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[ "# Number of zeros in decimal expansion What is the number of zeros in the decimal expansion of $11^{100}-1$? - According to python 12. Your number is 13780612339822270184118337172089636776264331200038466433146477552154985209552307‌​6769401159497458526446000. – P.. Oct 25 '12 at 18:44 how can it be solved using some theorems? – Vaolter Oct 25 '12 at 18:46 Or if that's binary, you have 3^4 - 1 = 80, so the answer is 1 zero in the decimal expansion. – Patrick87 Oct 25 '12 at 18:46 You can easily show that $11^{100} - 1$ has exactly $3$ trailing $0$'s. However, the $0$'s in the middle of the number are not really very amenable to analysis. – Robert Israel Oct 25 '12 at 19:05 ## 2 Answers 11^100 - 1 = 137806123398222701841183371720896367762643312000384664331464775521549852095523076769401159497458526446000 which has 12 zeros. - We ca see that, $$11^{100}-1=(10+1)^{100}-1=(10^{100}+100\\cdot10^{99}+99\\cdot50\\cdot 10^{98}+...+1)-1$$ Now, try to investigate the above. -", "# How do you expand 8m(m-6)? ##### 1 Answer Nov 17, 2016 $8 {m}^{2} - 48 m$ #### Explanation: Distribute $8 m$ across each of the terms within the parenthesis: $8 m \\cdot m - 8 m \\cdot 6 \\implies$ $8 {m}^{2} - 48 m$", "# How do you expand 7(4b-2)? Nov 19, 2016 $28 b - 14$ Distribute $7$ across each term in parenthesis: $\\left(7 \\cdot 4 b\\right) - \\left(7 \\cdot 2\\right) \\implies$ $28 b - 14$", "# How many zeros are there at the end of the product $33 \\times 175 \\times 180\\times 12 \\times 44 \\times 80 \\times 66$? Option 1) 6Option 2) Option 3) Option 4) Option 5) $33*175*180*12*44*80*66$ $211*54*....$ $4\\: zeros$", "the construction, all the numbers have no $1$s in their ternary expansion.", "# Decimal Expansion/Examples/1 over 6 $\\dfrac 1 6$ has a decimal expansion of $0 \\cdotp 1666 \\ldots$.", "# How do you expand 7(4b-2)? $28 b - 14$ Distribute $7$ across each term in parenthesis: $\\left(7 \\cdot 4 b\\right) - \\left(7 \\cdot 2\\right) \\implies$ $28 b - 14$", "## Elementary Algebra $25\\cdot1=25$ $25+1=26$ $25$ and $1$ are the only two numbers whose product is $25$ and sum is $26$. Therefore, we must use these numbers to fill in the blanks.", "Browse Questions # The number of non-zero terms in the expansion of $(1+ 3 \\sqrt {2}x)^9+(1-3 \\sqrt {2} x)^9$ is equal to : $(a)\\;0 \\quad (b)\\;5 \\quad (c)\\;9 \\quad (d)\\;10$ Can you answer this question? (b) 5 answered Nov 7, 2013 by", "Browse Questions # The number of non-zero terms in the expansion of $(1+ 3 \\sqrt {2}x)^9+(1-3 \\sqrt {2} x)^9$ is equal to : $(a)\\;0 \\quad (b)\\;5 \\quad (c)\\;9 \\quad (d)\\;10$ (b) 5", "Number Theory Factorials: Level 2 Challenges Which of these numbers is a perfect square? What is the product of all positive odd integers less than $10000$? $\\large 1 \\cdot 1! + 2\\cdot 2! + 3\\cdot 3! + \\cdots + 25\\cdot 25! = \\, ?$ Which is greater? $300! \\text{ or } 100^{300} ?$ Given that $13!= 13 \\times 12 \\times 11 \\times \\cdot \\cdot \\cdot \\times 2 \\times 1 = 6227020800$, it can be seen that $13!$ contains two trailing zeroes. How many trailing zeroes does $1000!$ contain? ×", "# Who is going to expand? $(1-x)(1-2x)(1-4x)\\cdots \\left(1-2^{101}x\\right)$ What is the coefficient of $x^{101}$ in the expansion of the above? ×", "# Fifty factorial 1 Find the number of trailing number of zeros can you find at the end of the decimal representation of $$50!$$. Notation: $$!$$ denotes the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "you the infinite expansion you found: $$0.101010\\ldots 1010\\ldots$$. • Wonderfully done. Very intuitive! – Graviton Aug 2 at 8:31", "# Number of zeros at the end of $10^{2}!+ 11^{2}!+12^{2}! \\cdots+99^{2}!$ How do I find the number of zeros at the end of the Integer $$10^{2}!+ 11^{2}!+12^{2}! \\cdots+99^{2}!$$ Answer provided for this question is $24$ • Can you answer a simpler question: how many zeros are at the end of the integer $100!$? Jul 16 '16 at 8:19 • HINT: This number is divisible by $10^2!$ but not by $10\\cdot10^2!$. Jul 16 '16 at 8:20 The number of zeroes at the end of all numbers $10^{2}!,\\ \\ 11^{2}!,\\ \\ 12^{2}!,\\ \\ \\cdots,\\ \\ 99^{2}!$, are equal or more than the one for $10^{2}!$. So, you just need to count the number of zeroes at the end of $10^{2}!$ Moreover, we know that the number of zeroes at the end of $100!$ is equal to $\\color{red}{\\lfloor {\\frac{100}{5}} \\rfloor+\\lfloor {\\frac{100}{25}} \\rfloor =24}$. Revise: As Hagen mentioned here, the number of zeroes at the end of all the numbers $11^{2}!,\\ \\ 12^{2}!,\\ \\ \\cdots,$ and $99^{2}!$ are more than $24$, that is the one for $10^2!$, and so the number of zeroes at the end of summation is equal to the number of zeroes at the end of $10^2!$. • @Ayushakj: This answer, unlike Hagen von Eitzen’s, is incomplete. If some of the terms $n^2!$ with $n>100$ also ended in $24$", "# Just ones and zeroes! How many ones are there in the expansion of $(101) \\times (10001) \\times (100000001) \\ .....\\times (100.... ( 2^7-1) zeroes ...0001) ?$ ×", "# How do you multiply 1.25*sqrt3 * 30? $37.5 \\sqrt{3}$ The radical is already in its simplest form; the only step is to multiply $1.25 \\cdot 30$ and leave as is.", "former case, the largest possible increase is $1/2^2 + 1/2^3 + \\cdots = 1/2$ minus $2/3^2 + 2/3^3 + \\cdots = 1/3$, which is $1/6$. In the latter case, $1/2^n - 1/3^n$ is exceeded by $1/2^{n-1} - 1/(2 \\cdot 3^{n-1})$, the total decrease from digit $2$ to digit $1$ from the $n$th place on. Thus, the minimum total increase from the increases exceeds the maximum total decrease from the decreases, which completes our proof of non-negativity for $x \\geq 1/2$.", "# Zeros How many trailing number of zeros does the following product end with? $1000\\times 999\\times 998\\times \\cdots \\times 3\\times 2\\times 1$ ×", "SQRT(9) = 16 OR 20 - (1 + SQRT(9)) = 16 17. 20 - 1 * SQRT(9) = 17 18. 20 + 1 - SQRT(9) = 18 OR (2 + 0) * 1 * 9 = 18 OR (2 + 0) / (1 / 9) = 18 19. 2 * 0 + 19 = 19 20. 2^0 + 19 = 20 21. 2 + 0 + 19 = 21 22. 20 - 1 + SQRT(9) = 22 23. 20 * 1 + SQRT(9) = 23 24. 20 + 1 + SQRT(9) = 24 25. 20 - 1 + (SQRT(9))! = 25 26. 20^1 + (SQRT(9))! = 26 27. 20 + 1 + (SQRT(9))! = 27 28. 20 - 1 + 9 = 28 29. 20 * 1 + 9 = 29 30. 20 + 1 + 9 = 30 • I really like how you gave \"or\" options. That said, you missed a few more that make 10. – sirjonsnow Apr 18 '18 at 14:11 • The ORs were as I found them, and not intended to be a complete solution. I have added a few for 10. – NetJohn Apr 18 '18 at 15:17 Here they are, as simple and neat as I could make them! $1-6:$ $$\\small\\begin{array}{c|c}1&2&3&4&5&6\\\\\\hline2\\cdot0+1^9&2+0\\cdot1\\cdot9&2\\cdot0\\cdot1+\\sqrt9&2+0!+1^9&-2-0!-1+9&-2+0-1+9\\end{array}$$ $7-12:$ $$\\small\\begin{array}{c|c}7&8&9&10&11&12\\\\\\hline-2+0\\cdot1+9&-2+0+1+9&2\\cdot0\\cdot1+9&2+0-1+9&2+0\\cdot1+9&2+0+1+9\\end{array}$$ $13-18:$ $$\\small\\begin{array}{c|c}13&14&15&16&17&18\\\\\\hline2+0!+1+9&(2+0!)!-1+9&20+1-\\sqrt9!&(2+0!)!+1+9&-2+0+19&2\\cdot(0+1)\\cdot9\\end{array}$$ $19-25:$ $$\\small\\begin{array}{c|c}19&20&21&22&23&24&25\\\\\\hline2\\cdot0+19&2^0+19&20+1^9&2+0!+19&20\\cdot1+\\sqrt9&20+1+\\sqrt9&(2+0!)!+19\\end{array}$$ $26-30:$ $$\\small\\begin{array}{c|c}26&27&28&29&30\\\\\\hline20+1\\cdot\\sqrt9!&20+1+\\sqrt9!&20-1+9&20\\cdot1+9&20+1+9\\end{array}$$ •" ]

[ "# Five times a number is 375. What is the number? Apr 7, 2016 75. #### Explanation: So the first part of the question: \"Five times a number is 375.\" is actually saying: \"Some number multiplied by 5 is equal to 375\", which looks like this: $x \\cdot 5 = 375$ Which is the same as: $5 x = 375$ The second part of the problem: \"What is the number?\" tells us that they want us to solve for $x$. We can do this by dividing both sides of our equation ($5 x = 375$) by 5, which looks like this: $5 x = 375$ =$\\frac{5 x}{5} = \\frac{375}{5}$ Lucky for us 375 is divisible by 5 (the divisibility rule for 5 states: any number is divisible by 5 if it ends in a 0 or 5) which means we will get a whole number as opposed to a fraction or decimal. We can solve this with mental math if we remember that: $5 \\cdot 20 = 100$ and that: $5 \\cdot 5 = 25$ If we add the 20 and the 5 together we get 25: $20 + 5 = 25$ Now if we multiply 25 by 5 we get 125 (This is the same as adding 100 and 25!). So this means that: $5 \\left(20 +", "# Factorial Trailing zeroes intuition I get that when we're trying to get the number of zeroes in 23!, we do 23/5 = 4 and we know there are 4 zeroes. My question is why only divide 23 by 5? What about 20, 15, 10, 5 all these numbers that are multiples of 5 and why does dividing 23 by 5 yield the number of 5's in the factorial? What's the mathematical intuition? Dividing $23$ by $5$ tells us how many multiples of $5$ there are less than (or equal to) $23$. That is how many multiples of $5$ occur in the product: $23\\times 22\\times \\cdots \\times 1$. Thus that product has $5^4$ as a factor. It also has $2^4$ as a factor, because $2$'s come much more frequently than $5$'s as we collect factors from $1$ to $23$. Thus the factorial has $10^4$ as a factor, hence $4$ zeros. It's not just a matter of counting the number of $5$'s though. If we look at $25!$, it ends with $6$ zeros, because when we reach $25$, we add not one but two factors of $5$, because $25$ is $5^2$. • Well, $5$ goes into $23$ four times, so if you count by $5$'s, you go through four multiples before getting there. Jul 27, 2017 at 5:37 •", "+0 # Factorial Help +1 157 1 The symbol 3! Means 3 x 2 x 1, which equals 6. Similarly 5! Means 5 x 4 x 3 x 2 x 1, which equals 120, and so on. Suppose that N! ends in exactly 3 zeros after fully multiplying out. What is the smallest value that N can have? Jun 13, 2019 #1 +195 +2 For a number end in $$x$$ zeros, it must be divisible by $$10^x$$ So for N! to have 3 zeros on the end of it, it must be divisible by 10*10*10 or $$5^3\\cdot2^3$$ To get 3 fives when multiplying out N! there must be 3 multiples of 5 less that or equal to N. 15!=15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 15, 10 and 5 are multiples of 5. There are also three 2s in 15!. $$\\boxed{15}$$ is the smallest value N can be because 15! will have 3 sets of 2s and 5s to make zeros. Jun 13, 2019 #1 +195 +2 For a number end in $$x$$ zeros, it must be divisible by $$10^x$$ So for N! to have 3 zeros on the end of it, it must be divisible by 10*10*10 or $$5^3\\cdot2^3$$ To get 3 fives when multiplying out N! there must be 3 multiples of 5 less that or equal to N. 15!=15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 15, 10", "precise mathematical statements about how many zeroes you'll get, search this site, where there are many variations on this question.) - It should be pretty obvious why $10!$ and all higher factorial must all have at least one zero at the end: they're all divisible by $10$. $$10! = \\mathbf{10} \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1$$ If you think about it a bit more, it's also pretty obvious that $20!$ and any factorials above it must haveat least two zeros at the end (because they're divisible by $10 \\times 20$) and that $30!$ and above must have at least three zeros (because they're divisible by $10 \\times 20 \\times 30$) and so on. Actually, this \"rule\" underestimates the number of zeros at the end of factorials by about a factor of $2$. Why? Because $2 \\times 5 = 10$, so $5! = 5 \\times 4 \\times 3 \\times 2 \\times 1$ already has one zero at the end, and every further multiple of $5$ adds yet another zero (there being plenty of even numbers to provide the multiples of $2$ needed to make up $10$). So $15!$ has three zeros at the end, not just one, and $20!$ actually has four, not two. Also, $25!$ actually", "352 views How many $0$’s are there at the end of $50!$? edited | 352 views We get a 0 at end of a number if the number is divisible by 10. In terms of prime numbers this means a number being divisible by each pair of (2, 5) we get a 0 at end of it. When we consider factorial, number of 2's will always be greater than number of 5's. So, we just need to count the no. of powers of 5. Number of $0$ at the end of $50!$ will be = $\\frac{50}{5}$ + $\\frac{50}{25}$ + $\\frac{50}{125}$ = 10 + 2 + 0 = 12 by edited by 0 Why u divided by 5,25,125?? 0 Why did you add videos here$?$ +1 Why you add videos here$?$ as we know for divisibilty of 10..any number must be divisible by 2 and 5...so whenever we want no of 0's at the end of any number..it will simply show how much time we can we can divide this number by 10...now The idea is to consider prime factor of a factorial n.If we can count the number of 5s and 2s, our task is done.Consider the following examples. n = 5: There is one 5 and 3 2s in prime factors of 5! =(2 * 2 *", "product, and the minimum of these two numbers gives you the number of factors $10$. In most cases, it boils down to counting the number of factors $5$ you find in the product. In your example the answer is $7$ because of the factors $5$ in $5, 10, 15, 20, 25, 25, 30$. Note that $25$ contains two factors $5$. Given positive integers $n_1, \\ldots, n_k$, let $a_i$ be the exponent of $2$ in the prime factorization of $n_i$, and $b_i$ be the exponent of $5$ in $n_i$. Then we have the number of zeroes at the end of $n_1 \\ldots n_k$ is equal to $\\min\\{ a_1 + \\ldots + a_k, b_1 + \\ldots + b_k\\}$. For $1 \\cdot 2 \\cdot \\ldots \\cdot 30$ we have 15 terms divisible by $2$, 7 terms divisible by $4$, 3 terms divisible by $8$, and 1 term divisible by $16$. So $a_1 + \\ldots + a_{30} = 26$. We have 6 terms divisible by $5$, and 1 term divisible by $25$, so $b_1 + \\ldots + b_{30} = 7$. The minimum of these is $7$. In general when looking at the number of zeroes at the ends of $n!$, we can argue its going to be $\\sum_{i=1}^{\\log_5(n)} \\lfloor \\frac{n}{5^i} \\rfloor$ Note that a $0$ is pretty much just a $5$ and", "of the integer x will all be zero? A) 21 B) 29 C) 38 D) 40 E) 100 We are given that x = a! and need to determine the least value for a that will create 8 trailing zeros in integer x. We must remember that a trailing zero is created with each 5-and-2 pair within the prime factorization of a!. Furthermore, since we know there are fewer 5s in a! (or any factorial) than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs. To determine the number of 5s within 21!, we can use the following shortcut in which we divide 21 by 5, and then divide the quotient of 21/5 by 5 and continue this process until we no longer get a nonzero quotient: 21/5 = 4 (we can ignore the remainder) Since 4/5 does not produce a nonzero quotient, we can stop. 21! contains four 5-and-2 pairs, and thus 21! does not contain 8 trailing zeros. Next we can test answer choice B: 29/5 = 5 (we can ignore the remainder) 5/5 =1 (we can ignore the remainder) Since 1/5 does not produce a nonzero quotient, we can stop. The final step is to add up our quotients; that sum represents the number of", "that $25 = 5\\cdot5$, so we must count double for each factor divisible by $25$, of which there are $5$; accordingly, our new answer is $30$. BUT WAIT... $125 = 5^3$, so we have to count it too, giving us a final answer of $31$ - The formula explaining above: $$\\text{Number of zeros in }n!=\\left[ \\frac {n}{5} \\right]+\\left[ \\frac {n}{5^2} \\right]+\\ldots$$ where [.] denotes the greatest integer function. - Note that if there is at least one zero after the last non-zero digit of $125!$, then $125!$ is divisible by $10$. Likewise, if there are two, then $125!$ is divisible by $100$, etc.. Thus, we need to check how many times $125!$ is divisible by $10$. So, we count the multiples of $5^1$, $5^2$, and $5^3 = 125$, in $125!$. It is easy to see that there are $25 = 125/5$ factors divisible by $5^1 = 5$, less than $125$. Similarly, there are $5 = 125/25$ factors divisible by $5^2 = 25$ in $125$. And finally, there is $1 = 125/125$ factors divisible by $5^3 = 125$. Thus, by the sum rule, there are $25 + 5 + 1 = 31$ such factors. But, this counts the number of times $125!$ is divisible by $10$, since $10 = 5 \\times 2$, and there is a $2$ in the", "# Math Help - factorial question 1. ## factorial question determine how many zeros are at the end of the numerals for the following numbers. a. 10! b. 100! c. 1000! 2. Hello, ihavvaquestion! Determine how many zeros are at the end of the numerals for the following numbers. . . $(a)\\;10! \\qquad (b)\\;100! \\qquad (c)\\;1000!$ Every factor of 5 (combined with an even number) creates a final zero. The question becomes: How many factors-of-5 are contained in $n!$ ? $(a)\\;10!$ We know that: . $10! \\:=\\:1\\cdot2\\cdot3\\cdots 10$ And every 5th number contains a factor-of-5: . $\\left[\\frac{10}{5}\\right] \\:=\\:2$ factors-of-5 Therefore: . $10!$ ends in two zeros. $(b)\\;100!$ Every 5th number contains a factor-of-5: . $\\left[\\frac{100}{5}\\right] \\:=\\:20$ factors-of-5. But every 25th number contains $5^2$ Each contributes an additional factor-of-5: . $\\left[\\frac{100}{25}\\right]\\:=\\:4$ more factors-of-5. Therefore, $100!$ ends in: . $20 + 4 \\:=\\:24$ zeros. $(c)\\;1000!$ Every fifth number contains a factor-of-5: . $\\left[\\frac{1000}{5}\\right] \\:=\\:200$ factors-of-5. But every 25th number contains $5^2$ Each contributes another factor-of-5: . $\\left[\\frac{1000}{25}\\right] \\:=\\:40$ more factors-of-5. And every 125th number contains $5^3$ Each contributes another factor-of-5: . $\\left[\\frac{1000}{125}\\right] \\:=\\:8$ more factors-of-5. And every 625th number contains $5^4$ Each contributes yet another factor-of-5\" . $\\left[\\frac{1000}{625}\\right] \\:=\\:1$ more factor-of-5. Therefore, $1000!$ ends in: . $200 + 40 + 8 + 1 \\:=\\:249$ zeros.", "is divisible by 2. From 1 and 2 : 25 has last digit 5 => 25 is divisible by 5 Jun 14, 2017 $25$ is a multiple of only $5$, not $2$ but see below for a better explanation. #### Explanation: Multiples of 5 always end with a $5$ or a $0$ and multiples of 2 always end with a even number or $0$. Since $25$ ends with $5$, it 25 is a multiple of 5! My source is my mind. I hope that helps you! -Moksha", "fifth number has a factor of 5. . . Hence, there are: . $\\frac{75}{5} \\:=\\:15$ factors of 5. But every twenty-fifth number has a factor of $5^2 = 25.$ . . Each of them contributes one more 5 to the total. And there are: . $\\frac{75}{25} \\:=\\:3$ of them. Hence, $75!$ contains: $15 + 3 \\:=\\: 18$ factors of 5 Therefore, $75!$ ends in eighteen zeros.", "is $$5$$ sums each equal to $$6$$, for a total of $$n(n+1) = 5 \\cdot 6 = 30$$. However, we’ve used each number from the sequence exactly twice, so we must divide by 2: $S_n = 10 = \\frac{n(n+1)}{2} = \\frac{4 \\cdot 5}{2} = 10$ Back to our puzzle. Given some $$n = 4k+1, k > 0$$, We can use the same trick of pairing numbers to cancel out every term but the last. Consider $$n = 9$$: $(-1) + 2 + (-3) + 4 + 5 + (-6) + 7 + (-8) + 9 = 9$ Here we’ve paired up 8 with 1, 7 with 2, and so on. We then pair up the pairs, and negate every other pair so all but the last number cancels to zero. Rearranging, this is more clear: $-(8 + 1) + (7 + 2) - (6 + 3) + (4 + 5) + 9 = 9$ $-9 + 9 - 9 + 9 + 9 = 9$ However, this only works for $$n = 4k + 1$$ because you need an even number of pairs to cancel out. ## Exponentiation William Kamovitch also notes that if exponentiation is allowed, there is a trivial solution available for all $$n$$: $1^{2 + ... + (n-1)} \\cdot n = n$ 1. We need", "View Single Post HW Helper P: 1,422 Quote by SteveRives Your question is the same as finding out how often 5 is a prime factor on the natural numbers. So I am guessing that 100! has 20 zeros, because in counting by fives, we get to 100 after 20 steps. You are WRONG, SteveRives. 24 zeros is the answer. (*)You will have a number which is divisible by 5 for every 5 successive integer. (*)You will have a number which is divisible by 25 (ie 5.5) for every 25 successive integer. (*)You will have a number which is divisible by 125 (ie 5.5.5) for every 125 successive integer, and so on... So you will have 20 numbers which is divisible by 5 (included numbers that is divisible by 25). And you will have 4 numbers which is divisible by 25. So you will have: 20 + 4 = 24 zeros in 100! In general, you will have: n! has: $$C = \\left[ \\frac{n}{5} \\right] + \\left[ \\frac{n}{5 ^ 2} \\right] + \\left[ \\frac{n}{5 ^ 3} \\right] + ... + \\left[ \\frac{n}{5 ^ q} \\right]$$ zeros at the end. You can stop the sum at any q such that: $$\\left[ \\frac{n}{5 ^ q} \\right] = 0$$ Where [...] denotes the integer part of a number. For example : [1.55]", "right. What does the 50/25 mean exactly in your sum when you say 50/5 + 50/25? That 25 is a factor of 50, right? But isn't it already accounted for in the 10 factors of five? Here is a list of multiples of five in 50 written in factored form. $5,\\;2 \\cdot 5,\\;3 \\cdot 5,\\;4 \\cdot 5,\\;5 \\cdot 5,\\;6 \\cdot 5,\\;7 \\cdot 5,\\;8 \\cdot 5,\\;9 \\cdot 5,\\;2 \\cdot 5 \\cdot 5$ those are $5,~10,~15,~20,~25,~30,~35,~40,~45,~50$. COUNT the number of fives in the first list. There are twelve factors of five. • Sep 28th 2013, 11:06 PM thevinh Re: Factorial Quote: Originally Posted by Natasha1 I have a question... Just one thing... 50! has 10 factors of five right. What does the 50/25 mean exactly in your sum when you say 50/5 + 50/25? That 25 is a factor of 50, right? But isn't it already accounted for in the 10 factors of five? To find out the number of trailing zeroes upon factorial expansion we need to count how many 5's there are in the product. Let's do a simple example, 16!. In this case you can see that there are three 5's in this product (1.2.3...5...10...15...16), hence you will have 3 trailing zeroes. Let's complicate thing a little bit, what about 47!? Well, 1...5...10...15...20...25...30...35...40...45.46.4 7, here you have", "# Why does the combination formula divide counts by two? This thing is confusing me for a long time and hence I am asking it here. We have a given set of numbers $$X = \\{1,2,3,4,5\\}$$ and we want to form a $4$- digit number of the form $aaac$(this is just symbolic; I want all the arrangements of this form for e.g. $aaca$ etc). (I know that such questions' answers do exist on MSE, but my doubt/ confusion is pertinent to something extraneous to such questions.) To make such a number the first approach that we can induce in our solution is that of using simple combination technique and choose two numbers out of the set of $5$ numbers. This can be done in the following way: $$\\binom{5}{2} \\cdot \\frac {4!}{3!}=40$$ That seems fine as far as the combination formula is confirmed. Let us now take a naive approach and pretend as if we don't know the combination formula. Here we use the gap method: Let us make one of our configuration as: $$\\fbox{a} \\fbox{a} \\fbox{a} \\fbox{c}$$ Here $a$ can be chosed in $5$ ways and $c$ can be chosen in $4$ ways Total ways to make $aaac$ configuration $=5 \\cdot 1\\cdot 1\\cdot 4 = 20$ ways Every such configuration now needs to be permuted since we want", "numbers are the numbers having factors $1$ and itself. A number is divisible by $5$ if the unit place digit is either $0$ or $5.$ Therefore, the pairs will be 1. $3 + 7 = 10$ 2. $2 + 3 = 5$ 3. $13 + 2 = 15$ 4. $13 + 7 = 20$ 5. $3 + 17 = 20$ 6. $5 + 5 = 10$ 15. Fill in the blanks: 1. A number which has only two factors is called a __________. Ans: We need to fill in the blanks with appropriate numbers or words. A number which has only two factors is called a prime number. 1. A number which has more than two factors is called a _________. Ans: We need to fill in the blanks with appropriate numbers or words. A number which has more than two factors is called a composite number. 1. $\\mathbf{1}$ neither ________ nor _________. Ans: We need to fill in the blanks with appropriate numbers or words. $1$ neither prime nor composite number. 1. The smallest prime number is ________. Ans: We need to fill in the blanks with appropriate numbers or words. The smallest prime number is $2.$ 1. The smallest composite number is ________. Ans: We need to fill in the blanks with appropriate numbers or words.", "# Number of zeros at the end of $10^{2}!+ 11^{2}!+12^{2}! \\cdots+99^{2}!$ How do I find the number of zeros at the end of the Integer $$10^{2}!+ 11^{2}!+12^{2}! \\cdots+99^{2}!$$ Answer provided for this question is $24$ • Can you answer a simpler question: how many zeros are at the end of the integer $100!$? Jul 16 '16 at 8:19 • HINT: This number is divisible by $10^2!$ but not by $10\\cdot10^2!$. Jul 16 '16 at 8:20 The number of zeroes at the end of all numbers $10^{2}!,\\ \\ 11^{2}!,\\ \\ 12^{2}!,\\ \\ \\cdots,\\ \\ 99^{2}!$, are equal or more than the one for $10^{2}!$. So, you just need to count the number of zeroes at the end of $10^{2}!$ Moreover, we know that the number of zeroes at the end of $100!$ is equal to $\\color{red}{\\lfloor {\\frac{100}{5}} \\rfloor+\\lfloor {\\frac{100}{25}} \\rfloor =24}$. Revise: As Hagen mentioned here, the number of zeroes at the end of all the numbers $11^{2}!,\\ \\ 12^{2}!,\\ \\ \\cdots,$ and $99^{2}!$ are more than $24$, that is the one for $10^2!$, and so the number of zeroes at the end of summation is equal to the number of zeroes at the end of $10^2!$. • @Ayushakj: This answer, unlike Hagen von Eitzen’s, is incomplete. If some of the terms $n^2!$ with $n>100$ also ended in $24$", "zero in $n!$ results in the number of factors of fives in $n!$. $26!$ is divisible by $5,~10,~15,~20,~\\&~25=5^2$ that is six factors of five. $50!$ has 12 trailing zeros. Fifty is has 10 factors of 5 and 2 factors of 25. • Mar 22nd 2013, 02:34 AM hitesh091 Re: factorial help @ max09 thanks i got it.. (Evilgrin)", "of the form $10 \\cdot n + 3$, and $10 \\cdot n + 5$ is always divisible by $5$. There is a similar example for case $4$. • But the OP also wants the \"in between\" number to be a multiple of $6$, if I read the question correctly. – David Oct 22 '15 at 6:13 • @David Correct. All four sets contain numbers adjacent to a multiple of six. – Tony Oct 22 '15 at 6:28", "it is n 2s all multiplied together. Like so: $2^8 = 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8$ In order for something to end in a zero it must be a multiple of 10, and the prime factorization of 10 is 5 * 2, and the prime factorization of $2^n$ will never contain a 5. Case closed. That was fun. That got me thinking about figuring out how many zeroes are at the end of a number if all you have is the prime factorization. Using my basic arithmetic skills I found out that: • $2 \\cdot 5 = 10$ • $2 \\cdot 2 \\cdot 5 = 20$ • $2 \\cdot 5 \\cdot 5 = 50$ • $2 \\cdot 2 \\cdot 5 \\cdot 5 = 10 \\cdot 10 = 100$ It appears (although isn’t proof) that the lowest quantity between twos and fives dictates how many zeroes are at the end of a number when you’re looking at its prime factorization. I tried this with many more combinations and it worked with every one of them. So what can we do with this information? A factorial is a number written as $n!$ where for any value $n$, $n! = 1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot n$ . For" ]

[ "right. What does the 50/25 mean exactly in your sum when you say 50/5 + 50/25? That 25 is a factor of 50, right? But isn't it already accounted for in the 10 factors of five? Here is a list of multiples of five in 50 written in factored form. $5,\\;2 \\cdot 5,\\;3 \\cdot 5,\\;4 \\cdot 5,\\;5 \\cdot 5,\\;6 \\cdot 5,\\;7 \\cdot 5,\\;8 \\cdot 5,\\;9 \\cdot 5,\\;2 \\cdot 5 \\cdot 5$ those are $5,~10,~15,~20,~25,~30,~35,~40,~45,~50$. COUNT the number of fives in the first list. There are twelve factors of five. • Sep 28th 2013, 11:06 PM thevinh Re: Factorial Quote: Originally Posted by Natasha1 I have a question... Just one thing... 50! has 10 factors of five right. What does the 50/25 mean exactly in your sum when you say 50/5 + 50/25? That 25 is a factor of 50, right? But isn't it already accounted for in the 10 factors of five? To find out the number of trailing zeroes upon factorial expansion we need to count how many 5's there are in the product. Let's do a simple example, 16!. In this case you can see that there are three 5's in this product (1.2.3...5...10...15...16), hence you will have 3 trailing zeroes. Let's complicate thing a little bit, what about 47!? Well, 1...5...10...15...20...25...30...35...40...45.46.4 7, here you have", "# Five times a number is 375. What is the number? Apr 7, 2016 75. #### Explanation: So the first part of the question: \"Five times a number is 375.\" is actually saying: \"Some number multiplied by 5 is equal to 375\", which looks like this: $x \\cdot 5 = 375$ Which is the same as: $5 x = 375$ The second part of the problem: \"What is the number?\" tells us that they want us to solve for $x$. We can do this by dividing both sides of our equation ($5 x = 375$) by 5, which looks like this: $5 x = 375$ =$\\frac{5 x}{5} = \\frac{375}{5}$ Lucky for us 375 is divisible by 5 (the divisibility rule for 5 states: any number is divisible by 5 if it ends in a 0 or 5) which means we will get a whole number as opposed to a fraction or decimal. We can solve this with mental math if we remember that: $5 \\cdot 20 = 100$ and that: $5 \\cdot 5 = 25$ If we add the 20 and the 5 together we get 25: $20 + 5 = 25$ Now if we multiply 25 by 5 we get 125 (This is the same as adding 100 and 25!). So this means that: $5 \\left(20 +", "# Factorial Trailing zeroes intuition I get that when we're trying to get the number of zeroes in 23!, we do 23/5 = 4 and we know there are 4 zeroes. My question is why only divide 23 by 5? What about 20, 15, 10, 5 all these numbers that are multiples of 5 and why does dividing 23 by 5 yield the number of 5's in the factorial? What's the mathematical intuition? Dividing $23$ by $5$ tells us how many multiples of $5$ there are less than (or equal to) $23$. That is how many multiples of $5$ occur in the product: $23\\times 22\\times \\cdots \\times 1$. Thus that product has $5^4$ as a factor. It also has $2^4$ as a factor, because $2$'s come much more frequently than $5$'s as we collect factors from $1$ to $23$. Thus the factorial has $10^4$ as a factor, hence $4$ zeros. It's not just a matter of counting the number of $5$'s though. If we look at $25!$, it ends with $6$ zeros, because when we reach $25$, we add not one but two factors of $5$, because $25$ is $5^2$. • Well, $5$ goes into $23$ four times, so if you count by $5$'s, you go through four multiples before getting there. Jul 27, 2017 at 5:37 •", "that $25 = 5\\cdot5$, so we must count double for each factor divisible by $25$, of which there are $5$; accordingly, our new answer is $30$. BUT WAIT... $125 = 5^3$, so we have to count it too, giving us a final answer of $31$ - The formula explaining above: $$\\text{Number of zeros in }n!=\\left[ \\frac {n}{5} \\right]+\\left[ \\frac {n}{5^2} \\right]+\\ldots$$ where [.] denotes the greatest integer function. - Note that if there is at least one zero after the last non-zero digit of $125!$, then $125!$ is divisible by $10$. Likewise, if there are two, then $125!$ is divisible by $100$, etc.. Thus, we need to check how many times $125!$ is divisible by $10$. So, we count the multiples of $5^1$, $5^2$, and $5^3 = 125$, in $125!$. It is easy to see that there are $25 = 125/5$ factors divisible by $5^1 = 5$, less than $125$. Similarly, there are $5 = 125/25$ factors divisible by $5^2 = 25$ in $125$. And finally, there is $1 = 125/125$ factors divisible by $5^3 = 125$. Thus, by the sum rule, there are $25 + 5 + 1 = 31$ such factors. But, this counts the number of times $125!$ is divisible by $10$, since $10 = 5 \\times 2$, and there is a $2$ in the", "# Number of such five digit numbers formed of $0,1,2,3,4,5$ which are divisible by $6$ without repetition I came across the question in the picture on the net. I couldn't figure out the solution given. But what I got is total number of such $5$ digit numbers using the given digits will be $5.5!$. Half of them will be divisible by $2$. And $1/3$ of them will be divisible by $3$. (Not sure though) What's next? I couldn't decipher what's written after that... Please try to use the permutations and combinations approach if possible: A $5$-digit number is formed by using the digits $0$, $1$, $2$, $3$, $4$ & $5$ without repetition. The probability that the number is divisible by $6$ is: (A) $8\\%$ (B) $17\\%$ (C) $18\\%$ (D) $36\\%$ Ans: (C) Hint: Number should be divisible by $2$ and $3$. $n(S) = 5 \\cdot 5!$; $n(A)$: reject '$0$' $= 2 \\cdot 4!$ reject '$3$' $= 4! + 2 \\cdot 3 \\cdot 3!$ Total $n(A) = 3 \\cdot 4! + 6 \\cdot 3! = 18 \\cdot 3!$ $\\therefore p = \\frac{18 \\cdot 3!}{5 \\cdot 5!} = 18\\%$ We try to count how many five-digit numbers formed with digits $0,1,2,3,4,5$ are divisible by six. To begin, we note three important facts: • To be a five-digit number, the first", "$3$ choices • Pick how the rest of the digits are arranged. $3!$ choices There are then a total of $2\\cdot 4!+4!+2\\cdot 3\\cdot 3! = 108$ possible five-digit numbers satisfying the conditions. The total number of five-digit numbers that can be created with those available digits will be $5\\cdot 5!=600$. (Arrange all six digits in a row, the final digit is unused. Remove the \"bad\" arrangements where zero is the leading digit for $6!-5! = 5\\cdot 5!$) The probability that a five-digit number formed from those digits is divisible by six is then $\\frac{108}{600}=0.18$ I. A number $n$ is divisible by $6$ if it is divisible by both $2$ and $3$. II. A number $n$ is divisible by $2$ if the rightmost digit is even. III. A number $n$ is divisible by $3$ if the sum of its digits is divisible by $3$. We have access to six digits, $\\{0,1,2,3,4,5\\}$, which sum to $15$. We need a $5$-digit number. This means we must remove one digit such that the resulting digit-sum is divisible by $3$. We can do this by either eliminating the $0$ (which results in the digit-sum $15$) or eliminating the $3$ (which results in the digit-sum $12$). This will allow us to create numbers that are divisible by $3$ no matter how we order the", "+0 # Factorial Help +1 157 1 The symbol 3! Means 3 x 2 x 1, which equals 6. Similarly 5! Means 5 x 4 x 3 x 2 x 1, which equals 120, and so on. Suppose that N! ends in exactly 3 zeros after fully multiplying out. What is the smallest value that N can have? Jun 13, 2019 #1 +195 +2 For a number end in $$x$$ zeros, it must be divisible by $$10^x$$ So for N! to have 3 zeros on the end of it, it must be divisible by 10*10*10 or $$5^3\\cdot2^3$$ To get 3 fives when multiplying out N! there must be 3 multiples of 5 less that or equal to N. 15!=15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 15, 10 and 5 are multiples of 5. There are also three 2s in 15!. $$\\boxed{15}$$ is the smallest value N can be because 15! will have 3 sets of 2s and 5s to make zeros. Jun 13, 2019 #1 +195 +2 For a number end in $$x$$ zeros, it must be divisible by $$10^x$$ So for N! to have 3 zeros on the end of it, it must be divisible by 10*10*10 or $$5^3\\cdot2^3$$ To get 3 fives when multiplying out N! there must be 3 multiples of 5 less that or equal to N. 15!=15*14*13*12*11*10*9*8*7*6*5*4*3*2*1 15, 10", "product, and the minimum of these two numbers gives you the number of factors $10$. In most cases, it boils down to counting the number of factors $5$ you find in the product. In your example the answer is $7$ because of the factors $5$ in $5, 10, 15, 20, 25, 25, 30$. Note that $25$ contains two factors $5$. Given positive integers $n_1, \\ldots, n_k$, let $a_i$ be the exponent of $2$ in the prime factorization of $n_i$, and $b_i$ be the exponent of $5$ in $n_i$. Then we have the number of zeroes at the end of $n_1 \\ldots n_k$ is equal to $\\min\\{ a_1 + \\ldots + a_k, b_1 + \\ldots + b_k\\}$. For $1 \\cdot 2 \\cdot \\ldots \\cdot 30$ we have 15 terms divisible by $2$, 7 terms divisible by $4$, 3 terms divisible by $8$, and 1 term divisible by $16$. So $a_1 + \\ldots + a_{30} = 26$. We have 6 terms divisible by $5$, and 1 term divisible by $25$, so $b_1 + \\ldots + b_{30} = 7$. The minimum of these is $7$. In general when looking at the number of zeroes at the ends of $n!$, we can argue its going to be $\\sum_{i=1}^{\\log_5(n)} \\lfloor \\frac{n}{5^i} \\rfloor$ Note that a $0$ is pretty much just a $5$ and", "# Prove that $2^{5n + 1} + 5^{n + 2}$ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2}$ is divisible by 27 for any positive integer. Work so far: (1) For n = 1: $2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$ Check if divisible by $27$: $189$ mod $27$ = $0$ As no remainder is left, the base case is divisible by $27$. (2) Assume $n = k$, then $2^{5k + 1} + 5^{k + 2} = 27k$ (3) Prove that this is true for n = k + 1: $$2^{5(k + 1) + 1} + 5^{(k + 1) + 2}$$ $$= 2^{5k + 5 + 1} + 5^{k + 1 + 2}$$ $$= 32 * 2^{5k + 1} + 5 * 5^{k + 2}$$ $$= ?$$ I know I am supposed to factor out 27 somehow, I just cant seem to figure it out. Any help would be appreciated. Use the fact that $32=5+27$ to get \\begin{align}2^{5(k+1)+1}+5^{k+1+2}&=2^{5k+6}+5^{k+3}\\\\&=32\\cdot2^{5k+1}+5\\cdot5^{k+2}\\\\&=5\\cdot2^{5k+1}+27\\cdot2^{5k+1}+5\\cdot5^{k+2}\\\\&=5(2^{5k+1}+5^{k+2})+27\\cdot2^{5k+1}\\\\&=27(5t+2^{5k+1})\\end{align} for some positive integer $t$. The result follows. • This is brilliant – Singh Chief Feb 8 '18 at 20:00 Because $$2^{5n+1}+5^{n+2}=2\\cdot32^n-2\\cdot5^n+27\\cdot5^n=2(32-5)(32^{n-1}+...+5^{n-1})+27\\cdot5^n,$$ which is divided by $27.$ It is: $$2^{5n+1}+5^{n+2}=2\\cdot 32^n+25\\cdot 5^n=$$", "answering this question and they are dubious in varying degrees because the real answer is “because I said so.” That is to say, the rule for multiplying negatives is a convention; adopted for good reasons, but a convention nonetheless. Those good reasons are mathematical: we want to make sure that when we extend multiplication and addition to negative numbers the properties of operations still apply. In particular, we want the distributive property to apply. Meditate on this: $$3\\cdot(5 + (-5)) = 3\\cdot5 + 3 \\cdot (-5).$$ The left side is really $3 \\cdot 0$, so it had better be zero. So the right side had better be zero as well. The first term on the right side is 15, so the other term had better be $-15$. So $3 \\cdot (-5) = -15$. We want the commutative law to hold, so we had better say $(-5)\\cdot 3 = -15$ as well. Now meditate on $$(-5)\\cdot(3 + (-3)) = (-5)\\cdot 3 + (-5)\\cdot(-3).$$ The same reasoning tells us that $(-5)\\cdot(-3) = 15$. Trouble is, all this is really hard to explain to middle schoolers, so people invent contexts. One context I’ve seen has something to do with sending out bills. If you receive 5 bills for 3 dollars then you have $5 \\cdot (-3) = -15$ dollars. Sending out", "# Math Help - factorial question 1. ## factorial question determine how many zeros are at the end of the numerals for the following numbers. a. 10! b. 100! c. 1000! 2. Hello, ihavvaquestion! Determine how many zeros are at the end of the numerals for the following numbers. . . $(a)\\;10! \\qquad (b)\\;100! \\qquad (c)\\;1000!$ Every factor of 5 (combined with an even number) creates a final zero. The question becomes: How many factors-of-5 are contained in $n!$ ? $(a)\\;10!$ We know that: . $10! \\:=\\:1\\cdot2\\cdot3\\cdots 10$ And every 5th number contains a factor-of-5: . $\\left[\\frac{10}{5}\\right] \\:=\\:2$ factors-of-5 Therefore: . $10!$ ends in two zeros. $(b)\\;100!$ Every 5th number contains a factor-of-5: . $\\left[\\frac{100}{5}\\right] \\:=\\:20$ factors-of-5. But every 25th number contains $5^2$ Each contributes an additional factor-of-5: . $\\left[\\frac{100}{25}\\right]\\:=\\:4$ more factors-of-5. Therefore, $100!$ ends in: . $20 + 4 \\:=\\:24$ zeros. $(c)\\;1000!$ Every fifth number contains a factor-of-5: . $\\left[\\frac{1000}{5}\\right] \\:=\\:200$ factors-of-5. But every 25th number contains $5^2$ Each contributes another factor-of-5: . $\\left[\\frac{1000}{25}\\right] \\:=\\:40$ more factors-of-5. And every 125th number contains $5^3$ Each contributes another factor-of-5: . $\\left[\\frac{1000}{125}\\right] \\:=\\:8$ more factors-of-5. And every 625th number contains $5^4$ Each contributes yet another factor-of-5\" . $\\left[\\frac{1000}{625}\\right] \\:=\\:1$ more factor-of-5. Therefore, $1000!$ ends in: . $200 + 40 + 8 + 1 \\:=\\:249$ zeros.", "because the real answer is “because I said so.” That is to say, the rule for multiplying negatives is a convention; adopted for good reasons, but a convention nonetheless. Those good reasons are mathematical: we want to make sure that when we extend multiplication and addition to negative numbers the properties of operations still apply. In particular, we want the distributive property to apply. Meditate on this: $$3\\cdot(5 + (-5)) = 3\\cdot5 + 3 \\cdot (-5).$$ The left side is really $3 \\cdot 0$, so it had better be zero. So the right side had better be zero as well. The first term on the right side is 15, so the other term had better be $-15$. So $3 \\cdot (-5) = -15$. We want the commutative law to hold, so we had better say $(-5)\\cdot 3 = -15$ as well. Now meditate on $$(-5)\\cdot(3 + (-3)) = (-5)\\cdot 3 + (-5)\\cdot(-3).$$ The same reasoning tells us that $(-5)\\cdot(-3) = 15$. Trouble is, all this is really hard to explain to middle schoolers, so people invent contexts. One context I’ve seen has something to do with sending out bills. If you receive 5 bills for 3 dollars then you have $5 \\cdot (-3) = -15$ dollars. Sending out is the opposite of receiving, so if you send out", "If we count the even divisors, that will result in some of the divisors being negative. (See thread.) Ok, lame post, sorry. Guys, do some of the earlier problems, like that staircase one ^_^ ## Variation of a previous number theory problem September 19th, 2009 Lol, just saw this on the math team forum, was posted by like last last year, but anyways… In the prime factorization of the expression $100! \\cdot 99! \\cdot 98! \\cdot \\ldots \\cdot 3! \\cdot 2! \\cdot 1!$ there is a $7^x$ term. Find x. ## Number Theory – Solutions September 18th, 2009 1) Written out in base 10, the number $345!$ ends in how many zeroes? Each zero in a number represents a factor of 10, or a factor each of five and two. Therefore, to find the number of zeros after a number, we need to find the number of factors of five and the number of factors of two in $345!$. How many factors of five are in $345!$? Simply divide 345 by five. $\\dfrac{345}{5} = 69$ However, we need to count the factors of five squared (25) twice, so we divide again by 25. $\\dfrac{345}{25} = 13$ And finally, let’s not forget the factors of five cubed which need to be counted three times. $\\dfrac{345}{125} = 2$ Summing these", "that $9^k-5^k$ is divisible by four, and you want to prove that $9^{k+1}-5^{k+1}$ is divisible by four. You noticed that you can write the last expression as $9\\cdot 9^k-5\\cdot 5^k$, and you said you're having trouble applying your assumption to it. What happens if you add $9\\cdot 5^k-9\\cdot 5^k$ to your expression? This is just zero, so you're not changing anything, but you can write $$9^{k+1}-5^{k+1} =9\\cdot 9^k-5\\cdot 5^k+9\\cdot 5^k-9\\cdot 5^k =9\\cdot (9^k-5^k)+(9-5)\\cdot 5^k.$$ By the assumption, the first term, $9\\cdot (9^k-5^k)$, is divisible by four, and by observation, the second term, $(9-5)\\cdot 5^k$, is divisible by four, so the sum of the two terms and therefore $9^{k+1}-5^{k+1}$ are divisible by four. This is what you wanted to show. You can always add zero (or multiply by one), and by writing zero (or one) in a particular way, you can sometimes make an argument easier to prove. Also, since your proof is by mathematical induction, you need to show that $9^n-5^n$ is divisible by four for, say, $n=0$ or $n=1$. You may have already done so and not asked about it in your question, but it's worth mentioning.", "Well there are still $a_k +1$ choices for all other $p_k$, but for $5$ we will only have $a_k$ choices, since there is no choice of putting $0$. - I don't understand fully what you have added, can you please clarify just a little the last portion? – Jeel Shah Jun 17 '13 at 16:16 @gekkostate For a divisor to be divisible by $5$, it needs at least one $5$ in its prime factorization, rather than $0$. – Tim Vermeulen Jun 17 '13 at 16:17 @gekkostate, Absolutely. The prime factorization of $4725$ is $7 \\cdot 5^2 \\cdot 3^3$. So we have $2$ ways to use the $7$ (one seven or no seven). For $5^2$, we have $2$ ways (one five or two five). We can't use zero fives, because it won't be divisible by $5$. We have $4$ ways to choose the $3$ (either $0, 1, 2$ or $3$ threes could be used). The answer is the product, or $2 \\cdot 3 \\cdot 4$. – George V. Williams Jun 17 '13 at 16:18 @GeorgeV.Williams Wow, that makes sense :D Thanks a bunch! – Jeel Shah Jun 17 '13 at 16:21 Using Divisor function, as $4725=7^1\\cdot5^2\\cdot3^3$ the number of divisors will be $(1+1)(2+1)(3+1)$ - Why do we add $1$ to each of number of exponents? Is it to include", "is a set of ordered pairs (with some restrictions)? There are no legal pairs for these restrictions, and so there are no legal subsets of these pairs...other than the empty set. So $\\emptyset$ -is- a function (it -is- a set of pairs (empty of course), all of whose pairs satisfy the function criteria). No other functions will work so $\\emptyset$ is the only function that works. So there is only 1 bijective function on a set of size 0. So $0! = 1$. Yes, this is weird, but it works. Negative numbers, complex numbers, they're all weird even when you just manipulate their properties. But you'll get over it. - Let's try a different approach from my other answer to this: To multiply a number $N$ by $6!$ is to multiply it by six factors and get $$N\\cdot1\\cdot2\\cdot3\\cdot4\\cdot5\\cdot6.$$ Similarly to multiply $N$ by $0!$ is to multiply it by $0$ numbers: $$N.$$ But that is the same as multiplying it by $1$ $$N\\cdot1.$$ Multiplying $N$ by no numbers at all is multiplying $N$ by $1$. (This answer doesn't apply only to factorials; it may be taken as a general explanation of why, when one multiplies no numbers, one gets $1$.) - You can define $\\exp(x)$ as $$1 + \\frac {x} {1!} + \\frac {x^2} {2!} + \\frac {x^3}", "is $$5$$ sums each equal to $$6$$, for a total of $$n(n+1) = 5 \\cdot 6 = 30$$. However, we’ve used each number from the sequence exactly twice, so we must divide by 2: $S_n = 10 = \\frac{n(n+1)}{2} = \\frac{4 \\cdot 5}{2} = 10$ Back to our puzzle. Given some $$n = 4k+1, k > 0$$, We can use the same trick of pairing numbers to cancel out every term but the last. Consider $$n = 9$$: $(-1) + 2 + (-3) + 4 + 5 + (-6) + 7 + (-8) + 9 = 9$ Here we’ve paired up 8 with 1, 7 with 2, and so on. We then pair up the pairs, and negate every other pair so all but the last number cancels to zero. Rearranging, this is more clear: $-(8 + 1) + (7 + 2) - (6 + 3) + (4 + 5) + 9 = 9$ $-9 + 9 - 9 + 9 + 9 = 9$ However, this only works for $$n = 4k + 1$$ because you need an even number of pairs to cancel out. ## Exponentiation William Kamovitch also notes that if exponentiation is allowed, there is a trivial solution available for all $$n$$: $1^{2 + ... + (n-1)} \\cdot n = n$ 1. We need", "precise mathematical statements about how many zeroes you'll get, search this site, where there are many variations on this question.) - It should be pretty obvious why $10!$ and all higher factorial must all have at least one zero at the end: they're all divisible by $10$. $$10! = \\mathbf{10} \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1$$ If you think about it a bit more, it's also pretty obvious that $20!$ and any factorials above it must haveat least two zeros at the end (because they're divisible by $10 \\times 20$) and that $30!$ and above must have at least three zeros (because they're divisible by $10 \\times 20 \\times 30$) and so on. Actually, this \"rule\" underestimates the number of zeros at the end of factorials by about a factor of $2$. Why? Because $2 \\times 5 = 10$, so $5! = 5 \\times 4 \\times 3 \\times 2 \\times 1$ already has one zero at the end, and every further multiple of $5$ adds yet another zero (there being plenty of even numbers to provide the multiples of $2$ needed to make up $10$). So $15!$ has three zeros at the end, not just one, and $20!$ actually has four, not two. Also, $25!$ actually", "it is n 2s all multiplied together. Like so: $2^8 = 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8 \\cdot 8$ In order for something to end in a zero it must be a multiple of 10, and the prime factorization of 10 is 5 * 2, and the prime factorization of $2^n$ will never contain a 5. Case closed. That was fun. That got me thinking about figuring out how many zeroes are at the end of a number if all you have is the prime factorization. Using my basic arithmetic skills I found out that: • $2 \\cdot 5 = 10$ • $2 \\cdot 2 \\cdot 5 = 20$ • $2 \\cdot 5 \\cdot 5 = 50$ • $2 \\cdot 2 \\cdot 5 \\cdot 5 = 10 \\cdot 10 = 100$ It appears (although isn’t proof) that the lowest quantity between twos and fives dictates how many zeroes are at the end of a number when you’re looking at its prime factorization. I tried this with many more combinations and it worked with every one of them. So what can we do with this information? A factorial is a number written as $n!$ where for any value $n$, $n! = 1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot n$ . For", "# Prove that for any $n \\in \\mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$? I have question Q Prove that for any $n \\in \\mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$? by using induction Thanks - $2^{n+1+2}3^{n+1}+5(n+1)-4=6\\cdot2^{n+2}3^{n}+5n+5-4=(2^{n+2}3^{n}+5n-4)+5(2^{n+‌​2}3^{n}+1)$. So it's enough to show that $2^{n+2}3^{n}+1$ is divisible by $5$. Again induction: $2^{n+1+2}3^{n+1}+1=6\\cdot2^{n+2}3^{n}+1=2^{n+2}3^{n}+1+5\\cdot(2^{n+2}3^{n}+1) \\dots$ – P.. Apr 26 '13 at 7:57 Let $f(n)=2^{n+2}3^n+5n-4=4\\cdot6^n+5n-4$ $f(m+1)-f(m)=4\\cdot6^{m+1}+5(m+1)-4-(4\\cdot6^m+5m-4)$ $=5\\{5+4(6^m-1)\\}$ Now, we know $(6^m-1)$ is divisible by $6-1=5$ $\\implies f(m+1)\\equiv f(m)\\pmod {25}$ Now, $f(1)=2^3\\cdot3+5\\cdot1-4=25\\implies f(1)$ divisible by $25$ Alternatively, $2^{n+2}3^n+5n-4=4\\cdot6^n+5n-4$ $=4(1+5)^n-5n-4$ $=4\\left(1+\\binom n1 5+\\binom n2 5^2+\\cdots +\\binom n{n-1}5^{n-1}+5^n\\right)+5n-4$ (Using Binomial Expansion) $= 4\\{1+5n+5^2\\left(\\binom n2 +\\binom n35 +\\binom n45^2+\\cdots+\\binom n{n-1}5^{n-3}+5^{n-2}\\right)\\}+5n-4$ As the Binomial coefficients are all integers, all the terms after $\\binom n1 5$ is divisible by $5^2=25,$ $5^2\\left(\\binom n25^2 +\\binom n35 +\\binom n45^2+\\cdots+\\binom n{n-1}5^{n-3}+5^{n-2}\\right)$ can bewritten as $25k$ where $k$ is an integer So, $2^{n+2}3^n+5n-4=4\\{1+5n+25k\\}+5n-4=25(n+4k)$ - can you explain more why 2^n 3^n=6^n and in step 4 how you get 4+4 5n from step 3 I know step 3 is Binomial Theorem, how you get step 3 can you explain more thank you – leena adam Apr 26 '13 at 6:34 @leenaadam, for real $a,b (ab)^n=a^n \\cdot b^n$. – lab bhattacharjee Apr 26 '13 at 6:37 how you get 4+4 5n from step 3 I know step 3 is Binomial Theorem, how you get step" ]

[ "their 5 26 Nov 2015, 06:33 1 How many different positive integers can be formed 5 24 Feb 2015, 11:42 How many different four-digit integers can be composed of digits 0, 2, 2 30 Sep 2014, 01:26 25 How many four-digit positive integers can be formed by using 20 22 May 2012, 09:35 1 What is the greatest number of four-digit integers whose 5 15 Sep 2008, 04:15 Display posts from previous: Sort by", "total number of different three-digit positive integers that can be formed by elements from Set B 25000 +道题目 139本备考书籍", "# Primes Again Given six distinct positive integers $a, b, c, d, e,$ and $f,$ what is the maximum number of distinct primes that can be formed from the pairwise sums? ×", "+0 # How many 11 -digit positive integers can be written using only the digits 1,2 and 3 and have exactly five 1’s and exactly two 2's as digits? 0 59 1 How many 11 -digit positive integers can be written using only the digits 1,2 and 3 and have exactly five 1’s and exactly two 2's as digits? Aug 15, 2022 #1 +2453 0 We know there must be five 1's, two 2's, and four 3's. There are $${11! \\over 2! \\times 5! \\times 4!} = \\color{brown}\\boxed{6930}$$ numbers that work. Aug 16, 2022", "+0 # Base Stuff +1 66 1 +770 How many two digit positive integers in the base-four representation are twice the two digit positive integers in the base-three representation? A) 5 B) 6 C) 7 D) 8 E) 9 The answer is apparently A) 5, but I am not sure how to get to it. Any help is appreciated. Thanks! - PM Jan 26, 2020 edited by PartialMathematician Jan 26, 2020 #1 +24133 +3 How many two digit positive integers in the base-four representation are twice the two digit positive integers in the base-three representation? I assume: $$\\begin{array}{|rl|c|c|c|c|c|c|l|} \\hline & & 10 & 11 & 12 & 20 & 21 & 22 & \\text{base}~3 \\\\ & & =3 & =4 & =5 & =6 & =7 & =8 & \\text{base}~10 \\\\ \\text{base}~4 & \\text{base}~10 & 6 & 8 & 10 & 12 & 14 & 16 & 2x \\\\ \\hline 10 & = 4 & & & & & & & \\\\ 11 & = 5 & & & & & & & \\\\ 12 & = 6 &6=6\\checkmark& & & & & & \\\\ 13 & = 7 & & & & & & & \\\\ 20 & = 8 & &8=8\\checkmark& & & & & \\\\ 21 & = 9 & & & & & &", "integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ? A. 400 B. 1728 C. 108 D. 216 E. 432 I am getting answer as 432, but the OA is D. Here is the approach I used. First number can be selected from 1 to 9 - in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1 so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks! XXYY can be arranged in $$\\frac{4!}{2!2!}=6$$ ways (# of arrangements of 4 letters out of which 2 X's and 2 Y's are identical); # of ways we can select 2 distinct digits out of 9 is $$C^2_9=36$$; Total # of integers that can be formed is 6*36=216. Hi Bunuel, I Want to know if the statement changes from \"How many four-digit positive integers can be formed by using the digits from 1 to 9\" to How many four-digit positive integers can be formed by using the", ". How many two-digit numbers can be formed from the numbers 1,2,3,4 is repetirion is allowed", "# Sum of two fifth powers! Find the nine-digit number of the form xxxxyyyyy that can be written as a sum of fifth powers of two positive integers. ×", "formed by 2,9,16 and you are done. -", "each 9 17 Dec 2012, 15:41 28 How many four-digit positive integers can be formed by using 20 22 May 2012, 08:35 19 In list L above, there are 3 positive integers, where each o 5 19 Jul 2011, 14:45 45 K and L are each four-digit positive integers with thousands 19 15 Nov 2009, 11:58 Display posts from previous: Sort by", "# Primes Again Logic Level 3 Given six distinct positive integers $$a, b, c, d, e,$$ and $$f,$$ what is the maximum number of distinct primes that can be formed from the pairwise sums? ×", "+0 # Counting 0 42 5 How many different positive, six-digit integers can be formed using the digits 2, 2, 5, 6, 8 and 9? Apr 14, 2022 #1 +1384 +1 $${6! \\over 2!} = \\color{brown}\\boxed{360}$$ . Apr 14, 2022 #3 +122 -1 Gm builderboi Kakashi Apr 14, 2022 #4 +1384 +1 gm to you as well :) BuilderBoi Apr 14, 2022 #2 +326 +2 We have 6 numbers in the set \"2, 2, 5, 6, 8, and 9\", so we have 6! to find the different positive 6 digit integers that can be formed. And we have two 2s... which means that we need to divide by 2! That will give us: 6!/2! which is 360 different positive 6 digit numbers Apr 14, 2022 #5 +122 -1 I had no clue how to do that bc i just forgot how to do some things i learn i the past Apr 14, 2022", "# High five Discrete Mathematics Level pending Suppose a 5-digit number is created from 5 distinct integers chosen among {1,2,3,4,5,6,12}. How many such 5-digit numbers include both 1 and 2 but neither begin nor end with 1? ×", "different numbers of nines when this would never happen for any finite number?", "# Multiples of 5 and 7 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "# multiples of five Level pending how many four digit numbers can be formed using 0,1,2,3,4 which are multiples of 5 ×", "9 + #1} & #1 + {1, 4, 9} &", "group of nine cells that form a 3x3 square must contain the numbers 1 through 9.", "0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?", "if each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30) #2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B) #3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer? #4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer? #5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer? #2: $$\\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8$$ 4.8 * 11111 =$$\\color{red}{53332.8}$$" ]

[ "at a time: $P(5,5) = \\frac{5!}{(5-5)!} = 5! = 120$ Example 2.2.11: Ordering of digits under different conditions. Consider only the digits 1, 2, 3, 4, and 5. (a) How many three-digit numbers can be formed if no repetition of digits can occur? (b) How many three-digit numbers can be formed if repetition of digits is allowed? (c) How many three-digit numbers can be formed if only non-consecutive repetition of digits are allowed? Solutions to (a): Solution 1: Using the rule of products. We have any one of five choices for digit one, any one of four choices for digit two, and three choices for digit three. Hence, 5 · 4 · 3 = 60 different three-digit numbers can be formed. Solution 2; Using the permutation formula. We want the total number of permutations of five digits taken three at a time: $P(5,3) = \\frac{5!}{(5-3)!} = 5 \\cdot 4 \\cdot 3 = 60$ Solution to (b): The definition of permutation indicates “...no two elements in each list are the same.” Hence the permutation formula cannot be used. However, the rule of products still applies. We have any one of five choices for the first digit, five choices for the second, and five for the third. So there are 5 · 5 · 5 = 125 possible different three-digit", "distinct bit strings can be formed from three 0’s and two 1’s? Solution: 5 total digits would give 5! permutations. But that is assuming the 0’s and 1’s are distinguishable (to make that explicit, let’s give each one a subscript). Here are the $3! \\cdot 2! = 12$ different ways that we could have arrived at the identical string \"01100\" if we thought of each 0 and 1 as unique. Since identical digits are indistinguishable, all the listed permutations are the same. For any given permutation, there are 3! ways of rearranging the 0’s and 2! ways of rearranging the 1’s (resulting in indistinguishable strings). We have over-counted. Using the formula for permutations of indistinct objects, we can correct for the over-counting: $$\\text{Total} = \\frac{5!}{3! \\cdot 2!} = \\frac{120}{6 \\cdot 2} = 10$$ Example: How many distinct orderings of characters are possible for the string \"MISSISSIPPI\"? Solution: In the case of the string \"MISSISSIPPI\", we should separate the characters into four distinct groups of indistinct characters: one \"M\", four \"I\"s, four \"S\"s, and two \"P\"s. The number of distinct orderings are: $$\\frac{11!}{1!4!4!2!} = 34,650$$ Example: Consider the 4-digit passcode smart-phone from before. How many distinct passcodes are possible if there are 3 smudges over 3 digits on the screen? Solution: One of 3 digits is repeated, but we", "new two-step process – or $(52\\cdot 51\\cdot 50)\\cdot 49 =52\\cdot 51\\cdot 50\\cdot 49$ ways. Finally, the number of ways to be dealt five cards, we have (the number of ways to be dealt the first four cards) $\\times$ (the number of ways to be dealt the fifth card) – here again, a two-step process – or $(52\\cdot 51\\cdot 50\\cdot 49)\\cdot 48 =52\\cdot 51\\cdot 50\\cdot 49\\cdot 48 =311,875,200$ possible outcomes for this process. This would be difficult to count without our technique. In what follows we will use the above counting technique in a variety of settings, examining its flexibility. Consider now the number of ways to order a collection of items. Such orderings are called permutations. ## Permutations Here is a model problem. Given a collection of four items, call them $A\\,$ , $\\, B\\,$ , $\\, C$ and $D\\,$ , in how many ways can we order them? That is, in how many ways can we pick a first item, then a second , then a third, and finally a fourth? This can be answered by considering it as a multi-step process. There are four ways to pick a first item, then three ways to pick a second, next two ways to pick a third, leaving the fourth item determined (i.e. only one way to pick the", "order, we divide the number of total ways of picking $$k$$ things in order by the number of ways of picking the same $$k$$ things in different orders. Example. To see this in action, check out $C(5,3) = \\frac{5\\cdot 4\\cdot 3}{3\\cdot 2\\cdot 1} = 10.$ The general class of situations. You use a binomial coefficient any time you want to count the number of ways of choosing $$k$$ things from $$n$$ things without regard to order. A particular example. Say you have five different sandwiches and you grab three of them. How many different outcomes does this experiment have? Well, since you're just grabbing them, you're obviously not making a point of checking what order you're taking them. So you're just choosing $$3$$ things from $$5$$ things without regard to order. How many ways are there of doing this? $$C(5,3) = 10.$$ Where could the process of understanding break down? You might have just skimmed this because you've already decided you won't ever understand math. If that's the case, I'd encourage you to go back and read it again. It's not as bad as you think it is. Perhaps your eyes glossed over at the symbol $$C(n,k)$$. If that's the case, remember: I told you what the symbol means. I don't expect you to remember it right away,", "or ${}_nP_r$. By the basic principle of counting we find that $$\\label{eq:perm}P(n,r)=\\prod_{j=0}^{r-1}(n-j)=n(n-1)(n-2)\\cdots(n-r+1)$$ The expression in \\eqref{eq:perm} can be written as $$\\label{eq:perm2}P(n,r)=\\frac{n!}{(n-r)!}$$ The sage command for calculating $P(n,r)$ is falling_factorial(n,r). Obviously, $P(n,n)=n!$. On the other hand, we have $P(n,n)=\\frac{n!}{0!}$ from \\eqref{eq:perm2}. Therefore, it is defined that $0!=1$. Using the formula \\eqref{eq:perm2} the answer to the question in the above example is simply $$P(25,3)=\\frac{25!}{(5-3)!}=25\\cdot 24\\cdot 23$$ Example. Consider the digits 1, 2, 3, 4, 5. 1. How many three-digit numbers can be formed if no repetition of digits can occur? 2. How many three-digit numbers can be formed if repetition is allowed? 3. How many three-digit numbers can be formed if only non-consecutive repetition of digits are allowed? Solution. 1. $P(5,3)=\\frac{5!}{(5-3)!}=5\\cdot 4\\cdot 3=60$. 2. $5\\cdot 5\\cdot 5=125$. 3. $5\\cdot 4\\cdot 4=80$. References: Not in particular order. [1] Applied Discrete Structures, Al Doerr and Ken Levasseur. This book is available for free under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 United States License. [2] A First Course in Probability, Sheldon Ross, 5th Edition, Prentice-Hall, 1998", "\\cdot 4 \\cdot 5=120$. This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the $k-th$ step involves $k+1$ numbers to choose from. The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally. The same goes for the blue ones. The numerator must equal $(1 \\cdot 2)^2$. Therefore, the probability for each of the orderings of $RRBB$ is $\\frac{4}{120}=\\frac{1}{30}$. There are 6 of these, so the total probability is $\\boxed{\\bf{(B)} \\frac{1}{5}}$. ~IceMatrix", "regard to order, we divide the number of total ways of picking $$k$$ things in order by the number of ways of picking the same $$k$$ things in different orders. Example. To see this in action, check out $C(5,3) = \\frac{5\\cdot 4\\cdot 3}{3\\cdot 2\\cdot 1} = 10.$ The general class of situations. You use a binomial coefficient any time you want to count the number of ways of choosing $$k$$ things from $$n$$ things without regard to order. A particular example. Say you have five different sandwiches and you grab three of them. How many different outcomes does this experiment have? Well, since you're just grabbing them, you're obviously not making a point of checking what order you're taking them. So you're just choosing $$3$$ things from $$5$$ things without regard to order. How many ways are there of doing this? $$C(5,3) = 10.$$ Where could the process of understanding break down? You might have just skimmed this because you've already decided you won't ever understand math. If that's the case, I'd encourage you to go back and read it again. It's not as bad as you think it is. Perhaps your eyes glossed over at the symbol $$C(n,k)$$. If that's the case, remember: I told you what the symbol means. I don't expect you to remember it", "be written with digits 1, 2, 3, 4, 5 i 6 if the digits: a) can’t be repeated; b) can be repeated; Solution: a) Of 6 offered digits we choose 3 keeping in mind that the digits can’t be repeated. So, NOT ALL digits are chosen, and their order IS important, therefore we have variations of 3 from the set of 6 elements without repeating ($n=6$ and $k=3$): $V_{6}^{3}=\\frac{6!}{(6-3)!}=\\frac{6!}{3!}=\\frac{6\\cdot 5\\cdot 4\\cdot 3!}{3!}=120$ We can write 3-digit 120 numbers. b) Here we have the same case the only difference being that this time digits can be repeated, so we have: $\\overline{V_{6}^{3}}={{6}^{3}}=216$ COMBINATIONS Suppose that in our group of 25 members we need to choose 2 to represent the group in a convention. In this case we chose 2 of 25 so, again, not all members are chosen. Difference from the variations is that this time the oreder of chosen members is not important (since both representatives have the same function). This are called combinations and the answers to our two questions are: NOT ALL elements from the set are chosen. The order of chosen elements IS NOT important. Combinations can also be with and without repeating. Combinations of k from the set of n elements without repeating $C_{n}^{k}$ is calculated using formula: $C_{n}^{k}=\\left( \\begin{matrix} n \\\\ k \\\\ \\end{matrix}", "# How many possible codes are there if the digits cannot be repeated? How many possible codes are there if the digits cannot be repeated? ## A Store-It-All storage facility requires users to enter a four digit code, using numbers 0-9, to get through the gate. You need to get a box of winter clothes from your storage container, but you forgot the code. How many possible codes are there if the digits cannot be repeated? What is the probability that you will guess the correct code on your first try? P(correct code) = __ $10 \\cdot 9 \\cdot 8 \\cdot 7 = 5040$ different codes. P(correct code)=$\\frac{1}{5040}$", "Solution. $25\\cdot 24\\cdot 23$ by the basic principle of counting. Definition. An ordered arrangement of $r$ elements selected from $n$ elements, $0\\leq r\\leq n$, where no two elements of the arrangement are the same, is called a permutation of $n$ objects taken $r$ at a time.The total number of such permutations is denoted by $P(n,r)$ or ${}_nP_r$. By the basic principle of counting we find that $$\\label{eq:perm}P(n,r)=\\prod_{j=0}^{r-1}(n-j)=n(n-1)(n-2)\\cdots(n-r+1)$$ The expression in \\eqref{eq:perm} can be written as $$\\label{eq:perm2}P(n,r)=\\frac{n!}{(n-r)!}$$ The sage command for calculating $P(n,r)$ is falling_factorial(n,r). Obviously, $P(n,n)=n!$. On the other hand, we have $P(n,n)=\\frac{n!}{0!}$ from \\eqref{eq:perm2}. Therefore, it is defined that $0!=1$. Using the formula \\eqref{eq:perm2} the answer to the question in the above example is simply $$P(25,3)=\\frac{25!}{(5-3)!}=25\\cdot 24\\cdot 23$$ Example. Consider the digits 1, 2, 3, 4, 5. 1. How many three-digit numbers can be formed if no repetition of digits can occur? 2. How many three-digit numbers can be formed if repetition is allowed? 3. How many three-digit numbers can be formed if only non-consecutive repetition of digits are allowed? Solution. 1. $P(5,3)=\\frac{5!}{(5-3)!}=5\\cdot 4\\cdot 3=60$. 2. $5\\cdot 5\\cdot 5=125$. 3. $5\\cdot 4\\cdot 4=80$. References: Not in particular order. [1] Applied Discrete Structures, Al Doerr and Ken Levasseur. This book is available for free under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 United States License. [2] A First Course in", "that you repeat (how many possibilities?), then choose where those two digits go in the number (how many possibilities?), and finally choose the order for the remaining $9$ digits (how many possibilities?) (The answer is $163,296,000$.) • I got $10$ ways to select the digit I want to repeat, $C(11,2)$ ways to place the repeated digit, and $9!$ ways for the remaining 9 digits. And the result was wrong. Where is my problem? – PHPIsTheBestLanguage Jun 29 '15 at 14:05 • You don't have C(11,2) ways to place the repeated digit because you can't place the repeated digit next to each other. – Amy B Jun 29 '15 at 14:09 • Oops you are right, I forgot it. :P – PHPIsTheBestLanguage Jun 29 '15 at 14:21 • Right: $10 \\cdot (_{11}C_2 - 10) \\cdot 9!$ – John Jun 29 '15 at 18:41 How many repeat digit $1$? There are $\\frac{11!}{2}$ orders total but $10!$ of those have both $1$'s together (put the two $1$'s together forming a single object). Hence there are $\\frac{11!}{2}-10!=\\frac{9\\cdot10!}{2}$ that repeat digit $1$, which implies there are $\\frac{10\\cdot9\\cdot10!}{2}=5\\cdot9\\cdot10!$ total. • I'm not quite understanding $11!/2$ orders for a repeat digit. How did you get that? – PHPIsTheBestLanguage Jun 29 '15 at 14:07 • There are $11!$ ways to order $11$ objects. But you have", "are $45$ such combinations with $2$ distinct digits and both digits repeated. There are $360$ such combinations with $3$ distinct digits ($120$ ways to pick the digits, $3$ ways to pick the one that repeats). Thus, there are $$1\\cdot 10+4\\cdot 90+6\\cdot 45+12\\cdot 360=4960$$ combinations with at least one repeat, leaving $5040$ combinations with no repeats, and so there are $$\\frac{5040}{24}=210$$ combinations in non-decreasing order with no repeats. In total, there are thus $10+90+45+360+210=715$ non-decreasing combinations that we need to try, and we can be sure to get it open by the end of that list. As for strategy, let's suppose that each of the $10000$ possible combinations is equally likely to be correct. Then we have \\begin{align}\\text{P(four distinct digits)} &= \\frac{24\\cdot 210}{10000}=50.4\\%\\\\\\text{P(three distinct digits)} &= \\frac{12\\cdot 360}{10000}=43.2\\%\\\\\\text{P(two distinct digits, both repeated)} &= \\frac{6\\cdot 45}{10000}=2.7\\%\\\\\\text{P(two distinct digits, only one repeated)} &= \\frac{4\\cdot 90}{10000}=3.6\\%\\\\\\text{P(single digit repeated four times)} &= \\frac{1\\cdot 10}{10000}=0.1\\%.\\end{align} Guessing a non-decreasing combination with 4 distinct digits is effectively worth $24$ guesses, and so on. Our expected value for a given kind of combinations--that is, how efficient a given nondecreasing combination will be as a guess--would then be \\begin{align}\\text{E(four distinct digits)} &= 24\\cdot50.4\\% = 12.096\\\\\\text{E(three distinct digits)} &= 12\\cdot43.2\\% = 5.184\\\\\\text{E(two distinct digits, both repeated)} &= 6\\cdot2.7\\% = 0.162\\\\\\text{E(two distinct digits, only one repeated)} &= 4\\cdot3.6\\% = 0.144\\\\\\text{E(single", "10 !}$ = $\\frac { 15 \\times 14 \\times 13 \\times 12 \\times 11}{5!}$ = 3003. There are 3003 ways of arranging a group such that 5 children accomodate each group. 2. ### In a lucky draw, 6 boxes are kept in which only 3 boxes contains gifts. In how many ways can the kids pick the gift boxes? Step 1 : given: n = 6, k = 3 Step 2 : Use the combination formula: nck = $\\frac{n!}{r! (n-k)!}$ = $\\frac{6!}{3! (6 - 3)!}$ = $\\frac{6!}{3! 3!}$ = $\\frac{6 \\times 5 \\times 4 \\times 3!}{3! 3!}$ = $\\frac{120}{36}$ = 20 ways.", "# Number of ways to choose numbers with repetition, order matters I wanted to ask if I got the answers to the following two (homework) questions right, and if not whether someone could give me a hint where I went wrong? We choose $10$ numbers with repetition, order matters, from $\\{ -9, -8, \\ldots ,0, \\ldots ,8,9 \\}$. 1. How many ways are there to do this when the sum of the last $3$ numbers are even? 2. How many ways are there to do this when exactly three numbers chosen are multiples of $3$? In (Q1.), I got $9^3+3 \\cdot 9 \\cdot 10^2$ total ways to choose the last three numbers (by counting the number of ways a sum of three ordered numbers can be even), and so my answer is $19^7 \\cdot (9^3+3 \\cdot 9 \\cdot 10^2)$. In (Q2.), I arrived at the answer: $12^7 \\cdot 7^3 \\cdot {8 \\choose 3}$ (first choose the $7$ numbers that aren't multiples of $3$, then choose the multiples of 3 and also locations to insert them twixt the first $7$). I hope this is considered a legitimate question! Thank you. - So your last factor is in fact $\\binom {10} 3$, not $\\binom 8 3$. Everything else is fine.", "the remaining 3 digits. So, that is $1\\cdot 3\\cdot 3 = 9$. Total 3-digit numbers with no repeated digits and only odd digits that are greater than 350: $36+9 = 45$. The 36 from the book does show up, but I am guessing your book is wrong. 5. ## Re: Permutations and Combinations Thanks for the help. 6. ## Re: Permutations and Combinations Hello, Fratricide! 1. Seven people are to be seated in a row. Calculate the number of ways in which this can be done so that two particular persons, A and B, always have exactly one of the others between them. We have: $A\\_\\;\\!B$ or $B\\_\\;\\!A$ . . . 2 choices. There are 5 choices for the \"middle\" person: $A\\,\\!X\\!B$ Now we have 5 \"people\" to permute: . $\\left\\{\\boxed{A\\,\\!X\\!B},\\, C,\\,D,\\,E,\\,F\\right\\}$ There are 5! permutations. Therefore, there are: . $2\\cdot5\\cdot 5! \\:=\\:1200$ ways. 2. Determine the number of possible permutations (from digits 1 through to 9) when the digits 1 and 2 are together, but not necessarily in that order. Duct-tape the \"1\" and \"2\" together. There are 2 possible orders: $12$ or $21.$ We have 8 \"numbers\" to permutate: . $\\left\\{\\boxed{12},\\,3,\\,4,\\,5,\\,6,\\,7,\\,8,\\,9 \\right\\}$ There are 8! ways. Therefore, there are: . $2\\cdot8! \\:=\\:80,640$ permutations. 3. There are 10 chairs in a row: a) In how many ways", "so the answer is $$5!$$. Each of these orders corresponds to two different orders in which $$A$$ and $$B$$ are adjacent, depending on whether $$A$$ or $$B$$ is first. So the $$6!$$ count is too high by $$2\\cdot5!$$ and the count we seek is $$6!-2\\cdot 5!=4\\cdot5!$$.", "number of possible options for that slot. #### Guided Practice 1. There are 20 hockey players on a pro NHL team, two of which are goalies. In how many different ways can 5 skaters and 1 goalie be on the ice at the same time? 2. In how many different ways could you score a 70% on a 10 question test where each question is weighted equally and is either right or wrong? 3. How many different 4 digit ATM passwords are there? Assume you can repeat digits. 1. The question asks for how many on the ice, implying that order does not matter. This is combination problem with two combinations. You need to choose 1 goalie out of a possible of 2 and choose 5 skaters out of a possible 18. $\\dbinom{2}{1}\\dbinom{18}{5}=2 \\cdot \\frac{18!}{5! \\cdot 13!}=17136$ 2. The order of the questions you got right does not matter, so this is a combination problem. $\\dbinom{10}{7}=\\frac{10!}{7!3!}=120$ 3. Order does matter. There are 10 digits and repetition is allowed. You can use a decision chart for each of the four options. $10 \\cdot 10 \\cdot 10 \\cdot 10=10,000$ #### Practice Simplify each of the following expressions so that they do not have a factorial symbol. 1. $\\frac{7!}{3!}$ 2. $\\frac{110!}{105!5!}$ 3. $\\frac{52!}{49!}$ 4. In how many ways can you choose", "to divide by 2 three times. More specifically, if you enumerate them, you'll be counting each of these combinations twice. If you doubt this, you could perhaps enumerate them for say 3 numbers instead of 10. Anyway, effectively you can only shuffle in $4! / 2^3 = 3$ ways. $q = {40 \\cdot 3 \\cdot 36 \\cdot 3 \\times 3 \\over 40 \\cdot 39 \\cdot 38 \\cdot 37}$ $q / p = 162$ Last edited: #### veronica1999 ##### Member Ah yes, you also forgot to take into account that you need a 2nd slip with the same number as the first for 3 choices, and also a 2nd slip with the same number as the second. I can see that this would be confusing, so let's take this apart a bit further. You have ignored ordering from the start, and you have tried to compensate for that. Let's keep the ordering. That is, let's assuming each slip can be distinguished. So we have 1a, 1b, 1c, 1d as the slips with the number 1. It means you have a total of 40 x 39 x 38 x 27 choices. If the 4 slips are the same, you have 40 choices for the first slip, 3 for the second, 2 for the third, and 1 for the last. This", "“...no two elements in each list are the same.” Hence the permutation formula cannot be used. However, the rule of products still applies. We have any one of five choices for the first digit, five choices for the second, and five for the third. So there are $$5 \\cdot 5\\cdot 5 = 125$$ possible different three-digit numbers if repetition is allowed. Solution to (c): Again, the rule of products applies here. We have any one of five choices for the first digit, but then for the next two digits we have four choices since we are not allowed to repeat the previous digit So there are $$5 \\cdot 4\\cdot 4 = 80$$ possible different three-digit numbers if only non-consecutive repetitions are allowed. ### Exercises2.2.2Exercises #### 1. If a raffle has three different prizes and there are 1,000 raffle tickets sold, how many different ways can the prizes be distributed? $$P(1000,3)$$ #### 2. 1. How many three-digit numbers can be formed from the digits 1, 2, 3 if no repetition of digits is allowed? List the three-digit numbers. 2. How many two-digit numbers can be formed if no repetition of digits is allowed? List them. 3. How many two-digit numbers can be obtained if repetition is allowed? #### 3. How many eight-letter words can be formed from the 26", "Question #52c10 2 Answers Feb 13, 2018 depends on what you're doing with them Explanation: if you are using different components of BEDMAS, then the order of the numbers is important. $6 + 5 - 8 \\cdot 2 = - 5$ $2 + 6 - 5 \\cdot 8 = - 32$ $8 + 2 - 6 \\cdot 5 = - 20$ and so on Feb 13, 2018 Order of operations matters. Explanation: Take for example: $3 \\times 2 + 6 = 12$ You do multiplication first, then addition. So if we just change the numbers, you get something like this: $6 \\times 2 + 3 = 15$ Order of operations matters!" ]

[ "\\binom{5}{1} \\cdot 5 \\cdot 5^4 = 4 \\binom{5}{1} 5^5$$ Case 2: The leading digit is odd. Two of the remaining five positions must be filled with even digits. There are $\\binom{5}{2}$ ways to select the positions of the even digits and five ways to fill each of those positions with an even digit. There are five ways to fill each of the four other positions with an odd digit. $$\\binom{5}{2} \\cdot 5^2 \\cdot 5^4 = \\binom{5}{2} 5^6$$ Total: Since the above cases are mutually exclusive and exhaustive, the number of six-digit positive integers that contain exactly two even digits is $$4 \\binom{5}{1} 5^5 + \\binom{5}{2} 5^6$$ In how many five-digit positive integers are there digits that appear more than once? Hint: Subtract the number of five-digit positive integers in which all the digits are distinct from the total number of five-digit positive integers. The total number of five-digit positive integers is $$9 \\cdot 10 \\cdot 10 \\cdot 10 \\cdot 10 = 90000$$ since the leading digit can be chosen in nine ways (as $0$ is prohibited) and each of the remaining digits can be selected in $10$ ways. Alternatively, the largest five-digit number is $99999$ and the largest number with fewer than five digits is $9999$, so the number of five-digit positive integers is $$99999 - 9999 =", "at a time: $P(5,5) = \\frac{5!}{(5-5)!} = 5! = 120$ Example 2.2.11: Ordering of digits under different conditions. Consider only the digits 1, 2, 3, 4, and 5. (a) How many three-digit numbers can be formed if no repetition of digits can occur? (b) How many three-digit numbers can be formed if repetition of digits is allowed? (c) How many three-digit numbers can be formed if only non-consecutive repetition of digits are allowed? Solutions to (a): Solution 1: Using the rule of products. We have any one of five choices for digit one, any one of four choices for digit two, and three choices for digit three. Hence, 5 · 4 · 3 = 60 different three-digit numbers can be formed. Solution 2; Using the permutation formula. We want the total number of permutations of five digits taken three at a time: $P(5,3) = \\frac{5!}{(5-3)!} = 5 \\cdot 4 \\cdot 3 = 60$ Solution to (b): The definition of permutation indicates “...no two elements in each list are the same.” Hence the permutation formula cannot be used. However, the rule of products still applies. We have any one of five choices for the first digit, five choices for the second, and five for the third. So there are 5 · 5 · 5 = 125 possible different three-digit", "distinct bit strings can be formed from three 0’s and two 1’s? Solution: 5 total digits would give 5! permutations. But that is assuming the 0’s and 1’s are distinguishable (to make that explicit, let’s give each one a subscript). Here are the $3! \\cdot 2! = 12$ different ways that we could have arrived at the identical string \"01100\" if we thought of each 0 and 1 as unique. Since identical digits are indistinguishable, all the listed permutations are the same. For any given permutation, there are 3! ways of rearranging the 0’s and 2! ways of rearranging the 1’s (resulting in indistinguishable strings). We have over-counted. Using the formula for permutations of indistinct objects, we can correct for the over-counting: $$\\text{Total} = \\frac{5!}{3! \\cdot 2!} = \\frac{120}{6 \\cdot 2} = 10$$ Example: How many distinct orderings of characters are possible for the string \"MISSISSIPPI\"? Solution: In the case of the string \"MISSISSIPPI\", we should separate the characters into four distinct groups of indistinct characters: one \"M\", four \"I\"s, four \"S\"s, and two \"P\"s. The number of distinct orderings are: $$\\frac{11!}{1!4!4!2!} = 34,650$$ Example: Consider the 4-digit passcode smart-phone from before. How many distinct passcodes are possible if there are 3 smudges over 3 digits on the screen? Solution: One of 3 digits is repeated, but we", "formed with the digits 1, 2, 3, . . . , 9, if no digit is repeated in any number? pls help on the problem above, thanks. $N=5 \\cdot \\frac{8!}{(8-2)!}=5 \\cdot 8 \\cdot 7 =280$", "---> $$\\frac{4!}{2!}= 12$$ ---> 1523|5: ---> $$\\frac{4!}{2!}= 12$$ So, we need to check those numbers whether they are divisible by 3 or not: ---> 1+1+2+3+5= 12 (divisible by 3) ---> 1+2+2+3+5=13 (not divisible by 3) ---> 1+3+2+3+5= 14 (not divisible by 3) ---> 1+5+2+3+5= 16 (not divisible by 3) In total, there are 12 numbers, which are multiple of 15. Re: How many five-digit positive integers can be formed using each of the [#permalink] 31 May 2020, 01:35", "## There are more numbers that contain digit 9 than numbers that do not contain digit 9 Remember from the last post that the sum of inverses of all strictly positive integers converges to infinity: $S_1 = \\displaystyle\\lim_{n \\to \\infty}\\displaystyle\\sum_{i=1}^{n}\\frac{1}{i} = \\infty$ I also proved that the sum of inverses of all strictly positive integers that don’t contain the digit 9 is finite. Denote this sum $S_2$. Let $S_3$ be the sum of all strictly positive integers that contain the digit 9: $S_3=\\frac{1}{ 9}+\\frac{1}{ 19}+\\frac{1}{ 29}+\\cdots+\\frac{1}{ 90}+\\cdots+\\frac{1}{ 99}+\\frac{1}{ 109}+\\cdots$ Then: $S_1 = S_2 + S_3$ Thus $S_3 = S_1 - S_2$, and it follows that $S_3=\\infty$ because $S_1=\\infty$ and $S_2$ is finite. In words: the sum of inverses of strictly positive integers that don’t contain digit 9 is finite and the sum of inverses of strictly positive integers that contain digit 9 is infinity.  This may be counterintuitive because apparently there are more numbers that don’t contain digit 9 than numbers that contain digit 9, but this is not actually true. There are $9^n-1$ strictly positive integers up to $n$ digits that don’t contain digit 9 (see here). And obviously the total number of strictly positive integers is $10^n-1$. This means that the number of strictly positive integers that contain digit 9 is: $10^n-9^n$. And here’s the", "Eight digit number is formed using all the digits: $1,1,2,2,3,3,4,5$ Eight digit number is formed using all the digits: $1,1,2,2,3,3,4,5$. Find 1. Total numbers in which no two identical digits appear together. 2. Exactly two pair of identical digits occur together. For first part, I thought of finding number of ways such that all three numbers are always together and then subtract it from total numbers that can be formed $(\\frac{8!}{2! 2! 2!})$ but it didn't work. How should I proceed? One way is to use Inclusion/Exclusion. You attempted a form of it, but something more elaborate is needed. You know that there are $\\frac{8!}{2!2!2!}$ possible sequences if we have no restriction. Now let us count the number that have the two $1$'s together. Tie them together, and think of them as a superletter. Now we have $7$ \"letters.\" These can be arranged in $\\frac{7!}{2!2!}$ ways. We get the same count for the $2$'s together, and for the $3$'s together. So our first estimate for the required number is $\\frac{8!}{2!2!2!}-3\\cdot \\frac{7!}{2!2!}$. But we have subtracted too much, for we have subtracted one too many times the patterns that have two $1$'s and two $2$'s together, as well as the ones in which we have two $2$'s and two $3$'s together, and so on. There are $\\frac{6!}{2!}$ of each", "eliminated. This leaves us with $$2480 - 75 = 2405$$ four-digit strings which contain only repeated odd digits. However, not all these strings are four-digit positive integers since the first digit may be zero. We must eliminate these. Three-digit positive integers in which a single odd digit is used in all three positions: There are $5$ such numbers since there are five odd digits. Three-digit odd positive integers in which exactly two digits are used and an odd digit is repeated: Notice that we have already eliminated numbers in which a zero appears since it would be a repeated even digit (as zero also occupies the thousands place). There are five ways of choosing the repeated digit, $\\binom{3}{2}$ ways of selecting two of the three locations for that digit, and eight ways to fill the remaining digit. Hence, there are $$5 \\cdot \\binom{3}{2} \\cdot 8 = 120$$ such numbers. Hence, there are $$\\frac{10^4 - 10 \\cdot 9 \\cdot 8 \\cdot 7}{2} - \\frac{5 \\cdot 5 \\cdot \\binom{4}{2}}{2} - 5 - 5 \\cdot \\binom{3}{2} \\cdot 8 = 2480 - 75 - 5 - 120 = 2280$$ four-digit positive integers in which the only repeated digits are odd, which agrees with the result we obtained above. Thus, there are a total of $$4536 + 2280 = 6816$$ four-digit positive integers", "all positive integers that are less than or equal to a given positive integer. For instance$3! = 3 \\times 2 \\times 1 =6$ and $5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120$. We also define $0! = 1! = 1$. #### Factorial Rule The factorial of any integer $n$ can be calculated from the factorial of the previous integer, i.e., $n-1$. The factorial rule states that $n! = n \\times (n-1)!$ #### Fundamental Counting Principle The fundamental counting principle states that if one event can occur in $A$ different ways and a second event can occur in $B$ different ways then the total number of ways in which both events can occur is $A \\times B$. For example, in a restaurant, if you can order three different types of deserts, four different main courses, and two different appetizers then the total number of possible meal combinations that you can order is $3 \\times 4 \\times 2 = 24$. Using factorial and the fundamental counting principle, we can derive a formula for combinations as described below ### Combinations Formula The possible combinations of $n$ different objects taken $r \\leq n$ at time is given as $\\large{C(n,r) = \\frac{n!}{r!(n-r)!}}$ Derivation Suppose that we have $n$ distinct digits for a code taken $r$ at a time.", "# Tuples with 4 consecutive integers from 1 to 50 How many ways are there to pick an ordered tuple of 5 elements from the set of positive integers between 1 and 50 inclusive, provided that there must be at least one consecutive sequence of 4 consecutive elements? So what I think is somehow we need to use inclusion exclusion principle ; So I proceed by calculating sequence with 5 consecutive element but then I have no idea how to proceed with this? • You have ruined your own question with that edit. It is now completely unintelligible! – TonyK Oct 17 '19 at 11:46 • @TonyK Oh sorry for that. – maths student Oct 17 '19 at 11:51 There are 47 sets of 4 consecutive elements, from $$\\{1,2,3,4\\}$$ to $$\\{47,48,49,50\\}.$$ For each of these, we can choose a fifth element for our 5-tuple in 46 ways, and this fifth element can be placed in the first or last position in the 5-tuple. So far our count is $$47 \\cdot 46 \\cdot 2$$ Notice this count would include the 5-tuples with five consecutive elements as well; however each of these 5-tuples with 5 consecutive elements would have been counted twice, either by choosing as our fifth element the first or the last of the five consecutive numbers. [e.g", "+0 0 111 1 Suppose 5 different integers are randomly chosen from between 20 and 69, inclusive. What is the probability that they each have a different tens digit? Guest Jun 27, 2018 #1 +964 +2 From 20 to 69, they are a total of 50 integers. To choose 5 of them, we do: $$\\binom{50}{5}$$ Now, we need to find out the number of successful outcomes. We need an integer between 20 and 29 Then another one from 30 to 39, 40 to 49, 50 to 59, and 60 to 69. There are 10 numbers in each category, and therefore $$10^5$$ ways of doing this. Your final answer is: $$\\boxed{\\frac{2500}{52969}}$$ I hope this helped, Gavin GYanggg Jun 27, 2018", "My answer to a combination problem is different from the textbook answer. The question in full goes as follows: How many $5$-digit numbers can be formed from the integers $1,2,\\dots,9$ if no digit can appear more than twice? (For instance, $41434$ is not allowed.) I put my answer as the sum of three subsets (no repetition, one pair, and two pairs). The number of possibilities for no reps is $9\\cdot8\\cdot7\\cdot6\\cdot5$. The # of possibilities for one pair is $\\binom94$. If there's a pair, then there are $4$ distinct digits in the $5$-digit number. Multiply this by $5!/2$, since each digit will need a place, which is calculated by $5!$. Divide by $2$ to remove redundancy of the digits in the pair. Likewise, the third set goes $\\binom93\\cdot5!/(2\\cdot2)$. The sum of the three subsets being: $$9\\cdot8\\cdot7\\cdot6\\cdot5 + \\binom94\\cdot\\frac{5!}2 + \\binom93\\cdot\\frac{5!}{2\\cdot2}$$ The answer at the back of the book goes: There are $\\binom52\\cdot8\\cdot7\\cdot6$ numbers in which only one specified digit appears twice, so there are $9\\cdot\\binom52\\cdot8\\cdot7\\cdot6$ numbers in which only a single digit appears twice. There are $7\\cdot5!/(2!\\cdot2!)$ numbers in which two specified digits appear twice, so there are $\\binom92\\cdot7\\cdot5!/(2!\\cdot2!)$ numbers in which two digits appear twice. Thus, the answer is $$9\\cdot8\\cdot7\\cdot6\\cdot5 + 9\\binom52\\cdot8\\cdot7\\cdot6 + \\binom92\\cdot7\\cdot\\frac{5!}{2\\cdot2}$$ The answer makes sense to me, but mine doesn't seem wrong either. Both answers", "# Math Help - digits 1. ## digits Of all the four-digit positive integers containing only digits from the set {2, 4, 6, 8}, what fraction of them have at least one of their digits repeated? Express your answer as a common fraction. 2. Originally Posted by sri340 Of all the four-digit positive integers containing only digits from the set {2, 4, 6, 8}, what fraction of them have at least one of their digits repeated? Express your answer as a common fraction. Hi sri340, We can ask \"how many have no repeated digits\" ? This is 4! as all 4 digits must be different. There are $4^4$ 4-digit numbers given the digits may be repeated. Hence $4^4-4!$ contain at least one repeated digit. The fraction is $\\frac{4^4-4!}{4^4}=\\frac{4(4^3-6)}{4(4^3)}=\\frac{4^3-6}{4^3}$", "can we write the 5 digits 1, 1, 1, 2, 2\". That's really a \"binomial\" problem: it is $\\frac{5!}{3!2!}= 10$. One way of seeing that is to imagine that each of those had a subscript: 11, 12, 13, 21, 22 so that they are distinguishable. There are 5! ways of ordering those 5 symbols. Now we not that 3! ways of rearranging just the 3 \"1\"s so that for each of the ways of rearranging those 5 \"distinguishable\" symbols there are 5 more that are just rearranging the different \"1\"s. Since they are NOT distinguishable, we don't want to count those as 6 different ways so we divide by 3!= 6. The same thing happens with the \"2\"s. In order NOT to count as different orders that just swap the two \"2\"s, we divide by 2!= 2. thanks again for the info... it help a lot on my understanding regarding permutations... can you check my other post, which is \"probability\".", "as a concatenation of two primes, and thus the estimate for the number of ways to write $$n$$ as a concatenation of more primes will be greater. For any $$n \\in \\mathbb{N}$$ there are $$n-1$$ many ways to see $$n$$ as a concatenation of two numbers. For example the 5-digit number 75319 can be seen as: 7-5319, 75-319, 753-19, or 7531-9 (this is a random example, in the work below each part of the concatenation is a prime number). Let me show how to estimate the number of ways to write a 5-digit number as a concatenation of primes, then the reader can see how I got the estimate for an $$n$$-digit number. For a 5-digit number, the first concatenation type is a single digit number followed by a 4-digit number. There are 4 single digit prime numbers, and there are approximately $$\\frac{9^4}{\\ln(10^4)}$$ primes which are 4-digits long. Thus there are approximately $$\\frac{4 \\cdot 9^4}{\\ln(10^4)}$$ many 5-digit numbers which are of the first concatenation type. There are just as many of the concatenation type which is a 4-digit number followed by a single digit number. For the concatenation type of a 5-digit number which is a 2-digit number followed by a 3-digit number, there are approximately $$\\frac{9^2 \\cdot 9^3}{\\ln(10^2) \\cdot \\ln(10^3)}$$ many 5-digit numbers of this concatenation type.", "including the digit zero, which can be placed anywhere in the remaining positions. This choice can be done in $$P(6,3) = \\frac{6!}{(6-3)!} = \\frac{6!}{3!}=6 \\cdot5 \\cdot4$$, different ways. Finally, with distinct digits, there is $$6 \\cdot6 \\cdot5 \\cdot4 =720$$ four-digit numbers to be constructed. We know that the amount of odd numbers plus the amount of even numbers equal the total amount of the 720 four-digit numbers. The odd numbers are 1, 3 and 5. The question to be asked is how many of the 720 are odd? The digit to fill the unit position can only be chosen from the digits 1, 3 or 5, and this choice can be made in $$P(3,1)=\\frac{3!}{(3-1)!}=\\frac{3!}{2!}=3$$, different ways. To choose the digit filling the thousand position, we have five valid digits to choose from, since the zero digit is not valid for this position. The choice can be made in $$P(5,1)=\\frac{5!}{(5-1)!}=\\frac{5!}{4!}=5$$, different ways. Now we are left with two positions to fill, the hundred and tenth position, and five digits to choose from, now including the zero digit. The choice for these two positions can be made in $$P(5,2)=\\frac{5!}{(5-2)!}=\\frac{5!}{3!}=5 \\cdot4=20$$, different ways. The total number of odd four-digit numbers, with distinct digits are $$5 \\cdot5 \\cdot4 \\cdot3 =300$$. Now, we can answer the question how many of these 720, four-digit", "integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ? A. 400 B. 1728 C. 108 D. 216 E. 432 I am getting answer as 432, but the OA is D. Here is the approach I used. First number can be selected from 1 to 9 - in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1 so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks! XXYY can be arranged in $$\\frac{4!}{2!2!}=6$$ ways (# of arrangements of 4 letters out of which 2 X's and 2 Y's are identical); # of ways we can select 2 distinct digits out of 9 is $$C^2_9=36$$; Total # of integers that can be formed is 6*36=216. Hi Bunuel, I Want to know if the statement changes from \"How many four-digit positive integers can be formed by using the digits from 1 to 9\" to How many four-digit positive integers can be formed by using the", "[#permalink] ### Show Tags 25 May 2013, 22:03 Bunuel wrote: Ramsay wrote: Sorry guys, Could someone please explain the following: \"There are 8C3 ways to determine where to place the separators\" I'm not familiar with this shortcut/approach. Ta Consider this: we have 5 $$d$$'s and 3 separators $$|$$, like: $$ddddd|||$$. How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$d$$'s and 3 $$|$$'s are identical, so $$\\frac{8!}{5!3!}=56$$. With these permutations we'll get combinations like: $$|dd|d|dd$$ this would be 3 digit number 212 OR $$|||ddddd$$ this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR $$ddddd|||$$ this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)... Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5. Hence the answer is $$\\frac{8!}{5!3!}=56$$. Answer: C (56). This can be done with direct formula as well: The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case)", "How many positive integers less than 10,000 are there in which the sum of the digits equals 5? A) 31 B) 51 C) 56 D) 62 E) 93 Consider this: we have 5 $$d$$'s and 3 separators $$|$$, like: $$ddddd|||$$. How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$d$$'s and 3 $$|$$'s are identical, so $$\\frac{8!}{5!3!}=56$$. With these permutations we'll get combinations like: $$|dd|d|dd$$ this would be 3 digit number 212 OR $$|||ddddd$$ this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR $$ddddd|||$$ this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)... Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5. Hence the answer is $$\\frac{8!}{5!3!}=56$$. This can be done with direct formula as well: The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is $${n+r-1}_C_{r-1}$$. In our case we'll get: $${n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\\frac{8!}{5!3!}=56$$ _________________ Manager Joined: 27 Apr", "Question 64 # A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is Solution The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once. The different possibilities include : Case 1:The four digits are ( 2, 2, 2, 3). Since the number 2 is repeated 3 times. The total number of arrangements are : $$\\frac{4!}{3!}$$ = 4. Case 2: The four digits are 2, 2, 3, 3. The total number of four-digit numbers formed using this are : $$\\frac{4!}{2!\\cdot2!}=\\ 6$$ Case 3: The four digits are 2, 3, 3, 3. The number of possible 4 digit numbers are : $$\\frac{4!}{3!}$$ = 4 Case4: The four digits are 2, 3, 3, 1. The number of possible 4 digit numbers are : $$\\frac{4!}{2!}=\\ 12$$ Case5: Using the digits 2, 2, 3, 1. The number of possible 4 digit numbers are : $$\\frac{4!}{2!}=\\ 12$$ Case 6: Using the digits 2, 3, 1, 1. The number of possible 4 digit numbers are : $$\\frac{4!}{2!}=\\ 12$$ A total of 12 + 12 + 12 + 4 + 6 + 4 = 50" ]

[ "a complete answer: Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3 • Now try for the values $38116532895986727945334202400$ and $49378235797073715747364762200$ ;) – Servaes Nov 18 '15 at 19:02 • @Servaes, give me a minute. – Will Jagy Nov 18 '15 at 19:05", "# Rows in Pascal's Triangle containing Numbers in Arithmetic Progression ## Theorem There are an infinite number of rows of Pascal's triangle which contain $3$ integers in arithmetic progression. ## Examples ### $7, 21, 35$ The integers: $7, 21, 35$ are in arithmetic progression and appear in row $7$ of Pascal's triangle. ### $1001, 2002, 3003$ The integers: $1001, 2002, 3003$ are in arithmetic progression and appear in row $14$ of Pascal's triangle. ### $490 \\,314, 817 \\, 190, 1 \\, 144 \\, 066$ The integers: $490 \\, 314, 817 \\, 190, 1 \\, 144 \\, 066$ are in arithmetic progression and appear in row $23$ of Pascal's triangle. ## Historical Note This result, along with Rows in Pascal's Triangle containing Numbers in Geometric Progression and Rows in Pascal's Triangle containing Numbers in Harmonic Progression, was apparently published by Theodore Samuel Motzkin in Volume $12$ of Scripta Mathematica, but details have not been established.", "Back ## Polygons in a Row $100$ regular triangles with side length $1$ are arranged in a row, as shown below: What is the perimeter of the figure? What would the perimeter be if there were $x$ triangles?", "14641$$ $$11^{5} = 161051 = 1|5|10|10|5|1$$ (See Exploding Dots!) Any guesses as to why these powers appear as rows of Pascal’s triangle? ## Books Take your understanding to the next level with easy to understand books by James Tanton. BROWSE BOOKS ## Guides & Solutions Dive deeper into key topics through detailed, easy to follow guides and solution sets. BROWSE GUIDES ## Donations Consider supporting G'Day Math! with a donation, of any amount. Your support is so much appreciated and enables the continued creation of great course content. Thanks!", "multiples of $p$ (except for $k=0$ and $k=p$). In particular, all the inner entries in row 11 of Pascal's triangle are divisible by 11, and all the entries in row 31 are divisible by 31. Adding a multiple of $341=11\\times31$ won't affect this divisibility property for those two rows.", "# 1000th number in a triangle Number Theory Level 3 A triangle of integers contains 1000 rows numbered 1 to 1000. In row $$i,$$ the integers $$1$$ to $$i$$ (inclusive) are written in order. If you start reading the numbers from the first row until the end in order, what is the 1000th number that is read? ×", "# Book:Blaise Pascal/Traité du Triangle Arithmétique/Historical Note Blaise Pascal wrote this book in $1653$, but did not publish it until $1665$, as he had turned his attention away from mathematics in order to concentrate on religion. Some give it as $1663$, while others give it as $1655$.", "# Hexagonal Numbers In the diagram above, we see a series of triangles and hexagons. How many dots are contained in the $$20^\\text{th}$$ hexagon? × Problem Loading... Note Loading... Set Loading...", "of Pascal triangle...", "## Algebra and Trigonometry 10th Edition $(a+6)^4=a^4+24a^3+216a^2+864a+1296$ Fourth row of Pascal's Triangle: $1~~4~~6~~4~~1$ $(a+6)^4=(1)a^4+(4)a^3(6)+(6)a^2(6^2)+(4)a(6)^3+(1)(6)^4=a^4+24a^3+216a^2+864a+1296$", "cube above, the numbers from $1$ to $3^3=27$ are arranged in a cube and the sum of the numbers on each column and each row is $42$. » Read more 1 7 8 9", "– 0 = 5\\\\a_{36} = a_{35} + a_{26} – a_{20} = 5 + 1 – 0 = 6\\\\a_{37} = a_{36} + a_{27} – a_{21} = 6 + 0 – 1 = 5\\\\a_{38} = a_{37} + a_{28} – a_{22} = 5 + 0 – 1 = 4\\\\a_{39} = a_{38} + a_{29} – a_{23} = 4 + 0 – 1 = 3$ Row 4 is less trivial, the first nine values being: $a_{41} = a_{40} + a_{31} – a_{3^-5} = 0 + 1 – 0 = 1\\\\ a_{42} = a_{41} + a_{32} – a_{3^-4} = 1 + 2 – 0 = 3\\\\a_{43} = a_{42} + a_{33} – a_{3^-3} = 3 + 3 – 0 = 6\\\\a_{44} = a_{43} + a_{34} – a_{3^-2} = 6 + 4 – 0 = 10\\\\a_{45} = a_{44} + a_{35} – a_{3^-1} = 10 + 5 – 0 = 15\\\\a_{46} = a_{45} + a_{36} – a_{30} = 15 + 6 – 0 = 21\\\\a_{47} = a_{46} + a_{37} – a_{31} = 21 + 5 – 1 = 25\\\\a_{48} = a_{47} + a_{38} – a_{32} = 25 + 4 – 2 = 27\\\\a_{49} = a_{48} + a_{39} – a_{33} = 27 + 3 – 3 = 27$ We can continue in this fashion. ### Generalizing the Pattern If we rewrite the Pascal’s Triangle to be left-aligned instead", "Comment Share Q) # $40,41,45,54,70, ?$ $\\begin{array}{1 1} 93 \\\\ 99 \\\\ 90 \\\\ 95 \\end{array}$ The pattern of the given series is $40,41,45,54,70$ $+1,+4,+9,+16,+25$ Missing number is $70+5^2$ =>$70+25$ =>$95$", "Numbers in a Row If we write the positive integers in a row (see it below) what number be the 1.000.000 digit in the sequence? $$12345678910111213141516...$$ ×", "# Hexagonal Numbers In the diagram above, we see a series of triangles and hexagons. How many dots are contained in the $$20^\\text{th}$$ hexagon? ×", "## Algebra and Trigonometry 10th Edition $_4C_2=6$ Fourth row of Pascal's triangle: $1~~4~~6~~4~~1$ It corresponds to: $_4C_0~~_4C_1~~_4C_2~~_4C_3~~_4C_4$ So, $_4C_2=6$", "Problem #219 219 In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3: 4: 5$? This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "Answer Comment Share Q) # The A.M of set of 50 numbers is 48. If two numbers 52 and 44 are discarded then A.M of the remaining set of numbers is $\\begin {array} {1 1} (A)\\;50 & \\quad (B)\\;52 \\\\ (C)\\;48 & \\quad (D)\\;44 \\end {array}$", "= 4 \\times 70$, $448 = 16 \\times 28$ and $30 = 30 \\times 1$, where $70$, $28$ and $1$ are weights for the Reed-Muller code $RM(2,3)$. The $4$, $16$ and $30$ appear in the bottom row of the Pascal triangle below. The number 759 is the summit of this triangle, which counts the number of blocks that (i) contain a point set with $n$ elements and (ii) are disjoint from another set with $m$ elements. The union cardinality $m + n$ is fixed in each row. So there are $253$ occurrences of any given object, and a complementary $506$ blocks that do not contain it. There are $42504$ ways to choose $5$ objects from $24$, and $42504 = 759 \\times 56$. To obtain the $8$ bits, after choosing the first $5$, there are $969$ possibilities (namely $3$ out of $19$) and we see that $969 \\times 759$ is indeed the binomial coefficient $B(24,8) = 735471$. Why the redundancy factor of $56$, the FTS dimension? This counts the number of ways of choosing $3$ objects from $8$, namely the $3$ that are omitted from the set of $5$. The Mathieu group $M_{24}$ has a 729 dimensional permutation representation with suborbit lengths given by the $1$, $30$, $280$ and $448$. Similarly, the Mathieu group $M_{12}$ is associated to the", "= 46 \"g}$" ]

[ "multiples of $p$ (except for $k=0$ and $k=p$). In particular, all the inner entries in row 11 of Pascal's triangle are divisible by 11, and all the entries in row 31 are divisible by 31. Adding a multiple of $341=11\\times31$ won't affect this divisibility property for those two rows.", "# Pascal's triangle curiosity I noticed the following pattern in the rows of Pascal's triangle: $$1 = 11^0\\\\ 11 = 11^1\\\\ 121= 11^2\\\\ 1331= 11^3\\\\ 14641=11^4$$ at this point I thought maybe this pattern would follow indefinitely, but the next row is $$15101051=7\\times2157293$$. The next two rows aren't powers of 11 either, they are $$1615201561=43\\times 37562827$$ and $$172135352171=29\\times5935701799$$. Are the first 5 rows the only rows in the Pascal triangle that are powers of 11? The following row 18285670562881 is a prime (at least according to Linux factor function). How often primes appear in these rows? This might be fun to show or investigate. I am just curious. I just wrote a Python script to generate a few more rows of the Pascal triangle. Each row was also factored up to row n=19: from scipy.special import binom from sympy import factorint for n in range(5, 20): row = '' print('n=%d' % n) for k in range(n+1): binom_nk = int(binom(n, k)) row = row+str(binom_nk) print(row) row_number = int(row) factors=factorint(row_number) print(factors) n=5 15101051 {7: 1, 2157293: 1} n=6 1615201561 {43: 1, 37562827: 1} n=7 172135352171 {29: 1, 5935701799: 1} n=8 18285670562881 {18285670562881: 1} n=9 193684126126843691 {5647: 1, 34298587945253: 1} n=10 1104512021025221012045101 {13: 1, 197: 1, 4649: 1, 92768668286052709: 1} n=11 1115516533046246233016555111 {11: 1, 101410593913295112092414101: 1} n=12 1126622049579292479249522066121 {523: 1, 4637:", "# Does there exist $a,n \\in \\mathbb{Z}^+$, where $n \\geq 2$, such that $a, an, an^2,an^3,\\ldots,an^5$ are all palindromes in base 10? Question: Does there exist $a,n \\in \\mathbb{Z}^+$, where $n \\geq 2$, such that $$a, an, an^2,an^3,\\ldots,an^5$$ are all palindromes in base 10? We see that $a=1$ and $n=11$ give rise to $$1, 11, 121, 1331, 14641$$ which is basically the first few rows of Pascal's triangle. There are infinitely many other examples of length 5, which we can generate by generalising the Pascal's triangle construction, giving $$11, 1111, 112211, 11333311, 1144664411$$ $$111, 111111, 111222111, 111333333111, 111444666444111$$ and so on. Some other examples are: $$147741, 13444431, 1223443221, 111333333111, 10131333313101$$ $$1478741, 134565431, 12245454221, 1114336334111, 101404606404101$$ However, I checked for $a,n \\leq 10^7+1$, and didn't find any of length greater than 5. -", "of such set. - Here's a variation on the theme of Didier's answer. Each number in Pascal's triangle gets added twice to the row below it. The first $1$ below gets added to the next row to get the $1$ at the end, and also gets added to the next row to contribute to the $9$. Then the $8$ gets added to the next row to contribute to the $9$, and also gets added to the next row to contribute to the $36$. And so on. $$\\begin{array}{ccccccccccccccccccc} & 1 & & 8 & & 28 & & 56 & & 70 & & 56 & & 28 & & 7 & & 1 \\\\ \\\\ 1 & & 9 & & 36 & & 84 & & 126 & & 126 & & 84 & & 36 & & 9 & & 1 \\end{array}$$ Since each number is added twice to the next row, the sum of the numbers in the next row is twice as big. -", "= 16$$ - 4 years ago That is quite natural because rth element of nth row of pascals triangle is nCr and sum of all those elements is equivalent to binomial expansion of $${(1+1)}^{n}$$ or $${2}^{n}$$. Also, $${2}^{n}$$ is equivalent to number of subsets of a set and vertices of n dimensional hypercube. Look at the diagrams of lattices based on inclusion operation, They form hypercubes. Basically, pascal's triangle, hypercubes in n-dimensions, binomial coefficients, power sets and their lattice and recurrence relation [T(n)=1+Sum of all previous elements {with T(0)=1}] are all related. Maybe i'll post all of their relations that i know with corresponding images sometime later. - 4 years ago I know another property: If in the row $$n$$ the second number is prime, all numbers in this row except the 1s on the sides, will be multiples of this number. Examples: Row 7: $$1\\quad 7\\quad 21\\quad 35\\quad 35\\quad 21\\quad 7 \\quad1$$ 7 is a prime and 21 and 35 are multiples of 7. Row 17: $$1\\quad17 \\quad136 \\quad 680 \\quad 2380 \\quad 6188\\quad 12376 \\quad19448\\quad 24310\\quad 24310 \\quad19448 \\quad12376 \\quad 6188 \\quad2380 \\quad 680 136 \\quad 17 \\quad 1$$ 17 is prime and 136, 680, 2380, 6188, 12376, 19448 and 24310 are multiples of 17. - 4 years ago Cool - 4 years ago If you", "# Rows in Pascal's Triangle containing Numbers in Arithmetic Progression ## Theorem There are an infinite number of rows of Pascal's triangle which contain $3$ integers in arithmetic progression. ## Examples ### $7, 21, 35$ The integers: $7, 21, 35$ are in arithmetic progression and appear in row $7$ of Pascal's triangle. ### $1001, 2002, 3003$ The integers: $1001, 2002, 3003$ are in arithmetic progression and appear in row $14$ of Pascal's triangle. ### $490 \\,314, 817 \\, 190, 1 \\, 144 \\, 066$ The integers: $490 \\, 314, 817 \\, 190, 1 \\, 144 \\, 066$ are in arithmetic progression and appear in row $23$ of Pascal's triangle. ## Historical Note This result, along with Rows in Pascal's Triangle containing Numbers in Geometric Progression and Rows in Pascal's Triangle containing Numbers in Harmonic Progression, was apparently published by Theodore Samuel Motzkin in Volume $12$ of Scripta Mathematica, but details have not been established.", "# Primey Pascal's Triangle Imagine that we have a triangle that starts with 2,3 and grows like Pascal's triangle but instead uses the smallest prime $\\geq$ to the sum of the above two primes. Visually: $$\\begin{array}{ccccccccc} &&& 2 && 3 \\\\ && 2 && 5 && 3 \\\\ & 2 && 7 && 11 && 3 \\\\ 2 && 11 && 19 && 17 && 3 \\end{array}$$ Obviously, all primes will be seen in the first diagonal on the left side. However, if we neglect that diagonal, then there are some primes that never appear. The sequence starts: $5,7,13,43,73,103,109,\\dots$ One might notice that all of these primes are second members of twin prime pairs. They would be correct. In fact, the diagonal on the right side does capture every prime except for those that are the second prime of a twin prime pair because $p+3$ skips over $p+2$. However...there ARE some hold-outs. There are primes that do not appear in the main body of the triangle, but DO appear in the right side diagonal despite not being a twin prime. In other words, they appear only twice in the entire triangle while having composite odd neighbors. The first few are $23, 53, 79, 83, 131, 157, \\dots$. Is there something special about these primes that explains this curious", "each base exceeds by unity the sum of all the preceding bases. In other words, $2^{n} - 1 = 2^{n-1} + 2^{n-2} + ... + 1.$ There are well known sequences of numbers Some of those sequences are better observed when the numbers are arranged in Pascal's form where because of the symmetry, the rows and columns are interchangeable. The first row contains only $1$s: $1, 1, 1, 1, \\ldots$ The second row consists of all counting numbers: $1, 2, 3, 4, \\ldots$ The third row consists of the triangular numbers: $1, 3, 6, 10, \\ldots$ The fourth row consists of tetrahedral numbers: $1, 4, 10, 20, 35, \\ldots$ The fifth row contains the pentatope numbers: $1, 5, 15, 35, 70, \\ldots$ \"Pentatope\" is a recent term. Regarding the fifth row, Pascal wrote that ... since there are no fixed names for them, they might be called triangulo-triangular numbers. Pentatope numbers exists in the $4D$ space and describe the number of vertices in a configuration of $3D$ tetrahedrons joined at the faces. In the standard configuration, the numbers $C^{2n}_{n}$ belong to the axis of symmetry. Numbers $\\frac{1}{n+1}C^{2n}_{n}$ are known as Catalan numbers. Every two successive triangular numbers add up to a square: $(n - 1)n/2 + n(n + 1)/2 = n^{2}.$ Hockey Stick Pattern In Pascal's words (and", "# This question involves Pascal's Triangle & Binomial Theorem. Full Question 1. Factorize the expression $P(n)=n^x-n$ for $x=2,3,4,5$ Determine if the expression is always divisible by the corresponding $x$. If divisible use mathematical induction to prove your results by showing whether $P(k+1)-P(k)$ is always divisible by $x$. Using appropriate technology, explore more cases, summarize your results and make a conjecture for when $n^x-n$is divisible by$x$. (Not divisible by$\\ 4$. Thus I concluded that $x$ is prime.) 2.Explain how to obtain the entries in Pascal's Triangle, and using appropriate technology, generate the first 15 rows. State the relationship between the expression $P(k+1)-P(k)$ and Pascal's Triangle. Reconsider your conjecture and revise if necessary. (Here, my previous conjecture appeared incorrect, because $x$=prime numbers did not work out for 11 and 13 from the Triangle). And on this part I am experiencing a bit of trouble as I am saying that $x=2,3,5,7$. But this does not work for all $r$...I decided that $k$ is a multiple of $x$ when $x=2, 3, 5, 7, 9$. However, $r$ is different for each of the $x$ ... Write an expression for the xth row of Pascal's Triangle. You will have noticed that ($x$ choose $r)=k$, $k$ is $N$. Determine when $k$ is a multiple of $x$. 1. make conclusions regarding the last result in part", "$3$ will weigh up to a maximum of $\\frac 1 2 \\left({3^{n + 1} - 1}\\right)$. ### Integer as Sum of 5 Non-Zero Squares $33$: Any integer greater than $33$ can be written as the sum of $5$ non-zero squares. [ Jackson, Masat, Mitchell, MM v61 41 ] ### Triplets of Products of Two Distinct Primes $33$: The triplet $33, 34, 35$ is the smallest in which each number is the product of $2$ distinct primes. ... The next such triplets are: $93, 94, 95$; $141, 142, 143$; $201, 202, 203$; $213, 214, 215$; $217, 218, 219$; ... ### Prime Factors of 35, 36, 4734 and 4735 $35$: $35$ and $4375$ have the same prime factors between them (namely $2$, $3$, $5$ and $7$) as have $36$ and $4374$. ### Element of Pascal's Triangle is Sum of Diagonal or Column starting above it going Upwards $35$: In Pascal's Triangle, each number is the sum of either of the diagonals starting immediately above it, and taking the long way to the edge: for example, $35 = 15 + 10 + 6 + 3 + 1$. ### Hilbert-Waring Theorem/Particular Cases/5 $37$: Every number is the sum of at most $37$ $5$th powers. ### Euler Lucky Number/Examples/41 $41$: [ $x^2 + x + 41$ ] has many further prime values, including", "thing I saw was that $11^2 = 121$ which are numbers in (zero indexed) first and second row of the triangle as if they were just single numbers instead. Trying to find a pattern I tried to take the square of $121$ hoping to get $1331$ but it gave me $14641$ instead which is still adaptable for a pattern. What was also interesting is that I was able to get $1331$ by multiplying $121$ and $11$ and then I realized what I had was: \\begin{align} n = 0:& 11^n = 1 \\\\ n = 1:& 11^n = 11 \\\\ n = 2:& 11^n = 121 \\\\ n = 3:& 11^n = 1331 \\\\ n = 4:& 11^n = 14641 \\\\ \\end{align} Is pascal's triangle just numbers $11^n$? \\begin{align} n = 5:& 11^n = 161051 \\\\ \\end{align} So the answer is no, well at least in simple terms. The obvious problem with the $5^{th}$ row is that there are numbers with multiple digits and it's ambiguous how to concatenate them, hence the obvious answer is that what if we had no multiple digits that is to say digits themselves do not overflow and form the next ones. Luckily I was doing the calculations on paper like this: \\begin{array}{cccccc} 1 & 4 & 6 & 4 & 1 \\\\ \\times", "45$ The three integers are $41 , 43 \\mathmr{and} 45$", "a complete answer: Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3 • Now try for the values $38116532895986727945334202400$ and $49378235797073715747364762200$ ;) – Servaes Nov 18 '15 at 19:02 • @Servaes, give me a minute. – Will Jagy Nov 18 '15 at 19:05", "then so does $p^2$. Such ought exist, but the program found none under $20\\cdot 120^4$. The trick was to see if $L_{2p}=3 \\pmod{p^2}$, where $L_n$ is the nth lucas number. It did find the Heron number $103$, where a triangle $n-1, n, n+1$ is a triangle with integer area, then if $103$ divides any of these numbers, so does $103^2$. -", "3 = $2^{2^0} + 1$ 101 = 5 = $2^{2^1} + 1$ 10001 = 17 = $2^{2^2} + 1$ 10000001 = 257 = $2^{2^3} + 1$ 1000000000000001 = 65537 = $2^{2^4} + 1$ The numbers listed above are all prime. Based on this evidence Fermat conjectured that all numbers of the form $2^{2^n} + 1$ are prime. But Euler crushed this dream by showing that the next Fermat number, $2^{2^5} + 1,$ is not prime. Indeed, even today, no other Fermat numbers are known to be prime! People have checked all of them up to $2^{2^{32}} + 1.$ They’ve even checked a few bigger ones, the largest being $2^{2^{3329780}} + 1$ which turns out to be divisible by $193 \\times 2^{3329782} + 1$ Here are some much easier challenges: Puzzle 1. Show that every row of Pascal’s triangle mod 2 corresponds to a product of distinct Fermat numbers: 1 = 1 11 = 3 101 = 5 1111 = 15 = 3 × 5 10001 = 17 110011 = 51 = 3 × 17 1010101 = 85 = 5 × 17 11111111 = 255 = 3 × 5 × 17 100000001 = 257 and so on. Also show that every product of distinct Fermat numbers corresponds to a row of Pascal’s triangle mod 2. What is the pattern?", "# 2008 AIME I Problem 6: Pascal's triangle? A triangular array of numbers has a first row consisting of the odd integers $1,3,5,\\ldots,99$ in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples of $67$? Except for writing down some numbers, I wasn't sure how to proceed on this question. Does anyone have any ideas? • don't they usually have solutions on their website? – Christopher Marley Jun 6 '18 at 3:34 • Yes, but the solution didn't make too much sense to me, so I figured I'd post it here – Dude156 Jun 6 '18 at 3:38 • In this case, you may want to post a link to the solution, and highlight which parts you are unable to understand. Of course, if the entire approach, while being correct, seems to come out of the blue, you may request a more intuitive understanding of the answer. – астон вілла олоф мэллбэрг Jun 6 '18 at 3:40 • no, why does he have to post other solutions? just because other", "that and I don't think it applies to me. – Max Apr 26 '17 at 2:11 Assuming you do not need rigorous, mathematical justification, you can simply do it in the following way: First, find $n = \\dfrac{10^9}{7}=142857142$. Then there will be in total of $T_n = \\dfrac{n(n+1)}{2}$ triangular block of consisting of $1's$. As you demonstrated in your code, each of those block has exactly $T_6 = 21$ $1's$, so there are $N = 21T_n$ numbers that are divisible by $7$ in the first $10^9$ rows of the Pascal triangle. Once you have found this, the answer to actual problem will simply be $1+2+...+10^9 - N$, obviously. Last but not least, notice that how the only rows that has no $1's$ are the rows numbered $6, 13, 20,...$. In fact, this pattern continues and $10^9$ th row will precisely be another zero row because $10^9= -1\\mod 7.$ Therefore, you do not need to worry about half triangle appearing at the very bottom. I am certain that the number $10^9$ was chosen for this particular purpose. One can prove all these assertions mathematically, but you need at least Lucas's theorem, or some equivalent machinery which in general assumes a very decent knowledge of number theory. The author of that article you mentioned did a very poor job of", "figure, shows an interesting pattern formed with the numbers those are placed diagonally. The first diagonal is just only 1s. The second diagonal has counting numbers i.e. 1, 2, 3, 4, 5, 6, 7 etc., which is a sequence of integers. The third diagonal of Pascal’s triangle has sequence of numbers 1, 3, 6, 10, 15, 21 etc. This series of numbers is also called triangular numbers. The fourth diagonal is also an interesting sequence of numbers 1, 4, 10, 20, 35 etc., which are called as tetrahedral numbers. ## Number pattern on rows of Pascal’s triangle The sum of numbers on rows of Pascal’s triangle makes up the power of 2. Pascal's triangle squares Let’s understand it from the above figure. $$1^{st}$$ row: $$1 \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\; = \\; \\; \\; \\; 1 = 1^1$$ $$2^{nd}$$ row: $$1 + 1 \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; = \\; \\; \\; \\;2 = 2^1$$ $$3^{rd}$$ row: $$1 + 2 + 1 \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\;", "$p$ except the endpoints. Divisiblity by $p$ is then automatic for all entries in a triangle descending and travelling inwards. Let me paste a picture and then identify an example... Well, here is the easy one to see: look at all the entries in rows 5,6,7,8 that are divisible by $5,$ a kind of upside down triangle with apex at $70,$ which is $8$ choose $4.$ • \"Divisibility by $p$ is then automatic for all entries in a triangle descending and traveling inwards.\" I don't understand this part. Please explain. – Janitha357 Jan 19 '17 at 3:30 • @Janitha357 in row 5, entries 1,2,3,4 are 5, 10, 10, 5. In row 6, entries 2,3,4 are 15, 20, 15. In row 7, entries 3,4 are 35, 35. Finally, in row 8, entry 4 is 70. – Will Jagy Jan 19 '17 at 3:35 • @Janitha357 a similar argument shows that there are three consecutive entries in row 14, those being 1001, 2002, 3003. That is, all three are divisible by 7, 11, and 13. The 7 works because 14 = 2 * 7 and we are not using the central position, while 7 is an odd prime. – Will Jagy Jan 19 '17 at 3:37 • Thank you. This is really nice. – Janitha357 Jan 19 '17 at 3:55", "a composite number. Why is 43 a prime number? Number 43 is prime because it doesn’t have proper factors. In other words, the only factors of 43 are 1 and itself. To be sure of it, we can use the following property. If 𝒏 is a natural number, and neither of the prime numbers less than $$\\sqrt n$$ divides 𝒏, then 𝒏 is a prime number. Notice that 43<49, thus $$\\sqrt{43}\\;<\\;\\sqrt{49}\\;=\\;7$$. Therefore, the prime umbers less than $$\\sqrt{43}$$ are 2, 3 and 5. Since 43 isn’t an even number, 2 doesn’t divide 43. Moreover, $$43 = (3 \\times 14) + 1$$ and $$43 = (5 \\times 8) + 3$$, meaning that neither 3 nor 5 divides 43. Then, by the property above, 43 is a prime number. On the other hand, a prime number of objects can’t be arranged into a rectangular grid with more than one column and more than one row. This is another way of verifying that 43 is a prime number: • For example, if we try to arrange 43 stars into a rectangular grid with four rows, one of the columns will be incomplete. The same happens if we try to arrange 43 stars into a rectangular grid with any number of rows and columns greater than one. • The only way of" ]

[ "= 16$$ - 4 years ago That is quite natural because rth element of nth row of pascals triangle is nCr and sum of all those elements is equivalent to binomial expansion of $${(1+1)}^{n}$$ or $${2}^{n}$$. Also, $${2}^{n}$$ is equivalent to number of subsets of a set and vertices of n dimensional hypercube. Look at the diagrams of lattices based on inclusion operation, They form hypercubes. Basically, pascal's triangle, hypercubes in n-dimensions, binomial coefficients, power sets and their lattice and recurrence relation [T(n)=1+Sum of all previous elements {with T(0)=1}] are all related. Maybe i'll post all of their relations that i know with corresponding images sometime later. - 4 years ago I know another property: If in the row $$n$$ the second number is prime, all numbers in this row except the 1s on the sides, will be multiples of this number. Examples: Row 7: $$1\\quad 7\\quad 21\\quad 35\\quad 35\\quad 21\\quad 7 \\quad1$$ 7 is a prime and 21 and 35 are multiples of 7. Row 17: $$1\\quad17 \\quad136 \\quad 680 \\quad 2380 \\quad 6188\\quad 12376 \\quad19448\\quad 24310\\quad 24310 \\quad19448 \\quad12376 \\quad 6188 \\quad2380 \\quad 680 136 \\quad 17 \\quad 1$$ 17 is prime and 136, 680, 2380, 6188, 12376, 19448 and 24310 are multiples of 17. - 4 years ago Cool - 4 years ago If you", "# Rows in Pascal's Triangle containing Numbers in Arithmetic Progression ## Theorem There are an infinite number of rows of Pascal's triangle which contain $3$ integers in arithmetic progression. ## Examples ### $7, 21, 35$ The integers: $7, 21, 35$ are in arithmetic progression and appear in row $7$ of Pascal's triangle. ### $1001, 2002, 3003$ The integers: $1001, 2002, 3003$ are in arithmetic progression and appear in row $14$ of Pascal's triangle. ### $490 \\,314, 817 \\, 190, 1 \\, 144 \\, 066$ The integers: $490 \\, 314, 817 \\, 190, 1 \\, 144 \\, 066$ are in arithmetic progression and appear in row $23$ of Pascal's triangle. ## Historical Note This result, along with Rows in Pascal's Triangle containing Numbers in Geometric Progression and Rows in Pascal's Triangle containing Numbers in Harmonic Progression, was apparently published by Theodore Samuel Motzkin in Volume $12$ of Scripta Mathematica, but details have not been established.", "# Does there exist $a,n \\in \\mathbb{Z}^+$, where $n \\geq 2$, such that $a, an, an^2,an^3,\\ldots,an^5$ are all palindromes in base 10? Question: Does there exist $a,n \\in \\mathbb{Z}^+$, where $n \\geq 2$, such that $$a, an, an^2,an^3,\\ldots,an^5$$ are all palindromes in base 10? We see that $a=1$ and $n=11$ give rise to $$1, 11, 121, 1331, 14641$$ which is basically the first few rows of Pascal's triangle. There are infinitely many other examples of length 5, which we can generate by generalising the Pascal's triangle construction, giving $$11, 1111, 112211, 11333311, 1144664411$$ $$111, 111111, 111222111, 111333333111, 111444666444111$$ and so on. Some other examples are: $$147741, 13444431, 1223443221, 111333333111, 10131333313101$$ $$1478741, 134565431, 12245454221, 1114336334111, 101404606404101$$ However, I checked for $a,n \\leq 10^7+1$, and didn't find any of length greater than 5. -", "# Pascal's triangle curiosity I noticed the following pattern in the rows of Pascal's triangle: $$1 = 11^0\\\\ 11 = 11^1\\\\ 121= 11^2\\\\ 1331= 11^3\\\\ 14641=11^4$$ at this point I thought maybe this pattern would follow indefinitely, but the next row is $$15101051=7\\times2157293$$. The next two rows aren't powers of 11 either, they are $$1615201561=43\\times 37562827$$ and $$172135352171=29\\times5935701799$$. Are the first 5 rows the only rows in the Pascal triangle that are powers of 11? The following row 18285670562881 is a prime (at least according to Linux factor function). How often primes appear in these rows? This might be fun to show or investigate. I am just curious. I just wrote a Python script to generate a few more rows of the Pascal triangle. Each row was also factored up to row n=19: from scipy.special import binom from sympy import factorint for n in range(5, 20): row = '' print('n=%d' % n) for k in range(n+1): binom_nk = int(binom(n, k)) row = row+str(binom_nk) print(row) row_number = int(row) factors=factorint(row_number) print(factors) n=5 15101051 {7: 1, 2157293: 1} n=6 1615201561 {43: 1, 37562827: 1} n=7 172135352171 {29: 1, 5935701799: 1} n=8 18285670562881 {18285670562881: 1} n=9 193684126126843691 {5647: 1, 34298587945253: 1} n=10 1104512021025221012045101 {13: 1, 197: 1, 4649: 1, 92768668286052709: 1} n=11 1115516533046246233016555111 {11: 1, 101410593913295112092414101: 1} n=12 1126622049579292479249522066121 {523: 1, 4637:", "multiples of $p$ (except for $k=0$ and $k=p$). In particular, all the inner entries in row 11 of Pascal's triangle are divisible by 11, and all the entries in row 31 are divisible by 31. Adding a multiple of $341=11\\times31$ won't affect this divisibility property for those two rows.", "# Primey Pascal's Triangle Imagine that we have a triangle that starts with 2,3 and grows like Pascal's triangle but instead uses the smallest prime $\\geq$ to the sum of the above two primes. Visually: $$\\begin{array}{ccccccccc} &&& 2 && 3 \\\\ && 2 && 5 && 3 \\\\ & 2 && 7 && 11 && 3 \\\\ 2 && 11 && 19 && 17 && 3 \\end{array}$$ Obviously, all primes will be seen in the first diagonal on the left side. However, if we neglect that diagonal, then there are some primes that never appear. The sequence starts: $5,7,13,43,73,103,109,\\dots$ One might notice that all of these primes are second members of twin prime pairs. They would be correct. In fact, the diagonal on the right side does capture every prime except for those that are the second prime of a twin prime pair because $p+3$ skips over $p+2$. However...there ARE some hold-outs. There are primes that do not appear in the main body of the triangle, but DO appear in the right side diagonal despite not being a twin prime. In other words, they appear only twice in the entire triangle while having composite odd neighbors. The first few are $23, 53, 79, 83, 131, 157, \\dots$. Is there something special about these primes that explains this curious", "each base exceeds by unity the sum of all the preceding bases. In other words, $2^{n} - 1 = 2^{n-1} + 2^{n-2} + ... + 1.$ There are well known sequences of numbers Some of those sequences are better observed when the numbers are arranged in Pascal's form where because of the symmetry, the rows and columns are interchangeable. The first row contains only $1$s: $1, 1, 1, 1, \\ldots$ The second row consists of all counting numbers: $1, 2, 3, 4, \\ldots$ The third row consists of the triangular numbers: $1, 3, 6, 10, \\ldots$ The fourth row consists of tetrahedral numbers: $1, 4, 10, 20, 35, \\ldots$ The fifth row contains the pentatope numbers: $1, 5, 15, 35, 70, \\ldots$ \"Pentatope\" is a recent term. Regarding the fifth row, Pascal wrote that ... since there are no fixed names for them, they might be called triangulo-triangular numbers. Pentatope numbers exists in the $4D$ space and describe the number of vertices in a configuration of $3D$ tetrahedrons joined at the faces. In the standard configuration, the numbers $C^{2n}_{n}$ belong to the axis of symmetry. Numbers $\\frac{1}{n+1}C^{2n}_{n}$ are known as Catalan numbers. Every two successive triangular numbers add up to a square: $(n - 1)n/2 + n(n + 1)/2 = n^{2}.$ Hockey Stick Pattern In Pascal's words (and", "– 0 = 5\\\\a_{36} = a_{35} + a_{26} – a_{20} = 5 + 1 – 0 = 6\\\\a_{37} = a_{36} + a_{27} – a_{21} = 6 + 0 – 1 = 5\\\\a_{38} = a_{37} + a_{28} – a_{22} = 5 + 0 – 1 = 4\\\\a_{39} = a_{38} + a_{29} – a_{23} = 4 + 0 – 1 = 3$ Row 4 is less trivial, the first nine values being: $a_{41} = a_{40} + a_{31} – a_{3^-5} = 0 + 1 – 0 = 1\\\\ a_{42} = a_{41} + a_{32} – a_{3^-4} = 1 + 2 – 0 = 3\\\\a_{43} = a_{42} + a_{33} – a_{3^-3} = 3 + 3 – 0 = 6\\\\a_{44} = a_{43} + a_{34} – a_{3^-2} = 6 + 4 – 0 = 10\\\\a_{45} = a_{44} + a_{35} – a_{3^-1} = 10 + 5 – 0 = 15\\\\a_{46} = a_{45} + a_{36} – a_{30} = 15 + 6 – 0 = 21\\\\a_{47} = a_{46} + a_{37} – a_{31} = 21 + 5 – 1 = 25\\\\a_{48} = a_{47} + a_{38} – a_{32} = 25 + 4 – 2 = 27\\\\a_{49} = a_{48} + a_{39} – a_{33} = 27 + 3 – 3 = 27$ We can continue in this fashion. ### Generalizing the Pattern If we rewrite the Pascal’s Triangle to be left-aligned instead", "of such set. - Here's a variation on the theme of Didier's answer. Each number in Pascal's triangle gets added twice to the row below it. The first $1$ below gets added to the next row to get the $1$ at the end, and also gets added to the next row to contribute to the $9$. Then the $8$ gets added to the next row to contribute to the $9$, and also gets added to the next row to contribute to the $36$. And so on. $$\\begin{array}{ccccccccccccccccccc} & 1 & & 8 & & 28 & & 56 & & 70 & & 56 & & 28 & & 7 & & 1 \\\\ \\\\ 1 & & 9 & & 36 & & 84 & & 126 & & 126 & & 84 & & 36 & & 9 & & 1 \\end{array}$$ Since each number is added twice to the next row, the sum of the numbers in the next row is twice as big. -", "a complete answer: Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3 • Now try for the values $38116532895986727945334202400$ and $49378235797073715747364762200$ ;) – Servaes Nov 18 '15 at 19:02 • @Servaes, give me a minute. – Will Jagy Nov 18 '15 at 19:05", "# This question involves Pascal's Triangle & Binomial Theorem. Full Question 1. Factorize the expression $P(n)=n^x-n$ for $x=2,3,4,5$ Determine if the expression is always divisible by the corresponding $x$. If divisible use mathematical induction to prove your results by showing whether $P(k+1)-P(k)$ is always divisible by $x$. Using appropriate technology, explore more cases, summarize your results and make a conjecture for when $n^x-n$is divisible by$x$. (Not divisible by$\\ 4$. Thus I concluded that $x$ is prime.) 2.Explain how to obtain the entries in Pascal's Triangle, and using appropriate technology, generate the first 15 rows. State the relationship between the expression $P(k+1)-P(k)$ and Pascal's Triangle. Reconsider your conjecture and revise if necessary. (Here, my previous conjecture appeared incorrect, because $x$=prime numbers did not work out for 11 and 13 from the Triangle). And on this part I am experiencing a bit of trouble as I am saying that $x=2,3,5,7$. But this does not work for all $r$...I decided that $k$ is a multiple of $x$ when $x=2, 3, 5, 7, 9$. However, $r$ is different for each of the $x$ ... Write an expression for the xth row of Pascal's Triangle. You will have noticed that ($x$ choose $r)=k$, $k$ is $N$. Determine when $k$ is a multiple of $x$. 1. make conclusions regarding the last result in part", "= 4 \\times 70$, $448 = 16 \\times 28$ and $30 = 30 \\times 1$, where $70$, $28$ and $1$ are weights for the Reed-Muller code $RM(2,3)$. The $4$, $16$ and $30$ appear in the bottom row of the Pascal triangle below. The number 759 is the summit of this triangle, which counts the number of blocks that (i) contain a point set with $n$ elements and (ii) are disjoint from another set with $m$ elements. The union cardinality $m + n$ is fixed in each row. So there are $253$ occurrences of any given object, and a complementary $506$ blocks that do not contain it. There are $42504$ ways to choose $5$ objects from $24$, and $42504 = 759 \\times 56$. To obtain the $8$ bits, after choosing the first $5$, there are $969$ possibilities (namely $3$ out of $19$) and we see that $969 \\times 759$ is indeed the binomial coefficient $B(24,8) = 735471$. Why the redundancy factor of $56$, the FTS dimension? This counts the number of ways of choosing $3$ objects from $8$, namely the $3$ that are omitted from the set of $5$. The Mathieu group $M_{24}$ has a 729 dimensional permutation representation with suborbit lengths given by the $1$, $30$, $280$ and $448$. Similarly, the Mathieu group $M_{12}$ is associated to the", "then so does $p^2$. Such ought exist, but the program found none under $20\\cdot 120^4$. The trick was to see if $L_{2p}=3 \\pmod{p^2}$, where $L_n$ is the nth lucas number. It did find the Heron number $103$, where a triangle $n-1, n, n+1$ is a triangle with integer area, then if $103$ divides any of these numbers, so does $103^2$. -", "$3$ will weigh up to a maximum of $\\frac 1 2 \\left({3^{n + 1} - 1}\\right)$. ### Integer as Sum of 5 Non-Zero Squares $33$: Any integer greater than $33$ can be written as the sum of $5$ non-zero squares. [ Jackson, Masat, Mitchell, MM v61 41 ] ### Triplets of Products of Two Distinct Primes $33$: The triplet $33, 34, 35$ is the smallest in which each number is the product of $2$ distinct primes. ... The next such triplets are: $93, 94, 95$; $141, 142, 143$; $201, 202, 203$; $213, 214, 215$; $217, 218, 219$; ... ### Prime Factors of 35, 36, 4734 and 4735 $35$: $35$ and $4375$ have the same prime factors between them (namely $2$, $3$, $5$ and $7$) as have $36$ and $4374$. ### Element of Pascal's Triangle is Sum of Diagonal or Column starting above it going Upwards $35$: In Pascal's Triangle, each number is the sum of either of the diagonals starting immediately above it, and taking the long way to the edge: for example, $35 = 15 + 10 + 6 + 3 + 1$. ### Hilbert-Waring Theorem/Particular Cases/5 $37$: Every number is the sum of at most $37$ $5$th powers. ### Euler Lucky Number/Examples/41 $41$: [ $x^2 + x + 41$ ] has many further prime values, including", "y) = 10^8 + (x+y)10^4 + xy$. I fiddled with this approach for a little while and didn’t get anywhere, so I thought, “hey, this is a contest math problem…if this approach seems laborious, then it’s probably the wrong approach…there has to be some ‘nice’ way of finding the factor…I must be missing some neat observation.” So I looked at the 9 digit number again and noticed that it amusingly looks a little like the 1, 4, 6, 4, 1 row of Pascal’s triangle…and then I realized how to do the problem because I know that the rows of Pascal’s triangle are the digits of powers of 11 (if you take a sufficiently high base, but here, 14641 is exactly $11^4$). That is, $104060401 = 101^4$. Therefore, $104060465 = 101^4 + 64 = 101^4 + 4(2^4)$, which is exactly in the form of one side of Sophie Germain‘s identity: $a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)$. Substituting $a = 101$ and $b= 2$, we find that $104060465 = (101^2 + 2(2^2) + 2(101)(2))(101^2 + 2(2^2) - 2(101)(2)) = (10613)(9805)$. Since we are told that there is a 5 digit prime factor, it must be 10613. Because this approach works so snugly and factoring directly would be tedious and time-consuming, I’m confident that", "# Sequences of sums of Pascal's triangle The sequence $$1,3,6,10,16,28,56,120,256,528,1056$$ is defined in OEIS as \"sum of every 4th entry of row n in Pascal's triangle, starting at \"n choose 2\"\". It satisfies the recurrence $$a(n) = 4a(n-1)-6a(n-2)+4a(n-3)$$ and can be explicitly calculated as $$\\sum_{k=0}^{\\lfloor\\frac{n+1}{2}\\rfloor}{n+1 \\choose 2k}\\frac{1-(-1)^k}{2}$$ I suspect that following sequences $$0,0,3,21,89,307,977,3031,9321,28479,86505,261615,788969,2375119$$ $$1,7,31,117,439,1729,7063,29201,120471,493617,2008567,8129265$$ $$0,0,10,122,906,5478,30274,160974,840602,4343142,22270674$$ also relate to sums of terms in Pascal's triangle but can't figure it out. For example, the first of above-mentioned sequences has minimal characteristic polynomial $$x^6 - 8x^5 + 27x^4 - 54x^3 + 70x^2 - 56x + 24$$ Can any relation with Pascal's triangle can be established for that sequences? Thanks! • Where do the three sequences come from? (How are you calculating them) – Andrew Szymczak Mar 16 '15 at 17:42 • By definition, I think a more intuitive explicit formula is $a(n) = {n \\choose 2} + {n \\choose 6} + {n \\choose 10} + \\ldots = \\sum_{k \\ge 0} {n \\choose 4k-2}$. Note the offset on the OEIS is wrong. With an offset 0 the sequence should start 0,0,1,3,... – Andrew Szymczak Mar 16 '15 at 18:59 • – Lucian Mar 16 '15 at 23:51 • @Lucian: can you please give an example? – user144765 Mar 17 '15 at 7:32 If you want to relate a sequence to Pascal's triangle,", "Numbers in Pascal's Triangle For each $n \\in \\mathbb{N}$ define $b(n)$ to be the smallest $m \\in \\mathbb{N}$ such that $n = {m \\choose k}$ for some $k$, i.e. $b(n)$ is the first row in Pascal's triangle containing $n$. Question, what are the $n$ such that $b(n) = n$? Ideas. I think that for any prime $p$, $b(p) = p$. I think it follows from the fact that $p$ is prime if and only if ${p \\choose k} \\equiv 0 \\pmod p$ for all $0 < k < p$. I also guess that powers of primes $n = p^k$ also satisfy $b(n) = n$ but not sure how to prove this. Other numbers satisfy the property by inspection, for instance $12$ and $14$ so I wonder what properties are shared by all these numbers. Perhaps it may be easier to consider for which $n$, $b(n) < n$. Any references to related results would be greatly appreciated. • Numbers not satisfying $b(n)=n$ are $\\{6,10,15,20,21,28,35,36,\\cdots\\}$. See oeis.org for some references. Apr 24 '17 at 9:36 • Your guesses are correct, if $n$ appears before row $n$ it will have the form $\\binom{m}{k}$ where $2\\le m\\le m-2$ so $n$ will be possible to factor into at least two distinct factors which excludes $p^l$ if $p$ is prime. Apr 24 '17 at", "3 = $2^{2^0} + 1$ 101 = 5 = $2^{2^1} + 1$ 10001 = 17 = $2^{2^2} + 1$ 10000001 = 257 = $2^{2^3} + 1$ 1000000000000001 = 65537 = $2^{2^4} + 1$ The numbers listed above are all prime. Based on this evidence Fermat conjectured that all numbers of the form $2^{2^n} + 1$ are prime. But Euler crushed this dream by showing that the next Fermat number, $2^{2^5} + 1,$ is not prime. Indeed, even today, no other Fermat numbers are known to be prime! People have checked all of them up to $2^{2^{32}} + 1.$ They’ve even checked a few bigger ones, the largest being $2^{2^{3329780}} + 1$ which turns out to be divisible by $193 \\times 2^{3329782} + 1$ Here are some much easier challenges: Puzzle 1. Show that every row of Pascal’s triangle mod 2 corresponds to a product of distinct Fermat numbers: 1 = 1 11 = 3 101 = 5 1111 = 15 = 3 × 5 10001 = 17 110011 = 51 = 3 × 17 1010101 = 85 = 5 × 17 11111111 = 255 = 3 × 5 × 17 100000001 = 257 and so on. Also show that every product of distinct Fermat numbers corresponds to a row of Pascal’s triangle mod 2. What is the pattern?", "## Fermat Primes and Pascal’s Triangle If you take the entries Pascal’s triangle mod 2 and draw black for 1 and white for 0, you get a pleasing pattern: The $2^n$th row consists of all 1’s. If you look at the triangle consisting of the first $2^n$ rows, and take the limit as $n \\to \\infty,$ you get a fractal called the Sierpinski gasket. This can also be formed by repeatedly cutting triangular holes out of an equilateral triangle: Something nice happens if you interpret the rows of Pascal’s triangle mod 2 as numbers written in binary: 1 = 1 11 = 3 101 = 5 1111 = 15 10001 = 17 110011 = 51 1010101 = 85 11111111 = 255 100000001 = 257 Notice that some of these rows consist of two 1’s separated by a row of 0’s. These give the famous ‘Fermat numbers‘: 11 = 3 = $2^{2^0} + 1$ 101 = 5 = $2^{2^1} + 1$ 10001 = 17 = $2^{2^2} + 1$ 10000001 = 257 = $2^{2^3} + 1$ 1000000000000001 = 65537 = $2^{2^4} + 1$ The numbers listed above are all prime. Based on this evidence Fermat conjectured that all numbers of the form $2^{2^n} + 1$ are prime. But Euler crushed this dream by showing that the next Fermat number, $2^{2^5}", "figure, shows an interesting pattern formed with the numbers those are placed diagonally. The first diagonal is just only 1s. The second diagonal has counting numbers i.e. 1, 2, 3, 4, 5, 6, 7 etc., which is a sequence of integers. The third diagonal of Pascal’s triangle has sequence of numbers 1, 3, 6, 10, 15, 21 etc. This series of numbers is also called triangular numbers. The fourth diagonal is also an interesting sequence of numbers 1, 4, 10, 20, 35 etc., which are called as tetrahedral numbers. ## Number pattern on rows of Pascal’s triangle The sum of numbers on rows of Pascal’s triangle makes up the power of 2. Pascal's triangle squares Let’s understand it from the above figure. $$1^{st}$$ row: $$1 \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\qquad \\; = \\; \\; \\; \\; 1 = 1^1$$ $$2^{nd}$$ row: $$1 + 1 \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; = \\; \\; \\; \\;2 = 2^1$$ $$3^{rd}$$ row: $$1 + 2 + 1 \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\; \\;" ]

[ "# AMC 12 questions #1 Probability Level 4 On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots? ×", "# Problem #1150 1150 On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots? $(\\mathrm {A}) \\ \\frac{5}{11} \\qquad (\\mathrm {B}) \\ \\frac{10}{21} \\qquad (\\mathrm {C})\\ \\frac{1}{2} \\qquad (\\mathrm {D}) \\ \\frac{11}{21} \\qquad (\\mathrm {E})\\ \\frac{6}{11}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "of the die. B is an Event: getting a perfect square on the upper surface of the die. view solution All the black face cards are removed from a pack of 52 playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a: (i) face card (ii) red card (iii) black card (iv) king view solution #### Reference Material Solution for concept: Probability Examples and Solutions. For the course 8th-10th CBSE S", "die is rolled (faces 1 and 6 have probability $$\\frac{1}{4}$$ each, and faces 2, 3, 4, 5 have probability $$\\frac{1}{8}$$ each). Find the mean and standard deviation of the die score. 1. $$\\frac{7}{2}$$ 2. $$1.8634$$", "# Why is the roll of a die considered random? I've been reading articles on pseudo-randomness in computing when generating a random value. They all state that the generated numbers are pseudo-random because we know all the factors that influence the outcome, and that the roll of a die is considered truly random. But I'm wondering why. Don't we know all the physical forces that influence the die when it's being rolled? Or is there too many of them? • Can you accurately predict every single nerve pulse running to the muscles controlling the hand that rolls the die? May 20, 2017 at 9:45 • \"It is shown that for the realistic values of the initial energy the probabilities of the die landing on the face which is the lowest one at the beginning is larger than the probabilities of landing on any other face. [...] In real experiment, the predictability is possible only for very small \\epsilon, i.e., an accuracy which in practice is extremely difficult to implement[...]\" -- The three-dimensional dynamics of the die throw, Kapitaniak, Strzalko, Grabski, Kapitaniak. May 20, 2017 at 22:08 • To go with all the great answers here, I'd like to point out your own phrasing, \"... the roll of a die is considered truly random.\" Whether or not there is", "drawing a ticket from a bag with your eyes closed are random and fair since each of the possibilities is equally likely. But the probability of things a little more complicated, like the sum of two dice, is not so easy---and that makes it fun! Suppose we have a single die. It has six faces, and as far as we know each is equally likely to be face up if we roll the die. We can now use our basic rule for computing probability to compute the probability of each possible outcome. The basic rule is to take the number of outcomes that represent the event (called X) that you're trying to compute the probability of, and divide it by the total number of possible outcomes. So, for rolling a die: Pr(\"getting a one\") = 1 face that has one dot / 6 faces = 1/6 Pr(\"getting a two\") = 1 face that has two dots / 6 faces = 1/6 Pr(\"getting a three\") = 1 face that has three dot / 6 faces = 1/6 Pr(\"getting a four\") = 1 face that has four dots / 6 faces = 1/6 Pr(\"getting a five\") = 1 face that has five dot / 6 faces = 1/6 Pr(\"getting a six\") = 1 face that has six dots / 6", "had on three faces a common colour, and on the other three faces a second colour. This randomisation machine operates like a coin - it will have some particular 2-state discrete probability distribution. Now imagine all possible combinations of six different colours written on the faces of dice. That's a lot of dice, each with its own number of states, with its own discrete probability distribution. Finally, imagine a die with no face markings. This represents the minimal 0-state machine which doesn't technically get to be called a randomisation machine, since there's no uncertainty in rolling it. But for completeness you can see how it fits in. Without knowing anything about the particular probability distribution of a die, you can see that the first die I mentioned, the one with 6 distinct faces, somehow provides the most randomness. That is, if you first build the die (and therefore fix its probability distribution), when you come to the decision of how to label its faces, there's something natural feeling about having all the available slots differently labelled. It is more efficient, less wasteful of capacity, more generative of randomness. In terms of information theory, it is the maximum entropy choice, given any particular probability distribution. The maximum entropy choice would have you use all available slots on this randomisation", "among the 18 equally-likely-to-be-selected faces . Your method neglected to include an indication of which die was chosen, and further the outcomes would not have equal probability weight.", "chosen uniformly at random from $S$. What is the probability that $\\left(\\frac34 + \\frac34i\\right)z$ is also in $S$? Two circles of radius 1 are to be constructed as follows. The center of circle $A$ is chosen uniformly and at random from the line segment joining $(0,0)$ and $(2,0)$. The center of circle $B$ is chosen uniformly and at random, and independently of the first choice, from the line segment joining $(0,1)$ to $(2,1)$. What is the probability that circles $A$ and $B$ intersect? A traffic light runs repeatedly through the following cycle: green for $30$ seconds, then yellow for $3$ seconds, and then red for $30$ seconds. Leah picks a random three-second time interval to watch the light. What is the probability that the color changes while she is watching? On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots? Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at", "# No face. But a 5 Level pending An experiment consists of randomly drawling a card from a standard deck of cards (no jokers) and rolling a regular dice (6 sides). Find the probability that the card is NOT a face card and the die shows a five. Note: The solution can be expressed as $$\\frac{a}{b}$$. Express your answer as $$a + b$$. ×", "HomeEnglishClass 10MathsChapterProbability If one die is rolled then find... # If one die is rolled then find the probability of each of the following events <br> Number on the upper face is prime Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free, (1)/(2)", "# Probability of getting maximum dots in n rolls of a die A fair die is rolled $n$ times. Let $X_i$ denote the number of spots that are on the up face on roll $i$, for $i=1, 2, \\dots, n$. Find the probability mass function of $Y=\\text{max}\\{X_1,X_2,\\dots,X_n\\}$. I've been trying various things but I know my answers aren't correct because the sum of the values of the pmf don't equal $1$. The probability that the maximum $Y$ is $\\le k$ is the probability that all the $X_i$ are $\\le k$. This is $\\left(\\frac{k}{6}\\right)^n$. The probability that $Y=k$ is the probability that $Y$ is $\\le k$ minus the probability that $Y$ is $\\le k-1$.", "Home > English > Class 10 > Maths > Chapter > Probability > A die is rolled. What is the p... # A die is rolled. What is the probability of getting a number less than 7 on the upper face ? Updated On: 27-06-2022 1/612/35/6", "If not, explain why not. 1. $$X$$ is the number of dots on the top face of fair die that is rolled. 2. $$X$$ is the number of hearts in a five-card hand drawn (without replacement) from a well-shuffled ordinary deck. 3. $$X$$ is the number of defective parts in a sample of ten randomly selected parts coming from a manufacturing process in which $$0.02\\%$$ of all parts are defective. 4. $$X$$ is the number of times the number of dots on the top face of a fair die is even in six rolls of the die. 5. $$X$$ is the number of dice that show an even number of dots on the top face when six dice are rolled at once. 2. Determine whether or not the random variable $$X$$ is a binomial random variable. If so, give the values of $$n$$ and $$p$$. If not, explain why not. 1. $$X$$ is the number of black marbles in a sample of $$5$$ marbles drawn randomly and without replacement from a box that contains $$25$$ white marbles and $$15$$ black marbles. 2. $$X$$ is the number of black marbles in a sample of $$5$$ marbles drawn randomly and with replacement from a box that contains $$25$$ white marbles and $$15$$ black marbles. 3. $$X$$ is the number of", "dots. But since we can’t cover “part” of a dot (in this problem) then that means that there exists a random placement of the coins that covers all 10 dots. So no matter the configuration of dots, we know that there is always a way to cover the dots with at most 10 coins 🙂", "they exhibit “stunning regularity”. ### ToplDice is Markovian Many problems in probability are solved by assuming independence of separate experiments. When we toss a coin, it is assumed that the outcome does not depend on the results of previous tosses. Similarly, each cast of a die is assumed to be independent of previous casts. However, this assumption is frequently invalid. Draw a card from a shuffled deck and reveal it. Then place it on the bottom and draw another card. The odds have changed: if the first card was an ace, the chances that the second is also an ace have diminished.", "higher or lower numbers than with a standard die. The coin flip bias is only at around the 1% level (I have no idea what it would be for a 20-sided die), so whether this is significant or not depends on what level of effects you care about. But casinos make a living on advantages on the level of a few percent. -", "Probability # Uniform Probability A die is made from a regular icosahedron so that each side labelled by a number from $1$ through $10,$ with some labels appearing multiple times. If the the probability that it lands on the number $8$ is $\\frac{1}{5},$ how many sides are labelled $8?$ Details and assumptions A regular icosahedron has $20$ faces. A dartboard is equally divided into fan-shaped sectors, each with a central angle of $20 ^\\circ.$ If exactly one of the sectors is painted red and a dart has equal probability of hitting every position in the dartboard, then the probability that the dart hits the red sector can be expressed as $\\frac{a}{b}$, where $a$ and $b$ are coprime positive integers. What is $a+b?$ Details and assumptions Assume that the dart never misses the dartboard. A dartboard is equally divided into fan-shaped sectors, each with a central angle of $6 ^\\circ$, and exactly one of the sectors is painted red. If the dart has equal probability of hitting every position in the dartboard, what is the probability that it hit the red sector? Details and assumptions Assume that the dart never misses the dartboard. Sophia rolls a fair standard 6-sided die with sides labeled $1$ through $6$. What is the probability that she rolls a $6$? Details and assumptions In", "Question Refer to Exercise 4.10. Assume that the die is a fair die, that is, each of the outcomes has a probability of 1/36. What is the probability of observing In exercise A die is rolled two times. Provide a list of the possible outcomes of the two rolls in this form: the result from the first roll and the result from the second roll. a. Event A: Exactly one dot appears on each of the two upturned faces? b. Event B: The sum of the dots on the two upturned faces is exactly 4? c. Event C: The sum of the dots on the two upturned faces is at most 4? Sales1 Views30 Comments0 • CreatedNovember 21, 2015 • Files Included Post your question 5000", "some unpleasantly complicated physics gets in the way. For example, even a standard six-sided die can be vulnerable to manipulation by a skilled cheat. The difficulty is that the fairness of a particular shape of die may depend on assumptions about how the die is rolled– what is the probability distribution from which we randomly draw the die’s initial position, velocity, and angular momentum? What surface does the die land on– can it slide and/or bounce, and if so, with what coefficients of friction and restitution, etc.? (For this discussion, we will focus on dice that are convex polyhedra, having flat polygonal faces with no holes or indentations. There are other interesting possibilities, though. For example, consider flipping a cylinder as a “three-sided coin,” which may land on heads, tails, or on its curved “edge.”) Symmetry One way to get around this dependence on physics modeling assumptions is to appeal to symmetry: if there is any shape of die that could possibly be considered fair, then a sufficient (but perhaps not necessary) condition for fairness is face-transitivity— the group of symmetries of the die should act transitively on its faces. That is, given any pair of faces $f_1$ and $f_2$, there must be a rigid transformation of the die into itself that maps $f_1$ to $f_2$. To see" ]

[ "# Probability of an Odd Number of Sixes ### Solution If $0\\le k\\le n,$ the probability of $k$ sixes turning up in a random toss of $n$ fair dice is $\\displaystyle {n\\choose k}\\left(\\frac{5}{6}\\right)^{n-k}\\left(\\frac{1}{6}\\right)^{k}.$ Hence, with $\\displaystyle a=\\frac{5}{6}$ and $\\displaystyle b=\\frac{1}{6},$ the required probability is \\displaystyle\\begin{align}P&=\\sum_{k\\text{ is odd}}{n\\choose k}\\left(\\frac{5}{6}\\right)^{n-k}\\left(\\frac{1}{6}\\right)^{k}\\\\ &=\\text{ sum of every second term in the expansion of }(a+b)^n\\\\ &=\\frac{1}{2}\\left[(a+b)^n-(a-b)^n\\right]\\\\ &=\\frac{1}{2}\\left[1-\\left(\\frac{2}{3}\\right)^n\\right]. \\end{align} In a similar manner, the probability of an even number of sixes equals \\displaystyle \\begin{align} Q&=\\frac{1}{2}\\left[(a+b)^n+(a-b)^n\\right]\\\\ &=\\frac{1}{2}\\left[1+\\left(\\frac{2}{3}\\right)^n\\right]. \\end{align} Naturally, $P+Q=1.$ Why $Q\\gt P?$ The graphics below may be suggestive. $0$ is an even number always lurking in the background. ### Acknowledgment This is a slight modification of problem 470 from Five Hundred Mathematical Challenges by E. J. Barbeau et all (MAA, 1995).", "have $d-(n-1)$ keys already drawn from a possible set of $d$, giving you a chance of $(d-(n-1))/d$ of not having a collision.) And therefore, the probability of having a collision is simply: $P(n,d)=1-Q(n,d)$. Whenever $d$ and $n$ are largish, we can safely approximate using: $\\displaystyle P(n,d)\\approx{}1-e^{-\\frac{n(n-1)}{2d}}$. So, knowing $d$ (for example $d=2^{128}$), how large does $n$ have to be to have a collision? OK, let us say only to have a probability $\\alpha$ of collision? Well, we just solve $P(n,d)=\\alpha$ for $n$. After a bit of pain, we find $\\displaystyle n=\\frac{1}{2}\\left(\\sqrt{1-8d\\ln(1-\\alpha)}\\right)$. * * * OK, let’s say now that we want a 50% chance of finding a collision, how large should $n$ be? Well, now we have $\\alpha=0.5$, $d=2^{128}$, and we solve for $n$, finding that $\\displaystyle n=\\frac{1}{2}\\left(\\sqrt{1-8d\\ln(1-\\alpha)}\\right)$ $\\displaystyle n=\\frac{1}{2}\\left(\\sqrt{1-8(2^{128})\\ln(\\frac{1}{2})}\\right)$ and finally $\\displaystyle n\\approx{}2.17\\times{}10^{19}$. We observe that $n\\approx 1.17 \\sqrt{d}$ (for $\\alpha=\\frac{1}{2}$, anyways). * * * Therefore, before reasonably expecting a collision with probability $\\alpha=\\frac{1}{2}$ using $d=2^{128}$, we would have to hash about $2.17\\times{}10^{19}$ keys, which… may take a while. 5 Responses to Finding Collisions 1. Interesting. This assumes, of course, that hash values are truly random over 128-bit. That’s a strong requirement. A slightly more realistic requirement would be to assume 32-bit hash values (as in, for example, Java). Your formula then gives me 77163. Not quite", "$P(n)$ on figure 1. Equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_approx_p} is very interesting since it makes it easy to compute a value of $n$ for which we will have a collision with probability larger than or equal to $p$: $$\\begin{eqnarray} p = 1 - e^{-n(n-1)/(2N)} &\\Longrightarrow& \\displaystyle\\frac{-n(n-1)}{2N} = \\log(1 - p) \\nonumber\\\\[5pt] &\\Longrightarrow& \\displaystyle\\frac{n(n-1)}{2N} = \\log\\left(\\displaystyle\\frac{1}{1-p}\\right) \\nonumber\\\\[5pt] &\\Longrightarrow& n^2 \\geq 2N\\log\\left(\\displaystyle\\frac{1}{1-p}\\right) \\end{eqnarray}$$ Therefore: $$\\boxed{ n \\geq \\sqrt{2N\\log\\left(\\displaystyle\\frac{1}{1-p}\\right)} } \\label{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n}$$ will yield a collision with probability larger than or equal to $p$. For $N = 365$ and $p = 0.5$, we get $n \\approx 22.49$, so indeed for $n = 23$ we will have collisions with at least $50\\%$ probability. The birthday paradox gets even stranger if we take $N = 10^6$ (a million) and $p = 0.5$: for those parameters, $n \\approx 1177.4$, so we will have a collision with more than $50\\%$ probability with as few as $1180$ elements! In general, to produce a collision with $50\\%$ probability for a given $N$, we can select a number of elements given by: $$\\boxed{ n \\geq 1.2\\sqrt{N} } \\label{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n_approx}$$ The equation above is obtained by setting $p = 0.5$ on equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n} and rounding up the constant factor. For $N = 365$, equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n_approx} yields $n \\geq 22.92$ (no surprises here). Equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n_approx} closes our discussion of the problem. To have a collision", "1. Equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_approx_p} is very interesting since it makes it easy to compute a value of $n$ for which we will have a collision with probability larger than or equal to $p$: $$\\begin{eqnarray} p = 1 - e^{-n(n-1)/(2N)} &\\Longrightarrow& \\displaystyle\\frac{-n(n-1)}{2N} = \\log(1 - p) \\nonumber\\\\[5pt] &\\Longrightarrow& \\displaystyle\\frac{n(n-1)}{2N} = \\log\\left(\\displaystyle\\frac{1}{1-p}\\right) \\nonumber\\\\[5pt] &\\Longrightarrow& n^2 \\geq 2N\\log\\left(\\displaystyle\\frac{1}{1-p}\\right) \\end{eqnarray}$$ Therefore: $$\\boxed{ n \\geq \\sqrt{2N\\log\\left(\\displaystyle\\frac{1}{1-p}\\right)} } \\label{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n}$$ will yield a collision with probability larger than or equal to $p$. For $N = 365$ and $p = 0.5$, we get $n \\approx 22.49$, so indeed for $n = 23$ we will have collisions with at least $50\\%$ probability. The birthday paradox gets even stranger if we take $N = 10^6$ (a million) and $p = 0.5$: for those parameters, $n \\approx 1177.4$, so we will have a collision with more than $50\\%$ probability with as few as $1180$ elements! In general, to produce a collision with $50\\%$ probability for a given $N$, we can select a number of elements given by: $$\\boxed{ n \\geq 1.2\\sqrt{N} } \\label{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n_approx}$$ The equation above is obtained by setting $p = 0.5$ on equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n} and rounding up the constant factor. For $N = 365$, equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n_approx} yields $n \\geq 22.92$ (no surprises here). Equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n_approx} closes our discussion of the problem. To have a collision with more than", "the probability that at least one student has the same birthday as any other student on any day is around 70% (for n = 30), from the formula $${\\displaystyle 1-{\\frac {365!}{(365-n)!\\cdot 365^{n}}}}$$. which can be rephrased in terms of the language in Cryptography Engineering: $$1 - \\frac{N!}{(N-k)! * N^k}$$ Is it supposed to equal to the following from Cryptography Engineering: $$(k(k-1))/(2N)$$ Why? The question asks how we go from $$\\displaystyle p=1 - \\frac{N!}{(N-k)!\\,N^k}$$ for the probability of collision of $$k$$ uniformly random values among $$N$$, to the approximation $$\\displaystyle p\\approx\\frac{k(k-1)}{2N}$$ (which assumes $$k$$ is small compared to $$\\sqrt N$$ ). First we get back to $$\\displaystyle1-p=\\prod_{j=0}^{k-1}{\\left(1-\\frac j N\\right)}$$, which is how $$p$$ was determined in the first place. Then we take the logarithm on both sides and use that $$u>0,v>0\\implies\\ln(u\\,v)=\\ln(u)+\\ln(v)$$ to get $$\\displaystyle\\ln(1-p)=\\sum_{j=0}^{k-1}{\\ln\\left(1-\\frac j N\\right)}$$ For small $$|x|$$, it hold $$\\ln(1+x)\\approx x$$. Applying this to $$x=p$$ on the left side and $$\\displaystyle x=\\frac j N$$ on the right side, we get $$\\displaystyle p\\approx\\sum_{j=0}^{k-1}\\frac j N$$. We rewrite this as $$\\displaystyle p\\approx\\frac 1 N\\sum_{j=0}^{k-1}j$$. Now we use that the sum of non-negative integers less than $$k$$ is $$\\displaystyle\\frac{k\\,(k-1)}2$$ and get the desired $$\\displaystyle p\\approx\\frac{k(k-1)}{2N}$$. Without proof: this approximation of $$p$$ is always by excess. It's off by less than $$+28\\%$$ when $$k\\le\\sqrt N$$, less than $$+14\\%$$ when $$k\\le\\sqrt{2N}$$, less", "number of trials $$\\displaystyle={13}$$ $$\\displaystyle{p}=$$ Probability of success $$\\displaystyle={\\frac{{{1}}}{{{2}}}}={0.5}$$ (Assuming heads and tails are equally likely) Since we are interested in the number of successes among a fixed number of independent trials with a constant probability of success, we can use binomial distribution. Now, we will find the binomial probability at $$\\displaystyle{k}={13}$$ $$\\displaystyle{P}{\\left({X}={13}\\right)}={{C}_{{{13}}}^{{{13}}}}\\cdot{\\left({0.5}\\right)}^{{{13}}}\\cdot{\\left({1}-{0.5}\\right)}^{{{13}-{13}}}$$ $$\\displaystyle={\\frac{{{13}!}}{{{13}!{\\left({13}-{13}\\right)}!}}}\\times{\\left({0.5}\\right)}^{{{13}}}\\times{\\left({1}-{0.5}\\right)}^{{{0}}}$$ $$\\displaystyle={0.5}^{{{13}}}$$ $$\\displaystyle={\\left({\\frac{{{1}}}{{{2}}}}\\right)}^{{{13}}}$$ $$\\displaystyle={\\frac{{{1}}}{{{2}^{{{13}}}}}}$$ $$\\displaystyle={\\frac{{{1}}}{{{8192}}}}\\approx{0.0001}$$", "2^5 + 2^5 + 2^5$ $\\vdots$ $+ 2^{17} + 2^{17} + 2^{17} + \\dots + 2^{17}$ Let’s now rearrange the terms of this sum. We will do this by adding along the diagonals instead of along the rows. In this way, the above sum can be rearranged as $2^3 + 2^4 + 2^5 + \\dots + 2^{17}$ $+ 2^4 + 2^5 + \\dots + 2^{17}$ $+ 2^5 + \\dots + 2^{17}$ $\\vdots$ $+ 2^{17}$ Each of these new rows (or the original diagonals) is a geometric series and can be calculated using the formula: $2^3 + 2^4 + 2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^3 (1-2^{15})}{1-2} = 2^{18} - 2^3$ $2^4 + 2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^4 (1-2^{14})}{1-2} = 2^{18} - 2^4$ $2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^5 (1-2^{13})}{1-2} = 2^{18} - 2^5$ $\\vdots$ $2^{17} = (2-1) \\cdot 2^{17} = 2^{18} - 2^{17}$ So, thus far in the calculation, we have established that $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2} = \\displaystyle \\sum_{n=3}^{17} \\left( 2^{18} - 2^n \\right)$. Simplifying, $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2}=\\displaystyle \\sum_{n=3}^{17} 2^{18} - \\sum_{n=3}^{17} 2^n$ The first sum on the right is the sum of a constant being added to itself 15 times: $\\displaystyle \\sum_{n=3}^{17} 2^{18} = 15 \\times 2^{18}$ The second sum on the right is yet another geometric", "of convergence of the pwer series $\\displaystyle\\sum_{n=1}^\\infty na_n(z-a)^{n-1}$ is also $R$; 2. $\\displaystyle\\left(\\sum_{n=0}^\\infty a_n(z-a)^n\\right)'=\\sum_{n=1}^\\infty na_n(z-a)^{n-1}$. • (I think this is the first time you answer one of my questions (I take a look to your answers and they are incredible)) Wow! amazings properties! Thank you. – manooooh Jul 4 '18 at 22:20 • @manooooh I'm glad I could help. – José Carlos Santos Jul 4 '18 at 22:47 In terms of finding $f''(-2)$ with $$f(x)=\\sum_{n=1}^\\infty{\\frac{{(-1)}^{n+1}}{n\\cdot3^n}{(x-3)}^n}$$ then consider the following. Method 1 Since \\begin{align} f(x) &= \\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n} \\, \\left(1 - \\frac{x}{3}\\right)^{n} = \\ln\\left(\\frac{x}{3}\\right) \\end{align} then \\begin{align} f'(x) &= \\frac{1}{x} \\\\ f''(x) &= - \\frac{1}{x^2} \\end{align} Method 2 \\begin{align} f(x) &= \\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n} \\, \\left(1 - \\frac{x}{3}\\right)^{n} \\\\ f'(x) &= \\frac{1}{3} \\, \\sum_{n=1}^{\\infty} \\left( \\frac{3 - x}{3} \\right)^{n} = \\frac{1}{3 \\, \\left(1 - \\left( 1 - \\frac{x}{3} \\right) \\right)} = \\frac{1}{x} \\\\ f''(x) &= - \\frac{1}{x^2}. \\end{align} The value sought can now be obtained by setting $x = -2$.", "can be seen by noting that the center of the standard simplex is ${\\textstyle \\left({\\frac {1}{n+1}},\\dots ,{\\frac {1}{n+1}}\\right)}$, and the centers of its faces are coordinate permutations of ${\\textstyle \\left(0,{\\frac {1}{n}},\\dots ,{\\frac {1}{n}}\\right)}$. Then, by symmetry, the vector pointing from ${\\textstyle \\left({\\frac {1}{n+1}},\\dots ,{\\frac {1}{n+1}}\\right)}$ to ${\\textstyle \\left(0,{\\frac {1}{n}},\\dots ,{\\frac {1}{n}}\\right)}$ is perpendicular to the faces. So the vectors normal to the faces are permutations of ${\\displaystyle (-n,1,\\dots ,1)}$, from which the dihedral angles are calculated. ### Simplices with an \"orthogonal corner\" An \"orthogonal corner\" means here that there is a vertex at which all adjacent edges are pairwise orthogonal. It immediately follows that all adjacent faces are pairwise orthogonal. Such simplices are generalizations of right triangles and for them there exists an n-dimensional version of the Pythagorean theorem: The sum of the squared (n − 1)-dimensional volumes of the facets adjacent to the orthogonal corner equals the squared (n − 1)-dimensional volume of the facet opposite of the orthogonal corner. ${\\displaystyle \\sum _{k=1}^{n}|A_{k}|^{2}=|A_{0}|^{2}}$ where ${\\displaystyle A_{1}\\ldots A_{n}}$ are facets being pairwise orthogonal to each other but not orthogonal to ${\\displaystyle A_{0}}$, which is the facet opposite the orthogonal corner. For a 2-simplex the theorem is the Pythagorean theorem for triangles with a right angle and for a 3-simplex it is de Gua's theorem for a tetrahedron with", "the original diagonals) is a geometric series and can be calculated using the formula: $2^3 + 2^4 + 2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^3 (1-2^{15})}{1-2} = 2^{18} - 2^3$ $2^4 + 2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^4 (1-2^{14})}{1-2} = 2^{18} - 2^4$ $2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^5 (1-2^{13})}{1-2} = 2^{18} - 2^5$ $\\vdots$ $2^{17} = (2-1) \\cdot 2^{17} = 2^{18} - 2^{17}$ So, thus far in the calculation, we have established that $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2} = \\displaystyle \\sum_{n=3}^{17} \\left( 2^{18} - 2^n \\right)$. Simplifying, $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2}=\\displaystyle \\sum_{n=3}^{17} 2^{18} - \\sum_{n=3}^{17} 2^n$ The first sum on the right is the sum of a constant being added to itself 15 times: $\\displaystyle \\sum_{n=3}^{17} 2^{18} = 15 \\times 2^{18}$ The second sum on the right is yet another geometric series. Indeed, it’s the same geometric series from the first diagonal above: $\\sum_{n=3}^{17} 2^n = \\displaystyle \\frac{2^3 (1-2^{15})}{1-2} = 2^{18} - 2^3$ Therefore, $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2} = 15 \\times 2^{18} - \\left(2^{18} - 2^3 \\right)$ $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2} = 14 \\times 2^{18} + 2^3$ $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2} = 3,670,024$ Not surprisingly, this matches the sum that was found via direct addition. # 2048 and algebra (Part 7) In this series of posts, I consider", "\\\\ &\\qquad\\qquad\\quad\\quad\\; =\\frac{1}{9}\\left( 1+\\frac{13}{18}\\ +\\ {{\\left( \\frac{13}{18} \\right)}^{2}}\\ +{{\\left( \\frac{13}{18} \\right)}^{3}}+\\ .....\\infty \\right) \\\\ &\\qquad\\qquad\\quad\\quad\\; =\\frac{1}{9}\\ \\cdot \\ \\frac{1}{1-\\frac{13}{18}} \\\\ &\\qquad\\qquad\\quad\\quad\\; =\\frac{2}{5} \\\\ \\end{align} In passing, note that the probability of obtaining a 7 before 5 is simply \\begin{align}& P\\left( 7\\text{ before }5 \\right)\\text{ }=\\text{ }1-\\text{ }P\\left( 5\\text{ before }7 \\right) \\\\ &\\qquad\\qquad\\qquad \\;=\\frac{3}{5} \\\\ \\end{align} Can you appreciate why obtaining a 7 before 5 is more likely than obtaining a 5 before 7? Example – 8 Two unit squares are chosen at random from a standard chessboard. What is the probability that the two squares have exactly one corner in common? Solution: The total number of ways of selecting two squares from a standard chessboard is simply $$^{64}{{C}_{2}}\\text{ }=\\text{ }2016$$. Now, let us find the number of ways of selecting a pair with exactly one corner in common. For that, consider the following figure. It is immediately evident that the number of favorable ways is 49 + 49 = 98. The required probability is \\begin{align}\\frac{98}{2016}=\\frac{7}{144}\\end{align} Example – 9 For two events A and B, we are given the following information: $P(A)=\\frac{1}{4},\\ \\ P(A/B)=\\frac{1}{4}\\ \\ \\text{and}\\ \\ P(B/A)=\\frac{1}{2}$ (a) Determine whether the two events are ME or independent, or neither. (b) Find $$P(\\bar{B}/\\bar{A})$$ . The notation $$\\bar{A}$$ stands for the complementary event of event A, i.e., $$\\bar{A}$$ occurs", "sampling $k$ elements $r_1 < r_2 < \\dots < r_k$ the probability due to step $3$ is $\\displaystyle \\frac{k}{r_1}\\frac{k-1}{r_2}\\dots\\frac{1}{r_k}$ That is the probability that each element is put into its own slot. The probability that $r_1$ doesn’t get overwritten by the upcoming elements (except for $r_2,\\dots,r_k$ which have already been picked by step $3$) is $\\displaystyle \\prod_{r_1 < s \\le n-k+1} \\frac{s-1}{s} = \\frac{r_1}{n-k+1}$ Thus, the probability of selecting $r_1 < \\dots < r_k$ according to the algorithm is $\\displaystyle \\frac{k}{r_1}\\frac{k-1}{r_2}\\dots\\frac{1}{r_k} \\times \\frac{r_1}{n-k+1}\\dots \\frac{r_k}{n} = \\frac{k!}{(n)_k}$ This entry was posted in statistics and tagged , , . Bookmark the permalink.", "this sum can be written in expanded form as $2^3$ $+ 2^4 + 2^4$ $+ 2^5 + 2^5 + 2^5$ $\\vdots$ $+ 2^{17} + 2^{17} + 2^{17} + \\dots + 2^{17}$ Let’s now rearrange the terms of this sum. We will do this by adding along the diagonals instead of along the rows. In this way, the above sum can be rearranged as $2^3 + 2^4 + 2^5 + \\dots + 2^{17}$ $+ 2^4 + 2^5 + \\dots + 2^{17}$ $+ 2^5 + \\dots + 2^{17}$ $\\vdots$ $+ 2^{17}$ Each of these new rows (or the original diagonals) is a geometric series and can be calculated using the formula: $2^3 + 2^4 + 2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^3 (1-2^{15})}{1-2} = 2^{18} - 2^3$ $2^4 + 2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^4 (1-2^{14})}{1-2} = 2^{18} - 2^4$ $2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^5 (1-2^{13})}{1-2} = 2^{18} - 2^5$ $\\vdots$ $2^{17} = (2-1) \\cdot 2^{17} = 2^{18} - 2^{17}$ So, thus far in the calculation, we have established that $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2} = \\displaystyle \\sum_{n=3}^{17} \\left( 2^{18} - 2^n \\right)$. Simplifying, $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2}=\\displaystyle \\sum_{n=3}^{17} 2^{18} - \\sum_{n=3}^{17} 2^n$ The first sum on the right is the sum of a constant being added to itself 15 times: $\\displaystyle \\sum_{n=3}^{17}", "that the probability that a dash is transmitted as a dot is likewise 1/10, then Bayes's rule can be used to calculate $\\displaystyle{ P(dot \\ received) }$. $\\displaystyle{ P(dot \\ received) = P(dot \\ received \\ \\cap \\ dot \\ sent ) + P(dot \\ received \\ \\cap \\ dash \\ sent) }$ $\\displaystyle{ P(dot \\ received) = P(dot \\ received \\mid dot \\ sent)P(dot \\ sent) + P(dot \\ received \\mid dash \\ sent)P(dash \\ sent) }$ $\\displaystyle{ P(dot \\ received) = \\frac{9}{10}\\times\\frac{3}{7} + \\frac{1}{10}\\times\\frac{4}{7} = \\frac{31}{70} }$ Now, $\\displaystyle{ P(dot \\ sent \\mid dot \\ received) }$can be calculated: $\\displaystyle{ P(dot \\ sent \\mid dot \\ received) = P(dot \\ received \\mid dot \\ sent) \\frac{P(dot \\ sent)}{P(dot \\ received)} = \\frac{9}{10}\\times \\frac{\\frac{3}{7}}{\\frac{31}{70}} = \\frac{27}{31} }$[16] ## Statistical independence Main page: Independence (probability theory) Events A and B are defined to be statistically independent if the probability of the intersection of A and B is equal to the product of the probabilities of A and B: $\\displaystyle{ P(A \\cap B) = P(A) P(B). }$ If P(B) is not zero, then this is equivalent to the statement that $\\displaystyle{ P(A\\mid B) = P(A). }$ Similarly, if P(A) is not zero, then $\\displaystyle{ P(B\\mid A) = P(B) }$ is also equivalent. Although the derived forms may seem more", "= P({s_2}) + P({s_3}) + P({s_7}) + P({s_9})$ • If we know the probability of the individual outcomes, we can compute the probability of any event by adding the probabilities of the relevant outcomes. • Core assumption: individual events in the collection are separate by convention. • Important special case (limit of most high school probability): Suppose that each of the $N$ outcomes is “equally likely” (recurring phrase we’ll hear a lot). In this case, each outcome has probability:$P({s_i}) = frac{1}{N} mbox{for } i = 1, 2, dots, N$ This means that with equally likely outcomes, figuring probabilities is basically just counting. • Example: $E = { s_1, s_2, dots, s_k}$ for $k$ outcomes in $E$, then $P(E) = P({s_1}) + P({s_2}) + dots + P({s_k})$ $P(E) = frac{1}{N} + frac{1}{N} + dots + frac{1}{N}$ $P(E) = frac{displaystyle k}{displaystyle N}= frac{displaystyle #(mbox{outcomes in E})}{ displaystyle#(mbox{outcomes in S})}$ • 3 of a kind example: Draw 5 cards from a standard deck of cards. What is $P(mbox{3 of a kind})$? A: $frac{displaystyle #(mbox{3 of a kind})}{displaystyle #(mbox{5 cards})} = frac{displaystyle formula , derived , previously}{displaystyle {52 choose 5}}$ • Counting example (Exercise 3.7.2 in book): Assume the probabilities of throwing an astragalus are: $P(1) = 0.1, P(3) = 0.4, P(4) = 0.4, P(6) = 0.1$ • So we have: $p_1", "= P({s_2}) + P({s_3}) + P({s_7}) + P({s_9})$ • If we know the probability of the individual outcomes, we can compute the probability of any event by adding the probabilities of the relevant outcomes. • Core assumption: individual events in the collection are separate by convention. • Important special case (limit of most high school probability): Suppose that each of the $N$ outcomes is “equally likely” (recurring phrase we’ll hear a lot). In this case, each outcome has probability:$P({s_i}) = frac{1}{N} mbox{for } i = 1, 2, dots, N$ This means that with equally likely outcomes, figuring probabilities is basically just counting. • Example: $E = { s_1, s_2, dots, s_k}$ for $k$ outcomes in $E$, then $P(E) = P({s_1}) + P({s_2}) + dots + P({s_k})$ $P(E) = frac{1}{N} + frac{1}{N} + dots + frac{1}{N}$ $P(E) = frac{displaystyle k}{displaystyle N}= frac{displaystyle #(mbox{outcomes in E})}{ displaystyle#(mbox{outcomes in S})}$ • 3 of a kind example: Draw 5 cards from a standard deck of cards. What is $P(mbox{3 of a kind})$? A: $frac{displaystyle #(mbox{3 of a kind})}{displaystyle #(mbox{5 cards})} = frac{displaystyle formula , derived , previously}{displaystyle {52 choose 5}}$ • Counting example (Exercise 3.7.2 in book): Assume the probabilities of throwing an astragalus are: $P(1) = 0.1, P(3) = 0.4, P(4) = 0.4, P(6) = 0.1$ • So we have: $p_1", "= P({s_2}) + P({s_3}) + P({s_7}) + P({s_9})$ • If we know the probability of the individual outcomes, we can compute the probability of any event by adding the probabilities of the relevant outcomes. • Core assumption: individual events in the collection are separate by convention. • Important special case (limit of most high school probability): Suppose that each of the $N$ outcomes is “equally likely” (recurring phrase we’ll hear a lot). In this case, each outcome has probability:$P({s_i}) = frac{1}{N} mbox{for } i = 1, 2, dots, N$ This means that with equally likely outcomes, figuring probabilities is basically just counting. • Example: $E = { s_1, s_2, dots, s_k}$ for $k$ outcomes in $E$, then $P(E) = P({s_1}) + P({s_2}) + dots + P({s_k})$ $P(E) = frac{1}{N} + frac{1}{N} + dots + frac{1}{N}$ $P(E) = frac{displaystyle k}{displaystyle N}= frac{displaystyle #(mbox{outcomes in E})}{ displaystyle#(mbox{outcomes in S})}$ • 3 of a kind example: Draw 5 cards from a standard deck of cards. What is $P(mbox{3 of a kind})$? A: $frac{displaystyle #(mbox{3 of a kind})}{displaystyle #(mbox{5 cards})} = frac{displaystyle formula , derived , previously}{displaystyle {52 choose 5}}$ • Counting example (Exercise 3.7.2 in book): Assume the probabilities of throwing an astragalus are: $P(1) = 0.1, P(3) = 0.4, P(4) = 0.4, P(6) = 0.1$ • So we have: $p_1", "\\dots + 2^{17}$ $\\vdots$ $+ 2^{17}$ Each of these new rows (or the original diagonals) is a geometric series and can be calculated using the formula: $2^3 + 2^4 + 2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^3 (1-2^{15})}{1-2} = 2^{18} - 2^3$ $2^4 + 2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^4 (1-2^{14})}{1-2} = 2^{18} - 2^4$ $2^5 + \\dots + 2^{17} = \\displaystyle \\frac{2^5 (1-2^{13})}{1-2} = 2^{18} - 2^5$ $\\vdots$ $2^{17} = (2-1) \\cdot 2^{17} = 2^{18} - 2^{17}$ So, thus far in the calculation, we have established that $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2} = \\displaystyle \\sum_{n=3}^{17} \\left( 2^{18} - 2^n \\right)$. Simplifying, $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2}=\\displaystyle \\sum_{n=3}^{17} 2^{18} - \\sum_{n=3}^{17} 2^n$ The first sum on the right is the sum of a constant being added to itself 15 times: $\\displaystyle \\sum_{n=3}^{17} 2^{18} = 15 \\times 2^{18}$ The second sum on the right is yet another geometric series. Indeed, it’s the same geometric series from the first diagonal above: $\\sum_{n=3}^{17} 2^n = \\displaystyle \\frac{2^3 (1-2^{15})}{1-2} = 2^{18} - 2^3$ Therefore, $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2} = 15 \\times 2^{18} - \\left(2^{18} - 2^3 \\right)$ $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2} = 14 \\times 2^{18} + 2^3$ $\\displaystyle \\sum_{n=1}^{15} n \\cdot 2^{n+2} = 3,670,024$ Not surprisingly, this matches the sum that was found via direct addition. #", "3. The probability of getting heads first and then tails: $\\scriptsize P(\\text{H;T)}=\\displaystyle \\frac{1}{4}$ 3. . 1. . \\scriptsize \\begin{align*}P(\\text{W)}&=\\displaystyle \\frac{6}{{14}}\\\\&=\\displaystyle \\frac{3}{7}\\end{align*} 2. . \\scriptsize \\begin{align*}P(\\text{R)}&=\\displaystyle \\frac{2}{{14}}\\\\&=\\displaystyle \\frac{1}{7}\\end{align*} 3. . $\\scriptsize \\begin{array}{l}P({R}')&=1-P(R)\\\\&=1-\\displaystyle \\frac{1}{7}\\\\&=\\displaystyle \\frac{6}{7}\\end{array}$ 4. . \\scriptsize \\begin{align*}P(\\text{W}\\cup \\text{R)}&=P(\\text{W)}+P(\\text{R)}\\\\&=\\displaystyle \\frac{3}{7}+\\displaystyle \\frac{1}{7}\\\\&=\\displaystyle \\frac{4}{7}\\end{align*} 5. . \\scriptsize \\begin{align*}P(\\text{neither W nor R})&=1-P(\\text{W or R)}\\\\&=1-\\displaystyle \\frac{4}{7}\\\\&=\\displaystyle \\frac{3}{7}\\end{align*} 4. . 1. . \\scriptsize \\begin{align*}P(\\text{G and G)}&=\\displaystyle \\frac{5}{{16}}\\cdot \\displaystyle \\frac{5}{{16}}\\\\&=\\displaystyle \\frac{{25}}{{256}}\\end{align*} 2. The probability of two green balls if the first ball is not replaced: \\scriptsize \\begin{align*}P(\\text{G and G)}&=\\displaystyle \\frac{5}{{16}}\\cdot \\displaystyle \\frac{4}{{15}}\\\\&=\\displaystyle \\frac{1}{{12}}\\end{align*} 3. . \\scriptsize \\displaystyle \\begin{align*}P(\\text{Y or O)}&=P(\\text{Y;O)}+P(\\text{O;Y)}\\\\&=\\left( {\\displaystyle \\frac{5}{{16}}\\cdot \\displaystyle \\frac{6}{{16}}} \\right)+\\left( {\\displaystyle \\frac{6}{{16}}\\cdot \\displaystyle \\frac{5}{{16}}} \\right)\\\\&=\\displaystyle \\frac{{15}}{{64}}\\end{align*} 5. . 1. . \\scriptsize \\begin{align*}P(\\text{M})&=\\displaystyle \\frac{{15}}{{35}}\\\\&=\\displaystyle \\frac{3}{7}\\end{align*} 2. . \\scriptsize \\begin{align*}P(\\text{F})&=\\displaystyle \\frac{{20}}{{35}}\\\\&=\\displaystyle \\frac{4}{7}\\end{align*} 3. . \\scriptsize \\begin{align*}P(\\text{M}\\cap \\text{F})&=P(\\text{M})\\cdot P(\\text{F})\\\\&=\\displaystyle \\frac{3}{7}\\cdot \\displaystyle \\frac{4}{7}\\\\&=\\displaystyle \\frac{{12}}{{49}}\\end{align*} Back to Exercise 1.3 ### Unit 1: Assessment 1. . 1. $\\scriptsize \\text{S}=\\{\\text{A; 2; 3; 4; 5; 6; 7; 8; 9; 10; J; K; Q}\\}$ 2. $\\scriptsize P(\\text{A})=\\displaystyle \\frac{1}{{13}}$ 3. $\\scriptsize P(2)+P(3)+P(5)+P(7)=\\displaystyle \\frac{4}{{13}}$ 4. $\\scriptsize P(\\text{A})+P(\\text{J})+P(\\text{K})+P(\\text{Q})=\\displaystyle \\frac{4}{{13}}$ 2. . 1. . \\scriptsize \\displaystyle \\begin{align*}P(\\text{A or B)}&=0.8\\\\P(\\text{A}\\cap \\text{B})&=0\\text{ since A and B are mutually exclusive}\\\\P(\\text{A or B)}&=P(\\text{A})+\\text{P(B)}\\\\0.3+k&=0.8\\\\\\therefore k&=0.5\\end{align*} 2. . \\scriptsize \\displaystyle \\begin{align*}P(\\text{A and B)}&=P(\\text{A)}\\cdot P(\\text{B) since A and B are independent}\\\\&=0.3\\times k\\\\P(\\text{A or B)}&=0.8\\\\P(\\text{A)}+P(\\text{B)}-P(\\text{A and B})&=0.8\\\\0.3+k-0.3k&=0.8\\\\0.7k&=0.5\\\\\\therefore k&=\\displaystyle \\frac{5}{7}\\end{align*} 3. The sum of the dice must equal eight, so it is", "heads, and write: $\\displaystyle P(X)-P(Y)=\\delta$ $\\displaystyle \\frac{1}{3^{50}}\\sum_{k=0}^{25}\\left[{50 \\choose 2k}2^{2k} \\right]-\\frac{1}{3^{50}}\\sum_{k=0}^{24}\\left[{50 \\choose 2k+1}2^{2k+1} \\right]=\\delta$ $\\displaystyle 3^{50}\\delta=\\sum_{k=0}^{50}\\left[{50 \\choose k}2^{k}(-1)^{50-k}\\right]$ By the binomial theorem, we have: $\\displaystyle 3^{50}\\delta=(2-1)^{50}=1$ and so: $\\displaystyle \\delta=\\frac{1}{3^{50}}$ Hence: $\\displaystyle P(X)-P(Y)=\\frac{1}{3^{50}}$ $\\displaystyle P(X)+P(Y)=1$ $\\displaystyle 2P(X)=1+\\frac{1}{3^{50}}$ and so finally, we have: $\\displaystyle P(X)=\\frac{1}{2}\\left(1+\\frac{1}{3^{50}} \\right)$ Thanks!!! Now it is absolutely clear. MarkFL Staff member I found this problem quite interesting and thought about it while I was out and so wanted to generalize a bit and cut out some of the unnecessary hand-waving... Let's say the probability of getting a heads is $\\displaystyle p$ and we flip the coin $\\displaystyle 2n$ times where $\\displaystyle n\\in\\mathbb{N}$. What is the probability that the total number of heads is even? Let $\\displaystyle P(X)$ be the probability that the total number of heads is even and $\\displaystyle P(Y)$ be the probability that the total number of heads is odd. $\\displaystyle P(X)+P(Y)=1$ $\\displaystyle P(X)-P(Y)=\\sum_{k=0}^{2n}\\left[{2n \\choose k}p^{2n-k}(p-1)^k \\right]=(2p-1)^{2n}$ $\\displaystyle 2P(X)=1+(2p-1)^{2n}$ $\\displaystyle P(X)=\\frac{1}{2}\\left(1+(2p-1)^{2n} \\right)$" ]

[ "# Rolling Defective Die ### Solution Originally, there are $1+2+3+4+5+6=21$ dots of which $1+3+5=9$ are on the odd numbered faces and $2+4+6$ are one the even numbered faces. After a dot was removed, there are either two even faces and four odd faces or four even faces and two odd faces. The probability of the first event is $\\displaystyle \\frac{12}{21};$ the probability of the second one is $\\displaystyle \\frac{9}{21}.$ In the first case the probability of having an odd face on top is $\\displaystyle \\frac{4}{6};$ it is $\\displaystyle \\frac{2}{6}$ for the second case. Since all the variants are mutually exclusive, the total probability is $\\displaystyle\\frac{4}{6}\\cdot\\frac{12}{21}+\\frac{2}{6}\\cdot\\frac{9}{21}=\\frac{2\\cdot 12+9}{3\\cdot 21}=\\frac{11}{21}.$ ### Acknowledgment This is problem 14 from the 2005 AMC 12A. That's a borrowing from J. D. Faires' First Steps for Math Olympians, MAA, 2006.", "and one spade $$= {}^{13}{C_3} \\times {}^{13}{C_1}$$ Thus, the probability of obtaining $$3$$ diamonds and one spade $$= \\frac{{{}^{13}{C_3} \\times {}^{13}{C_1}}}{{{}^{52}{C_4}}}$$ Chapter 16 Ex.16.ME Question 3 A die has two faces each with number $$'1'$$, three faces each with number $$'2'$$ and one face with number $$'3'$$. If die is rolled once, determine (i) $$P\\left( 2 \\right)$$ (ii) $$P\\left( {1\\;or\\;3} \\right)$$ (iii)$$P\\left( {not\\;3} \\right)$$ Solution Total number of faces $$= 6$$ Number of faces with number $$'2' = 3$$ Therefore, \\begin{align}P\\left( 2 \\right) &= \\frac{3}{6}\\\\&= \\frac{1}{2}\\end{align} Since, \\begin{align}P\\left( {1\\;or\\;3} \\right) &= P\\left( {not\\;2} \\right)\\\\&= 1 - P\\left( 2 \\right)\\\\&= 1 - \\frac{1}{2}\\\\&= \\frac{1}{2}\\end{align} Number of faces with number $$'3' = 1$$ Therefore, $P\\left( 3 \\right) = \\frac{1}{6}$ Thus, \\begin{align}P\\left( {not{\\rm{ }}3} \\right) &= 1 - P\\left( 3 \\right)\\\\&= 1 - \\frac{1}{6}\\\\&= \\frac{5}{6}\\end{align} Chapter 16 Ex.16.ME Question 4 In a certain lottery $${\\rm{1}}0,000$$ tickets are sold, and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket? (b) two tickets? (c) ten tickets? Solution (a) If we buy one ticket, then Total number of tickets sold $$= 10000$$ Number of prizes awarded $$= 10$$ \\begin{align}P\\left( \\text{getting a prize} \\right)& = \\frac{{10}}{{10000}}\\\\&= \\frac{1}{{1000}}\\end{align} Therefore, \\begin{align}P\\left( \\text{not getting a prize} \\right) &= 1 - \\frac{1}{{1000}}\\\\&= \\frac{{999}}{{1000}}\\end{align} (b) If we buy two", "red bead and then a black bead if the first bead is put back into the bag. 2. Calculate the probability of choosing two black beads if the first bead is NOT replaced into the bag. The full solutions are at the end of the unit. # Unit 1: Solutions ### Exercise 1.1 1. If you roll a die, the total possible outcomes are $\\scriptsize \\displaystyle \\left( {1;\\text{ }2;\\text{ }3;\\text{ }4;\\text{ }5;\\text{ }6} \\right)$. 1. The lowest possible score is one 2. The highest possible score is six. 3. Are all outcomes just as likely or will some happen more often? All outcomes are equally likely so all have a probability of $\\scriptsize \\displaystyle \\frac{1}{6}$ and no outcome is more likely than another to occur. 2. If you roll a die, what is the probability of throwing: 1. $\\scriptsize \\displaystyle P(4)=\\displaystyle \\frac{1}{6}$ 2. There are three prime numbers $\\scriptsize (2;\\text{ }3;\\text{ }5)$. \\scriptsize \\displaystyle \\begin{align*}P(\\text{prime number})&=\\displaystyle \\frac{3}{6}\\\\&=\\displaystyle \\frac{1}{2}\\end{align*} 3. There are three odd numbers $\\scriptsize (1;\\text{ }3;\\text{ }5)$. \\scriptsize \\displaystyle \\begin{align*}P(\\text{odd number})&=\\displaystyle \\frac{3}{6}\\\\&=\\displaystyle \\frac{1}{2}\\end{align*} 4. . \\scriptsize \\displaystyle \\begin{align*}P(8)&=\\displaystyle \\frac{0}{6}\\\\&=0\\end{align*} 5. There are four factors of $\\scriptsize \\displaystyle 6$, $\\scriptsize (1;\\text{ }2;\\text{ }3;\\text{ }6)$ . \\scriptsize \\displaystyle \\begin{align*}P(\\text{factor of }6)&=\\displaystyle \\frac{4}{6}\\\\&=\\displaystyle \\frac{2}{3}\\end{align*} 6. All the numbers are less than $\\scriptsize \\displaystyle 8$ so this is a certainty. \\scriptsize", "# Probability of an Odd Number of Sixes ### Solution If $0\\le k\\le n,$ the probability of $k$ sixes turning up in a random toss of $n$ fair dice is $\\displaystyle {n\\choose k}\\left(\\frac{5}{6}\\right)^{n-k}\\left(\\frac{1}{6}\\right)^{k}.$ Hence, with $\\displaystyle a=\\frac{5}{6}$ and $\\displaystyle b=\\frac{1}{6},$ the required probability is \\displaystyle\\begin{align}P&=\\sum_{k\\text{ is odd}}{n\\choose k}\\left(\\frac{5}{6}\\right)^{n-k}\\left(\\frac{1}{6}\\right)^{k}\\\\ &=\\text{ sum of every second term in the expansion of }(a+b)^n\\\\ &=\\frac{1}{2}\\left[(a+b)^n-(a-b)^n\\right]\\\\ &=\\frac{1}{2}\\left[1-\\left(\\frac{2}{3}\\right)^n\\right]. \\end{align} In a similar manner, the probability of an even number of sixes equals \\displaystyle \\begin{align} Q&=\\frac{1}{2}\\left[(a+b)^n+(a-b)^n\\right]\\\\ &=\\frac{1}{2}\\left[1+\\left(\\frac{2}{3}\\right)^n\\right]. \\end{align} Naturally, $P+Q=1.$ Why $Q\\gt P?$ The graphics below may be suggestive. $0$ is an even number always lurking in the background. ### Acknowledgment This is a slight modification of problem 470 from Five Hundred Mathematical Challenges by E. J. Barbeau et all (MAA, 1995).", "Assume the dots represent five observations. -a) 5/6 -b) 2/5 +c) 3/5 -d) 3/4 -e) 2/4=1/2 9) If a number is randomly selected from the set {2,3,4,5}, what is the probability that it is both even and prime? -a) 1 +b) 1/4 -c) 5/4 -d) 3/4 -e) 1/2 -f) 0 10) If a number is randomly selected from the set {2,3,4,5}, what is P(prime)+P(even), or the sum of the probability that it is even, plus the probability that it is prime? -a) 1/2 -b) 1 -c) 3/4 -d) 0 -e) 1/4 +f) 5/4 ### Bell:Bell2:V1 Bell152874216185 1) Suppose the referee selects neutral scoring with ${\\displaystyle Q={\\frac {4}{3}}\\left({\\frac {1-P_{S}}{P_{S}}}\\right).}$ What number does the penalty approach as the probability of asking the same question goes to 0? a) ${\\displaystyle 3}$ b) ${\\displaystyle 4/3}$ c) ${\\displaystyle 4}$ d) ${\\displaystyle 0}$ e) ${\\displaystyle \\infty }$ 2) Suppose the referee gives Alice and Bob receive question cards of the different suit (different questions). What are the best and worst possible outcomes for the partners? (Assume for this question that ${\\displaystyle Q>3}$) a) Best for partners: ${\\displaystyle +1}$ ... Worst: ${\\displaystyle -Q}$ b) Best for partners: ${\\displaystyle 0}$ ... Worst: ${\\displaystyle -3}$ c) None of these is correct d) Best for partners: ${\\displaystyle +1}$ ... Worst: ${\\displaystyle -3}$ e) Best for partners: ${\\displaystyle 0}$", "side.. Alternatively, If a chosen square is in a corner, then the probabilty that the 2nd chosen square is not sharing a side with it is $\\displaystyle\\frac{61}{63}$ hence the probability that the squares have one of them in the corner, with another not sharing a side is $\\displaystyle\\frac{4}{64}\\ \\frac{61}{63}$ There are 6(4)=24 remaining squares along the edge that are not in a corner. The probability of another square not sharing a side with one of these is $\\displaystyle\\frac{60}{63}$ hence the probability that the squares include one of the remaining edge squares, but the other does not share a side with it is $\\displaystyle\\frac{24}{64}\\ \\frac{60}{63}$ There are 36 squares not on an edge. Therefore the probability that the squares include one of these, and the 2nd one is not sharing a side with it is $\\displaystyle\\frac{36}{64}\\ \\frac{59}{63}$ Therefore, the probability that 2 squares are selected and do share a side is $\\displaystyle\\ 1-\\left(\\frac{4(61)+(24)(60)+(36)(59)}{(64)(63)}\\rig ht)=\\frac{(64)(63)-3808}{(64)(63)}=\\frac{4032-3808}{4032}=\\frac{224}{4032}=\\frac{1}{18}$", "heads, and write: $\\displaystyle P(X)-P(Y)=\\delta$ $\\displaystyle \\frac{1}{3^{50}}\\sum_{k=0}^{25}\\left[{50 \\choose 2k}2^{2k} \\right]-\\frac{1}{3^{50}}\\sum_{k=0}^{24}\\left[{50 \\choose 2k+1}2^{2k+1} \\right]=\\delta$ $\\displaystyle 3^{50}\\delta=\\sum_{k=0}^{50}\\left[{50 \\choose k}2^{k}(-1)^{50-k}\\right]$ By the binomial theorem, we have: $\\displaystyle 3^{50}\\delta=(2-1)^{50}=1$ and so: $\\displaystyle \\delta=\\frac{1}{3^{50}}$ Hence: $\\displaystyle P(X)-P(Y)=\\frac{1}{3^{50}}$ $\\displaystyle P(X)+P(Y)=1$ $\\displaystyle 2P(X)=1+\\frac{1}{3^{50}}$ and so finally, we have: $\\displaystyle P(X)=\\frac{1}{2}\\left(1+\\frac{1}{3^{50}} \\right)$ Thanks!!! Now it is absolutely clear. MarkFL Staff member I found this problem quite interesting and thought about it while I was out and so wanted to generalize a bit and cut out some of the unnecessary hand-waving... Let's say the probability of getting a heads is $\\displaystyle p$ and we flip the coin $\\displaystyle 2n$ times where $\\displaystyle n\\in\\mathbb{N}$. What is the probability that the total number of heads is even? Let $\\displaystyle P(X)$ be the probability that the total number of heads is even and $\\displaystyle P(Y)$ be the probability that the total number of heads is odd. $\\displaystyle P(X)+P(Y)=1$ $\\displaystyle P(X)-P(Y)=\\sum_{k=0}^{2n}\\left[{2n \\choose k}p^{2n-k}(p-1)^k \\right]=(2p-1)^{2n}$ $\\displaystyle 2P(X)=1+(2p-1)^{2n}$ $\\displaystyle P(X)=\\frac{1}{2}\\left(1+(2p-1)^{2n} \\right)$", "can be seen by noting that the center of the standard simplex is ${\\textstyle \\left({\\frac {1}{n+1}},\\dots ,{\\frac {1}{n+1}}\\right)}$, and the centers of its faces are coordinate permutations of ${\\textstyle \\left(0,{\\frac {1}{n}},\\dots ,{\\frac {1}{n}}\\right)}$. Then, by symmetry, the vector pointing from ${\\textstyle \\left({\\frac {1}{n+1}},\\dots ,{\\frac {1}{n+1}}\\right)}$ to ${\\textstyle \\left(0,{\\frac {1}{n}},\\dots ,{\\frac {1}{n}}\\right)}$ is perpendicular to the faces. So the vectors normal to the faces are permutations of ${\\displaystyle (-n,1,\\dots ,1)}$, from which the dihedral angles are calculated. ### Simplices with an \"orthogonal corner\" An \"orthogonal corner\" means here that there is a vertex at which all adjacent edges are pairwise orthogonal. It immediately follows that all adjacent faces are pairwise orthogonal. Such simplices are generalizations of right triangles and for them there exists an n-dimensional version of the Pythagorean theorem: The sum of the squared (n − 1)-dimensional volumes of the facets adjacent to the orthogonal corner equals the squared (n − 1)-dimensional volume of the facet opposite of the orthogonal corner. ${\\displaystyle \\sum _{k=1}^{n}|A_{k}|^{2}=|A_{0}|^{2}}$ where ${\\displaystyle A_{1}\\ldots A_{n}}$ are facets being pairwise orthogonal to each other but not orthogonal to ${\\displaystyle A_{0}}$, which is the facet opposite the orthogonal corner. For a 2-simplex the theorem is the Pythagorean theorem for triangles with a right angle and for a 3-simplex it is de Gua's theorem for a tetrahedron with", "1. Equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_approx_p} is very interesting since it makes it easy to compute a value of $n$ for which we will have a collision with probability larger than or equal to $p$: $$\\begin{eqnarray} p = 1 - e^{-n(n-1)/(2N)} &\\Longrightarrow& \\displaystyle\\frac{-n(n-1)}{2N} = \\log(1 - p) \\nonumber\\\\[5pt] &\\Longrightarrow& \\displaystyle\\frac{n(n-1)}{2N} = \\log\\left(\\displaystyle\\frac{1}{1-p}\\right) \\nonumber\\\\[5pt] &\\Longrightarrow& n^2 \\geq 2N\\log\\left(\\displaystyle\\frac{1}{1-p}\\right) \\end{eqnarray}$$ Therefore: $$\\boxed{ n \\geq \\sqrt{2N\\log\\left(\\displaystyle\\frac{1}{1-p}\\right)} } \\label{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n}$$ will yield a collision with probability larger than or equal to $p$. For $N = 365$ and $p = 0.5$, we get $n \\approx 22.49$, so indeed for $n = 23$ we will have collisions with at least $50\\%$ probability. The birthday paradox gets even stranger if we take $N = 10^6$ (a million) and $p = 0.5$: for those parameters, $n \\approx 1177.4$, so we will have a collision with more than $50\\%$ probability with as few as $1180$ elements! In general, to produce a collision with $50\\%$ probability for a given $N$, we can select a number of elements given by: $$\\boxed{ n \\geq 1.2\\sqrt{N} } \\label{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n_approx}$$ The equation above is obtained by setting $p = 0.5$ on equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n} and rounding up the constant factor. For $N = 365$, equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n_approx} yields $n \\geq 22.92$ (no surprises here). Equation \\eqref{post_c59e4c7dd11d1d61bf902136ee9cafcb_eq_n_approx} closes our discussion of the problem. To have a collision with more than", "calculator or a computer to calculate μ and σ to reduce rounding error. For some probability distributions, there are short-cut formulas for calculating μ and σ. ## Example 6 Toss a fair, six-sided die twice. Let X = the number of faces that show an even number. Construct a table and calculate the mean μ and standard deviation σ of X. Click here to show answer: Tossing one fair six-sided die twice has the same sample space as tossing two fair six-sided dice. The sample space has 36 outcomes: (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Use the sample space to complete the following table: Calculating μ and σ. x P(x) x ⋅ P(x) (xμ)2P(x) 0 $\\displaystyle\\frac{{9}}{{36}}$ 0 $\\displaystyle{({0}-{1})}^{{2}} \\cdot \\frac{{9}}{{36}}=\\frac{{9}}{{36}}$ 1 $\\displaystyle\\frac{{18}}{{36}}$ $\\displaystyle\\frac{{18}}{{36}}$ $\\displaystyle{({1}-{1})}^{{2}} \\cdot \\frac{{18}}{{36}}={0}$ 2 $\\displaystyle\\frac{{9}}{{36}}$ $\\displaystyle\\frac{{18}}{{36}}$ $\\displaystyle{({2}-{1})}^{{2}} \\cdot \\frac{{9}}{{36}}=\\frac{{9}}{{36}}$ Add the values in the third column to find the Expected Value: $\\displaystyle \\mu =\\frac{{36}}{{36}}={1}$. Add the values in the", "# Acting As a Team I ### Strategy 1 Each player chooses the color randomly. The probability of success is $\\displaystyle P=\\left(\\frac{1}{2}\\right)^3=\\frac{1}{8}.$ ### Strategy 2 There are eight possible combinations of hat colors. Each player has to choose between $R(ed),$ $B(lue),$ or $P(ass),$ making $27$ possible team answers. Each player makes a random choice, now between three possibilities. If all pass, the team loses. With two passes, there are $6$ possible answers, each with probability of $\\displaystyle \\frac{1}{2}$ of a correct guess. With one pass, there are $12$ possible answers, each with probability of $\\displaystyle \\frac{1}{4}$ of a correct guess. With no passes, there are $8$ possible answers, each with probability of $\\displaystyle \\frac{1}{8}$ of being right. The total comes to \\displaystyle\\begin{align}P&=\\frac{1}{27}\\left(0\\cdot 1+\\frac{1}{2}\\cdot 6+\\frac{1}{4}\\cdot 12+\\frac{1}{8}\\cdot 8\\right)\\\\ &=\\frac{3+3+1}{27}=\\frac{7}{27}\\gt \\frac{7}{28}=\\frac{1}{4}. \\end{align} ### Strategy 3 Strategies 1 and 2 are stupid: the fewer people make a guess, the higher is the probability of success. However, in order to win at least one guess ought to be made. The probability of success is $\\displaystyle \\frac{1}{2}.$ ### Elwin Berlekamp's Strategy 1. If you see two hats of the same color, guess the other color. 2. If you see two hats of different colors, pass. The results of the strategy are summarized in the table below: $\\begin{array}{ccccccccc} A&B&C&&A&B&C&&Outcome\\\\ \\hline R&R&R&&GBW&GBW&GBW&&lose\\\\ R&R&B&&pass&pass&GBC&&win\\\\ R&B&R&&pass&GBC&pass&&win\\\\ R&B&B&&GRC&pass&pass&&win\\\\ B&R&R&&GRC&pass&pass&&win\\\\", "3. The probability of getting heads first and then tails: $\\scriptsize P(\\text{H;T)}=\\displaystyle \\frac{1}{4}$ 3. . 1. . \\scriptsize \\begin{align*}P(\\text{W)}&=\\displaystyle \\frac{6}{{14}}\\\\&=\\displaystyle \\frac{3}{7}\\end{align*} 2. . \\scriptsize \\begin{align*}P(\\text{R)}&=\\displaystyle \\frac{2}{{14}}\\\\&=\\displaystyle \\frac{1}{7}\\end{align*} 3. . $\\scriptsize \\begin{array}{l}P({R}')&=1-P(R)\\\\&=1-\\displaystyle \\frac{1}{7}\\\\&=\\displaystyle \\frac{6}{7}\\end{array}$ 4. . \\scriptsize \\begin{align*}P(\\text{W}\\cup \\text{R)}&=P(\\text{W)}+P(\\text{R)}\\\\&=\\displaystyle \\frac{3}{7}+\\displaystyle \\frac{1}{7}\\\\&=\\displaystyle \\frac{4}{7}\\end{align*} 5. . \\scriptsize \\begin{align*}P(\\text{neither W nor R})&=1-P(\\text{W or R)}\\\\&=1-\\displaystyle \\frac{4}{7}\\\\&=\\displaystyle \\frac{3}{7}\\end{align*} 4. . 1. . \\scriptsize \\begin{align*}P(\\text{G and G)}&=\\displaystyle \\frac{5}{{16}}\\cdot \\displaystyle \\frac{5}{{16}}\\\\&=\\displaystyle \\frac{{25}}{{256}}\\end{align*} 2. The probability of two green balls if the first ball is not replaced: \\scriptsize \\begin{align*}P(\\text{G and G)}&=\\displaystyle \\frac{5}{{16}}\\cdot \\displaystyle \\frac{4}{{15}}\\\\&=\\displaystyle \\frac{1}{{12}}\\end{align*} 3. . \\scriptsize \\displaystyle \\begin{align*}P(\\text{Y or O)}&=P(\\text{Y;O)}+P(\\text{O;Y)}\\\\&=\\left( {\\displaystyle \\frac{5}{{16}}\\cdot \\displaystyle \\frac{6}{{16}}} \\right)+\\left( {\\displaystyle \\frac{6}{{16}}\\cdot \\displaystyle \\frac{5}{{16}}} \\right)\\\\&=\\displaystyle \\frac{{15}}{{64}}\\end{align*} 5. . 1. . \\scriptsize \\begin{align*}P(\\text{M})&=\\displaystyle \\frac{{15}}{{35}}\\\\&=\\displaystyle \\frac{3}{7}\\end{align*} 2. . \\scriptsize \\begin{align*}P(\\text{F})&=\\displaystyle \\frac{{20}}{{35}}\\\\&=\\displaystyle \\frac{4}{7}\\end{align*} 3. . \\scriptsize \\begin{align*}P(\\text{M}\\cap \\text{F})&=P(\\text{M})\\cdot P(\\text{F})\\\\&=\\displaystyle \\frac{3}{7}\\cdot \\displaystyle \\frac{4}{7}\\\\&=\\displaystyle \\frac{{12}}{{49}}\\end{align*} Back to Exercise 1.3 ### Unit 1: Assessment 1. . 1. $\\scriptsize \\text{S}=\\{\\text{A; 2; 3; 4; 5; 6; 7; 8; 9; 10; J; K; Q}\\}$ 2. $\\scriptsize P(\\text{A})=\\displaystyle \\frac{1}{{13}}$ 3. $\\scriptsize P(2)+P(3)+P(5)+P(7)=\\displaystyle \\frac{4}{{13}}$ 4. $\\scriptsize P(\\text{A})+P(\\text{J})+P(\\text{K})+P(\\text{Q})=\\displaystyle \\frac{4}{{13}}$ 2. . 1. . \\scriptsize \\displaystyle \\begin{align*}P(\\text{A or B)}&=0.8\\\\P(\\text{A}\\cap \\text{B})&=0\\text{ since A and B are mutually exclusive}\\\\P(\\text{A or B)}&=P(\\text{A})+\\text{P(B)}\\\\0.3+k&=0.8\\\\\\therefore k&=0.5\\end{align*} 2. . \\scriptsize \\displaystyle \\begin{align*}P(\\text{A and B)}&=P(\\text{A)}\\cdot P(\\text{B) since A and B are independent}\\\\&=0.3\\times k\\\\P(\\text{A or B)}&=0.8\\\\P(\\text{A)}+P(\\text{B)}-P(\\text{A and B})&=0.8\\\\0.3+k-0.3k&=0.8\\\\0.7k&=0.5\\\\\\therefore k&=\\displaystyle \\frac{5}{7}\\end{align*} 3. The sum of the dice must equal eight, so it is", "the total number of dots appearing is 7” c) C is the event “Toss a dice twice in a row so that the number of dots appearing on two rolls is the same” ### Solving exercises Lesson 4, page 45, Math textbook 10 Kite, episode 2 Roll a dice twice in a row. Calculate the probability of each of the following events: a) “The total number of dots appearing in two sows is not less than 10”; b) “The 1-dot face appears at least once”. Solution method Step 1: Calculate the number of elements of the pattern space “$$n\\left( \\Omega )$$” and the number of elements of the result in favor of the event “$$n\\left( A )$$,$$n\\left( B )$$” Step 2: The probability of the event is: $$P\\left( A ) = \\frac{{n\\left( A )}}{{n\\left( \\Omega )}};P\\left( B ) = \\frac{{n\\left( B )}}{{n\\left( \\Omega )}}$$ Solution guide +) The pattern space in the above game is the set $$\\Omega = {\\rm{ }}\\left\\{ {\\left( {i,j} ){\\rm{ | }}i,{\\rm{ }}j{\\rm{ }} = {\\rm{ }}1,{\\rm{ }}2,{\\rm{ }}3,{\\rm{ }}4,{\\rm{ }}5, {\\rm{ }}6} \\right\\}$$ where (i,j) is the result “The first time appears the dotted face, the second time appears the dotted face j”. So $$n\\left( \\Omega ) = 36$$ a) Let A be the event “The total number of dots appearing in two", "# Thread: Probability Q Again.. 1. ## Probability Q Again.. Hi >.< I got another Q: The number showing on the upper faces when 2 fair six-sided dice are tossed, is observed. Consider events A and B A- the total score is at most 5 B- at least one die achieves a score of 3 or 4 thanks 2. Hello, procrastinator! With \"dice\" problems, there are no neat formulas. Usually, we must list the outcomes (or visualize them). The number showing on the upper faces when two fair six-sided dice are tossed, is observed. Consider events A and B A: the total score is at most 5 B: at least one die achieves a score of 3 or 4 I assume we are to find $\\displaystyle P(A)$ and $\\displaystyle P(B)$. (1) There are: .$\\displaystyle 6 \\times 6 \\:=\\:36$ possible outcomes. There are 10 outcomes with a sum of 5 or less: . . $\\displaystyle (1,1),\\,(1,2),\\,(1,3),\\,(1,4),\\,(2,1),\\,(2,2),\\,(2 ,3),\\,(3,1),\\,(3,2),\\,(4,1)$ Therefore: .$\\displaystyle P(A) \\;=\\;\\frac{10}{36} \\;=\\;\\frac{5}{18}$ (2) I have a \"back door\" approach to the second part. The probability that a die does not show a 3 or 4 is: .$\\displaystyle \\frac{4}{6} \\,=\\,\\frac{2}{3}$ The probability that neither die shows a 3 or 4 is: .$\\displaystyle \\frac{2}{3}\\!\\cdot\\!\\frac{2}{3}\\:=\\:\\frac{4}{9}$ Therefore, the probability that at least one die shows a 3 or 4 is: . . $\\displaystyle", "that the probability that a dash is transmitted as a dot is likewise 1/10, then Bayes's rule can be used to calculate $\\displaystyle{ P(dot \\ received) }$. $\\displaystyle{ P(dot \\ received) = P(dot \\ received \\ \\cap \\ dot \\ sent ) + P(dot \\ received \\ \\cap \\ dash \\ sent) }$ $\\displaystyle{ P(dot \\ received) = P(dot \\ received \\mid dot \\ sent)P(dot \\ sent) + P(dot \\ received \\mid dash \\ sent)P(dash \\ sent) }$ $\\displaystyle{ P(dot \\ received) = \\frac{9}{10}\\times\\frac{3}{7} + \\frac{1}{10}\\times\\frac{4}{7} = \\frac{31}{70} }$ Now, $\\displaystyle{ P(dot \\ sent \\mid dot \\ received) }$can be calculated: $\\displaystyle{ P(dot \\ sent \\mid dot \\ received) = P(dot \\ received \\mid dot \\ sent) \\frac{P(dot \\ sent)}{P(dot \\ received)} = \\frac{9}{10}\\times \\frac{\\frac{3}{7}}{\\frac{31}{70}} = \\frac{27}{31} }$[16] ## Statistical independence Main page: Independence (probability theory) Events A and B are defined to be statistically independent if the probability of the intersection of A and B is equal to the product of the probabilities of A and B: $\\displaystyle{ P(A \\cap B) = P(A) P(B). }$ If P(B) is not zero, then this is equivalent to the statement that $\\displaystyle{ P(A\\mid B) = P(A). }$ Similarly, if P(A) is not zero, then $\\displaystyle{ P(B\\mid A) = P(B) }$ is also equivalent. Although the derived forms may seem more", "# Problem 8 from the Spring 2018 Mathcounts ### Solution If the probability of hitting a region is proportional to its area, then the odds of hitting one in the outer ring to those of hitting one in the middle ring to those in the central circle are $5:3:1.$ Summing up the probabilities for each number we get $\\displaystyle P(1)=\\frac{10}{72},$ $\\displaystyle P(2)=\\frac{14}{72},$ $\\displaystyle P(3)=\\frac{12}{72},$ $\\displaystyle P(4)=\\frac{14}{72},$ $\\displaystyle P(5)=\\frac{10}{72},$ $\\displaystyle P(6)=\\frac{10}{12}.$ As the dice is fair, the probability of coincidence is proportional to th probability of hitting a number on the dartboard. $2$ and $4$ have the highest probabilities of being hit so that they are most likely to hit the outcome of the die roll. For the first question, the distribution on the dartboard is inconsequential. Once a number was hit, there is the same probability of $\\displaystyle \\frac{1}{6}$ that it will be matched by the outcome of a die roll. ### Acknowledgment This problem is based on problem 8 from the Spring 2018 Mathcount competition.", "unfavorable. In both cases then, the probability that two girls have been selected to advance to the national finals is $\\displaystyle \\frac{2}{6} = \\frac{1}{3}.$ For the third question, you were as likely to cast your eyes on the first girl as on the second. By the Symmetry Principle, the probabilities of the two events are equal to $\\displaystyle \\frac{1}{2}$ whereas each of the girls had an individual probability of $\\displaystyle \\frac{1}{3}$ to be a winner. Thus the total probability of both girls is $\\displaystyle \\frac{1}{2}\\cdot\\frac{1}{3}+\\frac{1}{2}\\cdot\\frac{1}{3}=\\frac{1}{3}.$ For the fourth problem, there are $10$ possible outcomes, with only two favorable. The probability of having two lucky girls is then $\\displaystyle \\frac{2}{10} = \\frac{1}{5}.$ ### Solution 2 (NCTM) 1. If the girl that you saw on the stage was the first winner, then out of the three students left for the random drawing for a second winner, only one is a girl, and two are boys. The chances that the second one is also a girl will be $1$ in $3,$ or $\\displaystyle \\frac{1}{3}.$ However, if the girl that you saw on stage was actually the second winner, then the chance that the first winner was also a girl is $\\displaystyle \\frac{1}{2},$ because before the first girl was picked, there were two girls and two boys, so the chance of a", "4.9k views A six sided unbiased die with four green faces and two red faces is rolled seven times. Which of the following combinations is the most likely outcome of the experiment? 1. Three green faces and four red faces. 2. Four green faces and three red faces. 3. Five green faces and two red faces. 4. Six green faces and one red face edited | 4.9k views 0 The number of green faces is twice the number of red faces in the dice. So, green is obviously more likely to come up more. A eliminated. B is too close. Eliminated. D is too much out-of-proportion. Eliminated. Only A seems reasonable. We can calculate the probability of each of the options but by logic we can easily eliminate the ones with more number of ${\\color{Red} {red}}$ faces - option $A$ can be avoided without any calculation. $A = P(3 {\\color{Green} G},4{\\color{Red} R}) = {}^7C_3. {\\color{Green}{\\left(\\frac{4}{6}\\right)^3}}{\\color{Red}{\\left(\\frac{2}{6}\\right)^4}}$ $B = P(4 {\\color{Green} G},3{\\color{Red} R}) = {}^7C_4. {\\color{Green}{\\left(\\frac{4}{6}\\right)^4}}{\\color{Red}{\\left(\\frac{2}{6}\\right)^3}}$ $C = P(5 {\\color{Green} G},2{\\color{Red} R}) = {}^7C_5. {\\color{Green} {\\left(\\frac{4}{6}\\right)^5}}{\\color{Red} {\\left(\\frac{2}{6}\\right)^2}}$ $D = P(6 {\\color{Green} G},1{\\color{Red} R}) = {}^7C_6.{\\color{Green} { \\left(\\frac{4}{6}\\right)^6}}{\\color{Red}{\\left(\\frac{2}{6}\\right)^1}}$ Option $A$ is clearly smaller and hence eliminated. $\\frac{C}{B} = \\frac{{}^7C_5}{{}^7C_4}.\\frac{4}{6}. \\frac{6}{2} = \\frac{3}{5} . 2 > 1 .$ So, $C > B.$ $\\frac{C}{D} = \\frac{{}^7C_5}{{}^7C_6}.\\frac{6}{4}. \\frac{2}{6} = \\frac{21}{7}.0.5 > 1 .$ So,", "represented by$S = {s_1, s_2, dots, s_N}$ • For example: • $E = {s_2, s_3, s_7, s_9} = {s_2} cup {s_3} cup {s_7} cup {s_9}$ • So, $P(E) = ?$ • $P(E) = P({s_2}) + P({s_3}) + P({s_7}) + P({s_9})$ • If we know the probability of the individual outcomes, we can compute the probability of any event by adding the probabilities of the relevant outcomes. • Core assumption: individual events in the collection are separate by convention. • Important special case (limit of most high school probability): Suppose that each of the $N$ outcomes is “equally likely” (recurring phrase we’ll hear a lot). In this case, each outcome has probability:$P({s_i}) = frac{1}{N} mbox{for } i = 1, 2, dots, N$ This means that with equally likely outcomes, figuring probabilities is basically just counting. • Example: $E = { s_1, s_2, dots, s_k}$ for $k$ outcomes in $E$, then $P(E) = P({s_1}) + P({s_2}) + dots + P({s_k})$ $P(E) = frac{1}{N} + frac{1}{N} + dots + frac{1}{N}$ $P(E) = frac{displaystyle k}{displaystyle N}= frac{displaystyle #(mbox{outcomes in E})}{ displaystyle#(mbox{outcomes in S})}$ • 3 of a kind example: Draw 5 cards from a standard deck of cards. What is $P(mbox{3 of a kind})$? A: $frac{displaystyle #(mbox{3 of a kind})}{displaystyle #(mbox{5 cards})} = frac{displaystyle formula , derived , previously}{displaystyle {52 choose 5}}$", "1. I will now say QED just for extra asshattishness. QED PacWalker 03:37, 7 April 2015 (UTC) No proof, only foolhardy definition ${\\displaystyle 0.999\\dots ={\\frac {9}{10}}+{\\frac {9}{100}}+{\\frac {9}{1000}}+\\dots }$ ${\\displaystyle 0.999\\dots ={\\frac {9}{10}}+{\\frac {9}{10^{2}}}+{\\frac {9}{10^{3}}}+\\dots }$ ${\\displaystyle 0.999\\dots =\\sum _{i=1}^{\\infty }9\\cdot \\left({\\frac {1}{10}}\\right)^{i}=\\sum _{i=0}^{\\infty }{\\frac {9}{10}}\\cdot \\left({\\frac {1}{10}}\\right)^{i}}$ This is also known as the Eulerian Blunder, that is, S = Lim S where ${\\displaystyle S=0.999\\dots =\\sum _{i=1}^{\\infty }{\\frac {9}{10^{i}}}=lim_{n\\to \\infty }S=lim_{n\\to \\infty }\\sum _{i=1}^{n}{\\frac {9}{10^{i}}}=1}$ Notice there is no proof here of any sort, only a foolhardy definition of 0.999... (an ill-formed object) as 1.— Unsigned, by: Hot sauce / talk / contribs If they are distinct, give us a number between 0.9 recurring and 1. Show a property of 1 that 0.9 recurring does not satisfy other than appearance. 86.141.180.13 (talk) 04:18, 8 September 2016 (UTC) And if you have a number between 0.9 recurring and 1 (call it x), then you can construct a decimal of the format 0.9999...9 with a finite number of 9s that's greater than x but less than 0.9 recurring. Wow! Annquin (talk) 16:16, 8 September 2016 (UTC) Watch. {\\displaystyle {\\begin{aligned}{\\frac {1}{9}}&=0.111\\dots \\\\9\\times {\\frac {1}{9}}&=9\\times 0.111\\dots \\\\1&=0.999\\dots \\end{aligned}}} QED. Vive Liberté! 12:35, 21 May 2017 (UTC) ## P ?= NP This article seriously overstates the practical impact of P = NP. \"Polynomial" ]

[ "+0 # PLSSSSS HELPPPPP 0 52 1 A club has 30 members: 15 boys and 15 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl? Mar 21, 2022 #1 +117105 +1 A club has 30 members: 15 boys and 15 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl? 1- P(all girls) - P(all boys) =$$=1-2*\\frac{15C4}{30C4}\\\\ =1-2*\\frac{1365}{27405}\\\\ =1-\\frac{26}{261}\\\\ =\\frac{235}{261}$$ Mar 21, 2022", "25 Apr 2012, 10:59 1 A certain team has 12 members, including Joey. A three-membe 13 23 Oct 2011, 13:16 12 A certain club has 20 members. What is the ratio of the memb 18 25 Mar 2009, 23:04 Display posts from previous: Sort by", "# The ratio of boys to girls in the Science Club is 3:5. If there are 60 girls, how many boys are there? Nov 18, 2016 $36 \\text{ boys}$ #### Explanation: Express the ratio in fractional form. Let the number of boys be x. $\\Rightarrow \\frac{3}{5} = \\frac{x}{60}$ cross-multiply. $\\Rightarrow 5 x = 180$ $\\Rightarrow x = \\frac{180}{5} = 36$ There are 36 boys in the Science Club", "Jan 2008, 11:23 20 A certain club has 10 members, including Harry. One of the 19 08 Jan 2007, 23:22 Display posts from previous: Sort by", "# Math Help - Pri 5 maths percentage question 2 1. ## Pri 5 maths percentage question 2 There are 875 members in an astronomy club.350 of the members are females.What percentage of the members are males? 2. If there 875 members in an astronomy club (that's a lot of members ) and 350 of the members are females, how many members are male? Find that answer and write it as a fraction. (What do I mean by this? Well, if there were 256 members in an astronomy club and 176 of them are Asian, then the corresponding fraction would be $\\frac{176}{256}$.) Finally, setup a proportion to convert to percent, as I've explained in this thread: http://www.mathhelpforum.com/math-he...stion-1-a.html . 01", "trip. if 39 people went on that trip, how many are in the club...", "are 4 people in both clubs? Then there are 11 people in just the maths club, and 8 people just in the science club. That is a total of 19 people in only one club, which is 6 more than 13. So we could try half-way between 4 and 10 - what if there are 7 people in both clubs? Then there are 8 people in just the maths club, and 5 people in just the science club. That is a total of 13 people in only one club - perfect. Using trial on a venn diagram In this first Venn diagram, the 15 students in the maths club and the 12 students in the science club are all shown as dots, and all shown as being in only one club. Putting a student into the overlap is the same as moving one student from each side, since the student in the overlap is in both clubs - there are still 15 students in the maths club and 12 students in the science club. So for every student put into the overlap, the number of students in just one club goes down by 2. We can keep going until the number of people in one club is 13. There are 7 people in both clubs. Using algebra on", "Total members is 40>M>10, so 29 members. Now if each table has 6 members, that is 29/6 then the last table would have 5 (29-24) members on the table. Is this approach right? I did the problem in less than a minute with this approach. Thanks in advance. How did you get that there must be 29 members from 40>M>10? Does 29 members satisfy ANY of the conditions (3 members at one table and 4 members at each of the other tables and 3 members at one table and 5 members at each of the other tables)? _________________ IIMA, IIMC School Moderator Joined: 04 Sep 2016 Posts: 1340 Location: India WE: Engineering (Other) Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe [#permalink] ### Show Tags 27 Feb 2018, 09:14 Quote: Club X has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at", "+0 # Probability 0 37 1 The Grammar club has 20 members: 10 boys and 10 girls. A 4-person committee is chosen at random. What is the probability that the committee has 2 boys and 2 girls? Oct 7, 2021 #1 +785 0 $\\frac{\\binom{10}{2}}{\\binom{20}{2}} \\cdot \\frac{\\binom{10}{2}}{\\binom{18}{2}}$ Feel like this is wrong .. edited by MathProblemSolver101 Oct 7, 2021 edited by MathProblemSolver101 Oct 7, 2021", "topics Replies Last post Similar Topics: What is the number of members of the Booster Club who are under 21 5 02 Dec 2015, 05:54 A certain club has 20 members. How many of the members of the club are 3 28 Oct 2015, 06:01 3 Eleven members of Club A are also members of Club B, and five members 5 13 Feb 2015, 07:47 1 What is the number of members of Club X who are at least 35 2 23 Jul 2010, 19:11 11 The members of a club were asked whether they speak 19 05 Aug 2007, 03:32 Display posts from previous: Sort by", "# A chess club has 10 members, of whom 6 are men and 4 are women #### zx0xc0 A chess club has 10 members, of whom 6 are men and 4 are women. A team of 4 members is selected to play in a match. Find the number of different ways of selecting the team if Given that the 6 men include 2 brothers, find the total numberof ways in which the team can be selected if either ofthe brothers, but not both, must be included. #### 123qwerty Let's split the club into two parts: the brothers and everyone else. Then the total number of ways in which the team can be selection is $$\\displaystyle 2 \\times \\displaystyle{8 \\choose 3}$$.", "Question At Annville Junior High School, $30\\%$ of the students in the Math Club are in the Science Club, and $80\\%$ of the students in the Science Club are in the Math Club. There are $15$ students in the Science Club. How many students are in the Math Club 1. hongcuc2 The total number of students who are in Maths club is 12. If there are 15 students in Maths club and 80 percent of the students in the Science club are in Maths club. According to the given question. Total number of students in the science club = 15 30 percent of the students in maths club are in the science club. And, 80 percent of the students in the science club are in the maths club. ⇒ 80/100 × 15 = 12 ⇒ 12 students of the science club are in the Maths club. Therefore, the total number of students who are in Maths club is 12.", "(2 pts:): A computer science department has 50 faculty members_ (a) How many ways are there to choose an executive committee COIL- sisting of 6 members? (b) How many ways are there to choose the chair vice chair. director of undergraduate study; director of graduate study, director of computing facilities_ and director of public relations if no person can hold more than position? Problem (2 pts:): A computer science department has 50 faculty members_ (a) How many ways are there to choose an executive committee COIL- sisting of 6 members? (b) How many ways are there to choose the chair vice chair. director of undergraduate study; director of graduate study, director of comput...", "# 100,000 Members 1. May 29, 2008 ### Ivan Seeking Staff Emeritus 100,000 Members!!! We should break the 100,000 PF members mark in the next few days - we have been growing at a rate of 150 to 200 members per day but it does vary. As of right now we are at members = 99,565. I think this is a huge milestone for PF. Congratulations to the Admin and staff, our Science Advisors and Homework Helpers, our gold members who pay to keep PF alive, and the many bright and dedicated members who have helped to further the cause of education and to make PF the best on the web. If the typical internet forum is an internet community, then I think PF has achieved the status of being an internet city; a city of our best and brightest! When I think of PF, I think of a commercial... not sure who did it, but it showed a hall full of young academics, I think scientist and engineers, and asked [approximately] the question: What great things can be achieved when 100 of our best and brightest put their minds together? So when I think of PF, I wonder what great things can acheived when 100,000 of the worlds best and brightest put their minds together? I will", "### Inclusion Exclusion How many integers between 1 and 1200 are NOT multiples of any of the numbers 2, 3 or 5? ### Rain How many dry days did Aisha have on her holiday? # Clubs ##### Age 11 to 14 ShortChallenge Level Using words and trial Think about the people in both clubs. Suppose that there are 10 people in the maths club who are in both clubs. Then those 10 people are also in the science club. And there can't be any other people in the science club who are in both clubs, because if there were, then they would be in the maths club - but we have already counted the 10 people in the maths club who are in both clubs! So if there are 10 people in the maths club who are in both clubs, then there are also 10 people in the science club who are in both clubs - and they are the same people. If there are 10 people who are in both clubs, then there are 15$-$10 = 5 people in just the maths club, and 12$-$10 = 2 people in just the science club. That is a total of 7 people in only one club, which is 6 less than 13. 10$-$6 = 4, so what if there", "chart showing the number of students in each club is shown below. There are 50 children in Vidyarangam. Make a table showing the number of children in each club. Given data: Number of children in vidyarangam=50 Solution: From the pie chart, we can observe that science club is half of the vidyarangam club. Therefore, number of children in science club will be $$\\frac{1}{2}$$×50=25. Number of children in social and english club seems to be equal to the science club. Therefore, number of children in social and english club will be 25. Number of children in maths club are three times of the science club. Therefore, number of children in maths club will be 3×25=75. From our assumptions, we can make the below table. Piechart with computers Textbook Page No. 192 Pie charts can be easily drawn using a computer. Open Libre Office Calc program and type in the details as below: Click on any cell and choose Insert → chart → pie. This makes a pie-chart. Change the numbers and see what happens to the picture. Given data: From the above table, we get the number of students from each club and the total number of students belonging to all the clubs. Total number of students: 30+20+25+15+10=100 To draw a pie chart, we need to calculate the angle", "one cheese slice for each cracker. Patrick made the statement shown. If Andy doesn’t want any crackers or cheese slices left over, he needs to buy at least 432 of each. Is Patrick’s statement correct? Use numbers and words to explain why or why not. If Patrick’s statement is incorrect, what should he do to correct it? Explanation: Multiples of 18: 18,36,54,72 Multiples of 24: 24,48,72 LCM is 72 So the least packages he need to buy is 72. Question 13. There are 16 sixth graders and 20 seventh graders in the Robotics Club. For the first project, the club sponsor wants to organize the club members into equal-size groups. Each group will have only sixth graders or only seventh graders. Part A How many students will be in each group if each group has the greatest possible number of club members? Show your work. Answer: Each group will have 4 members, and 4 groups of sixth grade and 5 groups of seventh grade. Explanation: By distributive property 16+20 =(4×4)+(4×5) =4×(4+5) So each group will have 4 members, and 4 groups of sixth grade and 5 groups of seventh grade. Question 13. Part B If each group has the greatest possible number of club members, how many groups of sixth graders and how many groups of seventh graders", "# Word problem • July 8th 2012, 12:35 PM Ragnarok Word problem From a prealgebra problem book: 30% of the students in the Math Club are in the Science Club, and 80% of the students in the Science Club are in the Math Club. There are 15 students in the Science Club. How many students are in the Math Club? I'm getting confused about what my variables and equations should be. Could anyone help out? • July 8th 2012, 01:12 PM Furyan Re: Word problem '80% of the students in the Science Club are in the Math Club. There are 15 students in the Science Club'. You can find out how many students in the Science club are also in the Math club. 15 x 0.8... What can you do next? • July 8th 2012, 01:35 PM Soroban Re: Word problem Hello, Ragnarok! Quote: 30% of the students in the Math Club are in the Science Club, and 80% of the students in the Science Club are in the Math Club. There are 15 students in the Science Club. How many students are in the Math Club? Let $M$ = number of students in the Math Club. Let $S$ = number of students in the Science Club. $\\begin{array}{cccc}\\text{30\\% of the Math Club is in both clubs:}& 0.30M \\,=\\,\\text{both}", "Free Online Trial Hour Re: A certain club has 20 members. What is the ratio of the memb [#permalink] 06 Jun 2016, 08:41 Go to page 1 2 Next [ 22 posts ] Similar topics Replies Last post Similar Topics: 8 A local club has between 24 and 57 members. The members of the club ca 9 10 May 2017, 12:39 30 A certain club has exactly 5 new members at the end of its 12 13 Sep 2016, 11:51 4 A certain club has 10 members, including Harry. One of the 8 25 Jul 2012, 21:13 145 A certain club has 10 members, including Harry. One of the 33 31 May 2017, 22:01 1 A certain team has 12 members, including Joey. A three-membe 13 14 Dec 2011, 10:58 Display posts from previous: Sort by", "x + x = 55 ----> 45 + x = 55 ----> x = 10. Sufficient. Re: A club has 80 members, each of which practice one or more of tennis, [#permalink] 23 Jul 2019, 06:14 Display posts from previous: Sort by" ]

[ "Courses Courses for Kids Free study material Free LIVE classes More # There are three boys and two girls. A committee of two is to be formed. Find the probability of events that the committee contains one boy and one girl. Last updated date: 18th Mar 2023 Total views: 311.4k Views today: 8.91k Verified 311.4k+ views Hint: Here we will use the concept of combination. First find the selection of 2 people from the committee of 5 which gives us the total number of cases. Later find the selection of a boy and a girl from the committee which gives the favorable case. There are three boys and two girls and we have to select two person from the sample so, total number of ways of selecting them is ${}^5{C_2} = \\dfrac{{5!}}{{3! \\times 2!}} = 10$ Number of ways for selecting the committee of 1 girl & 1 boy =${}^2{C_1} \\times {}^3{C_1}$ $\\Rightarrow 2 \\times 3 = 6$ So, required probability is = $\\dfrac{favorable\\;ways}{total\\;ways}$ $\\Rightarrow \\dfrac{6}{{10}}$ $\\Rightarrow \\left( {\\dfrac{3}{5}} \\right)$ So, this is our required probability. NOTE: - In these types of questions always find the first total number of ways, then favorable ways, then divide favorable ways by total ways, we will get the required probability.", "the second ball is green is $$\\dfrac{1}{12}$$ Example 3 Mrs. Andrews has to select two students from 35 girls and 15 boys to be part of a club. What is the probability that both students are girls? Solution The total number of students $$= 35 + 15 = 50$$ Probability of choosing the first girl, P(girl 1) = $$\\dfrac { 35}{50}$$ Probability of choosing the second girl, P(girl 2) = $$\\dfrac { 34}{49}$$ Now, The probability that both students have chosen are girls $$\\text {P(first girl and second girl)}$$ $$= \\text{ P(first girl)} \\times \\text{P(second girl | first girl)}$$ $$= \\dfrac { 35}{50} \\times \\dfrac { 34}{49}$$ $$= \\dfrac{1190}{1666}$$ $$= \\dfrac{85}{119}\\$$ $$\\therefore$$The probability that both chosen students are girls is $$\\dfrac{85}{119}$$. Example 4 Joseph and David are playing a game of cards. Joseph drew a card at random without replacement. He asks David to help him determine the probability that the first card drawn was a queen and the second is a king. ### Solution As we understand that this probability is having a dependent event condition. P (drawing a queen in the first condition) =$$\\dfrac { 4}{52}$$ P (drawing a king in the second condition after a queen) = $$\\dfrac { 4}{51}$$ P (drawing a queen followed by a king) = $$\\dfrac{4}{52} \\times\\dfrac{4}{51}=\\dfrac{16}{2652}=\\dfrac{4}{663}$$ $$\\therefore$$The probability of drawing", "3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black? 1/4 1/2 1/2 5/8 2/3 3/4 http://gmatclub.com/forum/3-yellow-balls-and-5-black-balls-84530.html 47. Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary. What is the probability that only girls will be elected? What is the probability that only boys will be elected? What is the probability that at least one girl will be elected? http://gmatclub.com/forum/comb-none-56860.html 48. For one toss of a certain coin,the probability that the outcome is heads is 0.6.If this coin is tossed 5 times,which of the following is the probability that the outcome will be heads at least 4 times? A. (0.6)^5 B. 2(0.6)^4 C. 3(0.6)^4*(0.4) D. 4(0.6)^4*(0.4)+(0.6)^5 E. 5(0.6)^4*(0.4)+(0.6)^5 http://gmatclub.com/forum/probability-84565.html 49. A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women? 1/14 1/7 2/7 3/7 1/2 http://gmatclub.com/forum/probability-56037.html 50. The choir consists of 5 boys and 6 girls. In how many ways can the singers be arranged in a row, so that all the boys are together? (assume", "Question # A class has 20 girls and 16 boys. One student is selected at random. Find the probability that the selected student is a girl.A. $\\dfrac{1}{4}$B. $\\dfrac{5}{9}$C. $\\dfrac{3}{4}$D. $\\dfrac{4}{9}$ Hint: In this particular problem find the total number of students and then use the probability formula for finding probability of getting a girl on selection. Formula for probability of getting a favourable outcome is = $\\dfrac{{{\\text{Number of favourable outcomes}}}}{{{\\text{Total number of outcomes}}}}$ As we know from the concept of the probability that if there are n balls out of then x are of same type A then the probability of getting a ball of type A on random selection is equal to = $\\dfrac{{{\\text{Number of favourable outcomes}}}}{{{\\text{Total number of outcomes}}}}$. So, in this example favourable outcome will be ball of type A. So, probability = $\\dfrac{{{\\text{Number of balls of type A}}}}{{{\\text{Total number of balls}}}} = \\dfrac{x}{n}$ So, probability that the selected student is girl = $\\dfrac{{{\\text{Number of girls}}}}{{{\\text{Total number of students in class}}}} = \\dfrac{{20}}{{36}} = \\dfrac{5}{9}$", "The total probability is the product of these since the selections are independent, so $p = \\dfrac{16}{24}\\dfrac{15}{23}\\dfrac{8}{22} = \\dfrac {70}{253}$ one can observe that this number remains unchanged if we change the order that the students were selected. For example if we chose the girl first we'd have $p = \\dfrac{8}{24} \\dfrac{16}{23}\\dfrac{15}{22}$ and this is the same value as that calculated by assuming the selection was bbg. April 20th, 2017, 12:14 PM #3 Newbie Joined: Feb 2017 From: Denmark Posts: 11 Thanks: 1 Quote: Originally Posted by romsek we have to assume the selections are all independent of one another. Assume we chose 2 boys and then a girl Choosing the first boy has probability $\\dfrac{16}{24}$ as there are 16 boys to choose from 24 students. Similarly the probability of choosing the second boy is $\\dfrac{15}{23}$ as there are now 15 boys and 23 students to choose from. Finally the probability of choosing the girl is $\\dfrac{8}{22}$ The total probability is the product of these since the selections are independent, so $p = \\dfrac{16}{24}\\dfrac{15}{23}\\dfrac{8}{22} = \\dfrac {70}{253}$ one can observe that this number remains unchanged if we change the order that the students were selected. For example if we chose the girl first we'd have $p = \\dfrac{8}{24} \\dfrac{16}{23}\\dfrac{15}{22}$ and this is the same value as that calculated", "is the probability of having all the aces (4 in total) if you get dealt five cards in a row\". The second question just reduces the total number of possible outcomes in the denominator. By knowing, that one female and one male is already drawn the, remaining \"pool\" of possible outcomes comes from looking at what the problabiltiy of drawing the remaining female is. I.e., we look at (f,m,f),(f,f,m) and (m,f,f) and will have all possible outcomes as before, but, it is not possible to have (m,m,m). The probability that a randomly selected committee of three includes both females must equal the probability that the two individuals not selected for the committee are both male. The latter can be calculated as $\\frac{3}{5}\\cdot\\frac{2}{4} = 0.3$. For the second question, after selecting one female and one male there remain three unselected individuals, one female and two males. To obtain a committee with two females, the female must be selected from those three, which has probability $\\frac{1}{3}$.", "+0 # Probability with Combinations +1 48 3 +485 The Grammar club has 20 members: 10 boys and 10 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl? Apr 3, 2021 #1 +1 There is a probability of 1/2 of choosing a boy, then a 10/19 probability of choosing a girl. Then the other two people can be anyone, so the probability is 2*1/2*10/19 = 10/19. Apr 3, 2021 #2 +485 +1 That was incorrect, but I have one more try. RiemannIntegralzzz Apr 3, 2021 #3 +485 +1 Actually, I figured it out myself: You can use complementary counting/combinations here: The number of ways to choose a committee of all boys or all girls is $2*C(10,4)$(2 for the girls and boys)$=420$. The total number of committees is $C(20, 4)=4845$. So the answer is $1-\\frac{420}{4845}=\\frac{4425}{4845}=\\boxed{\\frac{295}{323}}$ Apr 3, 2021", "# Moderate Probability Solved QuestionAptitude Discussion Q. Four boys and three girls stand in queue for an interview. The probability that they stand in alternate positions is: ✔ A. 1/35 ✖ B. 1/34 ✖ C. 1/17 ✖ D. 1/68 Solution: Option(A) is correct Total number of possible arrangements for 4 boys and 3 girls in a queue $= 7!$ When they occupy alternate position the arrangement would be like: $B$ $G$ $B$ $G$ $B$ $G$ $B$ Thus, total number of possible arrangements for boys $= (4×3×2)$ Total number of possible arrangements for girls $= (3×2)$ Required probability $=\\dfrac{(4\\times 3\\times 2\\times 3\\times 2)}{7!}$ $=\\dfrac{1}{35}$ ## (6) Comment(s) Suman () IT can be like GBGBGBGBG like snehal said then 5C3*3! for girls and 4! for boys Ashok () The question specifically said that alternative position and choosing position other than BGBGBGB will result in at least 2 boys back to back which is not allowed. Now once this is understood, the 4 boys can shuffle themselves 4! and girls can 3!. so total favourable ways are 4!x3! and prob is divide it by 7! Snehal () Girls have 5 places to choose from ....at the start, between boys and a place after the last boy.....They can select the positions in 5C3 ways and then the girls can occupy the three", "is. Do not confuse yourself. You should know the probability $P(A)=\\dfrac{n(A)}{n(S)}$. You should be also familiar with the events such as we have an event $A$ that contains a committee of at least one girl.", "# Math Help - Simple Probabilities Plz Help 1. ## Simple Probabilities Plz Help Hey, Can someone plz answer the following questions and provide explanation. 1. P(E)=0.25 and 20 outcomes make up event E. Determine the number of outcomes? 2. There are 7 girls, 9 boys in a school camera club. The club needs to form a committee of 6 random people. Determine the outcome if: a) The committee has no boys. b) The committee has at least 1 girl. 3. You have a regular 6 side die. You roll it 2 times. Determine if: a) Probability that you roll a sum of 7. b) Probability that you roll a sum of 7 if the first roll is 5. Any help would be greatly appreciated. Thanx 2. 1. $20 \\times 0.25 = 5$ I'm not 100% sure about this one. 2. so there are a total of 16 students and out of them you are interested in choosing six. $\\binom{16}{6} = 8008$ possible picks. a) $P(X=0) = \\frac{\\binom{9}{0} \\binom{7}{6}}{8008} = \\frac{7}{8008}$ b) $P(Y \\geq 1) = 1- P(Y=0) = 1- \\frac{\\binom{9}{6} \\binom{7}{0}}{8008} = 1- \\frac{84}{8008} = \\frac{283}{286}$ 3. your sample would consist of $S= (1,1), (1,2), ... , (6,6)$ a) so out of those there are 6 possible ways to get a 7 and $36= 6\\times 6$ outcomes.", "+0 # If a committee of six students is chosen at random from a group of six boys and four girls, what is the probability that the committee conta 0 387 2 +201 If a committee of six students is chosen at random from a group of six boys and four girls, what is the probability that the committee contains the same number of boys and girls? Express your answer as a common fraction. May 19, 2021 #1 +177 +2 Number of ways for the committee to have 3 boys, 3 girls: $$\\binom{6}{3} \\cdot \\binom{4}{3} = 20 \\cdot 4 = 80.$$ Number of total ways: $$\\binom{10}{6} = 210$$ So, the answer is $$\\frac{8}{21}.$$ May 19, 2021 #2 +201 0 Thanks ! mathisopandcool May 19, 2021", "# Math Help - More Data .. Can you figure it out? 1. ## More Data .. Can you figure it out? A school has 480 girls and 520 boys. how many committees of 5 can u make if at least one boy must be on each committee. 2. There are two ways to do it: 1) Simpler Way (solving it indirectly): Find the total number of committees and then subtracts the number of committees which are all girls, as $\\binom{1000}{5}-\\binom{480}{5}$ OR 2) Solving it directly: $\\binom{520}{1}\\binom{480}{4}+\\binom{520}{2}\\binom{ 480}{3}+\\binom{520}{3}\\binom{480}{2}+\\binom{520}{4 }\\binom{480}{1}+\\binom{520}{5}$ 3. Hello, wikji! Who assigned this problem? . . In fact, who wrote it? A school has 480 girls and 520 boys. How many committees of five can be made if at least one boy must be on each committee? There are: . ${1000\\choose5} \\:\\approx\\:82.5\\times 10^{11}$ possible committes. The opposite of \"at least one boy\" is \"no boys\" (all girls). There are: . ${480\\choose5} \\;\\approx\\:2.1\\times10^{11}$ all-girl committees. There are about: . $8\\times 10^{12}$ (eight trillion) committees with at least one boy. 4. Originally Posted by Soroban Hello, wikji! Who assigned this problem? . . In fact, who wrote it? There are: . ${1000\\choose5} \\:\\approx\\:82.5\\times 10^{11}$ possible committes. The opposite of \"at least one boy\" is \"no boys\" (all girls). There are: . ${480\\choose5} \\;\\approx\\:2.1\\times10^{11}$ all-girl committees. There are about:", "How to find the probability of of simple events, multiple independent events, a union of two events. chocolate bars produced by a certain machine is normal with mean 8. However, if you mean the probability of there being at least one women in any committee makeup, then I'd recommend you use the complement. No, each student may have a different probability. Find the probability that the two have different party affiliations (that is, not both A, not both B, and not both independent). 2) The committee must contain at least three English teachers. B)Students outnumber the teachers on the committee if and only if at least 3 students are selected. Suppose that Alice, Betty, Cindy, and Dianne have their own private outdoors club, and that they need to select a president, a secretary, and a treasurer. The probability that a male develops some form of cancer in his lifetime is 0. Note that this is an aggregate result, and not necessarily true for any one particular piece of. When you want to find the probability of one event OR another occurring, you add their probabilities together. Ok so you have 10 students. SB 2955 has already cleared the Senate and has been transferred to the House. 4 Exactly two girls and a boy being selected. Its work is", "QUESTION # 4 boys and 2 girls occupy seats in a row at random. Then the probability that the girls occupy seat side by side is[a] $\\dfrac{1}{2}$[b] $\\dfrac{1}{4}$[c] $\\dfrac{1}{3}$[d] $\\dfrac{1}{6}$ Hint: Probability of event E = $\\dfrac{n(E)}{n(S)}=\\dfrac{\\text{Favourable cases}}{\\text{Total number of cases}}$ where S is called the sample space of the random experiment. Find n (E) and n (S) and use the above formula to find the probability. Use the grouping method to find the values of n(E) and use the fact that arranging n elements in a row is given by n!. Let E be the event: The two girls occupy seats side by side. Let us consider the girls as a single group. Hence we have four boys and one group to be arranged which can be done in 5! ways. But the girls inside the group can internally arrange themselves in 2! ways. Hence the total number of ways in which four boys and two girls can be seated so that the girls are seated side by side is given by 5!2! ways Hence, we have n (E) = 5!2! = 240. The total number of ways in which we can arrange four boys and two girls without any restriction is given by 6! =720. Hence, we have n (S) =720 Hence, P (E) = $\\dfrac{240}{720}=\\dfrac{1}{3}$", "sum of those, as nazz gives. I see, thanks. 9. ## Re: Probability What about part b) can someone shed more light on it please and thanks for all the replies 10. ## Re: Probability Originally Posted by nazz There are 4 girls and 5 boys would like to go on school trip but there are onlyy 2 seats available. b) if Sarah is one of the girls who want to go on the trip, what is the probabily of not choosing Sarah? How to solve it? The probability that Sarah is chosen equals $\\dfrac{\\binom{8}{1}}{\\binom{9}{2}}~.$ 11. ## Re: Probability Thanks for for all the replies you all have been great.", "# We are given a class consisting of 4 boys and 4 girls. We are given a class consisting of 4 boys and 4 girls. a committee that consists of a President, a Vice-President and a secretary is to be chosen among the 8 students of the class. Let a denote the number of ways of choosing the committee in such a way that the committee has at least one boy and atleast one one girl. Let b denote the number of ways of choosing the committee in such a way that the number of girls is greater than or equal to that of boys. Then what are the values of a and b? As Simon said in his answer above, $a$ can have two cases: $a.1:$ 2 boys 1 girl, which yields $\\binom{4}{2}\\binom{4}{1}$ choices for people and $3!$ arrangements for their positions which yields $6\\cdot4\\cdot6=144$ possibilities. $a.2:$ 1 boy 2 girls, which is the same as above and yields $144$. Thus $a=288$. $b$ also has two cases: $b.1:$ 1 boy 2 girls, which as above, yields $144$. $b.2:$ 3 girls, which yields $\\binom{4}{3}$ to choose the three girls and $3!$ arrangements which gives $4\\cdot6$ possibilities $24$. Thus $b=168$. Case a is either 2 boys 1 girl or 1 boy 2 girls, case b is either 1 boy", "Question # Six boys and six girls are to sit in a row at random. The probability that all the 6 girls sit together isA. $\\dfrac{{6!6!}}{{12!}}$B. $\\dfrac{{6!7!}}{{12!}}$C. $\\dfrac{{2 \\times 6!6!}}{{12!}}$D. $\\dfrac{{2 \\times 6!7!}}{{12!}}$ Hint: Here we will use the formula of permutation to find the total numbers of events and the favourable events and then we will find the probability. First, we will find the total number of events. Given that there are six boys and six girls so that there are a total 12 persons who are said to sit in a row. Now, using the formula ${}^n{P_r}$, we will find the total number of ways 12 persons can sit. So, total number of ways of sitting = ${}^n{P_r}$ =$\\dfrac{{n!}}{{(n - r)!}}$, where n is the number of persons and r is the total places. Here in this question, n = 12 and r = 12. So, applying the above formula. Total number of ways = ${}^{12}{P_{12}} = \\dfrac{{12!}}{{(12 - 12)!}} = \\dfrac{{12!}}{{0!}} = 12!$ Now, let a group be made in which all girls sit together. Let take this group as one element. So, we have a total 7 elements left out of which there are six boys and a group of girls. So, we have to find a number of ways in which we can", "# There Are Three Boys and Two Girls. a Committee of Two is to Be Formed. Find the Probability of the Following Events - Algebra There are three boys and two girls. A committee of two is to be formed. Find the probability of the following events: Event A: The committee contains at least one girl Event B: The committee contains one boy and one girl #### Solution Here, there are three boys B1, B2, B3 and two girls G1, G2. A committee of two is to be formed. Thus, the sample space (S) is given by S = {B1B2, B1B3, B2B3, B1G1, B1G2, B2G1, B2G2, B3G1, B3G2, G1G2} ∴n(S) = 10 A is the event that the committee contains at least one girl. Then, A = {B1G1, B1G2, B2G1, B2G2, B3G1, B3G2, G1G2} ∴n(A) = 7 therefore P(A)=(n(A))/(n(S)) therefore P(A)=7/10 B is the event that the committee contains one boy and one girl. Then, B = {B1G1, B1G2, B2G1, B2G2, B3G1, B3G2} therefore P(B)=(n(B))/(n(S)) therefore P(B)=6/10 therefore P(B)=3/5 Concept: Basic Ideas of Probability Is there an error in this question or solution?", "sum of numbers that turn up is 12.$\\\\$ Accordingly, B = {(6, 6)}$\\\\$ $\\therefore P(B)= \\dfrac{\\text{Number of outcomes favourable to B}}{\\text{Total number of possibleoutcomes}}=\\dfrac{n(B)}{n(S)}=\\dfrac{1}{52}$$\\\\ 28 There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a women? Solution : There are four men and six women on the city council.As one council member is to be selected for a committee at random, the sample space\\\\ contains 10 (4 + 6) elements.\\\\ Let A be the event in which the selected council member is a woman.\\\\ Accordingly, n(A) = 6\\\\ \\therefore P(A)= \\dfrac{\\text{Number of outcomes favourable to A}}{\\text{Total number of possibleoutcomes}}=\\dfrac{n(A)}{n(S)}=\\dfrac{6}{10}$$\\\\$ 29 A fair coin is tossed four times, and a person win Re 1 for each head and loss Rs1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts. Solution : Since the coin is tossed four time, there can be a maximum of 4 heads and tails.$\\\\$ When 4 heads turns up, Rs. 1+Rs. 1+Rs. 1+Rs. 1 = Rs. 4 is the gain.$\\\\$ When 3 heads and 1 tail turn up, Rs 1 + Rs", "among the nine, all of whom have equal probability to have been the first selected student. Clearly the conditional probability is $$5/9$$. If you wish to use Bayes' Rule to confirm: $$\\mathsf P(M_1\\mid M_2)=\\dfrac{\\mathsf P(M_1\\cap M_2)}{\\mathsf P(M_2)}=\\dfrac{\\tfrac 6{10}\\tfrac {5}{9}}{\\tfrac 6{10}\\tfrac 59+\\tfrac 4{10}\\tfrac 69}=\\dfrac 59$$ • Shuffle a deck of six black and four red cards. Every individual card has equal probability for being in any position. Pulling out the second card to see that it is black does not change that fact about the top card - each of the remaining nine cards has equal probability for being the top card, and five of them are black. – Graham Kemp Jun 25 at 11:08 I would have to disagree with Dirk. I actually think you are correct. Suppose we had 2 boys Arthur and Bob and 1 girl Clelia and we pick 2 at random. Our possibilities are $$AB,\\ AC,\\ BA,\\ BC,\\ CA,\\ CB$$. Now it is revealed to us that our second pick was a boy, leaving us with the options $$AB,\\ BA,\\ CA,\\ CB$$. In two of these cases our first pick was a boy, in two of these cases our first pick was a girl, leaving us with chance $$\\frac{1}{2}$$ our first pick was a boy, instead of the initial $$\\frac{2}{3}$$ chance. For bigger cases" ]

[ "# Math Help - More Data .. Can you figure it out? 1. ## More Data .. Can you figure it out? A school has 480 girls and 520 boys. how many committees of 5 can u make if at least one boy must be on each committee. 2. There are two ways to do it: 1) Simpler Way (solving it indirectly): Find the total number of committees and then subtracts the number of committees which are all girls, as $\\binom{1000}{5}-\\binom{480}{5}$ OR 2) Solving it directly: $\\binom{520}{1}\\binom{480}{4}+\\binom{520}{2}\\binom{ 480}{3}+\\binom{520}{3}\\binom{480}{2}+\\binom{520}{4 }\\binom{480}{1}+\\binom{520}{5}$ 3. Hello, wikji! Who assigned this problem? . . In fact, who wrote it? A school has 480 girls and 520 boys. How many committees of five can be made if at least one boy must be on each committee? There are: . ${1000\\choose5} \\:\\approx\\:82.5\\times 10^{11}$ possible committes. The opposite of \"at least one boy\" is \"no boys\" (all girls). There are: . ${480\\choose5} \\;\\approx\\:2.1\\times10^{11}$ all-girl committees. There are about: . $8\\times 10^{12}$ (eight trillion) committees with at least one boy. 4. Originally Posted by Soroban Hello, wikji! Who assigned this problem? . . In fact, who wrote it? There are: . ${1000\\choose5} \\:\\approx\\:82.5\\times 10^{11}$ possible committes. The opposite of \"at least one boy\" is \"no boys\" (all girls). There are: . ${480\\choose5} \\;\\approx\\:2.1\\times10^{11}$ all-girl committees. There are about:", "+0 # Probability with Combinations +1 48 3 +485 The Grammar club has 20 members: 10 boys and 10 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl? Apr 3, 2021 #1 +1 There is a probability of 1/2 of choosing a boy, then a 10/19 probability of choosing a girl. Then the other two people can be anyone, so the probability is 2*1/2*10/19 = 10/19. Apr 3, 2021 #2 +485 +1 That was incorrect, but I have one more try. RiemannIntegralzzz Apr 3, 2021 #3 +485 +1 Actually, I figured it out myself: You can use complementary counting/combinations here: The number of ways to choose a committee of all boys or all girls is $2*C(10,4)$(2 for the girls and boys)$=420$. The total number of committees is $C(20, 4)=4845$. So the answer is $1-\\frac{420}{4845}=\\frac{4425}{4845}=\\boxed{\\frac{295}{323}}$ Apr 3, 2021", "How to find the probability of of simple events, multiple independent events, a union of two events. chocolate bars produced by a certain machine is normal with mean 8. However, if you mean the probability of there being at least one women in any committee makeup, then I'd recommend you use the complement. No, each student may have a different probability. Find the probability that the two have different party affiliations (that is, not both A, not both B, and not both independent). 2) The committee must contain at least three English teachers. B)Students outnumber the teachers on the committee if and only if at least 3 students are selected. Suppose that Alice, Betty, Cindy, and Dianne have their own private outdoors club, and that they need to select a president, a secretary, and a treasurer. The probability that a male develops some form of cancer in his lifetime is 0. Note that this is an aggregate result, and not necessarily true for any one particular piece of. When you want to find the probability of one event OR another occurring, you add their probabilities together. Ok so you have 10 students. SB 2955 has already cleared the Senate and has been transferred to the House. 4 Exactly two girls and a boy being selected. Its work is", "+0 # PLSSSSS HELPPPPP 0 52 1 A club has 30 members: 15 boys and 15 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl? Mar 21, 2022 #1 +117105 +1 A club has 30 members: 15 boys and 15 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl? 1- P(all girls) - P(all boys) =$$=1-2*\\frac{15C4}{30C4}\\\\ =1-2*\\frac{1365}{27405}\\\\ =1-\\frac{26}{261}\\\\ =\\frac{235}{261}$$ Mar 21, 2022", "# There Are Three Boys and Two Girls. a Committee of Two is to Be Formed. Find the Probability of the Following Events - Algebra There are three boys and two girls. A committee of two is to be formed. Find the probability of the following events: Event A: The committee contains at least one girl Event B: The committee contains one boy and one girl #### Solution Here, there are three boys B1, B2, B3 and two girls G1, G2. A committee of two is to be formed. Thus, the sample space (S) is given by S = {B1B2, B1B3, B2B3, B1G1, B1G2, B2G1, B2G2, B3G1, B3G2, G1G2} ∴n(S) = 10 A is the event that the committee contains at least one girl. Then, A = {B1G1, B1G2, B2G1, B2G2, B3G1, B3G2, G1G2} ∴n(A) = 7 therefore P(A)=(n(A))/(n(S)) therefore P(A)=7/10 B is the event that the committee contains one boy and one girl. Then, B = {B1G1, B1G2, B2G1, B2G2, B3G1, B3G2} therefore P(B)=(n(B))/(n(S)) therefore P(B)=6/10 therefore P(B)=3/5 Concept: Basic Ideas of Probability Is there an error in this question or solution?", "Courses Courses for Kids Free study material Free LIVE classes More # There are three boys and two girls. A committee of two is to be formed. Find the probability of events that the committee contains one boy and one girl. Last updated date: 18th Mar 2023 Total views: 311.4k Views today: 8.91k Verified 311.4k+ views Hint: Here we will use the concept of combination. First find the selection of 2 people from the committee of 5 which gives us the total number of cases. Later find the selection of a boy and a girl from the committee which gives the favorable case. There are three boys and two girls and we have to select two person from the sample so, total number of ways of selecting them is ${}^5{C_2} = \\dfrac{{5!}}{{3! \\times 2!}} = 10$ Number of ways for selecting the committee of 1 girl & 1 boy =${}^2{C_1} \\times {}^3{C_1}$ $\\Rightarrow 2 \\times 3 = 6$ So, required probability is = $\\dfrac{favorable\\;ways}{total\\;ways}$ $\\Rightarrow \\dfrac{6}{{10}}$ $\\Rightarrow \\left( {\\dfrac{3}{5}} \\right)$ So, this is our required probability. NOTE: - In these types of questions always find the first total number of ways, then favorable ways, then divide favorable ways by total ways, we will get the required probability.", "+0 # Combination Probability +1 78 3 +485 There are 3 math clubs in the school district, with 5, 7, and 8 students respectively. Each club has two co-presidents. If I randomly select a club, and then randomly select three members of that club to give a copy of Introduction to Counting and Probability, what is the probability that two of the people who receive books are co-presidents? On numerous occasions, I have seen answers without work and have inputted them, and they are wrong most of the time. If you do not show work, I will not be using your answer. Apr 4, 2021 #1 +423 +1 Introduction to counting and probability is a great book ;) Let's first calculate the probability for each club then average them. 5 student club - There are $\\binom{5}{3}=10$ ways to choose the members. In $\\binom{2}{2}\\binom{3}{1}=3$ ways, the co-presidents are selected - the probability here is $\\frac{3}{10}$. 7 student and 8 student clubs are analogous. Try it, I believe in you! Apr 4, 2021 edited by thedudemanguyperson Apr 4, 2021 #2 +485 0 Thank you for your solution! I finished the problem for 7 and 8 students, and got $\\frac{11}{60}$, which was correct! Thanks again, this helped me a lot! RiemannIntegralzzz Apr 4, 2021 #3 +423 +1 No problem, keep the", "Question # A class has 20 girls and 16 boys. One student is selected at random. Find the probability that the selected student is a girl.A. $\\dfrac{1}{4}$B. $\\dfrac{5}{9}$C. $\\dfrac{3}{4}$D. $\\dfrac{4}{9}$ Hint: In this particular problem find the total number of students and then use the probability formula for finding probability of getting a girl on selection. Formula for probability of getting a favourable outcome is = $\\dfrac{{{\\text{Number of favourable outcomes}}}}{{{\\text{Total number of outcomes}}}}$ As we know from the concept of the probability that if there are n balls out of then x are of same type A then the probability of getting a ball of type A on random selection is equal to = $\\dfrac{{{\\text{Number of favourable outcomes}}}}{{{\\text{Total number of outcomes}}}}$. So, in this example favourable outcome will be ball of type A. So, probability = $\\dfrac{{{\\text{Number of balls of type A}}}}{{{\\text{Total number of balls}}}} = \\dfrac{x}{n}$ So, probability that the selected student is girl = $\\dfrac{{{\\text{Number of girls}}}}{{{\\text{Total number of students in class}}}} = \\dfrac{{20}}{{36}} = \\dfrac{5}{9}$", "the second ball is green is $$\\dfrac{1}{12}$$ Example 3 Mrs. Andrews has to select two students from 35 girls and 15 boys to be part of a club. What is the probability that both students are girls? Solution The total number of students $$= 35 + 15 = 50$$ Probability of choosing the first girl, P(girl 1) = $$\\dfrac { 35}{50}$$ Probability of choosing the second girl, P(girl 2) = $$\\dfrac { 34}{49}$$ Now, The probability that both students have chosen are girls $$\\text {P(first girl and second girl)}$$ $$= \\text{ P(first girl)} \\times \\text{P(second girl | first girl)}$$ $$= \\dfrac { 35}{50} \\times \\dfrac { 34}{49}$$ $$= \\dfrac{1190}{1666}$$ $$= \\dfrac{85}{119}\\$$ $$\\therefore$$The probability that both chosen students are girls is $$\\dfrac{85}{119}$$. Example 4 Joseph and David are playing a game of cards. Joseph drew a card at random without replacement. He asks David to help him determine the probability that the first card drawn was a queen and the second is a king. ### Solution As we understand that this probability is having a dependent event condition. P (drawing a queen in the first condition) =$$\\dfrac { 4}{52}$$ P (drawing a king in the second condition after a queen) = $$\\dfrac { 4}{51}$$ P (drawing a queen followed by a king) = $$\\dfrac{4}{52} \\times\\dfrac{4}{51}=\\dfrac{16}{2652}=\\dfrac{4}{663}$$ $$\\therefore$$The probability of drawing", "majors, a committee of 4 computer science majors, 4 mathematics majors, and 3 statistics majors is to be formed. How many distinct committees are there? }} To solve the problem, we will need to find the 3 constituting elements - 4 computer majors, 4 math majors, and 3 statistics majors and then, since they are elements of one set - multiply them. Jumping a little ahead, we will use the combinations formula and will get the following results for each group: 495, 210, and 84. Now, since for each computer science major there are a good number of math and stats majors, we multiply. The result we will get is $495*210*84 = 8,731,800$. ## Permutations and Combinations Besides the two problems explained in Kaplan, that you should know by heart, about the 3 out of 5 runners and 3 out of 8 committee members, there are few variations. For example, let's take a Permutation problem from high school textbook: {{#x:box| EXAMPLE 7. Given a selected committee of 8, in how many ways, can the members of the committee divide the responsibilities of a president, vice president, and secretary? }} The solution comes both from the permutations formula and from logic. (Permutations, not combinations formula is used because the order matters since the positions are unique). Scenario 1 -", "Question 118 The student mess committee of a reputed Engineering College has n members. Let P be the event that the Committee has students of both sexes and let Q be the event that there is at most one female student in the committee. Assuming that each committee member has probability 0.5 of being female, the value of n for which the events P and Q are independent is Solution Let's first estimate the sample space. In a committee of n members, there can be the following cases: (0 males, n females), (1 male, n-1 females), (2 males, n-2 females) ...(n males, 0 females). Thus, the sample space has n+1 cases. n(S) = n+1 cases Barring 2 cases from the sample space, event P occurs for all other cases. => n(P) = n+1-2 = n-1 cases => p(P) = (n-1)/(n+1) At most 1 female can be expressed as the sum of the cases exactly 0 females and exactly 1 female. => n(Q) = 2 cases => p(Q) = 2/(n+1) The set $$P \\cap Q$$ occurs when there is exactly one female member. Hence, n($$P \\cap Q$$) = 1 p($$P \\cap Q$$) = 1/(n+1) When two events are independent, the probability of both of them occurring is equal to the product of their individual probabilities. Thus, p($$P \\cap Q$$)", "+0 # Probability 0 37 1 The Grammar club has 20 members: 10 boys and 10 girls. A 4-person committee is chosen at random. What is the probability that the committee has 2 boys and 2 girls? Oct 7, 2021 #1 +785 0 $\\frac{\\binom{10}{2}}{\\binom{20}{2}} \\cdot \\frac{\\binom{10}{2}}{\\binom{18}{2}}$ Feel like this is wrong .. edited by MathProblemSolver101 Oct 7, 2021 edited by MathProblemSolver101 Oct 7, 2021", "Ed According to a recent study, having a role model at school who looks like you can have large and long-lasting effects. 4567 (Source: American Cancer Society). Out of 14 regions in the country, six will be chosen to be included in a survey. [Maximum mark: 4] Consider the grades of 11 students in a math test. Let Y be the random variable which represents the toss of a coin. There are still 18 balls. A committee of three people is selected at random from a set consisting of six teachers, eight parents of students, and five alumni. A committee of 4 persons has to be chosen from 8 boys and 6 girls consisting of at least one girl. the probability of choosing one whiteball is 9/18=1/2. Here is the probability distribution for X. What is the probability that both children are boys? This question is identical to question one, except that instead of specifying that the older child is a boy, it is specified that at least one of them is a boy. Calculate the number of ways that this committee can be formed if at least one geography teachers must be on the committee. Note that assigning the same people different tasks constitutes a different assignment. committee? (c) Among the 12 people there is one married", "# Math Help - Combinatorial probability question 1. ## Combinatorial probability question im sure this will be easy for most, i just am having trouble knowing where to start as im new to probability. the question is as follows: \"A science committee of 3 people is to be formed from a group of 17 scientists (7 biologists, 4 physicists and 6 chemists) with each possible committee equally likely. What is the probability the committee vote in favour of a proposal given: (a) all biologists in favour, everybody else against? (b) all chemists against, everybody else in favour? (c) all biologists in favour, all chemists against and all physicists abstain?\" thanks 2. In order to do this we must assume that a proposal passes on simple majority vote: 2 to 1 or 3 to 0. To do part (a) then we must have two biologists and one other or all three biologists. What is the probability of each of those committees? 3. Originally Posted by Plato In order to do this we must assume that a proposal passes on simple majority vote: 2 to 1 or 3 to 0. To do part (a) then we must have two biologists and one other or all three biologists. What is the probability of each of those committees? 7 choose 2 times", "the Science Club is in both clubs:}& 0.80S \\,=\\,\\text{both}\\end{array}$ $\\text{These quantities are equal: }\\:0.30M \\,=\\,0.8S \\quad\\Rightarrow\\quad M \\,=\\,\\tfrac{8}{3}S$ We are told that: . $S \\,=\\,15$ Therefore . . . 4. ## Re: Word problem Thank you so much! I was totally overlooking the fact that the numbers are equal. There are 40 students. : ) 5. ## Re: Word problem step 1-80% of 15=12 step2-now take 30% of a number, let say,x, i.e-30% of x =12 step 3- 12*100/30=40 (answer)", "# probability of chess clubs problem The chess clubs of two schools consists of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between the two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that (a) Rebecca and Elise will be paired? Can someone explain step in step? Because from the book, the solution is $$\\frac{\\binom{7}{3}\\binom{8}{3}3!}{\\binom{8}{4}\\binom{9}{4}4!}$$ I really don't know what is going on at all, especially $\\dfrac{3!}{4!}$??? • The expression, though correct, is kind of silly. The probability R is chosen to be on her school's team is $\\frac{4}{8}$. The corresponding probability for E is $\\frac{4}{9}$. And given they are both chosen, the probability they play against each other is $\\frac{1}{4}$. – André Nicolas Apr 30 '15 at 18:49 • how come the probability they play against each other is 1/4 , not 1/4 * 1/4 ? – user235829 Apr 30 '15 at 18:50 • Rebecca is equally likely to play against any of the $4$ members of the opposing team. – André Nicolas Apr 30 '15 at 18:51 • Please see", "them on a shelf we have to calculate number of permutations of four element set: bar(bar(Omega))=P_4=4! =1*2*3*4=24 The orders which are mentioned in the task are 1234 (increasing order) and 4321 (decreasing order), so bar(bar(A))=2 Finally we can calculate the probability: P(A)=bar(bar(A))/bar(bar(Omega))=2/24=1/12 Answer: The probability is 1/12 A problem which includes combinations would deal with choosing a subset from a larger set. For example: Before a science contest a team has to be chosen from a class. Assuming that there are 15 boys and 10 girls in how many ways can a team be choosen if it has to contain 2 girls and two boys ? Solution: In this task the order is not important. No matter if you choose Jack first and Ann next or the other way Ann first and next Jack, the team consists of the same people. So you have to use combinations to solve this task. To calculate the number we can assume that we choose girls first, then boys, so the number would be calculated as: n=C\"\"_10^2*C\"\"_15^2=(10!)/(2!*8!)(15!)/(2!*13!) n=15750 Answer There are 15750 ways to choose the team.", "jury foreman? 66. The United States Senate Committee on Commerce, Science, and Transportation consists of 23 members, 12 Republicans and 11 Democrats. The Surface Transportation and Merchant Marine Subcommittee consists of 8 Republicans and 7 Democrats. How many ways can members of the Subcommittee be chosen from the Committee? 67. You own 16 CDs. You want to randomly arrange 5 of them in a CD rack. What is the probability that the rack ends up in alphabetical order? 68. A jury pool consists of 27 people, 14 men and 13 women. Compute the probability that a randomly selected jury of 12 people is all male. 69. In a lottery game, a player picks six numbers from 1 to 48. If 5 of the 6 numbers match those drawn, they player wins second prize. What is the probability of winning this prize? 70. In a lottery game, a player picks six numbers from 1 to 48. If 4 of the 6 numbers match those drawn, they player wins third prize. What is the probability of winning this prize? 71. Compute the probability that a 5-card poker hand is dealt to you that contains all hearts. 72. Compute the probability that a 5-card poker hand is dealt to you that contains four Aces. 73. A bag contains 3 gold marbles,", "and B with B’ P (A’ U B’)’ = P (A ∩ B) P (A ∩ B) = 1/4 P (A’ U B’) = 1 – 1/4 = 3/4 Problem 11) A and B are two mutually exclusive events of an experiment. If P(A’) = 0.65 and P (A U B) = 0.65 and P(B) = p, find the value of p. Sol. P(A’) = 0.65 P(A) = 0.35 P (A U B) = 0.65 A and B are mutually exclusive events. Therefore, P (A U B) = P(A) + P(B) 0.65 = 0.35 + p P = 0.30 Problem 12) A committee of two persons is selected from two men and two women. What is the probability that the committee will have a) no man b) one man c) two men? Sol. Total number of persons = 4. Out of these 4 persons, two can be selected in 4Cways. a)There will be two women in the committee. Out of two women, two can be selected in 2C= 1 way Probability (no man) = 2C2/4C2 b)One man and one woman can be selected in 2C1 x 2C1 Probability (One man and one woman) = (2C1 x 2C1)/ (4C2) c)Out of two men, two can be selected in 2C= 1 way Probability (two-man) = 2C2/4C2 Problem 13) A carton", "have memberships at the gym. If one man and then one woman are selected at random, what is the probability that they are a married couple? A. 0 B. 1/14693 C. 1/3600 D. 1/1050 E. 1 OA:C Probability of Selecting Male (Married couple)$$= \\frac{300}{900}$$ Probability of Selecting female Married to male selected earlier$$= \\frac{1}{1200}$$ Final probability$$= \\frac{300}{900} * \\frac{1}{1200} = \\frac{1}{3600}$$ _________________ Good, good Let the kudos flow through you Manager Joined: 27 Dec 2016 Posts: 241 Re: A certain gym has 900 male members and 1200 female members. Three hund [#permalink] Show Tags 06 Sep 2018, 17:50 Princ wrote: Bunuel wrote: A certain gym has 900 male members and 1200 female members. Three hundred of the male members are married to women who have memberships at the gym. If one man and then one woman are selected at random, what is the probability that they are a married couple? A. 0 B. 1/14693 C. 1/3600 D. 1/1050 E. 1 OA:C Probability of Selecting Male (Married couple)$$= \\frac{300}{900}$$ Probability of Selecting female Married to male selected earlier$$= \\frac{1}{1200}$$ Final probability$$= \\frac{300}{900} * \\frac{1}{1200} = \\frac{1}{3600}$$ Hello, I was wondering could you please explain how you got 1/1200? 300 men married to 300 women. Shouldn't the probability of selecting a woman is 300/1200? Would greatly appreciate it if you" ]

[ "## Wednesday, February 7, 2007 ### Spacy. Topic: S&S/Sets. Level: AMC. Problem: (2007 AMC 12A - #25) Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\\{1, 2, 3, \\ldots, 12\\}$, including the empty set, are spacy? Solution: We will solve this problem with a recursion on $S_n$, which we will define as the number of spacy subsets of $\\{1, 2, 3, \\ldots, n\\}$. We can easily calculate $S_0 = 1$, $S_1 = 2$, $S_2 = 3$, and $S_3 = 4$. Now we will think about how to evaluate $S_n$ in terms of previous terms. Obviously, we can keep all the subsets in $S_{n-1}$. These will account for any space subsets that don't contain the element $n$. Now we just need to count the ones that do contain the element $n$. If a spacy subset contains $n$, it cannot have $n-1$ or $n-2$. So take all the spacy subsets of $\\{1, 2, \\ldots, n-3\\}$ and add the element $n$ to them to get all the spacy subsets that contain $n$. Since these are both of the cases, we obtain $S_n = S_{n-1}+S_{n-3}$. Some calculations yield $S_{12} = 129$. QED. -------------------- Comment: In all honesty, a pretty standard recursion problem, which is probably better off", "but is not quite the same. In fact, solving a similar equation for circumference = area, 2πr = πr2, we get r = 2, suggesting we should take half the radius of a circle as the natural measure and giving 4π for both the circumference and the area. (The Tauists might be glad to hear that.) Call a set of integers “spacy’ if it contains no more than one out of any three consecutive integers. How many subsets of {1, 2, 3, ……,12} are spacy? Let g(n) be the number of spacy subsets of $\\{1, 2, ..., n\\}$. Then clearly $g(0) = 1, g(1) = 2, g(2) = 3, g(3) = 4$ — in each case the spacy sets are just the empty set and the one-element subsets. Then for larger n, $g(n) = g(n-1) + g(n-3)$. How can you make a spacy subset of $\\{1, 2, \\ldots, n\\}$? Well, any spacy subset of $\\{ 1, 2, \\ldots, n-1 \\}$ is a spacy subset of $\\{ 1, 2, \\ldots, n \\}$ which does not contain$lates n$, and there are $g(n-1)$ of those. And any spacy subset of $\\{ 1, 2, \\ldots, n \\}$ which contains$n$must, when you remove $n$ from it, give a spacy subset of $\\{ 1, 2, \\ldots, n-3 \\}$. There are$g(n-3)$of those. Any spacy subset", "# Difference between revisions of \"2007 AMC 12A Problems/Problem 25\" ## Problem Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\\{1,2,3,\\ldots,12\\},$ including the empty set, are spacy? $$\\mathrm{(A)}\\ 121 \\qquad \\mathrm{(B)}\\ 123 \\qquad \\mathrm{(C)}\\ 125 \\qquad \\mathrm{(D)}\\ 127 \\qquad \\mathrm{(E)}\\ 129$$ ## Solution ### Solution 1 Let $S_{n}$ denote the number of spacy subsets of $\\{ 1, 2, ... n \\}$. We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$. The spacy subsets of $S_{n + 1}$ can be divided into two groups: • $A =$ those not containing $n + 1$. Clearly $|A|=S_{n}$. • $B =$ those containing $n + 1$. We have $|B|=S_{n - 2}$, since removing $n + 1$ from any set in $B$ produces a spacy set with all elements at most equal to $n - 2,$ and each such spacy set can be constructed from exactly one spacy set in $B$. Hence, $S_{n + 1} = S_{n} + S_{n - 2}$ From this recursion, we find that $S(0)$ $S(1)$ $S(2)$ $S(3)$ $S(4)$ $S(5)$ $S(6)$ $S(7)$ $S(8)$ $S(9)$ $S(10)$ $S(11)$ $S(12)$ 1 2 3 4 6 9 13 19 28 41 60 88 129 ### Solution 2 Since each of the elements of the subsets must be", "# How is this problem called? I have a set of integers $\\{8, 8, 6\\}$. I want to know if It is possible to get $21$ by adding a subset of them. I would like to know how is the problem called so that I can google it. - This is the Subset Sum Problem. - Your question is slightly flawed: $\\{8,8,6\\}$ is not a set of integers, because it contains a repetition. (Or rather, it is a set of integers, but it only contains the two elements $8$ and $6$.) If you really did mean \"set of integers\", then rattle's answer is correct. If you meant \"sequence of integers\", as suggested by your example, then this is the Knapsack Problem. The Subset Sum Problem is a special case of the Knapsack Problem. - This \"set\" is an example of what is sometimes called a \"multiset\": it allows one member (in this case 8) to have multiple memberships in it rather than only one. You can't get an odd number such as 21 by adding even numbers. -", "# Subsets Without Consecutive Integers How many subsets of $$\\{1,2,3, \\ldots, 14\\}$$ contain no pair of consecutive integers? Details and assumptions The empty set is a subset of every set. ×", "a revised proof that avoids that problem, I think: Let’s start by thinking about how many of the integers can be multiples of 25. If there are 3 or more multiples of 25, say 25a, 25b, and 25c, then at least 2 of a, b, and c must have the same parity, i.e. both even or both odd. Let’s say it’s a and b that have the same parity; then 25a – 25b is an even multiple of 25, i.e. a multiple of 50, and we’re done. So we may assume that at most 2 of the set are multiples of 25. Let’s discard those two integers (because they would cause problems in the proof below), so we still have at least 25 integers, none of which are a multiple of 25. Let’s say the remaining 25 integers are $x_1, x_2, \\dots , x_{25}$. We may assume without loss of generality that if x is in the set then –x is not; otherwise we would have x + (-x) = 0, which is divisible by 50, and we would be through. Consider the expanded set made by throwing in the negatives of the original integers, i.e. $x_1, x_2, \\dots , x_{25}, -x_1, -x_2, \\dots , -x_{25}$. This set contains 50 distinct integers. Now place the integers into pigeonholes", "# Fat Subsets Discrete Mathematics Level pending A finite set of positive integers is called fat if each of its members is at least as large as the number of elements in the set. (The empty set is considered to be $$fat$$.) Let $$a_n$$ denote the number of fat subsets of $$\\{1,2,\\ldots,n \\}$$ that contain no two consecutive integers, and let $$b_n$$ denote the number of subsets of $$\\{1,2,\\ldots,n \\}$$ in which any two elements differ by at least three. Find $$a_{1729}-b_{1729}+1$$. ×", "(just don't insist on points being consecutive integers) then the trace of such a system on a subset forms a similar system.", "# Show that for any subset of $10$ distinct integers there exists two disjoint subsets equal in sum The title abbreviates the following homework exercise on the Pigeonhole Principle. Show that for any set of $10$ distinct integers from $1 \\dots 60$ there exists two disjoint subsets of equal sum. (The $2$ disjoint sets may not include all $10$ integers. e.g. let a set of $10$ distinct integers contain $\\{1,2,3\\}$. Subsets $\\{1,2\\}, \\{3 \\}$ are disjoint and have equal sums.) Can anyone give a hint at how or why this problem is an application of the Pigeonhole Principle? Second, does the result to be proved hold for sets of integers $\\{1 \\dots n \\}$ where $n \\ne 60$? • For the last question no. Doesn't hold for $n=2$ – Guy Mar 12 '14 at 9:38 • @Sabyasachi: Yes it does, vacuously. Because there is no set of 10 distinct integers from 1...2, anything at all can be said to hold for all such sets. – TonyK Mar 12 '14 at 9:56 • @TonyK oh. I thought at most 10 elements. Not exactly 10. – Guy Mar 12 '14 at 9:57 • So at least, this is true for all $n \\le 60$. – Du Phan Mar 12 '14 at 9:58 • By the answer of @vonbrand, this is", "A set of integers Assume that there is a set of ordered integers initially containing number $1$ to a given $n$. At each step, the lowest number in the set is removed, if the number was odd, then we go to the next step and if it was even, half of that number is inserted into the set and the cycle repeats until the set is empty. My question is, how many steps does it take for a given number $n$ to finish the whole set? I think it should be something the form of $2k - x$ where $x$ itself is likely a complex expression but I just can't seem to figure it out. Any help would be appreciated. (I got that from trial and error, by the way, no logic or proof behind it, I know that the answer is $\\lfloor \\frac{n}{2} \\rfloor$+\"The number of times each even number can be divided by 2\", but I just can't find a closed formula) • It will have to do with the number of numbers in the original set that fall into each of the following categories: $(2k+1),(2k+1)2, (2k+1)2^2,(2k+1)2^3,\\dots$. Numbers from the first category are removed in a single step. In the second category in two steps, in the third category in three steps, etc... – JMoravitz Jan", "## Oblong numbers as a sum of two squares A pronic number is a number which is the product of two consecutive integers, that is, a number of the form $n(n+1)$. They are also called oblong numbers. Prove that there are infinitely many positive integers $n$ such that $n(n+1)$ can be expressed as a sum of two squares in at least two different ways", "# Subsets Without Consecutive Integers Probability Level 3 How many subsets of $\\{1,2,3, \\ldots, 14\\}$ contain no pair of consecutive integers? Details and assumptions The empty set is a subset of every set. ×", "from it, give a spacy subset of $\\{ 1, 2, \\ldots, n-3 \\}$. There are \\$g(n-3)\\$ of those. Any spacy subset of \\$\\latex \\{1, 2, \\ldots, n\\}\\$ either contains or does not contain $n$, and so $g(n) = g(n-1) + g(n-3)$. For example, the spacy subsets of \\$\\{ 1, 2 \\}\\$ are $\\emptyset, 1$, and \\$2\\$ (for notational ease I’m dropping the braces\\$. The spacy subsets of $\\{ 1, 2, 3, 4 \\}$ are $\\emptyset, 1, 2, 3, 4, 14$. So the spacy subsets of $\\{ 1, 2, 3, 4, 5 \\}$ are \\$\\emptyset, 1, 2, 3, 4, 14\\$ (from the spacy subsets of $\\{ 1, 2, 3, 4\\}$) and $5, 15, 25$ (from adding a 5 to each spacy subset of $\\{ 1, 2\\}$.) Interestingly, the winners in this contest all came out of the financial community: the affiliations mentioned in the post at the Times are Two Sigma Investments, Renaissance Technologies, Morgan Stanley, and Citadel. A few explanations for this present themselves: • people who can solve these sorts of problems quickly tend to end up in that industry; • this was a fundraising event and people in finance have lots of money; • this event took place in New York; you’d have different results if you tried this in, say, San Francisco The last of", "# Modular Subsets Discrete Mathematics Level pending For the set of integers from 1 to 30, inclusive, what is the total number of ordered subsets of the set, such that the sum of all the elements in the subset is equivalent to $$0 \\bmod{3}$$. Note: The empty set is one such selection. Do not exclude it from your answer. ×", "flavors in which flavors may be repeated is a combination with repetition. The number of solutions is the number of solutions of the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 3$ in the nonnegative integers, which is the number of ways of inserting four addition signs in a row of three ones, that is, $\\binom{3 + 4}{4} = \\binom{7}{4}$. – N. F. Taussig Jun 20 '15 at 10:09", "is denoted as (! To transfinite numbers, integers, composite numbers between 10 to 20 subsets of a correspondence..., such as 3 or 11 or 412, used in counting to indicate quantity but bijective.", "set of integers and natural numbers with which we are normally familiar. Definition 1.80 (Integers) The set of all integers is the set $\\ZZ = \\{\\dots, -3,-2,-1,0,1,2,3, \\dots \\}.$ Definition 1.81 (Natural numbers) The set of all natural numbers is the set $\\Nat = \\{1,2,3, \\dots \\}.$", "## Tarry-Escott Problem For each Positive Integer , there exists a Positive Integer and a Partition of , ..., as a disjoint union of two sets and , such that for , The results extended to three or more sets of Integers are called Prouhet's Problem. Hahn, L. The Tarry-Escott Problem.'' Problem 10284. Amer. Math. Monthly 102, 843-844, 1995.", "six integers as either being in the subset or not in the subset. Each integer can have 2 states. For a total of $2^6=64$ total states or subsets. Subtracting out the empty set, you are left with 63 subsets. We denote the size, or cardinality, of $S$ by $|S|=6$. Now for any nonempty set $X$ there are $2^{|X|}$ subsets of $X$ in total, these include $X$ itself and the empty set $\\varnothing$. So for your set there are $2^{|S|}=2^6$ subsets of $S$ in total. Since the empty set $\\varnothing$ is empty by definition you must subtract $1$ from $2^6$ to obtain the number of nonempty sets.", "up to the same number. Originally Posted by awkward The problem statement did not say the integers are distinct. So the minimum sum is 1+1+1+1 and the maximum is 50+50+50+50. I completely disagree with you on that. How do you account for “subsets of exactly four integers each”? I do not count the collection [1,1,1,1] as subset. 8. Originally Posted by Plato I completely disagree with you on that. How do you account for “subsets of exactly four integers each”? I do not count the collection [1,1,1,1] as subset. Yes, I guess you're right about that. Since the problem statement referred to a set of 10 integers and subsets of 4 integers, the implication is that set members are distinct. (The problem's conclusion still holds if the integers are not required to be distinct, however, but then to get the terminology correct the collections involved should be called multisets.)" ]

[ "# Difference between revisions of \"2007 AMC 12A Problems/Problem 25\" ## Problem Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\\{1,2,3,\\ldots,12\\},$ including the empty set, are spacy? $\\mathrm{(A)}\\ 121 \\qquad \\mathrm{(B)}\\ 123 \\qquad \\mathrm{(C)}\\ 125 \\qquad \\mathrm{(D)}\\ 127 \\qquad \\mathrm{(E)}\\ 129$ ## Solution ### Solution 1 Let $S_{n}$ denote the number of spacy subsets of $\\{ 1, 2, ... n \\}$. We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$. The spacy subsets of $S_{n + 1}$ can be divided into two groups: • $A =$ those not containing $n + 1$. Clearly $|A|=S_{n}$. • $B =$ those containing $n + 1$. We have $|B|=S_{n - 2}$, since removing $n + 1$ from any set in $B$ produces a spacy set with all elements at most equal to $n - 2,$ and each such spacy set can be constructed from exactly one spacy set in $B$. Hence, $S_{n + 1} = S_{n} + S_{n - 2}$ From this recursion, we find that $S(0)$ $S(1)$ $S(2)$ $S(3)$ $S(4)$ $S(5)$ $S(6)$ $S(7)$ $S(8)$ $S(9)$ $S(10)$ $S(11)$ $S(12)$ 1 2 3 4 6 9 13 19 28 41 60 88 129 And so the answer is $(E)$ $129$. ### Solution 2 Since each of", "from it, give a spacy subset of $\\{ 1, 2, \\ldots, n-3 \\}$. There are \\$g(n-3)\\$ of those. Any spacy subset of \\$\\latex \\{1, 2, \\ldots, n\\}\\$ either contains or does not contain $n$, and so $g(n) = g(n-1) + g(n-3)$. For example, the spacy subsets of \\$\\{ 1, 2 \\}\\$ are $\\emptyset, 1$, and \\$2\\$ (for notational ease I’m dropping the braces\\$. The spacy subsets of $\\{ 1, 2, 3, 4 \\}$ are $\\emptyset, 1, 2, 3, 4, 14$. So the spacy subsets of $\\{ 1, 2, 3, 4, 5 \\}$ are \\$\\emptyset, 1, 2, 3, 4, 14\\$ (from the spacy subsets of $\\{ 1, 2, 3, 4\\}$) and $5, 15, 25$ (from adding a 5 to each spacy subset of $\\{ 1, 2\\}$.) Interestingly, the winners in this contest all came out of the financial community: the affiliations mentioned in the post at the Times are Two Sigma Investments, Renaissance Technologies, Morgan Stanley, and Citadel. A few explanations for this present themselves: • people who can solve these sorts of problems quickly tend to end up in that industry; • this was a fundraising event and people in finance have lots of money; • this event took place in New York; you’d have different results if you tried this in, say, San Francisco The last of", "of$\\latex \\{1, 2, \\ldots, n\\}$either contains or does not contain $n$, and so $g(n) = g(n-1) + g(n-3)$. For example, the spacy subsets of$\\{ 1, 2 \\}$are $\\emptyset, 1$, and$2$(for notational ease I’m dropping the braces$. The spacy subsets of $\\{ 1, 2, 3, 4 \\}$ are $\\emptyset, 1, 2, 3, 4, 14$. So the spacy subsets of $\\{ 1, 2, 3, 4, 5 \\}$ are $\\emptyset, 1, 2, 3, 4, 14$ (from the spacy subsets of $\\{ 1, 2, 3, 4\\}$) and $5, 15, 25$ (from adding a 5 to each spacy subset of $\\{ 1, 2\\}$.) Interestingly, the winners in this contest all came out of the financial community: the affiliations mentioned in the post at the Times are Two Sigma Investments, Renaissance Technologies, Morgan Stanley, and Citadel. A few explanations for this present themselves: • people who can solve these sorts of problems quickly tend to end up in that industry; • this was a fundraising event and people in finance have lots of money; • this event took place in New York; you’d have different results if you tried this in, say, San Francisco The last of these explanations would be easy to check. # Kerrich, Ulam, and the scope of classes One of the classic stories of experimental probability comes from the internment", "2 added 110 characters in body Let $C(N,k)$ be the smallest positive integer $x$ such that $[1,x]\\subset \\mathbb{Z}$ contains $k$ disjoint intervals $I_1, ..., I_k$ of $N$ consecutive integers that are all composite. (For example, $C(2,2)=15$, with $I_1=[8,9]$ and $I_2=[14, 15]$.) I am interested in the asymptotic behavior of $C(N,k)$ for various fixed values of $k$. Clearly $C(N,1) \\leq (N+1)!+N+1$. Also if $M$ divides $N$, and we have $k$ disjoint intervals of $N$ consecutive composites, we can break each interval up into $N/M$ disjoint intervals of $M$ consecutive composites, giving a total of $kN/M$ intervals, and so $C(M, kN/M) \\leq C(N, k)$. So, we have $C(N,k) \\leq C(kN, 1) \\leq (kN+1)!+kN+1$. However, these bounds give $15=C(2,2)\\leq C(4,1)=27\\leq 5!+5=125$, which doesn't seem very tight. Can anyone come up with better bounds or asymptotics? The gap between primes is something like $O(\\log(N))$, and so maybe $C(N,k)$ grows sort of like $O(ke^N)$? 1 # Consecutive composite numbers Let $C(N,k)$ be the smallest positive integer $x$ such that $[1,x]\\subset \\mathbb{Z}$ contains $k$ disjoint intervals $I_1, ..., I_k$ of $N$ consecutive integers that are all composite. (For example, $C(2,2)=15$, with $I_1=[8,9]$ and $I_2=[14, 15]$.) I am interested in the asymptotic behavior of $C(N,k)$ for various fixed values of $k$. Clearly $C(N,1) \\leq (N+1)!+N+1$. Also if $M$ divides $N$, and we have $k$ disjoint", "# Difference between revisions of \"2007 AMC 12A Problems/Problem 25\" ## Problem Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\\{1,2,3,\\ldots,12\\},$ including the empty set, are spacy? $$\\mathrm{(A)}\\ 121 \\qquad \\mathrm{(B)}\\ 123 \\qquad \\mathrm{(C)}\\ 125 \\qquad \\mathrm{(D)}\\ 127 \\qquad \\mathrm{(E)}\\ 129$$ ## Solution ### Solution 1 Let $S_{n}$ denote the number of spacy subsets of $\\{ 1, 2, ... n \\}$. We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$. The spacy subsets of $S_{n + 1}$ can be divided into two groups: • $A =$ those not containing $n + 1$. Clearly $|A|=S_{n}$. • $B =$ those containing $n + 1$. We have $|B|=S_{n - 2}$, since removing $n + 1$ from any set in $B$ produces a spacy set with all elements at most equal to $n - 2,$ and each such spacy set can be constructed from exactly one spacy set in $B$. Hence, $S_{n + 1} = S_{n} + S_{n - 2}$ From this recursion, we find that $S(0)$ $S(1)$ $S(2)$ $S(3)$ $S(4)$ $S(5)$ $S(6)$ $S(7)$ $S(8)$ $S(9)$ $S(10)$ $S(11)$ $S(12)$ 1 2 3 4 6 9 13 19 28 41 60 88 129 ### Solution 2 Since each of the elements of the subsets must be", "# Number of compositions of $n$ such that each term is less than equal to $k.$ Let $$n$$ be an integer $$\\geq 1.$$ Then a partition of $$n$$ is a sequence of positive integers (greater than equal to $$1$$) such that their sum equals $$n.$$ So for instance if $$n=4$$ then $$[[4], [1, 3], [1, 1, 2], [1, 1, 1, 1], [1, 2, 1], [2, 2], [2, 1, 1], [3, 1]]$$ is a collection of all partitions of $$n$$ with order. Clearly for any $$n$$ the number of such ordered partitions is $$2^{n-1}.$$ However I want to count the number of paritions of $$n$$ where each integer in the parition is less than equal to some integer $$k.$$ So for instance if $$n=4$$ and $$k=2$$ then $$[[1, 1, 2], [1, 1, 1, 1], [1, 2, 1], [2, 2], [2, 1, 1]]$$ is a collection of all the $$2$$-partitions of $$4.$$ Is there a general formula for finding this or maybe an asymptotic expression? • See this, specifically $A$-restricted compositions. – MathematicsStudent1122 Dec 28 '18 at 21:48 • Well, for $k=2$ you get the Fibonacci sequence, so if there is a general formula, it will be complicated for sure. – SmileyCraft Dec 28 '18 at 21:48 • @MathematicsStudent1122 I saw this, but it does not give me the answer", "# Divisible subset sums Inspired by the recent 3Blue1Brown video Consider, for some positive integer $$\\n\\$$, the set $$\\\\{1, 2, ..., n\\}\\$$ and its subsets. For example, for $$\\n = 3\\$$, we have $$\\emptyset, \\{1\\}, \\{2\\}, \\{3\\}, \\{1,2\\}, \\{1,3\\}, \\{2,3\\}, \\{1,2,3\\}$$ If we take the sum of these subsets, we can then ask ourselves the following question: Of the sums of the subsets of $$\\\\{1, 2, ..., n\\}\\$$, how many are divisible by some given integer $$\\k\\$$? Again, using $$\\n = 3\\$$ as an example, we have our subset sums as $$0, 1, 2, 3, 3, 4, 5, 6$$ For $$\\k = 2\\$$, there are $$\\4\\$$ subsets whose sum is divisible by $$\\k\\$$: $$\\\\emptyset, \\{2\\}, \\{1,3\\}\\$$ and $$\\\\{1,2,3\\}\\$$. Given two positive integers $$\\n \\$$ and $$\\k\\$$, with $$\\2 \\le k \\le n\\$$, output the number of subsets of $$\\\\{1, 2, 3, ..., n-1, n\\}\\$$ such that their sum is divisible by $$\\k\\$$. You may take input and give output in any reasonable manner and format. This is , so the shortest code in bytes wins. ## Test cases n k out 3 2 4 5 5 8 13 8 1024 2 2 2 7 7 20 14 4 4096 15 10 3280 9 5 104 7 2 64 11 4 512 15 3 10944 16 7 9364 13", "answer? • What is the place of the non-wicked set $\\{1,3,4,5\\}$ in your classification? – Chen Wang May 12 '18 at 0:16 • To fix, need I/E. – vadim123 May 12 '18 at 0:19 There are two good ways to do this problem. 1. Let $a_n$ be the number of wicked subsets of $\\{1,2,\\dots,n\\}$. You can show that $$a_n=a_{n-1}+a_{n-2}+a_{n-3}$$ holds for all $n\\ge 3$. The $a_{n-1}$ term counts subsets where $n$ is absent, $a_{n-2}$ counts subsets where $n$ is present but $n-2$ is absent, and $a_{n-3}$ is subsets with $n$ and $n-1$ but without $n-2$. This recurrence, along with the bases cases $a_0=1,a_1=2,a_2=4$, allows you to iteratively compute $a_n$ from the ground up, starting with $a_3$ all the way up to $n=10$. 2. Instead, count the number of un-wicked (saintly) subsets. Thinking of subsets as strings of ten $0$s and $1$s, where $1$s correspond to the elements in the set, every saintly subset contains a $111$ somewhere. To eliminate the problem of overlaps, in any block of three or more ones, we will only consider the initial $111$, which corresponds to either a $111$ at the beginning or a $0111$ somewhere. There are $2^7$ strings starting with $111$, and $7\\cdot 2^6$ strings containing a $0111$ (the $7$ represents the choice of location of $0111$). Well, not quite; there", "\\} \\subseteq U$, $\\{ 2\\} \\subseteq U$, $\\{ 1 \\} \\subseteq \\{ 1 \\}$, $\\{ 2 \\} \\subseteq \\{ 2 \\}$. So, there are 7 such true relations out of 16. (3c) Now, let U be the set $\\{ 1,2,3\\}$. There are 8 subsets, namely, $\\phi$, $U$, $\\{ 1\\}$, $\\{ 2\\}$, $\\{ 3 \\}$, $\\{1,2\\}$, $\\{ 1,3\\}$, $\\{ 2,3\\}$. There are 64 possible relations of the form $A \\subseteq B$. Count the number of true ones. Answer: $\\phi \\subseteq U$, $\\phi \\subseteq \\{ 1 \\}$, $\\phi \\subseteq \\{ 2 \\}$, $\\phi \\subseteq \\{ 3 \\}$, $\\{ 1 \\} \\subseteq U$, $\\{ 2 \\} \\subseteq U$, $\\{ 3 \\} \\subseteq U$, $\\{ 1, 2\\} \\subseteq U$, $\\{ 1, 3\\} \\subseteq U$, $\\{ 2,3\\} \\subseteq U$, $\\{ 1\\} \\subseteq \\{ 1, 2 \\}$, $\\{ 1 \\} \\subseteq \\{ 1,3\\}$, $\\{ 2 \\} \\subseteq \\{ 1,2 \\}$, $\\{ 2\\} \\subseteq \\{ 2,3\\}$, $\\{ 3\\} \\subseteq \\{ 1,3\\}$, $\\{ 3\\} \\subseteq \\{ 2,3\\}$. And, also, except $\\phi$ and $U$, each is a subset of itself. Thus, there are such 30 true relations. (3d) Let U be the set $\\{ 1,2,3,\\ldots, n\\}$ for an arbitrary positive integer n. How many subsets are there ? Answer: $2^{n}$ subsets. How many possible relations of the form $A \\subseteq B$? Answer: $2^{n} \\times 2^{n}$, that is,", "each $$X_n$$ has two disjoint copies that embed in $$Z$$ (one in each copy of $$Y$$), so by the Lemma, $$Z$$ cannot be homeomorphic to $$X_n$$ for any $$n$$. Thus $$\\{X_n:n\\in\\mathbb{N}\\}$$ is not a complete list of all the countable compact subsets of $$\\mathbb{R}$$ up to homeomorphism. • So you did better than the exercise wanted: You showed that there exist uncountably many nonhomeomorphic countable compact subsets of $\\mathbb{R}$. Noice. – Questioner Oct 19 '20 at 21:11 • Nice. What about $\\mathfrak c$ nonhomeomorphic compact subsets of $\\mathbb R$? Can there be only $\\aleph_1$ types of compact subsets while $\\mathfrak c\\gt\\aleph_1$? – bof Oct 20 '20 at 1:41 • You could just assume that the sequence $X_n$ contains each element of the countable collection twice. – bof Oct 20 '20 at 1:45 • @bof: Hah, that was actually my original idea, but this way felt slightly easier to write up. – Eric Wofsey Oct 20 '20 at 2:18 • @bof: As for your question, both of us have in fact answered it before (the answer is there are always $\\mathfrak{c}$ of them): math.stackexchange.com/questions/1602978/…. (OK, that was about closed subsets, but the constructions in my answer in fact gives compact subsets, and your answer can easily be arranged to do so.) – Eric Wofsey Oct 20 '20 at 2:26", "# Collatz dropping times aperiodicity Let $a(n)$ for $n \\geq 1$ be the number of powers of two $2^m$ that lie between $3^{n-1}$ and $3^{n}$: $$a(n) = 1, 2, 1, 2, 1, 2, 2, 1, ...$$ It represents the increments between successive terms of allowable dropping times in the Collatz ($3x+1$) problem (see oeis A022921). Is there an easy way to prove that the sequence never becomes periodic? i.e. do not exist $n_0, k \\geq 1$ such that for all $n \\geq n_0$, $a(n) = a(n_0 + (n - n_0)\\bmod k))$ ? - It has an irrational Cesàro sum: $\\log_23$. – Emil Jeřábek Feb 28 '14 at 15:10 even more interesting: there are two types of partial sums, say $a=1,2$ and $b=1,2,2$. Then you can rewrite it $a_1(n)=a,a,a,b,a,a,a,b,a,a,b,a,a,a,b,...$ ,say. THis has again two types of \"words\",$c=a,a,a,b$ and $d=a,a,b$, say. Then one can rewrite this as $a_2(n) = c,d,d,d,c,d,d,c,d,d,d,....$. Then this has two types of \"words\"... and so on. It is a nice game to find the ever more compressed expression, and is also related to the continued fraction of log(3)/log(2) (or log(3/2), dan't have it at hand at the moment) – Gottfried Helms Mar 1 '14 at 0:42 (cont) I forgot to mention: the a-periodicity guarantees, that this does not run into a repeated sequence of only one", "# Count weighted integer compositions What is the asymptotic growth of the sequence $$a_n:=\\sum_{k\\geq 0} 3^k c_{n,k},$$ as $$n\\rightarrow\\infty$$, where $$c_{n,k}$$ denotes the number of integer compositions of $$n$$ with exactly $$k$$ many 2s? A composition of $$n$$ is a sum $$n=c_1+c_2+\\cdots+c_p$$, with all the $$c_i$$ positive. The first values of the sequence $$a_n$$ are $$1,1,4,8,22,52,135,\\ldots$$ (not in the OEIS). [Edit: As pointed out by Somos below, the value 135 is wrong, and must be corrected to 132, and then the sequence is in the OEIS.] So far, I was only able to prove the following bounds: As $$\\sum_{k\\geq 0}c_{n,k}=2^{n-1}$$, it follows that $$2^{n-1}\\leq a_n \\leq (2\\sqrt{3})^n=(3.464...)^n.$$ • Can we say $c_{n,k}=c_{n-2,k-1}k+1)$, and thus produce a recursion? Jun 21, 2019 at 17:05 • $c_{n,k}$ is the sequence A105114 in the OEIS. Jun 21, 2019 at 21:56 When corrected, $$a_n$$ is the OEIS sequence A052528 whose first nine values are $$\\,1,1,4,8,22,52,132,324,808,\\dots\\,$$ and it has a linear recurrence. The combinatorial recurrence from its combination definition leads immediately to $$a_n = 2a_{n-2} + \\sum_{k=0}^{n-1} a_k$$ where the $$\\, 2a_{n-2}\\,$$ comes from the combination part $$2.$$ The ordinary generating function is $$(1 - x)/(1 - 2 x - 2 x^2 + 2 x^3).$$ The growth rate of $$\\,a_n\\,$$ depends on the reciprocal of the smallest root $$\\,\\alpha\\,$$ of $$\\, 2x^3 -2x^2 -2x", "of the array $A$ and the number of queries respectively. The next line contains $N$ space separated integers $A_i$, the elements of the array $A$. The next $Q$ lines contains two types of queries: • $1\\space P\\space C$ • $2\\space L\\space R\\space K$ Constraints: $1\\leq N \\leq 20$ $1\\leq Q\\leq10^4$ $1\\leq A_i, C\\leq10^{9}$ $1\\leq P\\leq min(N,6)$ $1\\leq L\\leq R < 2^N$ $1\\leq K\\leq (R-L+1)$ ## Output For each of the second type of queries, print the value of the $K^{th}$ element in the range of $[L,R]$ indexes in a line as described in the statement. ## Sample InputOutput 3 5 81 93 93 2 4 7 2 2 1 5 5 1 3 5 2 1 5 5 2 4 7 2 93 267 179 86 Notes: For the first and second queries, Subsets in lexicographically ascending order will be: 1. Taking index $\\{1\\}$. Subset = $\\{81\\}$. Sum = $81$. 2. Taking indexes $\\{1,2\\}$. Subset = $\\{81,93\\}$. Sum = $174$. 3. Taking indexes $\\{1,2,3\\}$. Subset = $\\{81,93,93\\}$. Sum = $267$. 4. Taking indexes $\\{1,3\\}$. Subset = $\\{81,93\\}$. Sum = $174$. 5. Taking index $\\{2\\}$. Subset = $\\{93\\}$. Sum = $93$. 6. Taking indexes $\\{2,3\\}$. Subset = $\\{93,93\\}$. Sum = $186$. 7. Taking index $\\{3\\}$. Subset = $\\{93\\}$. Sum = $93$. So, the $B$ array is $\\{81, 174, 267,", "Partition of $\\{1,2,3,\\cdots,3n\\}$ into $n$ subsets, each with $3$ numbers, which have equal sum I want to show, that for every odd $n$ $(n\\ge3)$, there exists a partition of $\\{1,2,3,\\cdots,3n\\}$ into $n$ disjoint subsets, where each one has $3$ elements and equal sum. The first such number is $3$. For $3$ it is obvious. $\\{1,6,8\\}, \\{2,4,9\\}, \\{3,5,7\\}$. I tried to show this using induction, but it seems I have some trouble with it. Please help me, if you can. • Your example for $n=3$ doesn't work. The sums of the three subsets are $14,15$ and $16$. It should be $\\{3,5,7\\}, \\{2,4,9\\}$ and $\\{1,6,8\\}$. – Anurag A Feb 28 '16 at 8:42 • For $n=3$, either rows or columns of $3\\times 3$ magic square will do. – Ng Chung Tak Feb 28 '16 at 9:58 We let $k$ range from $1$ to $n$. Our sets are $$\\begin {cases} \\{k, \\frac{3n-1}2+k,3n+2-2k\\} &1\\le k \\le \\frac {n+1}2\\\\ \\{k,n+k-\\frac{n+1}2,4n-2k+2\\}&\\frac{n+1}2 \\lt k \\le n \\end {cases}$$ These can be seen to add to $\\frac {9n+3}2$ and to use the numbers $1$ to $n$ in the first entry, $n+1$ to $2n$ in the second and $2n+1$ to $3n$ in the third. They follow the pattern in Ng Chung Tak's answer for $n=5$ • slick !!!!!!!!! – Abr001am Feb 28 '16 at 16:30 • Still", "are all in the set $\\{-2,-1,0,1,2\\}$. Lemma. If $a_k\\in\\{-2,-1,0,1,2\\}$ for $k=0,\\dots,t$ and $a_t\\ne0$, then $\\sum_{k=0}^ta_k3^k\\ne0$. Proof. Without loss of generality assume that $a_t>0$. (Otherwise just multiply everything by $-1$.) Then $$\\sum_{k=0}^ta_k3^k\\ge 3^t-2\\sum_{k=0}^{t-1}3^k=3^t-2\\frac{3^t-1}{3-1}=3^t-\\left(3^t-1\\right)=1\\;.\\qquad\\qquad\\dashv$$ It follows from the lemma that the coefficients $d_k-c_k$ are all $0$ and hence that the two representations of $n$ are identical. To show that each $n\\in\\Bbb Z$ has at least one balanced ternary representation, it suffices to show that each positive integer has one: clearly if $n=\\sum_{k=0}^rc_k3^k$, then $-n=\\sum_{k=0}^r(-c_k)3^k$. Suppose, then, that there is at least one positive integer with no balanced ternary representation, and let $n$ be the smallest such integer. Clearly $n>1$, so by hypothesis there are a non-negative integer $r$ and $c_k\\in\\{-1,0,1\\}$ for $k=0,\\dots,r$ such that $n-1=\\sum_{k=0}^rc_k3^k$. If $c_0\\ne 1$, $$n=(c_0+1)+\\sum_{k=1}^rc_k3^k$$ is a valid representation of $n$, so we must have $c_0=1$. If some $c_k\\ne 1$, let $s$ be minimal such that $c_s\\ne1$. Then \\begin{align*} n&=\\sum_{k=s+1}^rc_k3^k+c_s3^s+\\sum_{k=0}^{s-1}3^k+1\\\\ &=\\sum_{k=s+1}^rc_k3^k+c_s3^s+\\frac{3^s-1}{3-1}+1\\\\ &=\\sum_{k=s+1}^rc_k3^k+c_s3^s+\\frac{3^s+1}2\\\\ &=\\sum_{k=s+1}^rc_k3^k+c_s3^s+\\left(3^s-\\frac{3^s-1}2\\right)\\\\ &=\\sum_{k=s+1}^rc_k3^k+c_s3^s+(c_s+1)3^s-\\sum_{k=0}^{s-1}3^k\\;. \\end{align*} Now $c_s\\ne 1$, so $c_s+1\\in\\{-1,0,1\\}$, Thus, if we let $d_k=-1$ for $k=0,\\dots,s-1$, $d_s=c_s+1$, and $d_k=c_k$ for $k=s+1,\\dots,r$, $$n=\\sum_{k=0}^rd_k3^k$$ is a valid balanced ternary representation of $n$. This contradiction shows that every positive integer has a balanced ternary representation and hence that every integer has one. - Well, the $-1$ as digit is not usual. Thus, you would like to see $5$", "are 8 subsets, namely, $\\phi$, $U$, $\\{ 1\\}$, $\\{ 2\\}$, $\\{ 3 \\}$, $\\{1,2\\}$, $\\{ 1,3\\}$, $\\{ 2,3\\}$. There are 64 possible relations of the form $A \\subseteq B$. Count the number of true ones. Answer: $\\phi \\subseteq U$, $\\phi \\subseteq \\{ 1 \\}$, $\\phi \\subseteq \\{ 2 \\}$, $\\phi \\subseteq \\{ 3 \\}$, $\\{ 1 \\} \\subseteq U$, $\\{ 2 \\} \\subseteq U$, $\\{ 3 \\} \\subseteq U$, $\\{ 1, 2\\} \\subseteq U$, $\\{ 1, 3\\} \\subseteq U$, $\\{ 2,3\\} \\subseteq U$, $\\{ 1\\} \\subseteq \\{ 1, 2 \\}$, $\\{ 1 \\} \\subseteq \\{ 1,3\\}$, $\\{ 2 \\} \\subseteq \\{ 1,2 \\}$, $\\{ 2\\} \\subseteq \\{ 2,3\\}$, $\\{ 3\\} \\subseteq \\{ 1,3\\}$, $\\{ 3\\} \\subseteq \\{ 2,3\\}$. And, also, except $\\phi$ and $U$, each is a subset of itself. Thus, there are such 30 true relations. (3d) Let U be the set $\\{ 1,2,3,\\ldots, n\\}$ for an arbitrary positive integer n. How many subsets are there ? Answer: $2^{n}$ subsets. How many possible relations of the form $A \\subseteq B$? Answer: $2^{n} \\times 2^{n}$, that is, $2^{2n}$ possible pairings. Can you make an informed guess as to how many of these are true? Answer (trial 1): The empty set $\\phi$ is paired with each of the remaining $2^{n}-1$ so these give $2^{n}-1$ pairings; each set which is", "• Yes, that's right. The hand {6♣,6♢,7♡} would count as one sequence just the same as {4♣,6♢,7♡}. – Joakim Sundh Aug 19 '19 at 11:36 Consider two consecutive numbers on the lower $$(1, 2)$$ and higher end $$(K, A)$$. In this case, combinations with one specific third value must be discarded ($$3$$ and $$Q$$, respectively). Assuming three different values, there are $${4 \\choose 1}{4 \\choose 1}{40 \\choose 1} = 640$$ valid combinations. Assuming two different values, there are $${2 \\choose 1}{4 \\choose 2}{4 \\choose 1} = 48$$ valid combinations. Consider two consecutive numbers $$(2, 3)$$ up to $$(Q, K)$$. In this case, combinations with two values must be discarded (e.g., $$1$$ and $$4$$). Assuming three different values, there are $${4 \\choose 1}{4 \\choose 1}{36 \\choose 1} = 576$$ valid combinations. Assuming two different values, there are again $${2 \\choose 1}{4 \\choose 2}{4 \\choose 1} = 48$$ valid combinations. Combining both together, we find: $$\\frac{2(640 + 48) + 10(576 + 48)}{52 \\choose 3} = \\frac{7616}{22100} \\approx 0.3446$$ This is confirmed by the following Python code: import itertools v = list(range(1, 14)) * 4 v.sort() j = 0 for c in itertools.combinations(v, 3): if c[0] == c[1] - 1 and c[1] != c[2] - 1: j += 1 if c[0] != c[1] - 1 and c[1] == c[2] - 1:", "2.5k views Choose the correct alternatives (more than one may be correct) and write the corresponding letters only. Let $r=1(1+0)^*, s=11^*0 \\text{ and } t=1^*0$ be three regular expressions. Which one of the following is true? 1. $L(s) \\subseteq L(r)$ and $L(s) \\subseteq L(t)$ 2. $L(r) \\subseteq L(s)$ and $L(s) \\subseteq L(t)$ 3. $L(s) \\subseteq L(t)$ and $L(s) \\subseteq L(r)$ 4. $L(t) \\subseteq L(s)$ and $L(s) \\subseteq L(r)$ 5. None of the above edited | 2.5k views +12 L(r) = $1(1+0)^*$ L(s) = $1^+0$ L(t) = $1^*0$ To know the answer let us check all the options: 1. $L(s)\\subseteq L(r)$: strings generated by $s$ are any numbers of $1$'s followed by one $0$, i.e., $10,110,1110,1110,\\dots$. Strings generated by $r$ are is $1$ followed by any combination of $0$ or $1$, i.e., $1,10,11,1110,101,110 \\dots$. This shows that all the strings that can be generated by $s$, can also be generated by $r$ it means $L(s)\\subseteq L(r)$ is true. $L(s)\\subseteq L(t)$: here strings generated by $t$ are any numbers of $1$ (here $1^*$ means we have strings as $\\epsilon,1,11,111,\\dots$) followed by only one $0$, i.e., $0,10,110,1110, \\dots$. So we can see that all the strings that are present in $s$ can also be generated by $t$, hence $L(s) \\subseteq L(t)$ which shows that option A is true. 2. $L(r)\\subseteq L(s)$:", "## Wednesday, February 7, 2007 ### Spacy. Topic: S&S/Sets. Level: AMC. Problem: (2007 AMC 12A - #25) Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\\{1, 2, 3, \\ldots, 12\\}$, including the empty set, are spacy? Solution: We will solve this problem with a recursion on $S_n$, which we will define as the number of spacy subsets of $\\{1, 2, 3, \\ldots, n\\}$. We can easily calculate $S_0 = 1$, $S_1 = 2$, $S_2 = 3$, and $S_3 = 4$. Now we will think about how to evaluate $S_n$ in terms of previous terms. Obviously, we can keep all the subsets in $S_{n-1}$. These will account for any space subsets that don't contain the element $n$. Now we just need to count the ones that do contain the element $n$. If a spacy subset contains $n$, it cannot have $n-1$ or $n-2$. So take all the spacy subsets of $\\{1, 2, \\ldots, n-3\\}$ and add the element $n$ to them to get all the spacy subsets that contain $n$. Since these are both of the cases, we obtain $S_n = S_{n-1}+S_{n-3}$. Some calculations yield $S_{12} = 129$. QED. -------------------- Comment: In all honesty, a pretty standard recursion problem, which is probably better off", "that do not contain a $$0$$, do not contain a $$2$$, do not contain a $$5$$ and do not contain an $$8$$. There are $$N_0=3^4=81$$ sequences that do not contain a $$0$$, and similarly $$N_2=N_5=N_8=81$$ sequences that do not contain a $$2$$, a $$5$$ or and $$8$$. Now we could compute the number of sequences that contain each digit as $$N-(N_0+N_2+N_5+N_8)=256-4\\times81=-68,$$ which is clearly wrong, because we have counted the sequences that do not contain a $$0$$ and do not contain a $$2$$ twice. 3. So next add the number of sequences that do not contain a $$0$$ and a $$2$$; there are $$N_{0,2}=2^4=16$$ such sequences, and similarly $$N_{0,5}=N_{0,8}=N_{2,5}=N_{2,8}=N_{5,8}=16.$$ Adding these back we get $$\\begin{eqnarray*} N&-&(N_0+N_2+N_5+N_8)+(N_{0,2}+N_{0,5}+N_{0,8}+N_{2,5}+N_{2,8}+N_{5,8})\\\\ &=&256-4\\times81+16\\times6=28. \\end{eqnarray*}$$ But now we have a problem with the sequences that are missing three digits; we first subtracted them three times, then added them back three times, so we should subtract them again. To clarify, for example the sequence $$8888$$ was counted in $$N_0$$ and $$N_2$$ and $$N_5$$ and subtracted, and then counted in $$N_{0,2}$$ and $$N_{0,5}$$ and $$N_{2,5}$$ and added again. 4. Finally we count the sequences that do not contain a $$0$$, a $$2$$ and a $$5$$; there is of course only $$N_{0,2,5}=1$$ such sequence which is $$8888$$. Similarly $$N_{0,2,8}=N_{0,5,8}=N_{2,5,8}=1$$. Then the total number of sequences is $$\\begin{eqnarray*} N&-&(N_0+N_2+N_5+N_8)+(N_{0,2}+N_{0,5}+N_{0,8}+N_{2,5}+N_{2,8}+N_{5,8})\\\\" ]

[ "# Difference between revisions of \"2007 AMC 12A Problems/Problem 25\" ## Problem Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\\{1,2,3,\\ldots,12\\},$ including the empty set, are spacy? $\\mathrm{(A)}\\ 121 \\qquad \\mathrm{(B)}\\ 123 \\qquad \\mathrm{(C)}\\ 125 \\qquad \\mathrm{(D)}\\ 127 \\qquad \\mathrm{(E)}\\ 129$ ## Solution ### Solution 1 Let $S_{n}$ denote the number of spacy subsets of $\\{ 1, 2, ... n \\}$. We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$. The spacy subsets of $S_{n + 1}$ can be divided into two groups: • $A =$ those not containing $n + 1$. Clearly $|A|=S_{n}$. • $B =$ those containing $n + 1$. We have $|B|=S_{n - 2}$, since removing $n + 1$ from any set in $B$ produces a spacy set with all elements at most equal to $n - 2,$ and each such spacy set can be constructed from exactly one spacy set in $B$. Hence, $S_{n + 1} = S_{n} + S_{n - 2}$ From this recursion, we find that $S(0)$ $S(1)$ $S(2)$ $S(3)$ $S(4)$ $S(5)$ $S(6)$ $S(7)$ $S(8)$ $S(9)$ $S(10)$ $S(11)$ $S(12)$ 1 2 3 4 6 9 13 19 28 41 60 88 129 And so the answer is $(E)$ $129$. ### Solution 2 Since each of", "# Difference between revisions of \"2007 AMC 12A Problems/Problem 25\" ## Problem Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\\{1,2,3,\\ldots,12\\},$ including the empty set, are spacy? $$\\mathrm{(A)}\\ 121 \\qquad \\mathrm{(B)}\\ 123 \\qquad \\mathrm{(C)}\\ 125 \\qquad \\mathrm{(D)}\\ 127 \\qquad \\mathrm{(E)}\\ 129$$ ## Solution ### Solution 1 Let $S_{n}$ denote the number of spacy subsets of $\\{ 1, 2, ... n \\}$. We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$. The spacy subsets of $S_{n + 1}$ can be divided into two groups: • $A =$ those not containing $n + 1$. Clearly $|A|=S_{n}$. • $B =$ those containing $n + 1$. We have $|B|=S_{n - 2}$, since removing $n + 1$ from any set in $B$ produces a spacy set with all elements at most equal to $n - 2,$ and each such spacy set can be constructed from exactly one spacy set in $B$. Hence, $S_{n + 1} = S_{n} + S_{n - 2}$ From this recursion, we find that $S(0)$ $S(1)$ $S(2)$ $S(3)$ $S(4)$ $S(5)$ $S(6)$ $S(7)$ $S(8)$ $S(9)$ $S(10)$ $S(11)$ $S(12)$ 1 2 3 4 6 9 13 19 28 41 60 88 129 ### Solution 2 Since each of the elements of the subsets must be", "## Wednesday, February 7, 2007 ### Spacy. Topic: S&S/Sets. Level: AMC. Problem: (2007 AMC 12A - #25) Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\\{1, 2, 3, \\ldots, 12\\}$, including the empty set, are spacy? Solution: We will solve this problem with a recursion on $S_n$, which we will define as the number of spacy subsets of $\\{1, 2, 3, \\ldots, n\\}$. We can easily calculate $S_0 = 1$, $S_1 = 2$, $S_2 = 3$, and $S_3 = 4$. Now we will think about how to evaluate $S_n$ in terms of previous terms. Obviously, we can keep all the subsets in $S_{n-1}$. These will account for any space subsets that don't contain the element $n$. Now we just need to count the ones that do contain the element $n$. If a spacy subset contains $n$, it cannot have $n-1$ or $n-2$. So take all the spacy subsets of $\\{1, 2, \\ldots, n-3\\}$ and add the element $n$ to them to get all the spacy subsets that contain $n$. Since these are both of the cases, we obtain $S_n = S_{n-1}+S_{n-3}$. Some calculations yield $S_{12} = 129$. QED. -------------------- Comment: In all honesty, a pretty standard recursion problem, which is probably better off", "from it, give a spacy subset of $\\{ 1, 2, \\ldots, n-3 \\}$. There are \\$g(n-3)\\$ of those. Any spacy subset of \\$\\latex \\{1, 2, \\ldots, n\\}\\$ either contains or does not contain $n$, and so $g(n) = g(n-1) + g(n-3)$. For example, the spacy subsets of \\$\\{ 1, 2 \\}\\$ are $\\emptyset, 1$, and \\$2\\$ (for notational ease I’m dropping the braces\\$. The spacy subsets of $\\{ 1, 2, 3, 4 \\}$ are $\\emptyset, 1, 2, 3, 4, 14$. So the spacy subsets of $\\{ 1, 2, 3, 4, 5 \\}$ are \\$\\emptyset, 1, 2, 3, 4, 14\\$ (from the spacy subsets of $\\{ 1, 2, 3, 4\\}$) and $5, 15, 25$ (from adding a 5 to each spacy subset of $\\{ 1, 2\\}$.) Interestingly, the winners in this contest all came out of the financial community: the affiliations mentioned in the post at the Times are Two Sigma Investments, Renaissance Technologies, Morgan Stanley, and Citadel. A few explanations for this present themselves: • people who can solve these sorts of problems quickly tend to end up in that industry; • this was a fundraising event and people in finance have lots of money; • this event took place in New York; you’d have different results if you tried this in, say, San Francisco The last of", "of$\\latex \\{1, 2, \\ldots, n\\}$either contains or does not contain $n$, and so $g(n) = g(n-1) + g(n-3)$. For example, the spacy subsets of$\\{ 1, 2 \\}$are $\\emptyset, 1$, and$2$(for notational ease I’m dropping the braces$. The spacy subsets of $\\{ 1, 2, 3, 4 \\}$ are $\\emptyset, 1, 2, 3, 4, 14$. So the spacy subsets of $\\{ 1, 2, 3, 4, 5 \\}$ are $\\emptyset, 1, 2, 3, 4, 14$ (from the spacy subsets of $\\{ 1, 2, 3, 4\\}$) and $5, 15, 25$ (from adding a 5 to each spacy subset of $\\{ 1, 2\\}$.) Interestingly, the winners in this contest all came out of the financial community: the affiliations mentioned in the post at the Times are Two Sigma Investments, Renaissance Technologies, Morgan Stanley, and Citadel. A few explanations for this present themselves: • people who can solve these sorts of problems quickly tend to end up in that industry; • this was a fundraising event and people in finance have lots of money; • this event took place in New York; you’d have different results if you tried this in, say, San Francisco The last of these explanations would be easy to check. # Kerrich, Ulam, and the scope of classes One of the classic stories of experimental probability comes from the internment", "2 added 110 characters in body Let $C(N,k)$ be the smallest positive integer $x$ such that $[1,x]\\subset \\mathbb{Z}$ contains $k$ disjoint intervals $I_1, ..., I_k$ of $N$ consecutive integers that are all composite. (For example, $C(2,2)=15$, with $I_1=[8,9]$ and $I_2=[14, 15]$.) I am interested in the asymptotic behavior of $C(N,k)$ for various fixed values of $k$. Clearly $C(N,1) \\leq (N+1)!+N+1$. Also if $M$ divides $N$, and we have $k$ disjoint intervals of $N$ consecutive composites, we can break each interval up into $N/M$ disjoint intervals of $M$ consecutive composites, giving a total of $kN/M$ intervals, and so $C(M, kN/M) \\leq C(N, k)$. So, we have $C(N,k) \\leq C(kN, 1) \\leq (kN+1)!+kN+1$. However, these bounds give $15=C(2,2)\\leq C(4,1)=27\\leq 5!+5=125$, which doesn't seem very tight. Can anyone come up with better bounds or asymptotics? The gap between primes is something like $O(\\log(N))$, and so maybe $C(N,k)$ grows sort of like $O(ke^N)$? 1 # Consecutive composite numbers Let $C(N,k)$ be the smallest positive integer $x$ such that $[1,x]\\subset \\mathbb{Z}$ contains $k$ disjoint intervals $I_1, ..., I_k$ of $N$ consecutive integers that are all composite. (For example, $C(2,2)=15$, with $I_1=[8,9]$ and $I_2=[14, 15]$.) I am interested in the asymptotic behavior of $C(N,k)$ for various fixed values of $k$. Clearly $C(N,1) \\leq (N+1)!+N+1$. Also if $M$ divides $N$, and we have $k$ disjoint", "# Pigeon Hole Principle: Six positive integers whose maximum is at most $14$ I was reading this example from my textbook: Let $S$ be a set of six positive integers whose maximum is at most $14$. Show that the sums of the elements in all the nonempty subsets of $S$ cannot all be distinct. For each nonempty subset $A$ of $S$ the sum of the elements in A denoted $S$ satisfies: $1 \\leq S \\leq 9 + 10 + 11 + 12 + 13 + 14 = 69$ And then there are $2^6 - 1 = 63$ non empty subsets of $S$. Could somebody please explain me the logic of $63$. How is this being calculated. Thanks, Rahul • Just for curiosity: since there are more possible results than subsets, how does the textbook finish the problem? (If it is long, don't bother). – ajotatxe Sep 19 '16 at 22:36 • The book is assuming you know that a finite set with n elements will have $2^n$ subsets including the empty set and the entire set itself. Do you know why? – fleablood Sep 19 '16 at 23:35 • Ajotaxte. Min sum is min elements. Max sum is min elements + 10+11+12+13+14. Max minus min is 60 < 63. – fleablood Sep 19 '16 at 23:45 Imagine the", "your sets were {2,5,11} and {13, 31} then your numbers would be 110 and 403. Petek Gold Member I think that Tinyboss has already given you the best advice: If you don't want sets that consist of individual primes, then create sets that consist of products of distinct primes. However, if you're looking for an algorithm to produce sets of pairwise coprime numbers, try this: Choose integers a and $S_0$ such that $S_0$ > a $\\geq$ 1 and $S_0$ and a are relatively prime. Then define $S_n$ recursively by $$S_n = a + S_{n-1}(S_{n-1} - a)$$ Then a and the $S_n$ will be a sequence of pairwise coprime integers. For a = 2 and $S_0$ = 3, you get the Fermat numbers. Petek I think that Tinyboss has already given you the best advice: If you don't want sets that consist of individual primes, then create sets that consist of products of distinct primes. However, if you're looking for an algorithm to produce sets of pairwise coprime numbers, try this: Choose integers a and $S_0$ such that $S_0$ > a $\\geq$ 1 and $S_0$ and a are relatively prime. Then define $S_n$ recursively by $$S_n = a + S_{n-1}(S_{n-1} - a)$$ Then a and the $S_n$ will be a sequence of pairwise coprime integers. For a = 2", "but is not quite the same. In fact, solving a similar equation for circumference = area, 2πr = πr2, we get r = 2, suggesting we should take half the radius of a circle as the natural measure and giving 4π for both the circumference and the area. (The Tauists might be glad to hear that.) Call a set of integers “spacy’ if it contains no more than one out of any three consecutive integers. How many subsets of {1, 2, 3, ……,12} are spacy? Let g(n) be the number of spacy subsets of $\\{1, 2, ..., n\\}$. Then clearly $g(0) = 1, g(1) = 2, g(2) = 3, g(3) = 4$ — in each case the spacy sets are just the empty set and the one-element subsets. Then for larger n, $g(n) = g(n-1) + g(n-3)$. How can you make a spacy subset of $\\{1, 2, \\ldots, n\\}$? Well, any spacy subset of $\\{ 1, 2, \\ldots, n-1 \\}$ is a spacy subset of $\\{ 1, 2, \\ldots, n \\}$ which does not contain$lates n$, and there are $g(n-1)$ of those. And any spacy subset of $\\{ 1, 2, \\ldots, n \\}$ which contains$n$must, when you remove $n$ from it, give a spacy subset of $\\{ 1, 2, \\ldots, n-3 \\}$. There are$g(n-3)$of those. Any spacy subset", "answer? • What is the place of the non-wicked set $\\{1,3,4,5\\}$ in your classification? – Chen Wang May 12 '18 at 0:16 • To fix, need I/E. – vadim123 May 12 '18 at 0:19 There are two good ways to do this problem. 1. Let $a_n$ be the number of wicked subsets of $\\{1,2,\\dots,n\\}$. You can show that $$a_n=a_{n-1}+a_{n-2}+a_{n-3}$$ holds for all $n\\ge 3$. The $a_{n-1}$ term counts subsets where $n$ is absent, $a_{n-2}$ counts subsets where $n$ is present but $n-2$ is absent, and $a_{n-3}$ is subsets with $n$ and $n-1$ but without $n-2$. This recurrence, along with the bases cases $a_0=1,a_1=2,a_2=4$, allows you to iteratively compute $a_n$ from the ground up, starting with $a_3$ all the way up to $n=10$. 2. Instead, count the number of un-wicked (saintly) subsets. Thinking of subsets as strings of ten $0$s and $1$s, where $1$s correspond to the elements in the set, every saintly subset contains a $111$ somewhere. To eliminate the problem of overlaps, in any block of three or more ones, we will only consider the initial $111$, which corresponds to either a $111$ at the beginning or a $0111$ somewhere. There are $2^7$ strings starting with $111$, and $7\\cdot 2^6$ strings containing a $0111$ (the $7$ represents the choice of location of $0111$). Well, not quite; there", "# Pigeonhole principle: prove that any set of $n+1$ integers from $\\{1,2,…,2n\\}$ has two consecutive integers differ by one. Let $n \\ge 1$ be an integer. Use the Pigeonhole Principle to prove that in any set of $n + 1$ integers from $\\{1, 2, . . . , 2n\\}$, there are two integers differ by one. I've attempt and come up with this: Let S be $\\{1,2,...,2n\\}$, $T_0$ be a subset of S with $n$ elements, $T_1$ is a subset of S with $n+1$ elements. We form $T_1$ by adding a number to $T_0$. If $T_0$ already has two integers differ by one, then we're done. If $T_0$ has no integers differ by one, how do we know $T_1$, having 1 more number, has two integers differ by one? • What does it mean by \"regard each entry in $T_0$ and $T_1$ as a box containing some integers\" please? – PSPACEhard Apr 26 '17 at 8:51 • I mean that each box in $T_0$ contains exactly 1 integer. $T_1$ is formed by adding one number to $T_0$, then one of the boxes in $T_1$ must have 2 integers – whatever Apr 26 '17 at 8:53 • So how \"box\" is defined here? – PSPACEhard Apr 26 '17 at 8:53 • Sets are unordered, so \"consecutive\" technically doesn't make", "# Subsets of $\\{1,2 \\dots n\\}$ with no consecutive integers How many subsets with cardinality k of $$\\{1, 2, \\dots n\\}$$ contain no consecutive integers? I know that there are $$F_{n+2}$$ subsets of $$\\{1, 2, \\dots n\\}$$ with no consecutive integers, but I do not know how to go about finding the number for a given $$k$$. This is equivalent to chooosing a sequence of $$k$$ ones and $$n-k$$ zeroes with no adjacent ones. An example with $$n=8$$ and $$k=3$$ is $$00101001, \\text{ corresponding to the set }\\{3,5,8\\}$$ To choose such a sequence, start with a string of $$n-k$$ zeroes, with $$n-k-1$$ spaces between the zeroes, plus two extra spaces before and after, for $$n-k+1$$ spaces total: $$\\;\\_\\; 0\\;\\_\\;0\\;\\_\\;0\\;\\_\\;0\\;\\_\\;0\\;\\_\\qquad,\\text{ with 8-3+1=6 gaps}.$$ Each of the $$k$$ $$1$$'s goes into exactly one gap. We need to choose $$k$$ of these gaps to put a $$1$$ in. This can be done in $$\\binom{n-k+1}{k}\\text{ ways.}$$ Without loss of generality, let your $$k$$ elements from a selected subset be $$x_1 Given such a $$k$$-tuple $$(x_1,x_2,\\dots,x_k)$$, construct the related $$(k+1)$$-tuple $$(x_1-1, x_2-x_1, x_3-x_2,\\dots, x_k-x_{k-1}, n-x_k)$$ describing the distance between each number and/or the boundaries in the case of the first and last numbers. Renaming those values in the $$(k+1)$$-tuple $$(y_1,y_2,\\dots,y_{k+1})$$ we can recognize that $$y_1+y_2+\\dots+y_{k+1} = n-1$$ as the sum telescopes. Now, consider", "integers”. So it isn’t the infinite-ness of one of the subsets that separates it from the other, but rather some sort of “simplicity of specification”. But this still isn’t quite right, because I can very easily specify the subset $\\{63\\}$ as “the number 63”, but this fact doesn’t make this subset any more interesting. There’s nothing cool about the subset $\\{63\\}$ whereas there seems to be something interesting about the set $\\{...,-6, -4, -2, 0, 2, 4, 6, ...\\}$. And whatever it is that is interesting about the subset $\\{..., -6, -4, -2, 0, 2, 4, 6, ...\\}$, the same thing will be interesting about the subset $\\{..., -12, -9, -6, -3, 0, 3, 6, 9, 12, ...\\}$, as well as the subset $\\{..., -30, -20, -10, 0, 10, 20, 30, ...\\}$. As a last piece of motivation for our next definition, let’s take a look at another group that we know and love—modular arithmetic. Let’s consider the set $\\{0, 1, 2, 3, 4, 5, 6, 7\\}$ with addition modulo 8, meaning that we continually “cycle back through” once we hit 8, in precisely the same way we do with the number 12 when we tell time (or the number 24 for the non-12-hour-clock readers). We note that for this group, the subset $\\{0, 2, 4, 6\\}$ is", "2n consisting of n X’s and n Y’s such that no initial segment of the string has more Y’s than X’s. E.g. XXXYYY, XYXXYY, XYXYXY, XXYYXY, XXYXYY If we have an ordered set of unique integers, the Dyck words represent the situations which we do NOT need to check the subset sums (X indicates a number in set #1, Y a number in set #2). Not surprisingly, the number of Dyck words of length 2n is equal to Catalan(n). The total number of words of length 2n with equal number of X’s and Y’s is Choose(2n, n)/2, so the number of non-Dyck words is: Choose(2n, n)/2 – Catalan(n) Finally, given a set of N integers, there are Choose(N, 2n) ways to select a set of 2n unique integers. We need to consider n up to floor(N/2). This leaves: # checks = sum(n=1..N/2, Choose(N,2n)*[Choose(2n,n)/2-Catalan(n)] ) Storing a Pascal’s triangle up to size N for computing the combinations makes this computation near instantaneous. Fangzhou To find the number of pairs of subsets, there’s a pretty easy formula using the principle of inclusion exclusion. On top of the 2 subsets, look at the remaining unused numbers. These numbers form a third subset. Therefore, each number can be in subset 1 (s1), subset 2 (s2), or the unused subset(s3). For a set", "# Show that in any set of $2n$ integers, there is a subset of $n$ integers whose sum is divisible by $n$. There was a problem in a recent programming competition which my friend solved by assuming the following conjecture: Show that for any set of $$2n$$ integers, there is a subset of $$n$$ integers whose sum is divisible by $$n$$. I have thought about this problem for a while but can't seem to prove it, but I couldn't come up with a counter-example either. A similar problem has a well-known solution: show that for any set of $$n$$ integers, there is a non-empty subset whose sum is divisible by $$n$$. The proof goes as follows. Suppose the set is $$\\{x_1, x_2, \\dots, x_n\\}$$ and hence define $$s_i = \\left(x_1 + x_2 + \\dots + x_i\\right)\\bmod n$$, with $$s_0 = 0$$. Then we have the set $$\\{s_0, s_1, \\dots, s_n\\}$$ with $$n+1$$ elements, but each $$s_i$$ can take only $$n$$ distinct values, so there are two $$i, j$$ with $$i\\neq j$$ such that $$s_i = s_j$$. Then $$s_j - s_i = x_{i+1} + x_{i+2} + \\dots + x_j$$ is divisible by $$n$$. However, this approach can't directly be applied to this problem since now we need to ensure that we choose exactly $$n$$ integers. • you can actually", "# Minimum sum of set whose average of subsets is positive integer A finite set of positive integers $A$ is called meanly if for each of its nonempty subsets the arithmetic mean of its elements is also a positive integer. In other words, $A$ is meanly if $\\frac{1}{k}(a_1 + \\dots + a_k)$ is an integer whenever $k \\ge 1$ and $a_1, \\dots, a_k \\in A$ are distinct. Given a positive integer $n$, what is the least possible sum of the elements of a meanly $n$-element set? Source: Middle European Math Competition For motivation, let's do some small ones by hand. For $n=1$, the set can be $\\{1\\}$ with sum $1$. For $n=2$, we can put $1$ in, but then we can't have $2$, which would give mean $\\frac 32$ for the whole set. But $\\{1,3\\}$ with sum $4$ works. It should be clear that if $n \\ge 2$ all the numbers have to have the same parity. For $n=3$, then, we can't use $4$, but $\\{1,3,5\\}$ with sum $9$ works. But if $n\\ge 4$, we can't have all of $1,3,5$, as whatever the fourth element $k$ is, either $\\{1,3,k\\}$ or $\\{1,5,k\\}$ will not have an integral mean. The best we can do is $\\{1,7,13,19\\}$ with sum $40$ This indicates that all the numbers in the set must be", "\\{ 1,2,\\dots,n \\}$. If $F$ is to have size $n$, then necessarily $F$ must consist of $n$-many copies of $\\{ 1,2,\\dots,n \\}$. Right? – Asher M. Kach Jul 2 '12 at 23:08 Yes, if $F$ has size $n$, then $F$ just contains $n$ copies of $[n]$, in other words, $F$ contains $n$ copies of the partition, which is just the whole set itself. In this sense, it contains $n$ partitions of $[n]$ in this case too. For example, for n=3, $F$ could be {{1,2,3},{1,2,3},{1,2,3}} or {{1},{2,3},{1,2},{3},{1,2,3}} or {{1},{2},{3},{1},{2},{3},{1},{2},{3}}, etc. – László Kozma Jul 3 '12 at 8:59 But $F$ is not necessarily of size $n$, it is just $\\geq n$ and $\\leq n^2$. – László Kozma Jul 3 '12 at 9:01", "# Number of compositions of $n$ such that each term is less than equal to $k.$ Let $$n$$ be an integer $$\\geq 1.$$ Then a partition of $$n$$ is a sequence of positive integers (greater than equal to $$1$$) such that their sum equals $$n.$$ So for instance if $$n=4$$ then $$[[4], [1, 3], [1, 1, 2], [1, 1, 1, 1], [1, 2, 1], [2, 2], [2, 1, 1], [3, 1]]$$ is a collection of all partitions of $$n$$ with order. Clearly for any $$n$$ the number of such ordered partitions is $$2^{n-1}.$$ However I want to count the number of paritions of $$n$$ where each integer in the parition is less than equal to some integer $$k.$$ So for instance if $$n=4$$ and $$k=2$$ then $$[[1, 1, 2], [1, 1, 1, 1], [1, 2, 1], [2, 2], [2, 1, 1]]$$ is a collection of all the $$2$$-partitions of $$4.$$ Is there a general formula for finding this or maybe an asymptotic expression? • See this, specifically $A$-restricted compositions. – MathematicsStudent1122 Dec 28 '18 at 21:48 • Well, for $k=2$ you get the Fibonacci sequence, so if there is a general formula, it will be complicated for sure. – SmileyCraft Dec 28 '18 at 21:48 • @MathematicsStudent1122 I saw this, but it does not give me the answer", "into Boxes Before we start, I want to introduce a few new terms. A $k$-partition of a set A is a collection of $k$ non-empty subsets of A. The number of $k$-partition of a set of $n$ elements is denoted by $S(n,k)$. This value is called a Stirling number of the second kind. (The forumula will be discussed later.) A $k$-partition of a positive integer $n$ is an integer solution to the equation $x_1+x_2+\\dots+x_k=n$, where $x_1\\geq x_2\\geq\\dots\\geq x_k\\geq 1$. The number of solutions is denoted by $P(n,k)$. (The closed formula is, however, not yet discovered.) In the four following cases, the number of objects is $k$ and the number of boxes is $n$. #### Different objects and Different boxes The boxes are numbered from 1 to n. The objects are numbered from 1 to k. Each object $i$ is assigned with a number $b_i$ from 1 to n, denoting the box into which this object will be put. An arrangement is the sequence $(1,b_1), (2,b_2),\\dots, (k, b_k)$. Count the number of arrangements. Condition # None $n^k$ Each box has $\\leq$ 1 object (only if $k\\leq n$) $\\frac{n!}{(n-k)!}$ Each box has $\\geq$ 1 object (only if $k\\geq n$) $n!\\times S(k,n)$ Box $i$ contains $n_i$ objects $\\frac{k!}{k_1!k_2!\\dots k_n!}$ #### Identical objects and Different boxes The boxes are numbered from 1", "# Show that $\\sum_{\\{a_1, a_2, \\dots, a_k\\}\\subseteq\\{1, 2, \\dots, n\\}}\\frac{1}{a_1*a_2*\\dots*a_k} = n$ Question: Show that $$\\sum_{\\{a_1, a_2, \\dots, a_k\\}\\subseteq\\{1, 2, \\dots, n\\}}\\frac{1}{a_1*a_2*\\dots*a_k} = n$$ (Here the sum is over all non-empty subsets of the set of the $n$ smallest positive integers.) I made an attempt and then encountered an inconsistency with the answer key which is detailed at the very bottom. My Attempt: $n=1$, there’s only one non-empty subset $\\{1\\}$, hence, $$\\frac{1}{1} = 1$$ Let the proposition in the question be $P(n)$. Suppose $P(n)$ is true, show that $P(n)\\rightarrow P(n+1)$ is true, $$P(n) = \\sum_{\\{a_1, a_2, \\dots, a_k\\}\\subseteq\\{1, 2, \\dots, n\\}}\\frac{1}{a_1*a_2*\\dots*a_k}=n$$ $P(n+1)$ would mean, I add another integer $n+1$ to the set, the change to the possible subsets are divided into 3 cases, 1. Subset of only $\\{n+1\\}$. 2. Subsets of all positive integers containing $\\{n+1\\}$. 3. Subsets of all positive integers not containing $\\{n+1\\}$. First case, just $$\\frac{1}{n+1}$$, Second case, is $$\\frac{P(n)}{n+1} = \\frac{n}{n+1}$$, Third case, by inductive hypothesis is just $n$, Thus, adding them all we get, $$\\frac{1}{n+1} + \\frac{n}{n+1} + n = n + 1$$, completing the proof. Problem: The following are cases in the answer key: 1. Just $\\{n+1\\}$ 2. a non-empty subsets of the first $n$ positive integers together with $n+1$. 3. a non-empty subset of the first $n$ positive integers. (I’ve rearrange" ]

[ "# Question #b3e6b Sep 14, 2016 $103$ toothpicks in the ${50}^{\\text{th}}$ figure. $2 n + 3$ toothpicks in the ${n}^{\\text{th}}$ figure. In each progressive figure, one toothpick is added to the top row and one is added to the bottom row. As the ${1}^{\\text{st}}$ figure has $1$ toothpick in the top row and $2$ in the bottom row, that means that the ${n}^{\\text{th}}$ figure will have $n$ toothpicks in the top row and $n + 1$ in the bottom row. Adding these to the $2$ toothpicks which constitute the left and right sides, we get the total for the ${n}^{\\text{th}}$ figure as $n + \\left(n + 1\\right) + 2 = 2 n + 3$. To figure out how many are in the ${50}^{\\text{th}}$ figure, then, we just let $n = 50$ to get $2 \\left(50\\right) + 3 = 103$.", "the pattern with the equation $t=5 n-3,$ where $t$ is the number of toothpicks in Stage $n .$a. Explain how each part of the equation is related to the toothpick pattern.b. How many toothpicks would Evan need for Stage 10$?$ For Stage 100$?$c. Evan used 122 toothpicks to make one stage of his pattern. Write and solve an equation to find the stage number.d. Is any stage of the pattern composed of 137 toothpicks? Why or why not?e. Is any stage of the Evan made this toothpick pattern. He described the pattern with the equation $t=5 n-3,$ where $t$ is the number of toothpicks in Stage $n .$ a. Explain how each part of the equation is related to the toothpick pattern. b. How many toothpicks would Evan need for Stage 10$?$ For Stage 100$?$ c. Eva...", "Problem #2045 2045 Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label(\"Figure\",(0.5,-1),S); label(\"0\",(0.5,-2.5),S); label(\"Figure\",(9.5,-1),S); label(\"1\",(9.5,-2.5),S); label(\"Figure\",(19.5,-1),S); label(\"2\",(19.5,-2.5),S); label(\"Figure\",(32.5,-1),S); label(\"3\",(32.5,-2.5),S); [/asy]$ $\\textbf{(A)}\\ 10401 \\qquad\\textbf{(B)}\\ 19801 \\qquad\\textbf{(C)}\\ 20201 \\qquad\\textbf{(D)}\\ 39801 \\qquad\\textbf{(E)}\\ 40801$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "85 ∗ 101 = ____________ b. 156 ∗ 9 = ____________ c. 48 ∗ 24 = ____________ Question 4. Rewrite each expression as a product by taking out a common factor a. 48 + 24 = _______ ∗ (_______ + _______) = _______ ∗ _______ b. 72 – 56 = _______ ∗ (_______ – _______) = _______ ∗ _______ c. (2y) + (3 ∗ y) = (_______ + _______) ∗ _______ = _______ ∗ _______ Practice Use <, >, or = to make the sentence true. Question 5. $$\\frac{2}{3}$$ _________ $$\\frac{2}{5}$$ Question 6. 0.7 _______ $$\\frac{4}{5}$$ Question 7. 0.3 ______ 0.23 Question 8. 1 $$\\frac{1}{4}$$ _______ 1.25 Building with Toothpicks Yaneli is building a pattern with toothpicks. The pattern grows in the following way: Question 1. How many toothpicks are needed for Design 5? _________ Question 2. How many toothpicks are needed for Design 10? __________ Question 3. Describe in words how you see the toothpick design growing. What stays the same from one figure to the next? What changes? Question 4. Write an expression to represent how many toothpicks are needed for Design n? Question 5. What toothpick design number could you build with exactly 82 toothpicks? ________ Question 6. Describe how you can figure out the number of toothpicks you need for any design number. Practice", "# Question 1 of 10 Helen uses some toothpicks to form the pattern below. \\begin{array}{|c|c|} \\hline \\mbox{Pattern} & \\mbox{Number of toothpick} \\\\ \\hline 1 & 6\\\\ \\hline 2 & 15\\\\ \\hline 3 & 25\\\\ \\hline 4 & 35\\\\ \\hline 5 & \\\\ \\hline \\end{array} (a) How many toothpicks will she need to form Pattern 5? (b) How many toothpick will she need to form Pattern 40? (c) Helen uses 4955 toothpicks to form a Pattern. Which Pattern is it? A (a) 45 (b) 40 (c) 496 B (a) 47 (b) 43 (c) 498 C (a) 49 (b) 45 (c) 500 D (a) 50 (b) 49 (c) 502 E None of the above", "the table below to determine the number of toothpicks needed for the pattern. Pattern number 1 2 3 4 5 ... ... 200 Number of toothpicks Completed, the table looks like: Pattern number 1 2 3 4 5 ... ... 200 Number of toothpicks 7 12 17 22 27 1002 The number of toothpicks in any pattern number is the result of multiplying the pattern number by 5 and adding 2 to the product. The number of toothpicks in pattern number 200 is: The data must be discrete. The graph would be dots representing the pattern number and the number of toothpicks. It is impossible to have a pattern number or a number of toothpicks that are not natural numbers. Therefore, the points would not be joined. The domain and range are: If the range is written in terms of a function, then the number system to which belongs must be designated in the range. ### Examples #### Example 1 Earlier, you were told that to avoid the road construction, Joseph decided to travel the gravel road. After driving for 20 minutes he was 62 miles away from work and after driving for 40 minutes he was 52 miles away from work. What would the graph look like for this problem? What would a suitable domain and range", "# Difference between revisions of \"2000 AMC 12 Problems/Problem 8\" The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page. ## Problem Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label(\"Figure\",(0.5,-1),S); label(\"0\",(0.5,-2.5),S); label(\"Figure\",(9.5,-1),S); label(\"1\",(9.5,-2.5),S); label(\"Figure\",(19.5,-1),S); label(\"2\",(19.5,-2.5),S); label(\"Figure\",(32.5,-1),S); label(\"3\",(32.5,-2.5),S); [/asy]$ $\\mathrm{(A)}\\ 10401 \\qquad\\mathrm{(B)}\\ 19801 \\qquad\\mathrm{(C)}\\ 20201 \\qquad\\mathrm{(D)}\\ 39801 \\qquad\\mathrm{(E)}\\ 40801$ ## Solution 1 We can divide up figure $n$ to get the sum of the sum of the first $n+1$ odd numbers and the sum of the first $n$ odd numbers. If you do not see this, here is the example for $n=3$: $[asy] draw((3,0)--(4,0)--(4,7)--(3,7)--cycle); draw((0,3)--(7,3)--(7,4)--(0,4)--cycle); draw((2,1)--(5,1)--(5,6)--(2,6)--cycle); draw((1,2)--(6,2)--(6,5)--(1,5)--cycle); draw((3,0)--(3,7)); [/asy]$ The sum of the first $n$ odd numbers is $n^2$, so for figure $n$, there are $(n+1)^2+n^2$ unit squares. We plug in $n=100$ to get $\\boxed{\\textbf{(C) }20201}$. ## Solution 2 Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$. Thus, the original sequence can", "2 3 4 5 Number of Toothpicks 4 10 18 28 40 Increase in Toothpicks +4 +6 +8 +10 +12 Number of Squares 1 3 6 10 15 Increase in Squares +1 +2 +3 +4 +5 My problem is that I can't get from the patterns to creating a formula that would work for any figure number without listing them all out consecutively #### MarkFL Staff member I see you have found that the difference in the number of toothpicks per iteration (where you state the increase in toothpicks) are the sequence of even numbers, beginning with 4. What is the second difference, that is, the difference of the differences. How much does the increase increase each time? #### Opalg ##### MHB Oldtimer Staff member Yeah I'm up to my neck in patterns; I made a table and everything. Figure Number 1 2 3 4 5 Number of Toothpicks 4 10 18 28 40 Increase in Toothpicks +4 +6 +8 +10 +12 Number of Squares 1 3 6 10 15 Increase in Squares +1 +2 +3 +4 +5 My problem is that I can't get from the patterns to creating a formula that would work for any figure number without listing them all out consecutively Let $x_n$ be the number of toothpicks in Fig. $n$. Your table shows", "\\ \\in \\ R \\} \\quad R=\\{y|y \\ge -3, y \\ \\in \\ R \\}$ #### Example B Whether a relation is discrete, continuous, or neither can often be determined without a graph. The domain and range can be determined without a graph as well. Examine the following toothpick pattern. Complete the table below to determine the number of toothpicks needed for the pattern. Pattern number $(n)$ 1 2 3 4 5 ... $n$ ... 200 Number of toothpicks $(t)$ Is the data continuous or discrete? Why? What is the domain? What is the range? Solution: Pattern number $(n)$ 1 2 3 4 5 ... $n$ ... 200 Number of toothpicks $(t)$ 7 12 17 22 27 $5n+2$ 1002 The number of toothpicks in any pattern number is the result of multiplying the pattern number by 5 and adding 2 to the product. The number of toothpicks in pattern number 200 is: $t&=5n+2\\\\t&=5({\\color{red}200})+2\\\\t&=1000+2\\\\t&=1002$ The data must be discrete. The graph would be dots representing the pattern number and the number of toothpicks. It is impossible to have a pattern number or a number of toothpicks that are not natural numbers. Therefore, the points would not be joined. The domain and range are: $D=\\{x|x \\ \\in \\ N\\} \\quad R=\\{y|y=5x+2, x \\ \\in \\ N\\}$ If the range is", "\\varepsilon \\ R \\} \\quad R=\\{y|y \\ge -3, y \\ \\varepsilon \\ R \\}$ #### Example B Whether a relation is discrete or continuous can often be determined without a graph. As well, the domain and range can also be determined. Let’s examine the following toothpick pattern. Complete the table below to determine the number of toothpicks needed for the pattern. Pattern number $(n)$ 1 2 3 4 5 ... $n$ ... 200 Number of toothpicks $(t)$ Is the data continuous or discrete? Why? What is the domain? What is the range? Solution: Pattern number $(n)$ 1 2 3 4 5 ... $n$ ... 200 Number of toothpicks $(t)$ 7 12 17 22 27 $5n+2$ 1002 The number of toothpicks in any pattern number is the result of multiplying the pattern number by 5 and adding 2 to the product. The number of toothpicks in pattern number 200 is: $t&=5n+2\\\\t&=5({\\color{red}200})+2\\\\t&=1000+2\\\\t&=1002$ The data must be discrete. The graph would be dots representing the pattern number and the number of toothpicks. It is impossible to have a pattern number or a number of toothpicks that are not natural numbers. Therefore, the points would not be joined. The domain and range are: $D=\\{x|x \\ \\varepsilon \\ N\\} \\quad R=\\{y|y=5x+2, x \\ \\varepsilon \\ N\\} \\ \\ OR \\ \\ R=\\{y|y \\ge", "Difference between revisions of \"2000 AMC 12 Problems/Problem 8\" The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page. Problem Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label(\"Figure\",(0.5,-1),S); label(\"0\",(0.5,-2.5),S); label(\"Figure\",(9.5,-1),S); label(\"1\",(9.5,-2.5),S); label(\"Figure\",(19.5,-1),S); label(\"2\",(19.5,-2.5),S); label(\"Figure\",(32.5,-1),S); label(\"3\",(32.5,-2.5),S); [/asy]$ $\\mathrm{(A)}\\ 10401 \\qquad\\mathrm{(B)}\\ 19801 \\qquad\\mathrm{(C)}\\ 20201 \\qquad\\mathrm{(D)}\\ 39801 \\qquad\\mathrm{(E)}\\ 40801$ Solution 1 We can divide up figure $n$ to get the sum of the sum of the first $n+1$ odd numbers and the sum of the first $n$ odd numbers. If you do not see this, here is the example for $n=3$: $[asy] draw((3,0)--(4,0)--(4,7)--(3,7)--cycle); draw((0,3)--(7,3)--(7,4)--(0,4)--cycle); draw((2,1)--(5,1)--(5,6)--(2,6)--cycle); draw((1,2)--(6,2)--(6,5)--(1,5)--cycle); draw((3,0)--(3,7)); [/asy]$ The sum of the first $n$ odd numbers is $n^2$, so for figure $n$, there are $(n+1)^2+n^2$ unit squares. We plug in $n=100$ to get $\\boxed{\\textbf{(C) }20201}$. Solution 2 Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$. Thus, the original sequence can be generated from a", "# Toothpick Squares Sequence #### marsman ##### New member Please help! I need to be able to find the number of total toothpicks used in any given figure in this sequence, in addition to total squares in any figure, using the figure number. The only way I can figure this out is by listing them all out one by one, which is TERRIBLE Thank you so much in advance!!!!! #### Opalg ##### MHB Oldtimer Staff member Please help! I need to be able to find the number of total toothpicks used in any given figure in this sequence, in addition to total squares in any figure, using the figure number.View attachment 1592 The only way I can figure this out is by listing them all out one by one, which is TERRIBLE Thank you so much in advance!!!!! Suggestion: In the six figures, count how many toothpicks you have to add to each figure in order to get the next one. Can you see a pattern there? #### marsman ##### New member Suggestion: In the six figures, count how many toothpicks you have to add to each figure in order to get the next one. Can you see a pattern there? Yeah I'm up to my neck in patterns; I made a table and everything. Figure Number 1", "So, there is a repeating pattern for every $30$ digits on the original list. As shown below, the digits erased in the first, second, and third rounds are colored in red, yellow, and green, respectively: $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,5)--(i+0.5,5)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,5)--(i+0.5,5)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,5)--(8,5)--(8,0)--cycle,green); fill((18,0)--(18,5)--(19,5)--(19,0)--cycle,green); fill((27,0)--(27,5)--(28,5)--(28,0)--cycle,green); for (real i=1; i<5; ++i) { for (real j=0; j<30; ++j) { label(\"\"+string(1+j%5)+\"\",(j+0.5,i+0.5)); } } for (real i=0; i<30; ++i) { label(\"\\vdots\",(i+0.5,2/3)); } add(grid(30,5,linewidth(1.25))); [/asy]$ As indicated by the white squares, each group of $30$ digits on the original list has $\\frac25\\cdot30=12$ digits remaining on the final list. Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,1)--(8,1)--(8,0)--cycle,green); fill((18,0)--(18,1)--(19,1)--(19,0)--cycle,green); fill((27,0)--(27,1)--(28,1)--(28,0)--cycle,green); for (real j=0; j<30; ++j) { label(\"\"+string(1+j%5)+\"\",(j+0.5,0.5)); } draw((3.5,-1.25)--(3.5,-0.2),linewidth(1.25),EndArrow); draw((6.5,-1.25)--(6.5,-0.2),linewidth(1.25),EndArrow); draw((9.5,-1.25)--(9.5,-0.2),linewidth(1.25),EndArrow); add(grid(30,1,linewidth(1.25))); [/asy]$ Therefore, the answer is $4+2+5=\\boxed{\\textbf{(D)}\\ 11}.$ ~MRENTHUSIASM ## Video Solution https://youtu.be/t6yjfKXpwDs 16:40 ~IceMatrix", "mm}\\diamondsuit \",(12,4.5),S); [/asy]$ $\\text{(A)}\\ 1 \\qquad \\text{(B)}\\ 2 \\qquad \\text{(C)}\\ 3 \\qquad \\text{(D)}\\ 4 \\qquad \\text{(E)}\\ 5$ ## Problem 25 How many different patterns can be made by shading exactly two of the nine squares? Patterns that can be matched by flips and/or turns are not considered different. For example, the patterns shown below are not considered different. $[asy] fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,gray); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle,gray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,linewidth(1)); draw((2,0)--(2,3),linewidth(1)); draw((0,1)--(3,1),linewidth(1)); draw((1,0)--(1,3),linewidth(1)); draw((0,2)--(3,2),linewidth(1)); fill((6,0)--(8,0)--(8,1)--(6,1)--cycle,gray); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle,linewidth(1)); draw((8,0)--(8,3),linewidth(1)); draw((6,1)--(9,1),linewidth(1)); draw((7,0)--(7,3),linewidth(1)); draw((6,2)--(9,2),linewidth(1)); fill((14,1)--(15,1)--(15,3)--(14,3)--cycle,gray); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle,linewidth(1)); draw((14,0)--(14,3),linewidth(1)); draw((12,1)--(15,1),linewidth(1)); draw((13,0)--(13,3),linewidth(1)); draw((12,2)--(15,2),linewidth(1)); fill((18,1)--(19,1)--(19,3)--(18,3)--cycle,gray); draw((18,0)--(21,0)--(21,3)--(18,3)--cycle,linewidth(1)); draw((20,0)--(20,3),linewidth(1)); draw((18,1)--(21,1),linewidth(1)); draw((19,0)--(19,3),linewidth(1)); draw((18,2)--(21,2),linewidth(1)); [/asy]$ $\\text{(A)}\\ 3 \\qquad \\text{(B)}\\ 6 \\qquad \\text{(C)}\\ 8 \\qquad \\text{(D)}\\ 12 \\qquad \\text{(E)}\\ 18$", "all possible maximum values of $S.$ $[asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label(\"\"+string(i+0.5)+\"\",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label(\"\"+string(i+42.5)+\"\",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label(\"\"+string(i+84.5)+\"\",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy]$ From the second row of the table above, we perform casework on the possible maximum value of $S:$ $$\\begin{array}{c||c|c|l} & & & \\\\ [-2.5ex] \\textbf{Max Value} & \\boldsymbol{S} & \\textbf{Valid?} & \\hspace{16.25mm}\\textbf{Reasoning/Conclusion} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 49 & \\{46,47,48,49\\} & & \\text{The student who gets } 46 \\text{ will reply yes.} \\\\ 50 & \\{47,48,49,50\\} & \\checkmark & \\text{Another possibility is } S=\\{89,90,91,92\\}. \\\\ 51 & \\{48,49,50,51\\} & & \\text{The student who gets } 51 \\text{ will reply yes.} \\\\ 56 & \\{53,54,55,56\\} & & \\text{The student who gets } 53 \\text{ will reply yes.} \\\\ 57 & \\{54,55,56,57\\} & & \\text{The student who gets } 57", "By observations, we circle all possible maximum values of $S.$ $[asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label(\"\"+string(i+0.5)+\"\",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label(\"\"+string(i+42.5)+\"\",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label(\"\"+string(i+84.5)+\"\",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy]$ From the second row of the table above, we perform casework on the possible maximum value of $S:$ $$\\begin{array}{c||c|c|l} & & & \\\\ [-2.5ex] \\textbf{Max Value} & \\boldsymbol{S} & \\textbf{Valid?} & \\hspace{16.25mm}\\textbf{Reasoning/Conclusion} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 49 & \\{46,47,48,49\\} & & \\text{The student who gets } 46 \\text{ will reply yes.} \\\\ 50 & \\{47,48,49,50\\} & \\checkmark & \\text{Another possibility is } S=\\{89,90,91,92\\}. \\\\ 51 & \\{48,49,50,51\\} & & \\text{The student who gets } 51 \\text{ will reply yes.} \\\\ 56 & \\{53,54,55,56\\} & & \\text{The student who gets } 53 \\text{ will reply yes.} \\\\ 57 & \\{54,55,56,57\\} & & \\text{The student", "We have a \"stack\" of rows of squares. The rows have consecutive odd numbers of squarea. The squares are made of toothpicks. How many toothpicks are used in the $$n^{th}$$ diagram? Code: n=4 n=3 * - * | | n=2 * - * * - * - * - * | | | | | | n=1 * - * * - * - * - * * - * - * - * - * - * | | | | | | | | | | | | * - * * - * - * - * * - * - * - * - * - * * - * - * - * - * - * - * - * | | | | | | | | | | | | | | | | | | | | * - * * - * - * - * * - * - * - * - * - * * - * - * - * - * - * - * - * 4 13 26 43 We have the sequence: .$$4,13,26,64,\\;\\text{ . . .}$$ Take the differences of consecutive terms, then take the differences of the differences, and so on. $$\\begin{array}{|c|ccccccccc|}\\hline \\text{Sequence} & 4 && 13 && 26", "Pie charts with LaTeX TikZ Define a new command to insert a pie slice: \\newcommand{\\slice}[4]{ \\pgfmathparse{0.5*#1+0.5*#2} \\let\\midangle\\pgfmathresult % slice \\draw[thick,fill=black!10] (0,0) -- (#1:1) arc (#1:#2:1) -- cycle; % outer label \\node[label=\\midangle:#4] at (\\midangle:1) {}; % inner label \\pgfmathparse{min((#2-#1-10)/110*(-0.3),0)} \\let\\temp\\pgfmathresult \\pgfmathparse{max(\\temp,-0.5) + 0.8} \\let\\innerpos\\pgfmathresult \\node at (\\midangle:\\innerpos) {#3}; } Then define the slices in the order you want to have them and with the percentages and labels. You can start at a different point in the circle by setting the counter ‘d’ to a different value before the loop, e.g. \\setcounter{d}{25}. \\begin{tikzpicture}[scale=3] \\newcounter{c} \\newcounter{d} \\foreach \\p/\\t in {66/, 17/Equative, 10/Difference, 7/} { \\setcounter{c}{\\value{d}}", "of getting four gumballs of the same color is $\\text{(A)}\\ 8 \\qquad \\text{(B)}\\ 9 \\qquad \\text{(C)}\\ 10 \\qquad \\text{(D)}\\ 12 \\qquad \\text{(E)}\\ 18$ ## Problem 22 The two wheels shown below are spun and the two resulting numbers are added. The probability that the sum is even is $[asy] draw(circle((0,0),3)); draw(circle((7,0),3)); draw((0,0)--(3,0)); draw((0,-3)--(0,3)); draw((7,3)--(7,0)--(7+3*sqrt(3)/2,-3/2)); draw((7,0)--(7-3*sqrt(3)/2,-3/2)); draw((0,5)--(0,3.5)--(-0.5,4)); draw((0,3.5)--(0.5,4)); draw((7,5)--(7,3.5)--(6.5,4)); draw((7,3.5)--(7.5,4)); label(\"3\",(-0.75,0),W); label(\"1\",(0.75,0.75),NE); label(\"2\",(0.75,-0.75),SE); label(\"6\",(6,0.5),NNW); label(\"5\",(7,-1),S); label(\"4\",(8,0.5),NNE); [/asy]$ $\\text{(A)}\\ \\dfrac{1}{6} \\qquad \\text{(B)}\\ \\dfrac{1}{4} \\qquad \\text{(C)}\\ \\dfrac{1}{3} \\qquad \\text{(D)}\\ \\dfrac{5}{12} \\qquad \\text{(E)}\\ \\dfrac{4}{9}$ ## Problem 23 If $X$, $Y$ and $Z$ are different digits, then the largest possible $3-$digit sum for $\\begin{tabular}{ccc} X & X & X \\\\ & Y & X \\\\ + & & X \\\\ \\hline \\end{tabular}$ has the form $\\text{(A)}\\ XXY \\qquad \\text{(B)}\\ XYZ \\qquad \\text{(C)}\\ YYX \\qquad \\text{(D)}\\ YYZ \\qquad \\text{(E)}\\ ZZY$ ## Problem 24 A $2$ by $2$ square is divided into four $1$ by $1$ squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares. $\\text{(A)}\\ 4 \\qquad \\text{(B)}\\ 6 \\qquad \\text{(C)}\\ 7 \\qquad \\text{(D)}\\ 8 \\qquad \\text{(E)}\\ 16$ ##", "fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); draw(arc((0,3),1,-15,-165)); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3)); draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3)); label(\"6\",.74*dir(-90)+(0,3)); label(\"5\",.74*dir(-60)+(0,3)); label(\"4\",.74*dir(-30)+(0,3)); [/asy]$ The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\\circ$ clockwise (i.e. 2 hours). $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150),EndArcArrow);label(\"30^\\circ\",2*dir(150),W); draw(1.8*dir(120)--2*dir(120)); draw(1.8*dir(90)--2*dir(90)); label(rotate(30)*\"12\",1.56*dir(120)); label(rotate(30)*\"1\",1.56*dir(90)); draw(arc((0,3),1,-15,-165),EndArcArrow);label(\"60^\\circ\",dir(-165)+(0,3),W); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3)); draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3)); label(rotate(-60)*\"6\",.74*dir(-150)+(0,3)); label(rotate(-60)*\"5\",.74*dir(-120)+(0,3)); label(rotate(-60)*\"4\",.74*dir(-90)+(0,3)); [/asy]$ However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\\circ$ clockwise to see what happens when the small cog rolls around it. $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); pair c=(1.5,sqrt(27)/2); draw(arc(c,1,0,-200),EndArcArrow);label(\"90^\\circ\",dir(-180)+c,W); draw(0.9*dir(-120)+c--dir(-120)+c); draw(0.9*dir(-150)+c--dir(-150)+c); draw(0.9*dir(-180)+c--dir(-180)+c); label(rotate(-90)*\"6\",.74*dir(-180)+c); label(rotate(-90)*\"5\",.74*dir(-150)+c); label(rotate(-90)*\"4\",.74*dir(-120)+c); [/asy]$ It turns out that, when the point of tangency moves $30^\\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\\boxed{\\textbf{(C) }\\text{4 o' clock}}$ ## Solution 5 We unfold the big clock and draw the small clock. We know that the small clocks arrow will be facing up again at 6 o clock on the unfolded big clock. We" ]

[ "Main content # Using inductive reasoning (example 2) ## Video transcript We're asked how many toothpicks will be needed to form the 50th figure in this sequence? And they've drawn the first four figures in the sequence. This is the first, the second, the third. Now, let's see what's happening here. So in this first figure, we have this nice little house that they've made out of toothpicks, and it has one, two, three, four, five, six toothpicks. So the first figure in our sequence has six toothpicks. Now, what happens in the second figure here? Well, it looks just like our first figure. We have our one, two, three, four, five, six toothpicks, but then they add some toothpicks. It looks like they added one, two, three, four, five, toothpicks. So it's 6 plus 5 or 11 toothpicks. So let me write this down. It is the 6 toothpicks plus 5 more, which is equal to 11 toothpicks. And then when we go to the third figure, it looks just like the second figure, so it has the 11 toothpicks, so the 11 toothpicks are one, two, three, four, five, six, seven, eight, nine, ten, eleven, just like the second figure. And then they add some more. They add one, two, three, four, five. So it looks", "place $$(3-1)$$ toothpicks because the dice take on 3 different values separated by 1. So the number of outcomes is: $\\binom {4 + (3-1)}{3-1}$", "to order the pills and on day 7 he will have to repeat one of the orders. Suppose he is taking 4 pills a day. There are 4 choices for the pill he takes on the first day. Regardless of which pill he chooses he has 3 pills left and he can order them in 6 ways. Thus he has $4 \\times 6 = 24$ ways to order the 3 pills. Hence there are 24 ways to order the pills and on day 25 he will have to repeat one of the orders. What about 5 pills or 6 pills? Penny", "= 33649,$$ so the number of sets that are not missing any of 3, 6, or 14 is $$33649 - 19810 = \\boxed{13839}.$$", "add all three cases: $1+17+243=\\boxed{261}$", "are 100 - 42 = 58 green gumballs left. The second phase thus lasts for 58/2 = 29 rounds, and so costs 29 pennies. Finally, the machine enters its third phase, wherein it dispenses 1 pink gumball for each penny, and continues to do this until there are no pink gumballs left. After the first round, there were 98 pink gumballs left, and this remained unchanged during the second round, which dispensed no pink gumballs, so the third phase will last for 98 rounds, and so cost 98 pennies. So, Putting all the phases together, we have 42 + 29 + 98 pennies to release all the gumballs, which is 169 pennies. 2) Let $N_p$ be the initial number of pink gumballs, $N_g$ be the initial number of green gumballs, $P$ be the percentage setting, $D_{p1}$ be the number of pink dispensed per penny in the pink-and-green phase, $D_{g1}$ be the number of green dispensed per penny in the pink-and-green phase, $D_{p2}$ be the number of pink dispensed per penny in the pink-only phase, and $D_{g2}$ be the number of green dispensed per penny in the green-only phase. If $N_p$ is 0, presumably the machine will skip the pink-and-green phase, go straight into the green-only phase, continue till all green gumballs are dispensed, and then skip the pink-only", "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021", "is 45, the least common multiple. This means buying 5 boxes of forks ($5\\boldcdot 9=45$) and 3 boxes of spoons ($3 \\boldcdot 15=45$).", "\\ | * - - - - - * Now there are $\\boxed{3}$ choices for plot #4. Hence, there are: . $5 \\times 4 \\times 3 \\times 3 \\:=$ 180 ways. Therefore, there are: . $80 + 180 \\:=$ 260 ways.", "money amount. The answer is that is costs $25,000 per second for a thirty second commercial. IV. Model and Solve Real-World Problems using Simple Whole-Number Equations You can construct models to help you solve a problem. You have already learned to use a verbal model to write and solve equations to solve problems. Creating a model for a problem may also include methods such as drawing a diagram or picture or making a table or chart. Example The triangles below were constructed using toothpicks. Determine the number of toothpicks needed to construct twenty triangles. As you can see, three toothpicks were needed to construct one triangle. Two more were needed to construct the second triangle. Therefore, five toothpicks were used to make two triangles. Continue to make more triangles along the row. Each time you construct a new triangle, record the number of toothpicks used on a chart. Triangle #: Toothpick # 1 3 2 5 3 7 4 9 5 11 6 13 7 15 8 17 9 19 10 21 Looking at the table, you can identify a pattern. You can see that two toothpicks are needed each time a new triangle is constructed. You can write a verbal model to express this amount. Total Number of Toothpicks Needed = Two Times the Number of Triangles", "# Sampling with replacement order doesn't matter problem At first you need distribute pizze to 22 students. Assume stars $*$ represent the pizza and each student like a box $\\\\|$ represent an end side of a box. $$\\underbrace{*\\ *\\ *\\ \\ \\ \\ \\ \\ *}_{15\\ balls}\\ \\ \\ \\underbrace{[ \\ \\vert \\ \\vert \\ \\vert \\ \\vert \\ \\ \\ \\ \\vert \\ \\vert \\ ]}_{22 \\ boxes}^{21\\ bars}$$ Take two of the bars as special, to represent left and right ends. Then the original problem may be reformulated : How many different combinations of these $15+22-1$ objects there are? This is $${(15+22-1)!\\over 15!\\cdot (21)!} = \\binom{36}{21}=C_{36}^{21}$$ For Coca-cola it is ${(24+22-1)!\\over 24!\\cdot (21)!} = \\binom{45}{21}=C_{45}^{21}$. And the common variants is the product $C_{36}^{21}C_{45}^{21}$.", "| \\ / | * We have \"nine\". | / \\ | / \\ | / \\ Code: | \\ / | \\ / | \\ / - + - - -* - - - \"Half of nine\" | / \\ | / \\ | / \\ Code: | \\ / | \\ / \"Four\" | \\ / | * ZOMG!!!! 8. Originally Posted by Soroban Hello, khotso! Finally . . . a corrected version of the problem! Use toothpicks and Roman numerals. Did you formally study toothpick analysis in school? 9. Actually, I majored in Toothpick Analysis . . with a minor in Matchstick Theory.", "3 + 4 = \\boxed{21}$", "us how many spoons we could buy, as shown here. • Forks: 9, 18, 27, 36, 45, 54, 63, 72, 90. . . • Spoons: 15, 30, 45, 60, 75, 90. . . If we want as many forks as spoons, our options are 45, 90, 135, and so on, but the smallest number of utensils we could buy is 45, the least common multiple. This means buying 5 boxes of forks ($$5\\boldcdot 9=45$$) and 3 boxes of spoons ($$3 \\boldcdot 15=45$$).", "1$$ Afterward, we do multiplication, then subtraction. $$-420-1$$ $$-421$$ Therefore, $14(4\\times 7-58)-1 = \\boxed{-421}$.", "30$$ We are looking for the first ratio 3x = 30*3= 90 Hope now is more clear to you. Regards Yes its more clear now. Thank you for your explanation. Re: QOTD#27 Keri, Neill, and Rich use toilet paper in their apar [#permalink] 10 Feb 2019, 08:03 Display posts from previous: Sort by", "about money in this problem. So our answer needs to be written as a money amount. The answer is that is costs 25,000 per second for a thirty second commercial. IV. Model and Solve Real-World Problems using Simple Whole-Number Equations You can construct models to help you solve a problem. You have already learned to use a verbal model to write and solve equations to solve problems. Creating a model for a problem may also include methods such as drawing a diagram or picture or making a table or chart. Example The triangles below were constructed using toothpicks. Determine the number of toothpicks needed to construct twenty triangles. As you can see, three toothpicks were needed to construct one triangle. Two more were needed to construct the second triangle. Therefore, five toothpicks were used to make two triangles. Continue to make more triangles along the row. Each time you construct a new triangle, record the number of toothpicks used on a chart. Triangle #: Toothpick # 1 3 2 5 3 7 4 9 5 11 6 13 7 15 8 17 9 19 10 21 Looking at the table, you can identify a pattern. You can see that two toothpicks are needed each time a new triangle is constructed. You can write a verbal model to express", "(16 - 1)^2 + 1 = \\boxed{406}$.", "Etc.? Solution to the previous Riddler Express Congratulations to 👏 Jay Lee 👏 of Portland, Oregon, winner of the previous Riddler Express! At the grocery store you work at, the evening manager assigns 15 grocery aisles to the five night stockers who work that shift. Assume that the aisles are assigned randomly. How many different ways can you be assigned three aisles out of the 15? How many different ways are there to assign the 15 aisles to the five stockers, assuming they are each assigned three aisles? And finally, if the manager can assign any number of aisles to any stocker, how many possibilities are there? First, there are 455 ways you might be assigned three aisles out of the 15. There are 15 possibilities for your first aisle, 14 for your second, and 13 for your third, giving (15)(14)(13) = 2,730. But we don’t care about the specific order in which you get assigned these aisles, so we can divide that by the number of orders in which these aisles might have been assigned, which is (3)(2)(1) = 6, giving us 2,730/6 = 455. Another way to put all this: It’s just “15 choose 3,” or $${15 \\choose 3}$$. Second, there are 168,168,000 ways the 15 aisles might have been assigned to the five stockers, assuming", "$2$s. The second $6$ is $3!$ for the number of different orders $21,51,81$ can go in (they have to go in the three $0$ places, but you can put them in those places in any order). Then there are $2$ ways to put in $31$ and $61$ (either $31$ goes in the first $1$ place and $61$ goes in the second, or vice versa), and similarly there are $2$ ways to put in $41$ and $71$ in the two $2$ places. Since all of these are independent choices you multiply them together. – Especially Lime May 23 '17 at 21:44" ]

[ "# Question #b3e6b Sep 14, 2016 $103$ toothpicks in the ${50}^{\\text{th}}$ figure. $2 n + 3$ toothpicks in the ${n}^{\\text{th}}$ figure. In each progressive figure, one toothpick is added to the top row and one is added to the bottom row. As the ${1}^{\\text{st}}$ figure has $1$ toothpick in the top row and $2$ in the bottom row, that means that the ${n}^{\\text{th}}$ figure will have $n$ toothpicks in the top row and $n + 1$ in the bottom row. Adding these to the $2$ toothpicks which constitute the left and right sides, we get the total for the ${n}^{\\text{th}}$ figure as $n + \\left(n + 1\\right) + 2 = 2 n + 3$. To figure out how many are in the ${50}^{\\text{th}}$ figure, then, we just let $n = 50$ to get $2 \\left(50\\right) + 3 = 103$.", "the pattern with the equation $t=5 n-3,$ where $t$ is the number of toothpicks in Stage $n .$a. Explain how each part of the equation is related to the toothpick pattern.b. How many toothpicks would Evan need for Stage 10$?$ For Stage 100$?$c. Evan used 122 toothpicks to make one stage of his pattern. Write and solve an equation to find the stage number.d. Is any stage of the pattern composed of 137 toothpicks? Why or why not?e. Is any stage of the Evan made this toothpick pattern. He described the pattern with the equation $t=5 n-3,$ where $t$ is the number of toothpicks in Stage $n .$ a. Explain how each part of the equation is related to the toothpick pattern. b. How many toothpicks would Evan need for Stage 10$?$ For Stage 100$?$ c. Eva...", "85 ∗ 101 = ____________ b. 156 ∗ 9 = ____________ c. 48 ∗ 24 = ____________ Question 4. Rewrite each expression as a product by taking out a common factor a. 48 + 24 = _______ ∗ (_______ + _______) = _______ ∗ _______ b. 72 – 56 = _______ ∗ (_______ – _______) = _______ ∗ _______ c. (2y) + (3 ∗ y) = (_______ + _______) ∗ _______ = _______ ∗ _______ Practice Use <, >, or = to make the sentence true. Question 5. $$\\frac{2}{3}$$ _________ $$\\frac{2}{5}$$ Question 6. 0.7 _______ $$\\frac{4}{5}$$ Question 7. 0.3 ______ 0.23 Question 8. 1 $$\\frac{1}{4}$$ _______ 1.25 Building with Toothpicks Yaneli is building a pattern with toothpicks. The pattern grows in the following way: Question 1. How many toothpicks are needed for Design 5? _________ Question 2. How many toothpicks are needed for Design 10? __________ Question 3. Describe in words how you see the toothpick design growing. What stays the same from one figure to the next? What changes? Question 4. Write an expression to represent how many toothpicks are needed for Design n? Question 5. What toothpick design number could you build with exactly 82 toothpicks? ________ Question 6. Describe how you can figure out the number of toothpicks you need for any design number. Practice", "Problem #2045 2045 Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label(\"Figure\",(0.5,-1),S); label(\"0\",(0.5,-2.5),S); label(\"Figure\",(9.5,-1),S); label(\"1\",(9.5,-2.5),S); label(\"Figure\",(19.5,-1),S); label(\"2\",(19.5,-2.5),S); label(\"Figure\",(32.5,-1),S); label(\"3\",(32.5,-2.5),S); [/asy]$ $\\textbf{(A)}\\ 10401 \\qquad\\textbf{(B)}\\ 19801 \\qquad\\textbf{(C)}\\ 20201 \\qquad\\textbf{(D)}\\ 39801 \\qquad\\textbf{(E)}\\ 40801$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Question 1 of 10 Helen uses some toothpicks to form the pattern below. \\begin{array}{|c|c|} \\hline \\mbox{Pattern} & \\mbox{Number of toothpick} \\\\ \\hline 1 & 6\\\\ \\hline 2 & 15\\\\ \\hline 3 & 25\\\\ \\hline 4 & 35\\\\ \\hline 5 & \\\\ \\hline \\end{array} (a) How many toothpicks will she need to form Pattern 5? (b) How many toothpick will she need to form Pattern 40? (c) Helen uses 4955 toothpicks to form a Pattern. Which Pattern is it? A (a) 45 (b) 40 (c) 496 B (a) 47 (b) 43 (c) 498 C (a) 49 (b) 45 (c) 500 D (a) 50 (b) 49 (c) 502 E None of the above", "# Difference between revisions of \"2000 AMC 12 Problems/Problem 8\" The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page. ## Problem Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label(\"Figure\",(0.5,-1),S); label(\"0\",(0.5,-2.5),S); label(\"Figure\",(9.5,-1),S); label(\"1\",(9.5,-2.5),S); label(\"Figure\",(19.5,-1),S); label(\"2\",(19.5,-2.5),S); label(\"Figure\",(32.5,-1),S); label(\"3\",(32.5,-2.5),S); [/asy]$ $\\mathrm{(A)}\\ 10401 \\qquad\\mathrm{(B)}\\ 19801 \\qquad\\mathrm{(C)}\\ 20201 \\qquad\\mathrm{(D)}\\ 39801 \\qquad\\mathrm{(E)}\\ 40801$ ## Solution 1 We can divide up figure $n$ to get the sum of the sum of the first $n+1$ odd numbers and the sum of the first $n$ odd numbers. If you do not see this, here is the example for $n=3$: $[asy] draw((3,0)--(4,0)--(4,7)--(3,7)--cycle); draw((0,3)--(7,3)--(7,4)--(0,4)--cycle); draw((2,1)--(5,1)--(5,6)--(2,6)--cycle); draw((1,2)--(6,2)--(6,5)--(1,5)--cycle); draw((3,0)--(3,7)); [/asy]$ The sum of the first $n$ odd numbers is $n^2$, so for figure $n$, there are $(n+1)^2+n^2$ unit squares. We plug in $n=100$ to get $\\boxed{\\textbf{(C) }20201}$. ## Solution 2 Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$. Thus, the original sequence can", "the table below to determine the number of toothpicks needed for the pattern. Pattern number 1 2 3 4 5 ... ... 200 Number of toothpicks Completed, the table looks like: Pattern number 1 2 3 4 5 ... ... 200 Number of toothpicks 7 12 17 22 27 1002 The number of toothpicks in any pattern number is the result of multiplying the pattern number by 5 and adding 2 to the product. The number of toothpicks in pattern number 200 is: The data must be discrete. The graph would be dots representing the pattern number and the number of toothpicks. It is impossible to have a pattern number or a number of toothpicks that are not natural numbers. Therefore, the points would not be joined. The domain and range are: If the range is written in terms of a function, then the number system to which belongs must be designated in the range. ### Examples #### Example 1 Earlier, you were told that to avoid the road construction, Joseph decided to travel the gravel road. After driving for 20 minutes he was 62 miles away from work and after driving for 40 minutes he was 52 miles away from work. What would the graph look like for this problem? What would a suitable domain and range", "# Difference between revisions of \"2002 AMC 8 Problems/Problem 7\" ## Problem The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E? $[asy] real[] r={6, 8, 4, 2, 5}; int i; for(i=0; i<5; i=i+1) { filldraw((4i,0)--(4i+3,0)--(4i+3,2r[i])--(4i,2r[i])--cycle, black, black); } draw(origin--(19,0)--(19,16)--(0,16)--cycle, linewidth(0.9)); for(i=1; i<8; i=i+1) { draw((0,2i)--(19,2i)); } label(\"0\", (0,2*0), W); label(\"1\", (0,2*1), W); label(\"2\", (0,2*2), W); label(\"3\", (0,2*3), W); label(\"4\", (0,2*4), W); label(\"5\", (0,2*5), W); label(\"6\", (0,2*6), W); label(\"7\", (0,2*7), W); label(\"8\", (0,2*8), W); label(\"A\", (0*4+1.5, 0), S); label(\"B\", (1*4+1.5, 0), S); label(\"C\", (2*4+1.5, 0), S); label(\"D\", (3*4+1.5, 0), S); label(\"E\", (4*4+1.5, 0), S); label(\"SWEET TOOTH\", (9.5,18), N); label(\"Kinds of candy\", (9.5,-2), S); label(rotate(90)*\"Number of students\", (-2,8), W);[/asy]$ $\\text{(A)}\\ 5 \\qquad \\text{(B)}\\ 12 \\qquad \\text{(C)}\\ 15 \\qquad \\text{(D)}\\ 16 \\qquad \\text{(E)}\\ 20$ ## Solution From the bar graph, we can see that $5$ students chose candy E. There are $6+8+4+2+5=25$ total students in Mrs. Sawyers class. The percent that chose E is $\\frac{5}{25} \\cdot 100 = \\boxed{\\text{(E)}\\ 20}$.", "2 3 4 5 Number of Toothpicks 4 10 18 28 40 Increase in Toothpicks +4 +6 +8 +10 +12 Number of Squares 1 3 6 10 15 Increase in Squares +1 +2 +3 +4 +5 My problem is that I can't get from the patterns to creating a formula that would work for any figure number without listing them all out consecutively #### MarkFL Staff member I see you have found that the difference in the number of toothpicks per iteration (where you state the increase in toothpicks) are the sequence of even numbers, beginning with 4. What is the second difference, that is, the difference of the differences. How much does the increase increase each time? #### Opalg ##### MHB Oldtimer Staff member Yeah I'm up to my neck in patterns; I made a table and everything. Figure Number 1 2 3 4 5 Number of Toothpicks 4 10 18 28 40 Increase in Toothpicks +4 +6 +8 +10 +12 Number of Squares 1 3 6 10 15 Increase in Squares +1 +2 +3 +4 +5 My problem is that I can't get from the patterns to creating a formula that would work for any figure number without listing them all out consecutively Let $x_n$ be the number of toothpicks in Fig. $n$. Your table shows", "Difference between revisions of \"2000 AMC 12 Problems/Problem 8\" The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page. Problem Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label(\"Figure\",(0.5,-1),S); label(\"0\",(0.5,-2.5),S); label(\"Figure\",(9.5,-1),S); label(\"1\",(9.5,-2.5),S); label(\"Figure\",(19.5,-1),S); label(\"2\",(19.5,-2.5),S); label(\"Figure\",(32.5,-1),S); label(\"3\",(32.5,-2.5),S); [/asy]$ $\\mathrm{(A)}\\ 10401 \\qquad\\mathrm{(B)}\\ 19801 \\qquad\\mathrm{(C)}\\ 20201 \\qquad\\mathrm{(D)}\\ 39801 \\qquad\\mathrm{(E)}\\ 40801$ Solution 1 We can divide up figure $n$ to get the sum of the sum of the first $n+1$ odd numbers and the sum of the first $n$ odd numbers. If you do not see this, here is the example for $n=3$: $[asy] draw((3,0)--(4,0)--(4,7)--(3,7)--cycle); draw((0,3)--(7,3)--(7,4)--(0,4)--cycle); draw((2,1)--(5,1)--(5,6)--(2,6)--cycle); draw((1,2)--(6,2)--(6,5)--(1,5)--cycle); draw((3,0)--(3,7)); [/asy]$ The sum of the first $n$ odd numbers is $n^2$, so for figure $n$, there are $(n+1)^2+n^2$ unit squares. We plug in $n=100$ to get $\\boxed{\\textbf{(C) }20201}$. Solution 2 Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$. Thus, the original sequence can be generated from a", "\\varepsilon \\ R \\} \\quad R=\\{y|y \\ge -3, y \\ \\varepsilon \\ R \\}$ #### Example B Whether a relation is discrete or continuous can often be determined without a graph. As well, the domain and range can also be determined. Let’s examine the following toothpick pattern. Complete the table below to determine the number of toothpicks needed for the pattern. Pattern number $(n)$ 1 2 3 4 5 ... $n$ ... 200 Number of toothpicks $(t)$ Is the data continuous or discrete? Why? What is the domain? What is the range? Solution: Pattern number $(n)$ 1 2 3 4 5 ... $n$ ... 200 Number of toothpicks $(t)$ 7 12 17 22 27 $5n+2$ 1002 The number of toothpicks in any pattern number is the result of multiplying the pattern number by 5 and adding 2 to the product. The number of toothpicks in pattern number 200 is: $t&=5n+2\\\\t&=5({\\color{red}200})+2\\\\t&=1000+2\\\\t&=1002$ The data must be discrete. The graph would be dots representing the pattern number and the number of toothpicks. It is impossible to have a pattern number or a number of toothpicks that are not natural numbers. Therefore, the points would not be joined. The domain and range are: $D=\\{x|x \\ \\varepsilon \\ N\\} \\quad R=\\{y|y=5x+2, x \\ \\varepsilon \\ N\\} \\ \\ OR \\ \\ R=\\{y|y \\ge", "\\ \\in \\ R \\} \\quad R=\\{y|y \\ge -3, y \\ \\in \\ R \\}$ #### Example B Whether a relation is discrete, continuous, or neither can often be determined without a graph. The domain and range can be determined without a graph as well. Examine the following toothpick pattern. Complete the table below to determine the number of toothpicks needed for the pattern. Pattern number $(n)$ 1 2 3 4 5 ... $n$ ... 200 Number of toothpicks $(t)$ Is the data continuous or discrete? Why? What is the domain? What is the range? Solution: Pattern number $(n)$ 1 2 3 4 5 ... $n$ ... 200 Number of toothpicks $(t)$ 7 12 17 22 27 $5n+2$ 1002 The number of toothpicks in any pattern number is the result of multiplying the pattern number by 5 and adding 2 to the product. The number of toothpicks in pattern number 200 is: $t&=5n+2\\\\t&=5({\\color{red}200})+2\\\\t&=1000+2\\\\t&=1002$ The data must be discrete. The graph would be dots representing the pattern number and the number of toothpicks. It is impossible to have a pattern number or a number of toothpicks that are not natural numbers. Therefore, the points would not be joined. The domain and range are: $D=\\{x|x \\ \\in \\ N\\} \\quad R=\\{y|y=5x+2, x \\ \\in \\ N\\}$ If the range is", "So, there is a repeating pattern for every $30$ digits on the original list. As shown below, the digits erased in the first, second, and third rounds are colored in red, yellow, and green, respectively: $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,5)--(i+0.5,5)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,5)--(i+0.5,5)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,5)--(8,5)--(8,0)--cycle,green); fill((18,0)--(18,5)--(19,5)--(19,0)--cycle,green); fill((27,0)--(27,5)--(28,5)--(28,0)--cycle,green); for (real i=1; i<5; ++i) { for (real j=0; j<30; ++j) { label(\"\"+string(1+j%5)+\"\",(j+0.5,i+0.5)); } } for (real i=0; i<30; ++i) { label(\"\\vdots\",(i+0.5,2/3)); } add(grid(30,5,linewidth(1.25))); [/asy]$ As indicated by the white squares, each group of $30$ digits on the original list has $\\frac25\\cdot30=12$ digits remaining on the final list. Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: $[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,1)--(8,1)--(8,0)--cycle,green); fill((18,0)--(18,1)--(19,1)--(19,0)--cycle,green); fill((27,0)--(27,1)--(28,1)--(28,0)--cycle,green); for (real j=0; j<30; ++j) { label(\"\"+string(1+j%5)+\"\",(j+0.5,0.5)); } draw((3.5,-1.25)--(3.5,-0.2),linewidth(1.25),EndArrow); draw((6.5,-1.25)--(6.5,-0.2),linewidth(1.25),EndArrow); draw((9.5,-1.25)--(9.5,-0.2),linewidth(1.25),EndArrow); add(grid(30,1,linewidth(1.25))); [/asy]$ Therefore, the answer is $4+2+5=\\boxed{\\textbf{(D)}\\ 11}.$ ~MRENTHUSIASM ## Video Solution https://youtu.be/t6yjfKXpwDs 16:40 ~IceMatrix", "# Toothpick Squares Sequence #### marsman ##### New member Please help! I need to be able to find the number of total toothpicks used in any given figure in this sequence, in addition to total squares in any figure, using the figure number. The only way I can figure this out is by listing them all out one by one, which is TERRIBLE Thank you so much in advance!!!!! #### Opalg ##### MHB Oldtimer Staff member Please help! I need to be able to find the number of total toothpicks used in any given figure in this sequence, in addition to total squares in any figure, using the figure number.View attachment 1592 The only way I can figure this out is by listing them all out one by one, which is TERRIBLE Thank you so much in advance!!!!! Suggestion: In the six figures, count how many toothpicks you have to add to each figure in order to get the next one. Can you see a pattern there? #### marsman ##### New member Suggestion: In the six figures, count how many toothpicks you have to add to each figure in order to get the next one. Can you see a pattern there? Yeah I'm up to my neck in patterns; I made a table and everything. Figure Number 1", "mm}\\diamondsuit \",(12,4.5),S); [/asy]$ $\\text{(A)}\\ 1 \\qquad \\text{(B)}\\ 2 \\qquad \\text{(C)}\\ 3 \\qquad \\text{(D)}\\ 4 \\qquad \\text{(E)}\\ 5$ ## Problem 25 How many different patterns can be made by shading exactly two of the nine squares? Patterns that can be matched by flips and/or turns are not considered different. For example, the patterns shown below are not considered different. $[asy] fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,gray); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle,gray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,linewidth(1)); draw((2,0)--(2,3),linewidth(1)); draw((0,1)--(3,1),linewidth(1)); draw((1,0)--(1,3),linewidth(1)); draw((0,2)--(3,2),linewidth(1)); fill((6,0)--(8,0)--(8,1)--(6,1)--cycle,gray); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle,linewidth(1)); draw((8,0)--(8,3),linewidth(1)); draw((6,1)--(9,1),linewidth(1)); draw((7,0)--(7,3),linewidth(1)); draw((6,2)--(9,2),linewidth(1)); fill((14,1)--(15,1)--(15,3)--(14,3)--cycle,gray); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle,linewidth(1)); draw((14,0)--(14,3),linewidth(1)); draw((12,1)--(15,1),linewidth(1)); draw((13,0)--(13,3),linewidth(1)); draw((12,2)--(15,2),linewidth(1)); fill((18,1)--(19,1)--(19,3)--(18,3)--cycle,gray); draw((18,0)--(21,0)--(21,3)--(18,3)--cycle,linewidth(1)); draw((20,0)--(20,3),linewidth(1)); draw((18,1)--(21,1),linewidth(1)); draw((19,0)--(19,3),linewidth(1)); draw((18,2)--(21,2),linewidth(1)); [/asy]$ $\\text{(A)}\\ 3 \\qquad \\text{(B)}\\ 6 \\qquad \\text{(C)}\\ 8 \\qquad \\text{(D)}\\ 12 \\qquad \\text{(E)}\\ 18$", "all possible maximum values of $S.$ $[asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label(\"\"+string(i+0.5)+\"\",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label(\"\"+string(i+42.5)+\"\",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label(\"\"+string(i+84.5)+\"\",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy]$ From the second row of the table above, we perform casework on the possible maximum value of $S:$ $$\\begin{array}{c||c|c|l} & & & \\\\ [-2.5ex] \\textbf{Max Value} & \\boldsymbol{S} & \\textbf{Valid?} & \\hspace{16.25mm}\\textbf{Reasoning/Conclusion} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 49 & \\{46,47,48,49\\} & & \\text{The student who gets } 46 \\text{ will reply yes.} \\\\ 50 & \\{47,48,49,50\\} & \\checkmark & \\text{Another possibility is } S=\\{89,90,91,92\\}. \\\\ 51 & \\{48,49,50,51\\} & & \\text{The student who gets } 51 \\text{ will reply yes.} \\\\ 56 & \\{53,54,55,56\\} & & \\text{The student who gets } 53 \\text{ will reply yes.} \\\\ 57 & \\{54,55,56,57\\} & & \\text{The student who gets } 57", "By observations, we circle all possible maximum values of $S.$ $[asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label(\"\"+string(i+0.5)+\"\",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label(\"\"+string(i+42.5)+\"\",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label(\"\"+string(i+84.5)+\"\",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy]$ From the second row of the table above, we perform casework on the possible maximum value of $S:$ $$\\begin{array}{c||c|c|l} & & & \\\\ [-2.5ex] \\textbf{Max Value} & \\boldsymbol{S} & \\textbf{Valid?} & \\hspace{16.25mm}\\textbf{Reasoning/Conclusion} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 49 & \\{46,47,48,49\\} & & \\text{The student who gets } 46 \\text{ will reply yes.} \\\\ 50 & \\{47,48,49,50\\} & \\checkmark & \\text{Another possibility is } S=\\{89,90,91,92\\}. \\\\ 51 & \\{48,49,50,51\\} & & \\text{The student who gets } 51 \\text{ will reply yes.} \\\\ 56 & \\{53,54,55,56\\} & & \\text{The student who gets } 53 \\text{ will reply yes.} \\\\ 57 & \\{54,55,56,57\\} & & \\text{The student", "We have a \"stack\" of rows of squares. The rows have consecutive odd numbers of squarea. The squares are made of toothpicks. How many toothpicks are used in the $$n^{th}$$ diagram? Code: n=4 n=3 * - * | | n=2 * - * * - * - * - * | | | | | | n=1 * - * * - * - * - * * - * - * - * - * - * | | | | | | | | | | | | * - * * - * - * - * * - * - * - * - * - * * - * - * - * - * - * - * - * | | | | | | | | | | | | | | | | | | | | * - * * - * - * - * * - * - * - * - * - * * - * - * - * - * - * - * - * 4 13 26 43 We have the sequence: .$$4,13,26,64,\\;\\text{ . . .}$$ Take the differences of consecutive terms, then take the differences of the differences, and so on. $$\\begin{array}{|c|ccccccccc|}\\hline \\text{Sequence} & 4 && 13 && 26", "answer is then $\\boxed{\\dfrac{145}{147}}$. ## Solution 5 $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); draw((0.3,0.3)--(3.6,0.3), dashed); draw((0.3,2.7)--(0.3,0.3), dashed); label(\"S\", (-0.05,-0.05), NE); draw((0.3,0.3)--(1.41,1.91)); draw((1.63,1.78)--(1.48,1.56)); draw((1.28,1.70)--(1.48,1.56)); label(\"4\", (2,0), S); label(\"3\", (0,1.5), W); label(\"\\frac{10}{3}\", (2,0.3), N); label(\"\\frac{5}{2}\", (0.3,1.5), E); label(\"2\", (1,1.2), E); draw((3.6,0)--(3.6,0.3), dashed); draw((0,2.7)--(0.3,2.7), dashed); label(\"\\small{l}\", (3.6,0.15), W); label(\"\\small{l}\", (0.15,2.7), S); label(\"\\small{l}\", (0.3,0.15), E); label(\"\\small{l}\", (0.15,0.3), N); [/asy]$ On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\\frac{5}{2}$ and $\\frac{10}{3}$. With $l$ being the side length of the square, we need to find an expression for $l$. Using the hypotenuse, we can see that $\\frac{3}{2}+\\frac{8}{3}+\\frac{5}{4}l+\\frac{5}{3}l=5$. Simplifying, $\\frac{35}{12}l=\\frac{5}{6}$, or $l=2/7$. A different calculation would yield $l+\\frac{3}{4}l+\\frac{5}{2}=3$, so $\\frac{7}{4}l=\\frac{1}{2}$. In other words, $l=\\frac{2}{7}$, while to check, $l+\\frac{4}{3}l+\\frac{10}{3}=4$. As such, $\\frac{7}{3}l=\\frac{2}{3}$, and $l=\\frac{2}{7}$. Finally, we get $A(\\Square S)=l^2=\\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\\frac{4}{49\\cdot6}=1-\\frac{2}{147}=\\frac{145}{147}$, so $\\boxed{\\textit D}$ is correct. ## Solution 6: Also Coordinate Geo Let the right angle be at $(0,0)$, the point $(x,x)$ be the far edge of the unplanted square and the hypotenuse be the line $y=-\\frac{3}{4}x+3$. Since the line from $(x,x)$ to the hypotenuse is the shortest possible distance, we know this line, call it line $\\l$, is perpendicular to the hypotenuse and therefore has", "fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); draw(arc((0,3),1,-15,-165)); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3)); draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3)); label(\"6\",.74*dir(-90)+(0,3)); label(\"5\",.74*dir(-60)+(0,3)); label(\"4\",.74*dir(-30)+(0,3)); [/asy]$ The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\\circ$ clockwise (i.e. 2 hours). $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150),EndArcArrow);label(\"30^\\circ\",2*dir(150),W); draw(1.8*dir(120)--2*dir(120)); draw(1.8*dir(90)--2*dir(90)); label(rotate(30)*\"12\",1.56*dir(120)); label(rotate(30)*\"1\",1.56*dir(90)); draw(arc((0,3),1,-15,-165),EndArcArrow);label(\"60^\\circ\",dir(-165)+(0,3),W); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3)); draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3)); label(rotate(-60)*\"6\",.74*dir(-150)+(0,3)); label(rotate(-60)*\"5\",.74*dir(-120)+(0,3)); label(rotate(-60)*\"4\",.74*dir(-90)+(0,3)); [/asy]$ However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\\circ$ clockwise to see what happens when the small cog rolls around it. $[asy] fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label(\"12\",1.56*dir(90)); label(\"1\",1.56*dir(60)); pair c=(1.5,sqrt(27)/2); draw(arc(c,1,0,-200),EndArcArrow);label(\"90^\\circ\",dir(-180)+c,W); draw(0.9*dir(-120)+c--dir(-120)+c); draw(0.9*dir(-150)+c--dir(-150)+c); draw(0.9*dir(-180)+c--dir(-180)+c); label(rotate(-90)*\"6\",.74*dir(-180)+c); label(rotate(-90)*\"5\",.74*dir(-150)+c); label(rotate(-90)*\"4\",.74*dir(-120)+c); [/asy]$ It turns out that, when the point of tangency moves $30^\\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\\boxed{\\textbf{(C) }\\text{4 o' clock}}$ ## Solution 5 We unfold the big clock and draw the small clock. We know that the small clocks arrow will be facing up again at 6 o clock on the unfolded big clock. We" ]

[ "attended the wedding, and averaged three drinks per person.", "gathering at least furnished a stepping-stone to better things. “the parties range from wine and snacks, to just full on parties with headliner-type bands,” said bryan parker, ceo of doubledutch.", "to the face during the process 1 guests", "in the wild, let alone at a party :-)).", "arrangement matters). And so while some attendees (the “core group”) might end up having fun, the party doesn’t really work for most participating parties. So, the next time you’re hosting a party, do yourself and your guests a favour and ensure that you don’t end up with between 6 and 10 people at the party. Either less or more is fine! You might want to read this other post I’ve written on coordinating guest lists for birthday parties. ## Meeting types There are essentially three kinds of meetings – those that are entirely “in person”, those that are entirely “on call” and hybrids. I argue that the quality of conversation in the third kind of meeting is significantly inferior to that of the first two types. In person meetings are those where all participants are in one room. These are perhaps the best kind of meetings (except when you know it’s likely to turn antagonistic), for you can maintain eye contact with the others and as long as the number of meeters is small, there is social pressure on meeters to not be distracted, and thus the meeting is likely to conclude its agenda productively and quickly. Conference calls allow you to multitask while you are on the meeting – the positive thing is that you can choose", "## Stats At a certain event, 16 people attend, and 5 will be chosen at random to receive door prizes. The prizes are all the same, so the order in which the people are chosen does not matter. How many different groups of 5 people can be chosen?", "259 Part Eight: How to Work a Party Like a Politician Works a Room: The Politician’s Six- Point Party Checklist . . . . . . . . . . . . . . 265 71 How to Avoid the Most Common Party Blooper . . . . 270 72 How to Make an Unforgettable Entrance . . . . . . . . . 272 73 How to Meet the People YOU Want to Meet . . . . . . 274 74 How to Subliminally Lure People to You at a Gathering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 75 How to Make ’Em Feel Like a Movie Star . . . . . . . . . 281 76 How to Amaze Them with What You Remember About Them . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 77 How to Make the Sale with Your Eyeballs . . . . . . . . . 288 Part Nine: How to Break the Most Treacherous Glass Ceiling of All: Sometimes People", "# When's the Listening Party? A listening party is an event where a bunch of people get together to listen to some music (like a viewing party, but for music). In the age of the internet you can do Listening Parties online with friends, where you just all hit play together. The problem is on the internet everyone is in different time-zones so you can't just say \"Let's start at 5:35\". As a solution we just type xs in the hours position. So xx:35 means whenever the minute hand is next at 35. If the minute hand is already at that value then the party happens in 1 hour. If you need to specify more than one hour ahead you can add 1+, 2+ etc. to the beginning. So 1+xx:35 is an hour after the next time the minute hand is at 35. Your task is to take the current local time, and a string representing when the listening party happens in the above format and output when the listening party will occur in local time. # IO You can take and output the local time either as a 24 hour string with the minutes and hours separated by a colon (e.g. 19:45), as a tuple of the minutes and hours or as some native time type when", "months each with 60 more attendees.", "trip. if 39 people went on that trip, how many are in the club...", "with x friends, how many cupcakes does each friend get?...", "5 people are to be seated around a 6 13 Jun 2011, 23:04 8 At a dinner party, 5 people are to be seated around a 11 22 Jul 2011, 14:10 10 At a dinner party, 5 people are to be seated around a circul 7 30 Dec 2013, 08:48 78 At a dinner party, 5 people are to be seated around a 12 08 Jul 2013, 07:41 Display posts from previous: Sort by", "some of them Will be a good 108 So party, I pressed 36 president too far. So they go on you. Could you want it? That's an answer", "place, people talk to strangers, adults and children build things together, and men and women interact as equals. The Congress is really unique. I can't recommend it enough. If you can free up those four days between Christmas and New Year's Eve at all, go there. You won't regret it!", "interesting - and very telling! 6 of 29 hold licenses (20%). 80% don't. I think that explains a lot of what I've read on this forum! 2", "party, he would have had a drink and it might be that if he had come to the party, he would have liked it. On this re-analysis, it’s dominant construal is the party, unlike (65).", "Last post Similar Topics: 6 At a dinner party, 5 people are to be seated around a 11 17 Mar 2010, 06:01 9 At a dinner party, 5 people are to be seated around a circul 7 30 Oct 2009, 11:26 1 At a dinner party , 5 people are to be seated around a 9 08 Jan 2008, 05:53 At a dinner party, 5 people are to be seated around a 4 05 Dec 2006, 16:53 At a dinner party, 5 people are to be seated around a 1 22 May 2006, 03:22 Display posts from previous: Sort by", "as many females as males [#permalink] 18 Oct 2016, 09:59 Similar topics Replies Last post Similar Topics: At a party, there were five times as many females as males. There were 1 10 Jan 2016, 08:45 In a certain town, there are four times as many people who were born 3 11 Dec 2015, 09:28 7 A certain college party is attended by both male and female students. 7 08 May 2015, 06:04 13 At a picnic there were 3 times as many adults as children 6 26 Nov 2012, 22:56 5 How many ways are there for 3 males and 3 females to sit 5 18 Dec 2010, 05:18 Display posts from previous: Sort by", "# The convention Logic Level 2 A certain convention was exactly attended by 100 politicians. We are given the following 2 facts. 1) Jon Huntsman is present. He is known to be honest. 2) If you choose any 2 politicians randomly at the convention, at least one of them will be crooked (not honest). Of the 100 politicians present, how many are crooked? ×", "of guests, visitors, or strangers. 5" ]

[ "$$\\binom {36}{13}$$ \"good\" hands", "To prove this particular case of the hockey stick identity, I just notice that the lefthand side of $(1)$ is yet a third way to count the $3$-element subsets of the set $S$ above: it’s what you get when you count them according to their largest element instead of their middle element. The largest element can be anything from $2$ through $n$. If $k$ is the largest element, the other two elements must be chosen from the $k$ members of $S$ less than $k$, and there are $\\binom{k}2$ ways to do that. Summing over $k$ then yields the identity $(1)$. If there are $N$ people at a party they exchange $\\binom{N}{2}$ handshakes among them. You can understand that as follows: everyone exchanges exactly one handshake with everybody else. So you can think of each handshake as a $2$-element subset of the set of $n$ people. Thus, counting the number of handshakes is equivalent to counting the number of $2$-element subsets. So, for your first party, since people leave one-after-the-other, the total number of handshakes is: $$\\sum_{k=2}^{N}\\binom{k}{2}=\\binom{2}{2}+\\binom{3}{2}+...+\\binom{N-1}{2}+\\binom{N}{2}$$ Now, use your result for the second party, solve for $N$ $$\\sum_{i=1}^{N-1}i(N-i)= \\\\ =1\\times 9+2\\times 8+3\\times 7+4\\times 6+5\\times 5 +6\\times 4+7\\times 3+8\\times2+9\\times1=165 \\Rightarrow \\\\ \\Rightarrow N=10$$ and plug it in the above formula to calculate $\\sum_{k=2}^{N}\\binom{k}{2}=\\sum_{k=2}^{10}\\binom{k}{2}$, to get the result for the", "of 12 people? We have: How many handshakes will take place between six people in a room when they each shakes hands with all the other people in the room one time? Here, Notice that no consideration is given to the order or arrangement of the items but simply the combinations. Another way of viewing combinations is as follows. Consider the number of combinations of 5 letters taken 3 at a time. This produces: Now assume you permute (arrange) the r = 3 letters in each of the 10 combinations in all possible ways. Each group would produce r! permutations. Letting x = 5C3 for the moment, we would therefore have a total of x(r!) different permutations. This total, however, represents all the possible permutations (arrangements) of n things taken r at a time, which is shown under arrangement numbers and defined as nPr. Therefore, Using the committee of 3 out of 12 people example from above, Consider the following: How many different ways can you enter a 4 door car? It is clear that there are 4 different ways of entering the car. Another way of expressing this is: If we ignore the presence of the front seats for the purpose of this example, how many different ways can you exit the car assuming that you do", "(5, 3), while he goes through cells (1, 4), (2, 5), (3, 4), (4, 5), and (5, 4). The two of them will then get to hold hands while on rows 1, 3, 4, and 5, for a total of 4 seconds. However, there's no way for both Alice and Bob to navigate the second sidewalk. They must start in cells (1, 1) and (1, 4), and no walkable cells on the second row can be reached from cell (1, 4).", "if this is not obvious). Now use the cross-products to obtain: 4m - 4 = 3m + 12 Solve for m and you'll see that there are 16 men. Substitute back into the second equation and you'll find the number of women. Using my versions of the equations (to avoid fractions), we might have chosen instead to eliminate m first, replacing m in the first equation with $$m = 4(w-1)$$ to obtain $$4(w-1)-1=3w$$. This simplifies to $$4w-5=3w$$, so that $$w=5$$, and $$m = 4(w-1) = 4(5-1) = 16$$. Therefore there are 16 men and 5 women (including the Li’s), for a total of 21 people at the party. Here I only have to show Mr. and Mrs. Li’s handshakes, with men in blue and women in red: Mr. Li shook hands with 15 men and 5 women (a 3:1 ratio), while Mrs. Li shook hands with 16 men and 4 women (a 4:1 ratio). It worked. You might also want to use this information to determine the TOTAL number of handshakes. You can find it by examining Pascal's Triangle. I had fun with this one! Thanks! The mention of Pascal’s Triangle refers to the fact that it contains binomial coefficients, which count combinations; in particular, we need $${{21}\\choose{2}} = \\frac{21\\cdot 20}{2\\cdot 1} = 210$$ handshakes in all. If", "answer is $6C2=\\frac{6!}{4!2!}=\\text{15}$ There are ten people at a party. If they all shake hands, how many hand-shakes are possible? Note that between any two people there is only one hand shake. Therefore, we have The shopping area of a town is in the shape of square that is 5 blocks by 5 blocks. How many different routes can a taxi driver take to go from one corner of the shopping area to the opposite cater-corner? Let us suppose the taxi driver drives from the point $A$ , the lower left hand corner, to the point $B$ , the upper right hand corner as shown in the figure below. To reach his destination, he has to travel ten blocks; five horizontal, and five vertical. So if out of the ten blocks he chooses any five horizontal, the other five will have to be the vertical blocks, and vice versa. Therefore, all he has to do is to choose 5 out of ten. The answer is 10C5, or 252. Alternately, the problem can be solved by permutations with similar elements. The taxi driver's route consists of five horizontal and five vertical blocks. If we call a horizontal block $H$ , and a vertical block a $V$ , then one possible route may be as follows. $\\text{HHHHHVVVVV}$ Clearly there are", "the index finger • Eleven → middle link of the index finger • Twelve → lower link of the index finger ### Two-handed counting up to 60 After the first dozen has been counted using the thumb as a pointer with the three phalanges of the remaining four fingers of the same hand (4 × 3 = 12), the counting capacity of one hand is initially exhausted. • The other hand is clenched in a fist. In order to remember that a dozen has been counted, one now extends a finger, e.g. B. the thumb. • Now you continue to count by starting again at one with your first hand . At twelve , the second dozen is full. • In order to remember that two dozen have been counted, one now extends the next finger of the other hand, e.g. B. after the thumb out the index finger. • With the five fingers of the first hand you can count five times a dozen, so 5 × 12 = 60. • Now you can count the next dozen again with the first hand, i.e. count up to 72 with two hands (12 on the first plus 60 on the other hand). This finger counting system still exists in parts of Turkey , Iraq , India and Indochina", "(iii) If the group contains the oldest person and excludes the youngest person, then the oldest person can go with 7 of the remaining 12. The number of ways this can happen is $\\binom{12}{7}$ hence $P=\\frac{\\binom{12}{7}}{\\binom{14}{8}}$", "proposition that says in a group of 25 people, it is possible for each to shake hands with exactly with 3 other people. I am pretty sure that it is true and that I have to use a directed graph to prove it but I don't know where to go after that. Can anyone give me a hint? Thanks in advance. 9. Measure of angle formed by minute and hour hands Hi, What is the measure of the angle formed by the minute and hour hands of a clock at 1:50? (A) 90^o (B) 95^o (C) 105^o (D) 115^o (E) 120^o Answer: D My guess was 90^o after mentally flipping it. My approach was worthless and remains worthless for questions of this type. I seek a better... 10. Number of card hands. Not entirely sure how to do this. Would appreciate some help. In a game of poker, five players are each dealt 5 cards from a 52-card deck. How many ways are there to deal the cards? 11. Blackjack hands How many hands are there in which the face-down card is an ace and the face-up card is not a heart? There are 4 ways to choose an ace times 39 ways to choose not a heart, then subtract out the combinations of two ace of spades,", "ways in which the people in the group are seated along a circle so that men and women sit in alternate positions and the number of ways in which the people in the group are seated in a row so that men and women sit in alternate positions is 1 : 10. (2) The number of handshakes possible separately among the men and separately among the women is equal and each is equal to 5/12 of the number of handshakes possible if the men and women shake hands only with each other such that every man shakes hands with every woman. [Reveal] Spoiler: OA _________________ Kudos [?]: 124569 [1], given: 12079 Intern Joined: 23 Apr 2017 Posts: 3 Kudos [?]: 2 [2], given: 155 Re: A group consists of a few men and a few women. What is the total numbe [#permalink] ### Show Tags 19 Jul 2017, 06:04 2 KUDOS Bunuel wrote: A group consists of a few men and a few women. What is the total number of men and women in the group? (1) The ratio of the number of ways in which the people in the group are seated along a circle so that men and women sit in alternate positions and the number of ways in which the people in the group are", "take to make 10 complete rotations? #### 11th Standard Maths Combinations and Mathematical Induction English Medium Free Online Test One Mark Questions 2020 - 2021 - by Question Bank Software - View & Read • 1) The sum of the digits at the 10th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is • 2) The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in physics, first in chemistry and first in English is • 3) The number of ways in which a host lady invite 8 people for a party of 8 out of 12 people of whom two do not want to attend the party together is • 4) Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is • 5) The product of r consecutive positive integers is divisible by _________ #### 11th Standard Maths Important Question - by Question Bank Software - View & Read • 1) By taking suitable sets A, B, C, verify the following results: C-(B-A) = (C$\\cap$ A) $\\cup$ (C$\\cap$B') • 2) Discuss the following relations for reflexivity, symmetricity", "not shake hands with himself, nor Mrs. Li with herself. Therefore the number of men should be equal to 3 times the number of women PLUS 1. The number of women therefore must be ONE MORE THAN 1/4 the number of men. Let's set w = no. of Women and m = no. of Men. Then we can set up two equations: m = 3w+1 and w = (m/4)+1 Here we’re assuming that the Li’s did shake hands with each other, taking what it says literally. (Otherwise, they would have shaken hands with exactly the same people, and what is described could not be true.) The first equation could have been written as $$m-1 = 3w$$, meaning that the number of men, minus Mr. Li, is three times the number of women; and the second equation could be $$m = 4(w-1)$$, since the number of men is four times the number of women, excluding Mrs. Li. From the first equation, we can solve for w to get: w = (m - 1)/3 and substitute this into our second equation to get: (m - 1)/3 = (m/4) + 1 Let's simplify the right side by changing the 1 to 4/4 and adding. Now we have: (m - 1)/3 = (m + 4)/4 (you'll want to work out the steps", "(Paul picks wrongly) = 23×23+13×13=59 10. 36 people {a1, a2… a36} meet and shake hands in a circular fashion. In other words, there are totally 36 handshakes involving the pairs, {a1, a2}, {a2, a3}, …, {a35, a36}, {a36, a1}. Then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is a) 12 b) 11 c) 13 d) 18 Ans: {a1, a2}, {a2, a3},{a3, a4}, {a4, a5},{a5, a6}, {a6, a7} …, {a35, a36}, {a36, a1} From the above arrangement, If we separate a3, a6, a9, …..a36. Total 12 persons the reamining persons must have shaked hand with atleast one person. So answer is 12. 11. There are two boxes, one containing 10 red balls and the other containing 10 green balls. You are allowed to move the balls between the boxes so that when you choose a box at random and a ball at random from the chosen box, the probability of getting a red ball is maximized. This maximum probability is If rearrangement is not allowed, then actual probability of picking up a red ball = 12(10)+12(0)=12 As we are allowed to move the ball, we keep only 1 red in the first box, and shirt the remaining 9 to the second. So = 12(1)+919(0)=1419", "# Greeting cards On Diwali, all the students of a class send greeting cards to one another . If the postmen deliver 1640 greeting cards to the students of this class,then the number of students can be $$\\text{__________}$$. ×", "of five to nine persons. There are so many good uses for paint swatches. Source: The Lettered Classroom Sole Mate – One of our all-time favorites we learned from the fine folks at Project Adventure way back in 2007. Gather your team in a video conference room, then divide your group into smaller teams. 5) Odd Birthdays. In my last school, every classroom had a theme. Here is a sample plan from a real-life scenario. In this decision game, the aim is to decide who will be in group A and group B. Optional: If you have a larger group, you can make this competitive by dividing the group into 2 (or more) pairs of teams. 3) Spades. 11.", "another group of $6$ people. This is nothing but choosing $6$ people from a group of $18$ people. This is nothing but $\\binom{18}{6} = 18564 \\Rightarrow 564$. ~coolmath_2018", "ready to go lingers, can that be! 9240 Ten ways to choose group 2 's that you cross your arms divide n identical objects among distinct. Choices for each of the $n = 6$ is special … this works... Answer site for people studying math at any level and professionals in related fields sizes matter not...", "take you. Say you go to a party and the host asks you to play a game by completing certain tasks. One task is to shake hands with two different people, and only two different people. If everyone who comes to the party is playing will everyone be able to shake the hands of exactly two other people? Or will someone be left out? The answer to this question is given by the Hand-Shaking Lemma, which is taught in MTH 438/538 (Theory and Application of Graphs). Have you ever seen the game Chomp before? if you are tired of losing to the computer MTH 437/537 (Game Theory and Related Topics) might help you understand the computer’s strategy. Did you play Dots and Boxes with your family and friends when you were younger? MTH 437/537 (Game Theory and Related Topics) might help you find a winning strategy. (Forget what it is, search online or click here:) Like games? Here’s one you might learn how to win in MTH 437/537 (Game Theory and Related Topics). It’s called Nim Want to figure out how much of a two different vitamin supplements to take in addition to your multivitamin sot that you don’t go over a healthy amount of any given vitamin but get the most of each possible? MTH 432/532 (Optimization)", "take you. Say you go to a party and the host asks you to play a game by completing certain tasks. One task is to shake hands with two different people, and only two different people. If everyone who comes to the party is playing will everyone be able to shake the hands of exactly two other people? Or will someone be left out? The answer to this question is given by the Hand-Shaking Lemma, which is taught in MTH 438/538 (Theory and Application of Graphs). Have you ever seen the game Chomp before? if you are tired of losing to the computer MTH 437/537 (Game Theory and Related Topics) might help you understand the computer’s strategy. Did you play Dots and Boxes with your family and friends when you were younger? MTH 437/537 (Game Theory and Related Topics) might help you find a winning strategy. (Forget what it is, search online or click here:) Like games? Here’s one you might learn how to win in MTH 437/537 (Game Theory and Related Topics). It’s called Nim Want to figure out how much of a two different vitamin supplements to take in addition to your multivitamin sot that you don’t go over a healthy amount of any given vitamin but get the most of each possible? MTH 432/532 (Optimization)", "bubbles: the ones branching off from the 58 people who took the train. The third bit of info in the question tells us that 71 people were late to the event. Given that 40 of those people took the bus, we get $\\text{travelled by train and late }=71-40=31$ Finally, if 31 of the 58 people who took the train were late, we must have that $\\text{travelled by train and on time }=58-31=27$ The completed frequency tree looks like what is shown below. #### Is this a topic you struggle with? Get help now. a) Firstly, there are 75 people in total and 63 are right-handed, so we get $\\text{left-handed people }=75-63=12$ Secondly, we’re told that of the left-handed people, one third are right-footed. So, we get $\\text{left-handed and right-footed people }=12\\div 3=4$ Furthermore, since 12 is the total number of left-handed people, we must have that $\\text{left-handed and left-footed people }=12-4=8$ So, we just have two remaining bubbles: the ones branching off from the 63 right-handed people. The last bit of info in the question tells us that 27 people are left-footed. Given that we now know 8 of the left-handed people are left-footed, we get $\\text{right-handed and left-footed people }=27-8=19$ Finally, if 19 of the 63 right-handed people are left-footed, we must have that $\\text{right-handed and right-footed" ]

[ "arrangement matters). And so while some attendees (the “core group”) might end up having fun, the party doesn’t really work for most participating parties. So, the next time you’re hosting a party, do yourself and your guests a favour and ensure that you don’t end up with between 6 and 10 people at the party. Either less or more is fine! You might want to read this other post I’ve written on coordinating guest lists for birthday parties. ## Meeting types There are essentially three kinds of meetings – those that are entirely “in person”, those that are entirely “on call” and hybrids. I argue that the quality of conversation in the third kind of meeting is significantly inferior to that of the first two types. In person meetings are those where all participants are in one room. These are perhaps the best kind of meetings (except when you know it’s likely to turn antagonistic), for you can maintain eye contact with the others and as long as the number of meeters is small, there is social pressure on meeters to not be distracted, and thus the meeting is likely to conclude its agenda productively and quickly. Conference calls allow you to multitask while you are on the meeting – the positive thing is that you can choose", "of 12 people? We have: How many handshakes will take place between six people in a room when they each shakes hands with all the other people in the room one time? Here, Notice that no consideration is given to the order or arrangement of the items but simply the combinations. Another way of viewing combinations is as follows. Consider the number of combinations of 5 letters taken 3 at a time. This produces: Now assume you permute (arrange) the r = 3 letters in each of the 10 combinations in all possible ways. Each group would produce r! permutations. Letting x = 5C3 for the moment, we would therefore have a total of x(r!) different permutations. This total, however, represents all the possible permutations (arrangements) of n things taken r at a time, which is shown under arrangement numbers and defined as nPr. Therefore, Using the committee of 3 out of 12 people example from above, Consider the following: How many different ways can you enter a 4 door car? It is clear that there are 4 different ways of entering the car. Another way of expressing this is: If we ignore the presence of the front seats for the purpose of this example, how many different ways can you exit the car assuming that you do", "× ## Graph Theory Any connected group of things can be represented as a graph: cities and roads, people and friendships, and more. Learn why an even number of people have an odd number of friends. # Level 3 What is the least number of cuts that must be made in order to completely cut through this rope fence? Note: The knots where multiple ropes meet are too thick to cut through. There is a beautiful way to solve this that is much more elegant than randomly attacking the fence with scissors. Which of the shapes above cannot be traced without lifting up your pencil and without tracing over the same edge twice? Alice invited seven of her friends to a party. At the party, several pairs of people shook hands, although no one shook hands with themselves or shook hands with the same person more than once. After the party, Alice asked each of her seven friends how many people they shook hands with during the party, and was surprised when they responded with seven distinct positive integers. Given that her friends were truthful, how many hands did Alice shake? The Queen plans to build a number of new cities in the wilderness, connected by a network of roads. She expects the annual tax revenues from each", "clear the room and then have the one plus five. Each coordinator must confirm which are the five. Could we do this expeditiously, please? Event is not active Filed under:", "## Stats At a certain event, 16 people attend, and 5 will be chosen at random to receive door prizes. The prizes are all the same, so the order in which the people are chosen does not matter. How many different groups of 5 people can be chosen?", "take to make 10 complete rotations? #### 11th Standard Maths Combinations and Mathematical Induction English Medium Free Online Test One Mark Questions 2020 - 2021 - by Question Bank Software - View & Read • 1) The sum of the digits at the 10th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is • 2) The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in physics, first in chemistry and first in English is • 3) The number of ways in which a host lady invite 8 people for a party of 8 out of 12 people of whom two do not want to attend the party together is • 4) Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is • 5) The product of r consecutive positive integers is divisible by _________ #### 11th Standard Maths Important Question - by Question Bank Software - View & Read • 1) By taking suitable sets A, B, C, verify the following results: C-(B-A) = (C$\\cap$ A) $\\cup$ (C$\\cap$B') • 2) Discuss the following relations for reflexivity, symmetricity", "# Near the end of a party, everyone shakes hands... If everyone shakes hands with each other person we essentially have a complete graph. If there are 12 original guests, then there are ${{12} \\choose 2} = 66$ handshakes. So what about the late comer?", "# At a party each man dances with 4 women and each woman dances with 3 men. If 9 men attended the party, how many women attended the party? At a party each man dances with 4 women and each woman dances with 3 men. If 9 men attended the party, how many women attended the party? I have no idea how to approach this problem. • How many ways can a women choose 3 men in a set of 9 ? that would help. – user451844 Jul 13 '17 at 13:22 If you assume that a man dances with a woman, how many dances will be performed? 9 men -> dance 9 * 4 times with a woman, so 36 dances Each woman performs 3 dances, so 12 women. Count dance pairs . . . Since there are $9$ men and each man dances with $4$ women (presumably this means each man dances exactly $4$ times), there are exactly $36$ dance pairs. Let $w$ be the number of women. Since each woman dances with $3$ men (presumably this means each woman dances exactly $3$ times), there are exactly $3w$ dance pairs. Thus, $3w=36$, so $w=12$. • Obviously, we agree, but I don't think once each really matters. 9 women can dance 3 times with the same men,", "# Party acquaintances There are $$n$$ people at a party. Suppose they are friendly people, and each person has 8 acquaintances. Now, Bob, a passer-by notices this special property: (i) Each pair of people who are mutually acquainted has exactly 4 common acquaintances. (ii) Each pair of people who are mutually unacquainted has exactly 2 common acquaintances Find the number of possible values of $$n$$.", "sit together. Formula: Treat the persons who sit together as one and arrange. Then, arrange the group internally. Calculation: Men can be treated as one group. So, five boys and 1 group of men can be arranged in 6! ways. Three men can be arranged in 3! ways. Total no. of cases = 6! × 3! = 720 × 6 = 4320 ways ∴ Required no. of ways = 4320 # There are 7 persons in a party. The number of hands shaken all together in the party is: 1. 21 2. 6 3. 10 4. 2 Option 1 : 21 ## Detailed Solution Given: Number of person in a party = 7 $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ Where, n = number of the person r = minimum number of person required for a handshake Calculation: $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ $$\\Rightarrow {7_{{C_2}}} = \\;\\frac{{7!}}{{2! \\times \\left( {7 - 2} \\right)!}}$$ $$\\Rightarrow \\frac{{7 \\times 6 \\times 5!}}{{2! \\times 5!}}$$ ⇒ 42/2 ⇒ 21 ∴ the number of hands shaken is 21. # A boy contains 5 pink , 3 blue & 6 black ball. How many solution of ball can be formed by taking at least one ball form the boy 1. 157 2. 187 3. 167 4. 197 5.", "in the second. How many chairs can be around one table? • Kate shares Kate shares a 64-ounce bottle of apple cider with five friends. Each person's serving will be the same number of ounces. Between what two whole number of ounces will each person's serving to be? Explain using division.", "multiple measures within each level of party.list ? I have also removed the objection to the binomial. – Robert Long Dec 16 '19 at 18:42 • Yes there are up to three measures of party.list in each ibge7 (one per year). Thanks for that response. That is my biggest concern, and your response helped me realize I could confirm it by ensuring (1 | ibge7) + (1 | party.list) is equal to (1 | ibge7) + (1 | ibge7:SIGLA_PARTIDO). – dcoy Dec 19 '19 at 20:53", "# When's the Listening Party? A listening party is an event where a bunch of people get together to listen to some music (like a viewing party, but for music). In the age of the internet you can do Listening Parties online with friends, where you just all hit play together. The problem is on the internet everyone is in different time-zones so you can't just say \"Let's start at 5:35\". As a solution we just type xs in the hours position. So xx:35 means whenever the minute hand is next at 35. If the minute hand is already at that value then the party happens in 1 hour. If you need to specify more than one hour ahead you can add 1+, 2+ etc. to the beginning. So 1+xx:35 is an hour after the next time the minute hand is at 35. Your task is to take the current local time, and a string representing when the listening party happens in the above format and output when the listening party will occur in local time. # IO You can take and output the local time either as a 24 hour string with the minutes and hours separated by a colon (e.g. 19:45), as a tuple of the minutes and hours or as some native time type when", "# Two persons with the same number of acquaintance in a party There are $n$ persons attending a party. Prove that there are two persons who have the same number of acquaintance among the people attending the party.", "from 6 boys and 4 girls. Solution: There are 6 boys and 4 girls. A team of 3 boys and 2 girls is to be selected. ∴ 3 boys can be selected from 6 boys in 6C3 ways. 2 girls can be selected from 4 girls in 4C2 ways. ∴ Number of ways the team can be selected = 6C3 × 4C2 = $$\\frac{6 !}{3 ! 3 !} \\times \\frac{4 !}{2 ! 2 !}$$ = $$\\frac{6 \\times 5 \\times 4 \\times 3 !}{3 \\times 2 \\times 1 \\times 3 !} \\times \\frac{4 \\times 3 \\times 2 !}{2 \\times 2 !}$$ = 20 × 6 = 120 ∴ The team of 3 boys and 2 girls can be selected in 120 ways. Question 9. After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting. Solution: Let there be n participants present in the meeting. A handshake occurs between 2 persons. ∴ Number of handshakes = nC2 Given 66 handshakes were exchanged. ∴ 66 = nC2 ∴ 66 = $$\\frac{n !}{2 !(n-2) !}$$ ∴ 66 × 2 = $$\\frac{\\mathrm{n}(\\mathrm{n}-1)(\\mathrm{n}-2) !}{(\\mathrm{n}-2) !}$$ ∴ 132 = n(n – 1) ∴ n(n – 1) = 12 × 11 Comparing on both sides, we get n = 12", "video calling has indeed given me an excuse to catch up “visually” with friends I haven’t seen in ages, I don’t see that much value from group video calls, after having participated in a few. The main problem is that there can, at a time, be only one channel of communication. A few years back I’d written about the “anti two pizza rule” for organising parties, where I said that if you have a party, you should either have five or fewer guests, or ten or more (or something of the sort). The idea was that five or fewer can indeed have one coherent conversation without anyone being left out. Ten or more means the group naturally splits into multiple smaller groups, with each smaller group able to have conversations that add value to them. In between (6-9 people) means it gets awkward – the group is too small to split, and too large to have one coherent conversation, and that makes for a bad party. Now take that online. Because we have only one audio channel, there can only be one conversation for the entire group. This means that for a group of 10 or above, any “cross talk” needs to be necessarily broadcast, and that interferes with the main conversation of the group. So however large", "• # question_answer On the eve of Holi festival, a group of 15 friends greeted every other friend by sending greeting cards. Find the number of cards purchased by group. A) 156B) 144C) 210D) 320 There being 15 friends in the group, each friend must have purchased $(15-1)$ i.e 14 cards for sending greeting to rest of his 14 friends. Thus total number of cards Purchased by all the friends together is $15\\times 14$ i.e. 210.", "# English question regarding pigeonhole principle classic question. Mr. and Mrs. Smith invited four couples to their home. Some guests were friends of Mr. Smith, and some others were friends of Mrs. Smith. When the guests arrived, people who knew each other beforehand shook hands, those who did not know each other just greeted each other. After all this took place, the observant Mr. Smith said \"How interesting. If you disregard me, there are no two people present who shook hands the same number of times\". How many times did Mrs. Smith shake hands? My question is regarding the use of \"some\" and \"guests\" Some is defined as being at least one —used to indicate that a logical proposition is asserted only of a subclass or certain members of the class denoted by the term which it modifies. But would \"guests\" imply more than one? • No; \"some elephants are blue\" means the same as \"some elephant is blue\". – mjqxxxx Oct 28 '13 at 23:00 • I suppose Mr. and Mr. Smith knew each other beforehand, so they shook hands with each other? Similarly, any two people who came as a couple shook hands with each other? – bof Oct 28 '13 at 23:13 • The story is obscure and confusing, but I think the intended mathematical", "the index finger • Eleven → middle link of the index finger • Twelve → lower link of the index finger ### Two-handed counting up to 60 After the first dozen has been counted using the thumb as a pointer with the three phalanges of the remaining four fingers of the same hand (4 × 3 = 12), the counting capacity of one hand is initially exhausted. • The other hand is clenched in a fist. In order to remember that a dozen has been counted, one now extends a finger, e.g. B. the thumb. • Now you continue to count by starting again at one with your first hand . At twelve , the second dozen is full. • In order to remember that two dozen have been counted, one now extends the next finger of the other hand, e.g. B. after the thumb out the index finger. • With the five fingers of the first hand you can count five times a dozen, so 5 × 12 = 60. • Now you can count the next dozen again with the first hand, i.e. count up to 72 with two hands (12 on the first plus 60 on the other hand). This finger counting system still exists in parts of Turkey , Iraq , India and Indochina", "Probability # Combinations: Level 2 Challenges At a party, everyone shook hands with everybody else exactly once. There were 66 handshakes. How many people were at the party? Details and Assumptions: • You can't shake your own hand. Robbie is sitting for 9 subject papers in the GCSE O level examinations. In order to pass the examination, the number of subjects that he gets a pass grade in must be more than the number of subjects that he gets a fail grade. The number of ways in which Robbie can pass his GCSE O level examinations is: I was on vacation in England, and wanted to visit the Tower of London. The roads were laid out on a grid map, and the castle was 4 blocks north and 5 blocks east. If I were to travel only north and east, how many routes did I have to get to the castle? I was on vacation in England, and wanted to visit the Tower of London. The roads were laid out on a grid map, and and the castle was 4 blocks north and 5 blocks east. Unfortunately, there was some police activity near the hotel, and that intersection was blocked off so I could not go through it. If I were to travel only north and east, how" ]

[ "+0 -1 261 1 +397 deleted. Nov 8, 2019 edited by sinclairdragon428 Nov 20, 2019 #1 +24378 +3 Our math club has 20 members, among which we have 3 officers: President, Vice President, and Treasurer. However, one member, Ali, hates another member, Brenda. How many ways can we fill the offices if Ali refuses to serve as an officer if Brenda is also an officer? (No person is allowed to hold more than one office.) $$(20\\times19\\times18)-(3\\times2\\times18) = 6840 - 108 = 6732$$ Nov 8, 2019", "2022-01-28 There are 9 people eligible for the positions of president, vice-president, and secretary of the math club. How many different ways can the positions be filled? alenahelenash Expert Number of eligible people = n = 9 The number of different ways the position of president, vice-president and sevretary of the math club can be filled: Using ${P}_{3}^{n}$, where 3 - positions ${P}_{3}^{9}=9×8×7=504$ - Answer Do you have a similar question?", "officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club? A. 6 B. 10 C. 65 D. 69 E. It cannot be determined from information given. Breakdown and solve - Quote: The ratio of men to women in a certain club with 150 members is m : w......... There are 75 men in the club..... Men = Women = 75 Quote: there are 16 officers in the club... Officers = 16 & Non Officers = 134 Quote: If two-fifteenths of female members are officers.... Female Officers = $$\\frac{2}{15}*75$$ => 10 Male Officers = 6 Finally we have - Officers = 16 ( Female Officers - 10 , Male Officers - 6) & Non Officers = 134 (Female Non Officer - 65 ; Male Non Officer = 69 ) So, Total number of male non-officers in the club is 69 , Answer is (D) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) VP Status: It's near - I can see.", "n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club? A. 6 B. 10 C. 65 D. 69 E. It cannot be determined from information given. [Reveal] Spoiler: OA _________________ examPAL Representative Joined: 07 Dec 2017 Posts: 301 Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink] ### Show Tags 01 Jan 2018, 03:04 Bunuel wrote: The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club? A. 6 B. 10 C. 65 D. 69 E. It cannot be determined from information given. As one of our solutions is 'it cannot be determined' we cannot use the answers in this question. Therefore, we'll go for a direct calculation, a Precise approach. Let's write down what we know, going from the start of the question: Men + Women = 150 Men = 75 --> Women = 150 - 75 = 75", "* 2/15 = 10 So men officers = 16-10 = 6 So men who are not officers = 75-6 = 69 Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 3399 Location: India GPA: 3.5 Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink] ### Show Tags 01 Jan 2018, 08:47 1 KUDOS Bunuel wrote: The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club? A. 6 B. 10 C. 65 D. 69 E. It cannot be determined from information given. Breakdown and solve - Quote: The ratio of men to women in a certain club with 150 members is m : w......... There are 75 men in the club..... Men = Women = 75 Quote: there are 16 officers in the club... Officers = 16 & Non Officers = 134 Quote: If two-fifteenths of female members are officers.... Female Officers = $$\\frac{2}{15}*75$$ => 10 Male Officers = 6 Finally we have - Officers =", "+0 # helppppp 0 238 1 +814 Our club has 25 members, and wishes to pick a president, secretary, and treasurer. In how many ways can we choose the officers, if individual members may hold more than one office? Dec 9, 2018 $$\\text{The list of officers can be considered a 3 digit base 25 number}\\\\ N = (25)^3 = 15625$$", "Office [Specify the timing (see next paragraph) of an annual election of SIGMAA Officers, describe the composition of the nominating committee, and indicate relevant information about the nomination and election process. Give the terms of office for elected officers or appointed members of the Executive Committee, and indicate limits on terms for officers and a method for filling vacancies in offices.] Note that election of officers of a SIGMAA must be conducted through an online balloting system by the MAA Office. Two one month windows of opportunity are available for the election process, namely, April 1 through April 30 and October 1 through October 31. Typically a SIGMAA that specifies the January Joint Mathematics Meeting as its general membership meeting will select the October 1 through October 31 window, while the April 1 through April 30 window will be selected by a SIGMAA whose general membership meeting is held at the August MathFest. [The following wording is required.] 1. Nomination and election of officers will take place by electronic ballot of the membership. By October 1 (or April 1) the MAA will distribute to the SIGMAA members instructions for voting by October 31 (or April 30). The electronic ballot will indicate that write-in votes are permitted. The winner will be the candidate receiving the most votes. Ties will", "+0 # helppppp 0 60 1 +809 Our club has 25 members, and wishes to pick a president, secretary, and treasurer. In how many ways can we choose the officers, if individual members may hold more than one office? Dec 9, 2018 #1 +3576 +1 $$\\text{The list of officers can be considered a 3 digit base 25 number}\\\\ N = (25)^3 = 15625$$ . Dec 9, 2018", "the officer election committee shall be conducted at the next regular meeting of the Union or electronically, if the next Union meeting will not take place within two weeks. Any tie results shall be resolved by a coin toss. 2. The officer election committee must: 1. Announce through an email to all members a procedure for receiving nominations for steering committee positions. 1. Any member in good standing may nominate another member (or self- nominate) for a steering committee position. A member may only appear once on the steering committee ballot. If nominated for more than one position, may choose which one to run for. 2. Members must have at least two weeks from the committee's initial notification of the opening of nominations to make nominations. 3. A member may only serve in a particular steering committee position for one term. 4. After nominations have closed, the committee will determine, in accordance with the Constitution, the eligibility of the nominees. 5. If no eligible nominee is found for particular steering committee positions, the committee shall email the membership and reopen nominations for a period of one week. If no eligible nominations are received, the newly elected steering committee may appoint a member to the vacant position for the next term. 2. Decide and publicize the date for the", "club has 10 members. One president and two vice-presidents in the Problem Solving forum “Number of options for president = 10. (Any of the 10 members.) From the remaining 9 members, the number of ways to choose 2 to serve as vice-president = 9C2 = (9*8)/(2*1) = 36. To combine these options, we multiply: 10*36 = 360. The correct answer is E.” November 28, 2018", "Jan 2008, 11:23 20 A certain club has 10 members, including Harry. One of the 19 08 Jan 2007, 23:22 Display posts from previous: Sort by", "## College Algebra (6th Edition) Number of members in club = 10 We have to choose three officers such that each office held by one person, then the number of ways to fill the office use the formula $n_{{P}{_{r}}}$ = $\\frac{n!}{(n - r)! }$ The number of ways to fill the office $10_{{P}{_{3}}}$ = $\\frac{10!}{(10 - 3)! }$ = $\\frac{10\\times9\\times8\\times7!}{7! }$ = $10\\times9\\times8$ = 720", "Question How many ways can a president, vice-president, secretary, and treasurer be chosen from a club with 9 members?", "in no more than 300 words, their suitability for those position(s). 5. The returning officer must give notice to all members of the name, nominated position(s), and statement of each candidate no later than one week before the annual general meeting. 6. If any position has more candidates than vacancies, the returning officer must conduct a secret ballot for each such position at the annual general meeting. 7. The returning officer must determine the elected candidate(s) for each contested position using a single transferable vote method (such as the Meek STV method in the OpenSTV voting software). 8. If one or more candidates would be elected to more than one position, they are each elected to the position for which they nominated the highest preference, and each lower-preferenced position is re-determined as though those candidates withdrew. This process is repeated as necessary. 9. Each elected candidate assumes their position two weeks after the conclusion of the annual general meeting. Any position not filled at an annual general meeting becomes vacant two weeks after the conclusion of the meeting. 10. If a majority of the trustee positions would become vacant in this way, Clubs Council may appoint those positions on an interim basis or dissolve the association. 11. If Clubs Council appoints positions in this way, the committee must", "for three years (with staggered retirement dates). If a member in either of these categories resigns from the committee before the end of the three-year term, the resulting casual vacancy shall be filled for the remainder of that term by nomination from the appropriate society. Note: In order to start the process, the Edinburgh Mathematical Society and the London Mathematical Society should each be invited to nominate a member to hold office for two years, a member to hold office for three years, a member to hold office for four years, subsequently nominating replacements as (normal or casual) vacancies arise.", "+0 # help 0 246 1 +630 My school's Physics Club has 22 members. It needs to select 3 officers: chairman, vice-chairman, and sergeant-at-arms. Each person can hold at most one office. Two of the members, Penelope and Quentin, will only be officers if the other one is also an officer. (In other words, either both Penelope and Quentin are officers, or neither are.) In how many ways can the club choose its officers? Oct 19, 2018 $$\\text{Let's count the combination that have neither Penelope or Quentin first}\\\\ \\text{there will be }\\dbinom{20}{3}3! =6840 \\text{ ways to assign the officers w/o P or Q}\\\\ \\text{now if we have P we have Q so that just leaves }\\dbinom{20}{1}=20 \\text{ combinations}\\\\ \\text{or }20\\cdot 3!=120 \\text{ different ways off assigning the officers using P and Q}\\\\ \\text{this gets us a total of }6960 \\text{ ways of assigning officers}$$", "Please read - Club consolidation c. Ballots shall be sent out to current regular members in good standing no later than seven days after the October meeting. d. At the December meeting the Chairman shall present the vote counts and announce the new presiding directors. The new officers will be installed as of January 1st. I've got a question on this. Has a vote method been determined? e.g. a member is presented with a ballot of 15 nominees and votes for... 1, 3, 6 of them? Jay W Jason Swindle Posts: 65 Joined: Thu May 15, 2008 11:26 am Club: GRA Car#: 5 ### Re: Everybody - Please read - Club consolidation Speaking as an avid participant who has \"mostly\" left the governing (and apparently) squabbling to others, I do see merit in both methods, club, and unified. I too was mildly confused by the whole \"points card\" / which club do you wanna join. I raced the first season with 600 #s and my SCCA card. But after that season, I saw that joining a club was more fun. There were club xmas parties, club members who had tire changers in their garage, and the friendly rivalry with other clubs. I think some of that fun has been lost of the past few years. There really isn't", "• mathmath333 There are 40 students .In how many ways can they be arranged to form a team with 1 captain and 4 vice-captains ? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "to be honest with you, I put it aside for a while because there were some issues I couldn't quite figure out. When you write a book where you're trying to explain things to others, you realize somebody has not done a very good job of explaining what they've done. So I put it aside. I actually just went back to it yesterday. I'm feeling a head of steam. It's probably about halfway done is my guess. I would've said two years. It's probably going to be more like three and a half. ZIERLER: The year before, when you were inducted into the National Academy of Sciences, at this late stage, given how you've been recognized elsewhere, I wonder if one of your reactions was, \"What took you so long?\" SIMON: Well, it wasn't just my reaction because for years, I was aware, and I actually was aware–I understand a lot more about the process, having now been in the Academy. The election process is rather chaotic. I heard a story that there was a group of people who were trying to push it. It's a very funny process. 50 years ago, there used to be five, six, or seven mathematicians elected each year. For whatever reason, until relatively recently, it went down to more like three. Percy", "16 ( Female Officers - 10 , Male Officers - 6) & Non Officers = 134 (Female Non Officer - 65 ; Male Non Officer = 69 ) So, Total number of male non-officers in the club is 69 , Answer is (D) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Director Status: It's near - I can see. Joined: 13 Apr 2013 Posts: 739 Location: India GMAT 1: 480 Q38 V22 GPA: 3.01 WE: Engineering (Consulting) The ratio of men to women in a certain club with 150 members is m : w [#permalink] ### Show Tags 01 Jan 2018, 09:51 Bunuel wrote: The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club? A. 6 B. 10 C. 65" ]

[ "the answer and set the problem up accordingly. Finally, you must perform the calculations. Some special agent exams will give you “none of these” as the fifth (E) choice, but the TEA has a fifth numerical choice. It is important that you perform all calculations carefully because an estimate or an approximation may lead you to choose the wrong answer. Reminder: Do all the practice exercise questions with pencil on scratch paper. You will not be permitted to use a calculator at the exam, so get used to doing arithmetic without one. Answer questions in the practice exercise by circling the letter of your choice. ## Solved Examples - Arithmetic Reasoning Example 1) A police department purchases badges at 16$each for all the graduates of the police training academy. The last training class graduated 10 new officers. What is the total amount of money the department will spend for badges for these new officers? (A)$70 (B) $116 (C)$160 (D) $180 (E)$200 Solution: The correct answer is (C). It can be obtained by computing the following: $16\\times 10\\:=\\:160$ The badges are priced at 16$each. The department must purchase 10 of them for the new officers. Multiplying the price of one badge (16$) by the number of graduates (10) gives the total price for all of the badges. Choice (A),", "# From 4 officers and 8 jawans in how many ways can 6 be chosen Question: From 4 officers and 8 jawans in how many ways can 6 be chosen (i) to include exactly one officer (ii) to include at least one officer? Solution: (i) From 4 officers and 8 jawans, 6 need to be chosen. Out of them, 1 is an officer. Required number of ways $={ }^{4} C_{1} \\times{ }^{8} C_{5}=4 \\times \\frac{8 !}{5 ! 3 !}=4 \\times \\frac{8 \\times 7 \\times 6 \\times 5 !}{5 ! \\times 6}=224$ (ii) From 4 officers and 8 jawans, 6 need to be chosen and at least one of them is an officer. Required number of ways = Total number of ways -\">- Number of ways in which no officer is selected $={ }^{12} C_{6}-{ }^{8} C_{6}$ $=\\frac{12 !}{6 ! 6 !}-\\frac{8 !}{6 ! 2 !}$ $=\\frac{12 \\times 11 \\times 10 \\times 9 \\times 8 \\times 7}{6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}-\\frac{8 \\times 7}{2}$ $=924-28$ = 896", "the number of officers is 15, then find the number of non-officer in the office? A. 450 B. 550 C. 510 D. 520 Explanation: Let the required number of non-officers = a Then, $$110a + 460 \\times 15$$ = 120 (15 + a) or, 120a – 110a = $$450 \\times 15 – 120 \\times 15$$ = 15 (460 – 120) or, 10a = $$15 \\times 340$$ a = $$15 \\times 34$$ = 510 Q3. Find the average of first 20 multiple of 7? A. 71.5 B. 73.5 C. 75.2 D. 76.6 Explanation: Average = $$\\frac { 7 (1 + 2 + 3 + ……. + 20)}{20} = 7 \\times 20 \\times \\frac {21} {20 \\times 2} = 73.5$$ Q4. The average of four consecutive ODD number is 28.Find the largest number. A. 25 B. 31 C. 13 D. 27 Explanation: $$\\frac {x + (x + 2) + (x + 4) + (x + 6)}{4}$$ = 28 4x +12 = $$28 \\times 4 = > 112$$ $$x = \\frac {(112 – 12)}{4} = \\frac {100}{4}$$ 24 Largest number $$x + 6 = 25 + 6 = 31$$ Q5. Find the average of first 60 natural numbers A. 30.5 B. 31 C. 31.5 D. 32 Explanation: Sum of First 60 Natural numbers = $$\\frac { n (n + 1)}{2}$$ $$\\frac", "he is not an officer). For f, you should write $|X|=20$, $|Y|=20$, $|X\\cup Y|=40$. - a) How many selections are there in which Connie is not an officer? $$\\binom{6-1}{3}\\cdot3!$$ b) How many selections are there in which neither Ben nor Francisco is an officer? $$\\binom{6-2}{3}\\cdot3!$$ c) How many selections are there in which both Ben and Francisco are officers? $$\\binom{6-2}{3-2}\\cdot3!$$ d) How many selections are there in which Dolph is an officer and Francisco is not an officer? $$\\binom{6-2}{3-1}\\cdot3!$$ e) How many selections are there in which either Dolph is chairperson or he is not an officer? $$\\binom{6-1}{3-1}\\cdot(3-1)!+\\binom{6-1}{3}\\cdot3!$$ f) How many selections are there in which Ben is either chairperson or treasurer? $$\\binom{6-1}{3-1}\\cdot(3-1)!+\\binom{6-1}{3-1}\\cdot(3-1)!$$ -", "to $924-28=896$. Note: There is a possibility that one may commit a mistake while finding the answer of the part (1). There is a possibility that one may add the two obtained numbers instead multiplying them. But since we have to select both 1 officer and 5 privates, we have to multiply the two obtained numbers.", "# Math Help - need help with this problem 1. ## need help with this problem ther are 1200 officers in the police servies. there are 854 in unifourms while the rest work in plain clothes the averge ratio for a police service is 1 in clothes officers for every 3 uniform oficersto meet this average ratio how many plain clothes officers must be trasferd to uniform how do i get this ansewer 2. Hello, 1200 are in service 854 are in uniform 1200-854=346 are not in uniform 'y' no. of officers are transfered FROM UNIFORM TO NON UNIFORM OFFICERS $ \\frac{346+y}{854-y}=\\frac{1}{3} $ Solve for x You will get", "${}^{8}{{C}_{5}}=\\dfrac{8!}{5!\\left( 3 \\right)!}=\\dfrac{8\\times 7\\times 6\\times 5!}{5!\\times 3!}=\\dfrac{8\\times 7\\times 6}{3\\times 2}=56$ ways to choose $5$ jawans from $8$ jawans. To calculate the total number of ways to choose $6$ people according to the given data, we will multiply both the values of choosing one officer from $4$ officers and $5$ jawans from $8$ jawans. Thus, the total number of ways to choose $6$ people according to the given data $=4\\times 56=224$. Hence, we can choose $6$ people such that it includes exactly one officer is $224$.", "##### Example 5a Sheldon Ross on Combinatorics A police department in a small city consists of 10 officers. If the department policy is to have 5 of the officers patrolling the streets, 2 of the officers working full time at the station, and 3 of the officers on reserve at the station, how many different divisions of the 10 officers into the 3 groups are possible? There are $10 ! \\over 5! 2! 3 !$ possible divisions.", "## Precalculus (6th Edition) Blitzer We know that the order in which the four officers are selected makes a difference as the designation for all the four officers would be different. The ordered arrangement in which the order of the arrangement makes a difference is solved using the concept of permutations. Four officers are to be selected from a club of fifteen members. So, $n=15,r=4$. Thus, \\begin{align} & _{15}{{P}_{4}}=\\frac{15!}{\\left( 15-4 \\right)!} \\\\ & =\\frac{15!}{11!} \\\\ & =\\frac{15\\times 14\\times 13\\times 12\\times 11!}{11!} \\\\ & =32,760 \\end{align}", "and we can chose the deviant box in $5$ ways. In all there are $1024-4-60=960$ admissible assignments.", "no restriction; (ii) two particular books are always selected; (iii) two particular books are never selected? (i) Required ways of selecting 4 books from 10 books without any restriction = (ii) Two particular books are selected from 10 books. So, 2 books need to be selected from 8 books. Required number of ways if 2 particular books are always selected = (iii) Two particulars books are never selected from 10 books. So, 4 books need to be selected from 8 books. Required number of ways if two particular books are never selected = #### Question 9: From 4 officers and 8 jawans in how many ways can 6 be chosen (i) to include exactly one officer (ii) to include at least one officer? (i) From 4 officers and 8 jawans, 6 need to be chosen. Out of them, 1 is an officer. Required number of ways = (ii) From 4 officers and 8 jawans, 6 need to be chosen and at least one of them is an officer. Required number of ways = Total number of ways $-$ Number of ways in which no officer is selected #### Question 10: A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20", "number of ways to pick one of those $k$ to be president, and $\\binom{n}{n-k}$ is the number of ways to fill out the committee with $n-k$ people from the non-eligible group. Then, as you say, you sum over the possible values of $k$. On the right, $n$ is the number of ways to choose the president, and $\\binom{2n-1}{n-1}$ is the number of ways to choose the rest of the committee. – Brian M. Scott Nov 28 '14 at 19:47", "Arrange 25 officers, each having one of five different ranks $a$, $b$, $c$, $d$ and $e$, and belonging to one of five different regiments $p$, $q$, $r$, $s$ and $t$, in a square formation 5 by 5, so that each row and each file contains just one officer of each rank and just one from each regiment.", "remaining 4 to be one of 25 choices can be done in $5 \\cdot 25^4$ ways, just as you said.", "& 3 civilans = 5C2 * 9C3 3 officers & 2 civilans = 5C3 * 9C2 total = 5C2 * 9C3 + 5C3 * 9C2 = 840 + 360 = 1200 shouldn't there one more case here, the case when exactly 2 officers and 2 civilians are selected $$C^5_2 *C^9_2 * C^{10}_1$$ = 3600 so total ways 2 officers 2 civilians = $$C^5_2 *C^9_2 * C^{10}_1$$ = 3600 3 officers 2 civilians = $$C^5_3 * C^9_2$$ = 840 2 officers 3 civilians = $$C^5_2 * C^9_3$$= 360 so 3600 +1200 = 4800 please do confirm if my solution is right or wrong? _________________ - Stne Kudos [?]: 88 [0], given: 484 Math Expert Joined: 02 Sep 2009 Posts: 42249 Kudos [?]: 132619 [2], given: 12326 ### Show Tags 09 Sep 2012, 05:31 2 KUDOS Expert's post 1 This post was BOOKMARKED stne wrote: bangalorian2000 wrote: gmatinfoonline wrote: Hi all, A friend is taking a GMAT prep class and passed along this question to me - - sorry that I don't have the original source or the answer choices. \"A group consists of 5 officers and 9 civilians. If you must select a committee of 5 people, and the committee must include at least 2 officers and at least 2 civilians, how many different committees are possible?\" 2 officers", "9 \\times 8 \\times 7 }{ 4 \\times 3 \\times 2 \\times 1 }$ $= 210$ ii) Number of ways in which $4$ books can be made from $10$different books if two particular books are always selected $= ^{8} \\rm C_{2} =$ $\\frac{8!}{2! 6!}$ $=$ $\\frac{ 8 \\times 7 }{ 2 \\times 1 }$ $= 28$ iii) Number of ways in which $4$ books can be made from $10$different books if two particular books are never selected $= ^{8} \\rm C_{4} =$ $\\frac{8!}{4! 4!}$ $=$ $\\frac{ 8 \\times 7 \\times 6 \\times 5 }{ 4 \\times 3 \\times 2 \\times 1 }$ $= 70$ $\\\\$ Question 9: From $4$ officers and $8$ soldiers in how many ways can $6$ be chosen (i) to include exactly one officer (ii) to include at least one officer? Number of ways in which from $4$ officers and $8$ soldiers in how many ways can 6 be chosen if there is no restriction $= ^{12} \\rm C_{6} =$ $\\frac{12!}{6! 6!}$ $=$ $\\frac{ 12 \\times 11 \\times 9 \\times 8 \\times 7}{ 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1 }$ $= 924$ i) Number of ways in which from $4$ officers and $8$ soldiers in how many ways can 6 be chosen to include exactly one officer $= ^{4} \\rm", "thirteen boys are not separate groups. • okay. got it. thank you so much.! you're a genius. Dec 4, 2016 at 11:30 There should be minimum $2$ perfects in the committee, there is no limit for maximum of perfects. So, The number of ways can be counted as: $1$. $2$ perfect and $6$ non perfect..... Number of ways become ${{5}\\choose{2}}\\times{{10}\\choose{6}}$. $2$. $3$ perfect and $5$ non perfect....... Number of ways become ${{5}\\choose{3}}\\times{{10}\\choose{5}}$. $3$. $4$ perfect and $4$ non perfect.....Number of ways become ${{5}\\choose{4}}\\times{{10}\\choose{4}}$. $4$. $5$ perfect and $3$ non perfect.....Number of ways become ${{5}\\choose{5}}\\times{{10}\\choose{3}}$. The total number of committees of $8$ is $\\binom{15}{8}$. The number of committees of $8$ with no prefects is equal to $\\binom{10}{8}$. The number of committees with exactly one prefect is equal to $\\binom{5}{1} \\cdot \\binom{10}{7}$. Then the answer to your question is $\\binom{15}{8} - \\binom{10}{8} - \\binom{5}{1} \\cdot \\binom{10}{7}=6435 - 45 - 5\\cdot 120 = 5790.$", "that's it? Aug 8, 2020 #3 +1082 +4 Not exactly, but you are getting closer! Let's think about it case by case. So, you are right that there is one chief, so that is 10. Now, don't go ahead and choose the officers first! First, choose your two supporting chiefs. So that is 9C2. Next, we choose our officers. It is not 5C2, since 1) we do not have 5 left (9-2=?) and 2) you need 4 officers, not 2 officers. Can you go from here? :) ilorty Aug 8, 2020 #4 +169 +1 I got it! Thanks! (also choosing what first doesn't matter, i just learned that :o) Inspirational Aug 9, 2020", "the nine officers (8 * 324 + 9) makes 2,601 which just happens to be 51 squared. We pressed on, because that's the sort of people we are, and found another solution when x = 105. In this case, each square of troops contains 11,025 men, and when combined with the nine officers (8 * 11025 + 9) makes 88,209 which just happens to be 297 squared. And there we stopped!", "it includes at least one officer. (1) In this part, we have to find the number of ways in which we can choose 6 persons such that it includes exactly one officer. So, we can say that among the 6 chosen persons, 1 will be an officer and other five will be privates. From formula $\\left( 1 \\right)$, the number of ways in which we can choose 1 officer from a total of 4 officers is equal to, \\begin{align} & {}^{4}{{C}_{1}}=\\dfrac{4!}{1!\\left( 4-1 \\right)!} \\\\ & \\Rightarrow \\dfrac{4\\times 3!}{3!} \\\\ & \\Rightarrow 4 \\\\ \\end{align} Also, from formula $\\left( 1 \\right)$, the number of ways in which we can choose 5 privates from a total of 8 privates is equal to, \\begin{align} & {}^{8}{{C}_{5}}=\\dfrac{8!}{5!\\left( 8-5 \\right)!} \\\\ & \\Rightarrow \\dfrac{8\\times 7\\times 6\\times 5!}{5!.3!} \\\\ & \\Rightarrow \\dfrac{8\\times 7\\times 6}{3\\times 2\\times 1} \\\\ & \\Rightarrow 56 \\\\ \\end{align} Since we have to choose 1 officer and 5 privates, the number of ways to do so will be given by multiplying the above two obtained numbers. So, the number of ways in which we can choose 6 persons such that it includes exactly one officer is equal to $56\\times 4=224$. (2) In this part, we have to find the number of ways in which we can choose 6 persons such that" ]

[ "+0 -1 261 1 +397 deleted. Nov 8, 2019 edited by sinclairdragon428 Nov 20, 2019 #1 +24378 +3 Our math club has 20 members, among which we have 3 officers: President, Vice President, and Treasurer. However, one member, Ali, hates another member, Brenda. How many ways can we fill the offices if Ali refuses to serve as an officer if Brenda is also an officer? (No person is allowed to hold more than one office.) $$(20\\times19\\times18)-(3\\times2\\times18) = 6840 - 108 = 6732$$ Nov 8, 2019", "the answer and set the problem up accordingly. Finally, you must perform the calculations. Some special agent exams will give you “none of these” as the fifth (E) choice, but the TEA has a fifth numerical choice. It is important that you perform all calculations carefully because an estimate or an approximation may lead you to choose the wrong answer. Reminder: Do all the practice exercise questions with pencil on scratch paper. You will not be permitted to use a calculator at the exam, so get used to doing arithmetic without one. Answer questions in the practice exercise by circling the letter of your choice. ## Solved Examples - Arithmetic Reasoning Example 1) A police department purchases badges at 16$each for all the graduates of the police training academy. The last training class graduated 10 new officers. What is the total amount of money the department will spend for badges for these new officers? (A)$70 (B) $116 (C)$160 (D) $180 (E)$200 Solution: The correct answer is (C). It can be obtained by computing the following: $16\\times 10\\:=\\:160$ The badges are priced at 16$each. The department must purchase 10 of them for the new officers. Multiplying the price of one badge (16$) by the number of graduates (10) gives the total price for all of the badges. Choice (A),", "he is not an officer). For f, you should write $|X|=20$, $|Y|=20$, $|X\\cup Y|=40$. - a) How many selections are there in which Connie is not an officer? $$\\binom{6-1}{3}\\cdot3!$$ b) How many selections are there in which neither Ben nor Francisco is an officer? $$\\binom{6-2}{3}\\cdot3!$$ c) How many selections are there in which both Ben and Francisco are officers? $$\\binom{6-2}{3-2}\\cdot3!$$ d) How many selections are there in which Dolph is an officer and Francisco is not an officer? $$\\binom{6-2}{3-1}\\cdot3!$$ e) How many selections are there in which either Dolph is chairperson or he is not an officer? $$\\binom{6-1}{3-1}\\cdot(3-1)!+\\binom{6-1}{3}\\cdot3!$$ f) How many selections are there in which Ben is either chairperson or treasurer? $$\\binom{6-1}{3-1}\\cdot(3-1)!+\\binom{6-1}{3-1}\\cdot(3-1)!$$ -", "## Precalculus (6th Edition) Blitzer We know that the order in which the four officers are selected makes a difference as the designation for all the four officers would be different. The ordered arrangement in which the order of the arrangement makes a difference is solved using the concept of permutations. Four officers are to be selected from a club of fifteen members. So, $n=15,r=4$. Thus, \\begin{align} & _{15}{{P}_{4}}=\\frac{15!}{\\left( 15-4 \\right)!} \\\\ & =\\frac{15!}{11!} \\\\ & =\\frac{15\\times 14\\times 13\\times 12\\times 11!}{11!} \\\\ & =32,760 \\end{align}", "+0 # helppppp 0 60 1 +809 Our club has 25 members, and wishes to pick a president, secretary, and treasurer. In how many ways can we choose the officers, if individual members may hold more than one office? Dec 9, 2018 #1 +3576 +1 $$\\text{The list of officers can be considered a 3 digit base 25 number}\\\\ N = (25)^3 = 15625$$ . Dec 9, 2018", "the number of officers is 15, then find the number of non-officer in the office? A. 450 B. 550 C. 510 D. 520 Explanation: Let the required number of non-officers = a Then, $$110a + 460 \\times 15$$ = 120 (15 + a) or, 120a – 110a = $$450 \\times 15 – 120 \\times 15$$ = 15 (460 – 120) or, 10a = $$15 \\times 340$$ a = $$15 \\times 34$$ = 510 Q3. Find the average of first 20 multiple of 7? A. 71.5 B. 73.5 C. 75.2 D. 76.6 Explanation: Average = $$\\frac { 7 (1 + 2 + 3 + ……. + 20)}{20} = 7 \\times 20 \\times \\frac {21} {20 \\times 2} = 73.5$$ Q4. The average of four consecutive ODD number is 28.Find the largest number. A. 25 B. 31 C. 13 D. 27 Explanation: $$\\frac {x + (x + 2) + (x + 4) + (x + 6)}{4}$$ = 28 4x +12 = $$28 \\times 4 = > 112$$ $$x = \\frac {(112 – 12)}{4} = \\frac {100}{4}$$ 24 Largest number $$x + 6 = 25 + 6 = 31$$ Q5. Find the average of first 60 natural numbers A. 30.5 B. 31 C. 31.5 D. 32 Explanation: Sum of First 60 Natural numbers = $$\\frac { n (n + 1)}{2}$$ $$\\frac", "+0 # helppppp 0 238 1 +814 Our club has 25 members, and wishes to pick a president, secretary, and treasurer. In how many ways can we choose the officers, if individual members may hold more than one office? Dec 9, 2018 $$\\text{The list of officers can be considered a 3 digit base 25 number}\\\\ N = (25)^3 = 15625$$", "## College Algebra (6th Edition) Number of members in club = 10 We have to choose three officers such that each office held by one person, then the number of ways to fill the office use the formula $n_{{P}{_{r}}}$ = $\\frac{n!}{(n - r)! }$ The number of ways to fill the office $10_{{P}{_{3}}}$ = $\\frac{10!}{(10 - 3)! }$ = $\\frac{10\\times9\\times8\\times7!}{7! }$ = $10\\times9\\times8$ = 720", "# Statistics Quiz Help Can anyone take a look and let me know if I'm doing this right? This year I am the advisor of a 29 member chapter of The Honor Society. 1. How many different 8 member teams can I create. 2. Elections must be 4 4 positions. All current members can be considered, how many different slates of officers can there be? Answer--I don't think the 29 students comes into play, I think it's just 4x3x2x1=24 3. Jake really wants to be treasurer, and he is the only one interested. How many different slates of officers will there be if Jake is Treasurer? Answer: Again I don't think the 29 comes into play, but the first number is not a 4, but a 3. So 3x3x2x1=18 - The first should be $\\binom{29}{8} \\gg 8!$: You can pick $1$ out of $29$ at first, then $1$ out of $28$, then $1$ out of $27$ etc. giving $29\\cdot 28\\cdot 27 \\cdot \\ldots \\cdot 22$ options. However, you counted each team in $8!$ different orders, so you should divide by $8!$ giving $\\binom{29}{8}$. Your other answers are wrong for similar reasons. (Note that for the first question, you can make way more different teams with $100000$ members than with $29$ members to choose from. Still your answer does", "# Permutations and combinations • Sep 14th 2012, 07:35 AM Shwems Permutations and combinations There are five members of the Math Club. In how many ways can the positions of officers, a president, and a treasurer, be chosen? I thinked it's just 5 factorial. But the answer is 20. Which means they divided by 3 factorial... Why ?(Crying) • Sep 14th 2012, 07:49 AM kalyanram Re: Permutations and combinations You have two offices (President, Treasurer). Now you have to choose 2 people from a group of 5 how many ways can this can be done. Spoiler: In $2.\\binom{5}{2} = 2.\\frac{5!}{2!3!} = 20$ ways. Multiplication by two is to allow permutations among them. Thanks to Evgeny (emakarov) for the correction. • Sep 14th 2012, 08:10 AM emakarov Re: Permutations and combinations Quote: Originally Posted by Shwems There are five members of the Math Club. In how many ways can the positions of officers, a president, and a treasurer, be chosen? I thinked it's just 5 factorial. But the answer is 20. Which means they divided by 3 factorial... Why ?(Crying) See k-permutations.", "# From 4 officers and 8 jawans in how many ways can 6 be chosen Question: From 4 officers and 8 jawans in how many ways can 6 be chosen (i) to include exactly one officer (ii) to include at least one officer? Solution: (i) From 4 officers and 8 jawans, 6 need to be chosen. Out of them, 1 is an officer. Required number of ways $={ }^{4} C_{1} \\times{ }^{8} C_{5}=4 \\times \\frac{8 !}{5 ! 3 !}=4 \\times \\frac{8 \\times 7 \\times 6 \\times 5 !}{5 ! \\times 6}=224$ (ii) From 4 officers and 8 jawans, 6 need to be chosen and at least one of them is an officer. Required number of ways = Total number of ways -\">- Number of ways in which no officer is selected $={ }^{12} C_{6}-{ }^{8} C_{6}$ $=\\frac{12 !}{6 ! 6 !}-\\frac{8 !}{6 ! 2 !}$ $=\\frac{12 \\times 11 \\times 10 \\times 9 \\times 8 \\times 7}{6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}-\\frac{8 \\times 7}{2}$ $=924-28$ = 896", "## Algebra and Trigonometry 10th Edition There are $5,586,853,480$ different ways to form this jury. The order in which the people are selected is not important. It is a combination of 12 people from a group of 40 people. $_{40}C_{12}=\\frac{40!}{(40-12)!~12!}=\\frac{40\\times39\\times38\\times37\\times36\\times35\\times34\\times33\\times32\\times31\\times30\\times29\\times28!}{28!~12!}=5,586,853,480$", "n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club? A. 6 B. 10 C. 65 D. 69 E. It cannot be determined from information given. [Reveal] Spoiler: OA _________________ examPAL Representative Joined: 07 Dec 2017 Posts: 301 Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink] ### Show Tags 01 Jan 2018, 03:04 Bunuel wrote: The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club? A. 6 B. 10 C. 65 D. 69 E. It cannot be determined from information given. As one of our solutions is 'it cannot be determined' we cannot use the answers in this question. Therefore, we'll go for a direct calculation, a Precise approach. Let's write down what we know, going from the start of the question: Men + Women = 150 Men = 75 --> Women = 150 - 75 = 75", "# Combinatorics Question - Permutations while fixing cases Here is the wording of my question: In how many ways can a class with 20 students (12 boys and 8 girls elect a class president, vice president, and secretary if each student is willing to serve in any of the offices and no student can be elected to more than one office, and with at least one boy and one girl each among the students selected? There are two ways that I am thinking to do this. The first way would be to take the total number of cases and subtract off the ones that only have boys and only have girls elected. So that would be: $$20\\cdot19\\cdot18-\\binom{8}{3}-\\binom{12}{3}$$ Which equals $6564$. The other way I thought to do it was to select first a boy, then a girl, then one final student from the pool left and multiply it by the 6 ways you can arrange the 3 positions, so: $$\\binom{12}{1}\\cdot\\binom{8}{1}\\cdot\\binom{18}{1}\\cdot3!$$ Which equals $10368$. In my mind, both ways of doing this should be equivalent. I have checked my computation several times over but it is possible I am continuing to make a mistake. Can anyone offer an explanation as to which way is the correct way and explain why the other way is wrong? Maybe both of them", "+0 # help 0 246 1 +630 My school's Physics Club has 22 members. It needs to select 3 officers: chairman, vice-chairman, and sergeant-at-arms. Each person can hold at most one office. Two of the members, Penelope and Quentin, will only be officers if the other one is also an officer. (In other words, either both Penelope and Quentin are officers, or neither are.) In how many ways can the club choose its officers? Oct 19, 2018 $$\\text{Let's count the combination that have neither Penelope or Quentin first}\\\\ \\text{there will be }\\dbinom{20}{3}3! =6840 \\text{ ways to assign the officers w/o P or Q}\\\\ \\text{now if we have P we have Q so that just leaves }\\dbinom{20}{1}=20 \\text{ combinations}\\\\ \\text{or }20\\cdot 3!=120 \\text{ different ways off assigning the officers using P and Q}\\\\ \\text{this gets us a total of }6960 \\text{ ways of assigning officers}$$", "# Math Help - need help with this problem 1. ## need help with this problem ther are 1200 officers in the police servies. there are 854 in unifourms while the rest work in plain clothes the averge ratio for a police service is 1 in clothes officers for every 3 uniform oficersto meet this average ratio how many plain clothes officers must be trasferd to uniform how do i get this ansewer 2. Hello, 1200 are in service 854 are in uniform 1200-854=346 are not in uniform 'y' no. of officers are transfered FROM UNIFORM TO NON UNIFORM OFFICERS $ \\frac{346+y}{854-y}=\\frac{1}{3} $ Solve for x You will get", "Discussion Forum Daily Magic Spell Discussion - January 6, 2017 Daily Magic Spell Discussion - January 6, 2017 Today, we will discuss about the Daily Magic Spell proposed on January 6, 2017. Problem: Suppose you have a club of $20$ members. The club chooses four officers: President, Vice-President, Treasurer, and Secretary. They also choose someone to be in charge of fundraising. The four officers must all be different, but the member in charge of fundraising can be one of the officers. How many ways can we choose the four officers in the club with no restrictions? Solution: We can first choose the President in the club. There are $20$ options. After the President is chosen, when choosing the Vice-President, there are $19$ options. We repeat this process until all of the positions are filled. Therefore, there are $20 \\times 19 \\times 18 \\times 17 = 116280$ ways to fill the four positions. Additionally, we wish to appoint someone in the club to take charge of fundraising. This can be any one of the $20$ members in the club. Therefore, we have: $116280 \\times 20 = 2325600$ ways to appoint four officers and a person in charge of fundraising. The common approach that most students who answered this question incorrectly is most likely interpreting the problem incorrectly. Some users", "+0 0 78 2 +69 A club has 15 members and needs to choose 2 members to be co-presidents. In how many ways can the club choose its co-presidents? rohitaop May 14, 2018 #1 +867 +2 This is a simple choosing problem. 15 choose 2 can be expressed as: $$\\binom{15}{2}=\\frac{15\\cdot14}{2\\cdot1}=105$$ i hope this helped, Gavin GYanggg May 14, 2018 #2 +2765 +1 This is just combinations, 15C2- $$\\frac{15!}{2!*13!}=\\frac{15*14}{2*1}=15*7=\\boxed{105}$$ tertre May 14, 2018", "to $924-28=896$. Note: There is a possibility that one may commit a mistake while finding the answer of the part (1). There is a possibility that one may add the two obtained numbers instead multiplying them. But since we have to select both 1 officer and 5 privates, we have to multiply the two obtained numbers.", "# Math Help - Permutations and combinations 1. ## Permutations and combinations There are five members of the Math Club. In how many ways can the positions of officers, a president, and a treasurer, be chosen? I thinked it's just 5 factorial. But the answer is 20. Which means they divided by 3 factorial... Why ? 2. ## Re: Permutations and combinations You have two offices (President, Treasurer). Now you have to choose 2 people from a group of 5 how many ways can this can be done. Spoiler: In $2.\\binom{5}{2} = 2.\\frac{5!}{2!3!} = 20$ ways. Multiplication by two is to allow permutations among them. Thanks to Evgeny (emakarov) for the correction. 3. ## Re: Permutations and combinations Originally Posted by Shwems There are five members of the Math Club. In how many ways can the positions of officers, a president, and a treasurer, be chosen? I thinked it's just 5 factorial. But the answer is 20. Which means they divided by 3 factorial... Why ? See k-permutations." ]

[ "to me why you are doing the Compute() at all, since \\$a1 is already a MathObject. There is no need to recompute it that I can think of. OOPS! I see Alex already answered. Sorry for the noise.", "Try to Solve Without a Calculator 2 $1045^{2089}-1044^{2089}\\equiv x \\pmod{2089}$ What is the minimum value of $$x$$, without using a calculator? ×", "of $$a^3 + \\dfrac{1}{27a^3}$$ without solving for $$a$$.", "# Try to Solve Without a Calculator 2 $1045^{2089}-1044^{2089}\\equiv x \\pmod{2089}$ What is the minimum value of $$x$$, without using a calculator? × Problem Loading... Note Loading... Set Loading...", "MathsGod1 Apr 1, 2015 #30 +92624 +5 I want you to do them WITHOUT the calculator please MG Here is some more - only do them if you want to. :) Hint: Don't make a big deal out of these - they are quite simple. $$\\\\1)\\;\\; 6-\\frac{2}{3}\\\\\\\\ 2) \\;\\;12-3\\frac{5}{9}\\\\\\\\$$ Melody Apr 2, 2015", "# Try to Solve Without a Calculator 2 Number Theory Level 2 $1045^{2089}-1044^{2089}\\equiv x \\pmod{2089}$ What is the minimum value of $$x$$, without using a calculator? ×", "× # How to solve huge exponential easily without calculator? I want to solve $$8^{2017}$$ without calculator so how Can I solve it? And what is the general method to solve similar equations? Note by Nafees Khabir 2 years, 1 month ago Sort by: Are you looking for the exact value of $$8^{2017} = 3408\\ldots165248$$ which is nearly 2000 digits long? Then there's no shortcut to it. Or are you looking for the last few digits of the number? Then this page might help. · 2 years, 1 month ago", "to compute, (and that only up to sign!). 6. Jul 6, 2016", "all very straightforward questions. what do you need help with? How can I do this without a scientific calc on hand? 4. ## Re: probability questions I need help with. Originally Posted by Antony519 How can I do this without a scientific calc on hand? looking at (1) $p(k)=\\dfrac{\\mu^k e^{-\\mu}}{k!} = \\dfrac{6^k e^{-6}}{k!}$ $p(6) = \\dfrac{6^6 e^{-6}}{6!}$ without a calculator I'd just leave it in that form. You've got access to a computer but not a scientific calculator?", "# Try to solve this without a calculator Simplify the expression below. $\\dfrac{999999\\times999999}{1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1}$ Try ×", "## Thinking Mathematically (6th Edition) $6^{-3}= \\dfrac{1}{6^3} = \\dfrac{1}{216}$ RECALL: For any non-zero real number $a$, $a^{-m} = \\dfrac{1}{a^m}$ Use the rule above to find: $6^{-3} \\\\=\\dfrac{1}{6^3} \\\\=\\dfrac{1}{216}$", "and simplify it. 3. Use the equation to compute $$99^{2}$$ quickly, without using a calculator. Explain how you did it.", "## Calculate Fractional Exponents Without Calculator? For example, 8^(4/3) is 16. How would I have known that without a calculator?", "### Home > CALC > Chapter 4 > Lesson 4.4.2 > Problem4-138 4-138. ≈ 0.42 It is important to recognize what skills you have and what skills you do not have. Think about why you are not expected to compute these limits without a calculator.", "with each steps without a calculator", "+ \\tfrac{1}{e} - e$ . . . . . . $=\\;-\\frac{16}{3} + \\frac{1-e^2}{e}$ 3. Re: Area under the shaded region answer check please? Haha yes I absolutely suck at doing that type of stuff without a calculator; no lies. I appreciate your help so much!!", "at 12:37 • The expoenential distributioin was not a good idea. I have made an edit. – callculus Jun 19 '15 at 13:52 • You don't have to compute $1000!$ to compute $1000 \\choose k$. For say $k=3$ it is just $1000*999*998/3!$. The factors in the denominator cancel most of the factors in the numerator. – BruceZ Jun 20 '15 at 0:07", "if $f(q)$ can be computed without the oracle) fails to compute $f$ on a $(\\frac{1}{2} - o(1))$-fraction of inputs in $H$.", "$a = 8;$b = Compute(\"\\$a\"); # creates a MathObject Real(8). Good luck and keep the questions coming! Paul Pearson", "Question Find the following values without a calculator. (a) 3 ÷ _____ = 13 (b) _____÷ 56 = 4 (c) 38 ÷ _____ = 116 (d) 16 ÷ _____ = 58 2 m Click button first when a symbol is required. X Find the following values without a calculator." ]

[ "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "\\dfrac{(x+1)(x+4)}{(x+3)(x+3)} \\cdot \\dfrac{(x+3)(x-3)}{x+1} = \\boxed{\\dfrac{x^2+x-12}{x+3}} $ Ah.....oops! Thank you! How did I forget that! Thank you again for your help earboth! drakmord", "$\\dfrac{1}{4 \\times x^4}$ $+$ $\\cdots\\Bigg)$ $\\implies$ $\\log_e{(y)}$ $\\,=\\,$ $x$ $-$ $\\Bigg(\\dfrac{x^2 \\times 1}{x}$ $-$ $\\dfrac{x^2 \\times 1}{2 \\times x^2}$ $+$ $\\dfrac{x^2 \\times 1}{3 \\times x^3}$ $-$ $\\dfrac{x^2 \\times 1}{4 \\times x^4}$ $+$ $\\cdots\\Bigg)$ $\\implies$ $\\log_e{(y)}$ $\\,=\\,$ $x$ $-$ $\\Bigg(\\dfrac{\\cancel{x^2} \\times 1}{\\cancel{x}}$ $-$ $\\dfrac{\\cancel{x^2} \\times 1}{2 \\times \\cancel{x^2}}$ $+$ $\\dfrac{\\cancel{x^2} \\times 1}{3 \\times \\cancel{x^3}}$ $-$ $\\dfrac{\\cancel{x^2} \\times 1}{4 \\times \\cancel{x^4}}$ $+$ $\\cdots\\Bigg)$ $\\implies$ $\\log_e{(y)}$ $\\,=\\,$ $x$ $-$ $\\Bigg(\\dfrac{x \\times 1}{1}$ $-$ $\\dfrac{1 \\times 1}{2 \\times 1}$ $+$ $\\dfrac{1 \\times 1}{3 \\times x}$ $-$ $\\dfrac{1 \\times 1}{4 \\times x^2}$ $+$ $\\cdots\\Bigg)$ $\\implies$ $\\log_e{(y)}$ $\\,=\\,$ $x$ $-$ $\\Bigg(\\dfrac{x}{1}$ $-$ $\\dfrac{1}{2}$ $+$ $\\dfrac{1}{3x}$ $-$ $\\dfrac{1}{4x^2}$ $+$ $\\cdots\\Bigg)$ $\\implies$ $\\log_e{(y)}$ $\\,=\\,$ $x$ $-$ $\\Bigg(x$ $-$ $\\dfrac{1}{2}$ $+$ $\\dfrac{1}{3x}$ $-$ $\\dfrac{1}{4x^2}$ $+$ $\\cdots\\Bigg)$ $\\implies$ $\\log_e{(y)}$ $\\,=\\,$ $x$ $-$ $x$ $+$ $\\dfrac{1}{2}$ $-$ $\\dfrac{1}{3x}$ $+$ $\\dfrac{1}{4x^2}$ $-$ $\\cdots$ $\\implies$ $\\log_e{(y)}$ $\\,=\\,$ $\\cancel{x}$ $-$ $\\cancel{x}$ $+$ $\\dfrac{1}{2}$ $-$ $\\dfrac{1}{3x}$ $+$ $\\dfrac{1}{4x^2}$ $-$ $\\cdots$ $\\implies$ $\\log_e{(y)}$ $\\,=\\,$ $\\dfrac{1}{2}$ $-$ $\\dfrac{1}{3x}$ $+$ $\\dfrac{1}{4x^2}$ $-$ $\\cdots$ ### Evaluate the Limit of the Algebraic equation The simplification of the logarithmic equation is completed. Now, let’s us evaluate the limit of the log equation as $x$ close to infinity. $\\implies$ $\\displaystyle \\large \\lim_{x \\,\\to\\, \\infty}{\\normalsize \\log_e{(y)}}$ $\\,=\\,$ $\\displaystyle \\large \\lim_{x \\,\\to\\, \\infty}{\\normalsize \\Bigg(\\dfrac{1}{2}}$ $-$ $\\dfrac{1}{3x}$ $+$ $\\dfrac{1}{4x^2}$ $-$ $\\cdots\\Bigg)$ Now, evaluate the limit of the infinite series as $x$ approaches infinity", "= \\dfrac{1}{n}$$ $\\dfrac{(n-1)!}{n!} = \\dfrac{\\cancel{(n-1)} \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}}{n \\times \\cancel{(n-1)} \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}} = \\dfrac{1}{n}$ $$\\dfrac{(n-2)!}{(n-1)!} = \\dfrac{1}{n-1} \\text{ for } n>1$$ $\\dfrac{(n-2)!}{(n-1)!} = \\dfrac{\\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}}{(n-1) \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}} = \\dfrac{1}{n-1}$", "5}{10 \\times 9 \\times 8 \\times 7 \\times 6 }$ 5 red out of 5: $\\dfrac{5 \\times 4 \\times 3 \\times 2 \\times 1}{5 \\times 4 \\times 3 \\times 2 \\times 1} \\times \\dfrac{5 \\times 4 \\times 3 \\times 2 \\times 1}{10 \\times 9 \\times 8 \\times 7 \\times 6 }$ Add those up and you get $\\frac{1}{2}$. Also add 0, 1 or 2 reds out of 5 and you get $1$. - add comment", "(4 \\times 3 \\times 2 \\times 1) - (3 \\times 2 \\times 1)= 720 + 24 -6 = \\text{738}$ $$\\dfrac{6! - 2!}{2!}$$ $\\dfrac{(6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1) - (2 \\times 1)}{2 \\times 1}= \\dfrac{720 - 2}{2} = \\text{359}$ $$\\dfrac{2! + 3!}{5!}$$ $\\dfrac{(2 \\times 1) + (3 \\times 2 \\times 1)}{5 \\times 4 \\times 3 \\times 2 \\times 1}= \\dfrac{2 + 6}{120} = \\dfrac{1}{15}$ $$\\dfrac{2! + 3! - 5!}{3! - 2!}$$ $\\dfrac{2 + 6 - 120}{6 - 2}= \\dfrac{-\\text{112}}{4} = -\\text{28}$ $$(3!)^{3}$$ $$6 \\times 6 \\times 6 = \\text{216}$$ $$\\dfrac{3! \\times 4!}{2!}$$ $\\dfrac{(3 \\times 2 \\times 1) \\times (4 \\times 3 \\times 2 \\times 1)}{2 \\times 1} = \\text{72}$ Calculate the following using a calculator: $$\\dfrac{12!}{2!}$$ $\\dfrac{(12 \\times 11 \\times 10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1)}{2 \\times 1} = \\text{239 500 800}$ $$\\dfrac{10!}{20!}$$ $$\\text{1,49} \\times \\text{10}^{-\\text{12}}$$ $$\\dfrac{10! + 12!}{5! + 6!}$$ $$\\text{574 560}$$ $$5!(2! + 3!)$$ $$\\text{960}$$ $$(4!)^{2}(3!)^{2}$$ $$\\text{20 736}$$ Show that the following is true: $$\\dfrac{n!}{(n-2)!} = n^{2} - n$$ \\begin{align*} \\dfrac{n!}{(n-2)!} & = \\dfrac{n \\times (n-1) \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}}{\\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}} \\\\ \\\\ &= n(n-1) = n^{2} - n \\end{align*} $$\\dfrac{(n-1)!}{n!}", "\\dfrac{x}{2} + \\dfrac{3}{8}x^{2} - \\cdots+(-1)^{n}\\dfrac{1\\times 3\\times\\dotsm\\times(2n-1)}{2\\times 4\\times\\dotsm\\times 2n} x^{n} + \\mathcal{O}\\left(x^{n+1}\\right) \\end{array} \\end{document}", "\\dfrac{{3 \\times 2!}}{{\\not 2!}}}}{{\\dfrac{{\\not 6 \\times 5 \\times 4 \\times \\not 3!}}{{\\not 3! \\times 3! \\times 2 \\times 1}}}}$ $= \\dfrac{9}{{20}} \\times \\dfrac{6}{{20}} = \\dfrac{{27}}{{200}}\\,\\,\\,\\,\\,\\,\\,.......(1)$ Now case II as the number of balls are same just colours are different, we have: $= \\dfrac{{{}^3{C_2} \\times {}^3{C_1}}}{{{}^6{C_3}}} \\times \\dfrac{{{}^1{C_1} \\times {}^2{C_1} \\times {}^3{C_1}}}{{{}^6{C_3}}}$ $= \\dfrac{{\\dfrac{{3!}}{{2!1!}} \\times \\dfrac{{3!}}{{2!1!}}}}{{\\dfrac{{6!}}{{3!3!}}}} \\times \\dfrac{{\\dfrac{{2!}}{{1!1!}} \\times 1 \\times \\dfrac{{3!}}{{2!1!}}}}{{\\dfrac{{6!}}{{3!3!}}}}$ $= \\dfrac{{\\dfrac{{3 \\times \\not 2!}}{{\\not 2!}} \\times \\dfrac{{3 \\times \\not 2!}}{{\\not 2!}}}}{{\\dfrac{{\\not 6 \\times 5 \\times 4 \\times \\not 3!}}{{3! \\times 3\\not ! \\times \\not 2 \\times 1}}}} \\times \\dfrac{{2! \\times \\dfrac{{3 \\times 2!}}{{\\not 2!}}}}{{\\dfrac{{\\not 6 \\times 5 \\times 4 \\times \\not 3!}}{{\\not 3! \\times 3! \\times 2 \\times 1}}}}$ $= \\dfrac{9}{{20}} \\times \\dfrac{6}{{20}} = \\dfrac{{27}}{{200}}\\,\\,\\,\\,\\,\\,\\,.......(2)$ $= \\dfrac{{27}}{{200}}.....(2)$ Probability of 3 later balls being all of different colours $= \\dfrac{{27}}{{200}} + \\dfrac{{27}}{{200}}$(from $e{q^n}$ 1 & 2) $= \\dfrac{{2 \\times 27}}{{200}}$$= \\dfrac{{27}}{{100}}$ Note: Cube root need to group digit in no. 3 and taking the unit place digit of the first group & ten’s from $I{I^{nd}}$. In mathematics, a cube root of a number x is a number y such that $\\;{y^3}\\; = \\;x$. All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube", "\\to \\infty} {n! \\over \\sqrt{2\\pi n} (n/e)^n} = 1,$$ if you rewrite the product as $(2n)!/(2^n n!)^2$. - Thanks to you two! Dear Michael s! – Bidgoli Mar 12 at 20:08 Here you need to find $\\lim A_n$, where $$A_n= \\frac{1\\times3\\times5\\times\\cdots\\times(2n-1)}{2\\times4\\times6\\times8\\times\\cdots\\times (2n)}.$$ Now note that, for $n\\gt1$, $2n-1\\gt n$. So, in the numerator, you get $$\\frac{1\\times 2\\times 3\\dots\\times n}{2\\times4\\times 6\\times\\dots\\times 2n}\\lt\\frac{1\\times3\\times5\\times\\cdots\\times(2n-1)}{2\\times4\\times6\\times8\\times\\cdots\\times (2n)}\\\\\\implies\\frac1{2^n}\\lt \\frac{1\\times3\\times5\\times\\cdots\\times(2n-1)}{2\\times4\\times6\\times8\\times\\cdots\\times (2n)}$$ Now, you use the identity $$\\frac n{n+1}\\lt\\frac {n+1}{n+2}$$ and define a sequence $$B_n=\\frac 23\\times \\frac 45\\times\\dots \\times \\frac {2n}{2n+1} .$$ So, now you have $$A_n\\lt B_n\\\\\\implies (A_n)^2\\lt A_nB_n=\\frac 1{2n+1}\\\\\\implies A_n\\lt \\frac 1{\\sqrt{2n+1}}.$$ So, you have $$\\frac 1{2^n}\\lt A_n\\lt \\frac 1{\\sqrt{2n+1}}\\\\\\implies\\lim \\frac 1{2^n}\\lt\\lim A_n\\lt\\lim \\frac 1{\\sqrt{2n+1}}.$$ So, by sandwitch theorem, $$\\lim A_n=0.$$ - What's wrong with this? – Subhadeep Dey Mar 11 at 19:06 Dear downvoter, are you going to explain the reason for downvoting? – Subhadeep Dey Mar 11 at 19:18 I didn't downvote, but the first half of the proof is unnecessary. For the sandwich theorem, all you need is $0 < A_n < 1/\\sqrt{2n+1}$, and $0 < A_n$ is obvious. – alephzero Mar 11 at 23:51 @alephzero, yes, I know. I have just showed that there are several ways to get the lower sequence. – Subhadeep Dey Mar 12 at 7:49 JUST BRILLIANT... – Bidgoli Mar 12 at 20:06 To paraphrase Wikipedia:", "### Factorials #### Definition A factorial of a positive number $n$ is defined to be the product of all positive whole numbers less than and equal to $n$. E.g. $n=5$ $5! = 5 \\times 4 \\times 3 \\times 2 \\times 1.$ Note: We define the factorial $0!=1$. #### Worked Examples ###### Example 1 Calculate $7!$. ###### Solution From the definition we have $7! = 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1.$ Hence $7! =5040.$ ###### Example 2 Simplify $\\dfrac{9!}{6!}$. ###### Solution Firstly we can expand the factorials using the definition $\\dfrac{9!}{6!} = \\dfrac{9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}{6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}.$ This can then be simplified by canceling terms on the top and bottom to give $\\dfrac{9!}{6!} = 9 \\times 8 \\times 7.$ Therefore, $\\dfrac{9!}{6!} = 504.$ ###### Example 3 Simplify $\\dfrac{16!}{13!\\cdot 8!}$ ###### Solution As above start by expanding the factorials $\\dfrac{16!}{13!\\cdot 8!} = \\dfrac{16 \\times 15 \\times \\dots \\times 3 \\times 2 \\times 1}{(13 \\times 12 \\times 11 \\times \\dots \\times 3 \\times 2 \\times 1 )( 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1)}.$ Then cancel out any terms that appear on the top", "is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$.", ". . . . . $=\\; \\dfrac{60}{9.8} \\;=\\;\\dfrac{300}{49} \\;=\\;6\\frac{6}{49}\\text{ seconds.}$", "– Dylan May 2 '13 at 5:32 No worries ;-) that is how we all learn. I still lose hours because of what I don't know. – Fixed Point May 2 '13 at 6:01 Hint: Use the Ratio Test. No modification of the expression is needed. - We have $$a_n = \\dfrac{n!}{(2n-1)!!} = \\dfrac{n!}{(2n)!} \\times 2^n n! = \\dfrac{2^n}{\\dbinom{2n}n}$$ Use ratio test now to get that $$\\dfrac{a_{n+1}}{a_n} = \\dfrac{2^{n+1}}{\\dbinom{2n+2}{n+1}} \\cdot \\dfrac{\\dbinom{2n}n}{2^n} = \\dfrac{2(n+1)(n+1)}{(2n+2)(2n+1)} = \\dfrac{n+1}{2n+1}$$ We can also use Stirling. From Stirling, we have $$\\dbinom{2n}n \\sim \\dfrac{4^n}{\\sqrt{\\pi n}}$$ Use this to conclude, about the convergence/divergence of the series. EDIT $$1 \\times 3 \\times 5 \\times \\cdots \\times(2n-1) = \\dfrac{\\left( 1 \\times 3 \\times 5 \\times \\cdots \\times(2n-1) \\right) \\times \\left(2 \\times 4 \\times \\cdots \\times (2n)\\right)}{ \\left(2 \\times 4 \\times \\cdots \\times (2n)\\right)}$$ Now note that $$\\left( 1 \\times 3 \\times 5 \\times \\cdots \\times(2n-1) \\right) \\times \\left(2 \\times 4 \\times \\cdots \\times (2n)\\right) = (2n)!$$ and $$\\left(2 \\times 4 \\times \\cdots \\times (2n)\\right) = 2^n \\left(1 \\times 2 \\times \\cdots \\times n\\right) = 2^n n!$$ Hence, $$1 \\times 3 \\times 5 \\times \\cdots \\times(2n-1) = \\dfrac{(2n)!}{2^n \\cdot n!}$$ - Yikes. It is not at all apparent how $\\frac{1}{(2n-1)!!}$ becomes $\\frac{2^{n}n!}{(2n)!}$ – Dylan May 2 '13 at 5:38 Soon it will be obvious. Put in the \"missing\" terms", "C}$ and $\\alpha_ \\text{brass} = \\dfrac{19 \\times 10^{-6}}{^\\circ C}$]", "# Do You Know The Rule? $\\large \\dfrac{1}{{8!}} + \\dfrac{1}{{9!}} = \\dfrac{x}{{10!}}$ Find the value of $$x$$ satisfying the equation above. Notation: $$!$$ denotes the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "= \\dfrac{{13 \\times 12! \\times 4 \\times 3! \\times 50! \\times 2 \\times 1}}{{52 \\times 51 \\times 50! \\times 12! \\times 3!}} \\\\ P = \\dfrac{{13 \\times 4 \\times 2}}{{52 \\times 51}} = \\dfrac{2}{{51}} \\\\$", "+ 2) = 9 \\times 10 + 9 \\times 2 = 108$ $9 \\times (110 + 2) = 9 \\times 110 + 9 \\times 2 = 1008$ So on. $$9 \\times 11\\cdots12 = 9 \\times (11\\cdots 11+1) = 99\\cdots99 + 9.$$ Write $1...12$ as $$\\overline{1...(\\times\\ n)...1} +1=\\frac{10^n-1}{9}+1$$ Hence $$9\\times (\\frac{10^n-1}{9}+1)=10^n-1+9=10^n+8$$", "\\dfrac{629988.3164 \\times 9 \\times 10^{-4}}{0.5}$ N = 1133.98 = 1134 T.", "47} &=& \\dfrac{3 \\times 5^2 \\times 11 \\times 13 \\times 17 \\times 29}{19 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{49} \\times p_{14}(29,20) &=& \\dfrac{29}{49} \\times \\dfrac{2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17}{37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11 \\times 13 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{50} \\times p_{15}(29,21) &=& \\dfrac{29}{50} \\times \\dfrac{2^2 \\times 3 \\times 5^3 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11 \\times 13 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{51} \\times p_{16}(29,22) &=& \\dfrac{29}{51} \\times \\dfrac{2^5 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^5 \\times 3 \\times 5 \\times 11 \\times 13 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{52} \\times p_{17}(29,23) &=& \\dfrac{29}{52} \\times \\dfrac{2^3 \\times 3 \\times 5 \\times 11^2 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{53} \\times p_{18}(29,24) &=& \\dfrac{29}{53} \\times \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2", "\\times 17 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 23 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{54} \\times p_{19}(29,25) &=& \\dfrac{29}{54} \\times \\dfrac{2^4 \\times 3^2 \\times 11^2 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^3 \\times 11^2 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} \\\\ \\dfrac{29}{55} \\times p_{20}(29,26) &=& \\dfrac{29}{55} \\times \\dfrac{2^5 \\times 5^3 \\times 11 \\times 17 \\times 19 \\times 23}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^5 \\times 5^2 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{56} \\times p_{21}(29,27) &=& \\dfrac{29}{56} \\times \\dfrac{2^4 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23}{3 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{57} \\times p_{22}(29,28) &=& \\dfrac{29}{57} \\times \\dfrac{2^2 \\times 3^2 \\times 5^2 \\times 11 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times" ]

[ "(4 \\times 3 \\times 2 \\times 1) - (3 \\times 2 \\times 1)= 720 + 24 -6 = \\text{738}$ $$\\dfrac{6! - 2!}{2!}$$ $\\dfrac{(6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1) - (2 \\times 1)}{2 \\times 1}= \\dfrac{720 - 2}{2} = \\text{359}$ $$\\dfrac{2! + 3!}{5!}$$ $\\dfrac{(2 \\times 1) + (3 \\times 2 \\times 1)}{5 \\times 4 \\times 3 \\times 2 \\times 1}= \\dfrac{2 + 6}{120} = \\dfrac{1}{15}$ $$\\dfrac{2! + 3! - 5!}{3! - 2!}$$ $\\dfrac{2 + 6 - 120}{6 - 2}= \\dfrac{-\\text{112}}{4} = -\\text{28}$ $$(3!)^{3}$$ $$6 \\times 6 \\times 6 = \\text{216}$$ $$\\dfrac{3! \\times 4!}{2!}$$ $\\dfrac{(3 \\times 2 \\times 1) \\times (4 \\times 3 \\times 2 \\times 1)}{2 \\times 1} = \\text{72}$ Calculate the following using a calculator: $$\\dfrac{12!}{2!}$$ $\\dfrac{(12 \\times 11 \\times 10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1)}{2 \\times 1} = \\text{239 500 800}$ $$\\dfrac{10!}{20!}$$ $$\\text{1,49} \\times \\text{10}^{-\\text{12}}$$ $$\\dfrac{10! + 12!}{5! + 6!}$$ $$\\text{574 560}$$ $$5!(2! + 3!)$$ $$\\text{960}$$ $$(4!)^{2}(3!)^{2}$$ $$\\text{20 736}$$ Show that the following is true: $$\\dfrac{n!}{(n-2)!} = n^{2} - n$$ \\begin{align*} \\dfrac{n!}{(n-2)!} & = \\dfrac{n \\times (n-1) \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}}{\\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}} \\\\ \\\\ &= n(n-1) = n^{2} - n \\end{align*} $$\\dfrac{(n-1)!}{n!}", "# Try to solve this without a calculator Simplify the expression below. $\\dfrac{999999\\times999999}{1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1}$ Try ×", "## 2013 DSE MATH (CORE) Question 19 Full Solution Posted: September 10, 2013 in Mathematics Tags: , , , , , (a) (i) The trick here is to recognize that you need to divide r by 100 and then add 1 to it. $9 \\times 10^6 \\cdot \\left(1+\\dfrac{r}{100} \\right) - 3 \\times 10^5$ $= 87 \\times 10^5 + 9 \\times 10^4 \\cdot r$ $= 8700000 + 90000r$ (a) (ii) similar to the last question, multiply the result from (i) by (1+r/100) and then minus 300000 $(8700000+90000r)\\left(1+\\dfrac{r}{100}\\right)-300000 = 1.026\\times10^7$ $87\\times10^5+9\\times10^4\\cdot r+87\\times10^3\\cdot r+9\\times10^2r^2-3\\times10^5-1026\\times10^4=0$ $9r^2+1770r-18600=0$ $3r^2+590r-6200=0$ $(3r+620)(r-10)=0$ $r=10$ or $r=-\\dfrac{620}{3}$ of course here we reject the negative answer for r since it can’t be negative and hence we have $r=10$ (b) (i) $((9\\times10^6\\cdot1.1-3\\times10^5)(1.1)-3\\times10^5)(1.1)\\ldots-3\\times10^5$ After 1st year, Area $=9\\times 10^6$ After 2nd year, Area $=9\\times10^6(1.1)-3\\times10^5$ After 3rd year, Area $=9\\times10^6(1.1)^2-3\\times10^5(1.1)-3\\times10^5$ After nth year, Area $=9\\times10^6(1.1)^{n-1}-3\\times10^5(1.1^{n-2}+\\ldots+1)$ $=9\\times10^6(1.1)^{n-1}-3\\times10^5\\left(\\dfrac{1.1^{n-1}-1}{1.1-1}\\right)$ $=9\\times10^6(1.1)^{n-1}-3\\times10^6(1.1^{n-1}-1)$ $=9\\times10^6(1.1)^{n-1}-3\\times10^6\\cdot1.1^{n-1}+3\\times10^6$ $=6\\times10^6\\times(1.1)^{n-1}+3\\times10^6$ (b) (ii) $6\\times10^6\\times(1.1)^{n-1}+3\\times10^6 > 4\\times10^7$ $6\\cdot1.1^{n-1}+3>40$ $1.1^{n-1} > \\dfrac{37}{6}$ $n > \\dfrac{\\log\\frac{37}{6}}{\\log 1.1}+1 \\approx 20.08\\ldots$ Since n is an integer therefore $n=21$. At the end of 21st year (c) Setup the system of equations by the given information in the table, $1.21a+b=10^7$ $1.21^2a+b=1.063\\times10^7$ $(1.21^2-1.21)a = 0.063\\times10^7$ $a=\\dfrac{63\\cdot10^4}{1.21(1.21-1)}=\\dfrac{63\\cdot10^4}{1.21\\cdot0.21}=\\dfrac{3\\cdot10^6}{1.21}$ $3\\cdot10^6+b=10^7$ $b=7\\cdot10^6$ Assume the claim is true, which means, $6\\cdot10^6\\cdot1.1^{n-1}+3\\cdot10^6>\\dfrac{3\\cdot10^6}{1.21}\\cdot1.21^n+7\\cdot10^6$ $6\\cdot1.1^{n-1}+3>\\dfrac{3}{1.1^2}\\cdot1.1^{2n}+7$ $0 > 3(1.1^{n-1})^2 - 6(1.1^{n-1}) + 4$ Now we need to show the RHS is", "## Algebra 1 $4 \\times 10^{-5}$ $\\dfrac{a^m}{a^n}=a^{m-n}, a\\ne0$ Divide the corresponding parts while using the rule above to find: $=\\dfrac{3.6}{9} \\times \\dfrac{10^{-10}}{10^{-6}} \\\\=0.4 \\times 10^{-10-(-6)} \\\\=0.4 \\times 10^{-10+6} \\\\=0.4 \\times 10^{-4} \\\\=(4 \\cdot 10^{-1}) \\times 10^{-4} \\\\=4 \\times 10^{-1+(-4)} \\\\=4 \\times 10^{-5}$", "– Dylan May 2 '13 at 5:32 No worries ;-) that is how we all learn. I still lose hours because of what I don't know. – Fixed Point May 2 '13 at 6:01 Hint: Use the Ratio Test. No modification of the expression is needed. - We have $$a_n = \\dfrac{n!}{(2n-1)!!} = \\dfrac{n!}{(2n)!} \\times 2^n n! = \\dfrac{2^n}{\\dbinom{2n}n}$$ Use ratio test now to get that $$\\dfrac{a_{n+1}}{a_n} = \\dfrac{2^{n+1}}{\\dbinom{2n+2}{n+1}} \\cdot \\dfrac{\\dbinom{2n}n}{2^n} = \\dfrac{2(n+1)(n+1)}{(2n+2)(2n+1)} = \\dfrac{n+1}{2n+1}$$ We can also use Stirling. From Stirling, we have $$\\dbinom{2n}n \\sim \\dfrac{4^n}{\\sqrt{\\pi n}}$$ Use this to conclude, about the convergence/divergence of the series. EDIT $$1 \\times 3 \\times 5 \\times \\cdots \\times(2n-1) = \\dfrac{\\left( 1 \\times 3 \\times 5 \\times \\cdots \\times(2n-1) \\right) \\times \\left(2 \\times 4 \\times \\cdots \\times (2n)\\right)}{ \\left(2 \\times 4 \\times \\cdots \\times (2n)\\right)}$$ Now note that $$\\left( 1 \\times 3 \\times 5 \\times \\cdots \\times(2n-1) \\right) \\times \\left(2 \\times 4 \\times \\cdots \\times (2n)\\right) = (2n)!$$ and $$\\left(2 \\times 4 \\times \\cdots \\times (2n)\\right) = 2^n \\left(1 \\times 2 \\times \\cdots \\times n\\right) = 2^n n!$$ Hence, $$1 \\times 3 \\times 5 \\times \\cdots \\times(2n-1) = \\dfrac{(2n)!}{2^n \\cdot n!}$$ - Yikes. It is not at all apparent how $\\frac{1}{(2n-1)!!}$ becomes $\\frac{2^{n}n!}{(2n)!}$ – Dylan May 2 '13 at 5:38 Soon it will be obvious. Put in the \"missing\" terms", "$p$ from the numerator to get that $$\\dbinom{p}{k} = \\dfrac{p \\times (p-1) \\times (p-2) \\times \\cdots \\times (p-k+2) \\times (p-k+1)}{k \\times (k-1) \\times (k-2) \\times \\cdots \\cdots \\times 2 \\times 1} = p \\times M$$ where $M \\in \\mathbb{Z}$.", "2. Expand the factorial notation: $\\dfrac{8!}{4!} = \\dfrac{8 \\times 7 \\times 6 \\times 5 \\times \\cancel{4} \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}}{\\cancel{4} \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}} = 8 \\times 7 \\times 6 \\times 5 = \\text{ RHS}$ 3. Expand the factorial notation:$\\dfrac{n!}{(n-1)!} = \\dfrac{n \\times \\cancel{(n-1)} \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}}{\\cancel{(n-1)} \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}} = n$ If $$n = 1$$, we get $$\\frac{1!}{0!}$$. This is a special case. Both $$1!$$ and $$0! =1$$, therefore $$\\frac{1!}{0!} = 1$$ so our identity still holds. # Don't get left behind Join thousands of learners improving their maths marks online with Siyavula Practice. ## Factorial notation Exercise 10.5 Work out the following without using a calculator: $$3!$$ $3 \\times 2 \\times 1 = \\text{6}$ $$6!$$ $6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1 = \\text{720}$ $$2!3!$$ $2 \\times 1 \\times 3 \\times 2 \\times 1 = \\text{12}$ $$8!$$ $8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1 = \\text{40 320}$ $$\\dfrac{6!}{3!}$$ $\\dfrac{6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}{3 \\times 2 \\times 1} = \\text{120}$ $$6! + 4! - 3!$$ $(6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1) +", "but ignore 2 with 1, 3 with 1, and 3 with 2 because these last 3 are duplicates of the first 3 respectively. $$C(n,r) = \\dfrac{n!}{( r! (n - r)! )}$$ $$C(n,2) = \\dfrac{n!}{( 2! (n - 2)! )}$$ expanding the factorials, $$= \\dfrac{1\\times2\\times3...\\times(n-2)\\times(n-1)\\times(n)}{( 2\\times1\\times(1\\times2\\times3...\\times(n-2)) )}$$ cancelling and simplifying, $$= \\dfrac{(n-1)\\times(n)}{2} = \\dfrac{n(n-1)}{2}$$ which is the same as the equation above. ## References [1] Zwillinger, Daniel (Editor-in-Chief). CRC Standard Mathematical Tables and Formulae, 31st Edition New York, NY: CRC Press, p. 206, 2003. Cite this content, page or calculator as: Furey, Edward \"Combinations Calculator (nCr)\"; CalculatorSoup, https://www.calculatorsoup.com - Online Calculators", "# Summing The Series Algebra Level 4 $\\dfrac{4\\times{3}\\times{2}\\times{1}}{3!+2!+1!}$+$\\dfrac{5\\times{4}\\times{3}\\times{2}}{4!+3!+2!}$+.......+$\\dfrac{2015\\times{2014}\\times{2013}\\times{2012}}{2014!+2013!+2012!}$= $a$ Find $\\lfloor{a}\\rfloor$ $\\lfloor...\\rfloor$ is the greatest integer less than equal to [$...$] ×", "All possible values of $x$ If $\\dfrac{13!}{2^x}$ is an integer, which of the following represents all possible values of $x?$ (where $x$ is a non-negative integer). A. $0 \\le x \\le 10$. B. $0 \\le x \\le 9$. C. $0 \\le x < 10$. D. $1 \\le x \\le 10$. E. $1 < x <10$. Notation: $!$ is the factorial notation. For example, $8! = 1\\times2\\times3\\times\\cdots\\times8$. ×", "\\dfrac{x}{2} + \\dfrac{3}{8}x^{2} - \\cdots+(-1)^{n}\\dfrac{1\\times 3\\times\\dotsm\\times(2n-1)}{2\\times 4\\times\\dotsm\\times 2n} x^{n} + \\mathcal{O}\\left(x^{n+1}\\right) \\end{array} \\end{document}", "& 6 & ^{1}4 & 9 \\\\ \\times & & & & 2 & 7 \\\\ \\hline & & 4 & 5 & 4 & 3 \\\\ & 1 & 2 & 9 & 8 & \\end{array}$ $\\begin{array}{rccccc} & & & ^{3}6 & ^{6}4 & 9 \\\\ \\times & & & & 2 & 7 \\\\ \\hline & & ^{1}4 & ^{1}5 & 4 & 3 \\\\ + & 1 & 2 & 9 & 8 & \\\\ \\hline & 1 & 7 & 5 & 2 & 3 \\end{array}$ The answer is $$27 \\times 649 = 17\\ 523$$. ## Worked Example 1.14: Multiplying whole numbers Calculate without using a calculator: $3\\ 487 \\times 93$ ### Write the two numbers in column, start with the bigger number. $\\begin{array}{rcccc} & 3 & 4 & 8 & 7 \\\\ \\times & & & 9 & 3 \\end{array}$ ### Multiply by the units of the second number first. $\\begin{array}{rccccc} & & ^{1}3 & ^{2}4 & ^{2}8 & 7 \\\\ \\times & & & & 9 & 3 \\\\ \\hline & 1 & 0 & 4 & 6 & 1 \\end{array}$ ### Multiply by the tens of the second number. $\\begin{array}{rcccccc} & & & ^{4}3 & ^{7}4 & ^{6}8 & 7 \\\\ \\times & & & & & 9 & 3 \\\\", "\\times 10^6) \\cdot (8.7 \\times 10^{11}) = 2.784 \\times 10^{18}$ (b) $(5.2 \\times 10^{-4}) \\cdot (3.8 \\times 10^{-19}) & = \\underbrace{ 5.2 \\times 3.8 }_{19.76} \\times \\underbrace{ 10^{-4} \\times 10^{-19} }_{10^{-23}} \\\\& = 1.976 \\times 10^1 \\times 10^{-23}$ Solution $(5.2\\times 10^{-4}) \\cdot (3.8 \\times 10^{-19}) = 1.976\\times 10^{-22}$ (c) $(1.7 \\times 10^6) \\cdot (2.7 \\times 10^{-11}) = \\underbrace { 1.7 \\times 2.7}_{4.59} \\times \\underbrace {10^6 \\times 10^{-11}}_{10^{-5}}$ Solution $(1.7 \\times 10^6) \\cdot (2.7 \\times 10^{-11}) = 4.59 \\times 10^{-5}$ Example 6 Evaluate the following expressions. Round to 3 significant figures and write your answer in scientific notation. (a) $(3.2 \\times 10^6) \\div (8.7 \\times 10^{11})$ (b) $(5.2 \\times 10^{-4}) \\div (3.8 \\times 10^{-19})$ (c) $(1.7 \\times 10^6) \\div (2.7 \\times 10^{-11})$ It will be easier if we convert to fractions and THEN separate out the powers of 10. (a) $(3.2 \\times 10^6) \\div (8.7 \\times 10^{11}) & = \\frac{3.2 \\times 10^6} {8.7 \\times 10^{11}} && \\text{Next we separate the powers of}\\ 10.\\\\& = \\frac{3.2} {8.7} \\times \\frac{10^6} {10^{11}} && \\text{Evaluate each fraction (round to}\\ 3\\ \\text{s.f.):}\\\\& = 0.368 \\times 10{(6 - 11)} && \\text{Remember how to write scientific notation!}\\\\& = 3.68 \\times 10^{-1} \\times 10^{-5}$ Solution $(3.2 \\times 10^6) \\div (8.7 \\times 10^{11}) = 3.86 \\times 10^{-6}$ (rounded to 3 significant figures) (b) $(5.2 \\times 10^{-4}) \\div (3.8", "### Factorials #### Definition A factorial of a positive number $n$ is defined to be the product of all positive whole numbers less than and equal to $n$. E.g. $n=5$ $5! = 5 \\times 4 \\times 3 \\times 2 \\times 1.$ Note: We define the factorial $0!=1$. #### Worked Examples ###### Example 1 Calculate $7!$. ###### Solution From the definition we have $7! = 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1.$ Hence $7! =5040.$ ###### Example 2 Simplify $\\dfrac{9!}{6!}$. ###### Solution Firstly we can expand the factorials using the definition $\\dfrac{9!}{6!} = \\dfrac{9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}{6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}.$ This can then be simplified by canceling terms on the top and bottom to give $\\dfrac{9!}{6!} = 9 \\times 8 \\times 7.$ Therefore, $\\dfrac{9!}{6!} = 504.$ ###### Example 3 Simplify $\\dfrac{16!}{13!\\cdot 8!}$ ###### Solution As above start by expanding the factorials $\\dfrac{16!}{13!\\cdot 8!} = \\dfrac{16 \\times 15 \\times \\dots \\times 3 \\times 2 \\times 1}{(13 \\times 12 \\times 11 \\times \\dots \\times 3 \\times 2 \\times 1 )( 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1)}.$ Then cancel out any terms that appear on the top", "# JEE Main 2016 (22) Algebra Level 5 $\\dfrac{25!}{x(x+1)(x+2)\\cdots(x+25)}=\\sum_{i=0}^{25}\\dfrac{A_{i}}{x+i}$ The equation above holds true for constants $$A_1, A_2, \\ldots , A_{25}$$. Find the sum of digits of the number $$A_{24}$$. Notation: $$!$$ denotes the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "can be reworked to $I = \\dfrac{f \\cdot T_{CPU}}{CPI}$ Try: - $100,000 \\times \\dfrac{750\\times 1.3}{600\\times 2.5}$ = 65,000", "This is not what you think! Algebra Level 4 $\\large\\dfrac{3}{1!+2!+3!} + \\dfrac{4}{2!+3!+4!} +\\cdots+ \\dfrac{2016}{2014!+2015!+2016!}$ If the above expression can be represented as$\\dfrac{1}{k!} - \\dfrac{1}{(k+a)!} \\; ,$ then find $$\\dfrac{a}{k}$$. Notation: $$!$$ denotes the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "+0 # somebody help 0 159 1 +36 Simplify the following expression as far as possible without using a calculator:5/6÷3/6×3/15? Siphe Aug 15, 2017 Sort: $$\\dfrac{5}{6}\\div \\dfrac{3}{6}\\times \\dfrac{3}{15}\\\\ =\\dfrac{5}{6}\\div \\dfrac{1}{2}\\times \\dfrac{1}{5}\\\\ =\\dfrac{5}{6}\\times 2 \\times \\dfrac{1}{5}\\\\ =\\dfrac{5\\times 2}{6\\times 5}\\\\ =\\dfrac{10}{30}\\\\ =\\dfrac{1}{3}$$", "\\to \\infty} {n! \\over \\sqrt{2\\pi n} (n/e)^n} = 1,$$ if you rewrite the product as $(2n)!/(2^n n!)^2$. - Thanks to you two! Dear Michael s! – Bidgoli Mar 12 at 20:08 Here you need to find $\\lim A_n$, where $$A_n= \\frac{1\\times3\\times5\\times\\cdots\\times(2n-1)}{2\\times4\\times6\\times8\\times\\cdots\\times (2n)}.$$ Now note that, for $n\\gt1$, $2n-1\\gt n$. So, in the numerator, you get $$\\frac{1\\times 2\\times 3\\dots\\times n}{2\\times4\\times 6\\times\\dots\\times 2n}\\lt\\frac{1\\times3\\times5\\times\\cdots\\times(2n-1)}{2\\times4\\times6\\times8\\times\\cdots\\times (2n)}\\\\\\implies\\frac1{2^n}\\lt \\frac{1\\times3\\times5\\times\\cdots\\times(2n-1)}{2\\times4\\times6\\times8\\times\\cdots\\times (2n)}$$ Now, you use the identity $$\\frac n{n+1}\\lt\\frac {n+1}{n+2}$$ and define a sequence $$B_n=\\frac 23\\times \\frac 45\\times\\dots \\times \\frac {2n}{2n+1} .$$ So, now you have $$A_n\\lt B_n\\\\\\implies (A_n)^2\\lt A_nB_n=\\frac 1{2n+1}\\\\\\implies A_n\\lt \\frac 1{\\sqrt{2n+1}}.$$ So, you have $$\\frac 1{2^n}\\lt A_n\\lt \\frac 1{\\sqrt{2n+1}}\\\\\\implies\\lim \\frac 1{2^n}\\lt\\lim A_n\\lt\\lim \\frac 1{\\sqrt{2n+1}}.$$ So, by sandwitch theorem, $$\\lim A_n=0.$$ - What's wrong with this? – Subhadeep Dey Mar 11 at 19:06 Dear downvoter, are you going to explain the reason for downvoting? – Subhadeep Dey Mar 11 at 19:18 I didn't downvote, but the first half of the proof is unnecessary. For the sandwich theorem, all you need is $0 < A_n < 1/\\sqrt{2n+1}$, and $0 < A_n$ is obvious. – alephzero Mar 11 at 23:51 @alephzero, yes, I know. I have just showed that there are several ways to get the lower sequence. – Subhadeep Dey Mar 12 at 7:49 JUST BRILLIANT... – Bidgoli Mar 12 at 20:06 To paraphrase Wikipedia:", "Thinking Mathematically (6th Edition) $n!=n(n-1)(n-2)\\cdots(3)(2)(1)$ and, by definition, $0!=1$ ----------------- Most calculators can display up to $69!\\approx 10^{98}$. Anything over results in an error. 900! and 899! are huge numbers. We employ a different approach: $900!=900\\times 899\\times 898\\times...\\times 2\\times 1$ $=900\\times 899!$ So, we write $\\displaystyle \\frac{900!}{899!}=\\frac{900\\times 899!}{899!}$ ... reduce the fraction by $899!$... $=900$" ]

[ "digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30) #2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B) #3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer? #4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer? #5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer? #1: 66660 #2: $$\\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8$$ 4.8 * 11111 =$$\\color{red}{53332.8}$$ #3: 133320 #4: 1333272 #5: 93324 Sunday, March 6, 2016 2016 Mathcounts State Prep : How to Avoid Making Careless Mistakes Why Can Some Kids Handle Pressure While Others Fall Apart? from the New York Times Relax! You'll Be", "if each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30) #2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B) #3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer? #4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer? #5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer? #2: $$\\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8$$ 4.8 * 11111 =$$\\color{red}{53332.8}$$", "# How many consecutive integers can you make using only four digits? In the spirit of the classic four fours, I wonder what's the optimal set of four numbers? Your goal is to make the most consecutive integers using four digits of your choice. Pick four: $0,1,2,3,4,5,6,7,8,9$ ( You can pick multiple instances of the same digit ) When constructing an integer: • all of your four candidates must be used exactly once (order/placing of digits is irrelevant) • You may use basic arithmetic operations $+,-,\\times,\\div$ and parentheses $()$ • You may use $a^b$ and $\\sqrt[a]{b}$ but at the expense of 2 numbers as you can see • You may not form new numbers, i.e. $ab$ is not allowed If we were to use four $4$s, the best we could do would be up to $9$: 0 = 4 ÷ 4 × 4 − 4 1 = 4 ÷ 4 + 4 − 4 2 = 4 −(4 + 4)÷ 4 3 = (4 × 4 − 4)÷ 4 4 = 4 + 4 ×(4 − 4) 5 = (4 × 4 + 4)÷ 4 6 = (4 + 4)÷ 4 + 4 7 = 4 + 4 − 4 ÷ 4 8 = 4 ÷ 4 × 4 + 4 9 = 4 ÷ 4 + 4", "+0 # Base Stuff +1 66 1 +770 How many two digit positive integers in the base-four representation are twice the two digit positive integers in the base-three representation? A) 5 B) 6 C) 7 D) 8 E) 9 The answer is apparently A) 5, but I am not sure how to get to it. Any help is appreciated. Thanks! - PM Jan 26, 2020 edited by PartialMathematician Jan 26, 2020 #1 +24133 +3 How many two digit positive integers in the base-four representation are twice the two digit positive integers in the base-three representation? I assume: $$\\begin{array}{|rl|c|c|c|c|c|c|l|} \\hline & & 10 & 11 & 12 & 20 & 21 & 22 & \\text{base}~3 \\\\ & & =3 & =4 & =5 & =6 & =7 & =8 & \\text{base}~10 \\\\ \\text{base}~4 & \\text{base}~10 & 6 & 8 & 10 & 12 & 14 & 16 & 2x \\\\ \\hline 10 & = 4 & & & & & & & \\\\ 11 & = 5 & & & & & & & \\\\ 12 & = 6 &6=6\\checkmark& & & & & & \\\\ 13 & = 7 & & & & & & & \\\\ 20 & = 8 & &8=8\\checkmark& & & & & \\\\ 21 & = 9 & & & & & &", "four-digit positive integer? (2003 Mathcounts Sprint #30) #2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B) #3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer? #4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer? #5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer? #2: $$\\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8$$ 4.8 * 11111 =$$\\color{red}{53332.8}$$", "+0 # Help! 0 326 1 1. The sequence 2, 3, 5, 6, 7, 10, 11, $\\ldots$ contains all the positive integers from least to greatest that are neither squares nor cubes. What is the $400^{\\mathrm{th}}$ term of the sequence? 2. The first digit of a string of 2002 digits is a 1. Any two-digit number formed by consecutive digits within this string is divisible by 19 or 31. What is the largest possible last digit in this string? Jun 3, 2018", "mersenneforum.org Classic problem: Four 4's Register FAQ Search Today's Posts Mark Forums Read 2005-09-28, 16:45 #1 Ken_g6 Jan 2005 Caught in a sieve 18B16 Posts Classic problem: Four 4's I found a version of the four 4's problem in an old book awhile back. Four those who haven't seen it befour, the challenge is to create fourmulas, using no more than four 4's, that evaluate to other integers. This version is more stringent on the operations allowed than most. The operations allowed are listed below. The book claims that all positive integers up to 119 can be created this way. I have been unable to create three of those integers, so I'd like to see what you can come up with. If the book is correct, though, all integers from 1 through at least 130 can be created. Allowed unary operations: (each example in parentheses uses one 4) Square (4^2) - this is the only place where 2's are allowed. Also note below that exponentiation other than this is not allowed. Factorial (4!) Decimal (.4) Allowed binary operations: (each example in parentheses uses two 4's) + (4+4) - (4-4) * (4*4) / (4/4) Concatenate (44); also before (4.4) or after (.44) a Decimal. Any number of parentheses are also allowed. Standard order of operations applies when parentheses are", "# All I see is fours Logic Level 2 $\\large{\\begin{eqnarray} 1 &=& 44 \\div 44 \\\\ 2 &=& 4 \\times 4 \\div (4 + 4 ) \\\\ 3 &=& (4 + 4 + 4) \\div 4 \\\\ 4 &=& 4 + (4\\times(4-4)) \\\\ 5 &=& (4 + (4\\times4)) \\div 4 \\end{eqnarray}}$ Above shows the first 5 positive integers formed by using the four mathematical operators ($$+ \\ - \\ \\times \\ \\div$$) only on the digit 4 four times. What is the smallest positive integer that cannot be represented using these conditions? Note: You are allowed to join the digits together: $$44 + 44$$. ×", "odd numbers from the set $\\left\\{ { - 8,{\\text{ }} - 7,\\, - 6,{\\text{ }} - 5,\\; - 4,\\; - 3,{\\text{ }} - 2,\\; - 1,\\,0,{\\text{ }}1,\\,2} \\right\\}$ can be derived by finding the numbers which leave a remainder $1$ when it is divided by $2$. They are $- 7,{\\text{ }} - 5,\\; - 3,\\, - 1,\\;1$. Therefore all the integers between -9 and 3 which are odd are $- 7,{\\text{ }} - 5,\\; - 3,\\, - 1,\\;1$. Note: Whole numbers: Whole numbers are simply the numbers in which zero is included $0,{\\text{ }}1,{\\text{ }}2,{\\text{ }}3,\\;{\\text{ }}4,\\,{\\text{ }}5,{\\text{ }}6,{\\text{ }} \\ldots ..$ Integers: Integers are like whole numbers, but they also include negative numbers but still no fractions allowed. So integers can be negative $\\left\\{ { - 1,{\\text{ }} - 2,{\\text{ }} - 3,\\; \\ldots \\ldots } \\right\\}$, positive $\\left\\{ {1,\\,{\\text{ }}2,\\,{\\text{ }}3,{\\text{ }}4,{\\text{ }} \\ldots \\ldots } \\right\\}$ or zero $\\left\\{ 0 \\right\\}$.", "# Sum of Some Primes The four values $$x$$, $$y$$, $$x - y$$ and $$x + y$$ are all positive prime integers. What is the sum of all the four integers? Details and assumptions Fact: 1 is not a prime number. ×", "each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30) #2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B) #3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer? #4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer? #5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer? #1: 66660 #2: $\\frac{1+ 3 + 5 + 7 + 8}{5} = 4.8 * 11111 = \\color{red}53332.8$ #3: 133320 #4: 1333272 #5: 93324 ## Friday, August 22, 2014 ### Learn How to Learn by BOGTRO from AoPS forum -- Thanks a bunch !! I love the following quotes : Insanity: doing the same thing over and", "Order the following numbers from greatest to least. $0.155$ $1.5\\times10^5$ $15{\\%}$ $\\frac{1}{5}$ Greatest Least", "In four-digit integer numbers from $1001$ to $9999$, the digit group $\\text{“37”}$ (in the same 1. $270$ 2. $279$ 3. $280$ 4. $299$", "# Each Digit Once in Powers of N $N$ is an integer. The numbers $N^3$ and $N^4$ together consist of all of the digits $0,1,2,3,4,5,6,7,8,9$ exactly once. What is the value of $N^2$? Details and assumptions The number $12 = 012$ does not contain the digit 0. Each digit (0 to 9) appears in either $N^3$ or $N^4$ exactly once, and doesn't appear in both of them. For example, if $N=2$, then $N^3$ and $N^4$ together consist of the digits $1, 6, 8$ exactly once. ×", "Eva lists the positive integers, but skips all numbers containing a digit divisible by $$3$$. For example, the first ten numbers she lists are $$1, 2, 4, 5, 7, 8, 11, 12, 14, 15$$. What is the $$100^{\\text{th}}$$ number in her list? This problem is shared by Muhammad A. Details and assumptions The number $$13$$ is not included in the list, since it contains the digit 3, which is divisible by 3. ×", "# All I see is fours Logic Level 2 \\large{\\begin{aligned} 1 &=& 44 \\div 44 \\\\ 2 &=& 4 \\times 4 \\div (4 + 4 ) \\\\ 3 &=& (4 + 4 + 4) \\div 4 \\\\ 4 &=& 4 + (4\\times(4-4)) \\\\ 5 &=& (4 + (4\\times4)) \\div 4 \\end{aligned}} Above shows the first 5 positive integers formed by using the four mathematical operators ($+ \\ - \\ \\times \\ \\div$) only on the digit 4 four times. What is the smallest positive integer that cannot be represented using these conditions? Note: You are allowed to join the digits together: $44 + 44$. ×", "with 0; xyz1^(positive integer) ends with 1; xyz5^(positive integer) ends with 5; xyz6^(positive integer) ends with 6; • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4; For example last digit of $$xyz3^{positive \\ integer}$$ repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... So cyclicity of the last digit of 3 in power is 4. • Integers ending with 4 and 9 have a cyclicity of 2. Now, last digit of $$xyz3^{positive \\ integer}$$ repeats in blocks of 4: {3, 9, 7, 1} - {3, 9, 7, 1} - ... means that: 3^1, 3^5, 3^9, 3^13, ..., 3^(4x+1), all will have the same last digit as 3^1 so 1; 3^2, 3^6, 3^10, 3^14, ..., 3^(4x+2), all will have the same last digit as 3^2 so 9; 3^3, 3^7, 3^11, 3^15, ..., 3^(4x+3), all will have the same last digit as 3^3 so 7; 3^4, 3^8, 3^12, 3^16, ..., 3^(4x), all will have the same last digit as 3^4 so 1; So to get the last digit of 3^x, (where x is a positive integer) you should divide x by 4 (cylcility) and look at the remainder: If remainder is 1 then the last digit will be the same as for 3^1 (so the first digit from", "least two states. I think we were stopped at 37 or thereabouts. Above, I used only the four basic operations. But they aren’t enough to get every numbers you’d like to get. Sooner or later the question arises, what operations are allowed? The next operations you might want to use (and within a group you get to decide on the restrictions) might be exponents ($$4^4$$, which in type we write as 4^4) and square roots ($$\\sqrt{4}$$). Often we allow concatenation of digits and decimal points, like $$44$$, $$.4$$, $$4.4$$. Doctor Chita leaps over those and suggests a “cheat”: I found that you could generate more of the more \"difficult\" numbers by knowing some specialized math functions. To my family's dismay and annoyance, I used the greatest integer function. That is, the greatest integer value of an integer is the integer itself. The greatest integer value of a decimal number is the greatest integer immediately to the left of the decimal number. For example: a. [3] = 3 b. [2.6] = 2 c. [-5.4] = -6 d. [-4] = -4 This function obviously helps to truncate a number. You could also introduce the rounding function, using a rule about lopping off decimals less than 0.5, and rounding up for decimals greater than or equal to 0.5. An example using", "# How do you write the verbal phrase given: n + 6? Having said that, $n$ usually denotes a positive integer: $\\left\\{1 , 2 , 3 , 4. . .\\right\\}$ But in this case it is not specified, and it could be $1 , - 1 , .5 , \\setminus \\pi$ among others.", "# From strict to non-strict Algebra Level 4 Let $$a$$ and $$b$$ be positive integers such that $$\\frac{ab}{a + b} > 2017.$$ Let $$C$$ be the greatest possible real number such that $$\\frac{ab}{a + b} \\geq C.$$ If $$C$$ can be written in the form $$\\frac{p}{q},$$ where $$p$$ and $$q$$ are coprime positive integers, find the last three digits of $$q.$$ ×" ]

[ "If we try $222$, $333$, and so on up to $888$, each of those attempts rule out $22$ possibilities as well. After these, we've ruled out $22 \\cdot 8 = 176$ possibilities. Which possibilities are left? Exactly those with no repeated digits. If we try $123$, we have an overlap with $121$, $122$, $113$, $133$, $223$, and $323$ being doubly ruled out, but we still rule out $22-6=16$ additional possibilities. We can get another $16$ ruled out by each of $456$, $781$, $234$, $567$, $812$, $345$, and $678$. Now we have a total of $512 - 22 \\cdot 8 - 16 \\cdot 8 = 208$ possibilities left to rule out. These are exactly the combinations $\\text{A}\\text{B}\\text{C}$ in which $\\text{B} - \\text{A} \\not\\equiv 0, 1 \\pmod{8}$ and $\\text{C} - \\text{B} \\not\\equiv 0, 1 \\pmod{8}$. If we try $135$, we have an overlap with $133$, $155$, $535$, $335$, $115$, $131$, $145$, $235$, and $125$ but we still rule out $22-9=13$ additional possibilities. Similarly, we try $246$, $357$, $468$, $571$, $682$, $713$, and $824$, each ruling out $13$. Now we have $512 - 22 \\cdot 8 - 16 \\cdot 8 - 13 \\cdot 8 = 104$ possibilities left, with $24$ attempted codes. If we try $147$, we have an overlap with $144$, $177$, $117$, $447$, $141$, $747$, $145$, $167$, $127$, $137$,", "\\pmod{5}$$. $$\\bullet$$ Case 2a) $$x \\equiv 2 \\pmod{4}$$ and $$x \\equiv 2 \\pmod{5} \\implies x \\equiv 2 \\pmod{20} \\implies x \\in \\{2,22,42,62,82\\}$$. $$\\bullet$$ Case 2b) $$x \\equiv 2 \\pmod{4}$$ and $$x \\equiv 3 \\pmod{5} \\implies x \\equiv 18 \\pmod{20} \\implies x \\in \\{18,38,58,78,98\\}$$. So, $$x \\in \\{ 2,4,16,18,22,24,36,38,42,44,56,58,62,64,76,78,82,84,96,98 \\}$$, which are indeed 20 values. · 4 years ago", "converting 7.478478.... into \\dfrac{q}{p} form, we get x = 7.478478.... ... (a) \\\\ 1000x = 7478.478478.... ... (b) \\\\ While, subtracting (a) from (b), we get \\\\ 999x = 7471\\ x \\dfrac{ 7471}{ 999} \\\\ Therefore, 7.478478.... is a rational number. \\\\ (v) 1.101001000100001.... \\\\ We can observe that the number 1.101001000100001.... is a non-terminating non- recurring decimal. Thus, non-terminating and non-recurring decimals cannot be converted into \\dfrac{q}{p} form.\\\\ Therefore, we conclude that 1.101001000100001.... is an irrational number. 17 Visualize 3.765 on the number line using successive magnification. ##### Solution : \\bullet Clearly, the value lies between 3 and 4 \\\\ \\bullet The numbers 3.7 and 3.8 lie between 3 and 4. \\\\ \\bullet The numbers 3.76 and 3.77 lie between 3.7 and 3.8. \\\\ \\bullet The numbers 3.764 and 3.766 lie between 3.76 and 3.77 \\\\ \\bullet The number 3.765 lies between 3.764 and 3.766. \\\\ Therefore, we need to use the successive magnification, after locating numbers 3 and 4 on the number line. 18 Visualize 4.\\bar{26} on the number line, up to 4 decimal places. ##### Solution : The number 4.\\bar{26} can also be written as 4.262.... .$$\\\\$ The number $4.2$ lie between $4$ and $5$ The numbers $4.26$ lie between $4.2$ and $4.3$ The numbers $4.262$ lie between $4.261$ and $4.263$ The number $4.2626$ lie", "even the normally imperturbable Humphreys gives it an exclamation mark. By the way, did you notice something funny about these numbers: $$1, 7, 11, 13, 17, 19, 23, 29 ?$$ Except for 1, they're all prime! 30 just happens to be the largest number for which every number besides 1 that's smaller than it and relatively prime to it is actually prime. I don't know if this is important here, but it's darn suspicious. For $$\\mathrm{E}_7$$ we have to work harder. The dimension of the Lie algebra $$\\mathrm{E}_7$$ is 133, so the number of roots is $$133 - 7 = 126$$, and the Coxeter number is $$126/7 = 18$$. Here are the numbers smaller than 18 that are relatively prime to 18: $$1, 5, 7, 11, 13, 17$$ Adding one, we get magic numbers for $$\\mathrm{E}_7$$: $$2, 6, 8, 12, 14, 18$$ Unfortunately we only get six of the seven magic numbers this way. Luckily, Humphreys also proved that the sequence of magic numbers must be 'symmetrical' as described in Puzzle 5. The only possibility is that the seventh magic number is 10: $$\\begin{array}{ccccccccccccccccc} 2 &&&& 6 &&8 && \\mathbf{10} && 12 &&14 &&&& 18 \\\\ \\bullet &\\circ &\\circ &\\circ & \\bullet &\\circ &\\bullet &\\circ &\\bullet &\\circ &\\bullet &\\circ &\\bullet &\\circ&\\circ &\\circ& \\bullet \\end{array}$$ By the way, did", "– Neal Oct 11 '18 at 19:11 • You meant the total number of possibilities is $6!$, not the total probability. – N. F. Taussig Oct 11 '18 at 19:25 • @N.F.Taussig Could you be more specific? I cannot properly obtain where the issue is. – Mark Oct 11 '18 at 19:26 All the ways to seat all of the people $$6!$$ At least $$2$$ girls together. We attach 2 girls, and allocate $$5$$ objects. Making for $$5!$$ allocations $$5!$$ But we can swap the order of the girls. And we have 3 girls who we can make the \"odd-man-out\" $$5!\\cdot 6 = 6!$$ But we have double counted all of the cases where we have all 3 girls together. $$4!3!$$ $$6! - (6! - 4!3!) = 4!3! = 144$$ You don't state it, but I'm assuming the boys are indistinguishable and the girls are indistinguishable. Please clarify this point! Consider three girls. There must be at least one boy separating pairs: $$GBGBG$$. The only question then remains where the last boy can go: $$\\bullet G \\bullet B \\bullet G \\bullet B \\bullet G \\bullet$$. Note that some of these cases are identical, so you can eliminate them How many distinguishable places all told? • I dont know how to calculate that. And is my assumption wrong? –", "### Av(1234, 1342, 1432, 2341, 2413, 3142, 3241, 4123, 4213) #### Permutation examples length 2: 12, 21 length 3: 123, 132, 213, 231, 312, 321 length 4: 1243, 1324, 1423, 2134, 2143, 2314, 2431, 3124, 3214, 3412, 3421, 4132, 4231, 4312, 4321 length 5: 13254, 21354, 21435, 21534, 23154, 24315, 31254, 32145, 32154, 34125, 34215, 35412, 35421, 41325, 42315, 43125, 43215, 45132, 45231, 45312, 45321, 52431, 53412, 53421, 54132, 54231, 54312, 54321 #### Conjectured Struct Cover $\\mathcal{A}$ $=$ $\\bullet$ $\\bullet$ $\\bullet$ \\ $\\bullet$ $\\bullet$ $\\bigsqcup$ $\\bullet$ $\\bullet$ $\\bullet$ $\\bullet$ $\\mathcal{B}$ $\\bigsqcup$ $\\bullet$ $\\bullet$ \\ $\\bullet$ $\\bullet$ $\\bigsqcup$ $\\bullet$ $\\bullet$ $\\bullet$ \\ $\\bullet$ $\\bullet$ $\\bigsqcup$ $\\bullet$ $\\bullet$ $\\bullet$ $\\bullet$ $\\bigsqcup$ $\\mathcal{B}$ $\\bullet$ $\\bullet$ $\\bullet$ $\\mathcal{C}$ $\\bigsqcup$ $\\bullet$ $\\mathcal{B}$ $\\bullet$ $\\bullet$ $\\bullet$ $\\mathcal{C}$ $\\bigsqcup$ $\\bullet$ $\\bullet$ $\\bullet$ $\\bullet$ \\ $\\bullet$ $\\bullet$ $\\bigsqcup$ $\\bullet$ $\\bullet$ \\ $\\bullet$ $\\bullet$ $\\bullet$ $\\mathcal{B}$ $\\bigsqcup$ $\\mathcal{D}$ #### Recurrence relation $a_{0} = 1,~a_{1} = 1,~a_{2} = 2,~a_{3} = 6,~a_{4} = 15,~a_{5} = 28,~a_{6} = 48$ $a_{n} = b_{n-4}+\\sum_{i=0}^{n-3}b_{i}c_{n-i-3}+\\sum_{i=0}^{n-4}b_{i}c_{n-i-4}+\\sum_{i=0}^{n-5}2b_{n-i-5}+d_{n}+5$", "+0 How many distinct numbers can be made (casework) +1 81 2 +59 An expression is formed using the numbers 7, 16, 25, and 27 according to the following rules. $\\bullet$ Each of the four numbers is used exactly once. $\\bullet$ The four numbers may be used in any order. $\\bullet$ Exactly three operations are used; each one is either $+$ or $\\times$. $\\bullet$ An unlimited number of parentheses may be used. No two distinct expressions have the same simplified value. The two expressions below are not distinct, and therefore must be counted as only one value. What is the greatest number of distinct values, including the one below, that can be obtained when building expressions following these rules? $(7+16+27) \\times 25 = 25 \\times (27+7+16)$ Jul 27, 2020 #1 -1 There are 7 possible distinct values. Jul 27, 2020 #2 +59 +2 I can already think of 8, and I know there's much more Williamjwu8 Jul 27, 2020", "of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$ Together, the answer is $3(12+12+12)-12=\\boxed{\\textbf{(A) } 96}.$ Solution 4 Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \\equiv 1\\pmod{3},$ and thus we need to count any $2$-digit number $\\equiv 2\\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \\boxed{\\textbf{(A) } 96}.$ Solution 5 We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\\cdot\\frac{1}{3}=150.$ Of these $150$ numbers, $\\frac{4}{5}$ $\\textbf{do not}$ have $3$ in their ones (units) digit, $\\frac{9}{10}$ $\\textbf{do not}$ have $3$ in their tens digit, and $\\frac{8}{9}$ $\\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$900\\cdot\\frac{1}{2}\\cdot\\frac{1}{3}\\cdot\\frac{4}{5}\\cdot\\frac{9}{10}\\cdot\\frac{8}{9}=\\boxed{\\textbf{(A) } 96}.$$ ~mathpro12345 Solution 6 We will start with the numbers that could work. This numbers include _ _ $1$, _ _ $5$, _ _ $7$, _", "zero doesn't introduce new solutions (case of Fermat's Last Thm.), allowing negative cubes does. For example $9^3 + (-1)^3 = 12^3 + (-10)^3$ is smaller than $1729$. – hardmath Jan 31 '13 at 14:36 +1 @hardmath . Noone knows this. – vikiiii Jan 31 '13 at 14:47 show 1 more comment There are a couple of code snippets there for you to work with. The sequence continues as follows: $1729 = (1^3 + 12^3)$ or $(9^3 + 10^3)$ $4104 = (2^3 + 16^3)$ or $(9^3 + 15^3)$ $13832 = (2^3 + 24^3)$ or $(18^3 + 20^3)$ $20683 = (10^3 + 27^3)$ or $(19^3 + 24^3)$ $32832 = (4^3 + 32^3)$ or $(18^3 + 30^3)$ $39312 = (2^3 + 34^3)$ or $(15^3 + 33^3)$ $40033 = (9^3 + 34^3)$ or $(16^3 + 33^3)$ $46683 = (3^3 + 36^3)$ or $(27^3 + 30^3)$ $64232 = (17^3 + 39^3)$ or $(26^3 + 36^3)$ $65728 = (12^3 + 40^3)$ or $(31^3 + 33^3)$ After that you have: $110656, 110808, 134379, 149389, 165464, 171288, 195841, 216027, 216125, 262656$, etc. You might also be interested in exploring: $\\bullet$ Diophantine Equation--3rd Powers $\\bullet$ Cubic Numbers $\\bullet$ Taxi Numbers in JavaScript Regards - Almost at a \"nice answer\" badge...+1 nudge closer ;-) – amWhy May 5 '13 at 0:20 The bruteforce approach is simple. Loop over", "knowing the room number and knowing the conversation, we cannot have two possible candidates $36$ and $48$ for the product. Hence $S\\ge13$ indeed is impossible, and the claim is proved. Claim 2: $S\\le11$ is impossible. Proof: Then for every choice of $n$ and $P$, there exists at most one list of $n$ positive integers $x_1,\\ldots,x_n$ with sum $S$ and product $P$. I have verified this statement by a computer program: I have generated all lists with sum $\\le11$, and it turns out that no two different such lists agree simultaneously in $n$ and in $P$ and in $S$. This collides with the statement \"No\" that the member makes after scribbling for some time. Claim 3: $S=12$ is possible. Proof: Consider the following lists: • The numbers $3$, $4$, $4$, $1$. • The numbers $2$, $2$, $2$, $6$. Both lists contain four numbers with sum $S=12$ and product $P=48$. My computer program shows that for $S=12$, these are the only two lists that simultaneously agree in $n$ and in $P$. • Thanks. I found the same solution, but not a proof of uniqueness. – Colonel Panic Apr 1 '16 at 14:22 • @Gamow why 6,8,8,2 and 4,4,4,12 are not okay? do I miss something? S would be 24, and product will be the same 768 – Oray Apr 1", "Questions & Answers Question Answers # The number of integer’s n with $100 \\leqslant n \\leqslant 999$and containing at most two distinct digits is:${\\text{a}}{\\text{. }}252 \\\\ {\\text{b}}{\\text{. }}280 \\\\ {\\text{c}}{\\text{. }}324 \\\\ {\\text{d}}{\\text{. }}360 \\\\$ Answer Verified 151.5k+ views Hint: - Calculate total numbers having distinct digits then subtract this from total numbers. Total three digit numbers between $100 \\leqslant n \\leqslant 999$ $999 - 99 = 900$, because 999 and 100 are included. Now, a three digit number is to be formed from the digits $0,1,2,3,4,5,6,7,8,9$ $\\bullet \\bullet \\bullet$ Since the left most place i.e. hundred’s place cannot have zero. So, there are 9 ways to fill hundred’s place. Since, we consider the number with distinct digits, therefore repletion is not allowed, so, ten’s place can be filled by 9 remaining ways. So, ten’s place can be filled in 9 ways. Similarly, to fill the unit's place, we have 8 digits remaining. So, the unit's place can be filled by 8 ways. So, the required number of ways in which three distinct digit number can be formed are $9 \\times 9 \\times 8 = 648$ So, the numbers having all the distinct digits $= 648$ Thus, the remaining numbers containing at most two distinct digits $=$ total numbers $-$numbers having all the distinct digits. $= 900 -", "\\pmod6$. The order of $10$ modulo $13$ is $6$ so the fraction $\\frac{1}{13}$ repeats with period $6$. Denote the repeating block of $13^{-1}$ as $r = 076923$. Our number is of the form $$2\\times \\frac{10^{6n}- 1}{13} = 2\\times\\underbrace{rr\\cdots r}_{n \\text{times}}$$ where the latter expression denotes the number formed by concatenating $r$ with itself $n$ times for some natural number $n$. Putting it all together, our number is of the form $153846$ appended with itself an arbitrary number of times, $$\\left\\{153846,\\ 153846153846,\\ 153846153846153846,\\ \\cdots\\right\\}$$ Besides EuYu's, another way to arrive at the answer is to say that the leading digit has to be $1$ or $2$, because if it were $3$ or greater there would be a carry and the product would have more digits than the original number. Let's try $1$. The multiplicand then starts with $1$ and the product ends with $1$. To have the product end with $1$, the multiplicand must end with $6$ as $6 \\cdot \\frac 72=21$. Then the product ends with $61$ and the multiplicand ends with $46$. We keep going until we have a $1$ at the front, giving $$153846 \\\\ \\underline{\\times \\ \\ \\ \\ 3.5} \\\\ 538461$$ Now we can see that multiplying by $2$ will carry, so will not work and this is the only primitive solution.", "Marking the first $3$ and the first $4$ after $3$, there are only $6$ cases: $$\\begin{array}{} 3&4&\\bullet&\\bullet&\\implies& 10\\cdot10 = 100\\\\ 3&\\bullet&4&\\bullet&\\implies& 9\\cdot 10 = 90\\\\ 3&\\bullet&\\bullet&4&\\implies& 9\\cdot9 =81\\\\ \\bullet&3&4&\\bullet&\\implies& 9\\cdot10 = 90\\\\ \\bullet&3&\\bullet&4&\\implies& 9\\cdot 9 = 81\\\\ \\bullet&\\bullet&3&4&\\implies& 9\\cdot9 = 81 \\end{array}$$ Adding up, $523$. Note: The first $4$ after the first $3$ has been marked, but this doesn't preclude a $4$ coming before the first $3$. - There are a little more, since some will have a $3$ before a $4$ and also a $4$ before a $3$. – André Nicolas Mar 14 at 5:18 Yea, thanks and (+1). Deleting for the nonce to examine if this route can lead somewhere ! – true blue anil Mar 14 at 5:48 The idea will work. Let $w$ be the number where a $3$ occurs before a $4$ but not the other way, and let $b$ be the number of ways both happen. Then we want $w+b$, and $2w+b$ is your $974$. If we can find $b$, we will be finished, and finding $b$ is not hard, one just has to be careful. – André Nicolas Mar 14 at 5:54 So this isn't correct? I feel like it is. – Ninjascroll86 Mar 14 at 7:29 523 is the correct answer as first given by Andre, I am also getting", "100.010 = 0.120012$$ The first number has only three significant digits, so that determines the number in our result: $$\\bf = 0.120$$ 2 $(1.002 \\times 10^{-3})(2.0230 \\times 10^4)$ Solution \\begin{align} (1.002 &\\times 10^{-3})(2.0230 \\times 10^4) \\\\ &= 20.27046 \\end{align} The first number has only four significant digits, so that determines the number in our result: $$\\bf = 20.27$$ 3 $1.00102 + 1.001$ Solution $$1.00102 + 1.001 = 2.00202$$ This is an addition, so we need to mindful of the rule. The second number has only three significant digits after the decimal. So must our result: $$\\bf = 2.002$$ 4 $(2.3 \\times 10^2) / 2.104$ Solution $$\\frac{2.3 \\times 10^2}{2.104} = 109.31558$$ The numerator has only two significant digits, so our result can have only two. In order to achieve that, we'll have to round 109 to 110. $$= \\bf 110$$ 5 $(10.9090)^2 · 9.92 \\times 10^4$ Solution \\begin{align} (10.9090)^2 &· 9.92 \\times 10^4\\\\ &= 1.18054 \\times 10^7 \\end{align} 9.92 has three significant digits, the least of the two numbers, so our result can have only three. $$= \\bf 1.18 \\times 10^7$$ 6 $0.0023 / (2.1 · 3.34159)$ Solution $$\\frac{0.0023}{2.1(3.34159)} = 3.27759 \\times 10^{-4}$$ 2.1 is the weak link here, with only two significant digits. So our result can have only two: $$= \\bf 3.3 \\times 10^{-4}$$ 7 $1.17 \\times", "so I apologize in advance. My thought is to try thinking about your placement of the $$110$$. For the case of $$n=3$$, you saw that there was only one place to put it - as the whole thing. Now consider the case of $$n=4$$. The approach will be to consider the possibilities which contain the string you are looking to avoid, find out how many such unique strings there are and subtract it from the total. There are two places to put the string $$110$$, namely the beginning $$110\\bullet$$ or the end $$\\bullet110$$. There are $$2^4$$ total four-digit bitstrings, and there are four which match these placement patterns $$1100, 1101$$ and $$0110, 1110$$. Just to check, we should make sure all four of these are different, which is clear for this many (just look at the third digit) but might not be so clear on a larger scale. So as I see it there are $$2^4 - 4 = 12$$ ways to do this. This is $$16-4=12 \\neq 8$$. I agree with your logic for finding $$S_3, S_2$$, and hence $$S_4 = 8$$. Do you see any errors in my logic? If you agree with my logic, is it possible that the question is not being asked the way you have asked it here? Alternatively are you sure", "(110,225),(115,235),(120,225),(125,235),(130,225),(135,235), (140,225),(145,235),(150,225),(155,235),(160,225),(165,235), (170,225),(175,235),(180,225),(185,235),(187.5,230),(200,230)} \\tlabel[tc](145,220){$R_\\mathrm{LC}$} \\lines{(200,230),(212.5,230),(215,235),(220,225),(225,235), (230,225),(235,235),(240,225),(245,235),(250,225),(255,235), (260,225),(265,235),(270,225),(275,235),(280,225),(285,235), (290,225),(295,235),(300,225),(305,235),(307.5,230),(320,230)} \\tlabel[tc](260,220){$R_\\mathrm{CR}$} \\lines{(320,230),(332.5,230),(335,235),(340,225),(345,235), (350,225),(355,235),(360,225),(365,235),(370,225),(372.5,230)} \\dashed\\lines{(372.5,230),(400,230)} \\tlabel[cc](80,230){$\\bullet$} \\tlabel[cc](200,230){$\\bullet$} \\tlabel[cc](320,230){$\\bullet$} \\arrow\\lines{(100,245),(180,245)} \\arrow\\lines{(220,245),(300,245)} \\tlabel[bc](140,255){$I_\\mathrm{LC}$} \\tlabel[bc](260,255){$I_\\mathrm{CR}$} \\tlabel[bc](80,255){$V_\\mathrm{L}$} \\tlabel[bc](200,255){$V_\\mathrm{C}$} \\tlabel[bc](320,255){$V_\\mathrm{R}$} % Build Right hand circuit \\lines{(320,230),(320,180),(300,180)} \\lines{(300,142.5),(295,140),(305,135),(295,130),(305,125), (295,120),(305,115),(295,110),(305,105),(295,100), (300,97.5),(300,90),(340,90),(340,130)} \\lines{(350,130),(330,130)} \\lines{(350,140),(330,140)} \\tlabel[bl](345,150){$c^{(m)}_\\mathrm{R}$} \\lines{(340,140),(340,180),(320,180)} \\lines{(320,90),(320,40)} \\lines{(310,40),(330,40)} \\lines{(312.5,35),(327.5,35)} \\lines{(315,30),(325,30)} \\lines{(317.5,25),(322.5,25)} \\dashed\\lines{(300,180),(260,180)} \\dashed\\lines{(300,90),(260,90)} \\lines{(300,180),(320,180)} \\lines{(300,180),(300,170)} % Battery \\lines{(295,170),(305,170)} \\lines{(295,162),(305,162)} \\lines{(295,154),(305,154)} \\lines{(255,150),(265,150)} \\lines{(255,166),(265,166)} \\lines{(255,158),(265,158)} \\pen{2pt} \\lines{(297.5,150),(302.5,150)} \\lines{(297.5,166),(302.5,166)} \\lines{(297.5,158),(302.5,158)} \\lines{(257.5,170),(262.5,170)} \\lines{(257.5,162),(262.5,162)} \\lines{(257.5,154),(262.5,154)} \\pen{1pt} \\lines{(300,150),(300,142.5)} \\lines{(260,180),(260,170)} \\lines{(260,150),(260,142.5),(255,140),(265,135),(255,130), (265,125),(255,120),(265,115),(255,110),(265,105),(255,100), (260,97.5),(260,90)} \\arrow\\lines{(290,135),(290,105)} \\arrow\\lines{(310,80),(310,50)} \\tlabel[cr](305,65){$I^{(m)}_\\mathrm{R}$} % Build Left hand circuit \\lines{(80,230),(80,180),(60,180)} \\lines{(60,142.5),(55,140),(65,135),(55,130),(65,125), (55,120),(65,115),(55,110),(65,105),(55,100),(60,97.5), (60,90),(100,90),(100,130)} \\lines{(110,130),(90,130)} \\lines{(110,140),(90,140)} \\tlabel[bl](105,150){$c^{(m)}_\\mathrm{L}$} \\lines{(100,140),(100,180),(80,180)} \\lines{(80,90),(80,40)} \\lines{(70,40),(90,40)} \\lines{(72.5,35),(87.5,35)} \\lines{(75,30),(85,30)} \\lines{(77.5,25),(82.5,25)} \\dashed\\lines{(60,180),(20,180)} \\dashed\\lines{(60,90),(20,90)} \\lines{(60,180),(80,180)} \\lines{(60,180),(60,170)} % Battery \\lines{(55,170),(65,170)} \\lines{(55,162),(65,162)} \\lines{(55,154),(65,154)} \\lines{(15,150),(25,150)} \\lines{(15,166),(25,166)} \\lines{(15,158),(25,158)} \\pen{2pt} \\lines{(57.5,150),(62.5,150)} \\lines{(57.5,166),(62.5,166)} \\lines{(57.5,158),(62.5,158)} \\lines{(17.5,170),(22.5,170)} \\lines{(17.5,162),(22.5,162)} \\lines{(17.5,154),(22.5,154)} \\pen{1pt} \\lines{(60,150),(60,142.5)} \\lines{(20,180),(20,170)} \\lines{(20,150),(20,142.5),(15,140),(25,135),(15,130), (25,125),(15,120),(25,115),(15,110),(25,105),(15,100), (20,97.5),(20,90)} \\arrow\\lines{(50,135),(50,105)} \\arrow\\lines{(70,80),(70,50)} \\tlabel[cr](65,65){$I^{(m)}_\\mathrm{L}$} % Build Middle circuit \\lines{(200,230),(200,180),(180,180)} \\lines{(180,142.5),(175,140),(185,135),(175,130),(185,125), (175,120),(185,115),(175,110),(185,105),(175,100), (180,97.5),(180,90),(220,90),(220,130)} \\lines{(230,130),(210,130)} \\lines{(230,140),(210,140)} \\tlabel[bl](225,150){$c^{(m)}_\\mathrm{C}$} \\lines{(220,140),(220,180),(200,180)} \\lines{(200,90),(200,40)} \\lines{(190,40),(210,40)} \\lines{(192.5,35),(207.5,35)} \\lines{(195,30),(205,30)} \\lines{(197.5,25),(202.5,25)} \\dashed\\lines{(180,180),(140,180)} \\dashed\\lines{(180,90),(140,90)} \\lines{(180,180),(200,180)} \\lines{(180,180),(180,170)} % Battery \\lines{(175,170),(185,170)} \\lines{(175,162),(185,162)} \\lines{(175,154),(185,154)} \\lines{(135,150),(145,150)} \\lines{(135,166),(145,166)} \\lines{(135,158),(145,158)} \\pen{2pt} \\lines{(177.5,150),(182.5,150)} \\lines{(177.5,166),(182.5,166)} \\lines{(177.5,158),(182.5,158)} \\lines{(137.5,170),(142.5,170)} \\lines{(137.5,162),(142.5,162)} \\lines{(137.5,154),(142.5,154)} \\pen{1pt} \\lines{(180,150),(180,142.5)} \\lines{(140,180),(140,170)} \\lines{(140,150),(140,142.5),(135,140),(145,135),(135,130), (145,125),(135,120),(145,115),(135,110),(145,105),(135,100), (140,97.5),(140,90)} \\arrow\\lines{(170,135),(170,105)} \\arrow\\lines{(190,80),(190,50)} \\tlabel[cr](185,65){$I^{(m)}_\\mathrm{C}$} % Far right \\dashed\\lines{(400,180),(380,180)} \\dashed\\lines{(400,90),(380,90)} \\lines{(380,150),(380,142.5),(375,140),(385,135),(375,130), (385,125),(375,120),(385,115),(375,110),(385,105),(375,100), (380,97.5),(380,90)} \\arrow\\lines{(50,135),(50,105)} \\arrow\\lines{(70,80),(70,50)} % Battery \\lines{(375,150),(385,150)} \\lines{(375,166),(385,166)} \\lines{(375,158),(385,158)} \\lines{(380,180),(380,170)} \\pen{2pt} \\lines{(377.5,170),(382.5,170)} \\lines{(377.5,162),(382.5,162)} \\lines{(377.5,154),(382.5,154)} \\pen{1pt} \\end{mfpic}} \\centering \\parbox{5.6in}{\\caption{\\label{circuit} A diagrammatic representation of two elemental circuits used to construct the compartmental model of a dendritic section. Axial current $I_\\mathrm{L}$ flows in the left hand compartment under the influence of the potential difference $(V_\\mathrm{L}-V_\\mathrm{C})$. Axial", "of double upon moving the last digit to the front. Let's introduce an arbitrary multiplying factor: $k$ So the solutions mentioned above and Talwalkar's puzzle are a particular case when $k=2$ For a different example, let's look at $k=5$ in base-7. The smallest solution is $$(5) \\cdot 1013042156536245_7 \\quad\\ \\ \\$$ $$||$$ $$5101304215653624_7$$ This solution is $16$ digits long. As we did before for base-10, we can write out a sequence of the number of digits in each base's solution starting with base-2 $$6,6,9,2,14,16,4,5,42,18,...$$ $$\\uparrow\\ \\$$ $$\\text{Base-}7\\ \\$$ Another search on http://oeis.org/ quickly shows that these numbers are the multiplicative order of $$5\\ (\\text{mod } 5\\cdot7-1)\\quad\\text{or rather}\\quad 5 \\ (\\text{mod }34)$$ $\\bullet\\textbf{ A Conjecture }\\bullet$ The pattern might be clear now. After checking other cases, it seems that the smallest positive number in any base $B$ - which is $k$ times the value gotten by moving it's last digit to the front - has a number of digits equal to the multiplicative order of $$k\\ (\\text{mod } Bk-1)$$ I can see how multiplicative order would be related to this problem. But I can't find an exact reason why this relationship should be so. Is there an intuitive reason? Given some solution in base $B$ - let's start by assigning the variable $x$ to the digit that get's moved", "# How is moving the last digit of a number to the front and multiplying related to multiplicative orders? There seems to be a relationship between multiplicative orders modulo $n$ and a puzzle Presh Talwalkar gave a few days ago at https://www.youtube.com/watch?v=1lHDCAIsyb8 I'm hoping someone can give a more rigorous explanation of the pattern I see. $\\bullet\\textbf{ The Puzzle }\\bullet$ He states the puzzle as: \"What Positive Number Doubles When Its Last Digit Becomes Its First Digit?\" For example, the smallest solution - assuming base-10 representation - is $105263157894736842$. Which means that $$(2)\\cdot105263157894736842 \\quad\\ \\ \\$$ $$||$$ $$210526315789473684$$ Naturally, one can proceede to find solutions in other bases. The base-5 solution is $$(2)\\cdot102342_5 \\quad\\ \\ \\$$ $$||$$ $$210234_5$$ $\\bullet\\textbf{ Multiplicative Order Connection }\\bullet$ The base-10 solutions is $18$ digits and the base-5 solution is $6$ digits. I wrote a program to generate the smallest solution in all bases and then counted the number of digits in each such solution. Here's the sequence, starting with base-2 $$2,4,3,6,10,12,4,8,18,6,11,20,...$$ $$\\uparrow \\quad\\quad\\quad\\quad\\ \\uparrow \\quad\\quad$$ $$\\text{Base-} 5 \\quad\\quad\\quad \\text{Base-} 10 \\quad\\quad$$ A quick search on http://oeis.org/ shows that these numbers are the multiplicative order of $$2\\ (\\text{mod } 2B-1)$$ where $B$ is the representation base we are working in. See http://oeis.org/A002326 This puzzle can be further generalized by finding numbers that triple instead", "of this: the ‘$\\,79\\,$’ in $\\,2.\\underline{79}845\\,$ becomes an ‘$\\,80\\,$’. EXAMPLES: $293.4991927\\,$, rounded to $\\,3\\,$ decimal places, is: $\\,293.499$ (The decider digit is $\\,1\\,$; round down.) $293.4699927\\,$, rounded to $\\,3\\,$ decimal places, is: $\\,293.470\\,$ (The decider digit is $\\,9\\,$; round up.) Notice that you must include the $\\,0\\,$ in the thousandths place. $293.4999927\\,$, rounded to $\\,3\\,$ decimal places, is: $\\,293.500\\,$ (The decider digit is $\\,9\\,$; round up.) Notice that you must include the two trailing zeros. $999.9999992\\,$, rounded to $\\,6\\,$ decimal places, is: $\\,999.999999\\,$ (The decider digit is $\\,2\\,$; round down.) $999.9999998\\,$, rounded to $\\,6\\,$ decimal places, is: $\\,1000.000000\\,$ (The decider digit is $\\,8\\,$; round up.) Notice that you must include all the trailing zeros. CAREFUL! When you are asked to round a number to $\\,n\\,$ decimal places, then your answer must have $\\,n\\,$ decimal places, even if there are zeros in those positions. MORE EXAMPLES: Round $\\,0.92746\\,$ to 3 decimal places. Answer: $\\,0.927$ If the decimal is less than $\\,1\\,$, then it is good style to put a zero in the ones place. In the web exercise below, you must put the zero in the ones place to be recognized as correct. Round $\\,2.999997\\,$ to $\\,4\\,$ decimal places. Answer: $\\,3.0000$ If you round to $\\,4\\,$ decimal places, then your answer must have $\\,4\\,$ decimal places, even if there", "of combinations of 3 digits from the 7 remaining. There are $\\displaystyle {7\\choose{3}} = 35$ combinations. That means the probability of having a number less than 4000 and greater than 3000 is $\\displaystyle \\frac{35}{70} = \\frac{1}{2}$. 3. Hello, johnsy123! $\\text{1)A bag contains five white, six red and seven blue balls.}$ $\\text{If three balls are selected at random, without replacement,}$ $\\text{find the probability they are all different colors.}$ There are: . ${18\\choose3} \\:=\\:816$ possible outcomes. There are: . ${5\\choose1}{6\\choose1}{7\\choose1} \\:=\\:5\\cdot6\\cdot7 \\:=\\:210$ ways to get one of each color. Therefore: . $P(\\text{all different colors}) \\;=\\;\\dfrac{210}{816} \\;=\\;\\dfrac{35}{136}$ $\\text{2) A four-digit number (no repetitions) is to be formed}$ $\\text{from the set of numbers (0,1,2,3,4,5,6,7).}$ Assuming that the number may not have a leading zero, . . there are: . $7\\cdot7\\cdot6\\cdot5 \\:=\\:1470$ possible 4-digit numbers. $\\text{(a) Find the probability that the number is even.}$ There are two cases to consider: . . The last digit is 0. . . The last digit is 2, 4 or 6. The last digit is 0: . $\\square\\:\\square\\;\\square\\;0$ . . The other 3 spaces can be filled in: . $7\\cdot6\\cdot5 \\:=\\:210$ ways. The last digit is 2, 4 or 6: .3 choices. The first must not be 0: .6 choices. The other 2 spaces can be filled in $6\\cdot5\\,=\\,30$ ways. . . There are: . $3\\cdot6\\cdot30 \\:=\\:540$" ]

[ "of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$ Together, the answer is $3(12+12+12)-12=\\boxed{\\textbf{(A) } 96}.$ Solution 4 Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \\equiv 1\\pmod{3},$ and thus we need to count any $2$-digit number $\\equiv 2\\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \\boxed{\\textbf{(A) } 96}.$ Solution 5 We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\\cdot\\frac{1}{3}=150.$ Of these $150$ numbers, $\\frac{4}{5}$ $\\textbf{do not}$ have $3$ in their ones (units) digit, $\\frac{9}{10}$ $\\textbf{do not}$ have $3$ in their tens digit, and $\\frac{8}{9}$ $\\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$900\\cdot\\frac{1}{2}\\cdot\\frac{1}{3}\\cdot\\frac{4}{5}\\cdot\\frac{9}{10}\\cdot\\frac{8}{9}=\\boxed{\\textbf{(A) } 96}.$$ ~mathpro12345 Solution 6 We will start with the numbers that could work. This numbers include _ _ $1$, _ _ $5$, _ _ $7$, _", "If we try $222$, $333$, and so on up to $888$, each of those attempts rule out $22$ possibilities as well. After these, we've ruled out $22 \\cdot 8 = 176$ possibilities. Which possibilities are left? Exactly those with no repeated digits. If we try $123$, we have an overlap with $121$, $122$, $113$, $133$, $223$, and $323$ being doubly ruled out, but we still rule out $22-6=16$ additional possibilities. We can get another $16$ ruled out by each of $456$, $781$, $234$, $567$, $812$, $345$, and $678$. Now we have a total of $512 - 22 \\cdot 8 - 16 \\cdot 8 = 208$ possibilities left to rule out. These are exactly the combinations $\\text{A}\\text{B}\\text{C}$ in which $\\text{B} - \\text{A} \\not\\equiv 0, 1 \\pmod{8}$ and $\\text{C} - \\text{B} \\not\\equiv 0, 1 \\pmod{8}$. If we try $135$, we have an overlap with $133$, $155$, $535$, $335$, $115$, $131$, $145$, $235$, and $125$ but we still rule out $22-9=13$ additional possibilities. Similarly, we try $246$, $357$, $468$, $571$, $682$, $713$, and $824$, each ruling out $13$. Now we have $512 - 22 \\cdot 8 - 16 \\cdot 8 - 13 \\cdot 8 = 104$ possibilities left, with $24$ attempted codes. If we try $147$, we have an overlap with $144$, $177$, $117$, $447$, $141$, $747$, $145$, $167$, $127$, $137$,", "knowing the room number and knowing the conversation, we cannot have two possible candidates $36$ and $48$ for the product. Hence $S\\ge13$ indeed is impossible, and the claim is proved. Claim 2: $S\\le11$ is impossible. Proof: Then for every choice of $n$ and $P$, there exists at most one list of $n$ positive integers $x_1,\\ldots,x_n$ with sum $S$ and product $P$. I have verified this statement by a computer program: I have generated all lists with sum $\\le11$, and it turns out that no two different such lists agree simultaneously in $n$ and in $P$ and in $S$. This collides with the statement \"No\" that the member makes after scribbling for some time. Claim 3: $S=12$ is possible. Proof: Consider the following lists: • The numbers $3$, $4$, $4$, $1$. • The numbers $2$, $2$, $2$, $6$. Both lists contain four numbers with sum $S=12$ and product $P=48$. My computer program shows that for $S=12$, these are the only two lists that simultaneously agree in $n$ and in $P$. • Thanks. I found the same solution, but not a proof of uniqueness. – Colonel Panic Apr 1 '16 at 14:22 • @Gamow why 6,8,8,2 and 4,4,4,12 are not okay? do I miss something? S would be 24, and product will be the same 768 – Oray Apr 1", "converting 7.478478.... into \\dfrac{q}{p} form, we get x = 7.478478.... ... (a) \\\\ 1000x = 7478.478478.... ... (b) \\\\ While, subtracting (a) from (b), we get \\\\ 999x = 7471\\ x \\dfrac{ 7471}{ 999} \\\\ Therefore, 7.478478.... is a rational number. \\\\ (v) 1.101001000100001.... \\\\ We can observe that the number 1.101001000100001.... is a non-terminating non- recurring decimal. Thus, non-terminating and non-recurring decimals cannot be converted into \\dfrac{q}{p} form.\\\\ Therefore, we conclude that 1.101001000100001.... is an irrational number. 17 Visualize 3.765 on the number line using successive magnification. ##### Solution : \\bullet Clearly, the value lies between 3 and 4 \\\\ \\bullet The numbers 3.7 and 3.8 lie between 3 and 4. \\\\ \\bullet The numbers 3.76 and 3.77 lie between 3.7 and 3.8. \\\\ \\bullet The numbers 3.764 and 3.766 lie between 3.76 and 3.77 \\\\ \\bullet The number 3.765 lies between 3.764 and 3.766. \\\\ Therefore, we need to use the successive magnification, after locating numbers 3 and 4 on the number line. 18 Visualize 4.\\bar{26} on the number line, up to 4 decimal places. ##### Solution : The number 4.\\bar{26} can also be written as 4.262.... .$$\\\\$ The number $4.2$ lie between $4$ and $5$ The numbers $4.26$ lie between $4.2$ and $4.3$ The numbers $4.262$ lie between $4.261$ and $4.263$ The number $4.2626$ lie", "Marking the first $3$ and the first $4$ after $3$, there are only $6$ cases: $$\\begin{array}{} 3&4&\\bullet&\\bullet&\\implies& 10\\cdot10 = 100\\\\ 3&\\bullet&4&\\bullet&\\implies& 9\\cdot 10 = 90\\\\ 3&\\bullet&\\bullet&4&\\implies& 9\\cdot9 =81\\\\ \\bullet&3&4&\\bullet&\\implies& 9\\cdot10 = 90\\\\ \\bullet&3&\\bullet&4&\\implies& 9\\cdot 9 = 81\\\\ \\bullet&\\bullet&3&4&\\implies& 9\\cdot9 = 81 \\end{array}$$ Adding up, $523$. Note: The first $4$ after the first $3$ has been marked, but this doesn't preclude a $4$ coming before the first $3$. - There are a little more, since some will have a $3$ before a $4$ and also a $4$ before a $3$. – André Nicolas Mar 14 at 5:18 Yea, thanks and (+1). Deleting for the nonce to examine if this route can lead somewhere ! – true blue anil Mar 14 at 5:48 The idea will work. Let $w$ be the number where a $3$ occurs before a $4$ but not the other way, and let $b$ be the number of ways both happen. Then we want $w+b$, and $2w+b$ is your $974$. If we can find $b$, we will be finished, and finding $b$ is not hard, one just has to be careful. – André Nicolas Mar 14 at 5:54 So this isn't correct? I feel like it is. – Ninjascroll86 Mar 14 at 7:29 523 is the correct answer as first given by Andre, I am also getting", "TL;DR On with TASK #2 from The Weekly Challenge #141. Enjoy! # The challenge You are given positive integers, $m and $n. Write a script to find total count of integers created using the digits of $m which is also divisible by $n. Repeating of digits are not allowed. Order/Sequence of digits can’t be altered. You are only allowed to use (n-1) digits at the most. For example, 432 is not acceptable integer created using the digits of 1. Also for 1234, you can only have integers having no more than three digits. Example 1: Input: $m = 1234,$n = 2 Output: 9 Possible integers created using the digits of 1234 are: 1, 2, 3, 4, 12, 13, 14, 23, 24, 34, 123, 124, 134 and 234. There are 9 integers divisible by 2 such as: 2, 4, 12, 14, 24, 34, 124, 134 and 234. Example 2: Input: $m = 768,$n = 4 Output: 3 Possible integers created using the digits of 768 are: 7, 6, 8, 76, 78 and 68. There are 3 integers divisible by 4 such as: 8, 76 and 68. # The questions It’s not entirely clear to me what would happen with a number having repeated digits inside. Let’s take 1223 as an example: the sub-sequence 123 can be generated in two", "# $$x^3 + 1$$ and Arithmetic Sequences While thinking up interesting problems, I stumbled upon this interesting fact: If $$x = 3n^2,$$ for some positive integer $$n,$$ then $$x^3 + 1$$ can be expressed as the product of three distinct positive integers that form an arithmetic sequence. For instance, when $$x = 12 = 3 \\times 2^2,$$ we have $$12^3 + 1 = 7 \\times 13 \\times 19.$$ When $$x = 75 = 3 \\times 5^2,$$ we have $$75^3 + 1 = 61 \\times 76 \\times 91.$$ This statement is relatively easy to prove with some algebraic manipulation, but I'm having a lot of trouble with the converse. If $$x^3 + 1$$ can be expressed as a product of three positive integers that form an arithmetic sequence, then must $$x = 3n^2$$ for some positive integer $$n$$? If not, what are the counterexamples? I'd appreciate it if anyone here can provide some information about this! Note by Steven Yuan 7 months, 1 week ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link >", "namely $6$, $7$, $8$, and $9$; if $6$, there are $3$ choices; if $7$, there are $2$ choices; and if $8$, there is $1$ choice. It follows that the total number of such integers is $5+4+3+2+1=\\boxed{\\textbf{(C) }15}$.", "three even squares and two odd squares. Because $4,16,36,64$ all appear, the two representations are different. $\\mathbf{k=3}$: All we need to do here is to apply the second bullet to $n-4-4$ and $n-16-16$, so we only need to assume $n\\ge35$. $\\mathbf{k=4}$: All we need to do here is to apply the second bullet to $n-16-1$ and $n-64-1$, which we are able to do provided $n\\ge 68$. Note that the two representations are distinct, because the sole even squares in them are different. $\\mathbf{k=5}$: If all the numbers $n-1-1$, $n-9-9$, $n-25-25$ are all positive, then by the second bullet they are all sums of three odd squares. Necessarily at least two of the resulting representations of $n$ as a sum of five squares are different, because $1,9,25$ cannot all appear twice among a set of five squares. This case is settled, if $n\\ge55$. $\\mathbf{k=6}$: By the last bullet if $n\\ge38$, then both $n-4-4$ and $n-16-16$ can be written as sums of an even square $\\equiv4\\pmod8$ and two odd squares. Again we get two different representations. $\\mathbf{k=7}$: Here we simply apply the second bullet to both $n-4-16$ and $n-36-16$. Again the resulting representations as sums of the prescribed two even squares and some three odd squares are different. This case is thus done, if $n\\ge59$. But the case $k=1$ gives", "I could understand it! Thank you very much for your guidance! – Andrej Hatzi Feb 11 at 11:12 • @AndrejHatzi: I preferred to let you think rather than giving a straight answer. – Yves Daoust Feb 11 at 11:22 You shall only choose the first and last digit from the two middle rows. That's because the bottom row contains only one number and for the top row, there's no row above to choose the second and third digit. Let's start by choosing two digits from the $$3$$rd row. We can choose and rearrange two items out of three in $$P(3,2)=\\frac{3!}{1!}=6$$ ways. Following the exact same reasoning, for the second and third digit, we have: $$P(3,2)=\\frac{3!}{1!}=6$$ ways for the middle digits. Thus, $$6$$ ways to choose first and last digits and $$6$$ ways to choose second and third. That's a total of $$36$$ ways to choose all $$4$$ digits. Also, we could repeat all these actions, choosing the first and last from the $$2$$nd row and the middle ones from the $$1$$st one. Again, that's a total of $$36$$ ways to choose all four digits. At last, since we can carry out this procedure either the first $$36$$ ways or the other $$36$$ ways, the total ways are $$36+36=72.$$", "a path to approach the more general problem. Staff - 4 years ago What are the last three digits of $$9^{9^9}$$? - 4 years ago We have $$9^{50} \\equiv 1$$ ($$1000$$) and $$9^9 \\equiv 39$$ ($$50$$). Hence $$9^{9^9} \\equiv 9^{39} \\equiv 289$$ ($$1000$$). - 4 years ago Here are a few easy ones I know about modular arithmetic: The fractional part of $$\\frac {n}{7}$$ is $$.142857$$ repeating, or some variant thereof. What is the minimal positive integer N such that the fractional part of $$\\frac{251}{7} \\times n$$ is $$.142857$$ repeating? - 4 years ago Sorry, I'm a new member, so I don't really know the formatting that well yet. We can start this problem by thinking about the number $$251$$, which can be written as $$251 \\equiv 6 \\pmod{7}$$. In order for $$\\frac {251}{7} \\times n$$ to be $$0.142857$$ repeating, $$251n \\equiv 1 \\pmod{7}$$. Looking at the first congruence, we can start by realizing that $$251n$$ will have a remainder of $$1$$ iff $$6n \\equiv 1 \\pmod{7}$$. We can see that the possible values for $$n$$ will be of the form $$7k - 1$$, where $$k$$ is a positive integer. Therefore, the minimum possible value for $$n$$ is $$6$$. - 4 years ago Your formatting is fine. Note, though, that you could simplify things a bit by making", "the number you are currently working with and bring down the next group of numbers behind it. This is your next group of numbers to work with. Step 7: Repeat steps 4 through 6 until you get an approximation with an acceptable number of significant digits. I understand everything but Step 5. In the picture above, for instance, how do they get that $$6$$ in order to form $$26$$? I had understood we should find the digit $$\\bullet$$ in $$2\\bullet$$ such that $$154/2\\bullet$$ would produce the least remainder. But if that were the case, one should have $$\\bullet=9$$ since $$154/29$$ has remainder $$0$$. Furtheremore, what if there were more than one digit $$\\bullet$$ which lead to the same remainder? Which would I pick? Thanks. • FYI, the algorithm is described (IMHO, somewhat better in at least certain ways, including the Step #$5$ as explained in Ross' answer below, with it being the second bullet point of its step #$2$) in the Decimal (base 10) section of Wikipedia's \"Methods of computing square roots\" article. Jun 16, 2020 at 4:26 • Also, FYI, the question at Derivation of Square-Root Computation Algorithm asks to have explained the Wikipedia algorithm section I link to in my comment above. Jun 16, 2020 at 4:50 In step $$5$$ you are going to multiply the", "two smallest numbers or the numbers which have 2 at the start, and check the combinations for the subtracted number, you will find that all combinations from that are negative. After that, you can just try all other combinations and you're done! Jamie, from Shrewsbury International School in Thailand, used a different method to systematically find the solutions. There are 4 numbers. In the first square, you are able to put 4 different numbers. Keep the 4. The second square can have 3 different numbers, since one number is already used. There are two numbers that can go into the third square. And the last square matches with the last number left. Multiplying $4 \\times 3 \\times 2 \\times 1 = 24$ gives the number of orders. This is also called $4$ factorial, written as $4!$. Zach was able to develop this to get to again show that there are 12 solutions: $4!$ gives 24 possible 4-digit numbers. These then form the two 2-digit numbers, but since the answer must be positive, $28$ and $57$ gives the same sum as $57$ and $28$. This means that half the possible 24 combinations are duplicates, as you always have to take the smaller number from the bigger number to get a positive result. Max and Ben, from St. Columba's Primary", "of this: the ‘$\\,79\\,$’ in $\\,2.\\underline{79}845\\,$ becomes an ‘$\\,80\\,$’. EXAMPLES: $293.4991927\\,$, rounded to $\\,3\\,$ decimal places, is: $\\,293.499$ (The decider digit is $\\,1\\,$; round down.) $293.4699927\\,$, rounded to $\\,3\\,$ decimal places, is: $\\,293.470\\,$ (The decider digit is $\\,9\\,$; round up.) Notice that you must include the $\\,0\\,$ in the thousandths place. $293.4999927\\,$, rounded to $\\,3\\,$ decimal places, is: $\\,293.500\\,$ (The decider digit is $\\,9\\,$; round up.) Notice that you must include the two trailing zeros. $999.9999992\\,$, rounded to $\\,6\\,$ decimal places, is: $\\,999.999999\\,$ (The decider digit is $\\,2\\,$; round down.) $999.9999998\\,$, rounded to $\\,6\\,$ decimal places, is: $\\,1000.000000\\,$ (The decider digit is $\\,8\\,$; round up.) Notice that you must include all the trailing zeros. CAREFUL! When you are asked to round a number to $\\,n\\,$ decimal places, then your answer must have $\\,n\\,$ decimal places, even if there are zeros in those positions. MORE EXAMPLES: Round $\\,0.92746\\,$ to 3 decimal places. Answer: $\\,0.927$ If the decimal is less than $\\,1\\,$, then it is good style to put a zero in the ones place. In the web exercise below, you must put the zero in the ones place to be recognized as correct. Round $\\,2.999997\\,$ to $\\,4\\,$ decimal places. Answer: $\\,3.0000$ If you round to $\\,4\\,$ decimal places, then your answer must have $\\,4\\,$ decimal places, even if there", "$4$ goes to $2$, so $4 \\to 2$. $5$ goes to $6$, and $6$ goes to $5$, so $5 \\to 5$. $6$ goes to $5$, and $5$ goes to $6$, so $6 \\to 6$. Putting this together gives $[(13)(24)]$, since $5$ and $6$ don't change. • Alright, that helped a lot! Do you know why it ends at [(1234)(56)]$^4$? Like why the person did not do [(1234)(56)]$^5$ and [(1234)(56)]$^6$? – Justin Feb 25 '15 at 1:51 • Also, if you do [(1234)(56)]$^4$, why is (1) rather than (1)(3,5,6) – Justin Feb 25 '15 at 1:55 • Because, for example, you have $1 \\to 2 \\to 3 \\to 4 \\to 1$. Similar for all the other elements. After 4 iterations, you end up with the identity map. As for $[(1234)(56)]^5$, the trick is that $[(1234)(56)]^5 = [(1234)(56)][(1234)(56)]^4 = [(1234)(56)] \\cdot \\text{id}$. So there are really only 3 different permutations in the list $[(1234)(56)], [(1234)(56)]^2, [(1234)(56)]^3, [(1234)(56)]^4, \\ldots$ – MarkG Feb 25 '15 at 1:57 • For [(1234)(56)]$^4$, I made a mistake, my fault. I see everything clearer now. thanks a lot! – Justin Feb 25 '15 at 2:13 • Is this a typo: $2$ goes to $3$, and $3$ goes to $4$, so $3 \\to 4$.? Should it not read, $2 \\to 4$? – Kevin Meredith May 20 '15", "\\to 1^2 + 0^2 = 1$ The index of the $8$th Mersenne prime after $2$, $3$, $5$, $7$, $13$, $17$, $19$: $M_{31} = 2^{31} - 1 = 2 \\, 147 \\, 483 \\, 647$ The $9$th lucky number: $1$, $3$, $7$, $9$, $13$, $15$, $21$, $25$, $31$, $\\ldots$ The $16$th positive integer which is not the sum of $1$ or more distinct squares: $2$, $3$, $6$, $7$, $8$, $11$, $12$, $15$, $18$, $19$, $22$, $23$, $24$, $27$, $28$, $31$, $\\ldots$ The $20$th positive integer after $2$, $3$, $4$, $7$, $8$, $9$, $10$, $11$, $14$, $15$, $16$, $19$, $20$, $21$, $24$, $25$, $26$, $29$, $30$ which cannot be expressed as the sum of distinct pentagonal numbers. The $21$st integer $n$ such that $2^n$ contains no zero in its decimal representation: $2^{31} = 2 \\, 147 \\, 483 \\, 648$ Expressible as the sum of successive powers starting from $1$ in in $2$ different ways: $31 = 1 + 5 + 5^2 = 1 + 2 + 2^2 + 2^3 + 2^4$", "## Divisible by 3 or 7 Find the number of four-digit positive integers divisible by $3$ or $7$. Source: NCTM Mathematics Teacher, August 2006 Solution Consider $A=\\{1002,1005,1008,\\cdots,9996,9999\\}$, the set of four-digit positive integers divisible by $3$ and $B=\\{1001,1008,1015,\\cdots,9989,9996\\}$, the set of four-digit positive integers divisible by $7$. The count of $A=(9999-1002)/3+1=3000$ and the count of $B=(9996-1001)/7+1=1286$. Some integers like $1008$ and $9996$ appear in both sets so we subtract them as duplicates from the total count. Since $1008+(7\\times 3)=1008+(3\\times 7)=1029$, starting from $1008$ the duplicates occur every $7$th integer in set $A$ or every $3$rd integer in set $B$. If we use set $A$, the count of duplicates = $(9996-1008)/(7\\times 3)+1=429$. The number of four-digit positive integers divisible by $3$ and/or $7$ is $3000+1286-429=3857$. Answer: $3857$ Alternative solution If we divide $9999$ by $3$, we get $3333$ which means that there are $3333$ integers $\\leq 9999$ that are divisible by $3$. But, if we divide $9999$ by $7$, we get $1428.43$ which is not an integer. So we say there are $1428$ integers $\\leq 9999$ that are divisible by $7$. To avoid this problem we use the floor function $\\lfloor x\\rfloor$ = the largest integer less than or equal to $x$ for some real number $x$. In our example, $\\lfloor 1428.43\\rfloor=1428$. Number of four-digit integers divisible by $3$", "9 $111222$ $20*20=400$ Case 10 $111222222$ $20$ Case 11 $11112$ $15*6=90$ Case 12 $11112222$ $15*15=225$ Case 13 $1111122$ $6*15=90$ Case 14 $1111122222$ $6*6=36$ Case 15 $111111$ $1$ Case 16 $111111222$ $20$ Case 17 $111111222222$ $1$ Total $1+20+1+36+90+225+90+20+400+20+90+225+90+36+1+20+1=484+360+450+72=1366$ $P=\\frac{1366}{2^{12}}=\\frac{683}{2^{11}}$ ANS=$\\boxed{683}$ By SpecialBeing2017 ## Solution 4 Consider the numbers $\\{1,4,7,10,13,16\\}$. Each of those are congruent to $1 \\pmod 3$. There is ${6 \\choose 0}=1$ way to choose zero numbers ${6 \\choose 1}=6$ ways to choose $1$ and so on. There ends up being ${6 \\choose 0}+{6 \\choose 3}+{6 \\choose 6} = 22$ possible subsets congruent to $0\\pmod 3$. There are $2^6=64$ possible subsets of these numbers. By symmetry there are $21$ subsets each for $1 \\pmod 3$ and $2 \\pmod3$. We get the same numbers for the subsets of $\\{2,5,8,11,14,17\\}$. For $\\{3,6,9,12,15,18\\}$, all $2^6$ subsets are $0 \\pmod3$. So the probability is: $\\frac{(22\\cdot22+21\\cdot21+21\\cdot21)\\cdot2^6}{2^{18}}=\\frac{683}{2^{11}}$ ## Solution 5 Notice that six numbers are $0\\pmod3$, six are $1\\pmod3$, and six are $2\\pmod3$. Having numbers $0\\pmod3$ will not change the remainder when $s(T)$ is divided by $3$, so we can choose any number of these in our subset. We ignore these for now. The number of numbers that are $1\\pmod3$, minus the number of numbers that are $2\\pmod3$, must be a multiple of $3$, possibly zero or negative. We can now split into cases", "# 226 Previous ... Next ## Number $226$ (two hundred and twenty-six) is: $2 \\times 113$ The $36$th happy number after $1$, $7$, $10$, $13$, $19$, $23$, $\\ldots$, $130$, $133$, $139$, $167$, $176$, $188$, $190$, $192$, $193$, $203$, $208$, $219$: $226 \\to 2^2 + 2^2 + 6^2 = 4 + 4 + 36 = 44 \\to 4^2 + 4^2 = 16 + 16 = 32 \\to 3^2 + 2^2 = 9 + 4 = 13 \\to 1^2 + 3^2 = 9 + 1 = 10 \\to 1^2 + 0^2 = 1$ The $14$th positive integer which cannot be expressed as the sum of a square and a prime: $1$, $10$, $25$, $34$, $58$, $64$, $85$, $91$, $121$, $130$, $169$, $196$, $214$, $226$, $\\ldots$ The smallest positive integer which can be expressed as the sum of $2$ distinct lucky numbers in $14$ different ways The $8$th non-square positive integer which cannot be expressed as the sum of a square and a prime: $10$, $34$, $58$, $85$, $91$, $130$, $214$, $226$, $\\ldots$", "of combinations of 3 digits from the 7 remaining. There are $\\displaystyle {7\\choose{3}} = 35$ combinations. That means the probability of having a number less than 4000 and greater than 3000 is $\\displaystyle \\frac{35}{70} = \\frac{1}{2}$. 3. Hello, johnsy123! $\\text{1)A bag contains five white, six red and seven blue balls.}$ $\\text{If three balls are selected at random, without replacement,}$ $\\text{find the probability they are all different colors.}$ There are: . ${18\\choose3} \\:=\\:816$ possible outcomes. There are: . ${5\\choose1}{6\\choose1}{7\\choose1} \\:=\\:5\\cdot6\\cdot7 \\:=\\:210$ ways to get one of each color. Therefore: . $P(\\text{all different colors}) \\;=\\;\\dfrac{210}{816} \\;=\\;\\dfrac{35}{136}$ $\\text{2) A four-digit number (no repetitions) is to be formed}$ $\\text{from the set of numbers (0,1,2,3,4,5,6,7).}$ Assuming that the number may not have a leading zero, . . there are: . $7\\cdot7\\cdot6\\cdot5 \\:=\\:1470$ possible 4-digit numbers. $\\text{(a) Find the probability that the number is even.}$ There are two cases to consider: . . The last digit is 0. . . The last digit is 2, 4 or 6. The last digit is 0: . $\\square\\:\\square\\;\\square\\;0$ . . The other 3 spaces can be filled in: . $7\\cdot6\\cdot5 \\:=\\:210$ ways. The last digit is 2, 4 or 6: .3 choices. The first must not be 0: .6 choices. The other 2 spaces can be filled in $6\\cdot5\\,=\\,30$ ways. . . There are: . $3\\cdot6\\cdot30 \\:=\\:540$" ]

[ "six jokers, but we do not take these into account here. Each card of a standard 52-card deck is represented by a string (str) that contains the rank of the card, followed by its suit: • ranks: 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) and A (ace) • suits: C (clubs; ), D (diamonds; ), H (hearts; ) and S (spades; ) As such, the ace of spades is for example represented as AS, the ten of hearts as 10H and the king of clubs as KC. A stack of cards is represented as a list (list) or a tuple (tuple) whose elements are the representations of the individual cards in the stack, listed from top to bottom. A stack of cards does not necessarily need to contain all 52 cards of a standard deck. Your task: • Write a function card_count that takes a stack of cards. The function also has an optional parameter color. If no argument is explicitly passed to the parameter color, the function must return the number of cards in the given stack. If the string red is passed to the parameter color, the function must return the number of red cards in the given stack. If the string black is passed to the parameter", "want to embed an alignged environment in one square of an array to make sure the hands line up pretty; I'd like the name of the player to be centered above the hands too) $$\\begin{array}{l l l} & North & \\\\ & \\clubsuit: AQ & \\\\ & \\diamondsuit: xx & \\\\ & \\heartsuit: Axx & \\\\ & \\spadesuit: JTxxxx & \\\\ West & & East \\\\ \\clubsuit: xxxx & & \\clubsuit: xxx \\\\ \\diamondsuit: Axx & & \\diamondsuit: xxxx \\\\ \\heartsuit: KQxxx & & \\heartsuit: xxxxx \\\\ \\spadesuit: 4 & & \\spadesuit: 6 \\\\ & South & \\\\ & \\clubsuit: KJxx & \\\\ & \\diamondsuit: KQxx & \\\\ & \\heartsuit: & \\\\ & \\spadesuit: AKQxx &$$ Play opens with west leading $K\\heartsuit$. Last edited: Nov 29, 2003 2. Nov 29, 2003 ### Loren Booda My mother often played a foursome with her girlfriend Andree, sometimes teaming up with bridge great Freddie Sheinwold. A particular hand was dealt while Andree was out of the room. When she picked it up, she found 13 of the same suit! My mother swore that there was no collusion, although the chance of such a deal is astronomical. 3. Dec 1, 2003 ### KLscilevothma Wow, coool !! You and your partner got total 29 points, and you got a void suit, perfect split!", "standard deck? hearts, diamonds, spades, clubs. Name the suits in a deck. 13. How many diamonds? 4 card suits. How many suits? 26. How many. 3. Wondrous Item, legendary. Usually found in a box or pouch, this deck contains a number of cards made of ivory or vellum. Most (75 percent) of these decks have only thirteen cards, but the rest have twenty-two. Before you draw a card, you must declare how many cards you intend to draw and then draw them randomly (you can use an altered deck of. 4. Answer: - There are 4 suits (Clubs, Hearts, Diamonds, and Spades) and there are 13 cards in each suit (Clubs/Spades are black, Hearts/Diamonds are red) - Without replacement means the card IS NOT put back into the deck 5. What is the probability that a card drawn at random from a pack of 52 cards either a king or a spade? [Hint 13 / 52 because there are 13 spades and 3 / 52 instead of 4 / 52 (there are four kings) because one king is already counted in spades.] 2). One card is drawn at random from a well-shuffled pack of 52 cards 6. A standard deck of playing cards consists of 52 cards. All cards are divided into 4 suits. There are two black", "+0 # help probability 0 34 2 A standard deck of 52 cards has 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King) and 4 suits such that there is exactly one card for any given rank and suit. Two of the suits are black and the other two suits are red. The deck is randomly arranged. What is the probability that the top card is a heart of a jack? Jul 3, 2022 #1 +2275 +1 There is exactly 1 heart of jack, and 52 cards, so the probability is $$\\color{brown}\\boxed{1 \\over 52}$$ Jul 4, 2022 #2 0 That problem is too easy. I think the OP made a typo and it should have asked \"..a heart or a jack?\" Also, \"a heart of a jack\" is an unusual way to say it. The expression would have been \"the jack of hearts.\" . Jul 4, 2022 edited by Guest Jul 4, 2022", "+0 # HALP ASAP 0 155 1 A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ($\\heartsuit$ and $\\diamondsuit$, called hearts' and diamonds') are red, the other two ($\\spadesuit$ and $\\clubsuit$, called spades' and clubs') are black. The cards in the deck are placed in random order (usually by a process called `shuffling'). What is the probability that the first two cards drawn from the deck are both red? Mar 29, 2021 #1 +207 +1 Hey there! From your definitions we know that there are 26 red cards in the deck. I'm assuming that there is no replacement when the first card is drawn. The chance of you drawing a red card first is $$\\frac{26}{52}$$ then, since a red card is removed from the deck, the chance of getting a second red card is $$\\frac{25}{51}$$. We then multiply this together to get the probability of this occurring: $$\\frac{26}{52}\\times \\frac{25}{51} \\approx 0.2451 \\approx 24.51\\%$$ Hope this helped :) Mar 29, 2021", "+0 # A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ( and , called `hearts' and -1 151 2 +49 A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ( and , called `hearts' and `diamonds') are red, the other two ( and , called `spades' and `clubs') are black. The cards in the deck are placed in random order (usually by a process called `shuffling'). What is the probability that the first two cards drawn from the deck are both red? off-topic Jun 9, 2020 edited by Guest Jun 9, 2020", "Red color and Spades and Clubs comes in Black color.Each 4 suits contains 13 cards : Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.A king, queen, or jack of a deck of playing cards are known as face cards.", "[Suit] suits = [Clubs ..] data Rank -- we also impose the aces-high order on ranks = Two | Three | Four | Five | Six | Seven | Eight | Nine | Ten | Jack | Queen | King | Ace deriving (Show, Eq, Ord, Enum) ranks :: [Rank] ranks = [Two ..] It will be convenient to distinguish between the “standard” deck of cards and the “full” deck which also includes the joker. So I’ll use two card types: Carte, the French word for card, will represent the standard cards. Card will represent either a standard card or a joker. data Carte = Carte { rank :: Rank , suit :: Suit } deriving (Show, Eq, Ord) standardDeck :: [Carte] standardDeck = [Carte r s | r <- ranks, s <- suits] data Card = Card Carte | Joker deriving (Show, Eq, Ord) fullDeck :: [Card] fullDeck = (map Card standardDeck) ++ [Joker] Sanity check: $> length fullDeck 53 Note the derived Ord instance for Card; we’ve cooked it up so that cards sort on rank and then suit. This will be handy later. We derived Show instances for the Suit, Rank, and Card types to make debugging easier. But it’d also be nice to have a more succinct string representation for cards. We’ll use the", "Hearts, and Spades. It's important to know that these suits come in two colors such as red and black. Suits —Clubs and Spades, come in black color while Diamonds and Hearts in red color. Each suite contains pip cards starting from 2-10 with face cards like Kings, Queens, and Jacks. In a standard deck of cards for spades, there are 4 Aces. The odds of picking any card and getting an ace are 4/52 (or around 8% probability if you prefer it said like that). This means that the probability of picking up the top value card as your next is exactly 1.92%! Multiply that by the 13 cards you have in your deck and it makes the. Deck One Cards On eBay - Deck One Cards On eBa 1. A standard deck contains 52 cards with four ace cards. Each suit; diamond, hearts, clubs, and spades, has its individual ace. This means that there are a total of four ace cards in a standard deck of cards. In a deck of playing cards, the highest value is given to the ace cards 2. Start studying Deck of Cards. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Home Match. Gravity. Created by. laughsara. Terms in this set (13) 52. How many cards in a", "face_names[13] = { \"Ace\", \"Two\", \"Three\", \"Four\", \"Five\", \"Seven\", \"Eight\", \"Nine\", \"Ten\", \"Jack\", \"Queen\", \"King\" }; std::cout << \"You have a \" << face_names[card % 13] << '\\n';} ##### Share on other sites An array of size 51 can hold only 51 elements. There are 52 cards in a set. As for naming cards, there is a different way to do it: #include <vector>#include <string>enum Suit{ DIAMOND, CLUB, SPADE, HEART};enum Rank{ ACE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT, NINE, TEN, JACK, QUEEN, KING};struct Card{ Suit suit; Rank rank; Card(Suit s, Rank r) : suit(s), rank(r) {} };std::ostream &operator<<(std::ostream &out, const Card &c){ static const std::string suits[] = { \"Diamond\", \"Club\", \"Spade\", \"Heart\" }; static const std::string ranks[] = { \"Ace\", \"Two\", \"Three\", \"Four\", \"Five\", \"Six\", \"Seven\", \"Eight\", \"Nine\", \"Ten\", \"Jack\", \"Queen\", \"King\" }; out << ranks[c.rank] << \" of \" << suits[c.suit]; return out;}typedef std::vector<Card> Deck;Deck makeDeck(){ Deck deck; for(int suit = 0 ; suit < 4 ; ++suit) { for(int rank = 0 ; rank < 13 ; ++rank) { deck.push_back(Card(Suit(suit),Rank(rank))); } } return deck;}int main(){ Deck deck = makeDeck(); Card card = deck[rand() % deck.size()]; std::cout << \"You have a \" << card << '\\n';} If you think that is a lot to digest, I agree [smile]. Why is this way \"better\" than yours? -", "# prealgebra A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ($\\heartsuit$ and $\\diamondsuit$, called 'hearts' and 'diamonds') are red, the other two ($\\spadesuit$ and $\\clubsuit$, called 'spades' and 'clubs') are black. The cards in the deck are placed in random order (usually by a process called 'shuffling'). In how many ways can we pick two different cards? (Order matters, thus ace of spades followed by jack of diamonds is different than jack of diamonds followed by ace of spades.) 1. 👍 0 2. 👎 0 3. 👁 632 1. since all the cards are different, this is just 52P2 = 52*51 = 2652 Think about it. There are 52 choices for the first card. For each of those choices there are 51 ways to pick the next card. 1. 👍 0 2. 👎 0 👨‍🏫 oobleck 2. ### math Consider a standard deck of 52 playing cards with 4 suits. What is the probability of randomly drawing 1 card that is both a red card and a face card? (Remember that face cards are jacks, queens, and kings.) Enter your answer as a 3. ### math Three cards are drawn at random without replacement from a standard deck of 52 playing cards. What is", "(`♠`), Hearts (`♥`), Diamonds (`♦`), and Clubs (`♣`). The ranks are Ace (A), 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack (J), Queen (Q), and King (K). Depending on the game that you are playing, an Ace may be higher than King or lower than 2. If we want to define a new object to represent a playing card, it is obvious what the attributes should be: rank and suit. It is not as obvious what type the attributes should be. One possibility is to use strings containing words like `\"Spade\"` for suits and `\"Queen\"` for ranks. One problem with this implementation is that it would not be easy to compare cards to see which had a higher rank or suit. An alternative is to use integers to encode the ranks and suits. In this context, “encode” means that we are going to define a mapping between numbers and suits, or between numbers and ranks. This kind of encoding is not meant to be a secret (that would be “encryption”). For example, this table shows the suits and the corresponding integer codes: • `♠` $\\mapsto$ 4 • `♥` $\\mapsto$ 3 • `♦` $\\mapsto$ 2 • `♣` $\\mapsto$ 1 This code makes it easy to compare cards; because higher suits map to higher numbers, we can compare", "# Math Help - i need help with this combination/probability thingy :( 1. ## i need help with this combination/probability thingy :( i know how to solve these questions, but i'm supposed to put them into categories but i don't know what kind of categories to put them into the teacher also says to write what these problems have in common and why you multiply to get the answers but i am very stupid so i don't know. 1. A deli has five types of meat, two types of cheese, and three types of bread. How many different sandwiches consisting of one type of meat, one type of cheese, and one type of bread, does the deli serve? 2. In a six-team division of a baseball league, each team plays each other 9 times. How many games will be played in the division? 3. A standard 52-card deck has four suits (hearts, clubs, diamonds, and spades) with 13 cards in each suit. How many five-card hands consist of four diamonds and one spade? 4. John has five shirts, three ties, and seven suits. How many possible outfits consisting of one shirt, one tie, and one suit does he have? 5. A locker combination system uses three digits from 0 to 9. How many different three-digit combinations with no", "create objects that contain lists (as attributes); we can create lists that contain lists; we can create objects that contain objects; and so on. In this chapter and the next, we will look at some examples of these combinations, using Card objects as an example. 11.4.2 Card objects If you are not familiar with common playing cards, now would be a good time to get a deck, or else this chapter might not make much sense. There are fifty-two cards in a deck, each of which belongs to one of four suits and one of thirteen ranks. The suits are Spades, Hearts, Diamonds, and Clubs (in descending order in bridge). The ranks are Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, and King. Depending on the game that we are playing, the rank of Ace may be higher than King or lower than 2. The rank is sometimes called the face-value of the card. If we want to define a new object to represent a playing card, it is obvious what the attributes should be: rank and suit. It is not as obvious what type the attributes should be. One possibility is to use strings containing words like &quot;Spade&quot; for suits and &quot;Queen&quot; for ranks. One problem with this implementation is that it would", "suits — spades (♠) and clubs (♣) and two red suits — hearts (♥) and diamonds (♦) In a standard deck of 52 cards there are 13 types. 1 - 10 (1's are called Aces) and three face cards: Jack, Queen and King. For each type there are 4 suites; clubs spades hearts and diamonds. Clubs and spades are black, and hearts and diamonds are red In light of this, how many black face are in a deck of 52 cards? Diamonds and Hearts are red cards (there are 26 total red cards) and Clubs and Spades are black cards (there are 26 total black cards). There are 26 black cards and 12 face cards in total. However, of those 26 black cards, there are 6 face cards How To Play Uno With Regular Playing Cards Card Art Jack Of Spades Playing Cards Design Steam Workshop Deck Of Many Things Deck Of Many Things Inspirational Cards Custom Decks Four Jacks Cards Card Games Game Lovers Amazing Jumping Jack Trick Card Tricks Cool Card Tricks Easy Card Trick The four suits — hearts, clubs, spades and diamonds — represent the four seasons. Meanwhile, the 13 cards in each suit represent the 13 phases of the lunar cycle. If so, it still might have escaped you that if you add", "+0 # I NEED HELP FAST 0 154 2 A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ( and , called 'hearts' and 'diamonds') are red, the other two ( and , called 'spades' and 'clubs') are black. The cards in the deck are placed in random order (usually by a process called 'shuffling'). In how many ways can we pick two different cards? (Order matters, thus ace of spades followed by jack of diamonds is different than jack of diamonds followed by ace of spades.) Sep 12, 2021 #1 0 The cards are ordered, so there are 52*51*2 = 5304 ways. Sep 12, 2021 #2 0 Wait why $$52 \\cdot 51 \\cdot 2?$$ Guest Sep 13, 2021", "say the 3 of Clubs and the Jack of Diamonds, use these commands: 1 2 three_of_clubs = Card(0, 3) card1 = Card(1, 11) In the first case above, for example, the first argument, 0, represents the suit Clubs. Save this code for later use . . . 198 Chapter 11. Classes and Objects How to Think Like a Computer Scientist: Learning with Python 3 Documentation, Release 3rd Edition In the next chapter we assume that we have save the Cards class, and the upcoming Deck class in a file called Cards.py. 11.4.3 Class attributes and the __str__ method In order to print Card objects in a way that people can easily read, we want to map the integer codes onto words. A natural way to do that is with lists of strings. We assign these lists to class attributes at the top of the class definition: 1 2 3 4 class Card: suits = [&quot;Clubs&quot;, &quot;Diamonds&quot;, &quot;Hearts&quot;, &quot;Spades&quot;] ranks = [&quot;narf&quot;, &quot;Ace&quot;, &quot;2&quot;, &quot;3&quot;, &quot;4&quot;, &quot;5&quot;, &quot;6&quot;, &quot;7&quot;, &quot;8&quot;, &quot;9&quot;, &quot;10&quot;, &quot;Jack&quot;, &quot;Queen&quot;, &quot;King&quot;] 5 6 7 8 def __init__(self, suit=0, rank=0): self.suit = suit self.rank = rank 9 10 11 def __str__(self): return (self.ranks[self.rank] + &quot; of &quot; + self.suits[self.suit]) A class attribute is defined outside of any method, and it can be accessed from any of", "as const. # End words Also, as you can see, I introduced assertion in the last code, paradoxically to the length of this post, i found code pretty well writes. It's time, I think, to adopt stronger concepts and methods. Assertions are one of those things. I hope not being too straight or rude, English isn't my primary language, so I miss some nuancies. # 10 days later... I wrote this few days ago, but didn't take time to add it before. Hope it will help. ## Cards Actually in your code, you compare cards with ranks, and then if equals, with suits. So the lowest card is the Ace of Spikes and the greatest the King of Hearts. Also, the Ace of Spikes is lower than the Ace of Clubs which is lower than all others. So, what's cards values? • 1: Ace of Spikes • 2: Ace of Clubs • 3: Ace of Diamonds ... • 51: King of Diamonds • 52: King of Hearts So, instead of dealing with suits and ranks, you can try to deal with values directly. The question is now: \"How retrieve suits and ranks from a card's value?\" Pretty simple! (with card values going from 0 to 51) //ensure card value < (number of suits * number of ranks), then:", "The flush is the easiest. If your Card data type has a selector to extract the suit field (I called mine cardSuit), then you can basically do S.map cardSuit hand. Since the result is a set, each distinct suit in the original hand appears only once. So if the final size of that set is 1, you have a flush! Here’s an illustration in pseudo-interpreter set notation: • S.map cardSuit $$\\{3\\heartsuit,J\\diamondsuit,8\\clubsuit\\}$$$$\\{\\heartsuit,\\diamondsuit,\\clubsuit\\}$$ (not flush) • S.map cardSuit $$\\{4\\heartsuit,A\\spadesuit,5\\heartsuit\\}$$$$\\{\\heartsuit,\\spadesuit\\}$$ (not flush) • S.map cardSuit $$\\{2\\clubsuit,Q\\clubsuit,5\\clubsuit\\}$$$$\\{\\clubsuit\\}$$ (flush!) Detecting a straight is a little more difficult. You can use S.map cardRank hand to extract only the ranks, and then S.toAscList to put them in ascending order. Then, figure out how to ensure there are three ranks with no gaps. • S.toAscList $S.map cardRank $$\\{4\\spadesuit,8\\spadesuit,5\\diamondsuit\\}$$[4,5,8] (not a straight) • S.toAscList$ S.map cardRank $$\\{2\\heartsuit,3\\clubsuit,2\\diamondsuit\\}$$[2,3] (not a 3-card straight) • S.toAscList $S.map cardRank $$\\{2\\heartsuit,4\\clubsuit,3\\heartsuit\\}$$[2,3,4] (straight!) Finally, I paired these two Boolean results into a custom data type like this: data Result = Straight | Flush | Both | Neither deriving (Show, Eq, Ord) scoreHand :: Hand -> Result scoreHand h = case (isStraight h, isFlush h) of (True, True) -> Both (True, False) -> Straight (False, True) -> Flush (False, False) -> Neither So you should be able to calculate: ghci> optimize (scoreHand <$> drawThreeCards)", "red and half are black. The 52 cards are divided evenly into 4 suits: spades, hearts, diamonds, and clubs. Each suit has three face cards (jack, queen, king), and an ace. Each suit also has 9 cards numbered from 2 to 10. Question 7. Dawn draws 1 card, replaces it, and draws another card. Is it more likely that she draws 2 red cards or 2 face cards? ______________ Explanation: There are 26 red cards in the deck and 12 face cards in the deck so it is more likely to draw two red cards than it is to draw two face cards. Question 8. Luis draws 1 card from a deck, 39 times. Predict how many times he draws an ace. _______ times Explanation: A standard deck of cards contains 52 cards, of which 26 are red and 26 are black, 13 are of each suit (hearts, diamonds, spades, clubs), and of which 4 are of each denomination (A, 2 to 10, J, Q, K). The face cards are the jacks J, queens Q, and kings K. There are 52 cards in the deck of cards and thus there 52 possible outcomes. possible outcomes = 52 4 of the 52 cards in a standard deck of cards area aces and thus there are 4 favorable outcomes. favorable" ]

[ "# Combinatorial Probability Question I came across the following probability question: There are $50$ cards of $5$ different colors. It comprises of $10$ Red cards, $10$ blue cards, $10$ orange cards, $10$ green cards and $10$ yellow cards. Each color will have the cards numbered between $1$ to $10$. You pick $2$ cards at random. What is the probability that they are not of same color and not of same number. So, initially, as a direct approach, I thought you would first choose the color of the first card. There are $5$ ways of doing this. Then you could choose the number of the first card. There are $10$ ways of doing this. Then you would choose the color of the second card. There are $4$ ways of doing this since it can't be the same color of the first card. Then you would choose the number of the second card. There are $9$ ways of doing this, since it cannot be the same number as the first card. Finally there are $\\binom{50}{2}$ ways of choosing the $2$ cards. Thus the answer would be $$\\frac{5 \\times 10 \\times 4 \\times 9}{\\binom{50}{2}} > 1$$ which is wrong since it is greater than $1$. The answer given is $$\\frac{\\binom{5}{2} \\times 10 \\times 9}{\\binom{50}{2}} = \\frac{36}{49}$$ My question with this answer", "# Difficult Probability Solved QuestionAptitude Discussion Q. Four unit squares are chosen at random on a chessboard. What is the probability that three of them are of one colour and fourth is of opposite colour? ✖ A. 80/427 ✖ B. 160/427 ✖ C. 320/1281 ✔ D. 640/1281 Solution: Option(D) is correct The number of ways of choosing 4 squares from 64 is $^{64}C_4$ The possible split ups as far as colour is concerned and the number of ways of selecting the square are listed below: White = 0, Black = 4, No. of Ways = ${^{32}C_0} × {^{32}C_4}$ White = 1, Black = 3, No. of Ways = ${^{32}C_1} × {^{32}C_3}$ White = 2, Black = 2, No. of Ways = ${^{32}C_2} × {^{32}C_2}$ White = 3, Black = 1, No. of Ways = ${^{32}C_3} × {^{32}C_1}$ White = 4, Black = 0, No. of Ways = ${^{32}C_4} × {^{32}C_0}$ The number of ways in which 3 squares are of one colour and fourth is of opposite colour is: $P(A) =2\\times {^{32}C_1} \\times {^{32}C_3}$ $=(2\\times 32)\\dfrac{32\\times 31\\times 30}{2\\times 3}$ The total number of ways of selecting 4 squares is: $= P(B)= {^{64}C_4}$ $=\\dfrac{64\\times 63\\times62\\times61}{2\\times3\\times4}$ The required probability $=\\dfrac{P(A)}{P(B)}$ $=\\dfrac{640}{1281}$ Edit: For an alternative solution, check comment by Laurence. ## (9) Comment(s) Siddharth () Let the combination, $WWWB$ be", "black. Ans. It is known that in the pack of $\\text{52}$ cards, there are $\\text{26}$ black cards. Let $\\text{P}\\left( \\text{A} \\right)$ denotes the probability that a black card is drawn in the first draw. Therefore, $\\text{P}\\left( \\text{A} \\right)\\text{=}\\dfrac{\\text{26}}{\\text{52}}\\text{=}\\dfrac{\\text{1}}{\\text{2}}$. Let suppose $\\text{P}\\left( \\text{B} \\right)$ denotes the probability that a black card is drawn in the second draw. Therefore, $\\text{P}\\left( \\text{B} \\right)\\text{=}\\dfrac{\\text{25}}{\\text{51}}$. Hence, the probability that both the cards drawn are black $\\text{=}\\dfrac{\\text{1}}{\\text{2}}\\text{ }\\!\\!\\times\\!\\!\\text{ }\\dfrac{\\text{25}}{\\text{51}}\\text{=}\\dfrac{\\text{25}}{\\text{102}}$. 3. Three oranges are chosen randomly without replacement from a box of oranges. The box is ready to go if all three oranges seem good, otherwise it is rejected. Determine the probability that a box having $\\mathbf{15}$ oranges out of which $\\mathbf{12}$ are good and $\\mathbf{3}$ are bad ones will be ready to go for sale. Ans. Let the events of choosing three oranges be $\\text{A,}\\,\\,\\text{B,}$ and $\\text{C}$ respectively. So, the probability that the orange drawn at first is good, $\\text{P}\\left( \\text{A} \\right)\\text{=}\\dfrac{\\text{12}}{\\text{15}}$. Since, that orange is not replaced so, there only $14$ oranges left. Similarly, the probability that the second orange chosen is good, $\\text{P}\\left( \\text{B} \\right)\\text{=}\\dfrac{\\text{11}}{\\text{14}}$. Also, the probability that the third orange drawn is good, $\\text{P}\\left( \\text{C} \\right)\\text{=}\\dfrac{\\text{10}}{\\text{13}}$. Now, it is described that, if all three oranges are good, then the whole box will be ready to go for sale. Therefore, the", "of choosing 1 black from 26 is 26; and the number of ways of choosing 2 red from 26 is $\\binom{26}{2}$. So the number of ways of choosing 1 black and 2 red is $26\\times\\binom{26}{2}=\\frac{26\\times26!}{2!24!}=\\f rac{26.26.25}{2}=26.13.25$ Therefore the probability that this occurs when 3 cards are chosen from 52 is $\\frac{26.13.25}{26.17.50}=\\frac{13}{34}\\approx 0.382$ I'm not quite sure what you mean by the 'opposite events'. Do you mean the probability that this doesn't happen? If so, then it's $1 - 0.382 = 0.618$. Do 1B and 1C in a similar way. Start B by saying \"The number of ways of choosing 3 reds from 26 is $\\binom{26}{3}$ ...\" And start C by saying \"The number of ways of choosing 1 club from 13 is 13... Can you complete these now? For question 2: The number of ways of choosing 4 balls from 32 is $\\binom{32}{4} =$ ... ? (a) The number of ways of choosing 2 white from 10 is ... ? The number of ways of choosing 2 black from 12 is ... ? So the number of ways of doing each of these things, one after the other is ... ? (b) (Hint: Multiply these two answers together.) Therefore the probability that this occurs is ... ? (Hint: divide the second answer (b) by the first, (a).) Can", "red is the the same as of getting 26 blacks: $(\\tfrac{18}{37})^{25} \\cdot \\tfrac{18}{37} = (\\tfrac{18}{37})^{26}$. The gamblers' chance of getting red or black was the same every time, no matter when they joined the table and what the previous outcomes were. The fact that today's casinos install an LED marquee at roulette tables showing which numbers have occurred recently is no contradiction to this. On the contrary, it is to the casino's advantage if people believe in the gambler's fallacy and keep playing even after a streak of losses or wins: The law of large numbers guarantees that any streak by a player will eventually be overcome by the parameters of the game—which are always in favour of the casino. ### Daily Kit Box The Tech card of the from Asphalt 8 can serve as another example. Every Daily Kit Box grants exactly one Tech card, and this card can only be Advanced Tech (AT) or Mid-Tech (MT). As there are only two possible outcomes, the experiment can be compared to a coin toss, but with two differences: • While the probabilites for a coin toss are known, there are no official drop rates for AT and MT in Daily Kit Boxes. Therefore, they have been inferred statistically from currently 254 boxes, showing an average ratio of 30", "there are $${4 \\choose 2} = 6$$ ways. In the fourth case, there are $${4 \\choose 2} \\times 2 = 12$$ ways. For the fifth case, there is only 1 way. So, there are $$1 + 4 + 6 + 12 + 1 = \\color{brown}\\boxed{24}$$ ways. Aug 28, 2022", "$P\\left( {\\dfrac{A}{{{S^1}}}} \\right)$ = $\\dfrac{{{}^{13}{C_2}}}{{{}^{51}{C_2}}}$ = $\\dfrac{{26}}{{425}}$ $P({S^1} \\cap A) = P({S^1}).P\\left( {\\dfrac{A}{{{S^1}}}} \\right)$ = $\\dfrac{3}{4} \\times \\dfrac{{26}}{{425}}$ The probability that the lost card is spade given that two spades are drawn = $P\\left( {\\dfrac{S}{A}} \\right)$ = $\\dfrac{{P(S \\cap A)}}{{P(A)}}$ = $\\dfrac{{P(S).P(A/S)}}{{P(S).P(A/S) + P({S^1}).P(A/{S^1})}}$ = $\\dfrac{{1/4 \\times 22/425}}{{1/4 \\times 22/425 + 3/4 \\times 26/425}}$ = 11/50 7. There are two bags containing white and black balls. In the first bag there are 8 white and 6 black balls and in the second bag, there are 4 white and 7 black balls. One ball is drawn at random from any of these two bags. Find the probability of this ball being black. a. 21/154 b. 7/54 c. 21/77 d. 41/77 Explanation: Probability = $\\dfrac{1}{2} \\times \\dfrac{{{}^6{C_1}}}{{{}^{14}{C_1}}} + \\dfrac{1}{2} \\times \\dfrac{{{}^7{C_1}}}{{{}^{11}{C_1}}}$ = $\\dfrac{{41}}{{77}}$ 8. A bag contains 1100 tickets numbered 1, 2, 3, ... 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it? a. 291/1100 b. 292/1100 c. 290/1100 d. 301/1100 Explanation: Numbers which dont have 2 from 1 to 9 = 8 Numbers which don't have 2 from 10 to 99: Let us take two places _ _. Now left most place is fixed in 8 ways. Units place is filled with", "2 ways to choose the word \"MATH\" (there are 2 T's), so the probability is $$\\color{brown}\\boxed{1 \\over 5000}$$ BuilderBoi Jul 27, 2022 #4 +12 0 There are $$10^4=10,000$$ ways to choose M, A, T, and H. There is only one way to choose M, A, and H and two ways to choose T in MATHCOUNTS. That means that there are $$1 \\cdot 1 \\cdot 1 \\cdot 2 = 2$$ ways to spell MATH. Then, the probability of getting MATH is $$\\dfrac{2}{10,000} = \\boxed{\\dfrac{1}{5,000}}.$$ And yes, HeWhoShallNotBeNamed (@Voldemort)s' answer was correct. Jul 28, 2022 edited by AngrySesame Jul 28, 2022", "No. of Red honor cards $= 8$ So, a red honor cards can be selected in $8$ ways So, required probability $= \\dfrac{8}{{52}} = \\dfrac{2}{{13}}$ So, the probability that a card drawn at random from a deck of $52$ cards is Red honor is$\\dfrac{2}{{13}}$. Note: There an ‘n’ no. of ways to choose from the deck of $52$ cards and then choosing $8$ red honor cards from that $26$ Red cards. Then probability $= \\dfrac{8}{{26}} \\times \\dfrac{{26}}{{52}}$ $= \\dfrac{2}{{13}}$ Here the probability of red honor cards from $26$ cards is $\\dfrac{8}{{26}}$. And probability of Red cards from a deck of $52$ cards is $\\dfrac{{26}}{{52}}$.", "cards. Let $V(b,r)$ be the expected value to you of state $(b,r)$. If you choose to draw a card in state $(b,r)\\ne (0,0)$, with probability $b/(r+b)$ you draw a black card, losing $1$ point, and then you are in state $(b-1,r)$, while with probability $r/(r+b)$ you draw a red card, gain $1$ point, and go to state $(b,r-1)$. The expected value is thus $$\\dfrac{b}{r+b} (-1 + V(b-1,r)) + \\dfrac{r}{r+b} (1 + V(b,r-1))$$ However, if this is negative, you should stop. Thus $$V(b,r) = \\max\\left(0, \\dfrac{b}{r+b} (-1 + V(b-1,r)) + \\dfrac{r}{r+b} (1 + V(b,r-1))\\right)$$ with $V(0,0) = 0$. I get $V(2,2) = 2/3$. With intelligent play, you never have a negative result (as you can always continue until all the cards have been dealt. The best play expectation is $\\frac12$ for the chance of the first card being good, plus $\\frac16$ for the chance that the first card is bad but the next two are good, for a total of $\\frac23$. • why are you considering only the cases where you win early? What about case where you choose Red Black Red? – Zack Mar 14 '16 at 2:14 • @Zack: I show that continuing after you draw red first doesn't change the expectation. – Ross Millikan Mar 14 '16 at 2:21 • Because after drawing red, you will", "we draw another card from the bag and lay it face up on the table. The face that is shown on the new card is white. Show that the probability that the first card, the one showing a black side, has black on its other side is now 0.75. Use the counting method, if you can. Hint: Treat this like the sequence of globe tosses, counting all the ways to see each observation, for each possible first card. As the hint suggests, let’s fill in the table below by thinking through each possible combination of first and second cards that could produce the observed data. If the first card was the first side of BB, then there would be 3 ways for the second card to show white (i.e., the second side of BW, the first side of WW, or the second side of WW). If the first card was the second side of BB, then there would be the same 3 ways for the second card to show white. So the total ways for the first card to be BB is $$3+3=6$$. If the first card was the first side of BW, then there would be 2 ways for the second card to show white (i.e., the first side of WW or the second side of WW;", "4}^2$$ There are $$4 \\choose 2$$ ways to choose which rolls are $$\\ge 10$$. Thus, the probability is: $$\\large{{1 \\over 4} \\times {1 \\over 4} \\times {3 \\over 4} \\times { 3 \\over 4} \\times {4 \\choose 2}} = \\color{brown}\\boxed{7 \\over 128}$$ BuilderBoi Apr 28, 2022", "third card. Bet$1.33 on the fourth certain card for a final total of $2.66 If the second card is red you will have$0.67 and there will be 2 black cards remaining in the deck. Bet everything on black twice for \\$2.68 Obviously, if the first card turns up black, a symmetric strategy will have the same result. All times are GMT -5. The time now is 10:23 PM.", "of ways we can choose 6 things out of a collection of 9. So we just need to know: what's 9 choose 6? We can work this out as before. There are 9 ways to pick one of the flips and draw a red H on it, then 8 ways left to pick another and draw a blue H on it, and 7 ways left to pick a third and draw a orange H on it. That sounds like 9 × 8 × 7. But we've overcounted! After all, we don't care about the colors. We don't care about the difference between this: (T, H, T, T, H, T, T, H, T) and this: (T, H, T, T, H, T, T, H, T) In fact we've counted each possibility 6 times! Why six? The first H could be red, green or blue — that's 3 choices. But then the second H could be either of the two remaining 2 colors... and for the third, we just have 1 choice. So there are 3 × 2 × 1 = 6 ways to permute the colors. So, the actual number of ways to get 6 heads out of 9 coin flips is $$\\displaystyle{ \\frac{9 \\times 8 \\times 7}{3 \\times 2 \\times 1} }$$ In other words: $$\\displaystyle{ \\binom{9}{6} = \\frac{9", "to make the later arithmetic simple. Then \"about\" $7700$ times she chooses the box with $3$ orange and $4$ black, and \"about\" $7700$ times she chooses the box with $5$ orange and $6$ black. Out of the about $7700$ times she chooses the first box, she gets orange \"about\" $\\frac{3}{7}(7700)$ times, that is, $3300$ times. And out of the roughly $7700$ times she chooses the second box, she gets orange about $\\frac{5}{11}(7700)$ times, so roughly $3500$ times. So she got orange about $6800$ times. And they came from the second box about $3500$ times. So we would expect that the probability the ball came from the second box given that it is orange should be $\\dfrac{3500}{6800}$. It is safer to use the machinery of conditional probabilities. But the above calculation may help give an informal idea of what is going on. In essence we are resricting the sample space to those situations where we got an orange ball. -", "This is not at all necessary. For example:$$ \\lim_{x\\to 0} \\frac {x^2}x =0, $$but \\left.\\dfrac{x^2}{x}\\right\\vert_{x=0} is undefined. If we do have that f is defined at a and \\lim\\limits_{x\\to a}f(x)=f(a), then f is called continuous at x=a. 0 Think of the color options as 4 slots with \"nothing\" being a possible subtype. Selecting that option in, say, the blue slot would simply mean that there was no blue bonus attached to that card. By the specified terms you could choose \"nothing\" in all 4 slots or in any subset. Viewed this way, there are 21 possible subtypes in each of the four slots ... 0 You have 3 possibility : you are looking the First card's side 1, you are looking the first card's side 2 or you are looking the third card's side red. Sot the probability that the second side is red is \\frac 2 3 1 Let B be the event that you selected the card with different sides, and R that you selected the red card. You are calculating$$P(\\text{down red} | \\text{up red})=P(B|\\text{red up})P(\\text{down red}|B)+P(R|\\text{red up})P(\\text{down red}|R)=\\frac{P(B\\cap\\text{red up})}{P(\\text{red up}}\\cdot 1+\\frac{P(R\\cap\\text{red up})}{P(\\text{red up}}\\cdot ... 4 The first card has sides $R_1$ and $R_2$, two red sides. The second has $B_1$ and $B_2$ (two black sides), and the last has $R_3$ and $B_3$. You see either", "still not it, I will be glad to give it another try. 5. Hello, jplondon! If I read the problem correctly, it is quite complicated. How many different combinations are there if i have 6 boxes with 4 different items in each box, and take 1 from each box If you don't shuffle the 24 cards before loading the six feeders, . . the problem is much simpler. The arrangement might look like this: . . $\\begin{array}{ccc}\\boxed{1,2,3,4} & \\boxed{5,6,7,8} & \\boxed{9,10,11,12} \\\\ \\\\ \\boxed{13,14,15,16} & \\boxed{17,18,19,20} & \\boxed{21,22,23,24} \\end{array}$ We take one from each feeder and for each there are 4 choices. Hence, there are: . $4^6 \\:=\\:4096$ possible four-card packs. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ If you do shuffle the 24 cards before leading the feeders, . . there will be far more possible outcomes. But there will so much duplication, that it is virtually impossible . . to come up with an accurate count. In the first arrangement, we could select: . $\\{4,8,12,16,20,24\\}$ In one of the shufflings, the distribution might be, for example: . . $\\begin{array}{ccc}\\boxed{1,{\\color{red}8},10,15} & \\boxed{5,{\\color{red}12},17,22} & \\boxed{{\\color{red}4},11,19,23} \\\\ \\\\ \\boxed{2,6,14,{\\color{red}20}} & \\boxed{7,13,18,{\\color{red}24}} & \\boxed{3,9,{\\color{red}16},21} \\end{array}$ And we", "# Which is the better deck of cards? You have two decks of cards: a 52 card deck (26 black , 26 red) and a 26 card deck (13 black, 13 red). You randomly draw two cards and win if both are the same color. Which deck would you prefer? What if the 26 card deck was randomly drawn from the 52 card deck? Which deck would you prefer then? The first part seemed easy since $\\dfrac{25}{51} > \\dfrac{12}{25}$. But the second part seems more complicated. Then the probabilities for drawing both reds or blacks would be different so how would we compare them? • The first question: You prefer the bigger deck, because it is \"closer\" to infinite - that is, the second draw is less skewed towards the other suit. The second is no difference - it is the same as picking two cards at random from the 52-card deck. – Thomas Andrews Nov 27 '15 at 23:56 • For the skewed deck, let there be $20$ cards that are red and $6$ cards that are black. The probability here is $\\dfrac{20}{26} \\cdot \\dfrac{19}{25}+\\dfrac{6}{26} \\cdot \\dfrac{5}{25} > \\dfrac{25}{51}$. This is greater than the probability of a greater deck so how is there no difference? Also, when we take the cards out of the $52$ card deck", "26, and how many ways are there to choose 13 black cards out of 26?", "pairs of cards from the previous step, there are 2 possible third cards, so there are a total of 24 possibilities for the first 3 cards: There is now only one card left, so we have no choice which card to use as the final card. This means that there are 24 possible permutations of the four cards, as shown in the first diagram above. For 4 cards the number of permutations is: $$4 \\times 3 \\times 2 \\times 1 = 24$$ This is, of course, 4 factorial. This is often written as 4! In general, the number of permutations of n unique items is n factorial: $$n \\times (n - 1) \\times ... \\times 2 \\times 1 = n!$$ ## Number of permutations of r items selected from n items Next, we will look at a slightly different case. We have n items, but we will only select r of those items. For example imagine we have 5 different coloured counters - red, yellow, green, blue, and black. We will select 2 of those counters. We might choose: • A red counter then a yellow counter. • A green counter then a white counter. • A yellow counter then a red counter. How many permutations of 2 counters are there? The first thing to be clear here" ]

[ "are blue. Three balls are chosen at random, without replacement. 1. Find the probability that the chosen balls are all the same color. 2. Find the probability that the chosen balls are all different colors. 1. $$\\frac{34}{455}$$ 2. $$\\frac{120}{455}$$ Suppose again that an urn contains 15 balls: 6 are red, 5 are green, and 4 are blue. Three balls are chosen at random, with replacement. 1. Find the probability that the chosen balls are all the same color. 2. Find the probability that the chosen balls are all different colors. 1. $$\\frac{405}{3375}$$ 2. $$\\frac{720}{3375}$$ #### Cards Recall that a standard card deck can be modeled by the product set $D = \\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k\\} \\times \\{\\clubsuit, \\diamondsuit, \\heartsuit, \\spadesuit\\}$ where the first coordinate encodes the denomination or kind (ace, 2-10, jack, queen, king) and where the second coordinate encodes the suit (clubs, diamonds, hearts, spades). Sometimes we represent a card as a string rather than an ordered pair (for example $$q \\heartsuit$$). Card games involve choosing a random sample without replacement from the deck $$D$$, which plays the role of the population. Thus, the basic card experiment consists of dealing $$n$$ cards from a standard deck without replacement; in this special context, the sample of cards is often", "want to embed an alignged environment in one square of an array to make sure the hands line up pretty; I'd like the name of the player to be centered above the hands too) $$\\begin{array}{l l l} & North & \\\\ & \\clubsuit: AQ & \\\\ & \\diamondsuit: xx & \\\\ & \\heartsuit: Axx & \\\\ & \\spadesuit: JTxxxx & \\\\ West & & East \\\\ \\clubsuit: xxxx & & \\clubsuit: xxx \\\\ \\diamondsuit: Axx & & \\diamondsuit: xxxx \\\\ \\heartsuit: KQxxx & & \\heartsuit: xxxxx \\\\ \\spadesuit: 4 & & \\spadesuit: 6 \\\\ & South & \\\\ & \\clubsuit: KJxx & \\\\ & \\diamondsuit: KQxx & \\\\ & \\heartsuit: & \\\\ & \\spadesuit: AKQxx &$$ Play opens with west leading $K\\heartsuit$. Last edited: Nov 29, 2003 2. Nov 29, 2003 ### Loren Booda My mother often played a foursome with her girlfriend Andree, sometimes teaming up with bridge great Freddie Sheinwold. A particular hand was dealt while Andree was out of the room. When she picked it up, she found 13 of the same suit! My mother swore that there was no collusion, although the chance of such a deal is astronomical. 3. Dec 1, 2003 ### KLscilevothma Wow, coool !! You and your partner got total 29 points, and you got a void suit, perfect split!", "+0 # HALP ASAP 0 155 1 A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ($\\heartsuit$ and $\\diamondsuit$, called hearts' and diamonds') are red, the other two ($\\spadesuit$ and $\\clubsuit$, called spades' and clubs') are black. The cards in the deck are placed in random order (usually by a process called `shuffling'). What is the probability that the first two cards drawn from the deck are both red? Mar 29, 2021 #1 +207 +1 Hey there! From your definitions we know that there are 26 red cards in the deck. I'm assuming that there is no replacement when the first card is drawn. The chance of you drawing a red card first is $$\\frac{26}{52}$$ then, since a red card is removed from the deck, the chance of getting a second red card is $$\\frac{25}{51}$$. We then multiply this together to get the probability of this occurring: $$\\frac{26}{52}\\times \\frac{25}{51} \\approx 0.2451 \\approx 24.51\\%$$ Hope this helped :) Mar 29, 2021", "six jokers, but we do not take these into account here. Each card of a standard 52-card deck is represented by a string (str) that contains the rank of the card, followed by its suit: • ranks: 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) and A (ace) • suits: C (clubs; ), D (diamonds; ), H (hearts; ) and S (spades; ) As such, the ace of spades is for example represented as AS, the ten of hearts as 10H and the king of clubs as KC. A stack of cards is represented as a list (list) or a tuple (tuple) whose elements are the representations of the individual cards in the stack, listed from top to bottom. A stack of cards does not necessarily need to contain all 52 cards of a standard deck. Your task: • Write a function card_count that takes a stack of cards. The function also has an optional parameter color. If no argument is explicitly passed to the parameter color, the function must return the number of cards in the given stack. If the string red is passed to the parameter color, the function must return the number of red cards in the given stack. If the string black is passed to the parameter", "The flush is the easiest. If your Card data type has a selector to extract the suit field (I called mine cardSuit), then you can basically do S.map cardSuit hand. Since the result is a set, each distinct suit in the original hand appears only once. So if the final size of that set is 1, you have a flush! Here’s an illustration in pseudo-interpreter set notation: • S.map cardSuit $$\\{3\\heartsuit,J\\diamondsuit,8\\clubsuit\\}$$$$\\{\\heartsuit,\\diamondsuit,\\clubsuit\\}$$ (not flush) • S.map cardSuit $$\\{4\\heartsuit,A\\spadesuit,5\\heartsuit\\}$$$$\\{\\heartsuit,\\spadesuit\\}$$ (not flush) • S.map cardSuit $$\\{2\\clubsuit,Q\\clubsuit,5\\clubsuit\\}$$$$\\{\\clubsuit\\}$$ (flush!) Detecting a straight is a little more difficult. You can use S.map cardRank hand to extract only the ranks, and then S.toAscList to put them in ascending order. Then, figure out how to ensure there are three ranks with no gaps. • S.toAscList $S.map cardRank $$\\{4\\spadesuit,8\\spadesuit,5\\diamondsuit\\}$$[4,5,8] (not a straight) • S.toAscList$ S.map cardRank $$\\{2\\heartsuit,3\\clubsuit,2\\diamondsuit\\}$$[2,3] (not a 3-card straight) • S.toAscList $S.map cardRank $$\\{2\\heartsuit,4\\clubsuit,3\\heartsuit\\}$$[2,3,4] (straight!) Finally, I paired these two Boolean results into a custom data type like this: data Result = Straight | Flush | Both | Neither deriving (Show, Eq, Ord) scoreHand :: Hand -> Result scoreHand h = case (isStraight h, isFlush h) of (True, True) -> Both (True, False) -> Straight (False, True) -> Flush (False, False) -> Neither So you should be able to calculate: ghci> optimize (scoreHand <$> drawThreeCards)", "# prealgebra A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ($\\heartsuit$ and $\\diamondsuit$, called 'hearts' and 'diamonds') are red, the other two ($\\spadesuit$ and $\\clubsuit$, called 'spades' and 'clubs') are black. The cards in the deck are placed in random order (usually by a process called 'shuffling'). In how many ways can we pick two different cards? (Order matters, thus ace of spades followed by jack of diamonds is different than jack of diamonds followed by ace of spades.) 1. 👍 0 2. 👎 0 3. 👁 632 1. since all the cards are different, this is just 52P2 = 52*51 = 2652 Think about it. There are 52 choices for the first card. For each of those choices there are 51 ways to pick the next card. 1. 👍 0 2. 👎 0 👨‍🏫 oobleck 2. ### math Consider a standard deck of 52 playing cards with 4 suits. What is the probability of randomly drawing 1 card that is both a red card and a face card? (Remember that face cards are jacks, queens, and kings.) Enter your answer as a 3. ### math Three cards are drawn at random without replacement from a standard deck of 52 playing cards. What is", "$\\frac{{39\\choose21}}{{52\\choose21}}\\left(1 - \\frac{{18\\choose2}}{{31\\choose2}} \\right)$ Suppose 1 $\\diamondsuit$ is removed. . This probability is: . $\\frac{{13\\choose1}{39\\choose20}}{{52\\choose21}}$ The deck contains 12 $\\diamondsuit$s and 19 Others. The probability that no $\\diamondsuit$ is drawn is: . $\\frac{{19\\choose2}}{{31\\choose2}}$ The probabiity that 1 $\\diamondsuit$ is removed and at least one $\\diamondsuit$ is drawn is: . $\\frac{{13\\choose1}{39\\choose20}}{{52\\choose21}}\\le ft(1 - \\frac{{19\\choose2}}{{31\\choose2}}\\right)$ . . . $\\vdots$ We must extend this list up to: Suppose 11 $\\diamondsuit$s are removed. . This probability is: . $\\frac{{13\\choose11}{39\\choose10}}{{52\\choose21}} $ The deck contains 2 $\\diamondsuit$s and 29 Others. The probability that no $\\diamondsuit$ is drawn is: . $\\frac{{29\\choose2}}{{31\\choose2}}$ The probability that 11 $\\diamondsuit$s are removed and at least one $\\diamondsuit$ is drawn is: . $\\frac{{13\\choose11}{39\\choose10}}{{52\\choose21}} \\left(1 - \\frac{{29\\choose2}}{{31\\choose2}}\\right)$ The final probability is the sum of these twelve probabilities. You go on ahead . . . I'll wait in the car. . 7. Originally Posted by Soroban Hello, khor! The second one has an incredibly long and tedious solution. Since it depends on how many diamonds were removed, . . we must consider all possible cases. Suppose no $\\diamondsuit$ is removed. . This probability is: . $\\frac{{39\\choose21}}{{52\\choose21}}$ The deck contains: 13 $\\diamondsuit$s and 18 Others. The probability that no $\\diamondsuit$ is drawn is: . $\\frac{{18\\choose2}}{{31\\choose2}}$ The probability that no $\\diamondsuit$ is removed and at least one $\\diamondsuit$ is drawn is: . $\\frac{{39\\choose21}}{{52\\choose21}}\\left(1 -", "\"Both in the second row.\" The word is BIBLE . . . and there are two B's. His cards are at those locations. Continue in this manner for each of the volunteers. 3. ## Another silly trick In the array below, [1] Touch any Red card. [2] Touch the screen at the nearest BLACK card . . .which is to the left or right of your chosen card. [3] Touch the screen at the nearest RED card . . .which is vertically up or down from the black card. [4] Touch the screen at the nearest BLACK card . . . which is diagonally adjacent from this card. [5] Touch the screen to the nearest RED card . . .which is down or to the right of this card. . . $\\begin{array}{ccccccc}\\boxed{3\\spadesuit}\\\\ \\\\ \\boxed{9\\heartsuit} \\\\ \\\\ \\boxed{Q\\diamondsuit} \\\\ \\\\ \\boxed{4\\heartsuit}\\end{array}$ $\\begin{array}{cccc}\\boxed{K\\diamondsuit} \\\\ \\\\ \\boxed{J\\spadesuit} \\\\ \\\\ \\boxed{4\\clubsuit} \\\\ \\\\ \\boxed{Q\\spadesuit}\\end{array}$ $\\begin{array}{cccc}\\boxed{8\\diamondsuit} \\\\ \\\\ \\boxed{2\\clubsuit} \\\\ \\\\ \\boxed{A\\diamondsuit} \\\\ \\\\ \\boxed{7\\clubsuit}\\end{array}$ $\\begin{array}{cccc}\\boxed{3\\heartsuit} \\\\ \\\\ \\boxed{6\\spadesuit} \\\\ \\\\ \\boxed{7\\heartsuit} \\\\ \\\\ \\boxed{2\\spadesuit}\\end{array}$ Concentrate on your final card. Your card is: (drag cursor between the asterisks) . . * The Ace of Diamonds * Page 2 of 2 First 12 , , , , , , ### shorttrick group in group throry Click on a term to search for related topics.", "# Assignment 10 due at 23:59 on +80 Question: Suppose we draw three cards from a shuffled deck of playing cards. What is the probability of getting a straight, a flush, or both? You should answer this using the probability distribution monad from class on 30 November, save your implementation as a10.hs, and submit it to the dropbox at https://www.dropbox.com/request/xz9LTmkMl6fCcCCLwZKR. # Definitions In case you’re not familiar, I’ll define all the situations needed here. Playing cards have both a suit and a rank. There are four suits, known as Hearts ($$\\heartsuit$$), Spades ($$\\spadesuit$$), Clubs ($$\\clubsuit$$), and Diamonds ($$\\diamondsuit$$). There are thirteen ranks, known as the integers 2 through 10, and then Jack, Queen, King, and Ace. Every combination of suit and rank appears in a deck, so that’s $$4\\times13=52$$ cards. When you have a small selection of cards (called a hand), you can recognize patterns in it. A flush means that all the cards in the hand have the same suit. So with three cards, if you drew $$3\\heartsuit, 9\\heartsuit, A\\heartsuit$$, that would be a flush. A straight means that all the cards in the hand have consecutive ranks. So with three cards, if you drew $$9\\diamondsuit, 10\\clubsuit, J\\spadesuit$$, that would be a straight. When you draw cards from a deck, you are selecting without replacement. So instead", "card2){ Card tempCard; tempCard = card1; card1 = card2; card 2 = tempCard;} It''s a little rough around the edges but it should get you a shuffled deck then all you need to do is to give both the player and the dealer 2 cards to start with. The dealer will usually stay if their hand is 18 or over. If not then they will hit. The player just needs to be asked if he''d like to hit. Then you count up the values and see who''s over 21 and BAM you''ve got blackjack! Share on other sites I''m bored, so the AP gets random suggestions: 1) The first enum. Try: enum { eSuit_Hearts, eSuit_Diamonds, eSuit_Clubs, eSuit_Spades, eSuit_Count}; This way, eSuit_Count will == the number of suits, no matter how many you add. 2) The second enum only causes confusion. Why would you want \"two\" valued at 0? And the face cards should all have the same value. And the aces are special cases. And... 3) Your deck fill-in is buggy, unless you wanted a deck of all Spades valued from 0 - 51. 4) There is not item #4 in this list. 5) I could mention the STL random_shuffle, or why using % with rand() isn''t the happiest of ideas in applications that matter, but this is", "as const. # End words Also, as you can see, I introduced assertion in the last code, paradoxically to the length of this post, i found code pretty well writes. It's time, I think, to adopt stronger concepts and methods. Assertions are one of those things. I hope not being too straight or rude, English isn't my primary language, so I miss some nuancies. # 10 days later... I wrote this few days ago, but didn't take time to add it before. Hope it will help. ## Cards Actually in your code, you compare cards with ranks, and then if equals, with suits. So the lowest card is the Ace of Spikes and the greatest the King of Hearts. Also, the Ace of Spikes is lower than the Ace of Clubs which is lower than all others. So, what's cards values? • 1: Ace of Spikes • 2: Ace of Clubs • 3: Ace of Diamonds ... • 51: King of Diamonds • 52: King of Hearts So, instead of dealing with suits and ranks, you can try to deal with values directly. The question is now: \"How retrieve suits and ranks from a card's value?\" Pretty simple! (with card values going from 0 to 51) //ensure card value < (number of suits * number of ranks), then:", ". Since we each win half the time, the bet is fair. The above explanations are comforting and reasonable . . . but wrong! . . There are six outcomes: . $(\\heartsuit\\,\\diamondsuit),\\;(\\heartsuit\\,\\spadesu it),\\;(\\heartsuit\\,\\clubsuit),\\;(\\diamondsuit\\,\\sp adesuit),\\;(\\diamondsuit\\,\\clubsuit),\\;(\\spadesuit \\,\\clubsuit)$ . . And in only two of them, $(\\heartsuit\\,\\diamondsuit),\\;(\\spadesuit\\,\\clubsui t)$, the colors match. Your probability of winning is: . $\\frac{2}{6}\\:=\\:\\frac{1}{3}$ 4. Originally Posted by Soroban This is a classic trick question . . . you know a lot of those, don't you? 5. Originally Posted by Soroban This is the basis for a \"Sucker Bet\". I have four playing cards, one of each suit, face down on the table. You pick any two of them. You will bet that the cards are of the same color. I will bet that they have opposite colors. . . And we bet \"even money\". Are you being hustled? Let's reason it out . . . There are only two outcomes: the colors match or they do not match. . . Since these outcomes are equally likely, the bet is \"fair\". Okay, there are four outcomes: (Red, Red), (Red, Black), (Black, Red), (Black, Black) . . Since we each win half the time, the bet is fair. The above explanations are comforting and reasonable . . . but wrong! . . There are six outcomes: . $(\\heartsuit\\,\\diamondsuit),\\;(\\heartsuit\\,\\spadesu", "+0 # help probability 0 34 2 A standard deck of 52 cards has 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King) and 4 suits such that there is exactly one card for any given rank and suit. Two of the suits are black and the other two suits are red. The deck is randomly arranged. What is the probability that the top card is a heart of a jack? Jul 3, 2022 #1 +2275 +1 There is exactly 1 heart of jack, and 52 cards, so the probability is $$\\color{brown}\\boxed{1 \\over 52}$$ Jul 4, 2022 #2 0 That problem is too easy. I think the OP made a typo and it should have asked \"..a heart or a jack?\" Also, \"a heart of a jack\" is an unusual way to say it. The expression would have been \"the jack of hearts.\" . Jul 4, 2022 edited by Guest Jul 4, 2022", "Hearts, and Spades. It's important to know that these suits come in two colors such as red and black. Suits —Clubs and Spades, come in black color while Diamonds and Hearts in red color. Each suite contains pip cards starting from 2-10 with face cards like Kings, Queens, and Jacks. In a standard deck of cards for spades, there are 4 Aces. The odds of picking any card and getting an ace are 4/52 (or around 8% probability if you prefer it said like that). This means that the probability of picking up the top value card as your next is exactly 1.92%! Multiply that by the 13 cards you have in your deck and it makes the. Deck One Cards On eBay - Deck One Cards On eBa 1. A standard deck contains 52 cards with four ace cards. Each suit; diamond, hearts, clubs, and spades, has its individual ace. This means that there are a total of four ace cards in a standard deck of cards. In a deck of playing cards, the highest value is given to the ace cards 2. Start studying Deck of Cards. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Home Match. Gravity. Created by. laughsara. Terms in this set (13) 52. How many cards in a", "of long hair: (55%*44%) + (45%*15%) = 0.2420 + 0.0675 = 0.3095 = 30.95% chance of female or long hair = 55% + 30.95% = 85.95% chance of female with long hair: 0.2420 = 24.2% 2. Hello, Sneaky! 4. Suppose a card is randomely chosen from a standard 52 card deck. What is the probability that the card is a Jack or a Club (or both) ? You started correctly, though . . . There are 4 Jacks: $J\\heartsuit,\\,J\\spadesuit,\\, J\\diamondsuit,\\,\\boxed{J\\clubsuit}$ . . $P(J) \\:=\\:\\dfrac{4}{52}$ There are 13 Clubs: $A\\clubsuit,\\,2\\clubsuit,\\,3\\clubsuit,\\,4\\clubsuit, \\,5\\clubsuit,\\,6\\clubsuit,\\,7\\clubsuit,\\,8\\clubsui t,\\,9\\clubsuit,\\,10\\clubsuit,\\,\\boxed{J\\clubsuit}, \\,Q\\clubsuit,\\,K\\clubsuit$ . . $P(\\clubsuit) \\:=\\:\\dfrac{13}{52}$ $\\text{But }\\,P(J\\text{ or }\\clubsuit) \\;=\\;\\dfrac{16}{52}$ There are only 16 cards that will \"win the bet\". You counted the $J\\clubsuit$ twice. 3. ok i understand why its 16/52. This also means that the other 4 questions were correct? But now i have some other concerns. This is a question from my book. a guy has 3 cards 1 red on both sides 1 black on both sides 1 red / black he puts 1 card on the table, and u only see the top side which is red. what are the chances of the other side being red my guess is 50/50, but the book says 2/3 using conditional probability. Why is it 2/3?? Also a question about number 5 Like u said", "say the 3 of Clubs and the Jack of Diamonds, use these commands: 1 2 three_of_clubs = Card(0, 3) card1 = Card(1, 11) In the first case above, for example, the first argument, 0, represents the suit Clubs. Save this code for later use . . . 198 Chapter 11. Classes and Objects How to Think Like a Computer Scientist: Learning with Python 3 Documentation, Release 3rd Edition In the next chapter we assume that we have save the Cards class, and the upcoming Deck class in a file called Cards.py. 11.4.3 Class attributes and the __str__ method In order to print Card objects in a way that people can easily read, we want to map the integer codes onto words. A natural way to do that is with lists of strings. We assign these lists to class attributes at the top of the class definition: 1 2 3 4 class Card: suits = [&quot;Clubs&quot;, &quot;Diamonds&quot;, &quot;Hearts&quot;, &quot;Spades&quot;] ranks = [&quot;narf&quot;, &quot;Ace&quot;, &quot;2&quot;, &quot;3&quot;, &quot;4&quot;, &quot;5&quot;, &quot;6&quot;, &quot;7&quot;, &quot;8&quot;, &quot;9&quot;, &quot;10&quot;, &quot;Jack&quot;, &quot;Queen&quot;, &quot;King&quot;] 5 6 7 8 def __init__(self, suit=0, rank=0): self.suit = suit self.rank = rank 9 10 11 def __str__(self): return (self.ranks[self.rank] + &quot; of &quot; + self.suits[self.suit]) A class attribute is defined outside of any method, and it can be accessed from any of", "it),\\;(\\heartsuit\\,\\clubsuit),\\;(\\diamondsuit\\,\\sp adesuit),\\;(\\diamondsuit\\,\\clubsuit),\\;(\\spadesuit \\,\\clubsuit)$ . . And in only two of them, $(\\heartsuit\\,\\diamondsuit),\\;(\\spadesuit\\,\\clubsui t)$, the colors match. Your probability of winning is: . $\\frac{2}{6}\\:=\\:\\frac{1}{3}$ [/size] i'll keep that in mind next time i go gambling, thanks 6. Hi, Jhevon! You know a lot of those, don't you? Yes, I do . . . Over the years, I've been surprised (or been had) by dozens of these \"sucker\" bets. . . The first was probably the \"Birthday Paradox\". A similar bet can be made with a friend while standing on a street corner. Bet your friend that, among the next 15 license plates that go by, . . some two will end in the same two-digit number. [You might agree to disregard those ending in letters.] Argument: \"Hey, there are a hundred different two-digit numbers. . . . . . . . . What are the chances?\" Truth: the odds are about 2-to-1 in your favor. Super Hustle: The same bet with 18 cars. . . . . . . . . . . The odds are about 4-to-1 in your favor. 7. Originally Posted by Soroban Hello, xfyz! This is a classic trick question . . . [size=3] When two coins are flipped, there are four possible outcomes: . $HH,\\:HT,\\:TH,\\:TT$ Since the student saw a Tail, there are only", "(`♠`), Hearts (`♥`), Diamonds (`♦`), and Clubs (`♣`). The ranks are Ace (A), 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack (J), Queen (Q), and King (K). Depending on the game that you are playing, an Ace may be higher than King or lower than 2. If we want to define a new object to represent a playing card, it is obvious what the attributes should be: rank and suit. It is not as obvious what type the attributes should be. One possibility is to use strings containing words like `\"Spade\"` for suits and `\"Queen\"` for ranks. One problem with this implementation is that it would not be easy to compare cards to see which had a higher rank or suit. An alternative is to use integers to encode the ranks and suits. In this context, “encode” means that we are going to define a mapping between numbers and suits, or between numbers and ranks. This kind of encoding is not meant to be a secret (that would be “encryption”). For example, this table shows the suits and the corresponding integer codes: • `♠` $\\mapsto$ 4 • `♥` $\\mapsto$ 3 • `♦` $\\mapsto$ 2 • `♣` $\\mapsto$ 1 This code makes it easy to compare cards; because higher suits map to higher numbers, we can compare", "at least one ace}) = \\dfrac{C^{25}_3 - C^{22}_3}{4960}$. #### MrGremlin ##### New member Thank you very much. So, b) P(A) - no face cards. there are 12 face cards in the deck.(including aces) 30-12 = 20 cards. $P(A)= \\dfrac {C^{20}_3} {4960}$ P(B)= the same number of colors. $P(\\clubsuit + \\diamondsuit+ \\heartsuit+ \\spadesuit)= 1 - P(\\overline{\\clubsuit + \\diamondsuit+ \\heartsuit+\\spadesuit})$, but they are incompatible events. So $P(\\overline{\\clubsuit + \\diamondsuit+ \\heartsuit+\\spadesuit})=0$ and P(B) = 1. - - - Updated - - - c)The number of cards by suits 7, 8, 8, 9. i.e. the number of successful samples = 7*8*8 +7*8*9 +7*8*9+8*8*9= 2032 P(A)= 2032/4960 #### Klaas van Aarsen ##### MHB Seeker Staff member b) choosing the same suit and there are no face cards Thank you very much. So, b) P(A) - no face cards. there are 12 face cards in the deck.(including aces) 30-12 = 20 cards. $P(A)= \\dfrac {C^{20}_3} {4960}$ P(B)= the same number of colors. $P(\\clubsuit + \\diamondsuit+ \\heartsuit+ \\spadesuit)= 1 - P(\\overline{\\clubsuit + \\diamondsuit+ \\heartsuit+\\spadesuit})$, but they are incompatible events. So $P(\\overline{\\clubsuit + \\diamondsuit+ \\heartsuit+\\spadesuit})=0$ and P(B) = 1. You can't lump them all together. It matters which suit you pick. So let's start with the first suit $\\clubsuit$. How many cards that are not facecards? Yielding how many ways to draw 3 cards in the", "you spot the $2 \\clubsuit$ then you know that 13 cards later (and If you reach the end of the deck, continue counting with the top card) is the $2 \\heartsuit$, 13 cards after that is the $2 \\spadesuit$, and 13 cards after that is the $2 \\diamondsuit$. Here are some rules this system satisfies: Rule 1 “Shuffling”: Never riff shuffle or mix the cards. Instead, split the deck in two, the “bottom half” as the left stack and the “top half” as the right stack. Take the left stack and place it on the right one. This has the effect of rotating the deck but preserving the ordering. Do this as many times as you like. Mathematically, each such cut adds an element of the group $G$ to each element of the deck. Some people call this a “false shuffle” of “false cut.” Rule 2 “Rank position”: The corresponding ranks of successive cards in the deck differs by 3. Rule 3 “Suit position”: Every card of the same denomination is 13 cards apart and runs in the same order of suits as in the CHaSeD mnemonic, namely, Clubs $\\clubsuit$, Hearts $\\heartsuit$, Spades $\\spadesuit$, Diamonds $\\diamondsuit$. At least, we can give a few simple card tricks based on this system. Trick 1: A player picks a card from" ]

[ "Browse Questions A coin and six faced die, both unbiassed, are thrown simultaneously. The probability of getting a head on the coin and an odd number on the die, is : $\\begin {array} {1 1} (a)\\frac{1}{2} & \\quad (b)\\;\\frac{3}{4} \\\\ (c)\\;\\frac{1}{4} & \\quad (d)\\;\\frac{2}{3} \\end {array}$ +1 vote $(c)\\;\\frac{1}{4}$", "pennies and 9 dimes. A coin is randomly taken and not replaced. A second coin is randomly taken. What is probability that both coins are pennies? -and- Two dice are rolled. What is the probability that the sum of the...", "one of the numbers rolled. (…)” Given this, we can have some insight into how the numbers on the 20-sided dice were chosen: they’re all products of two numbers between 1 and 9 inclusive, with the factors distributed somewhat uniformly across the interval: begin{array}{|c|c|c|} hline textbf{Die 3} & textbf{Die 4} & textbf{Die 5} \\ hline begin{aligned} 4 &= 1 times 4 = 2 times 2 \\ 5 &= 1 times 5 \\ 7 &= 1 times 7 \\ 10 &= 2 times 5 \\ 12 &= 2 times 6 = 3 times 4 \\ 15 &= 3 times 5 \\ 16 &= 2 times 8 = 4 times 4 \\ 18 &= 2 times 9 = 3 times 6 \\ 20 &= 4 times 5 \\ 21 &= 3 times 7 \\ 24 &= 3 times 8 = 4 times 6 \\ 28 &= 4 times 7 \\ 35 &= 5 times 7 \\ 36 &= 4 times 9 = 6 times 6 \\ 42 &= 6 times 7 \\ 49 &= 7 times 7 \\ 54 &= 6 times 9 \\ 56 &= 7 times 8 \\ 64 &= 8 times 8 \\ 72 &= 8 times 9 \\ end{aligned} & begin{aligned} 1 &= 1 times 1 \\ 2 &= 1 times 2 \\ 3 &= 1", "# Probability for rolling an odd number and tossing a coin on heads A coin is tossed and a die rolled. Find the probability of getting a head and an odd number. The answer is $\\frac{1}{4}$. My reasoning is that rolling an odd number is $\\frac{1}{2}$, and tossing a coin on heads is $\\frac{1}{2}$. So $0.5 \\times 0.5$ = $\\frac{1}{4}$. Is this basically it? • Yes. The outcome of the die is independent of the outcome of the coin: $P(d\\cap c)=P(d|c)\\cdot P(c)=P(d)\\cdot P(c)$ – callculus Jul 7 '15 at 10:19", "higher or lower numbers than with a standard die. The coin flip bias is only at around the 1% level (I have no idea what it would be for a 20-sided die), so whether this is significant or not depends on what level of effects you care about. But casinos make a living on advantages on the level of a few percent. -", "Together, these have a probability of $\\dfrac56 \\cdot \\dfrac18 = \\dfrac5{48}$. Now we add the probabilities from the separate cases. The probability that there is exactly one 6 is $\\frac{7}{48} + \\frac{5}{48} = \\frac{12}{48} = \\boxed{\\frac{1}{4}}.$ ## Similar Questions 1. ### math If a standard six-sided die is rolled twice, what is the probability that the result of the second roll is not less than the result of the first roll? 2. ### pre-algebra B Milli tosses a coin and a six-sided die.The numbers on the die are 1 through 6. What is the probability that Milli will toss a tail and roll an odd number? 3. ### math A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8? 4. ### counting and probability A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8? 5. ### probability - confused A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8? A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number", "arrange four different … 6. ### probability The digits 2, 3, 4, 7, and 8 are each used once in a random order to form a five-digit number. What is the probability that the resulting number is divisible by 4? 7. ### does anyone know probability???? The digits 2, 3, 4, 7, and 8 are each used once in a random order to form a five-digit number. What is the probability that the resulting number is divisible by 4? 8. ### probability - confused A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8?", "Independent probability Problem If you flip a coin and roll a 6-sided die, what is the probability that you will flip a heads and roll a 2?", "25 x 102 50 dimes per roll × 60 rolls = (5 x 6) x (10 × 10) = 30 x 102 50 dimes per roll × 70 rolls = (5 x 7) x (10 × 10) = 35 x 102 50 dimes per roll × 80 rolls = (5 x 8) x (10 × 10) = 40 x 102 50 dimes per roll × 90 rolls = (5 x 9) x (10 × 10) = 45 x 102 50 dimes per roll × 100 rolls = (5 x 10) x (10 × 10) = 50 x 102 Question 21. 1 roll = 40 quarters ; Think:40 quarters per roll × 20 rolls =(4 × 2) × (10 × 10) Type below: __________ Answer: Explanation: 1 roll = 40 quarters ; Think:40 quarters per roll × 20 rolls =(4 × 2) × (10 × 10) = 8 x 102 40 quarters per roll × 30 rolls =(4 × 3) × (10 × 10) = 12 x 102 40 quarters per roll × 40 rolls =(4 × 4) × (10 × 10) = 16 x 102 40 quarters per roll × 50 rolls =(4 × 5) × (10 × 10) = 20 x 102 40 quarters per roll × 60 rolls =(4 × 6) × (10 × 10) =", "# A Standard Die is rolled 5 times, How many Different outcomes are possible? Everytime you roll a die, 1 of 6 outcomes comes up. So for every die rolled there is 6 outcomes. Each roll is independent of the roll before, so for 5 rolls there are $6^5=7776$ outcomes.", "die is rolled (faces 1 and 6 have probability $$\\frac{1}{4}$$ each, and faces 2, 3, 4, 5 have probability $$\\frac{1}{8}$$ each). Find the mean and standard deviation of the die score. 1. $$\\frac{7}{2}$$ 2. $$1.8634$$", "coin once if the number on the die is even. If the number on the die is odd, the coin is flipped twice. Using the notation 4H, for example to denote an event that die comes … 6. ### DIsct MAth 1. Consider the \"coin flipping game\", and let X record the first time when the Head\" player has \\$1. What is the probability that X \u0014 7?", "There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "the probability of rolling the number when rolling a standard six-sided die? ### 6. What is the probability of drawing an even number from a set of cards numbered through ?", "1. ## probability 4. a) A single fair die is rolled. Find the probability of getting an odd number. b) A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is either a 10 or a diamond. 2. Originally Posted by bobby77 4. a) A single fair die is rolled. Find the probability of getting an odd number. All faces of the die are equally likely. The favourable outcomes in this case are: 1, 3, 5 out of a total of 6 possible outcomes. Therefore the probability of an odd number when the die is rolled is: [number of favourable outcomes]/[total number of outcomes] = 3/6 = 0.5 b) A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is either a 10 or a diamond. Here the favourable outcomes are A , 2 , 3 .., 10, J, Q, K of diamonds and 10 of hearts, 10 of Spades, 10 of clubs which is a total of 16 favourable cases. Total number of outcomes = 52, so the required probability is: 16/52=4/13 RonL", "# A Standard Die is rolled 5 times, How many Different outcomes are possible? • October 22nd 2009, 02:24 PM tjrustok A Standard Die is rolled 5 times, How many Different outcomes are possible? A Standard Die is rolled 5 times, How many Different outcomes are possible? The answer in the back of my textbook say's 7776, I have no clue as to how that is the answer. • October 22nd 2009, 04:43 PM Robb Everytime you roll a die, 1 of 6 outcomes comes up. So for every die rolled there is 6 outcomes. Each roll is independent of the roll before, so for 5 rolls there are $6^5=7776$ outcomes.", "the coin exactly four times, this would be $1/16$, as there is a unique outcome that this event corresponds to: $TTTH$.", "uniform distribution, in which all outcomes have an equal chance of occurring. A six-sided die has a uniform distribution. Each outcome has a probability of about 16.67% (1/6). Our plot below shows the solid line (so you can see it better), but keep in mind that this is a discrete distribution—you can't roll 2.5 or 2.11: Now, roll two dice together, as shown in the figure below, and the distribution is no longer uniform. It peaks at seven, which happens to have a 16.67% chance. In this case, all the other outcomes are less likely: Now, roll three dice together, as shown in the figure below. We start to see the effects of a most amazing theorem: the central limit theorem. The central limit theorem boldly promises that the sum or average of a series of independent variables will tend to become normally distributed, regardless of their own distribution. Our dice are individually uniform but combine them and—as we add more dice—almost magically their sum will tend toward the familiar normal distribution. ## Binomial Distribution The binomial distribution reflects a series of \"either/or\" trials, such as a series of coin tosses. These are called Bernoulli trials—which refer to events that have only two outcomes—but you don't need even (50/50) odds. The binomial distribution below plots a series of", "the product of two numbers on opposite sides of the multiplication symbol. Choose a number on each side of the multiplication symbol that when multiplied will equal one of the numbers rolled. (…)” Given this, we can have some insight into how the numbers on the 20-sided dice were chosen: they’re all products of two numbers between 1 and 9 inclusive, with the factors distributed somewhat uniformly across the interval: begin{array}{|c|c|c|} hline textbf{Die 3} & textbf{Die 4} & textbf{Die 5} \\ hline begin{aligned} 4 &= 1 times 4 = 2 times 2 \\ 5 &= 1 times 5 \\ 7 &= 1 times 7 \\ 10 &= 2 times 5 \\ 12 &= 2 times 6 = 3 times 4 \\ 15 &= 3 times 5 \\ 16 &= 2 times 8 = 4 times 4 \\ 18 &= 2 times 9 = 3 times 6 \\ 20 &= 4 times 5 \\ 21 &= 3 times 7 \\ 24 &= 3 times 8 = 4 times 6 \\ 28 &= 4 times 7 \\ 35 &= 5 times 7 \\ 36 &= 4 times 9 = 6 times 6 \\ 42 &= 6 times 7 \\ 49 &= 7 times 7 \\ 54 &= 6 times 9 \\ 56 &= 7 times 8 \\ 64 &= 8", "fair die is rolled 10 times. Find the probability that there is at least one 5. Solution: P(at least one 5) = 1 - P(none of the rolls equal 5) = 1 - P(all 10 rolls are not five) ⇒ P(5) = 1/6 P(not 5) = 5/6 = 1 - (5/6)10 = 0.838 2. A man purchases a box of 75 decorative light bulbs for his restaurant. If the probability that any light bulb is defective is 0.02, calculate the probability that at least one light bulb is defective in the box of 75. Solution: P(at least 1 light bulb is defective) = 1 - P(none of the light bulbs are defective) = 1 - P(all 75 light bulbs are good) ⇒ P(bad light bulb) = 0.02 P(good light bulb) = 0.98 = 1 - (0.98)75 = 0.7802" ]

[ "286 views An unbiased die is thrown $n$ times. The probability that the product of numbers would be even is 1. $\\dfrac{1}{(2n)}$ 2. $\\dfrac{1}{[(6n)!]}$ 3. $1 - 6^{-n}$ 4. $6^{-n}$ 5. None of the above. edited | 286 views Even number =$2,4,6$ odd number = $1,3,5$ Product will come even when even one time even number comes odd product will be if every time odd comes so, $P(Even) = 1- P(odd)$ $= 1- {^n}C{_n}\\times \\left(\\dfrac{3}{6}\\right)^{0}\\times \\left(\\dfrac{3}{6}\\right)^{n}$ $= 1- \\left(\\dfrac{1}{2}\\right)^{n}$ So i think option E edited 0 Prob. of product being even = Favorable outcome/Total outcomes = (6n - 3n)/6n= 1- 1/2n Total = 6n Favor. = 6n - [no. formed by only 1,3,5] =6n - 3n Here we need to find the probability that at least one time even number should come. Product of Even and Odd = Even Product of Even and Even =Even Product of Odd and Odd =Odd =1 - probability of coming only odds in \"n\" throws =1-(1/2)n reshown The product will come even when at least one time an even number comes in n rolls. P(even in 1 roll) =P(odd in 1 roll) =3/6 =1/2 P(atleast one time even in n rolls) = 1 - P(all times odd in n rolls) = 1 - 1/2n Hence ,the correct answer is 1 - 1/2", "craft Nov 5 '15 at 7:50 Calculate $1$ minus the probability of the complementary event: • The probability of getting an odd number in a roll is $\\frac36$ • The probability of getting an odd product in $n$ rolls is $\\left(\\frac36\\right)^n$ • The probability of getting an even product in $n$ rolls is $1-\\left(\\frac36\\right)^n$", "+0 # IMPOSSIBLE QUESTION 0 41 2 +103 3 six-sided dices are rolled. What is the probability that the product is even? MATHEXPERTISE Oct 21, 2018 #1 +2729 +2 $$P[\\text{product is even}]=1-P[\\text{product is odd}] = 1 - P[\\text{all three dice roll odd}]$$ $$P[\\text{1 die rolls odd}]=\\dfrac 1 2 \\\\ P[\\text{product is odd}]=P[\\text{all 3 dice roll odd}]=\\left(\\dfrac 1 2\\right)^3 = \\dfrac 1 8\\\\ P[\\text{product is even}]=1 - \\dfrac 1 8 = \\dfrac 7 8$$ Rom Oct 21, 2018 #2 +103 +1 thanx the answer is correct :D MATHEXPERTISE Oct 21, 2018", "+0 # IMPOSSIBLE QUESTION 0 141 2 +128 3 six-sided dices are rolled. What is the probability that the product is even? Oct 21, 2018 #1 +4402 +2 $$P[\\text{product is even}]=1-P[\\text{product is odd}] = 1 - P[\\text{all three dice roll odd}]$$ $$P[\\text{1 die rolls odd}]=\\dfrac 1 2 \\\\ P[\\text{product is odd}]=P[\\text{all 3 dice roll odd}]=\\left(\\dfrac 1 2\\right)^3 = \\dfrac 1 8\\\\ P[\\text{product is even}]=1 - \\dfrac 1 8 = \\dfrac 7 8$$ . Oct 21, 2018 #2 +128 +1 thanx the answer is correct :D MATHEXPERTISE Oct 21, 2018", "@fabulousgrizzly I think minji meant that in a set of five rolls, it doesn't matter what the first four rolls are. It only matters what the parity (oddness or evenness) of the sum of the first four rolls is. See, if the sum of the first four rolls is even (e.g. $$1 + 2 + 3 +4 = 10),$$ then we will want the fifth roll to be also even (e.g. $$10 + \\text{even} = \\text{even}$$), and if the sum of the first four rolls is odd (e.g. $$1 +2 + 3 + 3 = 9),$$ then we will want the fifth roll to be odd as well (since $$9 + \\text{odd} = \\text{even}$$). So since we don't care what the first four numbers are, there are $$6$$ choices for each of those, and since the last roll has to match the parity (oddness or evenness) of the running total, there are $$3$$ choices for that number. Actually, I have another way of solving this question! Each dice has three odd numbers and three even numbers. The chance of getting an even is $$\\frac{1}{2},$$ and the chance of getting an odd is $$\\frac{1}{2}.$$ How about not worrying about what the numbers are, and only keeping track of whether the number on the dice is even or odd? Then,", "and an unassigned even box. As such, the probability of having at least two even box/paper combinations equals 1 minus the probability of having only odd box/paper combinations: $$1-\\frac{(n!)^2}{(2n)!}$$", "\\frac 12 \\Rightarrow \\mathrm{(C)}$. Solution 3 The product will be a multiple of 3 if and only if at least one of the two rolls is a 3 or a 6. The probability that Juan rolls 3 or 6 is $2/8 = 1/4$. The probability that Juan does not roll 3 or 6, but Amal does is $(3/4) (1/3) = 1/4$. Thus, the probability that the product of the rolls is a multiple of 3 is $$\\frac{1}{4} + \\frac{1}{4} = \\boxed{\\frac{1}{2}}.$$ ~aopsav (Credit to AoPS Alcumus)", "# A dice is thrown, what is the probability that it is an odd prime number? Thank you! Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 27 Alan P. Share Nov 13, 2016 $\\frac{1}{3}$ or 33 1/3% #### Explanation: The odd prime numbers that could appear on a die roll are $3$ and $5$ (note that $1$ is not considered a prime). Since theer are $6$ possible roll outcomes, the probability of an odd prime is $2$ out of $6$ or $\\frac{2}{6} = \\frac{1}{3}$. (Often probability is expressed as a percentage in which case 1/3=33 1/3%) • 14 minutes ago • 14 minutes ago • 14 minutes ago • 14 minutes ago • 2 minutes ago • 4 minutes ago • 7 minutes ago • 10 minutes ago • 13 minutes ago • 14 minutes ago • 14 minutes ago • 14 minutes ago • 14 minutes ago • 14 minutes ago", "the probability the third, fourth, fifth, and sixth are different from the all previous rolls is is $\\frac{5}{6}$, $\\frac{4}{6}$, $\\frac{3}{6}$, and $\\frac{2}{6}$ respectively. The probability the first two are the same and the last four different is $\\frac{1}{6}\\cdot\\frac{1}{6}\\cdot\\frac{5}{6}\\cdot\\frac{4}{6}\\cdot\\frac{3}{6}\\cdot\\frac{2}{6} = \\frac{5!}{6^6}$. There are $\\binom{6}{2} = \\frac{6!}{2!\\ \\cdot\\ 4!} = 15$ different ways to order 6 things, two of which are the same so the probability of two rolls the same, the other 4 different in any order is $15\\cdot \\frac{5!}{6^6}$.", "# The probability of winning a peculiar dice game I'm a high school teacher, and students in my probability class created the following fun conundrum. We've been stuck on it for a couple of weeks: • Consider this dice game played with one fair $$n$$-sided dice. • On the first turn, a roll of $$n$$ wins, while a roll of $$1$$ loses. On any other result, the player rolls again. • On the 2nd roll, a roll of $$n$$ wins, while a roll of $$1$$ or $$2$$ loses. On any other roll, the game continues. • On roll $$k$$, the player wins with a roll of $$n$$ and loses with a roll of $$k$$ or below. • What is the probability of winning as $$n \\to \\infty$$ ?. The game must be won in no more than $$n - 1$$ turns, and for any given $$n$$ , $$\\mathrm{P}\\left(win\\right) = {1 \\over n} + \\sum_{i = 2}^{n - 1}\\frac{\\left(n - 2\\right)!}{\\left(n - i - 1\\right)!\\, n^{i}}$$ Here is where I'm stuck. Does: $$\\lim_{n \\to \\infty}\\mathrm{P}\\left(win\\right) = 0?$$ or does $$\\mathrm{P}\\left(win\\right)$$ converge on some other nonzero probability as $$n \\to \\infty$$ ?. How might one show this ?. • I think $P(win) \\lt \\dfrac{1}{\\sqrt{n}}$. If so, it does tend to $0$ but fairly slowly – Henry Jun 13 '19 at", "of a random variable. You can use this to work out the average score of a dice roll over 6 rolls. Given the outcomes=$(1, 2)$ and the probabilities=$(1/8, 1/4)$, the expected value $E[x]$ is: $E[x]$ = $1(1/8) + 2(1/4) = 0.625$. Interpretation: average in large number of independent repetitions. Caution: If we have an infinite sum, it needs to be well-defined. Rules -Linearity of Expected Value $E[F+G] = E[F] + E[G]$. -Multiplication Rule of Expected Value $E[F x G]$ does not equal $E[F] * E[G]$", "is odd. 3 even numbers. 3 odd numbers. First one odd and second even; first even and second odd = 3 × 3 × 2 = 18 outcomes The outcomes are (1, 2)(1, 4)(1, 6)(3, 2)(3, 4)(3, 6)(5, 2)(5, 4)(5, 6) (2, 1)(2, 3)(2, 5)(4, 1)(4, 3)(4, 5)(6, 1)(6, 3)(6, 5) (b) Both are even. 3 possibilities for the first and 3 for the second = 3 × 3 = 9 outcomes. The outcomes are (2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6) = 9 outcomes Total outcomes when product is even = 18 + 9 = 27 Total outcomes when two different dice are thrown = 6 × 6 = 36 Probability that product is even = $\\frac{\\text{number of outcomes when product is even}}{\\text{Total number of outcomes}}$ = $\\frac{27}{36}$ = $\\frac{3}{4}$ .", "Option (e) None of the above ,is the ans. edited by 0 only one possibility product will be odd when all the numbers is odd. total possiblities - 6^n prob that will be odd= 1/6^n so prob that even will be 1- prob that will be odd = 1-6^-n so answer c.. what is wrong with my approch", "4 views A fair dice is rolled twice. The probability that an odd number will follow an even number is 1. $\\frac{1}{2}$ 2. $\\frac{1}{6}$ 3. $\\frac{1}{3}$ 4. $\\frac{1}{4}$", "to $$\\dfrac12 bh$$ formula for area, we can see that the area is $$\\dfrac{42\\cdot20}2 = 420.$$ Thus, the correct answer is B. 7. Each of two boxes contains three chips numbered $$1,$$ $$2,$$ $$3.$$ A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even? $$\\dfrac{1}{9}$$ $$\\dfrac{2}{9}$$ $$\\dfrac{4}{9}$$ $$\\dfrac{1}{2}$$ $$\\dfrac{5}{9}$$ ###### Solution(s): First, let us try to find the probability that the product is odd. We know that the product will be odd if both boxes yield odd numbers. As there is a $$\\dfrac 23$$ chance of a box being odd, there is a $\\left(\\dfrac 23\\right)^2 = \\dfrac 49$ chance of the product being odd. The probability that a number is even is $$1$$ minus the probability that it is odd (as they are complements), so the odds of it being even is $1 - \\dfrac 49 = \\dfrac 59.$ Thus, the correct answer is E. 8. What is the smallest whole number larger than the perimeter of any triangle with a side of length $$5$$ and a side of length $$19?$$ $$24$$ $$29$$ $$43$$ $$48$$ $$57$$ ###### Solution(s): Let the third side be $$s.$$ By the triangle inequality, we know $$s < 5+19.$$ Therefore, the perimeter of the triangle: $s+5 +19 <", "answers. #### Choice B ##### The number of outcomes in which the sum of the digits that appear on the facing side is odd when a fair die rolled thrice. The total number of outcomes when a fair die is rolled thrice is 6 * 6 * 6 = 216. If you list down a few of the outcomes and observe the pattern, you will realize that half the outcomes have a sum that is odd and the other half has a sum that is even. Here is a sample of the first 6. {1, 1, 1, odd}, {1, 1, 2, even}, {1, 1, 3, odd}, {1, 1, 4, even}, {1, 1, 5, odd}, {1, 1, 6, even}. Therefore, ½ of 216 = 108 outcomes have an odd sum. Choice B is one of the answers where the number of outcomes is more than 50. #### Choice C ##### The number of outcomes in which the two cards drawn from a pack of well shuffled cards are both red and face cards. A pack of well shuffled cards has 52 cards. The break up of the pack on different parameters is listed in the table given below Color Suit Numbered cards Face cards Aces Red - 26 cards Diamond - 13 cards$2 to 10) 9 cards (J, Q,", "4}^2$$ There are $$4 \\choose 2$$ ways to choose which rolls are $$\\ge 10$$. Thus, the probability is: $$\\large{{1 \\over 4} \\times {1 \\over 4} \\times {3 \\over 4} \\times { 3 \\over 4} \\times {4 \\choose 2}} = \\color{brown}\\boxed{7 \\over 128}$$ BuilderBoi Apr 28, 2022", "in four rolls is the complement of at least one six in four rolls (thus they must sum to one), and used the rule of subtraction to compute the probability of interest: $P(\\text{at least one six in four rolls}) = 1 - \\bigg(\\frac{5}{6}\\bigg)^4=0.517$ de Méré’s gamble that he would throw at least one six in four rolls has a probability of greater than 0.5, explaning why de Méré made money on this bet on average. But what about de Méré’s second bet? Pascal used the same trick: $P(\\text{no double six in 24 rolls}) = \\bigg(\\frac{35}{36}\\bigg)^{24}=0.509$ $P(\\text{at least one double six in 24 rolls}) = 1 - \\bigg(\\frac{35}{36}\\bigg)^{24}=0.491$ The probability of this outcome was slightly below 0.5, showing why de Méré lost money on average on this bet. ## 3.3 Probability distributions We often want to be able to quantify the probability of any possible value in an experiment. For example, on Jan 20 2018, the basketball player Steph Curry hit only 2 out of 4 free throws in a game against the Houston Rockets. We know that Curry’s overall probability of hitting free throws across the entire season was 0.91, so it seems pretty unlikely that he would hit only 50% of his free throws in a game, but exactly how unlikely is it? We can determine this", "#### When a die is thrown, the probability of getting an odd number less than 3 is (A) $\\frac{1}{6}$ (B) $\\frac{1}{3}$ (C) $\\frac{1}{2}$ (D) 0 Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Total no. of cases = 6 odd number less than 3 = 1 Number of favorable cases = 1 Probability = $= \\frac{1}{6}$", "# Probability that the sum of 6 dice rolls is even Question: 6 unbiased dice are tossed together. What is the probability that the sum of all the dice is an even number? I think the answer would be 50%, purely by intuition. However, not sure if this is correct. How should I go about solving such a problem? Notice that whatever the sum of the first 5 rolls, whether the outcome is odd or even is totally determined by the last die. It is even or odd with equal probability, so the probability of an even sum is exactly the same as the probability of an odd sum. The first 5 dice don't matter. Statement: Let $X_1,\\ldots,X_n$ be independent, integer-valued random variables, and let some $X_k$ satisfy $P(X_k \\mbox{ odd}) = 0.5$. Then $P(\\sum X_i \\mbox{ odd}) = 0.5$. Proof: Without loss of generality let $k=n$, and let $S = \\sum_i^{n-1} X_i$. Since $S$ and $X_n$ are independent random variables, $$\\begin{eqnarray*} P\\left(\\sum X_i \\mbox{ odd}\\right) = P(S + X_n \\mbox{ odd}) &=& P(S \\mbox{ odd and } X_n \\mbox{ even}) + P(S \\mbox { even and } X_n \\mbox{ odd}) \\\\ &=& P(S \\mbox{ odd})P(X_n \\mbox{ even}) + P(S \\mbox{ even})P(X_n \\mbox{ odd}) \\\\ &=& P(S \\mbox{ odd})P(X_n \\mbox{ odd}) + P(S \\mbox{ even})P(X_n \\mbox{ odd}) \\\\" ]

[ "There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "# Probability for rolling an odd number and tossing a coin on heads A coin is tossed and a die rolled. Find the probability of getting a head and an odd number. The answer is $\\frac{1}{4}$. My reasoning is that rolling an odd number is $\\frac{1}{2}$, and tossing a coin on heads is $\\frac{1}{2}$. So $0.5 \\times 0.5$ = $\\frac{1}{4}$. Is this basically it? • Yes. The outcome of the die is independent of the outcome of the coin: $P(d\\cap c)=P(d|c)\\cdot P(c)=P(d)\\cdot P(c)$ – callculus Jul 7 '15 at 10:19", "# A dice is thrown, what is the probability that it is an odd prime number? Thank you! Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 27 Alan P. Share Nov 13, 2016 $\\frac{1}{3}$ or 33 1/3% #### Explanation: The odd prime numbers that could appear on a die roll are $3$ and $5$ (note that $1$ is not considered a prime). Since theer are $6$ possible roll outcomes, the probability of an odd prime is $2$ out of $6$ or $\\frac{2}{6} = \\frac{1}{3}$. (Often probability is expressed as a percentage in which case 1/3=33 1/3%) • 14 minutes ago • 14 minutes ago • 14 minutes ago • 14 minutes ago • 2 minutes ago • 4 minutes ago • 7 minutes ago • 10 minutes ago • 13 minutes ago • 14 minutes ago • 14 minutes ago • 14 minutes ago • 14 minutes ago • 14 minutes ago", "Browse Questions A coin and six faced die, both unbiassed, are thrown simultaneously. The probability of getting a head on the coin and an odd number on the die, is : $\\begin {array} {1 1} (a)\\frac{1}{2} & \\quad (b)\\;\\frac{3}{4} \\\\ (c)\\;\\frac{1}{4} & \\quad (d)\\;\\frac{2}{3} \\end {array}$ +1 vote $(c)\\;\\frac{1}{4}$", "286 views An unbiased die is thrown $n$ times. The probability that the product of numbers would be even is 1. $\\dfrac{1}{(2n)}$ 2. $\\dfrac{1}{[(6n)!]}$ 3. $1 - 6^{-n}$ 4. $6^{-n}$ 5. None of the above. edited | 286 views Even number =$2,4,6$ odd number = $1,3,5$ Product will come even when even one time even number comes odd product will be if every time odd comes so, $P(Even) = 1- P(odd)$ $= 1- {^n}C{_n}\\times \\left(\\dfrac{3}{6}\\right)^{0}\\times \\left(\\dfrac{3}{6}\\right)^{n}$ $= 1- \\left(\\dfrac{1}{2}\\right)^{n}$ So i think option E edited 0 Prob. of product being even = Favorable outcome/Total outcomes = (6n - 3n)/6n= 1- 1/2n Total = 6n Favor. = 6n - [no. formed by only 1,3,5] =6n - 3n Here we need to find the probability that at least one time even number should come. Product of Even and Odd = Even Product of Even and Even =Even Product of Odd and Odd =Odd =1 - probability of coming only odds in \"n\" throws =1-(1/2)n reshown The product will come even when at least one time an even number comes in n rolls. P(even in 1 roll) =P(odd in 1 roll) =3/6 =1/2 P(atleast one time even in n rolls) = 1 - P(all times odd in n rolls) = 1 - 1/2n Hence ,the correct answer is 1 - 1/2", "This puzzle came to me via Futility CLoset Two people are stranded on an island with only one banana to eat. To decide who gets it, they agree to play a game. Each of them will roll a fair 6-sided die. If the largest number rolled is a 1, 2, 3, or 4, then Player 1 gets the banana. If the largest number rolled is a 5 or 6, then Player 2 gets it. Which player has the better chance? I don’t think this is an especially difficult puzzle, but I think the key insight is nice. Spoilers below the line. The naive instinct, of course, is that player 1 has more options available, so their chance must be larger. But no! The probability of both rolling a 4 or lower is $\\left( \\frac23 \\right)^2$, or $\\frac{4}{9}$, which is less than a half! Player 2 is slightly more likely to win (a ratio of 5:4). As long as player 1’s win probability is smaller than $\\frac{1}{\\sqrt{2}}$, player 2 will (more likely) win.", "# You flip a coin and roll a die. What is the probability that you obtain a head on the coin and a 2 on the die? Feb 12, 2015 The probability that you obtain a head on the coin and a 2 on the die is $\\left(\\frac{1}{2}\\right) \\cdot \\left(\\frac{1}{6}\\right) = \\left(\\frac{1}{12}\\right)$ Assuming the coin and the die are unbiased: the probability of obtaining a head on the coin is $\\frac{1}{2}$ (50%) and the probability of obtaining a 2 on die is $\\frac{1}{6}$. These are independent events. What happens with the coin does not effect what happens with the die. Therefore the probability of the combined outcome is the product of their individual probabilities. From the above we can see that there are 12 (6 X 2) possible combined outcomes. The desired outcome (the intersection of the Die Outcome $= 2$ and the Coin Outcome $= H$) only occurs once. Therefore the probability of the desired outcome is $1$ out of $12$ (or, $\\frac{1}{12}$).", "# A six sided die is rolled six times. What is the probability that each side appears exactly once? Sep 5, 2017 The probability is approximately 1.54%. #### Explanation: On the first roll, there are no restrictions. The die is allowed to be any of the 6 equally likely values. Thus the probability of not duplicating any numbers so far after roll 1 is $\\frac{6}{6}$, or 100%. For each subsequent roll, the number of \"successful\" rolls decreases by 1. For instance, if our first roll was a $\\left[3\\right]$, then the second roll needs to be anything but $\\left[3\\right]$, meaning there are 5 \"successful\" outcomes (out of the 6 possible) for roll 2. So, since each roll is independent of the previous rolls, we multiply their \"success\" probabilities together. The probability of rolling no repeats after two rolls is $\\frac{6}{6} \\times \\frac{5}{6} = \\frac{5}{6} ,$ which is about 83.3%. Continuing this pattern, the third roll will have 4 \"successful\" outcomes out of 6, so we get Pr(\"3 unique rolls\") = 6/6 xx5/6 xx 4/6=55.6% and then Pr(\"4 unique rolls\") = 6/6 xx5/6 xx 4/6 xx3/6=27.8% Pr(\"5 unique rolls\") = 6/6 xx5/6 xx 4/6 xx3/6xx2/6=9.26% and finally Pr(\"6 unique rolls\") = (6 xx 5 xx 4 xx 3 xx 2 xx 1)/(6^6)=1.54%.", "Independent probability Problem If you flip a coin and roll a 6-sided die, what is the probability that you will flip a heads and roll a 2?", "Together, these have a probability of $\\dfrac56 \\cdot \\dfrac18 = \\dfrac5{48}$. Now we add the probabilities from the separate cases. The probability that there is exactly one 6 is $\\frac{7}{48} + \\frac{5}{48} = \\frac{12}{48} = \\boxed{\\frac{1}{4}}.$ ## Similar Questions 1. ### math If a standard six-sided die is rolled twice, what is the probability that the result of the second roll is not less than the result of the first roll? 2. ### pre-algebra B Milli tosses a coin and a six-sided die.The numbers on the die are 1 through 6. What is the probability that Milli will toss a tail and roll an odd number? 3. ### math A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8? 4. ### counting and probability A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8? 5. ### probability - confused A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8? A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number", "seen as an example of $$1-\\left(\\frac{n-1}{n}\\right)^n$$, which for increasing $$n$$ would tends towards $$1-\\frac1e \\approx 0.6321$$, getting further away from your normal distribution calculation • It is possible to find an artificial relationship using the standard normal density function at one standard deviation: $$1 - 2\\pi \\left(\\phi(1)\\right)^2 = 1-\\frac1e$$ too, but this expression has no natural interpretation The \" philosophical\" point is that the die does not have memory : when you throw it for the sixth time, it doesn't know what happened before, and does not care whether the ace already has appeared or not. Sixth draw is same as first. Yet the die is also \"fair\": it does not have preferences. And it is the combination of memoriless and fairness that produce \"randomness\". And randomness has only a \"regularity\" within statistics, as explained in the other answers. However, that the concept is difficult to grasp intuitively is demonstrated by all the people following the lotto numbers .. in delay. In this respect, consider the following situation. You launched the die four times without aces when a naughty boy steals your die and make a couple of draws getting two aces and returns the die to you: what are the chances you are left now ?", "1. ## probability 4. a) A single fair die is rolled. Find the probability of getting an odd number. b) A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is either a 10 or a diamond. 2. Originally Posted by bobby77 4. a) A single fair die is rolled. Find the probability of getting an odd number. All faces of the die are equally likely. The favourable outcomes in this case are: 1, 3, 5 out of a total of 6 possible outcomes. Therefore the probability of an odd number when the die is rolled is: [number of favourable outcomes]/[total number of outcomes] = 3/6 = 0.5 b) A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is either a 10 or a diamond. Here the favourable outcomes are A , 2 , 3 .., 10, J, Q, K of diamonds and 10 of hearts, 10 of Spades, 10 of clubs which is a total of 16 favourable cases. Total number of outcomes = 52, so the required probability is: 16/52=4/13 RonL", "uniform distribution, in which all outcomes have an equal chance of occurring. A six-sided die has a uniform distribution. Each outcome has a probability of about 16.67% (1/6). Our plot below shows the solid line (so you can see it better), but keep in mind that this is a discrete distribution—you can't roll 2.5 or 2.11: Now, roll two dice together, as shown in the figure below, and the distribution is no longer uniform. It peaks at seven, which happens to have a 16.67% chance. In this case, all the other outcomes are less likely: Now, roll three dice together, as shown in the figure below. We start to see the effects of a most amazing theorem: the central limit theorem. The central limit theorem boldly promises that the sum or average of a series of independent variables will tend to become normally distributed, regardless of their own distribution. Our dice are individually uniform but combine them and—as we add more dice—almost magically their sum will tend toward the familiar normal distribution. ## Binomial Distribution The binomial distribution reflects a series of \"either/or\" trials, such as a series of coin tosses. These are called Bernoulli trials—which refer to events that have only two outcomes—but you don't need even (50/50) odds. The binomial distribution below plots a series of", "# Flipping two strange coins 1. Jun 14, 2014 ### StrangeCoin These two coins are strange because the probability between their heads and tails can be adjusted from 0% to 100%. What is the probability of getting two heads or two tails, if for example one coin is set to flip heads 70% and the other to flip heads 20% of the time? Thanks. 2. Jun 14, 2014 ### pwsnafu Do you know what independence means? 3. Jun 14, 2014 ### StrangeCoin Not in terms of statistics and probability. I am only able to figure out the most simple scenarios, I don't know where to even begin with this one. 4. Jun 14, 2014 ### Fredrik Staff Emeritus Consider a normal six-sided die. The probability to roll a 6 AND then another 6 is $\\frac 1 6\\cdot\\frac 1 6=\\frac{1}{36}$. The probability to roll a 6 OR a 5 is $\\frac 1 6+\\frac 1 6=\\frac 1 3$. A good rule of thumb is that an \"and\" means that you need to multiply the probabilities, and an \"or\" means that you need to add them. Two events A and B are independent if knowing that one of them occurred doesn't affect the probability that the other one occurred. Suppose that you roll two dice, and see that the first one", "answers. #### Choice B ##### The number of outcomes in which the sum of the digits that appear on the facing side is odd when a fair die rolled thrice. The total number of outcomes when a fair die is rolled thrice is 6 * 6 * 6 = 216. If you list down a few of the outcomes and observe the pattern, you will realize that half the outcomes have a sum that is odd and the other half has a sum that is even. Here is a sample of the first 6. {1, 1, 1, odd}, {1, 1, 2, even}, {1, 1, 3, odd}, {1, 1, 4, even}, {1, 1, 5, odd}, {1, 1, 6, even}. Therefore, ½ of 216 = 108 outcomes have an odd sum. Choice B is one of the answers where the number of outcomes is more than 50. #### Choice C ##### The number of outcomes in which the two cards drawn from a pack of well shuffled cards are both red and face cards. A pack of well shuffled cards has 52 cards. The break up of the pack on different parameters is listed in the table given below Color Suit Numbered cards Face cards Aces Red - 26 cards Diamond - 13 cards$2 to 10) 9 cards (J, Q,", "faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "die is rolled 235 times, and 6 comes up 51 times. If the die is fair, we would expect 6 to come up ${\\displaystyle 235\\times 1/6=39.17}$ times. We have now observed that the number of 6s is higher than what we would expect on average by pure chance had the die been a fair one. But, is the number significantly high enough for us to conclude anything about the fairness of the die? This question can be answered by the binomial test. Our null hypothesis would be that the die is fair (probability of each number coming up on the die is 1/6). To find an answer to this question using the binomial test, we use the binomial distribution ${\\displaystyle B(N=235,p=1/6)}$ with pmf ${\\displaystyle f(k,n,p)=\\Pr(k;n,p)=\\Pr(X=k)={\\binom {n}{k}}p^{k}(1-p)^{n-k}}$ . As we have observed a value greater than the expected value, we could consider the probability of observing 51 6s or higher under the null, which would constitute a one-tailed test (here we are basically testing whether this die is biased towards generating more 6s than expected). In order to calculate the probability of 51 or more 6s in a sample of 235 under the null hypothesis we add up the probabilities of getting exactly 51 6s, exactly 52 6s, and so on up to probability of getting exactly 235 6s:", "# 4.5 Expectation ## Uneven Dice Say I have a six-sided die that lands on an even number with probability $$\\frac{1}{4}$$. If it lands on an even number, then it lands on each one with equal probability. The same is true for odd numbers. Find the expected value of this die. Please answer as a decimal with three significant figures. Exercise 1 Let $$R$$ be a random variable for the value of the die roll. From the description we know that $$\\Pr[R=i]=\\frac{1}{4}\\cdot\\frac{1}{3}=\\frac{1}{12}$$ for $$i\\in\\{2, 4, 6\\}$$ and $$\\Pr[R=i]=\\frac{3}{4}\\cdot\\frac{1}{3}=\\frac{1}{4}$$ for $$i\\in\\{1, 3, 5\\}$$. Then, $$E[R]=\\sum_{i=1}^6 \\Pr[R=i] = 1\\cdot\\frac{1}{4}+2\\cdot\\frac{1}{12}+3\\cdot\\frac{1}{4}+4\\cdot\\frac{1}{12}+5\\cdot\\frac{1}{4}+6\\cdot\\frac{1}{12}=\\frac{13}{4}$$", "# What is the probability you will flip a head and roll a four if you flip a coin and roll a die at the same time? Jul 27, 2018 $\\text{p(rolling a four and tossing a head)} = \\frac{1}{12}$ #### Explanation: Outcomes of tossing a coin: ie 2 outcomes • tail Outcomes of rolling a die: ie 6 outcomes • 1 • 2 • 3 • 4 • 5 • 6 $\\text{p(rolling a four and tossing a head)} = \\frac{1}{6} \\times \\frac{1}{2} = \\frac{1}{12}$", "fair die is rolled 10 times. Find the probability that there is at least one 5. Solution: P(at least one 5) = 1 - P(none of the rolls equal 5) = 1 - P(all 10 rolls are not five) ⇒ P(5) = 1/6 P(not 5) = 5/6 = 1 - (5/6)10 = 0.838 2. A man purchases a box of 75 decorative light bulbs for his restaurant. If the probability that any light bulb is defective is 0.02, calculate the probability that at least one light bulb is defective in the box of 75. Solution: P(at least 1 light bulb is defective) = 1 - P(none of the light bulbs are defective) = 1 - P(all 75 light bulbs are good) ⇒ P(bad light bulb) = 0.02 P(good light bulb) = 0.98 = 1 - (0.98)75 = 0.7802" ]

[ "math club for girls. This entry was posted in Contest Math and tagged , . Bookmark the permalink.", "• # question_answer Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six is and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If four members have to be girls, the members of the team are A) BCPQRS B) BFPRSTC) BCQRST D) BCPQRT Ans. Team having 4 girls 4 girls will be $\\to$ S, T, R, Q $\\downarrow$$\\downarrow$and 2 boys will be $\\to$ B C", "• # question_answer Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six is and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If, including P, the team has three girls, the members other than P are A) BCFQR B) ADEST C) ADBST D) BFRST Ans. This type of team can be formed having 3 boys and 3 girls. 3 boys will be B, C, F$\\downarrow$ $\\downarrow$and 3 girls will be R,Q, P $\\therefore$ Team should be BCFQR.", "2 people Math Team 2 people", "• mathmath333 There are 40 students .In how many ways can they be arranged to form a team with 1 captain and 4 vice-captains ? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "• Competition 15 boys and 10 girls are in the class. On school competition of them is selected 6-member team composed of 4 boys and 2 girls. How many ways can we select students?", "Competition 15 boys and 10 girls are in the class. On school competition of them is selected 6-member team composed of 4 boys and 2 girls. How many ways can we select students? • Points in plane The plane is given 12 points, 5 of which is located on a straight line. How many different lines could by draw from this points? • The confectionery The confectionery sold 5 kinds of ice cream. In how many ways can I buy 3 kinds if order of ice creams does not matter?", "Jebel Ali Primary School, Alex from Templars Primary, Alex from Maidstone and Beth from Torpoint who all sent clear solutions.", "one time can be six players per team. How many initial teams of this girls may trainer to choose?", "25 Apr 2012, 10:59 1 A certain team has 12 members, including Joey. A three-membe 13 23 Oct 2011, 13:16 12 A certain club has 20 members. What is the ratio of the memb 18 25 Mar 2009, 23:04 Display posts from previous: Sort by", "There are 14 girls and 11 boys in the class. How many ways can a four-member team be chosen so that there are exactly two boys in it? • Five girls Five girls traveling by bus. Each holds in each hand two baskets. In each basket are four cats. Each cat has three kittens. Two girls exits bus. How many legs are in the bus? • Classroom In a class are 32 pupils. Boys are 8 less than girls. How many boys and girls are in the classroom? • A library A library has 12,500 fiction books and 19,000 non fiction books. Currently 2/5 of the fiction books are checked out. Currently 2/5 of the non fiction books are checked out. Of the books checked out, only 1/10 are due back this week. How many books are due • Marbles Dave had 40 marbles. Junjun has 2 1/5 more than Dave’s marbles. How many marbles do they have altogether? • Equivalent fractions 2 Write the equivalent multiplication expression. 2 1/6÷3/4", "Question # Directions:- Study the following information carefully to answer the question given below. From a group of six boys M, N, O, P, Q and R and five girls G, H, I, J and K a team of six are to be selected. Some of the criteria of selections are as follows:- 1) M and J go together. 2) O cannot be placed with N. 3) I cannot go with J. 4) N goes with H. 5) P and Q have to be together. 6) K and R go together. If four members are boys, which of the following cannot constitute the team? A G, J, M, O, P and Q B J, K, M, N, O and R C H, J,M, N, P and Q D J, K, M, P, Q and R E None of these Solution ## The correct option is A J, K, M, N, O and RCannot be placed with N.Logical Reasoning Suggest Corrections 0 Similar questions View More People also searched for View More", "and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If, including P, the team has three girls, the members other than P are A) BCFQR D) BFRST Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six is and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If the team including C consists of four boys, the members of the team other B) ABDQR C) DEFAQ D) BEFRQ Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six", "9 _________________ ~ETERNAL~ Re: If there are g girls and b boys on a team, and two members of the team [#permalink] 10 Sep 2019, 13:16 Display posts from previous: Sort by", "Math Circle This year we are going to have a math circle time where we can address the various abilities of our unique students. We will find out the best fit for your child by considering their previous knowledge and skill level. Mixed ages will be split into groups where children can work with hands on activities and projects in tandem to share accordingly. Ms Suzanne and Ms Christy will be working with students in collaboration with each other to provide differentiated instruction. We are prepared to work online or in the classroom. 1 hr. Wednesdays & Fridays, 11:00am - 12:00pm", "### Home > CCAA > Chapter 9 Unit 8 > Lesson CC2: 9.1.1 > Problem9-19 9-19. The students at Dolan Middle School are competing in after-school activities in which they earn points for helping out around the school. Each team consists of the $30$ students in a homeroom. Halfway through the competition, the scores from the students in two of the teams were displayed in the plots at right. The champion team is the one with the most points when the scores of the $30$ students on the team are added. Which team would you rather be on? Explain. Homework Help ✎ What are the differences in the two plots? What do these differences tell you about the two teams? Use that information to make your decision.", "# Thread: Simple probability prolem :/ 1. ## Simple probability prolem :/ The Bridge club has 8 boys and 3 girls. Four players are randomly selected to play at one table. A) what is the probability of a table with 2 boys and 2 girls? B) what i the probability of a table with atleast 2 girls? 2. Originally Posted by needuhelp The Bridge club has 8 boys and 3 girls. Four players are randomly selected to play at one table. A) what is the probability of a table with 2 boys and 2 girls? B) what i the probability of a table with atleast 2 girls? A) $\\frac{{8 \\choose 2} \\cdot {3 \\choose 2}}{{11 \\choose 4}}$. B) At least two girls means either 2 girls and 2 boys or 3 girls and 1 boy or 4 girls and 0 boys. Calcultae each probability (use part A) as a guide) and then add them up.", "45 views +1 vote Given we have to form 4 mixed pairs for tournament i.e. we need 4 boys and 4 girls.. So we can select 4 boys from 5 in C(5,4) = 5 ways and also we can select 4 girls from 5 in 5 ways.. So by multiplication principle of counting , we have : No of ways of making mixed pair team = 5 * 5 = 25 Hence number of possible selections = 25 selected by I too did same. but answer given is $600$ Don't worry ..It is the ACE test series..Endowed with mistakes..:) +1 vote", "a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is ### Jee Advanced Class 12 | THREE DIMENSIONAL GEOMETRY For a> b>c>0, if the distance between (1,1) and the point of intersection of the line ax+by-c=0 is less than 2sqrt2 then, Latest from Doubtnut", "year. In addition, Eric, Schuyler, and Felix will all be accompanied by one or more younger siblings on this year's math circle teams. In our first year, we could not even field a full team. This year, we expect to have three teams. Our community has flourished and thrived across the years, with students building friendships and collaborative relationships that continue long after they have finished high school. It is delightful that many of our alumni return each year for our Memorial Day picnic. We will have one again this year. We look forward to celebrating a very special one in May." ]

[ ". Hence, there are: . $10\\cdot6 \\:=\\:{\\color{blue}60}$ ways to choose 2 boys and a girl. Therefore: . $P(\\text{2 boys, 1 girl}) \\:=\\:\\frac{60}{165} \\:=\\:\\boxed{\\frac{4}{11}}$", "team B = setup #2 team A. – barak manos Mar 8 '15 at 11:09 Number of pairs of teams where all boys are together: exactly $5$ (one for each girl). Number of possible pairs of teams: $\\frac{1}{2}\\binom{8}{4}=35$. The difference has to be the answer, i.e. $30$. The specific step at which you went wrong is when you preassigned the first two boys. This step excluded all arrangements in which those two boys are on the same team. As there are exactly three ways to put two boys on one team and one on the other team (when arrangements with the teams swapped are considered identical), your solution excluded exactly one-third of all possible arrangements, which is why you got only $20$ arrangements instead of $30$. The requirement that each team has at least one boy means that one team consists of two boys and two girls while the other team consists of one boy and three girls. Observe that selecting one of the teams completely determines who is on the other team. The number of ways of selecting two of the three boys and two of the five girls to be on one of the teams is $$\\binom{3}{2}\\binom{5}{2} = 3 \\cdot 10 = 30$$ Once we have chosen this team, there is only one way to select", "....$ 3 boys and 2 girls: ${17 \\choose 3} \\cdot {13 \\choose 2} = \\, ....$ Now add the two numbers together. 3. Originally Posted by Jones A team of 5 people are selected to participate in a chess tournament. The \"team\" is picked within a school class with 17 boys and 13 girls. The rules state that the team must consist of at least 2 boys and 2 girls. ${{17}\\choose{3}}{{13}\\choose{2}}+{{17}\\choose{2}}{ {13}\\choose{3}}$. That choose three boys and two girls or two boys and three girls. 4. Originally Posted by Plato ${{17}\\choose{3}}{{13}\\choose{2}}+{{17}\\choose{2}}{ {13}\\choose{3}}$. That choose three boys and two girls or two boys and three girls. A rare event - we both get the same answer (which means I've got a combinatorial question correct for once). 5. hah, it turned out to be really easy. Did i make a twat out of myself?", "• # question_answer Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six is and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If four members have to be girls, the members of the team are A) BCPQRS B) BFPRSTC) BCQRST D) BCPQRT Ans. Team having 4 girls 4 girls will be $\\to$ S, T, R, Q $\\downarrow$$\\downarrow$and 2 boys will be $\\to$ B C", "• # question_answer Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six is and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If, including P, the team has three girls, the members other than P are A) BCFQR B) ADEST C) ADBST D) BFRST Ans. This type of team can be formed having 3 boys and 3 girls. 3 boys will be B, C, F$\\downarrow$ $\\downarrow$and 3 girls will be R,Q, P $\\therefore$ Team should be BCFQR.", "# How many ways to assemble a team of 5 out of 15 girls and 10 boys with limitations? How many ways to assemble a team of 5 out of 15 girls and 10 boys, if the team must contain at least two boys and two girls? Why is it wrong to count the following way(?): 1. Choose 2 boys $= {10 \\choose 2}$ 2. Choose 2 girls $= {15 \\choose 2}$ 3. Choose one more boy/girl from the rest $= 21$ $${10 \\choose 2}{15 \\choose 2}21 = 99225$$ I quiet sure that this count has duplicates, if so, how do I eliminate them? The correct answer is: $${10 \\choose 2}{15 \\choose 3}+{10 \\choose 3}{15 \\choose 2}=33075$$ - The problem is that for the gender of which there are $3$, there are three ways of picking $2$ of the $3$ at first and then one later, so you're counting each team three times. You can either divide by that factor of three, or take the approach in the answer you gave: there must be either $2$ boys and $3$ girls or vice versa, so add the numbers of ways you can make those two selections.", "# Simple probability prolem :/ • Jan 18th 2010, 02:14 PM needuhelp Simple probability prolem :/ The Bridge club has 8 boys and 3 girls. Four players are randomly selected to play at one table. A) what is the probability of a table with 2 boys and 2 girls? B) what i the probability of a table with atleast 2 girls? • Jan 18th 2010, 02:19 PM mr fantastic Quote: Originally Posted by needuhelp The Bridge club has 8 boys and 3 girls. Four players are randomly selected to play at one table. A) what is the probability of a table with 2 boys and 2 girls? B) what i the probability of a table with atleast 2 girls? A) $\\frac{{8 \\choose 2} \\cdot {3 \\choose 2}}{{11 \\choose 4}}$. B) At least two girls means either 2 girls and 2 boys or 3 girls and 1 boy or 4 girls and 0 boys. Calcultae each probability (use part A) as a guide) and then add them up.", "# Thread: Simple probability prolem :/ 1. ## Simple probability prolem :/ The Bridge club has 8 boys and 3 girls. Four players are randomly selected to play at one table. A) what is the probability of a table with 2 boys and 2 girls? B) what i the probability of a table with atleast 2 girls? 2. Originally Posted by needuhelp The Bridge club has 8 boys and 3 girls. Four players are randomly selected to play at one table. A) what is the probability of a table with 2 boys and 2 girls? B) what i the probability of a table with atleast 2 girls? A) $\\frac{{8 \\choose 2} \\cdot {3 \\choose 2}}{{11 \\choose 4}}$. B) At least two girls means either 2 girls and 2 boys or 3 girls and 1 boy or 4 girls and 0 boys. Calcultae each probability (use part A) as a guide) and then add them up.", "1. ## Number of combinations Hi, I have this problem that i can't get right. A team of 5 people are selected to participate in a chess tournament. The \"team\" is picked within a school class with 17 boys and 13 girls. The rules state that the team must consist of at least 2 boys and 2 girls. In how many ways can the team be selected? So i thought this would be as simple as: Either 2 boys and 3 girls or 3 boys and 2 girls. $$(17*16*15/3) * (13*12/2)*(13*12*11/3)*(17*16/2)$$ That is incorrect though... Any ideas on how to do this? 2. Originally Posted by Jones Hi, I have this problem that i can't get right. A team of 5 people are selected to participate in a chess tournament. The \"team\" is picked within a school class with 17 boys and 13 girls. The rules state that the team must consist of at least 2 boys and 2 girls. In how many ways can the team be selected? So i thought this would be as simple as: Either 2 boys and 3 girls or 3 boys and 2 girls. $(17*16*15/3) * (13*12/2)*(13*12*11/3)*(17*16/2)$ That is incorrect though... Any ideas on how to do this? 2 boys and 3 girls: ${17 \\choose 2} \\cdot {13 \\choose 3} = \\,", "into three groups: the $14$ boys, the $2$ girls Roberta and Priya, and the other $15$ girls, with the requirement that committee consist of $3$ members of the first group, $1$ member of the second group, and $2$ members of the third group. The number of choices is thus $${14\\choose3}{2\\choose1}{15\\choose2}=364\\cdot2\\cdot105=76440$$ I'm not sure how the book got the answer $65520=364\\cdot180$, unless it mistakenly did some sort of inclusion-exclusion.", "group of 8 boys and 7 girls if: (a) There are no restrictions (b) There must be at least 2 boys chosen (c) There must be at least 1 girl chosen (a) There are 15 candidates. There are 15 choices for the President, . . 14 choices for the Vice President, . . 13 choices for the Treasure. Therefore, there are: . $15 \\cdot 14 \\cdot 13 \\:=\\:\\boxed{2730}$ ways. (b) \"at least 2 boys\" means \"2 boys or 3 boys\". If two boys (and one girl) are chosen, there are: . ${8\\choose2}{7\\choose1} = 196$ choices. Since each can hold a different office, there are $3! = 6$ assignments. . . Hence, there are: . $196 \\cdot 6 \\:=\\:1176$ ways for two boys. If three boys are chosen, there are: . $8 \\cdot 7 \\cdot 6 \\:=\\:336$ ways. Therefore, there are: . $1176 + 336 \\:=\\:\\boxed{1512}$ ways. (c) The opposite of \"at least one girl\" is \"no girls\" (or \"all boys\"). From (a) we know that there are $2730$ possible outcomes. From (b) we that there are $336$ outcomes with all boys. Therefore, \"at least one girl\" has: . $2730 - 336 \\:=\\:\\boxed{2394}$ ways.", "at least. 3) For the above condition, there will be K-4 different teams, containing different numbers of men and women in the team. In the first team, there would be 4 men and K-4 women, in second-team there would be 5 men and K-5 women, as we reach to end, the last team consists of K-1 men and 1 woman. 4) So, mathematically the total number of the team becomes : Total number of teams = $\\sum_{i=4}^{K-1}^{X}C_{i}\\times^{Y}C_{K-i}$ ### Example: Let’s assume, We have to form a team of size 6(K) and there needs to be atleast 4 men and 1 women out of 5(X) men and 2(Y) women. So we can form 2 teams, consisting Men and Women in following ways : 4 Men + 2 Women ( Total 5 ways ) 5 Men + 1 Women ( Total 2 ways ) Total number of ways to form a team of size 6 with 4 Men and 2 women : Assuming Men(M¹ M² M³ M⁴ M⁵) and Women (W¹ M²) are there to select, now we have to select team of 6, we need 2 women, so we’ll pick all women, and we need 4 men, so we’ll pick 4 men and then pair them with women. So possible ways are : M¹ M² M³ M⁴ + W¹", "$27$ ways of selecting a work partner. The next person in line who is not already on a team has $25$ ways of selecting a work partner. Continue. The number of teams that can be selected is $$27 \\cdot 25 \\cdot 23 \\cdot 21 \\cdot 19 \\cdot 17 \\cdot 15 \\cdot 13 \\cdot 11 \\cdot 9 \\cdot 7 \\cdot 5 \\cdot 3 \\cdot 1 = \\prod_{k = 1}^{14} (2k - 1) = 27!!$$ Observe that if we multiply the numerator and denominator of this answer by $2^{14}14!$, we obtain $$\\frac{28!}{2^{14}14!}$$", "Teams How many ways can divide 16 players into two teams of 8 member? Result n = 12870 Solution: $n=C_{{ 8}}(16) = \\dbinom{ 16}{ 8} = \\dfrac{ 16! }{ 8!(16-8)!} = \\dfrac{ 16 \\cdot 15 \\cdot 14 \\cdot 13 \\cdot 12 \\cdot 11 \\cdot 10 \\cdot 9 } { 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 12870$ Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! Following knowledge from mathematics are needed to solve this word math problem: Would you like to compute count of combinations? Next similar math problems: 1. Volleyball 8 girls wants to play volleyball against boys. On the field at one time can be six players per team. How many initial teams of this girls may trainer to choose? On the menu are 12 kinds of meal. How many ways can we choose four different meals into the daily menu? 3. Confectionery The village markets have 5 kinds of sweets, one weighs 31 grams. How many different ways a customer can buy 1.519 kg sweets. 4. Fish tank A fish tank at a pet store has 8 zebra fish. In how many different ways can George choose 2 zebra fish", "5 positions to the right of boys select either two different ($5\\choose 2$ possibilities) and add a chair, or select one (5 possibilities) and add two chairs. Then place the 6 girls on the empty chairs ($6!$). In total: $2\\cdot 4!\\cdot({5\\choose 2}+5)\\cdot 6!$ - Thanks for pointing out where I went wrong. Helps :) – user40143 Sep 17 '12 at 11:57", "# Choosing a 5 member team out of 12 girls and 10 boys We must choose a 5-member team from 12 girls and 10 boys. How many ways are there to make the choice so that there are no more than 3 boys on the team? The correct answer is $\\binom{22}{5} - \\binom{12}{1} \\binom{10}{4} - \\binom{10}{5}$. I understand the $\\binom{22}{5}$ part, but where I am confused at is the other two parts. I do not know how to get those parts. Can anyone help me understand how to get to the solution? - From the total number of ways to form a $5$ member team we subtract the numbers corresponding to the cases when exactly $4$ boys or exactly $5$ boys are chosen to form teams that have no more than $3$ boys. The first case happens when we choose $4$ boys and $1$ girl and the second happens when we choose $5$ boys and $0$ girls. This gives us $\\binom{12+10}{5}-\\binom{12}{1}\\binom{10}{4}-\\binom{12}{0}\\binom{10}{5}$. - Thank you. I didn't think it would be that simple. – Wooooop Dec 9 '12 at 7:53 That solution proceeds by starting with $\\binom{22}5$, the total number of possible $5$-person teams, and subtracting the $\\binom{12}1\\binom{10}4$ teams that have one girl and four boys and the $\\binom{10}5$ teams that have five boys. The problem could also be", "of them: $\\binom{11}{3}$ ways. There are 7 children; we choose 3 of them: $\\binom{7}{3}$ ways. Hence, there are: . $\\binom{11}{3}\\binom{7}{3}\\:=\\:165\\cdot35 \\:=\\:5,775$ ways to select team #1. Select team #2. There are 8 available adults; we choose 4 of them: $\\binom{8}{4}$ ways. There are 4 available children; we choose 2 of them: $\\binom{4}{2}$ ways. Hence, there are: . $\\binom{8}{4}\\binom{4}{2} \\:=\\:70\\cdot6\\:=\\:420$ ways to select team #2. The remaining 4 adults and 2 children will belong to team #3. Therefore, there are: . $5,775 \\times 420 \\:=\\:\\boxed{2,425,500}$ ways to select the teams. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ** If teams #2 and #3 are interchangeable, . . . then the \"420\" is twice as large as it should be. The number of ways would be: . $5,775 \\times 210 \\:=\\:\\boxed{1,212,750}$ 3. Thank you so much! You explained that in a very clear way.", "her \"Amy\") will be on one of the two teams. What is the probability that the other four girls will also be on her team? – Graham Kemp May 25 '16 at 0:04 • Your method is correct, and the answer you get is $2 / {10 \\choose 5} = 1/126$, which is also equal to $1/{9 \\choose 4}$. – svsring May 25 '16 at 6:39 Pick any one of the girls. She will be on one of the two teams but who cares which. What we are after is the probability that the other four girls, selected from the other nine players, will also be on her team. $$\\dfrac{1}{\\binom{9}{4}}$$ There are $\\frac{10\\cdot9\\cdot8\\cdot7\\cdot6}{5\\cdot4\\cdot3\\cdot2\\cdot1} = 252$ ways to select team A. Team B is whoever is not on team A. Team A could be all all girls or team A could be all boys (making team B all girls) $\\frac{2}{252} = \\frac{1}{126}$ If $R=\\frac{5}{10} + \\frac{4}{9} + \\frac{3}{8} + \\frac{2}{7} + \\frac{1}{6}$ then the answer is $2R$. But this problem is set up to use a hypergeometric distribution. Let $Y=$( # of girls on team A ). In general for a hypergeometric distribtion $P[Y=k]=\\dfrac{(_KC_k)(_{(N-K)}C_{(n-k)})}{_{N}C_n}$ . In this case $K$ represents the number of girls total (5 of them) and $N$ is the total amount of people (10 of them)", "+0 # greatest number of teams 0 171 1 There are 27 girls and 36 boys who want to participate in a math competition. If each team must have the same ratio of girls and boys what’s the greatest number of teams that can enter? Mar 3, 2020 #1 +6196 +1 $$27=3^3 = 3 \\cdot 9\\\\ 36=2^2 3^2 = 4 \\cdot 9\\\\ \\text{So if teams have 3 girls and 4 boys you can have 9 teams}$$ . Mar 3, 2020 $$27=3^3 = 3 \\cdot 9\\\\ 36=2^2 3^2 = 4 \\cdot 9\\\\ \\text{So if teams have 3 girls and 4 boys you can have 9 teams}$$", "have the boys seated, and $10!$ ways to have the girls seated. (All right, this part is about permutations.) So the number of solutions is $2\\cdot 10!\\cdot 10!$." ]

[ "• # question_answer Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six is and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If four members have to be girls, the members of the team are A) BCPQRS B) BFPRSTC) BCQRST D) BCPQRT Ans. Team having 4 girls 4 girls will be $\\to$ S, T, R, Q $\\downarrow$$\\downarrow$and 2 boys will be $\\to$ B C", "• # question_answer Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six is and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If, including P, the team has three girls, the members other than P are A) BCFQR B) ADEST C) ADBST D) BFRST Ans. This type of team can be formed having 3 boys and 3 girls. 3 boys will be B, C, F$\\downarrow$ $\\downarrow$and 3 girls will be R,Q, P $\\therefore$ Team should be BCFQR.", "# Thread: Simple probability prolem :/ 1. ## Simple probability prolem :/ The Bridge club has 8 boys and 3 girls. Four players are randomly selected to play at one table. A) what is the probability of a table with 2 boys and 2 girls? B) what i the probability of a table with atleast 2 girls? 2. Originally Posted by needuhelp The Bridge club has 8 boys and 3 girls. Four players are randomly selected to play at one table. A) what is the probability of a table with 2 boys and 2 girls? B) what i the probability of a table with atleast 2 girls? A) $\\frac{{8 \\choose 2} \\cdot {3 \\choose 2}}{{11 \\choose 4}}$. B) At least two girls means either 2 girls and 2 boys or 3 girls and 1 boy or 4 girls and 0 boys. Calcultae each probability (use part A) as a guide) and then add them up.", "### Home > CC2MN > Chapter 10 > Lesson 10.2.1 > Problem10-31 10-31. The students at Dolan Middle School are competing in after-school activities in which they earn points for helping out around the school. Each team consists of the $30$ students in a homeroom. Halfway through the competition, the scores from the students in two of the teams were displayed in the plots at right. The champion team is the one with the most points when the scores of the $30$ students on the team are added. Which team would you rather be on? Explain. What are the differences in the two plots? What do these differences tell you about the two teams? Use that information to make your decision.", "• # question_answer Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six is and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If four members including E have to be boys, the members other than E are A) ABCQR B) ADFSTC) BCFQR D) ACDFQ Correct Answer: B Solution : Ans. Team having 4 boys including E and 2 girls $\\to$ 4 boys Will be $\\to$ A, D, E, F and 2 girls will be $\\to$S T You need to login to perform this action. You will be redirected in 3 sec", "team B = setup #2 team A. – barak manos Mar 8 '15 at 11:09 Number of pairs of teams where all boys are together: exactly $5$ (one for each girl). Number of possible pairs of teams: $\\frac{1}{2}\\binom{8}{4}=35$. The difference has to be the answer, i.e. $30$. The specific step at which you went wrong is when you preassigned the first two boys. This step excluded all arrangements in which those two boys are on the same team. As there are exactly three ways to put two boys on one team and one on the other team (when arrangements with the teams swapped are considered identical), your solution excluded exactly one-third of all possible arrangements, which is why you got only $20$ arrangements instead of $30$. The requirement that each team has at least one boy means that one team consists of two boys and two girls while the other team consists of one boy and three girls. Observe that selecting one of the teams completely determines who is on the other team. The number of ways of selecting two of the three boys and two of the five girls to be on one of the teams is $$\\binom{3}{2}\\binom{5}{2} = 3 \\cdot 10 = 30$$ Once we have chosen this team, there is only one way to select", "1. ## Number of combinations Hi, I have this problem that i can't get right. A team of 5 people are selected to participate in a chess tournament. The \"team\" is picked within a school class with 17 boys and 13 girls. The rules state that the team must consist of at least 2 boys and 2 girls. In how many ways can the team be selected? So i thought this would be as simple as: Either 2 boys and 3 girls or 3 boys and 2 girls. $$(17*16*15/3) * (13*12/2)*(13*12*11/3)*(17*16/2)$$ That is incorrect though... Any ideas on how to do this? 2. Originally Posted by Jones Hi, I have this problem that i can't get right. A team of 5 people are selected to participate in a chess tournament. The \"team\" is picked within a school class with 17 boys and 13 girls. The rules state that the team must consist of at least 2 boys and 2 girls. In how many ways can the team be selected? So i thought this would be as simple as: Either 2 boys and 3 girls or 3 boys and 2 girls. $(17*16*15/3) * (13*12/2)*(13*12*11/3)*(17*16/2)$ That is incorrect though... Any ideas on how to do this? 2 boys and 3 girls: ${17 \\choose 2} \\cdot {13 \\choose 3} = \\,", "....$ 3 boys and 2 girls: ${17 \\choose 3} \\cdot {13 \\choose 2} = \\, ....$ Now add the two numbers together. 3. Originally Posted by Jones A team of 5 people are selected to participate in a chess tournament. The \"team\" is picked within a school class with 17 boys and 13 girls. The rules state that the team must consist of at least 2 boys and 2 girls. ${{17}\\choose{3}}{{13}\\choose{2}}+{{17}\\choose{2}}{ {13}\\choose{3}}$. That choose three boys and two girls or two boys and three girls. 4. Originally Posted by Plato ${{17}\\choose{3}}{{13}\\choose{2}}+{{17}\\choose{2}}{ {13}\\choose{3}}$. That choose three boys and two girls or two boys and three girls. A rare event - we both get the same answer (which means I've got a combinatorial question correct for once). 5. hah, it turned out to be really easy. Did i make a twat out of myself?", "# Choosing a 5 member team out of 12 girls and 10 boys We must choose a 5-member team from 12 girls and 10 boys. How many ways are there to make the choice so that there are no more than 3 boys on the team? The correct answer is $\\binom{22}{5} - \\binom{12}{1} \\binom{10}{4} - \\binom{10}{5}$. I understand the $\\binom{22}{5}$ part, but where I am confused at is the other two parts. I do not know how to get those parts. Can anyone help me understand how to get to the solution? - From the total number of ways to form a $5$ member team we subtract the numbers corresponding to the cases when exactly $4$ boys or exactly $5$ boys are chosen to form teams that have no more than $3$ boys. The first case happens when we choose $4$ boys and $1$ girl and the second happens when we choose $5$ boys and $0$ girls. This gives us $\\binom{12+10}{5}-\\binom{12}{1}\\binom{10}{4}-\\binom{12}{0}\\binom{10}{5}$. - Thank you. I didn't think it would be that simple. – Wooooop Dec 9 '12 at 7:53 That solution proceeds by starting with $\\binom{22}5$, the total number of possible $5$-person teams, and subtracting the $\\binom{12}1\\binom{10}4$ teams that have one girl and four boys and the $\\binom{10}5$ teams that have five boys. The problem could also be", "45 views +1 vote Given we have to form 4 mixed pairs for tournament i.e. we need 4 boys and 4 girls.. So we can select 4 boys from 5 in C(5,4) = 5 ways and also we can select 4 girls from 5 in 5 ways.. So by multiplication principle of counting , we have : No of ways of making mixed pair team = 5 * 5 = 25 Hence number of possible selections = 25 selected by I too did same. but answer given is $600$ Don't worry ..It is the ACE test series..Endowed with mistakes..:) +1 vote", "Question # Directions:- Study the following information carefully to answer the question given below. From a group of six boys M, N, O, P, Q and R and five girls G, H, I, J and K a team of six are to be selected. Some of the criteria of selections are as follows:- 1) M and J go together. 2) O cannot be placed with N. 3) I cannot go with J. 4) N goes with H. 5) P and Q have to be together. 6) K and R go together. If four members are boys, which of the following cannot constitute the team? A G, J, M, O, P and Q B J, K, M, N, O and R C H, J,M, N, P and Q D J, K, M, P, Q and R E None of these Solution ## The correct option is A J, K, M, N, O and RCannot be placed with N.Logical Reasoning Suggest Corrections 0 Similar questions View More People also searched for View More", "and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If, including P, the team has three girls, the members other than P are A) BCFQR D) BFRST Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six is and to de selected under the following conditions; A and D have be together. C cannot go with S. S and T have to be together. B cannot be termed with E. D cannot go with P. B and R have to be together. (7) C and Q have to be together. If the team including C consists of four boys, the members of the team other B) ABDQR C) DEFAQ D) BEFRQ Instructions Study the following information carefully and answer the questions given below it. From amongst six boys A, B, C, D, E and F and five girls P, Q, R, S and T, a team of six", "2 people Math Team 2 people", "# Simple probability prolem :/ • Jan 18th 2010, 02:14 PM needuhelp Simple probability prolem :/ The Bridge club has 8 boys and 3 girls. Four players are randomly selected to play at one table. A) what is the probability of a table with 2 boys and 2 girls? B) what i the probability of a table with atleast 2 girls? • Jan 18th 2010, 02:19 PM mr fantastic Quote: Originally Posted by needuhelp The Bridge club has 8 boys and 3 girls. Four players are randomly selected to play at one table. A) what is the probability of a table with 2 boys and 2 girls? B) what i the probability of a table with atleast 2 girls? A) $\\frac{{8 \\choose 2} \\cdot {3 \\choose 2}}{{11 \\choose 4}}$. B) At least two girls means either 2 girls and 2 boys or 3 girls and 1 boy or 4 girls and 0 boys. Calcultae each probability (use part A) as a guide) and then add them up.", "Competition 15 boys and 10 girls are in the class. On school competition of them is selected 6-member team composed of 4 boys and 2 girls. How many ways can we select students? • Points in plane The plane is given 12 points, 5 of which is located on a straight line. How many different lines could by draw from this points? • The confectionery The confectionery sold 5 kinds of ice cream. In how many ways can I buy 3 kinds if order of ice creams does not matter?", ". Hence, there are: . $10\\cdot6 \\:=\\:{\\color{blue}60}$ ways to choose 2 boys and a girl. Therefore: . $P(\\text{2 boys, 1 girl}) \\:=\\:\\frac{60}{165} \\:=\\:\\boxed{\\frac{4}{11}}$", "# Teams How many ways can divide 16 players into two teams of 8 member? n = 12870 ### Step-by-step explanation: $n={C}_{8}\\left(16\\right)=\\left(\\genfrac{}{}{0px}{}{16}{8}\\right)=\\frac{16!}{8!\\left(16-8\\right)!}=\\frac{16\\cdot 15\\cdot 14\\cdot 13\\cdot 12\\cdot 11\\cdot 10\\cdot 9}{8\\cdot 7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1}=12870$ We will be pleased if You send us any improvements to this math problem. Thank you! Tips to related online calculators Would you like to compute count of combinations? ## Related math problems and questions: • Volleyball 8 girls want to play volleyball against boys. On the field at one time can be six players per team. How many initial teams of these girls may trainer to choose? • Soccer teams Have to organize soccer teams. There are 3 age groups. How many different ways can you organize teams of ten for each age group? Is this a permutation or combination? • Divide How many different ways can three people divide 7 pears and 5 apples? • Tournament Determine how many ways can be chosen two representatives from 34 students to school tournament. • Examination The class is 21 students. How many ways can choose two to examination? • Football league In the football league is 16 teams. How many different sequence of results may occur at the end of the competition? • Bits, bytes Calculate how many different numbers can be", "Volleyball 8 girls wants to play volleyball against boys. On the field at one time can be six players per team. How many initial teams of this girls may trainer to choose? • Today in school There are 9 girls and 11 boys in the class today. What is the probability that Suzan will go to the board today? • Teams How many ways can divide 16 players into two teams of 8 member? • Confectionery The village markets have 5 kinds of sweets, one weighs 31 grams. How many different ways a customer can buy 1.519 kg sweets. • Chords How many 4-tones chords (chord = at the same time sounding different tones) is possible to play within 7 tones? • Blocks There are 9 interactive basic building blocks of an organization. How many two-blocks combinations are there? • PIN - codes How many five-digit PIN - code can we create using the even numbers? • Calculation of CN Calculate: ? • Balls The urn is 8 white and 6 black balls. We pull 4 randomly balls. What is the probability that among them will be two white? • One green In the container are 45 white and 15 balls. We randomly select 5 balls. What is the probability that it will be a maximum one green?", "# Teams How many ways can divide 16 players into two teams of 8 member? Correct result: n = 12870 #### Solution: $n={C}_{8}\\left(16\\right)=\\left(\\genfrac{}{}{0px}{}{16}{8}\\right)=\\frac{16!}{8!\\left(16-8\\right)!}=\\frac{16\\cdot 15\\cdot 14\\cdot 13\\cdot 12\\cdot 11\\cdot 10\\cdot 9}{8\\cdot 7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1}=12870$ We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! Tips to related online calculators Would you like to compute count of combinations? ## Next similar math problems: • Volleyball 8 girls wants to play volleyball against boys. On the field at one time can be six players per team. How many initial teams of this girls may trainer to choose? • Soccer teams Have to organize soccer teams. There are 3 age groups. How many different ways can you organize teams of ten for each age group? Is this a permutation or combination? • Divide How many different ways can three people divide 7 pears and 5 apples? • Tournament Determine how many ways can be chosen two representatives from 34 students to school tournament. • Examination The class is 21 students. How many ways can choose two to examination? • Football league In the football league is 16 teams. How many different sequence of results may occur at the end of the competition? • Bits,", "Frank T 2 There are two possible ways a team can be formed, either with only one boy or with no boys (at most one boy means \\geq 1 boy)Therefore we can find the number of ways with one boy: ^4 C_1 \\times ^6C_3 , and with no boys: ^6C_4 . Adding the two sets of possible teams results in 95. HOWEVER we must also account for all the possible captain selections. Multiplying by 4 accomplishes this and we have 380 for the final answer. Although this is not included in your answer options I am quite confident it is correct." ]

[ "$$n!$$ # of ways and out of these $$n!$$ ways only one, namely $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$ will be in ascending order. So 1 out of n!. $$P=\\frac{1}{n!}$$. Hence, according to the above, the only thing we need to know to answer the question is the size of the subset ($$n$$) we are selecting from set $$A$$. Hi, Got the first point but a doubt on the second one how can you say the subset of n numbers will be selected in n! ways. Suppose a situation where there are 12 distinct numbers and you select 5. Then the possible combinations will be 12C5 ways i.e 12!/(5!*7!) ways Say n = 5 and k = 12 (1, 2, 3, 4, ..., 12). Now, the number of 5-number groups out of 12 is indeed 12C5. Second point there says that the number of different sequences of 5 numbers is 5!. For example, if 5 numbers we are selecting are a, b, c, d, e, then the order of selections can be a, b, c, d, e, or e, d, c, b, a, or, e, a, b, c, d, ... Total = 5!. ok got it SO we can say Probability = possible outcomes/ total outcomes = 12C5/12P5 12C5: shows selection of 5 numbers and out of", "n numbers are selected from the set one-by-one, where n<=k, what is the probability that numbers will be selected in ascending order? (1) Set A consists of 12 even consecutive integers; (2) n=5. We should understand following two things: 1. The probability of selecting any n numbers from the set is the same. Why should any subset of n numbers have higher or lower probability of being selected than some other subset of n numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$, where $$x_1<x_2<...<x_n$$. We can select this subset of numbers in $$n!$$ # of ways and out of these n! ways only one, namely $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$ will be in ascending order. So 1 out of n!. $$P=\\frac{1}{n!}$$. Hence, according to the above the only thing we need to know to answer the question is the size of the subset (n) we are selecting from set A. \\\\ Bunuel, pls clarify how can there be n! ways to select n out of a set which has got \"k \" distinct numbers?? I m stupified.. _________________ The Mind is everything . What you think you become. - Lord Buddha Consider giving KUDOS if you appreciate my post !! Senior Manager", "be chosen (since the sum of all probabilities must be $$1$$, and each of the $$2^b$$ values has the same probability). Since $$S$$ is $$b$$-bit, $$S$$ is one of these $$2^b$$ values. Therefore $$\\Pr(S=S')\\,=\\,1/2^b\\,=\\,2^{-b}$$", "Given set $A=\\left\\{ 2, 3, 4, 5\\right\\}$ and set $B=\\left\\{ 11, 12, 13, 14, 15\\right\\}$, two numbers are randomly selected, one from each set. What is the probability that the sum of two numbers equal $16$? 1. $0.20$ 2. $0.25$ 3. $0.30$ 4. $0.33$", "telling you that the way numbers are selected at random from $S$ is such that you get a \"1\" in two thirds of the time (on average) and a \"2\" in two out of nine cases, etc...", "....: s = random_solution(5, 10) ....: assert sum(s[i] * (i + 1) for i in range(5)) == 10", "# Not a Chessboard Calvin fills in a $$7\\times7$$ grid with the numbers $$1$$ through $$49$$ in a random arrangement. He then erases his work and does the same thing again, to obtain two different random arrangements of the numbers in the grid. What is the expected number of pairs of numbers that occur in either the same row as each other or the same column as each other in both of the two arrangements? ×", "numbers will be selected in n! ways. Suppose a situation where there are 12 distinct numbers and you select 5. Then the possible combinations will be 12C5 ways i.e 12!/(5!*7!) ways Say n = 5 and k = 12 (1, 2, 3, 4, ..., 12). Now, the number of 5-number groups out of 12 is indeed 12C5. Second point there says that the number of different sequences of 5 numbers is 5!. For example, if 5 numbers we are selecting are a, b, c, d, e, then the order of selections can be a, b, c, d, e, or e, d, c, b, a, or, e, a, b, c, d, ... Total = 5!. _________________ Intern Joined: 04 Jul 2013 Posts: 16 ### Show Tags 14 Nov 2014, 04:59 2 Bunuel wrote: shyam593 wrote: Bunuel wrote: Official Solution: We should understand the following two things: 1. The probability of selecting any $$n$$ numbers from the set is the same. Why should any subset of $$n$$ numbers have higher or lower probability of being selected than some other subset of $$n$$ numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$, where $$x_1 \\lt x_2 \\lt ... \\lt x_n$$. We can select this subset of numbers in", "16 views | 16 views 0 2 numbers for eg :- 11+1 =12 0 I dont get it can u explain more Suppose we wrote $11$ numbers $1,2,3,4,5,6,7,8,9,10,11$ each on different piece of paper and then put all those papers in a box and we shuffle those pieces. Now how many minimum number of pieces we need to take out from that box to ensure that the number written on any $2$ of the pieces that we are taking out from the box results in sum $12$ without looking at the numbers that we are picking out from the box.. here the name of the box is $S$. i.e. box=set. The pairs that would result in sum of $12$ are $(1,11) ,(10,2), (3,9),(4,8),(5,7)$ only So we have to calculate the worst case of getting $12$ as sum of any $2$ numbers that we ae picking out. We pick $1$ then $2$ then $3$ then $4$ then $5$ then $6$ uptil here sum of none of the $2$ numbers that we picked results in $12$ Now whatever number we pick next from the remaining numbers in the set(box) we would definitely get $12$ as the sum.i.e. if we pick $7$ then we can get $5+7 =12$ if we pick $8$ then we can get $4+8 =12$ if we pick", "# Six in a Subset of S Level pending Let set $$S$$ = {$$1,2,3,4,5,6,7,8,9$$}. You randomly pick a subset of S from a box (not including the empty set). The probability that the subset will have the number $$6$$ in it can be expressed as $$\\frac{a}{b}$$. What is $$a + b$$? ×", "The four digits 5, 6, 7 and 8 are put at random in the spaces of the number below: 3 _ 1 _ 4 _ 0 _ 9 2 • Estimate the probability that the answer will be a multiple of 396.", "There's no reason that the sum will be uniform $\\mod n$. Last comment: As explained above, for the first hash, a probability of $1/n$ is over the choice of $a$.", "# Drawing two numbers from a set Two numbers $X_1$ and $X_2$ are drawn randomly from the set $\\{1,2,...,n\\}$ without replacement. Find $P(X_2 > X_1)$. Now I know that once we choose $X_2$ we have $X_2 - 1$ options for $X_1$ to hold the condition that $X_2 > X_1$ but the fact that we choose $X_1$ first kind of makes me confused.", "of $n$ possible positions. Taking $n=4,$ we have, for example, $$243\\mapsto\\{1243,2143,2413,2431\\}.$$ The probability that the insertion point is immediately in front of $2$ is $1/n.$", "Given Set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is the probability that the sum of the two numbers equals 16? Probability = Favorable Outcomes/Total Outcomes Probability = 4/20 = 0.20", "Subscribe to the weekly news from TrueShelf ## Minimum of Random Subsets Let $S =${$1,2,\\dots,n$}. Let $A,B$ be two random subsets of $S$. Let $\\min(A)$ denote the minimum number in the set $A$. • What is the probability that $\\min(A)= \\min(B)$ ? • Evaluate this probability as $n$ tends to infinity. 0 0 0 0 0 0 0 0 0 0", "The probability of selecting any $$n$$ numbers from the set is the same. Why should any subset of $$n$$ numbers have higher or lower probability of being selected than some other subset of $$n$$ numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$, where $$x_1 \\lt x_2 \\lt ... \\lt x_n$$. We can select this subset of numbers in $$n!$$ # of ways and out of these $$n!$$ ways only one, namely $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$ will be in ascending order. So 1 out of n!. $$P=\\frac{1}{n!}$$. Hence, according to the above, the only thing we need to know to answer the question is the size of the subset ($$n$$) we are selecting from set $$A$$. _________________ Intern Joined: 04 Jul 2013 Posts: 13 ### Show Tags 13 Nov 2014, 07:27 1 Bunuel wrote: Official Solution: We should understand the following two things: 1. The probability of selecting any $$n$$ numbers from the set is the same. Why should any subset of $$n$$ numbers have higher or lower probability of being selected than some other subset of $$n$$ numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$, where $$x_1", "# Let's re-share it! Two numbers $$x,y$$ with $$x>y$$ are chosen at random from the set$$\\left\\{1,2,3,.....,299,300\\right\\}.$$ If the probability that $$x^{2} - y^{2}$$ is divisible by 3 is $$\\frac{a}{b},$$ then what is the digit sum of $$a \\times b$$? Details and Assumptions: 1. Digit sum refers to the sum of all the digits of the number. For example, the digit sum of 123 is $$1+2+3=6.$$ 2. $$a$$ and $$b$$ are co-prime numbers. ×", "6), (6, 4), (5, 5) To get the sum as 11, possible outcomes = (5, 6), (6, 5) To get the sum as 12, possible outcomes = (6, 6) (ii)Probability of each of these sums will not be $\\frac{1}{11}$ as these sums are not equally likely.", "# How to calculate probability of these two pair-of-sums ($S_{n}$ and $T_{n}$) and ($SEvenF_{n}$ and $SOddF_{n}$) being the same? Say we have a sequence of $n$ positive integers, we can assume they're randomly chosen, let's call it $U_{n}$. Let $S_{n}$ = sum of $U_{n}$ from $1$ to $n$. Let $T_{n}$ = sum of $n$ from $1$ to $n$. In other words, $T_{n}$ is the triangle number $T_{n} = \\frac{1}{2}n(n+1)$. Define $F_{n} = U_{n} + n$. (note that $SF_{n}$, sum of $F_{n}$, is equal to $S_{n} + T_{n}$) Define two sets $EvenSet$ and $OddSet$ containing $F_{n}$ when it is even and odd respectively. Let $SEvenF_{n}$ be the sum of numbers in $EvenSet$ and $SOddF_{n}$ be the sum of numbers in $OddSet$. (note that $SEvenF_{n} + SOddF_{n} = SF_{n} = S_{n} + T_{n}$) What is the probability of the two pairs ($S_{n}$ and $T_{n}$) and ($SEvenF_{n}$ and $SOddF_{n}$) are the same pairs? In other words, either \"$S_{n} = SEvenF_{n}$ thus $T_{n} = SOddF_{n}$\" or \"$S_{n} = SOddF_{n}$ thus $T_{n} = SEvenF_{n}$\". How to calculate this probability? PS: I have a sample data with $n = 114$ in here http://goo.gl/k96FZ PS2: Is there a way or an algorithm to generate such sequence? PS3: @ZefChonoles has correctly pointed a concern about Probability Measure. Originally I intended this question to have no restriction" ]

[ "$12$ integers in $[1,100]$. From the chosen $55$ numbers, one of these classes must contain at least $7$ numbers. But this implies that two of them differ by exactly $9$. For $12$. Consider the classes $\\pmod {12}$. There are $12$ of these. As before, one of these must contain $5$ numbers and we get the same implication. For $13$. Consider the classes $\\pmod {13}$. There are $13$ of these. As before, one of these must contain $5$ numbers and we get the same implication. Taking lulu's method a step further, one can also show that there must be a pair in the set that differs by exactly $23$. Consider the classes $\\pmod{23}$, each containing either $4$ elements ($15$ classes) or $5$ elements ($8$ classes) of $[1,100]$. These allow $2\\times 15 + 3 \\times 8=54$ choices without an adjacent pair exactly $23$ apart. However we have $55$ chosen numbers so at least one of the classes has more elements than can be selected to stay non-adjacent in the class, so there must be a pair of numbers in the chosen set exactly $23$ apart.", "which is impossible as $r \\not = k$. This leaves us with the $16$ solutions $(2,34),(3,33),...,(17,19)$ To achieve set $A$, we eliminate $14$ elements from $B$. The maximum amount of pairs we can demolish is then $14$ (one element from each pair). This leaves us with at least $2$ pairs that sum up to $104$ in any set $A$, which of course means that there is two distinct integers in $A$ that sum up to $104$. Appreciate if someone had a look! Last edited: Feb 26, 2008 2. Feb 26, 2008 ### qspeechc Looks good to me-great stuff! What are these 'putnam problems'? They sound very interesting. 3. Feb 26, 2008 Thanks,", "the complement $S'$ of $S$. Note that the complement of the complement of $S$ is $S$, So the subsets of size $n$ can be divided into complementary pairs, and therefore the number of sets of size $n$ is even. If this is too abstract, let our initial set be $\\{1,2,3,4,5,6\\}$. Thus $n=3$. The set $\\{1,2,3\\}$ is paired with its complement, which is the set $\\{4,5,6\\}$. The set $\\{2,3,5\\}$ is paired with its complementary set $\\{1,4,6\\}$, and so on. This pairing divides the sets of size $3$ into \"couples.\" Thus the number of subsets is twice the number of couples, and in particular is even. Remark: This sort of pairing argument can be used to prove, for example, that if the positive integer $n$ is not a perfect square, then the number of (positive) divisors of $n$ is even. We illustrate the idea with $n=24$. Pair two divisors $a$ and $b$ of $n$ if $ab=n$. So $1$ is paired with $24$, $2$ is paired with $12$, $3$ is paired with $8$, $4$ is paired with $6$. Now the divisors of $24$ have been divided into couples, so $24$ has an even number of divisors. The same idea works for any non-square $n$. The method breaks down when $n$ is a perfect square, like $36$. For then there is nobody", "# Show that if seven integers are selected from the first 10 positive integers a) Show that if seven integers are selected from the first 10 positive integers, there must be at least two pairs of these integers with the sum 11. b) Is the conclusion in part (a) true if six integers are selected rather than seven? I don't know how should I show that. I know that in the worst situation we choose 1,2,...,7 that we have 7 + 4 = 11 and 5 + 6 = 11 but I don't know what should be the answer of this question Partition numbers from $1$ to $10$ into the following sets: $\\{1,10\\}$, $\\{2,9\\}$, $\\{3,8\\}$, $\\{4,7\\}$ and $\\{5,6\\}$. Now use pigeon hole principle. • It's a little tricky in that we are asked to show two pairs have sum $11$, and the pigeonhole principle gives one pair by immediate application (even if as few as six integers are selected). The difference between selecting seven and six integers from $1$ to $10$ is that we get (at least) two pairs with sum $11$ in the former case and as few as one pair in the latter case. – hardmath Dec 1 '13 at 16:58", "pairing up subsets of $A$ by complements, we see that exactly half the subsets of $A$ will have a sum greater than or equal to $28$. • How can we pairing the subsets of A with its complement? Can you show me an example to do that? It sounds unfamiliar to me. Thanks – akusaja Apr 23 '15 at 16:10 • Example: We form the complementary subset pair $\\{1, 3, 5, 6, 7, 8\\}$ and $\\{2,4, 9, 10 \\}$. In this case $1+3+5+6+7+8\\ge 28$; while $2+4+9+10 < 28$. – paw88789 Apr 23 '15 at 16:14 The correctness of paw88789's approach can perhaps be more clearly seen if we try a smaller set; that way, you can sit down and actually enumerate all the different possibilities. That can make a world of difference in terms of internalizing the approach. Suppose we ask how many subsets of $\\{1, 2, 3, 4, 5\\}$ add up to a number $\\geq 8$. The crucial idea is that we partition the set into two parts; these two parts are called complements of each other. Obviously, the sum of the two parts must add up to $15$. Exactly one of those parts is therefore $\\geq 8$. There must be at least one such part, because of the pigeonhole principle (specifically, two $7$'s are sufficient only", "at hand, given the observation that at least one of the pairs must be a square or twice a square? Then the density of such integers up to $N$ must be $\\ll \\sqrt{N}$, so the sum of reciprocals converges by partial summation. – Thomas Bloom Feb 6 '12 at 16:07", "# Question about the Pigeonhole Principle The question is: Let $A = \\{1,2,3,4,5,6,7,8\\}$. If five integers are selected from $A$, must at least one pair of the integers have a sum of $9$? The book explains the solution by dividing the number into $4$ disjoint subsets and pointing out that if the numbers in each set are added together, they result in a sum of $9$: $\\{1,8\\}, \\{2,7\\}, \\{3,6\\}, \\{4,5\\}$. Thus, if you pick $5$ numbers, you will inevitably find a pair, whose sum will equal to $9$. I don't get it. When the book asks to select $5$ numbers from $A$, I can pick $\\{1,2,3,4,5\\}$. Then from there, I can pick two integers like $1$ and $2$, whose sum does not equal to $9$. What am I not getting here? Thanks for any help. - its saying when you pick any 5 elements, you have some pair in that set of 5 elements whose sum is 9. In the set of 5 elements you picked, $4+5 = 9$ – Deven Ware Dec 15 '11 at 2:05 Of course some pairs, indeed many pairs, will not have sum $9$. But in your example, $4+5=9$. – André Nicolas Dec 15 '11 at 2:06 In that group of 5 you selected, the pair 4 and 5 sum to 9. Obviously,", "$x$ and $y$ must remain disjoint. 15.3.2 If a set omits $x$ and $y,$ and does not separate $x$ and $y,$ then no subset separates $x$ and $y,$ as enlarging the complementary component containing $x$ and $y$ cannot separate them.", "= 1$$ to $$12$$, replaced by $$N_{13}$$ the sum will again be even. Thus each $$N_j,\\, j = 1$$ to $$12$$ must be even. For (ii), any sequence of twelve integer with sum $$2P$$ has a subset of six integers with sum $$P$$. 'Mapping' the sequence to the newly formed sequences the sum would be $$2P - 12N$$ and the subset would be $$P - 6N$$ For (iii), similarly, any sequence of twelve now has sum $$2kP$$ and the subset of six would have sum $$kP$$. It is part (iv) I'm stuck on. I can demonstrate that the integers are either all odd, or all even using properties (i) and (ii) but can't see how property (iv) helps me here. Any thoughts? You descend through binary expansions. You can adapt your proof of iii to extend $$k$$ to include the reciprocal of the greatest common divisor of all the numbers. Use ii to subtract off the smallest one, leaving one of the numbers zero. i says they are all even, so all the original numbers had the same bit in the ones place in their binary expansions. Now all the numbers are even, so our extended iii says we can divide them all by $$2$$. Repeat the argument and you find that all the numbers have the same", "# Choosing $7$ numbers from $[1,2,…,11]$ will give us $2$ that have sum $12$. Choosing $7$ numbers from $[1,2,...,11]$ will give us $2$ that have sum $12$. I tried: There are only $5$ pairings possible: $(7,5),(8,4),(9,3),(10,2),(11,1)$ Suppose I pick $6$, and then not to be a pair I will need to draw exactly 1 number from each of the above pairs. Suppose I picked, $7,8,9,10,11$. I will use up $6$ numbers. Thus any number that is left belongs to some pair. How can I put this into mathematical terms and prove it? Prove that for any set of $7$ numbers chosen from the set $\\{1, 2, 3, ..., 11\\},$ there exists at least one pair of two numbers that sum to $12.$ We can formally prove this by contradiction. Imagine that we can find a set of $7$ numbers with no two elements with a sum of $12.$ We have $6$ boxes - one of which can hold up to $1$ number. Then our proposition is that each of $b_{1}, b_{2}, b_{3}, ..., b_{6}$ is less than or equal to $1.$ In other words, $b_{n} \\le 1.$ Then $$\\sum_{n = 1}^{6} b_{n} \\le 6.$$ • I meant that one box can hold up to 1 number. It can only hold $6.$ Because $6 + 6 = 12,$ and", "is also even. On the other hand, if we select an even integer and an odd integer, their sum is odd, so two integers is not sufficient. If we take three integers from set $S$, then at least two of them must have the same parity, which guarantees that we will have two integers such that their sum is even. What is the minimum number of integers selected from the set $S = \\{1, 2, 3, 4, 5, 6, 7, 8, 9\\}$ such that the difference of two of the integers is $5$? Consider the subsets $\\{1, 6\\}$, $\\{2, 7\\}$, $\\{3, 8\\}$, $\\{4, 9\\}$, $\\{5\\}$. The worst case scenario is that we select one number from each of these subsets. Therefore, how many numbers are required to guarantee that we have two numbers whose difference is $5$? • the answer would be 6? Since we can choose in the worst case a number from each pair [{1,6}, {2,7}, {3,8}, {4,9}] (4) and if the number 5 is chosen (1) it will not be possible for the difference between two numbers to result in 5. So another number (1) belonging to the correct pairs would have to be chosen. (4) + (1) + (1) = 6 – Lidy Monteiro Sep 2 '18 at 16:18 • That is correct. –", "$A$ and $B$ the sums of whose elements are equal. Then when you add four more integers in sequence, you can take the first and fourth new integers and append them to $A$, and you can take the second and third new integers and append them to $B$. Then their sums are still equal. • When $n = 3$, it works for $A=\\{1,2\\}$ and $B=\\{3\\}$. The sums of the elements of these sets is not even. – Eric Fisher Dec 6 '17 at 8:29 • I meant the sum of all numbers. $1+2+3=6$ which is quite obviously even. It is quite obvious that if that sum is not even, there's no way to divide the set into two subsets with equal sum; the not so obvious part is that the sum being even is already a sufficient condition. – celtschk Dec 6 '17 at 16:10", "# When does a set of sums uniquely determine a set of values? This is a follow up to a question about determining the set of values from a set of sums. We consider the following setup: Consider a vector $$A$$ (which you will not see) of $$n$$ positive integers. You are given the set of sums of the (contiguously indexed) subvectors of $$A$$. For example, say $$A = (3,2,1,2)$$ The subvectors are $$(3),(2),(1),(2), (3,2), (2,1), (1,2),(3,2,1), (2,1,2),(3,2,1,2)$$. We would be given the sums $$\\{1, 2, 3, 5, 6, 8\\}$$. Let us call this set of sums $$f(A)$$. It turns out it is not always possible to uniquely determine the set of integers in $$A$$ from the pair $$(f(A), n)$$? For example: $$A = (1,1,3)$$ and $$B = (1,2,2)$$ both give the same set of sums $$f(A) = f(B) = \\{1,2,3,4,5\\}$$. However in some cases it is possible. For example, if $$n=3$$ and $$f(C) = \\{1,2,3,5,6\\}$$ then we know that $$\\operatorname{set}(C) = \\{1, 2, 3\\}$$. This raises the following question: Is it possible to characterise which pairs $$(f(X), n)$$ uniquely determine the set of values in the array $$X$$? • Maybe it suffices to know that in the sum, there exist a one or a zero??? – user7427029 Oct 15 '18 at 22:50 • @user7427029 I don’t quite", "So here's the awkward question $$\\ldots$$ # What is a negative number anyway? Negative numbers are there to be the solutions to equations of the form $m+x = n$ when $$m>n$$. In particular, $$-n$$ is the solution to $$n+x=0$$, if this makes sense. So, let's make sense of it. We're going to do this a bit indirectly, thinking about pairs of integers $$(m,n)$$. Secretly we want to think of this pair as $$m-n$$, but we don't yet have a meaning for this when $$m \\lt n$$. So we're going to lump together certain pairs. We say that the pairs $$(m,n)$$ and $$(p,q)$$ are equivalent if $$m+q=n+p$$. A collection of pairs which are all equivalent to each other is called an equivalence class. This is actually quite a familiar idea: it's analogous to thinking of $$3/4$$ and $$6/8$$ as different ways of representing the same fraction. There are infinitely many pairs of integers $$(m,n)$$ which we can use to do it, namely all the ones satisfying $$3n=4m$$. We don't (often) talk explicitly about a fraction being an equivalence class of pairs of integers, but that's what's really going on. We're going to do the same trick, but with pairs where it's the difference than matters, not the ratio. We need to do three things now: 1. Define addition on", "So here's the awkward question $$\\ldots$$ # What is a negative number anyway? Negative numbers are there to be the solutions to equations of the form $m+x = n$ when $$m>n$$. In particular, $$-n$$ is the solution to $$n+x=0$$, if this makes sense. So, let's make sense of it. We're going to do this a bit indirectly, thinking about pairs of integers $$(m,n)$$. Secretly we want to think of this pair as $$m-n$$, but we don't yet have a meaning for this when $$m \\lt n$$. So we're going to lump together certain pairs. We say that the pairs $$(m,n)$$ and $$(p,q)$$ are equivalent if $$m+q=n+p$$. A collection of pairs which are all equivalent to each other is called an equivalence class. This is actually quite a familiar idea: it's analogous to thinking of $$3/4$$ and $$6/8$$ as different ways of representing the same fraction. There are infinitely many pairs of integers $$(m,n)$$ which we can use to do it, namely all the ones satisfying $$3n=4m$$. We don't (often) talk explicitly about a fraction being an equivalence class of pairs of integers, but that's what's really going on. We're going to do the same trick, but with pairs where it's the difference than matters, not the ratio. We need to do three things now: 1. Define addition on", "differs by $2$. Repeat until the number of elements in consecutive sequence is $P_s-1-2n =3$. Removing numbers with factor of $3$. There must be a pair of numbers where both of them are prime numbers. There must be infinite number of twin primes.", "contains $(2\\times N + 2)$ space separate integers. The first integer denotes $N.$ The next integer is $k.$ The next $N$ integers are $V[1],..., V[N].$ The last $N$ integers are $B[1],..., B[N].$ Constraints • $1 \\leq k \\leq 7$ • $-10^6 \\leq V[i] \\leq 10^6$, for all $i$ • $1 \\leq B[i] \\leq 2k$, for all $i$ Illustrated Examples • For the examples discussed here, let us assume that $k = 2$. The sequence 1, 1, 3 is not well-bracketed as one of the two 1's cannot be paired. The sequence 3, 1, 3, 1 is not well-bracketed as there is no way to match the second 1 to a closing bracket occurring after it. The sequence 1, 2, 3, 4 is not well-bracketed as the matched pair 2, 4 is neither completely between the matched pair 1, 3 nor completely outside of it. That is, the matched pairs cannot overlap. The sequence 1, 2, 4, 3, 1, 3 is well-bracketed. We match the first 1 with the first 3, the 2 with the 4, and the second 1 with the second 3, satisfying all the 3 conditions. If you rewrite these sequences using [, {, ], } instead of 1, 2, 3, 4 respectively, this will be quite clear. • Suppose $N = 6, k = 3,$", "Difference between revisions of \"2008 UNCO Math Contest II Problems/Problem 2\" Problem Let $S = \\left \\{a,b,c,d \\right \\}$ be a set of four positive integers. If pairs of distinct elements of $S$ are added, the following six sums are obtained: $5, 10, 11, 13, 14, 19.$ Determine the values of $a, b, c$, and $d.$ [Hint: there are two possibilities.] Solution The problem doesn't tell us that $a, b, c,$ and $d$ are distinct integers, but we can prove it easily. If we assume that these four positive integers are all distinct, then we will obtain six distinct pair-wise sums, since the number of ways to choose two objects out of four total where order does not matter is $(4 * 3)/2 = 6$. If any of the elements $a, b, c,$ or $d$ are equal to another, at least one of these distinct pair-wise sums would disappear, because at least two of the sums would be identical. However, we can see that the pair-wise sums we are given are all distinct, and that there are 6 of them, meaning that elements $a, b, c,$ and $d$ are all distinct. Before we move on to considering cases, it would help to consider what information might help save time. If we assume that $a < b < c", "6 = 12. 6 x 2 = 12. Question 2. The __________ of -6 is 6, because it is 6 units from zero on the number line. Answer: The absolute value of -6 is 6, because it is 6 units from zero on the number line. Explanation: In the above-given question, given that, The absolute value of -6 is 6, because it is 6 units from zero on the number line. for example: 6 and -6 are at the same distance from zero on the number line. Question 3. The number $$\\frac{5}{3}$$ is a _________ because 5 and 3 are integers and 3 ≠ 0. Answer: The number 5/3 is a rational number because 5 and 3 are integers and 3 is not equal to 0. Explanation: In the above-given question, given that, The number 5/3 is a rational number because 5 and 3 are integers and 3 is not equal to 0. for example: 3, 5, 6, and 7 are rational numbers. rational numbers are also integers. Question 4. The set of _________ consists of the counting numbers, their opposites, and zero. Answer: The set of integers consists of the counting numbers, their opposites, and zero. Explanation: In the above-given question, given that, The set of integers consists of the counting numbers, their opposites, and zero. Integers", "# Sums and differences of a set of numbers I am trying to prove that, given a set of $25$ positive numbers, it is always possible to choose a pair of them such that none of the other numbers equals either their sum or their difference. For instance, if you take the set to be the numbers from $1$ to $25$, taking $12$ and $24$ will give a sum of $36$ which is not in the set, and a difference of $12$ which is not a different element of the set than the two chosen. I feel like the proof of this is a lot simpler than I am making it. So far I have split the situation into two cases, being able to generalise for evenly spaced numbers by simply suggesting the first element chosen be a number $y \\in S$ such that $2ny \\in S$ but $3ny \\notin S$, where $n$ is the space between each number. Thus these two elements will always give a sum outside of the set and difference of $y$. However, this does not help at all for a random group of numbers and I feel there is a better argument that covers both cases. Considering the problem as a counting argument, I have determined that there are $300$ possible pairs within" ]

[ "16 views | 16 views 0 2 numbers for eg :- 11+1 =12 0 I dont get it can u explain more Suppose we wrote $11$ numbers $1,2,3,4,5,6,7,8,9,10,11$ each on different piece of paper and then put all those papers in a box and we shuffle those pieces. Now how many minimum number of pieces we need to take out from that box to ensure that the number written on any $2$ of the pieces that we are taking out from the box results in sum $12$ without looking at the numbers that we are picking out from the box.. here the name of the box is $S$. i.e. box=set. The pairs that would result in sum of $12$ are $(1,11) ,(10,2), (3,9),(4,8),(5,7)$ only So we have to calculate the worst case of getting $12$ as sum of any $2$ numbers that we ae picking out. We pick $1$ then $2$ then $3$ then $4$ then $5$ then $6$ uptil here sum of none of the $2$ numbers that we picked results in $12$ Now whatever number we pick next from the remaining numbers in the set(box) we would definitely get $12$ as the sum.i.e. if we pick $7$ then we can get $5+7 =12$ if we pick $8$ then we can get $4+8 =12$ if we pick", "# Choosing $7$ numbers from $[1,2,…,11]$ will give us $2$ that have sum $12$. Choosing $7$ numbers from $[1,2,...,11]$ will give us $2$ that have sum $12$. I tried: There are only $5$ pairings possible: $(7,5),(8,4),(9,3),(10,2),(11,1)$ Suppose I pick $6$, and then not to be a pair I will need to draw exactly 1 number from each of the above pairs. Suppose I picked, $7,8,9,10,11$. I will use up $6$ numbers. Thus any number that is left belongs to some pair. How can I put this into mathematical terms and prove it? Prove that for any set of $7$ numbers chosen from the set $\\{1, 2, 3, ..., 11\\},$ there exists at least one pair of two numbers that sum to $12.$ We can formally prove this by contradiction. Imagine that we can find a set of $7$ numbers with no two elements with a sum of $12.$ We have $6$ boxes - one of which can hold up to $1$ number. Then our proposition is that each of $b_{1}, b_{2}, b_{3}, ..., b_{6}$ is less than or equal to $1.$ In other words, $b_{n} \\le 1.$ Then $$\\sum_{n = 1}^{6} b_{n} \\le 6.$$ • I meant that one box can hold up to 1 number. It can only hold $6.$ Because $6 + 6 = 12,$ and", "\\lt x_2 \\lt ... \\lt x_n$$. We can select this subset of numbers in $$n!$$ # of ways and out of these $$n!$$ ways only one, namely $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$ will be in ascending order. So 1 out of n!. $$P=\\frac{1}{n!}$$. Hence, according to the above, the only thing we need to know to answer the question is the size of the subset ($$n$$) we are selecting from set $$A$$. Hi, Got the first point but a doubt on the second one how can you say the subset of n numbers will be selected in n! ways. Suppose a situation where there are 12 distinct numbers and you select 5. Then the possible combinations will be 12C5 ways i.e 12!/(5!*7!) ways Math Expert Joined: 02 Sep 2009 Posts: 55273 ### Show Tags 14 Nov 2014, 02:51 1 shyam593 wrote: Bunuel wrote: Official Solution: We should understand the following two things: 1. The probability of selecting any $$n$$ numbers from the set is the same. Why should any subset of $$n$$ numbers have higher or lower probability of being selected than some other subset of $$n$$ numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$, where $$x_1 \\lt x_2 \\lt ... \\lt", "# Question about the Pigeonhole Principle The question is: Let $A = \\{1,2,3,4,5,6,7,8\\}$. If five integers are selected from $A$, must at least one pair of the integers have a sum of $9$? The book explains the solution by dividing the number into $4$ disjoint subsets and pointing out that if the numbers in each set are added together, they result in a sum of $9$: $\\{1,8\\}, \\{2,7\\}, \\{3,6\\}, \\{4,5\\}$. Thus, if you pick $5$ numbers, you will inevitably find a pair, whose sum will equal to $9$. I don't get it. When the book asks to select $5$ numbers from $A$, I can pick $\\{1,2,3,4,5\\}$. Then from there, I can pick two integers like $1$ and $2$, whose sum does not equal to $9$. What am I not getting here? Thanks for any help. - its saying when you pick any 5 elements, you have some pair in that set of 5 elements whose sum is 9. In the set of 5 elements you picked, $4+5 = 9$ – Deven Ware Dec 15 '11 at 2:05 Of course some pairs, indeed many pairs, will not have sum $9$. But in your example, $4+5=9$. – André Nicolas Dec 15 '11 at 2:06 In that group of 5 you selected, the pair 4 and 5 sum to 9. Obviously,", "say the subset of n numbers will be selected in n! ways. Suppose a situation where there are 12 distinct numbers and you select 5. Then the possible combinations will be 12C5 ways i.e 12!/(5!*7!) ways Math Expert Joined: 02 Sep 2009 Posts: 52231 ### Show Tags 14 Nov 2014, 01:51 1 shyam593 wrote: Bunuel wrote: Official Solution: We should understand the following two things: 1. The probability of selecting any $$n$$ numbers from the set is the same. Why should any subset of $$n$$ numbers have higher or lower probability of being selected than some other subset of $$n$$ numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$, where $$x_1 \\lt x_2 \\lt ... \\lt x_n$$. We can select this subset of numbers in $$n!$$ # of ways and out of these $$n!$$ ways only one, namely $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$ will be in ascending order. So 1 out of n!. $$P=\\frac{1}{n!}$$. Hence, according to the above, the only thing we need to know to answer the question is the size of the subset ($$n$$) we are selecting from set $$A$$. Hi, Got the first point but a doubt on the second one how can you say the subset of n", "# Sums and differences of a set of numbers I am trying to prove that, given a set of $25$ positive numbers, it is always possible to choose a pair of them such that none of the other numbers equals either their sum or their difference. For instance, if you take the set to be the numbers from $1$ to $25$, taking $12$ and $24$ will give a sum of $36$ which is not in the set, and a difference of $12$ which is not a different element of the set than the two chosen. I feel like the proof of this is a lot simpler than I am making it. So far I have split the situation into two cases, being able to generalise for evenly spaced numbers by simply suggesting the first element chosen be a number $y \\in S$ such that $2ny \\in S$ but $3ny \\notin S$, where $n$ is the space between each number. Thus these two elements will always give a sum outside of the set and difference of $y$. However, this does not help at all for a random group of numbers and I feel there is a better argument that covers both cases. Considering the problem as a counting argument, I have determined that there are $300$ possible pairs within", "# NOSS - Editorial Author: AmirReza PoorAkhavan Tester: Danial Erfanian Editorialist: Michael Nematollahi DP, KNAPSACK ### PROBLEM: You are given a set of positive integers a. You are asked to replace any two of those numbers with any pair of positive integers so that the number of values you can obtain by adding up a subset of them (possibly empty) is maximized. You need to print this maximum number of values. ### QUICK EXPLANATION: The answer is \"the maximum NOSS over all subsets of size N-2\" imes 4. Let’s define dp* as the number of subsets whose sum is equal to i. dp can be calculated using the well-known knapsack problem. Now, for each unique pair of numbers in the set, try removing them and calculating the NOSS of the rest of the numbers. It can be proven that the number of unique pairs of numbers in the set is of O(S), where S is the sum of all of the numbers in the set. You can remove the pair from the dp array in O(S) and then count how many of the slots in dp are greater than 0 modulo some big prime MOD (like 10^9 + 7). The probability that those slots really contain 0 and not just 0 modulo MOD is very high. This way,", "($2n+1$ and $2n+2$) is chosen, we are done, since we must pick at least $n+1$ elements from the rest of the set $\\{1,...,2n\\}$ . A problem arises if BOTH of the two new numbers are chosen. Then we are required to prove that an arbitrary set of $n$ numbers from the set $\\{1,...,2n\\}$ contains two numbers, one of which divide another, or that the same arbitrary set of numbers must contain a number that divides either $2n+1$ or $2n+2$. If the arbitrary set of $n$ numbers contain either $1$, $2$, or $n+1$, then we are done. So we must prove that an arbitrary set of $n$ numbers from the set $\\{3,4,...,n-1,n,n+2,...,2n\\}$ contains two numbers, one of which divide another, or that the same arbitrary set of numbers must contain a number that divides either $2n+1$ or $2n+2$. How do I show this? • Similar: math.stackexchange.com/questions/769721/… also math.stackexchange.com/questions/1493253/… and probably many others. – Gerry Myerson Dec 31 '16 at 0:05 • The only numbers in if you choose any n+1 integers in 1...2n then they contain n integers and so one divides another. If you select 2n+1 but not 2n +2 then you must have also selected n and 2n. If you selected 2n+2 then you selected n+1. – fleablood Dec 31 '16 at 4:33 • Yes, yes I", "we have 12 numbers in set k={1,2,3...,11,12} All of the following are ascending order selection 1,2,3,4,5 6,7,8,9,10 4,7,8,9,12 1,3,5,6,11 .....and many more possible cases. Please elaborate the solution provided is not at all clear. Please got through the previous pages of discussion. _________________ Intern Joined: 23 Jun 2015 Posts: 29 ### Show Tags 06 Jul 2016, 04:49 1 Bunuel wrote: Official Solution: We should understand the following two things: 1. The probability of selecting any $$n$$ numbers from the set is the same. Why should any subset of $$n$$ numbers have higher or lower probability of being selected than some other subset of $$n$$ numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$, where $$x_1 \\lt x_2 \\lt ... \\lt x_n$$. We can select this subset of numbers in $$n!$$ # of ways and out of these $$n!$$ ways only one, namely $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$ will be in ascending order. So 1 out of n!. $$P=\\frac{1}{n!}$$. Hence, according to the above, the only thing we need to know to answer the question is the size of the subset ($$n$$) we are selecting from set $$A$$. Dear Bunuel .. I have a concern here . I think the answer is D .", "numbers will be selected in n! ways. Suppose a situation where there are 12 distinct numbers and you select 5. Then the possible combinations will be 12C5 ways i.e 12!/(5!*7!) ways Say n = 5 and k = 12 (1, 2, 3, 4, ..., 12). Now, the number of 5-number groups out of 12 is indeed 12C5. Second point there says that the number of different sequences of 5 numbers is 5!. For example, if 5 numbers we are selecting are a, b, c, d, e, then the order of selections can be a, b, c, d, e, or e, d, c, b, a, or, e, a, b, c, d, ... Total = 5!. _________________ Intern Joined: 04 Jul 2013 Posts: 16 ### Show Tags 14 Nov 2014, 04:59 2 Bunuel wrote: shyam593 wrote: Bunuel wrote: Official Solution: We should understand the following two things: 1. The probability of selecting any $$n$$ numbers from the set is the same. Why should any subset of $$n$$ numbers have higher or lower probability of being selected than some other subset of $$n$$ numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$, where $$x_1 \\lt x_2 \\lt ... \\lt x_n$$. We can select this subset of numbers in", "n numbers are selected from the set one-by-one, where n<=k, what is the probability that numbers will be selected in ascending order? (1) Set A consists of 12 even consecutive integers; (2) n=5. We should understand following two things: 1. The probability of selecting any n numbers from the set is the same. Why should any subset of n numbers have higher or lower probability of being selected than some other subset of n numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$, where $$x_1<x_2<...<x_n$$. We can select this subset of numbers in $$n!$$ # of ways and out of these n! ways only one, namely $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$ will be in ascending order. So 1 out of n!. $$P=\\frac{1}{n!}$$. Hence, according to the above the only thing we need to know to answer the question is the size of the subset (n) we are selecting from set A. \\\\ Bunuel, pls clarify how can there be n! ways to select n out of a set which has got \"k \" distinct numbers?? I m stupified.. _________________ The Mind is everything . What you think you become. - Lord Buddha Consider giving KUDOS if you appreciate my post !! Senior Manager", "173 views Selection of how many integers from the first ten positive integers (1, 2, ...) guarantees that there must be a pair of these integers with a sum equal to 11 ? my appr. Possible(x, y)solutions : (1, 10), (2, 9), (3, 8), (4, 7), (5, 6) so total 5 solution where i am wrong? In the worst case, we can select $5$ numbers such that sum of any pair of numbers is not equal to $11$, and selecting one more number will guarantee that sum of some pair of numbers is equal to $11$. $\\therefore answer = 6$. @shishir__roy can you give any example? It's simply pigeon hole principle If u choose randomly any 5 numbers and add 1 then surely there will be a pair whose sum will be 11 i.e. 1,2,3,4,5,6 there is (5,6) So ans will be 6 Among numbers 1 to 10, there are 5 pairs that corresponds to the sum 11. They are (10,1),(9,2),(8,3),(7,4) and (6,5). We can create a run of selections without any pair corresponding to the sum 11 upto a max of 5 numbers. For eg. the set {1,2,3,4,5} does not have any pairs summing up as 11. But in this set, if we add another number from the set {1,..10} – {1,2,3,4,5}, then there will be a", "$$n!$$ # of ways and out of these $$n!$$ ways only one, namely $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$ will be in ascending order. So 1 out of n!. $$P=\\frac{1}{n!}$$. Hence, according to the above, the only thing we need to know to answer the question is the size of the subset ($$n$$) we are selecting from set $$A$$. Hi, Got the first point but a doubt on the second one how can you say the subset of n numbers will be selected in n! ways. Suppose a situation where there are 12 distinct numbers and you select 5. Then the possible combinations will be 12C5 ways i.e 12!/(5!*7!) ways Say n = 5 and k = 12 (1, 2, 3, 4, ..., 12). Now, the number of 5-number groups out of 12 is indeed 12C5. Second point there says that the number of different sequences of 5 numbers is 5!. For example, if 5 numbers we are selecting are a, b, c, d, e, then the order of selections can be a, b, c, d, e, or e, d, c, b, a, or, e, a, b, c, d, ... Total = 5!. ok got it SO we can say Probability = possible outcomes/ total outcomes = 12C5/12P5 12C5: shows selection of 5 numbers and out of", "which is impossible as $r \\not = k$. This leaves us with the $16$ solutions $(2,34),(3,33),...,(17,19)$ To achieve set $A$, we eliminate $14$ elements from $B$. The maximum amount of pairs we can demolish is then $14$ (one element from each pair). This leaves us with at least $2$ pairs that sum up to $104$ in any set $A$, which of course means that there is two distinct integers in $A$ that sum up to $104$. Appreciate if someone had a look! Last edited: Feb 26, 2008 2. Feb 26, 2008 ### qspeechc Looks good to me-great stuff! What are these 'putnam problems'? They sound very interesting. 3. Feb 26, 2008 Thanks,", "###### back to index | new Let $A={1,2,3,4}$, and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$. Find the probability that the range of $f$ and the range of $g$ are disjoint. How many non-congruent triangles have vertices at three of the eight points in the array shown below? How many ways can all six numbers in the set $\\{4, 3, 2, 12, 1, 6\\}$ be ordered so that $a$ comes before $b$ whenever $a$ is a divisor of $b$? How many collections of six positive, odd integers have a sum of $18$? Note that $1 + 1 + 1 + 3 + 3 + 9$ and $9 + 1 + 3 + 1 + 3 + 1$ are considered to be the same collection.", "# Show that if seven integers are selected from the first 10 positive integers a) Show that if seven integers are selected from the first 10 positive integers, there must be at least two pairs of these integers with the sum 11. b) Is the conclusion in part (a) true if six integers are selected rather than seven? I don't know how should I show that. I know that in the worst situation we choose 1,2,...,7 that we have 7 + 4 = 11 and 5 + 6 = 11 but I don't know what should be the answer of this question Partition numbers from $1$ to $10$ into the following sets: $\\{1,10\\}$, $\\{2,9\\}$, $\\{3,8\\}$, $\\{4,7\\}$ and $\\{5,6\\}$. Now use pigeon hole principle. • It's a little tricky in that we are asked to show two pairs have sum $11$, and the pigeonhole principle gives one pair by immediate application (even if as few as six integers are selected). The difference between selecting seven and six integers from $1$ to $10$ is that we get (at least) two pairs with sum $11$ in the former case and as few as one pair in the latter case. – hardmath Dec 1 '13 at 16:58", "# When does a set of sums uniquely determine a set of values? This is a follow up to a question about determining the set of values from a set of sums. We consider the following setup: Consider a vector $$A$$ (which you will not see) of $$n$$ positive integers. You are given the set of sums of the (contiguously indexed) subvectors of $$A$$. For example, say $$A = (3,2,1,2)$$ The subvectors are $$(3),(2),(1),(2), (3,2), (2,1), (1,2),(3,2,1), (2,1,2),(3,2,1,2)$$. We would be given the sums $$\\{1, 2, 3, 5, 6, 8\\}$$. Let us call this set of sums $$f(A)$$. It turns out it is not always possible to uniquely determine the set of integers in $$A$$ from the pair $$(f(A), n)$$? For example: $$A = (1,1,3)$$ and $$B = (1,2,2)$$ both give the same set of sums $$f(A) = f(B) = \\{1,2,3,4,5\\}$$. However in some cases it is possible. For example, if $$n=3$$ and $$f(C) = \\{1,2,3,5,6\\}$$ then we know that $$\\operatorname{set}(C) = \\{1, 2, 3\\}$$. This raises the following question: Is it possible to characterise which pairs $$(f(X), n)$$ uniquely determine the set of values in the array $$X$$? • Maybe it suffices to know that in the sum, there exist a one or a zero??? – user7427029 Oct 15 '18 at 22:50 • @user7427029 I don’t quite", "there is an element $$m \\in S$$. But then $$m=m+0$$ and so $$0 \\in S$$. 1 cannot be in $$S,$$ otherwise it will contain all non-negative integers. By the division algorithm that if $$m, n$$ are in $$S$$ then so is their GCD. Therefore two coprime numbers cannot be in $$S .$$ Otherwise their GCD, which is $$1,$$ will be in $$S,$$ a contradiction. It follows that such sets $$S$$ are precisely those of the form $$n \\mathbb{Z} \\geq 0,$$ the set of all non-negative multiples of a fixed non-negative integer $$n .$$ So there are infinitely many such possible sets. ### Sample a divisor A4, 2017 Positive integers $$a$$ and $$b,$$ possibly equal, are chosen randomly from among the divisors of $$400 .$$ The numbers $$a, b$$ are chosen independently, each divisor being equally likely to be chosen. Find the probability that $$\\operatorname{gcd}(a, b)=1$$ and $$\\operatorname{lcm}(a, b)=400$$ Solution $$400=5^{2} \\times 2^{4}$$ has $$(2+1) \\times(4+1)=15$$ factors, so total number of pairs $$(a, b)$$ is $$225 .$$ For $$a, b$$ to be coprime, they should have no prime factor in common and then their lcm is just their product, which is required to be $$400 .$$ So there are only four allowed pairs: (1,400),(400,1),(25,16) and (16,25). The probability is $$\\frac{4}{225}$$. ### Carrom board math B6, 2018 Imagine the unit", "# Selecting from $\\{1,2,3,4,5,6,7,8,9\\}$ to guarantee at least one pair adds to $10$ How many numbers must be selected from the set $\\{1,2,3,4,5,6,7,8,9\\}$ to guarantee that at least one pair of these numbers add up to $10$? Justify your answer. Consider the two sets $\\{1,2,3,4\\}$ and $\\{6,7,8,9\\}$. In the worst case, one may pick the number $5$ and then all the numbers from one set and then a number from the other set. Therefore, $6$ numbers must be selected in order to guarantee that at least one pair of these numbers add up to $10$. Do you think my answer is correct? - Yes,now generalize it to other values.Given the numbers ${1,2,3. . .n}$, how many numbers should you pick to guarantee that a pair adds up to $n+1$ ? – rah4927 Apr 21 '14 at 7:44 Yes. Set $\\left\\{ 1,2,3,4,5,6,7,8,9\\right\\}$ is partitioned by the $5$ sets $\\left\\{ 1,9\\right\\}$, $\\left\\{ 2,8\\right\\}$, $\\left\\{ 3,7\\right\\}$, $\\left\\{ 4,6\\right\\}$, $\\left\\{ 5\\right\\}$. If $6$ numbers are picked out, then it is guaranteed that from one of these sets two distinct elements are picked out. – drhab Apr 21 '14 at 7:46 The argument is not fully persuasive, since conceivably we might pick, in addition to $5$, $3$ elements from one of our $4$-element sets and $2$ from another, and not get a sum", "where the absolute value of the larger number is greater than that of the smaller. With this restriction in mind, we have the following pairs of factors whose product is -48. {48, -1}, {24, -2}, {12, -4}, {8,-6} From these pairs, 8 and -6 has a sum of 2. Therefore, the factors are $(x + 8)(x - 6) = 0$ Equating to 0, we have $x + 8 = 0$, $x = -8$ $x - 6 = 0$, $x = 6$ Since we are looking for positive integers, we will take $x = 6$ and $x + 2 = 6 + 2 = 8$. Therefore, the two consecutive numbers are 6 and 8. Check: The numbers 6 and 8 are consecutive numbers and their product is 48. Therefore, we are correct." ]

[ "understand what $D(3,n)$ is. I assumed it was $D(3,n) = \\dbinom{n+2}{2}$. 10. ## Re: Help Generating Functions :) Originally Posted by SlipEternal I don't know what you mean by \"where do I attach importance here to a7\". You are looking for: $a_n = D(3,n) - ra_{n-1}-sa_{n-2}-ta_{n-3}$. You have: $a_n = \\dbinom{n+2}{2}-2a_{n-1}-2a_{n-2}-a_{n-3}$. What are $r,s,t$? I am not sure I understand what $D(3,n)$ is. I assumed it was $D(3,n) = \\dbinom{n+2}{2}$. Hey friend, I will try to explain myself more, I need to find three numbers (R, S, T) where the N is greater than or equal to 3, and then calculate A 7 using this formula (here I think it's only a placement and calculation) ... Hope more sense. thanks..... 11. ## Re: Help Generating Functions :) Ok, then $a_7 = \\dbinom{9}{2}-2a_6-2a_5-a_4$ $a_6 = \\dbinom{8}{2}-2a_5-2a_4-a_3$ $a_5 = \\dbinom{7}{2}-2a_4-2a_3-a_2 = \\dbinom{7}{2}-2a_4-2a_3-2$ $a_4 = \\dbinom{6}{2}-2a_3-2a_2-a_1 = \\dbinom{6}{2}-2a_3-5$ $a_3 = \\dbinom{5}{2}-2a_2-2a_1-a_0 = \\dbinom{5}{2}-7 = 3$ Plugging back in: $a_4 = \\dbinom{6}{2}-2a_3-5 = 15-6-5 = 4$ $a_5 = \\dbinom{7}{2}-2a_4-2a_3-2 = 21-8-6-2 = 5$ $a_6 = \\dbinom{8}{2}-2a_5-2a_4-a_3 = 28-10-8-3 = 7$ $a_7 = \\dbinom{9}{2}-2a_6-2a_5-a_4 = 36-14-10-4 = 8$", "\\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} & 360 \\cdot x^{44} & 360 \\cdot x^{45} & 120 \\cdot x^{46}\\\\ & & & &", "+0 # Help. I have been stuck 0 114 4 Given that $$\\binom{15}{8}=6435$$$$\\binom{16}{9}=11440$$, and $$\\binom{16}{10}=8008$$, find $$\\binom{15}{10}$$. I know that I should be able to do this but I can't figure it out. Aug 10, 2019 #1 +23295 +3 Given that $$\\dbinom{15}{8}=6435$$, $$\\dbinom{16}{9}=11440$$ , and $$\\dbinom{16}{10}=8008$$, find $$\\dbinom{15}{10}$$. Formula: $$\\dbinom{n}{k} + \\dbinom{n}{k+1} = \\dbinom{n+1}{k+1}$$ $$\\begin{array}{|lrcll|} \\hline & \\mathbf{\\dbinom{n}{k} + \\dbinom{n}{k+1}} &=& \\mathbf{\\dbinom{n+1}{k+1}} \\\\\\\\ (1) & \\dbinom{15}{8} + \\dbinom{15}{9} &=& \\dbinom{16}{9} \\\\ (2) & \\dbinom{15}{9} + \\dbinom{15}{10} &=& \\dbinom{16}{10} \\\\ \\hline (1)-(2): & \\dbinom{15}{8} + \\dbinom{15}{9} -\\left( \\dbinom{15}{9} + \\dbinom{15}{10}\\right) &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & \\dbinom{15}{8} - \\dbinom{15}{10} &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & 6435 - \\dbinom{15}{10} &=& 11440-11440 \\\\ & 6435 - \\dbinom{15}{10} &=& 3432 \\\\ & \\dbinom{15}{10} &=& 6435 - 3432 \\\\ & \\mathbf{\\dbinom{15}{10}} &=& \\mathbf{3003} \\\\ \\hline \\end{array}$$ Aug 11, 2019 #2 0 Thank you so much!! I can't keep all the formulas in my head sometimes. Aug 11, 2019 #3 +1655 +3 There are really too many formulas... tommarvoloriddle Aug 12, 2019 #4 +105540 +1 You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from. There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level. Most times this does", "$p$ from the numerator to get that $$\\dbinom{p}{k} = \\dfrac{p \\times (p-1) \\times (p-2) \\times \\cdots \\times (p-k+2) \\times (p-k+1)}{k \\times (k-1) \\times (k-2) \\times \\cdots \\cdots \\times 2 \\times 1} = p \\times M$$ where $M \\in \\mathbb{Z}$.", "can run the same method to compute $$a\\pmod{11!}$$.", "# Using generating functions to find the coefficient I was looking at an example in my textbook where it was asked to find in how many ways can a police captain distribute 24 rifle shells to four police officers so that each officer gets at least three shells, but not more than eight. Using generating functions: ${f(x)=(x^3+x^4+x^5+...+x^8)^4}$, and we will be looking to find the coefficient of ${x^{24}}$ in $f(x)$ We know that $f(x)= x^{12}((1-x^6)/(1-x))^4$ Therefore the answer will be the coefficient of $x^{12}$ in the expansion of $(1-x^6)^4*(1-x)^{-4}$ Which in turn is equal to $A$: $[1-\\dbinom{4}{1}x^6+\\dbinom{4}{2}x^{12}-\\dbinom{4}{3}x^{18}+x^{24}]*[\\dbinom{-4}{0}+\\dbinom{-4}{1}(-x)+\\dbinom{-4}{2}(-x)^2+...]$ Up to this point I perfectly understand the problem and the proposed solution; but the textbook goes on saying that the above expression is equivalent to $B$: $[\\dbinom{-4}{12}(-1)^{12}-\\dbinom{4}{1}\\dbinom{-4}{6}(-1)^6+\\dbinom{4}{2}\\dbinom{-4}{0}]$ I do not understand how they went from $A$ to $B$ I know that $C$: $[\\dbinom{-4}{12}(-1)^{12}]$ is the coefficient of $x^{24}$ in: $x^{12}[\\dbinom{-4}{0}+\\dbinom{-4}{1}(-x)+\\dbinom{-4}{2}(-x)^2+...]$ But I do not seem to make the link between $C$ and the following subtraction and addition, namely: $[C-\\dbinom{4}{1}\\dbinom{-4}{6}(-1)^6+\\dbinom{4}{2}\\dbinom{-4}{0}]$ • What are the ways to get an $x^{12}$ term in the product $(1 - ax^6 + bx^{12} -\\dotsc)\\cdot (1 - ux + vx^2 - wx^3 +\\dotsc)$? – Daniel Fischer Oct 12 '15 at 8:31 • @DanielFischer there will be $ax^6 * zx^6$ or $bx^{12}*1$ or $1 * mx^{12}$ – O.A.", "I get another hint? – jsluong Jul 23 '15 at 20:40 • Yep, the same recursion you have in your answer, only with different values $k=i$ and $n \\mapsto 1 + n + i$ – johannesvalks Jul 23 '15 at 20:42 • Should be $\\binom{n+i}{i+1}$ on the RHS. – xivaxy Jul 23 '15 at 20:47 • @Dr.MV - indead, I have corrected it. – johannesvalks Jul 23 '15 at 21:37 Oh, hey! I just figured it out. Funny how simply posting the question allows you re-evaluate the problem differently... So on Wikipedia apparently this is a thing (the recursive formula for computing the value of binomial coefficients): $$\\dbinom{n}{k} = \\dbinom{n-1}{k-1} + \\dbinom{n-1}{k}$$ On the RHS of the equation we have (let's call this equation 1): $$\\dbinom{2n+1}{n+1} = \\dbinom{2n}{n} + \\dbinom{2n-1}{n} + \\dbinom{2n-2}{n} + \\cdots + \\dbinom{n+1}{n} + \\dbinom{2n-(n-1)}{n+1}$$ The last term $\\dbinom{2n-(n-1)}{n+1}$ simplifies to $\\dbinom{n+1}{n+1}$ or just 1. Meanwhile on the LHS we have $\\sum\\limits_{i=0}^{n}\\dbinom{n+i}{i}$ which is also $\\sum\\limits_{i=0}^{n}\\dbinom{n+i}{n}$ because $\\dbinom{n}{k} = \\dbinom{n}{n-k}$. Written out that is (let's call this equation 2): $$\\sum\\limits_{i=0}^{n}\\dbinom{n+i}{n} = \\dbinom{n}{n} + \\dbinom{n+1}{n} + \\dbinom{n+2}{n} + \\cdots + \\dbinom{2n}{n}$$ The first term $\\dbinom{n}{n}$ simplifies to 1. Hey, look at that. In each written out sum there's a term in equation 1 and an equivalent in equation 2. And in each sum there's $n$ terms,", "we compute {\\$}T(\\mathbb{\\{}Z{\\}}{\\_}m \\times \\mathbb{\\{}Z{\\}}{\\_}n){\\$} for {\\$}2 \\leq m \\leq 7{\\$} and {\\$}2 \\leq n \\leq 19{\\$}.}, comment = {}, }", "let $N_1=N-\\dbinom{x_1}k$.\" Did you mean $N_1=N-\\dbinom{x_k}{k}$? That would be consistent with what you wrote in your next paragraph: $N_1=\\dbinom{x_{k-1}}{k-1}+\\ldots+\\dbinom{x_1}{1}$ – tychicus Aug 6 '13 at 7:21 • @tychicus: Yes, I did; I’ve fixed it now. (The same mistake was in the original version of your hint, and I must have unconsciously copied that.) – Brian M. Scott Aug 6 '13 at 12:02", "@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19 @AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22 Nope, that seems likely to be the most concise way to do it. Count ways to pick $n\\in\\{1,2,3,4\\}$ unique numbers, and to arrange them with $n-1$ \"$>$\" signs, to meet the criteria. $$\\begin{array}{|l:l|} \\hline \\rm a{=}a{=}a{=}a & \\dbinom 3 0 \\dbinom 6 1 \\\\\\hdashline\\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \\dbinom 3 1\\dbinom 6 2 \\\\\\hdashline\\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \\dbinom 3 2 \\dbinom 6 3 \\\\\\hdashline\\rm a{>}b{>}c{>}d & \\dbinom 3 3 \\dbinom 6 4 \\\\ \\hline\\end{array}$$ $$\\dfrac{\\sum_{n=1}^4 \\dbinom{3}{n-1}\\dbinom{6}{n}}{6^4} = \\dfrac{\\dbinom 6 1 + 3\\dbinom 6 2 + 3 \\dbinom 6 3 + \\dbinom 6 4}{6^4}$$ - The problem reduces to finding the cardinality of the following set $$A=\\{(a_1,a_2,a_3,a_4)\\in\\{1,\\ldots,6\\}^4: \\ a_1\\leq a_2\\leq a_3 \\leq a_4\\}.$$ Define $$B=\\{(a_1,a_2,a_3,a_4)\\}\\in\\{1,\\ldots 9\\}^4: \\ a_1 < a_2 < a_3 < a_4\\}$$ and $$f:A\\to B, \\quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \\choose 4}.$$ - @Shailesh This method actually does give the correct answer ($126/6^4$); If we", "# Why does $\\binom{10}{7} = \\frac{10!}{(10-7)!7!}$ We just learned that: $\\dbinom{10}{7}= \\frac{10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4}{7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}$, so that: If you throw a dice 10 times, the probability of getting $6$, $7$ of the times is: $\\dbinom{10}{7} \\times {\\frac{1}{6}}^7 \\times {\\frac{5}{6}}^3$, because $\\dbinom{10}{7}$ will give us the number of different ways you can get $7$ \"correct\" out of $10$. I wonder why this works. Why does $\\dbinom{10}{7}$ work as it does? (In my search I stumbled upon this way of writing it: $\\dbinom{10}{7} = \\frac{10!}{(10-7)!7!}$. It is a little bit different, but maybe it is more correct? 4 2022-07-25 20:42:53 Source Share", ")^{10} \\\\\\\\ && ( a + b + c )^{n} \\\\ &=& \\sum \\limits_{k=0}^{n} \\sum \\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} &", "## Intermediate Algebra for College Students (7th Edition) $x^{10}-5x^8+10x^6-10x^4+5x^2-1$ Apply Binomial Theorem. $(a+b)^n=\\sum_{r=0}^n\\dbinom{n}{r}(a)^{n-r}b^r$ $(x^2-1)^5=\\dbinom{5}{0}(x^2)^5(-1)^0+\\dbinom{5}{1}(x^2)^4(-1)^1+\\dbinom{5}{2}(x^2)^3(-1)^2+\\dbinom{5}{3}(x^2)^2(-1)^3+\\dbinom{5}{4}(x^2)^1(-1)^4+\\dbinom{5}{5}(x^2)^0(-1)^5$ or, $=x^{10}-5x^8+10x^6+10x^4(-1)+5x^2+(1)(1)(-1)$ or, $=x^{10}-5x^8+10x^6-10x^4+5x^2-1$", "(\\mathbf{A}\\mathbf{B})^{-1} \\dbinom12 = \\dbinom13 } \\\\ \\hline \\end{array}$$ Apr 17, 2020", "\\times \\mathbb P^2$ for $D$ some base-point free divisor on $C$. This gives an explicit Weierstraß equation, so most things you want to compute should be fairly explicit. Feb 24 '18 at 3:59", "# Summation this Summer! 3 Discrete Mathematics Level 4 $\\dbinom{2n}{1}^2 +2.\\dbinom{2n}{2}^2 +3.\\dbinom{2n}{3}^2 \\cdots \\cdots + 2n.\\dbinom{2n}{2n}^2$ Find the closed form of the above series. Notation: $$!$$ denotes the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "n + l\\times n\\times p$ multiplications; \\item Compute $A(BC)$ with $m\\times n\\times p + l\\times m\\times p$ multiplications. \\end{enumerate} Therefore, \\begin{enumerate} \\item If $lmn+lnp>mnp+lmp$, the product $ABC$ should be computed in the order of $A(BC)$. \\item If \\$lmn+lnp", "\\mathbf{b}^{(l)} ] \\times [ \\mathbf{A}^{(l)}; \\mathbf{1}]$$", "from stack } i+1), 0\\le t \\le l\\right\\}$$ We can compute $A[m,k]$ in time $O(k^2m)$.", "\\times 2$ matrices using only $7$ multiplications instead of $8$. …" ]

[ "20 }\\!\\!\\times\\!\\!\\text{ }\\dfrac{\\text{18}}{\\text{60 }\\!\\!\\times\\!\\!\\text{ 60}}\\text{ km = }\\dfrac{\\text{20 }\\!\\!\\times\\!\\!\\text{ 18}}{\\text{60 }\\!\\!\\times\\!\\!\\text{ 60}}\\times \\text{ 1000m} \\\\ \\end{align} d = 100m.", "h &= \\dfrac{4 \\pi^2Ed^4}{4 \\times 64L^2 \\times D_i^2\\times \\rho g}\\\\ h &= \\dfrac{4 \\pi^2 \\times 200 \\times 10^9 \\times (0.075)^4}{4 \\times 64 \\times \\times 4^2 \\times 1.5^2 \\times 10^4}\\\\ h &= 2.7\\ m\\end{aligned}", "( \\dfrac{1}{T_2^{2/3}}-\\dfrac{1}{T_1^{2/3}} \\right )\\\\ &=-\\left ( 2\\pi \\times 6.67\\times 10^{-11}\\times 6\\times 10^{24}\\right )^{2/3}\\times 50\\times \\left ( \\dfrac{1}{\\left ( 10.8\\times 10^{3} \\right )^{2/3}}-\\dfrac{1}{\\left (7.2\\times 10^{3} \\right )^{2/3}} \\right )\\\\ &=-\\left ( 2.51\\times 10^{15}\\right )^{2/3}\\times 50\\times \\left ( \\dfrac{1}{488}-\\dfrac{1}{373} \\right )\\\\ &=-1.36\\times 10^{10}\\times 50\\times \\left ( -6.32 \\times 10^{-4}\\right )\\\\ &=\\boxed{4.3\\times 10^{6}\\;\\rm J} \\end{align*} {/eq}", "{mg \\times \\left( {{R_c} - OA} \\right)} \\right) \\end{align*} {/eq} Substitute the given data in above equation, {eq}\\begin{align*} {M_a} &= \\left( {\\left( {\\dfrac{2}{{12}} \\times \\left( {\\dfrac{{50}}{2} - 10} \\right)} \\right) \\times {{\\left( {\\dfrac{{50}}{2} - 10} \\right)}^2}} \\right) + \\left( {2 \\times \\left( {\\dfrac{{50}}{2} - 10} \\right)} \\right) \\times 8 \\times 17.5 - \\left( {\\left( {2 \\times \\left( {\\dfrac{{50}}{2} - 10} \\right)} \\right)9.81 \\times \\left( {17.5 - 10} \\right)} \\right)\\\\ &= 2555.5\\;{\\rm{N}} \\cdot {\\rm{m}} \\end{align*} {/eq} Thus the moment at A is {eq}2555.5\\;{\\rm{N}} \\cdot {\\rm{m}} {/eq}", "10^7 \\\\ d &= 31.134 \\times 10^6 - 6.378 \\times 10^6 \\\\ d \\ &\\boxed{ = 24.756 \\times 10^6 } \\ \\rm m \\end{align*}", "\\times 17 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 23 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{54} \\times p_{19}(29,25) &=& \\dfrac{29}{54} \\times \\dfrac{2^4 \\times 3^2 \\times 11^2 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^3 \\times 11^2 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} \\\\ \\dfrac{29}{55} \\times p_{20}(29,26) &=& \\dfrac{29}{55} \\times \\dfrac{2^5 \\times 5^3 \\times 11 \\times 17 \\times 19 \\times 23}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^5 \\times 5^2 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{56} \\times p_{21}(29,27) &=& \\dfrac{29}{56} \\times \\dfrac{2^4 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23}{3 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{57} \\times p_{22}(29,28) &=& \\dfrac{29}{57} \\times \\dfrac{2^2 \\times 3^2 \\times 5^2 \\times 11 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times", "last row right. This works well, if one has an expression of the type a+b=c+d+e+f+g+h=j, where there's only one row between initial row and result. However, in my case, I have some mathematical transformation (meaning an additional row) and so I need the equal sign of the row: \"12\\times6+14\\times\\dfrac{48}{7}-6^2-\\left(\\dfrac{2}{7}-4\\right)^2-\\left(\\dfrac{48}{7}\\right)^2 -\\dfrac{2}{7}\\times\\dfrac{48}{7}-\\left(\\dfrac{2}{7}\\right)^2\" to be aligned under the first equal sign. So now I use this version: Code: Select all • Open in writeLaTeX \\begin{align*}\\pi^{M}(s^{M},f^{M},x^{M}) :=& 12s^{M}+14f^{M}-(s^{M})^2-(x^{M}-4)^2-(f^{M})^2\\\\ &-x^{M}f^{M}-(x^{M})^2\\pause\\\\ =&12\\times6+14\\times\\dfrac{48}{7}-6^2-\\left(\\dfrac{2}{7}-4\\right)^2-\\left(\\dfrac{48}{7}\\right)^2\\\\ &-\\dfrac{2}{7}\\times\\dfrac{48}{7}-\\left(\\dfrac{2}{7}\\right)^2\\pause\\\\ =&69,14\\pause\\end{align*} Does anybody have better ideas? Thank you avp3000 Posts: 49 Joined: Thu Nov 15th, 2007 ### Re: Too long equation You can try the split environment, which is also provided by the amsmath package. To get a proper alignment, there are some tricky things to do. Code: Select all • Open in writeLaTeX \\begin{align*} \\pi^{M}(s^{M},f^{M},x^{M}): \\begin{split} &= 12s^{M}+14f^{M}-(s^{M})^2-(x^{M}-4)^2 \\\\ &\\quad -(f^{M})^2-x^{M}f^{M}-(x^{M})^2 \\end{split}\\\\ \\begin{split} &= 12\\times 6+14\\times\\frac{48}{7}-6^2-\\left(\\frac{2}{7}-4\\right)^2 \\\\ &\\quad -\\left(\\frac{48}{7}\\right)^2-\\frac{2}{7}\\times\\frac{48}{7}-\\left(\\frac{2}{7}\\right)^2 \\end{split}\\\\ &=69,14\\end{align*} Note that the alignment as known from the align environment has to be put into the split environment. More hints about typesetting mathematical expressions are given in \"Math mode\". Best regards Thorsten LaTeX Community Moderator ¹ System: openSUSE 13.1 (Linux 3.11.10), TeX Live 2013 (vanilla), TeXworks 0.5 (r1351) ² Posting stopped indefinitely due to offenses localghost Site Moderator Posts: 9219 Joined: Fri Feb 2nd, 2007 Location: Braunschweig, Germany ###", "n-2 \\right)...(n-r)!$ \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5!}\\times \\left( \\dfrac{8\\times 7\\times 6}{3!} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6!}\\times \\left( \\dfrac{8\\times 7}{2!} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4!}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4!} \\right) \\\\ \\end{align} \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6}{3\\times 2\\times 1} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6\\times 5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7}{2\\times 1} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4\\times 3\\times 2\\times 1} \\right) \\\\ \\end{align} By cancelling the common terms in denominator and numerator we get $=\\left( 11\\times 9\\times 8 \\right)\\times \\left( 8\\times 7 \\right)+\\left( 11\\times 3\\times 4\\times 7 \\right)\\times \\left( 4\\times 7 \\right)+\\left( 11\\times 5\\times 9 \\right)\\times \\left( 5\\times 2\\times 7 \\right)$ $=792\\times 56+924\\times 28+495\\times 70$ = 44352 + 25872 + 34650 = 104874 Thus, the number of ways of selecting 10 students chosen for the competition = 104874 Note: The possibility of mistake can be the calculation mistake since the procedure of solving involves large calculations. Hence be very careful in simplifying and cancelling the terms to get the desired result.", "be a2 (in checkerboard coordinates) because $$b2/c2 \\ge 2$$ and at least two of those marked $$\\diamond$$ and at least one of those marked $$\\ast$$. Because only 8 small numbers are available a4 must be small and one of b4,c4. All others must be $$>8$$ (\"large\", marked $$\\Box$$): $$\\begin{matrix}\\bigcirc&+&\\Box\\bigcirc&+&\\Box\\bigcirc&=&\\Box\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&\\bigcirc&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&\\bigcirc&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&\\Box&=&\\Box\\end{matrix}$$ and we see that a4 must be $$1$$ Observe that $$d4 = d3/d2 + a1/b1 + c2\\times c3 + c4 \\ge 2 + 2 + 2\\times 3 + c4$$ Therefore c4 must be small and b4 large. $$\\begin{matrix}1&+&\\Box&+&\\bigcirc&=&13\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&\\bigcirc&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&\\bigcirc&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&9\\ldots 12&=&11\\ldots 14\\end{matrix}$$ Next, we see that $$2$$ must be in either c2 or c3: $$\\begin{matrix}1&+&9\\ldots 11&+&3\\ldots 6&=&14\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&2\\ldots 4&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&2\\ldots 4&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&10\\ldots 12&=&12\\ldots 14\\end{matrix}$$ We are almost done! Observe that in the products $$a2\\times b2 = c2\\times d2$$ the primes $$5$$ and $$7$$ cannot occur because they are available only once. $$\\begin{matrix}1&+&9\\ldots 10&+&5&=&15\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&2\\ldots 3&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&2\\ldots 3&=&\\bigcirc\\\\=&&=&&=&&=\\\\14&/&7&+&11&=&13\\end{matrix}$$ We can finish by filling in column a, $$\\begin{matrix}1&+&9&+&5&=&15\\\\\\times&&-&&+&&-\\\\10&+&\\Box&/&2\\ldots 3&=&\\Box\\\\+&&/&&\\times&&/\\\\4&\\times&\\bigcirc&/&2\\ldots 3&=&\\bigcirc\\\\=&&=&&=&&=\\\\14&/&7&+&11&=&13\\end{matrix}$$ row 3 and, finally, row 2. • Very well done and great explanation! I was getting worried that no one will solve this... – Dmitry Kamenetsky Oct 10 '20 at 0:46 Let the digits be a1,a2,a3,… d4, using standard algebraic chess notation (i.e. letters = columns, numbers = rows, a1 = lower left). Define a low digit = 1-8, high digit = 9-16. The cells,", "10^4\\times 9.46\\times 10^{15})^{1.5}}{\\sqrt{6.67\\times10^{-11}\\times 8\\times 10^{13}\\times 1.99\\times 10^{30}}}\\\\\\\\ &=8.25\\times 10^{14}\\, s\\\\\\\\ &=\\boxed{2.61\\times 10^7\\, yr} \\end{align} {/eq}", "47} &=& \\dfrac{3 \\times 5^2 \\times 11 \\times 13 \\times 17 \\times 29}{19 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{49} \\times p_{14}(29,20) &=& \\dfrac{29}{49} \\times \\dfrac{2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17}{37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11 \\times 13 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{50} \\times p_{15}(29,21) &=& \\dfrac{29}{50} \\times \\dfrac{2^2 \\times 3 \\times 5^3 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11 \\times 13 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{51} \\times p_{16}(29,22) &=& \\dfrac{29}{51} \\times \\dfrac{2^5 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^5 \\times 3 \\times 5 \\times 11 \\times 13 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{52} \\times p_{17}(29,23) &=& \\dfrac{29}{52} \\times \\dfrac{2^3 \\times 3 \\times 5 \\times 11^2 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{53} \\times p_{18}(29,24) &=& \\dfrac{29}{53} \\times \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2", "=9\\times \\left[ 4{{R}^{2}}-\\pi {{R}^{2}} \\right] \\\\ & =9\\times \\left( 4-\\pi \\right)\\times {{R}^{2}} \\\\ & =9\\times \\left( 4-3.14 \\right)\\times {{\\left( 7 \\right)}^{2}} \\\\ & =9\\times 0.86\\times 49 \\\\ & =397.26c{{m}^{2}} \\\\ \\end{align}", "19}{53 \\times 10} \\times \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^4 \\times 3^2 \\times 11^2 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ p_{20}(29,26) &=& \\dfrac{25 \\times 20}{54 \\times 11} \\times \\dfrac{2^4 \\times 3^2 \\times 11^2 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^5 \\times 5^3 \\times 11 \\times 17 \\times 19 \\times 23}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ p_{21}(29,27) &=& \\dfrac{26 \\times 21}{55 \\times 12} \\times \\dfrac{2^5 \\times 5^3 \\times 11 \\times 17 \\times 19 \\times 23}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^4 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23}{3 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ p_{22}(29,28) &=& \\dfrac{27 \\times 22}{56 \\times 13} \\times \\dfrac{2^4 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23}{3 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^2 \\times 3^2 \\times 5^2 \\times 11 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ p_{23}(29,29) &=& \\dfrac{28 \\times 23}{57 \\times 14} \\times \\dfrac{2^2 \\times 3^2 \\times 5^2 \\times", "3 Use the quotient rule to differentiate $\\displaystyle f(x)=\\dfrac{3x+1}{1-x}$. \\begin{align} \\displaystyle f^{\\prime}(x) &= \\dfrac{(3x+1)^{\\prime} \\times (1-x) – (3x+1) \\times (1-x)^{\\prime}}{(1-x)^2} \\\\ &= \\dfrac{3 \\times (1-x) – (3x+1) \\times (-1)}{(1-x)^2} \\\\ &= \\dfrac{3-3x +3x+1}{(1-x)^2} \\\\ &= \\dfrac{4}{(1-x)^2} \\\\ \\end{align} ### Example 4 Use the quotient rule to differentiate $\\displaystyle f(x)=\\dfrac{(2x+1)^3}{(4x-1)^4}$. \\begin{align} \\displaystyle f^{\\prime}(x) &= \\dfrac{\\big[(2x+1)^3\\big]^{\\prime} \\times (4x-1)^4 – (2x+1)^3 \\times \\big[(4x-1)^4\\big]^{\\prime}}{\\big[(4x-1)^4\\big]^2} \\\\ &= \\dfrac{3(2x+1)^{3-1} \\times (2x+1)^{\\prime} \\times (4x-1)^4 – (2x+1)^3 \\times 4(4x-1)^{4-1} \\times (4x-1)^{\\prime}}{(4x-1)^8} \\\\ &= \\dfrac{3(2x+1)^2 \\times 2 \\times (4x-1)^4 – (2x+1)^3 \\times 4(4x-1)^3 \\times 4}{(4x-1)^8} \\\\ &= \\dfrac{6(2x+1)^2 (4x-1)^4 – 16(2x+1)^3 (4x-1)^3}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3\\big[6(4x-1) – 16(2x+1)\\big]}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3(24x-6 – 32x-16)}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3(-8x-22)}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(-8x-22)}{(4x-1)^5} \\\\ \\end{align} ## Extension Examples These Extension Examples require to have some prerequisite skills including; \\begin{align} \\displaystyle \\dfrac{d}{dx}\\sin{x} &= \\cos{x} \\\\ \\dfrac{d}{dx}\\cos{x} &= -\\sin{x} \\\\ \\dfrac{d}{dx}e^x &= e^x \\\\ \\dfrac{d}{dx}\\log_e{x} &= \\dfrac{1}{x} \\\\ \\end{align} ### Example 5 Find $\\displaystyle \\dfrac{dy}{dx}$ of $\\displaystyle y=\\dfrac{\\sin{x}}{\\cos{x}}$, known that $\\dfrac{d}{dx}\\sin{x} = \\cos{x}$ and $\\dfrac{d}{dx}\\cos{x} = -\\sin{x}$. \\begin{align} \\displaystyle \\dfrac{dy}{dx} &= \\dfrac{\\dfrac{d}{dx}\\sin{x} \\times \\cos{x} – \\sin{x} \\times \\dfrac{d}{dx}\\cos{x}}{\\cos^2{x}} \\\\ &= \\dfrac{\\cos{x} \\times \\cos{x} – \\sin{x} \\times (-\\sin{x})}{\\cos^2{x}} \\\\ &= \\dfrac{\\cos^2{x} + \\sin^2{x}}{\\cos^2{x}} \\\\ \\end{align} Note that optionally $\\cos^2{x} + \\sin^2{x}$ can be simplified further to $1$, as $\\cos^2{x} + \\sin^2{x}=1$ in one of the trigonometric properties, however we do not cover this skill here.", "\\text{5}^{(2y-3) - (-5y+5)} \\times \\text{2}^{(4y+4) - (-5y+5)}\\\\ &= \\text{5}^{7y -\\text{8}} \\times \\text{2}^{9y-\\text{1}} \\end{align*} $$\\dfrac{6^4 \\times 12^3 \\times 4^5}{30^3 \\times 3^6}$$ \\begin{align*} \\frac{6^4 \\times 12^3 \\times 4^5}{30^3 \\times 3^6} & = \\frac{\\left(3^{4} \\times 2^{4}\\right) \\times \\left(3^{3} \\times 4^{3}\\right) \\times 4^{5}}{\\left(3^{3} \\times 10^{3}\\right) \\times 3^{6}} \\\\ & = \\frac{3^{4} \\times 2^{4} \\times 3^{3} \\times 2^{6} \\times 2^{10}}{3^{3} \\times 2^{3} \\times 5^{3} \\times 3^{6}} \\\\ & 3^{4 + 3 - 3 - 6} \\cdot 2^{4 + 6 + 10 - 3} \\cdot 5^{-3} \\\\ & = \\frac{2^{17}}{3^{2}5^{3}} \\end{align*} $$\\dfrac{9^3\\times 20^2}{4\\times 5^2 \\times 3^5}$$ \\begin{align*} \\frac{9^3\\times 20^2}{4\\times 5^2 \\times 3^5} & = \\frac{3^6 4^2 5^2}{4 \\times 5^2 3^5} \\\\ & = \\frac{3^6 2^4 5^2}{2^2 5^2 3^5} \\\\ & = 3^{6 - 5} 2^{4 - 2} 5^{2 - 2} \\\\ & = 3 \\times 2^2 \\\\ & = 12 \\end{align*} $$\\dfrac{7^{b} + 7^{b - 2}}{4 \\times 7^{b} + 3 \\times 7^{b}}$$ \\begin{align*} \\frac{7^{b} + 7^{b - 2}}{4 \\times 7^{b} - 3 \\times 7^{b}} &= \\frac{7^{b}(1 + 7^{-2})}{7^{b}(4 - 3)} \\\\ &= \\frac{(1 + 7^{-2})}{1} \\\\ &= \\frac{1 + \\frac{1}{49}}{1} \\\\ & = \\frac{51}{49} \\end{align*} $$\\dfrac{12^{y} - 96^{y}}{3^{y} + 6^{y}}$$ \\begin{align*} \\frac{12^{y} - 96^{y}}{3^{y} + 6^{y}} & = \\frac{(4 \\cdot 3)^{y} - \\left(2^{5} \\cdot 3\\right)^{y}}{3^{y} + \\left(2 \\cdot 3\\right)^{y}} \\\\ & = \\frac{3^{y}\\left(4^{y} - 2^{5y}\\right)}{3^{y}\\left(2 + 1\\right)} \\\\ &= \\frac{4^{y} - 2^{5y}}{3} \\end{align*}", "-\\dfrac{1}{3}x-\\dfrac{1}{8}y+\\dfrac{1}{6}z &= -\\dfrac{4}{3}\\\\ -\\dfrac{2}{3}x-\\dfrac{7}{8}y+\\dfrac{1}{3}z &= -\\dfrac{23}{3}\\\\ -\\dfrac{1}{3}x-\\dfrac{5}{8}y+\\dfrac{5}{6}z &= 0 \\end{align*} 35) \\begin{align*} -\\dfrac{1}{4}x-\\dfrac{5}{4}y+\\dfrac{5}{2}z &= -5\\\\ -\\dfrac{1}{2}x-\\dfrac{5}{3}y+\\dfrac{5}{4}z &= \\dfrac{55}{12}\\\\ -\\dfrac{1}{3}x-\\dfrac{1}{3}y+\\dfrac{1}{3}z &= \\dfrac{5}{3} \\end{align*} $$(-5,-5,-5)$$ 36) \\begin{align*} \\dfrac{1}{40}x+\\dfrac{1}{60}y+\\dfrac{1}{80}z &= \\dfrac{1}{100}\\\\ -\\dfrac{1}{2}x-\\dfrac{1}{3}y-\\dfrac{1}{4}z &= -\\dfrac{1}{5}\\\\ \\dfrac{3}{8}x+\\dfrac{3}{12}y+\\dfrac{3}{16}z &= \\dfrac{3}{20} \\end{align*} 37) \\begin{align*} 0.1x-0.2y+0.3z &= 2\\\\ 0.5x-0.1y+0.4z &= 8\\\\ 0.7x-0.2y+0.3z &= 8 \\end{align*} $$(10,10,10)$$ 38) \\begin{align*} 0.2x+0.1y-0.3z &= 0.2\\\\ 0.8x+0.4y-1.2z &= 0.1\\\\ 1.6x+0.8y-2.4z &= 0.2 \\end{align*} 39) \\begin{align*} 1.1x+0.7y-3.1z &= -1.79\\\\ 2.1x+0.5y-1.6z &= -0.13\\\\ 0.5x+0.4y-0.5z &= -0.07 \\end{align*} $$\\left ( \\dfrac{1}{2},\\dfrac{1}{5},\\dfrac{4}{5} \\right )$$ 40) \\begin{align*} 0.5x-0.5y+0.5z &= 10\\\\ 0.2x-0.2y+0.2z &= 4\\\\ 0.1x-0.1y+0.1z &= 2 \\end{align*} 41) \\begin{align*} 0.1x+0.2y+0.3z &= 0.37\\\\ 0.1x-0.2y-0.3z &= -0.27\\\\ 0.5x-0.1y-0.3z &= -0.03 \\end{align*} $$\\left ( \\dfrac{1}{2},\\dfrac{2}{5},\\dfrac{4}{5} \\right )$$ 42) \\begin{align*} 0.5x-0.5y-0.3z &= 0.13\\\\ 0.4x-0.1y-0.3z &= 0.11\\\\ 0.2x-0.8y-0.9z &= -0.32 \\end{align*} 43) \\begin{align*} 0.5x+0.2y-0.3z &= 1\\\\ 0.4x-0.6y+0.7z &= 0.8\\\\ 0.3x-0.1y-0.9z &= 0.6 \\end{align*} $$(2,0,0)$$ 44) \\begin{align*} 0.3x+0.3y+0.5z &= 0.6\\\\ 0.4x+0.4y+0.4z &= 1.8\\\\ 0.4x+0.2y+0.1z &= 1.6 \\end{align*} 45) \\begin{align*} 0.8x+0.8y+0.8z &= 2.4\\\\ 0.3x-0.5y+0.2z &= 0\\\\ 0.1x+0.2y+0.3z &= 0.6 \\end{align*} $$(1,1,1)$$ ### Extensions For the exercises 46-50, solve the system for $$x,y,$$ and $$z$$. 46) \\begin{align*} x+y+z &= 3\\\\ \\dfrac{x-1}{2}+\\dfrac{y-3}{2}+\\dfrac{z+1}{2} &= 0\\\\ \\dfrac{x-2}{3}+\\dfrac{y+4}{3}+\\dfrac{z-3}{3} &= \\dfrac{2}{3} \\end{align*} 47) \\begin{align*} 5x-3y-\\dfrac{z+1}{2} &= \\dfrac{1}{2}\\\\ 6x+\\dfrac{y-9}{2}+2z &= -3\\\\ \\dfrac{x+8}{2}-4y+z &= 4\\end{align*} $$\\left ( \\dfrac{128}{557},\\dfrac{23}{557},\\dfrac{428}{557} \\right )$$ 48) \\begin{align*} \\dfrac{x+4}{7}-\\dfrac{y-1}{6}+\\dfrac{z+2}{3} &= 1\\\\ \\dfrac{x-2}{4}+\\dfrac{y+1}{8}-\\dfrac{z+8}{2} &= 0\\\\ \\dfrac{x+6}{3}-\\dfrac{y+2}{3}+\\dfrac{z+4}{2} &= 3 \\end{align*} 49) \\begin{align*} \\dfrac{x-3}{6}+\\dfrac{y+2}{2}-\\dfrac{z-3}{3} &= 2\\\\ \\dfrac{x+2}{4}+\\dfrac{y-5}{2}+\\dfrac{z+4}{2} &= 1\\\\ \\dfrac{x+6}{2}-\\dfrac{y-3}{3}+z+1", "-\\dfrac{1}{3}x-\\dfrac{1}{8}y+\\dfrac{1}{6}z &= -\\dfrac{4}{3}\\\\ -\\dfrac{2}{3}x-\\dfrac{7}{8}y+\\dfrac{1}{3}z &= -\\dfrac{23}{3}\\\\ -\\dfrac{1}{3}x-\\dfrac{5}{8}y+\\dfrac{5}{6}z &= 0 \\end{align*} 35) \\begin{align*} -\\dfrac{1}{4}x-\\dfrac{5}{4}y+\\dfrac{5}{2}z &= -5\\\\ -\\dfrac{1}{2}x-\\dfrac{5}{3}y+\\dfrac{5}{4}z &= \\dfrac{55}{12}\\\\ -\\dfrac{1}{3}x-\\dfrac{1}{3}y+\\dfrac{1}{3}z &= \\dfrac{5}{3} \\end{align*} $$(-5,-5,-5)$$ 36) \\begin{align*} \\dfrac{1}{40}x+\\dfrac{1}{60}y+\\dfrac{1}{80}z &= \\dfrac{1}{100}\\\\ -\\dfrac{1}{2}x-\\dfrac{1}{3}y-\\dfrac{1}{4}z &= -\\dfrac{1}{5}\\\\ \\dfrac{3}{8}x+\\dfrac{3}{12}y+\\dfrac{3}{16}z &= \\dfrac{3}{20} \\end{align*} 37) \\begin{align*} 0.1x-0.2y+0.3z &= 2\\\\ 0.5x-0.1y+0.4z &= 8\\\\ 0.7x-0.2y+0.3z &= 8 \\end{align*} $$(10,10,10)$$ 38) \\begin{align*} 0.2x+0.1y-0.3z &= 0.2\\\\ 0.8x+0.4y-1.2z &= 0.1\\\\ 1.6x+0.8y-2.4z &= 0.2 \\end{align*} 39) \\begin{align*} 1.1x+0.7y-3.1z &= -1.79\\\\ 2.1x+0.5y-1.6z &= -0.13\\\\ 0.5x+0.4y-0.5z &= -0.07 \\end{align*} $$\\left ( \\dfrac{1}{2},\\dfrac{1}{5},\\dfrac{4}{5} \\right )$$ 40) \\begin{align*} 0.5x-0.5y+0.5z &= 10\\\\ 0.2x-0.2y+0.2z &= 4\\\\ 0.1x-0.1y+0.1z &= 2 \\end{align*} 41) \\begin{align*} 0.1x+0.2y+0.3z &= 0.37\\\\ 0.1x-0.2y-0.3z &= -0.27\\\\ 0.5x-0.1y-0.3z &= -0.03 \\end{align*} $$\\left ( \\dfrac{1}{2},\\dfrac{2}{5},\\dfrac{4}{5} \\right )$$ 42) \\begin{align*} 0.5x-0.5y-0.3z &= 0.13\\\\ 0.4x-0.1y-0.3z &= 0.11\\\\ 0.2x-0.8y-0.9z &= -0.32 \\end{align*} 43) \\begin{align*} 0.5x+0.2y-0.3z &= 1\\\\ 0.4x-0.6y+0.7z &= 0.8\\\\ 0.3x-0.1y-0.9z &= 0.6 \\end{align*} $$(2,0,0)$$ 44) \\begin{align*} 0.3x+0.3y+0.5z &= 0.6\\\\ 0.4x+0.4y+0.4z &= 1.8\\\\ 0.4x+0.2y+0.1z &= 1.6 \\end{align*} 45) \\begin{align*} 0.8x+0.8y+0.8z &= 2.4\\\\ 0.3x-0.5y+0.2z &= 0\\\\ 0.1x+0.2y+0.3z &= 0.6 \\end{align*} $$(1,1,1)$$ ### Extensions For the exercises 46-50, solve the system for $$x,y,$$ and $$z$$. 46) \\begin{align*} x+y+z &= 3\\\\ \\dfrac{x-1}{2}+\\dfrac{y-3}{2}+\\dfrac{z+1}{2} &= 0\\\\ \\dfrac{x-2}{3}+\\dfrac{y+4}{3}+\\dfrac{z-3}{3} &= \\dfrac{2}{3} \\end{align*} 47) \\begin{align*} 5x-3y-\\dfrac{z+1}{2} &= \\dfrac{1}{2}\\\\ 6x+\\dfrac{y-9}{2}+2z &= -3\\\\ \\dfrac{x+8}{2}-4y+z &= 4\\end{align*} $$\\left ( \\dfrac{128}{557},\\dfrac{23}{557},\\dfrac{428}{557} \\right )$$ 48) \\begin{align*} \\dfrac{x+4}{7}-\\dfrac{y-1}{6}+\\dfrac{z+2}{3} &= 1\\\\ \\dfrac{x-2}{4}+\\dfrac{y+1}{8}-\\dfrac{z+8}{2} &= 0\\\\ \\dfrac{x+6}{3}-\\dfrac{y+2}{3}+\\dfrac{z+4}{2} &= 3 \\end{align*} 49) \\begin{align*} \\dfrac{x-3}{6}+\\dfrac{y+2}{2}-\\dfrac{z-3}{3} &= 2\\\\ \\dfrac{x+2}{4}+\\dfrac{y-5}{2}+\\dfrac{z+4}{2} &= 1\\\\ \\dfrac{x+6}{2}-\\dfrac{y-3}{3}+z+1", "he received from B Jul 4, 2021 at 3:53 • I made an edit with the clarifications Jul 4, 2021 at 4:06 Here's my solution based on expectations. First, we calculate the expected value of the smallest, second-smallest, third-smallest... number on the 5 cards drawn. The calculation steps are shown below: $$E(\\text{smallest})=\\dfrac{1\\times \\dbinom{9}{4}+2\\times \\dbinom{8}{4}+\\cdots+6\\times \\dbinom{4}{4}}{\\dbinom{10}{5}}=\\dfrac{11}{6}.$$ Explanation: the numerator basically sums the smallest terms in all 5-card draws, and the denominator is the total number of cases. Similarly, we have $$E(\\text{second-smallest})=\\dfrac{2\\times \\dbinom{8}{3}\\times \\dbinom{1}{1}+3\\times \\dbinom{7}{3}\\times \\dbinom{2}{1}+\\cdots+7\\times \\dbinom{3}{3}\\times \\dbinom{6}{1}}{\\dbinom{10}{5}}=\\dfrac{11}{3};$$ $$E(\\text{third-smallest})=\\dfrac{3\\times \\dbinom{7}{2}\\times \\dbinom{2}{2}+4\\times \\dbinom{6}{2}\\times \\dbinom{3}{2}+\\cdots+8\\times \\dbinom{2}{2}\\times \\dbinom{7}{2}}{\\dbinom{10}{5}}=\\dfrac{11}{2};$$ $$E(\\text{fourth-smallest})=\\dfrac{4\\times \\dbinom{6}{1}\\times \\dbinom{3}{3}+5\\times \\dbinom{5}{1}\\times \\dbinom{4}{3}+\\cdots+9\\times \\dbinom{1}{1}\\times \\dbinom{8}{3}}{\\dbinom{10}{5}}=\\dfrac{22}{3};$$ $$E(\\text{fifth-smallest, or largest})=\\dfrac{5\\times \\dbinom{4}{4}+6\\times \\dbinom{5}{4}+\\cdots+10\\times \\dbinom{9}{4}}{\\dbinom{10}{5}}=\\dfrac{55}{6}.$$ As a quick double check, adding them all together yields $$27.5$$, which is the expected sum of the 5 cards. Next, let's think about A's strategy. In essence, he is discarding his $$N$$ smallest cards to get random cards. Thus, if the expected value of cards that he is discarding is smaller than the expected value of cards he is getting, then A is benefitting. The more A benefits, the higher his chances of winning. Then, clearly $$E(\\text{smallest})$$, $$E(\\text{second-smallest})$$ are smaller than $$5.5$$, the expected value on a random card. $$E(\\text{third-smallest})$$ is equal to $$5.5$$, and $$E(\\text{fourth-smallest})$$, $$E(\\text{fifth-smallest, or largest})$$ are greater than $$5.5$$. This means that discarding the smallest card and", "\\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17}{37 \\times 41 \\times 43 \\times 47}\\\\ p_{15}(29,21) &=& \\dfrac{20 \\times 15}{49 \\times 6} \\times \\dfrac{2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17}{37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^2 \\times 3 \\times 5^3 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ p_{16}(29,22) &=& \\dfrac{21 \\times 16}{50 \\times 7} \\times \\dfrac{2^2 \\times 3 \\times 5^3 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^5 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ p_{17}(29,23) &=& \\dfrac{22 \\times 17}{51 \\times 8} \\times \\dfrac{2^5 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^3 \\times 3 \\times 5 \\times 11^2 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ p_{18}(29,24) &=& \\dfrac{23 \\times 18}{52 \\times 9} \\times \\dfrac{2^3 \\times 3 \\times 5 \\times 11^2 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ p_{19}(29,25) &=& \\dfrac{24 \\times", "{2[(m + 1)} - 1}}}}{{{\\rm{1 - }}\\,{\\rm{Utilization}}}}} \\right] \\times \\left[ {\\dfrac{{{\\rm{CO}}{{\\rm{V}}_{\\rm{a}}}^2{\\rm{ + CO}}{{\\rm{V}}_{\\rm{p}}}^2}}{{\\rm{2}}}} \\right]\\\\ &= \\left[ {\\dfrac{{20}}{7}} \\right] \\times \\left[ {\\dfrac{{{{0.6}^3}}}{{1 - 0.6}}} \\right] \\times \\left[ {\\dfrac{{1 + 1}}{2}} \\right]\\\\ &= 2.8 \\times \\dfrac{{0.216}}{{0.4}} \\times 1\\\\ &= 2.8 \\times 0.54 \\times 1\\\\ &= 1.512 \\end{align*} {/eq}" ]

[ "quartets, but not the 13th: $\\dbinom{13}{12}\\dbinom{n}{12} \\left( \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{12} \\left(1 - \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{n-12}$ Then the probability that the next pull is the 13th quartet is: $\\dfrac{ \\dbinom{5}{4} }{ \\dbinom{52}{5} }$ So, the expected value is given by: \\displaystyle \\begin{align*} & \\sum_{n \\ge 12}\\dbinom{13}{12}\\dbinom{n}{12} \\left( \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{13} \\left(1 - \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{n-12}(n+1) \\\\ = & \\dbinom{13}{12} \\left( \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{13} \\sum_{n \\ge 12} \\dbinom{n}{12} \\left(1 - \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{n-12}(n+1) = 1830101\\end{align*} http://www.wolframalpha.com/input/?i...infinity%7D%5D Edit: I missed a factor of 48. 3. ## Re: Expectancy of number of trials I also got this expression but i suppose we have to multiply by 12! for permutations on the 12 quarters.For the same reason the probability for a quarter in one trial is 13*bin(5,4)*4!*(52-4)/(bin(52,5)*5!). Do you see a way of computing the expectancy by using indicators? 4. ## Re: Expectancy of number of trials Originally Posted by hedi I also got this expression but i suppose we have to multiply by 12! for permutations on the 12 quarters.For the same reason the probability for a quarter in one trial is 13*bin(5,4)*4!*(52-4)/(bin(52,5)*5!). Do you see a way of computing the expectancy by using indicators? I apologize. I missed a factor of 48 for", "he received from B Jul 4, 2021 at 3:53 • I made an edit with the clarifications Jul 4, 2021 at 4:06 Here's my solution based on expectations. First, we calculate the expected value of the smallest, second-smallest, third-smallest... number on the 5 cards drawn. The calculation steps are shown below: $$E(\\text{smallest})=\\dfrac{1\\times \\dbinom{9}{4}+2\\times \\dbinom{8}{4}+\\cdots+6\\times \\dbinom{4}{4}}{\\dbinom{10}{5}}=\\dfrac{11}{6}.$$ Explanation: the numerator basically sums the smallest terms in all 5-card draws, and the denominator is the total number of cases. Similarly, we have $$E(\\text{second-smallest})=\\dfrac{2\\times \\dbinom{8}{3}\\times \\dbinom{1}{1}+3\\times \\dbinom{7}{3}\\times \\dbinom{2}{1}+\\cdots+7\\times \\dbinom{3}{3}\\times \\dbinom{6}{1}}{\\dbinom{10}{5}}=\\dfrac{11}{3};$$ $$E(\\text{third-smallest})=\\dfrac{3\\times \\dbinom{7}{2}\\times \\dbinom{2}{2}+4\\times \\dbinom{6}{2}\\times \\dbinom{3}{2}+\\cdots+8\\times \\dbinom{2}{2}\\times \\dbinom{7}{2}}{\\dbinom{10}{5}}=\\dfrac{11}{2};$$ $$E(\\text{fourth-smallest})=\\dfrac{4\\times \\dbinom{6}{1}\\times \\dbinom{3}{3}+5\\times \\dbinom{5}{1}\\times \\dbinom{4}{3}+\\cdots+9\\times \\dbinom{1}{1}\\times \\dbinom{8}{3}}{\\dbinom{10}{5}}=\\dfrac{22}{3};$$ $$E(\\text{fifth-smallest, or largest})=\\dfrac{5\\times \\dbinom{4}{4}+6\\times \\dbinom{5}{4}+\\cdots+10\\times \\dbinom{9}{4}}{\\dbinom{10}{5}}=\\dfrac{55}{6}.$$ As a quick double check, adding them all together yields $$27.5$$, which is the expected sum of the 5 cards. Next, let's think about A's strategy. In essence, he is discarding his $$N$$ smallest cards to get random cards. Thus, if the expected value of cards that he is discarding is smaller than the expected value of cards he is getting, then A is benefitting. The more A benefits, the higher his chances of winning. Then, clearly $$E(\\text{smallest})$$, $$E(\\text{second-smallest})$$ are smaller than $$5.5$$, the expected value on a random card. $$E(\\text{third-smallest})$$ is equal to $$5.5$$, and $$E(\\text{fourth-smallest})$$, $$E(\\text{fifth-smallest, or largest})$$ are greater than $$5.5$$. This means that discarding the smallest card and", "{mg \\times \\left( {{R_c} - OA} \\right)} \\right) \\end{align*} {/eq} Substitute the given data in above equation, {eq}\\begin{align*} {M_a} &= \\left( {\\left( {\\dfrac{2}{{12}} \\times \\left( {\\dfrac{{50}}{2} - 10} \\right)} \\right) \\times {{\\left( {\\dfrac{{50}}{2} - 10} \\right)}^2}} \\right) + \\left( {2 \\times \\left( {\\dfrac{{50}}{2} - 10} \\right)} \\right) \\times 8 \\times 17.5 - \\left( {\\left( {2 \\times \\left( {\\dfrac{{50}}{2} - 10} \\right)} \\right)9.81 \\times \\left( {17.5 - 10} \\right)} \\right)\\\\ &= 2555.5\\;{\\rm{N}} \\cdot {\\rm{m}} \\end{align*} {/eq} Thus the moment at A is {eq}2555.5\\;{\\rm{N}} \\cdot {\\rm{m}} {/eq}", "# Division using Exponents (Indices) If we are given $a^8 \\div a^3$, we can also write this as $\\dfrac{a^8}{a^3}$, which means $\\dfrac{a \\times a \\times a \\times a \\times a \\times a \\times a \\times a}{a \\times a \\times a}$. As there are $8$ factors of $a$ on the top line (numerator), and $3$ factors of $a$ on the bottom line (denominator), we can cancel $3$ of them, giving us \\large \\begin{align} \\require{AMSsymbols} \\displaystyle \\require{cancel} &\\dfrac{\\cancel{a} \\times \\cancel{a} \\times \\cancel{a} \\times a \\times a \\times a \\times a \\times a}{\\cancel{a} \\times \\cancel{a} \\times \\cancel{a}} \\\\ &= a \\times a \\times a \\times a \\times a \\\\ &= a^5 \\end{align} ### Example 1 After first writing in factor form, simplify $\\dfrac{x^7}{x^4}$. \\begin{align} \\displaystyle \\require{AMSsymbols} \\require{cancel} \\dfrac{x^7}{x^4} &= \\dfrac{x \\times x \\times x \\times x \\times x \\times x \\times x}{x \\times x \\times x \\times x} \\\\ &= \\dfrac{\\cancel{x} \\times \\cancel{x} \\times \\cancel{x} \\times \\cancel{x} \\times x \\times x \\times x}{\\cancel{x} \\times \\cancel{x} \\times \\cancel{x} \\times \\cancel{x}} \\\\ &= x \\times x \\times x \\\\ &= x^3 \\end{align} An alternative solution can be formed by examining the result. We can see that the exponent (index) in the answer is the result of subtracting the exponents (indices) in the question. \\large \\begin{align} \\displaystyle \\require{cancel} a^8 \\div a^3 &= a^{8-3} \\\\", "n-2 \\right)...(n-r)!$ \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5!}\\times \\left( \\dfrac{8\\times 7\\times 6}{3!} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6!}\\times \\left( \\dfrac{8\\times 7}{2!} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4!}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4!} \\right) \\\\ \\end{align} \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6}{3\\times 2\\times 1} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6\\times 5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7}{2\\times 1} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4\\times 3\\times 2\\times 1} \\right) \\\\ \\end{align} By cancelling the common terms in denominator and numerator we get $=\\left( 11\\times 9\\times 8 \\right)\\times \\left( 8\\times 7 \\right)+\\left( 11\\times 3\\times 4\\times 7 \\right)\\times \\left( 4\\times 7 \\right)+\\left( 11\\times 5\\times 9 \\right)\\times \\left( 5\\times 2\\times 7 \\right)$ $=792\\times 56+924\\times 28+495\\times 70$ = 44352 + 25872 + 34650 = 104874 Thus, the number of ways of selecting 10 students chosen for the competition = 104874 Note: The possibility of mistake can be the calculation mistake since the procedure of solving involves large calculations. Hence be very careful in simplifying and cancelling the terms to get the desired result.", "### A result using complex numbers Suppose we expand out the expressions $$(1+1)^n$$, $$(1+i)^n$$, $$(1-1)^n$$, $$(1-i)^n$$ using the binomial theorem. This gives (for $$n>0$$), \\begin{align*} \\dbinom{n}{0} + \\dbinom{n}{1} + \\dbinom{n}{2} + \\dbinom{n}{3} + \\dots + \\dbinom{n}{n} &= 2^n\\\\ \\dbinom{n}{0} + i\\dbinom{n}{1} - \\dbinom{n}{2} - i\\dbinom{n}{3} + \\dots + i^n\\dbinom{n}{n} &= (1+i)^n\\\\ \\dbinom{n}{0} - \\dbinom{n}{1} + \\dbinom{n}{2} - \\dbinom{n}{3} + \\dots + (-1)^n\\dbinom{n}{n} &= 0^n\\\\ \\dbinom{n}{0} - i\\dbinom{n}{1} - \\dbinom{n}{2} + i\\dbinom{n}{3} + \\dots + (-i)^n\\dbinom{n}{n} &= (1-i)^n. \\end{align*} Now add these equations and divide by 4 to get $\\dbinom{n}{0} + \\dbinom{n}{4} + \\dbinom{n}{8} + \\dots = \\dfrac{1}{4}\\Big(2^n + (1-i)^n + (1+i)^n\\Big).$ It is easy to show, using the polar form, that $(1-i)^n + (1+i)^n = 2(\\sqrt{2})^n \\cos \\big(\\dfrac{n\\pi}{4}\\big).$ Thus, we have $\\dbinom{n}{0} + \\dbinom{n}{4} + \\dbinom{n}{8} + \\dots = \\dfrac{1}{4}\\Big(2^n + 2(\\sqrt{2})^n \\cos \\big(\\dfrac{n\\pi}{4}\\big)\\Big).$ This is a rather amazing and beautiful result. Next page - Links forward - Series for e", "y=\\dfrac{10!}{1!\\left( 10-1 \\right)!}\\times \\dfrac{10!}{2!\\left( 10-2 \\right)!}\\times \\dfrac{9!}{8!\\left( 9-8 \\right)!}\\times 8! \\\\ & y=\\dfrac{10\\times 9!}{9!}\\times \\dfrac{10\\times 9}{2}\\times \\dfrac{9\\times 8!}{8!}\\times 8! \\\\ & y=10\\times 45\\times 9\\times 8! \\\\ \\end{align} Now, we have all the required data to solve this problem. We have x = 10! And y = $10\\times 45\\times 9\\times 8!$ Then, $\\dfrac{y}{9x}=?$ ………………. (1) On substituting the value of x and y in equation (1), we get; \\begin{align} & \\dfrac{y}{9x}=\\dfrac{10\\times 45\\times 9\\times 8!}{9\\times 10!} \\\\ & =\\dfrac{{10}\\times 45\\times{9}\\times{8!}}{{9}\\times{10}\\times 9\\times {8!}} \\\\ & =\\dfrac{45}{9} \\\\ & =5 \\\\ \\end{align} Note: Please note that if repetition is not allowed then the total possibility will be n! but the repetition is allowed the total possibility will be ${{\\left( n \\right)}^{n}}$. Here in this question repetition is not allowed thus, we have the value of x = 10!.", "3 Use the quotient rule to differentiate $\\displaystyle f(x)=\\dfrac{3x+1}{1-x}$. \\begin{align} \\displaystyle f^{\\prime}(x) &= \\dfrac{(3x+1)^{\\prime} \\times (1-x) – (3x+1) \\times (1-x)^{\\prime}}{(1-x)^2} \\\\ &= \\dfrac{3 \\times (1-x) – (3x+1) \\times (-1)}{(1-x)^2} \\\\ &= \\dfrac{3-3x +3x+1}{(1-x)^2} \\\\ &= \\dfrac{4}{(1-x)^2} \\\\ \\end{align} ### Example 4 Use the quotient rule to differentiate $\\displaystyle f(x)=\\dfrac{(2x+1)^3}{(4x-1)^4}$. \\begin{align} \\displaystyle f^{\\prime}(x) &= \\dfrac{\\big[(2x+1)^3\\big]^{\\prime} \\times (4x-1)^4 – (2x+1)^3 \\times \\big[(4x-1)^4\\big]^{\\prime}}{\\big[(4x-1)^4\\big]^2} \\\\ &= \\dfrac{3(2x+1)^{3-1} \\times (2x+1)^{\\prime} \\times (4x-1)^4 – (2x+1)^3 \\times 4(4x-1)^{4-1} \\times (4x-1)^{\\prime}}{(4x-1)^8} \\\\ &= \\dfrac{3(2x+1)^2 \\times 2 \\times (4x-1)^4 – (2x+1)^3 \\times 4(4x-1)^3 \\times 4}{(4x-1)^8} \\\\ &= \\dfrac{6(2x+1)^2 (4x-1)^4 – 16(2x+1)^3 (4x-1)^3}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3\\big[6(4x-1) – 16(2x+1)\\big]}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3(24x-6 – 32x-16)}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3(-8x-22)}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(-8x-22)}{(4x-1)^5} \\\\ \\end{align} ## Extension Examples These Extension Examples require to have some prerequisite skills including; \\begin{align} \\displaystyle \\dfrac{d}{dx}\\sin{x} &= \\cos{x} \\\\ \\dfrac{d}{dx}\\cos{x} &= -\\sin{x} \\\\ \\dfrac{d}{dx}e^x &= e^x \\\\ \\dfrac{d}{dx}\\log_e{x} &= \\dfrac{1}{x} \\\\ \\end{align} ### Example 5 Find $\\displaystyle \\dfrac{dy}{dx}$ of $\\displaystyle y=\\dfrac{\\sin{x}}{\\cos{x}}$, known that $\\dfrac{d}{dx}\\sin{x} = \\cos{x}$ and $\\dfrac{d}{dx}\\cos{x} = -\\sin{x}$. \\begin{align} \\displaystyle \\dfrac{dy}{dx} &= \\dfrac{\\dfrac{d}{dx}\\sin{x} \\times \\cos{x} – \\sin{x} \\times \\dfrac{d}{dx}\\cos{x}}{\\cos^2{x}} \\\\ &= \\dfrac{\\cos{x} \\times \\cos{x} – \\sin{x} \\times (-\\sin{x})}{\\cos^2{x}} \\\\ &= \\dfrac{\\cos^2{x} + \\sin^2{x}}{\\cos^2{x}} \\\\ \\end{align} Note that optionally $\\cos^2{x} + \\sin^2{x}$ can be simplified further to $1$, as $\\cos^2{x} + \\sin^2{x}=1$ in one of the trigonometric properties, however we do not cover this skill here.", "and the bottom of the fraction $\\dfrac{16!}{13!\\cdot 8!} = \\dfrac{16 \\times 15 \\times 14}{8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2}.$ This can be canceled further if we factorize the numerator $\\dfrac{16!}{13!\\cdot 8!} = \\dfrac{8 \\times 2 \\times 5 \\times 3 \\times 2 \\times 7}{8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2}.$ Then finally we get \\begin{align} \\dfrac{16!}{13! \\cdot 8!} &= \\dfrac{1}{4 \\times 3},\\\\ &=\\dfrac{1}{12}. \\end{align} ###### Example 4 Simplify $\\dfrac{(n-1)!}{(n+1)!}.$ ###### Solution \\begin{align} \\dfrac{(n-1)!}{(n+1)!} &= \\dfrac{(n-1) \\times (n-2) \\times \\dots \\times 2 \\times 1}{ (n+1) \\times n \\times (n-1) \\times \\dots \\times 2 \\times 1},\\\\ &= \\dfrac{1}{(n+1) \\times n},\\\\ &= \\dfrac{1}{n^2 + n}. \\end{align}", "&= 15 \\times \\left[ {\\left( {-25} \\right) \\times \\left( {-4} \\right) \\times \\left( {-10} \\right)} \\right]\\\\ &= 15 \\times \\left[ {100 \\times \\left( {-10} \\right)} \\right]\\\\ &= 15 \\times \\left[ {1000} \\right]\\\\ &= -15000\\end{align} (d) $${\\rm{}}\\left( {-41} \\right){\\rm{ }} \\times {\\rm{ }}102$$ Using distributive law, we get \\begin{align}&=\\,{{(}} - {{41)}}\\,{{ \\times }}\\,{{(100}}\\,{{ + }}\\,{{2)}}\\dots\\,{{[a}}\\,{{ \\times }}\\,{{(b}}\\,{{ + }}\\,{{c)}}]=\\,{{(a}}\\,{{ \\times }}\\,{{b}}\\,{{ + }}\\,{{a}}\\,{{ \\times }}\\,{{c)]}}\\\\&={{{(}} - {{41)}}\\,{{ \\times }}\\,{{100}}\\,{{ + }}\\,{{(}}\\, - {{41) \\times }}\\,{{2}}}\\\\&= - {{4100}}\\, - {{82}}\\\\&=\\, - {{4182}}\\end{align} (e) $${{625}}\\,{{ \\times }}\\left( {{{-35}}} \\right){{ + }}\\left( {{{-}}\\,{{625}}} \\right){{ \\times }}\\,{{65}}$$ Using distributive property, we get \\begin{align} &=\\,{625}\\,\\times[{ (}-{35)}\\,{+}\\,{(}-{65) }]\\,{ }\\dots[{ a\\times b}\\,{+}\\,{a\\times c}\\,{=}\\,{a(b+c) }] \\\\ &=\\,{625}\\times[\\,-{35}\\,-{65 }] \\\\ &=\\,{625}\\times[-{100 }] \\\\ &={-62500} \\\\\\end{align} (f) $$7\\times \\left( 50-2 \\right)$$ Using distributive property, we get \\begin{align}&=7\\times { }\\,{(50}-{2)}\\,{ }\\dots[{ a\\times}\\,{ }{ }\\,{(b}-{c)}\\,{=}\\,{a }\\times{ b}-{a }\\times{ c }]{ } \\\\ &=7\\times { 50}-{7 }\\times { 2} \\\\ &=350-{14} \\\\ &=336\\end{align} (g) $$\\left( \\text{-17} \\right)\\text{ }\\!\\!\\times\\!\\!\\text{ }\\left( \\text{-29} \\right)$$ Using distributive property, we get \\begin{align} &= \\,( - 17)\\, \\times \\,\\,[( - 30)\\, + 1]\\dots[a\\, \\times \\,(b\\, + \\,c)={a\\times b}+{a \\times c}]\\\\ &= \\,( - 17)\\, \\times \\,( - 30)\\, + \\,( - 17)\\, \\times 1\\\\ &= \\,510\\, + ( - 17)\\\\ &= \\,510 - 17\\\\ &=\\,493\\end{align} (h)$${{}}\\left( {-57} \\right) \\times \\left( {-19} \\right) + 57$$ Using distributive property, we", "last row right. This works well, if one has an expression of the type a+b=c+d+e+f+g+h=j, where there's only one row between initial row and result. However, in my case, I have some mathematical transformation (meaning an additional row) and so I need the equal sign of the row: \"12\\times6+14\\times\\dfrac{48}{7}-6^2-\\left(\\dfrac{2}{7}-4\\right)^2-\\left(\\dfrac{48}{7}\\right)^2 -\\dfrac{2}{7}\\times\\dfrac{48}{7}-\\left(\\dfrac{2}{7}\\right)^2\" to be aligned under the first equal sign. So now I use this version: Code: Select all • Open in writeLaTeX \\begin{align*}\\pi^{M}(s^{M},f^{M},x^{M}) :=& 12s^{M}+14f^{M}-(s^{M})^2-(x^{M}-4)^2-(f^{M})^2\\\\ &-x^{M}f^{M}-(x^{M})^2\\pause\\\\ =&12\\times6+14\\times\\dfrac{48}{7}-6^2-\\left(\\dfrac{2}{7}-4\\right)^2-\\left(\\dfrac{48}{7}\\right)^2\\\\ &-\\dfrac{2}{7}\\times\\dfrac{48}{7}-\\left(\\dfrac{2}{7}\\right)^2\\pause\\\\ =&69,14\\pause\\end{align*} Does anybody have better ideas? Thank you avp3000 Posts: 49 Joined: Thu Nov 15th, 2007 ### Re: Too long equation You can try the split environment, which is also provided by the amsmath package. To get a proper alignment, there are some tricky things to do. Code: Select all • Open in writeLaTeX \\begin{align*} \\pi^{M}(s^{M},f^{M},x^{M}): \\begin{split} &= 12s^{M}+14f^{M}-(s^{M})^2-(x^{M}-4)^2 \\\\ &\\quad -(f^{M})^2-x^{M}f^{M}-(x^{M})^2 \\end{split}\\\\ \\begin{split} &= 12\\times 6+14\\times\\frac{48}{7}-6^2-\\left(\\frac{2}{7}-4\\right)^2 \\\\ &\\quad -\\left(\\frac{48}{7}\\right)^2-\\frac{2}{7}\\times\\frac{48}{7}-\\left(\\frac{2}{7}\\right)^2 \\end{split}\\\\ &=69,14\\end{align*} Note that the alignment as known from the align environment has to be put into the split environment. More hints about typesetting mathematical expressions are given in \"Math mode\". Best regards Thorsten LaTeX Community Moderator ¹ System: openSUSE 13.1 (Linux 3.11.10), TeX Live 2013 (vanilla), TeXworks 0.5 (r1351) ² Posting stopped indefinitely due to offenses localghost Site Moderator Posts: 9219 Joined: Fri Feb 2nd, 2007 Location: Braunschweig, Germany ###", "( \\dfrac{1}{T_2^{2/3}}-\\dfrac{1}{T_1^{2/3}} \\right )\\\\ &=-\\left ( 2\\pi \\times 6.67\\times 10^{-11}\\times 6\\times 10^{24}\\right )^{2/3}\\times 50\\times \\left ( \\dfrac{1}{\\left ( 10.8\\times 10^{3} \\right )^{2/3}}-\\dfrac{1}{\\left (7.2\\times 10^{3} \\right )^{2/3}} \\right )\\\\ &=-\\left ( 2.51\\times 10^{15}\\right )^{2/3}\\times 50\\times \\left ( \\dfrac{1}{488}-\\dfrac{1}{373} \\right )\\\\ &=-1.36\\times 10^{10}\\times 50\\times \\left ( -6.32 \\times 10^{-4}\\right )\\\\ &=\\boxed{4.3\\times 10^{6}\\;\\rm J} \\end{align*} {/eq}", "h &= \\dfrac{4 \\pi^2Ed^4}{4 \\times 64L^2 \\times D_i^2\\times \\rho g}\\\\ h &= \\dfrac{4 \\pi^2 \\times 200 \\times 10^9 \\times (0.075)^4}{4 \\times 64 \\times \\times 4^2 \\times 1.5^2 \\times 10^4}\\\\ h &= 2.7\\ m\\end{aligned}", "touchscreen computer, I had inadvertently downvoted instead of upvoting! The only fix I could think of was edit and upvote. – André Nicolas Jul 2 '14 at 7:04 • I think you can unvote, then upvote, anyway I cant see the difference after the edit, so thanks for the upvote. – Rene Schipperus Jul 2 '14 at 11:37 $\\dbinom{n}{k}\\dbinom{k}{m}$ counts the ways to divide $n$ items into three piles of size $n-k$, $k-m$, and $m$ by first removing $k$ items from and then dividing those items. $\\dbinom{n}{m}\\dbinom{n-m}{k-m}$ counts the ways to divide $n$ items into three piles of size $m$, $n-m$, and $n-k$ by first selecting $m$ items and then dividing the rest. \\begin{align}\\dbinom{n}{k}\\dbinom{k}{m}&=\\dfrac{n!}{k!\\, n-k!}\\dfrac{k!}{m!\\, (k-m)!}\\\\&=\\dfrac{n!}{(n-k)!\\,(k-m)!\\,m!}\\\\&=\\dfrac{n!}{m!\\,(n-m)!}\\dfrac{(n-m)!}{(n-k)!\\,(k-m)!}\\\\&=\\dbinom{n}{m}\\dbinom{n-m}{k-m}\\end{align}", "set of three elements can be arranged in six ways. We denote the number of ways of choosing $$r$$ objects from $$n$$ objects by $$\\dbinom{n}{r}$$, which is read as '$$n$$ choose $$r$$'. In the example above, we found that $\\dbinom{5}{3} = \\dfrac{{^5}P_3}{3!} = \\dfrac{60}{6} = 10.$ In general, we have $\\dbinom{n}{r} = \\dfrac{^nP_r}{r!} = \\dfrac{n!}{(n-r)!r!}.$ Consider the five-element set $$\\{a,b,c,d,e\\}$$ again. There are \\begin{alignat*}{2} \\dbinom{5}{1}&=5 \\text{ ways of choosing 1 letter},&\\qquad \\dbinom{5}{2}&=10 \\text{ ways of choosing 2 letters},\\\\ \\dbinom{5}{3}&=15 \\text{ ways of choosing 3 letters},&\\qquad \\dbinom{5}{4}&=10 \\text{ ways of choosing 4 letters},\\\\ \\dbinom{5}{5}&=1 \\text{ way of choosing 5 letters}. \\end{alignat*} Listing these as a row, including choosing no letters as the first entry: $\\dbinom{5}{0}=1, \\qquad \\dbinom{5}{1}=5, \\qquad \\dbinom{5}{2}=10, \\qquad \\dbinom{5}{3}=10, \\qquad \\dbinom{5}{4}=5, \\qquad \\dbinom{5}{5}=1.$ This is the fifth row of Pascal's triangle. #### Example Evaluate 1. $$\\dbinom{100}{2}$$ 2. $$\\dbinom{1000}{998}$$. #### Solution 1. $$\\dbinom{100}{2} = \\dfrac{100\\times99}{2} = 4950$$ 2. $$\\dbinom{1000}{998} = \\dfrac{1000\\times999}{2} = 499\\,500$$. #### Example In how many ways can you choose two people from a group of seven people? #### Solution There are $$\\dbinom{7}{2} = \\dfrac{7\\times6}{2} = 21$$ ways of choosing two people from seven. #### Example There are ten people in a basketball squad. Find how many ways: 1. the starting five can be chosen from the squad 2. the squad can be split into two", "7 \\times 100 + \\\\ 1 \\times 10 + 9 \\times 1\\end{array} \\right]\\\\&=\\! \\left[\\begin{array} = 1 \\times {10}^5 + 2 \\times {10}^4 + \\\\ 0 \\times {10}^3 + 7 \\times {10}^2 + \\\\ 1 \\times {10}^1 + 9 \\times {10}^0\\end{array} \\right]\\end{align} (v) $$20068$$ \\begin{align}&=\\left[\\begin{array} = 2 \\times 10000 + 0 \\times 1000 +\\\\ 0 \\times 100 + 6 \\times 10 + 8 \\times 1\\end{array} \\right]\\\\&=\\left[\\begin{array}=2 \\times {{10}^4} + 0 \\times {{10}^3} + 0 \\times\\\\ {{10}^2} + 6 \\times {{10}^1} + 8 \\times {{10}^0}\\end{array}\\right]\\end{align} ## Chapter 13 Ex.13.3 Question 2 Find the number from each of the following expanded forms: (a) \\left[ \\begin{align} & 8\\times {{10}^{4}}+6\\times {{10}^{3}}+0\\times \\\\ & {{10}^{2}}+4\\times {{10}^{1}}+5\\times {{10}^{0}} \\\\ \\end{align} \\right] (b) $$4 \\! \\times \\! 10^5 \\! + \\! 5 \\! \\times \\! 10^3 \\! + \\! 3 \\! \\times \\! 10^2 \\! + \\! 2 \\! \\times \\! 10^0$$ (c) $$3 \\times {{10}^4} + 7 \\times {{10}^2} + 5 \\times {{10}^0}$$ (d) $$9 \\times {{10}^5} + 2 \\times {{10}^2} + 3 \\times {{10}^1}$$ ### Solution Reasoning: Expressing a number means we have to expand the following powers of $$10.$$ (a) \\left[ \\begin{align} & 8\\times {{10}^{4}}+6\\times {{10}^{3}}+0\\times \\\\ & {{10}^{2}}+4\\times {{10}^{1}}+5\\times {{10}^{0}} \\\\ \\end{align} \\right] \\begin{align} & =\\left[ \\begin{array} & 8\\times 10000+6\\times 1000+ \\\\ 0\\times 100+4\\times 10+5\\times 1 \\\\ \\end{array} \\right] \\\\ & =80000+6000+0+40+5", "\\\\ &= (4 \\times 27) \\times x^{6+15} \\\\ &= 108x^{21} \\\\ \\end{align} ### Example 5 Simplify $\\Big(\\dfrac{2a^3}{b^2}\\Big)^3$. \\begin{align} \\displaystyle \\Big(\\dfrac{2a^3}{b^2}\\Big)^3 &= \\dfrac{(2a^3)^3}{(b^2)^3} \\\\ &= \\dfrac{2^3 a^{3 \\times 3}}{b^{2 \\times 3}} \\\\ &= \\dfrac{8 a^9}{b^6} \\\\ \\end{align}", "x \\times \\dfrac{d}{dx}\\sin{x} \\\\ &= 1 \\times \\sin{x} + x \\times \\cos{x} \\\\ &= \\sin{x} + x\\cos{x} \\\\ \\end{align} ### Example 5 Find $\\displaystyle \\dfrac{dy}{dx}$ of $y=\\sin{x}\\cos{x}$, known that $\\dfrac{d}{dx}\\sin{x} = \\cos{x}$ and $\\dfrac{d}{dx}\\cos{x} = -\\sin{x}$. \\begin{align} \\displaystyle \\dfrac{dy}{dx} &= \\dfrac{d}{dx}\\sin{x} \\times \\cos{x} + \\sin{x} \\times \\dfrac{d}{dx}\\cos{x} \\\\ &= \\cos{x} \\times \\cos{x} + \\sin{x} \\times (-\\sin{x}) \\\\ &= \\cos^2{x} – \\sin^2{x} \\\\ \\end{align} ### Example 6 Find $\\displaystyle \\dfrac{dy}{dx}$ of $y=x^2\\log_e{2x}$, known that $\\dfrac{d}{dx}\\log_e{x} = \\dfrac{1}{x}$. \\begin{align} \\displaystyle \\dfrac{dy}{dx} &= \\dfrac{d}{dx}x^2 \\times \\log_e{2x} + x^2 \\times \\dfrac{d}{dx}\\log_e{2x} \\\\ &= 2x \\times \\log_e{2x} + x^2 \\times \\dfrac{1}{2x} \\times \\dfrac{d}{dx}(2x) \\\\ &= 2x\\log_e{2x} + x^2 \\times \\dfrac{1}{2x} \\times 2 \\\\ &= 2x\\log_e{2x} + x \\\\ \\end{align} ### Example 7 Find $\\displaystyle \\dfrac{dy}{dx}$ of $y=e^{2x}\\sin{3x}$, known that $\\dfrac{d}{dx}\\sin{x} = \\cos{x}$ and $\\dfrac{d}{dx}e^x = e^x$. \\begin{align} \\displaystyle \\dfrac{dy}{dx} &= \\dfrac{d}{dx}e^{2x} \\times \\sin{3x} + e^{2x} \\times \\dfrac{d}{dx}\\sin{3x} \\\\ &= e^{2x} \\times \\dfrac{d}{dx}(2x) \\times \\sin{3x} + e^{2x} \\times \\cos{3x} \\times \\dfrac{d}{dx}(3x) \\\\ &= e^{2x} \\times 2 \\times \\sin{3x} + e^{2x} \\times \\cos{3x} \\times 3 \\\\ &= 3e^{2x}sin{3x} + 3e^{2x}\\cos{3x} \\\\ \\end{align}", "d(x^3+2x^2)}{\\mathrm dx} \\end{alignat} \\begin{align} \\text{i.e. }\\qquad3\\sin^2\\bigl(\\cos\\bigl(\\sqrt{x^3 +2x^2}\\bigr)\\bigr) \\times \\cos\\bigl(\\cos\\bigl(\\sqrt{x^3 +2x^2}\\bigr)\\bigr)&\\times\\bigl(-\\sin\\bigl(\\sqrt{x^3 +2x^2}\\bigr) \\\\ {}\\times \\frac 1{2\\sqrt{x^3 +2x^2}}&\\times (3x^2+4x). \\end{align}", "is: $\\set {\\dbinom {a \\ b} {1 \\ 2}, \\dbinom {a \\ b} {1 \\ 3}, \\dbinom {a \\ b} {2 \\ 1}, \\dbinom {a \\ b} {2 \\ 3}, \\dbinom {a \\ b} {3 \\ 1}, \\dbinom {a \\ b} {3 \\ 2} }$ thus demonstrating that there are $\\dfrac {3!} {\\paren {3 - 2}!} = \\dfrac {3!} {1!} = \\dfrac 6 1 = 6$ injections from $A$ to $B$." ]

[ "rolled repeatedly; different rolls are independent. N is the minimum number of rolls till each of the six faces of the die shows up at least once. Suppose that i ... 12 views Calculating the number of trials to get all four outcomes [duplicate] I was posed the following question recently: Suppose a brand of cereal has four different toys included in their cereal boxes, each with an equal probability of being found: On average, how many ... 18 views Combinatorics and statistics question regarding drawing every element [duplicate] Assuming a set of size X, on average how many attempts would it take to randomly draw every element in the set (returning it back after drawing)? I have a hard time formulating a title for this ... 20 views Expected number of tries before all elements of sample space have occurred. [duplicate] We have a bag with 4 balls, one white, one black, one blue and one red. We are drawing balls with replacement. Find the expected value of the number of tries needed to draw all four colours. More ... 16 views How many toys do I need to buy to get the whole set? [duplicate] I'm collecting toys, they come in blind packs and there are 9 of them. Therefore for the first toy I have", "# Colored balls A box contains 6 red, 4 white and 5 black balls. A person draws 4 balls from the box at random . Let P be the probability that among the balls drawn there is one ball of each color . Find 455P. ×", "many possible combinations can we create from 4 cups? • Components In the box are 8 white, 4 blue and 2 red components. What is the probability that we pull one white, one blue and one red component without returning?", "believed there was a 100% strategy - reached as follows: A guesses BLACK, and if possible switches boxes 2 and 3 so 2 would hold a black. B guesses BLACK, unless he sees two whites in 1 and 3. If possible and needed, he switches boxes 1 and 3 to put a white in box 3. C guesses WHITE. The box contents are as follows: Based on the explained logic, here are the guesses (red = mistakes) There is never more then one mistake, so always at least two correct answers, guaranteeing a 100% win. PS: also interesting variation: if box switching is allowed, how can they maximize odds of getting all three right. (I found a 3/8 strategy there) As pointed out below that is NOT correct, as the assumptions for option B aren't interpreted correctly in the table. However, if B assumes BLACK 100% of the time, winning chances are 7/8 • You wrote \"B guesses BLACK, unless he sees two whites in 1 and 3.\" But I see that two whites appear 3 times (first three cases) in box 1 and 3 for turn 2, so shouldn't B's guess have three whites? Aug 29, 2014 at 10:56 • Indeed there is an error in that table, this solution as described only get 75% wins. Concerning", "black, which gives seven possibilities, and color the rest white. This is of course the same as what Polya would produce.", "# A problem of distribution Discrete Mathematics Level pending You have 6 white balls and 6 black balls and you have 10 different boxes. The balls are all alike except for their colour. And each box can hold any number of balls. In how many ways can you distribute these balls in the boxes so that each box contains at least one ball. ×", "# A discrete mathematics problem by Tanishq Varshney Find the number of ways in which 4 different colored balls can be placed in four boxes whose colors are same of that of balls so that at least 2 balls go into the same colored boxes. Each box can only contain 1 ball. ×", "as the expected number of different colors in a single drawing. The only thing that changes when we go to a large number of drawings is that for a single drawing, it may be quite likely that you will see one or two colors more than expected or one or two colors fewer, but with a large number of drawings, the average number of colors is not so likely to be much different from the expected value. In other words, the expected value computed for a single drawing is the long-run average that you asked for.", "in a box. You draw a ball out of the box at random and note its color. Without replacing the first ball, you draw a second ball and then paint it to match the color of the first. Replace both balls, and repeat the process. The game ends when all four balls have become the same color. What is the expected number of turns to finish the game? Extra credit: What if there are more balls and more colors? Here is my solution to the first part (four balls): [Show Solution] Here is my solution to the general case with $N$ balls: [Show Solution] ## Pick a card! This Riddler puzzle is about a card game where the goal is to find the largest card. From a shuffled deck of 100 cards that are numbered 1 to 100, you are dealt 10 cards face down. You turn the cards over one by one. After each card, you must decide whether to end the game. If you end the game on the highest card in the hand you were dealt, you win; otherwise, you lose. What is the strategy that optimizes your chances of winning? How does the strategy change as the sizes of the deck and the hand are changed? Here is my solution: [Show Solution] ## Lucky", "Of these, seven are multicolored, and three are blue. Based on this information, there is a 7 out of 10 chance of pulling a multicolored toy from the box. Similarly, there is a 3 out of 10 chance of pulling a blue toy out of the box. However, if we randomly select two toys from the box, then what is the probability that we pull out a multicolored toy and then a blue toy without putting it back in the box? A common error while solving such problems is using the formula and then multiplying the probability of each toy together. As the toys are being taken out without putting back in the box, it means that the probability will change after every draw. This situation shows that an event is dependent on another event. Let's go back to the original situation. If we pull a multicolored toy out of the box that contains 10 toys, then the chances of the toy being multicolored are 7 out of 10. In the second event, the probability of pulling out a blue toy is, however, not 3 out of 10, as one multicolored toy is not put back in the box. The probability is now 3 out of 9 toys. To finish this problem, all we have to do is", "# A Difficult Choice! A box contains 4 balls. The color of each of the balls is one of the three: White, Black or Red. However, you don't know how many balls of each color are there. It might even be the case that all the balls are of the same color. You then draw two balls randomly from the box. The probability that both the drawn balls are red can be written as $$\\dfrac{p}{q}$$, where $$p$$ and $$q$$ are coprime positive integers. What is the value of $$p+q$$? ×", "# Permutations of Identical Objects (Part 1) So far we’ve talked about permutations of objects which were distinct. Things change when some (or all) of the objects to be arranged are identical. Suppose we have to arrange 5 balls in a row of which 3 are of the same color, and the other two are of different colors. How many different arrangements are possible? Let’s designate five boxes, each of which is to be filled by one ball. First, we’ll place the identical balls in any three of the five boxes. This can be done in 5C3 ways. Here’s one of them. Next, in the remaining two boxes, we’ll place the remaining two balls. This can be done in 2! ways. or Therefore the total number of arrangements will be 5C3 x 2! = 5!/3! = 20. Now let’s say 2 of the 5 balls are of the same color and the other 3 are of different colors. How many different arrangements are possible? Similar to the method used above, we’ll first place the identical balls in any 2 of the 5 boxes, in 5C2 ways. One of the arrangement is given below. Now in the remaining three boxes, the remaining three balls can be placed in 3! ways. Therefore the total number of arrangements will be 5C2", "# A box contains 30 bolts and 40 nuts .Half of the bolts and half of the nuts are rusted.If two items are drawn at random from the box,what is the probability that either both are rusted or both are bolts? This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com", "at the crux of the issue: \"pencils\" and \"boxes of pencils\" are different \"wholes.\" To take it a step further if they still have a hard time explaining it, have them draw three pencils plus two boxes of pencils and repeat the process. How many pencils are there? Why didn't 3 + 2 work? Draw 3 pencils + 2 pencils. How is this drawing different than 3 pencils + 2 boxes of pencils? • Also a great answer; maybe it would be even better if you just let the kids draw the boxes (\"Okay, now draw four pencil boxes. Draw five pencils in each of the boxes. How many pencils are there in total?\"). More advanced maths students/teachers tend to get too absorbed in the abstractness of maths - but in the end, it's all there for some final, real purpose - make it real, and it's much easier to understand. Aug 16, 2019 at 12:02 • @Luaan: I scanned though the comments and answers before writing the two comments I made under the question and I didn't notice that you had already described the drawing 4 boxes with 5 pencils in each idea that I gave in my first comment. Aug 16, 2019 at 13:23 Why you can multiply boxes and pencils, but cannot add? In this", "# Probability Probability Level 1 A box contains 4 blue, 6 yellow and 10 green marbles. If a marble is drawn, find the probability that it is either blue or yellow. ×", "to 31. For the dice game version, you'll also need 22 matchboxes, and counters in 6 different colors. You should have 16 counters in each of the 6 colors, for a total of 96 counters. You'll need to decide which color represents which number. For example, you might decide that red=1, blue=2, green=3, orange=4, white=5, and black=6. On 1 of the matchboxes, you're going to write “Move 1”. On 3 of the matchboxes, you're going to write “Move 3”. You'll also write “Move 5” on a different set of 3 matchboxes, and the same for “Move 7”, “Move 9”, “Move 11”, “Move 13”, and “Move 15”. In the single “Move 1” box, you're going to place counters of all 6 colors. This represents the fact that the matchbox array, which will always go first, can start the game by choosing any of the 6 numbers on the dice. On one of the “Move 3” boxes write “1 or 6 on top”, an only put in the counters with colors representing 2, 3, 4, and 5 in that box. Do the same for one of the “Move 5” boxes, one of the “Move 7” boxes, and so on. On another “Move 3” box, write “2 or 5 on top”, and only include the counters representing the numbers 1, 3,", "any number of \"even\" draws, we never know how many white or black ones we have, we can never be sure that their totals are equal...if we could, \"fabulous\" Steve would soon be \"broke\" Steve. Notice that after just 2 draws, there are 3 possibilities.....0W 2B, 1W 1B, 2W 0B ........Thus only one thing out of 3 wins for us. Notice that after just 4 draws, there are 5 possibilities.....0W 4B, 1W 3B , 2W 2B, 3W 1B, 4W 0B........Thus only one thing out of 5 wins for us And after 100 draws, again, only one thing out of 101 wins for us - 50W 50B, and 100 other combinations lose for us. But, consider the scenario in which two pearls remain in the bag......If we draw either a white or black one on the next draw and a white one remains, we win. Thus, (draw white)(white remains) or (draw black)(white remains) are two winning scenarios. But, if we draw either a black or white one and a black remains, we lose no matter what. Thus, out of the four remaining scenarios, 2 win for us and 2 lose, so we have a 50% chance of winning. And drawing all but one pearl out of the bag results in a greater chance of winning than stopping at any", "Distribute 12 balls into 5 boxes. If there are 12 balls, 3 red, 3 green, 3 black, and 3 white. How many ways can we distribute theses balls into 5 boxes such that each box contains exactly one ball? I was thinking to do this by cases. Case 1: 3 boxes contain one color (for example red). the other two boxes contain another color(for example white). There would be ${4 \\choose 1}{5 \\choose 3}{3 \\choose 1}$ many ways for this case. Case 2: Three boxes contain only one color (for example red). one boxes contain another different color(for example white) and the last box another different color(for example black). There would be ${4 \\choose 1}{5 \\choose 3}{3 \\choose 1}{2 \\choose 1}{2 \\choose 1}$ many ways for this case. Case 3: Two boxes contain only one color(for example red). Another two contain a different color. Last one contains a different color. there would be ${4 \\choose 1}{5 \\choose 2}{3 \\choose 1}{3 \\choose 2}{2 \\choose 1}$ many ways for this case. Case 4: Two boxes contain only one color(for example red). Another one contains a different color. Another one contains a different color. Last one contains a different color. there would be ${4 \\choose 1}{5 \\choose 2}{3 \\choose 1}{3 \\choose 1}{2 \\choose 1}{2 \\choose 1}{1 \\choose 1}{1 \\choose 1}$. Is", "# number of ways to put 4 black,4 white,4 red balls in 6 different boxes The question says:in how many ways we could put 4 black,4 white,4 red balls in 6 different boxes? boxes are distinguishable,black balls are identical,red balls are identical,and white balls are identical. one or more box can remain empty. i know number of ways to put k indistinguishable balls to n distinguishable boxes.but i don't know how to attack this problem because 3 different kind of balls. • Each color is independent from each other. Just calculate the possibilities for one color, and then multiply by each other color to get the total possibilities. Jul 24, 2014 at 4:49", "ways to put 5 different balls into 1 or 2 different boxes out of 3 (to leave at least 1 box empty). _________________ HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame Manager Joined: 11 Jul 2012 Posts: 52 GMAT 1: 650 Q49 V29 Re: Five balls of different colors are to be placed in three [#permalink] ### Show Tags 18 Oct 2012, 20:03 EvaJager wrote: avaneeshvyas wrote: Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done? A. 60 B. 90 C. 120 D. 150 E. 180 Please provide a small note of explanation for all the combinations used in the solution. Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box. For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box. Since in each box there must be at least one ball, we have the possibilities of" ]

[ "at first. Then: $$P(wbb)+P(bbw)+P(bwb)=\\frac59\\frac9{16}\\frac8{15}+\\frac49\\frac9{16}\\frac7{15}+\\frac49\\frac7{16}\\frac9{15}$$ where e.g. $bwb$ stands for the event that at first a black ball is drawn (from the first bag) secondly a white ball (from the second bag) and thirdly a black ball (from the second bag). • The answer given in my book is 1/6(case1)+7/30(case2)=2/5. – user517784 Feb 13 '18 at 13:08 • Can you just check it once. – user517784 Feb 13 '18 at 13:08 • That's also my answer (as you can check yourself). – drhab Feb 13 '18 at 13:14 • Yes thanks for answering – user517784 Feb 13 '18 at 13:15 • You are welcome. – drhab Feb 13 '18 at 13:18", "# Difficult Probability Solved QuestionAptitude Discussion Q. Four unit squares are chosen at random on a chessboard. What is the probability that three of them are of one colour and fourth is of opposite colour? ✖ A. 80/427 ✖ B. 160/427 ✖ C. 320/1281 ✔ D. 640/1281 Solution: Option(D) is correct The number of ways of choosing 4 squares from 64 is $^{64}C_4$ The possible split ups as far as colour is concerned and the number of ways of selecting the square are listed below: White = 0, Black = 4, No. of Ways = ${^{32}C_0} × {^{32}C_4}$ White = 1, Black = 3, No. of Ways = ${^{32}C_1} × {^{32}C_3}$ White = 2, Black = 2, No. of Ways = ${^{32}C_2} × {^{32}C_2}$ White = 3, Black = 1, No. of Ways = ${^{32}C_3} × {^{32}C_1}$ White = 4, Black = 0, No. of Ways = ${^{32}C_4} × {^{32}C_0}$ The number of ways in which 3 squares are of one colour and fourth is of opposite colour is: $P(A) =2\\times {^{32}C_1} \\times {^{32}C_3}$ $=(2\\times 32)\\dfrac{32\\times 31\\times 30}{2\\times 3}$ The total number of ways of selecting 4 squares is: $= P(B)= {^{64}C_4}$ $=\\dfrac{64\\times 63\\times62\\times61}{2\\times3\\times4}$ The required probability $=\\dfrac{P(A)}{P(B)}$ $=\\dfrac{640}{1281}$ Edit: For an alternative solution, check comment by Laurence. ## (9) Comment(s) Siddharth () Let the combination, $WWWB$ be", "boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, then the probability that 2 white and 1 black balls will be drawn is (a) $\\frac{13}{32}$ (b) $\\frac{1}{4}$ (c) $\\frac{1}{32}$ (d) $\\frac{3}{16}$ Q 2 | Page 103 A and B draw two cards each, one after another, from a pack of well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is (a) $\\frac{44}{85 \\times 49}$ (b) $\\frac{11}{85 \\times 49}$ (c) $\\frac{13 \\times 24}{17 \\times 25 \\times 49}$ (d) none of these Q 3 | Page 103 A and B are two events such that P (A) = 0.25 and P (B) = 0.50. The probability of both happening together is 0.14. The probability of both A and B not happening is (a) 0.39 (b) 0.25 (c) 0.11 (d) none of these Q 4 | Page 104 The probabilities of a student getting I, II and III division in an examination are $\\frac{1}{10}, \\frac{3}{5}\\text{ and } \\frac{1}{4}$respectively. The probability that the student fails in the examination is (a) $\\frac{197}{200}$ (b) $\\frac{27}{100}$ (c) $\\frac{83}{100}$ (d) none of these Q 5 | Page 104 India play two matches each with West Indies and Australia. In any match the probabilities of India getting 0,1 and", "multiple reasons have to treat these as 'dimer' - that is - units consisting of two 'subunits'. This means I can get white-white, white-black, black-white, and black-black dimers (white-black and black-white are different because I care about their orientation..) Now, when I wan to calculate then number of different combinations of these, I am effectively counting $\\frac{(ww+wb+bw+bb)!}{ww!wb!bw!bb!}$ summed over all the possible combinations of ww, wb, bw and bb such that w and b stays constant. A practical example - let's say B = 2 and W = 2, in this case $\\frac{(W+B)!}{W!B!} = \\frac{(2+2)!}{2!2!} = 6$ Now, if we treat them as dimers, we have 4 different scenarios: 1) ww = 1, bb = 1 number of combinations is $\\frac{(ww+wb+bw+bb)!}{ww!wb!bw!bb!}=\\frac{(1+0+0+1)!}{1!0!0!1!}= 2$ 2) wb = 2, bw = 0 number of combinations is $\\frac{(ww+wb+bw+bb)!}{ww!wb!bw!bb!}=\\frac{(0+2+0+0)!}{0!1!0!0!}= 1$ 3) wb = 1, bw = 1 number of combinations is $\\frac{(ww+wb+bw+bb)!}{ww!wb!bw!bb!}=\\frac{(0+1+1+0)!}{0!1!1!0!}= 2$ 4) wb = 0 bw = 2 number of combinations is $\\frac{(ww+wb+bw+bb)!}{ww!wb!bw!bb!}=\\frac{(0+0+2+0)!}{0!0!2!0!}= 1$ All of these sum to 6 which is the same result as from $\\frac{(W+B)!}{W!B!}$. This all actually intuitively makes sense, and I have proved this computationally from any arbitrary number of W and B, but was wondering if there was a mathematical proof for this for any general B and W (such that B+W is even so", "$WBBB$, $WWBB$, or $WWWB$. Notice we can't have 4 white balls since there are only a total of 3 white balls to choose from. So our $30$ configurations we counted previously must be split up between these 4 descriptions. So how many of each kind are there? $BBBB$: The only way this can happen is if we select two black from the first box and two black from the second box. So there are $\\binom{2}{2}\\binom{3}{2}=3$ ways for this to happen. $WBBB$: 1.) Select one white/one black from the first box and two black from the second box for a total of $2\\cdot \\binom{3}{2}=6$ ways. or 2.) Two black from the first and one white/one black from the second box for a total of $1\\cdot (3\\cdot 2)=6 ways. Thus there are a total of$6+6=12$ways for this to happen.$WWBB$: 1.) Select one white/one black from the first box and one white/one black from the second box for a total of$2\\cdot (3\\cdot 2)=12$ways. 2.) Two black from the first and two white from the second box for a total of$1\\cdot\\binom{2}{2}=1$ways. Thus there are a total of$12+1=13$ways for this to happen.$WWWB$: Select one white/one black from the first box and two white from the second box for a total of$2\\cdot \\binom{2}{2}=2$ways. From the above work we know that$P(BBBB)=\\frac{3}{30}$,$P(WBBB)=\\frac{12}{30}$,$P(WWBB)=\\frac{13}{30}$, and$P(WWWB)=\\frac{2}{30}\\$ So the probability", "arrange the 7 white balls first. _W_W_W_W_W_W_W_ 5 black balls have 8 spaces inbetween, so arrangements $$= \\frac{8*7*6*5*4}{5!} = 56$$ 3) B $$\\frac{13^4*48*47}{2}$$ .. we divide by two to avoid repetition of the last two cards i.e. ab or ba are the same in this context. 4) B we have to divide it into two cases i.e. 4-digit and 5-digit because 1034 and 10345 are both greater than 1000 and we can't put a condition that 0 can't be placed on 2nd position. 4-digit: 4*4*3*2 = 96 (0 can't be the first digit) 5-digit: 4*4*3*2*1 = 96 (0 can't be the first digit) total = 96*2 5) C at least one out of n is $$2^n-1$$. total mocktails = at least one whiskey * at least one soda * at least one juice $$= 2^3-1*2^4-1*2^5-1$$ $$=3255$$ _________________ press +1 Kudos to appreciate posts Manager Joined: 25 May 2011 Posts: 163 Followers: 2 Kudos [?]: 45 [0], given: 71 MBAhereIcome wrote: 1.1) i like to write is as $$\\frac{5!}{5} = 4! = 24$$ 1.2) AxB or BxA, so same neighbours are counted twice. that's why ans $$= \\frac{4!}{2} = 12$$ 2) A arrange the 7 white balls first. _W_W_W_W_W_W_W_ 5 black balls have 8 spaces inbetween, so arrangements $$= \\frac{8*7*6*5*4}{5!} = 56$$ 3) B $$\\frac{13^4*48*47}{2}$$ .. we divide by", "of choosing pairs of $$s_1$$ as white and $$s_2$$ as black square $$= |\\{ (s_1, s_2) | s_1 \\in W, s_2 \\in B \\}| \\\\ = |\\{ s_1| s_1 \\in W \\} \\times \\{ s_2| s_2 \\in B \\}| \\ [ \\ \\because W \\cap B = \\emptyset \\ ] \\\\ = |\\{ s_1| s_1 \\in W \\}| \\times |\\{ s_2| s_2 \\in B \\}| \\\\ = |W| \\times |B| = \\frac{|S|}{2} \\times \\frac{|S|}{2} = \\frac{|S|^2}{4}$$ Since order of choosing does not matter, Ways of choosing pairs of squares of different color = Ways of choosing pairs of $$s_1$$ as white and $$s_2$$ as black square $$= \\frac{|S|^2}{4} = \\frac{64^2}{4} = 1024$$ For each black square you can choose one of the 32 whites squares(32 possibilities), then, for the remaining 31 black and 31 white squares, you repeat the process (31 possibilities $$\\times$$ the first 32 possibilities) ... . So the result should be $$32!$$. Since there are 64 squares, 32 are black and 32 are white. You can choose a black square out of 32 in $${32}\\choose{1}$$ways and the same for the white squares. Then: $${32}\\choose{1}$$$${32}\\choose{1}=32$$$$32=1024$$ ways of choosing a pair of squares (one white and the other black) out of 64.", "as $$1+3$$, $$2+2$$, $$3+1$$ or $$4+0$$ (the $$0+4$$ case was the one where we already had a pair). The patterns for the four draws of other colours are $$AABB$$, $$ABAB$$ and $$ABBA$$ with equal probabilities. The $$AABB$$ is always a win, $$ABAB$$ is always a loss, and $$ABBA$$ is a win unless the groups were split $$2+2$$. So we take the $$2+2$$ case separate with a $$\\frac13$$ chance, and the three other splits contribute $$\\frac23$$ of their weight. All in all the chance to win is $$\\frac15+\\frac15\\times\\frac13+\\frac35\\times\\frac23=\\frac{3+1+6}{15}=\\frac23.$$ You can also brute-force it. If you call A the first color you see, B the second and C the third, there are only 15 possible patterns: AABBCC, AABCBC, AABCCB, ABABCC, ABACBC, ABACCB, ABBACC, ABCABC, ABCACB, ABBCAC, ABCBAC, ABCCAB, ABBCCA, ABCBCA, ABCCBA. The bold ones are those with 2 successive marbles of one color. They are 10 out of 15. That is a probability of 2/3. We're going to play a game. Here are the rules: • A game position is a pair $$(x,\\bar y)$$ of a natural number $$x$$ and a bag of positive numbers $$\\bar y$$. • To make a move, pick $$y \\in \\bar y$$. If $$x>0$$, the next game position is $$(y-1, (\\bar y\\backslash\\{y\\})\\cup\\{x\\})$$; otherwise don't insert $$x$$ and arrive at the next game position of $$(y-1,\\bar", "with \"BBW\" There are $\\binom 41=4$ equally probable arrangements like this ( you just have to decide where to put the second \"W\" in the 4 remaining places) To win on her thirrd selection the arrangement must start with \"BBBBW\" There are $\\binom 21=2$ equally probable arrangements like this ( you just have to decide where to put the second \"W\" in the 2 remaining places) so $$P(WIN) = \\frac{ \\binom 61+\\binom 41+\\binom 21 }{ \\binom 72}=\\frac{12}{21}$$ • This was great. Thank you very much – Pedro Alonso Jul 3 '18 at 1:59", "Sol: There are four possibilities First Ball Second Ball Event White White E1 White Black E2 Black White E3 Black Black E4 $$P(E_{1}) = \\frac{2}{4} * \\frac{1}{3} = \\frac{1}{6}$$ $$P(E_{2}) = \\frac{2}{4} * \\frac{2}{3} = \\frac{1}{3}$$ $$P(E_{3}) = \\frac{2}{4} * \\frac{2}{5} = \\frac{1}{5}$$ $$P(E_{4}) = \\frac{2}{4} * \\frac{3}{5} = \\frac{1}{10}$$ Let E denotes the event of drawing a black ball in the third attempt $$P(\\frac{E}{E_{1}}) = 1$$. Because there is no white ball left to be selected. $$P(\\frac{E}{E_{2}}) = \\frac{3}{4}$$.Because there are 3 black ball and 1 white ball left. $$P(\\frac{E}{E_{3}}) = \\frac{3}{4}$$. Because again there are 3 black and 1 white ball left. $$P(\\frac{E}{E_{4}}) = \\frac{4}{6} = \\frac{2}{3}$$. Because there are 4 black balls and 2 white balls. $$P(E) = P(E_{1}) P (\\frac{E}{E_{1}}) + P(E_{2}) P (\\frac{E}{E_{2}}) + P (E_{3}) P (\\frac{E}{E_{3}}) + P (E_{4}) P (\\frac{E}{E_{4}})$$ $$\\Rightarrow \\frac{1}{6} * 1 + \\frac{1}{3} * \\frac{3}{4} + \\frac{1}{5} * \\frac{3}{4} + \\frac{3}{10} + \\frac{2}{3}$$ $$\\Rightarrow \\frac{1}{6} + \\frac{1}{4} + \\frac{3}{20} + \\frac{1}{5}$$ $$\\Rightarrow \\frac{23}{30}$$ Therefore, the value of K is $$\\frac{23}{30}$$. Explore more such questions and answers at BYJU’S.", "and then a ball of the other color. In the case of a series of black balls the probability of such event is: $$\\frac{b}{w+b}\\frac{b-1}{w+b-1}\\cdots\\frac{b-k+1}{w+b-k+1}\\frac{w}{w+b-k}= \\frac{b!}{(b-k)!}\\frac{(w+b-k)!}{(w+b)!}\\frac{w}{w+b-k}$$ and similar expression for the series of white balls. Thus, we have for $$w,b>0$$: \\begin{align} P(w,b) &=\\sum_{k=1}^b\\frac{b!}{(b-k)!}\\frac{(w+b-k)!}{(w+b)!}\\frac{w}{w+b-k}P(w,b-k)\\\\ &+\\sum_{k=1}^{w}\\frac{w!}{(w-k)!}\\frac{(w+b-k)!}{(w+b)!}\\frac{b}{w+b-k}P(w-k,b), \\end{align} with $$P(w,0)=1$$, $$P(0,b)=0$$. We claim: $$\\forall w,b>0: P(w,b)=\\frac12.\\tag2$$ For $$P(1,1)$$ the claim is obviously valid. Assume it is valid for all $$P(w,b-k)$$ with $$0 and $$P(w-k,b)$$ with $$0. We have: \\begin{align} P(w,b) &\\stackrel{I.H.}=\\frac12\\sum_{k=1}^b\\frac{b!}{(b-k)!}\\frac{(w+b-k)!}{(w+b)!}\\frac{w}{w+b-k} +\\frac{b!}2\\frac{w!}{(w+b)!}\\\\ &+\\frac12\\sum_{k=1}^{w}\\frac{w!}{(w-k)!}\\frac{(w+b-k)!}{(w+b)!}\\frac{b}{w+b-k} -\\frac{w!}2\\frac{b!}{(w+b)!}\\\\ &\\stackrel{(1)}=\\frac12\\left(1-\\frac{b}{w+b}\\right)+\\frac12\\left(1-\\frac{w}{w+b}\\right)=\\frac12. \\end{align} Thus, the claim (2) is proved.", "is drawn from Box 2.}$ [color=beige]. . [/color]$\\text{The probabilities are: }\\:\\begin{Bmatrix}P(BB\\text{, Box 1})=&\\frac{{8\\choose2}}{{20\\choose2}}=&\\frac{28}{190} \\\\ \\\\ \\\\ \\\\ P(WB\\text{, Box 2})=&\\frac{{10\\choose1}{15\\choose1}}{{25\\choose2}}=&\\frac{150}{300}\\end{Bmatrix}=$ [color=beige]. . [/color]$\\text{Hence: }\\:P(BB|WB) \\:=\\:\\frac{28}{190}\\,\\cdot\\,\\frac{150}{300} \\:=\\:\\frac{35}{475}$ $\\text{Therefore: }\\:P(\\text{exactly 1 White}) \\;=\\;\\frac{84}{475}\\,+\\,\\frac{35}{475} \\;=\\;\\frac{119}{475}$ Chikis July 23rd, 2012 02:40 PM Re: The Probability That Exactly One Of The Four Balls Is Wh Don't be offended please, I cannot see the working clearly. Denis July 23rd, 2012 05:31 PM Re: The Probability That Exactly One Of The Four Balls Is Wh 1:12W,8B ; 2:10W,15B The White can be drawn 1st, 2nd, 3rd or 4th : so 4 different cases : Case1: W=12/20, B=8/19, B=15/25, B=14/24: 12/20 * 8/19 * 15/25 * 14/24 = 42/475 Case2: B=8/20, W=12/19, B=15/25, B=14/24: 8/20 * 12/19 * 15/25 * 14/24 = 42/475 Case3: B=8/20, B=7/19, W=10/25, B=15/24: 8/20 * 7/19 * 10/25 * 15/24 = 7/190 Case4: B=8/20, B=7/19, B=15/25, W=10/24: 8/20 * 7/19 * 15/25 * 10/24 = 7/190 42/475 + 42/475 + 7/190 + 7/190 = 119/475 (as per Sir Soroban!) Chikis July 24th, 2012 09:53 PM Re: The Probability That Exactly One Of The Four Balls Is Wh My thanks goes to soroban and Denis, for having intrest in my question, let me work with what you have given me. I will post more replies into the thread, if I found anything", "# A Followup on Lewis Carroll's Pillow Problem ### Solution 1 (Correct, short) Prior odds in favor of two white counters in the second bag is $w:b,$ but drawing $w$ is twice as likely under $ww,$ then under $bw$ so posterior odds in favor are $2w:b,$ i.e. $\\displaystyle Prob(ww|\\text{drawn }w)= \\frac{2w}{2w+b}.$ ### Solution 2 (Correct, long) We'll use Bayesian Odds. Introduce $\\displaystyle P(W)=\\frac{w}{w+b},$ probability of the counter removed from the first bag being white for the black counter it is $\\displaystyle P(W)=\\frac{b}{w+b};$ $P(W|W)$ and $P(B|W)$ are the probabilities that the remaining counter in the second bag is white and black, respectively; $\\overline{P}(W|W)$ and $P(W|B)$ are the probabilities of removing of a white counter from the second bag, provided that it contains two white or a black and a white counters. The Bayesian odds are expressed as $\\displaystyle \\frac{P(W|W)}{P(B|W)}=\\frac{\\overline{P}(W|W)}{P(W|B)}\\cdot\\frac{P(W)}{P(B)}.$ It is obvious that $\\displaystyle\\frac{\\overline{P}(W|W)}{P(W|B)}=2,$ implying $\\displaystyle \\frac{P(W|W)}{P(B|W)}=2\\cdot\\frac{w}{b}=\\frac{2w}{b}.$ Since $P(W|W)+P(B|W)=1,$ we derive $\\displaystyle P(W|W)=\\frac{2w}{2w+b}.$ ### Solution 3 (Correct, long) We'll use Bayes' Theorem. The notations are from the previous solution: \\displaystyle\\begin{align}P(W|W)&=\\frac{\\overline{P}(W|W)\\cdot P(W)}{\\overline{P}(W|W)\\cdot P(W)+P(W|B)\\cdot P(B)}\\\\ &=\\frac{\\displaystyle 1\\cdot \\frac{w}{w+b}}{\\displaystyle 1\\cdot \\frac{w}{w+b}+\\frac{1}{2}\\cdot \\frac{b}{w+b}}\\\\ &=\\frac{\\displaystyle \\frac{w}{w+b}}{\\displaystyle \\frac{1}{2}\\cdot\\frac{2w+b}{w+b}}=\\frac{2w}{2w+b}. \\end{align} ### Solution 4 (Incorrect) There's a $50%$ chance that the remaining ball is from the original bag, and so $\\displaystyle P = \\frac{w}{w+b}$ and another $50%$ chance that the remaining ball is the added \"white one\",", "yellow? I.e. we consider RY and YR to be the same event... Our event space therefore becomes: {R1Y1, R2Y1 Y1R1, Y1R2}. Thus, if we say event A is \"draw a red and a yellow marble\", $$P(A) = \\frac{N_A}{N} = \\frac{4}{30} = \\frac{2}{15}$$ We are essentially now dealing with combinations. Therefore in the above table, where we see things like... Selection 1 Selection 2 B1 B2 B2 B1 ...we can delete one of the rows. The set of all basic outcomes therefore becomes: Selection 1 Selection 2 Selection 1 Selection 2 B1 B2 B2 B1 B1 B3 B2 B3 B1 R1 B2 R1 B1 R2 B2 R2 B1 Y1 B2 Y1 B3 B1 B3 B2 B3 R1 B3 R2 B3 Y1 R1 R2 R2 R1 R1 B1 R2 B1 R1 B2 R2 B2 R1 B3 R2 B3 R1 Y1 R2 Y1 Y1 B1 Y1 B2 Y1 B3 Y1 R1 Y1 R2 Again a nightmare to figure out by hand, so maths to the resuce... \\begin{align} Combinations &= {_{n}P_r} = \\frac{n!}{r!(n-r)!} \\\\ &= {_{6}P_2} = \\frac{6!}{2!(6-2)!} \\\\ &= \\frac{6 \\times 5 \\times 4 \\times 3 \\times 2 }{4 \\times 3 \\times 2 \\times 2} \\\\ &= \\frac{6 \\times 5}{2} \\\\ &= 15 \\end{align} And, again, that is how many selections (not included the ones we've deleted using a strikethrough font)", ". . . Is this correct? . Yes! b) The probability of at least one black. The opposite of \"at least one black\" is \"no blacks\" (both white). You found: $P(\\text{2 whites}) = \\frac{1}{5}$ .in part (a). Therefore: . $P(\\text{at least one black}) \\;=\\;1-\\frac{1}{5} \\;=\\;\\frac{4}{5}$ c) The probability of at most one black. The opposite of \"at most one black\" is \"both black\". $P(\\text{both black}) \\:=\\:\\frac{2}{5}\\cdot\\frac{4}{6} \\:=\\:\\frac{4}{15}$ Therefore: . $P(\\text{at most one black}) \\;=\\;1-\\frac{4}{15} \\;=\\;\\frac{11}{15}$ d) The probability of a black followed by a white, or a white followed by a black. I did : . $\\frac{2}{5}\\cdot\\frac{2}{6} \\,=\\,\\frac{2}{15}\\:\\text{ or }\\:\\frac{3}{5}\\cdot\\frac{4}{6} \\,=\\,\\frac{2}{5}$ . . . Is this correct? . Um, not quite ... With an \"or\" statement, we add the probabilties. Therefore: . $P(\\text{BW }\\vee\\text{ WB}) \\;=\\;\\frac{2}{15} + \\frac{2}{5} \\;=\\;\\frac{8}{15}$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Here's an alternate approach to part (d). We want: . $P(\\text{opposite colors})$ In part (a), we found that: . $P(\\text{2 whites}) \\,=\\,\\frac{1}{5}$ In part (c), we found that: . $P(\\text{2 blacks}) \\,=\\,\\frac{4}{15}$ . . Hence: . $P(\\text{same color}) \\:=\\:\\frac{1}{5} + \\frac{4}{15} \\:=\\:\\frac{7}{15}$ Therefore: . $P(\\text{opposite colors}) \\;=\\;1 - \\frac{7}{15} \\;=\\;\\frac{8}{15}$ 3. ## Thank you! Thank you so much, I", "the second draw is then also blue is $\\frac 15$. Hence the probability $P(B_1\\cap B_2)=\\frac 1{15}$. What text is this? – lulu Jan 16 '17 at 12:52 • Just to stress: the probability that any given draw is blue is $\\frac 26$. There is nothing special about being the first, second, or sixth draw. It's fine if the author wants to carefully go through the law of total probability to confirm this...that's a good exercise. Just do it carefully. – lulu Jan 16 '17 at 12:55 • Oh, ok. Anybody can make a mistake at the board. – lulu Jan 16 '17 at 12:55 • As an alternate way of computing $P(B_1\\cap B_2)$: suppose the pens are numbered $b_1,b_2,r_1,r_2,y,g$. Then there are $6!$ equally probable ways of ordering them. There are $2\\times 4!$ ways of ordering them assuming the two blues come first (in some order). Thus the answer is $\\frac {2\\times 4!}{6!}=\\frac 2{6\\times 5}=\\frac 1{15}$. – lulu Jan 16 '17 at 12:58 $$\\boxed{P(B_2)=P(B_2|B_1)\\cdot P(B_1)+P(B_2|B_1^C)\\cdot P(B_1^C)}$$ $=\\frac15\\cdot \\frac{2}6+\\frac{2}{5}\\cdot \\frac46=\\frac2{30}+\\frac8{30}=\\frac{10}{30}=\\frac13$ Interpretation of $P(B_2|B_1)$: One blue pen is drawn first. $5$ pens are remaining. One of them is blue. Thus the probability that the second drawn pen is blue is $\\frac15$, given that the first drawn pen was blue.", "is R}) &=& \\frac{n}{3n} \\\\ \\\\[-3mm] P(\\text{2nd is W}) &=& \\frac{n}{n-1} \\\\ \\\\[-3mm] P(\\text{3rd is B}) &=& \\frac{n}{n-2} \\end{array}$ Hence: . $P(\\text{R-W-B}) \\;=\\;\\dfrac{n}{3n} \\times \\dfrac{n}{3n-1} \\times \\dfrac{n}{3n-2}$ What is the probability that the balls are drawn in the order Blue-Red-White? . . $\\begin{array}{cccccc} P(\\text{1st is B}) &=& \\frac{n}{3n} \\\\ \\\\[-3mm] P(\\text{2nd is R}) &=& \\frac{n}{n-1} \\\\ \\\\[-3mm] P(\\text{3rd is W}) &=& \\frac{n}{n-2} \\end{array}$ Hence: . $P(\\text{B-R-W}) \\;=\\;\\dfrac{n}{3n} \\times \\dfrac{n}{3n-1} \\times \\dfrac{n}{3n-2}$ What is the probability that the balls are drawn in the order White-Red-Blue? . . $\\begin{array}{cccccc} P(\\text{1st is W}) &=& \\frac{n}{3n} \\\\ \\\\[-3mm] P(\\text{2nd is R}) &=& \\frac{n}{n-1} \\\\ \\\\[-3mm] P(\\text{3rd is B}) &=& \\frac{n}{n-2} \\end{array}$ Hence: . $P(\\text{W-R-B}) \\;=\\;\\dfrac{n}{3n} \\times \\dfrac{n}{3n-1} \\times \\dfrac{n}{3n-2}$ . . . etc. . . . etc. • Oct 23rd 2010, 11:40 PM Lukybear Ok. The answer is actually infact (nC3)^3/3nC3. But i was confused here, since i thought it they picked one and didnt replace it, then it cant be computed using combinatorics. Namely, if the question stated, if it was picked and replaced, is the solution still (nC3)^3/3nC3? • Oct 24th 2010, 04:44 AM Plato Quote: Originally Posted by Lukybear Ok. The answer is actually infact (nC3)^3/3nC3. But i was confused here, since i thought it they picked one and didnt replace it, then it cant be computed using combinatorics.", "# Probability that two cards are black and one is red If I draw three cards at random (without replacement) from a standard 52-card deck, what is the probability that two of the cards will be black and one of them will be red? Thanks! - We have three possible sequences of cards that satisfy your conditions (two black, one red): BBR, BRB and RBB. Let $P(BBR),P(BRB),P(RBB)$ denote the probability that a randomly drawn sequence is of the form BBR, BRB and RBB respectively. The probability of three randomly drawn cards satisfying your condition is $P(BBR)+P(BRB)+P(RBB)$. I will show you how to calculate $P(BBR)$ and leave the rest to you. When we draw the first card, there are $26$ black cards and $52$ cards total, so there is a $\\frac{26}{52}=\\frac{1}{2}$ chance that this card is black. If the first card is black, there are $25$ black cards out of $51$ total when we draw the second card, so there is a $\\frac{25}{51}$ chance it is black. If the first two cards were both black, then there are $26$ red cards out of $50$ total when we draw the third card, so there is a $\\frac{26}{50}=\\frac{13}{25}$ chance it is red. Thus $P(BBR)=\\frac{1}{2}\\cdot\\frac{25}{51}\\cdot\\frac{13}{25}=\\frac{13}{102}$. There are $\\binom{52}{3}$ sets of 3 cards. $\\binom{26}{2}\\binom{26}{1}$ of them are as you described. The ratio of", "# Monthly Archives: February 2012 ## Probability musings From Futility Closet: A bag contains 16 billiard balls, some white and some black. You draw two balls at the same time. It is equally likely that the two will be the same color as different colors. What is the proportion of colors within the bag? A solution appears on the page, but it doesn’t suggest how it was determined or if the solution is unique. I’ll provide that below plus an extension. Assume there were w white balls originally, then the number of ways to draw two white is $\\displaystyle \\frac{w*(w-1)}{2!}$, two black is $\\displaystyle \\frac{(16-w)*(15-w)}{2!}$, and one of each is $\\displaystyle \\frac{w*(16-w)}{1!1!}$. As the likelihood of drawing one of each color is to be the same as drawing only one color, $\\displaystyle \\frac{w*(w-1)}{2}+ \\frac{(16-w)*(15-w)}{2}= w*(16-w)$ This is a quadratic equation with solutions $w=\\{6,10\\}$, verifying Futility Closet’s solution, and establishing those solutions as the only ones possible. Extension: For what total number(s) of white and black billiard balls is the likelihood for two drawn balls to be one of color the same as them being one of each? Let n be the total number of balls. Repeating the procedure above, the number of ways to draw two white is $\\displaystyle \\frac{w*(w-1)}{2!}$, two black is $\\displaystyle \\frac{(n-w)*((n-1)-w)}{2!}$, and one of", "* kudos for all correct solutions Hi We are given (w - # of white blls, b - # of black balls, x - # of white balls to be added): $$\\frac{w}{w+b} = \\frac{4}{5}$$ $$w = 4b$$ or $$w:b = 4:1$$ and $$\\frac{w + x}{w + b + x} = \\frac{7}{8}$$ $$\\frac{4b + x}{5b + x} = \\frac{7}{8}$$ $$x = 3b$$ Now let's lok at our options: (1) does not give us any additional information, we already know that w:b is 4:1. Insufficient. (2) w = b + 27, combinig this with w=4b we get b=9 and x=27. Sufficient. Hi, I have a question here. We already know that w:b=4:1 - agreed. But in statement 2, we use w=4b for the solution. As w=4b was also given in statement 1, I landed up opting choice C. I hope you understood my question. Please help Thanks, Uma Director Joined: 12 Nov 2016 Posts: 789 Re: A ball is randomly selected from a box containing white balls and [#permalink] ### Show Tags 23 Sep 2017, 23:33 GMATPrepNow wrote: A ball is randomly selected from a box containing white balls and black balls only. If the probability of randomly selecting a white ball is 4/5, how many white balls must be added to the box so that the probability of randomly" ]

[ "boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, then the probability that 2 white and 1 black balls will be drawn is (a) $\\frac{13}{32}$ (b) $\\frac{1}{4}$ (c) $\\frac{1}{32}$ (d) $\\frac{3}{16}$ Q 2 | Page 103 A and B draw two cards each, one after another, from a pack of well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is (a) $\\frac{44}{85 \\times 49}$ (b) $\\frac{11}{85 \\times 49}$ (c) $\\frac{13 \\times 24}{17 \\times 25 \\times 49}$ (d) none of these Q 3 | Page 103 A and B are two events such that P (A) = 0.25 and P (B) = 0.50. The probability of both happening together is 0.14. The probability of both A and B not happening is (a) 0.39 (b) 0.25 (c) 0.11 (d) none of these Q 4 | Page 104 The probabilities of a student getting I, II and III division in an examination are $\\frac{1}{10}, \\frac{3}{5}\\text{ and } \\frac{1}{4}$respectively. The probability that the student fails in the examination is (a) $\\frac{197}{200}$ (b) $\\frac{27}{100}$ (c) $\\frac{83}{100}$ (d) none of these Q 5 | Page 104 India play two matches each with West Indies and Australia. In any match the probabilities of India getting 0,1 and", "# Difficult Probability Solved QuestionAptitude Discussion Q. Four unit squares are chosen at random on a chessboard. What is the probability that three of them are of one colour and fourth is of opposite colour? ✖ A. 80/427 ✖ B. 160/427 ✖ C. 320/1281 ✔ D. 640/1281 Solution: Option(D) is correct The number of ways of choosing 4 squares from 64 is $^{64}C_4$ The possible split ups as far as colour is concerned and the number of ways of selecting the square are listed below: White = 0, Black = 4, No. of Ways = ${^{32}C_0} × {^{32}C_4}$ White = 1, Black = 3, No. of Ways = ${^{32}C_1} × {^{32}C_3}$ White = 2, Black = 2, No. of Ways = ${^{32}C_2} × {^{32}C_2}$ White = 3, Black = 1, No. of Ways = ${^{32}C_3} × {^{32}C_1}$ White = 4, Black = 0, No. of Ways = ${^{32}C_4} × {^{32}C_0}$ The number of ways in which 3 squares are of one colour and fourth is of opposite colour is: $P(A) =2\\times {^{32}C_1} \\times {^{32}C_3}$ $=(2\\times 32)\\dfrac{32\\times 31\\times 30}{2\\times 3}$ The total number of ways of selecting 4 squares is: $= P(B)= {^{64}C_4}$ $=\\dfrac{64\\times 63\\times62\\times61}{2\\times3\\times4}$ The required probability $=\\dfrac{P(A)}{P(B)}$ $=\\dfrac{640}{1281}$ Edit: For an alternative solution, check comment by Laurence. ## (9) Comment(s) Siddharth () Let the combination, $WWWB$ be", "at first. Then: $$P(wbb)+P(bbw)+P(bwb)=\\frac59\\frac9{16}\\frac8{15}+\\frac49\\frac9{16}\\frac7{15}+\\frac49\\frac7{16}\\frac9{15}$$ where e.g. $bwb$ stands for the event that at first a black ball is drawn (from the first bag) secondly a white ball (from the second bag) and thirdly a black ball (from the second bag). • The answer given in my book is 1/6(case1)+7/30(case2)=2/5. – user517784 Feb 13 '18 at 13:08 • Can you just check it once. – user517784 Feb 13 '18 at 13:08 • That's also my answer (as you can check yourself). – drhab Feb 13 '18 at 13:14 • Yes thanks for answering – user517784 Feb 13 '18 at 13:15 • You are welcome. – drhab Feb 13 '18 at 13:18", "47\\times 46}{3!} ways.$ $\\bg_white \\noindent So the answer is \\binom{4}{2}\\times \\binom{48}{3} = 6\\times \\frac{48\\times 47\\times 46}{3!} = 48\\times 47\\times 46 ways.$ #### fluffchuck ##### Active Member You've done your answers assuming there's an order that matters for the cards, but for a hand, the order doesn't matter, so we should take this into account. $\\bg_white \\noindent b) Pick the Ace: 4 ways.$ $\\bg_white \\noindent Pick the remaining four cards: We need to pick four cards from the remaining 48 non-Ace cards, where order is irrelevant, so this can be done in \\binom{48}{4}=\\frac{48\\times 47\\times 46\\times 45}{4!} ways.$ $\\bg_white \\noindent Hence the answer is 4\\times \\binom{48}{4} = 4\\times \\frac{48\\times 47\\times 46\\times 45}{4!}.$ $\\bg_white \\noindent c) Pick the two Aces (from a set of four Aces, with order irrelvant): \\binom{4}{2}=\\frac{4\\times 3}{2!}=6 ways.$ $\\bg_white \\noindent Pick the remaining three cards: We need to pick three cards from the remaining 48 non-Ace cards, where order is irrelevant, so this can be done in \\binom{48}{3}=\\frac{48\\times 47\\times 46}{3!} ways.$ $\\bg_white \\noindent So the answer is \\binom{4}{2}\\times \\binom{48}{3} = 6\\times \\frac{48\\times 47\\times 46}{3!} = 48\\times 47\\times 46 ways.$ Ooh very nice, I think you should explain to them what the binomial expressions mean (general doesn't cover it). #### InteGrand ##### Well-Known Member $\\bg_white \\noindent OK. For non-negative integers n and k\\leq n, we can define the", "46}{3!} = 48\\times 47\\times 46 ways.$ fluffchuck Active Member You've done your answers assuming there's an order that matters for the cards, but for a hand, the order doesn't matter, so we should take this into account. $\\bg_white \\noindent b) Pick the Ace: 4 ways.$ $\\bg_white \\noindent Pick the remaining four cards: We need to pick four cards from the remaining 48 non-Ace cards, where order is irrelevant, so this can be done in \\binom{48}{4}=\\frac{48\\times 47\\times 46\\times 45}{4!} ways.$ $\\bg_white \\noindent Hence the answer is 4\\times \\binom{48}{4} = 4\\times \\frac{48\\times 47\\times 46\\times 45}{4!}.$ $\\bg_white \\noindent c) Pick the two Aces (from a set of four Aces, with order irrelvant): \\binom{4}{2}=\\frac{4\\times 3}{2!}=6 ways.$ $\\bg_white \\noindent Pick the remaining three cards: We need to pick three cards from the remaining 48 non-Ace cards, where order is irrelevant, so this can be done in \\binom{48}{3}=\\frac{48\\times 47\\times 46}{3!} ways.$ $\\bg_white \\noindent So the answer is \\binom{4}{2}\\times \\binom{48}{3} = 6\\times \\frac{48\\times 47\\times 46}{3!} = 48\\times 47\\times 46 ways.$ Ooh very nice, I think you should explain to them what the binomial expressions mean (general doesn't cover it). InteGrand Well-Known Member $\\bg_white \\noindent OK. For non-negative integers n and k\\leq n, we can define the \\textsl{binomial coefficient} \\binom{n}{k} (read n choose k'', with one alternative notation being ^n C_k, which is why you will", "# Screws The box contains 8 iron, 6 brass, and 4 titanium screws. What is the probability that the randomly selected screw will not be brass? Correct result: p = 0.6667 #### Solution: $p=\\frac{8+4}{8+6+4}=\\frac{2}{3}=0.6667$ We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! Tips to related online calculators Would you like to compute count of combinations? ## Next similar math problems: • Balls The urn is 8 white and 6 black balls. We pull 4 randomly balls. What is the probability that among them will be two white? • Study results There are 7 students in the class with excellent results, 6 praiseworthy, 5 good, 4 sufficient and 3 insufficient students. What is the probability that it will be a good student when summoned? • Bulbs In the box are 6 bulbs with power 75 W, 14 bulbs with power 40 W and 15 with 60 W. Calculate probability that a randomly selected bulb is: • Rectangle In a rectangle with sides, 6 and 3 mark the diagonal. What is the probability that a randomly selected point within the rectangle is closer to the diagonal than to any side of the rectangle? • Dices We will throw two dice. What is the probability", "$WBBB$, $WWBB$, or $WWWB$. Notice we can't have 4 white balls since there are only a total of 3 white balls to choose from. So our $30$ configurations we counted previously must be split up between these 4 descriptions. So how many of each kind are there? $BBBB$: The only way this can happen is if we select two black from the first box and two black from the second box. So there are $\\binom{2}{2}\\binom{3}{2}=3$ ways for this to happen. $WBBB$: 1.) Select one white/one black from the first box and two black from the second box for a total of $2\\cdot \\binom{3}{2}=6$ ways. or 2.) Two black from the first and one white/one black from the second box for a total of $1\\cdot (3\\cdot 2)=6 ways. Thus there are a total of$6+6=12$ways for this to happen.$WWBB$: 1.) Select one white/one black from the first box and one white/one black from the second box for a total of$2\\cdot (3\\cdot 2)=12$ways. 2.) Two black from the first and two white from the second box for a total of$1\\cdot\\binom{2}{2}=1$ways. Thus there are a total of$12+1=13$ways for this to happen.$WWWB$: Select one white/one black from the first box and two white from the second box for a total of$2\\cdot \\binom{2}{2}=2$ways. From the above work we know that$P(BBBB)=\\frac{3}{30}$,$P(WBBB)=\\frac{12}{30}$,$P(WWBB)=\\frac{13}{30}$, and$P(WWWB)=\\frac{2}{30}\\$ So the probability", "\\\\ P(BB\\text{, Box 2})=&\\frac{{15\\choose2}}{{25\\choose2}}=&\\frac{105}{300}\\end{Bmatrix}=$ [color=beige]. . [/color]$\\text{Hence: }\\:P(WB|BB) \\:=\\:\\frac{96}{190}\\,\\cdot\\,\\frac{105}{300} \\:=\\:\\frac{84}{475}$ $[2]\\;BB\\text{ is drawn from Box 1, }\\,WB\\text{ is drawn from Box 2.}$ [color=beige]. . [/color]$\\text{The probabilities are: }\\:\\begin{Bmatrix}P(BB\\text{, Box 1})=&\\frac{{8\\choose2}}{{20\\choose2}}=&\\frac{28}{190} \\\\ \\\\ \\\\ \\\\ P(WB\\text{, Box 2})=&\\frac{{10\\choose1}{15\\choose1}}{{25\\choose2}}=&\\frac{150}{300}\\end{Bmatrix}=$ [color=beige]. . [/color]$\\text{Hence: }\\:P(BB|WB) \\:=\\:\\frac{28}{190}\\,\\cdot\\,\\frac{150}{300} \\:=\\:\\frac{35}{475}$ $\\text{Therefore: }\\:P(\\text{exactly 1 White}) \\;=\\;\\frac{84}{475}\\,+\\,\\frac{35}{475} \\;=\\;\\frac{119}{475}$ July 23rd, 2012, 02:40 PM #3 Senior Member Joined: Jan 2012 Posts: 579 Thanks: 3 Re: The Probability That Exactly One Of The Four Balls Is Wh Don't be offended please, I cannot see the working clearly. July 23rd, 2012, 05:31 PM #4 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 4,284 Thanks: 254 Re: The Probability That Exactly One Of The Four Balls Is Wh 1:12W,8B ; 2:10W,15B The White can be drawn 1st, 2nd, 3rd or 4th : so 4 different cases : Case1: W=12/20, B=8/19, B=15/25, B=14/24: 12/20 * 8/19 * 15/25 * 14/24 = 42/475 Case2: B=8/20, W=12/19, B=15/25, B=14/24: 8/20 * 12/19 * 15/25 * 14/24 = 42/475 Case3: B=8/20, B=7/19, W=10/25, B=15/24: 8/20 * 7/19 * 10/25 * 15/24 = 7/190 Case4: B=8/20, B=7/19, B=15/25, W=10/24: 8/20 * 7/19 * 15/25 * 10/24 = 7/190 42/475 + 42/475 + 7/190 + 7/190 = 119/475 (as per Sir Soroban!) July 24th, 2012, 09:53 PM #5 Senior Member Joined: Jan 2012 Posts: 579", "arrange the 7 white balls first. _W_W_W_W_W_W_W_ 5 black balls have 8 spaces inbetween, so arrangements $$= \\frac{8*7*6*5*4}{5!} = 56$$ 3) B $$\\frac{13^4*48*47}{2}$$ .. we divide by two to avoid repetition of the last two cards i.e. ab or ba are the same in this context. 4) B we have to divide it into two cases i.e. 4-digit and 5-digit because 1034 and 10345 are both greater than 1000 and we can't put a condition that 0 can't be placed on 2nd position. 4-digit: 4*4*3*2 = 96 (0 can't be the first digit) 5-digit: 4*4*3*2*1 = 96 (0 can't be the first digit) total = 96*2 5) C at least one out of n is $$2^n-1$$. total mocktails = at least one whiskey * at least one soda * at least one juice $$= 2^3-1*2^4-1*2^5-1$$ $$=3255$$ _________________ press +1 Kudos to appreciate posts Manager Joined: 25 May 2011 Posts: 163 Followers: 2 Kudos [?]: 45 [0], given: 71 MBAhereIcome wrote: 1.1) i like to write is as $$\\frac{5!}{5} = 4! = 24$$ 1.2) AxB or BxA, so same neighbours are counted twice. that's why ans $$= \\frac{4!}{2} = 12$$ 2) A arrange the 7 white balls first. _W_W_W_W_W_W_W_ 5 black balls have 8 spaces inbetween, so arrangements $$= \\frac{8*7*6*5*4}{5!} = 56$$ 3) B $$\\frac{13^4*48*47}{2}$$ .. we divide by", "of choosing pairs of $$s_1$$ as white and $$s_2$$ as black square $$= |\\{ (s_1, s_2) | s_1 \\in W, s_2 \\in B \\}| \\\\ = |\\{ s_1| s_1 \\in W \\} \\times \\{ s_2| s_2 \\in B \\}| \\ [ \\ \\because W \\cap B = \\emptyset \\ ] \\\\ = |\\{ s_1| s_1 \\in W \\}| \\times |\\{ s_2| s_2 \\in B \\}| \\\\ = |W| \\times |B| = \\frac{|S|}{2} \\times \\frac{|S|}{2} = \\frac{|S|^2}{4}$$ Since order of choosing does not matter, Ways of choosing pairs of squares of different color = Ways of choosing pairs of $$s_1$$ as white and $$s_2$$ as black square $$= \\frac{|S|^2}{4} = \\frac{64^2}{4} = 1024$$ For each black square you can choose one of the 32 whites squares(32 possibilities), then, for the remaining 31 black and 31 white squares, you repeat the process (31 possibilities $$\\times$$ the first 32 possibilities) ... . So the result should be $$32!$$. Since there are 64 squares, 32 are black and 32 are white. You can choose a black square out of 32 in $${32}\\choose{1}$$ways and the same for the white squares. Then: $${32}\\choose{1}$$$${32}\\choose{1}=32$$$$32=1024$$ ways of choosing a pair of squares (one white and the other black) out of 64.", "of black balls = x Then number of possible outcome = x + 30 Probability of drawing a black ball = $$\\frac { 2 }{ 5 }$$ of a white balls. Now, In case of black ball, Either x + 30 = 0, then x = -30 which is not possible or x – 12 = 0, then x = 12 Hence number of black balls = 12 Question 32. From a pack of 52 playing cards all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is (i) a face card (King, Jack or Queen) (ii) an even numbered red card ? (2011) Solution: No. of total cards = 52 cards removed of 4 colors = 3, 6, 9, 12 = 4 x 4 = 16 Remain using cards = 52 – 16 = 36 (i) No. of faces cards = 2 x 4 = 8 cards (excluding queen) ∴ Probebility P(E) = $$\\frac { 8 }{ 36 }$$ = $$\\frac { 2 }{ 9 }$$ (ii) An even number red cards = 2, 4, 8, 10 = 4×2 = 8 cards ∴ Probebility P(E) = $$\\frac { 8 }{ 36 }$$ = $$\\frac { 2 }{ 9 }$$ Question 33. A", "of black balls = x Then number of possible outcome = x + 30 Probability of drawing a black ball = $$\\frac { 2 }{ 5 }$$ of a white balls. Now, In case of black ball, Either x + 30 = 0, then x = -30 which is not possible or x – 12 = 0, then x = 12 Hence number of black balls = 12 Question 32. From a pack of 52 playing cards all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is (i) a face card (King, Jack or Queen) (ii) an even numbered red card ? (2011) Solution: No. of total cards = 52 cards removed of 4 colors = 3, 6, 9, 12 = 4 x 4 = 16 Remain using cards = 52 – 16 = 36 (i) No. of faces cards = 2 x 4 = 8 cards (excluding queen) ∴ Probebility P(E) = $$\\frac { 8 }{ 36 }$$ = $$\\frac { 2 }{ 9 }$$ (ii) An even number red cards = 2, 4, 8, 10 = 4×2 = 8 cards ∴ Probebility P(E) = $$\\frac { 8 }{ 36 }$$ = $$\\frac { 2 }{ 9 }$$ Question 33. A", "is drawn from Box 2.}$ [color=beige]. . [/color]$\\text{The probabilities are: }\\:\\begin{Bmatrix}P(BB\\text{, Box 1})=&\\frac{{8\\choose2}}{{20\\choose2}}=&\\frac{28}{190} \\\\ \\\\ \\\\ \\\\ P(WB\\text{, Box 2})=&\\frac{{10\\choose1}{15\\choose1}}{{25\\choose2}}=&\\frac{150}{300}\\end{Bmatrix}=$ [color=beige]. . [/color]$\\text{Hence: }\\:P(BB|WB) \\:=\\:\\frac{28}{190}\\,\\cdot\\,\\frac{150}{300} \\:=\\:\\frac{35}{475}$ $\\text{Therefore: }\\:P(\\text{exactly 1 White}) \\;=\\;\\frac{84}{475}\\,+\\,\\frac{35}{475} \\;=\\;\\frac{119}{475}$ Chikis July 23rd, 2012 02:40 PM Re: The Probability That Exactly One Of The Four Balls Is Wh Don't be offended please, I cannot see the working clearly. Denis July 23rd, 2012 05:31 PM Re: The Probability That Exactly One Of The Four Balls Is Wh 1:12W,8B ; 2:10W,15B The White can be drawn 1st, 2nd, 3rd or 4th : so 4 different cases : Case1: W=12/20, B=8/19, B=15/25, B=14/24: 12/20 * 8/19 * 15/25 * 14/24 = 42/475 Case2: B=8/20, W=12/19, B=15/25, B=14/24: 8/20 * 12/19 * 15/25 * 14/24 = 42/475 Case3: B=8/20, B=7/19, W=10/25, B=15/24: 8/20 * 7/19 * 10/25 * 15/24 = 7/190 Case4: B=8/20, B=7/19, B=15/25, W=10/24: 8/20 * 7/19 * 15/25 * 10/24 = 7/190 42/475 + 42/475 + 7/190 + 7/190 = 119/475 (as per Sir Soroban!) Chikis July 24th, 2012 09:53 PM Re: The Probability That Exactly One Of The Four Balls Is Wh My thanks goes to soroban and Denis, for having intrest in my question, let me work with what you have given me. I will post more replies into the thread, if I found anything", "which you have no information about. You draw two balls and both are white. What is the probability that all five of the balls are white ? We know there are at least 2 Whites among the 5 balls. I assume that the assortment colors of the balls are equally likely. $\\text{Let: }\\,W\\,=\\,\\text{white ball, }\\:A\\,=\\,\\text{another color}$ $\\text{Then: }\\;P(5W)\\:=\\:P(4W,\\,1A) \\:=\\:P(3W,\\,2A) \\:=\\:P(2W,\\,3A) \\:=\\:\\frac {1}{4}$ $\\text{Bayes' Theorem: }\\:P(5W\\,|\\,2W\\text{drawn}) \\;=\\;\\frac{P(5W\\,\\wedge\\,2W\\text{drawn})}{P(2W\\tex t{drawn})}\\;\\;[1]$ $\\text{Numerator:}$ [color=beige]. . [/color]$P(5W\\,\\wedge\\,2W\\text{drawn}) \\;=\\;\\left(\\frac{1}{4}\\right)(1) \\;=\\;\\frac{1}{4}\\;\\;[2]$ $\\text{Denominator: }\\;P(2W\\text{drawn})$ [color=beige]. . [/color]$P(5W\\,\\wedge\\,2W\\text{drawn})\\;=\\;\\left(\\frac{1}{4 }\\right)(1) \\;=\\;\\frac{1}{4}$ [color=beige]. . [/color]$P(4W,1A\\,\\wedge\\,2W\\text{drawn}) \\;=\\;\\left(\\frac{1}{4}\\right)\\left(\\frac{{4\\choose 2}{1\\choose1}}{{5\\choose2}}\\right) \\;=\\;\\left(\\frac{1}{4}\\right)\\left(\\frac{3}{5}\\rig ht) \\;=\\;\\frac{3}{20}$ [color=beige]. . [/color]$P(3W,2A\\,\\wedge\\,2W\\text{drawn}) \\;=\\;\\left(\\frac{1}{4}\\right)\\left(\\frac{{3\\choose 2}{2\\choose0}}{{5\\choose2}}\\right) \\;=\\;\\left(\\frac{1}{4}\\right)\\left(\\frac{3}{10}\\ri ght) \\;=\\;\\frac{3}{40}$ [color=beige]. . [/color]$P(2W,3A\\,\\wedge\\,2W\\text{drawn}) \\;=\\;\\left(\\frac{1}{4}\\right)\\left(\\frac{{2\\choose 2}{3\\choose0}}{{5\\choose2}}\\right) \\;=\\;\\left(\\frac{1}{4}\\right)\\left(\\frac{1}{10}\\ri ght) \\;=\\;\\frac{1}{40}$ $\\text{Hence: }\\:P(2W\\text{drawn}) \\;=\\;\\frac{1}{4}\\,+\\,\\frac{3}{20}\\,+\\,\\frac{3}{40} \\,+\\,\\frac{1}{40} \\;=\\;\\frac{30}{40} \\;=\\;\\frac{3}{4}\\;\\;[3]$ $\\text{Substitute [2] and [3] into [1]: }\\;P(5W\\,|\\,2W\\text{drawn}) \\;=\\;\\frac{\\frac{1}{4}}{\\frac{3}{4}} \\;=\\;\\frac{1}{3}$ January 19th, 2010, 12:51 PM #4 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Rule of succession problem Quote: Originally Posted by soroban $\\text{Then: }\\;P(5W)\\:=\\:P(4W,\\,1A) \\:=\\:P(3W,\\,2A) \\:=\\:P(2W,\\,3A) \\:=\\:\\frac {1}{4}$ This strikes me as a questionable assumption, though. Why not assume, say, that the prior probability that a ball is white is uniformly distributed on [0, 1]? That is, there is a universe with Bp white balls and B(1-p) nonwhite balls for B very large, and then the five balls are chosen with", "# Monthly Archives: February 2012 ## Probability musings From Futility Closet: A bag contains 16 billiard balls, some white and some black. You draw two balls at the same time. It is equally likely that the two will be the same color as different colors. What is the proportion of colors within the bag? A solution appears on the page, but it doesn’t suggest how it was determined or if the solution is unique. I’ll provide that below plus an extension. Assume there were w white balls originally, then the number of ways to draw two white is $\\displaystyle \\frac{w*(w-1)}{2!}$, two black is $\\displaystyle \\frac{(16-w)*(15-w)}{2!}$, and one of each is $\\displaystyle \\frac{w*(16-w)}{1!1!}$. As the likelihood of drawing one of each color is to be the same as drawing only one color, $\\displaystyle \\frac{w*(w-1)}{2}+ \\frac{(16-w)*(15-w)}{2}= w*(16-w)$ This is a quadratic equation with solutions $w=\\{6,10\\}$, verifying Futility Closet’s solution, and establishing those solutions as the only ones possible. Extension: For what total number(s) of white and black billiard balls is the likelihood for two drawn balls to be one of color the same as them being one of each? Let n be the total number of balls. Repeating the procedure above, the number of ways to draw two white is $\\displaystyle \\frac{w*(w-1)}{2!}$, two black is $\\displaystyle \\frac{(n-w)*((n-1)-w)}{2!}$, and one of", "1/3 * 2/5 ) / [ (1/3 * 2/5) + (2/3 * 3/4) ] = 4/19 We can solve this question with the help of probability tables $\\rightarrow$ $Red$ $Blue$ $P$ $\\color{red}{\\frac{1}{3} \\times \\frac{2}{5}}$ $\\frac{1}{3}\\times\\frac{3}{5}$ $\\frac{1}{3}$ $Q$ $\\frac{2}{3} \\times \\frac{3}{4}$ $\\frac{2}{3}\\times\\frac{1}{4}$ $\\frac{2}{3}$ $\\color{green}{\\frac{19}{30}}$ $\\frac{11}{30}$ $1$ Given that a ball selected in the above process is a red ball, the probability that it came from the box $P$ $= \\frac{\\color{red}{\\frac{1}{3} \\times \\frac{2}{5}}}{\\color{green}{\\frac{19}{30}}} = \\frac{4}{19}$ 1 2", "Solution: Two cards are drawn in succession from a deck of 52 cards. There are 6 denominations which are greater than 2 and less than 9. So there are 24 cards whose denominations are greater than 2 and less than 9. ∴ Their probability = $$\\frac{24}{52}$$ × $$\\frac{23}{51}$$. Question 16. From a bag containing 5 black and 7 white balls, 3 balls are drawn in succession. Find the probability that (i) all three are of the same colour. (ii) each colour is represented. Solution: From a bag containing 5 black and 7 white balls, 3 balls are drawn in succession. (i) The 3 balls drawn are of same colour. ∴ Probability of drawing 3 balls of black colour = $$\\frac{5}{12}$$ × $$\\frac{4}{11}$$ × $$\\frac{3}{10}$$ = $$\\frac{1}{22}$$ Probability of drawing 3 white balls = $$\\frac{7}{12}$$ × $$\\frac{6}{11}$$ × $$\\frac{5}{10}$$ = $$\\frac{7}{44}$$ ∴ Probability of drawing 3 balls of same colour = $$\\frac{5}{12}$$ × $$\\frac{4}{11}$$ × $$\\frac{3}{10}$$ + $$\\frac{7}{12}$$ × $$\\frac{6}{11}$$ × $$\\frac{5}{10}$$ = $$\\frac{9}{44}$$ (ii) Balls of both colour will be drawn. If B represents black ball and W represents the white ball. ∴ The possible draws are WWB, WBW, BWW. Question 17. A die is rolled until a 6 is obtained. What is the probability that (i) you end up in the second roll (ii) you end up", "# Screws The box contains 8 iron, 6 brass, and 4 titanium screws. What is the probability that the randomly selected screw will not be brass? p = 0.6667 ### Step-by-step explanation: $p=\\frac{8+4}{8+6+4}=\\frac{2}{3}=0.6667$ Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators Would you like to compute count of combinations? ## Related math problems and questions: • Balls The urn is 8 white and 6 black balls. We pull 4 randomly balls. What is the probability that among them will be two white? • Study results There are 7 students in the class with excellent results, 6 praiseworthy, 5 good, 4 sufficient, and 3 insufficient students. What is the probability that it will be a good student when summoned? • The box The box contains three light bulbs with a wattage of 40 W and two pieces of light bulbs with a wattage of 60 W. What is the probability of the event that two randomly selected light bulbs will both be 40 W? • Bulbs In the box are 6 bulbs with power 75 W, 14 bulbs with power 40 W and 15 with 60 W. Calculate probability that a randomly selected bulb is: • A box A box contains 6 red balls, 5 blue, and 4", "from thirty of which exactly twenty are defective. What is the probability that not all four are defective nor all four are non-defective\" you fully understood my question. It's ok I believe the given result is wrong (Y) 11. Originally Posted by javax Plato one more question. As I mentioned I tried to find it using $1-p_1^4*p2^4,$ where $p_1=\\frac{2}{3}$ and $p_2=\\frac{1}{3}$ which is quite close to your result. Is it ok finding it like this? Unless the trials, the events, are independent the binomial formula is not applicable. Selecting four objects from thirty is in no way independent. But say we have batch twenty black balls and ten white balls. We pick a random ball from that collection. Record its color and return it to the batch. We do that four times. Those outcomes are independent.", "PDA View Full Version : sat math probability permutations Jack 10-10-2013, 10:33 AM In how many different ways can 2 black boxes, 4 white boxes, 3 blue boxes and 2 brown boxes be placed in a row? (A) 48 (B) 576 (C) 1200 (D) 69300 Thanks miranda 10-10-2013, 10:34 AM In this case we have permutations since the order matters. Additionally we must divide the number of permutations by 2! , 4!, 3! and 2! Since there are 2 black boxes, 4 white boxes, 3 blue boxes and 2 brown boxes. Therefore the answer is \\frac{^11P_{11}}{(2!) \\cdot(4!) \\cdot(3!) \\cdot(2!)}= \\frac{11!}{576}=69300 (D). Jack 10-10-2013, 10:39 AM Thank you!!" ]

[ "Subset Subset A subset is a collection of numbers or objects within a larger set.", "think I can figure it out for a five digit based on your help. Thank you! or else if you have the time, can you explain it to me with a 5 digit? I get 5*5*C(8,3)*3! Make sense? Edit: I just learned about P(n, k) which gives the number of ordered k-subsets of {1,2,...,n}. This is in fact the same as C(n, k) * k! which I got above. But if you wanted to you could express as such: 5*5*P(8,3) Saves time", "+0 # I need help!!! 0 213 1 How many non-empty subsets of \\$\\{ 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \\}\\$ consist entirely of prime numbers? (We form a subset of the group of numbers by choosing some number of them, without regard to order. So, \\$\\{1,2,3\\}\\$ is the same as \\$\\{3,1,2\\}\\$.) Guest Sep 21, 2017 Sort: #1 +19084 +1 How many non-empty subsets of \\(\\{ 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \\}\\) consist entirely of prime numbers? (We form a subset of the group of numbers by choosing some number of them, without regard to order. So, \\(\\{1,2,3\\}\\) is the same as \\(\\{3,1,2\\}\\).) The set of prime numbers \\( \\{ 2, 3, 5, 7, 11 \\}\\) consist of 5 elements. The subsets with one element \\( = \\binom{5}{1} = 5 \\) subsets. The subsets with two elements \\(= \\binom{5}{2} = 10\\) subsets. The subsets with three elements \\(= \\binom{5}{3} = 10\\) subsets. The subsets with four elements \\(= \\binom{5}{4} = 5\\) subsets. The subsets with five elements \\(= \\binom{5}{5} = 1\\) subset. The sum is 5+10+10+5+1 = 31 subsets heureka Sep 21, 2017 ### 12 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic.", "\u0003 \u000e \u0015 \u0003 \u0007 \u0017 \b \u001a \u0014 \u0014 \u001a ( \u0017 \u0007 \u0002 \u000e \u0003 \u0007 \u0002 \u0006 \u0003 \u000e \u0015 \u001a \u0017 \u0003 \u0006 \u001b \u0002 \u0006 \u0007 \u0002 \u000e \u0013 # \u0015 \u001a \u0003 \u0006 \u0006 \u001a \u001b \u0007 \u000e \u001a \u0018 \u0002 \u0003 \u001b \u0017 \u0014 \u0018 \u0002 \b \u001b \u0018 \u001a \u0002 \u001b \u001a \u0006 \u0006 \u0003 \u001a \u000e \u0015 \u0003 \u000e \u0015 \u0003 \u000e \u0017 \u0007 \u0006 \u001a \u000e \u001a \u0018 \u0002 \u0007 \u0015 \u001a \u0002 ( \u0003 \u0019 \u001a \u0002 ( \u0003 5 \u000e \u001a \u0007 \u0003 \u001a \u0002 \u0003 \u0016 \u0017 \u0007 \u0002 \u0006 \u0018 \u0002 \u0007 \u0002 \u0006 \u0003 \u001a \u0002 \u001b \u0005 \u0006 \u0006 \u0003 \u0015 \u001a \u0002 ( \u0003 \u0019 \u001a \u0002 ( \u0003 , \u0018 \u0018 \u0006 \u0018 , , \u0017 \u001a \u0002 ( \u0003 \u000e \u0015 \u0003 + \u0005 \u0017 \u0002 \u001a \u0002 ( \u0003 \u0018 , \u0003 \u001a \u0002 \u001b \u0002 \u0007 \u0002 \u0006 \u0003 \u000e \u0018 * \u0002 \u0003 \u001b \u0005 \u0017 \u0017 \u0002 \u001b \b \u0003 \u001b \u0007 \u0017 \u0017 \u001a \u0006 \u0003 \u0018 \u0005 \u000e \u0003 \u001a \u0002 \u0002 \u0003 \u0004 \u0003 \u0005 \u0006 \u0007 \b \b \u0004 \u000e \u0006 \u0002 \u000f \u0003 \u0010 \u0003 \u0006 \u0011 \u0012 \u0013 \u000f \u0014 \u0003 \u000e \u0003 \u0006 \u0015 \u0003 \b \u0016 \u0017", "{0, 0, 3, 4, 5} -", "# Math Help - Subsets 1. ## Subsets This is my question, i can do the first part but then i need help with the rest please. How many ordered 5-subsets of {0,...,9} are there? How many of these start with 0? How many positive integers of exactly 5 digits, but with no two digits the same, are there? (In base 10.) 2. ## Re: Subsets Originally Posted by jonnyl How many ordered 5-subsets of {0,...,9} are there? How many of these start with 0? How many positive integers of exactly 5 digits, but with no two digits the same, are there? (In base 10.) These are permutations: $_NP_k=\\frac{N!}{(N-k)!}$ 3. ## Re: Subsets So how would i use this to do the rest of the question? 4. ## Re: Subsets Originally Posted by jonnyl So how would i use this to do the rest of the question? You should show some effort here. What answer do you get for the first part? 5. ## Re: Subsets I got 30240 for the first part, then 3024 for the second. But i dont really know how to do the last bit 6. ## Re: Subsets Originally Posted by jonnyl I got 30240 for the first part, then 3024 for the second. But i dont really know how to do the", "order. The number of subsets of $\\{n-1,\\ldots,0\\}$ is exactly $2^n$. • I really appreciate your experience. – Navjot Singh Mar 28 '18 at 15:40", "last bit Those are correct. 7. ## Re: Subsets so how would i go about doing the last part? 8. ## Re: Subsets Originally Posted by jonnyl so how would i go about doing the last part? It depends on how one understands the question. We all know that 70412 is a five digit number with all digits different. But would 09876 also count? It is five digits all of which are different. If we do not allow a leading zero then the answer is: $_{10}P_5 -~_9P_4$. 9. ## Re: Subsets But surely _{10}P_5-~_9P_4 is inclusive of numbers with repeated digits? 10. ## Re: Subsets Originally Posted by jonnyl But surely _{10}P_5-~_9P_4 is inclusive of numbers with repeated digits? No, it most certainly does not. If you do that calculation, you will see the result is just the answer in part 1 minus the answer in part 2.", "finite number of subsets.", "a subset of $(-1,1)$.", "# A Pre-RMO question! -22 Probability Level 1 If the number of subsets of $\\{ 1,2,3,4,5,6,7,8,9 \\}$ that are subsets of neither $\\{ 1,2,3,4,5 \\}$ nor $\\{ 4,5,6,7,8,9 \\}$ is $k$, find the sum of the digits of $k$. ×", "15$ $\\therefore$ The number of groups of three r more numbers $= 16+16+15 = 47.$ Correct Answer $: 47$ $\\textbf{Short method:}$ Given that, set of numbers $\\{1,2,3,4,5,6,7,8\\}$ From the above set, $\\{3,5\\}$ is taken, then the number of possible subsets containing at least three numbers $= 2^{6}-1 = 64-1 = 63$ From these $63,$ we need t remove the subsets which have $\\{3,5,7,8\\}.$ The remaining numbers are $\\{1,2,4,6\\}.$ For these numbers, we have two possibilities, either selected or not selected. So, the number of subsets possible $= 2^{4} = 16$ $\\therefore$ The total number of ways $= 63-16=47.$ 10.3k points 1 vote 1 297 views 1 vote", "always be 5, $\\E[5] = 5$.", "1. ## Combinatorics I'm stumped on this recreational problem: How many 5-digit numbers are there such that every next digit is smaller than the previous one? 2. Originally Posted by elninio I'm stumped on this recreational problem: How many 5-digit numbers are there such that every next digit is smaller than the previous one? $\\binom{10}{5}$ 3. Hmm, I didnt think it would be that simple! 4. Originally Posted by elninio Hmm, I didnt think it would be that simple! Well give me any subset of five digits. Say $\\{3,9,0,7,4\\}$. Arrange is desending order: $97430$. That's sort of number described. So it is just the number of five element subsets of the digits.", "# Which multiple of 5 is missing? This file contains the multiples of 5 between 5 and 955, inclusive. One of the multiples of 5 is missing though. Which number is missing?", "do not know what the OP means. Is that an exact wording or is it a translation? 5. It's a translation. Let me paraphrase. We have all the possible combinations of five numbers from 1 to 35 (35c5) contained in combinations of six numbers. for example: 1 2 3 4 5 6 (contains {1,2,3,4,5} {1,2,3,4,6} {1,2,3,5,6} {1,2,4,5,6} {1,3,4,5,6} {2,3,4,5,6}) What I want to know is how many of these -six- number combinations contain -five- specific numbers. Logically its 30. I'm confused however with the fact that every >1< combination of 6 contains 6 combinations of five!! 6. There are six combinations of six objects taken five at a time. $\\dbinom{6}{5}=\\dfrac{6!}{5!\\cdot 1!}=6$ The wording is confusing. All that emphasis on subsets of 5 numbers . . . Given are all possible 6-number subsets of numbers from 1-35, how many of these subsets contain 5 specific numbers? What do you mean by \"logically it's 30\"? .Is that the correct answer? Here's my reasoning . . . The is one way to get the five specific numbers. The sixth number can be any of the remaining 30 numbers.", "number of such sets $Y$ is $\\,\\binom52\\cdot 2^4$.", "Is it a Subset? Discrete Mathematics Level 1 $$A = \\{6,2,4,1,5\\}$$. Which of the options below is not a proper subset of $$A$$? • $$\\{2\\}$$ • $$\\{1,2,3\\}$$ • $$\\{6,5,2,1\\}$$ • $$\\{1,2,4,5\\}$$ ×", "Mathematics # Express the set in roster form. {x|x is a natural number multiple of 5} $\\{x|x\\in\\mathbb{N}\\ and\\ multiple\\ of\\ 5\\}=\\{0;\\ 5;\\ 10;\\ 15;\\ 20;...\\}=5k\\ (k\\in\\mathbb{N})$", "5 7 10 14 45 subsets evaluated in total ```" ]

[ "# The number of subset sum If we have that $S=\\{1,2,3,4,5,6,7,8,9,10\\}$,how to know the number of subset sum T from :S in which for every $x\\in T$ and for every $2x\\in S$ then: $2x \\in T$ - sorry $x$ and $2*x$ are member of $T$?or ? – dato datuashvili Aug 25 '12 at 21:28 ## 1 Answer The contribution from 1, 2, 4, 8 could be 0, 8, 12, 14, or 15; the contributiom from 3, 6 could be 0, 6, or 9; from 5, 10, could be 0, 10, or 15; then you have 0 or 7, and 0 or 9. So you pick one number from each of the groupings 0, 8, 12, 14, 15; 0, 6, 9; 0, 10, 15; 0, 7; 0, 9; and add the five numbers you've chosen, and see how many different sums you get (if I'm interpreting the problem correctly). The most you can get is 55; the least you can get, other than zero, is 6; there's no way to get 11; can you get every number between 12 and 55? No, there's no way to get 53. I don't see a clever way to see what you get and what you don't without actually trying all the possibilities. -", "the # of 5 digit numbers with only 1's and 2's: Such a number can have one 1, two 1's, three 1's, or four 1's. (All others being 2) How many ways can it have one 1? You have to choose 1 position out of 5 possible positions, so $\\binom{5}{1}=5$. How many ways can it have two 1's? You have to choose 2 positions out of 5 possible positions, so $\\binom{5}{2}=10$. Similarly for the others, we have 10 and 5. So there are 30. Do the same for all others, subtract, and you have your result. (150) If you don't like this approach, you can try counting them directly by splitting them into different cases. two 1's, two 2's, one 3 two 1's, two 3's, one 2 two 2's, two 3's, one 1 one 1, one 2's, three 3's one 2, one 3, three 1's one 1, one 3, three 2's. Of course, the first three are equal, and the last three are equal. I'm pretty sure the first way is the simplest way to do it though. I could be wrong.", "# Number of subsets of size 5 of a set of 25 numbers. The Set is S={1,2,3,.....,25}. We have to count the number of subsets of size 5 such that each has atleast one Odd number in it. Method 1: Count = (Total Subsets of size 5) - (Total subsets having all even numbers) $$= \\binom {25} {5} - \\binom {12} {5} = 53130 - 792 = 52338$$ Method 2: Consider the 5 element subsets as 5 distinct boxes, each can be filled with one number. 1st box can be filled by any of the 13 Odd numbers. 2nd by the remaining 24. 3rd by remaining 23. 4th by remaining 22. 5th by remaining 21. $$\\text{Total} = 13 \\times{24} \\times{23} \\times{22} \\times{21}=3315312$$ Why am I getting two answers? Which method is flawed and how? • Keep in mind that conpliment to 'each' is $\\exist$ at least 1. You can have two sets with no odd numbers. – Alex Jan 19 '17 at 9:34 First method is correct Second method is false because you are counting several solutions multiple times. For example the set {1, 2, 3, 4, 6 } is counted at least twice : • a first time when you consider putting 1 into the first box • a second time when you consider putting 3 into", "digit 4 times. – user Mar 14 at 20:45 • Are you sure? It's not given in the question that how many digits the formed number should have. – M. Kumar Mar 15 at 6:48 You could partition the problem into smaller problems composing numbers from the 6 digits: 1. How many different numbers contain neither $$2$$ nor $$3$$ 2. How many different numbers contain just one of $$2$$ and $$3$$ 3. How many different numbers contain both $$2$$ and $$3$$ Each of these numbers can be obtained by counting the solutions by the number of $$1$$'s they contain: 1. The composed string of digits will consist of one up to four $$1$$'s, there are $$4$$ ways to do so. 2. There are $$2$$ ways to decide between $$2$$ and $$3$$, and for $$k=0,...,4$$ there are $$k+1$$ ways to place that digit between $$k$$ $$1$$'s. So, counting the possibilities, we get $$2\\sum_{k=0}^{4}(k+1) = 2\\sum_{k=0}^{4}\\binom{k+1}{1} = 2\\binom{4+2}{2} = 30$$. 3. For $$k=0,...,4$$, starting with $$k$$ $$1$$'s, we have $$k+1$$ possible positions to insert the $$2$$ and then $$k+2$$ possible positions to place the $$3$$. Summing them up, we get $$\\sum_{k=0}^{4}(k+1)(k+2) = \\sum_{k=0}^{4}\\binom{k+2}{2}2! = 2\\binom{4+3}{3} = 70$$. In total, we get $$4+30+70 = 104$$ possible ways to compose the number. • I think this is the same thing as I", "# Math Help - [SOLVED] Subset 1. ## [SOLVED] Subset Consider a group of 4 boys and 6 girls. In how many ways can one choose a subset of 2 boys and 3 girls? I know there is a quicker way of doing this than manually but i don't know the method. 2. Do you know the combination formula? Like in the binomial formula? $\\begin{pmatrix}n \\\\k \\end{pmatrix} = \\frac{n!}{(n-k)!k!}$ That gives you the number of ways to choose k things from a set of n elements when order does not matter. Clearly in a subset, the order of choosing the things doesn't make a difference. Do you think you can take it from here? 3. thanks for the formula, does the fact that the boys and girls are seperate things make the calculation change? Or do i use the formula for both i.e. $\\frac{4!}{(4-2)!2!}$ and then add? to $\\frac{6!}{(6-3)!3!}$ ? 4. ok, i used some logic to work out that 20+6 = 26 and there are way more than 26, 120 to be precise which is 20*6 . Thanks for the formula and telling me it was a combinatorics question! 5. Yup looks good to me. That's what I got when I did it. When you have n ways to do one thing, and m ways to do", "thirteen boys are not separate groups. • okay. got it. thank you so much.! you're a genius. Dec 4, 2016 at 11:30 There should be minimum $2$ perfects in the committee, there is no limit for maximum of perfects. So, The number of ways can be counted as: $1$. $2$ perfect and $6$ non perfect..... Number of ways become ${{5}\\choose{2}}\\times{{10}\\choose{6}}$. $2$. $3$ perfect and $5$ non perfect....... Number of ways become ${{5}\\choose{3}}\\times{{10}\\choose{5}}$. $3$. $4$ perfect and $4$ non perfect.....Number of ways become ${{5}\\choose{4}}\\times{{10}\\choose{4}}$. $4$. $5$ perfect and $3$ non perfect.....Number of ways become ${{5}\\choose{5}}\\times{{10}\\choose{3}}$. The total number of committees of $8$ is $\\binom{15}{8}$. The number of committees of $8$ with no prefects is equal to $\\binom{10}{8}$. The number of committees with exactly one prefect is equal to $\\binom{5}{1} \\cdot \\binom{10}{7}$. Then the answer to your question is $\\binom{15}{8} - \\binom{10}{8} - \\binom{5}{1} \\cdot \\binom{10}{7}=6435 - 45 - 5\\cdot 120 = 5790.$", "Trickier 2013 Mathcounts State Sprint Round questions : Sprint #14: From Varun: Assume the term \"everything\" refers to all terms in the given set. 1 is a divisor of everything, so it must be first. Everything is a divisor of 12, so it must be last. The remaining numbers left are 2, 3, 4, and 6. 2 and 3 must come before 6, and 2 must come before 4. Therefore, we can list out the possibilities for the middle four digits: 2,3,4,6 2,3,6,4 3,2,4,6 3,2,6,4 2,4,3,6 There are 5 ways--therefore 5 is the answer. From Vinjai: Here's how I did #14: First, notice that 1 must be the first element of the set and 12 must be the last one. So that leaves only 2,3,4,6 to arrange. We can quickly list them out. The restrictions are that 2 must be before 4, 3 must be before 6, and 2 must be before 6: 2,3,4,6 3,2,4,6 4,2,3,6 6,2,3,4 2,3,6,4 3,2,6,4 4,2,6,3 6,2,4,3 2,4,3,6 3,4,2,6 4,3,2,6 6,3,2,4 2,4,6,3 3,4,6,2 4,3,6,2 6,3,4,2 2,6,3,4 3,6,2,4 4,6,2,3 6,4,2,3 2,6,4,3 3,6,4,2 4,6,3,2 6,4,3,2 Only the bold ones work. So, the answer is 5. #17: Common dimensional change problem $$\\overline {ZY}:\\overline {WV}=5:8$$ -- line ratio The volume ratio of the smaller cone to the larger cone is thus $$5^{3}: 8^{3}$$. The volume of the frustum", "the set) will contain the 2-element subset. There are $\\dbinom{5}{2}=10$ ways to choose the 2-element subset. Thus, there are $10$ ways to create these sets. Second, the subsets can have 3 and 4 elements. $3+4=7$ non-distinct elements. $7-5=2$ elements in the intersection. There are $\\dbinom{5}{3}=10$ ways to choose the 3-element subset. For the 4-element subset, two of the elements must be the remaining elements (not in the 3-element subset). The other two have to be a subset of the 3-element subset. There are $\\dbinom{3}{2}=3$ ways to choose these two elements, which means there are 3 ways to choose the 4-element subset. Therefore, there are $10\\cdot3=30$ ways to choose these sets. $10+30=40 \\implies \\boxed{\\textbf{(B) 40}}$ ### Solution 3 We label the subsets subset 1 and subset 2. Suppose the first subset has $k$ elements where $k<5.$ The second element has $5-k$ elements which the first subset does not contain (in order for the union to be the whole set). Additionally, the second set has 2 elements in common with the first subset. Therefore the number of ways to choose these sets is $\\binom{5}{k} \\cdot \\binom{k}{2}.$ Computing for $k<5$ we have $10+30+30+10=80.$ Divide by 2 as order does not matter to get $40.$ ### Solution 4 (Stars And Bars) There are $\\dbinom{5}{2}=10$ ways to choose the 2 shared elements. We", "()(()), (()()), ()()(). 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Can you arrange the numbers$1, 2, \\ldots, 9$along a circle such that the sum of two neighbors is never divisible by 3, 5, or 7? Can you arrange the numbers$1, 2, \\ldots, 9$along a circle such that the sum of two neighbors is never divisible by 3, 5, or 7? 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Choosing two distinct subsets In how many ways can you choose two distinct subsets from a collection of$N$items? The two subsets together need not include all$N$items. 1-2 3-4 5-6 7-8 9-10 11-12 13-14 In how many ways can you choose two distinct subsets from a collection of$N$items? The two subsets together need not include all$N$items. In how many ways can you choose two distinct subsets from a collection of$N$items? The two subsets together need not include all$N$items. 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Subsets with no consecutive elements How many subsets of the set$\\{1, 2, 3, \\ldots, N\\}$contain no two successive numbers? 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Distributing billiard balls into pockets How many ways are there to put seven white and two black billiard balls into nine pockets? Some of the pockets may be empty and the pockets are considered distinguishable. 1-2 3-4 5-6 7-8 9-10 11-12", "C 1? – Addison Mar 1 '13 at 14:52 19 choose 1 (you've selected 4 of the 23 when choosing the first four elements). Note that $\\binom{19}{1} = 19$ – amWhy Mar 1 '13 at 14:56 One more thing. What if I wanted all combinations of the 5 elements chosen? Wouldn't I just multiply by (5!) ? – Addison Mar 1 '13 at 15:52 ...after choosing the 5 elements as described, if order matters, then you could compute all possible arrangements (permutations) of those five elements by computing $5!$. – amWhy Mar 1 '13 at 16:07 Okay so, if {a1, a2, b1, b2, c1} is one set, and {a2, a1, b1, b2, c1} is a set as well, isn't that just (10 C 2 * 8 C 2 * 17 * 5!)? – Addison Mar 1 '13 at 16:09 If you are supposed to choose 2 elements from $s_1$, 2 elements from $s_2$, and one element from any of the sets, you can at first choose 2 elements from $s_1$ in $\\binom{10}{2}$ ways then you choose 2 elements from $s_2$ in $\\binom{8}{2}$ ways . Now after choosing these four elements you are left with $23-4=19$ elements from which you have to choose 1 element , which you can do in $\\binom{19}{1}$ ways . So the total no. of", "# Four different sets of objects contain 3, 5, 6, and 8 objects, respectively. How many unique combinations can be formed by picking one object from each set? $3 \\cdot 5 \\cdot 6 \\cdot 8 = 720$ #### Explanation: Let's first answer this by simplifying the question a bit and ask \"How many different combinations can be made from 2 different sets of objects - one containing 3 and the other containing 5.\" So that gives us Set A with 3 things and Set B with 5 things. If I pull out the first object from Set A, let's call that Object A1, I can then pull out Objects B1, B2, B3, B4, and B5 - and so can make 5 different unique combinations. And then I can pull out Object A2 and do it again (so make 5 more unique combinations), and then A3 and do it yet again (for yet another 5 unique combinations). In the end, we end up with 15 unique combinations. We can find this number by multiplying the number of objects in A and the number in B: $3 \\cdot 5 = 15$ We can do that with larger numbers of sets and groups as well. For instance, in the question above, we have 4 sets with 3, 5, 6, and 8", "inspection, a viable set must contain at least one element from both of the sets $\\{1, 2, 3, 4, 5\\}$ and $\\{4, 5, 6, 7, 8\\}$. Since 4 and 5 are included in both of these sets, then they basically don't matter, i.e. if set A is a subset of both of those two then adding a 4 or a 5 won't change that fact. Thus, we can count the number of ways to choose at least one number from 1 to 3 and at least one number from 6 to 8, and then multiply that by the number of ways to add in 4 and 5. The number of subsets of a 3 element set is $2^3=8$, but we want to exclude the empty set, giving us 7 ways to choose from $\\{1, 2, 3\\}$ or $\\{6, 7, 8\\}$. We can take each of these $7 \\times 7=49$ sets and add in a 4 and/or a 5, which can be done in 4 different ways (by adding both, none, one, or the other one). Thus, the answer is $49 \\times 4=\\boxed{196}$. ## Solution 3 This solution is very similar to Solution $2$. The set of all subsets of $\\{1,2,3,4,5,6,7,8\\}$ that are disjoint with respect to $\\{4,5\\}$ and are not disjoint with respect to the complements of sets", "# Basic Counting Method - Number of Arrangements For a $4$-digit string, where each digit can be from $\\{0,1,..,9\\}$, what is the probability that the number contains exactly two different digits (e.g. $3555$)? The answer is $\\binom{10}{2}\\cdot (2^4-2)/10^4$. I couldn't get anything close to that answer. I'm guessing I'm under-counting, but I don't see what I'm missing. - The answer here suggests you must have 1 digit of one kind and 3 of another. While the problem suggests that two of each kind is acceptable (Such as 5566). See my answer below for that case. – Guest 86 Jan 21 '13 at 16:07 ## 3 Answers There are $\\binom{10}2$ ways to choose the two digits. The smaller digit can appear $1,2$, or $3$ times. There are $\\binom41$ ways in which it can appear once, $\\binom42$ ways in which it can appear twice, and $\\binom43$ ways in which it can appear three times, for a total of $\\binom41+\\binom42+\\binom43=14=2^4-2$ ways. (The $2^4-2$ can also be obtained directly by starting with the $2^4$ strings of length $4$ and eliminating the two that have one digit repeated four times.) The total number of acceptable strings is therefore $\\binom{10}2(2^4-2)$, as claimed. - To create a four-digit string with two different digits, you first need to pick two digits, and then decide how to", "15$ $\\therefore$ The number of groups of three r more numbers $= 16+16+15 = 47.$ Correct Answer $: 47$ $\\textbf{Short method:}$ Given that, set of numbers $\\{1,2,3,4,5,6,7,8\\}$ From the above set, $\\{3,5\\}$ is taken, then the number of possible subsets containing at least three numbers $= 2^{6}-1 = 64-1 = 63$ From these $63,$ we need t remove the subsets which have $\\{3,5,7,8\\}.$ The remaining numbers are $\\{1,2,4,6\\}.$ For these numbers, we have two possibilities, either selected or not selected. So, the number of subsets possible $= 2^{4} = 16$ $\\therefore$ The total number of ways $= 63-16=47.$ 10.3k points 1 vote 1 297 views 1 vote", "phi_ftw1618 2015-10-07 19:55:07 Isn't it just$4C2+4C3+4C4$=$6+4+1$=11? gamjawon 2015-10-07 19:55:07 4 choose 2 + 4 choose 3 + 4 choose 4 mihirb 2015-10-07 19:55:13 there are${4 \\choose 2}$2 element ones${4 \\choose 3}$3 element ones and${4 \\choose 4}$4 element ones DPatrick 2015-10-07 19:55:17 Exactly. vsny23 2015-10-07 19:55:21 11 b/c 4C2 plus 4C3 plus 4C4 adas1 2015-10-07 19:55:21 4C2 + 4C3 + 4C4 = 6+4+1=11 DPatrick 2015-10-07 19:55:28 To count 2-element subsets, we have to choose two of the four elements to form the subset. One way to express this is as the combination$\\binom42$, which is 6. DPatrick 2015-10-07 19:55:38 Another way to think of it is that there are 4 choices for the first element of the subset, and then 3 choices for the second element of the subset. So that's$4 \\cdot 3 = 12$ways to choose two elements. DPatrick 2015-10-07 19:55:44 But subsets don't care about order, so this method of choosing counts each subset twice: once for either order of choosing the two elements. Thus, when counting 12, we've double-counted, and the actual number of subsets is$12/2 = 6$. DPatrick 2015-10-07 19:56:04 Similarly, we can count$\\binom43 = 4$ways to choose 3 elements out of 4. (But it's easier to just think of which element we decide to leave out: we have 4 choices for the missing element.) DPatrick", "multiple ways to create an adapted formula. One way would be \"Count of times a single box has a single ball and the other two balls are in the same box PLUS the number of ways to arrange a single ball in each box\". (There is no situation where only two boxes get a single ball.) So I would say $${3\\choose1}{3\\choose1}{2\\choose1} + 3!\\\\18+6$$ • I see, how can I adapt it to not account the duplicated values? Thank you. May 23 '19 at 14:16 You overcount. Say you want to put ball $$b_i$$ into box $$B_i$$. Then you count this... 3 times: $$\\binom{3}{1}$$ ways of choosing a ball, so say we choose ball 1 and put it in box 1. Then 1 way to fill in the rest. $$\\binom{3}{1}$$ ways of choosing a ball, so say we choose ball 2 and put it in box 2. Then 1 way to fill in the rest. $$\\binom{3}{1}$$ ways of choosing a ball, so say we choose ball 3 and put it in box 3. Then 3 way to fill in the rest. Thus you counted this combination 3 times in total, not once, and hence you overcounted.", "3, 4, 6, 8)$ then you'll get $(d_1, d_2, d_3, d_4, d_5) = (1, 3, 3, 5, 6)$ and your five-digit number will be $13356$. And since $c_1, c_2, c_3, c_4, c_5$ are all distinct and between 1 and 11, inclusive, there are ${11 \\choose 5}$ ways to choose them. (It's interesting that the answer key you have gives ${11 \\choose 6}$. This is equal to ${11 \\choose 5}$ but I don't see a \"natural\" way to get ${11 \\choose 6}$ as the answer since you're choosing five numbers, not six.) • This is \"Method 2\" from N. F. Taussig's answer, which I didn't realize when I answered the question. – Michael Lugo Jul 23 '17 at 20:41 Method 1: The leading digit must be at least $1$, so $d_1, d_2, d_3, d_4, d_5 \\in \\{1, 2, 3, 4, 5, 6, 7, 8, 9\\}$. There are three possibilities: 1. All the digits are different: There are $\\binom{9}{5}$ such cases since selecting five of the nine digits completely determines the number since $d_1 < d_2 < d_3 < d_4 < d_5$ 2. Four different digits are used: There are $\\binom{9}{4}$ such cases in which $d_2 = d_3$ since the repeated digit must be the second smallest digit selected. There are also $\\binom{9}{4}$ such cases in which $d_4 = d_5$", "asking. We need subsets without 0, which are: {empty}; {1}; ... {5} {1, 2} {1, 3} ... ... {1, 2, 3, 4, 5} _________________ Intern Joined: 24 Jun 2015 Posts: 46 ### Show Tags 03 Jul 2015, 11:38 Bunuel wrote: luisnavarro wrote: Bunuel wrote: Official Solution: How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0? A. $$16$$ B. $$27$$ C. $$31$$ D. $$32$$ E. $$64$$ Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is $$2^5=32$$. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0. Hi, I find it hard to understand why it should not be 5¡, because it is suppossed that there are 5 choices that you are combining; I think that answer 2^5=32 seems logic but It is hard to me to visualize why 5¡ does not work here. Thanks a lot. Regards Luis Navarro Looking for 700 5! gives different arrangements of {0, 1, 2, 3, 4, 5}: {1, 0, 2, 3, 4, 5}, {1, 2, 0, 3, 4, 5}, {1, 2,", "triple and a pair. The kind of card you have three of can be chosen in $\\binom{13}{1}$ ways, or, more simply, in $13$ ways. (By kind here we mean Ace, or King, or Queen, and so on.) For every choice of kind, there are $\\binom{4}{3}$ ways of choosing the actual cards of that kind. The number $\\binom{4}{3}$ is simply $4$, but I want to concentrate on the structure, so that you can apply similar reasoning to other problems. For every way of doing the above two tasks, there are $\\binom{12}{1}$ ways of choosing the kind of card you have two of. And for every such choice, there are $\\binom{4}{2}$ ways of choosing the actual two cards. Thus the total number of \"full house\" hands is $$\\binom{13}{1}\\binom{4}{3}\\binom{12}{1}\\binom{4}{2}.$$ Divide by $\\binom{52}{5}$ to find the probability. - Thank you so much! :) I assume that you reduce 1 from 13 as you can't have triplets, if you've already got a pair in your hand? – Zanathel Mar 21 '12 at 18:34 @Zanathel: Our pair cannot be of the same kind as what we have three of. One could have, however, counted another way, and said there are $13$ ways to choose the kind we have two of, and for each of these ways, there are $12$ ways to choose what we", "# How to approach this problem of combinatorics How many different numbers can be formed by the product of two or more of the numbers 3, 4, 4, 5, 5, 6, 7, 7, 7? The question doesn't even make sense to me because it says \"two or more,\" so I could pick $n$-many and there could be countably many. So that obviously doesn't make sense...But if I were to ignore that, what am I restricted to and why? How would I then proceed from there? - You have a multiset containing one $3$, two $4$’s, two $5$’s, one $6$, and three $7$’s, for a total of $9$ elements. You must pick at least $2$ of these $9$ elements to form your product, and the question is how many different products you can form in this way. A set of $9$ elements has $2^9$ subsets; one of them is empty, and $9$ have only one member each, so you’re down to $2^9-10$ subsets containing at least $2$ of these $9$ numbers. However, some of these subsets generate the same product. For instance, there are $\\binom32=3$ ways to pick two of the $7$’s, so there are $3$ ways to get the product $49$. There are $2\\cdot2=4$ ways to get the product $20$, since there are two $4$’s and two $5$’s." ]

[ "# The number of subset sum If we have that $S=\\{1,2,3,4,5,6,7,8,9,10\\}$,how to know the number of subset sum T from :S in which for every $x\\in T$ and for every $2x\\in S$ then: $2x \\in T$ - sorry $x$ and $2*x$ are member of $T$?or ? – dato datuashvili Aug 25 '12 at 21:28 ## 1 Answer The contribution from 1, 2, 4, 8 could be 0, 8, 12, 14, or 15; the contributiom from 3, 6 could be 0, 6, or 9; from 5, 10, could be 0, 10, or 15; then you have 0 or 7, and 0 or 9. So you pick one number from each of the groupings 0, 8, 12, 14, 15; 0, 6, 9; 0, 10, 15; 0, 7; 0, 9; and add the five numbers you've chosen, and see how many different sums you get (if I'm interpreting the problem correctly). The most you can get is 55; the least you can get, other than zero, is 6; there's no way to get 11; can you get every number between 12 and 55? No, there's no way to get 53. I don't see a clever way to see what you get and what you don't without actually trying all the possibilities. -", "# In how many ways can we partition a set into smaller subsets so the sum of the numbers in each subset is equal? Let $$A = \\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\\}$$.How many ways are there to partition this set into at least 2 subsets so the sum of the numbers in each subset is equal? Here's what I've tried: Let $$n$$ be the number of subsets and $$s$$ the sum of the numbers in a subset(Where all the sets are equal). Since $$n\\cdot{}s=45$$, $$n$$ has to divide $$45$$ and on the other hand, $$s$$ can't be smaller than $$9$$ so the only options for $$n$$ are $$3$$ and $$5$$. Now I know I can just count how many possible solutions there are given the fact that $$n$$ has only $$2$$ possible values but I was wondering is there a more elegant solution? P.S: This actually came in a math contest and I didn't have time to count all the possible solutions so if there aren't any \"better\" solution, what do you think I should have done since I only had around $$5$$ minutes for this question. • This may be cheating :) but may I ask: what is the level of the math contest? If it's highschool level perhaps you're just supposed", "digit 4 times. – user Mar 14 at 20:45 • Are you sure? It's not given in the question that how many digits the formed number should have. – M. Kumar Mar 15 at 6:48 You could partition the problem into smaller problems composing numbers from the 6 digits: 1. How many different numbers contain neither $$2$$ nor $$3$$ 2. How many different numbers contain just one of $$2$$ and $$3$$ 3. How many different numbers contain both $$2$$ and $$3$$ Each of these numbers can be obtained by counting the solutions by the number of $$1$$'s they contain: 1. The composed string of digits will consist of one up to four $$1$$'s, there are $$4$$ ways to do so. 2. There are $$2$$ ways to decide between $$2$$ and $$3$$, and for $$k=0,...,4$$ there are $$k+1$$ ways to place that digit between $$k$$ $$1$$'s. So, counting the possibilities, we get $$2\\sum_{k=0}^{4}(k+1) = 2\\sum_{k=0}^{4}\\binom{k+1}{1} = 2\\binom{4+2}{2} = 30$$. 3. For $$k=0,...,4$$, starting with $$k$$ $$1$$'s, we have $$k+1$$ possible positions to insert the $$2$$ and then $$k+2$$ possible positions to place the $$3$$. Summing them up, we get $$\\sum_{k=0}^{4}(k+1)(k+2) = \\sum_{k=0}^{4}\\binom{k+2}{2}2! = 2\\binom{4+3}{3} = 70$$. In total, we get $$4+30+70 = 104$$ possible ways to compose the number. • I think this is the same thing as I", "1. ## partioning sets... a set has 5 members.what is the the number ways of partioning it into 2 or more subsets??? 2. What things are you given and roughly what knowledge are you supposed to use? 3. all i have is the question. you should be able to answer it with only high school maths. 4. Originally Posted by earthboy a set has 5 members.what is the the number ways of partioning it into 2 or more subsets??? Originally Posted by earthboy all i have is the question. you should be able to answer it with only high school maths. I do not consider this to a high school level question. So I am giving you the technical answer. But I cannot explain to you. The fifth Bell number is $\\mathcal{B}(5)=52$. That is the number of ways to partition a set of five. But one of those is the set itself, so it does meet the condition of ‘two or more subsets’. Thus you answer must be $51$. 5. Originally Posted by earthboy all i have is the question. you should be able to answer it with only high school maths. You could try it using $n_C_r=\\frac{n!}{r!(n-r)!}$ You cannot have more than 4 members in a subset. A subset must contain at least one member. Therefore you", "# Math Help - [SOLVED] Subset 1. ## [SOLVED] Subset Consider a group of 4 boys and 6 girls. In how many ways can one choose a subset of 2 boys and 3 girls? I know there is a quicker way of doing this than manually but i don't know the method. 2. Do you know the combination formula? Like in the binomial formula? $\\begin{pmatrix}n \\\\k \\end{pmatrix} = \\frac{n!}{(n-k)!k!}$ That gives you the number of ways to choose k things from a set of n elements when order does not matter. Clearly in a subset, the order of choosing the things doesn't make a difference. Do you think you can take it from here? 3. thanks for the formula, does the fact that the boys and girls are seperate things make the calculation change? Or do i use the formula for both i.e. $\\frac{4!}{(4-2)!2!}$ and then add? to $\\frac{6!}{(6-3)!3!}$ ? 4. ok, i used some logic to work out that 20+6 = 26 and there are way more than 26, 120 to be precise which is 20*6 . Thanks for the formula and telling me it was a combinatorics question! 5. Yup looks good to me. That's what I got when I did it. When you have n ways to do one thing, and m ways to do", "15$ $\\therefore$ The number of groups of three r more numbers $= 16+16+15 = 47.$ Correct Answer $: 47$ $\\textbf{Short method:}$ Given that, set of numbers $\\{1,2,3,4,5,6,7,8\\}$ From the above set, $\\{3,5\\}$ is taken, then the number of possible subsets containing at least three numbers $= 2^{6}-1 = 64-1 = 63$ From these $63,$ we need t remove the subsets which have $\\{3,5,7,8\\}.$ The remaining numbers are $\\{1,2,4,6\\}.$ For these numbers, we have two possibilities, either selected or not selected. So, the number of subsets possible $= 2^{4} = 16$ $\\therefore$ The total number of ways $= 63-16=47.$ 10.3k points 1 vote 1 297 views 1 vote", "()(()), (()()), ()()(). 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Can you arrange the numbers$1, 2, \\ldots, 9$along a circle such that the sum of two neighbors is never divisible by 3, 5, or 7? Can you arrange the numbers$1, 2, \\ldots, 9$along a circle such that the sum of two neighbors is never divisible by 3, 5, or 7? 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Choosing two distinct subsets In how many ways can you choose two distinct subsets from a collection of$N$items? The two subsets together need not include all$N$items. 1-2 3-4 5-6 7-8 9-10 11-12 13-14 In how many ways can you choose two distinct subsets from a collection of$N$items? The two subsets together need not include all$N$items. In how many ways can you choose two distinct subsets from a collection of$N$items? The two subsets together need not include all$N$items. 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Subsets with no consecutive elements How many subsets of the set$\\{1, 2, 3, \\ldots, N\\}$contain no two successive numbers? 1-2 3-4 5-6 7-8 9-10 11-12 13-14 Distributing billiard balls into pockets How many ways are there to put seven white and two black billiard balls into nine pockets? Some of the pockets may be empty and the pockets are considered distinguishable. 1-2 3-4 5-6 7-8 9-10 11-12", "asking. We need subsets without 0, which are: {empty}; {1}; ... {5} {1, 2} {1, 3} ... ... {1, 2, 3, 4, 5} _________________ Intern Joined: 24 Jun 2015 Posts: 46 ### Show Tags 03 Jul 2015, 11:38 Bunuel wrote: luisnavarro wrote: Bunuel wrote: Official Solution: How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0? A. $$16$$ B. $$27$$ C. $$31$$ D. $$32$$ E. $$64$$ Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is $$2^5=32$$. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0. Hi, I find it hard to understand why it should not be 5¡, because it is suppossed that there are 5 choices that you are combining; I think that answer 2^5=32 seems logic but It is hard to me to visualize why 5¡ does not work here. Thanks a lot. Regards Luis Navarro Looking for 700 5! gives different arrangements of {0, 1, 2, 3, 4, 5}: {1, 0, 2, 3, 4, 5}, {1, 2, 0, 3, 4, 5}, {1, 2,", "the set) will contain the 2-element subset. There are $\\dbinom{5}{2}=10$ ways to choose the 2-element subset. Thus, there are $10$ ways to create these sets. Second, the subsets can have 3 and 4 elements. $3+4=7$ non-distinct elements. $7-5=2$ elements in the intersection. There are $\\dbinom{5}{3}=10$ ways to choose the 3-element subset. For the 4-element subset, two of the elements must be the remaining elements (not in the 3-element subset). The other two have to be a subset of the 3-element subset. There are $\\dbinom{3}{2}=3$ ways to choose these two elements, which means there are 3 ways to choose the 4-element subset. Therefore, there are $10\\cdot3=30$ ways to choose these sets. $10+30=40 \\implies \\boxed{\\textbf{(B) 40}}$ ### Solution 3 We label the subsets subset 1 and subset 2. Suppose the first subset has $k$ elements where $k<5.$ The second element has $5-k$ elements which the first subset does not contain (in order for the union to be the whole set). Additionally, the second set has 2 elements in common with the first subset. Therefore the number of ways to choose these sets is $\\binom{5}{k} \\cdot \\binom{k}{2}.$ Computing for $k<5$ we have $10+30+30+10=80.$ Divide by 2 as order does not matter to get $40.$ ### Solution 4 (Stars And Bars) There are $\\dbinom{5}{2}=10$ ways to choose the 2 shared elements. We", "How many $3$-subsets of $\\{1,2,\\ldots,10\\}$ contain at least one even and one odd integer? Find the number of three element subsets $A$ of $\\{1,2,...,10\\}$ such that $A$ contains at least one even and one odd integer. I think in this way: First we distribute one even and one odd number to 3 sets.Then there are 4 digits left of which 2 are even and 2 are odd.But how to distribute this 4 digits into 3 sets? • I don't know why you are distributing numbers to three sets. The question is about three-element sets, and how many different ones there are. – Gerry Myerson Apr 11 '13 at 13:30 • I am getting 3 three-element sets and one digit extra.So my question is how to get the three-element set?Is there any other way? – Germain Apr 11 '13 at 13:34 • @analysis89, I think you are misunderstanding the problem. Suppose the question didn't have the \"such that\" part. Then the answer would be the number of ways of choosing 3 things from 10, that is, the binomial coefficient 10-choose-3. We're not partitioning the set into 3-element subsets; we're counting the total number of 3-element subsets that exist. – Gerry Myerson Apr 11 '13 at 13:45 • The set $\\{1,2,2\\}$ does exist. It is a two element set. In", "the # of 5 digit numbers with only 1's and 2's: Such a number can have one 1, two 1's, three 1's, or four 1's. (All others being 2) How many ways can it have one 1? You have to choose 1 position out of 5 possible positions, so $\\binom{5}{1}=5$. How many ways can it have two 1's? You have to choose 2 positions out of 5 possible positions, so $\\binom{5}{2}=10$. Similarly for the others, we have 10 and 5. So there are 30. Do the same for all others, subtract, and you have your result. (150) If you don't like this approach, you can try counting them directly by splitting them into different cases. two 1's, two 2's, one 3 two 1's, two 3's, one 2 two 2's, two 3's, one 1 one 1, one 2's, three 3's one 2, one 3, three 1's one 1, one 3, three 2's. Of course, the first three are equal, and the last three are equal. I'm pretty sure the first way is the simplest way to do it though. I could be wrong.", "Trickier 2013 Mathcounts State Sprint Round questions : Sprint #14: From Varun: Assume the term \"everything\" refers to all terms in the given set. 1 is a divisor of everything, so it must be first. Everything is a divisor of 12, so it must be last. The remaining numbers left are 2, 3, 4, and 6. 2 and 3 must come before 6, and 2 must come before 4. Therefore, we can list out the possibilities for the middle four digits: 2,3,4,6 2,3,6,4 3,2,4,6 3,2,6,4 2,4,3,6 There are 5 ways--therefore 5 is the answer. From Vinjai: Here's how I did #14: First, notice that 1 must be the first element of the set and 12 must be the last one. So that leaves only 2,3,4,6 to arrange. We can quickly list them out. The restrictions are that 2 must be before 4, 3 must be before 6, and 2 must be before 6: 2,3,4,6 3,2,4,6 4,2,3,6 6,2,3,4 2,3,6,4 3,2,6,4 4,2,6,3 6,2,4,3 2,4,3,6 3,4,2,6 4,3,2,6 6,3,2,4 2,4,6,3 3,4,6,2 4,3,6,2 6,3,4,2 2,6,3,4 3,6,2,4 4,6,2,3 6,4,2,3 2,6,4,3 3,6,4,2 4,6,3,2 6,4,3,2 Only the bold ones work. So, the answer is 5. #17: Common dimensional change problem $$\\overline {ZY}:\\overline {WV}=5:8$$ -- line ratio The volume ratio of the smaller cone to the larger cone is thus $$5^{3}: 8^{3}$$. The volume of the frustum", "check, the number of subsets containing n elements is $\\begin{pmatrix}5 \\\\ n\\end{pmatrix}$. So there is $\\frac{5!}{0!5!}= 1$ subset with 0 elements, $\\frac{5!}{1!4!}= 5$ subsets with 1 element, $\\frac{5!}{2!3!}= 10$ subsets with two elements, $\\frac{5!}{3!2!}= 10$ subsets with three elements, $\\frac{5!}{4!1!}= 5$ subsets with 4 elements, and $\\frac{5!}{5!0!}= 1$ subsets with 5 elements (the entire set). For {5, 15, 25, 35, 45, 55, 65}, notice that 5= 1(5), 15= 3(5), 25= 5(5), 35= 7(5), 45= 9(5), 55= 11(5), 65= 13(5), odd multiples of 5. And those odd numbers can be written as 2n+1 for n from 0 to 6.", "solutions to $a_1 + a_2 + \\cdots + a_k = n$ in which the addends are indistinguishable). Then, for each partition, separately calculate the number of ways, and finally, add these results together. For example, if $(n,k) = (5,3)$, then our partitions are: $\\{5,0,0\\}$--this case has $1$ way. $\\{4,1,0\\}$--we choose one of $n$ to be the \"$1$\", so there are $5$ ways. $\\{3,2,0\\}$--we choose three objects to be the \"$3$'s\" (the rest are determined after this), so there are $\\binom53 = 10$ ways for this. $\\{3,1,1\\}$--again, we choose three objects to be the \"$3$'s\" (the rest are determined after this), so there are $\\binom53 = 10$ ways for this. $\\{2,2,1\\}$--first, we choose one object to be the \"$1$\", which has $5$ ways. Then, we can choose any two of the remaining four to be one of the \"$2$'s\", and there are $\\binom42 = 6$ ways for this. However, we must divide this by $2$, since the two \"$2$'s\" are interchangeable, and the total for this case is $5\\cdot6\\cdot\\frac12 = 15$. Adding up, we get $1 + 5 + 10 + 10 + 15 = 41$ ways. All of these problems are similar to this one in that you divide them up into smaller counting problems. ## Indistinguishable to indistinguishable This is part of the partition problem. Imagine that", "thirteen boys are not separate groups. • okay. got it. thank you so much.! you're a genius. Dec 4, 2016 at 11:30 There should be minimum $2$ perfects in the committee, there is no limit for maximum of perfects. So, The number of ways can be counted as: $1$. $2$ perfect and $6$ non perfect..... Number of ways become ${{5}\\choose{2}}\\times{{10}\\choose{6}}$. $2$. $3$ perfect and $5$ non perfect....... Number of ways become ${{5}\\choose{3}}\\times{{10}\\choose{5}}$. $3$. $4$ perfect and $4$ non perfect.....Number of ways become ${{5}\\choose{4}}\\times{{10}\\choose{4}}$. $4$. $5$ perfect and $3$ non perfect.....Number of ways become ${{5}\\choose{5}}\\times{{10}\\choose{3}}$. The total number of committees of $8$ is $\\binom{15}{8}$. The number of committees of $8$ with no prefects is equal to $\\binom{10}{8}$. The number of committees with exactly one prefect is equal to $\\binom{5}{1} \\cdot \\binom{10}{7}$. Then the answer to your question is $\\binom{15}{8} - \\binom{10}{8} - \\binom{5}{1} \\cdot \\binom{10}{7}=6435 - 45 - 5\\cdot 120 = 5790.$", "# Number of subsets of size 5 of a set of 25 numbers. The Set is S={1,2,3,.....,25}. We have to count the number of subsets of size 5 such that each has atleast one Odd number in it. Method 1: Count = (Total Subsets of size 5) - (Total subsets having all even numbers) $$= \\binom {25} {5} - \\binom {12} {5} = 53130 - 792 = 52338$$ Method 2: Consider the 5 element subsets as 5 distinct boxes, each can be filled with one number. 1st box can be filled by any of the 13 Odd numbers. 2nd by the remaining 24. 3rd by remaining 23. 4th by remaining 22. 5th by remaining 21. $$\\text{Total} = 13 \\times{24} \\times{23} \\times{22} \\times{21}=3315312$$ Why am I getting two answers? Which method is flawed and how? • Keep in mind that conpliment to 'each' is $\\exist$ at least 1. You can have two sets with no odd numbers. – Alex Jan 19 '17 at 9:34 First method is correct Second method is false because you are counting several solutions multiple times. For example the set {1, 2, 3, 4, 6 } is counted at least twice : • a first time when you consider putting 1 into the first box • a second time when you consider putting 3 into", "inspection, a viable set must contain at least one element from both of the sets $\\{1, 2, 3, 4, 5\\}$ and $\\{4, 5, 6, 7, 8\\}$. Since 4 and 5 are included in both of these sets, then they basically don't matter, i.e. if set A is a subset of both of those two then adding a 4 or a 5 won't change that fact. Thus, we can count the number of ways to choose at least one number from 1 to 3 and at least one number from 6 to 8, and then multiply that by the number of ways to add in 4 and 5. The number of subsets of a 3 element set is $2^3=8$, but we want to exclude the empty set, giving us 7 ways to choose from $\\{1, 2, 3\\}$ or $\\{6, 7, 8\\}$. We can take each of these $7 \\times 7=49$ sets and add in a 4 and/or a 5, which can be done in 4 different ways (by adding both, none, one, or the other one). Thus, the answer is $49 \\times 4=\\boxed{196}$. ## Solution 3 This solution is very similar to Solution $2$. The set of all subsets of $\\{1,2,3,4,5,6,7,8\\}$ that are disjoint with respect to $\\{4,5\\}$ and are not disjoint with respect to the complements of sets", "(and therefore not a subset of) $\\{1,2,3,4,5\\}$ and $\\{4,5,6,7,8\\}$ will be named $S$, which has $7\\cdot7=49$ members. The union of each member in $S$ and the $2^2=4$ subsets of $\\{4,5\\}$ will be the members of set $Z$, which has $49\\cdot4=\\boxed{196}$ members. $\\blacksquare$ Solution by a1b2 ## Solution 4 Consider that we are trying to figure out how many subsets are possible of $\\{1,2,3,4,5,6,7,8\\}$ that are not in violation of the two subsets $\\{1,2,3,4,5\\}$ and $\\{4,5,6,7,8\\}$. Assume that the number of numbers we pick from the subset $\\{1,2,3,4,5,6,7,8\\}$ is $n$. Thus, we can compute this problem with simple combinatorics: If $n=1$, $\\binom{8}{1}$ $-$ ($\\binom{5}{1}$ + $\\binom{5}{1}$ $- 2$) [subtract $2$ to eliminate the overcounting of the subset $\\{4\\}$ or $\\{5\\}$] = $8 - 8$ = $0$ If $n=2$, $\\binom{8}{2}$ $-$ ($\\binom{5}{2}$ + $\\binom{5}{2}$ $- 1$) [subtract $1$ to eliminate the overcounting of the subset $\\{4,5\\}$] = $28 - 19$ = $9$ If $n=3$, $\\binom{8}{3}$ $-$ ($\\binom{5}{3}$ + $\\binom{5}{3}$) = $56 - 20$ = $36$ If $n=4$, $\\binom{8}{4}$ $-$ ($\\binom{5}{4}$ + $\\binom{5}{4}$) = $70 - 10$ = $60$ If $n=5$, $\\binom{8}{5}$ $-$ ($\\binom{5}{5}$ + $\\binom{5}{5}$) = $56 - 2$ = $54$ If $n>5$, then the set $\\{1,2,3,4,5,6,7,8\\}$ is never in violation of the two subsets $\\{1,2,3,4,5\\}$ and $\\{4,5,6,7,8\\}$. Thus, If $n=6$, $\\binom{8}{6}$ = $28$ If $n=7$, $\\binom{8}{7}$ = $8$ If", "{1, 2, 3, 4, 5}, and in your explanation you said: 5! gives different arrangements of {0, 1, 2, 3, 4, 5}: {1, 0, 2, 3, 4, 5}, {1, 2, 0, 3, 4, 5}, {1, 2, 3, 0, 4, 5}, ... Which is not what the question is asking. But I was reffering to 5! without cero... I am confused, could you help me again? Thanks a lot Regards Luis Navarro Looking for 700 Yes, 5! gives different arrangements of {1, 2, 3, 4, 5}. 0 in my previous post was a typo. We need SUBSETS of {0, 1, 2, 3, 4, 5} without 0, not arrangements of {1, 2, 3, 4, 5}. Subsets of {0, 1, 2, 3, 4, 5} without 0 are: {empty}; {1}; ... {5} {1, 2} {1, 3} ... ... {1, 2, 3, 4, 5} Notice that arrangements of {1, 2, 3, 4, 5} give the same 5-element sets but arranged in different ways, while subsets give an empty set, 1-element sets, 2-elements set, ..., 5-element sets. Hope it's clear. Thanks, it is more clear know, I only have a little more doubt... What I inffer for this is that subsets orders by itself \"automaticly\", so then {1, 2, 3, 4, 5} only consider 1 posible combination, instead of mixing the order in the", "five ways: 4, 3+1, 2 + 1 + 1, 2 +2 and 1 + 1 + 1 + 1; and so 4 is said to have five partitions. What is the number of partitions for the number 7? #2: Extra: Try partition the number 5 and the number 8. Solution: #24: You can solve this problem using the same technique as counting coins: 7 6 5 4 3 2 1 1 1 way 1 1 way 1 1 2 ways ( 5 + 2 or 5 + 1 + 1) 1 1 1 way 1 1 2 ways ( 4 + 2 + 1 or 4 + 1 + 1 + 1) 2 0 1 1 ways 1 2 3 ways ( 3 + 2 + 2, 3 + 2 + 1 + 1 and 3 + 1 + 1 + 1 + 1) 3 4 ways (2 + 2 + 2 + 1, 2 + 2 + 3 ones, 2 + 5 ones and 7 ones.) Total 15 ways. The partitions of 5 are listed below (There are 7 ways total.): 5 4 3 2 1 1 1 way 1 1 way 1 1 2 ways (3 + 2 and 3 + 1 + 1) 2 3 ways (2 + 2 + 1, 2 + 1 +" ]

[ "### Home > CCG > Chapter 10 > Lesson 10.1.1 > Problem10-12 10-12. Multiple Choice: A penny, nickel, and dime are all flipped once. What is the probability that at least one coin comes up heads? 1. $\\frac {1} {3}$ 1. $\\frac {3} {8}$ 1. $1$ 1. $\\frac {7} {8}$ Draw a tree diagram that illustrates all the possible combinations that could occur. Then add up the probabilities of each combination that includes a coin coming up heads. Use the eTool below to support your ideas! Click on the link at right for the full eTool version: CCG 10-12 HW eTool", "I can offer this: In the question, the friend suggests coin flipping. Instead of the coin flipping scenario, he could have suggested that if you get an envelope with one dollar in it, then change envelopes. This is always correct and will increase above 1/2 your probability of ending up with the envelope with the larger amount. I think this is all the coin tossing does.", "### Home > GC > Chapter 8 > Lesson 8.1.4 > Problem8-42 8-42. Multiple Choice: A penny, nickel, and dime are all flipped once. What is the probability that at least one coin comes up heads? 1. $\\frac{1}{3}$ 1. $\\frac{3}{8}$ 1. $1$ 1. $\\frac{7}{8}$ Draw a tree diagram that illustrates all the possible combinations that could occur. Then add up the probabilities of each combination that includes a coin coming up heads. Use the eTool below to support your ideas! Click on the link at right for the full eTool version: GC 8-42 HW eTool", "coins were minted in 1933. The last 90% silver coins were minted in 1964, and the last 40% silver half dollar was minted in 1970. The United States Mint currently produces circulating coins at the Philadelphia and Denver Mints, and commemorative and proof coins for collectors at the San Francisco and West Point Mints. Mint mark conventions for these and for past mint branches are discussed in Coins of the United States dollar#Mint marks. The one-dollar coin has never been in popular circulation from 1794 to present, despite several attempts to increase their usage since the 1970s, the most important reason of which is the continued production and popularity of the one-dollar bill.[43] Half dollar coins were commonly used currency since inception in 1794, but has fallen out of use from the mid-1960s when all silver half dollars began to be hoarded. The nickel is the only coin whose size and composition (5 grams, 75% copper, and 25% nickel) is still in use from 1865 to today, except for wartime 1942-1945 Jefferson nickels which contained silver. Due to the penny's low value, some efforts have been made to eliminate the penny as circulating coinage. [44] [45] For a discussion of other discontinued and canceled denominations, see Obsolete denominations of United States currency#Coinage and Canceled denominations of United States", "At that time, it cost approximately 1.6 cents to mint each penny. In 1965, the United States stopped circulating the silver dollar since the material was worth more than a dollar. Most of the dollar coins that had been minted were then melted to recover the base metal. In history, coins did not always have a value imprinted on them. The purity of the metal that composed the coin could be tested on a touchstone, and the monetary value was calculated based on its weight and purity. Eventually gold and silver coins were stamped with their weights instead of a fixed monetary value. Governments would mint the coins as a way of guaranteeing the weight and purity of the material was correct.", "Fractions ### Anatomy of a fraction You probably know what fractions look like, and even what the parts are called, but do you really know what all of the parts mean? Let's start there. In the fraction below, the a is the numerator. To enumerate is to list or to count, so the numerator enumerates or counts. Specifically, it counts how many of the denominator there are. The denominator, b, is the size of the things that are being counted. Those things are the equal-sized parts into which the whole is divided. The denomination is the number of equal parts that would be required to make up the whole. In the fraction $\\frac{1}{4},$ four parts make up the whole, and we have one of them. A familiar example of denomination might be the denomination of money. The denomination of a coin or bill is its value in dollars (U.S. dollar-based system). A penny has a denomination of 1/100* because it's value is 1/100 of a dollar. A nickel has a denomination of 1/20 of a dollar, a dime 1/10 and a quarter 1/4 of a dollar (thus the name \"quarter\"). If we have four quarters, we have 4/4 of a dollar or one whole dollar. Two halves of a pie make a whole pie, if we have 87", "found different in step 1. Now, 3. You ask him to flip 3. Now for sure, you know that all of them are same. Initially, they are not all heads or tails. Thus, there are at least one heads and one tails, and the third coin is unknown. WLOG, we will assume the third coin is a heads. Thus, there are two heads and 1 tails. Lets say we get them to flip a single coin (say the nickel), hoping that this is the one coin different from the others. We are right 33% of the time and end the game with 3 heads. The rest of the time, we flip one of the heads, so we end up with two tails (one of which is the nickel) and one heads. If we didn't end the game (66% of the time), then we can get them to flip one of the remaining two coins (say the dime) hoping it is heads. If we are right (50% of the time) we have ended the game. Otherwise, the dime was tails and we now have two heads (dime and quarter) and one tails (nickel). The last move is to flip the nickel again. So, the simplified strategy is simply 1. Flip nickel 2. Flip dime 3. Flip nickel What ever", "that coin's $p_i$. Put more simply, it assumes variation in the probabilities of coins produced by a mint, and models that variation by way of the hierarchy. That's the sort of model described by your second interpretation. In your first interpretation, there's only one coin, and $x_4$ could represent the next, unseen flip.", "A coin flip decided which one of them to be the accepted answer.", "when you flip $n$ coins. -", "\"The penny came up heads\"$F$= \"The nickel came up heads\" For ordinary coins, it's ... 3 The two concepts of \"independent\" and \"mutually exclusive\" are different. Two events are \"mutually exclusive\" ( that is, $$P(E \\cap F) = 0$$) if they can't both happen at the same time. For example, if I roll a die and define E = \"I roll a 6\" and F = \"I roll a 3\", these can't both be true. If E happens, i know F didn't. Two events are \"independent\" (that ... 1 You're correct that if$E$and$F$are independent, then$P(E \\cap F)=P(E)P(F)$. If$E$and$F$are independent, then it means that the result of$E$has no influence on the result of$F$and vice versa. For example: I toss a quarter and a dime. The result of the quarter toss has no influence on the dime toss. However, if$P(E \\cap F)=0$, then they are ... 0 0.5 probability of player 2 winning means that if player 1 runs perpendicular to the line joining the initial positions of the players, then player 2 will catch him on the boundary of the circle. In that case the path of player 2 will be the hypotenuse of an isosceles right with side length$r$, the radius of the circle. the length of this path is$l=\\sqrt ... 1 The whole purpose of calculating the", "whole is, if not in error, then at least inconsistent in switching between two standards of value, namely, silver and wheat. Note In old English money 12 pence equals 1 shilling. Thus 6 shillings and 8 pence is almost a quarter of 28 shillings and exactly two-thirds of 10 shillings.", "will increase y, but if we look at both x's at the same time then if we hold x1 (total number of coins) constant, but increase x2 (the interpretation of the slope) then y would decrease because that means replacing quarters with smaller valued coins. The same general concept will hold for other datasets, including yours. - Thank you Greg Snow. – Bonnie Dec 31 '12 at 19:16", "after 1857, the us government started incorporating almost every other lower metals on combine. The fresh cent, being one cent, is actually perhaps not worthy of their lbs from inside the copper. Anybody you may melt copper cents and sell the latest copper for much more than the cents had been worth. Just after 1857, nickel are mixed with the larger copper. After 1864, the fresh escort services in Washington new penny is made from bronze. Tan try 95% copper and 5% zinc and you can tin. For 1 year, 1943, the brand new cent didn’t come with copper inside it as a result of the debts of the world War II. It had been simply zinc coated steel. Immediately after 1943 up to 1982, new penny experience episodes in which it absolutely was brass otherwise tan.", "# Flipping a Coin If you flip a fair coin, what is the difference of the percentage to get heads and the percentage to get tails? (Just write the number) ×", "the Bicentennial Quarter design which showed a date '1776 - 1976'. Unless you have a proof set that shows that it was released specifically in 1975 and 1976, you cannot tell the difference in the circulated coins. 5. Nickels minted from 1942-1945 during World War II contain 1.75 g (0.05626 oz) silver. The silver content of these \"war nickels\" as of October, 2007 is worth$0.77. 6. As of October 2007, the value of the metal in the nickel coin has reached 6.8 cents, a 36% premium over its face value, due to the rising costs of copper and nickel against a falling U.S. Dollar. In an attempt to avoid losing large quantities of circulating nickels to melting, the United States Mint introduced new interim rules on December 14, 2006 criminalizing the melting and export of pennies and nickels. Violators of these rules can be punished with a fine of up to \\$10,000, five years imprisonment, or both. • The Coinage Act of 1965, Pub.L. 89-81, 79 Stat. 254, enacted 1965-07-23, eliminated silver from the circulating dimes and quarter dollars of the United States, and diminished the silver content of the half dollar from 90% to 40%. This act was in response to coin shortages caused by the rising price of silver. ## References Note: Facts About United States", "1. ## confused how do i do this? A nickle, dime, and quarter are tossed. What is the probability that the nickel and dime will show heads and the quarter will show tails? What is the probability that only one of the coins will show tails? 2. Originally Posted by sevenlovesyou A nickle, dime, and quarter are tossed. What is the probability that the nickel and dime will show heads and the quarter will show tails? What is the probability that only one of the coins will show tails? A nickle, dime, and quarter are tossed. What is the probability that the nickel and dime will show heads and the quarter will show tails? $\\frac {1}{8}$ What is the probability that only one of the coins will show tails? $\\frac {3}{8}$ Make a table and count the possiblites. 3. Hey its easy identify all the possibilties fo all the coins. YUU WILL GET HHH HHT THH HTT TTT HTH THT TTH WHERE H= HEADS N T= TAILS... you can see there's only 1 possibility out of 8 that the first two will show tails n d last one a head n the probablity that only 1 coin show tails is 3/8", "U.S. penny. In 1982 the composition of a U.S. penny was changed from a brass alloy consisting of 95% w/w Cu and 5% w/w Zn, to a zinc core covered with copper.11 The pennies in Table 4.16, therefore, were drawn from different populations.", "once for each coworker. This feature is not available right now. But the very phrase tells us where we should classify that belief. Expected utility is an economic term summarizing the utility that an entity or aggregate economy is expected to reach under any number of circumstances. Palm Press with Magic Circle is a Pelvic Floor Exercise. I always weighed myself at the same time every week, wore more or less the same outfit and definitely the same shoes (which were ones I could slip off easily). It can't be done. Includes 83 gold, 16 silver and 1 bronze player. Current Weather. You may use it when selecting strike price for options trading. The 18 Karat Gold Melt Value Calculator, available below, can figure the total gold value of 18K gold items, measured by the weight unit of your choice. Sign in to comment. coin=TRUE shows a second plot with coin flip results (head or tail) Additional arguments from link{plot}. Transaction Value Median Transaction Value Tweets GTrends Active Addresses Top100ToTotal Fee in Reward. Last time we talked about independence of a pair of outcomes, but we can easily go on and talk about independence of a longer sequence of outcomes. Expected value is merely a weighted average. so for example if we have 2 coins, options will be", "× ## Decimals Coins are often used to express a fraction of a currency. In the US, a quarter is $0.25 and a dime is$0.10. Counting change is just one of many ways you can put decimals to use in everyday life. # Place Value Solving $$3.2\\times2.2,$$ which number do we get in the hundredths place? In the number $$0.0 8 3 5,$$ the digit $$3$$ is in which place? Round up $$0.004 0 9 5$$ to the nearest tenthousandth. Solving $$7.57-3.23,$$ which number do we have for the tenths place? What number is in the hundredths place of 662.373? ×" ]

[ "• anonymous MEDAL!! (bill and larry toss two coins. if both coins come up heads, bill pays larry $4. larry pays bill$1 otherwise.) What is the expected value of this game for bill? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "> t$$ coins. If the probabilities are constant at $$p$$ at each round, the expected value is $$1/p$$ which is like one coupon being collected in the coupon collector problem. Since the probability of heads is less than $$1/sqrt{t}$$ at each round, $$sqrt{t}$$ is a lower bound for the expectation. On the other hand, the probability is greater than $$1/sqrt{T}$$ in the bounded case, and so $$sqrt{T}$$ is an upper bound. The question is where in this regime $$[sqrt{t}, sqrt{T}]$$ does the expectation fall..", "# A bag contains 20 coins. If the probability that bag contains exactly 4 biased coin is 1/3 and that of exactly 5 biased coin is 2/3, then the probability that all the biased coin are sorted out from the bag in exactly 10 draws is Question : A bag contains 20 coins. If the probability that bag contains exactly 4 biased coin is 1/3 and that of exactly 5 biased coin is 2/3, then the probability that all the biased coin are sorted out from the bag in exactly 10 draws is (A) $\\dfrac{5}{33}\\dfrac{16C_{6}}{20C_{q}}+\\dfrac{1}{11}\\dfrac{15C_{5}}{20C_{9}}$ (B) $\\dfrac{2}{33}\\left[ \\dfrac{16C_{6}+15C_{5}}{20C_{9}}\\right]$ (C) $\\dfrac{5}{33}\\dfrac{16C_{7}}{20C_{9}}+\\dfrac{1}{11}\\dfrac{15_{C_{6}}}{20C_{9}}$ (D) none of these", "# Winning at Coins There are $10$ cards, labeled from $1$ to $10$, lying face down on a table. You pick a card, and if that card is a $10$, you win $\\20$. If the card is not $10$ you win $\\0$. What is the expected amount you would win (in dollars)? ×", "• anonymous A box contains nine $1 bills, eight$5 bills, two $10 bills, and four$20 bills. What is the expectation if one bill is selected? A.$6.48 B.$8.48 C.$5.48 D.$7.48 Probability Looking for something else? Not the answer you are looking for? Search for more explanations.", "Let $B$ be $1$ if both coins come up heads, and $0$ otherwise.", "be the sum of the stake ($1) and the expected value of winning), right?", "numbered from $1$ to $6$) is thrown. The probability that the face value is odd is $90\\%$ of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is ... one of the following options is closest to the probability that the face value exceeds $3$? $0.453$ $0.468$ $0.485$ $0.492$ 10 What is the probability that in a randomly chosen group of $r$ people at least three people have the same birthday? $1-\\dfrac{365-364 \\dots (365-r+1)}{365^{r}}$ ... $\\dfrac{365 \\cdot 364 \\dots (365-r+1)}{365^{r}}$ 11 Please explain how $P(A ∩ B) = P(A)P(B)$? If $A$ and $B$ are independent. 1 vote 12 Roll a die, then select at random, without replacement, as many cards from the deck as the number shown on the die. What is the probability that you get at least one Ace? 1 vote 13 Q32 [Mock 4]. Naveen's coin box contains 8 fair standard coins (heads and tails) and 1 coin which has heads on both sides. He selects a coin randomly and flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads? 1/3 2/3 ... )*(1/2)+(1/9)*(1) Probability of choosing fair coin and P(heads)+ P(unfair)*P(heads). Please help me understand this question.", "calculated similarly to how we calculated the probability of $\\dfrac{2}{5}$ above.", "## gunmate1 Group Title A box contains nine $1 bills, eight$5 bills, two $10 bills, and four$20 bills. What is the expectation if one bill is selected? one year ago one year ago • This Question is Open 1. texaschic101 Group Title The expectation is probably a $1 bill because there is more of them. There is a total of (9 + 8 + 2 + 4) = 23 bills. The chance of getting a$1 is 9/23, the chance of getting a $5 bill is 8/23, the chance of getting a$10 is 2/23, and the chance of getting a \\$20 is 4/23. 2. whpalmer4 Group Title The expectation (or expected value) is the sum of the probabilities multiplied by the payoffs. So $\\frac{9}{23} * 1 + \\frac{8}{23}*5 + ...$ 3. Skaternerd123 Group Title 1/24", "## anonymous one year ago a box contains five one dollar bills, nine five dollar bills, ten ten dollar bills, and nine twenty dollar bills. what is the expectation if one bill is selected. $10.00$9.00 $11.00$12.00 here we have to compute this: ${\\text{expectation}} = \\frac{{5 + 5 \\times 9 + 10 \\times 10 + 9 \\times 20}}{{33}} = \\frac{{330}}{{33}} = 10$", "Jenny places $100$ pennies on a table, with $30$ showing heads and $70$ showing tails. She chooses $40$ distinct pennies uniformly at random and turns them over. That is, if a chosen penny was showing heads, she turns it to show tails; if a chosen penny was showing tails, she turns it to show heads. After this process, what is the expected number of pennies showing heads?", "### Home > MC2 > Chapter 1 > Lesson 1.2.3 > Problem1-77 1-77. Remember, probability is the number of successful outcomes out of the total number of possible outcomes. How many nickels are in the bag? How many total coins are in the bag? $\\frac{4}{16}= \\frac{1}{4}$ What is the new number of nickels? What is the new total number of coins?", "are in the \"two-awakenings\" branch, the probability of it being a Tuesday or a Monday when the SB wakes up are both 50%. We can thus calculate the total probability of it being Tuesday when the SB wakes up as 0.5*0.5=0.25. Obviously then, the probability of it being Monday when she wakes up is 1-0.25=0.75 If the SB knew that she woke up on Tuesday, the probability of the coin having come up \"heads\" would have been zero. If she, however, knew that she woke up on Monday, the probability of the coin having come up \"heads\" would have been 50%. But we know that the probability of it being Monday is 0.75. So, to find out the total probability of the coin having come up \"heads\" we need to multiply 0.75*0.5=0.375 The answer is thus, the probability that the coin came up \"heads\" is 37.5% The above is just a suggestion. Please, point out, if you see flaws in my reasoning. • \"If she, however, knew that she woke up on Monday, the probability of the coin having come up \"heads\" would have been 50%.\" That is not right. The conditional probability of heads given Monday, or $P(H \\mid M)$, equals $P(H \\wedge M)/P(M)=P(H)/P(M)$. You end end up with 1/2. – Grassie Jan 5 '18 at 15:54", "There are three coins in a box. The first coin is two-headed. The second one is a fair coin. The third one is a biased coin that comes up heads $75\\%$ of the time. When one of the three coins was picked at random from the box and tossed, it landed heads.", "get exactly $3$ heads is $$abc+acd+abd+bcd.$$ If we want the \"at least\" case then we have to add $abcd$. In the case of $10$ coins the number of possibilities for exactly $3$ heads is ${10 \\choose 3}=120$ and all have to be listed like above. In the \"at least\" case we have much more possibilities to be listed: $\\sum_{i=3}^{10} {10\\choose i}$ and all have to be depicted.", "Question # A purse contains 4 coins which are either sovereigns or shillings; 2 coins are drawn and found to be shillings. If these are replaced, what is the chance that another drawing will give sovereign? Verified 160.8k+ views Hint: - This question may be interpreted in two ways, which we shall discuss separately. I) If we consider that all numbers of shillings are equally likely, we shall have three hypothesis: i. All the coins may be shillings ii. Three of them may be shillings iii. Only two of them may be shillings Here${P_1} = {P_2} = {P_3}$ Also${p_1} = 1,{p_2} = \\dfrac{1}{2},{p_3} = \\dfrac{1}{6}$ Hence probability of first hypothesis is${Q_1}$ $= 1 \\div \\left( {1 + \\dfrac{1}{2} + \\dfrac{1}{6}} \\right) = \\dfrac{6}{{10}}$ Probability of second hypothesis is${Q_2}$ $= \\dfrac{1}{2} \\div \\left( {1 + \\dfrac{1}{2} + \\dfrac{1}{6}} \\right) = \\dfrac{3}{{10}}$ Probability of third hypothesis is${Q_3}$ $= \\dfrac{1}{6} \\div \\left( {1 + \\dfrac{1}{2} + \\dfrac{1}{6}} \\right) = \\dfrac{1}{{10}}$ Therefore the probability that another drawing will give a sovereign is $= \\left( {{Q_1} \\times 0} \\right) + \\left( {{Q_2} \\times \\dfrac{1}{4}} \\right) + \\left( {{Q_3} \\times \\dfrac{2}{4}} \\right) \\\\ = \\left( {\\dfrac{6}{{10}} \\times 0} \\right) + \\left( {\\dfrac{3}{{10}} \\times \\dfrac{1}{4}} \\right) + \\left( {\\dfrac{1}{{10}} \\times \\dfrac{2}{4}} \\right) \\\\ = \\dfrac{3}{{40}} + \\dfrac{2}{{40}} \\\\ = \\dfrac{5}{{40}} = \\dfrac{1}{8} \\\\$ II) If", "with unfair coin which comes up heads $20 \\%$ of the time. If it comes up heads I will give you $4$ dollars if not, you will give me $1$ dollar. \" In the fair coin case your expected profit: $$2*0.5+(-2)*0.5 = 0$$ In the unfair coin case: $$4*0.2+(-1)*0.8=0$$ It is better not to waste your time with him because your expected profit is $0$.", "+0 0 42 4 There are three $20 bills, six$10 bills, seven $5 bills, and four$1 bills in your wallet. You select one bill at random. What is the expected value for this experiment? $_______ You roll a pair of six-sided number cubes, numbered 1 through 6. What is the probability of rolling two odd numbers? (Enter your probability as a fraction.) =__________ Guest Nov 9, 2018 #1 +13718 +2 Twenty bills 3/20 ($20) + 6/20 (10) + 7/20(5) + 4/20(1) = 7.95 expected value ElectricPavlov Nov 9, 2018 #2 +2793 +2 $$\\text{There are 20 total bills. Assume we choose any of them with equal probability.}\\\\ P[20] = \\dfrac{3}{20}\\\\ P[10]= \\dfrac{6}{20}=\\dfrac{3}{10}\\\\ P[5] = \\dfrac{7}{20}\\\\ P[1] = \\dfrac{4}{20} = \\dfrac{1}{5} \\\\ \\text{Let }X \\text{ be the value of the bill chosen.}\\\\ E[X] = 20\\cdot \\dfrac{3}{20} + 10\\cdot \\dfrac{3}{10} + 5 \\cdot \\dfrac{7}{20} + 1\\cdot \\dfrac{1}{5}\\\\ E[X] = \\dfrac{159}{20} = 7.95$$ Rom Nov 9, 2018 edited by Rom Nov 9, 2018 #3 +13718 +2 3/6 are odd for each die 3/6 x 3/6 = 9/36 = 1/4 chance of both being odd ElectricPavlov Nov 9, 2018 #4 +2793 0 deleted Rom Nov 9, 2018 edited by Rom Nov 9, 2018", "1 - $$\\dfrac {1} {8}$$= $$\\dfrac {7} {8}$$. #1: Amy tosses a nickel four times. What is the probability that she gets at least as many heads as tails ? #2: What is the probability that the product of the top faces on 2 standard die is even when rolled? #3: 3 numbers are selected at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction in lowest terms. #4: 9 fair coins are flipped. What is the probability that at least 4 are heads? #5: How many 3 digit numbers does not contain the digit 1 but have at least one digit that is 5? Answer: #1-$$\\dfrac {11} {16}$$ ; #2-$$\\dfrac {3} {4}$$ ; #3- $$\\dfrac {7} {8}$$ ; #4-$$\\dfrac {191} {256}$$ #5: 217 ## Tuesday, June 16, 2015 ### Problem Solving Strategies : Complementary Counting Check out Mathcounts here, the best competition math program for middle school students. Video to watch on complementary counting from \"Art of Problem Solving\" Part 1 Part 2 Question: How many two-digit numbers contain at least one 9? At beginning level, kids start to write down all the numbers that contain 9. However, this turns into impossible" ]

[ "simpler, consider the outcomes from flipping a pair of coins, one a quarter, then other a nickel. There are four possible outcomes: Both coins come up heads, the quarter comes up heads while the nickel comes up tails, the quarter comes up tails while the nickel comes up heads, and both come up tails. These are equiprobable events. Getting back to the problem of rolling the dice, look at rolling a sum of two versus rolling a sum of three. Both dice need to come up as one to roll a sum of two. The probability of this happening is 1/6*1/6, or 1/36. Now look at rolling a sum of three. It might help to imagine that the dice are colored, one red, the other green. You can roll a three by having the red die come up as one and the green die as two OR by having the red die come up as two and the green one as one. These are two independent events, each with probability 1/36, so the probability of rolling a sum of three with these colored dice is 2/36. Now suppose you used two white dice. Does the color of the dice change the probabilities one bit? 4. Dec 3, 2012 ### CAF123 No, the probabilities remain the same. So to", "### Home > CCG > Chapter 10 > Lesson 10.1.1 > Problem10-12 10-12. Multiple Choice: A penny, nickel, and dime are all flipped once. What is the probability that at least one coin comes up heads? 1. $\\frac {1} {3}$ 1. $\\frac {3} {8}$ 1. $1$ 1. $\\frac {7} {8}$ Draw a tree diagram that illustrates all the possible combinations that could occur. Then add up the probabilities of each combination that includes a coin coming up heads. Use the eTool below to support your ideas! Click on the link at right for the full eTool version: CCG 10-12 HW eTool", "\"The penny came up heads\"$F$= \"The nickel came up heads\" For ordinary coins, it's ... 3 The two concepts of \"independent\" and \"mutually exclusive\" are different. Two events are \"mutually exclusive\" ( that is, $$P(E \\cap F) = 0$$) if they can't both happen at the same time. For example, if I roll a die and define E = \"I roll a 6\" and F = \"I roll a 3\", these can't both be true. If E happens, i know F didn't. Two events are \"independent\" (that ... 1 You're correct that if$E$and$F$are independent, then$P(E \\cap F)=P(E)P(F)$. If$E$and$F$are independent, then it means that the result of$E$has no influence on the result of$F$and vice versa. For example: I toss a quarter and a dime. The result of the quarter toss has no influence on the dime toss. However, if$P(E \\cap F)=0$, then they are ... 0 0.5 probability of player 2 winning means that if player 1 runs perpendicular to the line joining the initial positions of the players, then player 2 will catch him on the boundary of the circle. In that case the path of player 2 will be the hypotenuse of an isosceles right with side length$r$, the radius of the circle. the length of this path is$l=\\sqrt ... 1 The whole purpose of calculating the", "### Home > GC > Chapter 8 > Lesson 8.1.4 > Problem8-42 8-42. Multiple Choice: A penny, nickel, and dime are all flipped once. What is the probability that at least one coin comes up heads? 1. $\\frac{1}{3}$ 1. $\\frac{3}{8}$ 1. $1$ 1. $\\frac{7}{8}$ Draw a tree diagram that illustrates all the possible combinations that could occur. Then add up the probabilities of each combination that includes a coin coming up heads. Use the eTool below to support your ideas! Click on the link at right for the full eTool version: GC 8-42 HW eTool", "Question 3 Explanation: Converting between fractions, decimals, and percents (Objective 0017). Question 4 #### If two fair coins are flipped, what is the probability that one will come up heads and the other tails? A $$\\large \\dfrac{1}{4}$$Hint: Think of the coins as a penny and a dime, and list all possibilities. B $$\\large \\dfrac{1}{3}$$Hint: This is a very common misconception. There are three possible outcomes -- both heads, both tails, and one of each -- but they are not equally likely. Think of the coins as a penny and a dime, and list all possibilities. C $$\\large \\dfrac{1}{2}$$Hint: The possibilities are HH, HT, TH, TT, and all are equally likely. Two of the four have one of each coin, so the probability is 2/4=1/2. D $$\\large \\dfrac{3}{4}$$Hint: Think of the coins as a penny and a dime, and list all possibilities. Question 4 Explanation: Topic: Calculate the probabilities of simple and compound events and of independent and dependent events (Objective 0026). Question 5 #### There are 15 students for every teacher. Let t represent the number of teachers and let s represent the number of students. Which of the following equations is correct? A $$\\large t=s+15$$Hint: When there are 2 teachers, how many students should there be? Do those values satisfy this equation? B $$\\large s=t+15$$Hint: When there", "get a dollar, or you can expect overall to realize half a dollar or $0.50 from flipping “heads.” The other half of the time, you can expect to lose a dollar, so your expectation has to include the possibility of flipping “tails” with an overall or average result of losing$0.50 or −\\$0.50. So you can expect 0.50 from one outcome and −0.50 from the other: altogether, you can expect 0.50 + −0.50 or 0 (which is why “flipping coins” is not a popular casino game.) The expected value (E(V)) of an event is the sum of each possible outcome’s probability multiplied by its result, or $E(V)=Σ( p n × r n ),$ where Σ means summation, p is the probability of an outcome, r is its result, and n is the number of outcomes possible. When faced with the uncertainty of an alternative that involves an independent event, it is often quite helpful to be able to at least calculate its expected value. Then, when making a decision, that expectation can be weighed against or compared to those of other choices. Figure 4.20 For example, Alice has projected four possible outcomes for her finances depending on whether she continues, gets a second job, wins in Vegas, or loses in Vegas, but there are really only three choices: continue,", "9/25 e) 17/25 My approach was calculating what the probability of nickel getting picked twice in a row is and subtracting that probability by 1. Probability of a nickel getting picked the first time is 2/5 and second time is 5/14. (2/5)*(5/14) = 1/7. So I was expecting 6/7 to show up but that choice is not given. I understand the OA and its methodology but what's wrong with my approach? Thanks! In this case it would be easier to calculate the probability in direct way: $$\\frac{non-nickel}{all}*\\frac{non-nickel -1}{all-1}=\\frac{9}{15}*\\frac{8}{14}=\\frac{12}{35}$$. The problem in your solution is that the opposite probability of two non-nickel coins in a row is not two nickel in a row, it's two nickels in a row plus either of two coins from two is nickel, (so basically the probability of at least one nickel from two picks): $$\\frac{6}{15}*\\frac{5}{14}+2*\\frac{6}{15}*\\frac{9}{14}=\\frac{23}{35}$$ $$P=1-\\frac{23}{35}=\\frac{12}{35}$$ Hi Bunuel, I tried a different approach but didn't end up with the same answer. Can you please clarify? Prob of pennies * Prob of Dimes * Permutation since order doesn't matter (5/15)(4/14)(2) = 4/21? Additionally, in your method above (9/15)(8/14) -- Why don't we factor in permutation? Math Expert Joined: 02 Sep 2009 Posts: 58335 In United States currency, a nickel is worth 5 cents [#permalink] ### Show Tags 27 May 2014, 01:08 russ9 wrote: Bunuel", "# Coin flipping game probability I have a 2 sided fair penny, if I flip heads 4 times in a row I win 10 dollars but it costs 1 dollar to play, should I play the game? Is the answer no because $$(10/2^4)-1$$ is negative? • you are correct. – Matthew Daly Nov 8 at 16:40 Win probability is (0.5)^4 and you'll have $10 Lose probability is 1-(0.5)^4 and it costs you$4 (10*0.0625) + (-4*0.9375) = -3.125 To put it in plain English, you will lose, on average, $3.125 for every $4 you will invest so if you came with a $400 to play 100 games, you will likely to leave with $87.5.", "A bag contains five chips numbered 2 through 6. Danya draws chips from the bag one at a time and sets them aside. After each draw, she totals the numbers on all the chips she has already drawn. What is the probability that at any point in this process her total will equal 10? Express your answer as a decimal to the nearest tenth. Answer: 1/5. The sequence starts with either a permutation of 2,3,5 or a permutation of 4,6. So the number of desired sequences is 3!2! + 2!3!, and the prob is (3!2!+2!3!)/5!. Jan 31: A drawer contains five socks: two green and three blue. What is the probability that two socks pulled out of the drawer at random will match? Express your answer as a common fraction. Answer: (2 choose 2)(3 choose 2) / (5 choose 2) = 3/10 Feb 1: A penny, a nickel and a dime are flipped. What is the probability that at least two coins land heads up and one of them is the nickel? Express your answer as a common fraction. Feb 2: When the circuit containing blinking lights A and B is turned on, lights A and B blink together. Then A blinks once every 5 seconds and B blinks once every 11 seconds. Lindsey looks at the two", "is in that range. Question 14 Explanation: Topics: Analyze and interpret various graphic and data representations, and use measures of central tendency (e.g., mean, median, mode) and spread to describe and interpret real-world data (Objective 0025). Question 15 #### If two fair coins are flipped, what is the probability that one will come up heads and the other tails? A $$\\large \\dfrac{1}{4}$$Hint: Think of the coins as a penny and a dime, and list all possibilities. B $$\\large \\dfrac{1}{3}$$Hint: This is a very common misconception. There are three possible outcomes -- both heads, both tails, and one of each -- but they are not equally likely. Think of the coins as a penny and a dime, and list all possibilities. C $$\\large \\dfrac{1}{2}$$Hint: The possibilities are HH, HT, TH, TT, and all are equally likely. Two of the four have one of each coin, so the probability is 2/4=1/2. D $$\\large \\dfrac{3}{4}$$Hint: Think of the coins as a penny and a dime, and list all possibilities. Question 15 Explanation: Topic: Calculate the probabilities of simple and compound events and of independent and dependent events (Objective 0026). Question 16 #### Tetrahedron Hint: All the faces of a tetrahedron are triangles. #### Triangular Prism Hint: A prism has two congruent, parallel bases, connected by parallelograms (since this is a right", "c) 13/35 d) 9/25 e) 17/25 My approach was calculating what the probability of nickel getting picked twice in a row is and subtracting that probability by 1. Probability of a nickel getting picked the first time is 2/5 and second time is 5/14. (2/5)*(5/14) = 1/7. So I was expecting 6/7 to show up but that choice is not given. I understand the OA and its methodology but what's wrong with my approach? Thanks! In this case it would be easier to calculate the probability in direct way: $$\\frac{non-nickel}{all}*\\frac{non-nickel -1}{all-1}=\\frac{9}{15}*\\frac{8}{14}=\\frac{12}{35}$$. The problem in your solution is that the opposite probability of two non-nickel coins in a row is not two nickel in a row, it's two nickels in a row plus either of two coins from two is nickel, (so basically the probability of at least one nickel from two picks): $$\\frac{6}{15}*\\frac{5}{14}+2*\\frac{6}{15}*\\frac{9}{14}=\\frac{23}{35}$$ $$P=1-\\frac{23}{35}=\\frac{12}{35}$$ Hi Bunuel, I tried a different approach but didn't end up with the same answer. Can you please clarify? Prob of pennies * Prob of Dimes * Permutation since order doesn't matter (5/15)(4/14)(2) = 4/21? Additionally, in your method above (9/15)(8/14) -- Why don't we factor in permutation? Hint: P(two non-nickel) = P(two pennies) + P(two dimes) + P(one nickel and one dime). As for your second question: (not nickel, not nickel) is one", "26, 2012 2. Jun 26, 2012 ### Villyer P(N) looks wrong, you cannot say that there are 5 total nickels and 10 total coins. Instead, you have to sum the probability of picking a nickel out of the right pocket and picking a nickel out of the left pocket. 3. Jun 27, 2012 ### HallsofIvy Yes, you can say \"there are 5 total nickels and 10 total coins\". That is given. What you cannot do is say \"there is a 50% chance I pick the right pocket\" when that is the probability you are asked to find. You have a total of 5 nickels, three in your left pocket and two in your right. If you take a nickel out of your pockets, the probability it came from your right pocket is 2/5. 4. Jun 27, 2012 ### Villyer You're right, what I meant to say was that you cannot use that fact alone to find his P(N), you need to know the distribution in each pocket. But the answer I got was 9/23. There is a 3/7 chance of picking a nickel out of the right pocket, and a 2/3 chance of picking a nickel out of the left pocket. So the probability of the nickel coming from the right pocket is $\\frac{\\frac{3}{7}}{\\frac{3}{7} + \\frac{2}{3}} = \\frac{9}{23}$.", "is a better choice from the perspective of risk-adjusted reward. ## Coin-Flip Casino: Reward Quiz To figure out which of these different scenarios is best, we need to consider reward and risk. When we talk about reward in this situation, we are talking about expected return. In the single-bet case, where we bet 1000 chips at once, our expected return is the chance of winning the bet times the amount that we would win plus the chance of losing the bet times the amount that we would lose. Since we are using a biased coin, we have a 51% chance of winning 1000 tokens and a 49% chance of losing 1000 tokens. If we multiply this out, we see that our expected return is 20 tokens. $\\mathbb{E} = 0.51 * 1000 + 0.49 * -1000$ $\\mathbb{E} = (0.51 - 0.49) * 1000$ $\\mathbb{E} = 0.02 * 1000 = 20$ What is the expected return in the multi-bet case, where we bet 1000 tokens across 1000 different tables? ## Coin-Flip Casino: Reward Quiz Solution In this case, we have the same chances of winning and losing - 51% and 49%, respectively - but the amount we stand to win or lose on any bet is only one dollar. Since we are placing 1000 bets, our overall expected value is", "# What is the optimal betting strategy to minimize expected waiting time to reach a goal in this coin flipping betting game? Suppose we play a game where we start with an initial amount of money $S$ and we wish to reach a larger target amount of money $T$. We a play a game in rounds as follows. There is a coin that has probability $p$ of coming up heads. We decide for the round how much to wager (less that or equal to our current amount of money) and if the coin flip is heads we win that amount, otherwise we lose that amount. Coin flips between different rounds are independent. Furthermore, if our current amount of money ever goes below $S$ then we are replenished back to $S$. We may as well assume $S = 1$. Then in terms of $T$ and $p$, and our current amount of money for a particular round, how much should we wager for that round? The goal is to minimize the expected number of rounds before our current amount of money is greater than or equal to $T$. I have a feeling that for $p$ around 0.75 or so, the optimal strategy might not be to wager everything each round. UPDATE: I ran some simulations for various $T$ with $p", "Jenny places $100$ pennies on a table, with $30$ showing heads and $70$ showing tails. She chooses $40$ distinct pennies uniformly at random and turns them over. That is, if a chosen penny was showing heads, she turns it to show tails; if a chosen penny was showing tails, she turns it to show heads. After this process, what is the expected number of pennies showing heads?", "even when rolled? #3: 3 numbers are selected at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction in lowest terms. #4: 9 fair coins are flipped. What is the probability that at least 4 are heads? #5: How many 3 digit numbers does not contain the digit 1 but have at least one digit that is 5? Answer: #1-$$\\dfrac {11} {16}$$ ; #2-$$\\dfrac {3} {4}$$ ; #3- $$\\dfrac {7} {8}$$ ; #4-$$\\dfrac {191} {256}$$ #5: 217", "once for each coworker. This feature is not available right now. But the very phrase tells us where we should classify that belief. Expected utility is an economic term summarizing the utility that an entity or aggregate economy is expected to reach under any number of circumstances. Palm Press with Magic Circle is a Pelvic Floor Exercise. I always weighed myself at the same time every week, wore more or less the same outfit and definitely the same shoes (which were ones I could slip off easily). It can't be done. Includes 83 gold, 16 silver and 1 bronze player. Current Weather. You may use it when selecting strike price for options trading. The 18 Karat Gold Melt Value Calculator, available below, can figure the total gold value of 18K gold items, measured by the weight unit of your choice. Sign in to comment. coin=TRUE shows a second plot with coin flip results (head or tail) Additional arguments from link{plot}. Transaction Value Median Transaction Value Tweets GTrends Active Addresses Top100ToTotal Fee in Reward. Last time we talked about independence of a pair of outcomes, but we can easily go on and talk about independence of a longer sequence of outcomes. Expected value is merely a weighted average. so for example if we have 2 coins, options will be", "possible outcomes = 12 We note that 5 of the 12 possible outcomes result in a number greater than 7 (that is 8, 9, 10, 11, 12) and thus there are 5 favorable outcomes. favorable outcomes = 5 The probability is the number of favorable outcomes divided by the number of possible outcomes: P(Greater than 7) = favorable outcomes/possible outcomes = $$\\frac{5}{12}$$ ≈ 0.4167 = 41.67% ### Page No. 426 Question 5. Christopher picked coins randomly from his piggy bank and got the numbers of coins shown in the table. Find each experimental probability. a. The next coin that Christopher picks is a quarter. $$\\frac{□}{□}$$ Answer: $$\\frac{6}{23}$$ Explanation: The table contains 7 pennies, 2 nickels, 8 dimes and 6 quarters, which are 7 + 2 + 8 + 6 = 23 coins in total and thus there are 23 possible outcomes. possible outcomes = 23 We note that 6 of the 23 coins are quarters and thus there are 6 favorable outcomes. favorable outcomes = 6 The probability is the number of favorable outcomes divided by the number of possible outcomes: P(Quarter) = favorable outcomes/possible outcomes = $$\\frac{6}{23}$$ ≈ 0.2609 = 26.09% Question 5. b. The next coin that Christopher picks is not a quarter. $$\\frac{□}{□}$$ Answer: $$\\frac{17}{23}$$ Explanation: The sum of the probabilities of an event and", "p(A=$10)5+p(A=$5)10=$7.50 which is identical to the expected content of the envelope that you already hold. If A=5, then P(A=10) is 0, but I think I understand what you're trying to say. You're just repeating the second calculation from my post (the one that says that the EV is 1/2*2X+1/2*X/2=3/2*X). Also note that the EV in dollars depends on the prior distribution, i.e. the method that was used to select the amount that went into the envelopes. (The EV in units of whatever the smaller amount is, is always 3/2, assuming that we don't know the value of A). Suppose for example that the person who prepared the envelopes flipped a coin until it came up heads and then put 2^n dollars and 2^(n+1) dollars, where n is the number of \"tails\" before the first \"heads\", into the envelopes. Then (assuming that we don't know the value of A) the EV in dollars is 1/2*1+1/4*2+1/8*4+...=1/2+1/2+1/2+... The possibility of infinite expectation values is probably the reason why this problem is so strange. • November 23rd 2006, 02:36 AM CaptainBlack Quote: Originally Posted by Fredrik The published articles about this do not dismiss it as nonsense. I think it probably is nonsense, but it's not easy to see why. It is nonsense because it is a calculation of the expectation for", "# Definition:Coin ## Definition $18$th century Irish penny A coin is a device for choosing a random number in the set $\\set {0, 1}$. It has been suggested that this device has other uses, but social anthropologists are divided as to what they might be. Perhaps they can resolve their differences by tossing a coin. ## Side of Coin The faces of a coin are known as its sides. They are distinguished one from another by means of different designs. The head of a coin is the side which traditionally bears an image of the head of state of the nation to which the coin is issued. ### Tails The tail of a coin is the opposite side to its head. It traditionally bears an image of cultural or historical significance to the nation to which the coin is issued. ## Biased Coin It is usual to consider that there is an equal probability for a coin to land either heads or tails. Such a coin is called a fair coin. However, discs of metal can crafted to look like coins, but with a higher probability of landing on one side rather than the other. Such a device is known as a biased coin. It is worth mentioning that as the manufacture of coins is such that the" ]

[ "Problem 21 Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even? $[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5)); label(\"A\",(-1,1)); label(\"B\",(2,1)); label(\"1\",(-.4,.4)); label(\"2\",(.4,.4)); label(\"3\",(.4,-.4)); label(\"4\",(-.4,-.4)); label(\"1\",(2.6,.4)); label(\"2\",(3.4,.4)); label(\"3\",(3,-.5)); [/asy]$ $\\textbf{(A)}\\ \\frac14\\qquad \\textbf{(B)}\\ \\frac13\\qquad \\textbf{(C)}\\ \\frac12\\qquad \\textbf{(D)}\\ \\frac23\\qquad \\textbf{(E)}\\ \\frac34$ ## Problem 22 At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is $\\frac25$. What fraction of the people in the room are married men? $\\textbf{(A)}\\ \\frac13\\qquad \\textbf{(B)}\\ \\frac38\\qquad \\textbf{(C)}\\ \\frac25\\qquad \\textbf{(D)}\\ \\frac{5}{12}\\qquad \\textbf{(E)}\\ \\frac35$ ## Problem 23 Tess runs counterclockwise around rectangular block $JKLM$. She lives at corner $J$. Which graph could represent her straight-line distance from home? $[asy] unitsize(5mm); pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2); draw(J--K--L--M--cycle); label(\"J\",J,NW); label(\"K\",K,SW); label(\"L\",L,SE); label(\"M\",M,NE); [/asy]$ ## Problem 24 In the figure, $ABCD$ is a rectangle and $EFGH$ is a parallelogram. Using the measurements given in the figure, what is the length $d$ of the segment that is perpendicular to $\\overline{HE}$ and $\\overline{FG}$? $[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair D=(0,0), C=(10,0), B=(10,8), A=(0,8); pair E=(4,8), F=(10,3), G=(6,0), H=(0,5); draw(A--B--C--D--cycle); draw(E--F--G--H--cycle); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"E\",E,N); label(\"F\",(10.8,3)); label(\"G\",G,S); label(\"H\",H,W); label(\"4\",A--E,N); label(\"6\",B--E,N);", "Also enter this probability in Table 1 (see Exercise 9). See the completed table in Exercise 9. Direct students to write their probabilities as fractions or as decimals. Fractions do not need to be reduced, as the unreduced fractions are meaningful based on the context. Exercise 9. Complete Table 1 by listing the outcomes for the other events and their probabilities based on the chance experiment for this scenario card. Event Outcomes Probability Odd number on Spinner 1 and a red card from the card bag Odd number on Spinner 1 Odd number on Spinner 1 and a blue card from the card bag Even number on Spinner 1 or a red card from the card bag Not picking a blue card from the card bag The following table organizes the responses to Exercises 7 – 9: Event Outcomes Probability Odd number on Spinner 1 and a red card from the card bag Note: If students struggle with the “and” language, discuss the outcomes with students. (, Red-E), (, Red-F), (, Red-E), (, Red-F) The probability is, which is approximately . Assign point to this event. Odd number on Spinner 1 Note: Students may indicate that the probability of this outcome is the same as the probability of just getting an odd number on the spinner without considering", "it sounds like P(T,circle) = \"probability of flipping a penny to land on tails AND picking a circle out of the bag\" 10. wio @wannabegurl is he right? 11. wannabegurl Yes @wio he is right thanks @jim_thompson5910 12. wio Then use the formula I wrote, since the flip is independent of the draw. $\\Pr(T,\\circ)=\\Pr(T)\\times \\Pr(\\circ)$ 13. wannabegurl Ok @wio if you say soo 14. jim_thompson5910 You could draw out a table to help you find the probability |dw:1361673237553:dw| 15. jim_thompson5910 Each cell would get the respective header from each row and column for example, this cell marked with an X |dw:1361673403499:dw| would have \"H, Square\" and this is the outcome where you flip a heads and pull out a square so you would enter that into the upper right cell to get |dw:1361673459247:dw|", "an odd number. $$\\text{1}$$; $$\\text{3}$$; $$\\text{5}$$ Is there an equal chance of getting an odd number and an even number? Explain. No, there is a higher chance of getting an odd number, because there are more possible outcomes for odd numbers. How could you use this spinner to design an unfair game of chance? The game would be unfair if a player has a small chance of winning, for example, if they win only if they get five. It would also be unfair if they win only if they get an even number. Draw a tree diagram with the first set of branches showing the possible outcomes for spinning the spinner in question $$\\text{3}$$. Then add the outcomes for a coin toss to each of the branches. How many possible outcomes are there altogether? Ten How many outcomes are there for getting a $$\\text{4}$$; H? Only one. What is the probability of getting a $$\\text{4}$$; H? One in ten or $$\\frac{\\text{1}}{\\text{10}}$$ How many outcomes are there for getting an even number and Heads? List them. Two: $$\\text{2}$$; H and $$\\text{4}$$; H What is the probability of getting an even number and Heads? The probability is $$\\text{2}$$ in $$\\text{10}$$, which simplifies to $$\\text{1}$$ in $$\\text{5}$$ or $$\\frac{\\text{1}}{\\text{5}}$$", "Probability is the ratio of favorable outcomes and all outcoms. P(E)=\"number of times event E was observed\"/\"total number of observations\" Part 1: Spinners Spinner 1 Spinner 2 Spinner 3 • Find a partner, then each select a different spinner. • Make a guess as whether your spinner might win more often over your partners spinner and state why. • Spin 20 times each. In a table, record the trial number, your spinner value, your partners spinner value, and whether you won or lost. Your table will have 4 columns and a row for each trial, i.e., at least 20 rows. • Use the results from your table to estimate the following probabilities 1. The probability of getting a ___ = 2. The probability of getting a ___ = 3. The probability of getting a ___ = 4. The probability that you win against your partner.", "that 25 isn’t evenly divisible by 7. We can think of our 25 possible outcomes from 2 calls to rand5 as a set of 25 “slots” in a list: results = [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ] Which we could then try to evenly distribute our 7 integers across: results = [ 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, ] It almost works . We could randomly pick an integer from the list, and the chances of getting any integer in 1…71…7 are pretty evenly distributed. Only problem is that extra 1, 2, 3, 4 in the last row. Any way we can sidestep this issue? What if we just “throw out” those extraneous results in the last row? 21 is divisible by 7. So if we just “throw out” our last 4 possible outcomes, we have a number of outcomes that are evenly divisible by 7. But what should we do if we get one of those 4 “throwaway” outcomes? We can just try the whole process again! Okay, this’ll work. But how do we translate our two", "to 4 as the first (tens) digit and 1 to 4 as the second (units) digit. Thus you resulting two-digit number can be any one of 11, 12, 13, 14, 21, ... 24, 31,..., 41, ... 44 This is the sample space and contains 16 elements. Since any single spin yields 1 to 4, equally likely, any of the 16 possible outcomes in the sample space are equally likely. For the first bullet in part (c) you want outcomes that are not in $E$ but are in $F.$ The sample space is $S$ so I would write the elements not in $E$ as $S - E.$ Your textbook or teacher may use some other notation. So for this bullet you want the outcomes in $S - E$ and in $F.$ This is the intersection of $S - E$ and $F$ which I would write $(S - E) \\cap F.$ Penny", "outcomes $=\\text{ }52$ $\\left( i \\right)$Number queens of red color $=\\text{ }2$ Number of favorable outcomes$~=\\text{ }2$ Hence, P(queen of red color)... read more ### A game consists of spinning arrow which comes to rest pointing at one of the numbers as shown below. If the outcomes are equally likely, find the probability that the pointer will point at: a number less than or equal to a number between and Solution: $\\left( v \\right)$ Favorable outcomes for a number less than or equal to $9\\text{ }are\\text{ }1,\\text{ }2,\\text{ }3,\\text{ }4,\\text{ }5,\\text{ }6,\\text{ }7,\\text{ }8,\\text{ }9$ So,... read more ### A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number, as shown in figure. What is the probability that it will point to:(iii) a number which is multiple of ? (iv) an even number? (iii) So, Favorable outcomes i.e. to get a multiple of $3$ are $3,6,9,$ and $12$ Therefore, total number of favorable outcomes i.e. to get a multiple of $3$ is $4$ We know that the Probability =... read more ### A game consists of spinning arrow which comes to rest pointing at one of the numbers as shown below. If the outcomes are equally likely, find the probability that the pointer will point", "outcomes. Find the probability that it will rest at (1) 8. (2) an odd number. (3) a number greater than 2. (4) a number less than 9 Total number of outcomes = 8 (1) Number of favourable outcomes (8) = 1 Therefore, required probability = $=\\frac{1}{8}$ Hence, the probability that the spinning arrow will rest at 8 is $\\frac{1}{8}$. (2) Number of favourable outcomes (odd number) = 1, 3, 5, 7 = 4 Therefore, required probability = $=\\frac{4}{8}\\phantom{\\rule{0ex}{0ex}}=\\frac{1}{2}$ Hence, the probability that the spinning arrow will rest at odd number is $\\frac{1}{2}$. (3) Number of favourable outcomes (a number greater than 2) = 3, 4, 5, 6, 7 and 8 = 6 Therefore, required probability = $=\\frac{6}{8}\\phantom{\\rule{0ex}{0ex}}=\\frac{3}{4}$ Hence, the probability that the spinning arrow will rest at a number greater than 2 is $\\frac{3}{4}$. (4) Number of favourable outcomes (a number less than 9) = 1, 2, 3, 4, 5, 6, 7, 8 = 8 Therefore, required probability = $=\\frac{8}{8}$ = 1 Hence, the probability that the spinning arrow will rest at a number less than 9 is 1. #### Question 11: There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that a card drawn shows, (1) a natural number. (2) a number", "your win rate past 85%. 7. Feb 5, 2016 ### DrChinese \"A\" would be the + outcomes of a polarization (spin) measurement at some angle a. \"Not A\" would be the - outcomes (note that + and - are arbitrary labels, you could also label it 1 and 0, or H and V). So \"B\" would be the + outcomes of a polarization (spin) measurement at some angle b. And so on. With an entangled pair, you can only measure 2 of the 3 angles at a time. So you would need to run a suitable number of trials of the 3 pairs: N(A, not B), N(B, not C) and N(A, not C). That would give you averages that can be plugged in. For a classical example: You have a room with 100 people in it. The A group are men, Not A are women. The B group are parents, Not B are people with no children. C are right handed, Not C are left handed. So we substitute into the formula N(A, not B) + N(B, not C) ≥ N(A, not C) and get: [1] N(men, not parents) + N(parents, left handed) >= N(men, lefthanded) The classical version will always be true, no matter how you select the people to go into the room. That is because", "in Figure 2.7 once and record the resulting $$(X, Y)$$ pair. (Recall that this spinner returns a pair of values.) Of course, the second method requires that the distribution of $$(X, Y)$$ is known. But in principle, there are always two ways of simulating a value $$x$$ of a random variable $$X$$. 1. (Simulate from the probability space.) Simulate an outcome $$\\omega$$ from the underlying probability space and set $$x = X(\\omega)$$. 2. (Simulate from the distribution.) Construct a spinner corresponding to the distribution of $$X$$ and spin it once to generate $$x$$. The second method requires that the distribution of $$X$$ is known. However, as we will see in many examples, it is common to specify the distribution of a random variable directly without defining the underlying probability space. Below is the Symbulate code for the second method, which corresponds to the spinner in Figure 2.7. Note that the probability space outcomes (the tickets in BoxModel) correspond to the possible $$(X, Y)$$ pairs, which are not equally likely (even though the 16 pairs of rolls are). We specify the probability of each outcome by using the probs option. To generate a single $$(X, Y)$$ pair, we spin the spinner once, and we draw one ticket from the box of pairs; this is why size = 1. xy_pairs", "the hotel Inverted Nine, each hotel room number is divisible by 6. How many rooms can we count with the three-digit number registered by digits 1,8,7,4,9? • Divisible by five How many different three-digit numbers divisible by five can we create from the digits 2, 4, 5? The digits can be repeated in the created number. • We roll We roll two dice A. - what is the probability that the sum of the falling numbers is at most 4 B. - is at least 10 C. - is divisible by 5? • What are 3 What are the two digits which when inserted in the blank spaces will make 234 _ _ divisible by 8? • Drawing from a hat When drawing numbers from a hat from 1 to 35, we select random given numbers. What is the probability that the drawn numbers will be divisible by 8 and 2? • The accompanying The accompanying table gives the probability distribution of the number of courses randomly selected student has registered Number of courses 1 2 3 4 5 6 7 Probability 0.02 0.03 0.1 0.3 0.4 - 0.01 respectively. a) Find the probability of a student registe • Three-digit numbers Use the numbers 4,5,8,9 to write all three-digit numbers without repetition. How many such numbers are there?", "result of rand5 as a digit in a two-digit base-5 integer. The first roll is the fives digit, and the second roll is the ones digit.) Can you adapt our function to use this math-based approach instead of the$results array? Because rand5 has only 5 possible outcomes, and we need 7 possible results for rand7, we need to call rand5 at least twice. When we call rand5 twice, we have 5*5=25 possible outcomes. If each of our 7 possible results has an equal chance of occurring, we'll need each outcome to occur in our set of possible outcomes the same number of times. So our total number of possible outcomes must be divisible by 7. 25 isn't evenly divisible by 7, but 21 is. So when we get one of the last 4 possible outcomes, we throw it out and roll again. So we roll twice and translate the result into a unique $outcomeNumber in the range 1..25. If the$outcomeNumber is greater than 21, we throw it out and re-roll. If not, we mod by 7 (and add 1). function rand7() { while (true) { // do our die rolls $roll1 = rand5();$roll2 = rand5(); $outcomeNumber = ($roll1-1) * 5 + ($roll2-1) + 1; // if we hit an extraneous // outcome we just re-roll if ($outcomeNumber >", "# RD Sharma Solutions Class 10 Mathematics Solutions for Probability Exercise 13.1 in Chapter 13 - Probability Question 65 Probability Exercise 13.1 A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number, 1, 2, 3, …., 12 as shown in figure. What is the probability that it will point to a number which is multiple of 3? Solution: When an experiment is repeated numerous times under identical circumstances, the proportion (or relative frequency) of times that the event is anticipated to occur is known as the probability of the event. Given: A game of chance consists of spinning an arrow which is equally likely to come to rest pointing number 1, 2, 3 ….12 Required to find: Probability of following Total numbers on the spin is 12 Favourable outcomes i.e. to get a multiple of 3 are 3, 6, 9, and 12 So, total number of favourable outcomes i.e. to get a multiple of 3 is 4 We know that, Probability = Number of favourable outcomes/ Total number of outcomes Thus, the probability of getting multiple of 3 = 4/12 = 1/3 Related Questions Lido Courses Teachers Book a Demo with us Syllabus Maths CBSE Maths ICSE Science CBSE Science ICSE English CBSE English ICSE", "# RD Sharma Solutions Class 10 Mathematics Solutions for Probability Exercise 13.1 in Chapter 13 - Probability Question 66 Probability Exercise 13.1 A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number, 1, 2, 3, …., 12 as shown in figure. What is the probability that it will point to an even number? Solution: The possibility that an event will occur is referred to as the probability of the event. In mathematics, this possibility of an event happening is represented by a number between 0 and 1. Given: A game of chance consists of spinning an arrow which is equally likely to come to rest pointing number 1, 2, 3 ….12 Required to find: Probability of following Total numbers on the spin is 12 Favourable outcomes i.e. to get an even number are 2, 4, 6, 8, 10, and 12 So, total number of favourable outcomes i.e. to get an even number is 6 We know that, Probability = Number of favourable outcomes/ Total number of outcomes Thus, the probability of getting an even number = 6/12 = 1/2 Related Questions Lido Courses Teachers Book a Demo with us Syllabus Maths CBSE Maths ICSE Science CBSE Science ICSE English CBSE English ICSE Coding Terms &", "1. Math 2. Statistics And Probability 3. let 0 1 3 that is all... # Question: let 0 1 3 that is all... ###### Question details Let Ω = {0, 1} 3 , that is, all possible (ordered) triples of zeros and ones. Suppose that all outcomes have equal probability. We define three random variables X1, X2, and X3 on this space representing the first, second, and third digit, respectively. We also define X = X1 + X2 + X3. compute: E(E(X|X1)|X2)", "below. For example, if the other team adopts the same algorithm, you would always tie. And I hate ties. Imagine playing the game with two slots and two spins of the wheel. On any given spin, you can get 0 through 9 with equal probability, meaning your expected payoff from a given spin is 4.5. So for this game your strategy is simple. If you get a 0 through 4 on the first roll, you’re likely to do better next time, so drop it in the one’s place. Otherwise, stick it in the tens place. There are 100 total possible orders of numbers that can come up in this two-turn game. The strategy described above fails only for numbers like 58 and 89 which have the first digit five or greater and the second digit bigger than the first. That’s ten total cases, so in the two-turn game you can play optimally 90% of the time. The probability distribution for your final outcome is that first digit will be 0 through 4 with probability $\\frac{1}{20}$ each, and 5 through 9 with probability $\\frac{3}{20}$ each. The expected value of this slot is 5.75. The units place is the same, but the probabilities for those two groups of digits are reversed, and its expected value is 3.25. Now we understand", "Krishna 0 Step 1: See the given figure and analyse the given information GIVEN: A game of chance spinning an arrow,which come to rest pointing any of the number 1,2,3,4,5,6,7,8 So, total number of outcomes = 8 Step 2: Determine the probability that it(arrow) will point at \"8\" Favourable outcomes of getting 8 = 1 The probability of pointing 8 = \\frac{\\text{Favourable outcome}}{\\text{total outcomes}} = \\frac{1}{8} Step 3: Find the probability that it will point at an odd number List out the odd number = {1, 3, 5, 7} Favourable outcomes of pointing an odd number = 4 The probability that arrow will point at an odd number = \\frac{\\text{Favourable outcome}}{\\text{total outcomes}} = \\frac{4}{8}", "3 letters followed by 3 digits Is this a true or false question? If so I think that it is false. ANSWERED AT 27/03/2020 - 12:45 PM QUESTION POSTED AT 27/03/2020 - 12:45 PM A spinner has four sections lettered A, B, C, and D. The table shows the results of several spins. Find the experimental probability of spinning each letter as a fraction in simplest form, a decimal, and a percent. Answer: Probability of getting A is given by =0.35=35% Probability of getting B is given by Probability of getting C is given by Probability of getting D is given by Step-by-step explanation: Since we have given that Number of possible outcomes of getting A = 14 Number of possible outcomes of getting B = 7 Number of possible outcomes of getting C = 11 Number of possible outcomes of getting D = 8 Total number of possible outcomes would be 14+7+11+8 =40 So, Probability of getting A is given by Probability of getting B is given by Probability of getting C is given by Probability of getting D is given by ANSWERED AT 27/03/2020 - 12:39 PM QUESTION POSTED AT 27/03/2020 - 12:39 PM April made \\$440 in tips last night waitressing. She had to give 20% of her tips to the boys who clean the", "## Geometry: Common Core (15th Edition) $\\frac{3}{8}$ To calculate probability, we need to divide the number of chances for a certain outcome to occur by the total number of outcomes. In this case, there are $3$ numbers greater than $5$ on the spinner, and $8$ numbers in total. When we divide $3$ by $8$, we get $\\frac{3}{8}$, which is the probability of getting a number greater than $5$." ]

[ "is even. In this case there are 4 possible digits (2, 4, 6, 8) since the first digit cannot be 0. There are again five possible digits for the other 5 digits and 10 ways to place the three odd digits in those 5 digts. There are $10(5^5)4= 125000$ ways in this case. Together there are 156250+ 125000= 281250 ways to do this. 3. Dec 21, 2012 ### rcgldr grinding this out with a simple program confirms the total is 281250: Code (Text): int main(int argc, char **argv) { int i, j, k, even, odd, total; total = 0; for(i = 100000; i < 1000000; i++){ j = i; even = odd = 0; for(k = 0; k < 6; k++){ if(j&1) odd += 1; else even += 1; j /= 10; } if((odd == 3) && (even == 3)) total += 1; } printf(\"%d\\n\", total); return(0); } 4. Dec 23, 2012 ### Vineeth T Then is it confirm that the answer is 281250. But the answer is given as 179550.(source:Problems In Mathematics With Hints And Solutions by V Govrov) Thanks for everyone. 5. Dec 23, 2012 ### rcgldr I think of this as the number of permutations for a multiset of 2 odds and 3 evens which is $$\\frac{5!}{2! \\ 3!} = 10$$ {OOEEE, OEOEE, OEEOE, OEEEO,", "Joined: 02 Oct 2017 Posts: 729 Re: If an integer n is to be chosen at random from the integers [#permalink] ### Show Tags 31 May 2018, 08:42 Total 96 Number 48 even and 48 odd For n=even. N+2=even Multiple of any two even is always divided by 8 so all 48 cases will be divided by 8 For odd It will include those cases in which multiple of 8 comes in between Eg 7*8*9 15*16*17 Etc 8,16,.......96 Total 12 cases So total cases= 48+12=60 Probability=60/96=5/8 Give kudos if it helps Posted from my mobile device _________________ Give kudos if you like the post Senior Manager Joined: 18 Jun 2018 Posts: 267 Re: If an integer n is to be chosen at random from the integers [#permalink] ### Show Tags 11 Aug 2018, 22:22 OA: C There are two different possible pattern Pattern 1: n= odd, then n+1=Even and n+2 = odd i.e odd*even*odd Number of such arrangement $$=\\frac{96}{2}=48$$ Pattern 2: n= Even, then n+1=Odd and n+2 = Even i.e even*odd*even Number of such arrangement $$=\\frac{96}{2}=48$$ In Pattern 1 : n+1 should be divisible by 8 as n and n+2 are odd $$n+1=8p ; n = 8p-1$$ Possible value of $$n = 7,15,23,31,39,47,55,63,71,79,87,95$$ Total possible values of n = 12 In pattern 2: all 48 are divisible by", "question. Kudos [?]: [0], given: 3 Math Expert Joined: 02 Sep 2009 Posts: 42606 Kudos [?]: 135613 [0], given: 12705 Re: If n is a positive even integer, what is the probability that the unit [#permalink] ### Show Tags 30 Oct 2017, 22:04 Expert's post 1 This post was BOOKMARKED gotonishu wrote: JeffTargetTestPrep wrote: nkmungila wrote: If n is a positive even integer, what is the probability that the unit’s digit of $$n^4$$ is 6? A. 1 B. $$\\frac{4}{5}$$ C. $$\\frac{3}{5}$$ D. $$\\frac{2}{5}$$ E. $$\\frac{1}{2}$$ Let’s write out all possible units digits of even numbers: 0^4 = 0 2^4 = 6 4^4 = 6 6^4 = 6 8^4 = 6 Thus, the probability is 4/5. But the question says n is positive. So we cant take 0 as a value for n. Also, where does it say that n has to be a single digit number. This seems like a bad question. Jeff there considers the units digits of n, not n itself and not single-digits. So, while n itself cannot be 0, its units digit can, for example n can be 10. _________________ Kudos [?]: 135613 [0], given: 12705 Re: If n is a positive even integer, what is the probability that the unit [#permalink] 30 Oct 2017, 22:04 Display posts from previous: Sort by", "+0 # help counting 0 77 1 How many even, three-digit positive integers have the property that exactly two of the integer's digits are even? Apr 16, 2022 #1 +2437 +1 There are 2 cases to consider: Even - Odd - Even Odd - Even - Even For the first case, there are $$4 \\times 5 \\times 5 = 100$$ numbers. (Remember, the leading digit can't be 0!!) For the second case, there are $$5 \\times 5 \\times 5 = 125$$ numbers. Thus, there are a total of $$100 + 125 = \\color{brown}\\boxed{225}$$ numbers that meet these criteria. Apr 16, 2022", "# Math Help - probability Help! 1. ## probability Help! Help in the probability.... How many 3-digits numbers can be formed from the digits 0,1,2,3,4,5 and 6 if each digit can be used only once? And how may of these are even numbers? Thanks for the help... Regards! 2. The number of 3-digits number (including the numbers that start with 0) is $A_6^3=6\\cdot 5\\cdot 4=120$ But we have to eliminate the numbers that star with 0, wich means $A_5^2=5\\cdot 4=20$ So, we have $120-20=100$ 3-digits numbers. To count the even numbers we put an even digit on the unit place. There are 4 possibilities: 0, 2, 4, 6. If the unit digit is 0, there are $A_5^2=20$ even numbers. If the unit digit is 2, 4 or 6, there are $A_5^2-A_4^1=16$ even numbers. So, there are $20+3\\cdot 16=20+48=68$ even numbers. 3. wow! this is so great... thanks a lot!", "7. Thanks 8. Hello, saberteeth! 1) Two dice are thrown simultaneously. Find the probability that the product of the two numbers on the two dice is an even number. There are $6^2=36$ possible outcomes. The product is odd if both numbers are odd. There are 3 choices of an odd number for the first die: $\\{1,3,5\\}$ . . and 3 choices of odd number for the second die. Hence, there are: . $3\\cdot3\\:=\\:9$ odd-product outcomes. So there are: . $36 - 9 \\:=\\:27$ even-product outcomes. Therefore: . $P(\\text{even product}) \\:=\\:\\frac{27}{36} \\;=\\;\\frac{3}{4}$ 3) What is the probability that the product of two consecutive non-negative integers will be a number ending with 0? Two consecutive integers can have the following pairs of units-digits: . . $(0,1), (1,2), (2,3),(3,4), (4,5), (5,6), (6,7), (7,8), (8,9), (9,0)$ The unit-digits of their product are: . $0,\\;2,\\;6,\\;2,\\;0,\\;0,\\;2,\\;6,\\;2,\\;0$ Got it? 9. Originally Posted by Soroban Hello, saberteeth! There are $6^2=36$ possible outcomes. The product is odd if both numbers are odd. There are 3 choices of an odd number for the first die: $\\{1,3,5\\}$ . . and 3 choices of odd number for the second die. Hence, there are: . $3\\cdot3\\:=\\:9$ odd-product outcomes. So there are: . $36 - 9 \\:=\\:27$ even-product outcomes. Therefore: . $P(\\text{even product}) \\:=\\:\\frac{27}{36} \\;=\\;\\frac{3}{4}$ Two consecutive integers can have the following", "+0 # Probability 0 26 4 A machine randomly generates one of the nine numbers 1, 2, 3, … , 9 with equal likelihood. What is the probability that when Tsuni uses this machine to generate four numbers their product is divisible by 4? Express your answer as a common fraction. Jul 22, 2022 #1 +2339 +1 Note that for the product to be divisible by 4, there need to be at least 2 even numbers. The probability of rolling only odd numbers is $${5 \\over 9}^4 = {1025 \\over 6561}$$ The probability of rolling exactly 1 even number is $$4 \\times 5^3 \\times {4 \\choose 3} = 2000$$ So, the probability is $$1 - ({1025 \\over 6561} + {2000 \\over 6561}) = \\text{_____}$$ Jul 22, 2022 #4 +2339 +1 BuilderBoi Jul 23, 2022 #2 0 BuilderBoi: A computer code gives the following numbers: 9^4 ==6,561 permutations. [This is based on the assumption that numbers are replaced at each draw] The computer calculates: The probability is; 4,936 / 6,561 Jul 22, 2022 #3 +2339 +1 REVISED ANSWER: There are $$9^4 = 6561$$ ways to get the numbers. The easiest way to solve this is to use complementary counting, in which we count what we don't want, and subtract that from 1 Note that the probability of getting an", "Whatever is in the tens place, its value will be a multiple of ten. Two is a factor of ten. Two is therefore a factor of any whole multiple of ten. Adding an even number (a multiple of two) to any number X can not change the evenness of X. So it makes no difference how many tens or hundreds or thousands we add; the units place alone determines whether the number is divisible by 2. This is true in all even bases, not just 10. We can illustrate this by writing 123 in bases 8, 6, 4 and 2. $123 = (173)_8 = (323)_6 = (1323)_4 = (1111011)_2$ In each of the above representations of 123, the digit in the units place is an odd number. That’s enough information to settle the question of evenness. It doesn’t matter that we don’t immediately recognize the that $(1232_4) = 123$; we can tell that it’s even just by checking the last digit. Now let’s see what happens when we rewrite 123 in a few odd bases. $123 = (146)_9 = (234)_7 = (443)_5 = (4120)_3$ The last digit can be even or odd but we know these numbers are all odd. Clearly the units place is not a reliable indicator of evenness. Why not? Because the radix is odd,", "Whatever is in the tens place, its value will be a multiple of ten. Two is a factor of ten. Two is therefore a factor of any whole multiple of ten. Adding an even number (a multiple of two) to any number X can not change the evenness of X. So it makes no difference how many tens or hundreds or thousands we add; the units place alone determines whether the number is divisible by 2. This is true in all even bases, not just 10. We can illustrate this by writing 123 in bases 8, 6, 4 and 2. $123 = (173)_8 = (323)_6 = (1323)_4 = (1111011)_2$ In each of the above representations of 123, the digit in the units place is an odd number. That’s enough information to settle the question of evenness. It doesn’t matter that we don’t immediately recognize the that $(1232_4) = 123$; we can tell that it’s even just by checking the last digit. Now let’s see what happens when we rewrite 123 in a few odd bases. $123 = (146)_9 = (234)_7 = (443)_5 = (4120)_3$ The last digit can be even or odd but we know these numbers are all odd. Clearly the units place is not a reliable indicator of evenness. Why not? Because the radix is odd,", "__ 28 __ 40 __ 42 __ 44 __ In each case, any of the 5 even digits (0, 2, 4, 6, 8) can take the units place. So acceptable cases = 5*5 = 25 Total number of integers = 200 Probability = 25/200 = 1/8 Nice approach. I didn't get your approach easily first May I know this approach more broadly ? Here is the way I thought about it: The number has to be chosen from 250 to 449. All digits have to be even. Let's consider the acceptable hundreds and tens digits : 26 __ - The units digit can be any even digit since all numbers in the range 260 to 269 are included. 28 __ - The units digit can be any even digit since all numbers in the range 280 to 289 are included. 3 __ __ - All such numbers are not acceptable since 3 is odd. Move on to 4 __ __ 40 __ - The units digit can be any even digit since all numbers in the range 400 to 409 are included. 42 __ - The units digit can be any even digit since all numbers in the range 420 to 429 are included. 44 __ - The units digit can be any even digit since all numbers", "be taken into account to build ... ### Module 4 - Multiple Logistic Regression - ReStore ... Multiple Logistic Regression ... our sample is 0.816. Odds and odds ... We don‟t actually have to calculate the odds directly from the numbers of students ... ### 1.1.5 Divisibility Rules - QED Infinity | Niagara ... The product of two odd numbers is odd, ... which is an even number, so $634\\;816$ is divisible by $2$. Deleting all but the last two digits leaves $16$, ... ### Mr. Number - Results for 1-405-816-XXXX Identify and stop mobile spam with Mr. Number. ... Phone numbers beginning with 405-816 are allocated to Oklahoma, in particular Oklahoma City in Cleveland County. ### Numeracy Starters. Put a circle around the EVEN numbers ... ... \"Numeracy Starters. Put a circle around the EVEN numbers: 932471642611 Put a circle around the ODD numbers: ... 2211 The difference between 22 and 816 ... ### An Encyclopedia of Probability - The Numbers - WSJ The Book of Odds, which launched Wednesday, hasn’t quite fulfilled all its goals. ... The Wall Street Journal examines numbers in the news, ... ### What is the prime factorization of 816 [SOLVED] What is the prime factorization of 816 ... Is 816 an odd number? ... Is 816 a composite number? Prime Factorization", "( 1,2,3,4,5) Again, the last number can be chosen in $2C1$ ways and the remaining 4 numbers can permute in $4!$ ways. The required number of ways for this case= $2*4!= 48$ Adding the number of ways in the two cases, 5 digit numbers that are divisible by 6= 108 Required Probability = $\\frac{108}{600} =$0.18$=$18%", "letters out of which \"g\" and \"o\" are represented twice. Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\\frac{9!}{4!3!2!}$$. So, the number of arrangements of four letters EEOO is 4!/(2!2!). Hope it's clear. _________________ Kudos [?]: 133076 [0], given: 12403 BSchool Forum Moderator Joined: 17 Jun 2016 Posts: 449 Kudos [?]: 215 [0], given: 199 Location: India GMAT 1: 720 Q49 V39 GMAT 2: 710 Q50 V37 GPA: 3.65 WE: Engineering (Energy and Utilities) Re: Gambling with 4 dice, what is the probability of getting an [#permalink] ### Show Tags 25 Jun 2017, 11:22 Gambling with 4 dice, what is the probability of getting an even sum? A. 3/4 B. 1/2 C. 2/3 D. 1/4 E. 1/3 What can a sum of two positive integers (or for that matter any integers) be ? It can only be EVEN or ODD.. there is no other possibility ...(note that I said INTEGERS - not just any numbers) So, a dice when rolled, can only produce integers .. So summing the results of any number of dice can only produce an ODD or an EVEN output. Hence, there are 50% chance that our output will be EVEN (and 50% chances that output will be ODD) So, Option B 1/2 _________________", "Browse Questions # If four digit numbers greater than 5000 are randomly formed from the digits 0,1,3,5 and 7,what is the probability of forming number divisible by 5 when repetition of digits is not allowed. $\\begin{array}{1 1}(A)\\;\\large\\frac{3}{8}\\\\(B)\\;\\large\\frac{3}{5}\\\\(C)\\;\\large\\frac{4}{7}\\\\(D)\\;\\large\\frac{7}{8}\\end{array}$ Toolbox: • Required probability=$\\large\\frac{n(E)}{n(S)}$ Step 1: 4 digit number greater than 5000 are to be formed Digits -0,1,3,5,7 Number to be divisible by 5 $\\therefore$ A number can be divisible by 5 only when its units digit is 0,5 Step 2: Since the number have to be greater than 5000.The thousand place will be 5 or 7 and the rest three digits will have options of 4,3,2 outcomes respectively. $\\therefore$ Total number of outcomes n(S)=24+24=48 Step 3: Number of favorable outcomes : For a number to be divisible by 5.The unit digit should be 0 or 5.The unit digit should be 0 or 5 and thousands digits are 5 or 7. $\\therefore$ Options for tens and hundreth digits are 3 and 2. $\\therefore$ Favorable outcomes n(E)=6+6+6=18 Step 4: $\\therefore$ Required probability =$\\large\\frac{n(E)}{n(S)}$ $\\Rightarrow \\large\\frac{18}{48}$ $\\Rightarrow \\large\\frac{3}{8}$ Hence (A) is the correct answer.", "note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits? 1) I have four slots _ _ _ _ that can contain only {0,2,3,5,6,7,8,9}. Four of them are even, four are odd. So the probability of having an even digit is $$4/8=1/2$$, as well having an odd digit. The solution can be calculated as: Probability of at least one even digit = Probability of one even digit + Probability of two even digits + Probability of three even digits + Probability of four even digits P1 = $$1/2$$ P2 = $$1/2^2=1/4$$ P3 = $$1/2^3=1/8$$ P4 = $$1/2^4=1/16$$ P = P1+P2+P3+P4 = $$15/16$$ 2) I have the following situation _ _ _ _ 1 1 1 Empty slots can contain only {2,3,4,5,6,7,8,9} Prime digits are {2,3,5,7}, not prime digits are {4,6,8,9}. So the probability of having a prime digit is $$4/8=1/2$$, as well having a non prime digit. I think I have the same situation of the previous problem: filling 4 empty slots with digits (even or prime) that have the same probability of being in the slot ($$1/2$$) or not being ($$1/2$$). So I", "(ii) When 4 or 6 are at unit place Number of numbers = 32 Total three digit even number = 20 + 32 = 52", "Aug 2013, 01:39 1 Bunuel wrote: If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8? A. 1/4 B. 3/8 C. 1/2 D. 5/8 E. 3/4 $$n(n + 1)(n + 2)$$ is divisible by 8 in two cases: A. $$n=even$$, in this case $$n+2=even$$ too and as $$n$$ and $$n+2$$ are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. $$n+1$$ is itself divisible by 8; (Notice that these two sets have no overlaps, as when $$n$$ and $$n+2$$ are even then $$n+1$$ is odd and when $$n+1$$ is divisible by 8 (so even) then $$n$$ and $$n+2$$ are odd.) Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8. Similar question: divisible-by-12-probability-121561.html Hope it helps. Hi, Can we use this formula for this problem # of multiples of x in the\\ range = frac{Last multiple of x in", "from the numbers 250 to 449, inclusiv [#permalink] Show Tags 07 Jun 2016, 22:54 AbdurRakib wrote: VeritasPrepKarishma wrote: DrAB wrote: If an integer is randomly chosen from the numbers 250 to 449, inclusive, what is the probability that all the digits of the number are even? (A) 3/5 (B) 17/50 (C)1/8 (D) 51/199 (E) 1/4 Another method is to enumerate the hundreds and tens digits which are acceptable: 26 __ 28 __ 40 __ 42 __ 44 __ In each case, any of the 5 even digits (0, 2, 4, 6, 8) can take the units place. So acceptable cases = 5*5 = 25 Total number of integers = 200 Probability = 25/200 = 1/8 Nice approach. I didn't get your approach easily first May I know this approach more broadly ? Here is the way I thought about it: The number has to be chosen from 250 to 449. All digits have to be even. Let's consider the acceptable hundreds and tens digits : 26 __ - The units digit can be any even digit since all numbers in the range 260 to 269 are included. 28 __ - The units digit can be any even digit since all numbers in the range 280 to 289 are included. 3 __ __ - All such numbers are not", "# Probability in Divisibility | AMC-10A, 2003 | Problem 15 Contents [hide] Try this beautiful problem from Probability based on divisibility. ## Probability in Divisibility - AMC-10A, 2003- Problem 15 What is the probability that an integer in the set ${1,2,3,…,100}$ is divisible by $2$ and not divisible by $3$? • $\\frac {33}{100}$ • $\\frac{1}{6}$ • $\\frac{17}{50}$ • $\\frac{1}{2}$ • $\\frac{18}{25}$ ### Key Concepts Number system Probability divisibility Answer: $\\frac{17}{50}$ AMC-10A (2003) Problem 15 Pre College Mathematics ## Try with Hints There are total number of integers are $100$.and numer of integers divisible by $2$ is $\\frac{100}{2}$=$50$. Now we have to find out divisible by $2$ and not divisible by $3$. so at first we have to find out the find out the numbers of integers which are divisible by $2$ and $3$ both....... can you finish the problem........ To be divisible by both $2$ and $3$, a number must be divisible by the lcm of $(2,3)=6$. Therefore numbers of integers which are divisible by $6$=$\\frac{100}{6}=16$ (between $1$ & $100$) can you finish the problem........ Therefore the number of integers wcich are divisible by $2$ and not divisible by $3$= $50 - 16=34$. So require probability= $\\frac{34}{100}=\\frac{17}{50}$ ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed. ###", "# Math Help - probability help 1. ## probability help The digits 2, 3, 4, 7 and 8 are each used once to form a five-digit number. What is the probability that the tens digit is odd and the number is divisible by 4? Express your 2. Hello, sri340! The digits 2, 3, 4, 7 and 8 are each used once to form a five-digit number. What is the probability that the tens digit is odd and the number is divisible by 4? With a little Thought, the solution is quite straight forward. We have a 5-digit number and its tens digit is odd. There are only 2 choices for the odd digit. The tens digit is 3: . _ _ _ 3 _ To be divisible by 4, the units digit must be 2. And the other 3 digits can be placed in $3! \\:=\\: 6$ ways. The tens digit is 7:_ _ _ 7 _ To be divisible by 4, the units digit must be 2. And the other 3 digits can be placed in $3! \\:=\\:6$ ways. Hence, there are: . $6 + 6 \\:=\\:12$ such numbers. Since there are $5! = 120$ possible 5-digit numbers, . . The probabiity is: . $\\frac{12}{120} \\:=\\:\\frac{1}{10}$" ]

[ "provide more opportunities for primary and secondary school students to experience international engagement. (sc : http://asmo2u.com/about-us) Berikut ini problems and solution ASMO 2019 grade 9 1. When two dice are rolled, find the probability of getting a sum that is divisible by 4. Mata dadu berjumlah $$4 : \\{(2, 2), (1, 3), (3, 1)\\}$$ ada $$3$$ Mata dadu berjumlah $$8 : \\{(4, 4), (2, 6), (3, 5), (5, 3), (6, 2)\\}$$ ada $$5$$ Mata dadu berjumlah $$12 : \\{(6, 6)\\}$$ ada $$1$$ $$\\frac{3+5+1}{36}=\\frac{9}{36}=\\frac{1}{4}$$ 2. The tens digit of a two-digit number is two more than the units digit. When this two-digit number is divided by the sum of its digits, the answer is 6 remainder 3. Determine the sum of the digits of the two-digit number. not yet available 3. The diagram shows a rectangle with length 9 cm and width 7 cm. One of the diagonals of the rectangle has been divided into seven equal parts. Determine the area of the shaded region. not yet available 4. If $$x + \\sqrt{xy} + y = 9$$ and $$x^2 + xy + y^2=27$$, then determine the value of $$x – \\sqrt{xy} + y$$. Misalkan $$𝑥 − \\sqrt{𝑥𝑦} + 𝑦 = 𝐴$$ Kurangkan persamaan $$𝑥 + \\sqrt{𝑥𝑦} + 𝑦 = 9$$ dan $$𝑥 − \\sqrt{𝑥𝑦} + 𝑦 = 𝐴$$, diperoleh", "Browse Questions # If four digit numbers greater than 5000 are randomly formed from the digits 0,1,3,5 and 7,what is the probability of forming number divisible by 5 when repetition of digits is not allowed. $\\begin{array}{1 1}(A)\\;\\large\\frac{3}{8}\\\\(B)\\;\\large\\frac{3}{5}\\\\(C)\\;\\large\\frac{4}{7}\\\\(D)\\;\\large\\frac{7}{8}\\end{array}$ Toolbox: • Required probability=$\\large\\frac{n(E)}{n(S)}$ Step 1: 4 digit number greater than 5000 are to be formed Digits -0,1,3,5,7 Number to be divisible by 5 $\\therefore$ A number can be divisible by 5 only when its units digit is 0,5 Step 2: Since the number have to be greater than 5000.The thousand place will be 5 or 7 and the rest three digits will have options of 4,3,2 outcomes respectively. $\\therefore$ Total number of outcomes n(S)=24+24=48 Step 3: Number of favorable outcomes : For a number to be divisible by 5.The unit digit should be 0 or 5.The unit digit should be 0 or 5 and thousands digits are 5 or 7. $\\therefore$ Options for tens and hundreth digits are 3 and 2. $\\therefore$ Favorable outcomes n(E)=6+6+6=18 Step 4: $\\therefore$ Required probability =$\\large\\frac{n(E)}{n(S)}$ $\\Rightarrow \\large\\frac{18}{48}$ $\\Rightarrow \\large\\frac{3}{8}$ Hence (A) is the correct answer.", "Courses Courses for Kids Free study material Free LIVE classes More # Cards marked with numbers 3, 4, 5, …...50 are placed in a box and mixed through. One card is drawn from the box. Find the probability that the number on the card is(i) Divisible by 6(ii) A single-digit number Last updated date: 17th Mar 2023 Total views: 304.2k Views today: 8.83k Verified 304.2k+ views Hint: To solve the question, we have to calculate the number of possible outcomes at the given conditions. To calculate probability, divide the obtained number of outcomes to the total number of outcomes obtained when a card is drawn. The numbers on the cards placed in a box given are 3, 4, 5, …...50 The number of possible outcomes when a card is drawn from the above-mentioned cards = 50 – 2 = 48 Since, only 1, 2 numbers have not marked the cards considering the series of natural numbers. The divisibility rule of 6 is that the number should be divisible by both numbers 2 and 3. The number is divisible by 2 if it is an even number. The number is divisible by 3 when the sum of all the digits of the number is a multiple of 3. Since, the marked numbers are in the range of the first", "Problem 21 Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even? $[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5)); label(\"A\",(-1,1)); label(\"B\",(2,1)); label(\"1\",(-.4,.4)); label(\"2\",(.4,.4)); label(\"3\",(.4,-.4)); label(\"4\",(-.4,-.4)); label(\"1\",(2.6,.4)); label(\"2\",(3.4,.4)); label(\"3\",(3,-.5)); [/asy]$ $\\textbf{(A)}\\ \\frac14\\qquad \\textbf{(B)}\\ \\frac13\\qquad \\textbf{(C)}\\ \\frac12\\qquad \\textbf{(D)}\\ \\frac23\\qquad \\textbf{(E)}\\ \\frac34$ ## Problem 22 At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is $\\frac25$. What fraction of the people in the room are married men? $\\textbf{(A)}\\ \\frac13\\qquad \\textbf{(B)}\\ \\frac38\\qquad \\textbf{(C)}\\ \\frac25\\qquad \\textbf{(D)}\\ \\frac{5}{12}\\qquad \\textbf{(E)}\\ \\frac35$ ## Problem 23 Tess runs counterclockwise around rectangular block $JKLM$. She lives at corner $J$. Which graph could represent her straight-line distance from home? $[asy] unitsize(5mm); pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2); draw(J--K--L--M--cycle); label(\"J\",J,NW); label(\"K\",K,SW); label(\"L\",L,SE); label(\"M\",M,NE); [/asy]$ ## Problem 24 In the figure, $ABCD$ is a rectangle and $EFGH$ is a parallelogram. Using the measurements given in the figure, what is the length $d$ of the segment that is perpendicular to $\\overline{HE}$ and $\\overline{FG}$? $[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair D=(0,0), C=(10,0), B=(10,8), A=(0,8); pair E=(4,8), F=(10,3), G=(6,0), H=(0,5); draw(A--B--C--D--cycle); draw(E--F--G--H--cycle); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"E\",E,N); label(\"F\",(10.8,3)); label(\"G\",G,S); label(\"H\",H,W); label(\"4\",A--E,N); label(\"6\",B--E,N);", "subsets divisible by 3 and 78 not. – kaine Nov 8 '13 at 15:08 • Edited solution is clearer. I appreciate that! – Satish Ramanathan Nov 8 '13 at 15:25 Note that our set has $34$ members equivalent to $1$ modulo $3$ (that is, with remainder $1$ upon division by $3$), $33$ equivalent to $2$, and $33$ equivalent to $0$. Suppose we draw three numbers at random to create our subset. In order to get a sum divisible by $3$, we could draw any of the following sets of remainders: $$2,2,2\\\\ 1,1,1\\\\ 0,0,0\\\\ 0,1,2 \\text{ (In each of }3! \\text{ arrangements)}$$ So, the number of possible drawings that will yield a multiple of $3$ (order matters here) is $$(34\\times 33\\times 32)+ (33\\times 32\\times 31)\\times 2+(33\\times 34 \\times 33)\\times (3\\times 2 \\times 1)$$ There are $100\\times 99\\times 98$ possible ways of drawing three numbers, our probability is $$\\frac{(34\\times 33\\times 32)+ (33\\times 32\\times 31)\\times 2+(33\\times 34 \\times 33)\\times (3\\times 2 \\times 1)}{100\\times 99 \\times 98}$$ Which, upon simplification, yields an answer of $\\frac{817}{2450}$. It is also possible to approach this with combinations instead of permutations. That is, noting we must have either all in one category or one from each category, our total number of possibilities is $$\\binom{33}{3} + \\binom{33}{3} + \\binom{34}{3} + 33\\times 34 \\times 33$$ producing a probability", "5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 10 11 b) Determine the probability that an even number was spun and a sum that is a multiple of four was obtained. Think: On the spinner there are two even numbers, $2$2 and $4$4. We now look down those columns for any sums that are a multiple of $4$4. Do: $\\frac{3}{30}$330 c) Determine the probability the sum was greater than $7$7 or a multiple of $4$4. Think: We need to look in the table for all sums greater than $7$7, or that are a multiple of $4$4, or both. I've highlighted those in the table below. Spinner 1 2 3 4 5 Dice 1 2 3 4 5 6 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 10 11 Do: $\\frac{13}{30}$1330 ## A Tree Diagram We've discussed tree diagrams in detail before, you can review that here if you'd like to. ## A Venn Diagram and a Two-Way Table Venn diagrams and Two-way tables are two different formats for representing the same ideas. You can review those here. #### Example 4: Sam", "$n$ are equal, then the probability ofobtaining that value is 1 ($100\\%$).\\par For a 3-spinner, I have 3 cases: • All three numbers are different. There is a probability of 1/3 ($33.33\\%$) of obtaining any of the three numbers. It corresponds to the polynomial $x^m + x^n + x^p$, with $m \\neq n \\neq p$ • Two of the three numbers are equal. There is a probability of 1/3 ($33.33\\%$) of obtaining the unique number, and a probability of $66.67\\%$ of obtaining the number which appears twice. It corresponds to the polynomial $x^m + 2x^n$, $m \\neq n$. • All 3 numbers are equal. The probability of obtaining this number is $100\\%$. Two ordinary dice are equivalent to 2 6-spinners. The theoretical probabilities and the simulated ones are: Number 2 3 4 5 6 7 8 9 10 11 12 Theoretical 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 Frequency 0.0281 0.0556 0.083 0.1123 0.1397 0.1674 0.1381 0.1093 0.0842 0.0539 0.028 This case corresponds to the polynomial $(x + x^2 + x^3 + x^4 + x^5 + x^6)^2$. The expansion of this polynomial is: $$x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + 6x^7 + 5x^8 + 4x^9 + 3x^{10} + 2x^{11} + x^{12}.$$ This polynomial could be factorised as follows: $$(x + x^2", "# Math Help - probability help 1. ## probability help The digits 2, 3, 4, 7 and 8 are each used once to form a five-digit number. What is the probability that the tens digit is odd and the number is divisible by 4? Express your 2. Hello, sri340! The digits 2, 3, 4, 7 and 8 are each used once to form a five-digit number. What is the probability that the tens digit is odd and the number is divisible by 4? With a little Thought, the solution is quite straight forward. We have a 5-digit number and its tens digit is odd. There are only 2 choices for the odd digit. The tens digit is 3: . _ _ _ 3 _ To be divisible by 4, the units digit must be 2. And the other 3 digits can be placed in $3! \\:=\\: 6$ ways. The tens digit is 7:_ _ _ 7 _ To be divisible by 4, the units digit must be 2. And the other 3 digits can be placed in $3! \\:=\\:6$ ways. Hence, there are: . $6 + 6 \\:=\\:12$ such numbers. Since there are $5! = 120$ possible 5-digit numbers, . . The probabiity is: . $\\frac{12}{120} \\:=\\:\\frac{1}{10}$", "Inverted nine In the hotel Inverted Nine, each hotel room number is divisible by 6. How many rooms can we count with the three-digit number registered by digits 1,8,7,4,9? • Divisible by five How many different three-digit numbers divisible by five can we create from the digits 2, 4, 5? The digits can be repeated in the created number. • We roll We roll two dice A. - what is the probability that the sum of the falling numbers is at most 4 B. - is at least 10 C. - is divisible by 5? • Drawing from a hat When drawing numbers from a hat from 1 to 35, we select random given numbers. What is the probability that the drawn numbers will be divisible by 8 and 2? • The accompanying The accompanying table gives the probability distribution of the number of courses randomly selected student has registered Number of courses 1 2 3 4 5 6 7 Probability 0.02 0.03 0.1 0.3 0.4 - 0.01 respectively. a) Find the probability of a student registe • Three digits number 2 Find the number of all three-digit positive integers that can be put together from digits 1,2,3,4 and which are subject to the same time has the following conditions: on one positions is one of the numbers", ", But we can tell that spinning a Pig is more likely than the other outcomes. It is more useful to think about the sample space instead, which has $6$6 sectors, and $2$2 of them have a Pig . Do: Probability $=$= $\\frac{2}{6}=\\frac{1}{3}$26=13 ##### Question 6 What is the probability of spinning a Star or an Apple on this spinner? Think: There are $10$10 different sectors, $3$3 of them have a Star and $3$3 of them have an Apple . These are the \"favorable outcomes\", and there are $3+3=6$3+3=6 all together. Do: Probability $=$= $\\frac{6}{10}=0.6$610=0.6 ##### Question 7 What is the probability of drawing a card from a standard deck of $52$52 cards, that is red and has an even number on it? What is the probability of drawing a Club that is not the Jack? Which is more likely? Think: For the first trial, the cards with even numbers are $2$2$4$4$6$6$8$8, and $10$10. Each of these numbers appear $4$4 times, once for each suit, and $2$2 of the suits are red. This means there are $10$10 cards we could draw corresponding to the event we want, the \"favorable outcomes\". For the second trial, there are $13$13 cards in each suit, and $12$12 of them are not the Jack. Do: For the first trial, Probability $=$= $\\frac{10}{52}$1052. For", "an odd number. $$\\text{1}$$; $$\\text{3}$$; $$\\text{5}$$ Is there an equal chance of getting an odd number and an even number? Explain. No, there is a higher chance of getting an odd number, because there are more possible outcomes for odd numbers. How could you use this spinner to design an unfair game of chance? The game would be unfair if a player has a small chance of winning, for example, if they win only if they get five. It would also be unfair if they win only if they get an even number. Draw a tree diagram with the first set of branches showing the possible outcomes for spinning the spinner in question $$\\text{3}$$. Then add the outcomes for a coin toss to each of the branches. How many possible outcomes are there altogether? Ten How many outcomes are there for getting a $$\\text{4}$$; H? Only one. What is the probability of getting a $$\\text{4}$$; H? One in ten or $$\\frac{\\text{1}}{\\text{10}}$$ How many outcomes are there for getting an even number and Heads? List them. Two: $$\\text{2}$$; H and $$\\text{4}$$; H What is the probability of getting an even number and Heads? The probability is $$\\text{2}$$ in $$\\text{10}$$, which simplifies to $$\\text{1}$$ in $$\\text{5}$$ or $$\\frac{\\text{1}}{\\text{5}}$$", "+ (n − 1)2 = 37 ⇒ 3 + 2n − 2 = 37 ⇒ 2n = 36 n = 18 Thus, total number of outcomes = 18. ​Let E be the event of getting a prime number. Out of these numbers, the prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37. Number of favourable outcomes = 11. ∴ P(getting a prime number) = P(E) = = $\\frac{11}{18}$ Thus, the probability that the number on the card is a prime number is $\\frac{11}{18}$. #### Question 3: Cards numbered 1 to 30 are put in a bag. A card is drawn at random from the bag. Find the probability that the number on the drawn card is (i) not divisible by 3, (ii) a prime number greater than 7, (iii) not a perfect square number. Total number of outcomes = 30. (i) ​Let E1 be the event of getting a number not divisible by 3. Out of these numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30. Number of favourable outcomes = 30 − 10 = 20 ∴ P(getting a number not divisible by 3) = P(E1) = = $\\frac{20}{30}=\\frac{2}{3}$ Thus, the probability that the number on the card is not divisible by 3 is", "# Chapter 8 - Section 8.5 - Counting and Introduction to Probability - Exercise Set: 22 $\\frac{2}{3}$ #### Work Step by Step There are 3 possible outcomes of a single spin: 1, 2, 3. The numbers that are not 3, 1 and 2, represents 2 of the 3 possible outcomes, so the probability is 2 out of 3. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "10 multiples of 6. We have to list the multiples of 6 below 50. Thus, we get 6, 12, ,18, 24, 30, 36, 42, 48 are the multiples of 6 that are marked on the given cards. Thus, the number of outcomes of getting a card marked number which is divisible by 6 when a card is drawn = 8. The probability of getting a card marked number which is divisible by 6 when a card is drawn = Ratio of the number of outcomes of getting a card marked number which is divisible by 6 when a card is drawn to the number of possible outcomes when a card is drawn from the above-mentioned cards $=\\dfrac{8}{48}$ $=\\dfrac{1}{6}$ $\\therefore$ The probability of getting a card marked number which is divisible by 6 when a card is drawn $=\\dfrac{1}{6}=0.16$ 3, 4, 5, 6, 7, 8, 9 are the single-digit numbers among the numbers marked on the cards when a card is drawn. Thus, the number of outcomes of getting a single-digit number marked card when a card is drawn = 7 The probability of getting a single-digit number marked card when a card is drawn = Ratio of the number of outcomes of getting a single-digit number marked card when a card is drawn, to the number of possible", "number, when each of the given numbers is equally likely to be selected. The possible outcomes are 1,2,3,4,................25. Number of all possible outcomes = 25 Out of these, the prime numbers are, 2, 3, 5, 7,11,13,17,19 and 23. Total prime numbers from 1 to 25 = 9 Thus, if the selected number is not prime, then number of favourable outcomes = 25- 9 = 16 P (getting a non-prime number) = $\\frac{16}{25}$ #### Question 17: There are 30 cards numbered from 1 to 30. One card is drawn at random. Find the probability that the number on the selected card is not divisible by 3. The possible outcomes are 1,2,3,4,................30. Number of all possible outcomes = 30 Out of these, the numbers divisible by 3 are 3, 6, 9,12,15,18, 21, 24, 27 and 30. Total numbers divisible by 3 = 10 Let E be the event of getting a number that is not divisible by 3. Then, number of favourable outcomes = 30 − 10 = 20 P (getting a number that is not divisible by 3) = P (E) = #### Question 18: A box contains 25 cards numbered from 1 to 25. A card is drawn from the box at random. Find the probability that the number on the drawn card is (i) even (ii) prime", "are $3$ even numbers, so there are $3$ different, equally likely ways of rolling an even number. • There are $6$ numbers on a die, so there are a total of $6$ possible outcomes. $\\textcolor{blue}{\\text{P(even)}} = \\dfrac{\\text{even numbers on a die}}{\\text{total numbers on a die}} = \\dfrac{3}{6} = 0.5$ Level 1-3 ## Probabilities Add Up to 1 If you consider all possible outcomes of an event (known as exhausting all options) then the probabilities must add up to $1$ Example: The chance of flipping a coin and landing on heads is one half, and the chance of landing on tails is the same, one half. These are the only two possibilities (we’ve exhausted all options). Adding these two up we get, $\\text{P(heads)}+\\text{P(tails)}=\\dfrac{1}{2}+\\dfrac{1}{2}=1$ Level 1-3 ## Listing Outcomes Listing outcomes is exactly what it sounds like – given a scenario, list every possible outcome. When there are two events taking place, we make use of a sample space diagram to help keep track of all the possible outcomes. This takes the form of a two way table Example: Megan spins two spinners numbered $1$ to $5$ and records the sum of the values each spinner lands on. a) List all possible outcomes that Megan could find. To do this we create and fill in a two way table to", "on the drawn card is: (2015D) (i) divisible by 3 or 5 (ii) a perfect square number. Solution: Total number of outcomes = 25 (i) Possible outcomes of numbers divisible by 3 or 5 in numbers 1 to 25 are (3, 6, 9, 12, 15, 18, 21, 24, 5, 10, 20, 25) = 12 ∴P(No. divisible by 3 or 5) = $\\frac{12}{25}$ (ii) Possible outcomes of numbers which are a perfect square = 5, i.e., (1, 4, 9, 16, 25) ∴ P(a perfect square no.) = $\\frac{5}{25}=\\frac{1}{5}$ Question 48. A game of chance consists of spinning an arrow on a 3 circular board, divided into / 4 8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, …, 8 (Figure), which are equally likely outcomes. What is the probability that the arrow will point at (i) an odd number (ii) a number greater than 3 (iii) a number less than 9. (2016 D) Solution: Total numbers = 8 (i) “Odd numbers” are 1, 3, 5, 7, i.e., 4 ∴ P(an odd number) = $\\frac{4}{8}=\\frac{1}{2}$ (ii) “nos. greater than 3” are 4, 5, 6, 7, 8, i.e., 5 ∴ P(a number > 3) = $\\frac{5}{8}$ (iii) “numbers less than 9” are 1, 2, 3, …8 i.e., 8 ∴ P(a number < 9) =", "+0 # Probability 0 26 4 A machine randomly generates one of the nine numbers 1, 2, 3, … , 9 with equal likelihood. What is the probability that when Tsuni uses this machine to generate four numbers their product is divisible by 4? Express your answer as a common fraction. Jul 22, 2022 #1 +2339 +1 Note that for the product to be divisible by 4, there need to be at least 2 even numbers. The probability of rolling only odd numbers is $${5 \\over 9}^4 = {1025 \\over 6561}$$ The probability of rolling exactly 1 even number is $$4 \\times 5^3 \\times {4 \\choose 3} = 2000$$ So, the probability is $$1 - ({1025 \\over 6561} + {2000 \\over 6561}) = \\text{_____}$$ Jul 22, 2022 #4 +2339 +1 BuilderBoi Jul 23, 2022 #2 0 BuilderBoi: A computer code gives the following numbers: 9^4 ==6,561 permutations. [This is based on the assumption that numbers are replaced at each draw] The computer calculates: The probability is; 4,936 / 6,561 Jul 22, 2022 #3 +2339 +1 REVISED ANSWER: There are $$9^4 = 6561$$ ways to get the numbers. The easiest way to solve this is to use complementary counting, in which we count what we don't want, and subtract that from 1 Note that the probability of getting an", "from a well-shuffled deck of 52 playing cards? (c) Probability of getting a red apple. (See figure below) Solution: (a) On the spinning wheel there are 5 sectors containing A, B, C, and D. Since, there is only one sector containing D, Number of possible outcome = 1 Number of equally likely outcomes = 5 ∴ Probability = $$\\frac{1}{5}$$ (b) Number of possible outcomes = 52 Since there are 4 aces in a pack of 52 cards, and out of the one ace can be obtained in 4 ways ∴ Probability of getting an ace = $$\\frac{4}{52}=\\frac{1}{13}$$ (c) There are 7 apples in all Possible number of ways = 7 Since there are 4 red apples No. of equally likely outcomes = 4 ∴ Probability of getting a red apple = $$\\frac{4}{7}$$ Question 4. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box, and mixed well. One slip is chosen from the box without looking into it. What is the probability of? (i) getting a number 6? (ii) getting a number less than 6? (iii) getting a number greater than 6? (iv) getting a 1-digit number? Solution: Since there are 10 slips, Total number of outcomes = 10 (i) We can get a slip containing the number ‘6’", "The outcome of each trial is independent of all... • Replies: 1 • Views: 1,802 09-29-2013, 05:49 PM 24. ### binomial distribution help / confirmation Q: A manufacturer produces 1000 computer chips for a mission-critical application. Each chip costs $100 to manufacture and sells for$2000, but has... • Replies: 1 • Views: 3,247 09-29-2013, 05:01 PM 25. ### Probability Question from Frederick Mosteller, #7 Hey all, I'm trying to improve my probability skills so hopefully the questions I'm asking aren't too amateur-ish. The following is a question... • Replies: 1 • Views: 2,094 09-27-2013, 01:01 PM 26. ### HELP! probability that a 7 digit number is divisible by 7? The seven digits {1; 2; 3; 4; 5; 6; 7} are written down in a random order. What is the chance that the resulting number is divisible by 7? For... • Replies: 8 • Views: 3,494 09-26-2013, 09:42 PM 27. ### Occupany problem This is a homework problem that deals with occupancy. Q: consider r indistinguishable balls and n cells where n > r. The r balls are placed at... • Replies: 6 • Views: 2,791 09-25-2013, 06:47 PM 28. ### Multiple comparisons in SEM Hello, I am using the Multiple Group Analysis function in AMOS to compare three groups. I am using pairwise comparisons to compare group 1" ]

[ "\"the expected value of t is 6\"... Tonio", "× Discrete Mathematics # Linearity of Expectation If you roll a six-sided die, the expected value for the number rolled is $$3.5.$$ If you roll 5 six-sided dice, what is the expected value for the sum of the numbers rolled? A standard deck of 52 cards is shuffled randomly, and then the cards are flipped over one-by-one. If the ace of spades is card number $$S$$ and the ace of hearts is card number $$H$$, what is the expected value of $$S+H?$$ Six friends are doing a \"secret santa gift exchange\". Their names are placed into a hat, and each person chooses a name randomly (without replacement) from the hat. However, this leads to the possibility that someone chooses their own name and doesn’t get a gift! What is the expected value for the number of people who will choose their own name? Patrick has a box with 4 yellow balls and 4 green balls. He randomly pulls out a ball, and if it is yellow, he paints it green. He then puts it back in the box and continues this process until all of the balls are green. Find the expected value for the number of times Patrick will pull out a ball. Andy the Ant is on a number line at the location 0. Each minute,", "6-sided die roll average value [duplicate] A die is rolled 100 times. Find the probability that the mean value would be between 3 and 3.5. I've found the expected value which is 3.5 without idea what to do next.", "Monday, November 22, 2010 Dice Statistics Observation -- The standard deviation of a normal polyhedral die (i.e., a discrete uniform distribution 1..N with small N) is approximately one-half of the mean. For the case of a 7-sided die, the standard deviation is exactly equal to one-half of the mean. For dice with fewer sides it is less, and for more sides it is greater. The d7 coincidentally also has the property that the variance exactly equals the mean. However, the two are not generally close for other sorts of dice (although still less for smaller dice, and more for bigger dice).", "= (30)(.5) = 15 (half of the questions). For a fair six-sided die rolled 60 times, the expected value of the number of times a “1” is tossed is np = (60)(1/6) = 10. The standard deviation for both of these would be, for the True-False test $$\\sqrt{30 \\times 0.5 \\times (1-0.5)}=\\sqrt{7.5}=2.74$$ and for the die $$\\sqrt{60 \\times \\frac{1}{6}\\times (1-\\frac {1}{6})}=\\sqrt{ \\frac{50}{6}}=2.89$$", "25 x 102 50 dimes per roll × 60 rolls = (5 x 6) x (10 × 10) = 30 x 102 50 dimes per roll × 70 rolls = (5 x 7) x (10 × 10) = 35 x 102 50 dimes per roll × 80 rolls = (5 x 8) x (10 × 10) = 40 x 102 50 dimes per roll × 90 rolls = (5 x 9) x (10 × 10) = 45 x 102 50 dimes per roll × 100 rolls = (5 x 10) x (10 × 10) = 50 x 102 Question 21. 1 roll = 40 quarters ; Think:40 quarters per roll × 20 rolls =(4 × 2) × (10 × 10) Type below: __________ Answer: Explanation: 1 roll = 40 quarters ; Think:40 quarters per roll × 20 rolls =(4 × 2) × (10 × 10) = 8 x 102 40 quarters per roll × 30 rolls =(4 × 3) × (10 × 10) = 12 x 102 40 quarters per roll × 40 rolls =(4 × 4) × (10 × 10) = 16 x 102 40 quarters per roll × 50 rolls =(4 × 5) × (10 × 10) = 20 x 102 40 quarters per roll × 60 rolls =(4 × 6) × (10 × 10) =", "expected value is a property of a model for your experiment (a die roll), as I have explained in various answers such as the one at stats.stackexchange.com/a/30330/919. – whuber Oct 6 '16 at 13:16 ## 2 Answers The expected value can be thought of as the long-run average. If you roll your die many, many times, and look at the average of the rolls, you will end up close to 3.5. This is not the value that you expect to see most often. This would be the mode of the distribution (in the discrete case - for a continuous distribution, any particular value has the same probability of zero). After all, you will never see a roll of 3.5 with your standard six-sided die. • How can I correct my definition in its current form? – user366312 Oct 6 '16 at 13:01 • The first sentence of the Wikipedia article is helpful: en.wikipedia.org/wiki/Expected_value – Stephan Kolassa Oct 6 '16 at 13:02 • Is my Continuous Case correct? – user366312 Oct 6 '16 at 13:06 • See whuber's comment. – Stephan Kolassa Oct 6 '16 at 14:41 I would like to help - you put $n$ in your definition, but this is exactly what we don't want. Since $n$ is never referenced elsewhere, its presence serves to inform the", "3 sides, +,0,-), '%' for a 100 sided die, or 'A' for an averaging die (with sides 2,3,3,4,4,5).", "a six-sided die an infinite amount of times, you see the average value equals 3. If this series does not converge absolutely, we say that the expected value of X does not exist. Expected Value for Multiple Events Of course, calculating expected value EV gets more complicated in real life. Perform the steps exactly as above. A fair six-sided die is tossed. Law of Large Numbers: Home Tables Binomial Distribution Table F Table PPMC Critical Values T-Distribution Table One Tail T-Distribution Table Two Tails Chi Squared Table Right Tail Z-Table Left of Curve Z-table Right of Curve Probability and Statistics Statistics Basics Probability Regression Analysis Hypothesis Testing Normal Distributions:", "# Expected Value and Standard Dev. #### HuskerJay ##### New Member Scenario: You pay $10 and roll a die. If you get a 6, you win$50. If not, you get to roll again. If you get a 6 this time, you get your \\$10 back.", "Use the example of rolling a six-sided die, probability = of. Dec, 19, 2020 0", "best way to cunduct this study is to join the results of several users and see and the fit is improved with increasing number of samples. The expected value for the average of N 6-sided dice is: $\\bar{\\mu} = \\frac{6N+N}{2}$", "# Different ways to calculate the expected value of getting all faces of a 6 sided die atleast once Problem Statement: You roll a fair six-sided die until all six numbers have been rolled at least once. What is the expected value of that ? Can you write different ways to solve this problem ? Without loss of generality let us focus on one of the numbers, say $1$. The probability that we see $1$ on the $n^{\\text{th}}$ attempt given that we did not observe $1$ on the previous $n-1$ tries is given by: ${(\\frac{5}{6})}^{n-1} \\cdot \\frac{1}{6}$ You can use the above to compute the expected value of the number of attempts to make to see $1$. We then multiply the above by $6$ to get the answer we seek.", "their expected values.", "+0 c&p halp polz +7 60 1 +258 A fair 6-sided die is rolled once. If I roll $n$, then I win $6-n$ dollars. What is the expected value of my win, in dollars? Feb 20, 2022 #1 +1384 +1 The expected value is the average of all the possibilities: You can win 0, 1, 2 ,3, 4, or 5 dollars, so the expected is $$\\color{brown}\\boxed{15\\over6}$$ Feb 20, 2022", "# Expected value for discrete (nominal) variable? I'm trying to understand the concept of the expected value. Especially, what bothers me is the expected value for discrete random variables. I will try to formulate it by examples: • The expected value when throwing a six-sided die is $E(X) = 3.5$, which as I see it, is not even part of the probability space $\\Omega = \\{1,2,3,4,5,6\\}$. Why is this not a problem at all? Is here anything wrong with my assumptions about the expected value? • Is there an expected value when throwing a coin? I cannot calculate a weighted mean on head and tail? Is there an expected value? Is it the maximum likely value, and what if there are multiple maximum likely values? I hope my questions have sufficiently revealed my (mis)conceptions of the topic. I'd appreciate any help and insights! • The expected value does not mean the value that you expect to see on the next roll of the die; it is what you would expect to be the average value of the outcomes on a large number of rolls. Roll the die $100$ times. You are likely to see a total of $350$ dots for an average of $350/100 = 3.5 = E[X]$. This is not guaranteed, you might get some number smaller", "Find the expected number of rolls of a fair six-sided die before the sequence of rolls contains $1, 2, 3, 4, 5, 6$ (in that order) as a subsequence.", "# Expected rolls until a 6 Suppose I roll a fair, six-sided die repeatedly until I get a 6. It is well-known that the expected number of rolls is exactly 6. Now, suppose instead that I roll a fair, six-sided die repeatedly until I get a 6, and I tell you that every time I rolled the die, an even number (2, 4, or 6) came up. What is the expected number of die rolls that I made? Clarification: For example, I could have rolled 2-4-4-2-4-6, 2-6, 4-4-6, or just 6. Source: Gil Kalai's blog ×", "the theoretically best normal variety of turn round, and that is $$\\frac\\log 10\\log 6 \\approx 1.2851,$$ afar greater than some of the processes depicted to another place in this particular twine.", "of six. The primary reason for going with an eight sided die instead of the standard die was three-fold. First it adds something unique to the game. Second, it speeds up the game as the average roll will be 4.5 as opposed to 3.5 for a six sided die. Finally, they are cool. Building the die movie is fairly simple. First we start with an image of the die in a finished position. On a separate layer we have text for the eight different values (labelled r1 to r8). We also have mid-roll die images. Figure \\ref{fig:die} shows the frames that make up the die. This is the minimal die animation approach. A more ambitious approach would have numbers on the between frames with the values shown. This would require many more frames of animation as you would have the 8 result frames and would need 18 additional between frames for 26 frames. Granted, the animation would look better so if you have the artist time to do this then it may be a good option. As with our other movies, we treat the die movie as a class. To handle the rolling of the die we will need some functions, but first we will initialize the die. This simply sets the value of the die to 8" ]

[ "# Rolling four fair dice and getting four 6s Probability There are 6 possible outcomes on a die and only one of them is a 6. The probability of rolling a 6 is then $$\\frac{1}{6}$$. The probability of rolling four 6s is then $$\\displaystyle\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{6}}=\\frac{{1}}{{1296}}$$", "fair die $$p=\\frac{1}{6}$$, for a die with two sixes $$p=\\frac{1}{3}$$ and for an unfair die we might have $$p=\\frac{1}{5}$$ or any other value $$0\\le p\\le1$$.", "# When do you add probabilities? For example, the probability of rolling a \"2\" on a six-sided die is $\\frac{1}{6}$, the probability of rolling a \"3\" is $\\frac{1}{6}$ and the probability of rolling a \"2\" or \"3\" is $\\frac{1}{6} + \\frac{1}{6} = \\frac{1}{3}$.", "× Discrete Mathematics # Linearity of Expectation If you roll a six-sided die, the expected value for the number rolled is $$3.5.$$ If you roll 5 six-sided dice, what is the expected value for the sum of the numbers rolled? A standard deck of 52 cards is shuffled randomly, and then the cards are flipped over one-by-one. If the ace of spades is card number $$S$$ and the ace of hearts is card number $$H$$, what is the expected value of $$S+H?$$ Six friends are doing a \"secret santa gift exchange\". Their names are placed into a hat, and each person chooses a name randomly (without replacement) from the hat. However, this leads to the possibility that someone chooses their own name and doesn’t get a gift! What is the expected value for the number of people who will choose their own name? Patrick has a box with 4 yellow balls and 4 green balls. He randomly pulls out a ball, and if it is yellow, he paints it green. He then puts it back in the box and continues this process until all of the balls are green. Find the expected value for the number of times Patrick will pull out a ball. Andy the Ant is on a number line at the location 0. Each minute,", "# What is the probability you will flip a head and roll a four if you flip a coin and roll a die at the same time? Jul 27, 2018 $\\text{p(rolling a four and tossing a head)} = \\frac{1}{12}$ #### Explanation: Outcomes of tossing a coin: ie 2 outcomes • tail Outcomes of rolling a die: ie 6 outcomes • 1 • 2 • 3 • 4 • 5 • 6 $\\text{p(rolling a four and tossing a head)} = \\frac{1}{6} \\times \\frac{1}{2} = \\frac{1}{12}$", "possible outcomes are equally likely, then the probability or chance of each outcome is $$\\displaystyle \\frac{1}{n}$$. Probability distribution A listing of all the possible outcomes of an experiment and their probabilities. Probability of an event If the event $$A$$ has $$k$$ outcomes, and $$\\Omega$$ has $$n$$ outcomes, and all the outcomes are equally likely, then the probability of $$A$$ is $$\\displaystyle \\frac{k}{n}$$. For example, consider tossing a coin twice. Then $$\\Omega = \\{\\text{HH, HT, TH, TT}\\}$$, where H denotes the coin landing heads, and T the coin landing tails. Let $$A$$ be the event that the coin lands the same way on both tosses, then $$A = \\{\\text{HH, TT}\\}$$. Therefore the probability of $$A = \\displaystyle \\frac{2}{4}$$. Let’s simulate rolling a die and counting how many times we see each face. If we roll it 6 times, we don’t really expect to see each face once, and as you can see below, in this particular instance of rolling the die six times, we didn’t see the face with one spot, but saw two spots twice. set.seed(12345) die <- 1:6 die_rolls <- sample(die, 6, replace = TRUE) data.frame(die_rolls) %>% group_by(die_rolls) %>% summarise(n = n()) # A tibble: 5 × 2 die_rolls n <int> <int> 1 2 2 2 3 1 3 4 1 4 5 1 5 6 1", "die is thrown then Possible outcomes = 6 Now probability of getting 6 is $$\\\\ \\frac { 1 }{ 6 }$$ (b) Question 6. Solution: A die is thrown Possible outcomes = 6 Now probability of getting an even number which are 2, 4, 6 = $$\\\\ \\frac { 3 }{ 6 }$$ = $$\\\\ \\frac { 1 }{ 2 }$$ (a) Question 7. Solution: One card is drawn from a well shuffled deck of 52 cards, possible out comes = 52 The probability of card which is a queen = $$\\\\ \\frac { 4 }{ 52 }$$ = $$\\\\ \\frac { 1 }{ 13 }$$ (c) Question 8. Solution: One card is drawn from a well-shuffled deck of 52 card, possible out comes = 52 Probability of a card being a black 6 (which are two) = $$\\\\ \\frac { 2 }{ 52 }$$ = $$\\\\ \\frac { 1 }{ 26 }$$ (b) Hope given RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24B are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.", "A die is thrown then Possible outcomes = 6 Now probability of getting 6 is $$\\\\ \\frac { 1 }{ 6 }$$ (b) Question 6. Solution: A die is thrown Possible outcomes = 6 Now probability of getting an even number which are 2, 4, 6 = $$\\\\ \\frac { 3 }{ 6 }$$ = $$\\\\ \\frac { 1 }{ 2 }$$ (a) Question 7. Solution: One card is drawn from a well shuffled deck of 52 cards, possible out comes = 52 The probability of card which is a queen = $$\\\\ \\frac { 4 }{ 52 }$$ = $$\\\\ \\frac { 1 }{ 13 }$$ (c) Question 8. Solution: One card is drawn from a well-shuffled deck of 52 card, possible out comes = 52 Probability of a card being a black 6 (which are two) = $$\\\\ \\frac { 2 }{ 52 }$$ = $$\\\\ \\frac { 1 }{ 26 }$$ (b) Hope given RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24B are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RS Aggarwal Class 8 Solutions Chapter 24 Probability Ex 24A These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS", "## Prealgebra (7th Edition) $\\frac{1}{3}$ There are 6 possible outcomes of tossing a single die: 1, 2, 3, 4, 5, 6. \"1 or 6\" represent 2 of the 6 possible outcomes, so the probability is 2 out of 6. $\\frac{2}{6}=\\frac{2\\div 2}{6\\div 2}=\\frac{1}{3}$", "H4, H5, H6, T6$$), so the probability is $$\\frac{7}{12}$$. How could we have found this from the individual probabilities? As we would expect, $$\\frac{1}{2}$$ of these outcomes have a head, and $$\\frac{1}{6}$$ of these outcomes have a 6 on the die. If we add these, $$\\frac{1}{2}+\\frac{1}{6}=\\frac{6}{12}+\\frac{2}{12}=\\frac{8}{12}$$, which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head and rolling a 6 is $$\\frac{1}{12}$$. If we subtract out this double count, we have the correct probability: $$\\frac{8}{12}-\\frac{1}{12}=\\frac{7}{12}$$. ##### $$P(A \\text { or } B)$$ The probability of either $$A$$ or $$B$$ occurring (or both) is $P(A \\text { or } B)=P(A)+P(B)-P(A \\text { and } B)$ ##### Example 10 Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King? Solution There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is: $P(\\text { King or Queen})=\\frac{8}{52}$ Note that in this case, there are no cards that are both a Queen and a King, so $$P(\\mathrm{King} \\text { and", "{1,2,3,4,5,6} Number of possible outcomes = 6 (ii) Let F be the event of getting a number lying between 2 and 6 Numbers lying between 2 and 6 on the die are = {3,4,5} Number of favorable outcomes = 3 $\\therefore P(F) = \\frac{favourable\\ outcomes}{total\\ outcomes} = \\frac{3}{6}$ $= \\frac{1}{2}$ Therefore, the probability of getting a number lying between 2 and 6 is $\\dpi{100} \\frac{1}{2}$ ## Q13 (iii) A die is thrown once. Find the probability of getting an odd number. Possible outcomes when a die is thrown = {1,2,3,4,5,6} Number of possible outcomes once = 6 (iii) Let O be the event of getting an odd number. Odd numbers on the die are = {1,3,5} Number of favorable outcomes = 3 $\\therefore P(O) = \\frac{favourable\\ outcomes}{total\\ outcomes} = \\frac{3}{6}$ $= \\frac{1}{2}$ Therefore, the probability of getting an odd number is $\\dpi{100} \\frac{1}{2}$. ## Q14 (i) One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting a king of red color. Total number of cards in a well-shuffled deck = 52 Hence, total possible outcomes = 52 (1) Let E be the event of getting a king of red color. There are only red color kings: Hearts and diamonds Hence, number of favorable outcomes = 2 $\\therefore P(E) = \\frac{favourable\\ outcomes}{total\\ outcomes}", "is rolled twice!) [6x6 = 36 possible outcomes] E = [ {(x,y): x,y S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)} Now, C: the sum is at least 7 and a multiple of 3 The sum can only be 9 or 12. C = [ {(a,b): (a,b) E, a+b>6 & a+b = 3k, k I} ]= {(3,6), (6,3), (5, 4), (4,5), (6,6)} 3. An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events: B: $2$ occurs on either die Sample space when a die is rolled: S = {1, 2, 3, 4, 5, 6} Let E = Event of rolling a pair of dice (= Event that a die is rolled twice!) [6x6 = 36 possible outcomes] E = [ {(x,y): x,y S } ] = {(1,1), (1,2)...(1,6),(2,1).....(6,5),(6,6)} Now, B: 2 occurs on either die Hence the number 2 can come on first die or second die or on both the die simultaneously. B = [ {(a,b): (a,b) E, a or b = 2 } ]=... 3. An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events: A: the sum is greater than $8$ Sample space when a die is rolled: S = {1, 2, 3, 4, 5, 6} Let E =", "\\frac{2}{3}.$$ Likewise $$P(B) = \\frac{2}{6} = \\frac{1}{3}.$$ Now the intersection of these sets is {4, 5}, for a probability of $\\frac{1}{3}.$ So we have: \\begin{align} P(A) P(B) &\\stackrel{?}{=} P(A \\cap B) \\\\[5pt] \\frac{2}{3} \\cdot \\frac{1}{3} &\\stackrel{?}{=} \\frac{1}{3} \\\\[5pt] \\frac{2}{9} &\\ne \\frac{1}{3} \\end{align} So these outcomes are not independent. We can also think about it using conditional probabilities. Consider the conditional probability P(A|B). If event B is obtained, then event A is guaranteed because both elements of B are in A. Once B is obtained, A is inevitable, so we have $$P(A|B) = 1.$$ which is not the same as P(A), so our system fails the independence test this way, too. Finally, if we take P(B|A) we find that the probability of obtaining B after A has been obtained is ½. That's not the same as P(B). ### Example 5 – another die roll Let's say we roll a six-sided die and we identify two outcomes: A = {4, 5} and B = {1, 3, 5}. Notice that the second outcome is like saying \"the value showing on the die is odd.\" Are these outcomes independent? Solution: The probabilities of events A and B are: \\begin{align} P(A) &= \\frac{2}{6} = \\frac{1}{3} \\\\[5pt] P(B) &= \\frac{3}{6} = \\frac{1}{2} \\end{align} Now these sets intersect only at 5, so $P(A \\cap B)", "? Free Version Easy # Basic Probability: Rolling a Die SATSTM-FL51Z0 What is the probability of rolling a six-sided die and getting a $5$? A $\\cfrac{1}{3}$ B $\\cfrac{1}{6}$ C $\\cfrac{1}{5}$ D $\\cfrac{5}{6}$ E $\\cfrac{1}{2}$", "Back to all chapters # Expected Value Know what outcome to expect when you're dealing with randomness. # A preview of \"Expected Value\"Join Brilliant Premium The expected value of a random variable $$X$$ is the average value that $$X$$ takes. More formally, $E[X] = \\sum_{x \\in S} x \\cdot P(X=x)$ where $$S$$ is the set of values that $$X$$ can take. For example, the expected value of rolling a fair, six-sided die is $\\frac{1}{6}(1)+\\frac{1}{6}(2)+\\frac{1}{6}(3)+\\frac{1}{6}(4)+\\frac{1}{6}(5)+\\frac{1}{6}(6) = 3.5.$ There are $$12$$ blue marbles and $$4$$ red marbles in a bag. You reach into the bag and pull $$5$$ marbles at random. What is the expected value of the number of blue marbles drawn? A fair coin is flipped until 2 heads are flipped in up to 3 successive flips. What is the expected value of the number of flips? ×", "Tails} or {H, T} Using the probability formula, we can find the probability of the event “getting tails”. The number of favorable outcomes associated with this event is 1, and the total number of possible outcomes is 2. P(T) = Probability coin lands on tails = $\\frac{\\text{Number of Favorable Outcomes}}{\\text{Total Number of Possible Outcomes}}$ = ½ or 0.5 The same goes for finding the probability of getting heads. P(H) = Probability coin lands on heads =$\\frac{\\text{Number of Favorable Outcomes}}{\\text{Total Number of Possible Outcomes}}$ = ½ or 0.5 Using the probability formula, see if you can find the probability of rolling heads or tails on a coin flip. Scroll to the bottom of this article to see the correct answer. ### Rolling Dice When rolling a die, there are six equally likely outcomes, so the sample space is: S = {1, 2, 3, 4, 5, 6} Using the probability formula, let’s find the probability of rolling a number that is less than 3. The only numbers on a die that are less than 3 are 1 and 2, so the probability of rolling a number that is less than 3 is the same thing as the probability of rolling a 1 or a 2. P(1 or 2) = Probability of rolling a number less than 3 = $\\frac{\\text{Number of", "probability of rolling a six given that I rolled an even number $\\displaystyle P(\\mathrm{6\\: and\\: even}) = P(6|\\mathrm{even}) \\times P(\\mathrm{even})= \\frac{1}{3} \\times \\frac{1}{2} = \\frac{1}{6}$, or equivalently the probability of rolling six multplied by the probability of rolling an even number given that I rolled a six $\\displaystyle P(\\mathrm{6\\: and\\: even}) = P(\\mathrm{even} | 6) \\times P(6) = 1 \\times \\frac{1}{6} = \\frac{1}{6}$. Reassuringly, we do get the same answer. This is a bit of a silly example, as we know that if we’ve rolled a six we have rolled an even number, so all we are doing if calculating the probability of rolling a six. We can use conditional probabilities for independent events: this is really easy as the conditional probability is just the straight probability. The probability of Ted surviving his surfeit of fudge given that I rolled a six is just the probability of him surviving, $P(\\mathrm{survive}|6) = P(\\mathrm{survive})$. Let’s try a more complicated example, let’s imagine that Ted is playing fudge roulette. This is like Russian roulette, except you roll a die and if it comes up six, then you have to eat the lethal dose of fudge. His survival probability now depends on the roll of the die. We want to calculate the probability that Ted will live to tomorrow. If Ted doesn’t", "of rolling a die and flipping a coin. The outcomes of the first event—say, rolling a die—can be represented as the first set of branches of the tree diagram. Now, the possible outcomes of the second event can be drawn as the sub-branch of each of the branches of the first event. Note that the order in which the events are considered is immaterial. That is, whether you consider the event of flipping the coin as first or second will not make any difference to the sample space. Example 2: What is the probability of tossing two coins simultaneously and getting exactly one tail? The possible outcomes for each of the coins are head and tail. Starting with the outcomes of one of the coins, we can draw the tree diagram as shown. Counting all the branch-endings of the tree diagram, the total number of possible outcomes is $4$ . Among these the favorable outcomes are the ones with one tail each. That is, the number of favorable outcomes is $2$ . Therefore, the probability of the compound event is $\\frac{2}{4}$ or $\\frac{1}{2}$ . A third way to represent a sample space is using a table. Let us again consider the compound event of rolling a die and flipping a coin. The outcomes of the event of rolling", "## Saturday, August 25, 2007 ### Expected value Expected value In probability theory the expected value (or mathematical expectation, or mean) of a discrete random variable is the sum of the probability of each possible outcome of the experiment multiplied by the outcome value (or payoff). Thus, it represents the average amount one \"expects\" as the outcome of the random trial when identical odds are repeated many times. Note that the value itself may not be expected in the general sense - the \"expected value\" itself may be unlikely or even impossible. For example, the expected value from the roll of an ordinary six-sided die is 3.5, found by, \\begin{align} \\operatorname{E}(X)& = 1 \\cdot \\frac{1}{6} + 2 \\cdot \\frac{1}{6} + 3 \\cdot \\frac{1}{6} + 4 \\cdot \\frac{1}{6} + 5 \\cdot \\frac{1}{6} + 6 \\cdot \\frac{1}{6}\\\\[6pt] & = \\frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5, \\end{align} which is not one of the possible outcomes. A common application of expected value is in gambling. For example, an American roulette wheel has 38 equally likely outcomes. A winning bet placed on a single number pays 35-to-1 (this means that you are paid 35 times your bet and your bet is returned, so you get 36 times your bet). So considering all 38 possible outcomes, the", "## A game of chance involves rolling a 16-sided die once. If a number from 1 to 3 comes up, you win 4 dollars. If the number 4 or 5 comes up, y Question A game of chance involves rolling a 16-sided die once. If a number from 1 to 3 comes up, you win 4 dollars. If the number 4 or 5 comes up, you win 7 dollars. If any other number comes up, you lose. If it costs 5 dollars to play, what is your expected net winnings in progress 0 5 months 2021-08-29T20:59:35+00:00 1 Answers 0 views 0 -$1.8 or loss of$1.8 The probability of winning $4 is 3/16 The probability of winning$7 is 2/16 The probability of losing $5 is 11/16 E(X) = (4 * 3/16) + (7 * 2/16) + (-5 * 11/16) = 12/16 + 14/16 – 55/16 = -29/16 = -1.8 Hence, your expected net “winnings” would be -$1.8. That’s an expected loss of \\$1.8." ]

[ "fair die $$p=\\frac{1}{6}$$, for a die with two sixes $$p=\\frac{1}{3}$$ and for an unfair die we might have $$p=\\frac{1}{5}$$ or any other value $$0\\le p\\le1$$.", "× Discrete Mathematics # Linearity of Expectation If you roll a six-sided die, the expected value for the number rolled is $$3.5.$$ If you roll 5 six-sided dice, what is the expected value for the sum of the numbers rolled? A standard deck of 52 cards is shuffled randomly, and then the cards are flipped over one-by-one. If the ace of spades is card number $$S$$ and the ace of hearts is card number $$H$$, what is the expected value of $$S+H?$$ Six friends are doing a \"secret santa gift exchange\". Their names are placed into a hat, and each person chooses a name randomly (without replacement) from the hat. However, this leads to the possibility that someone chooses their own name and doesn’t get a gift! What is the expected value for the number of people who will choose their own name? Patrick has a box with 4 yellow balls and 4 green balls. He randomly pulls out a ball, and if it is yellow, he paints it green. He then puts it back in the box and continues this process until all of the balls are green. Find the expected value for the number of times Patrick will pull out a ball. Andy the Ant is on a number line at the location 0. Each minute,", "expected value is a property of a model for your experiment (a die roll), as I have explained in various answers such as the one at stats.stackexchange.com/a/30330/919. – whuber Oct 6 '16 at 13:16 ## 2 Answers The expected value can be thought of as the long-run average. If you roll your die many, many times, and look at the average of the rolls, you will end up close to 3.5. This is not the value that you expect to see most often. This would be the mode of the distribution (in the discrete case - for a continuous distribution, any particular value has the same probability of zero). After all, you will never see a roll of 3.5 with your standard six-sided die. • How can I correct my definition in its current form? – user366312 Oct 6 '16 at 13:01 • The first sentence of the Wikipedia article is helpful: en.wikipedia.org/wiki/Expected_value – Stephan Kolassa Oct 6 '16 at 13:02 • Is my Continuous Case correct? – user366312 Oct 6 '16 at 13:06 • See whuber's comment. – Stephan Kolassa Oct 6 '16 at 14:41 I would like to help - you put $n$ in your definition, but this is exactly what we don't want. Since $n$ is never referenced elsewhere, its presence serves to inform the", "), then the weighted average turns into the simple average. This is intuitive: the expected value of a random variable is the average of all values it can take; thus the expected value is what one expects to happen on average. If the outcomes ${\\displaystyle x_{i}}$ are not equally probable, then the simple average must be replaced with the weighted average, which takes into account the fact that some outcomes are more likely than the others. The intuition however remains the same: the expected value of ${\\displaystyle X}$ is what one expects to happen on average. An illustration of the convergence of sequence averages of rolls of a die to the expected value of 3.5 as the number of rolls (trials) grows. #### ExamplesEdit • Let ${\\displaystyle X}$ represent the outcome of a roll of a fair six-sided die. More specifically, ${\\displaystyle X}$ will be the number of pips showing on the top face of the die after the toss. The possible values for ${\\displaystyle X}$ are 1, 2, 3, 4, 5, and 6, all equally likely (each having the probability of 1/6). The expectation of ${\\displaystyle X}$ is ${\\displaystyle \\operatorname {E} [X]=1\\cdot {\\frac {1}{6}}+2\\cdot {\\frac {1}{6}}+3\\cdot {\\frac {1}{6}}+4\\cdot {\\frac {1}{6}}+5\\cdot {\\frac {1}{6}}+6\\cdot {\\frac {1}{6}}=3.5.}$ If one rolls the die ${\\displaystyle n}$ times and computes the average (arithmetic", "## Prealgebra (7th Edition) $\\frac{1}{3}$ There are 6 possible outcomes of tossing a single die: 1, 2, 3, 4, 5, 6. \"1 or 6\" represent 2 of the 6 possible outcomes, so the probability is 2 out of 6. $\\frac{2}{6}=\\frac{2\\div 2}{6\\div 2}=\\frac{1}{3}$", "# What is the probability you will flip a head and roll a four if you flip a coin and roll a die at the same time? Jul 27, 2018 $\\text{p(rolling a four and tossing a head)} = \\frac{1}{12}$ #### Explanation: Outcomes of tossing a coin: ie 2 outcomes • tail Outcomes of rolling a die: ie 6 outcomes • 1 • 2 • 3 • 4 • 5 • 6 $\\text{p(rolling a four and tossing a head)} = \\frac{1}{6} \\times \\frac{1}{2} = \\frac{1}{12}$", "# Rolling four fair dice and getting four 6s Probability There are 6 possible outcomes on a die and only one of them is a 6. The probability of rolling a 6 is then $$\\frac{1}{6}$$. The probability of rolling four 6s is then $$\\displaystyle\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{6}}=\\frac{{1}}{{1296}}$$", "die is rolled (faces 1 and 6 have probability $$\\frac{1}{4}$$ each, and faces 2, 3, 4, 5 have probability $$\\frac{1}{8}$$ each). Find the mean and standard deviation of the die score. 1. $$\\frac{7}{2}$$ 2. $$1.8634$$", "Back to all chapters # Expected Value Know what outcome to expect when you're dealing with randomness. # A preview of \"Expected Value\"Join Brilliant Premium The expected value of a random variable $$X$$ is the average value that $$X$$ takes. More formally, $E[X] = \\sum_{x \\in S} x \\cdot P(X=x)$ where $$S$$ is the set of values that $$X$$ can take. For example, the expected value of rolling a fair, six-sided die is $\\frac{1}{6}(1)+\\frac{1}{6}(2)+\\frac{1}{6}(3)+\\frac{1}{6}(4)+\\frac{1}{6}(5)+\\frac{1}{6}(6) = 3.5.$ There are $$12$$ blue marbles and $$4$$ red marbles in a bag. You reach into the bag and pull $$5$$ marbles at random. What is the expected value of the number of blue marbles drawn? A fair coin is flipped until 2 heads are flipped in up to 3 successive flips. What is the expected value of the number of flips? ×", "Independent probability Problem If you flip a coin and roll a 6-sided die, what is the probability that you will flip a heads and roll a 2?", "# Expect value and standard deviation • Feb 17th 2013, 07:14 PM kkar Expect value and standard deviation Is there a way to calculate these two quantities if I'm throwing two twelve sided die and adding the results that's easier than going through all possible combination and their probabilities? Thanks. • Feb 17th 2013, 07:35 PM Plato Re: Expect value and standard deviation Quote: Originally Posted by kkar Is there a way to calculate these two quantities if I'm throwing two twelve sided die and adding the results that's easier than going through all possible combination and their probabilities? Look at this webpage Scroll down to the fully expanded form of the expression. That gives you all the impossibles. For example the term $11x^{14}$ tells us that there eleven ways to roll a sum of fourteen. So the probability is $\\frac{11}{144}$ of getting a sum of fourteen. • Feb 18th 2013, 11:25 AM kkar Re: Expect value and standard deviation Thanks Plato!", "## Saturday, August 25, 2007 ### Expected value Expected value In probability theory the expected value (or mathematical expectation, or mean) of a discrete random variable is the sum of the probability of each possible outcome of the experiment multiplied by the outcome value (or payoff). Thus, it represents the average amount one \"expects\" as the outcome of the random trial when identical odds are repeated many times. Note that the value itself may not be expected in the general sense - the \"expected value\" itself may be unlikely or even impossible. For example, the expected value from the roll of an ordinary six-sided die is 3.5, found by, \\begin{align} \\operatorname{E}(X)& = 1 \\cdot \\frac{1}{6} + 2 \\cdot \\frac{1}{6} + 3 \\cdot \\frac{1}{6} + 4 \\cdot \\frac{1}{6} + 5 \\cdot \\frac{1}{6} + 6 \\cdot \\frac{1}{6}\\\\[6pt] & = \\frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5, \\end{align} which is not one of the possible outcomes. A common application of expected value is in gambling. For example, an American roulette wheel has 38 equally likely outcomes. A winning bet placed on a single number pays 35-to-1 (this means that you are paid 35 times your bet and your bet is returned, so you get 36 times your bet). So considering all 38 possible outcomes, the", "+0 # expected value 0 52 1 In a game that costs $\\$3$to play, a fair die is rolled. If a five or a six is rolled the player receives$\\$6$. If any other number is rolled, the player receives nothing. What is the expected value of the game? Feb 19, 2021 If you roll a 1, 2, 3, or 4, you earn $3. If you roll a 5 or 6, you earn$6. So, we can write the following: $$E = \\frac{4}{6}(\\3) + \\frac{2}{6}(\\6) = \\boxed{\\4}$$", "expect to roll a $7$7? Round your answer to the nearest whole number. eight-sided die ##### Question 13 The following twelve-sided die is rolled. 1. What is the chance of rolling five or more? 2. What is the chance of rolling less than five? ### Outcomes #### MGSE7.SP.5 Understand that the probability of a chance event is a number between 0 and 1 that expresses the likelihood of the event occurring. Larger numbers indicate greater likelihood. A probability near 0 indicates an unlikely event, a probability around 1/2 indicates an event that is neither unlikely nor likely, and a probability near 1 indicates a likely event.", "that $$Q = 59.84.$$ However, if we actually have $$Q \\sim \\mathsf{Chisq}(5),$$ then this observed value of $$Q$$ seems very unlikely, because only 5% of values from $$\\mathsf{Chisq}(5)$$ should exceed the critical value $$c =11.07.$$ Put another way the probability that a value from this distribution exceeds $$59.84$$ is the P-value of the chi-squared test, which is much smaller than $$0.0001.$$ x = c(44, 97, 102, 99, 105, 153) q = sum((x-100)^2/1 0 0); q [1] 59.84 qchisq(.95, 5) [1] 11.0705 # critical value 1-pchisq(59.84, 5) [1] 1.311595e-11 # P-value The conclusion is that the data provide strong evidence that our die is unfair. [In fact, the values $$X_i$$ were simulated using probabilities $$(\\frac 1{12}, \\frac 1 6, \\frac 1 6, \\frac 1 6,\\frac 1 6,\\frac 1 4),$$ respectively, for the faces, instead of $$\\frac 1 6$$ for each face, as for a truly fair die. So the chi-squared test has been able to detect that the die is unfair.] Addendum: Shown below is a simulation of 100,000 values of $$Q,$$ each based on $$600$$ rolls of a fair die. Their histogram is plotted along with the density of $$\\mathsf{Chisq}(5)$$ in order to illustrate that is the the approximate distribution of such values of $$Q.$$ By way of explaining the code, one experiment with $$600$$ rolls of a", "# If a pair of die are rolled, what is the expected value of the sum of the rolls? Nov 6, 2015 Each die is independent , so the sum is the expected value of each separate die. #### Explanation: Each die has an expected value equal to: E(X) $= \\frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$ Expected value of sum $= 3.5 + 3.5 = 7$ hope that helped", "die is thrown then Possible outcomes = 6 Now probability of getting 6 is $$\\\\ \\frac { 1 }{ 6 }$$ (b) Question 6. Solution: A die is thrown Possible outcomes = 6 Now probability of getting an even number which are 2, 4, 6 = $$\\\\ \\frac { 3 }{ 6 }$$ = $$\\\\ \\frac { 1 }{ 2 }$$ (a) Question 7. Solution: One card is drawn from a well shuffled deck of 52 cards, possible out comes = 52 The probability of card which is a queen = $$\\\\ \\frac { 4 }{ 52 }$$ = $$\\\\ \\frac { 1 }{ 13 }$$ (c) Question 8. Solution: One card is drawn from a well-shuffled deck of 52 card, possible out comes = 52 Probability of a card being a black 6 (which are two) = $$\\\\ \\frac { 2 }{ 52 }$$ = $$\\\\ \\frac { 1 }{ 26 }$$ (b) Hope given RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24B are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.", "A die is thrown then Possible outcomes = 6 Now probability of getting 6 is $$\\\\ \\frac { 1 }{ 6 }$$ (b) Question 6. Solution: A die is thrown Possible outcomes = 6 Now probability of getting an even number which are 2, 4, 6 = $$\\\\ \\frac { 3 }{ 6 }$$ = $$\\\\ \\frac { 1 }{ 2 }$$ (a) Question 7. Solution: One card is drawn from a well shuffled deck of 52 cards, possible out comes = 52 The probability of card which is a queen = $$\\\\ \\frac { 4 }{ 52 }$$ = $$\\\\ \\frac { 1 }{ 13 }$$ (c) Question 8. Solution: One card is drawn from a well-shuffled deck of 52 card, possible out comes = 52 Probability of a card being a black 6 (which are two) = $$\\\\ \\frac { 2 }{ 52 }$$ = $$\\\\ \\frac { 1 }{ 26 }$$ (b) Hope given RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24B are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. ## RS Aggarwal Class 8 Solutions Chapter 24 Probability Ex 24A These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS", "# 4.5 Expectation ## Uneven Dice Say I have a six-sided die that lands on an even number with probability $$\\frac{1}{4}$$. If it lands on an even number, then it lands on each one with equal probability. The same is true for odd numbers. Find the expected value of this die. Please answer as a decimal with three significant figures. Exercise 1 Let $$R$$ be a random variable for the value of the die roll. From the description we know that $$\\Pr[R=i]=\\frac{1}{4}\\cdot\\frac{1}{3}=\\frac{1}{12}$$ for $$i\\in\\{2, 4, 6\\}$$ and $$\\Pr[R=i]=\\frac{3}{4}\\cdot\\frac{1}{3}=\\frac{1}{4}$$ for $$i\\in\\{1, 3, 5\\}$$. Then, $$E[R]=\\sum_{i=1}^6 \\Pr[R=i] = 1\\cdot\\frac{1}{4}+2\\cdot\\frac{1}{12}+3\\cdot\\frac{1}{4}+4\\cdot\\frac{1}{12}+5\\cdot\\frac{1}{4}+6\\cdot\\frac{1}{12}=\\frac{13}{4}$$", "# Expected value for discrete (nominal) variable? I'm trying to understand the concept of the expected value. Especially, what bothers me is the expected value for discrete random variables. I will try to formulate it by examples: • The expected value when throwing a six-sided die is $E(X) = 3.5$, which as I see it, is not even part of the probability space $\\Omega = \\{1,2,3,4,5,6\\}$. Why is this not a problem at all? Is here anything wrong with my assumptions about the expected value? • Is there an expected value when throwing a coin? I cannot calculate a weighted mean on head and tail? Is there an expected value? Is it the maximum likely value, and what if there are multiple maximum likely values? I hope my questions have sufficiently revealed my (mis)conceptions of the topic. I'd appreciate any help and insights! • The expected value does not mean the value that you expect to see on the next roll of the die; it is what you would expect to be the average value of the outcomes on a large number of rolls. Roll the die $100$ times. You are likely to see a total of $350$ dots for an average of $350/100 = 3.5 = E[X]$. This is not guaranteed, you might get some number smaller" ]

[ "P(X>Y) when X and Y are continuous uniform distribution Suppose $$X$$ and $$Y$$ are continuous uniform random variables. If $$X \\sim U[a,b]$$, $$Y \\sim U[c,d]$$ and $$[c,d] \\subset [a,b]$$ find the probability that a random $$X$$ value is greater than a random $$Y$$ value. I think maybe that's problem can resolve drawing a rectangule, but a i need help with that When you draw the arbitrary rectangle, with vertices $$(a,c), (a,d), (b,c), (b,d)$$ and draw in the line $$y=x$$, we want to find the area of the region below the line inside the rectangle. If you draw this region out, you find that the region is a trapezoid with vertices $$(c,c), (b,c), (d,d),$$ and $$(b,d)$$", "# rectangle • Nov 24th 2009, 03:33 PM Sampras rectangle Consider the unit circle C: $x^2+y^2 = 1$. Suppose 2 points are chosen randomly: (i) $p$ is chosen from the circumference and (ii) $q$ is chosen from the interior (chosen independently). Let $R$ be the rectangle with diagonal $pq$. What is the probability that no point of R lies outside of C? Wouldn't a rectangle with diagonal $pq$ always lie inside the circle? • Nov 24th 2009, 05:31 PM TKHunny The Diagonal pq will always be in or on the circle. Not so the other diagonal. Easy demonstration: $p = <0, 1>$ $q = <1-\\epsilon , 0>$ Where $0 < \\epsilon << 1$ p is the North West Vertex q is the South East Vertex Where is the North East vertex?", "Points in a box $$x$$, $$y$$, and $$z$$ are each chosen uniformly and at random in the interval $$[0,1]$$ What is the probability that $$x^2+y^2+z^2 < 1$$?", "# Random points in a sphere A point $$(x,y,z)$$ is chosen uniformly at random inside a unit sphere. What is the probability that $$|x|+|y|+|z| < 1$$?", "set of squares. Implying that $Y>X$ with probability $1$. See McKay's original paper:", "+0 # A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability t +5 542 4 +1766 A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability that $x < y$? Mellie Apr 27, 2015 ### Best Answer #3 +17711 +10 You're right -- I considered only the endpoints ... (I'm going to have to learn to read .... ) If you draw a diagonal line through the point (0,0) at a 45° angle, it will go through the point (4,4) and divide the square into two congruent parts: the points above the line will have x-values less than their y-values; the points below the line will have x-values greater than their y-values. Since the area above the diagonal line equals the area below the diagonal line, the probability will be 1/2. geno3141 Apr 27, 2015 Sort: ### 4+0 Answers #1 +17711 +5 Of the four points: (0,0), (4,0), (4,1), and (0,1) only the point (0,1) has an x-value smaller than the y-value. The probability that the x-value is less than the y-value is 1/4. geno3141 Apr 27, 2015 #2 +1766 +5 Thank you geno, but this was incorrect. Sorry!!!!!! $$The point (x,y) satisfies x <", "# $$n$$ Points $$A$$,$$B$$,$$C$$ and $$D$$ are the vertices of the rectangle. Let $$n\\geq 1$$ point(s) be marked on the rectangle in such a way that there are no 3 collinear points. 3 points are chosen at random. What is the probability of forming a triangle where point $$B$$ isn't one of the vertices? ×", "probability in the xy plane-OG 10-METHOD 4 29 May 2011, 04:51 Equation |x/2| + |y/2| = 5 encloses a certain region on 10 18 Oct 2008, 23:43 5 A rectangle is plotted on the standard coordinate plane, 13 02 Dec 2010, 12:13 Coordinate Plane 1 24 Nov 2010, 01:04 Display posts from previous: Sort by", "& \\text{elsewhere} \\end{cases}$ The continuous uniform distribution has a particularly simple representation, just as its discrete counterpart does. Nevertheless, this random variable has great practical and theoretical utility. We will explore this distribution in more detail in the following example and in the exercises. ### A Geometric Problem Consider the square in the xy-plane bounded by the lines x = 0, x = 1, y = 0 and y = 1. Now consider a vertical line with equation x = b, where 0 ≤ b ≤ 1 is fixed. Note that this line will intersect the unit square just defined. Suppose we select a point inside this square, uniformly at random. If we let X be the x-coordinate of this random point, what is the probability that X is in the interval [0 , b]? An illustration of our problem is given in the figure below. Graphically, we are trying to find the probability that a randomly selected point inside the square lies to the left of the red line. The region to the left of the red line is a rectangle with area equal to b. The probability that our random point lies inside this rectangle is proportional to the area of that rectangle, since the larger the area of the rectangle, the larger the probability is", "# Problem #1508 1508 Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube? $\\mathrm{(A)}\\ \\frac{1}{4} \\qquad \\mathrm{(B)}\\ \\frac{3}{8} \\qquad \\mathrm{(C)}\\ \\frac{4}{7} \\qquad \\mathrm{(D)}\\ \\frac{5}{7} \\qquad \\mathrm{(E)}\\ \\frac{3}{4}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "A token starts at the point $(0,0)$ of an $xy$-coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. Find the probability the token ends at a point on the graph of $|y|=|x|$.", "# Probability, Random points in rectangle There is a rectangle, the lower left is always fixed at co-ordinate $(0, 0)$. Let the width and height of the rectangle be $w$ and $h$. Let $P$ be a randomly chosen point from the rectangle with co-ordinates $(x,y)$, $x\\geq 0$ and $x\\leq w$, $y\\geq 0$ and $y\\leq h$. $x$ and $y$ can be any real number satisfying the constraint above. What is the probability that the area of the rectangle whose lower left is $(0,0)$ and upper right is $P(x, y)$ (Hence, the area is $x\\cdot y$) is greater than $A$? $w$, $h$ and $A$ will be provided. - By \"randomly chosen\", do you mean \"uniformly randomly chosen\"? Also, I'd delete the sentence beginning \"$x$ and $y$ can be ...\", since at best it doesn't add to what was already said and at worst it sounds as if $x$ and $y$ are given or can be arbitrarily chosen, in contrast to the preceding sentence which states that they're randomly chosen. Also, the notation $P(x,y)$ doesn't make much sense, since $P$ is itself $(x,y)$ and not defined as a function of $x$ and $y$. – joriki Feb 16 '13 at 10:09 The points that satisfy that are those where $xy>A$, that is, $y>\\frac{A}{x}$ (you can do that because $x$ is positive and", "The point $X$ moves around inside a rectangle of dimension $p$ units by $q$ units. The distances of $X$ from the vertices of the rectangle are $a$, $b$, $c$ and $d$ units. What are the least and the greatest values of $a^2 + b^2 + c^2 + d^2$ and where is the point $X$ when these values occur?", "+0 # help probability 0 85 1 A point (x,y) is randomly selected such that 0 <= x <= 3 and 0 <= y <= 6. What is the probability that x + y <= 5? Express your answer as a common fraction. Jun 26, 2022 #1 +2448 0 The blue area represents the successes (areas where the sum of x and y is less than or equal to 5) The big rectangle represents all the possibilities. So, the probability is $$\\text{area of blue shaded region} \\over \\text{area of total black rectangle}$$ Can you take it from here? The Jun 27, 2022", "any of the $m$ vertices. So the probability that the vertex we get to $w$ from is $v$ is $\\frac1m$. Combining these, we get a probability of $$\\frac1n + \\left(1 - \\frac1n\\right)\\frac1m = \\frac1n + \\frac1m - \\frac1{nm} = \\frac{m+n-1}{mn}.$$", "# How to check if a point is inside a rectangle? There is a point $(x,y)$, and a rectangle $a(x_1,y_1),b(x_2,y_2),c(x_3,y_3),d(x_4,y_4)$, how can one check if the point inside the rectangle? $M$ of coordinates $(x,y)$ is inside the rectangle iff", "Tag Info Here is the idea. Let the vertices be $v_0,\\ldots,v_n$, where $v_0$ is the origin and the vertices are arranged in clockwise order. The first step is either clockwise or counterclockwise — let's say clockwise. Then the probability you're after is the probability that you reach $v_n$ before you reach $v_0$, starting at $v_1$. This is random walk on a line, so ...", "Fall 2015, problem 1 The numbers $1,2,3,4,5,6,7,8$ are assigned to the vertices of a cube so that each vertex has a different number. Find the probability that no two consecutive numbers are written on vertices with a common edge, where $8$ and $1$ are considered consecutive.", "+0 # Probability 0 27 1 Four points W, X, Y, and Z are chosen at random on a circle. Find the probabiity that WX > YZ. Feb 23, 2023 #1 +2602 +1 I think that the probability $$WX > YZ$$ equals the probability of $$WX < YZ$$. This means that each has a probability of $$\\color{brown}\\boxed{1 \\over 2}$$ Feb 23, 2023", "closed rectangle with vertices at $(0,2/3,4/3)$, $(2/3,2/3,1)$, $(1,2/3,2/3)$, and $(4/3,2/3,0)$. For every point $Z$ in this region, there exist unique points $X$ and $Y$ such that $Z\\in XY$ and $ZY=2XZ$." ]

[ "have $x-y\\geq1$, $y-z\\geq1$. The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too. $[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green); draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red); draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed); [/asy]$ The volume of this region is $V_2=\\dfrac{(n-2)^3}{6}$. So the probability is $p=\\dfrac{V_2}{V_1}=\\dfrac{(n-2)^3}{n^3}$. Substituting $n$ with the values in the choices, we find that when $n=10$, $p=\\frac{512}{1000}>\\frac{1}{2}$, when $n=9$, $p=\\frac{343}{729}<\\frac{1}{2}$. So $n\\geq 10$. So the answer is $\\boxed{\\textbf{(D)}\\ 10}$. ## Solution 2 Because $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$, which means that $x$, $y$, and $z$ distributes uniformly and independently in the interval $[0,n]$. So the point $(x, y, z)$ distributes uniformly in the cubic $0\\leqslant x, y, z \\leqslant n$, as shown in the figure below. The volume of this cubic is $V_0=n^3$. As we want to find the probablity of the incident $A=\\big\\{ |x-y| \\geq 1, |y-z| \\geq1, |z-x| \\geq", "The combined area of those regions is 3/4, and the area of the entire square is 1, so the requested probability is $\\boxed{3/4}$. [asy] unitsize(2cm); draw((1.1,0)--(0,0)--(0,1.1),linewidth(1)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,gray(0.7)); fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white); fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white); label(\"$a$\",(1.1,0),E); label(\"$b$\",(0,1.1),N); label(\"1\",(1,0),S); label(\"1\",(0,1),W); label(\"0\",(0,0),SW); [/asy]", "The last case above splits the nonzero integration region into two rectangles and a triangle. It's also acceptable to draw a figure and say something intelligent w/o being explicit about all the details. 3. What is $f_X(x)$? $f_X(x)= \\int_0^{1-x}f_{XY}(x,y) dy = 3/2 (1-x)^2, 0\\le x\\le1$ Note that $\\int_0^1 f_X(x)=1$, which is correct. 4. Are X and Y independent? $f_X(x)= 3/2 (1-x)^2$ $f_Y(y)= 3/2 (1-y)^2$ $f_{XY}(x,y)= 3(x+y)\\ne f_X(x) f_Y(y)$ no. 5. What is $P[X\\le Y]$ ? Integrate: $\\int_0^1 \\int_0^{\\min{x,1-x)} f(x0,y0) dy0 dx0 = 1/2$ That's reasonable because X and Y are symmetric. The box around the above expression is a meaningless unwanted artifact. 6. What is $E[X]$? 1/8 7. What is $COV[X,Y]$? E[XY] = 1/10. E[Y] = E[X]. COV = 1/10 - 1/8 * 1/8 = .085. 8. What is $\\rho_{X,Y}$? OK to write the formula 9. What is $f_Y(y|x)$? In the nonzero triangle, $f_Y(y|x)=F_{XY}(x,y)/F_X(x)$ 10. What is $E[Y|x]$? $\\int yf_Y(y|x) dy$ 4. I'm comparing two types of widgets, red and blue. Assume that the probability of each widget dieing in a small interval dt, given that it was alive at the start, is independent of its age. Assume that the probability of the red widget dieing in the next hour is 0.1%, for the blue, it's 0.01%. 1. Give the pdf for the red widget's lifetime. (You have", "dot(\"E\", E, NW); dot(\"F\", F, SW); dot(\"G\", G, S); dot(\"H\", H, N); dot(\"I\", I, NE); label(\"X\", X,SE); dot(\"J\", J, SE);[/asy]$ First let $s=2$ (where $s$ is the side length of the squares) for simplicity. We can extend $\\overline{IJ}$ until it hits the extension of $\\overline{AB}$. Call this point $X$. The area of triangle $AXJ$ then is $\\dfrac{3 \\cdot 4}{2}$ The area of rectangle $BXIC$ is $2 \\cdot 1 = 2$. Thus, our desired area is $6-2 = 4$. Now, the ratio of the shaded area to the combined area of the three squares is $\\frac{4}{3\\cdot 2^2} = \\boxed{\\textbf{(C)}\\hspace{.05in}\\frac{1}{3}}$. Solution 2 $[asy] pair A,B,C,D,E,F,G,H,I,J,X; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); X= (1.25,1); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot(X,red); label(\"A\", A, NW); label(\"B\", B, NE); label(\"C\", C, NE); label(\"D\", D, NW); label(\"E\", E, NW); label(\"F\", F, SW); label(\"G\", G, S); label(\"H\", H, N); label(\"I\", I, NE); label(\"X\", X,SW,red); label(\"J\", J, SE);[/asy]$ Let the side length of each square be $1$. Let the intersection of $AJ$ and $EI$ be $X$. Since $[ABCD]=[GHIJ]$, $AD=IJ$. Since $\\angle IXJ$ and $\\angle AXD$ are vertical angles, they are congruent. We also have $\\angle JIH\\cong\\angle ADC$ by", "+0 # A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability t +5 542 4 +1766 A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability that $x < y$? Mellie Apr 27, 2015 ### Best Answer #3 +17711 +10 You're right -- I considered only the endpoints ... (I'm going to have to learn to read .... ) If you draw a diagonal line through the point (0,0) at a 45° angle, it will go through the point (4,4) and divide the square into two congruent parts: the points above the line will have x-values less than their y-values; the points below the line will have x-values greater than their y-values. Since the area above the diagonal line equals the area below the diagonal line, the probability will be 1/2. geno3141 Apr 27, 2015 Sort: ### 4+0 Answers #1 +17711 +5 Of the four points: (0,0), (4,0), (4,1), and (0,1) only the point (0,1) has an x-value smaller than the y-value. The probability that the x-value is less than the y-value is 1/4. geno3141 Apr 27, 2015 #2 +1766 +5 Thank you geno, but this was incorrect. Sorry!!!!!! $$The point (x,y) satisfies x <", "area bounded by $\\overline{A_i}{A_j}$ and the arc $A_iA_j$ is fixed, so we only need to consider the relevant triangles. $[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label(\"P\", P, dir(-75)); label(\"A_1\", dir(112.5), dir(112.5)); label(\"A_2\", dir(112.5-45), dir(112.5-45)); label(\"A_3\", dir(112.5-90), dir(112.5-90)); label(\"A_4\", dir(112.5-135), dir(112.5-135)); label(\"A_5\", dir(112.5-180), dir(112.5-180)); label(\"A_6\", dir(112.5-225), dir(112.5-225)); label(\"A_7\", dir(112.5-270), dir(112.5-270)); label(\"A_8\", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]$ Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\\tfrac{1}{7}-\\tfrac{1}{8}=\\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\\tfrac{1}{7}$. Similarly, $P$ was moved right by $\\tfrac{1}{8}-\\tfrac{1}{9}=\\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\\tfrac{1}{9}$. This means that $P$ has coordinates $(\\tfrac{1}{72},-\\tfrac{1}{56})$. Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\\tfrac{\\sqrt{2}}{2}(\\tfrac{1}{56}-\\tfrac{1}{72})=\\tfrac{\\sqrt{2}}{504}$. Remembering the definition of our unit, this yields a final area of", "lines. $\\setlength{\\unitlength}{2.5cm} \\begin{picture}(1,1) \\qbezier(0,0)(0,0)(1,0) \\qbezier(0,0)(0,0)(0,1) \\qbezier(1,0)(1,0)(1,1) \\qbezier(0,1)(0,1)(1,1) \\linethickness{1.3pt} \\qbezier(1,0)(1,0)(0,1) \\qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71) \\qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0) \\end{picture} $ 6. ## wow! +rep Originally Posted by JakeD Here's the picture. The probability is the area between the thicker lines. $\\setlength{\\unitlength}{2.5cm} \\begin{picture}(1,1) \\qbezier(0,0)(0,0)(1,0) \\qbezier(0,0)(0,0)(0,1) \\qbezier(1,0)(1,0)(1,1) \\qbezier(0,1)(0,1)(1,1) \\linethickness{1.3pt} \\qbezier(1,0)(1,0)(0,1) \\qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71) \\qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0) \\end{picture} $ wow! +rep 7. Originally Posted by JakeD Here's the picture. The probability is the area between the thicker lines. $\\setlength{\\unitlength}{2.5cm} \\begin{picture}(1,1) \\qbezier(0,0)(0,0)(1,0) \\qbezier(0,0)(0,0)(0,1) \\qbezier(1,0)(1,0)(1,1) \\qbezier(0,1)(0,1)(1,1) \\linethickness{1.3pt} \\qbezier(1,0)(1,0)(0,1) \\qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71) \\qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0) \\end{picture} $ Will you please explain how did you get that picture(how it is linked to the question)? 8. Originally Posted by malaygoel Will you please explain how did you get that picture(how it is linked to the question)? Ok, I understand it. I drew the graphs of $x + y = 1$ $x^2 + y^2 = 1$ and got the same picture as yours. I got the answer also from the picture. Did you also arrive at the picture using Analytic Geometry?If no, what method you adopted. KeepSmiling Malay 9. Originally Posted by malaygoel Ok, I understand it. I drew the graphs of $x + y = 1$ $x^2 + y^2 = 1$ and got the same picture as yours. I got the answer also", "draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(0,10),C=(10,0),P=(0,0),B=(13.660,13.660); dot(A); label(\"A\",A,W); dot(C); label(\"C\",C,S); dot(P); label(\"P_0\",P,SW); dot(B); label(\"B\",B,NE); draw(A--C--P--A); draw(A--B--C); draw(B--P,dotted); [/asy]$ By repeatedly applying the Midpoint Formula, we can determine the coordinates of $P_1$, $P_2$, $P_3$, and so on. We can also use the Distance Formula to calculate the distance of $P_0P_1$, $P_0P_2$, and so on. The values are shown in the below table. $n$ Coordinates of $P_n$ $P_0 P_n$ 1 $(0,2a)$ $2a$ 2 $(a+a\\sqrt{3},-a+a\\sqrt{3})$ $2a\\sqrt{2}$ 3 $(a-a\\sqrt{3},a-a\\sqrt{3})$ $a\\sqrt{6} - a\\sqrt{2}$ 4 $(-a+a\\sqrt{3},a+a\\sqrt{3})$ $2a\\sqrt{2}$ 5 $(2a,0)$ $2a$ 6 $(0,0)$ $0$ 7 $(0,2a)$ $2a$ 8 $(a+a\\sqrt{3},-a+a\\sqrt{3})$ $2a\\sqrt{2}$ Note that the coordinates of $P_n$ as well as the distance $P_0 P_n$ cycle after $n = 6$. Thus, $P_0P_n = \\frac{\\sqrt{6} - \\sqrt{2}}{2} \\cdot P_0P_1$ if $n \\equiv 3 \\pmod{6}$, $P_0P_n = 0$ if $n \\equiv 0 \\pmod{6}$, $P_0P_n = P_0P_1$ if $n \\equiv 1, 5 \\pmod{6}$, and $P_0P_n = \\sqrt{2} \\cdot P_0P_1$ if $n \\equiv 2, 4 \\pmod{6}$.", "a tricky probability question which has nothing to do with probability :) Posted: May 15, 2013 in Mathematics Here is the question: There are x green apples and y red apples in a box with $x>y$ and $x+y\\leq2008$ When two apples are drawn at random, the probability that the two apples are of the same color is exactly one-half. Please find the largest possible value of x. Solution: First of all, the probability of the getting same color is one-half. $P(\\mbox{same color}) = \\dfrac{x}{x+y}\\cdot\\dfrac{x-1}{x+y-1} + \\dfrac{y}{x+y}\\cdot\\dfrac{y-1}{x+y-1} = \\dfrac{1}{2}$ then do some algebraic manipulation, like cross multiplying and stuffs like that. $2x^2-2x+2y^2-2y = x^2+xy-x+xy+y^2-y$ $x^2+y^2-x-y-2xy=0$ $(x-y)^2-x-y=0$ $(x-y)^2=x+y\\leq2008$ we are good at this point, because we have found a relationship between $(x-y)$ and $(x+y)$, since x is greater than y so $(x-y)$ is positive and we can take the square root of the inequality below, and since x and y are integers we will take the integer part of the square root of 2008. $(x-y)^2\\leq2008$ $x-y \\leq 44$ and since $x+y = (x-y)^2$ then we have $x+y = (x-y)^2 \\leq 44^2 = 1936$ Now here comes the tricky part where most students can’t get to, let’s look at what the problem asks us to find. It says find the largest possible value of x. So we want to have this,", "ever want to talk about probabilities, then only use the word probability and avoid using the word odds. Also, @Vimath conditional probabilities are written with a vertical bar, not a slanted bar. It should be $P(A\\mid B')$, not $P(A/B')$ – JMoravitz Mar 23 at 13:59 • @JMoravitz I missed the syntax there , I'll try not to repeat it. – Vimath Mar 23 at 14:53 Guide: 1) Draw rectangle $$2\\le x\\le 4,$$ $$3\\le y\\le 5$$. 2) The area of the rectangle is $$4$$, so pdf is $$1/4$$. 3) Draw line $$y=x$$. 4) Find area of the rectangle above the line, which is $$7/2$$. 5) Finally, the required probability is $$7/2\\cdot 1/4=7/8$$. Here is the graph: $$\\hspace{2cm}$$ Bus $$A$$ arriving before bus $$B$$: $$A(2.5,4.5),B(3.5,4.5),C(2.5,3.5),D(3.4,3.7).$$ Bus $$A$$ arriving after bus $$B$$: $$E(3.7,3.3).$$ • do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify? – IrinaS Mar 23 at 5:15 • The graph is a good idea but you need to get rid of the block F/G. Obviously the intersection is 2 hrs by 2 hrs - a square. – Craig Hicks Mar 23 at 8:18 • Thanks, I rolled back to my first answer. – farruhota Mar 23 at 8:29 • But the graph was great! Bring", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\framebox{(D) 9/16}$. ## See Also 2009 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "the condition $x<y$? can you finish the problem…….. Now the line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.Therefore it will form a Triangle $\\triangle AOD$ (as shown above) whose $AO=1$ and $AD=1$.Therefore area of $\\triangle AOD=\\frac{1}{2}$ i.e (red region).Now can you find out the probability with the condition $x<y$? can you finish the problem…….. Therefore the required probability ($x<y$) is $\\frac{\\frac{1}{2}}{4}$=$\\frac{1}{8}$", "the red one) • $0< y_1 < 1/2 - x_1$ (the blue triangle is within the red one) • $x_2 > x_1$ (the green triangle is to the right of the blue) • $y_1< y_2 < 1/2 - x2$ (our Case A condition) So now we integrate the area under these conditions: Integrate[1/2 (-x2 + x1 + y1 - y2 + 1/2)^2, {x1, 0, 1/2}, {y1, 0, 1/2 - x1}, {x2, x1, 1/2 - y1}, {y2, y1, 1/2 - x2}] (* 1/7680 *) Integrate[ 1/2 (-x2 + x1 + y1 - y2 + 1/2)^2, {x1, 0, 1/2}, {y1, 0, 1/2 - x1}, {x2, x1, 1/2 - y1}, {y2, y1, 1/2 - x2}]/ Integrate[ 1, {x1, 0, 1/2}, {y1, 0, 1/2 - x1}, {x2, x1, 1/2 - y1}, {y2, y1, 1/2 - x2}] (* 1/20 *) Case B: Now the conditions are: • $0 < x_1 < 1/2$ (the blue triangle does not extend beyond the red one) • $x_2 > x_1$ (the green is to the right of the blue) • $x_1 < x_2 < 1/2$ (the blue triangle is within the red one) • $0 < y_2 < y_1$ (our Case B condition) • Re \"I'M POSTING AS I DERIVE:\", that's cool, that's like Vangelis composing and recording in real time (see YouTube documentary). Feb 25 '16", "is the side length of the squares) for simplicity. We can extend $\\overline{IJ}$ until it hits the extension of $\\overline{AB}$. Call this point $X$. The area of triangle $AXJ$ then is $\\dfrac{3 \\cdot 4}{2}$ The area of rectangle $BXIC$ is $2 \\cdot 1 = 2$. Thus, our desired area is $6-2 = 4$. Now, the ratio of the shaded area to the combined area of the three squares is $\\frac{4}{3\\cdot 2^2} = \\boxed{\\textbf{(C)}\\hspace{.05in}\\frac{1}{3}}$. ## Solution 2 $[asy] pair A,B,C,D,E,F,G,H,I,J,X; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); X= (1.25,1); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot(X,red); label(\"A\", A, NW); label(\"B\", B, NE); label(\"C\", C, NE); label(\"D\", D, NW); label(\"E\", E, NW); label(\"F\", F, SW); label(\"G\", G, S); label(\"H\", H, N); label(\"I\", I, NE); label(\"X\", X,SW,red); label(\"J\", J, SE);[/asy]$ Let the side length of each square be $1$. Let the intersection of $AJ$ and $EI$ be $X$. Since $[ABCD]=[GHIJ]$, $AD=IJ$. Since $\\angle IXJ$ and $\\angle AXD$ are vertical angles, they are congruent. We also have $\\angle JIH\\cong\\angle ADC$ by definition. So we have $\\triangle ADX\\cong\\triangle JIX$ by $\\textit{AAS}$ congruence. Therefore, $DX=JX$. Since $C$ and $D$ are midpoints of sides, $DH=CJ=\\dfrac{1}{2}$. This combined with", "$(x_0,y_0)$ and those of the second point be $(x_1, y_1)$. Then the coordinates of the other two points on the rectangle are $(x_0, y_1)$ and $(x_1, y_0)$. Let us suppose the first point must fall on the quarter-circle in the first quadrant. This does not affect the probability at all because of the symmetry of the circle. Then we have $y_0=\\sqrt{1-x_0^2}$. In order to have the rectangle fully inside of the circle, all of the following must be true: $$x_0^2+y_1^2 \\le 1$$ $$x_1^2+y_0^2 \\le 1$$ We can substitute into the second requirement $y_0=\\sqrt{1-x^2}$ to get $$x_1^2+1-x_0^2 \\le 1$$ $$x_1^2-x_0^2 \\le 0$$ $$x_0^2 \\ge x_1^2$$ $$|x_0| \\ge |x_1|$$ Similarly, the first requirement can be replaced by $$|y_0| \\ge |y_1|$$ and our requirements are $$|x_0| \\ge |x_1|$$ $$|y_0| \\ge |y_1|$$ Try and visualize this: Let $F$ be the event in which the rectangle fits, and $P(F|x_0)$ be the probability that the rectangle falls entirely inside of the circle given the value of $x_0$. All points $(x_1, y_1)$ for which the rectangle would fit in the circle are inside of the green rectangle shown with width $2x_0$ and height $2y_0$. This is now a spatial probability problem. Given $x_0$, we can now say that the probability of the rectangle fitting is the probability that $(x_1, y_1)$ lands in that green rectangle.", "label(\"A'\", AA, N); [/asy]$ From what we are given it easily follows that $ABC$ is an equilateral triangle, and its center $O$ is at the same time the foot of the height from $D$. We know that $AD=5$ and $OD=4$, from the Pythagorean theorem it follows that $OA=3$. The segment $OA$ is $2/3$ of the segment $AA'$. Then $AA'$ has length $9/2$. $AA'$ is the height in the triangle $ABC$, hence we have $AA' = \\dfrac{BC\\cdot \\sqrt 3}2$, and therefore $BC=3\\sqrt 3$. We can now compute the area of the triangle $ABC$ as $\\dfrac{BC\\cdot AA'}2 = \\dfrac{27\\sqrt 3}{4}$, and finally the volume of the tetrahedron $ABCD$ as $\\dfrac{ABC\\cdot DO}3 = 9\\sqrt 3$. Why did we compute all this stuff? Consider what happens when the leg $BD$ breaks in the described place $E$: import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple BB=0.8*B + 0.2*D; triple AC=(A+C)/2; filldraw(A--BB--C--cycle, lightgray, black ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(BB--D, black+1); draw(B--BB); draw(C--D, black+1); draw(A--O, dashed); draw(B--AC, dashed); draw(BB--AC, dashed); draw(C--O, dashed); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$E$\", BB, E); label(\"$C$\", C, W); label(\"$O$\", O, SE); label(\"$D$\", D, Z); label(\"$B'$\", AC, W); (Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)') The triangle $AEC$ will now be the base of the tripod, and our goal is", "+0 # do theese 3 points lie on the same lines? (-4,-2) (2,2.5) (8,7) or (-10,4) (-3,2.8) (-17,6.8) pleeeease respond 0 301 4 do theese 3 points lie on the same lines? (-4,-2) (2,2.5) (8,7) or (-10,4) (-3,2.8) (-17,6.8) pleeeease respond Guest Dec 3, 2014 #3 +18843 +10 3 points lie on the same line, if the area of the triangle from 3 points is 0: $$\\\\ P_1 = (x_1,y_1) \\\\ P_2 = (x_2,y_2) \\\\ P_3 = (x_3,y_3) \\\\ \\small{\\text{ The area of a triangle with 3 Points is the determinant } = \\left| \\begin{array}{cc} (x_1-x_2) & (x_3-x_2) \\\\ (y_1-y_2) & (y_3-y_2) \\\\ \\end{array} \\right|\\\\\\\\ \\boxed{ \\text{ area } = (x_1-x_2) *(y_3-y_2) - (y_1-y_2)*(x_3-x_2)} }$$ I do theese 3 points lie on the same lines? (-4,-2) (2,2.5) (8,7) $$\\\\P_1(x_1 = -4,\\ y_1 = -2 ) \\\\ P_2(x_2 = 2,\\ y_2 = 2.5 ) \\\\ P_3(x_3 = 8,\\ y_3 = 7 ) \\\\ \\text{ area } = (x_1-x_2) *(y_3-y_2) - (y_1-y_2)*(x_3-x_2)} \\\\ \\text{ area } = (-4-2) *(7-2.5) - ( (-2)-2.5)*(8-2)} \\\\ \\text{ area } = (-6) *(4.5) - ( -4.5)*(6)} \\\\ \\text{ area } = (-27) - ( -27)} \\\\ \\text{ area } = -27 + 27 = 0 }$$ area = 0: The 3 Points lie on the same line. II do theese 3 points lie on", "(1947), which shows that the aforementioned trio of properties is necessary and sufficient for the representability of $$\\succeq$$ by a function $$U$$ such that $U(P_f)=\\sum\\limits_{i=1}^{n}P_f(x_i)u(x_i),$ where $$u : \\mathcal{X}\\mapsto \\mathbb{R}$$ is a consequence utility function unique up to positive linear transformation. 1.3 The probability triangle The probability triangle (aka “Marschak-Machina triangle”) offers a helpful visual representation of preferences over the space of lotteries over $$\\{x_1, x_2, x_3\\}$$, with $$x_3\\succ x_2 \\succ x_1$$. Since, for any $$P\\in \\mathcal{P}$$, $$P(x_2)= 1- P(x_1)-P(x_3)$$, one can represent the situation two-dimensionally, with lotteries appearing as points in a unit triangle in which the horizontal axis gives us $$P(x_1)$$ and the vertical one gives us $$P(x_3)$$. The northwestern, southwestern and southeastern corners respectively correspond to the lotteries yielding $$x_3, x_2$$ and $$x_1$$ for sure. Now, as is easily demonstrated, SEU is committed to Stochastic Dominance For all acts $$f$$ and $$g$$: if, for any outcome $$x$$, the probability according to $$P_f$$ of obtaining an outcome that is weakly preferred to $$x$$ is at least as great as the corresponding probability according to $$P_g$$ (in other words: $$\\sum_{\\{y\\in\\mathcal{X}: y\\succeq x\\}} P_f(y)$$ $$\\geq$$ $$\\sum_{\\{y\\in\\mathcal{X}: y\\succeq x\\}} P_g(y)$$), then $$P_f\\succeq P_g$$. Indeed, the above principle follows from Independence and is in fact equivalent to Savage’s Eventwise Monotonicity condition, given the other conditions in place (Grant 1995). Therefore" ]

[ "have $x-y\\geq1$, $y-z\\geq1$. The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too. $[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green); draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red); draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed); [/asy]$ The volume of this region is $V_2=\\dfrac{(n-2)^3}{6}$. So the probability is $p=\\dfrac{V_2}{V_1}=\\dfrac{(n-2)^3}{n^3}$. Substituting $n$ with the values in the choices, we find that when $n=10$, $p=\\frac{512}{1000}>\\frac{1}{2}$, when $n=9$, $p=\\frac{343}{729}<\\frac{1}{2}$. So $n\\geq 10$. So the answer is $\\boxed{\\textbf{(D)}\\ 10}$. ## Solution 2 Because $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$, which means that $x$, $y$, and $z$ distributes uniformly and independently in the interval $[0,n]$. So the point $(x, y, z)$ distributes uniformly in the cubic $0\\leqslant x, y, z \\leqslant n$, as shown in the figure below. The volume of this cubic is $V_0=n^3$. As we want to find the probablity of the incident $A=\\big\\{ |x-y| \\geq 1, |y-z| \\geq1, |z-x| \\geq", "The combined area of those regions is 3/4, and the area of the entire square is 1, so the requested probability is $\\boxed{3/4}$. [asy] unitsize(2cm); draw((1.1,0)--(0,0)--(0,1.1),linewidth(1)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,gray(0.7)); fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white); fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white); label(\"$a$\",(1.1,0),E); label(\"$b$\",(0,1.1),N); label(\"1\",(1,0),S); label(\"1\",(0,1),W); label(\"0\",(0,0),SW); [/asy]", "and $y$ is either $0$ or $1$. Case 4: $x$ is is chosen from the interval $[0,1]$, and $y$ is also chosen from the interval $[0,1]$. Each case has a $\\frac{1}{4}$ chance of occurring (as it requires two coin flips). For Case 1, we need $x$ and $y$ to be different. Therefore, the probability for success in Case 1 is $\\frac{1}{2}$. For Case 2, if $x$ is 0, we need $y$ to be in the interval $\\left(\\frac{1}{2}, 1\\right]$. If $x$ is 1, we need $y$ to be in the interval $\\left[0, \\frac{1}{2}\\right)$. Regardless of what $x$ is, the probability for success for Case 2 is $\\frac{1}{2}$. By symmetry, Case 3 has the same success rate as Case 2. For Case 4, we must use geometric probability because there are an infinite number of pairs $(x, y)$ that can be selected, whether they satisfy the inequality or not. Graphing $|x-y| > \\tfrac{1}{2}$ gives us the following picture where the shaded area is the set of all the points that fulfill the inequality: $[asy] filldraw((0,0)--(0,1)--(1/2,1)--(0,1/2)--cycle,black+linewidth(1)); filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+linewidth(1)); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); [/asy]$ The shaded area is $\\frac{1}{4}$, which means the probability for success for case 4 is $\\frac{1}{4}$ (since the total area of the bounding square, containing all possible pairs, is $1$). Adding up the success rates from each case, we get: $\\left(\\frac{1}{4}\\right) \\cdot", "$\\ 12 \\choose 2$=$66$ ways to pick to of the colours… can you finish the problem…….. Now given condition is that the sum will be $7$.So $7$ can be obtained by $1 +6$,$2+5$,$3+4$ and Each number in the bag has two different colors, Therefore each of these three options corresponds to four pairs of colors.SO $7$ comes from $3.4$=$12$ pairs….. can you finish the problem…….. So our required probability will be $\\frac{12}{66}=\\frac{2}{11}$. Categories ## Probability in Coordinates | AMC-10A, 2003 | Problem 12 Try this beautiful problem from Probability based on Coordinates. ## Probability in Coordinates – AMC-10A, 2003- Problem 12 A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability that $x<y$? • $\\frac{1}{8}$ • $\\frac{1}{6}$ • $\\frac{2}{3}$ ### Key Concepts Number system Cube Answer: $\\frac{1}{8}$ AMC-10A (2003) Problem 12 Pre College Mathematics ## Try with Hints The given vertices are $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$.if we draw a figure using the given points then we will get a rectangle as shown above.Clearly lengtht of $OC$= $4$ and length of $AO$=$1$.Therefore area of the rectangle is $4 \\times 1=4$.now we have to find out the probability that $x<y$.so we draw a line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.can you find out the area with", "+0 # A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability t +5 542 4 +1766 A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability that $x < y$? Mellie Apr 27, 2015 ### Best Answer #3 +17711 +10 You're right -- I considered only the endpoints ... (I'm going to have to learn to read .... ) If you draw a diagonal line through the point (0,0) at a 45° angle, it will go through the point (4,4) and divide the square into two congruent parts: the points above the line will have x-values less than their y-values; the points below the line will have x-values greater than their y-values. Since the area above the diagonal line equals the area below the diagonal line, the probability will be 1/2. geno3141 Apr 27, 2015 Sort: ### 4+0 Answers #1 +17711 +5 Of the four points: (0,0), (4,0), (4,1), and (0,1) only the point (0,1) has an x-value smaller than the y-value. The probability that the x-value is less than the y-value is 1/4. geno3141 Apr 27, 2015 #2 +1766 +5 Thank you geno, but this was incorrect. Sorry!!!!!! $$The point (x,y) satisfies x <", "The last case above splits the nonzero integration region into two rectangles and a triangle. It's also acceptable to draw a figure and say something intelligent w/o being explicit about all the details. 3. What is $f_X(x)$? $f_X(x)= \\int_0^{1-x}f_{XY}(x,y) dy = 3/2 (1-x)^2, 0\\le x\\le1$ Note that $\\int_0^1 f_X(x)=1$, which is correct. 4. Are X and Y independent? $f_X(x)= 3/2 (1-x)^2$ $f_Y(y)= 3/2 (1-y)^2$ $f_{XY}(x,y)= 3(x+y)\\ne f_X(x) f_Y(y)$ no. 5. What is $P[X\\le Y]$ ? Integrate: $\\int_0^1 \\int_0^{\\min{x,1-x)} f(x0,y0) dy0 dx0 = 1/2$ That's reasonable because X and Y are symmetric. The box around the above expression is a meaningless unwanted artifact. 6. What is $E[X]$? 1/8 7. What is $COV[X,Y]$? E[XY] = 1/10. E[Y] = E[X]. COV = 1/10 - 1/8 * 1/8 = .085. 8. What is $\\rho_{X,Y}$? OK to write the formula 9. What is $f_Y(y|x)$? In the nonzero triangle, $f_Y(y|x)=F_{XY}(x,y)/F_X(x)$ 10. What is $E[Y|x]$? $\\int yf_Y(y|x) dy$ 4. I'm comparing two types of widgets, red and blue. Assume that the probability of each widget dieing in a small interval dt, given that it was alive at the start, is independent of its age. Assume that the probability of the red widget dieing in the next hour is 0.1%, for the blue, it's 0.01%. 1. Give the pdf for the red widget's lifetime. (You have", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\framebox{(D) 9/16}$. ## See Also 2009 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 10\" ## Problem On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\ \\frac{1}{16}\\qquad\\textbf{(B)}\\ \\frac{7}{16}\\qquad\\textbf{(C)}\\ \\frac{1}2\\qquad\\textbf{(D)}\\ \\frac{9}{16}\\qquad\\textbf{(E)}\\ \\frac{49}{64}$ ## Solution There are $8^2=64$ total squares. There are $(8-1)(4)=28$ unit squares on the perimeter and therefore $64-28=36$ NOT on the perimeter. The probability of choosing one of those squares is $\\frac{36}{64} = \\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$. ## Solution 2 The squares that don't touch the border are in a 6 by 6 square. Therefore, the number of squares that don't touch the border is $6*6=36$. There are 64 choices total. Therefore, the answer is $36/64$, which simplifies to $\\boxed{\\textbf{(D)}\\ \\frac{9}{16}}$.", "lines. $\\setlength{\\unitlength}{2.5cm} \\begin{picture}(1,1) \\qbezier(0,0)(0,0)(1,0) \\qbezier(0,0)(0,0)(0,1) \\qbezier(1,0)(1,0)(1,1) \\qbezier(0,1)(0,1)(1,1) \\linethickness{1.3pt} \\qbezier(1,0)(1,0)(0,1) \\qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71) \\qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0) \\end{picture} $ 6. ## wow! +rep Originally Posted by JakeD Here's the picture. The probability is the area between the thicker lines. $\\setlength{\\unitlength}{2.5cm} \\begin{picture}(1,1) \\qbezier(0,0)(0,0)(1,0) \\qbezier(0,0)(0,0)(0,1) \\qbezier(1,0)(1,0)(1,1) \\qbezier(0,1)(0,1)(1,1) \\linethickness{1.3pt} \\qbezier(1,0)(1,0)(0,1) \\qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71) \\qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0) \\end{picture} $ wow! +rep 7. Originally Posted by JakeD Here's the picture. The probability is the area between the thicker lines. $\\setlength{\\unitlength}{2.5cm} \\begin{picture}(1,1) \\qbezier(0,0)(0,0)(1,0) \\qbezier(0,0)(0,0)(0,1) \\qbezier(1,0)(1,0)(1,1) \\qbezier(0,1)(0,1)(1,1) \\linethickness{1.3pt} \\qbezier(1,0)(1,0)(0,1) \\qbezier(1.0, 0.0)(1.0, 0.41)(0.71, 0.71) \\qbezier(0.71, 0.71)(0.41, 1.0)(0.0, 1.0) \\end{picture} $ Will you please explain how did you get that picture(how it is linked to the question)? 8. Originally Posted by malaygoel Will you please explain how did you get that picture(how it is linked to the question)? Ok, I understand it. I drew the graphs of $x + y = 1$ $x^2 + y^2 = 1$ and got the same picture as yours. I got the answer also from the picture. Did you also arrive at the picture using Analytic Geometry?If no, what method you adopted. KeepSmiling Malay 9. Originally Posted by malaygoel Ok, I understand it. I drew the graphs of $x + y = 1$ $x^2 + y^2 = 1$ and got the same picture as yours. I got the answer also", "# Probability: 2 random numbers such that $x > 2y$ This is probably an easy one, but I'm new to probability and really don't know how to solve this question. Select two numbers randomly from interval $[0, 1]$ call them $x$ and $y$, what is the probability that $x$, (the first number) is greater that $2y$ ? Hint: Draw the part $T$ of the square $[0,1]\\times[0,1]$ defined by the inequation $x\\gt2y$. (Sub-hint: $T$ is a triangle.) Measure the area of $T$. Conclude. • Just for my understanding, what do you mean with T? What's the definition? – Hans Mar 20 '13 at 7:20 • @Hans see my picture. $x>2y$ is the portion of the rectangle with $(x,y)$ coordinates below the line $y=\\frac{1}{2}x$ – Rustyn Mar 20 '13 at 7:59 • What's the definition? Please read: the part T of the square [0,1]×[0,1] defined by the inequation x>2y. – Did Mar 20 '13 at 8:10 • @RustynYazdanpour Thanks for the picture. – Did Mar 20 '13 at 8:11 • @Did No problemo – Rustyn Mar 20 '13 at 8:13 $T$ is the shaded part of this picture: $x>2y$ is the portion of the rectangle with $(x,y)$ coordinates below the line $y=\\frac{1}{2}x$ If $y = t$, then valid $x$ points are located above the ${x}\\gt{2t}$ (see yellow area in image).", "【全网首发】2021AIME I 真题及答案 【全网首发】2021AIME I 真题及答案 Problem1 Zou and Chou are practicing their 100-meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\\frac23$ if they won the previous race but only $\\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ Problem2 In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.$[asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, W); label(\"B\", B, W); label(\"C\", C, (1,0)); label(\"D\", D, (1,0)); label(\"F\", F, N); label(\"E\", E, S); [/asy]$ Problem3 Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$ Problem4 Find the number of ways $66$ identical coins can be separated into three nonempty piles so that", "$(x_0,y_0)$ and those of the second point be $(x_1, y_1)$. Then the coordinates of the other two points on the rectangle are $(x_0, y_1)$ and $(x_1, y_0)$. Let us suppose the first point must fall on the quarter-circle in the first quadrant. This does not affect the probability at all because of the symmetry of the circle. Then we have $y_0=\\sqrt{1-x_0^2}$. In order to have the rectangle fully inside of the circle, all of the following must be true: $$x_0^2+y_1^2 \\le 1$$ $$x_1^2+y_0^2 \\le 1$$ We can substitute into the second requirement $y_0=\\sqrt{1-x^2}$ to get $$x_1^2+1-x_0^2 \\le 1$$ $$x_1^2-x_0^2 \\le 0$$ $$x_0^2 \\ge x_1^2$$ $$|x_0| \\ge |x_1|$$ Similarly, the first requirement can be replaced by $$|y_0| \\ge |y_1|$$ and our requirements are $$|x_0| \\ge |x_1|$$ $$|y_0| \\ge |y_1|$$ Try and visualize this: Let $F$ be the event in which the rectangle fits, and $P(F|x_0)$ be the probability that the rectangle falls entirely inside of the circle given the value of $x_0$. All points $(x_1, y_1)$ for which the rectangle would fit in the circle are inside of the green rectangle shown with width $2x_0$ and height $2y_0$. This is now a spatial probability problem. Given $x_0$, we can now say that the probability of the rectangle fitting is the probability that $(x_1, y_1)$ lands in that green rectangle.", "# probability question - help! posted by . Two people agree to meet at a coffee shop. They each independently pick a random moment in time between 8 a.m. and 9 a.m. and show up exactly at their selected time. But they are very impatient, and only stay for 10 minutes after when they arrive. What is the probability that they meet? Express your answer as a common fraction. • probability question - help! - • probability question - help! - Gee you guys always seem to provide wrong answers to easy problems. By using geometric probability, we find the answer to be 11/36. • probability question - help! - 11/36 • probability question - help! - mathcounts and yea are correct We solve the problem by graphing. Let $x$ and $y$ be the arrival times of the two people in minutes after 8 a.m., so $0 \\le x,y \\le 60$. Then the two people meet if and only if $|x - y| \\le 10$. We graph the set of points $(x,y)$ such that $|x - y| \\le 10$. [asy] unitsize(0.08 cm); filldraw((0,0)--(10,0)--(60,50)--(60,60)--(50,60)--(0,10)--cycle,gray(0.7),invisible); draw((0,0)--(60,0)--(60,60)--(0,60)--cycle); draw((10,0)--(60,50)); draw((0,10)--(50,60)); label(\"$x$\", (60,0), E); label(\"$y$\", (0,60), N); dot(\"$(0,0)$\", (0,0), SW); dot(\"$(60,0)$\", (60,0), S); dot(\"$(60,60)$\", (60,60), NE); dot(\"$(0,60)$\", (0,60), W); dot(\"$(10,0)$\", (10,0), S); dot(\"$(60,50)$\", (60,50), E); dot(\"$(50,60)$\", (50,60), N); dot(\"$(0,10)$\", (0,10), W); [/asy] The", "ever want to talk about probabilities, then only use the word probability and avoid using the word odds. Also, @Vimath conditional probabilities are written with a vertical bar, not a slanted bar. It should be $P(A\\mid B')$, not $P(A/B')$ – JMoravitz Mar 23 at 13:59 • @JMoravitz I missed the syntax there , I'll try not to repeat it. – Vimath Mar 23 at 14:53 Guide: 1) Draw rectangle $$2\\le x\\le 4,$$ $$3\\le y\\le 5$$. 2) The area of the rectangle is $$4$$, so pdf is $$1/4$$. 3) Draw line $$y=x$$. 4) Find area of the rectangle above the line, which is $$7/2$$. 5) Finally, the required probability is $$7/2\\cdot 1/4=7/8$$. Here is the graph: $$\\hspace{2cm}$$ Bus $$A$$ arriving before bus $$B$$: $$A(2.5,4.5),B(3.5,4.5),C(2.5,3.5),D(3.4,3.7).$$ Bus $$A$$ arriving after bus $$B$$: $$E(3.7,3.3).$$ • do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify? – IrinaS Mar 23 at 5:15 • The graph is a good idea but you need to get rid of the block F/G. Obviously the intersection is 2 hrs by 2 hrs - a square. – Craig Hicks Mar 23 at 8:18 • Thanks, I rolled back to my first answer. – farruhota Mar 23 at 8:29 • But the graph was great! Bring", "+0 # Probability 0 61 3 Hi Max: Please take a look at this question: https://web2.0calc.com/questions/pls-halp_27 Apr 16, 2022 #1 +9369 +2 This problem is equivalent to the probability of choosing two real numbers x, y with 0 < x < y < 5 such that x < 3, y - x < 3, and 5 - y < 3. We can view this geometrically. Let $$D$$ be the square with vertices (0, 0), (5, 0), (0, 5), (5, 5). Let $$\\Omega$$ be the region bounded by 0 < x < 3, y - x < 3, 5 - y < 3, x < y and $$D$$. Basically, the probability is equal to $$\\dfrac{\\text{area}(\\Omega)}{\\text{area}(D)} = \\dfrac{9 - 2 - \\dfrac12}{25} = \\dfrac{13}{50}$$. I believe that is the answer. Edit: Edited. This should be correct now. Apr 16, 2022 edited by MaxWong Apr 16, 2022 edited by MaxWong Apr 16, 2022 #1 +9369 +2 This problem is equivalent to the probability of choosing two real numbers x, y with 0 < x < y < 5 such that x < 3, y - x < 3, and 5 - y < 3. We can view this geometrically. Let $$D$$ be the square with vertices (0, 0), (5, 0), (0, 5), (5, 5). Let $$\\Omega$$ be the region bounded by", "Select Page Try this beautiful problem from Probability based on Coordinates. ## Probability in Coordinates – AMC-10A, 2003- Problem 12 A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability that $x<y$? • $\\frac{1}{8}$ • $\\frac{1}{6}$ • $\\frac{2}{3}$ ### Key Concepts Number system Cube But try the problem first… Answer: $\\frac{1}{8}$ Source AMC-10A (2003) Problem 12 Pre College Mathematics ## Try with Hints First hint The given vertices are $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$.if we draw a figure using the given points then we will get a rectangle as shown above.Clearly lengtht of $OC$= $4$ and length of $AO$=$1$.Therefore area of the rectangle is $4 \\times 1=4$.now we have to find out the probability that $x<y$.so we draw a line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.can you find out the area with the condition $x<y$? can you finish the problem…….. Second Hint Now the line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.Therefore it will form a Triangle $\\triangle AOD$ (as shown above) whose $AO=1$ and $AD=1$.Therefore area of $\\triangle AOD=\\frac{1}{2}$ i.e (red region).Now can you find out the probability with the condition $x<y$? can you finish the problem…….. Final Step Therefore the required probability ($x<y$) is $\\frac{\\frac{1}{2}}{4}$=$\\frac{1}{8}$", "$b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$? $\\textbf{(A) }20\\qquad\\textbf{(B) }30\\qquad\\textbf{(C) }60\\qquad\\textbf{(D) }120\\qquad \\textbf{(E) }180$ ## Problem 21 A top hat contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn? $\\textbf{(A) }\\dfrac{3}{10}\\qquad\\textbf{(B) }\\dfrac{2}{5}\\qquad\\textbf{(C) }\\dfrac{1}{2}\\qquad\\textbf{(D) }\\dfrac{3}{5}\\qquad \\textbf{(E) }\\dfrac{7}{10}$ ## Problem 22 Rectangle $DEFA$ below is a $3 \\times 4$ rectangle with $DC=CB=BA$. What is the area of the \"bat wings\" (shaded area)? $[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label(\"A\",(3.05,4.2)); label(\"B\",(2,4.2)); label(\"C\",(1,4.2)); label(\"D\",(0,4.2)); label(\"E\", (0,-0.2)); label(\"F\", (3,-0.2)); [/asy]$ $\\textbf{(A) }2\\qquad\\textbf{(B) }2 \\frac{1}{2}\\qquad\\textbf{(C) }3\\qquad\\textbf{(D) }3 \\frac{1}{2}\\qquad \\textbf{(E) }4$ ## Problem 23 Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$. The circles intersect at two points, one of which is $E$. What is the degree measure of $\\angle CED$? $\\textbf{(A) }90\\qquad\\textbf{(B) }105\\qquad\\textbf{(C) }120\\qquad\\textbf{(D) }135\\qquad \\textbf{(E) }150$ ## Problem 24 The digits $1$, $2$, $3$, $4$, and $5$ are each used", "second point. Therefore the total probability is $1\\times\\frac{1}{3}\\times\\frac{1}{4}=\\frac{1}{12}$ or $\\boxed{\\text{(D)}}$ Solution 2 We will use geometric probability. The first point can be anywhere. Each point must be $\\frac{\\pi}{3}$ or less away from each other. Define $x$ be the amount of radians away the second point is from the first. We limit $x$ to be in the interval $[-\\pi, \\pi]$. Define $y$ be the amount of radians away the third point is from the first. We limit $y$ to be in the interval $[-\\pi, \\pi]$. Now, we can deduct that: $$|x| \\le \\frac{\\pi}{3},$$ $$|y| \\le \\frac{\\pi}{3},$$ and $$|x-y| \\le \\frac{\\pi}{3}.$$ We now begin plotting these on the coordinate grid. First note that the area of the points that $x$ and $y$ can be (ignoring the conditions) is $(2\\pi)^2$ (remember what we restricted $x$ and $y$ to). Now, we can graph the equations we deduced on the coordinate grid. That should look like this: $[asy] fill((40, 0)--(40, 40)--(0, 40)--(-40, 0)--(-40, -40)--(0, -40)--cycle, green); draw((-50, 0)--(50, 0)); draw((0, -50)--(0, 50)); draw((-50, -10)--(10, 50),red); draw((-10, -50)--(50, 10),red); draw((-40, 40)--(40, 40),red); draw((-40, -40)--(40, -40),red); draw((40, 40)--(40, -40),red); draw((-40, 40)--(-40, -40),red); label(\"\\frac{\\pi}{3}\", (40, 0), SE); label(\"\\frac{\\pi}{3}\", (-40, 0), SW); label(\"\\frac{\\pi}{3}\", (0, 40), NE); label(\"\\frac{\\pi}{3}\", (0, -40), NW); [/asy]$ The area of the shaded region can be calculated in many ways. Eventually, you will find", "the same way, until you construct an octagon. How many sides does the resulting polygon have? $[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]$ $\\textbf{(A)}21\\qquad\\textbf{(B)}23\\qquad\\textbf{(C)}25\\qquad\\textbf{(D)}27\\qquad\\textbf{(E)}29$ Problem 10 On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board? $[asy] unitsize(10); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle); draw((1,8)--(1,0)); draw((7,8)--(7,0)); draw((6,8)--(6,0)); draw((5,8)--(5,0)); draw((4,8)--(4,0)); draw((3,8)--(3,0)); draw((2,8)--(2,0)); draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle,black); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle,black); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle,black); fill((6,2)--(7,2)--(7,3)--(6,3)--cycle,black); fill((0,4)--(1,4)--(1,5)--(0,5)--cycle,black); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle,black); fill((4,4)--(5,4)--(5,5)--(4,5)--cycle,black); fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); fill((0,6)--(1,6)--(1,7)--(0,7)--cycle,black); fill((2,6)--(3,6)--(3,7)--(2,7)--cycle,black); fill((4,6)--(5,6)--(5,7)--(4,7)--cycle,black); fill((6,6)--(7,6)--(7,7)--(6,7)--cycle,black); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,black); fill((5,1)--(6,1)--(6,2)--(5,2)--cycle,black); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle,black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle,black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle,black); fill((5,3)--(6,3)--(6,4)--(5,4)--cycle,black); fill((7,3)--(8,3)--(8,4)--(7,4)--cycle,black); fill((1,5)--(2,5)--(2,6)--(1,6)--cycle,black); fill((3,5)--(4,5)--(4,6)--(3,6)--cycle,black); fill((5,5)--(6,5)--(6,6)--(5,6)--cycle,black); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); fill((1,7)--(2,7)--(2,8)--(1,8)--cycle,black); fill((3,7)--(4,7)--(4,8)--(3,8)--cycle,black); fill((5,7)--(6,7)--(6,8)--(5,8)--cycle,black); fill((7,7)--(8,7)--(8,8)--(7,8)--cycle,black);[/asy]$ $\\textbf{(A)}\\frac{1}{16}\\qquad\\textbf{(B)}\\frac{7}{16}\\qquad\\textbf{(C)}\\frac{1}2\\qquad\\textbf{(D)}\\frac{9}{16}\\qquad\\textbf{(E)}\\frac{49}{64}$ Problem 11 The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$. How many more sixth graders than seventh graders bought a pencil? $\\textbf{(A)}\\ 1\\qquad\\textbf{(B)}\\ 2\\qquad\\textbf{(C)}\\ 3\\qquad\\textbf{(D)}\\ 4\\qquad\\textbf{(E)}\\ 5$ Problem 12 The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime? $[asy] unitsize(30); draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label(\"1\",(0,.5)); label(\"3\",((cos(pi/6))/2,(-sin(pi/6))/2)); label(\"5\",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]$ $[asy] unitsize(30);" ]

[ "124 views At a family group meeting of 30 women, 17 are descended from George, 16 are descended from John, and 5 are not descended from George or John. How many of the 30 women are descended from both George and John? Total Women = 30 Descended from John ($f(J)$) = $16$ Descended from George ($f(G)$) = $17$ Not descended from either John or George = $5$ $\\therefore$ Descended from either John or George ($f(J) \\cup f(G)$ = $25$ Women are descended from both George and John = $f(G) \\cap f(J)$ $f(G) \\cup f(J) = f(G) + f(J) – f(J)\\cap f(g)$ $25 = 17 + 16 – f(J)\\cap f(G)$ $f(J) \\cap f(G) = 33-25 = 8$ by 👌", "# How many children did Mr. and Mrs. Crachit have? How many children did Mr. and Mrs. Crachit have?", "# Math Help - Word problem 1. ## Word problem John was born 1980. In 2025 he will be 45 years old. These numbers are related because 45² = 2025. John's great grand mother was also 'n' years old in the year n². How old was she when John was born? I know this is probably simple, but I just can't seem to wrap my brain around it. thanks 2. Originally Posted by jacs John was born 1980. In 2025 he will be 45 years old. These numbers are related because 45² = 2025. John's great grand mother was also 'n' years old in the year n². How old was she when John was born? I know this is probably simple, but I just can't seem to wrap my brain around it. thanks She would have been either 44 in 1936, or 43 in 1849, I think the earlier possibilities are implausible. If she was 44 in 1936, she was born in 1892, so she was 88 in 1980. You can work the other case yourself (and probably reject it as implausible). RonL 3. Please post the explanation and answer if you know it, I'm also interested in this question. -NineZeroFive 4. i used n + n² = 1980 to get the 44 just didnt know what to", "have a two-month old baby. john feels that after the baby’s birth, love and sexual intimacy between him and lisa John and lisa have a two-month old baby. john feels that after the baby’s birth, love and sexual intimacy between him and lisa has diminished. what is the action john should take? a. john should nag lisa for not paying attention to him. b. john should ignore lisa and spend time out with friends. c...", "of the female offspring.", "# Mr. and Mrs. Mustard have six daughters and each daughter has one brother Riddle: Mr. and Mrs. Mustard have six daughters and each daughter has one brother. How many people are in the Mustard family? Solution: There are nine Mustards in the family. Since each daughter shares the same brother, there are six girls, one boy, and Mr. and Mrs. Mustard.", "# Question John and Jane are married. The probability that John watches a certain television show is .4. The probability that Jane watches the show is .5. The probability that John watches the show, given that Jane does, is .7. a. Find the probability that both John and Jane watch the show. b. Find the probability that Jane watches the show, given that John does. c. Do John and Jane watch the show independently of each other? Justify your answer. Sales0 Views34", "# John along with his wife,parents and four children of which two are boys and two are girls,put up at a rest house with four bed rooms for a night. John along with his wife,parents and four children of which two are boys and two are girls,put up at a rest house with four bed rooms for a night.There are beds for two persons in each room.If either a couple or two persons of the same sex sleep in a room,what is the chance that Mr and Mrs John will sleep together? I tried to solve the problem. Chance that Mr and Mrs John will sleep together=favorable conditions/total conditions=2/total no of ways. Here,i put 2,because there are 2 couples.But the total number of ways is elusive.Its answer is 1/37. There are 4 males and 4 females. If arranged by gender, there are ${4\\choose2}{4\\choose2} = 36$ ways to make 4 groups If John and wife sleep together, so do his parents, sons and daughters, which means just $1$ way $$\\text{Thus}\\;\\;Pr = \\frac{1}{37}$$ Edit User264781's comments are valid, I got carried away by the \"matching\" answer ! The number of genderwise groups will be $3*3 = 9$ and Pr = $1/10$ • You seem to say there are $\\binom{4}{2}$ ways of choosing how the males are paired up. But this", "John was 1 07 May 2012, 04:07 4 A town's oldest inhabitant is x years older than the sum of the ages 6 17 Jul 2011, 06:12 Display posts from previous: Sort by", "a loving father to his children and doting grandfather of ten and great-grandfather of six. and", "the Godhead. Among these fathers – 8 percent are raising three or more of their own children under 18 years old. Numbers 13:31-33 King James Version (KJV). Conservative scholars remain cautious about assigning too much importance to the meaning of numbers in the Bible. The number 3 is also associated with certain of God’s mighty acts.", "or independent 12. Large family I have as many brothers as sisters and each my brother has twice as many sisters as brothers. How many children do parents have?", "plus five times Claire’s age is 204. Nine times John’s age minus three times Claire’s age is also 204. How old are John and Claire? ## Date Created: Feb 23, 2012 Apr 02, 2015 You can only attach files to section which belong to you If you would like to associate files with this section, please make a copy first.", "case A or B, Chris may or may not have the most money. Re: Three children, Alice, Brian, and Chris have a total of [#permalink] 29 Mar 2012, 02:58 Similar topics Replies Last post Similar Topics: 2 Three Children Alice, Brain & Chris have a total of 4 07 Aug 2010, 21:30 3 10 children have ordered a total of 17 hamburgers, and each 12 07 Jul 2006, 07:44 Three children A,B and C have a total of$1.20 between them. 3 06 Apr 2006, 23:25 Display posts from previous: Sort by", "## How Many Sons Jane has two children. She has at least one son. What’s the chance that she has two sons? You can assume that boys and girls are equally likely to be born. Solution", "John and his dad were calculating the ratios of their ages. Currently [#permalink] 25 Nov 2019, 19:15 Display posts from previous: Sort by", "I don't think you are in any danger! 6. ## Re: Is this answer correct? Alternatively the money is divided up between John, mother and father in the ratio 2:3:2 Hence the mother receives 3/7 of the money.", "Family has 4 children. Ondra is 3 years older than Matthew and Karlos 5 years older than the youngest Jane. We know that they are together 30 years and 3 years ago they were together 19 years. Determine how old the children are.", "more as of 2011: ..Over the years, she watched the number of children in her son’s group grow. And grow. Today there are 150 children, all conceived with sperm from one donor, in this group of half siblings, and more are on the way…While Ms. Daily’s group is among the largest, many others comprising 50 or more half siblings are cropping up on Web sites and in chat groups, where sperm donors are tagged with unique identifying numbers. …No one knows how many children are born in this country each year using sperm donors. Some estimates put the number at 30,000 to 60,000, perhaps more. …Sperm donors, too, are becoming concerned. “When I asked specifically how many children might result, I was told nobody knows for sure but that five would be a safe estimate,” said a sperm donor in Texas who asked that his name be withheld because of privacy concerns. “I was told that it would be very rare for a donor to have more than 10 children.” He later discovered in the Donor Sibling Registry that some donors had dozens of children listed…Ms. Kramer, the registry’s founder, said that one sperm donor on her site learned that he had 70 children. He now keeps track of them all on an Excel spreadsheet.2 An article on", "given to babies born in 1800 were Mary and John, with 24% of female babies and 22% of male babies receiving those names, respectively. In contrast, the corresponding statistics for in England and Wales in 1994 were Emily and James, with 3% and 4% of names, respectively. Not only have Mary and John gone out of favor in the English speaking world, also the overall distribution of names has changed [substantially] over the last 100 years for females, but not for males. (The female trend has continued through to 2010: “The 1,000 top girl names accounted for only 67% of all girl names last year, down from 91% in 1960 and compared with 79% for boys last year.”29) The theory could probably be rescued by saying that the advantage of having an unique given name (and thus a relatively unique full name) goes that far back, but then we would need to explain why the advantage would be there for women, but not men. On the other hand, Social Security data seems to indicate both a 2-century long decline in the popularity of the top ten names and also a convergence of top-ten name rarity; from Andrew Gelman: • Pop culture is known to have a very strong influence on baby names (cf. the popularity of Star Wars" ]

[ "# John along with his wife,parents and four children of which two are boys and two are girls,put up at a rest house with four bed rooms for a night. John along with his wife,parents and four children of which two are boys and two are girls,put up at a rest house with four bed rooms for a night.There are beds for two persons in each room.If either a couple or two persons of the same sex sleep in a room,what is the chance that Mr and Mrs John will sleep together? I tried to solve the problem. Chance that Mr and Mrs John will sleep together=favorable conditions/total conditions=2/total no of ways. Here,i put 2,because there are 2 couples.But the total number of ways is elusive.Its answer is 1/37. There are 4 males and 4 females. If arranged by gender, there are ${4\\choose2}{4\\choose2} = 36$ ways to make 4 groups If John and wife sleep together, so do his parents, sons and daughters, which means just $1$ way $$\\text{Thus}\\;\\;Pr = \\frac{1}{37}$$ Edit User264781's comments are valid, I got carried away by the \"matching\" answer ! The number of genderwise groups will be $3*3 = 9$ and Pr = $1/10$ • You seem to say there are $\\binom{4}{2}$ ways of choosing how the males are paired up. But this", "or human-meaningful. You have to compromise on some aspect. • There’s already a decline in popular names, according to Wikipedia: Since about 1800 in England and Wales and in the U.S., the popularity distribution of given names has been shifting so that the most popular names are losing popularity. For example, in England and Wales, the most popular female and male names given to babies born in 1800 were Mary and John, with 24% of female babies and 22% of male babies receiving those names, respectively. In contrast, the corresponding statistics for in England and Wales in 1994 were Emily and James, with 3% and 4% of names, respectively. Not only have Mary and John gone out of favor in the English speaking world, also the overall distribution of names has changed [substantially] over the last 100 years for females, but not for males. (The female trend has continued through to 2010: “The 1,000 top girl names accounted for only 67% of all girl names last year, down from 91% in 1960 and compared with 79% for boys last year.”22) The theory could probably be rescued by saying that the advantage of having a unique given name (and thus a relatively unique full name) goes that far back, but then we would need to explain why the advantage", "have a two-month old baby. john feels that after the baby’s birth, love and sexual intimacy between him and lisa John and lisa have a two-month old baby. john feels that after the baby’s birth, love and sexual intimacy between him and lisa has diminished. what is the action john should take? a. john should nag lisa for not paying attention to him. b. john should ignore lisa and spend time out with friends. c...", "will help him decide large nurber of births to see if the data morc than half of all babies born are male? hypothesis test for J populatiori proportion Ahypothesis test comparing two population proportions A hypothesis test for a population meanAhypothesis test Question 10 1pe Many people believe that half of all than half of all babies babies born are male and half are female born are actually male. He decided rescarcher believed that more supported his belief. Which to look at type of inference will help him decide large nurber of births to see if the d...", "I don't think you are in any danger! 6. ## Re: Is this answer correct? Alternatively the money is divided up between John, mother and father in the ratio 2:3:2 Hence the mother receives 3/7 of the money.", "or exactly two sons, or is childless\"? Well, it seems to be a disjunction about $x$, so split it up into cases: if you know the whole thing is a disjunction, you can tackle each disjunct separately and then put it all together with $\\vee$s at the end. So you just need to analyze \"$x$ has only daughters\", \"$x$ has exactly two sons\", and \"$x$ is childless\". Hopefully, things are clear enough that you can do these on your own.", "given to babies born in 1800 were Mary and John, with 24% of female babies and 22% of male babies receiving those names, respectively. In contrast, the corresponding statistics for in England and Wales in 1994 were Emily and James, with 3% and 4% of names, respectively. Not only have Mary and John gone out of favor in the English speaking world, also the overall distribution of names has changed [substantially] over the last 100 years for females, but not for males. (The female trend has continued through to 2010: “The 1,000 top girl names accounted for only 67% of all girl names last year, down from 91% in 1960 and compared with 79% for boys last year.”29) The theory could probably be rescued by saying that the advantage of having an unique given name (and thus a relatively unique full name) goes that far back, but then we would need to explain why the advantage would be there for women, but not men. On the other hand, Social Security data seems to indicate both a 2-century long decline in the popularity of the top ten names and also a convergence of top-ten name rarity; from Andrew Gelman: • Pop culture is known to have a very strong influence on baby names (cf. the popularity of Star Wars", "were Emily and James, with 3% and 4% of names, respectively. Not only have Mary and John gone out of favor in the English speaking world, also the overall distribution of names has changed [substantially] over the last 100 years for females, but not for males. (The female trend has continued through to 2010: “The 1,000 top girl names accounted for only 67% of all girl names last year, down from 91% in 1960 and compared with 79% for boys last year.”2) The theory could probably be rescued by saying that the advantage of having a unique given name (and thus a relatively unique full name) goes that far back, but then we would need to explain why the advantage would be there for women, but not men. On the other hand, Social Security data seems to indicate both a 2-century long decline in the popularity of the top ten names and also a convergence of top-ten name rarity; from Andrew Gelman: • Pop culture is known to have a very strong influence on baby names (cf. the popularity of Star Wars and the subsequent massive spike in ‘Luke’). The counter-arguments to The Long Tail marketing theory say that pop culture is becoming ever more monolithic and hit-driven. The fewer hits, and the more mega-hits, the more we", "matter of perspective. Also, it is assumed that the daughters can only have integer ages, meaning that for example all 10 year olds are the same age. – Skydiver Jun 3 '16 at 21:39", "also a simple hystorical reason for men to have female second names: There was once a common tradition to give the child its mother's name if she died giving birth, even if the child was a boy. That is at least the case with Karl Maria Weber. June 14, 2012 @ 1:44 pm [...] What would Jesús do? — Language Log on, among other things, cross-cultural naming conventions. [...]", "might have, everyone is either married or unmarried. Male or female makes no difference. – DJClayworth Aug 20 '16 at 4:06 • Agree with DJClayworth, except I'm not sure it's full marks for political correctness, either. The sexes of Jack, Anne, and George are not given, only their names. – 6005 Aug 21 '16 at 9:12 Answer: Yes. Because: George is the child of Jack and Anne. Married mother is looking at unmarried child. • This is not a lateral-thinking puzzle. What makes you think George is the child of the other two? – elias Aug 20 '16 at 1:27", "# Math Help - Word problem 1. ## Word problem John was born 1980. In 2025 he will be 45 years old. These numbers are related because 45² = 2025. John's great grand mother was also 'n' years old in the year n². How old was she when John was born? I know this is probably simple, but I just can't seem to wrap my brain around it. thanks 2. Originally Posted by jacs John was born 1980. In 2025 he will be 45 years old. These numbers are related because 45² = 2025. John's great grand mother was also 'n' years old in the year n². How old was she when John was born? I know this is probably simple, but I just can't seem to wrap my brain around it. thanks She would have been either 44 in 1936, or 43 in 1849, I think the earlier possibilities are implausible. If she was 44 in 1936, she was born in 1892, so she was 88 in 1980. You can work the other case yourself (and probably reject it as implausible). RonL 3. Please post the explanation and answer if you know it, I'm also interested in this question. -NineZeroFive 4. i used n + n² = 1980 to get the 44 just didnt know what to", "therefore have that 2 (Joe), 4, 5, 7, and 9 (Michael) are males, so by process of elimination, 3 must be female and so Jolene must be 3. Jolene says that Robert and Jacob sit next to each other, so they must be 4 and 5 in some order; 2 lies and says Jacob doesn't sit next to Mary, so Jacob must be 5 and Robert must be 4. This leaves Christian as 7. Summary of Characters: 1: Anna (Sometimer); 2: Joe (Liar); 3: Jolene (Truthteller); 4: Robert (Truthteller); 5: Jacob (Liar); 6: Mary (Truthteller); 7: Christian (Sometimer); 8: Susanna (Liar); 9: Michael (Sometimer). Then we have (number in brackets) 1: TRUE (9) 2: TRUE (93) 3: FALSE (93) 4: FALSE (93) 5: FALSE (93) 6: TRUE (932) 7: TRUE (9324) 8: TRUE (93248) 9: FALSE (93248) The middle number is therefore 2 and so your Secret Santa is Joe.", "124 views At a family group meeting of 30 women, 17 are descended from George, 16 are descended from John, and 5 are not descended from George or John. How many of the 30 women are descended from both George and John? Total Women = 30 Descended from John ($f(J)$) = $16$ Descended from George ($f(G)$) = $17$ Not descended from either John or George = $5$ $\\therefore$ Descended from either John or George ($f(J) \\cup f(G)$ = $25$ Women are descended from both George and John = $f(G) \\cap f(J)$ $f(G) \\cup f(J) = f(G) + f(J) – f(J)\\cap f(g)$ $25 = 17 + 16 – f(J)\\cap f(G)$ $f(J) \\cap f(G) = 33-25 = 8$ by 👌", "biological sex of a baby---is it the mother, father, or both? Why? 9. Consider the below Punnett square cross. Explain how you could use it to determine which parentdetermines the biological sex of a baby---is it the mother, father, or both? Why?... 1 answer ### What investment decisions destabilized the economy during the 1920's What investment decisions destabilized the economy during the 1920's... -- 0.056232--", "he is 1. unmarried 2. male since to state \"John is a bachelor\" implies John has each of those three additional predicates. Example 2: For the whole numbers greater than two, being odd is necessary to being prime, since two is the only whole number that is both even and prime. Example 3: Consider thunder, in the technical sense, the acoustic quality demonstrated by the shock wave that inevitably results from any lightning bolt in the atmosphere. It may fairly be said that thunder is necessary for lightning, since lightning cannot occur without thunder, too, occurring. That is, if lightning does occur, then there is thunder. Example 4: Being at least 30 years old is necessary of serving in the U.S. Senate. If you are under 30 years old then it is impossible for you to be a senator. That is, if you are a senator, it follows that you are at least 30 years old. Example 5: In algebra, in order for some set S together with an operation $\\star$ to form a group, it is necessary that $\\star$ be associative. It is also necessary that S include a special element e such that for every x in S it is the case that e $\\star$ x and x $\\star$ e both equal x. It is also", "siblings, all biological children of his adoptive parents. brian's math skills are far better than those of his adoptive parents or his adoptive siblings. this provides evidence that _____....", "his central themes, in particular that being male is far more complex than has been realized and that “man, and not woman, is the second sex.” (However, his choice of value-laden terms, “first” and “second sex,” may pique some readers, as might a reference to crime as “a hobby of half the population.”) He argues his thesis with reference to a compelling array of facts, anecdotes, and speculations. We learn, for example, that genes older than the SRY gene (a trigger of testicular development in mammals) “rule the world of maleness.” One in several thousand children is born with a mix of testicles and ovaries, and most such children are XX. Jones poses the puzzle of how embryonic humans who lack a Y can grow a testis, usually on the right side. He finds the answer in the developmental asymmetry of hermaphrodites, which is caused by early accelerated growth of the embryo's right half, growth triggered by these more ancient genes. Jones also challenges the notion that older mothers, not fathers, are mostly responsible for producing children with genetic defects. He notes that many sperm fail to mature or are abnormal in some way. Furthermore, DNA in each sperm undergo multiple replication cycles, which leads to more errors in male than female gametes. In fact, for some conditions", "mother's egg that became you. From what I know, eggs (and also sperms) contain a random sample of ~50% of your mother's genes. So, could it be possible (though unlikely) that this random sample contains exactly those 50% that you previously inherited from your mother? This way, you and your mother/daughter only differ by the 50% of genes that the two fathers contributed to each of you. Genetically, you are half siblings. 2. Denote the two fathers F1 and F2. Also, denote yourself M1 and your daughter/mother M2. Via some bootstrap paradox mechanism (see below), it just so happens that your genetic fingerprint contains 1/3 of F1's genes and 2/3 of F2's genes, or in other words M1 = 1/3 F1 + 2/3 F2. If now M1 has a child with F1, it will be 1/2 M1 + 1/2 F1 = 1/6 F1 + 1/3 F2 + 1/2 F1 = 2/3 F1 + 1/3 F2 =: M2 and if this M2 has a child with F2, it will be 1/2 M2 + 1/2 F2 = 1/3 F1 + 1/6 F2 + 1/2 F2 = 1/3 F1 + 2/3 F2 = M1 My concern with this is that, even though the ratios of genes match, it's still unlikely that the randomly sampled egg of your daughter happens to contain", "# survivors list grammar question Obituaries often include lists with this grammatical structure: John Doe is survived by his children, Steve Doe and his wife, June; Will Doe and his wife, Janet; Susan Richards and her husband, Walter....\" It seems to me that the above is incorrect because, despite phrases like \"his wife\" and \"her husband,\" the wives and husbands still fall under the rubric of \"his [John Doe’s] children.\" So my first question is, am I wrong about this? According to obituary convention, you could write, \"his children, Steve (June) Doe, Will (Janet) Doe, and Susan (Walter) Richards.\" However, many families do not like how this looks. Would \"His children and their spouses\" followed by their names be right? I have some doubts about this because, without the word \"respectively,\" it’s potentially ambiguous. This brings me to my second question: Other than using parentheses, what would be the correct way to write this list? - Why not imply that the respective husbands and wives are also his children? \"I'm not losing a daughter, I'm gaining a son\".. – TsSkTo Sep 6 '13 at 11:11 Interesting title, but why? – Kris Sep 7 '13 at 14:31 See linguistics.SE: Grammatical rules governing extended survivors lists – hippietrail Sep 7 '13 at 16:25 I don't read obituaries, so I'm not" ]

[ "# John along with his wife,parents and four children of which two are boys and two are girls,put up at a rest house with four bed rooms for a night. John along with his wife,parents and four children of which two are boys and two are girls,put up at a rest house with four bed rooms for a night.There are beds for two persons in each room.If either a couple or two persons of the same sex sleep in a room,what is the chance that Mr and Mrs John will sleep together? I tried to solve the problem. Chance that Mr and Mrs John will sleep together=favorable conditions/total conditions=2/total no of ways. Here,i put 2,because there are 2 couples.But the total number of ways is elusive.Its answer is 1/37. There are 4 males and 4 females. If arranged by gender, there are ${4\\choose2}{4\\choose2} = 36$ ways to make 4 groups If John and wife sleep together, so do his parents, sons and daughters, which means just $1$ way $$\\text{Thus}\\;\\;Pr = \\frac{1}{37}$$ Edit User264781's comments are valid, I got carried away by the \"matching\" answer ! The number of genderwise groups will be $3*3 = 9$ and Pr = $1/10$ • You seem to say there are $\\binom{4}{2}$ ways of choosing how the males are paired up. But this", "124 views At a family group meeting of 30 women, 17 are descended from George, 16 are descended from John, and 5 are not descended from George or John. How many of the 30 women are descended from both George and John? Total Women = 30 Descended from John ($f(J)$) = $16$ Descended from George ($f(G)$) = $17$ Not descended from either John or George = $5$ $\\therefore$ Descended from either John or George ($f(J) \\cup f(G)$ = $25$ Women are descended from both George and John = $f(G) \\cap f(J)$ $f(G) \\cup f(J) = f(G) + f(J) – f(J)\\cap f(g)$ $25 = 17 + 16 – f(J)\\cap f(G)$ $f(J) \\cap f(G) = 33-25 = 8$ by 👌", "# Math Help - Word problem 1. ## Word problem John was born 1980. In 2025 he will be 45 years old. These numbers are related because 45² = 2025. John's great grand mother was also 'n' years old in the year n². How old was she when John was born? I know this is probably simple, but I just can't seem to wrap my brain around it. thanks 2. Originally Posted by jacs John was born 1980. In 2025 he will be 45 years old. These numbers are related because 45² = 2025. John's great grand mother was also 'n' years old in the year n². How old was she when John was born? I know this is probably simple, but I just can't seem to wrap my brain around it. thanks She would have been either 44 in 1936, or 43 in 1849, I think the earlier possibilities are implausible. If she was 44 in 1936, she was born in 1892, so she was 88 in 1980. You can work the other case yourself (and probably reject it as implausible). RonL 3. Please post the explanation and answer if you know it, I'm also interested in this question. -NineZeroFive 4. i used n + n² = 1980 to get the 44 just didnt know what to", "have a two-month old baby. john feels that after the baby’s birth, love and sexual intimacy between him and lisa John and lisa have a two-month old baby. john feels that after the baby’s birth, love and sexual intimacy between him and lisa has diminished. what is the action john should take? a. john should nag lisa for not paying attention to him. b. john should ignore lisa and spend time out with friends. c...", "for the MAXIMUM POSSIBLE number of girls who had chocolate ice cream, so THAT also implies that there were additional children beyond the 17 described. Fact 1: 30 children attended the party. Here we have an actual number to work with. 30 total children - the 17 mentioned in the prompt = 13 additional children. We don't know if they're boys or girls and we don't know whether they ate chocolate or strawberry ice cream. All of that is fine - the question is asking for the maximum POSSIBLE number of girls who COULD have had chocolate ice cream. IF those 13 extra children were ALL girls and IF they ALL had chocolate ice cream.....then that would make the maximum possible = 13. Fact 1 is SUFFICIENT. Fact 2: Fewer than half the children had strawberry ice cream. Since we know that 9 had strawberry and 8 had chocolate, this Fact also implies that there were additional children. Unfortunately, we don't know how many. There would have to be AT LEAST 2 extra children, but it could be dozens or hundreds and we have no idea about whether they're male or female or what type of ice cream each ate. Fact 2 is INSUFFICIENT. [Reveal] Spoiler: A GMAT assassins aren't born, they're made, Rich _________________ # Rich Cohen", "given to babies born in 1800 were Mary and John, with 24% of female babies and 22% of male babies receiving those names, respectively. In contrast, the corresponding statistics for in England and Wales in 1994 were Emily and James, with 3% and 4% of names, respectively. Not only have Mary and John gone out of favor in the English speaking world, also the overall distribution of names has changed [substantially] over the last 100 years for females, but not for males. (The female trend has continued through to 2010: “The 1,000 top girl names accounted for only 67% of all girl names last year, down from 91% in 1960 and compared with 79% for boys last year.”29) The theory could probably be rescued by saying that the advantage of having an unique given name (and thus a relatively unique full name) goes that far back, but then we would need to explain why the advantage would be there for women, but not men. On the other hand, Social Security data seems to indicate both a 2-century long decline in the popularity of the top ten names and also a convergence of top-ten name rarity; from Andrew Gelman: • Pop culture is known to have a very strong influence on baby names (cf. the popularity of Star Wars", "will help him decide large nurber of births to see if the data morc than half of all babies born are male? hypothesis test for J populatiori proportion Ahypothesis test comparing two population proportions A hypothesis test for a population meanAhypothesis test Question 10 1pe Many people believe that half of all than half of all babies babies born are male and half are female born are actually male. He decided rescarcher believed that more supported his belief. Which to look at type of inference will help him decide large nurber of births to see if the d...", "case A or B, Chris may or may not have the most money. Re: Three children, Alice, Brian, and Chris have a total of [#permalink] 29 Mar 2012, 02:58 Similar topics Replies Last post Similar Topics: 2 Three Children Alice, Brain & Chris have a total of 4 07 Aug 2010, 21:30 3 10 children have ordered a total of 17 hamburgers, and each 12 07 Jul 2006, 07:44 Three children A,B and C have a total of$1.20 between them. 3 06 Apr 2006, 23:25 Display posts from previous: Sort by", "There is a boy named Adam in each of the three children subgroup and a girl named Beata in each quartet (four-member subgroup). How many children can be in such a group and what are their names in that case? • Apples and pears Mom divided 24 apples and 15 pears to children. Each child received the same number of apples and pears - same number as his siblings. How many apples (j=?) and pears (h=?) received each child? • At the At the presentation of the travelers came three times as many men than women. When eight men left with their partners, there were five times more men than women at the presentation. How many were men and women originally? • Children The group has 42 children. There are 4 more boys than girls. How many boys and girls are in the group? • The sum 8 The sum of two numbers is 21. If three times the smaller numbers is two less than twice the larger number, find the two numbers. • Math logic There are 20 children in the group, each two children have a different name. Alena and John are among them. How many ways can we choose 8 children to be among the selected A) was John B) was John and Alena C)", "# survivors list grammar question Obituaries often include lists with this grammatical structure: John Doe is survived by his children, Steve Doe and his wife, June; Will Doe and his wife, Janet; Susan Richards and her husband, Walter....\" It seems to me that the above is incorrect because, despite phrases like \"his wife\" and \"her husband,\" the wives and husbands still fall under the rubric of \"his [John Doe’s] children.\" So my first question is, am I wrong about this? According to obituary convention, you could write, \"his children, Steve (June) Doe, Will (Janet) Doe, and Susan (Walter) Richards.\" However, many families do not like how this looks. Would \"His children and their spouses\" followed by their names be right? I have some doubts about this because, without the word \"respectively,\" it’s potentially ambiguous. This brings me to my second question: Other than using parentheses, what would be the correct way to write this list? - Why not imply that the respective husbands and wives are also his children? \"I'm not losing a daughter, I'm gaining a son\".. – TsSkTo Sep 6 '13 at 11:11 Interesting title, but why? – Kris Sep 7 '13 at 14:31 See linguistics.SE: Grammatical rules governing extended survivors lists – hippietrail Sep 7 '13 at 16:25 I don't read obituaries, so I'm not", "Three children, Alice, Brian, and Chris have a total of [#permalink] ### Show Tags 27 Mar 2012, 22:11 I have to disagree. In the case where each have 40c ie you have a set (40, 40, 40) they each individually have the most (ie the highest value = 40) and it so happens they each individually have the least, again 40. Since this provides two cases, Brian having the most when they all share the most (similar to tied for 1st place - they are equally best) and Brian not having the most when any other values are chosen, one requires both (1) and (2) to determine if Chris does/doesn't have the most. Clearly this is a definition debate around \"most\" and ties for most, and the question would likely (hopefully) be thrown out by the gmac folks! Math Expert Joined: 02 Sep 2009 Posts: 34475 Followers: 6288 Kudos [?]: 79775 [1] , given: 10022 Re: Three children, Alice, Brian, and Chris have a total of [#permalink] ### Show Tags 28 Mar 2012, 00:59 1 KUDOS Expert's post bschool83 wrote: Three children, Alice, Brian, and Chris have a total of \\$1.20 between them. Does Chris have the most money? (1) Alice has 35 cents. (2) Chris has 40 cents. This is a poor quality question because of its", "quite sure of the structure, but I agree that the way you state it implies the spouses are his children as well. My initial feeling is that I don't even think it's necessary to include spouses, but I would rewrite it as: John Doe is survived by his children, Steve Doe, married to June; Will Doe, married to Janet; and Susan Richards, married to Walter. Or something like that. Your second option, children and their spouses, makes more sense. I feel like the reader would assume his children are the ones mentioned first, but it is ambiguous. How do the obituaries deal with grandchildren? Do they mention them? - Personally, I see nothing significantly wrong with your original suggestion, and it contains no ambiguity: John Doe is survived by his children, Steve Doe and his wife, June; Will Doe and his wife, Janet; Susan Richards and her husband, Walter....\" His children are the first persons mentioned in each individual list item, so you can read it as: John Doe is survived by his children: • Steve Doe (and his wife, June); • Will Doe (and his wife, Janet); • Susan Richards (and her husband, Walter); • .... I'm not suggesting that you should include the brackets - I've put them in only to make it clear that the", "at least n and at most n (Geurts & Nouwen 2007; Cohen & Krifka 2014; Spector 2014) as well as approximative constructions such as approximately n and about n (Rodríguez 2008; Spector 2014; Solt 2018). (8) a. Lisa has at least 40 sheep. b. *Lisa doesn’t have at least 40 sheep. (9) a. Lisa has about / roughly / approximately 40 sheep. b. *Lisa doesn’t have about / roughly / approximately 40 sheep. Here too, one has the feeling that the (b) examples would be improved if the numerical value were salient in the context. Perhaps bare numerals should in some way be aligned to this class. Secondly, it has long been recognized that negative utterances more generally tend to be odd discourse initially and in neutral contexts, but instead require a context in which the “positive counterpart” has been previously asserted or implied, or is at least in some way under consideration (see Horn 1989 for extensive discussion and references). For example, (10) (based on Ducrot 1973) would be strange if it had not been earlier claimed that Pierre was Marie’s cousin; similarly, (11) (from Givón 1978) would be odd if the possibility of the speaker’s wife being pregnant had not in any way been raised. (10) Pierre isn’t Marie’s cousin. (11) Oh, my wife’s not pregnant.", "more as of 2011: ..Over the years, she watched the number of children in her son’s group grow. And grow. Today there are 150 children, all conceived with sperm from one donor, in this group of half siblings, and more are on the way…While Ms. Daily’s group is among the largest, many others comprising 50 or more half siblings are cropping up on Web sites and in chat groups, where sperm donors are tagged with unique identifying numbers. …No one knows how many children are born in this country each year using sperm donors. Some estimates put the number at 30,000 to 60,000, perhaps more. …Sperm donors, too, are becoming concerned. “When I asked specifically how many children might result, I was told nobody knows for sure but that five would be a safe estimate,” said a sperm donor in Texas who asked that his name be withheld because of privacy concerns. “I was told that it would be very rare for a donor to have more than 10 children.” He later discovered in the Donor Sibling Registry that some donors had dozens of children listed…Ms. Kramer, the registry’s founder, said that one sperm donor on her site learned that he had 70 children. He now keeps track of them all on an Excel spreadsheet.2 An article on", "clients who had 4 children, homemakers with 4 boys and no girl accounted for 4.1%, however, no boy and 4 girls 1.8% of the total. Among 52 clients, who had 5 children, woman with 5 boys and no girl comprised 3.9%, no boy and 5 girls 0%. Among 7 cases who had 6 children, there were 3 cases who had 3 boys and 3 girls, but only 1 cases had 1 boy and 5 girls and so on. These results showed a strong trend of male preference in Korea and this could be one of the inhibit factors for family planning. TOOLS Share : METRICS • 40 View Related articles in Clin Exp Reprod Med A Study on the Morphological Analysis of Sperm.1997 August;24(2)", "# Mr. and Mrs. Mustard have six daughters and each daughter has one brother Riddle: Mr. and Mrs. Mustard have six daughters and each daughter has one brother. How many people are in the Mustard family? Solution: There are nine Mustards in the family. Since each daughter shares the same brother, there are six girls, one boy, and Mr. and Mrs. Mustard.", "know, I was married three times, and I have two sons, from two different fathers. These fathers were also married several times and had other children. My two sons are half-brothers, and they also have half-siblings through their fathers. Thus a half-sister of a half-brother became a quarter-sister. Inventing this term was quite logical for a son of two mathematicians. As you can imagine, our family tree is complicated. One day Sergei pointed out that our tree doesn't look like a standard tree and called it a family bush. ## A Splashy Math Problem A problem from the 2021 Moscow Math Olympiad went viral on Russian math channels. The author is Dmitry Krekov. Problem. Does there exist a number A so that for any natural number n, there exists a square of a natural number that differs from the ceiling of An by 2? ## I Miss John Conway Yesterday I stumbled upon a picture of John Conway I completely forgot I had: it was saved in a wrong folder. The photo was taken at the MOVES conference in 2015. There is a third person in the original shot, but I do not remember her. I decided to cut her out as I didn't know how to contact her for permission to post this photo. ## The Anniversary", "that the Kim dynasty has had above-average fertility: Kim Jong-Un is believed to have 3 children as of early 2018, Kim Jong-Il had 5 children, and Kim Il-Sung had 6 (known) children. In comparison, Jong-Un’s South Korean contemporaries, President Park, herself the only child of a previous president, has no children, and President Moon Jae-in likewise has no children. One of the other anomalous monarchies that comes to mind is the Saudi monarchy, which has a truly excessive number of heirs and princelings, and in one of its most radical developments in decades, the succession has been overturned in favor of handing most power to the young and ambitious Prince Mohammad bin Salman, who has launched major military campaigns into Yemen and attempted ambitious economic reforms. 32. Incidentally, the difference in trait means between large & small or family-owned businesses is expected from greater competition and hence more extreme order statistics. 33. Adams et al 2015 helpfully interprets the respective strength of the correlations between odds of becoming CEO and IQ/height; for an increased height of 5cm, that is almost as good as an increase of a third of a standard deviation in intelligence: To get a better sense of what the numbers mean, one can do the following thought experiment. Following the convention that one standard deviation", "at which she served both chocolate and strawberry ice cream. There were 8 boys who had chocolate ice cream, and nine girls who had strawberry. Everybody there had some ice cream, but nobody tried both. What is the maximum possible number of girls who had some chocolate ice cream? (1) Exactly thirty children attended the party. (2) Fewer than half the children had strawberry ice cream. [Reveal] Spoiler: OA VP Joined: 05 Jul 2008 Posts: 1431 Followers: 40 Kudos [?]: 312 [0], given: 1 ### Show Tags 13 Sep 2008, 13:44 ssandeepan wrote: IMO E.. Total number of girls is not known from any of the statements..also this question looks awkward.. Yepp. I chose E. Total number of children is a level above the distinction of boys and girls. Children includes both. Senior Manager Joined: 31 Jul 2008 Posts: 306 Followers: 1 Kudos [?]: 35 [0], given: 0 ### Show Tags 13 Sep 2008, 14:18 A as we need the maximum no. of girls VP Joined: 17 Jun 2008 Posts: 1397 Followers: 8 Kudos [?]: 227 [0], given: 0 ### Show Tags 14 Sep 2008, 00:37 dancinggeometry wrote: Recently Mary gave a birthday party for her daughter at which she served both chocolate and strawberry ice cream. There were 8 boys who had chocolate ice cream, and nine", "# How many siblings? Algebra Level 2 Henry and Alice are siblings. Henry has twice as many sisters as brothers. Alice has the same number of brothers as sisters. How many siblings are there all together (including Henry and Alice)? Note: Henry is a boy, and Alice is a girl. ×" ]

[ "can run the same method to compute $$a\\pmod{11!}$$.", "{b}^{3}} \\sqrt{6}$", "@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19 @AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22 Nope, that seems likely to be the most concise way to do it. Count ways to pick $n\\in\\{1,2,3,4\\}$ unique numbers, and to arrange them with $n-1$ \"$>$\" signs, to meet the criteria. $$\\begin{array}{|l:l|} \\hline \\rm a{=}a{=}a{=}a & \\dbinom 3 0 \\dbinom 6 1 \\\\\\hdashline\\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \\dbinom 3 1\\dbinom 6 2 \\\\\\hdashline\\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \\dbinom 3 2 \\dbinom 6 3 \\\\\\hdashline\\rm a{>}b{>}c{>}d & \\dbinom 3 3 \\dbinom 6 4 \\\\ \\hline\\end{array}$$ $$\\dfrac{\\sum_{n=1}^4 \\dbinom{3}{n-1}\\dbinom{6}{n}}{6^4} = \\dfrac{\\dbinom 6 1 + 3\\dbinom 6 2 + 3 \\dbinom 6 3 + \\dbinom 6 4}{6^4}$$ - The problem reduces to finding the cardinality of the following set $$A=\\{(a_1,a_2,a_3,a_4)\\in\\{1,\\ldots,6\\}^4: \\ a_1\\leq a_2\\leq a_3 \\leq a_4\\}.$$ Define $$B=\\{(a_1,a_2,a_3,a_4)\\}\\in\\{1,\\ldots 9\\}^4: \\ a_1 < a_2 < a_3 < a_4\\}$$ and $$f:A\\to B, \\quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \\choose 4}.$$ - @Shailesh This method actually does give the correct answer ($126/6^4$); If we", "\\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} & 360 \\cdot x^{44} & 360 \\cdot x^{45} & 120 \\cdot x^{46}\\\\ & & & &", "+0 # Help. I have been stuck 0 114 4 Given that $$\\binom{15}{8}=6435$$$$\\binom{16}{9}=11440$$, and $$\\binom{16}{10}=8008$$, find $$\\binom{15}{10}$$. I know that I should be able to do this but I can't figure it out. Aug 10, 2019 #1 +23295 +3 Given that $$\\dbinom{15}{8}=6435$$, $$\\dbinom{16}{9}=11440$$ , and $$\\dbinom{16}{10}=8008$$, find $$\\dbinom{15}{10}$$. Formula: $$\\dbinom{n}{k} + \\dbinom{n}{k+1} = \\dbinom{n+1}{k+1}$$ $$\\begin{array}{|lrcll|} \\hline & \\mathbf{\\dbinom{n}{k} + \\dbinom{n}{k+1}} &=& \\mathbf{\\dbinom{n+1}{k+1}} \\\\\\\\ (1) & \\dbinom{15}{8} + \\dbinom{15}{9} &=& \\dbinom{16}{9} \\\\ (2) & \\dbinom{15}{9} + \\dbinom{15}{10} &=& \\dbinom{16}{10} \\\\ \\hline (1)-(2): & \\dbinom{15}{8} + \\dbinom{15}{9} -\\left( \\dbinom{15}{9} + \\dbinom{15}{10}\\right) &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & \\dbinom{15}{8} - \\dbinom{15}{10} &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & 6435 - \\dbinom{15}{10} &=& 11440-11440 \\\\ & 6435 - \\dbinom{15}{10} &=& 3432 \\\\ & \\dbinom{15}{10} &=& 6435 - 3432 \\\\ & \\mathbf{\\dbinom{15}{10}} &=& \\mathbf{3003} \\\\ \\hline \\end{array}$$ Aug 11, 2019 #2 0 Thank you so much!! I can't keep all the formulas in my head sometimes. Aug 11, 2019 #3 +1655 +3 There are really too many formulas... tommarvoloriddle Aug 12, 2019 #4 +105540 +1 You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from. There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level. Most times this does", "}{ \\log(3) }$", "6 )}^{C}=\\empty$ lol", "\\, 2!} = 15$$ (already calculated above) $$\\text {coef} (x^3y^3) =\\dbinom {6} {3} =\\dfrac {6!} {3! \\, 3!} = 20$$ $$\\text {coef} (x^2y^4) =\\dbinom {6} {2} =\\dfrac {6!} {2! \\, 4!} = 15$$ $$\\text {coef} (xy^5) =\\dbinom {6} {1} =\\dfrac {6!} {1! \\, 5!} = 6$$ $$\\text {coef} (y^6) =\\dbinom {6} {0} =\\dfrac {6!} {0! \\, 6!} = 1$$ Therefore, $$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$$ Notice that some coefficients are equal : $$x^2y^4$$ and $$x^4y^2$$ , $$xy^5$$ and $$x^5y$$. This symmetry is easily deduced from the binomial coefficient formula. This simplifies the calculations. Applied to $$(x+y) ^8$$ polynom, we get, $$\\text {coef} (x^8) = \\text {coef} (y^8) = 1$$ For the other coefficients, we only calculate the coefficients of $$xy^7$$,$$x^2y^6$$,$$x^3y^5$$ and $$x^4y^4$$. The other coefficients are deduced symetrically.", ")^{10} \\\\\\\\ && ( a + b + c )^{n} \\\\ &=& \\sum \\limits_{k=0}^{n} \\sum \\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} &", "let $N_1=N-\\dbinom{x_1}k$.\" Did you mean $N_1=N-\\dbinom{x_k}{k}$? That would be consistent with what you wrote in your next paragraph: $N_1=\\dbinom{x_{k-1}}{k-1}+\\ldots+\\dbinom{x_1}{1}$ – tychicus Aug 6 '13 at 7:21 • @tychicus: Yes, I did; I’ve fixed it now. (The same mistake was in the original version of your hint, and I must have unconsciously copied that.) – Brian M. Scott Aug 6 '13 at 12:02", "db \"eASSWzRD<11$\"", "\\text{dm}^3}$", "# Thread: dbl int and desmos graoh 1. ## dbl int and desmos graoh $\\tiny{15.3.53}\\\\ \\displaystyle \\int_{0}^{\\frac{\\pi}{6}} \\int_{0}^{\\sec{\\theta}} 6r^{3} drd\\theta$ just want to see if this desmos graph is correct wanted to shade the wedge area but ??? also got $0.96225$ for calulated answer but they wanted exact", "(\\mathbf{A}\\mathbf{B})^{-1} \\dbinom12 = \\dbinom13 } \\\\ \\hline \\end{array}$$ Apr 17, 2020", "$$\\PageIndex{6}$$).", "## Intermediate Algebra for College Students (7th Edition) $x^{10}-5x^8+10x^6-10x^4+5x^2-1$ Apply Binomial Theorem. $(a+b)^n=\\sum_{r=0}^n\\dbinom{n}{r}(a)^{n-r}b^r$ $(x^2-1)^5=\\dbinom{5}{0}(x^2)^5(-1)^0+\\dbinom{5}{1}(x^2)^4(-1)^1+\\dbinom{5}{2}(x^2)^3(-1)^2+\\dbinom{5}{3}(x^2)^2(-1)^3+\\dbinom{5}{4}(x^2)^1(-1)^4+\\dbinom{5}{5}(x^2)^0(-1)^5$ or, $=x^{10}-5x^8+10x^6+10x^4(-1)+5x^2+(1)(1)(-1)$ or, $=x^{10}-5x^8+10x^6-10x^4+5x^2-1$", "$\\dbinom n k = \\dbinom n {k + 1}$ So: $k \\ge \\dfrac {n - 1} 2$ By the Symmetry Rule for Binomial Coefficients, we also have that: $\\dfrac {n - 1} 2 < k \\le n \\iff \\dbinom n k > \\dbinom n {k + 1}$ and so for $\\dbinom n k = \\dbinom n {k + 1}$ it is not the case that $\\dfrac {n - 1} 2 < k \\le n$. So: $k \\le \\dfrac {n - 1} 2$ Hence that means: $k = \\dfrac {n - 1} 2$ which means: $n = 2 k + 1$ and so $n$ is odd. $\\blacksquare$", "}{ 3 }\\sqrt{3})$", "}_{3}/{\\alpha }_{2}}}$$ (5)", "## Numbers Greater Than 5,000,000 Using all the digits $3,4,5,5,5,6$ and $6$, how many distinct integers greater than $5,\\!000,\\!000$ can be formed? Source: NCTM Mathematics Teacher, January 2006 Solution The first digit must be a $5$ or a $6$. Case 1: $5$ _ _ _ _ _ _ Since copies of the same digits cannot be distinguished from one another, once we have positions to put them in we can copy them into the positions in any order. We construct an ordering for the digits $3,4,5,5,6,6$ in a four-step process 1) choose a subset of $2$ positions for the two $6$’s 2) choose a subset of $2$ positions for the two $5$’s 3) choose a subset of $1$ position for the $4$ 4) choose a subset of $1$ position for the $3$ $\\dbinom{6}{2}\\dbinom{4}{2}\\dbinom{2}{1}\\dbinom{1}{1}=180$ Case 2: $6$ _ _ _ _ _ _ Similarly, we construct an ordering for the digits $3,4,5,5,5,6$ 1) choose a subset of $1$ position for the $6$ 2) choose a subset of $3$ positions for the $5$’s 3) choose a subset of $1$ position for the $4$ 4) choose a subset of $1$ position for the $3$ $\\dbinom{6}{1}\\dbinom{5}{3}\\dbinom{2}{1}\\dbinom{1}{1}=120$ $180+120=300$ $300$ distinct integers greater than $5,\\!000,\\!000$ can be formed using all the digits $3,4,5,5,5,6$, and $6$. Answer: $300$" ]

[ "20 }\\!\\!\\times\\!\\!\\text{ }\\dfrac{\\text{18}}{\\text{60 }\\!\\!\\times\\!\\!\\text{ 60}}\\text{ km = }\\dfrac{\\text{20 }\\!\\!\\times\\!\\!\\text{ 18}}{\\text{60 }\\!\\!\\times\\!\\!\\text{ 60}}\\times \\text{ 1000m} \\\\ \\end{align} d = 100m.", "\\text{5}^{(2y-3) - (-5y+5)} \\times \\text{2}^{(4y+4) - (-5y+5)}\\\\ &= \\text{5}^{7y -\\text{8}} \\times \\text{2}^{9y-\\text{1}} \\end{align*} $$\\dfrac{6^4 \\times 12^3 \\times 4^5}{30^3 \\times 3^6}$$ \\begin{align*} \\frac{6^4 \\times 12^3 \\times 4^5}{30^3 \\times 3^6} & = \\frac{\\left(3^{4} \\times 2^{4}\\right) \\times \\left(3^{3} \\times 4^{3}\\right) \\times 4^{5}}{\\left(3^{3} \\times 10^{3}\\right) \\times 3^{6}} \\\\ & = \\frac{3^{4} \\times 2^{4} \\times 3^{3} \\times 2^{6} \\times 2^{10}}{3^{3} \\times 2^{3} \\times 5^{3} \\times 3^{6}} \\\\ & 3^{4 + 3 - 3 - 6} \\cdot 2^{4 + 6 + 10 - 3} \\cdot 5^{-3} \\\\ & = \\frac{2^{17}}{3^{2}5^{3}} \\end{align*} $$\\dfrac{9^3\\times 20^2}{4\\times 5^2 \\times 3^5}$$ \\begin{align*} \\frac{9^3\\times 20^2}{4\\times 5^2 \\times 3^5} & = \\frac{3^6 4^2 5^2}{4 \\times 5^2 3^5} \\\\ & = \\frac{3^6 2^4 5^2}{2^2 5^2 3^5} \\\\ & = 3^{6 - 5} 2^{4 - 2} 5^{2 - 2} \\\\ & = 3 \\times 2^2 \\\\ & = 12 \\end{align*} $$\\dfrac{7^{b} + 7^{b - 2}}{4 \\times 7^{b} + 3 \\times 7^{b}}$$ \\begin{align*} \\frac{7^{b} + 7^{b - 2}}{4 \\times 7^{b} - 3 \\times 7^{b}} &= \\frac{7^{b}(1 + 7^{-2})}{7^{b}(4 - 3)} \\\\ &= \\frac{(1 + 7^{-2})}{1} \\\\ &= \\frac{1 + \\frac{1}{49}}{1} \\\\ & = \\frac{51}{49} \\end{align*} $$\\dfrac{12^{y} - 96^{y}}{3^{y} + 6^{y}}$$ \\begin{align*} \\frac{12^{y} - 96^{y}}{3^{y} + 6^{y}} & = \\frac{(4 \\cdot 3)^{y} - \\left(2^{5} \\cdot 3\\right)^{y}}{3^{y} + \\left(2 \\cdot 3\\right)^{y}} \\\\ & = \\frac{3^{y}\\left(4^{y} - 2^{5y}\\right)}{3^{y}\\left(2 + 1\\right)} \\\\ &= \\frac{4^{y} - 2^{5y}}{3} \\end{align*}", "( \\dfrac{1}{T_2^{2/3}}-\\dfrac{1}{T_1^{2/3}} \\right )\\\\ &=-\\left ( 2\\pi \\times 6.67\\times 10^{-11}\\times 6\\times 10^{24}\\right )^{2/3}\\times 50\\times \\left ( \\dfrac{1}{\\left ( 10.8\\times 10^{3} \\right )^{2/3}}-\\dfrac{1}{\\left (7.2\\times 10^{3} \\right )^{2/3}} \\right )\\\\ &=-\\left ( 2.51\\times 10^{15}\\right )^{2/3}\\times 50\\times \\left ( \\dfrac{1}{488}-\\dfrac{1}{373} \\right )\\\\ &=-1.36\\times 10^{10}\\times 50\\times \\left ( -6.32 \\times 10^{-4}\\right )\\\\ &=\\boxed{4.3\\times 10^{6}\\;\\rm J} \\end{align*} {/eq}", "-\\dfrac{1}{3}x-\\dfrac{1}{8}y+\\dfrac{1}{6}z &= -\\dfrac{4}{3}\\\\ -\\dfrac{2}{3}x-\\dfrac{7}{8}y+\\dfrac{1}{3}z &= -\\dfrac{23}{3}\\\\ -\\dfrac{1}{3}x-\\dfrac{5}{8}y+\\dfrac{5}{6}z &= 0 \\end{align*} 35) \\begin{align*} -\\dfrac{1}{4}x-\\dfrac{5}{4}y+\\dfrac{5}{2}z &= -5\\\\ -\\dfrac{1}{2}x-\\dfrac{5}{3}y+\\dfrac{5}{4}z &= \\dfrac{55}{12}\\\\ -\\dfrac{1}{3}x-\\dfrac{1}{3}y+\\dfrac{1}{3}z &= \\dfrac{5}{3} \\end{align*} $$(-5,-5,-5)$$ 36) \\begin{align*} \\dfrac{1}{40}x+\\dfrac{1}{60}y+\\dfrac{1}{80}z &= \\dfrac{1}{100}\\\\ -\\dfrac{1}{2}x-\\dfrac{1}{3}y-\\dfrac{1}{4}z &= -\\dfrac{1}{5}\\\\ \\dfrac{3}{8}x+\\dfrac{3}{12}y+\\dfrac{3}{16}z &= \\dfrac{3}{20} \\end{align*} 37) \\begin{align*} 0.1x-0.2y+0.3z &= 2\\\\ 0.5x-0.1y+0.4z &= 8\\\\ 0.7x-0.2y+0.3z &= 8 \\end{align*} $$(10,10,10)$$ 38) \\begin{align*} 0.2x+0.1y-0.3z &= 0.2\\\\ 0.8x+0.4y-1.2z &= 0.1\\\\ 1.6x+0.8y-2.4z &= 0.2 \\end{align*} 39) \\begin{align*} 1.1x+0.7y-3.1z &= -1.79\\\\ 2.1x+0.5y-1.6z &= -0.13\\\\ 0.5x+0.4y-0.5z &= -0.07 \\end{align*} $$\\left ( \\dfrac{1}{2},\\dfrac{1}{5},\\dfrac{4}{5} \\right )$$ 40) \\begin{align*} 0.5x-0.5y+0.5z &= 10\\\\ 0.2x-0.2y+0.2z &= 4\\\\ 0.1x-0.1y+0.1z &= 2 \\end{align*} 41) \\begin{align*} 0.1x+0.2y+0.3z &= 0.37\\\\ 0.1x-0.2y-0.3z &= -0.27\\\\ 0.5x-0.1y-0.3z &= -0.03 \\end{align*} $$\\left ( \\dfrac{1}{2},\\dfrac{2}{5},\\dfrac{4}{5} \\right )$$ 42) \\begin{align*} 0.5x-0.5y-0.3z &= 0.13\\\\ 0.4x-0.1y-0.3z &= 0.11\\\\ 0.2x-0.8y-0.9z &= -0.32 \\end{align*} 43) \\begin{align*} 0.5x+0.2y-0.3z &= 1\\\\ 0.4x-0.6y+0.7z &= 0.8\\\\ 0.3x-0.1y-0.9z &= 0.6 \\end{align*} $$(2,0,0)$$ 44) \\begin{align*} 0.3x+0.3y+0.5z &= 0.6\\\\ 0.4x+0.4y+0.4z &= 1.8\\\\ 0.4x+0.2y+0.1z &= 1.6 \\end{align*} 45) \\begin{align*} 0.8x+0.8y+0.8z &= 2.4\\\\ 0.3x-0.5y+0.2z &= 0\\\\ 0.1x+0.2y+0.3z &= 0.6 \\end{align*} $$(1,1,1)$$ ### Extensions For the exercises 46-50, solve the system for $$x,y,$$ and $$z$$. 46) \\begin{align*} x+y+z &= 3\\\\ \\dfrac{x-1}{2}+\\dfrac{y-3}{2}+\\dfrac{z+1}{2} &= 0\\\\ \\dfrac{x-2}{3}+\\dfrac{y+4}{3}+\\dfrac{z-3}{3} &= \\dfrac{2}{3} \\end{align*} 47) \\begin{align*} 5x-3y-\\dfrac{z+1}{2} &= \\dfrac{1}{2}\\\\ 6x+\\dfrac{y-9}{2}+2z &= -3\\\\ \\dfrac{x+8}{2}-4y+z &= 4\\end{align*} $$\\left ( \\dfrac{128}{557},\\dfrac{23}{557},\\dfrac{428}{557} \\right )$$ 48) \\begin{align*} \\dfrac{x+4}{7}-\\dfrac{y-1}{6}+\\dfrac{z+2}{3} &= 1\\\\ \\dfrac{x-2}{4}+\\dfrac{y+1}{8}-\\dfrac{z+8}{2} &= 0\\\\ \\dfrac{x+6}{3}-\\dfrac{y+2}{3}+\\dfrac{z+4}{2} &= 3 \\end{align*} 49) \\begin{align*} \\dfrac{x-3}{6}+\\dfrac{y+2}{2}-\\dfrac{z-3}{3} &= 2\\\\ \\dfrac{x+2}{4}+\\dfrac{y-5}{2}+\\dfrac{z+4}{2} &= 1\\\\ \\dfrac{x+6}{2}-\\dfrac{y-3}{3}+z+1", "-\\dfrac{1}{3}x-\\dfrac{1}{8}y+\\dfrac{1}{6}z &= -\\dfrac{4}{3}\\\\ -\\dfrac{2}{3}x-\\dfrac{7}{8}y+\\dfrac{1}{3}z &= -\\dfrac{23}{3}\\\\ -\\dfrac{1}{3}x-\\dfrac{5}{8}y+\\dfrac{5}{6}z &= 0 \\end{align*} 35) \\begin{align*} -\\dfrac{1}{4}x-\\dfrac{5}{4}y+\\dfrac{5}{2}z &= -5\\\\ -\\dfrac{1}{2}x-\\dfrac{5}{3}y+\\dfrac{5}{4}z &= \\dfrac{55}{12}\\\\ -\\dfrac{1}{3}x-\\dfrac{1}{3}y+\\dfrac{1}{3}z &= \\dfrac{5}{3} \\end{align*} $$(-5,-5,-5)$$ 36) \\begin{align*} \\dfrac{1}{40}x+\\dfrac{1}{60}y+\\dfrac{1}{80}z &= \\dfrac{1}{100}\\\\ -\\dfrac{1}{2}x-\\dfrac{1}{3}y-\\dfrac{1}{4}z &= -\\dfrac{1}{5}\\\\ \\dfrac{3}{8}x+\\dfrac{3}{12}y+\\dfrac{3}{16}z &= \\dfrac{3}{20} \\end{align*} 37) \\begin{align*} 0.1x-0.2y+0.3z &= 2\\\\ 0.5x-0.1y+0.4z &= 8\\\\ 0.7x-0.2y+0.3z &= 8 \\end{align*} $$(10,10,10)$$ 38) \\begin{align*} 0.2x+0.1y-0.3z &= 0.2\\\\ 0.8x+0.4y-1.2z &= 0.1\\\\ 1.6x+0.8y-2.4z &= 0.2 \\end{align*} 39) \\begin{align*} 1.1x+0.7y-3.1z &= -1.79\\\\ 2.1x+0.5y-1.6z &= -0.13\\\\ 0.5x+0.4y-0.5z &= -0.07 \\end{align*} $$\\left ( \\dfrac{1}{2},\\dfrac{1}{5},\\dfrac{4}{5} \\right )$$ 40) \\begin{align*} 0.5x-0.5y+0.5z &= 10\\\\ 0.2x-0.2y+0.2z &= 4\\\\ 0.1x-0.1y+0.1z &= 2 \\end{align*} 41) \\begin{align*} 0.1x+0.2y+0.3z &= 0.37\\\\ 0.1x-0.2y-0.3z &= -0.27\\\\ 0.5x-0.1y-0.3z &= -0.03 \\end{align*} $$\\left ( \\dfrac{1}{2},\\dfrac{2}{5},\\dfrac{4}{5} \\right )$$ 42) \\begin{align*} 0.5x-0.5y-0.3z &= 0.13\\\\ 0.4x-0.1y-0.3z &= 0.11\\\\ 0.2x-0.8y-0.9z &= -0.32 \\end{align*} 43) \\begin{align*} 0.5x+0.2y-0.3z &= 1\\\\ 0.4x-0.6y+0.7z &= 0.8\\\\ 0.3x-0.1y-0.9z &= 0.6 \\end{align*} $$(2,0,0)$$ 44) \\begin{align*} 0.3x+0.3y+0.5z &= 0.6\\\\ 0.4x+0.4y+0.4z &= 1.8\\\\ 0.4x+0.2y+0.1z &= 1.6 \\end{align*} 45) \\begin{align*} 0.8x+0.8y+0.8z &= 2.4\\\\ 0.3x-0.5y+0.2z &= 0\\\\ 0.1x+0.2y+0.3z &= 0.6 \\end{align*} $$(1,1,1)$$ ### Extensions For the exercises 46-50, solve the system for $$x,y,$$ and $$z$$. 46) \\begin{align*} x+y+z &= 3\\\\ \\dfrac{x-1}{2}+\\dfrac{y-3}{2}+\\dfrac{z+1}{2} &= 0\\\\ \\dfrac{x-2}{3}+\\dfrac{y+4}{3}+\\dfrac{z-3}{3} &= \\dfrac{2}{3} \\end{align*} 47) \\begin{align*} 5x-3y-\\dfrac{z+1}{2} &= \\dfrac{1}{2}\\\\ 6x+\\dfrac{y-9}{2}+2z &= -3\\\\ \\dfrac{x+8}{2}-4y+z &= 4\\end{align*} $$\\left ( \\dfrac{128}{557},\\dfrac{23}{557},\\dfrac{428}{557} \\right )$$ 48) \\begin{align*} \\dfrac{x+4}{7}-\\dfrac{y-1}{6}+\\dfrac{z+2}{3} &= 1\\\\ \\dfrac{x-2}{4}+\\dfrac{y+1}{8}-\\dfrac{z+8}{2} &= 0\\\\ \\dfrac{x+6}{3}-\\dfrac{y+2}{3}+\\dfrac{z+4}{2} &= 3 \\end{align*} 49) \\begin{align*} \\dfrac{x-3}{6}+\\dfrac{y+2}{2}-\\dfrac{z-3}{3} &= 2\\\\ \\dfrac{x+2}{4}+\\dfrac{y-5}{2}+\\dfrac{z+4}{2} &= 1\\\\ \\dfrac{x+6}{2}-\\dfrac{y-3}{3}+z+1", "3 Use the quotient rule to differentiate $\\displaystyle f(x)=\\dfrac{3x+1}{1-x}$. \\begin{align} \\displaystyle f^{\\prime}(x) &= \\dfrac{(3x+1)^{\\prime} \\times (1-x) – (3x+1) \\times (1-x)^{\\prime}}{(1-x)^2} \\\\ &= \\dfrac{3 \\times (1-x) – (3x+1) \\times (-1)}{(1-x)^2} \\\\ &= \\dfrac{3-3x +3x+1}{(1-x)^2} \\\\ &= \\dfrac{4}{(1-x)^2} \\\\ \\end{align} ### Example 4 Use the quotient rule to differentiate $\\displaystyle f(x)=\\dfrac{(2x+1)^3}{(4x-1)^4}$. \\begin{align} \\displaystyle f^{\\prime}(x) &= \\dfrac{\\big[(2x+1)^3\\big]^{\\prime} \\times (4x-1)^4 – (2x+1)^3 \\times \\big[(4x-1)^4\\big]^{\\prime}}{\\big[(4x-1)^4\\big]^2} \\\\ &= \\dfrac{3(2x+1)^{3-1} \\times (2x+1)^{\\prime} \\times (4x-1)^4 – (2x+1)^3 \\times 4(4x-1)^{4-1} \\times (4x-1)^{\\prime}}{(4x-1)^8} \\\\ &= \\dfrac{3(2x+1)^2 \\times 2 \\times (4x-1)^4 – (2x+1)^3 \\times 4(4x-1)^3 \\times 4}{(4x-1)^8} \\\\ &= \\dfrac{6(2x+1)^2 (4x-1)^4 – 16(2x+1)^3 (4x-1)^3}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3\\big[6(4x-1) – 16(2x+1)\\big]}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3(24x-6 – 32x-16)}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3(-8x-22)}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(-8x-22)}{(4x-1)^5} \\\\ \\end{align} ## Extension Examples These Extension Examples require to have some prerequisite skills including; \\begin{align} \\displaystyle \\dfrac{d}{dx}\\sin{x} &= \\cos{x} \\\\ \\dfrac{d}{dx}\\cos{x} &= -\\sin{x} \\\\ \\dfrac{d}{dx}e^x &= e^x \\\\ \\dfrac{d}{dx}\\log_e{x} &= \\dfrac{1}{x} \\\\ \\end{align} ### Example 5 Find $\\displaystyle \\dfrac{dy}{dx}$ of $\\displaystyle y=\\dfrac{\\sin{x}}{\\cos{x}}$, known that $\\dfrac{d}{dx}\\sin{x} = \\cos{x}$ and $\\dfrac{d}{dx}\\cos{x} = -\\sin{x}$. \\begin{align} \\displaystyle \\dfrac{dy}{dx} &= \\dfrac{\\dfrac{d}{dx}\\sin{x} \\times \\cos{x} – \\sin{x} \\times \\dfrac{d}{dx}\\cos{x}}{\\cos^2{x}} \\\\ &= \\dfrac{\\cos{x} \\times \\cos{x} – \\sin{x} \\times (-\\sin{x})}{\\cos^2{x}} \\\\ &= \\dfrac{\\cos^2{x} + \\sin^2{x}}{\\cos^2{x}} \\\\ \\end{align} Note that optionally $\\cos^2{x} + \\sin^2{x}$ can be simplified further to $1$, as $\\cos^2{x} + \\sin^2{x}=1$ in one of the trigonometric properties, however we do not cover this skill here.", "last row right. This works well, if one has an expression of the type a+b=c+d+e+f+g+h=j, where there's only one row between initial row and result. However, in my case, I have some mathematical transformation (meaning an additional row) and so I need the equal sign of the row: \"12\\times6+14\\times\\dfrac{48}{7}-6^2-\\left(\\dfrac{2}{7}-4\\right)^2-\\left(\\dfrac{48}{7}\\right)^2 -\\dfrac{2}{7}\\times\\dfrac{48}{7}-\\left(\\dfrac{2}{7}\\right)^2\" to be aligned under the first equal sign. So now I use this version: Code: Select all • Open in writeLaTeX \\begin{align*}\\pi^{M}(s^{M},f^{M},x^{M}) :=& 12s^{M}+14f^{M}-(s^{M})^2-(x^{M}-4)^2-(f^{M})^2\\\\ &-x^{M}f^{M}-(x^{M})^2\\pause\\\\ =&12\\times6+14\\times\\dfrac{48}{7}-6^2-\\left(\\dfrac{2}{7}-4\\right)^2-\\left(\\dfrac{48}{7}\\right)^2\\\\ &-\\dfrac{2}{7}\\times\\dfrac{48}{7}-\\left(\\dfrac{2}{7}\\right)^2\\pause\\\\ =&69,14\\pause\\end{align*} Does anybody have better ideas? Thank you avp3000 Posts: 49 Joined: Thu Nov 15th, 2007 ### Re: Too long equation You can try the split environment, which is also provided by the amsmath package. To get a proper alignment, there are some tricky things to do. Code: Select all • Open in writeLaTeX \\begin{align*} \\pi^{M}(s^{M},f^{M},x^{M}): \\begin{split} &= 12s^{M}+14f^{M}-(s^{M})^2-(x^{M}-4)^2 \\\\ &\\quad -(f^{M})^2-x^{M}f^{M}-(x^{M})^2 \\end{split}\\\\ \\begin{split} &= 12\\times 6+14\\times\\frac{48}{7}-6^2-\\left(\\frac{2}{7}-4\\right)^2 \\\\ &\\quad -\\left(\\frac{48}{7}\\right)^2-\\frac{2}{7}\\times\\frac{48}{7}-\\left(\\frac{2}{7}\\right)^2 \\end{split}\\\\ &=69,14\\end{align*} Note that the alignment as known from the align environment has to be put into the split environment. More hints about typesetting mathematical expressions are given in \"Math mode\". Best regards Thorsten LaTeX Community Moderator ¹ System: openSUSE 13.1 (Linux 3.11.10), TeX Live 2013 (vanilla), TeXworks 0.5 (r1351) ² Posting stopped indefinitely due to offenses localghost Site Moderator Posts: 9219 Joined: Fri Feb 2nd, 2007 Location: Braunschweig, Germany ###", "(2) &\\div (1) \\\\ \\dfrac{ar^5}{ar^2} &= \\dfrac{-3072}{48} \\\\ r^3 &= -64 \\\\ r^3 &= (-4)^3 \\\\ r &= -4 \\\\ \\text{Substitute } r &= 4 \\text{ into } T_3 = ar^2 = 48 \\cdots (1)\\\\ a \\times 4^2 &= 48 \\\\ a &= 3 \\\\ \\therefore T_n &= 3 \\times (-4)^{n-1} \\end{aligned}", "10^4\\times 9.46\\times 10^{15})^{1.5}}{\\sqrt{6.67\\times10^{-11}\\times 8\\times 10^{13}\\times 1.99\\times 10^{30}}}\\\\\\\\ &=8.25\\times 10^{14}\\, s\\\\\\\\ &=\\boxed{2.61\\times 10^7\\, yr} \\end{align} {/eq}", "\\times 17 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 23 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{54} \\times p_{19}(29,25) &=& \\dfrac{29}{54} \\times \\dfrac{2^4 \\times 3^2 \\times 11^2 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^3 \\times 11^2 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} \\\\ \\dfrac{29}{55} \\times p_{20}(29,26) &=& \\dfrac{29}{55} \\times \\dfrac{2^5 \\times 5^3 \\times 11 \\times 17 \\times 19 \\times 23}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^5 \\times 5^2 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{56} \\times p_{21}(29,27) &=& \\dfrac{29}{56} \\times \\dfrac{2^4 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23}{3 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{57} \\times p_{22}(29,28) &=& \\dfrac{29}{57} \\times \\dfrac{2^2 \\times 3^2 \\times 5^2 \\times 11 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times", "{mg \\times \\left( {{R_c} - OA} \\right)} \\right) \\end{align*} {/eq} Substitute the given data in above equation, {eq}\\begin{align*} {M_a} &= \\left( {\\left( {\\dfrac{2}{{12}} \\times \\left( {\\dfrac{{50}}{2} - 10} \\right)} \\right) \\times {{\\left( {\\dfrac{{50}}{2} - 10} \\right)}^2}} \\right) + \\left( {2 \\times \\left( {\\dfrac{{50}}{2} - 10} \\right)} \\right) \\times 8 \\times 17.5 - \\left( {\\left( {2 \\times \\left( {\\dfrac{{50}}{2} - 10} \\right)} \\right)9.81 \\times \\left( {17.5 - 10} \\right)} \\right)\\\\ &= 2555.5\\;{\\rm{N}} \\cdot {\\rm{m}} \\end{align*} {/eq} Thus the moment at A is {eq}2555.5\\;{\\rm{N}} \\cdot {\\rm{m}} {/eq}", "be a2 (in checkerboard coordinates) because $$b2/c2 \\ge 2$$ and at least two of those marked $$\\diamond$$ and at least one of those marked $$\\ast$$. Because only 8 small numbers are available a4 must be small and one of b4,c4. All others must be $$>8$$ (\"large\", marked $$\\Box$$): $$\\begin{matrix}\\bigcirc&+&\\Box\\bigcirc&+&\\Box\\bigcirc&=&\\Box\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&\\bigcirc&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&\\bigcirc&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&\\Box&=&\\Box\\end{matrix}$$ and we see that a4 must be $$1$$ Observe that $$d4 = d3/d2 + a1/b1 + c2\\times c3 + c4 \\ge 2 + 2 + 2\\times 3 + c4$$ Therefore c4 must be small and b4 large. $$\\begin{matrix}1&+&\\Box&+&\\bigcirc&=&13\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&\\bigcirc&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&\\bigcirc&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&9\\ldots 12&=&11\\ldots 14\\end{matrix}$$ Next, we see that $$2$$ must be in either c2 or c3: $$\\begin{matrix}1&+&9\\ldots 11&+&3\\ldots 6&=&14\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&2\\ldots 4&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&2\\ldots 4&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&10\\ldots 12&=&12\\ldots 14\\end{matrix}$$ We are almost done! Observe that in the products $$a2\\times b2 = c2\\times d2$$ the primes $$5$$ and $$7$$ cannot occur because they are available only once. $$\\begin{matrix}1&+&9\\ldots 10&+&5&=&15\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&2\\ldots 3&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&2\\ldots 3&=&\\bigcirc\\\\=&&=&&=&&=\\\\14&/&7&+&11&=&13\\end{matrix}$$ We can finish by filling in column a, $$\\begin{matrix}1&+&9&+&5&=&15\\\\\\times&&-&&+&&-\\\\10&+&\\Box&/&2\\ldots 3&=&\\Box\\\\+&&/&&\\times&&/\\\\4&\\times&\\bigcirc&/&2\\ldots 3&=&\\bigcirc\\\\=&&=&&=&&=\\\\14&/&7&+&11&=&13\\end{matrix}$$ row 3 and, finally, row 2. • Very well done and great explanation! I was getting worried that no one will solve this... – Dmitry Kamenetsky Oct 10 '20 at 0:46 Let the digits be a1,a2,a3,… d4, using standard algebraic chess notation (i.e. letters = columns, numbers = rows, a1 = lower left). Define a low digit = 1-8, high digit = 9-16. The cells,", "= \\dfrac{{13 \\times 12! \\times 4 \\times 3! \\times 50! \\times 2 \\times 1}}{{52 \\times 51 \\times 50! \\times 12! \\times 3!}} \\\\ P = \\dfrac{{13 \\times 4 \\times 2}}{{52 \\times 51}} = \\dfrac{2}{{51}} \\\\$", "\\\\ &= (4 \\times 27) \\times x^{6+15} \\\\ &= 108x^{21} \\\\ \\end{align} ### Example 5 Simplify $\\Big(\\dfrac{2a^3}{b^2}\\Big)^3$. \\begin{align} \\displaystyle \\Big(\\dfrac{2a^3}{b^2}\\Big)^3 &= \\dfrac{(2a^3)^3}{(b^2)^3} \\\\ &= \\dfrac{2^3 a^{3 \\times 3}}{b^{2 \\times 3}} \\\\ &= \\dfrac{8 a^9}{b^6} \\\\ \\end{align}", "E=13.6\\left[ \\dfrac{1}{{{1}^{2}}}-\\dfrac{1}{{{5}^{2}}} \\right] \\\\ & \\Delta E=\\dfrac{hc}{\\lambda } \\\\ & \\dfrac{hc}{\\lambda }=13.6\\left[ \\dfrac{1}{{{1}^{2}}}-\\dfrac{1}{{{5}^{2}}} \\right]\\text{e}\\text{.V} \\\\ & \\dfrac{1240}{\\lambda }=13.6\\times \\dfrac{24}{25}\\approx 95\\text{nm} \\\\ \\end{align} So the correct option is A.", "n-2 \\right)...(n-r)!$ \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5!}\\times \\left( \\dfrac{8\\times 7\\times 6}{3!} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6!}\\times \\left( \\dfrac{8\\times 7}{2!} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4!}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4!} \\right) \\\\ \\end{align} \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6}{3\\times 2\\times 1} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6\\times 5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7}{2\\times 1} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4\\times 3\\times 2\\times 1} \\right) \\\\ \\end{align} By cancelling the common terms in denominator and numerator we get $=\\left( 11\\times 9\\times 8 \\right)\\times \\left( 8\\times 7 \\right)+\\left( 11\\times 3\\times 4\\times 7 \\right)\\times \\left( 4\\times 7 \\right)+\\left( 11\\times 5\\times 9 \\right)\\times \\left( 5\\times 2\\times 7 \\right)$ $=792\\times 56+924\\times 28+495\\times 70$ = 44352 + 25872 + 34650 = 104874 Thus, the number of ways of selecting 10 students chosen for the competition = 104874 Note: The possibility of mistake can be the calculation mistake since the procedure of solving involves large calculations. Hence be very careful in simplifying and cancelling the terms to get the desired result.", "be given by: {eq}\\displaystyle{\\begin{align*} J &= \\dfrac{V}{R \\ A}\\\\ &= \\dfrac{V}{R\\times \\pi r^2}\\\\ &= \\dfrac{(20 \\ \\rm V)}{(10^3 \\ \\Omega)\\times \\pi \\times (10^{-4} \\ \\rm m)^2}\\\\ &\\approx \\boxed{{63661.98 \\ \\rm A/m^2}}\\\\ \\end{align*}} {/eq}", "=9\\times \\left[ 4{{R}^{2}}-\\pi {{R}^{2}} \\right] \\\\ & =9\\times \\left( 4-\\pi \\right)\\times {{R}^{2}} \\\\ & =9\\times \\left( 4-3.14 \\right)\\times {{\\left( 7 \\right)}^{2}} \\\\ & =9\\times 0.86\\times 49 \\\\ & =397.26c{{m}^{2}} \\\\ \\end{align}", "or 6. Thus, the probability of E occurring is: \\begin{align}P( E ) &= \\dfrac{\\text{number of favorable outcomes}}{\\text{Number of total outcomes}}\\\\ &= \\dfrac{2}{6}\\\\ &= \\dfrac{1}{3}\\end{align} $$\\therefore$$ the probability= $$\\dfrac{1}{3}$$ Example 2 Shawn throws a die 400 times and he records the score of getting 5 as 30 times. What could be the probability of a) getting a score of 5 b) getting a score under 5 Solution a) P(getting a score of 5) \\begin{align}&=\\dfrac{\\text{number of times getting 5}}{\\text{total times}}\\\\\\\\&=\\dfrac{30}{400}\\\\\\\\&= \\dfrac{3}{40}\\end{align} b) P(getting a score under 5) \\begin{align}&=\\dfrac{\\text{number of times getting under 5}}{\\text{total times}}\\\\\\\\&=\\dfrac{370}{400}\\\\\\\\&= \\dfrac{37}{40}\\end{align} $$\\therefore$$ a) P(getting 5) $$= \\dfrac{3}{40}$$ b) P(getting under 5) $$= \\dfrac{37}{40}$$ Example 3 Patrick rolls a 6-sided die. What will be the probability that he will get a prime number? Solution The total possible outcomes in our sample space is 6 The numbers in the sample space is {1,2,3,4,5,6} The prime numbers among these are {2,3, 5} Probability of getting prime numbers \\begin{align}&= \\dfrac{\\text{favorable outcomes}}{\\text{total possible outcomes}}\\\\ &= \\dfrac{3}{ 6}\\\\&=\\dfrac{1}{2}\\end{align} $$\\therefore$$ Probability(getting a prime number) $$=\\dfrac{1}{2}$$ ## Interactive Questions Here are a few activities for you to practice. Select/Type your answer and click the \"Check Answer\" button to see the result. Think Tank 1. You have rolled a die 3 times in a row. You make a 3 digit number. The second digit is", "47} &=& \\dfrac{3 \\times 5^2 \\times 11 \\times 13 \\times 17 \\times 29}{19 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{49} \\times p_{14}(29,20) &=& \\dfrac{29}{49} \\times \\dfrac{2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17}{37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11 \\times 13 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{50} \\times p_{15}(29,21) &=& \\dfrac{29}{50} \\times \\dfrac{2^2 \\times 3 \\times 5^3 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11 \\times 13 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{51} \\times p_{16}(29,22) &=& \\dfrac{29}{51} \\times \\dfrac{2^5 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^5 \\times 3 \\times 5 \\times 11 \\times 13 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{52} \\times p_{17}(29,23) &=& \\dfrac{29}{52} \\times \\dfrac{2^3 \\times 3 \\times 5 \\times 11^2 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{53} \\times p_{18}(29,24) &=& \\dfrac{29}{53} \\times \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2" ]

[ "@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19 @AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22 Nope, that seems likely to be the most concise way to do it. Count ways to pick $n\\in\\{1,2,3,4\\}$ unique numbers, and to arrange them with $n-1$ \"$>$\" signs, to meet the criteria. $$\\begin{array}{|l:l|} \\hline \\rm a{=}a{=}a{=}a & \\dbinom 3 0 \\dbinom 6 1 \\\\\\hdashline\\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \\dbinom 3 1\\dbinom 6 2 \\\\\\hdashline\\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \\dbinom 3 2 \\dbinom 6 3 \\\\\\hdashline\\rm a{>}b{>}c{>}d & \\dbinom 3 3 \\dbinom 6 4 \\\\ \\hline\\end{array}$$ $$\\dfrac{\\sum_{n=1}^4 \\dbinom{3}{n-1}\\dbinom{6}{n}}{6^4} = \\dfrac{\\dbinom 6 1 + 3\\dbinom 6 2 + 3 \\dbinom 6 3 + \\dbinom 6 4}{6^4}$$ - The problem reduces to finding the cardinality of the following set $$A=\\{(a_1,a_2,a_3,a_4)\\in\\{1,\\ldots,6\\}^4: \\ a_1\\leq a_2\\leq a_3 \\leq a_4\\}.$$ Define $$B=\\{(a_1,a_2,a_3,a_4)\\}\\in\\{1,\\ldots 9\\}^4: \\ a_1 < a_2 < a_3 < a_4\\}$$ and $$f:A\\to B, \\quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \\choose 4}.$$ - @Shailesh This method actually does give the correct answer ($126/6^4$); If we", "he received from B Jul 4, 2021 at 3:53 • I made an edit with the clarifications Jul 4, 2021 at 4:06 Here's my solution based on expectations. First, we calculate the expected value of the smallest, second-smallest, third-smallest... number on the 5 cards drawn. The calculation steps are shown below: $$E(\\text{smallest})=\\dfrac{1\\times \\dbinom{9}{4}+2\\times \\dbinom{8}{4}+\\cdots+6\\times \\dbinom{4}{4}}{\\dbinom{10}{5}}=\\dfrac{11}{6}.$$ Explanation: the numerator basically sums the smallest terms in all 5-card draws, and the denominator is the total number of cases. Similarly, we have $$E(\\text{second-smallest})=\\dfrac{2\\times \\dbinom{8}{3}\\times \\dbinom{1}{1}+3\\times \\dbinom{7}{3}\\times \\dbinom{2}{1}+\\cdots+7\\times \\dbinom{3}{3}\\times \\dbinom{6}{1}}{\\dbinom{10}{5}}=\\dfrac{11}{3};$$ $$E(\\text{third-smallest})=\\dfrac{3\\times \\dbinom{7}{2}\\times \\dbinom{2}{2}+4\\times \\dbinom{6}{2}\\times \\dbinom{3}{2}+\\cdots+8\\times \\dbinom{2}{2}\\times \\dbinom{7}{2}}{\\dbinom{10}{5}}=\\dfrac{11}{2};$$ $$E(\\text{fourth-smallest})=\\dfrac{4\\times \\dbinom{6}{1}\\times \\dbinom{3}{3}+5\\times \\dbinom{5}{1}\\times \\dbinom{4}{3}+\\cdots+9\\times \\dbinom{1}{1}\\times \\dbinom{8}{3}}{\\dbinom{10}{5}}=\\dfrac{22}{3};$$ $$E(\\text{fifth-smallest, or largest})=\\dfrac{5\\times \\dbinom{4}{4}+6\\times \\dbinom{5}{4}+\\cdots+10\\times \\dbinom{9}{4}}{\\dbinom{10}{5}}=\\dfrac{55}{6}.$$ As a quick double check, adding them all together yields $$27.5$$, which is the expected sum of the 5 cards. Next, let's think about A's strategy. In essence, he is discarding his $$N$$ smallest cards to get random cards. Thus, if the expected value of cards that he is discarding is smaller than the expected value of cards he is getting, then A is benefitting. The more A benefits, the higher his chances of winning. Then, clearly $$E(\\text{smallest})$$, $$E(\\text{second-smallest})$$ are smaller than $$5.5$$, the expected value on a random card. $$E(\\text{third-smallest})$$ is equal to $$5.5$$, and $$E(\\text{fourth-smallest})$$, $$E(\\text{fifth-smallest, or largest})$$ are greater than $$5.5$$. This means that discarding the smallest card and", "is: $\\set {\\dbinom {a \\ b} {1 \\ 2}, \\dbinom {a \\ b} {1 \\ 3}, \\dbinom {a \\ b} {2 \\ 1}, \\dbinom {a \\ b} {2 \\ 3}, \\dbinom {a \\ b} {3 \\ 1}, \\dbinom {a \\ b} {3 \\ 2} }$ thus demonstrating that there are $\\dfrac {3!} {\\paren {3 - 2}!} = \\dfrac {3!} {1!} = \\dfrac 6 1 = 6$ injections from $A$ to $B$.", "set of three elements can be arranged in six ways. We denote the number of ways of choosing $$r$$ objects from $$n$$ objects by $$\\dbinom{n}{r}$$, which is read as '$$n$$ choose $$r$$'. In the example above, we found that $\\dbinom{5}{3} = \\dfrac{{^5}P_3}{3!} = \\dfrac{60}{6} = 10.$ In general, we have $\\dbinom{n}{r} = \\dfrac{^nP_r}{r!} = \\dfrac{n!}{(n-r)!r!}.$ Consider the five-element set $$\\{a,b,c,d,e\\}$$ again. There are \\begin{alignat*}{2} \\dbinom{5}{1}&=5 \\text{ ways of choosing 1 letter},&\\qquad \\dbinom{5}{2}&=10 \\text{ ways of choosing 2 letters},\\\\ \\dbinom{5}{3}&=15 \\text{ ways of choosing 3 letters},&\\qquad \\dbinom{5}{4}&=10 \\text{ ways of choosing 4 letters},\\\\ \\dbinom{5}{5}&=1 \\text{ way of choosing 5 letters}. \\end{alignat*} Listing these as a row, including choosing no letters as the first entry: $\\dbinom{5}{0}=1, \\qquad \\dbinom{5}{1}=5, \\qquad \\dbinom{5}{2}=10, \\qquad \\dbinom{5}{3}=10, \\qquad \\dbinom{5}{4}=5, \\qquad \\dbinom{5}{5}=1.$ This is the fifth row of Pascal's triangle. #### Example Evaluate 1. $$\\dbinom{100}{2}$$ 2. $$\\dbinom{1000}{998}$$. #### Solution 1. $$\\dbinom{100}{2} = \\dfrac{100\\times99}{2} = 4950$$ 2. $$\\dbinom{1000}{998} = \\dfrac{1000\\times999}{2} = 499\\,500$$. #### Example In how many ways can you choose two people from a group of seven people? #### Solution There are $$\\dbinom{7}{2} = \\dfrac{7\\times6}{2} = 21$$ ways of choosing two people from seven. #### Example There are ten people in a basketball squad. Find how many ways: 1. the starting five can be chosen from the squad 2. the squad can be split into two", "2\\dbinom 21}{\\dbinom 52}=\\dfrac{3}{5}$$", "### A result using complex numbers Suppose we expand out the expressions $$(1+1)^n$$, $$(1+i)^n$$, $$(1-1)^n$$, $$(1-i)^n$$ using the binomial theorem. This gives (for $$n>0$$), \\begin{align*} \\dbinom{n}{0} + \\dbinom{n}{1} + \\dbinom{n}{2} + \\dbinom{n}{3} + \\dots + \\dbinom{n}{n} &= 2^n\\\\ \\dbinom{n}{0} + i\\dbinom{n}{1} - \\dbinom{n}{2} - i\\dbinom{n}{3} + \\dots + i^n\\dbinom{n}{n} &= (1+i)^n\\\\ \\dbinom{n}{0} - \\dbinom{n}{1} + \\dbinom{n}{2} - \\dbinom{n}{3} + \\dots + (-1)^n\\dbinom{n}{n} &= 0^n\\\\ \\dbinom{n}{0} - i\\dbinom{n}{1} - \\dbinom{n}{2} + i\\dbinom{n}{3} + \\dots + (-i)^n\\dbinom{n}{n} &= (1-i)^n. \\end{align*} Now add these equations and divide by 4 to get $\\dbinom{n}{0} + \\dbinom{n}{4} + \\dbinom{n}{8} + \\dots = \\dfrac{1}{4}\\Big(2^n + (1-i)^n + (1+i)^n\\Big).$ It is easy to show, using the polar form, that $(1-i)^n + (1+i)^n = 2(\\sqrt{2})^n \\cos \\big(\\dfrac{n\\pi}{4}\\big).$ Thus, we have $\\dbinom{n}{0} + \\dbinom{n}{4} + \\dbinom{n}{8} + \\dots = \\dfrac{1}{4}\\Big(2^n + 2(\\sqrt{2})^n \\cos \\big(\\dfrac{n\\pi}{4}\\big)\\Big).$ This is a rather amazing and beautiful result. Next page - Links forward - Series for e", "&=& \\dbinom{2}{\\frac{11}{3}}\\\\\\\\ \\dbinom{1.5}{\\frac{13}{2}} + \\dbinom{2}{\\frac{14}{3}} &=& \\dbinom{3.5}{\\frac{67}{6}} \\end{array}$$ The vertices of the image are $$\\begin{array}{rcll} A' (0.5,-\\frac{5}{6} ) \\qquad B' ( 2, \\frac{11}{3}) \\qquad C' ( 3.5, \\frac{67}{6} ) \\end{array}$$ . Feb 16, 2016", "# Division using Exponents (Indices) If we are given $a^8 \\div a^3$, we can also write this as $\\dfrac{a^8}{a^3}$, which means $\\dfrac{a \\times a \\times a \\times a \\times a \\times a \\times a \\times a}{a \\times a \\times a}$. As there are $8$ factors of $a$ on the top line (numerator), and $3$ factors of $a$ on the bottom line (denominator), we can cancel $3$ of them, giving us \\large \\begin{align} \\require{AMSsymbols} \\displaystyle \\require{cancel} &\\dfrac{\\cancel{a} \\times \\cancel{a} \\times \\cancel{a} \\times a \\times a \\times a \\times a \\times a}{\\cancel{a} \\times \\cancel{a} \\times \\cancel{a}} \\\\ &= a \\times a \\times a \\times a \\times a \\\\ &= a^5 \\end{align} ### Example 1 After first writing in factor form, simplify $\\dfrac{x^7}{x^4}$. \\begin{align} \\displaystyle \\require{AMSsymbols} \\require{cancel} \\dfrac{x^7}{x^4} &= \\dfrac{x \\times x \\times x \\times x \\times x \\times x \\times x}{x \\times x \\times x \\times x} \\\\ &= \\dfrac{\\cancel{x} \\times \\cancel{x} \\times \\cancel{x} \\times \\cancel{x} \\times x \\times x \\times x}{\\cancel{x} \\times \\cancel{x} \\times \\cancel{x} \\times \\cancel{x}} \\\\ &= x \\times x \\times x \\\\ &= x^3 \\end{align} An alternative solution can be formed by examining the result. We can see that the exponent (index) in the answer is the result of subtracting the exponents (indices) in the question. \\large \\begin{align} \\displaystyle \\require{cancel} a^8 \\div a^3 &= a^{8-3} \\\\", "\\dbinom{5}{1}=\\frac{5!}{1!4!}=5; \\ \\dbinom{5}{2}=\\frac{5!}{2!3!}=10; \\ \\dbinom{5}{3}=\\frac{5!}{3!2!}=10; \\ \\dbinom{5}{4}=\\frac{5!}{4!1!}=5; \\ \\dbinom{5}{5}=\\frac{5!}{5!0!}=1$ These are the elements of the $6^{th}$ row of Pascal’s Triangle: $1 \\ \\ 5 \\ \\ 10 \\ \\ 10 \\ \\ 5 \\ \\ 1$ The Binomial Theorem allows us to determine the coefficients of the terms in the expansion without having to extend the triangle to the appropriate row. #### Example A Use the Binomial Theorem to expand $(x+2y)^6$ Solution: First, in this example, $a=x$ , $b=2y$ and $n=6$ . Now we can substitute into the rule. $(x+2y)^6=\\dbinom{6}{0}x^6(2y)^0+\\dbinom{6}{1}x^5(2y)^1+\\dbinom{6}{2}x^4(2y)^2+\\dbinom{6}{3}x^3(2y)^3+\\dbinom{6}{4}x^2(2y)^4+\\dbinom{6}{5}=x^1(2y)^5+\\dbinom{6}{6}x^0(2y)^6$ Now we can simplify: $&=(1)x^6(1)+(6)x^5(2y)+(15)x^4(4y^2)+(20)x^3(8y^3)+(15)x^2(16y^4)+(6)x(32y^5)+(1)(1)(64y^6) \\\\&=x^6+12x^5y+30x^4y^2+160x^3y^3+240x^2y^4+192xy^5+64y^6$ ### More Guidance What if we just want to find a single term in the expansion? We can use the following rule to represent the $(r+1)^{st}$ term in the expansion: $\\dbinom{n}{r}a^{n-r}b^r$ . The rule is for the $(r+1)^{st}$ term because if we want the $1^{st}$ term, then $r=0$ (refer back to the Binomial Theorem expansion rule). The value of $r$ in the expansion is always one less than the term number. #### Example B Find the $4^{th}$ term in the expansion of $(3x-5)^8$ . Solution: Since we want the $4^{th}$ term, $r=3$ . Now we can set up the formula with $a=3x$ , $b=-5$ , $n=8$ and $r=3$ and evaluate: $\\dbinom{8}{3}(3x)^{8-3}(-5)^3=(56)(243x^5)(-125)=-1,701,000x^5$ #### Example C Find the coefficient of the term containing $y^5$", "\\div 3 = 2(5 + 6) - 2\\\\ &&& 12 + 2 \\times 3 \\times 2^2 \\div 3 = 2\\times 11 - 2\\\\ &&& 12 + 2 \\times 3 \\times 4 \\div 3 = 2\\times 11 - 2\\\\ &&& 12 + 8 = 22 - 2\\\\ &&& 20 = 20\\end{align*} 2. Follow the order of operations and show each step. What is the value of the variable? \\begin{align*}10^2 - 6(4 + 6) - 3^2 - 3 \\times 4 - 1 = d + 50 \\div 5^2\\end{align*} We can use the problem solving steps to help us with the order of operations. \\begin{align*}& \\mathbf{Describe:} && \\text{The equation has parentheses, exponents, multiplication, subtraction, and addition.}\\\\ &&& d \\ \\text{is the variable.}\\\\ \\\\ & \\mathbf{My \\ Job:} && \\text{Apply the order of operations rule to figure out the value of}\\ d.\\\\ \\\\ & \\mathbf{Solve:} && 10^2 - 6(4 + 6) - 3^2 - 3 \\times 4 - 1 = d + 50 \\div 5^2\\\\ &&& \\mathbf{Parentheses} \\qquad \\qquad \\quad 10^2 - 6\\times 10 - 3^2 - 3 \\times 4 - 1 = d + 50 \\div 5^2\\\\ &&& \\mathbf{Exponents} \\qquad \\qquad \\quad \\ 100 - 6\\times 10 - 9 - 3 \\times 4 - 1 = d + 50 \\div 25\\\\ &&& \\mathbf{Multiplication/} \\qquad \\quad 100 - 60 - 9 -", "# Factorial Notation $n!$ is the product of the first $n$ positive integers for $n\\ge 1$. $n!=n(n-1)(n-2)(n-3) \\cdots \\times 3 \\times 2 \\times 1$ $n!$ is read “$n$ factorial”. For example, the product $5 \\times 4 \\times 3 \\times 2 \\times 1$ can be written as $5!$. Notice that $5 \\times 4 \\times 3$ can be written using factorial numbers only as $5 \\times 4 \\times 3 = \\dfrac{5 \\times 4 \\times 3 \\times 2 \\times 1}{2 \\times 1} = \\dfrac{5!}{2!}$ You can also notice that: \\begin{align} n! &= n(n-1)(n-2)(n-3) \\cdots \\times 3 \\times 2 \\times 1 \\\\ &= n(n-1)! \\\\ &= n(n-1)(n-2)! \\\\ &= n(n-1)(n-2)(n-3)! \\\\ &= \\cdots \\\\ \\end{align} Using this rule of factorial: \\begin{align} 1! &= 1 \\times 0! \\\\ 0! &= 1 \\\\ \\end{align} ## Example 1 Evaluate $5!$. \\begin{align} \\displaystyle 5! &= 5 \\times 4 \\times 3 \\times 2 \\times 1 \\\\ &= 120 \\end{align} ## Example 2 Evaluate $\\dfrac{6!}{4!}$. \\begin{align} \\displaystyle \\dfrac{6!}{4!} &= \\dfrac{6 \\times 5 \\times 4!}{4!} \\\\ &= 6 \\times 5 \\\\ &= 30 \\end{align} ## Example 3 Evaluate $\\dfrac{7!}{4! \\times 3!}$. \\begin{align} \\displaystyle \\dfrac{7!}{4! \\times 3!} &= \\dfrac{7 \\times 6 \\times 5 \\times 4!}{4! \\times 3!} \\\\ &= \\dfrac{7 \\times 6 \\times 5}{3!} \\\\ &= \\dfrac{7 \\times 6 \\times 5}{3 \\times 2 \\times 1} \\\\ &= 35 \\end{align}", "3 Use the quotient rule to differentiate $\\displaystyle f(x)=\\dfrac{3x+1}{1-x}$. \\begin{align} \\displaystyle f^{\\prime}(x) &= \\dfrac{(3x+1)^{\\prime} \\times (1-x) – (3x+1) \\times (1-x)^{\\prime}}{(1-x)^2} \\\\ &= \\dfrac{3 \\times (1-x) – (3x+1) \\times (-1)}{(1-x)^2} \\\\ &= \\dfrac{3-3x +3x+1}{(1-x)^2} \\\\ &= \\dfrac{4}{(1-x)^2} \\\\ \\end{align} ### Example 4 Use the quotient rule to differentiate $\\displaystyle f(x)=\\dfrac{(2x+1)^3}{(4x-1)^4}$. \\begin{align} \\displaystyle f^{\\prime}(x) &= \\dfrac{\\big[(2x+1)^3\\big]^{\\prime} \\times (4x-1)^4 – (2x+1)^3 \\times \\big[(4x-1)^4\\big]^{\\prime}}{\\big[(4x-1)^4\\big]^2} \\\\ &= \\dfrac{3(2x+1)^{3-1} \\times (2x+1)^{\\prime} \\times (4x-1)^4 – (2x+1)^3 \\times 4(4x-1)^{4-1} \\times (4x-1)^{\\prime}}{(4x-1)^8} \\\\ &= \\dfrac{3(2x+1)^2 \\times 2 \\times (4x-1)^4 – (2x+1)^3 \\times 4(4x-1)^3 \\times 4}{(4x-1)^8} \\\\ &= \\dfrac{6(2x+1)^2 (4x-1)^4 – 16(2x+1)^3 (4x-1)^3}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3\\big[6(4x-1) – 16(2x+1)\\big]}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3(24x-6 – 32x-16)}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3(-8x-22)}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(-8x-22)}{(4x-1)^5} \\\\ \\end{align} ## Extension Examples These Extension Examples require to have some prerequisite skills including; \\begin{align} \\displaystyle \\dfrac{d}{dx}\\sin{x} &= \\cos{x} \\\\ \\dfrac{d}{dx}\\cos{x} &= -\\sin{x} \\\\ \\dfrac{d}{dx}e^x &= e^x \\\\ \\dfrac{d}{dx}\\log_e{x} &= \\dfrac{1}{x} \\\\ \\end{align} ### Example 5 Find $\\displaystyle \\dfrac{dy}{dx}$ of $\\displaystyle y=\\dfrac{\\sin{x}}{\\cos{x}}$, known that $\\dfrac{d}{dx}\\sin{x} = \\cos{x}$ and $\\dfrac{d}{dx}\\cos{x} = -\\sin{x}$. \\begin{align} \\displaystyle \\dfrac{dy}{dx} &= \\dfrac{\\dfrac{d}{dx}\\sin{x} \\times \\cos{x} – \\sin{x} \\times \\dfrac{d}{dx}\\cos{x}}{\\cos^2{x}} \\\\ &= \\dfrac{\\cos{x} \\times \\cos{x} – \\sin{x} \\times (-\\sin{x})}{\\cos^2{x}} \\\\ &= \\dfrac{\\cos^2{x} + \\sin^2{x}}{\\cos^2{x}} \\\\ \\end{align} Note that optionally $\\cos^2{x} + \\sin^2{x}$ can be simplified further to $1$, as $\\cos^2{x} + \\sin^2{x}=1$ in one of the trigonometric properties, however we do not cover this skill here.", "\\begin{align*} \\dbinom{x+y}{n}=\\sum_{i\\in\\mathbb{N}}\\dbinom{x}{i}\\dbinom{y}{n-i}. \\end{align*} Theorem 4 (trinomial revision). Let $$n,a,b\\in\\mathbb{R}$$. Then, \\begin{align*} \\dbinom{n}{a}\\dbinom{a}{b}=\\dbinom{n}{b}\\dbinom{n-b}{a-b}. \\end{align*} Theorem 5 (symmetry of binomial coefficients). Let $$n\\in\\mathbb{N}$$ and $$k\\in\\mathbb{R}$$. Then, \\begin{align*} \\dbinom{n}{k}=\\dbinom{n}{n-k}. \\end{align*} Theorem 6 (sum of a row of Pascal's triangle). Let $$n\\in\\mathbb{N}$$. Then, \\begin{align*} \\sum_{p=0}^{n}\\dbinom{n}{p}=2^{n}. \\end{align*} I omit the proofs of these results, as they are well-known. (For the sake of completeness: Theorems 2, 3, 4, 5 and 6 appear as Proposition 2.6.13, Theorem 2.6.1, Proposition 1.3.35, Theorem 1.3.11 and Corollary 1.3.27 in my Enumerative Combinatorics, 4 May 2021, respectively. It should not be hard to find them in any other text on enumerative combinatorics as well, although possibly with more restrictions on some of the variables.) We will use a minor variant of Theorem 6: Lemma 7. Let $$q, k \\in \\mathbb{N}$$. Then, \\begin{align*} \\dbinom{q}{k} \\sum_{p=0}^q \\dbinom{q-k}{p} = \\dbinom{q}{k} 2^{q-k}. \\end{align*} Proof of Lemma 7. If $$q < k$$, then $$\\dbinom{q}{k} = 0$$ (since $$q \\in \\mathbb{N}$$), and therefore both sides of the equality in question equal $$0$$ (since they both contain a factor of $$\\dbinom{q}{k}$$). Thus, for the rest of this proof, we WLOG assume that we don't have $$q < k$$. Hence, $$q \\geq k$$ and therefore $$q - k \\in \\mathbb{N}$$. Therefore, we have $$\\dbinom{q-k}{p} = 0$$ whenever $$p > q-k$$. Thus, $$\\sum_{p=0}^q \\dbinom{q-k}{p} \\mathrel{\\overset{0}{=}} \\sum_{p=0}^{q-k} \\dbinom{q-k}{p} =", "3^2 - 2^3 + 4^1 = 3^2 \\times 3 - 2 \\times 3\\\\ &&& 4\\times 4 + 0 + 9 - 8 + 4 = 9 \\times 3 - 2 \\times 3\\\\ &&& 16 + 0 + 9 - 8 + 4 = 27 - 6\\\\ &&& 21 = 21\\end{align*} #### Earlier Problem Revisited \\begin{align*}6 \\times 3^2 \\div 2 + z + 4(7 - 3) + 2^3 \\div 4 = 3^2 \\times 6 + 1\\end{align*} We can use the problem solving steps to help us with the order of operations. \\begin{align*}& \\mathbf{Describe:} && \\text{The equation has parentheses, exponents, multiplication, division, and addition.}\\\\ &&& z \\ \\text{is the variable.}\\\\ \\\\ & \\mathbf{My \\ Job:} && \\text{Apply the order of operations rule to figure out the value of}\\ z.\\\\ \\\\ & \\mathbf{Solve:} && 6 \\times 3^2 \\div 2 + z + 4(7 - 3) + 2^3 \\div 4 = 3^2 \\times 6 + 1\\\\ &&& \\mathbf{Parentheses} \\qquad \\qquad \\quad 6 \\times 3^2 \\div 2 + z + 4 \\times 4 + 2^3 \\div 4 = 3^2 \\times 6 + 1\\\\ &&& \\mathbf{Exponents} \\qquad \\qquad \\quad \\ 6 \\times 9 \\div 2 + z + 4 \\times 4 + 8 \\div 4 = 9 \\times 6 + 1\\\\ &&& \\mathbf{Multiplication/} \\qquad \\quad 27 + z + 16 + 2 = 54", "and the bottom of the fraction $\\dfrac{16!}{13!\\cdot 8!} = \\dfrac{16 \\times 15 \\times 14}{8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2}.$ This can be canceled further if we factorize the numerator $\\dfrac{16!}{13!\\cdot 8!} = \\dfrac{8 \\times 2 \\times 5 \\times 3 \\times 2 \\times 7}{8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2}.$ Then finally we get \\begin{align} \\dfrac{16!}{13! \\cdot 8!} &= \\dfrac{1}{4 \\times 3},\\\\ &=\\dfrac{1}{12}. \\end{align} ###### Example 4 Simplify $\\dfrac{(n-1)!}{(n+1)!}.$ ###### Solution \\begin{align} \\dfrac{(n-1)!}{(n+1)!} &= \\dfrac{(n-1) \\times (n-2) \\times \\dots \\times 2 \\times 1}{ (n+1) \\times n \\times (n-1) \\times \\dots \\times 2 \\times 1},\\\\ &= \\dfrac{1}{(n+1) \\times n},\\\\ &= \\dfrac{1}{n^2 + n}. \\end{align}", "= \\dfrac{1}{n+1}\\dbinom{2n}{n} \\\\ \\hline \\end{array}$$ $$\\begin{array}{rcll} \\mathbf{C_5} &=& \\dfrac{1}{5+1}\\dbinom{2\\cdot 5}{5} \\\\ &=& \\dfrac{1}{6}\\dbinom{10}{5} \\\\ &=& \\dfrac{1}{6}\\cdot \\dfrac{10}{5} \\cdot \\dfrac{9}{4}\\cdot \\dfrac{8}{3}\\cdot \\dfrac{7}{2}\\cdot \\dfrac{6}{1} \\\\ &=& 2\\cdot 3\\cdot 7 \\\\ &\\mathbf{=}& \\mathbf{42} \\\\ \\end{array}$$ A convex heptagon can be divided into five triangles in 42 ways (with no lines crossing). heureka Jan 22, 2018 edited by heureka Jan 22, 2018 ### 19 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details", "\\text{5}^{(2y-3) - (-5y+5)} \\times \\text{2}^{(4y+4) - (-5y+5)}\\\\ &= \\text{5}^{7y -\\text{8}} \\times \\text{2}^{9y-\\text{1}} \\end{align*} $$\\dfrac{6^4 \\times 12^3 \\times 4^5}{30^3 \\times 3^6}$$ \\begin{align*} \\frac{6^4 \\times 12^3 \\times 4^5}{30^3 \\times 3^6} & = \\frac{\\left(3^{4} \\times 2^{4}\\right) \\times \\left(3^{3} \\times 4^{3}\\right) \\times 4^{5}}{\\left(3^{3} \\times 10^{3}\\right) \\times 3^{6}} \\\\ & = \\frac{3^{4} \\times 2^{4} \\times 3^{3} \\times 2^{6} \\times 2^{10}}{3^{3} \\times 2^{3} \\times 5^{3} \\times 3^{6}} \\\\ & 3^{4 + 3 - 3 - 6} \\cdot 2^{4 + 6 + 10 - 3} \\cdot 5^{-3} \\\\ & = \\frac{2^{17}}{3^{2}5^{3}} \\end{align*} $$\\dfrac{9^3\\times 20^2}{4\\times 5^2 \\times 3^5}$$ \\begin{align*} \\frac{9^3\\times 20^2}{4\\times 5^2 \\times 3^5} & = \\frac{3^6 4^2 5^2}{4 \\times 5^2 3^5} \\\\ & = \\frac{3^6 2^4 5^2}{2^2 5^2 3^5} \\\\ & = 3^{6 - 5} 2^{4 - 2} 5^{2 - 2} \\\\ & = 3 \\times 2^2 \\\\ & = 12 \\end{align*} $$\\dfrac{7^{b} + 7^{b - 2}}{4 \\times 7^{b} + 3 \\times 7^{b}}$$ \\begin{align*} \\frac{7^{b} + 7^{b - 2}}{4 \\times 7^{b} - 3 \\times 7^{b}} &= \\frac{7^{b}(1 + 7^{-2})}{7^{b}(4 - 3)} \\\\ &= \\frac{(1 + 7^{-2})}{1} \\\\ &= \\frac{1 + \\frac{1}{49}}{1} \\\\ & = \\frac{51}{49} \\end{align*} $$\\dfrac{12^{y} - 96^{y}}{3^{y} + 6^{y}}$$ \\begin{align*} \\frac{12^{y} - 96^{y}}{3^{y} + 6^{y}} & = \\frac{(4 \\cdot 3)^{y} - \\left(2^{5} \\cdot 3\\right)^{y}}{3^{y} + \\left(2 \\cdot 3\\right)^{y}} \\\\ & = \\frac{3^{y}\\left(4^{y} - 2^{5y}\\right)}{3^{y}\\left(2 + 1\\right)} \\\\ &= \\frac{4^{y} - 2^{5y}}{3} \\end{align*}", "this, we get \\begin{aligned}\\dfrac{1}{6}-\\dfrac{4}{9} &= \\dfrac{1\\times 3}{18}+\\dfrac{4\\times 2}{18} \\\\ &=\\dfrac{3}{18} - \\dfrac{8}{18} \\\\ &=-\\dfrac{5}{18}\\end{aligned} #### Is this a topic you struggle with? Get help now. Writing $9$ as $\\frac{9}{1}$, the calculation becomes $\\dfrac{7}{4}-\\dfrac{9}{1}$ Using $1\\times 4=4$ as a common denominator, we get \\begin{aligned}\\dfrac{7}{4}-\\dfrac{9}{1} &= \\dfrac{7}{4}-\\dfrac{9\\times 4}{4} \\\\ &=\\dfrac{7}{4} - \\dfrac{36}{4} \\\\ &=-\\dfrac{29}{4}\\end{aligned} ### Learning resources you may be interested in We have a range of learning resources to compliment our website content perfectly. Check them out below.", "n-2 \\right)...(n-r)!$ \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5!}\\times \\left( \\dfrac{8\\times 7\\times 6}{3!} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6!}\\times \\left( \\dfrac{8\\times 7}{2!} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4!}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4!} \\right) \\\\ \\end{align} \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6}{3\\times 2\\times 1} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6\\times 5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7}{2\\times 1} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4\\times 3\\times 2\\times 1} \\right) \\\\ \\end{align} By cancelling the common terms in denominator and numerator we get $=\\left( 11\\times 9\\times 8 \\right)\\times \\left( 8\\times 7 \\right)+\\left( 11\\times 3\\times 4\\times 7 \\right)\\times \\left( 4\\times 7 \\right)+\\left( 11\\times 5\\times 9 \\right)\\times \\left( 5\\times 2\\times 7 \\right)$ $=792\\times 56+924\\times 28+495\\times 70$ = 44352 + 25872 + 34650 = 104874 Thus, the number of ways of selecting 10 students chosen for the competition = 104874 Note: The possibility of mistake can be the calculation mistake since the procedure of solving involves large calculations. Hence be very careful in simplifying and cancelling the terms to get the desired result.", "& h = 4 \\\\ \\hline A_1 & 1 & 1 & 1 & 1 & z, x^2, y^2, z^2 \\\\ A_2 & 1 & 1 & -1 & -1 & xy, R_z \\\\ B_1 & 1 & -1 & 1 & -1 & x, xz, R_y \\\\ B_2 & 1 & -1 & -1 & 1 & y, yz, R_x \\\\ \\hline \\end{array}$ As before, we use Equation (15.20) to find out the number of times each irreducible representation appears. $a_k = \\dfrac{1}{h}\\sum_C n_C \\chi(g) \\chi_k(g) \\label{23.4}$ We have $\\begin{array}{rcll} a(A_1) & = & \\dfrac{1}{4}(1 \\times 6 \\times 1 + 1 \\times 0 \\times 1 + 1\\times 4\\times 1 + 1\\times 2\\times 1) & = 3 \\\\ a(A_2) & = & \\dfrac{1}{4}(1\\times 6\\times 1 + 1\\times 0\\times 1 + 1\\times 4\\times -1 + 1\\times 2\\times -1) & = 0 \\\\ a(B_1) & = & \\dfrac{1}{4}(1\\times 6\\times 1 + 1\\times 0\\times -1 + 1\\times 4\\times 1 + 1\\times 2\\times -1) & = 2 \\\\ a(B_2) & = & \\dfrac{1}{4}(1\\times 6\\times 1 + 1\\times 0\\times -1 + 1\\times 4\\times -1 + 1\\times 2\\times 1) & = 1 \\end{array} \\label{23.5}$ so the basis spans $$3A_1 + 2B_1 + B_2$$. Now we use the projection operators applied to each basis function $$f_i$$ in turn to determine the SALCs $$\\phi_i = \\Sigma_g" ]

[ "the nearest tenth. More Similar Questions", "getting closer and closer to $3$, so it looks like $x=3$. If we got to $3$ in our sequence, then the next number in the sequence would be $\\sqrt{3+2\\times 3}$, which is $\\sqrt{3+6}=\\sqrt{9}=3$. So if the sequence ever gets to $3$, it will stay at $3$ forever, which strongly suggests that $x=3$. You can find more short problems, arranged by curriculum topic, in our short problems collection.", "# Digit Supreme! A number is called digit supreme if it is divisible by some natural number $$n$$, and it's repeated digit sum is $$n$$. Find the percentage of numbers below 100, which are digit supreme numbers. Give answer to two decimal places. Repeated digit sum is the continuous application of digits sums, until a constant number is reached. ×", "closer to $3$, so it looks like $x=3$. If we got to $3$ in our sequence, then the next number in the sequence would be $\\sqrt{3+2\\times 3}$, which is $\\sqrt{3+6}=\\sqrt{9}=3$. So if the sequence ever gets to $3$, it will stay at $3$ forever, which strongly suggests that $x=3$. You can find more short problems, arranged by curriculum topic, in our short problems collection.", "1, so you repeat maybe $$O(1)$$ times).", "digits $a$ can be anything except $0$ (since they include the first digit), and the last digit can be anything except $a$, so you get $3*9*9$. Then your answer is $9*9+3*9*9=4*9*9$. • Thank you for explaining. This one is very close to my original approach. – Snailwalker Oct 3 '17 at 3:43", "tenths digit in decimal representation of x [#permalink] Show Tags 12 Feb 2017, 07:00 HKD1710 wrote: what is the tenths digit in decimal representation of $$x$$? (1) $$x$$ is greater than $$\\frac{3}{8}$$ (2) $$x$$ is less than $$\\frac{3}{7}$$ This is a copy of the following OG question: https://gmatclub.com/forum/what-is-the- ... 39476.html _________________ Re: what is the tenths digit in decimal representation of x [#permalink] 12 Feb 2017, 07:00 Display posts from previous: Sort by", "consecutive equal digits equals $9^n$.", "I would want to use 10^10^10 as x? I usually just use 999 or anything random. Yeah, Let $x>y$ Then obviously $f(x)$ is going to be a better approximation than $f(y)$ because $x$ is closer to $\\infty$ than $y$ 13. Ok that is good to know.", "don't allow decimals beginning with $0$ we can just multiply by $(9/10)^2$. There are two places that we might not allow a $0$, the first digit and the digit immediately following $E$. Although perhaps $E0$ should be allowed and not, say, $E01$? But in any case, the magnitude of the answer doesn't depend on this condition. Nov 3 '17 at 16:42", "problem tells you you don't need to go past $x^4$.", "down\" the next digit of $p$. That is, set $s$ to be $(s - cq)$ concatenated with the next digit of $p$. 4. If $s < q$, then it's your answer, stop. Else, go back to step $3$ and repeat. Here's an image from Wikipedia, for the example of $q = 37$. -", "can't begin with $0$. For the follow up question, find the number of $8$ digit numbers with no repeated digits. (i.e. look at the complement).", "100 is 10, hence, we have: $9 < \\sqrt {95} < 10$ Hence, $\\sqrt {95}$ lies between the consecutive numbers 9 and 10. Hence, the correct answer is option (c). Note: You can also find the value of $\\sqrt {95}$ by using long division method to find the value of $\\sqrt {95}$ up to one decimal place and find the two consecutive numbers between which $\\sqrt {95}$ lies.", "for any confusion I may have caused you, but what I meant by that phrase is that the answer to the question is suppose to be 24. Jul 2 '19 at 16:32 • No problem. It would be nice of you to reward the question and the title to reflect what you truly mean for future users. Thanks. Jul 2 '19 at 20:41 I assume you want all the pairs $$(y,x)$$ such that $$y$$ is 5 digits, $$x$$ is 4 digits, $$y-x=11111$$ and every digit from $$1$$ to $$9$$ appears either in $$y$$ nor $$x$$. This implies that there is no digit $$0$$ in $$y$$ nor in $$x$$, so the first digit of $$y$$ must be $$1$$. Write $$x=abcd$$ in decimal notation. Then $$y=1(a+1)(b+1)(c+1)(d+1)$$ in decimal notation. Note that there is no digit $$0$$ so the digit $$9$$ is not in $$x$$ and there is no carry-over. Moreover, note that the difference between two digits of $$x$$ must be at least $$2$$ because if it was $$1$$, then the same digit would be in $$y$$ and $$x$$ and this is impossible. This implies that all the even digits are in $$x$$. The even digits are $$\\left\\lbrace 2,4,6,8\\right\\rbrace$$ and the number of arrengements of this is $$24$$. • Thank you for this answer, if it won't be any trouble", "is x? 17. Unknown number Identify unknown number which 1/5 is 40 greater than one tenth of that number.", "of $(1 + \\sqrt 2)$ seem to be getting closer and closer to integers. • In particular, they seem to alternate between being just under an integer (for even powers) and just over an integer (for odd powers). If this is true, the decimal expansion of $(1 + \\sqrt 2)^{500}$ must be of the form $n.99999999\\dots$ for some big integer $n$ and some number of $9$s after the decimal point. And it seems reasonable that if Colin is posing this question, it must have more than 99 nines, which means the answer would be 9. But why does this happen? Do the powers really keep alternating being just over and under an integer? And how close do they get—how do we know for sure that $(1 + \\sqrt 2)^{500}$ is close enough to an integer that the 99th digit will be a 9? This is what I want to explore in a series of future posts—and as should come as no surprise it will take us on a tour of some fascinating mathematics!", "# Getting Easy $11! - 1 = x$ What is the unit digit of $x$? Note : $!$ means factorial. For example, $3! = 1 \\times 2\\times 3 = 6$ ×", "## Thinking Mathematically (6th Edition) The decimals either $\\underline{\\text{repeat}}$ or $\\underline{\\text{terminate}}$. Decimals either repeat or terminate if an equivalent fraction can be expressed in the form $a/b$, where both a and b are integers, and b does not equal 0. Thus, we know that repeating or terminating decimals are rational numbers.", "have (say) one occurrence of $X + Y = 4$ in several thousand realizations." ]

[ "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910} \\$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys.", "\\;(\\text{-}10)\\,\\frac{1,\\!000,\\!000,\\!001}{11} \\;=\\;(\\text{-}10)(90,\\!909,\\!091) \\;=\\;\\boxed{-909,\\!090,\\!910}$ 4. Yes, that's what I meant...hard to use that program so I had to copy/paste. Thanks guys.", "of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$ Together, the answer is $3(12+12+12)-12=\\boxed{\\textbf{(A) } 96}.$ Solution 4 Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \\equiv 1\\pmod{3},$ and thus we need to count any $2$-digit number $\\equiv 2\\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \\boxed{\\textbf{(A) } 96}.$ Solution 5 We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\\cdot\\frac{1}{3}=150.$ Of these $150$ numbers, $\\frac{4}{5}$ $\\textbf{do not}$ have $3$ in their ones (units) digit, $\\frac{9}{10}$ $\\textbf{do not}$ have $3$ in their tens digit, and $\\frac{8}{9}$ $\\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$900\\cdot\\frac{1}{2}\\cdot\\frac{1}{3}\\cdot\\frac{4}{5}\\cdot\\frac{9}{10}\\cdot\\frac{8}{9}=\\boxed{\\textbf{(A) } 96}.$$ ~mathpro12345 Solution 6 We will start with the numbers that could work. This numbers include _ _ $1$, _ _ $5$, _ _ $7$, _", "= 9$, you have $4\\cdot8 = 32$ positions without revealing the correct information. You can do even better, if you choose to start with $b$. Then you can extend your number of settings by moving $(a,b)$, $(b,c)$, $(a,b,c)$, $(b,c,d)$, and $(a,b,c,d)$ supplying $(1 + 2 + 2 + 1)\\cdot8 = 48$ settings in total. (Of course the order you choose does not matter.) You would get the same opportunity with $c$ instead of $b$. $d$ however, like $a$, would supply only 32 possible positions. The maximum number of different positions, before the thief has discoverd the correct one, is for $n > 2$ digits and an even number $m$ of dials: $$(n-2)\\cdot\\frac{m}{2}\\cdot\\frac{m+2}{2}.$$ For an odd number $m$ of dials you get $$(n-2)\\cdot(\\frac{m+1}{2})^2.$$ If no single dial is moved, we have only 48 settings: 3 pairs, 2 triples and 1 quadruple in 8 positions each. But if pair $(a,b)$ has been moved twice, pair $(c,d)$ cannot be moved without revealing the secret. Hence, we get only 40 settings. That is less than the constructed 48. So, in order to maximize the number of the secret-maintaining settings, we have to move also at least one single dial. But having moved it twice, we can no longer move any other single dial or the pair not containing the first. This subtracts", "refer to the symbols \"0\", \"1\", \"2\", \"3\", \"4\", \"5\", \"6\", \"7\", \"8\", and \"9\". There are a total of ten digits. The problem told us that we cannot have the first digit of the code be zero -- this means there are 9 total choices for the first digit of the lock box code. For the second and third digits of the lock box code, we are not told a restriction, and so there are 10 choices for each. For the fourth digit of the code, we may pick any odd digit: there are 5 total choices here. This means there are a total of $$9 \\cdot 10 \\cdot 10 \\cdot 5=4500$$ choices for a code.", "digits of $$\\left(\\frac{10^{1992}-1}{9}\\right)^2 = \\frac{1}{81}(10^{3984} - 2\\cdot 10^{1992} +1)$$ To write this out as an integer, note that $$\\frac{1}{81} = \\frac{12,345,679}{999,999,999} \\quad \\text{ and thus} \\quad \\frac{2}{81} = \\frac{24,691,358}{999,999,999}$$ This tells us the repeating 9-digit patterns in $\\frac{10^{3984} }{81}$ and $\\frac{2\\cdot 10^{1992}}{81}$. To take the difference, note that the two leading terms are $1$'s in the $10^{3982}$ place and in the $10^{1990}$ place, respectively. The first subtraction then occurs at the latter digit, which corresponds to the digit $4$ in the expansion of $\\frac{10^{3984} }{81}$, as $3982-3 = 3980 \\equiv 1990 \\mod 9$. The difference therefore has the repeating pattern unchanged from digits in the $10^{1991}$ place to the $10^{3982}$ place, and for lesser digits, the repeating pattern is now $$456,790,123-246,913,580 = 209,876,543$$ This pattern repeats indefinitely, until the decimal point, at which point we add $1/81$ with the effect of rounding $...320.\\overline{976543210}$ up to $...321$. Note that there are $221$ nine-digit sequences in each pattern, in addition to the leading three digits and the trailing two digits. The sum of the digits is then $1+2+3+221(44) + 221(37) + 2 + 1 = 9 + 221\\cdot 81 = 17910$. • ah, beat me to it as I was typing it out. This is the correct answer – qwr Apr 17 '15 at 1:30 This is not particularly pretty", "Question # Consider all four-digit numbers that can be created from the digits 0-9 where the first and last digits must be even and no digit can repeat. Probability and combinatorics Consider all four-digit numbers that can be created from the digits 0-9 where the first and last digits must be even and no digit can repeat. Assume that numbers can start with 0. What is the probability of choosing a random number that starts with 8 from this group? Enter a fraction or round your answer to 4 decimal places, if necessary. 2021-09-05 The first and last digits should be even. Hence, Total number of numbers possible $$\\displaystyle={5}\\cdot{4}\\cdot{8}\\cdot{7}={1120}$$ Total number that starts with $$\\displaystyle{2}={4}\\cdot{8}\\cdot{7}={224}$$ Hence, P(Stars with 2) $$\\displaystyle=\\frac{224}{{1120}}=\\frac{1}{{5}}$$", "\\right\\}$ there $9^6-1$ numbers in $A_1$, we subtract the 1 to account for 000000000. 7. Hello, shaoen01! Plato's approach to #2 is correct . . . I would further assume that an integer does not begin with zero. 2) How many integers from 1 through 999999 contain each of the digits 1,2 and 3 at least once? The solution asked me to use the inclusion-exclusion method, so how do i know when to use this? It is used when counting the items which are not in the set is easier than counting the items that are in the set. There are 999,999 integers. How many do not contain at least {1,2,3} ? Let $n(i')$ = number of integers that do not contain an $i.$ We want: $n(1' \\,\\cup \\,2' \\,\\cup \\,3') \\;=\\;n(1') \\,+\\, n(2') \\,+ \\,n(3') \\,-\\, n(1' \\,\\cap \\,2') \\,- \\,n(2' \\,\\cap \\,3') \\,- \\,n(1' \\,\\cap \\,3') \\,+$ $n(1' \\cap 2' \\cap 3')$ Consider $n(1')$ 1-digit numbers: 8 choices . (It must not be 0 or 1.) 2-digit numbers: $8\\cdot9$ choices. 3-digit numbers: $8\\cdot9^2$ choices. 4-digit numbers: $8\\cdot9^3$ choices. 5-digit numbers: $8\\cdot9^4$ choices. 6-digit numbers: $8\\cdot9^5$ choices. There are: . $8(1 + 9 + 9^2 + \\hdots + 9^5) \\;=\\;8\\,\\frac{9^5-1}{9-1} \\;=\\;59,048$ such numbers. The same reasoning holds for $n(2')\\text{ and }n(3')$ . . Hence: . $n(1') \\:=\\:n(2')", "$\\left\\lfloor \\frac{2001}{3\\cdot4\\cdot5} \\right\\rfloor$ yields the numbers $133$, $100$, and $33$. The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$. The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$. Subtracting this number from the original $1001$ numbers procures $\\boxed{\\textbf{(B)}\\ 801}$. ### Solution 3 First find the number of such integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of $3$. There are $\\frac45\\cdot2000=1600$ numbers that are not multiples of $5$. $\\frac23\\cdot\\frac34\\cdot1600=800$ are not multiples of $3$ or $4$, so $800$ numbers are. $800+1=\\boxed{\\textbf{(B)}\\ 801}$ ### Solution 4 Take a good-sized sample of consecutive integers; for example, the first $25$ positive integers. Determine that the numbers $3, 4, 6, 8, 9, 12, 16, 18, 21,$ and $24$ exhibit the properties given in the question. $25$ is a divisor of $2000$, so there are $\\frac{10}{25}\\cdot2000=800$ numbers satisfying the given conditions between $1$ and $2000$. Since $2001$ is a multiple of $3$, add $1$ to $800$ to get $800+1=\\boxed{\\textbf{(B)}\\ 801}$. ~ mathmagical ### Solution 5 By PIE, there are $1001$ numbers that are multiples of $3$ or $4$ and less than or equal to $2001$.", "I could understand it! Thank you very much for your guidance! – Andrej Hatzi Feb 11 at 11:12 • @AndrejHatzi: I preferred to let you think rather than giving a straight answer. – Yves Daoust Feb 11 at 11:22 You shall only choose the first and last digit from the two middle rows. That's because the bottom row contains only one number and for the top row, there's no row above to choose the second and third digit. Let's start by choosing two digits from the $$3$$rd row. We can choose and rearrange two items out of three in $$P(3,2)=\\frac{3!}{1!}=6$$ ways. Following the exact same reasoning, for the second and third digit, we have: $$P(3,2)=\\frac{3!}{1!}=6$$ ways for the middle digits. Thus, $$6$$ ways to choose first and last digits and $$6$$ ways to choose second and third. That's a total of $$36$$ ways to choose all $$4$$ digits. Also, we could repeat all these actions, choosing the first and last from the $$2$$nd row and the middle ones from the $$1$$st one. Again, that's a total of $$36$$ ways to choose all four digits. At last, since we can carry out this procedure either the first $$36$$ ways or the other $$36$$ ways, the total ways are $$36+36=72.$$", "0 and 1 repeated as well so we have 3*12=36 (not 18 ). For 4), I became interested in this type of thing because at my place of employment, in order to log in remotely, we have to enter a code that is chosen at random, 6 each of the digits 0-9. I noticed that it was relatively rare to have no repeated digits so I tried to figure out the odds of getting exactly two repeating digits (among other things). I wrote programs in Ruby (which I will gladly share but I doubt anyone uses Ruby here) to run random numbers and looking at the percentage, and also to calculate exactly by running through all of the combinations and looking at the percentage. I am pretty sure that I used reasoning similar to that given earlier in the thread but the calculated results did not match the programs' outputs until I used the methods given above. Caveat: My 'exactly two matching numbers' routine may be flawed but for base 3 with 4 symbols taken, it does give 36, in agreement with the result above. For base 10 and 6 numbers taken it gives 453600 which agrees with: $\\displaystyle \\large 10 \\cdot 1 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot \\frac{6!}{2! \\cdot 4!}$ My random number", "of combinations of 3 digits from the 7 remaining. There are $\\displaystyle {7\\choose{3}} = 35$ combinations. That means the probability of having a number less than 4000 and greater than 3000 is $\\displaystyle \\frac{35}{70} = \\frac{1}{2}$. 3. Hello, johnsy123! $\\text{1)A bag contains five white, six red and seven blue balls.}$ $\\text{If three balls are selected at random, without replacement,}$ $\\text{find the probability they are all different colors.}$ There are: . ${18\\choose3} \\:=\\:816$ possible outcomes. There are: . ${5\\choose1}{6\\choose1}{7\\choose1} \\:=\\:5\\cdot6\\cdot7 \\:=\\:210$ ways to get one of each color. Therefore: . $P(\\text{all different colors}) \\;=\\;\\dfrac{210}{816} \\;=\\;\\dfrac{35}{136}$ $\\text{2) A four-digit number (no repetitions) is to be formed}$ $\\text{from the set of numbers (0,1,2,3,4,5,6,7).}$ Assuming that the number may not have a leading zero, . . there are: . $7\\cdot7\\cdot6\\cdot5 \\:=\\:1470$ possible 4-digit numbers. $\\text{(a) Find the probability that the number is even.}$ There are two cases to consider: . . The last digit is 0. . . The last digit is 2, 4 or 6. The last digit is 0: . $\\square\\:\\square\\;\\square\\;0$ . . The other 3 spaces can be filled in: . $7\\cdot6\\cdot5 \\:=\\:210$ ways. The last digit is 2, 4 or 6: .3 choices. The first must not be 0: .6 choices. The other 2 spaces can be filled in $6\\cdot5\\,=\\,30$ ways. . . There are: . $3\\cdot6\\cdot30 \\:=\\:540$", "including the digit zero, which can be placed anywhere in the remaining positions. This choice can be done in $$P(6,3) = \\frac{6!}{(6-3)!} = \\frac{6!}{3!}=6 \\cdot5 \\cdot4$$, different ways. Finally, with distinct digits, there is $$6 \\cdot6 \\cdot5 \\cdot4 =720$$ four-digit numbers to be constructed. We know that the amount of odd numbers plus the amount of even numbers equal the total amount of the 720 four-digit numbers. The odd numbers are 1, 3 and 5. The question to be asked is how many of the 720 are odd? The digit to fill the unit position can only be chosen from the digits 1, 3 or 5, and this choice can be made in $$P(3,1)=\\frac{3!}{(3-1)!}=\\frac{3!}{2!}=3$$, different ways. To choose the digit filling the thousand position, we have five valid digits to choose from, since the zero digit is not valid for this position. The choice can be made in $$P(5,1)=\\frac{5!}{(5-1)!}=\\frac{5!}{4!}=5$$, different ways. Now we are left with two positions to fill, the hundred and tenth position, and five digits to choose from, now including the zero digit. The choice for these two positions can be made in $$P(5,2)=\\frac{5!}{(5-2)!}=\\frac{5!}{3!}=5 \\cdot4=20$$, different ways. The total number of odd four-digit numbers, with distinct digits are $$5 \\cdot5 \\cdot4 \\cdot3 =300$$. Now, we can answer the question how many of these 720, four-digit", "\\:=\\:n(3') \\:=\\:59,048$ Consider $n(1' \\cap 2')$ 1-digit numbers: 7 choices. .(It must not be 0, 1, or 2.) 2-digit numbers: $7\\cdot8$ choices. 3-digit numbers: $7\\cdot8^2$ choices. 4-digit numbers: $7\\cdot8^3$ choices. 5-digit numbers: $7\\cdot8^4$ choices. 6-digit numbers: $7\\cdot8^5$ choices. There are: . $7(1 + 8 + 8^2 +\\hdots+8^5) \\:=\\:7\\,\\frac{8^5-1}{8-1} \\:=\\:32,767$ such numbers. The same reasoning holds for $n(2'\\cap3')\\text{ and }n(1'\\cap3')$ . . Hence: . $n(1'\\cap2') \\:=\\:n(2' \\cap 3') \\:=\\:n(1' \\cap 3') \\:=\\:32,767$ Consider $n(1' \\cap 2' \\cap 3')$ 1-digit numbers: 6 choices. .(It must not be 0, 1, 2, or 3.) 2-digit numbers: $6\\cdot7$ choices. 3-digit numbers: $6\\cdot7^2$ choices. 4-digit numbers: $6\\cdot7^3$ choices. 5-digit numbers: $6\\cdot7^4$ choices. 6-digit numbers: $6\\cdot7^5$ choices. There are: . $6(1 + 7 + 7^2 + \\hdots + 7^5) \\:=\\:6\\,\\frac{7^5-1}{7-1} \\:=\\:16,806$ such numbers. . . Hence: . $n(1' \\cap 2' \\cap 3') \\:=\\:16,806$ Then: . $n(1' \\cup 2' \\cup 3') \\:=\\:3(59,048) - 3(32,767) + 16,806 \\:=\\:95,649$ Therefore: . $n(1 \\cap 2\\cap 3) \\;=\\;999,999 - 95,649 \\;=\\;\\boxed{904,350}$ But check my work and my reasoning . . . please! . 8. Originally Posted by Soroban I would further assume that an integer does not begin with zero. But check my work and my reasoning . . . please! But in the case \"assume that an integer does not begin with zero\", how would account for 234? That", "since there are three sets of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$ Together, the answer is $3(12+12+12)-12=\\boxed{\\textbf{(A) } 96}.$ ## Solution 4 Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \\equiv 1\\pmod{3},$ and thus we need to count any $2$-digit number $\\equiv 2\\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \\boxed{\\textbf{(A) } 96}.$ ## Solution 5 We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\\cdot\\frac{1}{3}=150.$ Of these $150$ numbers, $\\frac{4}{5}$ $\\textbf{do not}$ have $3$ in their ones (units) digit, $\\frac{9}{10}$ $\\textbf{do not}$ have $3$ in their tens digit, and $\\frac{8}{9}$ $\\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$900\\cdot\\frac{1}{2}\\cdot\\frac{1}{3}\\cdot\\frac{4}{5}\\cdot\\frac{9}{10}\\cdot\\frac{8}{9}=\\boxed{\\textbf{(A) } 96}.$$ ~mathpro12345 ~ pi_is_3.14", "5 choices. . . .The 4th digit has 4 choices. There are: . $4\\cdot6\\cdot5\\cdot4 \\:=\\:{\\color{blue}480}$ other numbers greater than 3400. Therefore, there are: . $80 + 480 \\:=\\:\\boxed{560}$ numbers greater than 3400. Quote: 2) How many numbers of different digits lying between 100 and 1000 can be formed with the digits {0,2,3,4,8,9}? These are all 3-digit numbers. The first digit can be: {2,3,4,8,9} . . . 5 choices. The second digit has 5 choices. The third digit has 4 choices. Therefore, there are: . $5\\cdot5\\cdot4 \\:=\\:\\boxed{100}$ such numbers. • October 23rd 2009, 10:12 PM saberteeth Cool, thanks :) I got it.", "are $45$ such combinations with $2$ distinct digits and both digits repeated. There are $360$ such combinations with $3$ distinct digits ($120$ ways to pick the digits, $3$ ways to pick the one that repeats). Thus, there are $$1\\cdot 10+4\\cdot 90+6\\cdot 45+12\\cdot 360=4960$$ combinations with at least one repeat, leaving $5040$ combinations with no repeats, and so there are $$\\frac{5040}{24}=210$$ combinations in non-decreasing order with no repeats. In total, there are thus $10+90+45+360+210=715$ non-decreasing combinations that we need to try, and we can be sure to get it open by the end of that list. As for strategy, let's suppose that each of the $10000$ possible combinations is equally likely to be correct. Then we have \\begin{align}\\text{P(four distinct digits)} &= \\frac{24\\cdot 210}{10000}=50.4\\%\\\\\\text{P(three distinct digits)} &= \\frac{12\\cdot 360}{10000}=43.2\\%\\\\\\text{P(two distinct digits, both repeated)} &= \\frac{6\\cdot 45}{10000}=2.7\\%\\\\\\text{P(two distinct digits, only one repeated)} &= \\frac{4\\cdot 90}{10000}=3.6\\%\\\\\\text{P(single digit repeated four times)} &= \\frac{1\\cdot 10}{10000}=0.1\\%.\\end{align} Guessing a non-decreasing combination with 4 distinct digits is effectively worth $24$ guesses, and so on. Our expected value for a given kind of combinations--that is, how efficient a given nondecreasing combination will be as a guess--would then be \\begin{align}\\text{E(four distinct digits)} &= 24\\cdot50.4\\% = 12.096\\\\\\text{E(three distinct digits)} &= 12\\cdot43.2\\% = 5.184\\\\\\text{E(two distinct digits, both repeated)} &= 6\\cdot2.7\\% = 0.162\\\\\\text{E(two distinct digits, only one repeated)} &= 4\\cdot3.6\\% = 0.144\\\\\\text{E(single", "may be either a letter or a digit. 4) Repetition is allowed. 5) They are not case sensitive. The first must be a letter: 26 choices. The last must be a digit: 10 choices. The others must be a letter or a digit: 36 choices. $\\begin{array}{c|c} \\text{Length} & \\text{Choices} \\\\ \\hline 2 & 26\\cdot10 \\\\ 3 & 26\\cdot36\\cdot10 \\\\ 4 & 26\\cdot36\\cdot36\\cdot10 \\\\ 5 & 26\\cdot36\\cdot36\\cdot36\\cdot10 \\\\ 6 & 26\\cdot36\\cdot36\\cdot36\\cdot36\\cdot10 \\end{array}$ Total: . $260 + 260\\cdot36 + 260\\cdot36^2 + 260\\cdot36^3 + 260\\cdot36^4$ . . $= \\;260\\left(1 + 36 + 36^2 + 36^3 + 36^4\\right) \\;=\\;260\\,\\frac{36^5-1}{36-1}$ . . $=\\;260\\,(1,\\!727,\\!605) \\;=\\;\\boxed{449,\\!177,\\!300}$ • Nov 2nd 2008, 07:13 PM ezwind72 Quote: Originally Posted by Soroban Hello, ezwind72! These are not combinations! With a password, the order of the characters is important. The first must be a letter: 26 choices. The last must be a digit: 10 choices. The others must be a letter or a digit: 36 choices. $\\begin{array}{c|c} \\text{Length} & \\text{Choices} \\\\ \\hline 2 & 26\\cdot10 \\\\ 3 & 26\\cdot36\\cdot10 \\\\ 4 & 26\\cdot36\\cdot36\\cdot10 \\\\ 5 & 26\\cdot36\\cdot36\\cdot36\\cdot10 \\\\ 6 & 26\\cdot36\\cdot36\\cdot36\\cdot36\\cdot10 \\end{array}$ Total: . $260 + 260\\cdot36 + 260\\cdot36^2 + 260\\cdot36^3 + 260\\cdot36^4$ . . $= \\;260\\left(1 + 36 + 36^2 + 36^3 + 36^4\\right) \\;=\\;260\\,\\frac{36^5-1}{36-1}$ . . $=\\;260\\,(1,\\!727,\\!605) \\;=\\;\\boxed{449,\\!177,\\!300}$ I understand where you are coming from with the first", "= 1 - (5^4)/9000. – fleablood Sep 3 '18 at 0:03 As you say, if there is at least one even digit the product will be even, so count how many four digit numbers have all odd digits. There are five odd digits, so there are $5^4=625$ of them. The fraction of even products is then $\\frac {9000-625}{9000}$", "Hence, there are: . $2\\!\\cdot\\!30 \\:=\\:60$ numbers between 300 and 500. Therefore: . $P(\\text{between 300 and 500}) \\;=\\;\\frac{60}{210} \\;=\\;\\boxed{\\frac{2}{7}}$ (d) Between 300 and 600 The first digit must be 3, 4, or 5: 3 choices. The last two digits are chosen from the remaining 6 digits; . $6\\!\\cdot\\!5\\:=\\:{\\color{blue}30}$ ways. . . Hence, there are: . $3\\!\\cdot\\!30 \\:=\\:90$ numbers bewteen 300 and 600. Therefore: . $P(\\text{between 300 and 6 00}) \\:=\\;\\frac{90}{210} \\;=\\;\\boxed{\\frac{3}{7}}$ 3. t ty" ]

[ "0 and 1 repeated as well so we have 3*12=36 (not 18 ). For 4), I became interested in this type of thing because at my place of employment, in order to log in remotely, we have to enter a code that is chosen at random, 6 each of the digits 0-9. I noticed that it was relatively rare to have no repeated digits so I tried to figure out the odds of getting exactly two repeating digits (among other things). I wrote programs in Ruby (which I will gladly share but I doubt anyone uses Ruby here) to run random numbers and looking at the percentage, and also to calculate exactly by running through all of the combinations and looking at the percentage. I am pretty sure that I used reasoning similar to that given earlier in the thread but the calculated results did not match the programs' outputs until I used the methods given above. Caveat: My 'exactly two matching numbers' routine may be flawed but for base 3 with 4 symbols taken, it does give 36, in agreement with the result above. For base 10 and 6 numbers taken it gives 453600 which agrees with: $\\displaystyle \\large 10 \\cdot 1 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot \\frac{6!}{2! \\cdot 4!}$ My random number", "of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$ Together, the answer is $3(12+12+12)-12=\\boxed{\\textbf{(A) } 96}.$ Solution 4 Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \\equiv 1\\pmod{3},$ and thus we need to count any $2$-digit number $\\equiv 2\\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \\boxed{\\textbf{(A) } 96}.$ Solution 5 We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\\cdot\\frac{1}{3}=150.$ Of these $150$ numbers, $\\frac{4}{5}$ $\\textbf{do not}$ have $3$ in their ones (units) digit, $\\frac{9}{10}$ $\\textbf{do not}$ have $3$ in their tens digit, and $\\frac{8}{9}$ $\\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$900\\cdot\\frac{1}{2}\\cdot\\frac{1}{3}\\cdot\\frac{4}{5}\\cdot\\frac{9}{10}\\cdot\\frac{8}{9}=\\boxed{\\textbf{(A) } 96}.$$ ~mathpro12345 Solution 6 We will start with the numbers that could work. This numbers include _ _ $1$, _ _ $5$, _ _ $7$, _", "Question # Consider all four-digit numbers that can be created from the digits 0-9 where the first and last digits must be even and no digit can repeat. Probability and combinatorics Consider all four-digit numbers that can be created from the digits 0-9 where the first and last digits must be even and no digit can repeat. Assume that numbers can start with 0. What is the probability of choosing a random number that starts with 8 from this group? Enter a fraction or round your answer to 4 decimal places, if necessary. 2021-09-05 The first and last digits should be even. Hence, Total number of numbers possible $$\\displaystyle={5}\\cdot{4}\\cdot{8}\\cdot{7}={1120}$$ Total number that starts with $$\\displaystyle{2}={4}\\cdot{8}\\cdot{7}={224}$$ Hence, P(Stars with 2) $$\\displaystyle=\\frac{224}{{1120}}=\\frac{1}{{5}}$$", "digits of $$\\left(\\frac{10^{1992}-1}{9}\\right)^2 = \\frac{1}{81}(10^{3984} - 2\\cdot 10^{1992} +1)$$ To write this out as an integer, note that $$\\frac{1}{81} = \\frac{12,345,679}{999,999,999} \\quad \\text{ and thus} \\quad \\frac{2}{81} = \\frac{24,691,358}{999,999,999}$$ This tells us the repeating 9-digit patterns in $\\frac{10^{3984} }{81}$ and $\\frac{2\\cdot 10^{1992}}{81}$. To take the difference, note that the two leading terms are $1$'s in the $10^{3982}$ place and in the $10^{1990}$ place, respectively. The first subtraction then occurs at the latter digit, which corresponds to the digit $4$ in the expansion of $\\frac{10^{3984} }{81}$, as $3982-3 = 3980 \\equiv 1990 \\mod 9$. The difference therefore has the repeating pattern unchanged from digits in the $10^{1991}$ place to the $10^{3982}$ place, and for lesser digits, the repeating pattern is now $$456,790,123-246,913,580 = 209,876,543$$ This pattern repeats indefinitely, until the decimal point, at which point we add $1/81$ with the effect of rounding $...320.\\overline{976543210}$ up to $...321$. Note that there are $221$ nine-digit sequences in each pattern, in addition to the leading three digits and the trailing two digits. The sum of the digits is then $1+2+3+221(44) + 221(37) + 2 + 1 = 9 + 221\\cdot 81 = 17910$. • ah, beat me to it as I was typing it out. This is the correct answer – qwr Apr 17 '15 at 1:30 This is not particularly pretty", "are $45$ such combinations with $2$ distinct digits and both digits repeated. There are $360$ such combinations with $3$ distinct digits ($120$ ways to pick the digits, $3$ ways to pick the one that repeats). Thus, there are $$1\\cdot 10+4\\cdot 90+6\\cdot 45+12\\cdot 360=4960$$ combinations with at least one repeat, leaving $5040$ combinations with no repeats, and so there are $$\\frac{5040}{24}=210$$ combinations in non-decreasing order with no repeats. In total, there are thus $10+90+45+360+210=715$ non-decreasing combinations that we need to try, and we can be sure to get it open by the end of that list. As for strategy, let's suppose that each of the $10000$ possible combinations is equally likely to be correct. Then we have \\begin{align}\\text{P(four distinct digits)} &= \\frac{24\\cdot 210}{10000}=50.4\\%\\\\\\text{P(three distinct digits)} &= \\frac{12\\cdot 360}{10000}=43.2\\%\\\\\\text{P(two distinct digits, both repeated)} &= \\frac{6\\cdot 45}{10000}=2.7\\%\\\\\\text{P(two distinct digits, only one repeated)} &= \\frac{4\\cdot 90}{10000}=3.6\\%\\\\\\text{P(single digit repeated four times)} &= \\frac{1\\cdot 10}{10000}=0.1\\%.\\end{align} Guessing a non-decreasing combination with 4 distinct digits is effectively worth $24$ guesses, and so on. Our expected value for a given kind of combinations--that is, how efficient a given nondecreasing combination will be as a guess--would then be \\begin{align}\\text{E(four distinct digits)} &= 24\\cdot50.4\\% = 12.096\\\\\\text{E(three distinct digits)} &= 12\\cdot43.2\\% = 5.184\\\\\\text{E(two distinct digits, both repeated)} &= 6\\cdot2.7\\% = 0.162\\\\\\text{E(two distinct digits, only one repeated)} &= 4\\cdot3.6\\% = 0.144\\\\\\text{E(single", "$10^4 - (9 \\cdot 9 \\cdot 8 \\cdot 7)$ by $2$. In addition, we also have to notice that there are some $4$-digit integers that contain both odd repeated digits and even repeated digits. Those integers have this form: \"aabb\", \"abba\", \"abab\", \"baba\". To be consistent, we'll allow these integers to start with $0$ as well, but this is not a problem since we'll ignore the integers that contain repeated even digits. Each \"a\" can be one of $5$ odd digits and each \"b\" can be one of $5$ even digits. This means we have an extra $(5 \\cdot 5) \\cdot 4$, $4$-digit integers with even digits repeated. We've already removed half of them (when we divided by $2$) so we only need to remove another half. So the final result is $$9 \\cdot 9 \\cdot 8 \\cdot 7 - \\frac{10^4 - 9 \\cdot 9 \\cdot 8 \\cdot 7}{2} - \\frac{5 \\cdot 5 \\cdot 4}{2} = 7218$$ My first question is: Given my potentially flawed assumption, is this solution correct? And if it is correct, can you think of an easier way to go about it? The solution for what I think the author actually asked (If the integers contain repeated digits all the digits in the integer must be odd): There are $9 \\cdot 9 \\cdot 8", "\\cdot 7$ $4$-digit integers that don't contain any repeated digits. We have $5^4$ $4$-digit integers that can have odd digits repeated. There is a set that intersects the previously mentioned sets. That is the set formed by all the $4$-digit integers with odd digits that don't repeat. So the final result is: $9 \\cdot 9 \\cdot 8 \\cdot 7 + 5^4 - 5 \\cdot 4 \\cdot 3 \\cdot 2 = 5041$. As for my second question: Is this solution correct and is this what you understood from the problem statement as well? Mention: This problem comes from The Book of Proof (Section 3.5 problem 2) • I understand your confusion. The meaning of the word \"all\" is unclear in this context. Also, please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Mar 22 '17 at 15:47 • Another interpretation seems viable considering the topic is PIE: \"Count positive 4-digit integers that have no repeat digits, or, do contain repeat digits but, if so, are odd integers.\" You can have a set $A_{\\text{no repeats}}$ of 4-digit numbers with no repeat digits, a set of odd 4-digit numbers $A_{\\text{odd}}$ and the overlap is those odd 4-digit numbers with no repeat digits. Trying to use PIE for your interpretations seems harder than the rest", "\\right\\}$ there $9^6-1$ numbers in $A_1$, we subtract the 1 to account for 000000000. 7. Hello, shaoen01! Plato's approach to #2 is correct . . . I would further assume that an integer does not begin with zero. 2) How many integers from 1 through 999999 contain each of the digits 1,2 and 3 at least once? The solution asked me to use the inclusion-exclusion method, so how do i know when to use this? It is used when counting the items which are not in the set is easier than counting the items that are in the set. There are 999,999 integers. How many do not contain at least {1,2,3} ? Let $n(i')$ = number of integers that do not contain an $i.$ We want: $n(1' \\,\\cup \\,2' \\,\\cup \\,3') \\;=\\;n(1') \\,+\\, n(2') \\,+ \\,n(3') \\,-\\, n(1' \\,\\cap \\,2') \\,- \\,n(2' \\,\\cap \\,3') \\,- \\,n(1' \\,\\cap \\,3') \\,+$ $n(1' \\cap 2' \\cap 3')$ Consider $n(1')$ 1-digit numbers: 8 choices . (It must not be 0 or 1.) 2-digit numbers: $8\\cdot9$ choices. 3-digit numbers: $8\\cdot9^2$ choices. 4-digit numbers: $8\\cdot9^3$ choices. 5-digit numbers: $8\\cdot9^4$ choices. 6-digit numbers: $8\\cdot9^5$ choices. There are: . $8(1 + 9 + 9^2 + \\hdots + 9^5) \\;=\\;8\\,\\frac{9^5-1}{9-1} \\;=\\;59,048$ such numbers. The same reasoning holds for $n(2')\\text{ and }n(3')$ . . Hence: . $n(1') \\:=\\:n(2')", "In reality you are approximating, but mathematics is more than reality so it prefer preciseness whenever possible. Here $\\frac{1}{3}\\times100=33.333\\cdots$ and $3\\times 33.333\\cdots=99.999\\cdots$. I think this is what confused you. But $100=99.999\\cdots$ Let $$x={\\Large99.999\\cdots}\\\\\\implies 10x=999.999\\cdots=900+99.999\\cdots=900+x\\\\\\implies10x-x=900\\\\\\implies9x=900\\\\\\implies x=\\frac{900}{9}={\\Large100}.$$ Here, in a practical way – I understand all your concepts, sirs – I was using a simple calculator on my computer, and looks like mine's got a 10 decimal digit precision. Which make this possible: 9 decimal \"3s\" 33.333333333 * 3 = 99.999999999 10 decimal \"3s\" 33.3333333333 * 3 = 100 If I divide 100 by 3, it'll give me 10 decimal \"3s\". And this expression was valid: 100/3 * 3 = 100 • That's simply because your calculator is doing an approximation. Mar 31 '14 at 1:12 • No offense, but your answer is kind of dumb. As comeAndGo says, it's only an approximation. But in practice, approximations aren't actually bad. Apr 2 '14 at 18:28 • Well, I thought the OP was trying to solve a problem instead of get to know all the concepts of divisibility. I should go back to stackoverflow, maybe :) Apr 3 '14 at 21:50 In base 10, no, not as long as you only allow finite digits. In base 3, we certainly can (or any other base divisible by three). The number 100 as", "# Is there any mathematical reason for this “digit-repetition-show”? The number has a crazy decimal representation : Is there any mathematical reason for so many repetitions of the digits ? A long block containing only a single digit would be easier to understand. This could mean that there are extremely good rational approximations. But here we have many long one-digit-blocks , some consecutive, some interrupted by a few digits. I did not calculate the probability of such a “digit-repitition-show”, but I think it is extremely small. Does anyone have an explanation ? The architect’s answer, while explaining the absolutely crucial fact that $$√308642≈5000/9=555.555…,\\sqrt{308642}\\approx 5000/9=555.555\\ldots,$$ didn’t quite make it clear why we get several runs of repeating decimals. I try to shed additional light to that using a different tool. $$√1+x=1+x2−x28+x316−5x4128+7x5256−21x61024+⋯ \\sqrt{1+x}=1+\\frac x2-\\frac{x^2}8+\\frac{x^3}{16}-\\frac{5x^4}{128}+\\frac{7x^5}{256}-\\frac{21x^6}{1024}+\\cdots$$ If we plug in $$x=2/(5000)2=8⋅10−8x=2/(5000)^2=8\\cdot10^{-8}$$, we get $$M:=√1+8⋅10−8=1+4⋅10−8−8⋅10−16+32⋅10−24−160⋅10−32+⋯. M:=\\sqrt{1+8\\cdot10^{-8}}=1+4\\cdot10^{-8}-8\\cdot10^{-16}+32\\cdot10^{-24}-160\\cdot10^{-32}+\\cdots.$$ √308462=50009M=50009+200009⋅10−8−400009⋅10−16+1600009⋅10−24+⋯=59⋅103+29⋅10−4−49⋅10−12+169⋅10−20+⋯. \\begin{aligned} \\sqrt{308462}&=\\frac{5000}9M=\\frac{5000}9+\\frac{20000}9\\cdot10^{-8}-\\frac{40000}9\\cdot10^{-16}+\\frac{160000}9\\cdot10^{-24}+\\cdots\\\\ &=\\frac{5}9\\cdot10^3+\\frac29\\cdot10^{-4}-\\frac49\\cdot10^{-12}+\\frac{16}9\\cdot10^{-20}+\\cdots. \\end{aligned} This explains both the runs, their starting points, as well as the origin and location of those extra digits not part of any run. For example, the run of $$5+2=75+2=7$$s begins when the first two terms of the above series are “active”. When the third term joins in, we need to subtract a $$44$$ and a run of $$33$$s ensues et cetera.", "since there are three sets of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$ Together, the answer is $3(12+12+12)-12=\\boxed{\\textbf{(A) } 96}.$ ## Solution 4 Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \\equiv 1\\pmod{3},$ and thus we need to count any $2$-digit number $\\equiv 2\\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \\boxed{\\textbf{(A) } 96}.$ ## Solution 5 We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\\cdot\\frac{1}{3}=150.$ Of these $150$ numbers, $\\frac{4}{5}$ $\\textbf{do not}$ have $3$ in their ones (units) digit, $\\frac{9}{10}$ $\\textbf{do not}$ have $3$ in their tens digit, and $\\frac{8}{9}$ $\\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$900\\cdot\\frac{1}{2}\\cdot\\frac{1}{3}\\cdot\\frac{4}{5}\\cdot\\frac{9}{10}\\cdot\\frac{8}{9}=\\boxed{\\textbf{(A) } 96}.$$ ~mathpro12345 ~ pi_is_3.14", "Thinking Mathematically (6th Edition) $n!=n(n-1)(n-2)\\cdots(3)(2)(1)$ and, by definition, $0!=1$ ----------------- Most calculators can display up to $69!\\approx 10^{98}$. Anything over results in an error. 900! and 899! are huge numbers. We employ a different approach: $900!=900\\times 899\\times 898\\times...\\times 2\\times 1$ $=900\\times 899!$ So, we write $\\displaystyle \\frac{900!}{899!}=\\frac{900\\times 899!}{899!}$ ... reduce the fraction by $899!$... $=900$", "an important role in the representation of the number as a whole. Since the decimal system we use is a positional numeric system, all of the digits in a decimal number is termed according to their position in respect to the decimal point and it is important to name the decimal places properly. The entire decimal system is completely based on number$10$and all of the digits, before and after the decimal point, are defined in terms of ten because of that. The digit placed furthest to the right of the decimal point has the smallest value. Hence, in the number$28.531$, the digit 5 is placed furthest from the decimal point and hence has the smallest value. The entire number can be defined as$ 2cdot 10+8 cdot 1+5 cdot frac<1><10>+3 cdot frac<1><100>+1 cdot frac<1><1000>$. The number after the decimal point can be collectively pronounced as$6$tenths,$4$hundredths and$5$thousandths or, more simply, as$645$thousandths. All the place values of the numbers depend on position on the left or right side of a decimal point. Look at the example with more digits. Let’s take a look at, for example the number$1,987,654,321.123456$, The first digit before the decimal point represents the ones (number$1$), -the second stands for the tens (number$2$), the third for the hundreds (number$3$), -the fourth for the thousands (number$4$, after the comma),", "$$5 \\cdot 3 \\cdot 9 \\cdot 8 = 1080$$ such numbers. In total, there are $960 + 1080 = 2040$ four-digit positive integers using exactly three digits in which one odd digit is repeated. Total: Adding these cases yields $$5 + 175 + 60 + 2040 = 2280$$ four-digit positive integers containing repeated digits in which the only digits that are repeated are odd digits. Your tried to use the Inclusion-Exclusion Principle to do this. Let's see what went wrong. As you noted, there are $10^4$ four-digit strings. However, the number of such strings with no repeated digits is $10 \\cdot 9 \\cdot 8 \\cdot 7 = 5040$. Hence, there are $$10^4 - 10 \\cdot 9 \\cot 8 \\cdot 7 = 1000 - 5040 = 4960$$ such strings with repeated digits. You attempted to use symmetry to eliminate those that contain repeated even digits by dividing by $2$. Doing so yields $2480$. As you realized, we are still counting strings with two even and two odd digits. There are actually $$5 \\cdot 5 \\cdot \\binom{4}{2} = 150$$ such strings since there are five choices of even digit, five choices of odd digit, and $\\binom{4}{2}$ ways of selecting the positions of the even digit. Half of these were eliminated when we divided by $2$, leaving $75$ more to be", "7345.345 \\cdots - 7.345 \\cdots$ which results to $9990x = 7338$. Now, $x =\\displaystyle\\frac{7338}{9990}$ which is a fraction. Therefore, $0.7345345 = \\frac{7338}{9990}$ is rational. You may also want to read more examples of converting non-terminating, repeating decimals in Irrational Numbers as Decimals.", "# Difference between revisions of \"1970 AHSME Problems/Problem 10\" ## Problem Let $F=.48181\\cdots$ be an infinite repeating decimal with the digits $8$ and $1$ repeating. When $F$ is written as a fraction in lowest terms, the denominator exceeds the numerator by $\\text{(A) } 13\\quad \\text{(B) } 14\\quad \\text{(C) } 29\\quad \\text{(D) } 57\\quad \\text{(E) } 126$ ## Solution Multiplying by $100$ gives $100F = 48.181818...$. Subtracting the first equation from the second gives $99F = 47.7$, and all the other repeating parts cancel out. This gives $F = \\frac{47.7}{99} = \\frac{477}{990} = \\frac{159}{330} = \\frac{53}{110}$. Subtracting the numerator from the denominator gives $\\fbox{D}$.", "is $$3.230\\cdots$$. Step 5: Compare both numbers $$3.666\\cdots$$ and $$3.230\\cdots$$. On comparing both numbers $$3.666\\cdots$$ and $$3.230\\cdots$$, the conclusion is that $$3.666\\cdots$$ is greater than $$3.230\\cdots$$, i.e., $$3.666{\\cdots} > 3.230{\\cdots}$$. So, $$\\frac{11}{3} > \\frac{42}{13}$$. ## E1.5B: Recognise equivalence and convert between these forms. The non-recurring numbers are the simple decimal numbers that do not end with zero and do not repeat the same interval. But the recurring numbers do not end with zero. The numbers are continuous till infinity and are periodic. The recurring numbers can be converted into fractions, whereas non-recurring numbers cannot be converted into fractions. ### Worked example Example 1: Convert $$\\frac{1}{6}$$ into a recurring decimal number. Step 1: Divide $$1$$ by $$6$$. $$1$$ divided by $$6$$ is $$0$$, so the remainder is $$1$$ carrying $$1$$ to the tenth column. Now, $$10$$ divided by $$6$$ by $$1$$ times, then the remainder will be $$4$$. Similarly, $$40$$ divided by $$6$$ times, then the remainder will be $$4$$, and so on. Step 2: Collect the quotient from the above step. The quotient is $$0.1666{\\cdots}$$. So, the recurring decimal is $$0.1\\bar{6}$$.", "\\frac{9}{100} + \\frac{9}{1000}$ But by what right can we write down something like $.999\\ldots = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000} + \\cdots$ where the sum just keeps going and going? How can we add infinitely many numbers? Does it even make sense to talk about something like that? To answer this, let’s consider the sequence of partial sums: $.9 = \\frac{9}{10} \\;\\;\\;\\;\\;\\; .99 = \\frac{9}{10} + \\frac{9}{100} \\;\\;\\;\\;\\;\\; .999 = \\frac{9}{10} + \\frac{9}{100} + \\frac{9}{1000}$ and so on. Notice that each term in the sequence is a bit larger than the preceding, and that all are less than 1: $.9 < .99 < .999 < .9999 < \\cdots < 1$ Modern mathematics defines the meaning of the infinite decimal expansion as follows: the value of the expansion is the number x such that, for any positive number ϵ, no matter how small, our sequence of partial sums is eventually in between x – ϵ and x. Put in another way, the value is x if, for any small number ϵ, the difference between x and the nth partial sum is smaller than ϵ if n is large enough. So, you see, the admission that the difference between 1 and .999… is a number “smaller than every other number” means that .999… is in fact equal to 1. #", "less. It means that the digits with respect to $q$ are the same - except that there are $n-1$ zeroes in between each of the digits. We have that $\\frac{10203}{100}=102 \\cdot 100+3$ and so $10203=1023$ with respect to $100$ and $\\frac{10203}{10}=1020 \\cdot 10+3$ and so $10203=10203$ with respect to $10$, right? More accurately, we have $10203=1\\cdot 100^2+2\\cdot 100^1+3\\cdot 100^0$ with respect to $p=100$. And we have $10203=1\\cdot 10^4+0\\cdot 10^3 + 2\\cdot 10^2+0\\cdot 10^1 + 3\\cdot 10^0$ with respect to $q=10$. #### evinda ##### Well-known member MHB Site Helper Didn't we just show that? We effectively wrote the same number with respect to the system with basis $q$, didn't we? Could it be that's a mistake in the problem statement? Consider for example that $p=100$ and $q=10$. So $n=2$. We can represent the number $10203$ with respect to $p$ as $123$, and we can represent the same number with respect to $q$ as $10203$. In this case we need $m=3$ digits to represent the number and $n=2$ digits does not suffice to represent the number. Is it maybe meant that each digit of the number is expressed with $n$ digits of the system $q$ ? For example in this case $3$ is written as $03$, $2$ as $02$ and $1$ to $01=1$. So since $p=q^n$ it means that each", "# Find the E(X) where X be the number of different numbers drawn during the first four week. In a certain lottery 5 numbers are drawn without replacement out of the set {1,2,...,90} randomly each week. Let X be the number of different numbers drawn during the first four weeks of the year. (For example, if the draws are {1,2,3,4,5}, {2,3,4,5,6}, {3,4,5,6,7} and {4,5,6,7,8} then X=8.) Find E(X). My work: • I am confused about how you can come up the P(Ii=1) where there is without replacement part(days) and with the replacement part(week). – Michelle Feb 26 '19 at 20:57 Your approach seems good. Basically, you want to calculate $$P(I_i = 1)$$. For a given round, there are $$5$$ numbers drawn uniformly at random (uar) without replacement. So the probability of a certain number not being drawn in a given round is $$p_5 = \\frac{89}{90} \\cdot \\frac{88}{89} \\cdot \\frac{87}{88} \\cdot \\frac{86}{87} \\cdot \\frac{85}{86} = \\frac{85}{90} = 1 - \\frac{5}{90}.$$ (The subscript $$5$$ denotes that we draw $$5$$ numbers; in general it's a similar product of length $$k$$ if there are $$k$$ numbers drawn in one round.) Why is this? Let's assume that we don't want to hit number $$90$$. Draw the first one: what's the probability that it's not $$90$$? Exactly $$89/90$$. Now draw the second: probability $$88/89$$." ]

[ "can run the same method to compute $$a\\pmod{11!}$$.", "+0 # Help. I have been stuck 0 114 4 Given that $$\\binom{15}{8}=6435$$$$\\binom{16}{9}=11440$$, and $$\\binom{16}{10}=8008$$, find $$\\binom{15}{10}$$. I know that I should be able to do this but I can't figure it out. Aug 10, 2019 #1 +23295 +3 Given that $$\\dbinom{15}{8}=6435$$, $$\\dbinom{16}{9}=11440$$ , and $$\\dbinom{16}{10}=8008$$, find $$\\dbinom{15}{10}$$. Formula: $$\\dbinom{n}{k} + \\dbinom{n}{k+1} = \\dbinom{n+1}{k+1}$$ $$\\begin{array}{|lrcll|} \\hline & \\mathbf{\\dbinom{n}{k} + \\dbinom{n}{k+1}} &=& \\mathbf{\\dbinom{n+1}{k+1}} \\\\\\\\ (1) & \\dbinom{15}{8} + \\dbinom{15}{9} &=& \\dbinom{16}{9} \\\\ (2) & \\dbinom{15}{9} + \\dbinom{15}{10} &=& \\dbinom{16}{10} \\\\ \\hline (1)-(2): & \\dbinom{15}{8} + \\dbinom{15}{9} -\\left( \\dbinom{15}{9} + \\dbinom{15}{10}\\right) &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & \\dbinom{15}{8} - \\dbinom{15}{10} &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & 6435 - \\dbinom{15}{10} &=& 11440-11440 \\\\ & 6435 - \\dbinom{15}{10} &=& 3432 \\\\ & \\dbinom{15}{10} &=& 6435 - 3432 \\\\ & \\mathbf{\\dbinom{15}{10}} &=& \\mathbf{3003} \\\\ \\hline \\end{array}$$ Aug 11, 2019 #2 0 Thank you so much!! I can't keep all the formulas in my head sometimes. Aug 11, 2019 #3 +1655 +3 There are really too many formulas... tommarvoloriddle Aug 12, 2019 #4 +105540 +1 You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from. There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level. Most times this does", "\\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} & 360 \\cdot x^{44} & 360 \\cdot x^{45} & 120 \\cdot x^{46}\\\\ & & & &", "db \"eASSWzRD<11$\"", ")^{10} \\\\\\\\ && ( a + b + c )^{n} \\\\ &=& \\sum \\limits_{k=0}^{n} \\sum \\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} &", "(\\mathbf{A}\\mathbf{B})^{-1} \\dbinom12 = \\dbinom13 } \\\\ \\hline \\end{array}$$ Apr 17, 2020", "let $N_1=N-\\dbinom{x_1}k$.\" Did you mean $N_1=N-\\dbinom{x_k}{k}$? That would be consistent with what you wrote in your next paragraph: $N_1=\\dbinom{x_{k-1}}{k-1}+\\ldots+\\dbinom{x_1}{1}$ – tychicus Aug 6 '13 at 7:21 • @tychicus: Yes, I did; I’ve fixed it now. (The same mistake was in the original version of your hint, and I must have unconsciously copied that.) – Brian M. Scott Aug 6 '13 at 12:02", "with $n\\le 9$ which has been done exhaustively by computer.", "this algorithm is $$(^{n}_{10})*n = O(n^{11})$$.", "## Intermediate Algebra for College Students (7th Edition) $x^{10}-5x^8+10x^6-10x^4+5x^2-1$ Apply Binomial Theorem. $(a+b)^n=\\sum_{r=0}^n\\dbinom{n}{r}(a)^{n-r}b^r$ $(x^2-1)^5=\\dbinom{5}{0}(x^2)^5(-1)^0+\\dbinom{5}{1}(x^2)^4(-1)^1+\\dbinom{5}{2}(x^2)^3(-1)^2+\\dbinom{5}{3}(x^2)^2(-1)^3+\\dbinom{5}{4}(x^2)^1(-1)^4+\\dbinom{5}{5}(x^2)^0(-1)^5$ or, $=x^{10}-5x^8+10x^6+10x^4(-1)+5x^2+(1)(1)(-1)$ or, $=x^{10}-5x^8+10x^6-10x^4+5x^2-1$", "process $P_{2}$ which is $9$ ms. 1 2", "this case trivially):$2^3 = 8$The$2$nd of the$1$st pair of consecutive integers whose product is a primorial:$1 \\times 2 = 2 = 2 \\#$$3$rd Term The$3$rd integer$m$after$0$,$1$such that$m! + 1$(its factorial plus$1$) is prime:$2! + 1 = 2 + 1 = 3$The$3$rd integer after$0$,$1$such that its double factorial plus$1$is prime:$2!! + 1 = 3$The$3$rd integer after$0$,$1$which is (trivially) the sum of the increasing powers of its digits taken in order:$2^1 = 2$The$3$rd subfactorial after$0$,$1$:$2 = 3! \\paren {1 - \\dfrac 1 {1!} + \\dfrac 1 {2!} - \\dfrac 1 {3!} }$The$3$rd integer$n$after$-1$,$0$such that$\\dbinom n 0 + \\dbinom n 1 + \\dbinom n 2 + \\dbinom n 3 = m^2$for integer$m$:$\\dbinom 2 0 + \\dbinom 2 1 + \\dbinom 2 2 + \\dbinom 2 3 = 2^2$The$3$rd integer$m$after$0$,$1$such that$m^2 = \\dbinom n 0 + \\dbinom n 1 + \\dbinom n 2 + \\dbinom n 3$for integer$n$:$2^2 = \\dbinom 2 0 + \\dbinom 2 1 + \\dbinom 2 2 + \\dbinom 2 3$The$3$rd palindromic integer after$0$,$1$which is the index of a palindromic triangular number$T_2 = 3$The$3$rd integer$n$after$0$,$1$such that$2^n$contains no zero in its decimal representation:$2^2 = 4$The$3$rd integer$n$after$0$,$1$such that$5^n$contains no zero in its decimal representation:$5^2 = 25$The$3$rd integer$n$after$0$,$1$such that both$2^n$and$5^n$have no zeroes:$2^2 = 4, 5^2 = 25$The$3$rd integer after$0$,$1$which is palindromic in both decimal and ternary:$2_{10} = 2_3$The$3$rd Fibonacci number after$1$,$1$:$2 =", "\\mathbf{d}$$...", "polynomial-time algorithm to compute $$\\chi'(G)$$.", "${9}$ (5)", "...9\\} = {9 \\choose 4}$.", "@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19 @AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22 Nope, that seems likely to be the most concise way to do it. Count ways to pick $n\\in\\{1,2,3,4\\}$ unique numbers, and to arrange them with $n-1$ \"$>$\" signs, to meet the criteria. $$\\begin{array}{|l:l|} \\hline \\rm a{=}a{=}a{=}a & \\dbinom 3 0 \\dbinom 6 1 \\\\\\hdashline\\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \\dbinom 3 1\\dbinom 6 2 \\\\\\hdashline\\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \\dbinom 3 2 \\dbinom 6 3 \\\\\\hdashline\\rm a{>}b{>}c{>}d & \\dbinom 3 3 \\dbinom 6 4 \\\\ \\hline\\end{array}$$ $$\\dfrac{\\sum_{n=1}^4 \\dbinom{3}{n-1}\\dbinom{6}{n}}{6^4} = \\dfrac{\\dbinom 6 1 + 3\\dbinom 6 2 + 3 \\dbinom 6 3 + \\dbinom 6 4}{6^4}$$ - The problem reduces to finding the cardinality of the following set $$A=\\{(a_1,a_2,a_3,a_4)\\in\\{1,\\ldots,6\\}^4: \\ a_1\\leq a_2\\leq a_3 \\leq a_4\\}.$$ Define $$B=\\{(a_1,a_2,a_3,a_4)\\}\\in\\{1,\\ldots 9\\}^4: \\ a_1 < a_2 < a_3 < a_4\\}$$ and $$f:A\\to B, \\quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \\choose 4}.$$ - @Shailesh This method actually does give the correct answer ($126/6^4$); If we", "-1 \\\\ k&=& 3 \\\\ \\dfrac{\\dbinom{14}{3}} {\\dbinom{14}{4}} &=& \\dfrac{4}{11} \\\\ \\hline \\end{array}$$ Mar 26, 2019", "$\\mathbf{u}$.", "\\right)^9 - 1}} {i} $" ]

[ "is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$.", "= 0$ $r = 1, 9$ Quite obviously $r > 1$, so $r = 9 \\boxed{(D)}$.", ". . . . . $=\\; \\dfrac{60}{9.8} \\;=\\;\\dfrac{300}{49} \\;=\\;6\\frac{6}{49}\\text{ seconds.}$", "Die kombinatorische Analysis, als Vorbereitungslehre zum Studium der theoretischen höheren Mathematik. It appears to have become the de facto standard in recent years. As a result, $\\dbinom n k$ is frequently voiced the binomial coefficient $n$ over $k$. Other notations include: $C \\left({n, k}\\right)$ ${}^n C_k$ ${}_n C_k$ $C^n_k$ ${C_n}^k$ all of which can cause a certain degree of confusion. ## Examples ### $2$ from $5$ The number of ways of choosing $2$ objects from a set of $5$ is: $\\dbinom 5 2 = \\dfrac {5 \\times 4} {2 \\times 1} = \\dfrac {5!} {2! \\, 3!} = 10$ ### $3$ from $7$ The number of ways of choosing $3$ objects from a set of $7$ is: $\\dbinom 7 3 = \\dfrac {7 \\times 6 \\times 5} {3 \\times 2 \\times 1} = \\dfrac {7!} {3! \\, 4!} = 35$ ### $3$ from $8$ The number of ways of choosing $3$ objects from a set of $8$ is: $\\dbinom 8 3 = \\dfrac {8 \\times 7 \\times 6} {3 \\times 2 \\times 1} = \\dfrac {8!} {3! \\, 5!} = 56$ ### $4$ from $52$ The number of ways of choosing $4$ objects from a set of $52$ (for example, cards from a deck) is: $\\dbinom {52} 4 = \\dfrac {52 \\times 51 \\times 50 \\times 49} {4", "2\\dbinom 21}{\\dbinom 52}=\\dfrac{3}{5}$$", "is: $\\set {\\dbinom {a \\ b} {1 \\ 2}, \\dbinom {a \\ b} {1 \\ 3}, \\dbinom {a \\ b} {2 \\ 1}, \\dbinom {a \\ b} {2 \\ 3}, \\dbinom {a \\ b} {3 \\ 1}, \\dbinom {a \\ b} {3 \\ 2} }$ thus demonstrating that there are $\\dfrac {3!} {\\paren {3 - 2}!} = \\dfrac {3!} {1!} = \\dfrac 6 1 = 6$ injections from $A$ to $B$.", "\\dfrac{(x+1)(x+4)}{(x+3)(x+3)} \\cdot \\dfrac{(x+3)(x-3)}{x+1} = \\boxed{\\dfrac{x^2+x-12}{x+3}} $ Ah.....oops! Thank you! How did I forget that! Thank you again for your help earboth! drakmord", "6 is $\\boxed{\\textbf{(D) } 13}$. - mathleticguyyy", "\\dfrac{629988.3164 \\times 9 \\times 10^{-4}}{0.5}$ N = 1133.98 = 1134 T.", "5}{10 \\times 9 \\times 8 \\times 7 \\times 6 }$ 5 red out of 5: $\\dfrac{5 \\times 4 \\times 3 \\times 2 \\times 1}{5 \\times 4 \\times 3 \\times 2 \\times 1} \\times \\dfrac{5 \\times 4 \\times 3 \\times 2 \\times 1}{10 \\times 9 \\times 8 \\times 7 \\times 6 }$ Add those up and you get $\\frac{1}{2}$. Also add 0, 1 or 2 reds out of 5 and you get $1$. - add comment", "gives us $\\dbinom{n+1}{2}$ ways, for a total of $3\\dbinom{n+1}{2}$ ways. The last case gives us $\\dbinom{n+1}{1}=n+1$ ways. Therefore, the total number of possible ways where there are no isolated men is $\\[\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1).\\]$ The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or $\\[2\\dbinom{n+1}{2}+(n+1).\\]$ Thus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that $\\[\\dfrac{2\\dbinom{n+1}{2}+(n+1)}{\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1)}\\le\\dfrac{1}{100}.\\]$ After simplification, we arrive at $\\[\\dfrac{6(n+1)}{n^2+8n+6}\\le\\dfrac{1}{100}.\\]$ Simplifying again, we see that we seek the smallest positive integer value of $n$ such that $n(n-592)\\ge594$. Clearly $n>592$, or the left side will not even be positive; we quickly see that $n=593$ is too small but $n=\\boxed{594}$ satisfies the inequality.", "# Binomial Coefficient/Examples ## Examples of Binomial Coefficients ### $2$ from $5$ The number of ways of choosing $2$ objects from a set of $5$ is: $\\dbinom 5 2 = 10$ ### $5$ from $2$ $\\dbinom 2 5 = 0$ ### $2$ from $-5$ $\\dbinom {-5} 2 = 15$ ### $5$ from $-2$ $\\dbinom {-2} 5 = -6$ ### $3$ from $7$ The number of ways of choosing $3$ objects from a set of $7$ is: $\\dbinom 7 3 = \\dfrac {7 \\times 6 \\times 5} {3 \\times 2 \\times 1} = \\dfrac {7!} {3! \\, 4!} = 35$ ### $3$ from $8$ The number of ways of choosing $3$ objects from a set of $8$ is: $\\dbinom 8 3 = \\dfrac {8 \\times 7 \\times 6} {3 \\times 2 \\times 1} = \\dfrac {8!} {3! \\, 5!} = 56$ ### $4$ from $52$ The number of ways of choosing $4$ objects from a set of $52$ (for example, cards from a deck) is: $\\dbinom {52} 4 = \\dfrac {52 \\times 51 \\times 50 \\times 49} {4 \\times 3 \\times 2 \\times 1} = \\dfrac {52!} {48! \\, 4!} = 270 \\, 725$ ### Number of Bridge Hands The total number $N$ of possible different hands for a game of bridge is: $N = \\dfrac {52!} {13! \\,", "C}$ and $\\alpha_ \\text{brass} = \\dfrac{19 \\times 10^{-6}}{^\\circ C}$]", "answer is $\\dfrac{8}{20}=\\boxed{\\textbf{(B)} \\dfrac{2}{5}}$. ~JH. L ## Video Solution 1 ~Education, the Study of Everything ~savannahsolver", "= \\dfrac{{13 \\times 12! \\times 4 \\times 3! \\times 50! \\times 2 \\times 1}}{{52 \\times 51 \\times 50! \\times 12! \\times 3!}} \\\\ P = \\dfrac{{13 \\times 4 \\times 2}}{{52 \\times 51}} = \\dfrac{2}{{51}} \\\\$", "he received from B Jul 4, 2021 at 3:53 • I made an edit with the clarifications Jul 4, 2021 at 4:06 Here's my solution based on expectations. First, we calculate the expected value of the smallest, second-smallest, third-smallest... number on the 5 cards drawn. The calculation steps are shown below: $$E(\\text{smallest})=\\dfrac{1\\times \\dbinom{9}{4}+2\\times \\dbinom{8}{4}+\\cdots+6\\times \\dbinom{4}{4}}{\\dbinom{10}{5}}=\\dfrac{11}{6}.$$ Explanation: the numerator basically sums the smallest terms in all 5-card draws, and the denominator is the total number of cases. Similarly, we have $$E(\\text{second-smallest})=\\dfrac{2\\times \\dbinom{8}{3}\\times \\dbinom{1}{1}+3\\times \\dbinom{7}{3}\\times \\dbinom{2}{1}+\\cdots+7\\times \\dbinom{3}{3}\\times \\dbinom{6}{1}}{\\dbinom{10}{5}}=\\dfrac{11}{3};$$ $$E(\\text{third-smallest})=\\dfrac{3\\times \\dbinom{7}{2}\\times \\dbinom{2}{2}+4\\times \\dbinom{6}{2}\\times \\dbinom{3}{2}+\\cdots+8\\times \\dbinom{2}{2}\\times \\dbinom{7}{2}}{\\dbinom{10}{5}}=\\dfrac{11}{2};$$ $$E(\\text{fourth-smallest})=\\dfrac{4\\times \\dbinom{6}{1}\\times \\dbinom{3}{3}+5\\times \\dbinom{5}{1}\\times \\dbinom{4}{3}+\\cdots+9\\times \\dbinom{1}{1}\\times \\dbinom{8}{3}}{\\dbinom{10}{5}}=\\dfrac{22}{3};$$ $$E(\\text{fifth-smallest, or largest})=\\dfrac{5\\times \\dbinom{4}{4}+6\\times \\dbinom{5}{4}+\\cdots+10\\times \\dbinom{9}{4}}{\\dbinom{10}{5}}=\\dfrac{55}{6}.$$ As a quick double check, adding them all together yields $$27.5$$, which is the expected sum of the 5 cards. Next, let's think about A's strategy. In essence, he is discarding his $$N$$ smallest cards to get random cards. Thus, if the expected value of cards that he is discarding is smaller than the expected value of cards he is getting, then A is benefitting. The more A benefits, the higher his chances of winning. Then, clearly $$E(\\text{smallest})$$, $$E(\\text{second-smallest})$$ are smaller than $$5.5$$, the expected value on a random card. $$E(\\text{third-smallest})$$ is equal to $$5.5$$, and $$E(\\text{fourth-smallest})$$, $$E(\\text{fifth-smallest, or largest})$$ are greater than $$5.5$$. This means that discarding the smallest card and", "\\dfrac{{3 \\times 2!}}{{\\not 2!}}}}{{\\dfrac{{\\not 6 \\times 5 \\times 4 \\times \\not 3!}}{{\\not 3! \\times 3! \\times 2 \\times 1}}}}$ $= \\dfrac{9}{{20}} \\times \\dfrac{6}{{20}} = \\dfrac{{27}}{{200}}\\,\\,\\,\\,\\,\\,\\,.......(1)$ Now case II as the number of balls are same just colours are different, we have: $= \\dfrac{{{}^3{C_2} \\times {}^3{C_1}}}{{{}^6{C_3}}} \\times \\dfrac{{{}^1{C_1} \\times {}^2{C_1} \\times {}^3{C_1}}}{{{}^6{C_3}}}$ $= \\dfrac{{\\dfrac{{3!}}{{2!1!}} \\times \\dfrac{{3!}}{{2!1!}}}}{{\\dfrac{{6!}}{{3!3!}}}} \\times \\dfrac{{\\dfrac{{2!}}{{1!1!}} \\times 1 \\times \\dfrac{{3!}}{{2!1!}}}}{{\\dfrac{{6!}}{{3!3!}}}}$ $= \\dfrac{{\\dfrac{{3 \\times \\not 2!}}{{\\not 2!}} \\times \\dfrac{{3 \\times \\not 2!}}{{\\not 2!}}}}{{\\dfrac{{\\not 6 \\times 5 \\times 4 \\times \\not 3!}}{{3! \\times 3\\not ! \\times \\not 2 \\times 1}}}} \\times \\dfrac{{2! \\times \\dfrac{{3 \\times 2!}}{{\\not 2!}}}}{{\\dfrac{{\\not 6 \\times 5 \\times 4 \\times \\not 3!}}{{\\not 3! \\times 3! \\times 2 \\times 1}}}}$ $= \\dfrac{9}{{20}} \\times \\dfrac{6}{{20}} = \\dfrac{{27}}{{200}}\\,\\,\\,\\,\\,\\,\\,.......(2)$ $= \\dfrac{{27}}{{200}}.....(2)$ Probability of 3 later balls being all of different colours $= \\dfrac{{27}}{{200}} + \\dfrac{{27}}{{200}}$(from $e{q^n}$ 1 & 2) $= \\dfrac{{2 \\times 27}}{{200}}$$= \\dfrac{{27}}{{100}}$ Note: Cube root need to group digit in no. 3 and taking the unit place digit of the first group & ten’s from $I{I^{nd}}$. In mathematics, a cube root of a number x is a number y such that $\\;{y^3}\\; = \\;x$. All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube", "Oct 28 '14 at 21:20 Here is an alternative solution to the (very good) answer already given here: The number of ways that you can choose $2$ out of $6$ values is $\\dbinom{6}{2}=15$. Given $2$ values A and B, the number of combinations that you can get in $6$ rolls is: • Value A appearing $1$ time and value B appearing $5$ times: $\\dbinom{6}{1}=6$ • Value A appearing $2$ times and value B appearing $4$ times: $\\dbinom{6}{2}=15$ • Value A appearing $3$ times and value B appearing $3$ times: $\\dbinom{6}{3}=20$ • Value A appearing $4$ times and value B appearing $2$ times: $\\dbinom{6}{4}=15$ • Value A appearing $5$ times and value B appearing $1$ time : $\\dbinom{6}{5}=6$ So you can get $6+15+20+15+6=62$ combinations containing A and B. And you can get $15\\cdot62=930$ combinations containing any $2$ out of $6$ values. The total number of combinations that you can get in $6$ rolls is simply $6^6=46656$. Therefore, the probability of having exactly $2$ values in $6$ rolls is $\\dfrac{930}{46656}\\approx0.0199$. • Just a note: the $62$ calculated here through sumation is precisely the $2^6 - 2$ calculated in my answer. This is due to the fact that $\\sum_{i=1}^6{6\\choose i} = 2^6$. – 5xum Oct 28 '14 at 12:12 • @5xum: I agree (except for the fact that you need to", "= \\dfrac{5 \\times 10^{-9} \\mathrm{~ C}}{(8.85 \\times 10^{-12} \\,\\text{F/m}) (0.0225 \\mathrm{~m^2})} \\\\ \\boxed{E_i = 2.5 \\times 10^{4} \\,\\text{N/C}} \\end{gather*}", "roll ten dice. I like to reduce the case approach to two multinomial coefficients for [choose] $\\times$ [place] to make it more mechanical and thus less error prone. $5-1-1-1-1-1: \\dbinom6{1,5}\\dbinom{10}{5,1,1,1,1,1}$ $4-2-1-1-1-1: \\dbinom6{1,1,4}\\dbinom{10}{4,2,1,1,1,1}$ $3-3-1-1-1-1: \\dbinom6{2,4}\\dbinom{10}{3,3,1,1,1,1}$ $3-2-2-1-1-1: \\dbinom6{1,2,3}\\dbinom{10}{3,2,2,1,1,1}$ $2-2-2-2-1-1: \\dbinom6{4,2}\\dbinom{10}{2,2,2,2,1,1}$ Add up and divide by $6^{10}$ PS Or you might like to write it as [choose] $\\times$ [permute], viz. $5-1-1-1-1-1: \\dbinom6{1,5}\\times\\dfrac{10!}{5!}$ $4-2-1-1-1-1: \\dbinom6{1,1,4}\\times\\dfrac{10!}{4!2!}$ $3-3-1-1-1-1: \\dbinom6{2,4}\\times\\dfrac{10!}{3!3!}$ $3-2-2-1-1-1: \\dbinom6{1,2,3}\\times\\dfrac{10}{3!2!2!}$ $2-2-2-2-1-1: \\dbinom6{4,2}\\times\\dfrac{10!}{2!2!2!2!}$ Based on the comments below, I try a different approach. I consider $1-$the number of cases where all numbers are not there. Choose 5 out of 6 numbers. Each dice can be any number = $5^{10}$ So, the total number of ways is $6^{10} - 6*5^{10}$ • Double Counting! Suppose you pick the first die and its a one, there's another one in the six you didn't pick. Suppose you picked that place, and its a one, and the one among the six you didn't pick is in first place. You've counted these at least twice. – Graham Kemp Feb 16 '16 at 1:37 • This method double counts just as I warned above in the comments. – Gregory Grant Feb 16 '16 at 1:37 • @RobertMingHao Be careful this is a wrong solution, it double counts many possibilities. You should convince yourself of why this is wrong there's an important" ]

[ "he received from B Jul 4, 2021 at 3:53 • I made an edit with the clarifications Jul 4, 2021 at 4:06 Here's my solution based on expectations. First, we calculate the expected value of the smallest, second-smallest, third-smallest... number on the 5 cards drawn. The calculation steps are shown below: $$E(\\text{smallest})=\\dfrac{1\\times \\dbinom{9}{4}+2\\times \\dbinom{8}{4}+\\cdots+6\\times \\dbinom{4}{4}}{\\dbinom{10}{5}}=\\dfrac{11}{6}.$$ Explanation: the numerator basically sums the smallest terms in all 5-card draws, and the denominator is the total number of cases. Similarly, we have $$E(\\text{second-smallest})=\\dfrac{2\\times \\dbinom{8}{3}\\times \\dbinom{1}{1}+3\\times \\dbinom{7}{3}\\times \\dbinom{2}{1}+\\cdots+7\\times \\dbinom{3}{3}\\times \\dbinom{6}{1}}{\\dbinom{10}{5}}=\\dfrac{11}{3};$$ $$E(\\text{third-smallest})=\\dfrac{3\\times \\dbinom{7}{2}\\times \\dbinom{2}{2}+4\\times \\dbinom{6}{2}\\times \\dbinom{3}{2}+\\cdots+8\\times \\dbinom{2}{2}\\times \\dbinom{7}{2}}{\\dbinom{10}{5}}=\\dfrac{11}{2};$$ $$E(\\text{fourth-smallest})=\\dfrac{4\\times \\dbinom{6}{1}\\times \\dbinom{3}{3}+5\\times \\dbinom{5}{1}\\times \\dbinom{4}{3}+\\cdots+9\\times \\dbinom{1}{1}\\times \\dbinom{8}{3}}{\\dbinom{10}{5}}=\\dfrac{22}{3};$$ $$E(\\text{fifth-smallest, or largest})=\\dfrac{5\\times \\dbinom{4}{4}+6\\times \\dbinom{5}{4}+\\cdots+10\\times \\dbinom{9}{4}}{\\dbinom{10}{5}}=\\dfrac{55}{6}.$$ As a quick double check, adding them all together yields $$27.5$$, which is the expected sum of the 5 cards. Next, let's think about A's strategy. In essence, he is discarding his $$N$$ smallest cards to get random cards. Thus, if the expected value of cards that he is discarding is smaller than the expected value of cards he is getting, then A is benefitting. The more A benefits, the higher his chances of winning. Then, clearly $$E(\\text{smallest})$$, $$E(\\text{second-smallest})$$ are smaller than $$5.5$$, the expected value on a random card. $$E(\\text{third-smallest})$$ is equal to $$5.5$$, and $$E(\\text{fourth-smallest})$$, $$E(\\text{fifth-smallest, or largest})$$ are greater than $$5.5$$. This means that discarding the smallest card and", "## Pascal’s Triangle Combinations – product of 6 integers $\\dbinom{0}{0}$ $\\dbinom{1}{0}$ $\\dbinom{1}{1}$ $\\dbinom{2}{0}$ $\\dbinom{2}{1}$ $\\dbinom{2}{2}$ $\\dbinom{3}{0}$ $\\dbinom{3}{1}$ $\\dbinom{3}{2}$ $\\dbinom{3}{3}$ $\\dbinom{4}{0}$ $\\dbinom{4}{1}$ $\\dbinom{4}{2}$ $\\dbinom{4}{3}$ $\\dbinom{4}{4}$ $\\dbinom{5}{0}$ $\\dbinom{5}{1}$ $\\dbinom{5}{2}$ $\\dbinom{5}{3}$ $\\dbinom{5}{4}$ $\\dbinom{5}{5}$ $\\dbinom{6}{0}$ $\\dbinom{6}{1}$ $\\dbinom{6}{2}$ $\\dbinom{6}{3}$ $\\dbinom{6}{4}$ $\\dbinom{6}{5}$ $\\dbinom{6}{6}$ Take, for example, $\\dbinom{5}{3}$ and the six integers surronding it, $\\dbinom{4}{2}$ $\\dbinom{4}{3}$ $\\dbinom{5}{4}$ $\\dbinom{6}{4}$ $\\dbinom{6}{3}$ $\\dbinom{5}{2}$ The product of these 6 integers is: $\\dbinom{4}{2} \\cdot \\dbinom{4}{3} \\cdot \\dbinom{5}{4} \\cdot \\dbinom{6}{4} \\cdot \\dbinom{6}{3} \\cdot \\dbinom{5}{2}$ $= 6 \\times 4 \\times 5 \\times 15 \\times 20 \\times 10$ $= \\; 360000$ $= \\; 600^2$ The product of 6 integers surrounding a number interior is always a square number $\\dbinom{n}{k} \\; = \\; n! \\,/ \\,(k! \\, (n-k)!)$ $\\dbinom{n}{k} \\; = \\; \\dbinom{n-1}{k-1} \\; + \\; \\dbinom{n-1}{k}$ We have $\\dbinom{n}{k}$ surrounded by 6 integers: ………. $\\dbinom{n-1}{k-1}$ $\\dbinom{n-1}{k}$ ….. $\\dbinom{n}{k-1}$ $\\dbinom{n}{k}$ $\\dbinom{n}{k+1}$ ………. $\\dbinom{n+1}{k}$ $\\dbinom{n+1}{k+1}$ $\\dbinom{n-1}{k-1} \\cdot \\dbinom{n-1}{k} \\cdot \\dbinom{n}{k+1} \\cdot \\dbinom{n+1}{k+1} \\cdot \\dbinom{n+1}{k} \\cdot \\dbinom{n}{k-1}$ $\\dbinom{n-1}{k-1} \\; = \\; (n-1)! \\,/ \\,((k-1)! \\, (n-k)!)$ $\\dbinom{n-1}{k} \\; = \\; (n-1)! \\,/ \\,(k! \\, (n-k-1)!)$ $\\dbinom{n}{k+1} \\; = \\; n! \\,/ \\,((k+1)! \\, (n-k-1)!)$ $\\dbinom{n+1}{k+1} \\; = \\; (n+1)! \\,/ \\,((k+1)! \\, (n-k)!)$ $\\dbinom{n+1}{k} \\; = \\; (n+1)! \\,/ \\,(k! \\, (n-k+1)!)$ $\\dbinom{n}{k-1} \\; = \\; n! \\,/ \\,((k-1)! \\, (n-k+1)!)$ numerator: $(n-1)! \\, (n-1)! \\, n! \\, (n+1)! \\, (n+1)! \\, n! \\; = \\;", "roll ten dice. I like to reduce the case approach to two multinomial coefficients for [choose] $\\times$ [place] to make it more mechanical and thus less error prone. $5-1-1-1-1-1: \\dbinom6{1,5}\\dbinom{10}{5,1,1,1,1,1}$ $4-2-1-1-1-1: \\dbinom6{1,1,4}\\dbinom{10}{4,2,1,1,1,1}$ $3-3-1-1-1-1: \\dbinom6{2,4}\\dbinom{10}{3,3,1,1,1,1}$ $3-2-2-1-1-1: \\dbinom6{1,2,3}\\dbinom{10}{3,2,2,1,1,1}$ $2-2-2-2-1-1: \\dbinom6{4,2}\\dbinom{10}{2,2,2,2,1,1}$ Add up and divide by $6^{10}$ PS Or you might like to write it as [choose] $\\times$ [permute], viz. $5-1-1-1-1-1: \\dbinom6{1,5}\\times\\dfrac{10!}{5!}$ $4-2-1-1-1-1: \\dbinom6{1,1,4}\\times\\dfrac{10!}{4!2!}$ $3-3-1-1-1-1: \\dbinom6{2,4}\\times\\dfrac{10!}{3!3!}$ $3-2-2-1-1-1: \\dbinom6{1,2,3}\\times\\dfrac{10}{3!2!2!}$ $2-2-2-2-1-1: \\dbinom6{4,2}\\times\\dfrac{10!}{2!2!2!2!}$ Based on the comments below, I try a different approach. I consider $1-$the number of cases where all numbers are not there. Choose 5 out of 6 numbers. Each dice can be any number = $5^{10}$ So, the total number of ways is $6^{10} - 6*5^{10}$ • Double Counting! Suppose you pick the first die and its a one, there's another one in the six you didn't pick. Suppose you picked that place, and its a one, and the one among the six you didn't pick is in first place. You've counted these at least twice. – Graham Kemp Feb 16 '16 at 1:37 • This method double counts just as I warned above in the comments. – Gregory Grant Feb 16 '16 at 1:37 • @RobertMingHao Be careful this is a wrong solution, it double counts many possibilities. You should convince yourself of why this is wrong there's an important", "is: $\\set {\\dbinom {a \\ b} {1 \\ 2}, \\dbinom {a \\ b} {1 \\ 3}, \\dbinom {a \\ b} {2 \\ 1}, \\dbinom {a \\ b} {2 \\ 3}, \\dbinom {a \\ b} {3 \\ 1}, \\dbinom {a \\ b} {3 \\ 2} }$ thus demonstrating that there are $\\dfrac {3!} {\\paren {3 - 2}!} = \\dfrac {3!} {1!} = \\dfrac 6 1 = 6$ injections from $A$ to $B$.", "2\\dbinom 21}{\\dbinom 52}=\\dfrac{3}{5}$$", "@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19 @AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22 Nope, that seems likely to be the most concise way to do it. Count ways to pick $n\\in\\{1,2,3,4\\}$ unique numbers, and to arrange them with $n-1$ \"$>$\" signs, to meet the criteria. $$\\begin{array}{|l:l|} \\hline \\rm a{=}a{=}a{=}a & \\dbinom 3 0 \\dbinom 6 1 \\\\\\hdashline\\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \\dbinom 3 1\\dbinom 6 2 \\\\\\hdashline\\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \\dbinom 3 2 \\dbinom 6 3 \\\\\\hdashline\\rm a{>}b{>}c{>}d & \\dbinom 3 3 \\dbinom 6 4 \\\\ \\hline\\end{array}$$ $$\\dfrac{\\sum_{n=1}^4 \\dbinom{3}{n-1}\\dbinom{6}{n}}{6^4} = \\dfrac{\\dbinom 6 1 + 3\\dbinom 6 2 + 3 \\dbinom 6 3 + \\dbinom 6 4}{6^4}$$ - The problem reduces to finding the cardinality of the following set $$A=\\{(a_1,a_2,a_3,a_4)\\in\\{1,\\ldots,6\\}^4: \\ a_1\\leq a_2\\leq a_3 \\leq a_4\\}.$$ Define $$B=\\{(a_1,a_2,a_3,a_4)\\}\\in\\{1,\\ldots 9\\}^4: \\ a_1 < a_2 < a_3 < a_4\\}$$ and $$f:A\\to B, \\quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \\choose 4}.$$ - @Shailesh This method actually does give the correct answer ($126/6^4$); If we", "-1 \\\\ k&=& 3 \\\\ \\dfrac{\\dbinom{14}{3}} {\\dbinom{14}{4}} &=& \\dfrac{4}{11} \\\\ \\hline \\end{array}$$ Mar 26, 2019", "# Why does $\\binom{10}{7} = \\frac{10!}{(10-7)!7!}$ We just learned that: $\\dbinom{10}{7}= \\frac{10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4}{7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}$, so that: If you throw a dice 10 times, the probability of getting $6$, $7$ of the times is: $\\dbinom{10}{7} \\times {\\frac{1}{6}}^7 \\times {\\frac{5}{6}}^3$, because $\\dbinom{10}{7}$ will give us the number of different ways you can get $7$ \"correct\" out of $10$. I wonder why this works. Why does $\\dbinom{10}{7}$ work as it does? (In my search I stumbled upon this way of writing it: $\\dbinom{10}{7} = \\frac{10!}{(10-7)!7!}$. It is a little bit different, but maybe it is more correct? 4 2022-07-25 20:42:53 Source Share", "$\\dbinom n k = \\dbinom n {k + 1}$ So: $k \\ge \\dfrac {n - 1} 2$ By the Symmetry Rule for Binomial Coefficients, we also have that: $\\dfrac {n - 1} 2 < k \\le n \\iff \\dbinom n k > \\dbinom n {k + 1}$ and so for $\\dbinom n k = \\dbinom n {k + 1}$ it is not the case that $\\dfrac {n - 1} 2 < k \\le n$. So: $k \\le \\dfrac {n - 1} 2$ Hence that means: $k = \\dfrac {n - 1} 2$ which means: $n = 2 k + 1$ and so $n$ is odd. $\\blacksquare$", "+0 # binomial question 0 250 3 (a) Simplify $$\\dfrac{\\dbinom{n}{k}}{\\dbinom{n}{k - 1}}.\\$$ (b) For some positive integer $$n$$ the expansion of $$(1 + x)^n$$ has three consecutive coefficients $$a,b,c$$ that satisfy $$a:b:c = 1:7:35.$$ What must $$n$$ be? Jul 20, 2021 #1 +26226 +2 (a) Simplify $$\\dfrac{\\dbinom{n}{k}}{\\dbinom{n}{k - 1}}$$. $$\\begin{array}{|rcll|} \\hline \\dbinom{n}{k - 1} &=& \\dfrac{n!}{(k-1)!\\Big( n-(k-1)\\Big)!} \\\\\\\\ &=& \\dfrac{n!}{(k-1)!\\Big( n-k+1)\\Big)!} \\\\\\\\ && \\boxed{ (k-1)!k=k!\\\\ (k-1)! =\\dfrac {k!}{k} } \\\\\\\\ &=& \\dfrac{n!*k}{k!\\Big( n-k+1)\\Big)!} \\\\\\\\ && \\boxed{ \\Big( n-k+1)\\Big)!=(n-k)!(n-k+1) } \\\\\\\\ &=& \\dfrac{n!*k}{k!(n-k)!(n-k+1) } \\\\\\\\ &=& \\dfrac{n!}{k!(n-k)! } *\\dfrac{k}{n-k+1} \\\\\\\\ && \\boxed{ \\dbinom{n}{k}=\\dfrac{n!}{k!(n-k)! } } \\\\\\\\ &=& \\dbinom{n}{k} *\\dfrac{k}{n-k+1} \\\\\\\\ \\hline \\dfrac{\\dbinom{n}{k}}{\\dbinom{n}{k - 1}} &=& \\dfrac{\\dbinom{n}{k}}{\\dbinom{n}{k} *\\dfrac{k}{n-k+1}} \\\\\\\\ \\dfrac{\\dbinom{n}{k}}{\\dbinom{n}{k - 1}} &=& \\dfrac{1}{\\dfrac{k}{n-k+1}} \\\\\\\\ \\dfrac{\\dbinom{n}{k}}{\\dbinom{n}{k - 1}} &=& \\dfrac{n-k+1}{k}\\\\ \\hline \\end{array}$$ Jul 21, 2021 #3 0 Thank you so much for answering my question but could you please elaborate on how you got $$(n-k+1)!=(n-k)!(n-k+1)$$ Guest Jul 28, 2021 #2 +26226 +1 (b) For some positive integer $$n$$ the expansion of $$(1 + x)^n$$ has three consecutive coefficients $$a,~b,~c$$ that satisfy $$a:b:c = 1:7:35$$. What must $$n$$ be? $$\\text{Let a=\\dbinom{n}{k - 1} } \\\\ \\text{Let b=\\dbinom{n}{k} } \\\\ \\text{Let c=\\dbinom{n}{k+1} }$$ $$\\begin{array}{|rcll|} \\hline \\dfrac{b}{a} = \\dfrac{7}{1} &=& \\dfrac{\\dbinom{n}{k}}{\\dbinom{n}{k - 1}} \\\\\\\\ \\dfrac{7}{1} &=& \\dfrac{\\dbinom{n}{k}}{\\dbinom{n}{k - 1}} \\\\\\\\ \\dfrac{7}{1} &=& \\dfrac{n-k+1}{k} \\\\\\\\ 7k &=& n-k+1 \\\\ 8k &=& n+1 \\\\ \\mathbf{n}", "= \\dfrac{5 \\times 10^{-9} \\mathrm{~ C}}{(8.85 \\times 10^{-12} \\,\\text{F/m}) (0.0225 \\mathrm{~m^2})} \\\\ \\boxed{E_i = 2.5 \\times 10^{4} \\,\\text{N/C}} \\end{gather*}", "is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$.", "(4 \\times 3 \\times 2 \\times 1) - (3 \\times 2 \\times 1)= 720 + 24 -6 = \\text{738}$ $$\\dfrac{6! - 2!}{2!}$$ $\\dfrac{(6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1) - (2 \\times 1)}{2 \\times 1}= \\dfrac{720 - 2}{2} = \\text{359}$ $$\\dfrac{2! + 3!}{5!}$$ $\\dfrac{(2 \\times 1) + (3 \\times 2 \\times 1)}{5 \\times 4 \\times 3 \\times 2 \\times 1}= \\dfrac{2 + 6}{120} = \\dfrac{1}{15}$ $$\\dfrac{2! + 3! - 5!}{3! - 2!}$$ $\\dfrac{2 + 6 - 120}{6 - 2}= \\dfrac{-\\text{112}}{4} = -\\text{28}$ $$(3!)^{3}$$ $$6 \\times 6 \\times 6 = \\text{216}$$ $$\\dfrac{3! \\times 4!}{2!}$$ $\\dfrac{(3 \\times 2 \\times 1) \\times (4 \\times 3 \\times 2 \\times 1)}{2 \\times 1} = \\text{72}$ Calculate the following using a calculator: $$\\dfrac{12!}{2!}$$ $\\dfrac{(12 \\times 11 \\times 10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1)}{2 \\times 1} = \\text{239 500 800}$ $$\\dfrac{10!}{20!}$$ $$\\text{1,49} \\times \\text{10}^{-\\text{12}}$$ $$\\dfrac{10! + 12!}{5! + 6!}$$ $$\\text{574 560}$$ $$5!(2! + 3!)$$ $$\\text{960}$$ $$(4!)^{2}(3!)^{2}$$ $$\\text{20 736}$$ Show that the following is true: $$\\dfrac{n!}{(n-2)!} = n^{2} - n$$ \\begin{align*} \\dfrac{n!}{(n-2)!} & = \\dfrac{n \\times (n-1) \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}}{\\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}} \\\\ \\\\ &= n(n-1) = n^{2} - n \\end{align*} $$\\dfrac{(n-1)!}{n!}", "= \\dfrac{1}{n}$$ $\\dfrac{(n-1)!}{n!} = \\dfrac{\\cancel{(n-1)} \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}}{n \\times \\cancel{(n-1)} \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}} = \\dfrac{1}{n}$ $$\\dfrac{(n-2)!}{(n-1)!} = \\dfrac{1}{n-1} \\text{ for } n>1$$ $\\dfrac{(n-2)!}{(n-1)!} = \\dfrac{\\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}}{(n-1) \\times \\cancel{(n-2)} \\times \\cancel{(n-3)} \\times \\ldots \\times \\cancel{3} \\times \\cancel{2} \\times \\cancel{1}} = \\dfrac{1}{n-1}$", "\\dfrac{{3 \\times 2!}}{{\\not 2!}}}}{{\\dfrac{{\\not 6 \\times 5 \\times 4 \\times \\not 3!}}{{\\not 3! \\times 3! \\times 2 \\times 1}}}}$ $= \\dfrac{9}{{20}} \\times \\dfrac{6}{{20}} = \\dfrac{{27}}{{200}}\\,\\,\\,\\,\\,\\,\\,.......(1)$ Now case II as the number of balls are same just colours are different, we have: $= \\dfrac{{{}^3{C_2} \\times {}^3{C_1}}}{{{}^6{C_3}}} \\times \\dfrac{{{}^1{C_1} \\times {}^2{C_1} \\times {}^3{C_1}}}{{{}^6{C_3}}}$ $= \\dfrac{{\\dfrac{{3!}}{{2!1!}} \\times \\dfrac{{3!}}{{2!1!}}}}{{\\dfrac{{6!}}{{3!3!}}}} \\times \\dfrac{{\\dfrac{{2!}}{{1!1!}} \\times 1 \\times \\dfrac{{3!}}{{2!1!}}}}{{\\dfrac{{6!}}{{3!3!}}}}$ $= \\dfrac{{\\dfrac{{3 \\times \\not 2!}}{{\\not 2!}} \\times \\dfrac{{3 \\times \\not 2!}}{{\\not 2!}}}}{{\\dfrac{{\\not 6 \\times 5 \\times 4 \\times \\not 3!}}{{3! \\times 3\\not ! \\times \\not 2 \\times 1}}}} \\times \\dfrac{{2! \\times \\dfrac{{3 \\times 2!}}{{\\not 2!}}}}{{\\dfrac{{\\not 6 \\times 5 \\times 4 \\times \\not 3!}}{{\\not 3! \\times 3! \\times 2 \\times 1}}}}$ $= \\dfrac{9}{{20}} \\times \\dfrac{6}{{20}} = \\dfrac{{27}}{{200}}\\,\\,\\,\\,\\,\\,\\,.......(2)$ $= \\dfrac{{27}}{{200}}.....(2)$ Probability of 3 later balls being all of different colours $= \\dfrac{{27}}{{200}} + \\dfrac{{27}}{{200}}$(from $e{q^n}$ 1 & 2) $= \\dfrac{{2 \\times 27}}{{200}}$$= \\dfrac{{27}}{{100}}$ Note: Cube root need to group digit in no. 3 and taking the unit place digit of the first group & ten’s from $I{I^{nd}}$. In mathematics, a cube root of a number x is a number y such that $\\;{y^3}\\; = \\;x$. All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube", "us $\\dbinom{n+1}{2}$ ways, for a total of $3\\dbinom{n+1}{2}$. The last case gives us $\\dbinom{n+1}{1}=n+1$ ways. Therefore, the total number of possible ways where there are no isolated men is $\\[\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1).\\]$ The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or $\\[2\\dbinom{n+1}{2}+(n+1).\\]$ Thus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that $\\[\\dfrac{\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1)}{2\\dbinom{n+1}{2}+(n+1)}\\le\\dfrac{1}{100}.\\]$ the numerator simplifies to $(n+1)^2$. The denominator simplifies to $\\dfrac{(n+6)(n+2)(n+1)}{6}$ so the whole faction simplifies to $\\dfrac{6(n+1)}{(n+6)(n+2)}$ Since $\\dfrac{n+1}{n+2}$ is slightly less than 1 when $n$ is large, $\\dfrac{6}{n+6}$ will be close to $\\dfrac{1}{100}$. They equal each other when $n = 594$. If we let $n= 595$ or $593$, we will notice that the answer is $\\boxed{594}$", "&=& \\mathbf{8k-1} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\dbinom{n}{k + 1} &=& \\dfrac{n!}{(k+1)!\\Big( n-(k+1)\\Big)!} \\\\\\\\ &=& \\dfrac{n!}{(k+1)!\\Big( n-k-1)\\Big)!} \\\\\\\\ && \\boxed{ (k+1)!=k!(k+1)} \\\\\\\\ &=& \\dfrac{n!}{k!(k+1)( n-k-1)!} \\\\\\\\ && \\boxed{ ( n-k-1)!(n-k)=(n-k)! \\\\ ( n-k-1)!=\\dfrac{(n-k)!}{n-k} } \\\\\\\\ &=& \\dfrac{n!*(n-k)}{k!(k+1)(n-k)!} \\\\\\\\ &=& \\dfrac{n!}{k!(n-k)! } *\\dfrac{(n-k)}{(k+1)} \\\\\\\\ && \\boxed{ \\dbinom{n}{k}=\\dfrac{n!}{k!(n-k)! } } \\\\\\\\ &=& \\dbinom{n}{k} *\\dfrac{(n-k)}{(k+1)} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\dfrac{c}{b} = \\dfrac{35}{7} &=& \\dfrac{\\dbinom{n}{k+1}}{\\dbinom{n}{k}} \\\\\\\\ \\dfrac{35}{7} &=& \\dfrac{\\dbinom{n}{k} *\\dfrac{(n-k)}{(k+1)}}{\\dbinom{n}{k}} \\\\\\\\ \\dfrac{35}{7} &=&\\dfrac{(n-k)}{(k+1)} \\\\\\\\ 35(k+1) &=&7(n-k) \\\\ 42k &=& 7n-35 \\\\ \\mathbf{k} &=& \\mathbf{ \\dfrac{7n-35}{42}} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline n &=& 8k-1 \\quad | \\quad \\mathbf{k=\\dfrac{7n-35}{42}} \\\\\\\\ n &=& 8\\left(\\dfrac{7n-35}{42}\\right)-1 \\\\\\\\ n &=& \\dfrac{4(7n-35)}{21}-1 \\\\\\\\ n &=& \\dfrac{4(7n-35)-21}{21} \\\\\\\\ 21n &=&4(7n-35)-21\\\\ 21n &=&28n-140-21\\\\ 7n &=& 161 \\quad | \\quad :7 \\\\ \\mathbf{n} &=& \\mathbf{23} \\\\ \\hline k &=& \\dfrac{7n-35}{42} \\\\\\\\ k &=& \\dfrac{7*23-35}{42} \\\\\\\\ \\mathbf{k} &=& \\mathbf{3} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline a:b:c =\\dbinom{n}{k - 1}:\\dbinom{n}{k}:\\dbinom{n}{k + 1} &=& 1:7:35 \\\\\\\\ \\dbinom{n}{k - 1}:\\dbinom{n}{k}:\\dbinom{n}{k + 1} &=& 1:7:35 \\\\\\\\ \\dbinom{23}{3 - 1}:\\dbinom{23}{3}:\\dbinom{23}{3 + 1} &=& 1:7:35 \\\\\\\\ \\dbinom{23}{2}:\\dbinom{23}{3}:\\dbinom{23}{4} &=& 1:7:35 \\\\\\\\ 253:1771:8855 &=& 1:7:35 \\\\ \\hline \\end{array}$$ Jul 21, 2021", "\\epsilon _0Gm_pm_p}$$ $$\\dfrac{F_e}{F_g}=\\dfrac{9\\times 10^9\\times (1.6\\times 10^{-19})^2}{6.67\\times 10^{-11}\\times \\left(1.67\\times 10^{-27}\\right)^2}$$ $$\\dfrac{F_e}{F_g}=1.3\\times 10^{36}$$", "5}{10 \\times 9 \\times 8 \\times 7 \\times 6 }$ 5 red out of 5: $\\dfrac{5 \\times 4 \\times 3 \\times 2 \\times 1}{5 \\times 4 \\times 3 \\times 2 \\times 1} \\times \\dfrac{5 \\times 4 \\times 3 \\times 2 \\times 1}{10 \\times 9 \\times 8 \\times 7 \\times 6 }$ Add those up and you get $\\frac{1}{2}$. Also add 0, 1 or 2 reds out of 5 and you get $1$. - add comment", "one of mine: this could be made on any of the four choices. The other three choices follow the pattern above. The choices of the matching digit in the terms below is the first in the first line, the second in the second line, etc. $$P_2 = \\dfrac{4}{10} \\times \\dfrac{6}{9} \\times \\dfrac{5}{8} \\times \\dfrac{4}{7}$$ $$+ \\dfrac{6}{10} \\times \\dfrac{4}{9} \\times \\dfrac{5}{8} \\times \\dfrac{4}{7}$$ $$+ \\dfrac{6}{10} \\times \\dfrac{5}{9} \\times \\dfrac{4}{8} \\times \\dfrac{4}{7}$$ $$+ \\dfrac{6}{10} \\times \\dfrac{5}{9} \\times \\dfrac{4}{8} \\times \\dfrac{4}{7}$$ $$P_2 = \\dfrac{1}{1} \\times \\dfrac{2}{3} \\times \\dfrac{1}{1} \\times \\dfrac{1}{7}$$ $$+ \\dfrac{2}{1} \\times \\dfrac{1}{3} \\times \\dfrac{1}{1} \\times \\dfrac{1}{7}$$ $$+ \\dfrac{1}{1} \\times \\dfrac{1}{3} \\times \\dfrac{1}{1} \\times \\dfrac{2}{7}$$ $$+ \\dfrac{1}{1} \\times \\dfrac{1}{3} \\times \\dfrac{1}{1} \\times \\dfrac{2}{7}$$ $$P_2 = \\dfrac{4*2}{3*7} = \\dfrac{8}{21}$$ That's the Case II probability. Add those two: $$P_{1,2} = \\dfrac{1}{14} + \\dfrac{8}{21} = \\dfrac{3}{42} + \\dfrac{16}{42} = \\dfrac{19}{42}$$ That's the complete probability of condition of the question not being satisfied, so we subtract that from 1 to answer the question. $$P = 1 - \\dfrac{19}{42} = \\dfrac{23}{42}$$ That's the answer to the question. This would be an extremely challenging PS. So far as I can tell, there's not a significantly short solution than the one I have shown. Does all this make sense? Mike _________________ Mike McGarry Magoosh Test Prep Education is not the filling of a pail, but the lighting" ]

[ "# Marbles in the Bag You have three bags, each containing two marbles. Bag A contains two white marbles, Bag B contains two black marbles, and Bag C contains one white marble and one black marble. You pick a random bag and take out one marble. It is a white marble. What is the probability that the remaining marble from the same bag is also white? Probability is $$\\dfrac{a}{b}$$, answer as $$a+b$$ ×", "$n_{1,1}$ balls from the second bag. number of combinations without repetition with limited supply The above question helps explain how to find the number of combinations of $n_{1,1}$ marbles from bag two. Essentially you analyze the following polynomial for the coefficients: $$P = \\prod_{j=1}^{c_2}{\\sum_{k=0}^{n_{2,j}}x^k}$$ For the question example: $$P=(1+x)(1+x)(1+x+x^2+x^3) = 1+3x+4x^2+4x^3+3x^4+x^5$$ and the answer is 4 because the coefficient of $x^2$ is 4. This makes sense as the potential cases can be described as the colors paired with blue (black/black, red/black, white/black, and red/white). b) The problem becomes more complicated when each bag contains more than 2 types of marbles. For the case of 3 colors in bag 1, the simular case requires that one consider both an x and a y. Define a polynomial: $$F = \\prod_{j=1}^{c_2}{\\sum_{k=0}^{n_{2,j}}(\\sum_{l=0}^{k} x^ly^{k-l})}$$ Convert it into an expanded form and note that the coefficient of $x^{n_{1,1}}y^{n_{1,2}}$ is the number of relevant combinations. For examples, if $n=5$, $c_1=3$, $c_2=3$, $n_{2,1}=2$, $n_{2,2}=2$, $n_{2,3}=1$ then: $$F=(1+x+y+x^2+x*y+y^2)^2*(1+x+y)$$ $$F=x^5+3 x^4 y+3 x^4+5 x^3 y^2+8 x^3 y+5 x^3+5 x^2 y^3+11 x^2 y^2+11 x^2 y+5 x^2+3 x y^4+8 x y^3+11 x y^2+8 x y+3 x+y^5+3 y^4+5 y^3+5 y^2+3 y+1$$ If bag 2 contains 2 black, 2 red, and 1 green (for instance) there will be 11 combinations. c) For the final case of an arbitrary number of colors in", "learned prior – how strongly we expect any newly encountered bag to conform to the distribution for the population prototype beta. For instance, suppose that we observe that bag1 consists of all blue marbles, bag2 consists of all green marbles, bag3 all red, and so on. This doesn’t tell us to expect a particular color in future bags, but it does suggest that bags are very regular—that all bags consist of marbles of only one color. var colors = ['black', 'blue', 'green', 'orange', 'red']; var observedData = [ {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag4', draw: 'orange'}] var predictives = Infer({method: 'MCMC', samples: 30000}, function(){ var alpha = gamma(2,2) var beta = dirichlet(ones([5, 1])) var getBag = mem(function(bagName) { var colorProbs = T.toScalars(dirichlet(beta.mul(alpha))); return Categorical({ vs: colors, ps: colorProbs }) }) map(function(d) { observe(getBag(d.bag), d.draw) }, observedData) return {bag1: sample(getBag('bag1')), bag2: sample(getBag('bag2')), bag3: sample(getBag('bag3')), bag4: sample(getBag('bag4')), bagN: sample(getBag('bagN')), alpha: alpha} });", "yellow tiles and 22 black tiles, what i [#permalink] 19 Sep 2018, 10:07 1 KUDOS Expert's post Carcass wrote: If a bag contains 18 yellow tiles and 22 black tiles, what is the probability that a tile selected at random from the bag will be yellow? A. $$\\frac{1}{10}$$ B. $$\\frac{9}{20}$$ C. $$\\frac{11}{20}$$ D. $$\\frac{9}{11}$$ E. $$\\frac{11}{9}$$ Number of YELLOW tiles in bag = 18 TOTAL number of tiles in bag = 18 + 22 = 40 P(selected tile is yellow) = (# of possible outcomes in which the selected tile is yellow)/(TOTAL # of possible outcomes) = 18/40 = 9/20 Cheers, Brent _________________ Brent Hanneson – Creator of greenlighttestprep.com Re: If a bag contains 18 yellow tiles and 22 black tiles, what i [#permalink] 19 Sep 2018, 10:07 Display posts from previous: Sort by", "newly encountered bag to conform to the distribution for the population prototype phi. For instance, suppose that we observe that bag1 consists of all blue marbles, bag2 consists of all green marbles, bag3 all red, and so on. This doesn’t tell us to expect a particular color in future bags, but it does suggest that bags are very regular—that all bags consist of marbles of only one color. var colors = ['black', 'blue', 'green', 'orange', 'red']; var observedData = [ {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag4', draw: 'orange'}] var predictives = Infer({method: 'MCMC', samples: 30000}, function(){ // the global prototype mixture: var phi = dirichlet(ones([colors.length, 1])) // regularity parameters: how strongly we expect the global prototype to project // (ie. determine the local prototypes): var alpha = gamma(2,2) var prototype = T.mul(phi, alpha) var makeBag = mem(function(bag){ var colorProbs = dirichlet(prototype) return Categorical({vs: colors, ps: colorProbs}) }) var obsFn =", "Markers (broad point) Black Sharpie Marker - 1 fine pt& 1 regular Expo Dry Erase Marker (dark color) Expo Dry Erase Marker 1 ERASER 2015-2016 GR4 5 see not 2 see not [email protected]\" see note soft bag K 12 GRI 2pks see note box GR2 2pks see note see note. pricefrom $31. In order to get the correct answer we need to include only blue, yellow and white balls because red and green balls are less than 9. If the outcome of the roll is 1 or 2, two white beads are added to the bag, if the outcome of the roll is a 3 or 4, four white beads are added to the bag, and if the roll is a 5 or 6, six white beads are added to the bag. Question 1129622: a bag contains 5 white and 8 red balls. New! Ziploc ® Brand Bags featuring Frozen 2 Designs. What is the probability that (i) all of them are red (ii) two of them would be red and t. The probability of randomly drawing a red marble is 1 6 6, or 3 8. A ball is now drawn at random from the 2nd urn. Blue Black Red White Silver Grey Turquoise Green Gold Purple Pink Yellow Orange Copper Magenta Teal Lime Sky Blue", "green shirts, 4 white shirts, and 2 red shirts in his shelf. Which of the following questions about these shirts could you use probability to solve? a. If Bill takes 1 shirt from the shelf, how many shirts will be left in the shelf? b. How many shirts did Bill have all together? c. If Bill picks 1 shirt from the shelf without looking, what color will it most likely be? d. How many more green shirts than red shirts did Bill have in the shelf? 218. Sheila packed 10 apples, 25 oranges, 36 strawberries and 5 pineapples in her picnic basket. Which of the following questions about these fruits could you use probability to solve? a. What is the total number of apples and oranges in the basket? b. If a fruit is picked at random, what kind of fruit is least likely to be picked? c. How many more apples than pineapples are there in the basket? d. How many fruits did Sheila pack in all? 219. A bag contains 11 blue marbles, 5 red marbles, and 4 green marbles. Pam draws one marble from the bag, records its color, and then puts it back in the bag. Then, he draws another marble and records its color. What is the probability of drawing 2 red marbles?", "* 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility. Could you please explain why we ignored 1-1-1 combination? Hey jhabib, You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed. Dear Karishma .. I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong . All the approaches are considering 3 bags as identical , but in problem they are distinct . My approach to this problem : I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 . So first distribute 3 balls one ball each to each bag . select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways . so total 60 ways . Now we are left with 2 balls and 3 bags each containing one ball each . for first ball we have 3 options and second ball again 3", "balls, and Bag B contains 2 red and 5 white balls. What is the probability that (i) all of them are red (ii) two of them would be red and t. Then an object is drawn at random from the bag. It is also considered for the case of conditional probability. find the probablity that the transferred ball is balck. The ratio of green balls to blue balls is 1 : 2 Bag 3 contains:. If one ball is drawn from the bag, find the probability that it will be either white or green. Question: A bag contains 8 red marbles, 6 white marbles, and 8 blue marbles. From the above formula we have get answer 47 because 6 + 8 + 10 + 12 + 15- 5 + 1 = 47 But it is not correct. S M: A special Poké Ball that appears in your Bag out of nowhere in the Entree Forest. Penn Tour Extra Duty Tennis Balls 24 Can Case. Bag A contains$3$white and$4$red balls whereas bag B contains$4$white and$3$red balls. Image: Shutterstock. A black marble is chosen without replacement. Three balls are drawn at random from the bag. 75”, available with set-up fee of$120 (V) and additional run charge of $40 (V). All outcomes must be shown from each node. A bag contains", "disk in a bag is either blue, yellow or green. &nbs [#permalink] 09 Sep 2018, 08:24 Display posts from previous: Sort by", "shows, mem is particularly useful when writing hierarchical models because it allows us to associate arbitrary random draws with categories across entire runs of the program. In this case it allows us to associate a particular mixture of marble colors with each bag. The mixture is drawn once, and then remains the same thereafter for that bag. Intuitively, you can see how each sample is sufficient to learn a lot about what that bag is like; there is typically a fair amount of similarity between the empirical color distributions in each of the four samples from bagA. In contrast, you should see a different distribution of samples from bagB. Now let’s explore how this model learns about the contents of different bags. We represent the results of learning in terms of the posterior predictive distribution for each bag: a single hypothetical draw from the bag. We will also draw a sample from the posterior predictive distribution on a new bag, for which we have had no observations. var colors = ['black', 'blue', 'green', 'orange', 'red']; var observedData = [ {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'black'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag2', draw: 'blue'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'blue'}, {bag: 'bag2', draw:", "1. ## Marbles in Bags Bag A contains 3 black marbles and 1 white marble and Bag B contains 1 black marble and 3 white marbles. If two marbles are randomly taken from Bag A and placed in Bag B, then a marble is randomly selected from Bag B, what is the probability that the marble selected from Bag B is black? I've tried to draw a tree, but I must be doing something wrong cause I can't get the correct answer. Could someone give me some guidance? 2. Hello mneox Originally Posted by mneox Bag A contains 3 black marbles and 1 white marble and Bag B contains 1 black marble and 3 white marbles. If two marbles are randomly taken from Bag A and placed in Bag B, then a marble is randomly selected from Bag B, what is the probability that the marble selected from Bag B is black? I've tried to draw a tree, but I must be doing something wrong cause I can't get the correct answer. Could someone give me some guidance? The number of ways of selecting $2$ marbles from the four in bag A is $\\binom42=6$. The number of ways of selecting the white marble and one of the black marbles from bag A $= 3$. Therefore the probability that", "a bag of 3 blue and 4 red marbles without replacement is right at about 0. SECTION D 26. asked • 04/01/17 A bag contains 5 white balls and 9 red balls. Stussy Mock Neck Sweat Olive. Find the expectation of a man who is allowed to draw two balls from the bag and who is to get one rupee for each black ball and two rupee for each white ball drawn Solution [] 03. More details… Tea Bag Squeezer. !!5 of the counters are red!3 of the counters are purple!2 of the counters are white!Sharon chooses a counter at random, records the colour, then replaces it. A bag contains 3 black beads, 5 red beads and 2 green beads. A bag contains 8 white balls and 6 black balls. One ball is drawn from the bag. A bag contains 5 red and 7 white balls. total number of Sample space = 12C5. To ask Unlimited Maths doubts download Doubtnut from - https://goo. Roll into tiny balls, dust with fine granulated sugar and press into rubber or plastic molds. Question 1102302: A bag contains 8 red and 5 white balls. Visit bowlingball. White Cells & Granulocytes. a bag contains seven red b***s numbered 2,4,5,6,7,8,10 and three white b***s numbered 1,3, and 9. 4 red, 2 white and 3", "# Probably not! A bag contains exactly three red marbles, five yellow marbles and two blue marbles. If three marbles are to be drawn from the bag at the same time, what is the probability that all three will be the same color? If your answer is a/b where a and b are coprime integers, express your answer as a+B. × Problem Loading... Note Loading... Set Loading...", "of candy. A bag contains 10 red marbles, 8 blue marbles and 2 yellow marbles. Notice that “red” appears twice and so “red” actually has twice the probability of any of the other colors. The answer is 16. Dns66 ダウンロード. What is the sample space for 2 spins of the first spinner? b. Find the probability that the spinner will stop on purple and the cube will show a six. tossing two fair coins and having both land tails up _____ 6. In $6$ spins, two land on red while the third lands on either blue or yellow. Spinning a number less than 7 20. This Probability Worksheet produces problems using a spinner. A spinning wheel is divided into 9 equal sections and each section is labeled with a number. A spinner is divided into five colored sections that are not of equal size: red, blue, green, yellow, and purple. For example, when you flip a coin, the probability that it will turn into a bottle of ketchup is 0. ANYONE GOOD IN PROBABILITY?D. The larger the section, the more probable it will be to land on it. The possible combinations are: * Red + Blue * Red +. If I get a 3 on the first spinner, then I must get a 1 or 2 on the", "Question A bag contains 8 red marbles, 4 blue marbles and 2 green marbles. If two marbles are drawn out of the bag, what is the exact probability that both marbles drawn will be red?", "bag. 7. Two cards are drawn from a deck of cards. Determine the probability of each of the following events: 1. $P$ (heart or club) 2. $P$ (heart and club) 3. $P$ (red or heart) 4. $P$ (jack or heart) 5. $P$ (red or ten) 6. $P$ (red queen or black jack) 1. A box contains 5 purple and 8 yellow marbles. What is the probability of successfully drawing, in order, a purple marble and then a yellow marble? {Hint: In order means they are not replaced.} 2. A bag contains 4 yellow, 5 red, and 6 blue marbles. What is the probability of drawing, in order, 2 red, 1 blue, and 2 yellow marbles? 3. A card is chosen at random. What is the probability that the card is black and is a 7? Mixed Review 1. A circle is inscribed within a square, meaning the circle's diameter is equal to the square’s side length. The length of the square is 16 centimeters. Suppose you randomly threw a dart at the figure. What is the probability the dart will land in the square, but not in the circle? 2. Why is $7-14x^4+7xy^5-1x^{-1}=8x^2 y^3$ not considered a polynomial? 3. Factor $72b^5 m^3 w^9-6(bm)^2 w^6$ . 4. Simplify $2^5-7^3 a^3 b^7+3^5 a^3 b^7-2^3$ . 5. Bleach breaks down cotton", "Bag A has Red and Green marbles and Bag B contains Blue and Yellow. If you want to calculate of picking a red marble, you calculate the probability of choosing from bag A instead of B and then multiply by the probability of getting Red from Bag A. Contents of bag B would not matter. One way of checking if the events $$A$$ and $$B$$ are independent is calculate the conditional probability of one of them, let's say $$P(A)$$, and see if is the same as $$P(A|B)$$, or probability of $$A$$ occurs when $$B$$ already happened (actually you should check the in-dependency of two events by the very way you're using to calculate $$P(A\\cap B)$$). By this you easily conclude that, if $$A$$: the individual is male; $$B$$: the individual is infected: $$P(A)=\\frac{43}{81}\\neq \\frac{36}{59}=P(A|B)$$ So the events aren't independent.", "the same time, player two gets a point. If one red and one green marble are pulled at the same time, both players get a point. Is this a fair game?", "# Math Help - Probility 1. ## Probility A bag contains 3 orange, 3 red, and 2 black balls. Two balls are to be selected at random, and the first ball selected will be placed back in bag before second marble is selected. What is the probablility that both balls selected will be black? 1/8 or 1/16. I think 1/8 because it is a total of 8 balls. Am I correct? 2. ## Re: Probility Originally Posted by tahirakw A bag contains 3 orange, 3 red, and 2 black balls. Two balls are to be selected at random, and the first ball selected will be placed back in bag before second marble is selected. What is the probablility that both balls selected will be black? 1/8 or 1/16. I think 1/8 because it is a total of 8 balls. Am I correct? Replacing the ball after it is drawn makes the events independent. Therefore, multiply: $\\left(\\frac{2}{8}\\right)^2$. 3. ## Re: Probility Thanks. so that gives me 4/64 when I reduce its 1/16?" ]

[ "this. :) I appreciate the direction. Apr 10 '17 at 1:48 • I got the answer 0.5724 Is this correct? Apr 10 '17 at 2:16 • It's correct :) Apr 10 '17 at 2:18 • Yay! Thanks for your help. ^.^ Apr 10 '17 at 2:20 • No problem. Glad to help! Apr 10 '17 at 2:20 If we chose from box $A$, and that has $0.55$ probability, then the probability we got a yellow marble is $\\frac{5}{15}$. However if we chose from box $B$, and that has $0.45$ probability, then the probability we got a yellow is $\\frac{6}{11}$. Hence the probably of getting a yellow by the law of total probability is, $$(0.55)(\\frac{5}{15})+(0.45)( \\frac{6}{11})$$ $$\\frac{(0.45)( \\frac{6}{11})}{(0.55)(\\frac{5}{15})+(0.45)( \\frac{6}{11})}$$", "random without replacement then what’s the probability of 3 yellow, I figured if you picked up a marble out of a bag for the first time your chance of getting a yellow marble would be 3/13 since there’s no replacement it would decrease the second time and the third. So the first chance would be 3/13 the second would be 2/12 and the third would be 1/11 – J.sant Dec 4 '18 at 16:58 HINT 1. Find the probability that none are yellow. 2. Remember that if $$A$$ is any event, then $$\\mathbb{P}[A] = 1-\\mathbb{P}[\\bar{A}]$$ UPDATE 1. The first time you grab a yellow ball with a chance of 3/13. You want the negation of this, so the chance of that is $$1-3/13 = 10/30$$. 2. The second time, you grab a yellow (assuming first time you did not) with a chance 3/12. Third time it will be 3/11. 3. You want to draw yellow in neither of the tries, the probability of that is $$\\left[1 - \\frac{3}{13}\\right] \\times \\left[1 - \\frac{3}{12}\\right] \\times \\left[1 - \\frac{3}{11}\\right] = \\frac{10 \\cdot 9 \\cdot 8}{13 \\cdot 12 \\cdot 11} = \\frac{60}{143}$$ 4. Finally you want the probability that any draw results in a yellow, which is $$1 - \\frac{60}{143} = \\frac{83}{143} \\approx 58\\%.$$ • This is what I got. 1-(3/13)*(2/12)*(1/11). Is", "mutually exclusive paths (M1) e.g. $$\\left( {\\frac{1}{3} \\times \\frac{7}{8}} \\right) + \\left( {\\frac{2}{3} \\times \\frac{3}{8}} \\right)$$ , $$\\left( {\\frac{1}{3} \\times \\frac{1}{8}} \\right) + \\left( {\\frac{2}{3} \\times \\frac{5}{8}} \\right)$$ $${\\rm{P}}(F) = \\frac{{13}}{{24}}$$ A1 N2 [4 marks] a(i) and (ii). (i) $$\\frac{1}{3} \\times \\frac{1}{8}$$ (A1) $$\\frac{1}{{24}}$$ A1 (ii) recognizing this is $${\\rm{P}}(E|F)$$ (M1) e.g. $$\\frac{7}{{24}} \\div \\frac{{13}}{{24}}$$ $$\\frac{{168}}{{312}}$$ $$\\left( { = \\frac{7}{{13}}} \\right)$$ A2 N3 [5 marks] b(i) and (ii). A2A1 N3 [3 marks] c. correct substitution into $${\\rm{E}}(X)$$ formula (M1) e.g. $$0 \\times \\frac{1}{9} + 3 \\times \\frac{4}{9} + 6 \\times \\frac{4}{9}$$ , $$\\frac{{12}}{9} + \\frac{{24}}{9}$$ $${\\rm{E}}(X) = 4$$ (euros) A1 N2 [2 marks] d. ## Question A box contains six red marbles and two blue marbles. Anna selects a marble from the box. She replaces the marble and then selects a second marble. Write down the probability that the first marble Anna selects is red. [1] a. Find the probability that Anna selects two red marbles. [2] b. Find the probability that one marble is red and one marble is blue. [3] c. Answer/Explanation ## Markscheme Note: In this question, method marks may be awarded for selecting without replacement, as noted in the examples. $${\\rm{P}}(R) = \\frac{6}{8}\\left( { = \\frac{3}{4}} \\right)$$ A1 N1 [1 mark] a. attempt to find $${\\rm{P(Red)}} \\times {\\rm{P(Red)}}$$ (M1) e.g. $${\\rm{P(}}R{\\rm{)}} \\times {\\rm{P(}}R{\\rm{)}}$$ ,", "was 12.2% He solved like this:[color=beige] .[/color]$\\frac{_{20}C_2\\,\\times\\,_{40}C_2\\,\\times\\,_{10}C_1 }{_{70}C_5}$ I know he was right, but I don't understand it. He reasoned it out like this . . . $\\text{W\\!e draw 5 marbles from the available 70 marbles.}$ [color=beige]. . [/color]$\\text{There are: }\\:_{70}C_5\\text{ possible outcomes.}$ We want 2 White from the available 20 Whites. [color=beige]. . [/color]$\\text{There are: }\\:_{20}C_2\\text{ ways.}$ We want 2 Red from the available 40 Reds. [color=beige]. . [/color]$\\text{There are: }\\:_{40}C_2\\text{ ways.}$ We want 1 Black from the available 10 Blacks. [color=beige]. . [/color]$\\text{There are: }\\:_{10}C_2\\text{ ways.}$ $\\text{Hence, there are: }\\:_{20}C_2\\,\\times\\,_{40}C_2\\,\\times\\,_{10}C_2\\,\\ text{ ways to get 2 White, 2 Red, and 1 Black.}$ $\\text{Therefore: }\\:P(\\text{2 White, 2 Red, 1 Black}) \\;=\\;\\frac{_{20}C_2\\,\\times\\,_{40}C_2\\,\\times\\,_{1 0}C_2}{_{70}C_5}$ Can you give me more explaination to help me understand the relation between my and his way...... I can't make sense how related them are...... June 11th, 2011, 09:11 PM #4 Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: Probablity !!!! Confusing WOW. This is the THIRD posting of this same problem. I've stopped attempting to read it cause I don't feel like jumping back and forth from thread to thread to get all the various pieces. Have you EVER thought to post again on your same thread so that everything is in one place, so as not to confuse everyone? Sigh.", "(8/13) = 0.3043 - P[3B|2R] = ((10C2 * 6C2)/(16C4)) * (8/13) = 0.2282 - P[3B|3R] = ((10C2 * 7C3)/(17C5)) * (8/13) = 0.15663 - P[3B|4R] = ((10C2 * 8C4)/(18C6)) * (8/13) = 0.1044 Basically it is (combination without repetition -for blues) * (combination with repetition - for reds) / total combination) * (8/13). Then I would muliply the number of marbles * probability to get expected number of total balls drawn per game. But this way won't work as the probabilities (although diminishing) are going to total over 1. Any one have any thoughts on this? Probably something very simple I'm overlooking. I'm stumped all the same! Thanks very much for any and all help. 2. ## Re: Marble question - hypergeometric/binomial hybrid You draw a sea of reds, with blues on the i-th, j-th, and n-th (terminating) draws 1<=i < j < n Ex: i = 4, j = 10, n = 15: R R R B R R R R R B R R R R B The probabilitiy of such an $(i, j, n)$ string, $Prob(i, j, n)$ is: $\\left(\\frac{5}{15}\\right)^{i-1} \\left(\\frac{10}{15}\\right) \\left(\\frac{5}{14}\\right)^{j-i-1} \\left(\\frac{9}{14}\\right) \\left(\\frac{5}{13}\\right)^{n-j-1} \\left(\\frac{8}{13}\\right)$ $= \\left(\\frac{(10)(9)(8)}{(15)(14)(13)}\\right) \\left( \\frac{(5^{i-1})(5^{j-i-1})(5^{n-j-1})}{({15}^{i-1})({14}^{j-i-1})({13}^{n-j-1})} \\right)$ $= \\left(\\frac{24}{91}\\right) \\left \\frac{(5^{n-3})}{({15}^{i-1})({14}^{j-i-1})({13}^{n-j-1})} \\right$ $= \\left(\\frac{24}{91}\\right) \\left \\frac{(5^{n-3})}{({15}^{i-1})({14}^{j-i-1})({13}^{n-j-1})} \\right$ $= \\left(\\frac{24}{91}\\right) (5^{n-3}) \\left( (15)(14)\\left(\\frac{14}{15}\\right)^i \\left(\\frac{13}{14}\\right)^j \\left(\\frac{1}{13}\\right)^{n-1} \\right)$ $= \\left(\\frac{(24)(15)(14)(13)}{(91)(5^3)}\\right) \\left(\\frac{14}{15}\\right)^i \\left(\\frac{13}{14}\\right)^j \\left(\\frac{5}{13}\\right)^{n}$ $= \\left(\\frac{144}{25}\\right)", "drew another card. What is the probability that the first card was an E and the second was a B? a. $\\frac{1}{13}$ b. $\\frac{1}{676}$ c. $\\frac{4}{473}$ d. $\\frac{1}{26}$ 197. What is the probability that a card randomly picked from a deck of playing cards is a diamond? a. $\\frac{1}{2}$ b. $\\frac{1}{52}$ c. $\\frac{1}{4}$ d. $\\frac{5}{52}$ 198. Clara has 4 blue marbles, 5 green marbles, and 3 yellow marbles. Find the probability of drawing either a green or a yellow marble. a. $\\frac{5}{48}$ b. $\\frac{3}{5}$ c. $\\frac{3}{4}$ d. $\\frac{2}{3}$ 199. A pair of dice is rolled. What is the probability that the sum of the numbers rolled is either 8 or 12? a. $\\frac{1}{6}$ b. $\\frac{5}{36}$ c. $\\frac{1}{5}$ d. $\\frac{2}{9}$ a. $\\frac{1}{3}$ b. $\\frac{3}{4}$ c. $\\frac{1}{4}$ d. $\\frac{1}{2}$", "David K Dec 30 '18 at 15:39 • Your left hand column has the six cases with \"no two of the marbles drawn have the same color\", but your right hand column does not: $$R,B,G$$ are all different but $$R,G,R$$ has two $$R$$s • The probability of drawing $$R,B,G$$ in that order is $$\\frac{30}{60} \\times \\frac{10}{60} \\times \\frac{20}{60} = \\frac1{36}$$. Each of the others in the left hand column have the same probability and adding these up gives $$6 \\times \\frac1{36}= \\frac16$$, which I would expect to be the answer", "ways to select the gender and the day of the week the child was born on. There are 132 = 169 ways which do not have a girl born on Monday and which 196 – 169 = 27 which do, so P(Y) = $$\\frac{27}{196}$$. This gives is that P(X|Y) = $$\\frac{\\frac{13}{19}*\\frac{1}{4}}{\\frac{27}{196}} = \\frac{13}{27}$$. 7. Suppose a fair eight-sided die is rolled once. If the value on the die is 1, 3, 5 or 7 the die is rolled a second time. Determine the probability that the sum of values that turn up is at least 8? a) $$\\frac{32}{87}$$ b) $$\\frac{12}{43}$$ c) $$\\frac{6}{13}$$ d) $$\\frac{23}{64}$$ Explanation: Sample space consists of 8*8=64 events. While (8) has $$\\frac{1}{8}$$ probability of occurrence, (1,7) has only $$\\frac{1}{64}$$ probability. So, the required probability = $$\\frac{1}{6} + (9 * \\frac{1}{64}) = \\frac{69}{192} = \\frac{23}{64}$$. 8. A jar containing 8 marbles of which 4 red and 4 blue marbles are there. Find the probability of getting a red given the first one was red too. a) $$\\frac{4}{13}$$ b) $$\\frac{2}{11}$$ c) $$\\frac{3}{7}$$ d) $$\\frac{8}{15}$$ Explanation: Suppose, P (A) = getting a red marble in the first turn, P (B) = getting a black marble in the second turn. P (A) = $$\\frac{4}{8}$$ and P (B) = $$\\frac{3}{7}$$ and P (A and B) = $$\\frac{4}{8}*\\frac{3}{7} = \\frac{3}{14}$$ P(B/A)", "Contest Duration: - (local time) (120 minutes) Back to Home ## How to find the number of black stones in the end from the colors of squares and $$s$$ If the squares at both ends have the same color, all $$n$$ stones will be of that color. Otherwise, let $$S_B$$ and $$S_W$$ be the connected components of black and white squares at the ends, respectively. Then, • If the distance from $$s$$ to $$S_W$$ is less than the distance to $$S_B$$, there will be $$|S_B|$$ black stones; • otherwise, there will be $$n - |S_W|$$ black stones. ### Why? We can prove that the white stones always form a connected component, and so do the black stones. Additionally, once a stone is placed at either end, the color of that stone never changes. Thus, if the squares at both ends have the same color, all stones will be of that color. Let us move on to the case the squares at the ends have different colors. We will consider two cases: the case $$S_W$$ gets a stone first, and the case $$S_B$$ gets a stone first. If $$S_B$$ gets a stone first, at the moment just before $$S_W$$ gets a stone later, the leftmost and rightmost stones already placed are both black, so all the stones placed are", "isn't yellow. Therefore, there's a $$\\frac45\\cdot\\frac34\\cdot\\frac23\\cdot\\frac12=\\frac15$$ chance that none of the four marbles drawn is yellow, so there's a $$1-\\frac15=\\frac45$$ chance that one of the four marbles is yellow. As a third approach (which I'll discuss in more detail), since there's only one yellow marble, then to get the probability that the yellow marble was chosen, we need only add the probability of the following distinct events: (i) the yellow marble was chosen first, (ii) the yellow marble was chosen second, (iii) the yellow marble was chosen third, (iv) the yellow marble was chosen fourth. Hopefully, you see why these events have no overlap (mutually exclusive), and why together they comprise all the possible ways that the yellow marble could be chosen in this circumstance. We already know that $$P(\\text{yellow first})=\\frac15.\\tag{i}$$ If yellow is chosen second, then some other marble was chosen first--there are ${}_4C_1=4$ ways this can happen out of ${}_5C_1=5$ possibilities--leaving $1$ yellow marble out of a total of $4$ remaining, so $$P(\\text{yellow second})=\\frac45\\cdot\\frac14=\\frac15.\\tag{ii}$$ If yellow is chosen third, then two non-yellow marbles were chosen first--there are ${}_4C_2=6$ ways this can happen out of ${}_5C_2=10$ possibilities--leaving $1$ yellow marble out of a total of $3$ remaining, so $$P(\\text{yellow third})=\\frac{6}{10}\\cdot\\frac13=\\frac15.\\tag{iii}$$ If the yellow marble is chosen fourth, then three non-yellow marbles were chosen first--there are ${}_4C_3=4$ ways", "of black marbles is also reduced by $1$. The probability $P (1)$ of drawing a black marble in the first draw is $\\frac{4}{13}$ and $P (2)$, the same for the second draw is now $\\frac{3}{12}$ or $\\frac{1}{4}$ Therefore the probability of drawing black marble on both occasions is $P (1) \\times P (2)$ = $(\\frac{4}{13})$ $\\times$ $(\\frac{1}{4})$ = $\\frac{1}{13}$ Example 4: Two fair dies are thrown at the same time. What is the probability of rolling on the same number on the both the dies? Solution: The events are independent because two separate dies are rolled. Now, the probability of rolling on a particular number with the first die is $\\frac{1}{6}$ and the probability of rolling on the same number with the second die is also $\\frac{1}{6}$ So the combined probability of getting one particular number is $(\\frac{1}{6})$ $\\times$ $(\\frac{1}{6})$ = $\\frac{1}{36}$ But there are 6 possible ways to roll on the same numbers with both the dies. That is it can be $\\{1, 1\\},\\ \\{2, 2\\},\\ \\{3, 3\\},\\ \\{4, 4\\},\\ \\{5, 5\\}$ and $\\{6, 6\\}$. The ultimate probability of rolling on the same numbers on both the dies is $6\\ \\times$ $(\\frac{1}{36})$ = $\\frac{1}{6}$", "orange $y = \\sqrt{x}$ Lime green $y = x$ Turquoise $y = \\frac{x^2+x}{2}$ Purple $y = x^2$ There are an infinite number of curves that can be found that will intersect only at the two end points. One potential form is $y = \\frac{x^2+ax}{1+a}$ for $a \\neq 1$", "bead from the bag and record its colour. 1. What is the probability that the first bead is yellow? 2. What is the probability that the second bead is black? 3. What is the probability that the first bead is yellow and the second bead is black? 4. Are the first bead being yellow and the second bead being black independent events? ### Probability of a yellow bead first Since there is a total of $$\\text{7}$$ beads, of which $$\\text{3}$$ are yellow, the probability of getting a yellow bead is $P(\\text{first bead yellow}) = \\frac{3}{7}$ ### Probability of a black bead second The problem states that the first bead is placed back into the bag before we take the second bead. This means that when we draw the second bead, there are again a total of $$\\text{7}$$ beads in the bag, of which $$\\text{4}$$ are black. Therefore the probability of drawing a black bead is $P(\\text{second bead black}) = \\frac{4}{7}$ ### Probability of yellow first and black second When drawing two beads from the bag, there are $$\\text{4}$$ possibilities. We can get We want to know the probability of the second outcome, where we have to get a yellow bead first. Since there are $$\\text{3}$$ yellow beads and $$\\text{7}$$ beads in total, there are $$\\frac{3}{7}$$ ways to get", "marble is red given that the first is yellow. I have to also attach a tree diagram with this problem and don't understand the multipliang and adding acrooss the diagram. 1]There are $_5C_2=10$ ways to chose to yellow marbles. There are $_{11}C_2=55$ ways to chose two marbles. Thus, the ratio is the probability which is $\\frac{10}{55}=\\frac{2}{11}$. 2]Thus, either RR or YR. Probability of chosing a red first is $\\frac{6}{11}$ and then another red is $\\frac{5}{10}=\\frac{1}{2}$ thus, the probability of RR is $\\frac{6}{11}\\cdot\\frac{1}{2}=\\frac{3}{11}$. The probability of chosing a yellow first is $\\frac{5}{11}$ and then choosing a red is $\\frac{6}{10}=\\frac{3}{5}$ thus the probability of YR is $\\frac{5}{11}\\cdot\\frac{3}{5}=\\frac{3}{11}$. Thus, the probability of RR or YR is $\\frac{3}{11}+\\frac{3}{11}=\\frac{6}{11}$.", "# Probability Worksheets - Page 21 Probability Worksheets • Page 21 201. A die is rolled. Find the probability of rolling a 2 or 4. a. $\\frac{1}{2}$ b. $\\frac{1}{6}$ c. $\\frac{1}{3}$ d. $\\frac{2}{3}$ 202. What is the probability of selecting a head when a coin is tossed, if the probability of selecting a tail is $\\frac{1}{2}$? a. 1 b. $\\frac{1}{2}$ c. $\\frac{1}{4}$ d. $\\frac{2}{6}$ 203. What is the probability of not selecting a number 4 when a number cube is rolled, if the probability of selecting a number 4 is $\\frac{1}{6}$? a. 1 b. $\\frac{5}{6}$ c. $\\frac{2}{6}$ d. $\\frac{1}{6}$ 204. Which of the following are complementary events? a. getting a number that is a multiple of 2, or a number that is a multiple of 3, when a number cube is rolled b. getting a number greater than 2 or a number greater than 3, when a number cube is rolled c. picking a marble from a bag containing only red and yellow marbles d. picking a Jack or a Spade from a pack of playing cards 205. Terry is choosing a marble from a bag containing 1 black marble, 1 brown marble, and 1 white marble. What is the probability that he will not choose a brown marble? a. $\\frac{2}{3}$ b. $\\frac{3}{2}$ c. $\\frac{1}{3}$ d. 1 206.", "B2 then R1. That means my population is: Selection 1 Selection 2 Selection 1 Selection 2 B1 B2 B2 B1 B1 B3 B2 B3 B1 R1 B2 R1 B1 R2 B2 R2 B1 Y1 B2 Y1 B3 B1 B3 B2 B3 R1 B3 R2 B3 Y1 R1 R2 R2 R1 R1 B1 R2 B1 R1 B2 R2 B2 R1 B3 R2 B3 R1 Y1 R2 Y1 Y1 B1 Y1 B2 Y1 B3 Y1 R1 Y1 R2 Which is a real nightmare to compute by trying to figure out all the permutations by hand, and imagine, this is only a small set! Maths to the rescue... \\begin{align} Permutations &= {_{n}P_r} = \\frac{n!}{(n-r)!} \\\\ &= {_{6}P_2} = \\frac{6!}{(6-2)!} \\\\ &= \\frac{6 \\times 5 \\times 4 \\times 3 \\times 2 }{4 \\times 3 \\times 2} \\\\ &= 6 \\times 5 \\\\ &= 30 \\end{align} And that is how many permutations we have in the above table (thankfully!). So, if my question is what is the probability of drawing a red then a yellow marble, my event space is the set {R1Y1, R2Y1}. Thus, if we say event A is \"draw a red then a yellow marble\", $$P(A) = \\frac{N_A}{N} = \\frac{2}{30} = \\frac{1}{15}$$ What if we don't care about the order. What is the probability of drawing a red and a", "add the probability of the event “select a triangle” to the probability of the event “select a yellow figure”? There are 5 triangles and 5 yellow figures, so after add them up we get 10 figures, while we need 7 figures. The three excess figures occurs, because they are common and we have three yellow triangles, that belong to the group “triangles” and to the group “yellow figures” at the same time. Notice that after adding we count common figures two times: first, when we count triangles and second, when we count yellow figures: Thus we need to subtract the number of common figures. The probability of the event “randomly select a triangle or a yellow figure” is: $P(Triangle~ or~ Yellow~figure)=P(Triangle)+P(Yellow~figure)- \\\\-P(Triangle~ and~Yellow~figure)=\\frac{5}{12}+\\frac{5}{12}-\\frac{3}{12}=\\frac{7}{12}.$ ## Basic Probability, part 3 ### August 23, 2009 Dice are very old instruments for games, thus they had been found in Egyptian tombs, dated to 2000 BC. Dice gamble were greatly popular in ancient Greek, Rome, Asia, then in medieval Europe and it has been being popular today. In ancient times the throw of dice was believed to be controlled by the gods. Famous was the game with throwing two dice and guessing the sum of the numbers. People noticed that the number 7 as the sum turns out more frequently and they thought", "black, similarly to the argument above. The colors of stones never change from then on, so there will be $$n-|S_W|$$ black stones in the end. In the same way, if $$S_W$$ gets a stone first, there will be $$|S_B|$$ black stones in the end. Therefore, Kuro will put stones trying to reach $$S_B$$ as fast as possible, and Shiro will try to reach $$S_W$$ as fast as possible, so the result depends on the distance from $$s$$ to $$S_B$$ and $$S_W$$, in the way stated above. ## The counting Let $$W$$ and $$B$$ be the expected number of white and black stones, respectively. Then, $$W+B=n$$, so we can find $$W$$ if we can find $$W-B$$. Now, consider the effect of inverting the colors of squares on the result. In almost all cases, inverting a game that ends with $$x$$ black stones results in a game that ends with $$x$$ white stones. Thus, those cases do not contribute to $$W-B$$, so we will ignore them. The only other cases are the cases the distances from $$s$$ to $$S_B$$ and $$S_W$$ are equal. Consider when $$S_B$$ is to the left, and $$s$$ is to the left of the middle. The difference between the final numbers of black and white stones is $$4s-2|S_B|-n+2$$ with probability $$2^{-n+2s-2|S_B|-1}$$. Thus, we need to efficiently", "C has a black stone, if possible. • B always guesses white. • C always rearranges so that B has a white stone, if possible. • C always guesses black. Here is a table, showing the initial layout, the layout after A leaves the room, the layout after B leaves, the final layout after C leaves, what the three guesses were, and the tally of results. $$\\begin{matrix} \\underline{\\text{Initial}} & \\underline{\\text{A}} & \\underline{\\text{B}} & \\underline{\\text{C (final)}} & \\underline{\\text{Guessses}} & \\underline{\\text{Result}}\\\\ www & www & www & www & wwb & ++- \\\\ wwb & wwb & wwb & wwb & bwb & -++ \\\\ wbw & wwb & wwb & wwb & bwb & -++ \\\\ wbb & wbb & wbb & bwb & bwb & +++ \\\\ bww & bww & wwb & wwb & wwb & +++ \\\\ bwb & bwb & bwb & bwb & bwb & +++ \\\\ bbw & bwb & bwb & bwb & bwb & +++ \\\\ bbb & bbb & bbb & bbb & bwb & +-+ \\end{matrix}$$ As the table shows: • C only guesses incorrectly if all the stones are white. • B only guesses incorrectly if all the stones are black. • A only guesses incorrectly if there are exactly two white stones and A started with one", "inside my fist and remove a blue stone. What is the probability that the other stone in my hand is also blue? The problem is small enough to LIST all the outcomes. The four stones are:. $W,\\,Y,\\,B_1,\\,B_2$ There are six possible outcomes: . . $\\{WY,\\:WB_1,\\:WB_2,\\:YB_1,\\:YB_2,\\:B_1B_2\\}$ We are told that one of the two stones is blue. The sample space is reduced to five outcomes: . . $\\{WB_1,\\:WB_2,\\:YB_1,\\:YB_2,\\:B_1B_2\\}$ The stones are both blue in exactly one case. Therefore: . $P(\\text{both blue}\\,|\\,\\text{one is blue}) \\;=\\;\\frac{1}{5}$ 3. ## Re: A simple puzzle (with a lot of controversy) Thank You Soroban! That was the official solution. Some people convinced themselves somehow that we must know whether he chooses one of the stones from his fist at random and then presents it to us (in which case we must include more outcomes in the sample space and the answer is 1/3) or whether he always picks the blue one first. The simplest explanation that I gave was that it is irrelevant once we know one of the stones is blue. I am a layman and I don't know mathematical jargon and I i think I failed to explain it to them. If anyone is interested in modifying the puzzle to accommodate different answers or explaining properly why 1/3 is definitely wrong in this" ]

[ "$n_{1,1}$ balls from the second bag. number of combinations without repetition with limited supply The above question helps explain how to find the number of combinations of $n_{1,1}$ marbles from bag two. Essentially you analyze the following polynomial for the coefficients: $$P = \\prod_{j=1}^{c_2}{\\sum_{k=0}^{n_{2,j}}x^k}$$ For the question example: $$P=(1+x)(1+x)(1+x+x^2+x^3) = 1+3x+4x^2+4x^3+3x^4+x^5$$ and the answer is 4 because the coefficient of $x^2$ is 4. This makes sense as the potential cases can be described as the colors paired with blue (black/black, red/black, white/black, and red/white). b) The problem becomes more complicated when each bag contains more than 2 types of marbles. For the case of 3 colors in bag 1, the simular case requires that one consider both an x and a y. Define a polynomial: $$F = \\prod_{j=1}^{c_2}{\\sum_{k=0}^{n_{2,j}}(\\sum_{l=0}^{k} x^ly^{k-l})}$$ Convert it into an expanded form and note that the coefficient of $x^{n_{1,1}}y^{n_{1,2}}$ is the number of relevant combinations. For examples, if $n=5$, $c_1=3$, $c_2=3$, $n_{2,1}=2$, $n_{2,2}=2$, $n_{2,3}=1$ then: $$F=(1+x+y+x^2+x*y+y^2)^2*(1+x+y)$$ $$F=x^5+3 x^4 y+3 x^4+5 x^3 y^2+8 x^3 y+5 x^3+5 x^2 y^3+11 x^2 y^2+11 x^2 y+5 x^2+3 x y^4+8 x y^3+11 x y^2+8 x y+3 x+y^5+3 y^4+5 y^3+5 y^2+3 y+1$$ If bag 2 contains 2 black, 2 red, and 1 green (for instance) there will be 11 combinations. c) For the final case of an arbitrary number of colors in", "yellow tiles and 22 black tiles, what i [#permalink] 19 Sep 2018, 10:07 1 KUDOS Expert's post Carcass wrote: If a bag contains 18 yellow tiles and 22 black tiles, what is the probability that a tile selected at random from the bag will be yellow? A. $$\\frac{1}{10}$$ B. $$\\frac{9}{20}$$ C. $$\\frac{11}{20}$$ D. $$\\frac{9}{11}$$ E. $$\\frac{11}{9}$$ Number of YELLOW tiles in bag = 18 TOTAL number of tiles in bag = 18 + 22 = 40 P(selected tile is yellow) = (# of possible outcomes in which the selected tile is yellow)/(TOTAL # of possible outcomes) = 18/40 = 9/20 Cheers, Brent _________________ Brent Hanneson – Creator of greenlighttestprep.com Re: If a bag contains 18 yellow tiles and 22 black tiles, what i [#permalink] 19 Sep 2018, 10:07 Display posts from previous: Sort by", "# Probability Worksheets - Page 21 Probability Worksheets • Page 21 201. A die is rolled. Find the probability of rolling a 2 or 4. a. $\\frac{1}{2}$ b. $\\frac{1}{6}$ c. $\\frac{1}{3}$ d. $\\frac{2}{3}$ 202. What is the probability of selecting a head when a coin is tossed, if the probability of selecting a tail is $\\frac{1}{2}$? a. 1 b. $\\frac{1}{2}$ c. $\\frac{1}{4}$ d. $\\frac{2}{6}$ 203. What is the probability of not selecting a number 4 when a number cube is rolled, if the probability of selecting a number 4 is $\\frac{1}{6}$? a. 1 b. $\\frac{5}{6}$ c. $\\frac{2}{6}$ d. $\\frac{1}{6}$ 204. Which of the following are complementary events? a. getting a number that is a multiple of 2, or a number that is a multiple of 3, when a number cube is rolled b. getting a number greater than 2 or a number greater than 3, when a number cube is rolled c. picking a marble from a bag containing only red and yellow marbles d. picking a Jack or a Spade from a pack of playing cards 205. Terry is choosing a marble from a bag containing 1 black marble, 1 brown marble, and 1 white marble. What is the probability that he will not choose a brown marble? a. $\\frac{2}{3}$ b. $\\frac{3}{2}$ c. $\\frac{1}{3}$ d. 1 206.", "of a bag with 12 marbles in it? 2. What is the probability of a spinner landing on “6” if there are 6 equally spaced points on the spinner? 3. What is the probability of pulling a red card at random from a standard deck? 4. What is the experimental probability of heads in an experiment where Scott flipped a coin 50 times and got heads 21 times? 5. What is the probability of shaking the hand of a female student if you randomly shake the hand of one person in a room with 23 female students and 34 male students? Solutions: 1. $P(red)=\\frac{\\text{1 red marble}}{\\text{12 total marbles}}=\\frac{1}{12} \\ or \\ 8.3 \\%$ 2. $P(6)=\\frac{\\text{1 number 6}}{\\text{6 total numbers}}=\\frac{1}{6} \\ or \\ 16.7 \\%$ 3. $P(red)=\\frac{\\text{26 red cards}}{\\text{52 total cards}}=\\frac{26}{52}=\\frac{1}{2} \\ or \\ 50 \\%$ 4. $P(heads)=\\frac{\\text{21 heads}}{\\text{50 flips}}=\\frac{21}{50} \\ or \\ 42 \\%$ 5. $P(female)=\\frac{\\text{23 females}}{\\text{57 students }}=\\frac{23}{57} \\ or \\ 40.4 \\%$ Practice Questions 1-10, find the probability: 1. Rolling a 4 on a standard die 2. Pulling a King from a standard deck 3. Pulling a green candy from an opaque bag with 5 red, 3 yellow, 3 blue, and 6 green candies. 4. Getting a 5 from one spin on a spinner numbered 1-8 (equally spaced) 5. Rolling an even number on a 20-sided", "Posts: 190 Location: India GMAT 1: 500 Q47 V15 GPA: 3.4 WE: Information Technology (Computer Software) Re: If a bag contains 18 yellow tiles and 22 black tiles, what is the prob [#permalink] ### Show Tags 17 Sep 2018, 02:11 Bunuel wrote: If a bag contains 18 yellow tiles and 22 black tiles, what is the probability that a tile selected at random from the bag will be yellow? A. $$\\frac{1}{10}$$ B. $$\\frac{9}{20}$$ C. $$\\frac{11}{20}$$ D. $$\\frac{9}{11}$$ E. $$\\frac{11}{9}$$ . Total number of tiles in a bag is $$18 + 22 = 40$$. Probability = $$\\frac{Favorable Outcome}{Total number of Outcome}$$ Probability = $$\\frac{18}{40}$$ Probability = $$\\frac{9}{20}$$ Ans - B _________________ If it helps you please press Kudos! Thank You Sudhanshu Re: If a bag contains 18 yellow tiles and 22 black tiles, what is the prob [#permalink] 17 Sep 2018, 02:11 Display posts from previous: Sort by", "cards with 4 aces, what is the probability that an ace will not show up? Let, P(A) = Ace appears • $$1-P(A)$$ • $$1-\\frac 1 {13}$$ ## Card Problem #03 Two cards are drawn with replacement; What is the probability that they are 1. Kings of same color 2. Kings of different color 3. Not Kings at all 1. $$P(BUR) =P(B)+P(R)$$ • $$\\frac{^2C_1 \\times ^2C_1}{^{52}C_1\\times ^{52}C_1}+\\frac{^2C_1 \\times ^2C_1}{^{52}C_1\\times ^{52}C_1}$$ Why not $$^{52}C_2$$, $$^4C_2$$ 1. $$1-P(B \\cup R)$$ • $$P(K)= \\frac{^4C_1 \\times ^4C_1}{^{52}C_1\\times ^{52}C_1}$$ 1. $$1-P(K)$$ ## Card Problem #04 A card is drawn from a pack of 52 cards. What is the probability that it is 1. an Ace 3. A Hearts or a King # Box ## Box Problem #01 In a box, there are 5 blue marbles, 7 green marbles, and 8 yellow marbles. If two marbles are randomly selected, what is the probability that both will be green or yellow, if taken 1. with replacement 2. without replacement • Correct or not: $$\\frac{^7C_2}{^{20}C_2}+\\frac{^8C_2}{^{20}C_2}$$ • $$\\frac{^7C_1 \\times ^7C_1}{^{20}C_1 \\times ^{20}C_1}+\\frac{^8C_1 \\times ^8C_1}{^{20}C_1 \\times ^{20}C_1}$$ • Without replacement: $$\\frac{^7C_1 \\times ^6C_1}{^{20}C_1 \\times ^{20}C_1}+\\frac{^8C_1 \\times ^7C_1}{^{20}C_1 \\times ^{20}C_1}$$ ## Box Problem #02 There are some balls in a box as below Color # Balls White 3 Black 6 Red 7 Green 5 Yellow 4 Violet 9 Blue 8", "learned prior – how strongly we expect any newly encountered bag to conform to the distribution for the population prototype beta. For instance, suppose that we observe that bag1 consists of all blue marbles, bag2 consists of all green marbles, bag3 all red, and so on. This doesn’t tell us to expect a particular color in future bags, but it does suggest that bags are very regular—that all bags consist of marbles of only one color. var colors = ['black', 'blue', 'green', 'orange', 'red']; var observedData = [ {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag4', draw: 'orange'}] var predictives = Infer({method: 'MCMC', samples: 30000}, function(){ var alpha = gamma(2,2) var beta = dirichlet(ones([5, 1])) var getBag = mem(function(bagName) { var colorProbs = T.toScalars(dirichlet(beta.mul(alpha))); return Categorical({ vs: colors, ps: colorProbs }) }) map(function(d) { observe(getBag(d.bag), d.draw) }, observedData) return {bag1: sample(getBag('bag1')), bag2: sample(getBag('bag2')), bag3: sample(getBag('bag3')), bag4: sample(getBag('bag4')), bagN: sample(getBag('bagN')), alpha: alpha} });", "bag. 7. Two cards are drawn from a deck of cards. Determine the probability of each of the following events: 1. $P$ (heart or club) 2. $P$ (heart and club) 3. $P$ (red or heart) 4. $P$ (jack or heart) 5. $P$ (red or ten) 6. $P$ (red queen or black jack) 1. A box contains 5 purple and 8 yellow marbles. What is the probability of successfully drawing, in order, a purple marble and then a yellow marble? {Hint: In order means they are not replaced.} 2. A bag contains 4 yellow, 5 red, and 6 blue marbles. What is the probability of drawing, in order, 2 red, 1 blue, and 2 yellow marbles? 3. A card is chosen at random. What is the probability that the card is black and is a 7? Mixed Review 1. A circle is inscribed within a square, meaning the circle's diameter is equal to the square’s side length. The length of the square is 16 centimeters. Suppose you randomly threw a dart at the figure. What is the probability the dart will land in the square, but not in the circle? 2. Why is $7-14x^4+7xy^5-1x^{-1}=8x^2 y^3$ not considered a polynomial? 3. Factor $72b^5 m^3 w^9-6(bm)^2 w^6$ . 4. Simplify $2^5-7^3 a^3 b^7+3^5 a^3 b^7-2^3$ . 5. Bleach breaks down cotton", "2 \\ 3 \\ 4 \\ 5 \\ 6 \\ \\ 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \\ \\ 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6\\end{align*} The red numbers 3 and 6 represent the two favorable outcomes out of 36 total, therefore \\begin{align*}P(2 \\ and \\ (3 \\ or \\ 6))=\\frac{2}{36} \\ or \\ \\frac{1}{18} \\ or \\ 5.5\\overline{5} \\%\\end{align*} Example 3 What is the probability of pulling a red and then a black card at random from a standard deck (replacing the first card after drawing)? There are 26 black and 26 red cards in the deck, so the probability on the first pull is \\begin{align*}P(red)=\\frac{26 \\ \\text{red cards}}{52 \\ \\text{total cards}}=\\frac{26}{52}=\\frac{1}{2} \\ or \\ 50 \\%\\end{align*} On the second pull, we again have a 50% chance of favorable outcome, but that 50% only applies to the half of the first pulls that were favorable. Therefore: \\begin{align*}P(\\text{red} \\ then \\ \\text{black})=50\\% \\ of \\ 50\\%=25\\%\\end{align*} Example 4 What probability of picking a red and then a green marble from a bag with 5 red and 1 green marbles in it (replacing the first marble after the draw)? Make a chart: \\begin{align*}& \\underline{\\text{first pull:} \\qquad \\quad r \\quad \\qquad \\ r \\qquad \\quad r \\qquad \\quad r \\qquad \\quad", "Bag A has Red and Green marbles and Bag B contains Blue and Yellow. If you want to calculate of picking a red marble, you calculate the probability of choosing from bag A instead of B and then multiply by the probability of getting Red from Bag A. Contents of bag B would not matter. One way of checking if the events $$A$$ and $$B$$ are independent is calculate the conditional probability of one of them, let's say $$P(A)$$, and see if is the same as $$P(A|B)$$, or probability of $$A$$ occurs when $$B$$ already happened (actually you should check the in-dependency of two events by the very way you're using to calculate $$P(A\\cap B)$$). By this you easily conclude that, if $$A$$: the individual is male; $$B$$: the individual is infected: $$P(A)=\\frac{43}{81}\\neq \\frac{36}{59}=P(A|B)$$ So the events aren't independent.", "Explanation: There are 52 cards in the standard deck of cards and thus there are 52 possible outcomes. possible outcomes = 52 We note that 4 of the 52 cards are 7’s and thus there are 4 favorable outcomes. favorable outcomes = 4 The probability is the number of favorable outcomes divided by the number of possible outcomes. P(7) = favorable outcomes/possible outcomes 4/52 = $$\\frac{1}{13}$$ ≈ 0.0769 = 7.69% Question 3. Picking a blue marble from a bag of 4 red marbles, 6 blue marbles, and 1 white marble. $$\\frac{□}{□}$$ Answer: $$\\frac{6}{11}$$ Explanation: The bag contains 4 red, 6 blue, and 1 white marble, thus the bag contains 4 + 6 + 1 = 11 marbles in total and thus there are 11 possible outcomes. possible outcomes = 11 We note that 6 of the 11 marbles in the bag are blue and thus there are 6 favorable outcomes. favorable outcomes = 6 The probability is the number of favorable outcomes divided by the number of possible outcomes. P(blue) = favorable outcomes/possible outcomes = $$\\frac{6}{11}$$ ≈ 0.5455 = 54.55% Question 4. Rolling a number greater than 7 on a 12-sided number cube. $$\\frac{□}{□}$$ Answer: $$\\frac{5}{12}$$ Explanation: A 12 side number cube has 12 possible outcomes: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.", "770/1631 = 0.47 d. Louisiana The probability of the ship landing in Louisiana: 397/1631 = 0.24 Question 15. DRAWING CONCLUSIONS You roll a six-sided die 60 times. The table shows the results. For which number is the experimental probability of rolling the number the same as the theoretical probability? Question 16. DRAWING CONCLUSIONS A bag contains 5 marbles that are each a different color. A marble is drawn, its color is recorded, and then the marble is placed back in the hag. This process is repeated until 30 marbles have been drawn. The table shows the results. For which marble is the experimental probability of drawing the marble the same as the theoretical probability? Total number of marbles = 30 6/30 = 1/5 For black marble the experimental probability of drawing the marble the same as the theoretical probability. Question 17. REASONING Refer to the spinner shown. The spinner is divided into sections with the same area. a. What is the theoretical probability that the spinner stops on a multiple of 3? b. You spin the spinner 30 times. If stops on a multiple of 3 twenty times. What is the experimental probability of Stopping on a multiple of 3? c. Explain why the probability you found in part (b) is different than the probability you found in", "every 13 times you don't draw a Answers and Explanations. A black card that isn’t queen if they can occur together, like the outcome of picking a queen or a club. Each card in a set of cards has one of the letters from the word mathematics. Two balls need to be drawn. Thus the probability of drawing two red marbles is r(r-1)/(15 16)) Two marbles are chosen without replacement There are two types of Based on deck of cards I am looking for a way to sum the results (how many of each color marble were selected out of x trials, etc I am looking for a way to sum the results (how many of each color marble were The red cards are either the two suits: hearts and diamonds. View solution > An urn contains 2 white and 2 black balls. There are 52 cards in the deck; 13 in each of the four suits, and two of the suits are red. Let. What is the probability that we will draw a card that is a six and a spade? There is only 16 of spades. e] at least one red marble. $\\endgroup$ – Multiple-choice. What is the probability of drawing a black card? The probability of drawing a black card is (Type an integer or", "other number that adds to 11. gl/9WZjCW A fair six-faced die is rolled 12 times. Getting an odd number and getting a number less than 4 4. If you meant to say 1,2,3,4,5, Or 6, then the answer would be 1/6. If a die is rolled twice, the outcome of the second trial is not affected by For a class exercise, students conduct a probability experiment is conducted in which they toss a coin 5 times in a row. If at each step the recipient of the rumor is chosen at. So, using the platonic solids we can have dice with 4, 6, 8, 12 or 20 faces! But with some imagination we can actually make fair dice with any number of faces we want. Let's take it up another notch. What is the probability of getting a number greater than 4 ? Total outcomes that can occur are 1, 2, 3, 4, 5, 6 Total number of. datingadvice. If a marble is drawn from the jar at random, what is the probability that this marble is white? Solution We first construct a table of frequencies that gives the marbles color. Probability. Find the probability that the ball remove from Box A is red and the ball remove from Box B is blue. Country Profiles. Willy will carve", "drawing a black card or a ten in a deck of cards. There are 4 tens in a deck of cards P(10) = 4/52 There are 26 black cards P(black) = 26/52 There are 2 black tens P(black and 10) = 2/52 $P&space;(black\\&space;or\\&space;ten)&space;=&space;\\frac{4}{52}&space;+&space;\\frac{26}{52}&space;-&space;\\frac{2}{52}&space;=&space;\\frac{30}{52}&space;-&space;\\frac{2}{52}&space;=&space;\\frac{28}{52}&space;=&space;\\frac{7}{13}$ ## How to Convert Odds to Probability Determining the likelihood of this event actually occurring is referred to as the odds. The odds, or chance, of something happening, depends on the probability. The odds take the probability of an event occurring and divide it by the probability of the event not occurring. Example: Colored Marbles. Figure out the probability of drawing a white marble (of which there are 11) out of the total pot of marbles (which contains 20). 1. Set the odds as a ratio with a positive outcome as a numerator. There are 11 white and 9 non-white marbles, you’ll write the odds as the ratio 11:9 2. Add the numbers together to convert the odds to probability. The event that you’ll draw a white marble is 11; the event another color will be drawn is 9. The total number of outcomes is 11 + 9, or 20. 3. Find the odds as if you were calculating the probability of a single event. The probability of drawing a white marble is 11/20. Divide", "that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails? (A) 1/8 (B) 1/2 (C) 3/4 (D) 7/8 (E) 15/16 P(at last 1 tails) = 1 - P(all heads) = 1 - (1/2)^3 = 7/8. Bunuel, would you mind explaining how you find the power to which you have to raise? What if e.g. there is a bag with three marbles, blue, red and yellow. Now the question is e.g. \"What is the probabilty to get a blue marble on at least 1 try if you try 4 times\" (putting the marbles back all the time). Would it be P(at least 1 blue marble) = 1 - P(none blue) = 1 - (2/3)^4 ?? Yes, that's correct, the power must be the number of tries. For the original question: P(at last 1 tails) = 1 - P(all heads) = 1 - (1/2*1/2*1/2)= 1 - (1/2)^3 = 7/8. For your example: P(at least 1 blue) = 1 - P(no blue) = 1- (2/3*2/3*2/3*2/3) = 1- (2/3)^4. Hope it's clear. _________________ Manager Joined: 21 Oct 2013 Posts: 194 Location: Germany GMAT 1: 660 Q45 V36 GPA: 3.51 Followers: 0 Kudos [?]:", "drawing a blue ball from the bag is twice that of a red ball,find the number of blue balls in the bag. (ii) if probability of drawing a blue ball from the bag is four times that of a red ball,find the number of blue balls in the bag. Let number of blue balls Total number of balls Probability of red ball Probability of blue ball By given condition, (i) No. of blue balls (ii) Here, Hence, number of blue balls Question 9: A box contains 3 blue marbles, 2 white marbles. If a marble is taken out at random from the box, what is the probability that it will be a white one? Blue one? Red one? Total no. of possible outcomes No. of favourable outcomes for white marbles Required probability No. of favourable outcomes for blue marbles Required probability No. of favourable outcomes for red marbles Required probability", "green shirts, 4 white shirts, and 2 red shirts in his shelf. Which of the following questions about these shirts could you use probability to solve? a. If Bill takes 1 shirt from the shelf, how many shirts will be left in the shelf? b. How many shirts did Bill have all together? c. If Bill picks 1 shirt from the shelf without looking, what color will it most likely be? d. How many more green shirts than red shirts did Bill have in the shelf? 218. Sheila packed 10 apples, 25 oranges, 36 strawberries and 5 pineapples in her picnic basket. Which of the following questions about these fruits could you use probability to solve? a. What is the total number of apples and oranges in the basket? b. If a fruit is picked at random, what kind of fruit is least likely to be picked? c. How many more apples than pineapples are there in the basket? d. How many fruits did Sheila pack in all? 219. A bag contains 11 blue marbles, 5 red marbles, and 4 green marbles. Pam draws one marble from the bag, records its color, and then puts it back in the bag. Then, he draws another marble and records its color. What is the probability of drawing 2 red marbles?", "newly encountered bag to conform to the distribution for the population prototype phi. For instance, suppose that we observe that bag1 consists of all blue marbles, bag2 consists of all green marbles, bag3 all red, and so on. This doesn’t tell us to expect a particular color in future bags, but it does suggest that bags are very regular—that all bags consist of marbles of only one color. var colors = ['black', 'blue', 'green', 'orange', 'red']; var observedData = [ {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag1', draw: 'blue'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag2', draw: 'green'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag3', draw: 'red'}, {bag: 'bag4', draw: 'orange'}] var predictives = Infer({method: 'MCMC', samples: 30000}, function(){ // the global prototype mixture: var phi = dirichlet(ones([colors.length, 1])) // regularity parameters: how strongly we expect the global prototype to project // (ie. determine the local prototypes): var alpha = gamma(2,2) var prototype = T.mul(phi, alpha) var makeBag = mem(function(bag){ var colorProbs = dirichlet(prototype) return Categorical({vs: colors, ps: colorProbs}) }) var obsFn =", "from their bag each round. If both marbles are blue $A$ wins $\\3$ and if both are red $A$ wins $\\1$. If the two colors are different, $B$ wins $\\2$. If you can play this game 100 times, who would you play as? Solution. Since there is an equal chance of picking marbles with different colors as picking two marbles that are the same color, the expected value of playing player $A$ versus player $B$ is the same (EV $= \\1$). Then, minimizing risk, we can calculate the variance of the payoff for each player: \\begin{aligned} \\sigma^2_A &= (1-1)^2 \\cdot \\frac{1}{4} + (3-1)^2 \\cdot \\frac{1}{4} + (0-1)^2 \\cdot \\frac{1}{2} = \\frac{3}{2}\\\\ \\sigma^2_B &= (0-1)^2 \\cdot \\frac{1}{2} + (2-1)^2 \\cdot \\frac{1}{2} = 1 \\end{aligned} $A$’s payoff has larger variance and he holds more risk so it’s advantageous to play as player $B$ when risk-averse. However, since $n$ is relatively large, the actual payoff approaches the expected payoff of playing, minimizing the risk of playing as player $A$. Problem 5. You are trapped in a dark cave with three indistinguishable exits on the walls. One of the exits takes you 3 hours to travel and takes you outside. One of the other exits takes 1 hour to travel and the other takes 2 hours, but both drop you back in" ]

[ "+0 # Equation 0 88 1 Find the value of n that satisfies $2(n+1)!+6n!=10$, where $n! = n\\cdot (n-1)\\cdot (n-2) \\cdots 2\\cdot 1$. Jun 11, 2022 #1 +2448 0 Just plug in n = 1, n = 2, and n = 3. You should find it pretty quickly. Jun 11, 2022", "3)}} - \\underbrace{(n - 1)}_{\\text{Previous factor of n}}$ $[(n + 4) - (n -1)]t_n = n(n + 1)(n + 2)(n + 3)[(n + 4) - (n - 1)] \\\\ 5t_n = n(n + 1)(n + 2)(n + 3)(n + 4) - (n - 1)n(n + 1)(n + 2)(n + 3)$ Now put $n = 1$, we will get $5t_1 = 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 - 0$ Now put $n = 2$, we will get $5t_2 = 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 - 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5$ Now put $n = 3$, we will get $5t_3 = 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 - 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6$ And so on, now put $n = (n - 1)$, we will get $5t_{n - 1} = (n - 1) \\cdot n \\cdot (n + 1) \\cdot (n + 2) \\cdot (n + 3) - (n - 2) \\cdot (n - 1) \\cdot n \\cdot (n + 1) \\cdot (n + 2)$ And now put $n = n$, we will get $5t_n = n \\cdot (n + 1) \\cdot (n + 2) \\cdot (n + 3) \\cdot (n + 4) - (n - 1) \\cdot n \\cdot (n +", "3)}} - \\underbrace{(n - 1)}_{\\text{Previous factor of n}}$ $[(n + 4) - (n -1)]t_n = n(n + 1)(n + 2)(n + 3)[(n + 4) - (n - 1)] \\\\ 5t_n = n(n + 1)(n + 2)(n + 3)(n + 4) - (n - 1)n(n + 1)(n + 2)(n + 3)$ Now put $n = 1$, we will get $5t_1 = 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 - 0$ Now put $n = 2$, we will get $5t_2 = 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 - 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5$ Now put $n = 3$, we will get $5t_3 = 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 - 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6$ And so on, now put $n = (n - 1)$, we will get $5t_{n - 1} = (n - 1) \\cdot n \\cdot (n + 1) \\cdot (n + 2) \\cdot (n + 3) - (n - 2) \\cdot (n - 1) \\cdot n \\cdot (n + 1) \\cdot (n + 2)$ And now put $n = n$, we will get $5t_n = n \\cdot (n + 1) \\cdot (n + 2) \\cdot (n + 3) \\cdot (n + 4) - (n - 1) \\cdot n \\cdot (n +", "$$y_3 = 2\\cdot 6\\cdot 8= 96$$, $$y_4 = 2\\cdot 96\\cdot 98 = 18616$$, and $$y_5 = 2\\cdot 18616\\cdot 18618 = 693185376$$. If we can show that $$n | y_n$$, then $$(n, x_n) = 1$$ since, if $$d|n$$ and $$d|x_n$$ and $$d \\ge 2$$ then $$d | n | y_n$$ so $$d | (x_n-1)$$ implies $$d | 1$$. Unfortunately, this doesn't work as $$y_5$$ shows. Next try: see if can find $$a_n, b_n$$ such that $$na_n-x_nb_n = 1$$. This will show that $$(n, x_n) = 1$$. If $$na_n-x_nb_n = 1$$ and $$(n+1)a_{n+1}-x_{n+1}b_{n+1} = 1$$ then $$\\begin{array}\\\\ 1 &=(n+1)a_{n+1}-(2x_n^2-1)b_{n+1}\\\\ &=na_{n+1}+a_{n+1}-2x_n^2b_{n+1}+b_{n+1}\\\\ &=na_{n}+n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1})+b_{n+1}\\\\ &=x_nb_n+1+n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1})+b_{n+1}\\\\ &=1+n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1}+b_{n})+b_{n+1}\\\\ \\text{so}\\\\ 0 &=n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1}+b_{n})+b_{n+1}\\\\ \\end{array}$$ and I don't know where to go from here. Hope someone else can make use of these. This is the pigeonhole principle. Among odd primes $$p,$$ there are solutions to $$2t^2 - 1 \\equiv 0 \\pmod p$$ if and only if $$p \\equiv \\pm 1 \\pmod 8.$$ For such a prime $$p,$$ there are some $$\\frac{p+1}{2}$$ distinct values $$\\pmod p$$ of $$2x^2 - 1.$$ Therefore, taking $$p \\geq 17,$$ if we take, say, $$p-5$$ steps of your sequence, there are just two possibilities: either (I) there is a repetition of some value $$\\pmod p$$ that is not one of $$-1,0,1.$$ In this case, the sequence repeats again and again, forever. Thus, this prime", "be a multiple of $n^n = 169^{169} = 13^{338}$, so $m > 338$. Then $k = m + n > 169 + 338 = 507$. All other possible values for $n$ have $n \\ge 242$, and in this case $m > n \\ge 242$, so $k \\ge 2 \\cdot 242 = 484$. Hence no greater values of $n$ can produce lesser values for $k$, and the least possible $k$ is indeed $407$.", "recognisable Using not-exactly-standard notation, $b_n=[n-1](8\\dots8)9$ where the number of 8's is n-1. $\\{b_n\\}=9,189,2889,38889,488889,\\dots$ In other words, $b_n = (n-1)\\cdot10^n \\ \\ \\ +\\ \\ \\ (10^n-10)\\cdot\\frac{8}{9}\\ \\ \\ +\\ \\ \\ 9$ So there's no need to solve for n, we can tell just by looking. 4. Wow this is amazing. Thank you for looking at this", "= 0 = a_1$. Thus $a_n = 2 \\cdot 3^n - 3 \\cdot 2^n$. • Case $n \\ge 2$ Since $% $, by induction hypothesis $a_{n-1} = 2 \\cdot 3^{n-1} - 3 \\cdot 2^{n-1}$. Similarly since $% $, by induction hypothesis we get $a_{n-2} = 2 \\cdot 3^{n-2} - 3 \\cdot 2^{n-2}$. Thus $5a_{n-1} - 6a_{n-2} = 5(2 \\cdot 3^{n-1} - 3 \\cdot 2^{n-1}) - 6(2 \\cdot 3^{n-2} - 3 \\cdot 2^{n-2})$ $= 10 \\cdot 3^{n-1} - 12 \\cdot 3^{n-2} - 15 \\cdot 2^{n-1} + 18 \\cdot 2^{n-2}$ $= 10 \\times 3 \\cdot 3^{n-2} - 12 \\cdot 3^{n-2} - 15 \\times 2 \\cdot 2^{n-2} + 18 \\cdot 2^{n-2}$ $= 30 \\cdot 3^{n-2} - 12 \\cdot 3^{n-2} - 30 \\cdot 2^{n-2} + 18 \\cdot 2^{n-2}$ $= 18 \\cdot 3^{n-2} - 12 \\cdot 2^{n-2}$ $= 2 \\cdot 3^n - 3 \\cdot 2^n$ $= a_n$ Thus $a_n = 2 \\cdot 3^n - 3 \\cdot 2^n$. Thus $a_n = 2 \\cdot 3^n - 3 \\cdot 2^n$ from all the cases. Soln11 Suppose $n$ is an arbitrary natural number. Suppose $% $. Thus we have following possible cases: • Case $n \\le 2$ Clearly from the given values $a_0, a_1, a_2$ are fibonacci numbers. • Case $n \\ge 3$ We have $a_n = \\frac {a_{n-3}} 2 + \\frac {3a_{n-2}} 2 + \\frac {a_{n-1}}", "# Finding the $n$-th derivative of $a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_1x+a_0$ Let $f(x)= a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_1x+a_0$. I am trying to find $f^n(x)$. By applying the power rule $n$ times, I get this $$f^n(x)=a_{n}(n\\cdot n)x^{n-n}+\\cdots+ a_1$$ which I think can be simplified to $$f^n(x)=a_{n}n^2+\\cdots+ a_1$$ However, I don't think I have the correct answer, as my exercise book is telling me the answer is. $$a_n\\,n\\cdot(n-1)\\cdot\\,\\cdots\\,\\cdot 1$$ What did I do wrong? I'm under the impression I have done multiple mistakes. • Start with $\\,\\left(x^n\\right)' = n x^{n-1}\\,$, $\\,\\left(x^n\\right)'' = n(n-1) x^{n-2}\\,$, $\\,\\left(x^n\\right)''' = n(n-1)(n-2) x^{n-3} \\,\\ldots\\,$ – dxiv Jul 30 '18 at 0:34 • Hint: $(x^n)' = n x^{n-1}$, but $(nx^{n-1})' = n(n-1) x^{n-2}$, and so on. Repeated application of the power rule give a falling factorial coefficient. – Xander Henderson Jul 30 '18 at 0:35 • The exercise book should say the answer is $a_nn(n-1)...1$. – 高田航 Jul 30 '18 at 0:39 • You are right, that is the answer in the book. I badly transcribed it, I modified my question. Thank you – Cedric Martens Jul 30 '18 at 0:41 Recall that the power rule for differentiation states: Let $f(x)=x^n$ for some $n\\in\\mathbb{R}\\backslash\\lbrace 0\\rbrace$. Then, $$\\frac{d}{dx}f(x)=nx^{n-1}$$ Also recall that differentiation is linear in the sense that $$\\frac{d}{dx}[f_1(x)+f_2(x)+\\cdots+f_k(x)]=\\frac{d}{dx}f_1(x)+\\frac{d}{dx}f_2(x)+\\cdots+\\frac{d}{dx}f_k(x)$$ Now consider your function $f(x)=a_nx^n+a_{n-1}x^{n-1}+\\cdots+a_1x+a_0$. Differentiating and applying linearity, we obtain $$\\frac{d}{dx}f(x)=\\frac{d}{dx}a_nx^n+\\frac{d}{dx}a_{n-1}x^{n-1}+\\cdots+\\frac{d}{dx}a_1x+\\frac{d}{dx}a_0$$ The constant term at", "- 2$ and $\\sum_1^n P\\cdot k! = {\\left(n^{3} - 3 n + 3\\right)} \\left(n + 1\\right)! - 3$ • $P = k^5 + 9$ and $\\sum_1^n P\\cdot k! = {\\left(n^{4} - 4 n^{2} + 6 n + 4\\right)} \\left(n + 1\\right)! - 4$ • $P = k^6 - 9$ and $\\sum_1^n P\\cdot k! = {\\left(n^{5} - 5 n^{3} + 10 n^{2} + 5 n - 30\\right)} \\left(n + 1\\right)! + 30$ • $P = k^7 - 50$ and $\\sum_1^n P\\cdot k! = {\\left(n^{6} - 6 n^{4} + 15 n^{3} + 4 n^{2} - 66 n + 55\\right)} \\left(n + 1\\right)! - 55$ • $P = k^8 + 267$ and $\\sum_1^n P\\cdot k! = {\\left(n^{7} - 7 n^{5} + 21 n^{4} - 119 n^{2} + 175 n + 126\\right)} \\left(n + 1\\right)! - 126$ All above results are computed and displayed via sage.", "= n + (n-2) + (n-4) + \\cdots + 4 + 2 =$$ $$2 ( 1 + 2 + \\cdots + \\frac{n}{2}) = 2\\frac{\\frac{n}{2}(\\frac{n}{2}+1)}{2} = \\frac{n(n+2)}{4} = \\Theta(n^2)$$", "$m=n$, I assert that $$N(n,m)\\approx n!\\cdot e^{m/n-1}$$ This approximation is by excess when $m>0$. The error made is less than $4.0\\%$ for $n\\ge4$; less than $1.8\\%$ for $n\\ge8$; less than $1.0\\%$ for $n\\ge14$. More rigourously stated: $$\\forall r\\in\\mathbb R, 0\\le r\\le1\\implies\\lim_{n\\rightarrow\\infty}{n!\\cdot e^{r-1}\\over N(n,\\lfloor r\\cdot n\\rfloor)}=1$$ -", "With this assumption, we will see what could be happened. $2^{N+1} = 2^N \\times 2 = 2^N + 2^N > 2N \\ge N + 1$. (By our assumption) This is $n = N+1$ case, so we proved the statement. Now the statement is true for $n = 2$, $n=3$, $\\cdots$.", "# How do you write the expression for the nth term of the sequence given 2, -4, 6, -8, 10,...? May 4, 2017 detail in explanation #### Explanation: case 1 you know how much is n 1.n is odd number then $A n = 2 + \\left(n - 1\\right) \\cdot 2$ 2.n is even number then $A n = - 4 + \\left(n - 2\\right) \\cdot \\left(- 2\\right)$ case 2. n is unknown $A n = \\left(2 + 2 \\cdot \\left(n - 1\\right)\\right) \\cdot {\\left(- 1\\right)}^{n - 1}$", "then $n!=1, \\text{when}\\ n=0$ $n!=n\\times (n-1)!\\ \\text{when} \\ n>1$ In other words, if $$n$$ is an integer greater than $$0$$, then $$n!=n\\times (n-1)\\times (n-2)\\times \\cdots \\times 2\\times 1$$. And $$0!=1$$ by definition. I hope this helps! - 4 years, 1 month ago", "= n(n+1)(n+2)(n+3)/4$. Thus we have $0 \\cdot 1 \\cdot 2 + 1 \\cdot 2 \\cdot 3 + 2 \\cdot 3 \\cdot 4 + \\cdot \\cdot \\cdot + n(n+1)(n+2) + (n+1)(n+1+1)(n+1+2)$. $= n(n+1)(n+2)(n+3)/4 + (n+1)(n+1+1)(n+1+2)$. $= n(n+1)(n+2)(n+3)/4 + (n+1)(n+2)(n+3)$. $= (n(n+1)(n+2)(n+3) + 4(n+1)(n+2)(n+3))/4$. $= (n+1)(n+2)(n+3)(n+4)/4$. $= (n+1)(n+1+1)(n+1+2)(n+1+3)/4$. Thus $P(n+1)$ is also true if $P(n)$ is true. Soln7 By checking the example mentioned and with some hit and try, it seems that formulae is $(3^{n+1} -1)/2$. Base Case: For $n = 0$, $3^0 = (3^{0+1} -1)/2 = 1$. Thus $P(0)$ is true. Induction Step: Suppose $P(n)$ is true. Thus $3^0 + 3^1 + 3^2 + 3^3 + \\cdot \\cdot \\cdot + 3^n = (3^{n+1} -1)/2$. Thus we have $3^0 + 3^1 + 3^2 + 3^3 + \\cdot \\cdot \\cdot + 3^n + 3^{n+1}$. $= (3^{n+1} -1)/2 + 3^{n+1}$. $= ((3^{n+1} -1) + (2 \\cdot 3^{n+1}))/2$. $= (3 \\cdot 3^{n+1} - 1)/2$. $= (3^{n+1+1} - 1)/2$. Thus $P(n+1)$ is also true if $P(n)$ is true. Soln8 First we can observe that on LHS there are $2n$ terms and on RHS there are $n$ terms. By Mathematical induction: Base Case: For $n = 1$. LHS will contain two terms = $\\frac 1 {2 \\cdot 1 - 1} - \\frac 1 {2 \\cdot 1} = 1 - \\frac 1 2$. And RHS will", "(n -1) \\cdot (n-1)! + (n+1 -1) \\cdot n! $ $= (2! - 1!) + (3! - 2!) + (4! - 3!) + \\cdots + (n! - (n-1)!) + ((n+1)! - n!)$ $ = (n+1)! - 1 $ so here $n = 50$ so the answer $S = 51!-1$. • April 30th 2009, 05:16 PM adhyeta Quote: Originally Posted by pickslides $ S_{50} = \\frac{50}{2}(2(-3)+(50-1)(-4))$ $ S_{50} = 25(-6+(49)(-4))$ $ S_{50} = 25(-6+-196)$ $ S_{50} = 25(-202)$ $ S_{50} = -5050$ Worked for me! hey! so sorry. i think i went wrong with the calc.(-calculation-) (Happy) • May 3rd 2009, 07:16 AM adhyeta Quote: Originally Posted by danny arrigo Here's the second one. Let $S_n = 1 \\cdot 1! + 2 \\cdot 2! + 3 \\cdot 3! + \\cdots + (n-1) \\cdot (n-1)! + n \\cdot n!$ $= (2-1) 1! + (3-1)2! + (4-1)3! + \\cdots + (n -1) \\cdot (n-1)! + (n+1 -1) \\cdot n! $ $= (2! - 1!) + (3! - 2!) + (4! - 3!) + \\cdots + (n! - (n-1)!) + ((n+1)! - n!)$ $ = (n+1)! - 1 $ so here $n = 50$ so the answer $S = 51!-1$. hey! can we similarily prove the sum $n^{2}.n!$???", "# Solve for n 72n + 432 n= ? Jul 9, 2016 $504 n$ see explanation #### Explanation: $72 n + 432 n = 504 n$ Unless I read the question wrong and it was $72 n + 432$, in which case you would leave it the same since there is nothing to simplify.", "$F_n$ with the following formula: $$F_n = 0.723607\\cdot 1.618034^n$$ Indeed, this formula does a great job approximating $F_n$, predicting its value exactly (provided one rounds to the nearest integer) all the way up to $n=29$. Further, had we better approximated the apparent constant $1.618034...$, the resulting formula would have worked all that much longer. This begs the question, however -- what is the exact value of the constant $1.618034...$ that we seek, and is there anything significant about it? Hmmm....", "# How do you find the indicated term of the arithmetic sequence a_1=7, d=3, n=14? Mar 17, 2018 ${a}_{14} = 46$ #### Explanation: ${a}_{1} = 7 , d = 3 , n = 14$ ${a}_{2} = {a}_{1} + \\left(2 - 1\\right) \\cdot d = 7 + 3 = 10$ ${a}_{3} = {a}_{1} + \\left(3 - 1\\right) \\cdot d = 7 + 6 = 13$ ${a}_{n} = {a}_{1} + \\left(n - 1\\right) \\cdot d$ $\\therefore {a}_{14} = {a}_{1} + \\left(14 - 1\\right) \\cdot d = 7 + 13 \\cdot 3 = 7 + 39 = 46$", "but the correct interpretation does not repeat the $3$ and the $2$. When $n=2$, the literal reading is even worse, since the factor $(n-2)$ is then $0$, but the intended meaning is still \"start at $n$ and continue to $1$\", so you get $2\\cdot 1$ (not $2\\cdot1\\cdot0\\cdots3\\cdot2\\cdot1$). Going down to $3, 2, 1$ is saying that it will keep multiplying until it reaches $1$, which is basically saying it will keep multiplying until it reaches $n - \\left( n - 1 \\right) = 1$. What you are saying is, for $n=5$, it goes till $n-4$. In general, it goes till $n-(n-1)$. But this is just $1$. It's equivalent. Concisely, it is $$n! = \\displaystyle\\prod_{k=1}^{n} k.$$" ]

[ "# Factorial Notation Factorial Notations $n!$ is the product of the first $n$ positive integers for $n\\ge 1$. $$n!=n(n-1)(n-2)(n-3) \\cdots \\times 3 \\times 2 \\times 1$$ $n!$ is read \"$n$ factorial\". For example, the product $5 \\times 4 \\times 3 \\times 2 \\times 1$ can be written as $5!$. Notice that $5 \\times 4 \\times 3$ can be written using factorial numbers only as $$5 \\times 4 \\times 3 = \\dfrac{5 \\times 4 \\times 3 \\times 2 \\times 1}{2 \\times 1} = \\dfrac{5!}{2!}$$ You can also notice that: \\begin{align} n! &= n(n-1)(n-2)(n-3) \\cdots \\times 3 \\times 2 \\times 1 \\\\ &= n(n-1)! \\\\ &= n(n-1)(n-2)! \\\\ &= n(n-1)(n-2)(n-3)! \\\\ &= \\cdots \\\\ \\end{align} Using this rule of factorial: \\begin{align} 1! &= 1 \\times 0! \\\\ 0! &= 1 \\\\ \\end{align} ### Example 1 Evaluate $5!$. ### Example 2 Evaluate $\\dfrac{6!}{4!}$. ### Example 3 Evaluate $\\dfrac{7!}{4! \\times 3!}$.", "&= 1 + a(16) + a(6) = 1 + 3 + 2 = 6 \\end{align*} Thus the answer is N = 15+6+8+6+4+3+2+2+1 = \\boxed{047}. 1 Like", "If $$n$$ is an integer such that $$n \\ge 0$$ then $$n$$ factorial is defined as, \\begin{align*}n! & = n\\left( {n - 1} \\right)\\left( {n - 2} \\right) \\cdots \\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right) & \\hspace{0.15in} & {\\mbox{if }}n \\ge 1\\\\ 0! & = 1 & \\hspace{0.15in} & {\\mbox{by definition}}\\end{align*} Let’s compute a couple real quick. \\begin{align*}& 1! = 1\\\\ & 2! = 2\\left( 1 \\right) = 2\\\\ & 3! = 3\\left( 2 \\right)\\left( 1 \\right) = 6\\\\ & 4! = 4\\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right) = 24\\\\ & 5! = 5\\left( 4 \\right)\\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right) = 120\\end{align*} In the last computation above, notice that we could rewrite the factorial in a couple of different ways. For instance, \\begin{align*}5! & = 5\\underbrace {\\left( 4 \\right)\\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right)}_{4!} = 5 \\cdot 4!\\\\ 5! & = 5\\left( 4 \\right)\\underbrace {\\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right)}_{3!} = 5\\left( 4 \\right) \\cdot 3!\\end{align*} In general, we can always “strip out” terms from a factorial as follows. \\begin{align*}n! & = n\\left( {n - 1} \\right)\\left( {n - 2} \\right) \\cdots \\left( {n - k} \\right)\\left( {n - \\left( {k + 1} \\right)} \\right) \\cdots \\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right)\\\\ & = n\\left( {n - 1} \\right)\\left( {n -", "every $n = 2, 3, \\ldots, N$, find the prime factorization of $n!$. This is not as nasty as it sounds: if you know that the prime factorization of $n!$ is $$n! = \\prod_i p_i^{e_i}$$ then the prime factorization of $(n + 1)!$ is simply that and the prime factorization of $n + 1$. For example, \\begin{alignat*}{4} 4! &= 2 \\cdot 3 \\cdot 4 &&= 2^3 \\cdot 3 \\\\ 5! &= 4! \\cdot 5 &&= (2^3 \\cdot 3) \\cdot 5 &&= 2^3 \\cdot 3 \\cdot 5 \\\\ 6! &= 5! \\cdot 6 &&= (2^3 \\cdot 3 \\cdot 5)(2 \\cdot 3) &&= 2^4 \\cdot 3^2 \\cdot 5\\\\ \\end{alignat*} Next, you need to partition $n!$ into factors $a$ and $b$. To ensure that $a$ and $b$ have no common factors, you simply have to ensure that each prime factor $p_i$ is attributed to either $a$ or $b$, along with all of its exponents $e_i$. For example, $4!$ can be partitioned as \\begin{alignat*}{3} (a, b) &= (1&&,\\ 2^3 \\cdot 3) \\\\ (a, b) &= (3&&,\\ 2^3) \\end{alignat*} but not as $$(a, b) = (2 \\cdot 3,\\ 2^2)$$ since 2 appears in both $a$ and $b$, making $\\frac{a}{b}$ reducible. So, the problem reduces to counting the power sets of the distinct prime factors of $n!$. Half of those partitions will meet the", "$$a_n=\\left(\\prod_{k=1}^nk\\right)^2.$$ \\begin{align} a_n&=n^2 a_{n-1}\\\\ &=n^2(n-1)^2a_{n-2}\\\\ &=n^2(n-1)^2(n-2)^2a_{n-3}\\\\ &\\vdots\\\\ &=n^2(n-1)^2(n-2)^2\\cdots 2^2\\overbrace{a_1}^{=1}\\\\ &=[n(n-1)(n-2)\\cdots 2\\cdot 1]^2\\\\ &=(n!)^2\\qquad\\blacksquare\\end{align}", "... + b_m = n$. We will now expand the lefthand side of the equation. We first expand the first two factors to get: (4) \\begin{align} \\quad \\binom{n}{b_1} \\cdot \\binom{n - b_1}{b_2} = \\frac{n!}{b_1! \\cdot (n - b_1)!} \\cdot \\frac{(n - b_1)!}{b_2! \\cdot (n - b_1 - b_2)!} = \\frac{n!}{b_1! \\cdot b_2! \\cdot (n - b_1 - b_2)!} \\end{align} • We then expand this with the third factor to get: (5) \\begin{align} \\quad \\binom{n}{b_1} \\cdot \\binom{n - b_1}{b_2} \\cdot \\binom{n - b_1 - b_2}{b_3} = \\frac{n!}{b_1! \\cdot b_2! \\cdot (n - b_1 - b_2)!} \\cdot \\binom{n - b_1 - b_2}{b_3} \\\\ = \\frac{n!}{b_1! \\cdot b_2! \\cdot (n - b_1 - b_2)!} \\cdot \\frac{(n - b_1 - b_2)!}{b_3! \\cdot (n - b_1 - b_2 - b_3)!} = \\frac{n!}{b_1! \\cdot b_2! \\cdot b_3! \\cdot (n - b_1 - b_2 - b_3)!} \\end{align} • Inductively, by expanding the $m^{\\mathrm{th}}$ factor of the product and noting that $n - b_1 - b_2 - ... - b_m = 0$, we see that: (6) \\begin{align} \\quad \\binom{b_1 + b_2 + ... + b_m}{b_1} \\cdot \\binom{b_2 + b_3 + ... + b_m}{b_2} \\cdot ... \\cdot \\binom{b_{m-1}}{b_m} \\\\ = \\binom{b_1 + b_2 + ... + b_m}{b_1, b_2, ..., b_m} = \\frac{n!}{b_1! \\cdot b_2! \\cdot ... \\cdot b_{m-1}! (n - b_1 - b_2 - ... - b_m)!}", "factors until we cannot divide any further. You may have seen a factor tree before: The factor tree for 90 looks like this: \\begin{align*}90 & = 9 \\cdot 10 \\\\ 9 & = 3 \\cdot 3 \\\\ 10 & = 5 \\cdot 2 \\\\ 90 & = 3 \\cdot 3 \\cdot 5 \\cdot 2\\end{align*} The factor tree for 126 looks like this: \\begin{align*}126 & = 9 \\cdot 14 \\\\ 9 & = 3 \\cdot 3 \\\\ 14 & = 7 \\cdot 2\\\\ 126 & = 3 \\cdot 3 \\cdot 7 \\cdot 2 \\end{align*} The LCM for 90 and 126 is made from the smallest possible collection of primes that enables us to construct either of the two numbers. We take only enough of each prime to make the number with the highest number of factors of that prime in its factor tree. Prime Factors in 90 Factors in 126 We Take 2 1 1 1 3 2 2 2 5 1 0 1 7 0 1 1 So we need: one 2, two 3's, one 5 and one 7. In other words: \\begin{align*}2 \\cdot 3 \\cdot 3 \\cdot 5 \\cdot 7 = 630\\end{align*} • The lowest common multiple of 90 and 126 is 630. The LCD for our calculation is 630. 90 divides into 630 seven times (notice", "that \\begin{align*} 0! &= 1 \\end{align*} The first few factorials are \\begin{align*} 1! &=1 & 2! &=2 & 3! &=6\\\\ 4! &=24 & 5! &=120 & 6! &=720 \\end{align*} Now consider $T_n(x)$ and its derivatives: \\begin{alignat*}{4} T_n(x) &=& c_0 &+ c_1(x-a) & + c_2 (x-a)^2 & + c_3(x-a)^3 &+ \\cdots+ & c_n (x-a)^n\\\\ T_n'(x) &=& &c_1 & + 2 c_2 (x-a) & + 3c_3(x-a)^2 &+ \\cdots +& n c_n (x-a)^{n-1}\\\\ T_n''(x) &=& & & 2 c_2 & + 6c_3(x-a) &+ \\cdots +& n(n-1) c_n (x-a)^{n-2}\\\\ T_n'''(x) &=& & & & 6c_3 &+ \\cdots + & n(n-1)(n-2) c_n (x-a)^{n-3}\\\\ & \\vdots\\\\ T_n^{(n)}(x) &=& & & & & & n! \\cdot c_n \\end{alignat*} Now notice that when we substitute $x=a$ into the above expressions only the constant terms survive and we get \\begin{align*} T_n(a) &= c_0\\\\ T_n'(a) &= c_1\\\\ T_n''(a) &= 2\\cdot c_2\\\\ T_n'''(a) &= 6 \\cdot c_3\\\\ &\\vdots\\\\ T_n^{(n)}(a) &= n! \\cdot c_n \\end{align*} So now if we want to set the coefficients of $T_n(x)$ so that it agrees with $f(x)$ at $x=a$ then we need \\begin{align*} T_n(a) &= c_0 = f(a) & c_0 &= f(a) = \\frac{1}{0!} f(a)\\\\ \\end{align*} We also want the first $n$ derivatives of $T_n(x)$ to agree with the derivatives of $f(x)$ at $x=a\\text{,}$ so \\begin{align*} T_n'(a) &= c_1 = f'(a) & c_1 &= f'(a)", "we’ll choose to eliminate $$2^{k+1}$$, but we could do it the other way, and it would be fine. \\begin{align} & P(k)=2^{k+1} + 5\\cdot 9^{k} \\\\ & 2^{k+1} = P(k) - 5\\cdot 9^{k} \\end{align} We can now apply this to $$P(k+1)$$, by taking out a factor of $$2$$ from $$2^{k+1+1}$$ and then substituting. \\begin{align} P(k+1) &= 2^{k+1+1} + 5\\cdot 9^{k+1} \\\\ &= 2\\cdot 2^{k+1} + 5\\cdot 9^{k+1} \\\\ &= 2\\left (P(k) - 5\\cdot 9^{k}\\right ) + 5\\cdot 9^{k+1} \\end{align} Now we can factorise the expression, and hopefully find a factor of $$7$$ lurking somewhere! \\begin{align} P(k+1) &= 2\\left (P(k) - 5\\cdot 9^{k}\\right ) + 5\\cdot 9^{k+1} \\\\ &= 2P(k) - 10\\cdot 9^{k} + 5\\cdot 9^{k+1} \\\\ &= 2P(k) - 10\\cdot 9^{k} + 5\\cdot 9\\cdot 9^{k} \\\\ &= 2P(k) + 35\\cdot 9^{k} \\end{align} This means that $$P(k)$$ being divisible by $$7$$ implies that $$P(k+1)$$ is also divisible by $$7$$. Now onto the completion step! This is pretty much the same every time: “ Thus $$P(n)$$ implies $$P(n+1)$$ and as $$P(1)$$ is true, $$P(n)$$ holds for all $$n \\in \\mathbb {N}$$ where $$n \\geqq 1$$. ”", "4! + 3\\cdot 3! + 2\\cdot 2! + 0\\cdot 1! & = \\\\ 5\\cdot120+4\\cdot24 + 3\\cdot6 + 2\\cdot 2 + 0\\cdot 1 & =\\\\ 600 + 96 + 18 + 4 + 0 & = 718.\\end{align}\\$ This has actual applications; for example it is a useful way to represent a permutation of a list. -", "F_0 \\cdots F_{k - 1} + 2$. Now observe that \\begin{align*} F_{k + 1} & = 2^{2^{k + 1}} + 1 \\\\ & = \\left(\\left(2^{2^{k}}\\right)^2 + 2 \\cdot 2^{2^k} + 1\\right) - 2\\cdot 2^{2^k} \\\\ & = (F_k)^2 - 2F_k + 2 \\\\ & = (F_0 \\cdot F_1\\cdots F_{k - 1} + 2)\\cdot F_k - 2F_k + 2 \\\\ & = F_0 \\cdot F_1 \\cdots F_{k - 1} \\cdot F_k + 2, \\end{align*} as desired. It follows that if $m < n$ that $\\gcd(F_m, F_n) = \\gcd(F_m, 2) = 1$, so $F_m$ and $F_n$ are relatively prime, as claimed.", "c &= 3^2 + 1^2 = 10. \\end{align*} Furthermore, $c = 2\\cdot 5 \\cdot 1$, so letting $m = 5$ and $n = 1$ gives \\begin{align*} d &= 5^2 - 1^2 = 24\\\\ e &= 5^2 + 1^2 = 26. \\end{align*} And again, $e = 2\\cdot 13 \\cdot 1$, so letting $m = 13$ and $n = 1$ gives \\begin{align*} f &= 13^2 - 1^2 = 168\\\\ g &= 13^2 + 1^2 = 170. \\end{align*} Then \\begin{align*} g^2 = 170^2 = 26^2 + 168^2 = 10^2 + 24^2 + 168^2 = 6^2 + 8^2 + 24^2 + 168^2 \\end{align*} Going one more step we can write $170 = 2\\cdot 17 \\cdot 5$ so $m = 17$ and $n = 5$ gives \\begin{align*} h &= 17^2 - 5^2 = 264\\\\ i &= 17^2 + 5^2 = 314, \\end{align*} in which $314 = 2\\cdot 157 \\cdot 1$. It seems I may be on the right track, but I'm not sure if this is true in general. Maybe it can be proven that this process can be continued arbitrarily given we start with two distinct primes $m$ and $n$ in the first step? Any thoughts or references or anything would be wonderful. Thanks in advance! - Yes, your process can always be continued if you start with $m_1>n_1$ both odd. $$a_1", "= \\prod_{i=1}^k (a_i+1)$. Proof: To see that this is true, notice that since $\\nu$ is multiplicative we only need to know how to evaluate $\\nu$ on prime powers. But notice that for a prime power $p^a$, the only divisors of $p^a$ are $1,p,p^2,\\cdots,p^a$. There are $a+1$ many divisors, and so we have $\\nu(p^a) = a+1$. Hence we can compute (5) \\begin{align} \\nu(n) = \\nu(p_1^{a_1}\\cdots p_k^{a_k}) = \\nu(p_1^{a_1})\\cdots \\nu(p_k^{a_k}) = (a_1+1)\\cdots(a_k+1) = \\prod_{i=1}^k(a_i+1). \\end{align} $\\square$ # The Product of Divisors of a Function At the end of class you all split up into groups and tried to come up with a simple formula for (6) \\begin{align} \\prod_{d \\mid n} d. \\end{align}", "2^3 \\cdot 6 \\cdot 5^{n-3} + 2^4 \\cdot a_{n-3} \\end{align} Hence, in general (you need to prove the claim below using induction) $$a_{n+1} = 2^{j+1}a_{n-j} + 6 \\left( \\sum_{k=0}^{j} 2^k 5^{n-k} \\right)$$ Setting $j=n$, we get that \\begin{align} a_{n+1} & = 2^{n+1}a_0 + 6 \\left( \\sum_{k=0}^{n} 2^k 5^{n-k} \\right) = 2^{n+1}a_0 + 6 \\cdot 5^n \\left( \\sum_{k=0}^{n} \\left(\\dfrac25 \\right)^k \\right)\\\\ & = 2^{n+1}a_0 + 6 \\cdot 5^n \\cdot \\dfrac{1-(2/5)^{n+1}}{1-2/5}\\\\ & = 2^{n+1}a_0 + 6 \\cdot \\dfrac{5^{n+1} -2^{n+1}}3\\\\ & = 2^{n+1}a_0 + 2 \\cdot (5^{n+1} -2^{n+1}) \\end{align} EDIT You can verify if it is true by plugging the value for $a_n$ and $a_{n+1}$ and computing $a_{n+1} - 2a_n$. $$a_n = 2^{n}a_0 + 2 \\cdot (5^{n} -2^{n})$$ Hence, $$2a_n = 2^{n+1}a_0 + 4 \\cdot (5^{n} -2^{n})$$ Hence, \\begin{align} a_{n+1} - 2a_n & = 2^{n+1}a_0 + 2 \\cdot (5^{n+1} -2^{n+1}) - \\left( 2^{n+1}a_0 + 4 \\cdot (5^{n} -2^{n})\\right)\\\\ & = 2 \\cdot (5^{n+1} -2^{n+1}) - 4 \\cdot (5^n - 2^n)\\\\ & = 2 \\cdot 5^{n+1} -2^{n+2} - 4 \\cdot 5^n + 4 \\cdot 2^n\\\\ & = 10 \\cdot 5^{n} - 4 \\cdot 5^n\\\\ & = 6 \\cdot 5^n \\end{align} - @WillJagy I verified it and I think what I have is correct unless I have made some error in the verification process. – user17762 Nov 23 '12 at 23:51 The answer", "## Wesley 5 years ago Factor each polynominal 44a^2+11a Can anyone show me how to work this? Well, the first thing in these cases is always to see that a * a = a^2, so you have: \\begin{align} 44a^2 &+ 11a\\\\ 44a\\cdot a &+ 11a \\end{align} In that case, you can factor an $$a$$ out: $a(44a + 11)$ Does that make sense? 2. Wesley Yes 3. anonymous You can also factor out an 11: $11a(4a+1)$", "1$ and $n$ be a non-zero integer. We have \\begin{align*}a^n-1=(a-1)(a^{n-1}+a^{n-2}+\\cdots+a+1).\\end{align*} Let us give a simple proof of this identity: We set \\begin{align*} A=a^{n-1}+a^{n-2}+\\cdots+a+1.\\end{align*}We have \\begin{align*}(a-1)A&=a A-A\\cr &= (a^{n}+a^{n-1}+\\cdots+a^2+a)-(a^{n-1}+a^{n-2}+\\cdots+a+1)\\cr &= a^n-1.\\end{align*} Let now $a$ and $b$ be two numbers such that $a\\neq b$. Then \\begin{align*}a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\\cdots+ab^{n-2}+b^{n-1}).\\end{align*} Preuve: We assume that $a$ is non-zero. Then we factor by $a^n$, we get \\begin{align*}a^n-b^n=a^n\\left(1-\\left(\\frac{b}{a}\\right)^n\\right)\\end{align*}By using the first higher remarkable identity, we have\\begin{align*} \\left(1-\\left(\\frac{b}{a}\\right)^{n}\\right)=\\left(1-\\frac{b}{a}\\right)\\left( \\left(\\frac{b}{a}\\right)^{n-1}+\\left(\\frac{b}{a}\\right)^{n-2}+\\cdots+\\frac{b}{a}+1 \\right) \\end{align*}On the other hand, we write $a^n=a a^{n-1}$. We obtain \\begin{align*}a^n-b^n&=a\\left(1-\\frac{b}{a}\\right)a^{n-1}\\left( \\left(\\frac{b}{a}\\right)^{n-1}+\\left(\\frac{b}{a}\\right)^{n-2}+\\cdots+\\frac{b}{a}+1 \\right)\\cr & =(a-b)(a^{n-1}+a^{n-2}b+\\cdots+ab^{n-2}+b^{n-1}). \\end{align*}", "$(2)$, multiply both sides by $\\rm x$ and then set $\\rm x=2$. - Your last term in $(2)$ should be $nx^{n-1}$ instead – Kirthi Raman Mar 14 '12 at 16:07 Although you got great answers, I will add another way to look at the expression you are looking for Let us denote $$S = \\sum_{k=1}^{n} k \\cdot 2^k$$ This can be expanded as \\begin{align*} S &= 2 + 2 . 2^2 + 3 . 2^3 + \\cdots n 2^n\\\\ 2S &= \\hspace{8pt}+1.2^2+2.2^3 + \\cdots (n-1) . 2^n+n . 2^{n+1} \\end{align*} If you notice, by multiplying by $2$ and writing under similar terms and then subtracting, we get \\begin{align*} S = n . 2^{n+1} - \\left(2+2^2+2^3+\\cdots+2^n \\right) &= n .2^{n+1}-2^{n+1}+2\\\\ &= 2(n . 2^n-2^n+1) \\end{align*} -", "• h0pe Let S be the set of numbers of the form $$n(n + 1)(n + 2)(n + 3)(n + 4),$$ where n is any positive integer. The first few terms of S are \\begin{align*} 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 &= 120, \\\\ 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 &= 720, \\\\ 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 &= 2520, \\end{align*} and so on. What is the GCD of the elements of S? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "\\begin{align}1+10=&100-1 \\\\ 1+10+100=&1000-1 \\\\ \\vdots\\end{align} So in general, for base $r$ it holds that $$(r-1)+r(r-1)+r^2(r-1)+\\cdots r^n(r-1)=r^{n+1}-1$$ Therefore $$(r-1)(1+r+\\cdots r^n)=r^{n+1}-1\\implies 1+r+\\cdots r^n=\\frac{r^{n+1}-1}{r-1}$$ If there is a constant in front, just factor it out to get $$a_0\\frac{r^{n+1}-1}{r-1}$$", "+0 # help 0 82 3 Consider the following two values: \\begin{align} A &= 4^{5000} \\cdot 4^{5000} \\\\ B &= 2^{19999} + 2^{19999} \\end{align} Which is bigger? Jul 1, 2020 #1 +732 +1 i think they are equal since if you simplify, $$A=4^{5000}\\cdot4^{5000}=2^{10000}\\cdot2^{10000}=\\boxed{2^{20000}}$$ $$B=2^{19999}+2^{19999}=2(2^{19999})=\\boxed{2^{20000}}$$ I'm not sure if this is right, so please let me know if it is! :) Jul 1, 2020 #2 0 Yes, you are right. They are EQUAL. Jul 1, 2020 #3 +732 0 oh wow! yay! :D" ]

[ "c &= 3^2 + 1^2 = 10. \\end{align*} Furthermore, $c = 2\\cdot 5 \\cdot 1$, so letting $m = 5$ and $n = 1$ gives \\begin{align*} d &= 5^2 - 1^2 = 24\\\\ e &= 5^2 + 1^2 = 26. \\end{align*} And again, $e = 2\\cdot 13 \\cdot 1$, so letting $m = 13$ and $n = 1$ gives \\begin{align*} f &= 13^2 - 1^2 = 168\\\\ g &= 13^2 + 1^2 = 170. \\end{align*} Then \\begin{align*} g^2 = 170^2 = 26^2 + 168^2 = 10^2 + 24^2 + 168^2 = 6^2 + 8^2 + 24^2 + 168^2 \\end{align*} Going one more step we can write $170 = 2\\cdot 17 \\cdot 5$ so $m = 17$ and $n = 5$ gives \\begin{align*} h &= 17^2 - 5^2 = 264\\\\ i &= 17^2 + 5^2 = 314, \\end{align*} in which $314 = 2\\cdot 157 \\cdot 1$. It seems I may be on the right track, but I'm not sure if this is true in general. Maybe it can be proven that this process can be continued arbitrarily given we start with two distinct primes $m$ and $n$ in the first step? Any thoughts or references or anything would be wonderful. Thanks in advance! - Yes, your process can always be continued if you start with $m_1>n_1$ both odd. $$a_1", "# Factorial Notation Factorial Notations $n!$ is the product of the first $n$ positive integers for $n\\ge 1$. $$n!=n(n-1)(n-2)(n-3) \\cdots \\times 3 \\times 2 \\times 1$$ $n!$ is read \"$n$ factorial\". For example, the product $5 \\times 4 \\times 3 \\times 2 \\times 1$ can be written as $5!$. Notice that $5 \\times 4 \\times 3$ can be written using factorial numbers only as $$5 \\times 4 \\times 3 = \\dfrac{5 \\times 4 \\times 3 \\times 2 \\times 1}{2 \\times 1} = \\dfrac{5!}{2!}$$ You can also notice that: \\begin{align} n! &= n(n-1)(n-2)(n-3) \\cdots \\times 3 \\times 2 \\times 1 \\\\ &= n(n-1)! \\\\ &= n(n-1)(n-2)! \\\\ &= n(n-1)(n-2)(n-3)! \\\\ &= \\cdots \\\\ \\end{align} Using this rule of factorial: \\begin{align} 1! &= 1 \\times 0! \\\\ 0! &= 1 \\\\ \\end{align} ### Example 1 Evaluate $5!$. ### Example 2 Evaluate $\\dfrac{6!}{4!}$. ### Example 3 Evaluate $\\dfrac{7!}{4! \\times 3!}$.", "4! + 3\\cdot 3! + 2\\cdot 2! + 0\\cdot 1! & = \\\\ 5\\cdot120+4\\cdot24 + 3\\cdot6 + 2\\cdot 2 + 0\\cdot 1 & =\\\\ 600 + 96 + 18 + 4 + 0 & = 718.\\end{align}\\$ This has actual applications; for example it is a useful way to represent a permutation of a list. -", "$(2)$, multiply both sides by $\\rm x$ and then set $\\rm x=2$. - Your last term in $(2)$ should be $nx^{n-1}$ instead – Kirthi Raman Mar 14 '12 at 16:07 Although you got great answers, I will add another way to look at the expression you are looking for Let us denote $$S = \\sum_{k=1}^{n} k \\cdot 2^k$$ This can be expanded as \\begin{align*} S &= 2 + 2 . 2^2 + 3 . 2^3 + \\cdots n 2^n\\\\ 2S &= \\hspace{8pt}+1.2^2+2.2^3 + \\cdots (n-1) . 2^n+n . 2^{n+1} \\end{align*} If you notice, by multiplying by $2$ and writing under similar terms and then subtracting, we get \\begin{align*} S = n . 2^{n+1} - \\left(2+2^2+2^3+\\cdots+2^n \\right) &= n .2^{n+1}-2^{n+1}+2\\\\ &= 2(n . 2^n-2^n+1) \\end{align*} -", "# How do you factor n^2-10n+9? Apr 10, 2018 $\\left(n - 1\\right) \\left(n - 9\\right)$ #### Explanation: ${n}^{2} - 10 n + 9 = \\left(n - 1\\right) \\left(n - 9\\right)$ $n \\cdot n = {n}^{2}$ $n \\cdot - 9 = - 9 n$ $n \\cdot - 1 = - n$ $- 9 \\cdot - 1 = 9$ When you combine those you get: ${n}^{2} - 10 n + 9$", "calculate the value of $f(2)$ from the value of $f(1)$, which itself is calculated from $f(0)$ -- which was given to us. $$f(1)=9\\implies f(2)=2\\cdot f(1)+3=2\\cdot 9+3$$ - Thank you very much this made it much clearer. I didn't realize f(1) was referencing the answer to the equation. – John Smith Jul 12 '13 at 2:28 $$f(0)=0$$ $$f(n+1)=n+1$$ \\begin{align} f(n=1)&=f(n=0)+1=0+1=1\\\\ f(n=2)&=f(n=1)+1=(f(n=0)+1)+1=(0+1)+1=2\\\\ f(n=3)&=f(n=2)+1=(f(n=1)+1)+1=((f(n=0)+1)+1)+1=((0+1)+1)+1=3\\\\ \\end{align}", "of $3+6+12+\\cdots$ to $8$ terms. \\begin{align} \\displaystyle u_1 &= 3 \\\\ r &= 2 \\\\ n &= 8 \\\\ S_8 &= \\dfrac{3(2^8-1)}{2-1} \\\\ &= 765 \\end{align} ## Example 2 Find the sum of $0.2+0.02+0.002+\\cdots$ of the first $n$ terms. \\begin{align} \\displaystyle u_1 &= 0.2 \\\\ r &= 0.1 \\\\ n &= n \\\\ S_n &= \\dfrac{0.2(0.1^n-1)}{0.1-1} \\\\ &= \\dfrac{2(1-0.1^n)}{9} \\end{align}", "+ (n+1)) \\\\ & = -1 +(n+1)! \\cdot (n+2) \\\\ & = -1 + (n+2)! \\end{align*} • I did like one of the answers which got deleted, it said let $y=(n+1)!$, then, $-1+(y+y(n+1))=-1+y(1+(n+1))$ – Elis Jones Jan 20 '15 at 12:52 $n \\cdot n! = (n+1-1)n!=(n+1)n!-n!=(n+1)!-n!$ Now, sum of all this taking $n=1,2,3,...$ $1 \\cdot 1!+2 \\cdot 2! + 3 \\cdot 3! + ... + n \\cdot n!$$=(2!-1!)+(3!-2!)+(4!-3!)+...+((n+1)!-n!)=(n+1)!-1! • Your post is overlapping to the right and I don't see how it answers the question at hand. – AlexR Jan 20 '15 at 12:57 • Sorry, I've just provided a new method to you. I didn't read whole the question, I understood you need a general method. – Hardey Pandya Jan 20 '15 at 12:59 • That should actually be an answer to the linked question, not this one. – AlexR Jan 20 '15 at 13:00 For example, to simplify 6!, substitute 6 for n in the following equation:$$n!=n\\cdot(n-1)\\cdot(n-2)\\cdot(n-3)\\cdot...6!=6\\cdot(6-1)\\cdot(6-2)\\cdot(6-3)\\cdot(6-4)\\cdot(6-5)$$and now simplify:$$6!=6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1$$but you don't have to including the 1 because it's the same thing as not using it:$$6!=6\\cdot5\\cdot4\\cdot3\\cdot2$$Now evaluate it:$$30\\cdot4\\cdot3\\cdot2120\\cdot3\\cdot2360\\cdot2720$\\$Use the regular equation to help you with not just this, but any whole-number factorial.", "that \\begin{align*} 0! &= 1 \\end{align*} The first few factorials are \\begin{align*} 1! &=1 & 2! &=2 & 3! &=6\\\\ 4! &=24 & 5! &=120 & 6! &=720 \\end{align*} Now consider $T_n(x)$ and its derivatives: \\begin{alignat*}{4} T_n(x) &=& c_0 &+ c_1(x-a) & + c_2 (x-a)^2 & + c_3(x-a)^3 &+ \\cdots+ & c_n (x-a)^n\\\\ T_n'(x) &=& &c_1 & + 2 c_2 (x-a) & + 3c_3(x-a)^2 &+ \\cdots +& n c_n (x-a)^{n-1}\\\\ T_n''(x) &=& & & 2 c_2 & + 6c_3(x-a) &+ \\cdots +& n(n-1) c_n (x-a)^{n-2}\\\\ T_n'''(x) &=& & & & 6c_3 &+ \\cdots + & n(n-1)(n-2) c_n (x-a)^{n-3}\\\\ & \\vdots\\\\ T_n^{(n)}(x) &=& & & & & & n! \\cdot c_n \\end{alignat*} Now notice that when we substitute $x=a$ into the above expressions only the constant terms survive and we get \\begin{align*} T_n(a) &= c_0\\\\ T_n'(a) &= c_1\\\\ T_n''(a) &= 2\\cdot c_2\\\\ T_n'''(a) &= 6 \\cdot c_3\\\\ &\\vdots\\\\ T_n^{(n)}(a) &= n! \\cdot c_n \\end{align*} So now if we want to set the coefficients of $T_n(x)$ so that it agrees with $f(x)$ at $x=a$ then we need \\begin{align*} T_n(a) &= c_0 = f(a) & c_0 &= f(a) = \\frac{1}{0!} f(a)\\\\ \\end{align*} We also want the first $n$ derivatives of $T_n(x)$ to agree with the derivatives of $f(x)$ at $x=a\\text{,}$ so \\begin{align*} T_n'(a) &= c_1 = f'(a) & c_1 &= f'(a)", "we’ll choose to eliminate $$2^{k+1}$$, but we could do it the other way, and it would be fine. \\begin{align} & P(k)=2^{k+1} + 5\\cdot 9^{k} \\\\ & 2^{k+1} = P(k) - 5\\cdot 9^{k} \\end{align} We can now apply this to $$P(k+1)$$, by taking out a factor of $$2$$ from $$2^{k+1+1}$$ and then substituting. \\begin{align} P(k+1) &= 2^{k+1+1} + 5\\cdot 9^{k+1} \\\\ &= 2\\cdot 2^{k+1} + 5\\cdot 9^{k+1} \\\\ &= 2\\left (P(k) - 5\\cdot 9^{k}\\right ) + 5\\cdot 9^{k+1} \\end{align} Now we can factorise the expression, and hopefully find a factor of $$7$$ lurking somewhere! \\begin{align} P(k+1) &= 2\\left (P(k) - 5\\cdot 9^{k}\\right ) + 5\\cdot 9^{k+1} \\\\ &= 2P(k) - 10\\cdot 9^{k} + 5\\cdot 9^{k+1} \\\\ &= 2P(k) - 10\\cdot 9^{k} + 5\\cdot 9\\cdot 9^{k} \\\\ &= 2P(k) + 35\\cdot 9^{k} \\end{align} This means that $$P(k)$$ being divisible by $$7$$ implies that $$P(k+1)$$ is also divisible by $$7$$. Now onto the completion step! This is pretty much the same every time: “ Thus $$P(n)$$ implies $$P(n+1)$$ and as $$P(1)$$ is true, $$P(n)$$ holds for all $$n \\in \\mathbb {N}$$ where $$n \\geqq 1$$. ”", "\\text{Age of boiler now} \\\\ B_t & \\text{Age of boiler then} \\\\ t & \\text{Time between then and now} \\end{array}$ This leaves us with the following equations to solve: \\begin{align} S_n & = S_t + t \\tag{a} \\\\ B_n & = B_t + t \\tag{b} \\\\ S_n & = 2 \\cdot B_t \\tag{c} \\\\ S_t & = B_n \\tag{d} \\\\ S_n + B_n &= 30 \\tag{e} \\\\ \\end{align} Combining (a) and (d) yields $S_n = B_n + t$ or alternatively $t = S_n - B_n$ which is more to the point. Combining this with (b) and (c) gives us \\begin{align} S_n & = 2 \\cdot B_t \\\\ S_n & = 2 \\cdot (B_n - t) \\\\ S_n & = 2 \\cdot (B_n - (S_n - B_n)) \\\\ S_n & = 2 \\cdot (2 \\cdot B_n - S_n) \\\\ S_n & = 4 \\cdot B_n - 2 \\cdot S_n \\\\ 3 \\cdot S_n & = 4 \\cdot B_n \\end{align} Combining this with (e), we can solve it: \\begin{align} 3 \\cdot S_n & = 4 \\cdot B_n \\\\ 3 \\cdot S_n & = 4 \\cdot (30 - S_n) \\\\ 3 \\cdot S_n & = 120 - 4 \\cdot S_n \\\\ 7 \\cdot S_n & = 120 \\\\ S_n & = \\frac{120}{7} \\end{align} Remembering that they total to 30 years, we", "the sequence $\\left \\{ \\frac{n!}{n} \\right \\}_{n=1}^{\\infty}$ is increasing. Recall that $n! = n \\cdot (n-1) \\cdot (n-2) \\cdot ...2 \\cdot 1$, so then $\\left \\{ \\frac{n!}{n} \\right \\}_{n=1}^{\\infty} = \\left \\{ (n - 1) \\cdot (n - 2) \\cdot ... \\cdot 2 \\cdot 1 \\right \\}_{n=1}^{\\infty} = \\left \\{(n-1)! \\right \\}_{n=1}^{\\infty}$. So we must show that $a_n < a_{n+1}$ for all $n \\in \\mathbb{N}$ to show this sequence is increasing. We note that $a_n = (n - 1)! = (n-1) \\cdot (n-2) \\cdot ... \\cdot 2 \\cdot 1$ while $a_{n+1} = (n + 1 - 1)! = n! = n \\cdot (n-1) \\cdot (n-2) \\cdot ... \\cdot 2 \\cdot 1$. Therefore: (2) \\begin{align} (n - 1)! ≤ n! \\\\ \\quad (n-1) \\cdot (n-2) \\cdot ... \\cdot 2 \\cdot 1 ≤ n \\cdot (n-1) \\cdot (n-2) \\cdot ... \\cdot 2 \\cdot 1 \\\\ 1 ≤ n \\end{align} Since $n > 1$, $a_{1} = 0! = 1$ while $a_{2} = 1! = 1$. So for $n > 1$, our sequence is increasing.", "every $n = 2, 3, \\ldots, N$, find the prime factorization of $n!$. This is not as nasty as it sounds: if you know that the prime factorization of $n!$ is $$n! = \\prod_i p_i^{e_i}$$ then the prime factorization of $(n + 1)!$ is simply that and the prime factorization of $n + 1$. For example, \\begin{alignat*}{4} 4! &= 2 \\cdot 3 \\cdot 4 &&= 2^3 \\cdot 3 \\\\ 5! &= 4! \\cdot 5 &&= (2^3 \\cdot 3) \\cdot 5 &&= 2^3 \\cdot 3 \\cdot 5 \\\\ 6! &= 5! \\cdot 6 &&= (2^3 \\cdot 3 \\cdot 5)(2 \\cdot 3) &&= 2^4 \\cdot 3^2 \\cdot 5\\\\ \\end{alignat*} Next, you need to partition $n!$ into factors $a$ and $b$. To ensure that $a$ and $b$ have no common factors, you simply have to ensure that each prime factor $p_i$ is attributed to either $a$ or $b$, along with all of its exponents $e_i$. For example, $4!$ can be partitioned as \\begin{alignat*}{3} (a, b) &= (1&&,\\ 2^3 \\cdot 3) \\\\ (a, b) &= (3&&,\\ 2^3) \\end{alignat*} but not as $$(a, b) = (2 \\cdot 3,\\ 2^2)$$ since 2 appears in both $a$ and $b$, making $\\frac{a}{b}$ reducible. So, the problem reduces to counting the power sets of the distinct prime factors of $n!$. Half of those partitions will meet the", "If $$n$$ is an integer such that $$n \\ge 0$$ then $$n$$ factorial is defined as, \\begin{align*}n! & = n\\left( {n - 1} \\right)\\left( {n - 2} \\right) \\cdots \\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right) & \\hspace{0.15in} & {\\mbox{if }}n \\ge 1\\\\ 0! & = 1 & \\hspace{0.15in} & {\\mbox{by definition}}\\end{align*} Let’s compute a couple real quick. \\begin{align*}& 1! = 1\\\\ & 2! = 2\\left( 1 \\right) = 2\\\\ & 3! = 3\\left( 2 \\right)\\left( 1 \\right) = 6\\\\ & 4! = 4\\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right) = 24\\\\ & 5! = 5\\left( 4 \\right)\\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right) = 120\\end{align*} In the last computation above, notice that we could rewrite the factorial in a couple of different ways. For instance, \\begin{align*}5! & = 5\\underbrace {\\left( 4 \\right)\\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right)}_{4!} = 5 \\cdot 4!\\\\ 5! & = 5\\left( 4 \\right)\\underbrace {\\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right)}_{3!} = 5\\left( 4 \\right) \\cdot 3!\\end{align*} In general, we can always “strip out” terms from a factorial as follows. \\begin{align*}n! & = n\\left( {n - 1} \\right)\\left( {n - 2} \\right) \\cdots \\left( {n - k} \\right)\\left( {n - \\left( {k + 1} \\right)} \\right) \\cdots \\left( 3 \\right)\\left( 2 \\right)\\left( 1 \\right)\\\\ & = n\\left( {n - 1} \\right)\\left( {n -", "$$a_n=\\left(\\prod_{k=1}^nk\\right)^2.$$ \\begin{align} a_n&=n^2 a_{n-1}\\\\ &=n^2(n-1)^2a_{n-2}\\\\ &=n^2(n-1)^2(n-2)^2a_{n-3}\\\\ &\\vdots\\\\ &=n^2(n-1)^2(n-2)^2\\cdots 2^2\\overbrace{a_1}^{=1}\\\\ &=[n(n-1)(n-2)\\cdots 2\\cdot 1]^2\\\\ &=(n!)^2\\qquad\\blacksquare\\end{align}", "## zmudz one year ago Find a closed form for $$S_n = 1 \\cdot 1! + 2 \\cdot 2! + \\ldots + n \\cdot n!.$$ for integer $$n \\geq 1.$$ Your response should have a factorial. 1. anonymous Notice that $n\\cdot n!=(n+1-1)\\cdot n!=(n+1)\\cdot n!-n!=(n+1)!-n!$ Now, \\begin{align*}S_n&=1\\cdot1!+2\\cdot2!+\\cdots+(n-1)\\cdot(n-1)!+n\\cdot n!\\\\[1ex] &=(2!-1!)+(3!-2!)+\\cdots+(n!-(n-1)!)+((n+1)!-n!)\\\\[1ex] &=\\cdots \\end{align*} 2. anonymous clever telescoping", "2^3 \\cdot 6 \\cdot 5^{n-3} + 2^4 \\cdot a_{n-3} \\end{align} Hence, in general (you need to prove the claim below using induction) $$a_{n+1} = 2^{j+1}a_{n-j} + 6 \\left( \\sum_{k=0}^{j} 2^k 5^{n-k} \\right)$$ Setting $j=n$, we get that \\begin{align} a_{n+1} & = 2^{n+1}a_0 + 6 \\left( \\sum_{k=0}^{n} 2^k 5^{n-k} \\right) = 2^{n+1}a_0 + 6 \\cdot 5^n \\left( \\sum_{k=0}^{n} \\left(\\dfrac25 \\right)^k \\right)\\\\ & = 2^{n+1}a_0 + 6 \\cdot 5^n \\cdot \\dfrac{1-(2/5)^{n+1}}{1-2/5}\\\\ & = 2^{n+1}a_0 + 6 \\cdot \\dfrac{5^{n+1} -2^{n+1}}3\\\\ & = 2^{n+1}a_0 + 2 \\cdot (5^{n+1} -2^{n+1}) \\end{align} EDIT You can verify if it is true by plugging the value for $a_n$ and $a_{n+1}$ and computing $a_{n+1} - 2a_n$. $$a_n = 2^{n}a_0 + 2 \\cdot (5^{n} -2^{n})$$ Hence, $$2a_n = 2^{n+1}a_0 + 4 \\cdot (5^{n} -2^{n})$$ Hence, \\begin{align} a_{n+1} - 2a_n & = 2^{n+1}a_0 + 2 \\cdot (5^{n+1} -2^{n+1}) - \\left( 2^{n+1}a_0 + 4 \\cdot (5^{n} -2^{n})\\right)\\\\ & = 2 \\cdot (5^{n+1} -2^{n+1}) - 4 \\cdot (5^n - 2^n)\\\\ & = 2 \\cdot 5^{n+1} -2^{n+2} - 4 \\cdot 5^n + 4 \\cdot 2^n\\\\ & = 10 \\cdot 5^{n} - 4 \\cdot 5^n\\\\ & = 6 \\cdot 5^n \\end{align} - @WillJagy I verified it and I think what I have is correct unless I have made some error in the verification process. – user17762 Nov 23 '12 at 23:51 The answer", "$$y_3 = 2\\cdot 6\\cdot 8= 96$$, $$y_4 = 2\\cdot 96\\cdot 98 = 18616$$, and $$y_5 = 2\\cdot 18616\\cdot 18618 = 693185376$$. If we can show that $$n | y_n$$, then $$(n, x_n) = 1$$ since, if $$d|n$$ and $$d|x_n$$ and $$d \\ge 2$$ then $$d | n | y_n$$ so $$d | (x_n-1)$$ implies $$d | 1$$. Unfortunately, this doesn't work as $$y_5$$ shows. Next try: see if can find $$a_n, b_n$$ such that $$na_n-x_nb_n = 1$$. This will show that $$(n, x_n) = 1$$. If $$na_n-x_nb_n = 1$$ and $$(n+1)a_{n+1}-x_{n+1}b_{n+1} = 1$$ then $$\\begin{array}\\\\ 1 &=(n+1)a_{n+1}-(2x_n^2-1)b_{n+1}\\\\ &=na_{n+1}+a_{n+1}-2x_n^2b_{n+1}+b_{n+1}\\\\ &=na_{n}+n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1})+b_{n+1}\\\\ &=x_nb_n+1+n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1})+b_{n+1}\\\\ &=1+n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1}+b_{n})+b_{n+1}\\\\ \\text{so}\\\\ 0 &=n(a_{n+1}-a_n)+a_{n+1}-x_n(2x_nb_{n+1}+b_{n})+b_{n+1}\\\\ \\end{array}$$ and I don't know where to go from here. Hope someone else can make use of these. This is the pigeonhole principle. Among odd primes $$p,$$ there are solutions to $$2t^2 - 1 \\equiv 0 \\pmod p$$ if and only if $$p \\equiv \\pm 1 \\pmod 8.$$ For such a prime $$p,$$ there are some $$\\frac{p+1}{2}$$ distinct values $$\\pmod p$$ of $$2x^2 - 1.$$ Therefore, taking $$p \\geq 17,$$ if we take, say, $$p-5$$ steps of your sequence, there are just two possibilities: either (I) there is a repetition of some value $$\\pmod p$$ that is not one of $$-1,0,1.$$ In this case, the sequence repeats again and again, forever. Thus, this prime", "Mathematics OpenStudy (anonymous): So, let's start at the beginning. $$2(n + 1.5) = 4.5$$ You can distribute the 2: \\begin{align} 2n + 2 \\cdot (1.5) &= 4.5\\\\ 2n + 3 &= 4.5\\\\ \\end{align} Can you see how that does not lead to $$2n = 6$$, and therefore your final answer isn't right?", "3)}} - \\underbrace{(n - 1)}_{\\text{Previous factor of n}}$ $[(n + 4) - (n -1)]t_n = n(n + 1)(n + 2)(n + 3)[(n + 4) - (n - 1)] \\\\ 5t_n = n(n + 1)(n + 2)(n + 3)(n + 4) - (n - 1)n(n + 1)(n + 2)(n + 3)$ Now put $n = 1$, we will get $5t_1 = 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 - 0$ Now put $n = 2$, we will get $5t_2 = 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 - 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5$ Now put $n = 3$, we will get $5t_3 = 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 - 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6$ And so on, now put $n = (n - 1)$, we will get $5t_{n - 1} = (n - 1) \\cdot n \\cdot (n + 1) \\cdot (n + 2) \\cdot (n + 3) - (n - 2) \\cdot (n - 1) \\cdot n \\cdot (n + 1) \\cdot (n + 2)$ And now put $n = n$, we will get $5t_n = n \\cdot (n + 1) \\cdot (n + 2) \\cdot (n + 3) \\cdot (n + 4) - (n - 1) \\cdot n \\cdot (n +" ]

[ "have his picnic. What is the chance that Bob will have his picnic? Manager Joined: 20 Dec 2010 Posts: 140 Location: Stockholm, Sweden ### Show Tags Updated on: 23 Jan 2011, 05:37 If you har having trouble with probabilities I'd recommend drawing probability tree. In this case we just multiply .3 * .5 and adds .7 = 85% Originally posted by Mackieman on 22 Jan 2011, 09:29. Last edited by Mackieman on 23 Jan 2011, 05:37, edited 1 time in total. Math Expert Joined: 02 Sep 2009 Posts: 60646 There is a 30 % chance of rain and a 70% chance of shine. If [#permalink] ### Show Tags 22 Jan 2011, 09:43 ajit257 wrote: There is a 30 % chance of rain and a 70% chance of shine. If it rains there is a 50% chance that Bob will cancel his picnic but if the sun is shining he will definitely have his picnic. What is the chance that Bob will have his picnic? 50% chance that Bob will cancel his picnic in case of a rain means that there is 50% chance that Bob will NOT cancel his picnic in case of a rain. So, $$P=0.3*0.5+0.7*1=0.85$$. Or $$P(picnic)=1-P(no \\ picnic)=1-0.3*0.5=0.85$$. _________________ Manager Joined: 28 Aug 2010 Posts: 145 ### Show Tags 22 Jan 2011, 11:19 why are", "## anonymous one year ago Tara makes the following statement: \"There is a 70% chance of sunny weather tomorrow and a 60% chance I will go to the beach.\" What is the probability that it will be sunny tomorrow and Tara will go to the beach? 42% 22% 11% 13% 1. anonymous @amoodarya 2. anonymous @heretohelpalways 3. anonymous $\\frac{ 7 }{ 10 }*\\frac{ 6 }{ 10 }$ 4. anonymous 42/100 5. anonymous so 42%", "1. ## Probability problem 5 5.43 A rock concert promoter has scheduled an outdoor concert on July 4th. If it does not rain, the promoter will make $30,000. If it does rain, the promoter will lose$15,000 in guarantees made to the band and other expenses. The probability of rain on the 4th is A. a. What is the promoter's expected profit? Is the expected profit a reasonable decision criterion? Explain. b. How much should an insurance company charge to insure the promoter's full losses?", "Every morning Mae has toast for breakfast. Each day she either chooses honey or jam to spread on her toast, with equal chance of choosing either one. 1. Draw a tree diagram for three consecutive days of Mae’s breakfast choices. 2. What is the probability that on the fourth day Mae chooses honey for her toast? 3. What is the probability that Mae chooses jam for her toast three days in a row? ##### QUestion 2 Based on historical data, it rains$19$19 days out of$31$31 in July in Bangkok, Thailand. 1. What is the chance it will rain in Bangkok on the 7th July 2027? Give your answer as a fraction. 2. What is the probability, to the nearest percent, it will rain two days in a row? 3. What is the probability, to the nearest percent, that from a Monday to a Wednesday in July, it only rained on the Monday? 4. What is the probability, to the nearest percent, that from a Monday to a Wednesday in July, it rained on only one of the days? ##### Question 3 Lucy has a box of Favourites chocolates. In this box there are$30$30 chocolates,$5\\$5 of which are Picnics. Lucy takes and eats a chocolate without looking until she gets a Picnic. 1. What is the probability she only", "of rain is higher that night. Since the probability of 75% is greater than 50%, it is likely that it will rain. Question 8. b. Tara would like to go camping for the next 3 nights, but will not go if it is likely to rain on all 3 nights. Should she go? Use probability to justify your answer. Type below: ____________ The probability that it will rain all three nights is 0.2 × 0.2 × 0.75 = 0.03 = 3%. It is unlikely that it will rain all 3 nights since the probability is 3% so she should go. Question 9. Sinead tossed 4 coins at the same time. She did this 50 times, and 6 of those times, all 4 coins showed the same result (heads or tails). a. Find the experimental probability that all 4 coins show the same result when tossed. $$\\frac{□}{□}$$ Answer: $$\\frac{3}{25}$$ Explanation: The 4 coins were tossed 50 times and thus there are 50 possible outcomes. possible outcomes = 50 The result showed that all 4 coins have the same result on 6 of the 50 tosses. favorable outcomes = 6 The probability is the number of favorable outcomes divided by the number of possible outcomes: P(Same result) = favorable outcomes/possible outcomes = $$\\frac{6}{50}$$ = $$\\frac{3}{25}$$ Question 9. b. Can you", "<meta http-equiv=\"refresh\" content=\"1; url=/nojavascript/\"> Complement Rule for Probability % Progress Practice Complement Rule for Probability Progress % Complement Rule for Probability Have you ever played a basketball game at an amusement park? Well, Jeff is going to do just that. Jeff stopped going on rides to play a basketball game at the amusement park. In this game, Jeff had ten chances to make a basket. Kyle stopped by to support Jeff in his attempt. \"I have to make four out of ten,\" Jeff told Kyle. \"Well, I'm pretty good at basketball. I bet I have a 75% chance of getting the four in.\" If Jeff has a 75% chance of making the basket, what is the probability that he won't make the basket? This is an situation with complementary events. You will learn how to answer this question during Concept. Guidance What happens when we know the likelihood that something will happen? Well, we can determine or base our actions on that event happening. If there is a 10% chance of rain, what is the probability that it will be sunny? We can say that there is a 90% chance that it will be sunny. If someone only knew that there was a 10% chance that it would be rainy, and that if it wasn't rainy the only", "us now. If she took that flight yesterday, she is somewhere in town today. If she took that flight yesterday, we will see her tomorrow. A condition clause (protasis) in the present tense refers to a future event, a current event which may be true or untrue, or an event which could be verified in the future. The result can be in the past, present, or future: If it's raining here now, then it was raining on the West Coast this morning. If it's raining now, then your laundry is getting wet. If it's raining now, there will be mushrooms to be picked next week. If it rains this afternoon, then yesterday's weather forecast was wrong. If it rains this afternoon, your garden party is doomed. If it rains this afternoon, everybody will stay home. If I become President, I'll lower taxes. Certain modal auxiliary verbs (mainly will, may, might, and could) are not usually used in the condition clause (protasis) in English: *If it will rain this afternoon, … *If it may have rained yesterday, … There are exceptions, however, in which will is used exactly as in the first example, namely when the action in the if clause takes place after that in the main clause: (The weather forecast says it's going to rain.) Well, if", "probability from the previous 192 rolls is low. When meteorologists predict the weather, they are largely using empirical probabilities. When a weatherman says there is a 70% chance of rain tomorrow, that number isn’t arbitrary. Part of what it means is that certain weather conditions are expected tomorrow that could lead to rain. Historically, when similar conditions have been present, it has rained about 70% of the time. The likelihood of the outcome of a future event is based on the outcomes of similar past events. I hope that this video increased your understanding of empirical probability and how it can be used! See you next time! Practice Questions Question #1: Empirical Probability is also known as _____________. Conditional Probability Theoretical Probability Experimental Probability Subjective Probability Empirical probability is also referred to as experimental probability because it is based on an actual experiment. Experiments do not have fixed results, so the outcomes may vary. For example, the theoretical probability of a flipped coin landing on heads is $$\\frac{1}{2}$$. However, when an actual experiment is conducted, tails could be flipped five out of five times. Empirical probability, or experimental probability, is based on the data produced from an experiment, which does not always match the theoretical probability for the scenario. Question #2: True or False: Over time, as an", "we shall encounter will be nondogmatic, that is, neither zero nor one. An event with probability $$0$$ has no chance of occurring; an event of probability $$1$$ is certain to occur. It is hard to think of any examples of interest to statistics in which the probability is either $$0$$ or $$1$$. (Even the probability that the Sun will come up tomorrow is less than $$1$$.) The following example illustrates our attitude about probabilities. Suppose you wish to know what the weather will be like next Saturday because you are planning a picnic. You turn on your radio, and the weather person says, “There is a 10% chance of rain.” You decide to have the picnic outdoors and, lo and behold, it rains. You are furious with the weather person. But was she wrong? No, she did not say it would not rain, only that rain was unlikely. She would have been flatly wrong only if she said that the probability is $$0$$ and it subsequently rained. However, if you kept track of her weather predictions over a long period of time and found that it rained on $$50\\%$$ of the days that the weather person said the probability was $$0.10$$, you could say her probability assessments are wrong. So when is it accurate to say that the", "50% chance of rain tomorrow\" means that the probability of rain. On September 28, 2017, the NBA's Board of Governors approved a reform of the lottery system, with the changes taking effect at the 2019 lottery. Our lottery odds calculator automatically displays calculations in real-time. How is Overall Odds Calculated? We need to take a step back and first see how does a state lottery calculate the overall odds of winning. You can use it to calculate any lotto games such as Powerball, Mega Million. Media: You may freely use any information on this page, but you must credit www. Calculate Expected Gross Winnings of a Lottery Ticket. 3 Powerballs and bonus balls. Next Jackpot: 130,000,000. 000000072$$ It is difficult to appreciate how small this number is and how unlikely it is that an event with such a low probability will occur. On the other hand if casino games are what you are after. Lotto Odds Calculator This copyrighted, unique, mathematically correct, interactive calculator is the only resource in the world that identifies the all-important facts to lotto players. Overround of 50/50 odds. You can use it to calculate any lotto games such as Powerball, Mega Million. \\there is a 50% chance of rain tomorrow\" means that the probability of rain. In smaller lottery set ups, for example,", "In a bag with 5 white marbles and 5 black marbles, Sanjay pulls out a white marble. B: Without returning the marble to the bag, Sanjay pulls out a second marble. 3. A: Eddie chooses the color blue for his new bike. B: Eddie chooses lasagne from the dinner menu. 4. A: The probability that it will rain tomorrow. B: The probability that the Red Wings hockey team will win their game tomorrow. 5. A: From a deck of cards, the probability of one player drawing a heart from the deck. B: On the next player’s turn, the probability of drawing another heart. 6. A: The probability of a spinner landing on blue 6 times in a row. B: The probability of the spinner landing on blue on the next spin. 7. A: The probability that the Rockies will be in the playoffs. B: The probability that the Rockies will win the World Series. 8. A: The probability that tomorrow will be sunny. B: The probability that tomorrow will be a full moon. 9. A The probability that tomorrow will be sunny. B: The probability that tomorrow will be cloudy. 10. A: The probability that it will be a half-moon today. B: The probability that it will be a full moon tomorrow. Directions: Solve the problems. 1. A", "## Python, Random Numbers and Probability ### Introduction \"Every American should have above average income, and my Administration is going to see they get it.\" This saying is attributed to Bill Clinton on umpteen websites. Usually, there is no context given, so it is not clear, if he might have meant it as a \"joke\". Whatever his intentions might have been, we quoted him to show a \"real\" life example of statistics. Statistics and probability calculation is all around us in real-life situations. We have to cope with it whenever we have to make a decision from various options. Can we go for a hike in the afternoon or will it rain? The weather forecast tells us, that the probability of precipitation will be 30 %. So what now? Will we go for a hike? Another situation: Every week you play the lottery and dream of a far away island. What is the likelihood of winning the Jackpot so that you will never have to work again and live in \"paradise\"? Now imagine that you right on this island of your dreams. Most probably not because you won the jackpot, but rather because you booked your time as an all-inclusive holiday package. You are on holiday on a paradisal island far from home. Suddenly, you meet your neighbor,", "## Calculating the Odds of an Event Here you’ll calculate odds by using outcomes or probability. Have you ever thought about the likelihood of an event happening? Take a look at this dilemma: Telly and Carey were already hard at work when Ms. Kelley came into the bike shop on Thursday morning. It was three days before the big race and there was still a lot of work to be done. “I can’t believe it!” Ms. Kelley exclaimed as she came into the shop. “There is a 4 to 5 chance that it is going to rain on Saturday. I just heard the weather report,” Ms. Kelley said sighing. “Well, there is still a chance that it won’t,” Telly said trying to cheer her up. When we think about chances and odds, we can calculate the likelihood that an event will or won’t occur. In this case, there are odds that it will rain and odds that it won’t. We can also express those odds as a fraction or a percentage. Learn about odds in this reading, and you can work on the odds of the rainstorm at the end. ## Guidance You’ve seen that the probability of an event is defined as a ratio that compares the favorable out comes to the total outcomes. We can write", "in the shade (presuming it isn't raining). Rain Plans: There is a 40% chance of rain on Friday, and a 30% chance of rain on Saturday.  Should it be really raining (not drizzling), we'll meet in section 30 of the main arena, although that may move. Frequencies: There will be a talk-in frequency circulating via Twitter via DM.  It'll be somewhere on 70cm. Click for a larger and more expanded picture. Direct link to large map (right-click to save, sorry it isn't a PDF). -73- and see you there! Category: QSTs ## First PIC Project: A Soil Moisture Monitor I decided a few weeks ago to bring some life into my office with a plant.  Since I was getting a few things at the local 24-hour megalowmart, I decided what the heck.  One plant, bag of dirt, and pot (that was on sale), I had this... It's called a \"Little Herman Ivy\" I decided I wanted something to help me keep this thing alive.  I didn't want to go the route of a twitter reminder, as I'm in here nearly every day.  A nice, visual indicator should do well. Since I'm a traffic engineer, I thought something with a traffic signal would be appropriate.  A green for \"you're good\", a yellow", "push us all the way to Probabilism. Consider Dima and Esther. They both have minimal confidence---i.e. 0---that it won't rain tomorrow. But Dima has credence 0.01 that it will rain, while Esther has credence 0.99 that it will. If we permit only actions that maximise expected utility, then Dima and Esther are required to pay exactly the same prices for bets on rain---that is, Dima will be required to pay a price exactly when Esther is. After all, if £$S$ is the payoff when it rains, £0 is the payoff when it doesn't, and $x$ is a proposed price, then $0.01\\times (S- x) + 0 \\times (0-x) \\geq 0$ iff $0.99 \\times (S-x) + 0 \\times (0-x) \\geq 0$ iff $S \\geq x$. So, according to MSEU, Dima and Esther are rationally required to pay anything up to the stake of the bet for such a bet. But this is surely wrong. It is surely at least permissible for Dima to refuse to pay a price that Esther accepts. It is surely permissible for Esther to pay £99 for a bet on rain that pays £100 if it rains and £0 if it doesn't, while Dima refuses to pay anything more than £1 for such a bet, in line with Ramsey's Thesis. Suppose Dima were offered such a", "### Rain or Shine Predict future weather using the probability that tomorrow is wet given today is wet and the probability that tomorrow is wet given that today is dry. ### Squash If the score is 8-8 do I have more chance of winning if the winner is the first to reach 9 points or the first to reach 10 points? ### Snooker A player has probability 0.4 of winning a single game. What is his probability of winning a 'best of 15 games' tournament? # Dicey Dice ##### Stage: 5 Challenge Level: This problem follows naturally from X-Dice, although it may be attempted independently. A company wishes to produce a set of three different 6-sided dice coloured Apple Green, Bright Pink and Cool Grey, called $A, B$ and $C$ respectively. They are to be made with the following properties: 1. The faces are to be numbered using only whole numbers 1 to 6. 2. Some of the numbers 1 to 6 can be left out or repeated as desired on each dice. 3. Apple Green is expected to beat Bright Pink on a single roll 4. Bright Pink is expected to beat Cool Grey on a single roll. 5. Cool Grey is expected to beat Apple Green on a single roll. Invent a set of such dice.", "it doesn't rain there is a $$90 \\%$$ chance that they play. What is the probability that students play football on February $$15^{\\text{th}}$$ ? 2. Cathy occasionally likes to drink soda. She mainly drinks it if she eats fried chicken on Tuesdays (\"special deal day\"). The probability that she eats fried chicken on any Tuesday is $$0.8$$. If she eats fried chicken the probability of her drinking soda is $$0.4$$, otherwise it is $$0.05$$. Calculate the probability that Cathy will drink soda next Tuesday. 3. In a grade $$12$$ class, the probability that a students studies Higer Level mathematics (HL) is $$0.3$$, altenratively students will study Standard Level Mathematics (SL). $$80 \\%$$ of students studying HL study Two Sciences, whereas only $$40 \\%$$ of the SL students study Two Sciences. A student is selected at random from the class. What is the probability that this student studies Two Sciences? 4. On any given day there is a $$80 \\%$$ chance that John has to take an object away from his two year old daughter, Charlotte. If he takes an object away from her there is a $$90\\%$$ chance she cries. If he doesn't take an object away from her there is still a $$30\\%$$ chance that she cries that day. What is the probability that Charlotte cries tomorrow?", "korporasidn 2021-11-12 1.The probability of Christine getting a new job or going on vacation (or both) is 0.85. If the probability of Christine going on vacation is 0.10 and the probability of Christine getting a new job is 0.75, are the events Christine gets a new job and Christine goes on vacation mutually exclusive? 2. The probability that Isabelle will go trick-or-treating is 0.4. The probability that she will go trick-or-treating or watch a scary movie is 0.65. The probability of Isabelle going trick-or-treating and watching a scary movie is 0.20. A. What is the probability that Isabelle will not watch a scary movie? B. What is the probability that Isabelle will watch a scary movie but does not go trick-or-treating? Nicole Keller Step 1 1.Given that The probability of Christine getting a new job $=0.75$ the probability of Christine going on vacation $=0.10$ The probability that getting a new job or going on vacation $=0.85$ Are the events mutually exclusive 2. The probability that Isabelle will go trick-or-treating $=0.4$ The probability that she will trick or treating and watching a scary movie $=0.20$ The probability that she will trick and treating and watching a scary movie $=0.65$ A. Find The probability that she will not watch a scary movie B. Find the probability that she will watch", "Consider Dima and Esther. They both have minimal confidence---i.e. 0---that it won't rain tomorrow. But Dima has credence 0.01 that it will rain, while Esther has credence 0.99 that it will. If we permit only actions that maximise expected utility, then Dima and Esther are required to pay exactly the same prices for bets on rain---that is, Dima will be required to pay a price exactly when Esther is. After all, if £$S$ is the payoff when it rains, £0 is the payoff when it doesn't, and $x$ is a proposed price, then $0.01\\times (S- x) + 0 \\times (0-x) \\geq 0$ iff $0.99 \\times (S-x) + 0 \\times (0-x) \\geq 0$ iff $S \\geq x$. So, according to MSEU, Dima and Esther are rationally required to pay anything up to the stake of the bet for such a bet. But this is surely wrong. It is surely at least permissible for Dima to refuse to pay a price that Esther accepts. It is surely permissible for Esther to pay £99 for a bet on rain that pays £100 if it rains and £0 if it doesn't, while Dima refuses to pay anything more than £1 for such a bet, in line with Ramsey's Thesis. Suppose Dima were offered such a bet for the price of £99, and", "Suppose she is only 50% confident that she will beat you (her personal probability of winning is p=0.5). What is her expected return now? (Report your answer without the$ symbol.) Q6. Chess: • Now assuming your friend will only agree to fair bets (expected return of $0), find her personal probability that she will win. Report your answer as a simplified fraction. Hint: Use the expected return of her proposed bet. Preview will appear here.. Q7. For Questions 7-8, consider the following “Dutch book” scenario: Suppose your friend offers a pair of bets: (i) if it rains or is overcast tomorrow, you pay him$4, otherwise he pays you $6; (ii) if it is sunny you pay him$5, otherwise he pays you $5. • Suppose rain, overcast, and sunny are the only events in consideration. If you make both bets simultaneously, this is called a “Dutch book,” as you are guaranteed to win money. How much do you win regardless of the outcome? (Report your answer without the$ symbol.) Q8. Dutch book: Apparently your friend doesn’t understand the laws of probability. Let’s examine the bets he offered. 1. For bet (i) to be fair, his probability that it rains or is overcast must be .6 (you can verify this by calculating his expected return and setting it equal to" ]

[ "UK Secondary (7-11) Law of Total Probability ## Interactive practice questions At a particular school, there are $270$270 students in year 9 and $180$180 students in year 10. $20%$20% of the year 9 students and $45%$45% of the year 10 students like maths. a What is the probability that a randomly chosen student is in year 9 and does not like maths? b What is the overall proportion of students who do not like maths? Easy Approx 6 minutes The probability that it will rain in Penrith on a Saturday is $0.27$0.27. If it does rain on a Saturday, then the probability that it will rain on the Sunday is $0.78$0.78. But if it doesn't rain on a Saturday, the probability that it will rain on the Sunday is $0.4$0.4. In the month of June, it rains on average $17$17 days of the month in Perth. It if rains, the chance that Carl catches the bus to work is $\\frac{3}{10}$310. If it doesn’t rain, the chance that Carl catches the bus to work is $\\frac{9}{10}$910. The probability that Jo starts a conversation is $70%$70%. If Jo starts talking, the chance that Em responds is $80%$80%. If Jo doesn’t start a conversation, the chance that Em does is $10%$10%.", "## Algebra 2 (1st Edition) Since the odds of it not raining on Friday are $=1-.05=.95$, this is the highest probability.", "Dora´s chances of traveling to Israel next holidays, if she knows there is a 70% probability for the raining season in Jerusalem to begin? (A) 15% (B) 29% (C) 35% (D) 42% (E) 46% _________________ Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our high-level \"quant\" preparation starts here: https://gmath.net Intern Joined: 25 Dec 2018 Posts: 2 If the raining season in Jerusalem begins, Dora estimates a 20% probab [#permalink] ### Show Tags 08 Mar 2019, 09:49 2 There are two possibilities of Dora travelling 1. If it rains and she travels- p(rains) x p(travel)= 0.7 x 0.2 = 0.14 2. If it does not rain and she travels - p (doesn't rain) x p (travels when it doesn't rain) Now, the question says that probability of travelling when it does not rain is higher by 250% means that, 250% of 20% = 0.5 therefore, probability of travelling when it doesn't rain is 0.2+0.5= 0.7 So, p (doesn't rain) x p (travels when it doesn't rain)= 0.3 (given in question) x 0.7 = 0.21 P (dora travels) = P (rains and travels) OR P (doesn't rain and travels) = 0.14 + 0.21 = 0.35 Hope this is correct. GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 935 Re: If the raining season in Jerusalem", "the probability of at least one of the events occurring is $0.44$. The probability of no events occurring is then one minus that, i.e. $1-0.44=0.56$. Alternatively, if $A$ and $B$ are independent, then $A^c$ and $B^c$ are also independent. We have then $Pr(A^c\\cap B^c)=Pr(A^c)Pr(B^c)=(1-0.3)(1-0.2)=0.7\\cdot 0.8 = 0.56$ Your approach double-counts some possibilities. The $0.3$ covers all possibility of rain - so, it includes both \"rain and quiz\" and \"rain and no quiz\". The $0.2$ covers all possibility of a quiz, so it includes both \"rain and quiz\" and \"quiz and no rain\". So $0.3 + 0.2$ covers \"rain and quiz\", \"rain and no quiz\", \"rain and quiz\" again, and \"quiz and no rain\". To cut out one of the countings of \"rain and quiz\", we need to subtract the probability of that event, which is $0.3 \\cdot 0.2 = 0.06$. So the probability of \"rain or quiz or both\" is $0.2 + 0.3 - 0.06 = 0.44$. The probability of neither is then $1 - 0.44 = 0.56$.", "korporasidn 2021-11-12 1.The probability of Christine getting a new job or going on vacation (or both) is 0.85. If the probability of Christine going on vacation is 0.10 and the probability of Christine getting a new job is 0.75, are the events Christine gets a new job and Christine goes on vacation mutually exclusive? 2. The probability that Isabelle will go trick-or-treating is 0.4. The probability that she will go trick-or-treating or watch a scary movie is 0.65. The probability of Isabelle going trick-or-treating and watching a scary movie is 0.20. A. What is the probability that Isabelle will not watch a scary movie? B. What is the probability that Isabelle will watch a scary movie but does not go trick-or-treating? Nicole Keller Step 1 1.Given that The probability of Christine getting a new job $=0.75$ the probability of Christine going on vacation $=0.10$ The probability that getting a new job or going on vacation $=0.85$ Are the events mutually exclusive 2. The probability that Isabelle will go trick-or-treating $=0.4$ The probability that she will trick or treating and watching a scary movie $=0.20$ The probability that she will trick and treating and watching a scary movie $=0.65$ A. Find The probability that she will not watch a scary movie B. Find the probability that she will watch", "Every morning Mae has toast for breakfast. Each day she either chooses honey or jam to spread on her toast, with equal chance of choosing either one. 1. Draw a tree diagram for three consecutive days of Mae’s breakfast choices. 2. What is the probability that on the fourth day Mae chooses honey for her toast? 3. What is the probability that Mae chooses jam for her toast three days in a row? ##### QUestion 2 Based on historical data, it rains$19$19 days out of$31$31 in July in Bangkok, Thailand. 1. What is the chance it will rain in Bangkok on the 7th July 2027? Give your answer as a fraction. 2. What is the probability, to the nearest percent, it will rain two days in a row? 3. What is the probability, to the nearest percent, that from a Monday to a Wednesday in July, it only rained on the Monday? 4. What is the probability, to the nearest percent, that from a Monday to a Wednesday in July, it rained on only one of the days? ##### Question 3 Lucy has a box of Favourites chocolates. In this box there are$30$30 chocolates,$5\\$5 of which are Picnics. Lucy takes and eats a chocolate without looking until she gets a Picnic. 1. What is the probability she only", "## Stats: Data and Models (3rd Edition) $25$ % chance of raining means that it would rain $1$ out of $4$ times and that it is $3$ times more likely not to rain than to rain.", "• 2008:$3 million (lump-sum payoff). Probability: 1/909,000. • 2010: \\$10 million (lump-sum payoff). Probability: 1/1,200,000. The probability of two or more independent events all happening is the product of their individual probabilities. Compare the probability of Joan winning all four lotteries with the probability of picking, at random, a grain of sand painted red that is hidden somewhere within the Sahara desert. < Previous Section Home Next Section >", "correctly describe the fact that $$P(\\mbox{umbrella})=1$$. What two numbers should we put in the empty cells? Again, let’s not worry about the maths and instead use our intuitions. We worked out that the joint probability of “rain and umbrella” was 7.05%, and the joint probability of “dry and umbrella” was 3.83%. But notice that both of these possibilities are consistent with the fact that he actually is carrying an umbrella. So what we expect to see in our final table are some numbers that preserve the fact that “rain and umbrella” is more plausible than “dry and umbrella” while still ensuring that the numbers in the table add up. Something like this, perhaps? Umbrella No umbrella Rainy day 0.648 0 Dry day 0.352 0 Total 1.000 0 This table infers that, after being told that Attila is carrying an umbrella, you believe that there is a 64.8% chance that today will be a rainy day and a 35.2% chance that it will not. That is the answer to our problem! The posterior probability of rain $$P(h|d)$$ given that Attila is carrying an umbrella is 64.8%. How did we calculate these numbers? You can probably guess: take the 0.0705 probability of “rain and umbrella” and divide it by the 0.1088 chance of “umbrella”. This produces a table that satisfies", "push us all the way to Probabilism. Consider Dima and Esther. They both have minimal confidence---i.e. 0---that it won't rain tomorrow. But Dima has credence 0.01 that it will rain, while Esther has credence 0.99 that it will. If we permit only actions that maximise expected utility, then Dima and Esther are required to pay exactly the same prices for bets on rain---that is, Dima will be required to pay a price exactly when Esther is. After all, if £$S$ is the payoff when it rains, £0 is the payoff when it doesn't, and $x$ is a proposed price, then $0.01\\times (S- x) + 0 \\times (0-x) \\geq 0$ iff $0.99 \\times (S-x) + 0 \\times (0-x) \\geq 0$ iff $S \\geq x$. So, according to MSEU, Dima and Esther are rationally required to pay anything up to the stake of the bet for such a bet. But this is surely wrong. It is surely at least permissible for Dima to refuse to pay a price that Esther accepts. It is surely permissible for Esther to pay £99 for a bet on rain that pays £100 if it rains and £0 if it doesn't, while Dima refuses to pay anything more than £1 for such a bet, in line with Ramsey's Thesis. Suppose Dima were offered such a", "have his picnic. What is the chance that Bob will have his picnic? Manager Joined: 20 Dec 2010 Posts: 140 Location: Stockholm, Sweden ### Show Tags Updated on: 23 Jan 2011, 05:37 If you har having trouble with probabilities I'd recommend drawing probability tree. In this case we just multiply .3 * .5 and adds .7 = 85% Originally posted by Mackieman on 22 Jan 2011, 09:29. Last edited by Mackieman on 23 Jan 2011, 05:37, edited 1 time in total. Math Expert Joined: 02 Sep 2009 Posts: 60646 There is a 30 % chance of rain and a 70% chance of shine. If [#permalink] ### Show Tags 22 Jan 2011, 09:43 ajit257 wrote: There is a 30 % chance of rain and a 70% chance of shine. If it rains there is a 50% chance that Bob will cancel his picnic but if the sun is shining he will definitely have his picnic. What is the chance that Bob will have his picnic? 50% chance that Bob will cancel his picnic in case of a rain means that there is 50% chance that Bob will NOT cancel his picnic in case of a rain. So, $$P=0.3*0.5+0.7*1=0.85$$. Or $$P(picnic)=1-P(no \\ picnic)=1-0.3*0.5=0.85$$. _________________ Manager Joined: 28 Aug 2010 Posts: 145 ### Show Tags 22 Jan 2011, 11:19 why are", "= .01$ $P(T|C) = .8$ $P(T|\\neg C) = .096$ Now the problem statement: A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer? We have to infer $P(C|T)$. Note how this is a reversal of the conditional statements we encounted in the information given about the test. Calculation Now comes the calculation. A good place to start when thinking about conditional probability is the ratio formula for the probability of a condititional event: $P(B|A) = \\frac{P(A \\& B) }{P(A)}$ Take an interpretatation of “If it is raining, then I have an umbrella” as the conditional event expression: I have an umbrella | it is raining The probability of this is the probability that I have an umbrella and it is raining, divided by the probability that it is raining. This can easily be rewritten to $P(A \\& B) = P(B|A) P(A)$ So if you know the probability of rain, and the probability that I have an umbrella when it rains, then you can multiply them to infer the probability that it is raining and I have an umbrella. One step towards Bayes’ rule begins with: 1. $P(B|A) = P(A \\& B) / P(A)$ 2. $P(A|B) = P(A \\& B) / P(B)$ [$A \\& B", "the coins were connected in some sneaky way, maybe each time one landed heads up, the other would land tails up. Then the answer to our question would be zero. Of course this seems silly. But it’s good to be very clear about this issue… because sometimes one event does affect another! For example, suppose there’s a 5% probability of rain each day in the winter in Riverside. What’s the probability that it rains two days in a row? Remember that 5% is 0.05. So, you might guess the answer is $0.05 \\times 0.05 = 0.0025$ But this is wrong, because if it rains one day, that increases the probability that it will rain the next day. In other words, these events aren’t independent. But if two events are independent, there’s an easy way to figure out the probability that they both happen: just multiply their probabilities! For example, if the chance that it will rain today in Riverside is 5% and the chance that it will rain tomorrow in Singapore is 60%, the chance that both these things will happen is $0.05 \\times 0.6 = 0.03$ or 3%, if these events are independent. I could try to persuade that this is a good rule, and maybe I will… but for now let’s just state it in a", "of rain is higher that night. Since the probability of 75% is greater than 50%, it is likely that it will rain. Question 8. b. Tara would like to go camping for the next 3 nights, but will not go if it is likely to rain on all 3 nights. Should she go? Use probability to justify your answer. Type below: ____________ The probability that it will rain all three nights is 0.2 × 0.2 × 0.75 = 0.03 = 3%. It is unlikely that it will rain all 3 nights since the probability is 3% so she should go. Question 9. Sinead tossed 4 coins at the same time. She did this 50 times, and 6 of those times, all 4 coins showed the same result (heads or tails). a. Find the experimental probability that all 4 coins show the same result when tossed. $$\\frac{□}{□}$$ Answer: $$\\frac{3}{25}$$ Explanation: The 4 coins were tossed 50 times and thus there are 50 possible outcomes. possible outcomes = 50 The result showed that all 4 coins have the same result on 6 of the 50 tosses. favorable outcomes = 6 The probability is the number of favorable outcomes divided by the number of possible outcomes: P(Same result) = favorable outcomes/possible outcomes = $$\\frac{6}{50}$$ = $$\\frac{3}{25}$$ Question 9. b. Can you", "it doesn't rain there is a $$90 \\%$$ chance that they play. What is the probability that students play football on February $$15^{\\text{th}}$$ ? 2. Cathy occasionally likes to drink soda. She mainly drinks it if she eats fried chicken on Tuesdays (\"special deal day\"). The probability that she eats fried chicken on any Tuesday is $$0.8$$. If she eats fried chicken the probability of her drinking soda is $$0.4$$, otherwise it is $$0.05$$. Calculate the probability that Cathy will drink soda next Tuesday. 3. In a grade $$12$$ class, the probability that a students studies Higer Level mathematics (HL) is $$0.3$$, altenratively students will study Standard Level Mathematics (SL). $$80 \\%$$ of students studying HL study Two Sciences, whereas only $$40 \\%$$ of the SL students study Two Sciences. A student is selected at random from the class. What is the probability that this student studies Two Sciences? 4. On any given day there is a $$80 \\%$$ chance that John has to take an object away from his two year old daughter, Charlotte. If he takes an object away from her there is a $$90\\%$$ chance she cries. If he doesn't take an object away from her there is still a $$30\\%$$ chance that she cries that day. What is the probability that Charlotte cries tomorrow?", "My Math Forum \"And\" and \"Or\" Probability Probability and Statistics Basic Probability and Statistics Math Forum December 3rd, 2016, 09:28 AM #1 Newbie Joined: Dec 2016 From: United Kingdom Posts: 2 Thanks: 0 \"And\" and \"Or\" Probability Sheila either walks or cycles to school. The probability that she walks to school is 0.65. If she walks, the probability that she will be late is 0.4. If she cycles, the probability that she will be late is 0.1. Work out the probability that on any one day Sheila will not be late for school. December 3rd, 2016, 10:13 AM #2 Senior Member Joined: Sep 2015 From: USA Posts: 1,980 Thanks: 1027 For any two events $A,~B$ $P[A \\cap B] = P[A | B] P[B]$ if $C$ and $D$ are mutually exclusive events (such as walking and cycling) then $P[C \\cup D] = P[C] + P[D]$ For two mutually exclusive events, $E,~F$ whose union is the entire sample space (again like walking and cycling). $P[F] = 1 - P[E]$ from these you can work out this problem. , , , , , Dan either walks or cycles to school Click on a term to search for related topics. Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post johnmath Math Books 6 January 27th, 2013", "Consider Dima and Esther. They both have minimal confidence---i.e. 0---that it won't rain tomorrow. But Dima has credence 0.01 that it will rain, while Esther has credence 0.99 that it will. If we permit only actions that maximise expected utility, then Dima and Esther are required to pay exactly the same prices for bets on rain---that is, Dima will be required to pay a price exactly when Esther is. After all, if £$S$ is the payoff when it rains, £0 is the payoff when it doesn't, and $x$ is a proposed price, then $0.01\\times (S- x) + 0 \\times (0-x) \\geq 0$ iff $0.99 \\times (S-x) + 0 \\times (0-x) \\geq 0$ iff $S \\geq x$. So, according to MSEU, Dima and Esther are rationally required to pay anything up to the stake of the bet for such a bet. But this is surely wrong. It is surely at least permissible for Dima to refuse to pay a price that Esther accepts. It is surely permissible for Esther to pay £99 for a bet on rain that pays £100 if it rains and £0 if it doesn't, while Dima refuses to pay anything more than £1 for such a bet, in line with Ramsey's Thesis. Suppose Dima were offered such a bet for the price of £99, and", "1. ## probability being late The probability that James is late for school on normal days is 0.01 .However , the probability that James is late for school on a rainy day is 0.02 . The probability that it would rain on Monday is 0.005 .What would be the probability that James would be late on Monday ? 2. Originally Posted by thereddevils The probability that James is late for school on normal days is 0.01 .However , the probability that James is late for school on a rainy day is 0.02 . The probability that it would rain on Monday is 0.005 .What would be the probability that James would be late on Monday ? $R^c$ means it does not rain. $P(L)=P(L\\cap R)+P(L\\cap R^c)=P(L|R)P(R) + P(L|R^c)P(R^c)$.", "once is 100% minus the odds that it never happened. The odds of something never happening are the chance of it not happening each hour, multiplied by the other chances of it not happening. If there’s a 10% chance of rain at 2PM, there’s a 90% chance of no rain. That’s pretty clear, right? But if there’s a 10% chance of rain at 2PM and a 15% chance of rain at 3PM, what are the odds of no rain between 2 and 3PM? $$(1-.1)\\cdot(1-.15)$$ That works out to roughly 77% chance of it not raining. That means there’s a 23% chance it raining between those two hours, even though the highest chance of rain it 15% at 3PM! Thus, even low percentages quickly add up. Assuming that every hour has a 10% chance of rain, the chance of it never raining is $(1-.1)^{24}$ (90% multiplied by itself 24 times), or roughly .08, or 8%. Thus, the chance of it raining at least once tomorrow would be 92%. Even though each hour only has a 10% chance of rain, I still need to send my daughter to school with an umbrella.", "want $\\Pr(R|L)$. There are some useful formulas that come into play. The most basic one comes from the definition of conditional probability. We have $$\\Pr(R|L)\\Pr(L)=\\Pr(R\\cap L).$$ If we can calculate $\\Pr(L)$ and $\\Pr(R\\cap L)$ we will be finished. We first compute $\\Pr(R\\cap L)$. The probability it rains is $\\frac{1}{4}$. Given that it rains, the probability she is late is $\\frac{2}{3}$. So the probability it rains and she is late is $\\frac{1}{4}\\cdot\\frac{2}{3}$. Next we compute $\\Pr(L)$. She can be late in two ways: (i) It rains and she is late or (ii) It doesn't rain and she is late. We already computed the probability of (i). In the same way, we find that the probability it doesn't rain and she is late is $\\frac{3}{4}\\cdot\\frac{1}{5}$. To find $\\Pr(L)$, add together the probabilities of (i) and of (ii). Remark: Here is an imprecise but useful way to view the calculation. Imagine we track what happens over $300$ days. It will rain about $75$ days. On two-thirds of these days she will be late, so she will be late on about $50$ days because of rain. It will not rain about $225$ days, and she will be late on about one-fifth of these days, so $45$ days. So she is late a total of $50+45$ days. Let's confine attention (cut down the" ]

[ "have his picnic. What is the chance that Bob will have his picnic? Manager Joined: 20 Dec 2010 Posts: 140 Location: Stockholm, Sweden ### Show Tags Updated on: 23 Jan 2011, 05:37 If you har having trouble with probabilities I'd recommend drawing probability tree. In this case we just multiply .3 * .5 and adds .7 = 85% Originally posted by Mackieman on 22 Jan 2011, 09:29. Last edited by Mackieman on 23 Jan 2011, 05:37, edited 1 time in total. Math Expert Joined: 02 Sep 2009 Posts: 60646 There is a 30 % chance of rain and a 70% chance of shine. If [#permalink] ### Show Tags 22 Jan 2011, 09:43 ajit257 wrote: There is a 30 % chance of rain and a 70% chance of shine. If it rains there is a 50% chance that Bob will cancel his picnic but if the sun is shining he will definitely have his picnic. What is the chance that Bob will have his picnic? 50% chance that Bob will cancel his picnic in case of a rain means that there is 50% chance that Bob will NOT cancel his picnic in case of a rain. So, $$P=0.3*0.5+0.7*1=0.85$$. Or $$P(picnic)=1-P(no \\ picnic)=1-0.3*0.5=0.85$$. _________________ Manager Joined: 28 Aug 2010 Posts: 145 ### Show Tags 22 Jan 2011, 11:19 why are", "Every morning Mae has toast for breakfast. Each day she either chooses honey or jam to spread on her toast, with equal chance of choosing either one. 1. Draw a tree diagram for three consecutive days of Mae’s breakfast choices. 2. What is the probability that on the fourth day Mae chooses honey for her toast? 3. What is the probability that Mae chooses jam for her toast three days in a row? ##### QUestion 2 Based on historical data, it rains$19$19 days out of$31$31 in July in Bangkok, Thailand. 1. What is the chance it will rain in Bangkok on the 7th July 2027? Give your answer as a fraction. 2. What is the probability, to the nearest percent, it will rain two days in a row? 3. What is the probability, to the nearest percent, that from a Monday to a Wednesday in July, it only rained on the Monday? 4. What is the probability, to the nearest percent, that from a Monday to a Wednesday in July, it rained on only one of the days? ##### Question 3 Lucy has a box of Favourites chocolates. In this box there are$30$30 chocolates,$5\\$5 of which are Picnics. Lucy takes and eats a chocolate without looking until she gets a Picnic. 1. What is the probability she only", "of rain is higher that night. Since the probability of 75% is greater than 50%, it is likely that it will rain. Question 8. b. Tara would like to go camping for the next 3 nights, but will not go if it is likely to rain on all 3 nights. Should she go? Use probability to justify your answer. Type below: ____________ The probability that it will rain all three nights is 0.2 × 0.2 × 0.75 = 0.03 = 3%. It is unlikely that it will rain all 3 nights since the probability is 3% so she should go. Question 9. Sinead tossed 4 coins at the same time. She did this 50 times, and 6 of those times, all 4 coins showed the same result (heads or tails). a. Find the experimental probability that all 4 coins show the same result when tossed. $$\\frac{□}{□}$$ Answer: $$\\frac{3}{25}$$ Explanation: The 4 coins were tossed 50 times and thus there are 50 possible outcomes. possible outcomes = 50 The result showed that all 4 coins have the same result on 6 of the 50 tosses. favorable outcomes = 6 The probability is the number of favorable outcomes divided by the number of possible outcomes: P(Same result) = favorable outcomes/possible outcomes = $$\\frac{6}{50}$$ = $$\\frac{3}{25}$$ Question 9. b. Can you", "1. ## Probability problem 5 5.43 A rock concert promoter has scheduled an outdoor concert on July 4th. If it does not rain, the promoter will make $30,000. If it does rain, the promoter will lose$15,000 in guarantees made to the band and other expenses. The probability of rain on the 4th is A. a. What is the promoter's expected profit? Is the expected profit a reasonable decision criterion? Explain. b. How much should an insurance company charge to insure the promoter's full losses?", "korporasidn 2021-11-12 1.The probability of Christine getting a new job or going on vacation (or both) is 0.85. If the probability of Christine going on vacation is 0.10 and the probability of Christine getting a new job is 0.75, are the events Christine gets a new job and Christine goes on vacation mutually exclusive? 2. The probability that Isabelle will go trick-or-treating is 0.4. The probability that she will go trick-or-treating or watch a scary movie is 0.65. The probability of Isabelle going trick-or-treating and watching a scary movie is 0.20. A. What is the probability that Isabelle will not watch a scary movie? B. What is the probability that Isabelle will watch a scary movie but does not go trick-or-treating? Nicole Keller Step 1 1.Given that The probability of Christine getting a new job $=0.75$ the probability of Christine going on vacation $=0.10$ The probability that getting a new job or going on vacation $=0.85$ Are the events mutually exclusive 2. The probability that Isabelle will go trick-or-treating $=0.4$ The probability that she will trick or treating and watching a scary movie $=0.20$ The probability that she will trick and treating and watching a scary movie $=0.65$ A. Find The probability that she will not watch a scary movie B. Find the probability that she will watch", "push us all the way to Probabilism. Consider Dima and Esther. They both have minimal confidence---i.e. 0---that it won't rain tomorrow. But Dima has credence 0.01 that it will rain, while Esther has credence 0.99 that it will. If we permit only actions that maximise expected utility, then Dima and Esther are required to pay exactly the same prices for bets on rain---that is, Dima will be required to pay a price exactly when Esther is. After all, if £$S$ is the payoff when it rains, £0 is the payoff when it doesn't, and $x$ is a proposed price, then $0.01\\times (S- x) + 0 \\times (0-x) \\geq 0$ iff $0.99 \\times (S-x) + 0 \\times (0-x) \\geq 0$ iff $S \\geq x$. So, according to MSEU, Dima and Esther are rationally required to pay anything up to the stake of the bet for such a bet. But this is surely wrong. It is surely at least permissible for Dima to refuse to pay a price that Esther accepts. It is surely permissible for Esther to pay £99 for a bet on rain that pays £100 if it rains and £0 if it doesn't, while Dima refuses to pay anything more than £1 for such a bet, in line with Ramsey's Thesis. Suppose Dima were offered such a", "# Does this break down adding independent probabilities? Suppose there is a jar with a red, blue, green, and yellow ball. The probability of picking a red ball is 25%. The probability of picking a red ball, putting it back, AND then picking a green ball is 12.5% (25% * 25%). The probability of picking either a red OR a green ball on the first try is 50% (25% + 25%). These are independent events. Now lets take another set of independent events. 1. The weather man said there is an 80% chance of rain. This means it should rain 80% of the time I assume. If you looked back at the past 100 days with an 80% rain chance it should rain on ~80 of them. 2. A girl hooked up with 5 of our 10 mutual friends. I have a 50% chance of hooking up with her too. 3. Rolling an even number on a dice carries a 50% probability. If it rains OR I hook up with the girl OR I roll an even number on the dice then I win. If any one of these things happen I win, just like if I pick a red or a green ball, it doesn't matter if its red or green. With that being said the probability", "## anonymous one year ago Tara makes the following statement: \"There is a 70% chance of sunny weather tomorrow and a 60% chance I will go to the beach.\" What is the probability that it will be sunny tomorrow and Tara will go to the beach? 42% 22% 11% 13% 1. anonymous @amoodarya 2. anonymous @heretohelpalways 3. anonymous $\\frac{ 7 }{ 10 }*\\frac{ 6 }{ 10 }$ 4. anonymous 42/100 5. anonymous so 42%", "50% chance of rain tomorrow\" means that the probability of rain. On September 28, 2017, the NBA's Board of Governors approved a reform of the lottery system, with the changes taking effect at the 2019 lottery. Our lottery odds calculator automatically displays calculations in real-time. How is Overall Odds Calculated? We need to take a step back and first see how does a state lottery calculate the overall odds of winning. You can use it to calculate any lotto games such as Powerball, Mega Million. Media: You may freely use any information on this page, but you must credit www. Calculate Expected Gross Winnings of a Lottery Ticket. 3 Powerballs and bonus balls. Next Jackpot: 130,000,000. 000000072$$ It is difficult to appreciate how small this number is and how unlikely it is that an event with such a low probability will occur. On the other hand if casino games are what you are after. Lotto Odds Calculator This copyrighted, unique, mathematically correct, interactive calculator is the only resource in the world that identifies the all-important facts to lotto players. Overround of 50/50 odds. You can use it to calculate any lotto games such as Powerball, Mega Million. \\there is a 50% chance of rain tomorrow\" means that the probability of rain. In smaller lottery set ups, for example,", "Consider Dima and Esther. They both have minimal confidence---i.e. 0---that it won't rain tomorrow. But Dima has credence 0.01 that it will rain, while Esther has credence 0.99 that it will. If we permit only actions that maximise expected utility, then Dima and Esther are required to pay exactly the same prices for bets on rain---that is, Dima will be required to pay a price exactly when Esther is. After all, if £$S$ is the payoff when it rains, £0 is the payoff when it doesn't, and $x$ is a proposed price, then $0.01\\times (S- x) + 0 \\times (0-x) \\geq 0$ iff $0.99 \\times (S-x) + 0 \\times (0-x) \\geq 0$ iff $S \\geq x$. So, according to MSEU, Dima and Esther are rationally required to pay anything up to the stake of the bet for such a bet. But this is surely wrong. It is surely at least permissible for Dima to refuse to pay a price that Esther accepts. It is surely permissible for Esther to pay £99 for a bet on rain that pays £100 if it rains and £0 if it doesn't, while Dima refuses to pay anything more than £1 for such a bet, in line with Ramsey's Thesis. Suppose Dima were offered such a bet for the price of £99, and", "Joined: 02 Sep 2009 Posts: 60646 ### Show Tags 03 Feb 2011, 14:03 1 VeritasPrepKarishma wrote: ajit257 wrote: why are we adding .7..how do I figure out whether it is OR/AND solution to see if it has to be added or multiplied To figure out whether you need OR/AND, focus on how you speak about it. You say, \"I need a shirt AND a trouser\" - AND so multiplication \"I need a shirt OR a T-shirt\" - OR so addition Here you say: It will Rain OR Shine Rains - probability of party is 15% Shines - probability of party is 70% Since it will Rain OR Shine, total probability = 15% + 70% = 85% You can directly translate the question into probability: Bob will have his picnic if: there is no rain (0.7) AND he goes (1) OR there is a rain (0.3) AND he still goes (0.5). Now, put * for AND and + for OR: 0.7*1+0.3*0.5. _________________ Retired Moderator Joined: 20 Dec 2010 Posts: 1533 ### Show Tags 04 Feb 2011, 05:45 Rain: 3/10; 1/2 chance. Shine: 7/10; 1 $$\\frac{3}{10}*\\frac{1}{2}+\\frac{7}{10}*\\frac{1}{1}$$ $$\\frac{17}{20}$$ $$85\\%$$ Senior Manager Joined: 08 Nov 2010 Posts: 264 ### Show Tags 05 Feb 2011, 08:52 what lvl is that question? Senior Manager Joined: 08 Nov 2010 Posts: 264 ### Show Tags 06", "Dora´s chances of traveling to Israel next holidays, if she knows there is a 70% probability for the raining season in Jerusalem to begin? (A) 15% (B) 29% (C) 35% (D) 42% (E) 46% _________________ Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our high-level \"quant\" preparation starts here: https://gmath.net Intern Joined: 25 Dec 2018 Posts: 2 If the raining season in Jerusalem begins, Dora estimates a 20% probab [#permalink] ### Show Tags 08 Mar 2019, 09:49 2 There are two possibilities of Dora travelling 1. If it rains and she travels- p(rains) x p(travel)= 0.7 x 0.2 = 0.14 2. If it does not rain and she travels - p (doesn't rain) x p (travels when it doesn't rain) Now, the question says that probability of travelling when it does not rain is higher by 250% means that, 250% of 20% = 0.5 therefore, probability of travelling when it doesn't rain is 0.2+0.5= 0.7 So, p (doesn't rain) x p (travels when it doesn't rain)= 0.3 (given in question) x 0.7 = 0.21 P (dora travels) = P (rains and travels) OR P (doesn't rain and travels) = 0.14 + 0.21 = 0.35 Hope this is correct. GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 935 Re: If the raining season in Jerusalem", "we shall encounter will be nondogmatic, that is, neither zero nor one. An event with probability $$0$$ has no chance of occurring; an event of probability $$1$$ is certain to occur. It is hard to think of any examples of interest to statistics in which the probability is either $$0$$ or $$1$$. (Even the probability that the Sun will come up tomorrow is less than $$1$$.) The following example illustrates our attitude about probabilities. Suppose you wish to know what the weather will be like next Saturday because you are planning a picnic. You turn on your radio, and the weather person says, “There is a 10% chance of rain.” You decide to have the picnic outdoors and, lo and behold, it rains. You are furious with the weather person. But was she wrong? No, she did not say it would not rain, only that rain was unlikely. She would have been flatly wrong only if she said that the probability is $$0$$ and it subsequently rained. However, if you kept track of her weather predictions over a long period of time and found that it rained on $$50\\%$$ of the days that the weather person said the probability was $$0.10$$, you could say her probability assessments are wrong. So when is it accurate to say that the", "it doesn't rain there is a $$90 \\%$$ chance that they play. What is the probability that students play football on February $$15^{\\text{th}}$$ ? 2. Cathy occasionally likes to drink soda. She mainly drinks it if she eats fried chicken on Tuesdays (\"special deal day\"). The probability that she eats fried chicken on any Tuesday is $$0.8$$. If she eats fried chicken the probability of her drinking soda is $$0.4$$, otherwise it is $$0.05$$. Calculate the probability that Cathy will drink soda next Tuesday. 3. In a grade $$12$$ class, the probability that a students studies Higer Level mathematics (HL) is $$0.3$$, altenratively students will study Standard Level Mathematics (SL). $$80 \\%$$ of students studying HL study Two Sciences, whereas only $$40 \\%$$ of the SL students study Two Sciences. A student is selected at random from the class. What is the probability that this student studies Two Sciences? 4. On any given day there is a $$80 \\%$$ chance that John has to take an object away from his two year old daughter, Charlotte. If he takes an object away from her there is a $$90\\%$$ chance she cries. If he doesn't take an object away from her there is still a $$30\\%$$ chance that she cries that day. What is the probability that Charlotte cries tomorrow?", "## Python, Random Numbers and Probability ### Introduction \"Every American should have above average income, and my Administration is going to see they get it.\" This saying is attributed to Bill Clinton on umpteen websites. Usually, there is no context given, so it is not clear, if he might have meant it as a \"joke\". Whatever his intentions might have been, we quoted him to show a \"real\" life example of statistics. Statistics and probability calculation is all around us in real-life situations. We have to cope with it whenever we have to make a decision from various options. Can we go for a hike in the afternoon or will it rain? The weather forecast tells us, that the probability of precipitation will be 30 %. So what now? Will we go for a hike? Another situation: Every week you play the lottery and dream of a far away island. What is the likelihood of winning the Jackpot so that you will never have to work again and live in \"paradise\"? Now imagine that you right on this island of your dreams. Most probably not because you won the jackpot, but rather because you booked your time as an all-inclusive holiday package. You are on holiday on a paradisal island far from home. Suddenly, you meet your neighbor,", "= .01$ $P(T|C) = .8$ $P(T|\\neg C) = .096$ Now the problem statement: A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer? We have to infer $P(C|T)$. Note how this is a reversal of the conditional statements we encounted in the information given about the test. Calculation Now comes the calculation. A good place to start when thinking about conditional probability is the ratio formula for the probability of a condititional event: $P(B|A) = \\frac{P(A \\& B) }{P(A)}$ Take an interpretatation of “If it is raining, then I have an umbrella” as the conditional event expression: I have an umbrella | it is raining The probability of this is the probability that I have an umbrella and it is raining, divided by the probability that it is raining. This can easily be rewritten to $P(A \\& B) = P(B|A) P(A)$ So if you know the probability of rain, and the probability that I have an umbrella when it rains, then you can multiply them to infer the probability that it is raining and I have an umbrella. One step towards Bayes’ rule begins with: 1. $P(B|A) = P(A \\& B) / P(A)$ 2. $P(A|B) = P(A \\& B) / P(B)$ [$A \\& B", "100 other prizes of $25 each. You buy a single ticket costing$3.50 a. what is the probability that you win the grand prize? b. what is the probability that you do NOT win any prize? c. what is the expected value of a single ticket to the individual purchasing the ticket? 6.) John figures that if he stays home for the weekend and studies for his Math exam he will have a 70% change of getting an A. If he does not stay home he figures he still has a 20% chance for an A. John's problem is that he plans to attend a beach party that weekend if it doesn't rain. The weather forecast shows a 40% chance of rain over the weekend. Draw a probability tree diagram for John and answer the following questions. (hint: 1st selection home or beach, 2nd selection A or not A) a. what is John's chance of getting an A? b. what is John's chance of not getting an A? c. if John decides to go to the beach regardless of the weather, what is his chance of an A? 7.) In a game of yahtzee you are left with (1,2,2,2,3) on your second roll. Show the procedure and calculate the \"best\" probability for tossing each of the following on your", "$0.44 \\div2 = 0.22$ b) From part a), we know that the probability of Jimmy watching a sci-fi movie is $0.22$. The probability of Jimmy watching a romantic comedy is $0.56$. In order to calculate the probability of Jimmy watching either a romantic comedy or a sci-fi movie we need to add the probabilities of each, since this is an either / or scenario: $0.22 + 0.56 = 0.78$ If Macy chooses black trousers, then she has $3$ choices for the colour of her jumper, so the $3$ possible outcomes are: BO, BY, BW If she chooses navy trousers, then the $3$ possible outcomes are: NO, NY, NW Finally, if she chooses purple trousers, then the remaining possible outcomes are: PO, PY, PW BY was already given in the question, so the full list of other possible outcomes is: BO, BW, NO, NY, NW, PO, PY, PW a) We know that the probability of selecting a red, blue or green bead must add up to $1$. We know that the probability of selecting a red bead is $0.25$, so the probability of selecting either a blue or a green bead must be: $1 - 0.25 = 0.75$ The problem in this question is that the probability of selecting a blue bead and the probability of selecting a green", "On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.2, what is the probability that Malachi will return home at the end of the day on the following Monday? A. 0.008 B. 0.128 C. 0.488 D. 0.512 E. 0.640 We are looking for the probability of the following even NNR: no rain on first day, no rain on second day, rain on third day (Monday). $$P(NNR)=0.8*0.8*0.2=0.128$$. OPEN DISCUSSION OF THIS QUESTION IS HERE: on-saturday-morning-malachi-will-begin-a-camping-vacation-100297.html _________________ Non-Human User Joined: 09 Sep 2013 Posts: 7711 Re: On Saturday morning, Malachi will begin a camping vacation a [#permalink] ### Show Tags 30 Jul 2018, 03:49 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as", "# What is the probability that the pass is a daily pass? 60.8% There are $8 + 23 + 5 + 15 = 51$ passes in total. However, only $8 + 23 = 31$ passes are daily passes. Since none of the passes are more likely to be chosen, the probability of choosing a daily pass is 31/51\\approx60.78%\\approx60.8%." ]

[ "+0 # Help +1 145 1 +335 A palindrome is an integer that reads the same forwards and backward. How many positive 3-digit palindromes are multiples of 3? Dec 14, 2018 #1 +22181 +10 A palindrome is an integer that reads the same forwards and backward. How many positive 3-digit palindromes are multiples of 3? $$\\begin{array}{|r|r|r|r|} \\hline & n & \\text{palindrome } a(n) & \\text{divisible by } 3 \\\\ \\hline & 20& 1 01 \\\\ 1. & 21& 1 11 & \\checkmark \\\\ & 22& 1 21 \\\\ & 23& 1 31 \\\\ 2. & 24& 1 41 &\\checkmark \\\\ & 25& 1 51 \\\\ & 26& 1 61 \\\\ 3.& 27& 1 71 &\\checkmark \\\\ & 28& 1 81 \\\\ & 29& 1 91 \\\\ \\hline & 30& 2 02 \\\\ & 31& 2 12 \\\\ 4. & 32& 2 22 &\\checkmark \\\\ & 33& 2 32 \\\\ & 34& 2 42 \\\\ 5. & 35& 2 52 &\\checkmark \\\\ & 36& 2 62 \\\\ & 37& 2 72 \\\\ 6.& 38& 2 82 &\\checkmark \\\\ & 39& 2 92 \\\\ \\hline 7. & 40& 3 03 &\\checkmark \\\\ & 41& 3 13 \\\\ & 42& 3 23 \\\\ 8. & 43& 3 33 &\\checkmark \\\\ & 44& 3 43 \\\\ & 45& 3 53 \\\\ 9.", "(2016-02-19). \"Every positive integer is a sum of three palindromes\". Mathematics of Computation. arXiv:1602.06208. (arXiv preprint)", "# Special Palindromes A palindrome is a positive integer that reads the same forwards and backwards; e.g., 101 and 2552 are both palindromes. What is the smallest $$x$$ for which $$x$$ and $$x+2016$$ are both palindromes? × Problem Loading... Note Loading... Set Loading...", "and backward, the number 121 is a palindrome.", "# Palindromic Primes What is the number of palindromic primes $$p$$ less than $$10^5$$? Hint: The answer is a prime number. ×", "Every positive integer is a sum of three palindromes Article by Javier Cilleruelo and Florian Luca and Lewis Baxter In collections: Easily explained, Fun maths facts For integer $g\\ge 5$, we prove that any positive integer can be written as a sum of three palindromes in base $g$. URL: arxiv.org/abs/1602.06208v2 PDF: arxiv.org/pdf/1602.06208v2 Entry: read.somethingorotherwhatever. · tooter · · · A Mastodon instance for maths people. The kind of people who make $\\pi z^2 \\times a$ jokes. Use $ and $ for inline LaTeX, and $ and $ for display mode.", "# Palindromic Primes Computer Science Level 3 What is the number of palindromic primes $$p$$ less than $$10^5$$? Hint: The answer is a prime number. ×", "“PWC 164 › Palindromic Primes”", "non-prime numbers. There are only three: 1, 4 and 6. Aha! So, not only are all invalid palindromes above 6 odd, they are all prime as well! (Assuming this continues beyond 10,000. I have no proof it does, but so far it works.)", "form the palindromes 1, 121, 12321, 1234321, ... The only known non-palindromic number whose cube is a palindrome is 2201, and it is a conjecture the fourth root of all the palindrome fourth powers are a palindrome with 100000...000001 (10n + 1). G. J. Simmons conjectured there are no palindromes of form ''n''''k'' for ''k'' > 4 (and ''n'' > 1). # Other bases Palindromic numbers can be considered in numeral system A numeral system (or system of numeration) is a writing system for expressing numbers; that is, a mathematical notation for representing number A number is a mathematical object used to count, measure, and label. The original example ... s other than decimal The decimal numeral system (also called the base-ten positional numeral system and denary or decanary) is the standard system for denoting integer and non-integer numbers. It is the extension to non-integer numbers of the Hindu–Arabic numeral ... . For example, the binary Binary may refer to: Science and technology Mathematics * Binary number A binary number is a number expressed in the base-2 numeral system or binary numeral system, a method of mathematical expression which uses only two symbols: typical ... palindromic numbers are those with the binary representations: :0, 1, 11, 101, 111, 1001, 1111, 10001, 10101, 11011, 11111, 100001, ... or in", "# Palindromic Primes What is the number of palindromic primes $$p$$ less than $$10^5$$? Hint: The answer is a prime number. ##### this problem is a part of set Prime Crimes via Computer Science × Problem Loading... Note Loading... Set Loading...", "palindromic integers in base 5) + (the only positive integer that is five times the sum of its digits) • (the only Fibonacci number that is a double of a prime) + (the only prime p such that p! has p digits) − (the only fixed point of look-and-say operation) • (the only number whose concatenation with itself is prime) • (the only positive integer that that differs by 1 from a square and a nonsquare cube) − (the largest number such that its divisors are each 1 less than a prime) • (the smallest evil untouchable number) ***** • (the alphanumeric value of MANIC SAGES) + (the sum of all three-digit numbers you can get by permuting digits 1, 2, and 3) + (the number of two-digit integers divisible by 9) − (the number of rectangles whose sides are composed of edges of squares of a chess board) • (the integer whose standard Roman numeral representation is alphabetically later than all others) − (the number you get if you divide a three digit number with identical digits by the sum of the digits) • (the largest even integer that is not a sum of two abundant numbers) − (the digit in the first position where e and π have the same digit) • (the number formed by", "mostly because my iPod had packed up on me, leaving me with a good half an hour with nothing better to do. A palindrome is a word like LEVEL, which reads the same forwards and backwards. Specifically, I was thinking about 5-digit numbers, so my palindromes were things like 01410. I’d seen one like this at work the previous day, which got me thinking-what were the chances of that happening? This turns out not to be a very interesting question to ask—for a random 5-digit number, the answer is simply 1 in 100. The first 3 digits can be whatever they like, and then the remaining two must match up with a 1-in-10 chance for each. # Primes à la Euler Anyone with a more than a passing interest in numbers knows that there are infinitely many primes.If we highlight them in a grid like this, it is clear that the last digit of any prime must be either 1, 3, 7 or 9 (except for 2 and 5 of course). But there seems to be no good reason for primes to prefer any one of these four possibilities over any other. It is natural therefore to conjecture that There are infinitely many primes whose last digit is $b$ for $b$ equal to any of 1, 3, 7,", "# 0110110110 How many binary strings of length 10 are palindromes? Details and assumptions A palindrome is a number that reads the same forward as it does backward. A binary string is a sequence of integers that are $$1$$ or $$0$$. The length of a string refers to the number of integers in it. As an explicit example, the binary string of length $$3$$ are $$000, 001, 010, 011$$, $$100, 101, 110, 111$$. ×", "= 7 7 = 1 + 5 + 1 7 = 2 + 3 + 2 7 = 3 + 1 + 3 7 = 1 + 1 + 3 + 1 + 1 7 = 2 + 1 + 1 + 1 + 2 7 = 1 + 2 + 1 + 2 + 1 7 = 1 + 1 + 1 + 1 + 1 + 1 + 1 In total, there are 8 palindromic partitions.", "roads running along 7 30 Jul 2015, 11:15 57 A palindrome is a number that reads the same forward and bac 12 07 Oct 2013, 01:31 1 Dan completes a 3 part exam as follows 3 21 Feb 2013, 08:05 Display posts from previous: Sort by", "will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: A palindrome number reads the same backward and forward [#permalink] 13 Nov 2014, 10:35 Similar topics Replies Last post Similar Topics: 7 A palindrome is a number that reads the same forward and backward, 2 06 Sep 2014, 10:40 28 A palindrome is a number that reads the same forward and bac 9 07 Oct 2013, 00:31 17 A palindrome is a number that reads the same forward and 4 30 Mar 2012, 04:00 5 A palindrome is a number that reads the same forward and backward, suc 6 26 Aug 2011, 05:42 2 A palindrome is a number that reads the same forward and backward, suc 5 06 Aug 2011, 12:40 Display posts from previous: Sort by", "# 32. Longest Double Palindrome ## I'm a slow walker, but I never walk backwards. We define a palindrome as a sequence of at least one integer that reads the same either forward or backward. For example 1 3 5 3 1 and 1 2 2 1 are both palindromes. To the extreme case 1 is also a palindrome. We also define a double palindrome as the concatenation of two palindromes. For example 1 3 5 3 1 1 2 2 1 is a double palindrome. To the extreme case 1 3 5 3 1 1 is also a double palindromes. Now given a sequence of at least two integers, please find the longest double palindrome. If there are multiple longest double palindromes, print the one that appears last . ## Limits The number of integers in the given sequence is no more than 100. ## Input Format There is one line in the input. The first line has the integers from which we want to find the longest double palindrome. ## Output Format There is one lines in the output. The first line has all the integers of the longest double palindrome. number of ways as in the description. ## Sample Input 7 1 3 5 3 1 1 ## Sample Output 1 3 5 3 1", "#### Quantity A The number of primes that are divisible by 9 #### Quantity B The number of primes that are divisible by 19 If a, b, x, and y are positive integers, and $13^a \\times 13^b=(13^x)^y=13^{13}$, what is the average (arithmetic mean) of a, b, x, and y? _____ A rectangle is drawn in a standard xy-coordinate plane. If the sides of the rectangle are not parallel to the axes, what is the product of the slope of the four sides? In a certain state, each license plate consists of either three digits (between 0 and 9, inclusive) followed by two letters or three letters followed by two digits. For example, 055-XY, 123-PP, and AAA-70 are all acceptable plates. How many different license plates can the state issue? _____ n is an even integer #### Quantity A The number of prime factors of n #### Quantity B The number of prime factors of $\\frac{n}{2}$ A positive integer is a palindrome if it reads exactly the same from right to left as it does from left to right. For example, 5 and 66 and 373 are all palindromes. How many palindromes are there between 1 and 1,000, inclusive? _____ Line l passes through points in both quadrants II and III. Which of the following statements are true? Indicate", "x2 Introduction = # Palindromic Numbers In English, a \"palindrome\" is a sentence that reads the same backwards as forwards. For example, \"Able was I ere I saw Elba.\"1. Mathematicians use the word in almost exactly the same sense to describe numbers." ]

[ "there are $90$ palindromes. You can choose the first digit $9$ ways and the middle digit $10$ ways. Generally, for $n$ digit numbers there are $$\\begin {cases} 9\\cdot 10^{\\frac {n-2}2} & n \\text { even} \\\\ 9\\cdot 10^{\\frac {n-1}2} & n \\text { odd} \\end {cases}$$ palindromes. Again, you can choose the first digit $9$ ways and the rest of the first half of the number (rounded up for odd numbers of digits) $10$ ways . To get $66$ with reverse and add, you can have $15,24,33,42,51$ as starting numbers. For $5556555$ you can certainly have five choices $(1-5)$ for the first digit, six $(0-5)$for the next two, and one choice $(3)$for the middle digit. Then the lower three digits are determined by the top three. This gives $180$ numbers. There might be more, as I have avoided carrying. - I believe if you ever have some carrying in the summation, then the resulting number won't be palindromic. Yet to be proven though :) – Dolma Apr 11 '13 at 17:09 You think it is possible to make the following statement: In the interval [a,b] we have a specific number of palindroms, what can we say about the numbers which appear when we reverse the reverse-and-add algorithm. – Babla Apr 11 '13 at 17:10 For exmaple: There are", "where $w$ will appear four times, $x$ will appear three times, and $y$ and $z$ will appear twice each. As you observed, an $x$ has to go in the middle; once that’s been taken care of, we have to have two $w$’s, one $x$, one $y$, and one $z$ on each side of that middle $x$. Since we’re building a palindrome, the five letters to the right of the centre $x$ must be the mirror image of the five on the left, so there will be one palindrome for every way of arranging two $w$’s, one $x$, one $y$, and one $z$. I can put the $x$ in any of $5$ places. Once that’s done, the $y$ can go in any of $4$ places, and then the $z$ can go in any of $3$ places. The $2$ $w$’s will necessarily go in the remaining two slots, so there are $5 \\cdot 4 \\cdot 3$ ways to arrange the five letters and therefore $5 \\cdot 4 \\cdot 3$ palindromes that can be made from the $11$ letters that we chose. \\begin{align*}26\\cdot 25 \\binom{24}{2} &= \\frac{26\\cdot 25\\cdot 24\\cdot 23}{2} \\cdot 5 \\cdot 4 \\cdot 3\\\\ &= \\frac{26\\cdot 25\\cdot 24\\cdot 23\\cdot 5 \\cdot 4 \\cdot 3}{2} \\end{align*}", "For five-digit palindromes, we start with our three-digit palindromes (90 of them) and we can insert one of ten digits into the middle. This gives us 900. The comparison of 6-digit to 5-digit is the same as 4 to 3. That is, we add two digits but they must be the same, so we’ll end up with 900 again. The pattern becomes obvious pretty quickly • 2 digits – 9 palindromic numbers • 3 digits – 90 • 4 digits – 90 • 5 digits – 900 • 6 digits – 900 • 7 digits – 9000 • 8 digits – 9000 This pattern is $x=9\\times10^{\\frac{d-2}{2}}$ if d is even or $x=9\\times10^{\\frac{d-1}{2}}$ if d is odd. (You nerds out there, tell me if there’s a better way of writing different-but-related odd/even functions) So substituting $d=351$ into $x=9\\times10^{\\frac{d-1}{2}}$ gives • $x=9\\times10^{\\frac{351-1}{2}}$ • $x=9\\times10^{\\frac{350}{2}}$ • $x=9\\times10^{175}$ That’s 90,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000 numbers. How far apart are the closest ones? Well, if we change any number in the first 175 digits, we have to make one on the symmetrical digit. The only way we can change just one digit is if it’s the middle one. In a more simple situation, lets look at some five digit numbers. • 55155 • 55255 • 55355 • 55455 These numbers are all 100 apart, and", "get $\\boxed{550}$ ~$\\alpha b \\alpha$ ## Solution 5 (Extremely Fast Solution) The average values of the first and last digits are each $5,$ and the average value of the middle digit is $4.5,$ so the average of all three-digit palindromes is $5\\cdot 10^2+4.5\\cdot 10+5=\\boxed{550}.$ ~MathIsFun286 ## Remark Visit the Discussion Page for questions and further generalizations. ~MRENTHUSIASM", "40 & 32 & 24 & 16 & 8 \\\\ 7 & & & 49 & 42 & 35 & 28 & 21 & 14 & 7 \\\\ 6 & & & & 36 & 30 & 24 & 18 & 12 & 6 \\\\ 5 & & & & & 25 & 20 & 15 & 10 & 5 \\\\ 4 & & & & & & 16 & 12 & 8 & 4 \\\\ 3 & & & & & & & 9 & 6 & 3 \\\\ 2 & & & & & & & & 4 & 2 \\\\ 1 & & & & & & & & & 1 \\\\ \\end{array}$ The first thing to jump out at me is that the vast majority of these products are not palindromes. That’s not really surprising I guess – there are $$9 \\cdot 10^{2t-1}$$ different $$2t$$-digit numbers, but only $$9 \\cdot 10^{t-1}$$ of them are palindromes, so to a first approximation the probability that an arbitrary $$2t$$-digit number is a palindrome is about $$10^{-t}$$. So maybe instead of searching among the products of $$t$$-digit numbers for palindromes, it would be faster to search among the palindromes for products of $$t$$-digit numbers. ## A Digression on Palindromes The rarity of palindromes highlights a disturbing possibility. The", "- 1 2 - 5 8 - 8. At $AB:CD:EF$, $A<3; C,E<6$. If $A<2$, there are 2 values available for $A$, with whichever chosen also used for $F$. $B$ and $C$ can be 0, 1, 2 or 5 ($2*4*4=32$ combinations). If $A=2$, then $B=0,1$ or $2$, and $C=0,1,2$ or $5$ ($3*4=12$ combinations). There are 44 in total. Edit: the last one wasn't what the OP meant. • What does vertical symmetry have to do with palindromicity? – Rand al'Thor Jan 10 '17 at 13:35 It's all down to the minutes - it needs a palindromic number of minutes. 00, 11, 22, 33, 44, 55 So for every hour, there are six possible palindromes. Then work out which hours can have a reversed number of seconds (less than 60): 01, 02, 03, 04, 05 (06, 07, 08, 09) 10, 11, 12, 13, 14, 15 (16, 17, 18, 19) 20, 21, 22, 24, 24. 16 possibles. 16 x 6 = 96 In the am/pm scenario, the preceding zero would never be included, therefore the number of palindromes possible would be significantly reduced below 96. Also, since the AM and PM symbols cannot be part of the palindrome, the first half, and the second half, of the 24 hour period would be indistinguishable, and therefore not add to the total. •", "once again there are 90. For five-digit palindromes, we start with our three-digit palindromes (90 of them) and we can insert one of ten digits into the middle. This gives us 900. The comparison of 6-digit to 5-digit is the same as 4 to 3. That is, we add two digits but they must be the same, so we’ll end up with 900 again. The pattern becomes obvious pretty quickly • 2 digits – 9 palindromic numbers • 3 digits – 90 • 4 digits – 90 • 5 digits – 900 • 6 digits – 900 • 7 digits – 9000 • 8 digits – 9000 This pattern is $x=9\\times10^{\\frac{d-2}{2}}$ if d is even or $x=9\\times10^{\\frac{d-1}{2}}$ if d is odd. (You nerds out there, tell me if there’s a better way of writing different-but-related odd/even functions) So substituting $d=351$ into $x=9\\times10^{\\frac{d-1}{2}}$ gives • $x=9\\times10^{\\frac{351-1}{2}}$ • $x=9\\times10^{\\frac{350}{2}}$ • $x=9\\times10^{175}$ That’s 90,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000 numbers. How far apart are the closest ones? Well, if we change any number in the first 175 digits, we have to make one on the symmetrical digit. The only way we can change just one digit is if it’s the middle one. In a more simple situation, lets look at some five digit numbers. • 55155 • 55255 • 55355 • 55455 These numbers", "mostly because my iPod had packed up on me, leaving me with a good half an hour with nothing better to do. A palindrome is a word like LEVEL, which reads the same forwards and backwards. Specifically, I was thinking about 5-digit numbers, so my palindromes were things like 01410. I’d seen one like this at work the previous day, which got me thinking-what were the chances of that happening? This turns out not to be a very interesting question to ask—for a random 5-digit number, the answer is simply 1 in 100. The first 3 digits can be whatever they like, and then the remaining two must match up with a 1-in-10 chance for each. # Primes à la Euler Anyone with a more than a passing interest in numbers knows that there are infinitely many primes.If we highlight them in a grid like this, it is clear that the last digit of any prime must be either 1, 3, 7 or 9 (except for 2 and 5 of course). But there seems to be no good reason for primes to prefer any one of these four possibilities over any other. It is natural therefore to conjecture that There are infinitely many primes whose last digit is $b$ for $b$ equal to any of 1, 3, 7,", "1. ## Palindromes - Hi MHF Here's my question: There are 90 four-digit palindromic numbers. How many of these four-digit palindromic numbers are divisible by 7. So i have no idea how to start or what to do, i'm unable to obtain a pattern of any sort, i assume there are 10 palindromes between each thousand number bracket (1000-1999, 2000-2999 etc) 2. Originally Posted by 99.95 Hi MHF Here's my question: There are 90 four-digit palindromic numbers. How many of these four-digit palindromic numbers are divisible by 7. So i have no idea how to start or what to do, i'm unable to obtain a pattern of any sort, i assume there are 10 palindromes between each thousand number bracket (1000-1999, 2000-2999 etc) Hello 99.95 Yes there are \"10 palindromes between each thousand number bracket 1000-1999, 2000-2999 etc\" And also there are only two palindromes in each thousand number bracket for e.g 1001 & 1771, 2002 & 2772,.....9009 & 9779 So in total it makes $9*2=18$ four-digit palindromic numbers that are divisible by $7$. Hope this helps 3. Yes that helps, but i would like to know how you found that there are only two palindromes in each thousand number bracket which are divisble by 7. is there some sort of rule/theorem, or did you happen to use", "if a large one. Hence, we explored 100:999 in greater detail. Figure 1 shows a plot of $f[100:999]$ Figure 1. y-axis: $\\log_{10}(f[n])$ Of these our experiments suggested that the following 13 numbers in the range 100:999 never converge to a palindrome (marked by red dots in Figure 1): 196, 295, 394, 493, 592, 689, 691, 788, 790, 879, 887, 978, 986. One can see that barring 790 all of them come in pairs. The maximum value attained by $f[n]$ in this range is the 13 digit number: 8813200023188. This value is attained when $n$ is 187, 286, 385, 484, 583, 682, 781, 869, 880, 968. Comparable to the non-converging cases, here all ten cases come as pairs. There is also an interesting pattern for both the maxima and the non-converging cases: we observe that in each century it comes one number earlier but at some point a “new line of descent” emerges which then perpetuates the same pattern along with the older one. The same maximum value is attained for $n$ in the range 1:99, with $f[89], f[98]=16668488486661$. Omitting the single digit numbers, 11 is the primordial palindromic number and its multiples often tend to have a palindromic structure. Its multiples for $k=1:9$ are the only double digit palindromes. Multiples of 11 are also found among the triple", "+0 # Help +1 145 1 +335 A palindrome is an integer that reads the same forwards and backward. How many positive 3-digit palindromes are multiples of 3? Dec 14, 2018 #1 +22181 +10 A palindrome is an integer that reads the same forwards and backward. How many positive 3-digit palindromes are multiples of 3? $$\\begin{array}{|r|r|r|r|} \\hline & n & \\text{palindrome } a(n) & \\text{divisible by } 3 \\\\ \\hline & 20& 1 01 \\\\ 1. & 21& 1 11 & \\checkmark \\\\ & 22& 1 21 \\\\ & 23& 1 31 \\\\ 2. & 24& 1 41 &\\checkmark \\\\ & 25& 1 51 \\\\ & 26& 1 61 \\\\ 3.& 27& 1 71 &\\checkmark \\\\ & 28& 1 81 \\\\ & 29& 1 91 \\\\ \\hline & 30& 2 02 \\\\ & 31& 2 12 \\\\ 4. & 32& 2 22 &\\checkmark \\\\ & 33& 2 32 \\\\ & 34& 2 42 \\\\ 5. & 35& 2 52 &\\checkmark \\\\ & 36& 2 62 \\\\ & 37& 2 72 \\\\ 6.& 38& 2 82 &\\checkmark \\\\ & 39& 2 92 \\\\ \\hline 7. & 40& 3 03 &\\checkmark \\\\ & 41& 3 13 \\\\ & 42& 3 23 \\\\ 8. & 43& 3 33 &\\checkmark \\\\ & 44& 3 43 \\\\ & 45& 3 53 \\\\ 9.", "questions like “what five-digit palindrome starting with 28 is divisible by 11?” There is at most one $(2k+1)$-digit palindrome with any first $k$-digit number which is divisible by 11. Consider the numbers $28082, 28182, \\ldots, 28982.$ Each of these differs by 100 from the last one, which is one more than a multiple of 11, so when we reduce mod 11 we get $k, k+1, \\ldots, k+9$ which are all different. Unless 28082 happens to be one more than a multiple of 11, one of these is a multiple of 11. As it turns out, 28082 is one {\\it less} than a multiple of 11, and 28182 is a multiple of 11: $28182 = 11 \\times 2562$. So there is either zero or one palindrome of each of the forms $10x01, 11x11, 12x21, \\ldots, 99x99$. When is there no such palindrome? Observe that, for example, $12000 - 21 = (10000 + 2000) - (20 + 1) = (10000 - 1) + 10 \\times 2 \\times (100 - 1)$$ and so 12000 and 21 differ by a multiple of 11. So $12x21$ and $24x00$ are congruent mod 11. To get $24x00$ to be a multiple of 11 we need $24x$ (that is, $240 + x$) to be a multiple of 11 – so we take $x = 2$. Indeed,", "+0 # Question 0 112 1 +182 Determine the number of palindromes with 7 digits. Apr 22, 2018 #1 +971 +4 There are two methods to do this problem. Method 1: Pattern Finding There are 9 palindromes with 2 digits There are 90 palindromes with 3 digits There are 90 palindromes with 4 digits There are 900 palindromes with 5 digits So 1 digit: 9 2 digits: 9 3 digits: 90 4 digits: 90 5 digits: 900 6 digits: 900 7 digits: 9000 9000 Method 2: Counting The 7 digit palindrome is in this format: ABCDCBA There are 9 choices for A 10 for B 10 for C 10 for D $$9\\cdot{10}\\cdot{10}\\cdot{10}=\\boxed{9000}$$ I hope this helped, Gavin . Apr 22, 2018 edited by GYanggg Apr 22, 2018", "(when given integer inputs) is prime, and furthermore, every prime is obtained in this way. (Note: $$Z$$ spits out many non-positive values as well.) Unfortunately, this polynomial $$Z$$ is not very practical. For one thing, its degree is about $$38!$$. (Yes, I mean 38 factorial.) If we consider digits instead of letters or words, then numbers such as 676, 100020001, and 12345678987654321 are palindromes. These are in fact quite special palindromes, as they are all squares of integers: $$26^2 = 676$$, $$111111111^2 = 12345678987654321$$, and $$10001^2 = 100020001$$. Are there infinitely many palindromic squares? Yes: the last example above hints that $$10\\ldots01^2 = 10\\ldots020\\ldots01$$ for however many 0s you put in the middle. By contrast, if we ask for $$n^5$$ to be palindromic (or any higher power), then no solutions are currently known! You might ask for other types of palindromic numbers, such as palindromic primes. A few examples include 7 (this is a palindrome!), 101, 19391, and 1000000008000000001. Are there infinitely many? We don’t know. But the largest palindromic prime currently known is $$10^{200000} + 47960506974 \\cdot 10^{99995} + 1$$, according to Wikipedia.", "+0 # Question 0 96 1 +182 Determine the number of palindromes with 7 digits. wiskdls Apr 22, 2018 #1 +963 +4 There are two methods to do this problem. Method 1: Pattern Finding There are 9 palindromes with 2 digits There are 90 palindromes with 3 digits There are 90 palindromes with 4 digits There are 900 palindromes with 5 digits So 1 digit: 9 2 digits: 9 3 digits: 90 4 digits: 90 5 digits: 900 6 digits: 900 7 digits: 9000 9000 Method 2: Counting The 7 digit palindrome is in this format: ABCDCBA There are 9 choices for A 10 for B 10 for C 10 for D $$9\\cdot{10}\\cdot{10}\\cdot{10}=\\boxed{9000}$$ I hope this helped, Gavin GYanggg Apr 22, 2018 edited by GYanggg Apr 22, 2018", "total. Kudos [?]: 41 [2], given: 46 Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 627 Kudos [?]: 1387 [7], given: 136 ### Show Tags 06 Sep 2013, 06:04 7 KUDOS 2 This post was BOOKMARKED arakban99 wrote: A palindrome is a number which reads the same when read forward as it it does when read backward. So, how many 5 digit palindromes are there? (A)720 (B)800 (C)890 (D)900 (E)950 A 5 digit palindrome would look like this : $$abcba.$$ Thus, the first digit can be filled in 9 ways(1-9), the second digit can be filled in 10 ways (0-9) and the 3rd digit can again be filled in 10 ways. The last 2 digits would just mirror the already selected digit. Thus, the no of ways : 9*10*10*1*1 = 900. D. _________________ Kudos [?]: 1387 [7], given: 136 Intern Joined: 03 Aug 2013 Posts: 15 Kudos [?]: 41 [1], given: 46 ### Show Tags 06 Sep 2013, 06:04 1 KUDOS Here is the official solution: there five digit places __ __ __ __ __ The last two depend on the first two (because the last to is the flipped version of the first two digits i.e 54145 - 45 is the flipped version of 54 so the number of ways to select the last two is 1*1", "know $$A+C \\ge 9$$ from the highest digit and $$A+C$$ ends in 1 from the lowest digit, so $$A+C=11$$. Then the minimum possible sum is 1111 where $$B=D=0$$, which happens to be a palindrome. One possible sum is $$505 + 606 = 1111$$. The only other sum that meets the condition is 1221, where $$B+D = 11$$.", "possibilities for each a and 10 possibilities for each b. Thus, the total number of palindromes of digit 5 are $$9\\times10\\times10=900.$$ Finally, the type of the 6 digit palindromes are given as: ab00ba,ab11ba,ab22ba,...,ab99ba There are 9 possibilities for each a and 10 possibilities for each b. Thusm the total number of palindromes of digit 6 are $$9\\times10\\times10=900$$ Thus, the total number of palindromic numbers between 1000 and 1000000 are $$90+900+900=1890.$$ Not exactly what you’re looking for? • Questions are typically answered in as fast as 30 minutes", "1. Palindromes The product of two positive three-digit palindromes is 436995. What is their sum? By the way, a palindrome is number that is read the same backwards and forwards: example: 1331, 7557, 131, 767, 454, etc.. This is what I have tried: $[100a+10b+a][100c+10d+c]=436995$ $[101a+10b][101c+10d]=436995$ I realize that either a or c must be 5, for the product to end in a 5. Also I see that one or both of these numbers must be a factor of 9 and 3, but i'm not sure what to make of these observations. I'm not exactly sure where to go from here. Help is appreciated. 2. Here's a simple method: since the product of the two numbers is 436,995 (which ends with 5), one of the two numbers should end with 5.And since it is a palindrome, it should also begin with 5. So, this number is is $5a5$ ;where a can be any number from 0 to 9. Now you can divide the number 436,995 by 5a5 till the result get a palindrome.. 436,995/505= 865.33 436,995/515=848.53 436,995/525=832.37 436,995/535=816.81 436,995/545=801.82 436,995/555=787.37 436,995/565=773.44 436,995/575=759.99 436,995/585=747 ...There you go! 3. Hello, rtblue! The product of two positive three-digit palindromes is 436995. What is their sum? I found no satisfactory algebraic approach. So I factored 436,995: . $3^4\\cdot5\\cdot13\\cdot83$ . . and found two", "increase 1 “zero” in denominator, the answer would be correct. This is a case where I would use permutations, but not in the same way as for cards. The numerator will be the number of ways to form a number consisting of 5 different digits, which means a permutation of 5 of the ten possible digits. That is 10P5 = 10*9*8*7*6 = 10!/5! = 30,240. The denominator, again, is the number of ways to choose 5 (non-distinct) digits, which is just 10^5 = 100,000, since there are 10 independent ways to choose each. So the probability is 30,240/100,000 = 0.3024. That is, $$P(\\text{all different})=\\frac{_{10}\\text{P}_5}{10^5}=\\frac{10\\cdot9\\cdot8\\cdot7\\cdot6}{100,000}=0.3024=30.24\\%.$$" ]

[ "+0 # Help +1 145 1 +335 A palindrome is an integer that reads the same forwards and backward. How many positive 3-digit palindromes are multiples of 3? Dec 14, 2018 #1 +22181 +10 A palindrome is an integer that reads the same forwards and backward. How many positive 3-digit palindromes are multiples of 3? $$\\begin{array}{|r|r|r|r|} \\hline & n & \\text{palindrome } a(n) & \\text{divisible by } 3 \\\\ \\hline & 20& 1 01 \\\\ 1. & 21& 1 11 & \\checkmark \\\\ & 22& 1 21 \\\\ & 23& 1 31 \\\\ 2. & 24& 1 41 &\\checkmark \\\\ & 25& 1 51 \\\\ & 26& 1 61 \\\\ 3.& 27& 1 71 &\\checkmark \\\\ & 28& 1 81 \\\\ & 29& 1 91 \\\\ \\hline & 30& 2 02 \\\\ & 31& 2 12 \\\\ 4. & 32& 2 22 &\\checkmark \\\\ & 33& 2 32 \\\\ & 34& 2 42 \\\\ 5. & 35& 2 52 &\\checkmark \\\\ & 36& 2 62 \\\\ & 37& 2 72 \\\\ 6.& 38& 2 82 &\\checkmark \\\\ & 39& 2 92 \\\\ \\hline 7. & 40& 3 03 &\\checkmark \\\\ & 41& 3 13 \\\\ & 42& 3 23 \\\\ 8. & 43& 3 33 &\\checkmark \\\\ & 44& 3 43 \\\\ & 45& 3 53 \\\\ 9.", "there are $90$ palindromes. You can choose the first digit $9$ ways and the middle digit $10$ ways. Generally, for $n$ digit numbers there are $$\\begin {cases} 9\\cdot 10^{\\frac {n-2}2} & n \\text { even} \\\\ 9\\cdot 10^{\\frac {n-1}2} & n \\text { odd} \\end {cases}$$ palindromes. Again, you can choose the first digit $9$ ways and the rest of the first half of the number (rounded up for odd numbers of digits) $10$ ways . To get $66$ with reverse and add, you can have $15,24,33,42,51$ as starting numbers. For $5556555$ you can certainly have five choices $(1-5)$ for the first digit, six $(0-5)$for the next two, and one choice $(3)$for the middle digit. Then the lower three digits are determined by the top three. This gives $180$ numbers. There might be more, as I have avoided carrying. - I believe if you ever have some carrying in the summation, then the resulting number won't be palindromic. Yet to be proven though :) – Dolma Apr 11 '13 at 17:09 You think it is possible to make the following statement: In the interval [a,b] we have a specific number of palindroms, what can we say about the numbers which appear when we reverse the reverse-and-add algorithm. – Babla Apr 11 '13 at 17:10 For exmaple: There are", "where $w$ will appear four times, $x$ will appear three times, and $y$ and $z$ will appear twice each. As you observed, an $x$ has to go in the middle; once that’s been taken care of, we have to have two $w$’s, one $x$, one $y$, and one $z$ on each side of that middle $x$. Since we’re building a palindrome, the five letters to the right of the centre $x$ must be the mirror image of the five on the left, so there will be one palindrome for every way of arranging two $w$’s, one $x$, one $y$, and one $z$. I can put the $x$ in any of $5$ places. Once that’s done, the $y$ can go in any of $4$ places, and then the $z$ can go in any of $3$ places. The $2$ $w$’s will necessarily go in the remaining two slots, so there are $5 \\cdot 4 \\cdot 3$ ways to arrange the five letters and therefore $5 \\cdot 4 \\cdot 3$ palindromes that can be made from the $11$ letters that we chose. \\begin{align*}26\\cdot 25 \\binom{24}{2} &= \\frac{26\\cdot 25\\cdot 24\\cdot 23}{2} \\cdot 5 \\cdot 4 \\cdot 3\\\\ &= \\frac{26\\cdot 25\\cdot 24\\cdot 23\\cdot 5 \\cdot 4 \\cdot 3}{2} \\end{align*}", "For five-digit palindromes, we start with our three-digit palindromes (90 of them) and we can insert one of ten digits into the middle. This gives us 900. The comparison of 6-digit to 5-digit is the same as 4 to 3. That is, we add two digits but they must be the same, so we’ll end up with 900 again. The pattern becomes obvious pretty quickly • 2 digits – 9 palindromic numbers • 3 digits – 90 • 4 digits – 90 • 5 digits – 900 • 6 digits – 900 • 7 digits – 9000 • 8 digits – 9000 This pattern is $x=9\\times10^{\\frac{d-2}{2}}$ if d is even or $x=9\\times10^{\\frac{d-1}{2}}$ if d is odd. (You nerds out there, tell me if there’s a better way of writing different-but-related odd/even functions) So substituting $d=351$ into $x=9\\times10^{\\frac{d-1}{2}}$ gives • $x=9\\times10^{\\frac{351-1}{2}}$ • $x=9\\times10^{\\frac{350}{2}}$ • $x=9\\times10^{175}$ That’s 90,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000 numbers. How far apart are the closest ones? Well, if we change any number in the first 175 digits, we have to make one on the symmetrical digit. The only way we can change just one digit is if it’s the middle one. In a more simple situation, lets look at some five digit numbers. • 55155 • 55255 • 55355 • 55455 These numbers are all 100 apart, and", "mostly because my iPod had packed up on me, leaving me with a good half an hour with nothing better to do. A palindrome is a word like LEVEL, which reads the same forwards and backwards. Specifically, I was thinking about 5-digit numbers, so my palindromes were things like 01410. I’d seen one like this at work the previous day, which got me thinking-what were the chances of that happening? This turns out not to be a very interesting question to ask—for a random 5-digit number, the answer is simply 1 in 100. The first 3 digits can be whatever they like, and then the remaining two must match up with a 1-in-10 chance for each. # Primes à la Euler Anyone with a more than a passing interest in numbers knows that there are infinitely many primes.If we highlight them in a grid like this, it is clear that the last digit of any prime must be either 1, 3, 7 or 9 (except for 2 and 5 of course). But there seems to be no good reason for primes to prefer any one of these four possibilities over any other. It is natural therefore to conjecture that There are infinitely many primes whose last digit is $b$ for $b$ equal to any of 1, 3, 7,", "1,570 views A palindrome is a word that reads the same forwards and backwards. In a game of words, a player has the following two plates painted with letters. From the additional plates given in the options, which one of the combinations of additional plates would allow the player to construct a five-letter palindrome. The player should use all the five plates exactly once. The plates can be rotated in their plane. A. B. C. D. Logic : Player needs to make a 5-Letter Palindrome word. In any odd-letters palindrome word, the middle letter must be present odd number of times. Every other letter(which is Not the middle letter) must be present even number of times. Using this logic, we can see that in a 5-letter word, Exactly one letter must appear odd number of times(this letter will become the middle letter), and every other letter must appear even number of times. Also, if we have the following collection of letters then we can definitely make some palindrome from them : “Exactly one letter appears odd number of times(this letter will become the middle letter), and every other letter appears even number of times”. So, we have a “if and only if theorem” from these observations : $\\color{blue}{\\text{“Let S be a collection of letters. There exists a palindrome", "appear as palindromes on a digital clock. You may need to ignore the colon ($:$) in order to regard it a palindrome. For example, $\\text{2:52}$ is read backwards $\\text{25:2}$, but ignoring the colon gives you the palindrome number $252$. $$\\begin{array}{|c|c|c|c|} \\hline \\text{0:00} & ... 5 I think completely random movement is sub optimal. I think a better strategy is to pick a direction (any will do), stick to it and randomise your speed. If both parties do this their paths will cross twice each orbit (unless they are on the same orbit in which case they will meet sooner due to randomised speed) if you go full random changing direction as ... 5$$ \\begin{align} &1+2+\\cdots+2015+2016+\\cdots+4030\\\\ =\\:&S+(2015+1)+(2015+2)+\\cdots+(2015+2015)\\\\ =\\:&S+(1+2+\\cdots+2015)+(\\underbrace{2015+2015+\\cdots+2015}_{2015\\text{ times}})\\\\ =\\:&S+S+2015\\cdot2015\\\\ =\\:&2S+2015^2 \\end{align} $$If you want to find the value of S, you can do the following additional pairing:$$ \\begin{align} ... 4 Your equations are correct, so you apparently just made a mistake in solving the system. I’d multiply the second equation by $10$ to get $200=b+0.2g$ and then subtract that from the first equation to get $200=0.8g$. Then $g=250$, so $b=400-250=150$. 4 $$\\def\\r{\\color{red}{1}}\\def\\b{\\color{blue}{2}}\\def\\g{\\color{green}{3}} \\matrix{\\r\\cr \\r&\\r\\cr \\r&\\r&\\r\\cr \\r&\\r&\\r&\\r\\cr \\r&\\r&\\r&\\g&\\g\\cr \\r&\\r&\\r&\\g&\\g&\\g\\cr \\b&\\b&\\b&\\g&\\g&\\g&\\g\\cr \\b&\\b&\\b&\\g&\\g&\\g&\\g&\\g\\cr}$$ can be rearranged as ... 3 Are you allowed to leave your 'footsteps'? I believe it makes the problem more interesting (otherwise the solution of MvG seems to", "non-prime numbers. There are only three: 1, 4 and 6. Aha! So, not only are all invalid palindromes above 6 odd, they are all prime as well! (Assuming this continues beyond 10,000. I have no proof it does, but so far it works.)", "total. Kudos [?]: 41 [2], given: 46 Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 627 Kudos [?]: 1387 [7], given: 136 ### Show Tags 06 Sep 2013, 06:04 7 KUDOS 2 This post was BOOKMARKED arakban99 wrote: A palindrome is a number which reads the same when read forward as it it does when read backward. So, how many 5 digit palindromes are there? (A)720 (B)800 (C)890 (D)900 (E)950 A 5 digit palindrome would look like this : $$abcba.$$ Thus, the first digit can be filled in 9 ways(1-9), the second digit can be filled in 10 ways (0-9) and the 3rd digit can again be filled in 10 ways. The last 2 digits would just mirror the already selected digit. Thus, the no of ways : 9*10*10*1*1 = 900. D. _________________ Kudos [?]: 1387 [7], given: 136 Intern Joined: 03 Aug 2013 Posts: 15 Kudos [?]: 41 [1], given: 46 ### Show Tags 06 Sep 2013, 06:04 1 KUDOS Here is the official solution: there five digit places __ __ __ __ __ The last two depend on the first two (because the last to is the flipped version of the first two digits i.e 54145 - 45 is the flipped version of 54 so the number of ways to select the last two is 1*1", "form the palindromes 1, 121, 12321, 1234321, ... The only known non-palindromic number whose cube is a palindrome is 2201, and it is a conjecture the fourth root of all the palindrome fourth powers are a palindrome with 100000...000001 (10n + 1). G. J. Simmons conjectured there are no palindromes of form ''n''''k'' for ''k'' > 4 (and ''n'' > 1). # Other bases Palindromic numbers can be considered in numeral system A numeral system (or system of numeration) is a writing system for expressing numbers; that is, a mathematical notation for representing number A number is a mathematical object used to count, measure, and label. The original example ... s other than decimal The decimal numeral system (also called the base-ten positional numeral system and denary or decanary) is the standard system for denoting integer and non-integer numbers. It is the extension to non-integer numbers of the Hindu–Arabic numeral ... . For example, the binary Binary may refer to: Science and technology Mathematics * Binary number A binary number is a number expressed in the base-2 numeral system or binary numeral system, a method of mathematical expression which uses only two symbols: typical ... palindromic numbers are those with the binary representations: :0, 1, 11, 101, 111, 1001, 1111, 10001, 10101, 11011, 11111, 100001, ... or in", "#### Quantity A The number of primes that are divisible by 9 #### Quantity B The number of primes that are divisible by 19 If a, b, x, and y are positive integers, and $13^a \\times 13^b=(13^x)^y=13^{13}$, what is the average (arithmetic mean) of a, b, x, and y? _____ A rectangle is drawn in a standard xy-coordinate plane. If the sides of the rectangle are not parallel to the axes, what is the product of the slope of the four sides? In a certain state, each license plate consists of either three digits (between 0 and 9, inclusive) followed by two letters or three letters followed by two digits. For example, 055-XY, 123-PP, and AAA-70 are all acceptable plates. How many different license plates can the state issue? _____ n is an even integer #### Quantity A The number of prime factors of n #### Quantity B The number of prime factors of $\\frac{n}{2}$ A positive integer is a palindrome if it reads exactly the same from right to left as it does from left to right. For example, 5 and 66 and 373 are all palindromes. How many palindromes are there between 1 and 1,000, inclusive? _____ Line l passes through points in both quadrants II and III. Which of the following statements are true? Indicate", "= 7 7 = 1 + 5 + 1 7 = 2 + 3 + 2 7 = 3 + 1 + 3 7 = 1 + 1 + 3 + 1 + 1 7 = 2 + 1 + 1 + 1 + 2 7 = 1 + 2 + 1 + 2 + 1 7 = 1 + 1 + 1 + 1 + 1 + 1 + 1 In total, there are 8 palindromic partitions.", "# Decimal/hex palindromes: why multiples of 53? A previous question (371 = 0x173 (Decimal/hexidecimal palindromes?)) described numbers whose decimal and hex representations are reversed of each other. Other than the trivial one-digit numbers, there are only 5 numbers with this property. My question: why are they all multiples of 53? $$53 = 35_{16} = 53 \\times 1$$ $$371 = 173_{16} = 53 \\times 7$$ $$5141 = 1415_{16} = 53 \\times 97$$ $$99481 = 18499_{16} = 53 \\times 1877$$ $$8520280 = 0820258_{16} = 53 \\times 160760$$ The palindrome condition given is a special case of a sequence of positive integers $\\{a_i\\}_{i=0}^n$ that satisfies $$N = \\sum_{i=0}^n a_i 10^i = \\sum_{i=0}^n a_i 16^{n-i}$$ In this case it follows that $$a_0 = \\frac{\\sum_{i=1}^n a_i\\left(10^i-16^{n-i}\\right)}{16^n-1}$$ Then we can write \\begin{align} N & = a_0 + \\sum_{i=1}^n a_i 10^i \\\\ & = \\sum_{i=1}^n a_i \\left(10^i+\\frac{10^i-16^{n-i}}{16^n-1}\\right) \\\\ & = \\frac{1}{16^n-1} \\sum_{i=1}^n a_i \\left(16^n10^i-16^{n-i}\\right) \\\\ & = \\frac{1}{16^n-1} \\sum_{i=1}^n a_i 16^{n-i}\\left(160^i-1\\right) \\end{align} Since $159 \\mid 160^i-1$ for every $i\\ge 0$, and $\\operatorname{ord}_{53} 16 = \\operatorname{ord}_{53} 10 = 13$, then if $13 \\nmid n$ we must have $53 \\mid N$. Of course this generalizes to bases $B_1,B_2$. If $$N = \\sum_{i=0}^n a_i B_1^i = \\sum_{i=0}^n a_i B_2^{n-i}$$ then we can write $$N = \\frac{1}{B_2^n-1}\\sum_{i=1}^n a_i B_2^{n-i} \\left((B_1B_2)^i-1\\right)$$ and similarly $$N = \\frac{1}{B_1^n-1}\\sum_{i=0}^{n-1} a_i B_1^i \\left((B_1B_2)^{n-i}-1\\right)$$ and", "trial division into $${\\color{black}{n/2^{k}\\cdot p_2^{k_2} \\cdots \\cdot p_{i-1}^{k_{i-1}}}}$$, repeating until $${\\color{black}{p_i}}$$ fails to divide $${\\color{black}{n/2^k\\cdot p_2^{k_2} \\cdots \\cdot p_{i-1}^{k_{i-1}}p_i^{k_i}}}$$. This process is continued for all primes $${\\color{black}{i < \\sqrt{n}}}$$. Applying this algorithm to 600851475143, we find the prime factors 71, 839, 1471, and 6857. And so, the largest prime factor of 600851475143 is $${\\color{black}{ \\boxed{6857.} }}$$ #### 4. Largest palindrome product A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 $${\\color{black}{\\times}}$$ 99. Find the largest palindrome made from the product of two 3-digit numbers. This is exercise is another computational problem. The product of two 3-digit numbers is either 5- or 6-digits. The code is easy: the interesting part is figuring out how to access the individual digits of 5- and 6-digit numbers to check for palindromicity (is that a word?) Consider 5-digit numbers in the form $${\\color{black}{n = abcde}}$$. The first digit $${\\color{black}{a}}$$ can be gotten as follows: divide $${\\color{black}{n}}$$ by $${\\color{black}{10^4}}$$ so that $${\\color{black}{n/10^4 = a.bcde}}$$. By making use of the floor function in C, we can round this down to the nearest integer: floor$${\\color{black}{(n/10^4) = a}}$$. The second digit, $${\\color{black}{b}}$$, can be gotten similarly but with an additional step. Consider floor$${\\color{black}{(n/10^3) = }}$$floor$${\\color{black}{(ab.cde) = ab}}$$. Then, from above, $${\\color{black}{10 \\times}}$$ floor$${\\color{black}{(n/10^4)", "and last) is either the sum or the difference of the digits either side of it. (6) • 17 The sum of this number’s digits is the square of its first digit. (5) • 19 2 more than a palindrome. (4) • 20 The sum of this number and 33A is greater than 10,000. (4) • 22 A non-prime number with only one (distinct) prime factor. (3) • 24 This number contains only two different digits. (6) • 27 A multiple of 13D. (5) • 28 Each digit of this number (except the first) is one less than the previous digit. (4) • 29 8 more than a palindrome. (4) • 30 A multiple of the final digit of 22A. (8) • 31 The sum of this number’s digits is 7. (3) • 36 A multiple of 9 that is less than 4D. (6) • 38 33A less than 39D. (6) • 39 Equal to 2D. (6) • 42 Equal to 18A. (4) • 44 1 more than a palindrome. (4) • 46 A factor of 14A. (2) • 49 The first digit of this number is one of the digits of 45A. (2) • 50 More than 8D. (2) • ### Rewatch the Issue 12 virtual launch event Now available to catch up on YouTube • ###", "once again there are 90. For five-digit palindromes, we start with our three-digit palindromes (90 of them) and we can insert one of ten digits into the middle. This gives us 900. The comparison of 6-digit to 5-digit is the same as 4 to 3. That is, we add two digits but they must be the same, so we’ll end up with 900 again. The pattern becomes obvious pretty quickly • 2 digits – 9 palindromic numbers • 3 digits – 90 • 4 digits – 90 • 5 digits – 900 • 6 digits – 900 • 7 digits – 9000 • 8 digits – 9000 This pattern is $x=9\\times10^{\\frac{d-2}{2}}$ if d is even or $x=9\\times10^{\\frac{d-1}{2}}$ if d is odd. (You nerds out there, tell me if there’s a better way of writing different-but-related odd/even functions) So substituting $d=351$ into $x=9\\times10^{\\frac{d-1}{2}}$ gives • $x=9\\times10^{\\frac{351-1}{2}}$ • $x=9\\times10^{\\frac{350}{2}}$ • $x=9\\times10^{175}$ That’s 90,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000 numbers. How far apart are the closest ones? Well, if we change any number in the first 175 digits, we have to make one on the symmetrical digit. The only way we can change just one digit is if it’s the middle one. In a more simple situation, lets look at some five digit numbers. • 55155 • 55255 • 55355 • 55455 These numbers", "base to have a palindrome containing all the digits in its base, for bases up to 100. All base 2 palindromes contain both of its 2 numerals, as do, it appears, about 22% of base 3 palendromes, but after base 5, there don’t seem to be any for small (<10000) integers. The numerological specialness of numeral representation of [imath]\\left( \\frac{1}{137} \\right) _{10}[/imath] and [imath]\\left( \\frac{1}{898} \\right) _5[/imath] depends as much on the number 10 and 5 as they do on the numbers 137 and 898. Since the choice of base for a given numeral system, or even the choice of regular-exponent-of-a-base numeration systems, is arbitrary and cultural (in our case, almost certainly due to us having 10 fingers), the numerological specialness of numbers like these has as much to do with our culture as with fundamental qualities of nature or numbers. ##### Share on other sites Quoting CraigD, Without that thrill akin to love that comes from intuiting patterns and formally teasing/proving them out.... Yeah! If there is no challenge, then life quickly becomes very boring. This is true even when it comes to movies and other forms of entertainment. Imagine a James Bond flick in which .007 finally retires and the entire film is about him just relaxing by the pool and drinking vodka martinis \"shaken, not", "# 32. Longest Double Palindrome ## I'm a slow walker, but I never walk backwards. We define a palindrome as a sequence of at least one integer that reads the same either forward or backward. For example 1 3 5 3 1 and 1 2 2 1 are both palindromes. To the extreme case 1 is also a palindrome. We also define a double palindrome as the concatenation of two palindromes. For example 1 3 5 3 1 1 2 2 1 is a double palindrome. To the extreme case 1 3 5 3 1 1 is also a double palindromes. Now given a sequence of at least two integers, please find the longest double palindrome. If there are multiple longest double palindromes, print the one that appears last . ## Limits The number of integers in the given sequence is no more than 100. ## Input Format There is one line in the input. The first line has the integers from which we want to find the longest double palindrome. ## Output Format There is one lines in the output. The first line has all the integers of the longest double palindrome. number of ways as in the description. ## Sample Input 7 1 3 5 3 1 1 ## Sample Output 1 3 5 3 1", "the non-palindromes finally generate using the method of division by largest prime factor, a palindrome other than 1? - The remainder of dividing a number by 9 is the same as dividing the sum of its digits by 9. This is one of the classic divisibility rules. The sum of digits is the same after you reverse the numbers, so the difference of the sums of digits will be zero. The difference of the numbers is therefore divisible by 9. - The key fact is that $9 \\vert 10^a - 10^b$ for all $a$ and $b$. In particular, $9 \\vert (10^k-1)$ for all $k$. Any $n+1$ digit number is of the form $$s= a_n 10^n + a_{n-1} 10^{n-1} + \\cdots + a_0$$ The new number is given by $$s_{\\text{new}} = a_0 10^n + a_{n-1} 10^{n-1} + \\cdots + a_{1} 10 + a_n$$ Hence, \\begin{align} s - s_{\\text{new}} & = a_n (10^n-1) + a_0 (1-10^n) = (10^{n}-1)(a_n-a_0) \\end{align} Hence, $9 \\vert (s-s_{\\text{new}})$. Note that $\\vert a_n - a_0 \\vert = d \\in \\{0,1,2,\\ldots,9\\}$ and $\\dfrac{10^n-1}{9} = \\underbrace{111\\ldots 11}_{n \\text{ ones}}$. Hence, if $\\vert a_n - a_0 \\vert = d$, then $$\\vert s - s_{\\text{new}} \\vert = \\underbrace{ddd\\ldots dd}_{n \\text{ d's}}$$ which is indeed more than a palindrome. EDIT In fact, the first half of your question is true in", "So this the total of ways to select (combination) the digits: 9 (digits 1-9, remember 0 at the start will make the number a four digit number) * 10 (0-9) * 10 (0-9) * 1 *1 = 900. Hope it helped _____________ Comments and KUDOS are deeply appreciated Last edited by arakban99 on 06 Sep 2013, 06:07, edited 1 time in total. Kudos [?]: 41 [1], given: 46 Intern Joined: 03 Aug 2013 Posts: 15 Kudos [?]: 41 [0], given: 46 ### Show Tags 06 Sep 2013, 06:06 mau5 wrote: arakban99 wrote: A palindrome is a number which reads the same when read forward as it it does when read backward. So, how many 5 digit palindromes are there? (A)720 (B)800 (C)890 (D)900 (E)950 A 5 digit palindrome would look like this : $$abcba.$$ Thus, the first digit can be filled in 9 ways(1-9), the second digit can be filled in 10 ways (0-9) and the 3rd digit can again be filled in 10 ways. The last 2 digits would just mirror the already selected digit. Thus, the no of ways : 9*10*10*1*1 = 900. D. Yep, that's my solution too. Kudos [?]: 41 [0], given: 46 Math Expert Joined: 02 Sep 2009 Posts: 42257 Kudos [?]: 132703 [1], given: 12335 ### Show Tags 06 Sep 2013, 07:33" ]

[ "# Find the Coefficient Find the coefficient of $$x^4$$ in the expansion of $\\large (2-x+3x^2)^6.$ ×", "Find the coefficient of $$x^{9}$$ in the expansion of $(2+x)^{10}(1+x)^0+(2+x)^{9}(1+x)+(2+x)^{8}(1+x)^{2}+ \\ldots + (2+x)^0(1+x)^{10}.$", "+0 # What is the coefficient of x^3 in the expansion of (x+2*sqrt3)^7? 0 48 1 What is the coefficient of x^3 in the expansion of (x+2*sqrt3)^7? Jul 16, 2021 #1 +114112 +1 $$(x+2\\sqrt3)^7\\\\~\\\\ 7C3*(2\\sqrt3)^4(x)^3$$ That is the x^3 tern. You can take it from there. Jul 16, 2021", "# Coefficient In A Septic Expansion What is the coefficient of $$x^3$$ in the expansion of $$(x+2)^7$$? ×", "# Challenge you What is the coefficient of $$x^{7}y^{3}z^{2}$$ in the expansion of $$(x+2y+3z)^{12}$$. ×", "Free Version Moderate # Coefficient in $(5x-2y)^{7}(5x+2y)^{7}$ COMBIN-QZDN7G What is the coefficient of $x^8y^{6}$ in the expansion of: $$(5x-2y)^{7}(5x+2y)^{7}?$$ A $25,000,000$ B $875,000,000$ C $14,000,000,000$ D $56,000,000,000$ E none of the above F", "# Coefficient play The number of terms in the expansion of $$(3x-4y+6z)^4$$ is $$m$$ and the coefficient of $$xy^2z$$ in the same expansion is $$n$$. Find $$m+n$$. ×", "+0 # Hapl Asap!!! 0 30 1 What is the coefficient of $x^3$ in the expansion of $$(x+2\\sqrt3)^7?$$ Apr 26, 2021 ### 1+0 Answers #1 +274 +3 5040 since when we expand it it is 5040 Apr 26, 2021 edited by Dennis070sinneD Apr 26, 2021", "# The coefficient of $x^3y^4z^5$ in the expansion of $(xy+yz+xz)^6$ is : $\\begin {array} {1 1} (a)\\;70 & \\quad (b)\\;60 \\\\ (c)\\;50 & \\quad (d)\\;none\\;of\\;these \\end {array}$ (b) 60", "× # Expansion What would be the coefficient of $$x^t$$ in the expansion of: $$(1+x+x^2+\\cdots+x^r)^n$$ if $$t>r$$? Is there any general formulae? Note by Rahul Vernwal 1 year, 3 months ago", "+0 # Hapl Asap!!! 0 137 1 What is the coefficient of $x^3$ in the expansion of $$(x+2\\sqrt3)^7?$$ Apr 26, 2021", "# The coefficient of the $8^{th}$ term in expansion of $(1+x)^{10}$ is equal to : ( A ) $210$ ( B ) $120$ ( C ) $7$ ( D ) $^{10}C_3$", "coefficient of x^3.", "Free Version Moderate # Coefficient in $(2a+b-3c)^5$ COMBIN-1VSGQG What is the coefficient of $a^2bc^2$ in the expansion of: $$(2a+b-3c)^5?$$ A $0$ B $243$ C $1024$ D $1080$ E $7776$ F none of the above", "# Submitted by Mohammad ElKammash Find the coefficient of $$x^{9}$$ in the expansion of $(2+x)^{10}(1+x)^0+(2+x)^{9}(1+x)+(2+x)^{8}(1+x)^{2}+ \\ldots + (2+x)^0(1+x)^{10}.$ ×", "in the expansion is $\\binom{2015}{2014} x (x^2+1)^{2014}$ and hence the coefficient is $2015+1=2016$", "+0 # Hapl Asap!!! 0 85 1 What is the coefficient of $x^3$ in the expansion of $$(x+2\\sqrt3)^7?$$ Apr 26, 2021", "# The coefficient of x^17in the expansion of(x-1)(x-2)(x-3)….(x-18) is Updated On: 17-04-2022 Get Answer to any question, just click a photo and upload the photo and get the answer completely free,", "# Expand? What is the coefficient of $$x^{5}$$ in the polynomial expansion $$(2-x-x^{2})^{4}$$ ? ×", "The coefficient of x in the expansion of Question: The coefficient of $x$ in the expansion of $(x+3)^{3}$ is (a) 1 (b) 9 (c) 18 (d) 27 Solution: $(x+3)^{3}$ $=x^{3}+3^{3}+9 x(x+3)$ $=x^{3}+27+9 x^{2}+27 x$ So, the coefficient of $x$ in $(x+3)^{3}$ is 27 . Hence, the correct answer is option (d)." ]

[ "$\\sqrt{51} = \\sqrt{49+2} = 7\\sqrt{1+(2/49)}$. Use the binomial expansion of $(1+x)^{1/2}$ with x = 2/49. The terms will get small quite rapidly, so you should not need very many to get within 4dp accuracy. (Don't forget to multiply by 7 at the end to get the correct answer.)", "## Algebra 2 (1st Edition) $$3125x^5-3125x^4+1250x^3-250x^2+25x-1$$ Using the binomial expansion theorem, we can expand the binomial without multiplying it out. Doing this, we find: $$3125x^5-3125x^4+1250x^3-250x^2+25x-1$$", "Comment Share Q) # Find the expansion of $(3x^2-2ax+3a^2)^3$ using Binomial theorem Comment A) Toolbox: • $(a-b)^n=nC_0a^n-nC_1a^{n-1}b+nC_2a^{n-2}b^2-....+(-1)^nnC_ra^{n-r}b^r+.....+nC_n(-b)^n$ $[3x^2-a(2x-3a)]^3=(3x^2)^3-3C_1(3x^2)^2.a(2x-3a)+3C_2(3x^2)a^2(2x-3a)^2-a^3(2x-3a)^3$ $\\Rightarrow 27x^6-27x^4a(2x-3a)+9x^2a^2(4x^2-12ax+9a^2)-a^3(8x^3-3\\times 4\\times x^2\\times 3a+3\\times 2x\\times 9a^2-27a^3)$ $\\Rightarrow 27x^6-54x^5a+81a^2x^4-108a^3x^3+81a^4x^2-8a^3x^3+36a^4x^2-54a^5x+27a^6$ $\\Rightarrow 27x^6-54x^5a+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6$", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $7,838,208\\sqrt2u^{5}$ The $r+1$ th term in the binomial expansion of $(x+y)^n$ is: $\\binom{n}{r}x^{n-r}y^r$ Here $x=2u$, $y=3\\sqrt{2}$, $n=10$: There are 11 terms in this expansion, the middle one is the 6th. $r+1=6$ $r=5$ The indicated term is: $\\binom{10}{5}(2u)^{10-5}(3\\sqrt{2})^5=252\\times2^5\\times u^{5}\\times 972\\sqrt2=7,838,208\\sqrt2u^{5}$", "probability that a binomial($10,.3$) is 5 larger than a binomial($10,.5$) – Macro Aug 24 '11 at 17:34", "Browse Questions Find the expansion of $(3x^2-2ax+3a^2)^3$ using Binomial theorem Toolbox: • $(a-b)^n=nC_0a^n-nC_1a^{n-1}b+nC_2a^{n-2}b^2-....+(-1)^nnC_ra^{n-r}b^r+.....+nC_n(-b)^n$ $[3x^2-a(2x-3a)]^3=(3x^2)^3-3C_1(3x^2)^2.a(2x-3a)+3C_2(3x^2)a^2(2x-3a)^2-a^3(2x-3a)^3$ $\\Rightarrow 27x^6-27x^4a(2x-3a)+9x^2a^2(4x^2-12ax+9a^2)-a^3(8x^3-3\\times 4\\times x^2\\times 3a+3\\times 2x\\times 9a^2-27a^3)$ $\\Rightarrow 27x^6-54x^5a+81a^2x^4-108a^3x^3+81a^4x^2-8a^3x^3+36a^4x^2-54a^5x+27a^6$ $\\Rightarrow 27x^6-54x^5a+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6$", "$x$, if you're really going to use it to obtain the Maclaurin expansion (by expanding the binomial and integrating termwise). – Hans Lundmark Nov 23 '17 at 10:53", "# Find the expansion of $(3x^2-2ax+3a^2)^3$ using Binomial theorem Toolbox: • $(a-b)^n=nC_0a^n-nC_1a^{n-1}b+nC_2a^{n-2}b^2-....+(-1)^nnC_ra^{n-r}b^r+.....+nC_n(-b)^n$ $[3x^2-a(2x-3a)]^3=(3x^2)^3-3C_1(3x^2)^2.a(2x-3a)+3C_2(3x^2)a^2(2x-3a)^2-a^3(2x-3a)^3$ $\\Rightarrow 27x^6-27x^4a(2x-3a)+9x^2a^2(4x^2-12ax+9a^2)-a^3(8x^3-3\\times 4\\times x^2\\times 3a+3\\times 2x\\times 9a^2-27a^3)$ $\\Rightarrow 27x^6-54x^5a+81a^2x^4-108a^3x^3+81a^4x^2-8a^3x^3+36a^4x^2-54a^5x+27a^6$ $\\Rightarrow 27x^6-54x^5a+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6$", "knew the coefficient of the $x^2$ term. Using this expansion suggests that we should choose x so that $2-\\frac{x}{8}=1.995$, that is, $0.005=\\frac{x}{8}$ or x=0.04. Substituting x=0.04 into the expansion gives $(1.995)^8\\approx256-256\\times 0.04+28\\times 0.04^2=245.8048$ The actual answer to 7 decimal places, using a calculator, is 250.9245767, so not a great approximation. ## Binomial Expansion Board Question Note that C2 was a module when A-Level Maths was delivered in a modular fashion. Binomial Expansion is now an AS-Level topic. ## Statistics Example Consider the binomially distributed random variable $X\\sim B(7,x)$. Find the probability $P(X=3)$ in terms of x. Write your answer as a polynomial in x. Using the formula: $P(X=3)={{7}\\choose{3}}x^3(1-x)^4$ We can use Example 2 above to expand $(1-x)^4$: $P(X=3)=35x^3\\left(1-4x+6x^2-4x^3+x^4\\right)$ $=35x^3-140x^4+210x^5-140x^4+35x^3$ Click here to find Questions by Topic and scroll down to all past BINOMIAL EXPANSION exam questions to practice some more.", "16 x4 + 96x3 +216x2 + 216x + 81 Example 4 Find the expansion of (2x − y)4 Solution (2x − y)4 = (2x) + (−y)4 = (2x)4 + 4(2x)3 (−y) + 6(2x)2(−y)2 + 4(2x) (−y)3+ (−y)4 = 16x4 − 32x3y + 24x2y2 − 8xy3 + y4 Example 5 Use the Binomial Theorem to expand (2 + 3x)3 Solution By comparing with the Binomial formula, a = 2; b = 3x and n = 3 ⟹ (2 + 3x) 3 = 23 + (31) 22(3x)1 + (32) 2(3x)2 + (3x)3 = 8 + 36x + 54x2 + 27x3 Example 6 Expand (x2 + 2)6 Solution (x2 +2)6 = 6C(x2)6(2)0 + 6C1(x2)5(2)1 + 6C2(x2)4(2)2 + 6C(x2)3(2)3 + 6C(x2)2(2)4 + 6C(x2)1(2)5 + 6C(x2)0(2)6 = (1) (x12) (1) + (6) (x10) (2) + (15) (x8) (4) + (20) (x6) (8) + (15) (x4) (16) + (6) (x2) (32) + (1)(1) (64) = x12 + 12 x10 + 60 x8 + 160 x6 + 240 x4 + 192 x2 + 64 Example 7 Expand the expression (√2 + 1)5 + (√2 − 1)5 using the Binomial formula. Solution (x + y)5 + (x – y)5 = 2[5C0 x5 + 5C2 x3 y2 + 5C4 xy4] = 2(x+ 10 x3 y+ 5xy4) = (√2 + 1)+ (√2 − 1)= 2[(√2)+ 10(√2)3(1)+ 5(√2) (1)4]", "the expansion = #### Question 19: If ${T}_{2}/{T}_{3}$ in the expansion of in the expansion of ${\\left(a+b\\right)}^{n+3}$ are equal, then n = (a) 3 (b) 4 (c) 5 (d) 6 (c) 5 #### Question 20: The coefficient of $\\frac{1}{x}$ in the expansion of is (a) (b) (c) (d) none of these (b) #### Question 21: If the sum of the binomial coefficients of the expansion ${\\left(2x+\\frac{1}{x}\\right)}^{n}$ is equal to 256, then the term independent of x is (a) 1120 (b) 1020 (c) 512 (d) none of these (a) 1120 #### Question 22: If the fifth term of the expansion does not contain 'a'. Then n is equal to (a) 2 (b) 5 (c) 10 (d) none of these (c) 10 #### Question 23: The coefficient of ${x}^{-3}$ in the expansion of ${\\left(x-\\frac{m}{x}\\right)}^{11}$ is (a) (b) (c) (d) (d) #### Question 24: The coefficient of the term independent of x in the expansion of ${\\left(ax+\\frac{b}{x}\\right)}^{14}$ is (a) (b) (c) (d) $\\frac{14!}{{\\left(7!\\right)}^{3}}{a}^{7}{b}^{7}$ (c) #### Question 25: The coefficient of x5 in the expansion of (a) 51C5 (b) 9C5 (c) 31C621C6 (d) 30C5 + 20C5 (c) 31C621C6 #### Question 26: The coefficient of x8y10 in the expansion of (x + y)18 is (a) 18C8 (b) 18p10 (c) 218 (d) none of these (a) 18C8 #### Question 27: If the coefficients of", "# Binomal Expansion ## Homework Statement Expand in ascending powers of x upto and including the term x³: $$3(1-3x)^{-1} + 4(2+x)^{-2}$$ ## The Attempt at a Solution Err never done it with 2 terms like above before. Any suggestions/hints Thanks :) expand $(1-3x)^{-1}$ up to the term in $x^3$ and expand $(2+x)^{-2}$ as well up to the term in $x^3$, the just add them.", "expression, especially when the powers are high e.g. (1 − 2x)14 #### Tekan Revision Exercise Find the term independent of x in the expansion . #### [+] B. Expand & Substitute Instead of ‘Expand & Multiply’ describe above, this time you expand and substitute a suitable value/expression into your expansion in order to expand a more complicated expression, estimate a value, and achieve world peace. Expand (1 − p)5. Use this result to find the expansion of (1 − x + x2)5 as far as the term in x3. Use a suitable value of x to estimate (1.11)5 from this expansion. ANS: Using the Binomial Expansion formula given in your formula sheet, (1 − p)5 = 15 + 5C1(14)(−p) + 5C2(13)(−p)2 + 5C3(12)(−p)3 + 5C4(12)(−p)4 + (−p)5 = 1 − 5p + 10p2 − 10p3 + 5p4p5 ##### Do the Substitution! To make (1 − x + x2)5 be like (1 − p)5, we equate 1 − p with 1 − x + x2: ⇒ 1 − p = 1 − x + x2 p = xx2 → our substitute expression! So sub p = xx2 into the expansion: = 1 − 5(xx2) + 10(xx2)2 − 10(xx2)3 + 5(xx2)4 − (xx2)5 → we’re only interested in terms up to x3, so we ignore all combos that result in", "expansion looks nice. Seems like we are dealing with product of terms here? Shall we compress it again? 10$$\\equiv \\prod_{i=0}^{x}(1+X)^{n_iP^i} \\mod P$$Not bad! We are getting somewhere. We know that $a^{bc} = \\left(a^b\\right)^c$ 11$$\\equiv \\prod_{i=0}^{x}\\left((1+X)^{P^i}\\right)^{n_i} \\mod P$$Hey! Is that $(1+X)^{P^i}$ ? Great! We can apply “Extended Lucas Polynomial Expansion Lemma” here 12$$\\equiv \\prod_{i=0}^{x}\\left(1+X^{P^i}\\right)^{n_i} \\mod P$$So far so good. How about we let $Y = X^{P^i}$ and try to expand again! 13$$\\equiv \\prod_{i=0}^{x}\\left(1+Y\\right)^{n_i} \\mod P$$Well, we know that $(1+Y)^a = \\sum_{j=0}^{a}\\binom{a}{j}Y^j$ 14$$\\equiv \\prod_{i=0}^{x}\\left( \\sum_{j=0}^{n_i}\\binom{n_i}{j}Y^j \\right)\\mod P$$ Now, in order to move forward we have to understand an important fact: what is $\\sum_{j=0}^{n_i}\\binom{n_i}{j}Y^j$ ? It’s just the polynomial expansion of $(1+Y)^{n_i}$ right? Let’s write it down in more simplified form. $\\binom{n_i}{0}Y^0+\\binom{n_i}{1}Y^1+ \\dots + \\binom{n_i}{n_i}Y^{n_i}$ Now, what is $n_i$? I hope you didn’t forget that. $n_i$ is simply a digit of $N$ is its $P$-base form. Since it’s a digit of a number in $P$ base form, we know that $0 \\leq n_i < P$. $\\therefore P-1 \\geq n_i$ and there is no doubt right? Then can we say that… $$\\sum_{j=0}^{n_i}\\binom{n_i}{j}Y^j = \\sum_{j=0}^{P-1}\\binom{n_i}{j}Y^j$$ Note that only the upper limit of the summation has changed from $n_i$ to $P-1$. Does that change anything at all? Previously in the original summation, $j$ was suppose to stop at $n_i$, but now, if $P-1$ is", "in your first step). You get $(-1)^s \\binom{k+s}{s}$. 3. This is called Generalized Binomial expansion: $(1+a)^{-r} = \\sum_{k=0}^{\\infty} \\binom{-r}{k} a^k$. This is valid for $|a|<1$.", "# Binomal Expansion 1. Jun 9, 2008 ### thomas49th 1. The problem statement, all variables and given/known data Expand in ascending powers of x upto and including the term x³: $$3(1-3x)^{-1} + 4(2+x)^{-2}$$ 2. Relevant equations 3. The attempt at a solution Err never done it with 2 terms like above before. Any suggestions/hints Thanks :) 2. Jun 9, 2008 ### rock.freak667 expand $(1-3x)^{-1}$ up to the term in $x^3$ and expand $(2+x)^{-2}$ as well up to the term in $x^3$, the just add them.", "Browse Questions # Find the expansion of $(3x^2-2ax+3a^2)^3$ using Binomial theorem Can you answer this question? Toolbox: • $(a-b)^n=nC_0a^n-nC_1a^{n-1}b+nC_2a^{n-2}b^2-....+(-1)^nnC_ra^{n-r}b^r+.....+nC_n(-b)^n$ $[3x^2-a(2x-3a)]^3=(3x^2)^3-3C_1(3x^2)^2.a(2x-3a)+3C_2(3x^2)a^2(2x-3a)^2-a^3(2x-3a)^3$ $\\Rightarrow 27x^6-27x^4a(2x-3a)+9x^2a^2(4x^2-12ax+9a^2)-a^3(8x^3-3\\times 4\\times x^2\\times 3a+3\\times 2x\\times 9a^2-27a^3)$ $\\Rightarrow 27x^6-54x^5a+81a^2x^4-108a^3x^3+81a^4x^2-8a^3x^3+36a^4x^2-54a^5x+27a^6$ $\\Rightarrow 27x^6-54x^5a+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6$ answered Jun 24, 2014", "326 c) 1320 d) 456 Explanation: The coefficient of the 8th term is 11C8 = 165. Hence, the 8th term of the expansion is 165 * 23 * x8 = 1320x8, where the coefficient is 1320. 8. Determine the coefficient of the x5y7 term in the polynomial expansion of (m+n)12. a) 792 b) 439 c) 382 d) 630 Explanation: Note that, the “x” in the binomial has to be chosen 5 times out of 12. Thus, the coefficient of the term x5y7 must be equal to the number of combinations of 5 objects out of 12: 12C5 = 792. 9. The last digit of the number (($$\\sqrt{51}$$ + 1)51 – $$\\sqrt{51}$$ – 1)51 is _______ a) 32 b) 8 c) 51 d) 1 Explanation: Consider the binomial expansion of (m+1)71 and (m-1)71 which gives these two expressions below respectively: 1) m51 + 51C1m50 + 51C2m49 + 51C3m48 + … + 51C50m1 + 51C51m0 2) m5151C1m50 + 51C2m4951C3m48 + … + 51C50m151C51m0 . By taking the difference we have, 2(51C1m50 + 51C3m4851C5m46 + … + 51C50m251C51m0 ). In this case, m = $$\\sqrt{51}$$ and 2(51C1m50 + 51C3m4851C5m46 + … + 51C50m251C51m0 ). Consider, module 10 on the powers(for any natural number n): (51)n ≡ (51 mod 10n) ≡ 1 gives 2(51C1 + 51C3 + 51C5 + … + 51C50", "$(3-x)^{4}$. Hence find the coefficient of $x^{2}$ in the expansion of $(1+2 x)^{5}(3-x)^{4}$. $\\mbox{ (5 marks)}$ 2.Using binomial theorem, find the coefficient of $x^{2}$ in the expansion of $\\left(3+x-2 x^{2}\\right)^{5}$. $\\mbox{ (5 marks)}$ 3.If the coefficient of $x^{2}$ in the expansion of $(4+k x)(2-x)^{6}$ is 384 , find the value of $k$ and the coefficient of $x^{3}$ in that expansion. $\\mbox{ (5 marks)}$ 4.Find the term independent of $x$ in the expansion of $\\left(5 x+\\frac{5}{x}\\right)^{6}\\left(5 x-\\frac{5}{x}\\right)^{6}$. $\\mbox{ (5 marks)}$ 5.In the expansion of $\\left(x^{2}-\\frac{2}{x}\\right)^{n}$ in descending power of $x$, the $5^{\\text {th }}$ term is independent of $x .$ Find the value of $n$ and the $4^{\\text {th }}$ term. $\\mbox{ (5 marks)}$ 6.In the binomial expansion of $\\left(1+\\frac{1}{4}\\right)^{n}$, the third term is twice the fourth term. Calculate the value of $n$. $\\mbox{ (5 marks)}$ 7.In the expansion of $\\left(1+\\frac{1}{3}\\right)^{n}$, the third term and the fourth term are in the ratio $3: 2$. Find the value of $n$ and the middle term of that expansion. $\\mbox{ (5 marks)}$ 8.If the $2^{\\text {nd }}$ and the $3^{r d}$ terms in $(a+b)^{n}$ are in the same ratio as the $3^{\\text {rd }}$ and the $4^{\\text {th }}$ terms in $(a+b)^{n+3}$, find the value of $n .$ Calculate also the middle term of $(a+b)^{n+3}$. $\\mbox{ (5 marks)}$ $\\quad\\;$$\\, 1.(1+2 x)^{5}=1+10 x+40", "there in this expansion? (b) Find the constant term in this expansion. (Total 6 marks) 18.) Find the coefficient of $\\D x^3$ in the expansion of $\\D (2- x)^5.$ (Total 6 marks) 19.) Use the binomial theorem to complete this expansion. $\\D (3x + 2y)^4 = 81x^4 + 216x^3 y + \\cdots$ (Total 4 marks) 20.) Consider the binomial expansion . $(1+x)^4=1+\\iixi{4}{1}x+ \\iixi{4}{2}x^2 +\\iixi{4}{3}x^3+x^4.$ (a) By substituting $\\D x = 1$ into both sides, or otherwise, evaluate $\\D \\iixi{4}{1}+ \\iixi{4}{2} +\\iixi{4}{3}$ (b) Evaluate $\\D \\iixi{9}{1}+ \\iixi{9}{2}+ \\iixi{9}{3}+ \\iixi{9}{4}+ \\iixi{9}{5}+ \\iixi{9}{6}+ \\iixi{9}{7}+ \\iixi{9}{8}.$ (Total 4 marks) 21.) Determine the constant term in the expansion of $\\D \\left(x-\\frac{2}{x^2}\\right)^9.$ (Total 4 marks) 22.) Find the coefficient of $\\D a^5b^7$ in the expansion of $\\D (a + b)^{12}.$ (Total 4 marks) 23.) Find the coefficient of $\\D x^5$ in the expansion of $\\D (3x - 2)^8.$ (Total 4 marks) 24.) Find the coefficient of $\\D a^3b^4$ in the expansion of $(5a + b)^7.$ (Total 4 marks) 1. 12,28160x^2 2. 16 32 24 8 1 25x^2 3. 1080x^4 4. n=10,a=p,b=2q,10C5p^5(2q)^5 5. x^3+3x^2h+3xh^2+h^3 6. -4032x^3 7. x^4-8x^3+24x^2-32x+16;40x^3 8. 7;-540x^{12} 9. 180 10. e^4+4e^2+6+4/e^2+1/e^4 2e^4+12+2?e^4 11. 6;A=-80 12. p=90,q=34 13. 3 14. -48384x^3 15. 16800x^{10} 16. ...+24a^2x^2+8a^3x^3+a^4x^4 17. 10;2268 18. -40 19. ...+216x^2y^2+96xy^3+16y^4 20. 14;510 21. -672 22. 792 23. -108864 24. 4375" ]

[ "+0 # What is the coefficient of x^3 in the expansion of (x+2*sqrt3)^7? 0 48 1 What is the coefficient of x^3 in the expansion of (x+2*sqrt3)^7? Jul 16, 2021 #1 +114112 +1 $$(x+2\\sqrt3)^7\\\\~\\\\ 7C3*(2\\sqrt3)^4(x)^3$$ That is the x^3 tern. You can take it from there. Jul 16, 2021", "# The coefficient of $x^3y^4z^5$ in the expansion of $(xy+yz+xz)^6$ is : $\\begin {array} {1 1} (a)\\;70 & \\quad (b)\\;60 \\\\ (c)\\;50 & \\quad (d)\\;none\\;of\\;these \\end {array}$ (b) 60", "# Find the Coefficient Find the coefficient of $$x^4$$ in the expansion of $\\large (2-x+3x^2)^6.$ ×", "in the expansion is $\\binom{2015}{2014} x (x^2+1)^{2014}$ and hence the coefficient is $2015+1=2016$", "Free Version Moderate # Coefficient in $(5x-2y)^{7}(5x+2y)^{7}$ COMBIN-QZDN7G What is the coefficient of $x^8y^{6}$ in the expansion of: $$(5x-2y)^{7}(5x+2y)^{7}?$$ A $25,000,000$ B $875,000,000$ C $14,000,000,000$ D $56,000,000,000$ E none of the above F", "# Coefficient play The number of terms in the expansion of $$(3x-4y+6z)^4$$ is $$m$$ and the coefficient of $$xy^2z$$ in the same expansion is $$n$$. Find $$m+n$$. ×", "# The coefficient of $x^{32}$ in the expansion of $(x^4-\\large\\frac{1}{x^3})^{15}$ is $\\begin{array}{1 1}(A)\\;-15C_3\\\\(B)\\;15C_4\\\\(C)\\;-15C_5\\\\(D)\\;15C_2\\end{array}$", "Find the coefficient of $$x^{9}$$ in the expansion of $(2+x)^{10}(1+x)^0+(2+x)^{9}(1+x)+(2+x)^{8}(1+x)^{2}+ \\ldots + (2+x)^0(1+x)^{10}.$", "× # Expansion What would be the coefficient of $$x^t$$ in the expansion of: $$(1+x+x^2+\\cdots+x^r)^n$$ if $$t>r$$? Is there any general formulae? Note by Rahul Vernwal 1 year, 3 months ago", "# The coefficient of the $8^{th}$ term in expansion of $(1+x)^{10}$ is equal to : ( A ) $210$ ( B ) $120$ ( C ) $7$ ( D ) $^{10}C_3$", "# Challenge you What is the coefficient of $$x^{7}y^{3}z^{2}$$ in the expansion of $$(x+2y+3z)^{12}$$. ×", "of the corresponding three terms of $$(1-x)^{-6}$$ viz., $$x^{20}, x^{11}, x^2$$ respectively. Finally the coefficient is given by $$\\binom{25} {5} - 6*\\binom{16} {5} + 15*\\binom{7} {5} = 27237$$ • Thanks PTDS for helping! – Vasu090 Jul 27 at 5:50", "The coefficient of the third term in the expansion of $(x^2-\\large\\frac{1}{4})^n$,when expanded in descending powers of $x$ is $31$.Then n is equal to $\\begin{array}{1 1}(A)\\;16\\\\(B)\\;30\\\\(C)\\;32\\\\(D)\\;\\text{None of these}\\end{array}$", "marks)}$ 3. Find the coefficient of $x^{2}$ in the expansion of $\\left(2 x^{2}-x-3\\right)^{6}$. $\\quad\\mbox{ (5 marks)}$ 4. Find the coefficient of $x^{2}$ and $x^{3}$ in the expansion of $\\left(x^{2}-x-2\\right)^{7}$. $\\quad\\mbox{ (5 marks)}$ 5. Evaluate the coefficients of $x^{5}$ and $x^{4}$ in the binomial expansion of $\\left(\\frac{x}{3}-3\\right)^{7}$. $\\quad\\mbox{ (5 marks)}$ Hence evaluate the coefficient of $x^{5}$ in the expansion of $\\left(\\frac{x}{3}-3\\right)^{7}(x+6)$. $\\quad\\mbox{ (5 marks)}$ 6. In the expansion of $(1+2 x)^{11}$, the coefficient of $x^{3}$ is $k$ times the coefficient of $x^{2}$. Find $k$. $\\quad\\mbox{ (5 marks)}$ 7. If the coefficient of $x^{4}$ in the expansion of $(3+2 x)^{6}$ is equal to the coefficient of $x^{4}$ in the expansion of $(2 k+3 x)^{6}$, find $k$. $\\quad\\mbox{ (5 marks)}$ 8. The ratio of the coefficient of $x^{4}$ in the expansion of $(3-2 x)^{6}$ and the coefficient of $x^{4}$ in the expansion of $(k+2 x)^{7}$ is $1: 7.$ Find the value of $k$. $\\quad\\mbox{ (5 marks)}$ 9. The first three terms in the binomial expasion of $(a+b)^{n}$, in ascending power of $b$, are denoted by $p, q$ and $r$ respectively. Show that $\\frac{q^{2}}{p r}=\\frac{2 n}{n-1}$. Given that $p=4, q=32$ and $r=96$, evaluate $n$. $\\quad\\mbox{ (5 marks)}$ 10. In the binomial expansion of $(1+x)^{n}$ the first three terms are $1+3+4+\\ldots$ Calculate the numerical values of $n$ and $x$, and the value of", "+0 # Hapl Asap!!! 0 137 1 What is the coefficient of $x^3$ in the expansion of $$(x+2\\sqrt3)^7?$$ Apr 26, 2021", "# The coefficient if $x^{10}$ in the expansion $(1+x^2-x^3)^8$ is ? $\\begin {array} {1 1} (a)\\; 476 \\\\ (b)\\; 496 \\\\ \\ (c) \\;506 \\\\ \\ (d) \\;528 \\end {array}$", "# The coefficient if $x^n$ in the expansion of $(1+x+x^2+...)^{-n}$ is $\\begin{array}{1 1}(A)\\;1\\\\(B)\\;(-1)^n\\\\(C)\\;n\\\\(D)\\;n+1\\end{array}$", "# The largest coefficient in the expansion of $(1+x)^{30}$ is __________ $\\begin{array}{1 1}(A)\\;30C_{15}\\\\(B)\\;30C_{10}\\\\(C)\\;30C_{14}\\\\(D)\\;30C_5\\end{array}$", "Free Version Easy # Coefficient in $(5x-2y)^7$ COMBIN-XIIW4G What is the coefficient of $x^3y^4$ in the expansion of $(5x-2y)^7$? A $725$ B $32$ C $126$ D $252$ E $1024$", "The coefficient of x in the expansion of Question: The coefficient of $x$ in the expansion of $(x+3)^{3}$ is (a) 1 (b) 9 (c) 18 (d) 27 Solution: $(x+3)^{3}$ $=x^{3}+3^{3}+9 x(x+3)$ $=x^{3}+27+9 x^{2}+27 x$ So, the coefficient of $x$ in $(x+3)^{3}$ is 27 . Hence, the correct answer is option (d)." ]

[ "+0 +1 193 2 +553 Coach Grunt is preparing the 5-person starting lineup for his basketball team, the Grunters. There are 12 players on the team. Two of them, Ace and Zeppo, are league All-Stars, so they'll definitely be in the starting lineup. How many different starting lineups are possible? (The order of the players in a basketball lineup doesn't matter.) ant101 Feb 18, 2018 #1 +88898 +1 Well ....two of the positions are \"set\" So..... we just want to choose any 3 of the remaining 10 players = C(10,3) = 120 different lineups CPhill Feb 18, 2018 #2 +139 +1 Coach Grunt has to choose 3 players from the 10 players that are remaining after Ace and Zeppo have been placed in the lineup. The order in which the players are chosen doesn't matter, so the answer is $$\\binom{10}{3} = \\frac{10 \\times 9 \\times 8}{3 \\times 2 \\times 1} = \\boxed{120}.$$ azsun Feb 18, 2018", "the coach chooses a team that includes both John and Pete = 35/126 = 5/18 Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Re: A basketball coach will select the members of a five-player [#permalink] 29 Jan 2018, 07:47 Display posts from previous: Sort by", "### Home > CCG > Chapter 10 > Lesson 10.3.3 > Problem10-143 10-143. Akio coaches the girls volleyball team. He needs to select players for the six different starting positions from his roster of $16$ players. On Akio’s teams, each position has its own special responsibility: setter, front left-side and middle hitters, back right- and left-side passers, and libero. 10-143 HW eTool (Desmos). Homework Help ✎ 1. If he chooses randomly, how many ways can Akio form his starting lineup? $_{16}P_6$ 2. How many of those teams have Sidney playing in the libero position? 3. If Akio chooses starting teams randomly, what is the probability (in percent) that Sidney gets chosen as the starting libero? $37.5$%", "burgers.They are known to make 20 32 20 Aug 2009, 15:22 12 A coach will make 3 substitutions. The team has 11 players, 16 19 Dec 2007, 13:58 Display posts from previous: Sort by", "+0 0 104 2 Corey and Peter on a basketball team. There are 10 players on the team. (a) How many starting lineups (of five players) include both Corey and Peter? (b) How many starting lineups (of five players) include Corey, Peter, or both? Mar 10, 2020 #1 +321 +2 (a) We automatically put Corey and Peter on the team, so we have 3 choices left to make with 8 people to choose from, so the answer is 8C3 = 56. (b) We want to count these cases seperately. With just Corey we have 8C4 = 70 ways to choose a team and the same goes for Peter. We add 70+70+56 = 196 lineups. Mar 10, 2020 #2 +28 0 a) A starting lineup of five players that includes Corey and Peter is: Corey, Pater, and three other players. These three must be chosen from the eight that are not Corey and Peter. Choosing three people out of eight when order doesn't matter makes: $${{8 \\cdot 7 \\cdot 6} \\over {3 \\cdot 2 \\cdot 1}} = 56$$ starting lineups. a) ANSWER: $$56$$ starting lineups. b) A starting lineup that includes Corey but not Peter is: Corey and four other players who are not Peter. These four must again be chosen from the eight that are not Corey and", "3 substitutions. The team has 11 players, 6 29 Oct 2007, 09:01 7 basketball teams with five players each are at a 1 10 Oct 2006, 09:58 Display posts from previous: Sort by", "Home > CCA2 > Chapter 11 > Lesson 11.1.1 > Problem11-7 11-7. Coach Kenadt has $12$ players on his basketball team and needs to select one player to be the team representative to the athletic student council. He wants the selection to be fair, meaning each player has an equal chance of being selected. Rummaging through his gym bag, he finds a single six-sided die and a nickel. How could he use the nickel and die to select the team representative fairly? There are $6$ numbers of the die and heads and tails on the nickel. How many combinations of numbers and H/T can you make?", "Filed under: Kentucky students are changing shirts mid-game based on which platoon is in Kentucky has so many good players that you can't bring only one shirt to the game. Kentucky famously has so many good players that John Calipari has divided his squad into two five-man units, the starters -- the \"blue platoon\" -- and the 6th-through-10th men -- the \"white platoon.\" Fans can't just show up to support one, right? So this is happening: Sure enough, here's Kentucky's student section in blue: And here they are in white:", "# Combine all of This. In how many ways can a basketball team with 9 members can be chosen from 12 players? Note: that the formula of Combination is nCr = n!/r!(n-r)! ×", "and not all who wish to join will make the cut. The team will be limited to 3 people for a probationary period so we don't get dumped in the deep end right away, and it will never have more than 5 people. Once appointed by the council, the team lead will choose applicants for the rest of the team to forward on for council approval. \\chapter{Stabilization}", "of people you're dealing with. Which is better, the best basketball team that you can put together from people born in Pennsylvania or the best basketball team that you can put together from people born in Delaware? Probably the Pennsylvania team, since there are about 13x as many people in that state so you get to draw from a larger pool. If there were no other relevant differences between the states then you'd expect 13 of the best 14 players to be Pennsylvanians, and probably the two neighboring states are similar enough so that Delaware can't overcome that population gap. Now, imagine you're picking the best 10 basketball players from the 1,000 tallest basketball-aged Americans (20-34 year-olds), and you're putting together another group consisting of the best 10 basketball players from the next 100,000 tallest basketball-aged Americans. Which is a better group of basketball players? In this case it's not obvious - getting to pick from a pool of 100x as many people is an obvious advantage, but that height advantage could matter a lot too. That's the tails coming apart - the very tallest don't necessarily give you the very best basketball players, because \"the very tallest\" is a much smaller set than the \"also really tall but not quite as tall\". (I ran some numbers and", "teams that I could create from a list of non where I chose three University of Louisville", "5 12 May 2014, 02:25 9 basketball players are trying out to be on a newly formed basketball 3 11 Aug 2010, 20:30 62 John and Peter are among the nine players a basketball coach 32 14 May 2016, 12:46 Display posts from previous: Sort by", "have 15 players on their roster. Rudy Gobert and Donovan Mitchell both tested positive for COVID. Every single player on the team was exposed often. (Here is an opinion, not fact, I will share: Had Rudy Gobert not been a star athlete for an NBA team, both he and Donovan Mitchell would likely have been in Group 4, but due to the power and wealth of the NBA, a minor sniffle as Rudy Gobert called it was enough to get the whole team tested. This is good for our stats as we have some numbers we can use to extrapolate data from.) Can we be certain that all 15 players on the team were exposed to COVID? Yes. How often? A lot (sorry, this number is hard to get, so lets use numbers we have). Rosters have 15 players and 1 head coach. There are assistant coaches, staff, medical professionals, trainers, etc. Let’s only count the 15 players and 1 head coach and 4 assistant coaches for a total of 20 individuals. • Number exposed: 20 • Number who caught COVID: 2 • Number who were asymptomatic: 1 • Number with minor symptoms: 1 • Number exposed but didn’t catch it: 18 So extrapolating from the Utah Jazz experience, only 2 in 20 exposed caught it. That means", "second example: For formation 4-4-2, he can put players 1, 2, 9 and 10 as defense, players 4, 5, 6 and 7 as midfielders, and players 3 and 8 as offense. For formation 3-5-2, he can put players 4, 9 and 10 as defense, players 1, 2, 5, 6 and 7 as midfielders, and players 3 and 8 as offense. Formation 4-3-3 can’t be set up because the coach has only 2 offensive players.", "good enough to hire the best players. As any football fan knows it, an equilibrium in the team must be met, and complementarities between players must be found. In mathematical words, these nonlinear aspects make the lineup design nontrivial. In more straightforward terms, we need to be smart about how to design the starting eleven. This is where column generation enters in play. How does that work? The trouble is that the set of combinations of eleven players is extremely huge. To face this issue, column generation proposes to divide the job of searching for good players (let’s call that recruitment and give it to a recruiter) and the job of selecting among recruited players the best starting eleven (let’s call that the lineup and have a coach doing it). Crucially, because the coach picks up players out of a small set of recruited players, it will be a much easier job for him to find the best starting eleven. Can’t the coach have a say in recruitment? Not only he can. He should. Given the weaknesses of his starting eleven, the coach can direct the recruiter towards the profiles of players that would definitely improve his team. Typically, if a quality defender is missing, the coach should say so, hence signaling the recruiter that he should focus", "coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter? a) 1/9 b) 1/6 c) 2/9 d) 5/18 e) 1/3 OPEN DISCUSSION OF THIS QUESTION IS HERE: john-and-peter-are-among-the-nine-players-a-basketball-coach-68186.html Remember - if it is certain that X HAS TO BE chosen then he can be chosen only in one way. because it is certain. so we need to choose 3 people from remaining 7 people because John an peter is included. Probability of choosing a team that includes John and Peter = (1 * 1 * 7C3) / 10C5 = 5/18 _________________ --------------------------------------------------------------- Target - 720-740 http://gmatclub.com/forum/information-on-new-gmat-esr-report-beta-221111.html http://gmatclub.com/forum/list-of-one-year-full-time-mba-programs-222103.html Re: John and Peter are among the nine players a basketball coach [#permalink] 09 Apr 2016, 02:06 Similar topics Replies Last post Similar Topics: 6 Coach John has to form a team of 4 players to be selected from a pool 5 01 Apr 2016, 13:57 17 A basketball coach will select the members of a five-player 16 20 Aug 2016, 04:49 1 A basketball coach will select the members of a five-player 5 12 May 2014, 02:25 9 basketball players are trying out to be on a newly formed basketball 3 11 Aug 2010, 20:30 62 John and Peter are", "Free Version Moderate # Permutations and Combinations 15 ALGEBR-MPWKCE A college has $10$ basketball players. A $5$-member team will be selected out of those $10$ players. Then $1$ player from the remaining team members will be chosen to serve as captain. How many different selections can be made in filling the roster and captain? A $120$ B $252$ C $1260$ D $151,200$", "Next? Now with a solid example under our belt, we can get into the fun stuff. In part 3, I’ll take data regarding fantasy-football players and find some of the best lineups to use for FanDuel competitions. Better grab a beer! Until then, see the full example in GitHub.", "way. +Furthermore, the team's lead will be appointed by the council to further +ensure a cohesive CoC interpretation.++It is expected that membership on this team will be highly selective and +not all who wish to join will make the cut. The team will be limited to +3 people for a probationary period so we don't get dumped in the deep +end right away, and it will never have more than 5 people. Once +appointed by the council, the team lead will choose applicants for the +rest of the team to forward on for council approval. \\ No newline at end of file" ]

[ "+0 +1 193 2 +553 Coach Grunt is preparing the 5-person starting lineup for his basketball team, the Grunters. There are 12 players on the team. Two of them, Ace and Zeppo, are league All-Stars, so they'll definitely be in the starting lineup. How many different starting lineups are possible? (The order of the players in a basketball lineup doesn't matter.) ant101 Feb 18, 2018 #1 +88898 +1 Well ....two of the positions are \"set\" So..... we just want to choose any 3 of the remaining 10 players = C(10,3) = 120 different lineups CPhill Feb 18, 2018 #2 +139 +1 Coach Grunt has to choose 3 players from the 10 players that are remaining after Ace and Zeppo have been placed in the lineup. The order in which the players are chosen doesn't matter, so the answer is $$\\binom{10}{3} = \\frac{10 \\times 9 \\times 8}{3 \\times 2 \\times 1} = \\boxed{120}.$$ azsun Feb 18, 2018", "into cases: • A) The designated hitter is a catcher • B) The designated hitter is an infielder • C) The designated hitter is a pitcher • D) The designated hitter is an outfielder Looking at one of these cases more closely, looking at A again: Break it up via multiplication principle as before. Since the D.H. is a catcher, that means our lineup will consist of 2 catchers, Jones, 2 other outfielders, and 4 infielders. • Pick which two of the catchers are in the lineup ($\\binom{2}{2}=1$ choice since there are two catchers available and we want to choose two of them to appear in our lineup) • Pick which two of the other outfielders are in the lineup ($\\binom{5}{2} = 10$ choices since there are $\\color{red}{\\star}$five$\\color{red}{\\star}$ outfielders available and we want to choose two of them to appear in our lineup) (since we already know Jones is in the lineup) • Pick which four of the infielders are in the lineup ($\\binom{5}{4} = 5$ choices since there are five infielders available and we want to choose four of them to appear in our lineup) • Pick which zero of the pitchers are in the lineup ($\\binom{12}{0} = 1$ choice since there are twelve pitchers available and we don't want to choose any of them, i.e. the", "ordering that would make any sense. And it’s clear from the wording of (b) that the batting order is not determined by the starting lineup, so the starting lineup must refer just to the set of players. – Brian M. Scott Feb 2 '13 at 21:34 Here one has to know the culture of baseball to work on a math question! – Maesumi Feb 2 '13 at 22:07 @Maesumi: Could be worse. There are $20$ cricketers in the village of Little Piddle in the Marsh. Four are fast bowlers. Six others have soft hands and exceptional reflexes. Only one is willing to keep wicket. In how many ways can the captain field two fast bowlers, three slips, and a gully? – Brian M. Scott Feb 2 '13 at 23:07 They are apparently treating a lineup as an ordered list of the nine starting players. This is unusual; perhaps they’re thinking that the list starts out as a list of the $9$ playing positions (catcher, pitcher, first base, etc.) which then gets filled in with some set of $9$ names; in that case listing Joe as pitcher would be different from listing Joe as catcher. Thus, they arrive at $\\binom{15}99!=\\frac{15!}{6!}$ ways to fill out the lineup card. They then treat the batting order as a distinct ordering of the", "\\times \\;46\\; \\times \\;45\\; \\times \\;44! }{ 4\\; \\times 3\\; \\times 2\\; \\times 1\\; \\times \\;44! } \\times \\frac{ 4\\; \\times \\;3! }{ 3! }$ = 48×47×46×454×3×2×1×4$\\frac{ 48\\; \\times \\;47\\; \\times \\;46\\; \\times \\;45 }{ 4\\; \\times \\;3\\; \\times \\;2\\; \\times \\;1 } \\times 4$ = 778320 number of combinations. Q-7: Find the number of ways by which one can select a cricket team of 11 players from a bunch of 17 players among which only 5 players can bowl if every cricket team of 11 players must have exactly 4 bowlers. Solution: Among 17 players, 5 players can bowl. Now, we need to select a cricket team of 11 players where there are exactly 4 players who can bowl. So, Among 5, 4 bowlers can be selected in 5C4$^{ 5 }C_{ 4 }$ number of ways and the remaining 7 players for the team of 11 players will be selected from the remaining 12 players in 12C7$^{ 12 }C_{ 7 }$ number of ways. Therefore, by the principle of multiplication, number of ways required to select a cricket team: = 5C4×12C7$^{ 5 }C_{ 4 } \\times ^{ 12 }C_{ 7 }$ = 5!4!1!×12!7!5!$\\frac{ 5! }{ 4!\\; 1! } \\times \\frac{ 12! }{ 7!\\; 5! }\\\\$ = 5×4!4!×12×11×10×9×8×7!7!×5×4×3×2×1$\\\\\\frac{ 5\\; \\times \\;4! }{ 4! } \\times \\frac{ 12\\; \\times", "Author Message TAGS: ### Hide Tags Intern Joined: 28 Aug 2016 Posts: 17 GMAT 1: 560 Q44 V23 Coach Jackson will choose at least two players for his team from those [#permalink] ### Show Tags Updated on: 15 Dec 2016, 03:03 16 00:00 Difficulty: 95% (hard) Question Stats: 23% (01:42) correct 77% (01:34) wrong based on 220 sessions ### HideShow timer Statistics Coach Jackson will choose at least two players for his team from those who try out on Saturday. How many players will Coach Jackson choose? (1) Coach Jackson could choose exactly 20 different teams. (2) At least two players at the tryout will not be chosen. Originally posted by Amby02 on 15 Dec 2016, 03:01. Last edited by Bunuel on 15 Dec 2016, 03:03, edited 1 time in total. Renamed the topic. Senior Manager Joined: 13 Oct 2016 Posts: 367 GPA: 3.98 Coach Jackson will choose at least two players for his team from those [#permalink] ### Show Tags 15 Dec 2016, 05:43 4 2 Very interesting and, I'd like to say, not an easy question. Now let's go to our options. (1) Coach Jackson could choose exactly 20 different teams That means nCr = 20, where n is our total number of players and r is # of layers we need to choose. The only", "the coach chooses a team that includes both John and Pete = 35/126 = 5/18 Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Re: A basketball coach will select the members of a five-player [#permalink] 29 Jan 2018, 07:47 Display posts from previous: Sort by", "2009, 01:55 PM Soroban [size=3]Hello, skeske1234! Quote: A baseball team has 13 members. If a batting line-up consists of 9 players, how many different batting line-ups are possible? The answer in my textbook for this question uses permutations to derive the answer. However, I believe that it's wrong. You should be using combinations to find the answer. I think combinations because in this question, its asking for a line-up of batting players. In this case, order doesn't matter... Does it? Well, of course, order matters! Do you understand what a line-up is? It looks like this: . . $\\boxed{\\begin{array}{cc}1. &\\text{Hoo bats first.} \\\\ 2.&\\text{Watt bats second.} \\\\ 3. & \\text{Idunno bats third.} \\\\ \\vdots & \\vdots\\end{array} }$", "roster no longer needs any more people at a qualified position. 1. By position, remove every remaining candidate from the pool once the roster requirement is satisfied at that position. 8. Beer myself. After some noodling, I ended up finding that filtering the set of all players for 99th weight and 25th height yielded this pool of candidates: This was the furthest I could push the filter before I started losing positional representation. 2B and 1B were the bottlenecks. From these, grouping players by position and sorting by descending weight yields this as the inaugural roster of the Wonkaville Huskies. ### And Here’s How They Compare to the Aforementioned 20,000-player Historical Means: Alright. So now we’ve got our roster. How good is this team? ## WAR — What is it good for? Absolutely Everything. When you start any sort of comparative analysis between baseball players, the good news is that you’ve got about a zillion ways to see how they measure up. Some of these are super simple. How many homers did someone hit in a given time period? And some of these are not simple: $SIERA = 6.145 - 16.896 \\times \\frac{SO}{PA} + 11.434 \\times \\frac{BB}{PA} - 1.858 \\times \\frac{GB-FB-PU}{PA} + \\\\7.653 \\times (\\frac{SO}{PA})^2 \\pm 6.664 \\times (\\frac{GB-FB-PU}{PA})^2 + 10.130 \\times \\frac{SO}{PA} \\times \\\\\\frac{GB - FB -", "Browse Questions # In how many ways can we select a cricket eleven from 17 players in which only 5 players can bowl if each cricket eleven must include exactly 4 bowlers? $\\begin{array}{1 1}(A)\\;3860\\\\(B)\\;3960\\\\(C)\\;3660\\\\(D)\\;3760\\end{array}$ Four bowlers can be selected from the five players and seven players can be selected from 12 players (17-5)=12 is the number of ways of selecting the cricket eleven,then $P=C(5,4)\\times C(12,7)$ $\\;\\;\\;=\\large\\frac{5!}{4!(5-4)!}\\times\\frac{ 12!}{7!(12-7)!}$ $\\;\\;\\;=\\large\\frac{5\\times 4!}{1!4!}\\times\\frac{ 12\\times 11\\times 10\\times 9\\times 8\\times 7!}{7!5!}$ $\\;\\;\\;=\\large\\frac{5}{1}\\times \\frac{12\\times 11\\times 10\\times 9\\times 8}{5\\times 4\\times 3\\times 2\\times 1}$ $\\;\\;\\;=3960$ Hence (B) is the correct answer.", "with the following adjustment: since we want to count the number of cases Jones is not in the lineup, redo the calculations for the first part as though Jones weren't on the team at all (I.e. remove Jones from the list of available outfielders, making it so that there are only $5$ outfielders available). We again break it into the same cases as before: A) d.h. is a catcher, B) d.h. is an infielder, C) d.h. is an outfielder, D) d.h. is a pitcher. Let us examine the first case in detail again: • Pick which two of the catchers are in the lineup ($\\binom{2}{2}=1$ choice since there are two catchers available and we want to choose two of them to appear in our lineup) • Pick which three of the outfielders are in the lineup ($\\binom{5}{3} = 10$ choices since there are $\\color{red}{\\star}$five$\\color{red}{\\star}$ outfielders available and we want to choose three of them to appear in our lineup) (since we don't want Jones) • Pick which four of the infielders are in the lineup ($\\binom{5}{4} = 5$ choices since there are five infielders available and we want to choose four of them to appear in our lineup) • Pick which zero of the pitchers are in the lineup ($\\binom{12}{0} = 1$ choice since there are twelve pitchers", "in 12C4 ways Required number of ways = 12C4 $$=\\frac{12 \\times 11 \\times 10 \\times 9}{4 \\times 3 \\times 2 \\times 1}=495$$ (iv) There are 26 red cards and 26 black cards. ∴ We have select 2 red and 2 black cards. This can be done in 26C2 x 26C2 ways. \\begin{aligned} &=\\frac{26 \\times 25}{2 \\times 1} \\times \\frac{26 \\times 25}{2 \\times 1}\\\\ &=325 \\times 325=105625 \\end{aligned} ∴ Required number of ways = 26C2 x 26C2 \\begin{aligned}&=\\frac{26 \\times 25}{2 \\times 1} \\times \\frac{26 \\times 25}{2 \\times 1}\\\\&=325 \\times 325=105625\\end{aligned} (v) There are 26 red and 26 black cards ∴ 4 red cards can be selected in 26 C4 ways 4 black cards can be selected in 26 C4 ways ∴ Required number of ways = 26 C4 +26C4 $$=2 x^{26} C_{4}=\\frac{26 \\times 25 \\times 24 \\times 26}{4 \\times 3 \\times 2 \\times 1}=29900$$ Question 10. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl, if each cricket team of 11 must include exactly 4 bowlers? We have, Number of players = 17 Number of players to be select for team =11 Number of bowlers = 5 Number of bowlers to be select = 4 Number of players other than bowlers =12 4 bowlers out of", "2 goes to team 1 given student 1 went to team 1 plus the probability student 1 went to team 2 times the probability student 2 goes to team 2 given student 1 went to team 2. – Jack Moody Nov 14 '18 at 16:15 Suppose that one team of three students is chosen at random from the class. Two given students A and B are in one team if they go to chosen team or if stay at the rest part of $$7$$ students. The total number of possibilities to choose $$3$$ students is $$\\binom{3}{10}=120$$, the number of favorable combinations is $$\\binom{1}{8}+\\binom{3}{8}=8+56=64,$$ and the probability is $$\\frac{64}{120}=\\frac{8}{15}$$. Here $$\\binom{1}{8}$$ is the number of variants to choose three students with A and B and one extra student chosen from the rest $$8$$ students. The summand $$\\binom{3}{8}$$ calculates the number of variants to choose $$3$$ students so that A and B stay in the second team.", "for position three, etc. The alphabetical ordering of the players of a baseball team is one permutation of the set of players. Other orderings of the players' names might be done by batting average, age, or height. The information that determines the ordering is called the key. We would expect that each key would give a different permutation of the names. If there are twenty-five players on the team, there are $$25 \\cdot 24 \\cdot 23 \\cdot \\cdots \\cdot 3 \\cdot 2 \\cdot 1$$ different permutations of the players. This number of permutations is huge. In fact it is 15511210043330985984000000, but writing it like this isn't all that instructive, while leaving it as a product as we originally had makes it easier to see where the number comes from. We just need to find a more compact way of writing these products. We now develop notation that will be useful for permutation problems. #### Definition2.2.5.Factorial. If $$n$$ is a positive integer then $$n$$ factorial is the product of the first $$n$$ positive integers and is denoted $$n!\\text{.}$$ Additionally, we define zero factorial, $$0!\\text{,}$$ to be 1. The first few factorials are \\begin{equation*} \\begin{array}{ccccccccc} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\\ n! & 1 & 1 & 2", "Q # In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers? Q.7. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers? Views Out off, 17 players, 5 are bowlers. A cricket team of 11 is to be selected such that there are exactly 4 bowlers. 4 bowlers can be selected in $^5C_4$ ways and 7 players can be selected in $^1^2C_7$ ways. Thus, using multiplication priciple, number of ways of selecting the team $=^5C_4 .^1^2C_7$ $=\\frac{5!}{1!4!}\\times \\frac{12!}{5!7!}$ $=5 \\times \\frac{12\\times 11\\times 10\\times 9\\times 8}{5\\times 4\\times 3\\times 2}$ $=3960$ Exams Articles Questions", "$k$ names of the players that were selected by the $i^\\textrm {th}$ owner. The $n$ teams should be in the original order of owners and players should be listed in the order in which they were drafted following the rules above. Sample Input 1 Sample Output 1 2 2 0 0 6 Shoresy Jonesy Reilly Sholtzy Fisky Yorkie Shoresy Reilly Jonesy Sholtzy Sample Input 2 Sample Output 2 2 2 2 Reilly Shoresy 2 Shoresy Reilly 6 Shoresy Jonesy Reilly Sholtzy Fisky Yorkie Reilly Jonesy Shoresy Sholtzy", "empty choice) • Pick which order the nine people chosen appear in the lineup ($\\color{red}{\\star}8! = 40320\\color{red}{\\star}$ choices since Jones is already occupying the fourth slot, and we want to arrange the other eight players around him) Do so similarly for the other cases to find the number of ways that Jones appears in the fourth batting position. Complete by applying the definition of probability (num outcomes for event / num outcomes regardless).", "# In how many ways can one select a cricket team of eleven from 17 players in which only 5 persons can bowl if each cricket team of 11 must include exactly 4 bowlers? Question: In how many ways can one select a cricket team of eleven from 17 players in which only 5 persons can bowl if each cricket team of 11 must include exactly 4 bowlers? Solution: Out of 17 players, 11 need to be selected. There are 5 bowlers, of which four must be selected in the team. So, we have to choose 7 players from the remaining 12 players. Required number of ways $={ }^{5} C_{4} \\times{ }^{12} C_{7}=5 \\times \\frac{12 !}{7 ! 5 !}=5 \\times 792=3960$", "100, 3 times in Up To, and 15 times in Must Have • WR – AJ Green – Appear 1 times in Top 100, 5 times in Up To, and 20 times in Must Have Players Without Limits • DST – Packers – Appear 3 times in Top 100, 25 times in Up To, and 25 times in Must Have • QB – Russell Wilson – Appear 3 times in Top 100, 0 times in Up To, and 2 times in Must Have • RB – Matt Forte – Appear 5 times in Top 100, 19 times in Up To, and 16 times in Must Have • TE – Rob Gronkowski – Appear 8 times in Top 100, 29 times in Up To, and 22 times in Must Have • WR – Doug Baldwin – Appear 97 times in Top 100, 100 times in Up To, and 100 times in Must Have Stats By Lineup Type Notes on the above worksheet: • The complete worksheet can be viewed by clicking on the button at the lower right of the embedded sheet above. • The most important thing to note is the difference in the lineup rankings for each method. For Top 100 the best lineup rank is 1 and the worst is 100, for Up To the best", "# Baseball Combinations Problem Two part question (My work below). For both questions will use the orioles current roster: -Current orioles roster: 12 pitchers, 2 catchers, 5 in-fielders, and 6 out-fielders: Similar to the list here but not identical http://www.foxsports.com/mlb/baltimore-orioles-team-roster?season=2015 In the American League, the pitcher in the game does not bat. Each of the other 9 players on the team are put into a “batting order”. (Somebody bats first, somebody bats second, etc.). Below is a typical batting order: -1. De Aza (outfielder) -2. Paredes (designated hitter) -3. Young (outfielder) -4. Jones (outfielder) -5. Davis (infielder) -7. Joseph (catcher) -8. Cabrera (infielder) -9. Navarro (infielder) Question A: How many different batting orders are possible for the Orioles? Hints: -You have to have 1 catcher, three outfielders, four infielders, and a designated hitter (who could be anyone, even those players who are listed under the “pitchers” section of the roster). My answer: I'm choosing 1 catcher, 3 out-fielders, 4 in-fielders, and 1 designated hitter of what is left of the roster. So: $${2 \\choose 1}*{6 \\choose 3}*{5 \\choose 4}*{17 \\choose 1} = 3400$$ Question B.: If the batting order is selected randomly (assume each viable batting order from the previous question is equally likely), then what is the probability that Jones will be batting in the fourth position?", "available and we don't want to choose any of them, i.e. the empty choice) • Pick which order the nine people chosen appear in the lineup ($9! = 362880$ choices since every way of ordering the players counts as \"different\") For a total of $\\binom{2}{2}\\binom{5}{3}\\binom{5}{4}\\binom{12}{0}9!$ different outcomes for the first case. Calculate the other cases similarly. Now that we have a tally of how many lineups do not have Jones on the roster, take the answer from the first half of the problem and subtract the number just calculated here to get the number of different ways that Jones IS on the starting lineup (as per a simple application of inclusion-exclusion ). Take this newest number and divide by the answer found in the first half to get the probability that Jones is in the starting lineup as per the definition of probability in an equiprobable sample space: $$Pr(A) := \\frac{|A|}{|S|}$$ Complete the problem by noting that, given that Jones is in the starting lineup, he is equally likely to be in each of the spots and apply the multiplication rule (probability). $$Pr(\\text{Jones bats}~4^{th}) = Pr(\\text{Jones in lineup})\\cdot Pr(\\text{Jones bats}~4^{th}~|~\\text{Jones in lineup})$$ There is another equally acceptable approach to solving the second part. Directly count the number of ways that Jones bats fourth. Do so by breaking it" ]

[ "+0 +1 193 2 +553 Coach Grunt is preparing the 5-person starting lineup for his basketball team, the Grunters. There are 12 players on the team. Two of them, Ace and Zeppo, are league All-Stars, so they'll definitely be in the starting lineup. How many different starting lineups are possible? (The order of the players in a basketball lineup doesn't matter.) ant101 Feb 18, 2018 #1 +88898 +1 Well ....two of the positions are \"set\" So..... we just want to choose any 3 of the remaining 10 players = C(10,3) = 120 different lineups CPhill Feb 18, 2018 #2 +139 +1 Coach Grunt has to choose 3 players from the 10 players that are remaining after Ace and Zeppo have been placed in the lineup. The order in which the players are chosen doesn't matter, so the answer is $$\\binom{10}{3} = \\frac{10 \\times 9 \\times 8}{3 \\times 2 \\times 1} = \\boxed{120}.$$ azsun Feb 18, 2018", "### Home > CCG > Chapter 10 > Lesson 10.3.3 > Problem10-143 10-143. Akio coaches the girls volleyball team. He needs to select players for the six different starting positions from his roster of $16$ players. On Akio’s teams, each position has its own special responsibility: setter, front left-side and middle hitters, back right- and left-side passers, and libero. 10-143 HW eTool (Desmos). Homework Help ✎ 1. If he chooses randomly, how many ways can Akio form his starting lineup? $_{16}P_6$ 2. How many of those teams have Sidney playing in the libero position? 3. If Akio chooses starting teams randomly, what is the probability (in percent) that Sidney gets chosen as the starting libero? $37.5$%", "the coach chooses a team that includes both John and Pete = 35/126 = 5/18 Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Re: A basketball coach will select the members of a five-player [#permalink] 29 Jan 2018, 07:47 Display posts from previous: Sort by", "burgers.They are known to make 20 32 20 Aug 2009, 15:22 12 A coach will make 3 substitutions. The team has 11 players, 16 19 Dec 2007, 13:58 Display posts from previous: Sort by", "Home > CCA2 > Chapter 11 > Lesson 11.1.1 > Problem11-7 11-7. Coach Kenadt has $12$ players on his basketball team and needs to select one player to be the team representative to the athletic student council. He wants the selection to be fair, meaning each player has an equal chance of being selected. Rummaging through his gym bag, he finds a single six-sided die and a nickel. How could he use the nickel and die to select the team representative fairly? There are $6$ numbers of the die and heads and tails on the nickel. How many combinations of numbers and H/T can you make?", "+0 0 104 2 Corey and Peter on a basketball team. There are 10 players on the team. (a) How many starting lineups (of five players) include both Corey and Peter? (b) How many starting lineups (of five players) include Corey, Peter, or both? Mar 10, 2020 #1 +321 +2 (a) We automatically put Corey and Peter on the team, so we have 3 choices left to make with 8 people to choose from, so the answer is 8C3 = 56. (b) We want to count these cases seperately. With just Corey we have 8C4 = 70 ways to choose a team and the same goes for Peter. We add 70+70+56 = 196 lineups. Mar 10, 2020 #2 +28 0 a) A starting lineup of five players that includes Corey and Peter is: Corey, Pater, and three other players. These three must be chosen from the eight that are not Corey and Peter. Choosing three people out of eight when order doesn't matter makes: $${{8 \\cdot 7 \\cdot 6} \\over {3 \\cdot 2 \\cdot 1}} = 56$$ starting lineups. a) ANSWER: $$56$$ starting lineups. b) A starting lineup that includes Corey but not Peter is: Corey and four other players who are not Peter. These four must again be chosen from the eight that are not Corey and", "Free Version Easy Permutation or Combination? ALGEBR-8GSHQ1 A baseball coach must choose 9 starters from a group of 14 players on his team. If order doesn't matter, how many different ways can he choose? A $605,048$ B $2,002$ C $1760$ D $240,240$", "Free Version Moderate # Permutations and Combinations 15 ALGEBR-MPWKCE A college has $10$ basketball players. A $5$-member team will be selected out of those $10$ players. Then $1$ player from the remaining team members will be chosen to serve as captain. How many different selections can be made in filling the roster and captain? A $120$ B $252$ C $1260$ D $151,200$", "1/3 [Reveal] Spoiler: OA Kudos [?]: 162 [2], given: 15 Math Expert Joined: 02 Sep 2009 Posts: 43304 Kudos [?]: 139237 [1], given: 12781 Re: A basketball coach will select the members of a five-player [#permalink] ### Show Tags 07 May 2012, 07:46 1 KUDOS Expert's post 3 This post was BOOKMARKED Galiya wrote: A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter? A. 1/9 B. 1/6 C. 2/9 D. 5/18 E. 1/3 Total # of five-player teams possible is $$C^5_9=126$$; # of teams that include both John and Peter is $$C^2_2*C^3_7=35$$; $$P=\\frac{35}{126}=\\frac{5}{18}$$. _________________ Kudos [?]: 139237 [1], given: 12781 Intern Joined: 04 Jan 2012 Posts: 4 Kudos [?]: 18 [4], given: 3 Re: A basketball coach will select the members of a five-player [#permalink] ### Show Tags 07 May 2012, 07:51 4 KUDOS 1 This post was BOOKMARKED There are 3 spots left after selecting John and Peter in the team. There are 7 players available to fill in the 3 positions. It can be done in 7C3 ways (favorable event) The total number of ways of selecting 5 out of 9 players", "Joined: 02 Sep 2009 Posts: 44566 Re: A basketball coach will select the members of a five-player [#permalink] ### Show Tags 07 May 2012, 08:46 1 KUDOS Expert's post 3 This post was BOOKMARKED Galiya wrote: A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter? A. 1/9 B. 1/6 C. 2/9 D. 5/18 E. 1/3 Total # of five-player teams possible is $$C^5_9=126$$; # of teams that include both John and Peter is $$C^2_2*C^3_7=35$$; $$P=\\frac{35}{126}=\\frac{5}{18}$$. _________________ Intern Joined: 04 Jan 2012 Posts: 4 Re: A basketball coach will select the members of a five-player [#permalink] ### Show Tags 07 May 2012, 08:51 4 KUDOS 1 This post was BOOKMARKED There are 3 spots left after selecting John and Peter in the team. There are 7 players available to fill in the 3 positions. It can be done in 7C3 ways (favorable event) The total number of ways of selecting 5 out of 9 players is given by 9C5 ways. 7C3 / 9C5 = 5/18 Manager Status: mba here i come! Joined: 07 Aug 2011 Posts: 243 ### Show Tags 19 Aug 2012, 01:30 total", "Filed under: Kentucky students are changing shirts mid-game based on which platoon is in Kentucky has so many good players that you can't bring only one shirt to the game. Kentucky famously has so many good players that John Calipari has divided his squad into two five-man units, the starters -- the \"blue platoon\" -- and the 6th-through-10th men -- the \"white platoon.\" Fans can't just show up to support one, right? So this is happening: Sure enough, here's Kentucky's student section in blue: And here they are in white:", "second example: For formation 4-4-2, he can put players 1, 2, 9 and 10 as defense, players 4, 5, 6 and 7 as midfielders, and players 3 and 8 as offense. For formation 3-5-2, he can put players 4, 9 and 10 as defense, players 1, 2, 5, 6 and 7 as midfielders, and players 3 and 8 as offense. Formation 4-3-3 can’t be set up because the coach has only 2 offensive players.", "Total cost: 34800, Total Projected Points: 140.09 You can see above each lineup, the team we are stacking 4 players from, and the opposing pitcher listed that we are trying to target. This is just the first 3 lineups created, but there were 7 more like them. You could easily loop through all the pitchers and create any number of lineups. Maybe you just want the top one, or maybe you want all available options. In theory, this should give us a list of 10 lineups with decent “boom or bust” potential. What this creates for us is a list of optimized lineups that we can choose from on any given day, any given slate. And a system that we can use to create lineups for historical contests and slates and see how they would have done, as well as future contests to speed up our lineup creating process when we want to play. ## Limitations This is a great start as this should be a list of mathematically optimized lineups. The fact that we treated it as an optimization problem, means that we selected the 9 players with the highest number of projected points as possible, while making sure it’s still a legal lineup. In theory, we should get the maximum bang for our buck with these", "to define the value of coalitions with more than 5 players. For this case we can aggregate the results from any lineup consisting exclusively of players in the coalition at hand. For example, if we want to estimate the value of the characteristic function for a coalition of 7 players, we will calculate the average net rating for any 5-man lineup played consisting only of players from this set of 7 players. Let’s see an example here. Of course, I will choose one of Pitt’s games and let’s go with the first conference game of the season against Miami. Despite the fact that Miami had only 6 or 7 scholarship players available for the game, they fought well (partially due to early foul troubles for Champagnie and Johnson) before Pitt eventually got away in the latter part of the second half with a 70-55 win. Using the excellent R package from Jake Flancer, bigballR, I was able to obtain the lineup information needed to calculate the Shapley Player Value (SPV) for that game (code is available here). These SPV values essentially consider all the different lineups that played for Pitt in that game and how they performed, and assigned a net rating to each player. According to SPV, Toney was the game MVP for Pitt (which also passes", "Author Message TAGS: ### Hide Tags Intern Joined: 28 Aug 2016 Posts: 17 GMAT 1: 560 Q44 V23 Coach Jackson will choose at least two players for his team from those [#permalink] ### Show Tags Updated on: 15 Dec 2016, 03:03 16 00:00 Difficulty: 95% (hard) Question Stats: 23% (01:42) correct 77% (01:34) wrong based on 220 sessions ### HideShow timer Statistics Coach Jackson will choose at least two players for his team from those who try out on Saturday. How many players will Coach Jackson choose? (1) Coach Jackson could choose exactly 20 different teams. (2) At least two players at the tryout will not be chosen. Originally posted by Amby02 on 15 Dec 2016, 03:01. Last edited by Bunuel on 15 Dec 2016, 03:03, edited 1 time in total. Renamed the topic. Senior Manager Joined: 13 Oct 2016 Posts: 367 GPA: 3.98 Coach Jackson will choose at least two players for his team from those [#permalink] ### Show Tags 15 Dec 2016, 05:43 4 2 Very interesting and, I'd like to say, not an easy question. Now let's go to our options. (1) Coach Jackson could choose exactly 20 different teams That means nCr = 20, where n is our total number of players and r is # of layers we need to choose. The only", "2009, 01:55 PM Soroban [size=3]Hello, skeske1234! Quote: A baseball team has 13 members. If a batting line-up consists of 9 players, how many different batting line-ups are possible? The answer in my textbook for this question uses permutations to derive the answer. However, I believe that it's wrong. You should be using combinations to find the answer. I think combinations because in this question, its asking for a line-up of batting players. In this case, order doesn't matter... Does it? Well, of course, order matters! Do you understand what a line-up is? It looks like this: . . $\\boxed{\\begin{array}{cc}1. &\\text{Hoo bats first.} \\\\ 2.&\\text{Watt bats second.} \\\\ 3. & \\text{Idunno bats third.} \\\\ \\vdots & \\vdots\\end{array} }$", "roster no longer needs any more people at a qualified position. 1. By position, remove every remaining candidate from the pool once the roster requirement is satisfied at that position. 8. Beer myself. After some noodling, I ended up finding that filtering the set of all players for 99th weight and 25th height yielded this pool of candidates: This was the furthest I could push the filter before I started losing positional representation. 2B and 1B were the bottlenecks. From these, grouping players by position and sorting by descending weight yields this as the inaugural roster of the Wonkaville Huskies. ### And Here’s How They Compare to the Aforementioned 20,000-player Historical Means: Alright. So now we’ve got our roster. How good is this team? ## WAR — What is it good for? Absolutely Everything. When you start any sort of comparative analysis between baseball players, the good news is that you’ve got about a zillion ways to see how they measure up. Some of these are super simple. How many homers did someone hit in a given time period? And some of these are not simple: $SIERA = 6.145 - 16.896 \\times \\frac{SO}{PA} + 11.434 \\times \\frac{BB}{PA} - 1.858 \\times \\frac{GB-FB-PU}{PA} + \\\\7.653 \\times (\\frac{SO}{PA})^2 \\pm 6.664 \\times (\\frac{GB-FB-PU}{PA})^2 + 10.130 \\times \\frac{SO}{PA} \\times \\\\\\frac{GB - FB -", "with 3 linemen and 4 linebackers. A professional derivative in the 1970s of the earlier Oklahoma or 50 defense, which had 5 linemen and 2 linebackers. The 3-4 outside linebackers resemble \"stand-up ends\" in the older defense. It is sometimes pronounced thirty-four defense. The 3-4 also was spun off from the Miami Dolphins \"53 defense named for the jersey number worn by linebacker Bob Matheson, who was often used by the Dolphins as a fourth linebacker in passing situations. 4–3 defense A defensive formation with 4 linemen and 3 linebackers. Several variations are employed. First used by coach Joe Kuharich[citation needed] and Tom Landry.[1][2] It is sometimes pronounced forty-three defense. 4–4–4 Defense Illegal participation (name so derived from the fact that 4+4+4=12 men on the field; each team is limited to 11). Coined by coach and color commentator John Madden. (However, this formation is legal in Canadian football, as there are 12 players on the Field) 46 defense Usually pronounced forty-six defense, a formation of the 4-3 defense (four linemen and three linebackers) featuring several dramatic shifts of personnel. The line is heavily shifted toward the offense's weak side; both outside linebackers tend to play on the strong side outside of the defensive linemen; and three defensive backs (the two cornerbacks and the strong safety) crowd the line", "# Baseball Roster Optimization I'm trying to programmatically optimize a Fantasy Baseball Roster that requires a fixed number of players at position (2 Catchers, 5 Outfielders, etc.) and has a salary constraint (total draft price cannot exceed n dollars). The success of the roster is based on the combined team production in batting (or pitching) categories. I have projections of each player's estimated statistical contributions - from which, I'm able to determine (using Standard Scoring) their approximate value. For this exercise, we can assume that I have priced each player appropriately. For this particular league, there are six batting categories (HR, RBI, Stolen Bases, etc)... and the projections I have allow me determine how much each player will contribute within each statistical category. Given 154 draftable players, and requiring 14 batters per team - gives an astounding 26,327,386,978,706,200,000 number of different combinations. Obviously, I don't want to attempt a brute-force method of testing each possible combination of players to determine an optimal roster. (I'd have a lot of rosters that are too expensive, and a lot of rosters worth $14.00 (or less)). Clearly, I need to be smarter about this and I'm looking for some degree of direction in order to get started. What I've tried: My first attempt at optimizing the roster was to select the BEST", "OpenKattis Baylor University # 2, 4, 6, Greaaat Against her wishes, Rainbow Dash was put in charge of coach the buckball halftime cheerleading squad. While she doesn’t have the slightest care for cheerleading (she’d much rather coach the team), she’s determined to give it her all to show everypony in the squad how much she cares about them. One of the most important things in cheerleading is regulating the crowd’s level of enthusiasm. By instructing the squad to execute different cheers, she can raise or lower this level; for example, less lively cheers temper the level of enthusiasm, while other, more exciting cheers raise it. At the beginning, the crowd’s level of enthusiasm is zero. To prime the crowd for the second half of the game, she wants the level of enthusiasm after all the cheers to be exactly $T$. Note that it is possible for the crowd’s level of enthusiasm to dip below zero or exceed $T$ in between the cheers. There are $N+1$ cheers at Rainbow Dash’s disposal. The $i^\\text {th}$ of these cheers changes the level of enthusiasm by $S_ i$, and has a level of difficulty $D_ i$—a higher difficulty requires more practice and is harder to execute correctly. The first of these cheers is always a standard cheer with $S_1 = 1$ and" ]

[ "# Thread: Probabilities and Odds Problems 1. ## Probabilities and Odds Problems Hey, Can someone help me on this plz and explain your solution. A confident and boastful coach claims that on the next game the odds of her team winning are 3:1, the odds against losing are 5:1, and the odds against tieing are 7:1. Can these odds be right? Show why or why not these claims make sense. Thanx Ballin 2. Originally Posted by ballin_sensation Hey, Can someone help me on this plz and explain your solution. A confident and boastful coach claims that on the next game the odds of her team winning are 3:1, the odds against losing are 5:1, and the odds against tieing are 7:1. Can these odds be right? Show why or why not these claims make sense. Thanx Ballin I find it easiest to work these problems by rewriting the odds as probabilities. The probability of winning is 3/4. Similarly, the probability of not losing is 5/6, and the probability of not tying is 7/8. But the probability of tying should be the probability of not losing minus the probability of winning, which we find is: $\\frac{5}{6} - \\frac{3}{4} = \\frac{20}{24} - \\frac{18}{24} = \\frac{2}{24} = \\frac{1}{12}$, whence the probability of not tying is 11/12. So you would conclude that", "Probability of winning a game $A$ and $B$ play a game. • The probability of $A$ winning is $0.55$. • The probability of $B$ winning is $0.35$. • The probability of a tie is $0.10$. The winner of the game is the person who first wins two rounds. What is the probability that $A$ wins? The answer is $0.66$. I don't know how it's coming $0.66$. EDIT : The right combinations according to me are {null,T,TT,TTT....}A{null,T,TT,TTT....}A {null,T,TT,TTT....}A{null,T,TT,TTT....}B{null,T,TT,TTT....}A {null,T,TT,TTT....}B{null,T,TT,TTT....}A{null,T,TT,TTT....}A - What have you tried? – TMM Apr 30 '13 at 15:17 I edited it to show you – Rajarshi Sarkar Apr 30 '13 at 15:19 Ties don't count, don't record them. So in effect we are playing a game in which A has probability $p=\\frac{0.55}{0.90}$ of winning a game, and B has probability $1-p$ of winning a game. Now there are several ways to finish. The least thinking one is that A wins with the pattern AA, or the patterns ABA, or BAA. - They will not play exactly $3$ games in which there is a winner. If A wins the first two non-tie games, the series is over. If A loses one of the first two non-tie games, she wins the series precisely if she wins the third non-tie game. Hence the $3$ cases. Alternately, we can take", "# probability that Janani will win the game? Two people are playing a coin toss game with a fair penny. Manu gets a point if the penny lands on heads. Janani gets a point if the penny lands on tails. The score is Janani 9, Manu 7, in a game to 10 points. What is the probability that Janani will win the game? - ## 3 Answers Hint: Janini will win if Manu loses. Manu wins only if the next three results are heads. - Janini will win only if Manu loses. Probability that Manu wins the game: since Manu has 3 chances is $\\frac{1}{8}\\$. Thus probability that Janini wins is 1- $\\frac{1}{8}\\$ = $\\frac{7}{8}\\$. - The most number of tosses that can happen before someone wins for sure is 2 (if 2 heads are tossed). Thus after 3 tosses, it is impossible for the game to not yet be over. It is clear that Manu needs all of these tosses to be heads, which occurs with probability $\\frac{1}{8}$. Thus, Janani wins with probability $\\frac{7}{8}$. An analogous way to think about this is to sum up the probabilities of getting 1, 2, and 3 tails in 3 tosses, all of which will lead to Janani winning. Even though Janani will win after just 1 tail, if we pretend", "possible outcomes. The probability of a player having arthritis is then the total number of players with arthritis divided by the sample size: P(arthritis)=# of favorable outcomes/# of possible outcomes=(10+9+24)/(10+61+9+206+24+548)=48/858 ~ 0.0501=5.01%", "the probability that you win. What's left to figure out is the probability that you tie. If $p_t$ is the probability of a tie, then the probability that you win is $\\frac{1}{2}(1-p_t)$. – TravisJ Jun 22 '15 at 19:56", "decks, namely, $\\frac12$. The only effect of not replacing the first player’s card is to decrease the expected number of tied rounds before the game is won or lost. - All orderings of the cards are equally likely. It follows that the probability the first player wins is equal to the probability the second player wins. -", "# Probability of Game Outcomes (with ties!) Let's say I'm playing 10 \"games\". For each game, I know the probability of winning, the probability of tying, and the probability of losing. From these values, I can calculate the probability of winning X games, the probability of losing X games, and the probability of tying X games (for X = 0 to 10). I'm just trying to figure out the probability of each outcome after playing all 10 games. For example, what is the probability of winning 7, losing 2, and tying 1? Any ideas? Thanks! Here's some example data for 2 games: Game 1: • win: 23.3% • tie: 1.1% • lose: 75.6% Game 2: • win: 29.5% • tie: 3.2% • lose: 67.3% Based on this, we can calculate the probability after playing 2 games of: • 0 wins: 54.0% • 1 win: 39.1% • 2 wins: 6.9% • 0 ties: 95.8% • 1 tie: 4.2% • 2 ties: 0.0% • 0 losses: 8.0% • 1 loss: 41.1% • 2 losses: 50.9% Based on these numbers, what is the probability of each outcome? The possible outcomes (W-L-T) would be: • 2-0-0 • 1-1-0 • 1-0-1 • 0-1-1 • 0-2-0 • 0-0-2 Obviously this is pretty trivial with 2 games, but is there a generic formula for finding", "## If the probability of losing a game is 1 /5 , then find the probability of winning the game Question If the probability of losing a game is 1 /5 , then find the probability of winning the game in progress 0 1 month 2021-08-14T04:53:49+00:00 1 Answer 0 views 0 ## Answers ( ) 4/5 Step-by-step explanation: See... when the probability of losing the game is given I. e, we need to find the probability of winning the game.. Now we know that the whole part given is 5 So, Simply we’ll subtract the probability of losing the game from the whole part.. =51/5", "Want to ask us a question? Click here Browse Questions Ad 0 votes # The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is _____________ $\\begin{array}{1 1}(A)\\;0.15\\\\(B)\\;0.10\\\\(C)\\;0.25\\\\(D)\\;0.05\\end{array}$ Can you answer this question? ## 1 Answer 0 votes 0.15 Hence (A) is the correct answer. answered Jul 9, 2014 0 votes 1 answer 0 votes 1 answer 0 votes 1 answer 0 votes 1 answer 0 votes 1 answer 0 votes 1 answer 0 votes 1 answer", "‘Black ball drawn from the first bag’ P(D) = 7/14 Probability of this ball being black = P(A). P(C) + P(B). P(D) = (1/2). (6/14) + (1/2). (7/14) = (1/2). (13/14) = 13/28 Problem 8) The probability that Ajeet will solve a problem is 1/5. What is probability that a) he does not solve a single problem out of ten problems b) he solves at least one problem out of ten problems Sol. Let A be an event ‘Ajeet will solve a problem’ P(A) = 1/5 P(A)’ = 4/5 He solves no problems i.e. he does not solve the first problem and he does not solve the second problem……….. and he does not solve the tenth problem. Probability that he does not solve a single problem out of ten problems = (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) = (1/5)10 b) Let B be an event ‘Ajeet does not solve a single problem out of ten problems’ P(B) = (1/5)10 P(B)’ = 1 – (1/5)10 Problem 9) In a four-game series between Radha and Anand, the probability that Anand wins a particular game is 2/5 and that of Radha winning a game is 3/5. Assuming that there is no probability of a draw in an individual", "53 [1], given: 0 Intern Joined: 03 Oct 2012 Posts: 10 Kudos [?]: 6 [1], given: 23 GMAT 1: 620 Q39 V38 GMAT 2: 700 Q49 V38 Re: 4 games tournament - probability [#permalink] ### Show Tags 05 Dec 2012, 13:16 1 KUDOS Marco83 wrote: Two equally skilled teams play a four games tournament. What is the probability of the tournament ending at the fourth game with a winner? [bonus question: how about a 40 games tournament ending on the 40th game with a winner?] First, we have to determine what we're looking for. If we are looking for a probability that after playing 4 games there will be a winner than we must consider a case of a tie. The statement does not tell us that the teams should either win or lose. It implies there might be ties. Therefore: 1) The total possible options are WLT: 3*3*3*3=81 2) In order to be a winner, a team can do 4W, 3W+L&T, 2W+L+T, 2W+2T, 1W+3T a)4W=1 b)3W=4!/3!=4 and we have T or L as options=4*2=8 c)2W+L+T=4!/2!=12 d)2W+2T=4!/2!*2!=6 e)1W+3T=4!/3!=4 3)Total for one team to become a winner is 1+8+12+6+4=31 4)Total for any of two to become a winner is 31*2=62 5)The probability is 62/81 Kudos [?]: 6 [1], given: 23 Intern Joined: 11 Dec 2009 Posts: 2 Kudos [?]:", "# Efficient Method for Calculating the Probability of a Set of Outcomes? Let's say I'm playing 10 different games. For each game, I know the probability of winning, the probability of tying, and the probability of losing (each game has different probabilities). From these values, I can calculate the probability of winning X games, the probability of losing X games, and the probability of tying X games (for X = 0 to 10). I'm just trying to figure out the probability of winning W games, tying T games, and losing L games after playing all 10 games... and hopefully do better than O(3^n). For example, what is the probability of winning 7, losing 2, and tying 1? Any ideas? Thanks! Edit - here's some example data for if there were only 2 games: Game 1: • win: 23.3% • tie: 1.1% • lose: 75.6% Game 2: • win: 29.5% • tie: 3.2% • lose: 67.3% Based on this, we can calculate the probability after playing 2 games of: • 0 wins: 54.0% • 1 win: 39.1% • 2 wins: 6.9% • 0 ties: 95.8% • 1 tie: 4.2% • 2 ties: 0.0% • 0 losses: 8.0% • 1 loss: 41.1% • 2 losses: 50.9% Based on these numbers, is there a generic formula for finding the probability of", "Why is this wrong? Since the prompt says the winner of the series has to win the last game, and we're told team A wins in 5 games, we're asked to find the probability of it losing the first game under these terms. Thanks! • It's worth noting that you answered the question \"what is the probability that team A will win the series in 5 games...\". However the question you were asked was \"given that team A won series in 5 games...\". The question, \"What is the probability of Team A losing the first game?\" is confusing because it is a question about the past asked in a way that suggests a future outcome. I've found that a number of math problems on MSE and elsewhere revolve around fairly subtle English constructions. – Χpẘ Jul 28 '17 at 19:46 Since we know the series lasted only five games, there are four possible outcomes: $$1. LWWWW\\\\ 2. WLWWW\\\\ 3. WWLWW\\\\ 4. WWWLW$$ Thus, the probability that they Team A lost game one is $1/4$. • But if team A lost game one, how can you have arrangements 2, 3, and 4? – AleksandrH Jul 28 '17 at 18:38 • @AleksandrH No, we don;t know that A lost game one ... we just know they lost one game –", "# Game Probability Question #### system2016 ##### New Member Hi, I am playing a game. I win with an 88% chance and lose with 12% probability. If I have won 20 games in a row what is the probabilty that I will lose the next game. If you can figure out the answer and provide also the calculation logic that would be great. Thanks B #### Buckeye ##### Active Member Is each game independent from other games? There's no opportunity for a tie? Last edited: #### system2016 ##### New Member Is each game independent from other games? There's no opportunity for a tie? No tie is possible its either win or lose. Would be interested to see how the numbers come out for 1) games are independent 2) games are not independent. Thanks #### ab-stats ##### New Member Hi, I am playing a game. I win with an 88% chance and lose with 12% probability. If I have won 20 games in a row what is the probabilty that I will lose the next game. The wording of your question implies that you wish to find $$P(G_{21}=L|G_1=W , G_2=W , ... , G_{20}=W)$$, the probability of losing the 21st game, conditional on having won all of the first 20. If the probability of winning/losing is the same for", "probability that you lose every point. One minus this is the probability that you don’t. 1. Andrew Dale says: That makes sense. Thanks for the explanation", "+0 # Probability 0 87 1 In a certain Algebra 2 class of 30 students, 7 of them play basketball and 11 of themplay baseball. There are 5 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball? Jun 12, 2022 #1 +2448 0 We know that in total, $$7 + 11 - 5 = 13$$ students play either basketball or baseball. This means that the probability is $$\\color{brown}\\boxed{13 \\over 30}$$ Jun 13, 2022", "is the... 0 answers and 6 views An urn contains 4 white and 4 black balls. We randomly choose 4 balls. If 2 of them are white and 2 are black, we stop. If not, we replace the balls in the urn and again randomly select 4 balls. This continues until exactly 2 of the 4 chosen are white. What is the probability that we shall make exactly n selections? 1 answers and 7 views Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the teams is stronger than the other and wins each game with probability .6, independently of the outcomes of the other games. Find the probability, for i = 4, 5, 6, 7, that the stronger team wins the series in exactly i games. Compare the probability that the stronger team wins with the probability that it would win a 2-out of-3 series. 1 answers and 32 views Consider a roulette wheel consisting of 38 numbers 1 through 36, 0, and double 0. If Smith always bets that the outcome will be one of the numbers 1 through 12, what is the probability that (a) Smith will lose his first 5 bets; (b) his first win will occur on his", "Browse Questions # The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is _____________ $\\begin{array}{1 1}(A)\\;0.15\\\\(B)\\;0.10\\\\(C)\\;0.25\\\\(D)\\;0.05\\end{array}$ Can you answer this question? 0.15 Hence (A) is the correct answer. answered Jul 9, 2014", "# Probability Of Winning A Game A series of $$2014$$ games is being played between two players $$A$$ and $$B$$. There are no ties-- exactly one of the players wins in a game and the other loses. The first game is won by $$A$$, and $$B$$ wins the second game. In the consequent games, the probability of $$A$$ winning is equal to $$\\dfrac{P}{P+Q}$$, where $$P$$ and $$Q$$ denote the number of games won by $$A$$ and $$B$$ respectively so far (for example, the probability of $$A$$ winning the third game is $$\\dfrac{1}{2},$$ if $$A$$ wins the third game, the probability of $$A$$ winning the fourth game is $$\\dfrac{2}{3},$$ etc). The probability that the series is tied, i.e. $$A$$ and $$B$$ both win exactly $$\\dfrac{2014}{2}$$ games., can be expressed as $$\\dfrac{a}{b},$$ where $$a$$ and $$b$$ are coprime positive integers. Find $$a+b+1$$. Details and assumptions • This problem is not original. ×", "Probability Grumps The probability that Danny wins a game is twice the probability that Arin wins a game. If each game is independent (that is, the outcome of one game does not affect another), the probability that Danny wins exactly two games out of five is $\\frac{a}{b}$ where $$a$$ and $$b$$ are relatively prime positive integers. Find $$a+b.$$ ×" ]

[ "types. The formula of the probability of an event is: $\\text{probability}=\\dfrac{\\text{number of desired outcomes}}{\\text{total number of favourable outcomes}}$ Or $P(A)=\\dfrac{n(A)}{n(S)}$ Where, $P(A)$ is the probability of an event $A$ $n(A)$ is the number of favourable outcomes And $n(S)$ is the total number of possible outcomes of a set. If the probability of occurring an event is $P(A)$ then the probability of not occurring an event is $P(A')=1-PA$ Now, according to the given question: There are two conditions for the event to be happen: Case1: India wins the toss and wins the match. Case2: India loses the toss and wins the match. Now we will find out the probability of the events: Probability of India wins the toss:$\\dfrac{3}{4}$ Probability of India wins the toss and got victory : $\\dfrac{4}{5}$ If the probability of occurring an event is $P(A)$ then the probability of not occurring an event is $P(A')=1-PA$ Hence, Probability of India loses the toss: $1-\\dfrac{3}{4}$ $=\\dfrac{4-3}{4}$ $=\\dfrac{1}{4}$ Probability of India loses the toss and got victory:$\\dfrac{1}{2}$ Now, $\\text{P(victory)=P(India wins the toss and got victory)+P(India loses the toss and got victory)}$ $\\text{P(victory)=(}\\dfrac{3}{4}\\times \\dfrac{4}{5}\\text{)+(}\\dfrac{\\text{1}}{4}\\times \\dfrac{\\text{1}}{2}\\text{)}$ $\\text{=}\\dfrac{3}{5}+\\dfrac{1}{8}$ $=\\dfrac{24+5}{40}$ $=\\dfrac{29}{40}$ So, the correct answer is “Option 4”. Note: In probability Null set $\\phi$ and sample space $S$ also represent events because both are the subsets of $S$. Here $\\phi$ represents an impossible", "denote the probability over loss sequences conditioned on . We have maxk∈VE[LA,T−Lk,T] =1α(G", "of course that the sum of these probabilities is $1$, as it should be. What I needed to use was:", "Rs 6.00, i.e., Rs 6.00 is the loss. There are $2^{4}=16$ elements in the sample space $S$, which is given by: S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT} $\\therefore n(S)=16$ The person wins Rs 4.00 when 4 heads turn up, i.e., when the event {HHHH} occurs. $\\therefore$ Probability (of winning Rs $4.00$ ) $=\\frac{1}{16}$ The person wins Rs 1.50 when 3 heads and one tail turn up, i.e., when the event {HHHT, HHTH, HTHH, THHH} occurs. $\\therefore$ Probability (of winning Rs $1.50)=\\frac{4}{16}=\\frac{1}{4}$ The person loses Re 1.00 when 2 heads and 2 tails turn up, i.e., when the event {HHTT, HTTH, TTHH, HTHT, THTH, THHT} occurs. $\\therefore$ Probability (of losing Re $1.00)=\\frac{6}{16}=\\frac{3}{8}$ The person loses Rs 3.50 when 1 head and 3 tails turn up, i.e., when the event {HTTT, THTT, TTHT, TTTH} occurs. Probability (of losing Rs $3.50$ ) $=\\frac{4}{16}=\\frac{1}{4}$ The person loses Rs 6.00 when 4 tails turn up, i.e., when the event {TTTT} occurs. Probability (of losing Rs $6.00$ ) $=\\frac{1}{16}$", "win this match. India may win or lose. Values of probability: Probabilities can be expressed in terms of fractions, decimals or percent. The values of probability range from $$0$$ to $$1$$. If the particular event is certain to happen, then that probability will be $$1$$. If the event is cannot possibly occur, then the probability for that event is $$0$$.", "the probability assigned to any subset of the values taken on by $X$ must be $0$, contradicting the fact that the sum of the probabilities over the range of $X$ must be $1$.", "and $C$ are $\\frac{4}{7},\\frac{2}{7}$ and $\\frac{1}{7}$respectively. Note: Sum of probabilities of mutually exclusive events is always $1.$So in case if there were more than three horses, the sum of respective probabilities of winning of each horse would also have been $1.$", "P(A)$ • Option 3) $P(A\\mid B)=1$ • Option 4) $P(A\\mid B)=P(B)-P(A)$ and non null events (given) Since Option 1) Option 2) Option 3) Option 4) In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs.50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is : • Option 1) $0$ • Option 2) $\\frac{400}{3}gain$ • Option 3) $\\frac{400}{9}loss$ • Option 4) $\\frac{400}{3}loss$ Multiplication Theorem of Probability - If A and B are any two events then - wherein where The law of Total Probability - Let S be the sample space and E1, E2, ......En be n mutually exclusive and exhaustive events associated with a random experiment. - wherein where A is any event which occurs with E1, E2, E3......En. p (success)= p ( 5 or 6 ) =1/3 Option 1) Option... Let S = {1,2,.........,20}. A subset B of S is said to be ''nice'' , if the sum of the elements of B is 203. Then the probability that a randomly chosen subset of S is ''nice'' is: • Option 1) $\\frac{7}{2^{20}}$ • Option", "• Event 5 doesn't happen: $7/10$ Since the events are independent, the probability of all these new events occurring simultaneously is just their product, which is $133/1000$. Based on the calculation above $$\\Pr(\\text{at least one event}) = 1 - \\Pr(\\text{none of the events}) = 1 - \\frac{133}{1000} = \\frac{867}{1000} = 86.7\\%.$$ -", "that, The sum of these probabilities is 1.", "• 2008:$3 million (lump-sum payoff). Probability: 1/909,000. • 2010: \\$10 million (lump-sum payoff). Probability: 1/1,200,000. The probability of two or more independent events all happening is the product of their individual probabilities. Compare the probability of Joan winning all four lotteries with the probability of picking, at random, a grain of sand painted red that is hidden somewhere within the Sahara desert. < Previous Section Home Next Section >", "### Expected Value of a Discrete Function • Understand conditions for discrete probability distribution function. • Calculate the expected value of a discrete function. $E(X) = \\Sigma x_i P(x_i)$ ### Solved Example: #### 9-2-01 You pick 3 different numbers between 1 and 9. If you pick all the numbers correctly you win \\$50. What are your expected earnings if it costs \\$1 to play? Solution: Probability of winning, $P(\\mathrm{Winning\\ \\$49}) = \\dfrac{1}{({}^{9}C_{3})} = \\dfrac{1}{84}$Probability of losing,$P(\\mathrm{Losing\\ \\$1}) = 1- \\dfrac{1}{84} = \\dfrac{83}{84}$", "0.9) [1] 97967 > prob_lose_by_kth_turn[97967] [1] 0.8995897 > prob_lose_by_kth_turn[97968] [1] 0.9001809 Similarly, to have a probability less than $0.1$ of the job failing, you need to aim for $96771$ or fewer numbers in the set.", "to the anticipated likelihood of a particular event occurring and represent the winnings you get back (plus your stake) if a particular event happens. So 2/1 (2 to 1) fractional odds say: if the event happens, you’ll get 2 back for every 1 you placed, plus your stake back. If I bet 1 unit at 2/1 and win, I get 3 back: my original 1 plus 2 more. If I bet 3, I get 9 back: my original 3 plus 2 for every 1 I placed. Since I placed 3 1s, I get back 3 x 2 = 6 in winnings. Plus my original 3, which gives me 9 back on 3 staked, a profit of 6. Odds are related (loosely) to the likelihood of an event happening. 2/1 odds represent a likelihood (probability) that an event will happen (1/3) = 0.333… of the time (to cur down on confusion between fractional odds and fractional probabilities, I’ll try to remember to put the fractional probabilities in brackets; so 1/2 is fractional odds of 2 to 1 on, and (1/2) is a probability of one half). To see how this works, consider an evens bet, fractional odds of 1/1, such as someone might make for tossing a coin. The probability of getting heads on a single toss is (1/2);", "= \\dfrac{9}{36}$$ Add the values in the third column to find the expected value: $$\\mu$$ = $$\\dfrac{36}{36}$$ = 1. Use this value to complete the fourth column. Add the values in the fourth column and take the square root of the sum: $\\sigma = \\sqrt{\\dfrac{18}{36}} \\approx 0.7071.$ Example $$\\PageIndex{6}$$ On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Iran was about 21.42%. Suppose you make a bet that a moderate earthquake will occur in Iran during this period. If you win the bet, you win $50. If you lose the bet, you pay$20. Let X = the amount of profit from a bet. $$P(\\text{win}) = P(\\text{one moderate earthquake will occur}) = 21.42%$$ $$P(\\text{loss}) = P(\\text{one moderate earthquake will not occur}) = 100% – 21.42%$$ If you bet many times, will you come out ahead? Explain your answer in a complete sentence using numbers. What is the standard deviation of $$X$$? Construct a table similar to Table and Table to help you answer these questions. $$x$$ $$P(x)$$ $$xP(x)$$ $$(x – \\mu^{2})P(x)$$ win 50 0.2142 10.71 [50 – (–5.006)]2(0.2142) = 648.0964 loss –20 0.7858 –15.716 [–20 – (–5.006)]2(0.7858) = 176.6636 Mean = Expected Value = 10.71 + (–15.716) = –5.006. If you make this bet", "multiplied 9/4 as you did in the last equation. Can you please explain why have you done that? – Sumit Kumar Aug 27 '17 at 6:11 • We are only concerned with expected value $E(N|A)$ of coin flips given that $A$ wins. $A$ wins if $N$ is odd, and loses if $N$ is even. We know how to easily get the total expectation $E(N)$ and the probability of $A$ winning. It is known that if $N$ is a random variable and $A$ is any event, then $E(N|A)$ is given as the weighted arithmetic mean of $E(N|X=x)$ with weights being probabilities of those outcomes $X=x$ that make $A$ happen. Recall that in the discrete case, the weighted mean is $\\frac{w_1x_1+\\cdots+w_nx_n}{w_1+\\cdots+w_n}$ – Roman Chokler Aug 27 '17 at 13:40", "# Calculate probability of pattern of indipendent events I have a long list of independent events. Of these, $$71\\%$$ are WINS and $$29\\%$$ of them are LOSSES. I have calculated the probability of losses with this formula : \\begin{align} 0.29^2 &= P(\\text{two losses}) \\\\ 0.29^3 &= P(\\text{three losses}) \\\\ 0.29^4 &= P(\\text{four losses}) \\end{align} How I can calculate a probability of this specific ordination pattern : loss, loss, win, loss, loss, loss? As they are independent events, The probability of any pattern is simply multiplication of events' individual probabilities. So, here pattern is loss, loss, win, loss, loss, loss $$P = 0.29 * 0.29 * 0.71 * 0.29 * 0.29 * 0.29$$ The easiest way to make sense of this question is to assume you are talking about particular patterns of length $$k$$ on the next $$k$$ trials. Then, the probability of $$A = \\{LL \\text{ on next } 2\\}$$ has $$P(A) = (.29)^2 = 0.0841,$$ as you say. Similarly, $$B = \\{LLL \\text{ on next } 3\\}$$ has $$P(B) = (.29)^3 = 0.0841 = 0.024389.$$ Notice that $$B \\subset A$$ and $$P(B) \\le P(A),$$ as required. Then letting $$M = \\{LLWLLL \\text{ on next } 6\\},$$ we have $$P(M) = (.29)^5(.71) = 0.0015.$$ This is the answer to your question (with the interpretation that we are talking", "chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is: A) A gain of Rs. 27 B) A loss of Rs. 37 C) A loss of Rs. 27 D) A gain of Rs. 37 Answer & Explanation Answer: B) A loss of Rs. 37 Explanation: As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order). The overall resultant will remain same. So final amount with the person will be (in all cases): 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27 Hence the final result is: 64 − 27 37 A loss of Rs.37 Report Error 19 1526", "the probability is 2/4=1/2. D $$\\large \\dfrac{3}{4}$$Hint: Think of the coins as a penny and a dime, and list all possibilities. Question 1 Explanation: Topic: Calculate the probabilities of simple and compound events and of independent and dependent events (Objective 0026). There is 1 question to complete. If you found a mistake or have comments on a particular question, please contact me (please copy and paste at least part of the question into the form, as the numbers change depending on how quizzes are displayed). General comments can be left here.", "a given sequence is $P(W)^{\\text{Number of wins}}*P(L)^{\\text{Number of losses}}$. So the probability that any of these sequences occurs is $\\sum_{k=1}^9 2(0.4)^k(0.6)^{10-k}$." ]

[ "# Thread: Probabilities and Odds Problems 1. ## Probabilities and Odds Problems Hey, Can someone help me on this plz and explain your solution. A confident and boastful coach claims that on the next game the odds of her team winning are 3:1, the odds against losing are 5:1, and the odds against tieing are 7:1. Can these odds be right? Show why or why not these claims make sense. Thanx Ballin 2. Originally Posted by ballin_sensation Hey, Can someone help me on this plz and explain your solution. A confident and boastful coach claims that on the next game the odds of her team winning are 3:1, the odds against losing are 5:1, and the odds against tieing are 7:1. Can these odds be right? Show why or why not these claims make sense. Thanx Ballin I find it easiest to work these problems by rewriting the odds as probabilities. The probability of winning is 3/4. Similarly, the probability of not losing is 5/6, and the probability of not tying is 7/8. But the probability of tying should be the probability of not losing minus the probability of winning, which we find is: $\\frac{5}{6} - \\frac{3}{4} = \\frac{20}{24} - \\frac{18}{24} = \\frac{2}{24} = \\frac{1}{12}$, whence the probability of not tying is 11/12. So you would conclude that", "not win). Thus his probalities are $.1*.1=.01$ that he loses $0-2$. The probability is $.1*.9=.09$ that he loses the first game and ties the second game to lose $.5 - 1.5$. The probability is $.9*5/9 = .5$ that he ties the first game and loses the second to lose $.5 - 1.5$ and the probability is $.9*4/9 = .4$ that he ties the first and wins the second to win. So a probability of winning at $.4$ and a probability of losing at $.6$ and an expected score of $.09*0 + (.09 + .5)*.5 + .4*1.5$ to $.09*2 + (.09+.5)*1.5 + .4*.5 = .895$ to $1.105$. If he plays boldly the first game the probailities are $\\frac 59*.1 = .05555...$ that he loses $0-2$. The probability is $\\frac 59*.9 = .5$ that he loses $.5-1.5$. The probability is $\\frac 49*.1 = .044444....$ that hee ties $1-1$ and the probability is $\\frac 49*.9 = .4$ that he wins $1.5-5$. With the expected score of $.05555*0 + .5*.5 + .0444444*1 + .4*1.5$ to $.055555*2 + .5*1.5 + .0444444*1 =$ .89444444....$to$1.10555555....$. So best strategy is timid first, then bold if he ties or timid if he loses. ...... Meanwhile we can figure out other strategies if the goal is to minimize probability of losing, or winning first, not losing second, maximizing", "Probability of winning a game $A$ and $B$ play a game. • The probability of $A$ winning is $0.55$. • The probability of $B$ winning is $0.35$. • The probability of a tie is $0.10$. The winner of the game is the person who first wins two rounds. What is the probability that $A$ wins? The answer is $0.66$. I don't know how it's coming $0.66$. EDIT : The right combinations according to me are {null,T,TT,TTT....}A{null,T,TT,TTT....}A {null,T,TT,TTT....}A{null,T,TT,TTT....}B{null,T,TT,TTT....}A {null,T,TT,TTT....}B{null,T,TT,TTT....}A{null,T,TT,TTT....}A - What have you tried? – TMM Apr 30 '13 at 15:17 I edited it to show you – Rajarshi Sarkar Apr 30 '13 at 15:19 Ties don't count, don't record them. So in effect we are playing a game in which A has probability $p=\\frac{0.55}{0.90}$ of winning a game, and B has probability $1-p$ of winning a game. Now there are several ways to finish. The least thinking one is that A wins with the pattern AA, or the patterns ABA, or BAA. - They will not play exactly $3$ games in which there is a winner. If A wins the first two non-tie games, the series is over. If A loses one of the first two non-tie games, she wins the series precisely if she wins the third non-tie game. Hence the $3$ cases. Alternately, we can take", "# probability that Janani will win the game? Two people are playing a coin toss game with a fair penny. Manu gets a point if the penny lands on heads. Janani gets a point if the penny lands on tails. The score is Janani 9, Manu 7, in a game to 10 points. What is the probability that Janani will win the game? - ## 3 Answers Hint: Janini will win if Manu loses. Manu wins only if the next three results are heads. - Janini will win only if Manu loses. Probability that Manu wins the game: since Manu has 3 chances is $\\frac{1}{8}\\$. Thus probability that Janini wins is 1- $\\frac{1}{8}\\$ = $\\frac{7}{8}\\$. - The most number of tosses that can happen before someone wins for sure is 2 (if 2 heads are tossed). Thus after 3 tosses, it is impossible for the game to not yet be over. It is clear that Manu needs all of these tosses to be heads, which occurs with probability $\\frac{1}{8}$. Thus, Janani wins with probability $\\frac{7}{8}$. An analogous way to think about this is to sum up the probabilities of getting 1, 2, and 3 tails in 3 tosses, all of which will lead to Janani winning. Even though Janani will win after just 1 tail, if we pretend", "should expect 1/2 because the number of desired outcomes are equal in both cases. 8. Re: Probability question please help Originally Posted by HeroWise $You mean \\, \\frac{25}{36} and how is it half? Like it means she eventually wins so we dont include her losses but continues and wins in different situations$ You've made a silly mistake in evaluating the probability of continuing the game; it should be 26/36 instead of 25/36. When using this, the limiting sum will be equal to 1/2, hence the probability of winning the game is 1/2 Thread Information Users Browsing this Thread There are currently 1 users browsing this thread. (0 members and 1 guests) Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •", "‘Black ball drawn from the first bag’ P(D) = 7/14 Probability of this ball being black = P(A). P(C) + P(B). P(D) = (1/2). (6/14) + (1/2). (7/14) = (1/2). (13/14) = 13/28 Problem 8) The probability that Ajeet will solve a problem is 1/5. What is probability that a) he does not solve a single problem out of ten problems b) he solves at least one problem out of ten problems Sol. Let A be an event ‘Ajeet will solve a problem’ P(A) = 1/5 P(A)’ = 4/5 He solves no problems i.e. he does not solve the first problem and he does not solve the second problem……….. and he does not solve the tenth problem. Probability that he does not solve a single problem out of ten problems = (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) = (1/5)10 b) Let B be an event ‘Ajeet does not solve a single problem out of ten problems’ P(B) = (1/5)10 P(B)’ = 1 – (1/5)10 Problem 9) In a four-game series between Radha and Anand, the probability that Anand wins a particular game is 2/5 and that of Radha winning a game is 3/5. Assuming that there is no probability of a draw in an individual", "Why is this wrong? Since the prompt says the winner of the series has to win the last game, and we're told team A wins in 5 games, we're asked to find the probability of it losing the first game under these terms. Thanks! • It's worth noting that you answered the question \"what is the probability that team A will win the series in 5 games...\". However the question you were asked was \"given that team A won series in 5 games...\". The question, \"What is the probability of Team A losing the first game?\" is confusing because it is a question about the past asked in a way that suggests a future outcome. I've found that a number of math problems on MSE and elsewhere revolve around fairly subtle English constructions. – Χpẘ Jul 28 '17 at 19:46 Since we know the series lasted only five games, there are four possible outcomes: $$1. LWWWW\\\\ 2. WLWWW\\\\ 3. WWLWW\\\\ 4. WWWLW$$ Thus, the probability that they Team A lost game one is $1/4$. • But if team A lost game one, how can you have arrangements 2, 3, and 4? – AleksandrH Jul 28 '17 at 18:38 • @AleksandrH No, we don;t know that A lost game one ... we just know they lost one game –", "in the 3 to 1 ratio, but we can say that the odds are 3 to 1 in favor of the Astros not winning. 3. The probability that the Yankees win is 0.2 and that they don’t win is 0.8, and $$0.8/0.2 = 4$$). So the odds are 4 to 1 against the Yankees winning (“against” because the Yankees are less likely to win than to not win). 4. The probability that the Braves win is 0.08 and that they don’t win is 0.92, and $$0.92/0.08 = 11.5$$). So the odds are 11.5 to 1, or 23 to 2, against the Braves winning (“against” because the Braves are less likely to win than to not win). • The odds of an event is a ratio involving the probability that the event occurs and the probability that the event does not occur \\begin{aligned} \\text{odds in favor} & = \\frac{\\text{probability that the event occurs}}{\\text{probability that the event does not occur}} \\\\ & \\\\ \\text{odds against} & = \\frac{\\text{probability that the event does not occur}}{\\text{probability that the event occurs}}\\end{aligned} • In many situations (e.g., gambling) odds are implicitly reported as odds against. • Odds are usually expressed as whole numbers, e.g., 11 to 1, 7 to 2. Ron and Leslie make the following bet. If Boston wins, Leslie will pay Ron", "# Efficient Method for Calculating the Probability of a Set of Outcomes? Let's say I'm playing 10 different games. For each game, I know the probability of winning, the probability of tying, and the probability of losing (each game has different probabilities). From these values, I can calculate the probability of winning X games, the probability of losing X games, and the probability of tying X games (for X = 0 to 10). I'm just trying to figure out the probability of winning W games, tying T games, and losing L games after playing all 10 games... and hopefully do better than O(3^n). For example, what is the probability of winning 7, losing 2, and tying 1? Any ideas? Thanks! Edit - here's some example data for if there were only 2 games: Game 1: • win: 23.3% • tie: 1.1% • lose: 75.6% Game 2: • win: 29.5% • tie: 3.2% • lose: 67.3% Based on this, we can calculate the probability after playing 2 games of: • 0 wins: 54.0% • 1 win: 39.1% • 2 wins: 6.9% • 0 ties: 95.8% • 1 tie: 4.2% • 2 ties: 0.0% • 0 losses: 8.0% • 1 loss: 41.1% • 2 losses: 50.9% Based on these numbers, is there a generic formula for finding the probability of", "+0 # Probability 0 87 1 In a certain Algebra 2 class of 30 students, 7 of them play basketball and 11 of themplay baseball. There are 5 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball? Jun 12, 2022 #1 +2448 0 We know that in total, $$7 + 11 - 5 = 13$$ students play either basketball or baseball. This means that the probability is $$\\color{brown}\\boxed{13 \\over 30}$$ Jun 13, 2022", "# Probability of Game Outcomes (with ties!) Let's say I'm playing 10 \"games\". For each game, I know the probability of winning, the probability of tying, and the probability of losing. From these values, I can calculate the probability of winning X games, the probability of losing X games, and the probability of tying X games (for X = 0 to 10). I'm just trying to figure out the probability of each outcome after playing all 10 games. For example, what is the probability of winning 7, losing 2, and tying 1? Any ideas? Thanks! Here's some example data for 2 games: Game 1: • win: 23.3% • tie: 1.1% • lose: 75.6% Game 2: • win: 29.5% • tie: 3.2% • lose: 67.3% Based on this, we can calculate the probability after playing 2 games of: • 0 wins: 54.0% • 1 win: 39.1% • 2 wins: 6.9% • 0 ties: 95.8% • 1 tie: 4.2% • 2 ties: 0.0% • 0 losses: 8.0% • 1 loss: 41.1% • 2 losses: 50.9% Based on these numbers, what is the probability of each outcome? The possible outcomes (W-L-T) would be: • 2-0-0 • 1-1-0 • 1-0-1 • 0-1-1 • 0-2-0 • 0-0-2 Obviously this is pretty trivial with 2 games, but is there a generic formula for finding", "round play each other. Sportscasters predict that in the first round, NE has a 2/3 probability of beating IA and MN has a 3/5 probability of beating WI. In the second round, they predict that NE would have a 1/2 probability of winning against MN, and a 7/10 probability of winning against WI. Suppose that no game can end in a tie. 1. Are the results of the two first round games dependent or independent events? 2. You trust the sportscasters and fill out a bracket predicting that NE and MN will win the first round. What is the probability that at least part of your prediction turns out to be wrong? 3. Are Nebraska's results in the first and second games dependent or independent events? 4. Given that NE wins the first game, what is the probability that NE wins the second game? Are Nebraska's results in the first and second games dependent or independent events? 5. What is the probability that NE wins the tournament? (That is, what is the probability that NE wins the first game and then wins the second game?) 11. A group of people is introducing themselves to each other at the beginning of a conference. Assume throughout the problem that everyone has a name spelled using the 26 letters of the", "two events, $$A$$ and $$B$$, are independent if and only if: $\\Pr(A\\mid B)=\\Pr(A)$ An example of two independent events would be rolling a 5 with one die after having rolled a 3 with another die. The two events are independent, so the probability of rolling a 5 is always 0.17 regardless of what we rolled on the first die.11 But what if we want to know the probability of some event occurring that requires that multiple events first to occur? For instance, let’s say we’re talking about the Cleveland Cavaliers winning the NBA championship. In 2016, the Golden State Warriors were 3–1 in a best-of-seven playoff. What had to happen for the Warriors to lose the playoff? The Cavaliers had to win three in a row. In this instance, to find the probability, we have to take the product of all marginal probabilities, or $$\\Pr(\\cdot)^n$$, where $$\\Pr(\\cdot)$$ is the marginal probability of one event occurring, and $$n$$ is the number of repetitions of that one event. If the unconditional probability of a Cleveland win is 0.5, and each game is independent, then the probability that Cleveland could come back from a 3–1 deficit is the product of each game’s probability of winning: $\\text{Win probability} =\\Pr\\big(W,W,W\\big)= (0.5)^3= 0.125$ Another example may be helpful. In Texas Hold’em poker, each player", "will happen in theory. If all outcomes are equally likely, it is calculated as the number of favourable outcomes divided by the total number of possible outcomes. For example, if two teams play a sport against each other 10 times (and we assume that there will not be a tie), we expect Team A to win five games and Team B to win five games because the probability of each team winning is 50%, $\\frac{1}{2}$, or 0.5 as both teams have equal chances to win. The theoretical probability is determined by dividing the number of favourable (desired) outcomes by the total number of possible outcomes. $\\text{Theoretical Probability}=\\frac{\\text{Number of favourable (desired) outcomes}}{\\text{Total number of possible outcomes}}$ Experimental probability is the measurement of the likelihood of an event happening based on the results gathered from running many trials of an experiment. For example, we conducted an experiment by observing two teams play 10 times (and we assume that there will not be a tie). We record four wins for Team A and six wins for Team B. The experimental probability of Team A winning is and the experimental probability of Team B winning is $\\frac{6}{10}.$ ## Task 2: Five option probability Imagine a spinner divided into five equal sections or an electronic game that provides the same options. The event", "is the... 0 answers and 6 views An urn contains 4 white and 4 black balls. We randomly choose 4 balls. If 2 of them are white and 2 are black, we stop. If not, we replace the balls in the urn and again randomly select 4 balls. This continues until exactly 2 of the 4 chosen are white. What is the probability that we shall make exactly n selections? 1 answers and 7 views Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the teams is stronger than the other and wins each game with probability .6, independently of the outcomes of the other games. Find the probability, for i = 4, 5, 6, 7, that the stronger team wins the series in exactly i games. Compare the probability that the stronger team wins with the probability that it would win a 2-out of-3 series. 1 answers and 32 views Consider a roulette wheel consisting of 38 numbers 1 through 36, 0, and double 0. If Smith always bets that the outcome will be one of the numbers 1 through 12, what is the probability that (a) Smith will lose his first 5 bets; (b) his first win will occur on his", "Race is run between Mahesli and John ∴ P(E) + P($$\\bar { E }$$ )=1 Where P(E) is the probability of lose the race by John and P($$\\bar { E }$$ ) is the probability of not losing or winning the race by Mahesh But P( $$\\bar { E }$$ ) = 0.54 then 0.54 + P(E) = 1 ⇒ P(E) = 1 – 0.54 = 0.46 ∴ Probability of winning the race by John = 0.46 . Question 9. (i) Write the probability of a sure event. (ii) Write the probability of an event which is impossible. (iii) For an event E. write a relation representing the range of values of P(E). Solution: (i) We know that if the probability of an event is 1 then the probability is called a certain event or a sure event Hence probability of a sure event = 1 (ii) The probability of an event which is impossible = 0 (iii) Probability of no event can be less than 0 and more than 1 and E be any event then 0 ≤ P(E) ≤ 1 Question 10. In a single throw of a die, find the probability of getting : (i) 5 (ii) 8 (in) a number less than 8. (iv) a prime number. Solution: On a die the numbers are", "possible outcomes. The probability of a player having arthritis is then the total number of players with arthritis divided by the sample size: P(arthritis)=# of favorable outcomes/# of possible outcomes=(10+9+24)/(10+61+9+206+24+548)=48/858 ~ 0.0501=5.01%", "Question Bob and Meena play a two-person game which is won by the first person to accumulate 10 points. At each turn Bob gains a point with probability of $\\frac{1}{3}$ . If he doesn’t get a point, then Meena gets a point. Meena is now ahead 9 to 8. What is the probability that Meena will win 1. the probability that Meena will win = 8/9 ### What is the probability? Mathematical explanations of the likelihood that an event will occur or that a statement is true are referred to as probabilities. A value between 0 and 1 represents the likelihood of an event, with 0 basically denoting impossibility and 1 generally indicating certainty. ### According to the given information: Meena must triumph in the subsequent turn or the one after it in order to win; else, she loses. The likelihood of that is determined by deducting the likelihood of the complimentary event, Bob winning in the following two turns, from one. The likelihood of Bob winning the following two turns may be calculated by taking the square of the probability of him winning on one turn because each turn is independent of the others; hence, it is = (1/3)^2 = 1/9 the probability that Meena will win = 1- 1/9 = 8/9 To know more about probability", "below (taken from http://khamel83.tripod.com/intro.htm), players don’t have set positions in the triangle offense. Rather, there are regions based on optimality and spacing: Many examples can be found from teams playing in the triangle offense system of guards posting up, big men coming out to shoot threes, etc… ## An Analysis of The 2015 NBA Finals Matchup The NBA finals are exactly five days away, and I wanted to present an analysis breaking down the matchup between The Golden State Warriors and Cleveland Cavaliers. I used machine and statistical learning techniques to generate the most probable scenarios for the outcome of each game, and this is what I found. Note that the probabilities listed above are not the probabilities for a team to win a specific game, they are the probabilities of a specific scenario occurring. Also, multiple scenarios can occur in a single game, so the probability of multiple scenarios occurring would be the sum of the individual ones. The Model Results So Far (Updated: June 11, 2015) Game 1: Scenario Outcomes: 1 and 2 – GSW win Game 2: Scenario Outcome: 9 – CLE win Game 3: Scenario Outcomes: 5, 8 – CLE win Thoughts so far: Despite GSW being down right now 2-1, I still believe that Cleveland’s wins were statistical anomalies. Cleveland’s Game 2 and", "card is drawn. Find the probability that the card drawn will: be a face card. From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will be a face card of red colour. If A and B are two complementary events then what is the relation between P(A) and P(B)? If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A? In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of: winning of Geeta In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of: not winning of Ritu In a race between Mahesh and John, the probability that John will lose the race is 0.54. Find the probability of: (i) winning of Mahesh (ii) winning of John Write the probability of a sure event Write the probability of an event when impossible For an event E, write a relation representing the range of values of P(E) In a single throw of die, find the probability of getting: 5 In a single throw of die, find the probability of getting: In a single throw of die, find" ]

[ "can run the same method to compute $$a\\pmod{11!}$$.", "db \"eASSWzRD<11$\"", "\\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} & 360 \\cdot x^{44} & 360 \\cdot x^{45} & 120 \\cdot x^{46}\\\\ & & & &", "+0 # Help. I have been stuck 0 114 4 Given that $$\\binom{15}{8}=6435$$$$\\binom{16}{9}=11440$$, and $$\\binom{16}{10}=8008$$, find $$\\binom{15}{10}$$. I know that I should be able to do this but I can't figure it out. Aug 10, 2019 #1 +23295 +3 Given that $$\\dbinom{15}{8}=6435$$, $$\\dbinom{16}{9}=11440$$ , and $$\\dbinom{16}{10}=8008$$, find $$\\dbinom{15}{10}$$. Formula: $$\\dbinom{n}{k} + \\dbinom{n}{k+1} = \\dbinom{n+1}{k+1}$$ $$\\begin{array}{|lrcll|} \\hline & \\mathbf{\\dbinom{n}{k} + \\dbinom{n}{k+1}} &=& \\mathbf{\\dbinom{n+1}{k+1}} \\\\\\\\ (1) & \\dbinom{15}{8} + \\dbinom{15}{9} &=& \\dbinom{16}{9} \\\\ (2) & \\dbinom{15}{9} + \\dbinom{15}{10} &=& \\dbinom{16}{10} \\\\ \\hline (1)-(2): & \\dbinom{15}{8} + \\dbinom{15}{9} -\\left( \\dbinom{15}{9} + \\dbinom{15}{10}\\right) &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & \\dbinom{15}{8} - \\dbinom{15}{10} &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & 6435 - \\dbinom{15}{10} &=& 11440-11440 \\\\ & 6435 - \\dbinom{15}{10} &=& 3432 \\\\ & \\dbinom{15}{10} &=& 6435 - 3432 \\\\ & \\mathbf{\\dbinom{15}{10}} &=& \\mathbf{3003} \\\\ \\hline \\end{array}$$ Aug 11, 2019 #2 0 Thank you so much!! I can't keep all the formulas in my head sometimes. Aug 11, 2019 #3 +1655 +3 There are really too many formulas... tommarvoloriddle Aug 12, 2019 #4 +105540 +1 You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from. There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level. Most times this does", ")^{10} \\\\\\\\ && ( a + b + c )^{n} \\\\ &=& \\sum \\limits_{k=0}^{n} \\sum \\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} &", "(\\mathbf{A}\\mathbf{B})^{-1} \\dbinom12 = \\dbinom13 } \\\\ \\hline \\end{array}$$ Apr 17, 2020", "## Intermediate Algebra for College Students (7th Edition) $x^{10}-5x^8+10x^6-10x^4+5x^2-1$ Apply Binomial Theorem. $(a+b)^n=\\sum_{r=0}^n\\dbinom{n}{r}(a)^{n-r}b^r$ $(x^2-1)^5=\\dbinom{5}{0}(x^2)^5(-1)^0+\\dbinom{5}{1}(x^2)^4(-1)^1+\\dbinom{5}{2}(x^2)^3(-1)^2+\\dbinom{5}{3}(x^2)^2(-1)^3+\\dbinom{5}{4}(x^2)^1(-1)^4+\\dbinom{5}{5}(x^2)^0(-1)^5$ or, $=x^{10}-5x^8+10x^6+10x^4(-1)+5x^2+(1)(1)(-1)$ or, $=x^{10}-5x^8+10x^6-10x^4+5x^2-1$", "this algorithm is $$(^{n}_{10})*n = O(n^{11})$$.", "$10^{-8}$", "{10}^{7}$", "10 \\}$$", "let $N_1=N-\\dbinom{x_1}k$.\" Did you mean $N_1=N-\\dbinom{x_k}{k}$? That would be consistent with what you wrote in your next paragraph: $N_1=\\dbinom{x_{k-1}}{k-1}+\\ldots+\\dbinom{x_1}{1}$ – tychicus Aug 6 '13 at 7:21 • @tychicus: Yes, I did; I’ve fixed it now. (The same mistake was in the original version of your hint, and I must have unconsciously copied that.) – Brian M. Scott Aug 6 '13 at 12:02", "\\;k=10c.}$", "$\\boxed {L = 8 r }$", "Documentation Compute logarithm to the base ten(10). More... ## Modules log10 Compute logarithm to the base ten(10). ## Detailed Description Compute logarithm to the base ten(10). Performs a base ten(10) logarithm on each element of the array.", "polynomial-time algorithm to compute $$\\chi'(G)$$.", "\\mathbf{d}$$...", "10^{-6}$", "{ 10 }^{ n }-1 } + \\ldots$ ×", "@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19 @AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22 Nope, that seems likely to be the most concise way to do it. Count ways to pick $n\\in\\{1,2,3,4\\}$ unique numbers, and to arrange them with $n-1$ \"$>$\" signs, to meet the criteria. $$\\begin{array}{|l:l|} \\hline \\rm a{=}a{=}a{=}a & \\dbinom 3 0 \\dbinom 6 1 \\\\\\hdashline\\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \\dbinom 3 1\\dbinom 6 2 \\\\\\hdashline\\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \\dbinom 3 2 \\dbinom 6 3 \\\\\\hdashline\\rm a{>}b{>}c{>}d & \\dbinom 3 3 \\dbinom 6 4 \\\\ \\hline\\end{array}$$ $$\\dfrac{\\sum_{n=1}^4 \\dbinom{3}{n-1}\\dbinom{6}{n}}{6^4} = \\dfrac{\\dbinom 6 1 + 3\\dbinom 6 2 + 3 \\dbinom 6 3 + \\dbinom 6 4}{6^4}$$ - The problem reduces to finding the cardinality of the following set $$A=\\{(a_1,a_2,a_3,a_4)\\in\\{1,\\ldots,6\\}^4: \\ a_1\\leq a_2\\leq a_3 \\leq a_4\\}.$$ Define $$B=\\{(a_1,a_2,a_3,a_4)\\}\\in\\{1,\\ldots 9\\}^4: \\ a_1 < a_2 < a_3 < a_4\\}$$ and $$f:A\\to B, \\quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \\choose 4}.$$ - @Shailesh This method actually does give the correct answer ($126/6^4$); If we" ]

[ "## Pascal’s Triangle Combinations – product of 6 integers $\\dbinom{0}{0}$ $\\dbinom{1}{0}$ $\\dbinom{1}{1}$ $\\dbinom{2}{0}$ $\\dbinom{2}{1}$ $\\dbinom{2}{2}$ $\\dbinom{3}{0}$ $\\dbinom{3}{1}$ $\\dbinom{3}{2}$ $\\dbinom{3}{3}$ $\\dbinom{4}{0}$ $\\dbinom{4}{1}$ $\\dbinom{4}{2}$ $\\dbinom{4}{3}$ $\\dbinom{4}{4}$ $\\dbinom{5}{0}$ $\\dbinom{5}{1}$ $\\dbinom{5}{2}$ $\\dbinom{5}{3}$ $\\dbinom{5}{4}$ $\\dbinom{5}{5}$ $\\dbinom{6}{0}$ $\\dbinom{6}{1}$ $\\dbinom{6}{2}$ $\\dbinom{6}{3}$ $\\dbinom{6}{4}$ $\\dbinom{6}{5}$ $\\dbinom{6}{6}$ Take, for example, $\\dbinom{5}{3}$ and the six integers surronding it, $\\dbinom{4}{2}$ $\\dbinom{4}{3}$ $\\dbinom{5}{4}$ $\\dbinom{6}{4}$ $\\dbinom{6}{3}$ $\\dbinom{5}{2}$ The product of these 6 integers is: $\\dbinom{4}{2} \\cdot \\dbinom{4}{3} \\cdot \\dbinom{5}{4} \\cdot \\dbinom{6}{4} \\cdot \\dbinom{6}{3} \\cdot \\dbinom{5}{2}$ $= 6 \\times 4 \\times 5 \\times 15 \\times 20 \\times 10$ $= \\; 360000$ $= \\; 600^2$ The product of 6 integers surrounding a number interior is always a square number $\\dbinom{n}{k} \\; = \\; n! \\,/ \\,(k! \\, (n-k)!)$ $\\dbinom{n}{k} \\; = \\; \\dbinom{n-1}{k-1} \\; + \\; \\dbinom{n-1}{k}$ We have $\\dbinom{n}{k}$ surrounded by 6 integers: ………. $\\dbinom{n-1}{k-1}$ $\\dbinom{n-1}{k}$ ….. $\\dbinom{n}{k-1}$ $\\dbinom{n}{k}$ $\\dbinom{n}{k+1}$ ………. $\\dbinom{n+1}{k}$ $\\dbinom{n+1}{k+1}$ $\\dbinom{n-1}{k-1} \\cdot \\dbinom{n-1}{k} \\cdot \\dbinom{n}{k+1} \\cdot \\dbinom{n+1}{k+1} \\cdot \\dbinom{n+1}{k} \\cdot \\dbinom{n}{k-1}$ $\\dbinom{n-1}{k-1} \\; = \\; (n-1)! \\,/ \\,((k-1)! \\, (n-k)!)$ $\\dbinom{n-1}{k} \\; = \\; (n-1)! \\,/ \\,(k! \\, (n-k-1)!)$ $\\dbinom{n}{k+1} \\; = \\; n! \\,/ \\,((k+1)! \\, (n-k-1)!)$ $\\dbinom{n+1}{k+1} \\; = \\; (n+1)! \\,/ \\,((k+1)! \\, (n-k)!)$ $\\dbinom{n+1}{k} \\; = \\; (n+1)! \\,/ \\,(k! \\, (n-k+1)!)$ $\\dbinom{n}{k-1} \\; = \\; n! \\,/ \\,((k-1)! \\, (n-k+1)!)$ numerator: $(n-1)! \\, (n-1)! \\, n! \\, (n+1)! \\, (n+1)! \\, n! \\; = \\;", "+0 # Help. I have been stuck 0 114 4 Given that $$\\binom{15}{8}=6435$$$$\\binom{16}{9}=11440$$, and $$\\binom{16}{10}=8008$$, find $$\\binom{15}{10}$$. I know that I should be able to do this but I can't figure it out. Aug 10, 2019 #1 +23295 +3 Given that $$\\dbinom{15}{8}=6435$$, $$\\dbinom{16}{9}=11440$$ , and $$\\dbinom{16}{10}=8008$$, find $$\\dbinom{15}{10}$$. Formula: $$\\dbinom{n}{k} + \\dbinom{n}{k+1} = \\dbinom{n+1}{k+1}$$ $$\\begin{array}{|lrcll|} \\hline & \\mathbf{\\dbinom{n}{k} + \\dbinom{n}{k+1}} &=& \\mathbf{\\dbinom{n+1}{k+1}} \\\\\\\\ (1) & \\dbinom{15}{8} + \\dbinom{15}{9} &=& \\dbinom{16}{9} \\\\ (2) & \\dbinom{15}{9} + \\dbinom{15}{10} &=& \\dbinom{16}{10} \\\\ \\hline (1)-(2): & \\dbinom{15}{8} + \\dbinom{15}{9} -\\left( \\dbinom{15}{9} + \\dbinom{15}{10}\\right) &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & \\dbinom{15}{8} - \\dbinom{15}{10} &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & 6435 - \\dbinom{15}{10} &=& 11440-11440 \\\\ & 6435 - \\dbinom{15}{10} &=& 3432 \\\\ & \\dbinom{15}{10} &=& 6435 - 3432 \\\\ & \\mathbf{\\dbinom{15}{10}} &=& \\mathbf{3003} \\\\ \\hline \\end{array}$$ Aug 11, 2019 #2 0 Thank you so much!! I can't keep all the formulas in my head sometimes. Aug 11, 2019 #3 +1655 +3 There are really too many formulas... tommarvoloriddle Aug 12, 2019 #4 +105540 +1 You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from. There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level. Most times this does", "= \\boxed{\\textbf{(B)}\\ 24}.$", "i.e. numbering $52$ Thus $Pr =$ $\\dbinom{52}{2} \\over \\dbinom{157}{2}$", "- 36 = \\boxed{\\textbf{(E)}\\; 64}$", "is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$.", "\\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} & 360 \\cdot x^{44} & 360 \\cdot x^{45} & 120 \\cdot x^{46}\\\\ & & & &", "}8)^2 < 81$ and $(9\\text{ to }80)^2 > 80$). Thus $m = \\boxed{243}$. ~Shreyas S", "= \\boxed{\\textbf{(C) }\\frac{85}{8}\\pi}$.", "+ 8 + 4 + 10 + 5 = \\boxed{\\textbf{(C) }45}$ ~sugar_rush", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "= 0$ $r = 1, 9$ Quite obviously $r > 1$, so $r = 9 \\boxed{(D)}$.", "60 out of all positions in search order. So the answer is $\\dbinom{7}{3}$ i.e. 35. more less 30 Nov 2015 - 6:58pm These are best explained in Kurose and Ross book of Computer Networking.Please go through it once. more less", "for $1 \\le n \\le 31$), and so our answer is $\\boxed{\\textbf{(B) } 31}$", "gives us $\\dbinom{n+1}{2}$ ways, for a total of $3\\dbinom{n+1}{2}$ ways. The last case gives us $\\dbinom{n+1}{1}=n+1$ ways. Therefore, the total number of possible ways where there are no isolated men is $\\[\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1).\\]$ The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or $\\[2\\dbinom{n+1}{2}+(n+1).\\]$ Thus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that $\\[\\dfrac{2\\dbinom{n+1}{2}+(n+1)}{\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1)}\\le\\dfrac{1}{100}.\\]$ After simplification, we arrive at $\\[\\dfrac{6(n+1)}{n^2+8n+6}\\le\\dfrac{1}{100}.\\]$ Simplifying again, we see that we seek the smallest positive integer value of $n$ such that $n(n-592)\\ge594$. Clearly $n>592$, or the left side will not even be positive; we quickly see that $n=593$ is too small but $n=\\boxed{594}$ satisfies the inequality.", ")^{10} \\\\\\\\ && ( a + b + c )^{n} \\\\ &=& \\sum \\limits_{k=0}^{n} \\sum \\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} &", "females must also be picked, for a total of $\\dbinom{12}{12}$. Therefore, these cases can be combined to $$\\dbinom{11}{0} \\cdot \\left(\\dbinom{12}{1} + \\dbinom{12}{12}\\right)$$ Since $\\dbinom{12}{12} = \\dbinom{12}{0}$, and $\\dbinom{12}{0} + \\dbinom{12}{1} = \\dbinom{13}{1}$, we can further simplify this to $$\\dbinom{11}{0} \\cdot \\dbinom{13}{1}$$ All other cases proceed similarly. For example, the case with one men or ten men is equal to $\\dbinom{11}{1} \\cdot \\dbinom{13}{2}$. Now, if we factor out a $13$, then all cases except the first two have a factor of $121$, so we can factor this out too to make our computation slightly easier. The first two cases (with $13$ factored out) give $1+66=67$, and the rest gives $121(10+75+270+504) = 103,939$. Adding the $67$ gives $104,006$. Now, we can test for prime factors. We know there is a factor of $2$, and the rest is $52,003$. We can also factor out a $7$, for $7,429$, and the rest is $17 \\cdot 19 \\cdot 23$. Adding up all the prime factors gives $2+7+13+17+19+23 = \\boxed{081}$. ### Video Solution: (Solves using both methods - Casework and Vandermonde's Identity) ## Solution 3 (Vandermonde's identity) Applying Vandermonde's identity by setting $m=12$, $n=11$, and $r=11$, we obtain $\\binom{23}{11}\\implies$ $\\boxed{081}$. ~Lcz ### Short Proof Consider the following setup: $[asy] size(1000, 100); for(int i=0; i<23; ++i){ dot((i, 0)); } draw((10.5, -1.5)--(10.5, 1.5), dashed); [/asy]$ The", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19 @AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22 Nope, that seems likely to be the most concise way to do it. Count ways to pick $n\\in\\{1,2,3,4\\}$ unique numbers, and to arrange them with $n-1$ \"$>$\" signs, to meet the criteria. $$\\begin{array}{|l:l|} \\hline \\rm a{=}a{=}a{=}a & \\dbinom 3 0 \\dbinom 6 1 \\\\\\hdashline\\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \\dbinom 3 1\\dbinom 6 2 \\\\\\hdashline\\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \\dbinom 3 2 \\dbinom 6 3 \\\\\\hdashline\\rm a{>}b{>}c{>}d & \\dbinom 3 3 \\dbinom 6 4 \\\\ \\hline\\end{array}$$ $$\\dfrac{\\sum_{n=1}^4 \\dbinom{3}{n-1}\\dbinom{6}{n}}{6^4} = \\dfrac{\\dbinom 6 1 + 3\\dbinom 6 2 + 3 \\dbinom 6 3 + \\dbinom 6 4}{6^4}$$ - The problem reduces to finding the cardinality of the following set $$A=\\{(a_1,a_2,a_3,a_4)\\in\\{1,\\ldots,6\\}^4: \\ a_1\\leq a_2\\leq a_3 \\leq a_4\\}.$$ Define $$B=\\{(a_1,a_2,a_3,a_4)\\}\\in\\{1,\\ldots 9\\}^4: \\ a_1 < a_2 < a_3 < a_4\\}$$ and $$f:A\\to B, \\quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \\choose 4}.$$ - @Shailesh This method actually does give the correct answer ($126/6^4$); If we", "&\\ldots \\\\ &\\dbinom{n}{0}^2 + \\dbinom{n}{1}^2 + \\cdots + \\dbinom{n}{n}^2 &=& \\dbinom{2n}{n} \\\\ \\hline \\end{array}$$ Apr 11, 2019 #2 +1 thanks! Apr 11, 2019" ]

[ "## Pascal’s Triangle Combinations – product of 6 integers $\\dbinom{0}{0}$ $\\dbinom{1}{0}$ $\\dbinom{1}{1}$ $\\dbinom{2}{0}$ $\\dbinom{2}{1}$ $\\dbinom{2}{2}$ $\\dbinom{3}{0}$ $\\dbinom{3}{1}$ $\\dbinom{3}{2}$ $\\dbinom{3}{3}$ $\\dbinom{4}{0}$ $\\dbinom{4}{1}$ $\\dbinom{4}{2}$ $\\dbinom{4}{3}$ $\\dbinom{4}{4}$ $\\dbinom{5}{0}$ $\\dbinom{5}{1}$ $\\dbinom{5}{2}$ $\\dbinom{5}{3}$ $\\dbinom{5}{4}$ $\\dbinom{5}{5}$ $\\dbinom{6}{0}$ $\\dbinom{6}{1}$ $\\dbinom{6}{2}$ $\\dbinom{6}{3}$ $\\dbinom{6}{4}$ $\\dbinom{6}{5}$ $\\dbinom{6}{6}$ Take, for example, $\\dbinom{5}{3}$ and the six integers surronding it, $\\dbinom{4}{2}$ $\\dbinom{4}{3}$ $\\dbinom{5}{4}$ $\\dbinom{6}{4}$ $\\dbinom{6}{3}$ $\\dbinom{5}{2}$ The product of these 6 integers is: $\\dbinom{4}{2} \\cdot \\dbinom{4}{3} \\cdot \\dbinom{5}{4} \\cdot \\dbinom{6}{4} \\cdot \\dbinom{6}{3} \\cdot \\dbinom{5}{2}$ $= 6 \\times 4 \\times 5 \\times 15 \\times 20 \\times 10$ $= \\; 360000$ $= \\; 600^2$ The product of 6 integers surrounding a number interior is always a square number $\\dbinom{n}{k} \\; = \\; n! \\,/ \\,(k! \\, (n-k)!)$ $\\dbinom{n}{k} \\; = \\; \\dbinom{n-1}{k-1} \\; + \\; \\dbinom{n-1}{k}$ We have $\\dbinom{n}{k}$ surrounded by 6 integers: ………. $\\dbinom{n-1}{k-1}$ $\\dbinom{n-1}{k}$ ….. $\\dbinom{n}{k-1}$ $\\dbinom{n}{k}$ $\\dbinom{n}{k+1}$ ………. $\\dbinom{n+1}{k}$ $\\dbinom{n+1}{k+1}$ $\\dbinom{n-1}{k-1} \\cdot \\dbinom{n-1}{k} \\cdot \\dbinom{n}{k+1} \\cdot \\dbinom{n+1}{k+1} \\cdot \\dbinom{n+1}{k} \\cdot \\dbinom{n}{k-1}$ $\\dbinom{n-1}{k-1} \\; = \\; (n-1)! \\,/ \\,((k-1)! \\, (n-k)!)$ $\\dbinom{n-1}{k} \\; = \\; (n-1)! \\,/ \\,(k! \\, (n-k-1)!)$ $\\dbinom{n}{k+1} \\; = \\; n! \\,/ \\,((k+1)! \\, (n-k-1)!)$ $\\dbinom{n+1}{k+1} \\; = \\; (n+1)! \\,/ \\,((k+1)! \\, (n-k)!)$ $\\dbinom{n+1}{k} \\; = \\; (n+1)! \\,/ \\,(k! \\, (n-k+1)!)$ $\\dbinom{n}{k-1} \\; = \\; n! \\,/ \\,((k-1)! \\, (n-k+1)!)$ numerator: $(n-1)! \\, (n-1)! \\, n! \\, (n+1)! \\, (n+1)! \\, n! \\; = \\;", "+0 # Help. I have been stuck 0 114 4 Given that $$\\binom{15}{8}=6435$$$$\\binom{16}{9}=11440$$, and $$\\binom{16}{10}=8008$$, find $$\\binom{15}{10}$$. I know that I should be able to do this but I can't figure it out. Aug 10, 2019 #1 +23295 +3 Given that $$\\dbinom{15}{8}=6435$$, $$\\dbinom{16}{9}=11440$$ , and $$\\dbinom{16}{10}=8008$$, find $$\\dbinom{15}{10}$$. Formula: $$\\dbinom{n}{k} + \\dbinom{n}{k+1} = \\dbinom{n+1}{k+1}$$ $$\\begin{array}{|lrcll|} \\hline & \\mathbf{\\dbinom{n}{k} + \\dbinom{n}{k+1}} &=& \\mathbf{\\dbinom{n+1}{k+1}} \\\\\\\\ (1) & \\dbinom{15}{8} + \\dbinom{15}{9} &=& \\dbinom{16}{9} \\\\ (2) & \\dbinom{15}{9} + \\dbinom{15}{10} &=& \\dbinom{16}{10} \\\\ \\hline (1)-(2): & \\dbinom{15}{8} + \\dbinom{15}{9} -\\left( \\dbinom{15}{9} + \\dbinom{15}{10}\\right) &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & \\dbinom{15}{8} - \\dbinom{15}{10} &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & 6435 - \\dbinom{15}{10} &=& 11440-11440 \\\\ & 6435 - \\dbinom{15}{10} &=& 3432 \\\\ & \\dbinom{15}{10} &=& 6435 - 3432 \\\\ & \\mathbf{\\dbinom{15}{10}} &=& \\mathbf{3003} \\\\ \\hline \\end{array}$$ Aug 11, 2019 #2 0 Thank you so much!! I can't keep all the formulas in my head sometimes. Aug 11, 2019 #3 +1655 +3 There are really too many formulas... tommarvoloriddle Aug 12, 2019 #4 +105540 +1 You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from. There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level. Most times this does", "\\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} & 360 \\cdot x^{44} & 360 \\cdot x^{45} & 120 \\cdot x^{46}\\\\ & & & &", "@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19 @AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22 Nope, that seems likely to be the most concise way to do it. Count ways to pick $n\\in\\{1,2,3,4\\}$ unique numbers, and to arrange them with $n-1$ \"$>$\" signs, to meet the criteria. $$\\begin{array}{|l:l|} \\hline \\rm a{=}a{=}a{=}a & \\dbinom 3 0 \\dbinom 6 1 \\\\\\hdashline\\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \\dbinom 3 1\\dbinom 6 2 \\\\\\hdashline\\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \\dbinom 3 2 \\dbinom 6 3 \\\\\\hdashline\\rm a{>}b{>}c{>}d & \\dbinom 3 3 \\dbinom 6 4 \\\\ \\hline\\end{array}$$ $$\\dfrac{\\sum_{n=1}^4 \\dbinom{3}{n-1}\\dbinom{6}{n}}{6^4} = \\dfrac{\\dbinom 6 1 + 3\\dbinom 6 2 + 3 \\dbinom 6 3 + \\dbinom 6 4}{6^4}$$ - The problem reduces to finding the cardinality of the following set $$A=\\{(a_1,a_2,a_3,a_4)\\in\\{1,\\ldots,6\\}^4: \\ a_1\\leq a_2\\leq a_3 \\leq a_4\\}.$$ Define $$B=\\{(a_1,a_2,a_3,a_4)\\}\\in\\{1,\\ldots 9\\}^4: \\ a_1 < a_2 < a_3 < a_4\\}$$ and $$f:A\\to B, \\quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \\choose 4}.$$ - @Shailesh This method actually does give the correct answer ($126/6^4$); If we", ")^{10} \\\\\\\\ && ( a + b + c )^{n} \\\\ &=& \\sum \\limits_{k=0}^{n} \\sum \\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} &", "this case trivially):$2^3 = 8$The$2$nd of the$1$st pair of consecutive integers whose product is a primorial:$1 \\times 2 = 2 = 2 \\#$$3$rd Term The$3$rd integer$m$after$0$,$1$such that$m! + 1$(its factorial plus$1$) is prime:$2! + 1 = 2 + 1 = 3$The$3$rd integer after$0$,$1$such that its double factorial plus$1$is prime:$2!! + 1 = 3$The$3$rd integer after$0$,$1$which is (trivially) the sum of the increasing powers of its digits taken in order:$2^1 = 2$The$3$rd subfactorial after$0$,$1$:$2 = 3! \\paren {1 - \\dfrac 1 {1!} + \\dfrac 1 {2!} - \\dfrac 1 {3!} }$The$3$rd integer$n$after$-1$,$0$such that$\\dbinom n 0 + \\dbinom n 1 + \\dbinom n 2 + \\dbinom n 3 = m^2$for integer$m$:$\\dbinom 2 0 + \\dbinom 2 1 + \\dbinom 2 2 + \\dbinom 2 3 = 2^2$The$3$rd integer$m$after$0$,$1$such that$m^2 = \\dbinom n 0 + \\dbinom n 1 + \\dbinom n 2 + \\dbinom n 3$for integer$n$:$2^2 = \\dbinom 2 0 + \\dbinom 2 1 + \\dbinom 2 2 + \\dbinom 2 3$The$3$rd palindromic integer after$0$,$1$which is the index of a palindromic triangular number$T_2 = 3$The$3$rd integer$n$after$0$,$1$such that$2^n$contains no zero in its decimal representation:$2^2 = 4$The$3$rd integer$n$after$0$,$1$such that$5^n$contains no zero in its decimal representation:$5^2 = 25$The$3$rd integer$n$after$0$,$1$such that both$2^n$and$5^n$have no zeroes:$2^2 = 4, 5^2 = 25$The$3$rd integer after$0$,$1$which is palindromic in both decimal and ternary:$2_{10} = 2_3$The$3$rd Fibonacci number after$1$,$1$:$2 =", "(\\mathbf{A}\\mathbf{B})^{-1} \\dbinom12 = \\dbinom13 } \\\\ \\hline \\end{array}$$ Apr 17, 2020", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "females must also be picked, for a total of $\\dbinom{12}{12}$. Therefore, these cases can be combined to $$\\dbinom{11}{0} \\cdot \\left(\\dbinom{12}{1} + \\dbinom{12}{12}\\right)$$ Since $\\dbinom{12}{12} = \\dbinom{12}{0}$, and $\\dbinom{12}{0} + \\dbinom{12}{1} = \\dbinom{13}{1}$, we can further simplify this to $$\\dbinom{11}{0} \\cdot \\dbinom{13}{1}$$ All other cases proceed similarly. For example, the case with one men or ten men is equal to $\\dbinom{11}{1} \\cdot \\dbinom{13}{2}$. Now, if we factor out a $13$, then all cases except the first two have a factor of $121$, so we can factor this out too to make our computation slightly easier. The first two cases (with $13$ factored out) give $1+66=67$, and the rest gives $121(10+75+270+504) = 103,939$. Adding the $67$ gives $104,006$. Now, we can test for prime factors. We know there is a factor of $2$, and the rest is $52,003$. We can also factor out a $7$, for $7,429$, and the rest is $17 \\cdot 19 \\cdot 23$. Adding up all the prime factors gives $2+7+13+17+19+23 = \\boxed{081}$. ### Video Solution: (Solves using both methods - Casework and Vandermonde's Identity) ## Solution 3 (Vandermonde's identity) Applying Vandermonde's identity by setting $m=12$, $n=11$, and $r=11$, we obtain $\\binom{23}{11}\\implies$ $\\boxed{081}$. ~Lcz ### Short Proof Consider the following setup: $[asy] size(1000, 100); for(int i=0; i<23; ++i){ dot((i, 0)); } draw((10.5, -1.5)--(10.5, 1.5), dashed); [/asy]$ The", "# Using generating functions to find the coefficient I was looking at an example in my textbook where it was asked to find in how many ways can a police captain distribute 24 rifle shells to four police officers so that each officer gets at least three shells, but not more than eight. Using generating functions: ${f(x)=(x^3+x^4+x^5+...+x^8)^4}$, and we will be looking to find the coefficient of ${x^{24}}$ in $f(x)$ We know that $f(x)= x^{12}((1-x^6)/(1-x))^4$ Therefore the answer will be the coefficient of $x^{12}$ in the expansion of $(1-x^6)^4*(1-x)^{-4}$ Which in turn is equal to $A$: $[1-\\dbinom{4}{1}x^6+\\dbinom{4}{2}x^{12}-\\dbinom{4}{3}x^{18}+x^{24}]*[\\dbinom{-4}{0}+\\dbinom{-4}{1}(-x)+\\dbinom{-4}{2}(-x)^2+...]$ Up to this point I perfectly understand the problem and the proposed solution; but the textbook goes on saying that the above expression is equivalent to $B$: $[\\dbinom{-4}{12}(-1)^{12}-\\dbinom{4}{1}\\dbinom{-4}{6}(-1)^6+\\dbinom{4}{2}\\dbinom{-4}{0}]$ I do not understand how they went from $A$ to $B$ I know that $C$: $[\\dbinom{-4}{12}(-1)^{12}]$ is the coefficient of $x^{24}$ in: $x^{12}[\\dbinom{-4}{0}+\\dbinom{-4}{1}(-x)+\\dbinom{-4}{2}(-x)^2+...]$ But I do not seem to make the link between $C$ and the following subtraction and addition, namely: $[C-\\dbinom{4}{1}\\dbinom{-4}{6}(-1)^6+\\dbinom{4}{2}\\dbinom{-4}{0}]$ • What are the ways to get an $x^{12}$ term in the product $(1 - ax^6 + bx^{12} -\\dotsc)\\cdot (1 - ux + vx^2 - wx^3 +\\dotsc)$? – Daniel Fischer Oct 12 '15 at 8:31 • @DanielFischer there will be $ax^6 * zx^6$ or $bx^{12}*1$ or $1 * mx^{12}$ – O.A.", "= 0$ $r = 1, 9$ Quite obviously $r > 1$, so $r = 9 \\boxed{(D)}$.", "6 is $\\boxed{\\textbf{(D) } 13}$. - mathleticguyyy", "&=& (10+1)^{2011} \\\\ &=& \\dbinom{2011}{0} 10^{2011}1^{0}+\\dbinom{2011}{1} 10^{2010}1^{1} + \\cdots + \\dbinom{2011}{2010} 10^11^{2010} + \\dbinom{2011}{2011} 1^{2011} \\end{aligned} However, all powers of 10 above $10^2$ are divisible by $1000$, so are $0 \\pmod{1000}$. Therefore, we can reduce them to $0$. All terms at the beginning are nullified, and we are left with $\\dbinom{2011}{2009} 10^2 1^{2009} + \\dbinom{2011}{2010} 10^11^{2010} + \\dbinom{2011}{2011} 1^{2011} = 100 \\cdot \\dbinom{2011}{2} + 10 \\cdot \\dbinom{2011}{1} + 1 \\cdot \\dbinom{2011}{0}$ because of the property that $\\dbinom{2011}{2009} = \\dbinom{2011}{2}$, $\\dbinom{2011}{2010} = \\dbinom{2011}{1}$, and $\\dbinom{2011}{2011} = \\dbinom{2011}{0}$. We can write this as \\begin{aligned}100 \\cdot \\dfrac{2011 \\cdot 2010}{2} + 10 \\cdot 2011 + 1 &\\equiv& 100 \\cdot \\dfrac{11 \\cdot 10}{2}+ 10 \\cdot 11 + 1 \\\\ &=& 5500 + 110 + 1 \\\\ &\\equiv& \\boxed{611} \\pmod{1000}\\end{aligned} $\\blacksquare$ 5 years, 11 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$.", "x= \\boxed{\\mathbf{\\pm 4}}$$", "is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$.", "let $N_1=N-\\dbinom{x_1}k$.\" Did you mean $N_1=N-\\dbinom{x_k}{k}$? That would be consistent with what you wrote in your next paragraph: $N_1=\\dbinom{x_{k-1}}{k-1}+\\ldots+\\dbinom{x_1}{1}$ – tychicus Aug 6 '13 at 7:21 • @tychicus: Yes, I did; I’ve fixed it now. (The same mistake was in the original version of your hint, and I must have unconsciously copied that.) – Brian M. Scott Aug 6 '13 at 12:02", "can run the same method to compute $$a\\pmod{11!}$$.", "+ 8 + 4 + 10 + 5 = \\boxed{\\textbf{(C) }45}$ ~sugar_rush", "no men are picked can be grouped with the case where all men are picked. When all men are picked, all women must also be picked, for a total of $\\dbinom{12}{12}$. Therefore, these cases can be combined to $$\\dbinom{11}{0} \\cdot \\left(\\dbinom{12}{1} + \\dbinom{12}{12}\\right)$$ Since $\\dbinom{12}{12} = \\dbinom{12}{0}$, and $\\dbinom{12}{0} + \\dbinom{12}{1} = \\dbinom{13}{1}$, we can further simplify this to $$\\dbinom{11}{0} \\cdot \\dbinom{13}{1}$$ All other cases proceed similarly. For example, the case with one men or ten men is equal to $\\dbinom{11}{1} \\cdot \\dbinom{13}{2}$. Now, if we factor out a $13$, then all cases except the first two have a factor of $121$, so we can factor this out too to make our computation slightly easier. The first two cases (with $13$ factored out) give $1+66=67$, and the rest gives $121(10+75+270+504) = 103,939$. Adding the $67$ gives $104,006$. Now, we can test for prime factors. We know there is a factor of $2$, and the rest is $52,003$. We can also factor out a $7$, for $7,429$, and the rest is $17 \\cdot 19 \\cdot 23$. Adding up all the prime factors gives $2+7+13+17+19+23 = \\boxed{081}$. ## Solution 3 (Vandermonde's identity) Applying Vandermonde's identity by setting $m=12$, $n=11$, and $r=11$, we obtain $\\binom{23}{11}\\implies$ $\\boxed{081}$. ~Lcz ### Short Proof Consider the following setup: $[asy] size(1000, 100); for(int i=0; i<23; ++i){" ]

[ "fair coin had been tossed and landed heads four times in a row, the odds of getting a heads in the next two flips is still $3/4$. The preceding flips don't matter either. – Tavrock Feb 28 '17 at 6:30", "## Saturday, November 25, 2017 ### Three Coin Flips The following game has a $10 entry fee. You are to flip a fair coin three times. The first time it comes up heads you are paid$5. The second time it comes up heads you're paid an additional $7. The third time it comes up heads you're paid$9 more, for a possible maximum prize of $21. Would you pay the$10 entry fee to play? If not, what would be a fair price for this game? Click below for the answer.", "with unfair coin which comes up heads $20 \\%$ of the time. If it comes up heads I will give you $4$ dollars if not, you will give me $1$ dollar. \" In the fair coin case your expected profit: $$2*0.5+(-2)*0.5 = 0$$ In the unfair coin case: $$4*0.2+(-1)*0.8=0$$ It is better not to waste your time with him because your expected profit is $0$.", "# Problem 5 A fair coin is tossed $9$ times. The probability that at least $5$ consecutive head occurs", "# Possible Six-Flip Results When $6$ indistinguishable fair coins are thrown, how many different outcomes are there? Details and Assumptions: • Two outcomes are the same if they contain the same number of heads. ×", "fair coin will be flipped 3 times. What is the probability that the coin will land on tails exactly once? $1/8$ $1/3$ $3/8$ $5/8$ 1 2 3 4 ... 610 You need to have at least 5 reputation to vote a question down. Learn How To Earn Badges.", "# Is a coin fair? [duplicate] This question already has an answer here: Is a coin fair ? In other words, does a coin come up $50$% heads ? If a coin is unfair, then it would not come up $50$% heads. My thoughts : let's first identify the population and the parameter", "# Coin flipping game probability I have a 2 sided fair penny, if I flip heads 4 times in a row I win 10 dollars but it costs 1 dollar to play, should I play the game? Is the answer no because $$(10/2^4)-1$$ is negative? • you are correct. – Matthew Daly Nov 8 at 16:40 Win probability is (0.5)^4 and you'll have $10 Lose probability is 1-(0.5)^4 and it costs you$4 (10*0.0625) + (-4*0.9375) = -3.125 To put it in plain English, you will lose, on average, $3.125 for every $4 you will invest so if you came with a $400 to play 100 games, you will likely to leave with $87.5.", "number $5$ and the fair coin are important.", "# Problem 5 Probability Level 5 A fair coin is tossed $9$ times. The probability that at least $5$ consecutive head occurs For more problems click here × Problem Loading... Note Loading... Set Loading...", "even when rolled? #3: 3 numbers are selected at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction in lowest terms. #4: 9 fair coins are flipped. What is the probability that at least 4 are heads? #5: How many 3 digit numbers does not contain the digit 1 but have at least one digit that is 5? Answer: #1-$$\\dfrac {11} {16}$$ ; #2-$$\\dfrac {3} {4}$$ ; #3- $$\\dfrac {7} {8}$$ ; #4-$$\\dfrac {191} {256}$$ #5: 217", "# Possible Six-Flip Results When $$6$$ indistinguishable fair coins are thrown, how many different outcomes are there? Details and assumptions Two outcomes are the same if they contain the same number of heads. ×", "that the probability of a fair coin coming up 9 or 10 heads in 10 flips is 0.0107, one would expect that in flipping 100 fair coins ten times each, to see a particular (i.e., pre-selected) coin come up heads 9 or 10 times would still be very unlikely, but seeing some coin behave that way, without concern for which one, would be more likely than not. Precisely, the likelihood that all 100 fair coins are identified as fair by this criterion is $(1-0.0107)^{100} \\approx 0.34$. Therefore the application of our single-test coin-fairness criterion to multiple comparisons would more likely than not falsely identify at least one fair coin as unfair. Similar Posts: Your email is never published nor shared.", "697 views If a fair coin is tossed $5$ times and comes up heads four times out of five. Then if the coin is tossed a sixth time under the same conditions, the probability of it turning up heads is: 1. $80\\%$ 2. $125\\%$ 3. $50\\%$ 4. $20\\%$ Coin being fair and events being independent, when the coin is tossed a sixth time, probability of head is 50%. by 212 points 1 vote 1", "$100$, and is considering taking a gamble in which the consumer flips a coin, and gets $20$ she ...", "# You have $5 and your opponent has$10. You flip a fair coin and if heads comes up, your opponent... 1 answer below » You have $5 and your opponent has$10. You flip a fair coin and if heads comes up, your opponent pays you $1. If tails comes up, you pay your opponent$1. The game is finished when one player has all the money or after 100 tosses, whichever comes first. Use simulation to estimate the probability that you end up with all the money and the probability that neither of you goes broke in 100 tosses. PALNATI N Looking for Something Else? Ask a Similar Question", "# How many possible outcomes are there when flipping a coin 10 times? Possible Outcomes flipping fair coin 10 times $= {2}^{10} = 1024$", "is fair, and flip it. Therefore, do not assume that the card is fair in case 2. You need a proof that it is ... Top 50 recent answers are included", "the coin exactly four times, this would be $1/16$, as there is a unique outcome that this event corresponds to: $TTTH$.", "Free Version Easy # Expected Value: Coin Flips APSTAT-KX1LVA Suppose a fair coin is flipped $1,000$ times. Of the $1,000$ trials, $505$ were heads and $495$ were tails. If this same coin was flipped $1,000$ more times, how many heads would be expected? A $495$ B $505$ C $500$ D $502$ E $498$" ]

[ "posting 6 different letters in 2 different post boxes such that at least one letter is posted in each of the boxes. 6. The number of ways of selecting at least one Indian and at least one American for a debate from a group comprising 3 Indians and 4 Americans and no one else. Video explanation will be added soon. #### Choice A ##### The number of outcomes in which at least three heads appears in 6 consecutive tosses of a fair coin. A coin tossed 6 times consecutively could result in outcomes ranging from 0 heads to 6 heads. We have to count the outcomes in which at least three heads appear What does at least three heads mean? 1. Number of outcomes with 3 heads: 6 choose 3 or 6C3 = $$frac{6!}{3! \\times 3!}$ = 20. 2. Number of outcomes with 4 heads: 6 choose 4 or 6C4 = $\\frac{6!}{4! \\times 2!}$ = 15. 3. Number of outcomes with 5 heads: 6 choose 5 or 6C5 = $\\frac{6!}{5! \\times 1!}$ = 6. 4. Number of outcomes with 6 heads: 6 choose 6 or 6C6 = 1 way. So, number of outcomes in which at least 3 heads will appear = 20 + 15 + 6 + 1 = 42 Choice A is NOT one of the", "each of the boxes. 6. The number of ways of selecting at least one Indian and at least one American for a debate from a group comprising 3 Indians and 4 Americans and no one else. ## GMAT Live Online Classes ### Explanatory Answer | GRE Quant SOMA Question 10 #### Choice A: The number of outcomes in which at least three heads appears in 6 consecutive tosses of a fair coin. A coin tossed 6 times consecutively could result in outcomes ranging from 0 heads to 6 heads. We have to count the outcomes in which at least three heads appear; What does at least three heads mean? 1. Number of outcomes with 3 heads: 6 choose 3 or 6C3 = $$frac{6!}{3! × 3!}$ = 20. 2. Number of outcomes with 4 heads: 6 choose 4 or 6C4 = $\\frac{6!}{4! × 2!}$ = 15. 3. Number of outcomes with 5 heads: 6 choose 5 or 6C5 = $\\frac{6!}{5! \\times 1!}$ = 6. 4. Number of outcomes with 6 heads: 6 choose 6 or 6C6 = 1 way. So, number of outcomes in which at least 3 heads will appear = 20 + 15 + 6 + 1 = 42 Choice A is NOT one of the answers. #### Choice B: The number of outcomes in which the", "$0$ successive heads to $2$ successive heads is obviously not possible in one toss, so this probability is $0$. Similarly, you can read off the other two rows. Judging by the last row, this game ends when you get $2$ successive heads, since from here it can never go back down to $0$. I suppose this must have been in the question. EDIT: In an attempt to explain this better, I will also go through the second row. Firstly, we must recognise what the columns and rows mean. The row represents the state in which we start - so the first row represents starting in the state of having $0$ successive heads, etc. The column represents the state in which we finish after $1$ toss of a coin. So the second column represents finishing with $1$ successive head, which means you used to have none, and you have just rolled a heads. Now, the second row is $$\\left(\\begin{matrix}\\frac12 & 0 & \\frac12 \\end{matrix}\\right)$$ The first element in this represents the probability of going from the state of $1$ successive heads to $0$ successive heads (row $2$ corresponds to starting with $1$ successive heads, column $1$ corresponds to ending with $0$ successive heads). Clearly, to go from having $1$ successive heads to having none means we rolled a tails,", "3 = $$\\frac{1}{12}$$. (ii) The sum 12 will occur, when (6, 6) appears, i.e., 6 on the coin as well as 6 on the die. ∴ No. of favourable cases = 1. ∴ Probability of getting the sum 12 is $$\\frac{1}{12}$$. Question 6. There are four men and six women on the city council, If one council member is selected for a committee at random, how likely is it that it is a woman? Solution: There are 10 members of the council. Any one may be selected for a committee. ∴ No. of exhaustive cases = 10. One woman out of 6 may be selected in 6 ways. ∴ No. of favourable cases = 6. ∴ Probability of selection of a woman = $$\\frac{6}{10}$$ = 0.6. Question 7. A fair coin is tossed four times and a person wins ₹ 1 for each head and lose ₹ 1.50 for each tail that turns up. From the sample space, calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts. Solution: There are five ways in which heads and tails appear. (i) No head and 4 tails, appear Money lost = ₹ 4 × 1.50 = ₹ 6.00 There is only 1 way when TTTT occurs. No. of", "the same result. You have $3$ heads to place into $8$ slots, the remaining $5$ must be tails. The number of ways to choose where the heads go is given by $${8\\choose 3}=\\frac{8!}{3!5!}.$$ Note that we could have chosen where we want to place the tails, in ${8\\choose 5}$ ways, which gives the same thing. This is where Brian says it is not really permutations, but combinations.", "I first tried to think about putting 2 unique people in a line of 6. Obviously the different amount of combinations would be $6\\cdot5=30$. Now if this were a circular permutation as in now 2 people sit in a circle of 6 chairs, each unique layout can be put into 6 different perspectives (because you can rotate), giving $\\frac{6\\cdot5}{6}$. Now each bullet is not unique unlike the persons in chairs, so I would do $\\frac{6\\cdot5}{6\\cdot2!} = 2.5$ which intuitively makes no sense... Anyways, I assumed there were 3 cases as outlined above that the bullets could be placed, and each would happen with the same probability so the probability of you living given you didn't spin is $$\\frac{2}{3}\\frac{1}{2} + \\frac{1}{3}\\frac{3}{4} = \\frac{7}{12}$$ I was wondering if this approach was reasonable and how exactly to mathematically deal with the different probabilities of arrangements of the bullets. Thanks - Your $3$ cases are not equally likely. – André Nicolas May 22 '12 at 21:18 Given the first shot is blank, if you don't spin then the probability that the next shot is a bullet is $\\frac{2}{5}$ with two bullets in five places. If you do it is $\\frac26=\\frac13$. So you should spin. The possible patterns are *BB*** **BB** ***BB* ****BB *B*B** *B**B* *B***B **B*B* **B**B ***B*B If the two bullets", "Total: Since the three cases are disjoint, the number of sequences with at least five consecutive heads is $5 + 2 + 1 = 8$. • It appears the solution is 9/128 but your approach seems completely valid. Wonder why... – Ian Limarta Aug 5 '16 at 17:40 • @IanLimarta I suspect that there is a typographical error in the answer key. – N. F. Taussig Aug 6 '16 at 0:03 First of all, the total number of cases are $$128$$, because each coin flips head or tail, and seven such flips are happening, so $$2^7=128$$. Now, suppose five consecutive heads do appear. Let us number the tosses from $$1$$ to $$7$$. the following cases arise, and we will eliminate overlaps when they happen: 1) Tosses $$1$$ to $$5$$ are heads, and tosses $$6$$ and $$7$$ are whatever. Then we have four total events: HHHHHHH,HHHHHTT,HHHHHTH,HHHHHHT. 2) Tosses $$2$$ to $$6$$ are heads, and tosses $$1$$ and $$7$$ are whatever. Then we have the following cases: HHHHHHH,THHHHHH,THHHHHT,HHHHHHT. Two of these cases are covered above, namely HHHHHHH and HHHHHHT. So we have $$4+2$$ = six favourable cases. 3) Tosses $$3$$ to $$7$$ are heads, and tosses $$1$$ and $$2$$ are whatever. Then we have the following cases: HHHHHHH,TTHHHHH,THHHHHH,HTHHHHH. Of these, HHHHHHH and THHHHHH are the repeated cases. So we have", "we would need $$6$$ pennies. Now we have $$4$$ coins for $$60$$ cents. Note that we need at least one quarter, since otherwise the maximum we could make is $$40$$ cents with $$4$$ dimes. One quarter leaves $$35$$ cents, which cannot be accomplished with $$3$$ coins ($$3$$ dimes would only achieve $$30$$ cents). This means that there are $$2$$ quarters, leaving $$10$$ cents. We can see that $$2$$ nickels can make this amount. Therefore, as only the first one was used, we must only have one dime. Thus, A is the correct answer. 21. Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is $$\\dfrac{1}{4}$$ $$\\dfrac{3}{8}$$ $$\\dfrac{1}{2}$$ $$\\dfrac{2}{3}$$ $$\\dfrac{3}{4}$$ ###### Solution(s): They can each either get one or zero heads. The probability that Keiko gets one head is $$\\dfrac{1}{2}.$$ The probability that Ephraim gets one head is $\\dfrac{1}{2} \\cdot \\dfrac{1}{2} + \\dfrac{1}{2} \\cdot \\dfrac{1}{2} = \\dfrac{1}{2}$(heads then tails or tails than heads). This means the probability that they both get one head is $\\dfrac{1}{2} \\cdot \\dfrac{1}{2} = \\dfrac{1}{4}.$ The probably that Keiko gets zero heads is $$\\dfrac{1}{2}.$$ The probability that Ephraim gets zero heads is if both flips are tails: $\\dfrac{1}{2} \\cdot \\dfrac{1}{2} = \\dfrac{1}{4}.$ Therefore, the probability that both flip zero heads is $\\dfrac{1}{2}", "flipping two coins, the number of possible outcomes are four. Let, H1 and T1 be the head and tail of the first coin and H2 and T2 be the head and tail of the second coin respectively and the sample space can be written as Sample Space, S = { (H1, H2), (H1, T2), (T1, H2), (T1, T2) } In general, if you have “n” coins, then the possible number of outcomes will be 2n. Example: If you toss 3 coins, “n” is taken as 3. Therefore, the possible number of outcomes will be 23 = 8 outcomes Sample space for tossing three coins is written as Sample space S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} ## A Die is Thrown When a single die is thrown, it has 6 outcomes since it has 6 faces. Therefore, the sample is given as S = { 1, 2, 3, 4, 5, 6} ## Two Dice are Thrown When two dice are thrown together, we will get 36 pairs of possible outcomes. Each face of the first die can fall with all the six faces of the second die. As there are 6 x 6 possible pairs, it becomes 36 outcomes. The 36 outcome pairs are written as: $\\begin{Bmatrix} (1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6) \\\\ (2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6) \\\\ (3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\\\ (4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6) \\\\", "# A coin is tossed $5$ times. How many possible outcomes contain AT MOST $3$ heads There are $$32$$ possible outcomes in total when a coin is tossed $$5$$ times. I have found that there are 10 possible outcomes that contain exactly$$3$$ heads by using $$5C3=5!/3!2!$$, but how do I find out how many possibilities contain at most $$3$$ heads? Thanks! • Add up the number of possibilities that contain 1,2 or 3 heads. – Noe Blassel May 4 at 7:50 • @NoeBlassel what about $0$ heads? – Mohammad Zuhair Khan May 4 at 7:54 • Right, those too – Noe Blassel May 4 at 8:39 This would be modeled with a sum of binomial coefficients. If you want precisely $$k$$ events to occur out of $$n$$ events, where event can only have success or failure, then the number of corresponding outcomes is $$n$$ choose $$k$$, i.e. $$\\binom n k$$. If you want at most some number of events, you can sum over the corresponding $$k$$ values, the $$k$$ values you deem valid. For example, if you want at most $$3$$ successes, you take the sum of $$\\binom n k$$ for $$k=0,1,2,3$$. In your case, $$n=5$$ and thus your result is $$\\sum_{k=0}^3 \\binom 5 k$$ It is $$\\displaystyle \\frac{32}{2}+\\binom 5 3$$. (Half of the possibilities are for getting", "– Neal Oct 11 '18 at 19:11 • You meant the total number of possibilities is $6!$, not the total probability. – N. F. Taussig Oct 11 '18 at 19:25 • @N.F.Taussig Could you be more specific? I cannot properly obtain where the issue is. – Mark Oct 11 '18 at 19:26 All the ways to seat all of the people $$6!$$ At least $$2$$ girls together. We attach 2 girls, and allocate $$5$$ objects. Making for $$5!$$ allocations $$5!$$ But we can swap the order of the girls. And we have 3 girls who we can make the \"odd-man-out\" $$5!\\cdot 6 = 6!$$ But we have double counted all of the cases where we have all 3 girls together. $$4!3!$$ $$6! - (6! - 4!3!) = 4!3! = 144$$ You don't state it, but I'm assuming the boys are indistinguishable and the girls are indistinguishable. Please clarify this point! Consider three girls. There must be at least one boy separating pairs: $$GBGBG$$. The only question then remains where the last boy can go: $$\\bullet G \\bullet B \\bullet G \\bullet B \\bullet G \\bullet$$. Note that some of these cases are identical, so you can eliminate them How many distinguishable places all told? • I dont know how to calculate that. And is my assumption wrong? –", "\\right\\}$. Some simple sample spaces are discussed below. A coin is tossed When a coin is tossed, it has two possible outcomes: heads and tails. To be brief, heads is denoted by $H$ and tails is denoted by $T$. Thus the sample space consists of heads and tails. In set theory notation, we can write $S$ as: Two coins are tossed When two coins are tossed, there are four possible outcomes. Let ${H_1}$and ${T_1}$ denote the head and tail on the first coin and ${H_2}$ and ${T_2}$ denote the head and tail on the second coin. The sample space $S$ can be written as It may be noted that the sample space of the throw of two coins has $4$ possible points. A sample space of $3$ coins will have ${2^3} = 8$ possible points and for $n$ coins, the number of possible points will be${2^n}$. A die is thrown An ordinary die has six faces. These six faces contain $1,2,3,4,5,6$ dots on them. Thus for a single throw of a die, the sample space has $6$ possible outcomes, which are: Two dice are thrown A die has six faces. Each face of the first die can occur with all six faces of the second die. Thus there are $6 \\times 6 = 36$ possible pairs or points", "that now, in every case last turn will have only $$1$$ option, as it is same as the first one. Hence everytime i will be multiplying by 1 showing last turn has 1 option only. Case - 1: First 2 turns are same. $$4$$ ways.(every possible way will satisfy) Case-2 : First 2 turns are different and third turn is same as first. $$4 \\times 3 \\times 1 = 12$$ ways Case - 3: First 3 turns are different and fourth one is same as the first one : $$4 \\times 3 \\times 2 \\times 1 = 24$$ ways. Case-4 : First four turns are different and fifth turn is same as first turn : $$4 \\times 3 \\times 2 \\times 1 \\times 1 = 24$$ ways. Hence total number of favorable ways = $$64$$. Hence probability = $$\\frac{64}{196} = \\boxed{\\frac{16}{49}} .$$ 5 years ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they", "being replaced. In this case, the head of the string is simply concatenated with the result of converting the rest of the string. Figure 12.3.14 shows an example of its execution. #### ExercisesSelf-Study Exercise ##### Activity12.3.3.Tracing convert(). Use Codelens to step through the convert() method, taking care to notice the different state of the method on each recursive call. It should match Figure 12.3.14. Write a recursive method, addBlanks() that changes each blank in a string into two consecutive blanks, leaving the rest of the string unchanged. ### Subsection12.3.5Printing all Possible Outcomes when Tossing N Coins Suppose that a student who is studying probability wishes to have a Java program that, for any positive integer $$N\\text{,}$$ will print out a list of all possible outcomes when $$N$$ coins are tossed. For purposes of analyzing the problem, it is assumed that the coins are numbered $$1$$ through $$N$$ and that they are tossed one at a time. An outcome will be represented by a string of H's and T's corresponding to heads and tails. Thus, if $$N = 2\\text{,}$$ the string HT represents a head on the first coin and a tail on the second coin. What we need is a method which, when given the number of coins, will print out all strings of this type. In case", "$$g_m(s)=s^m\\sum_w\\frac{c_w}{w-s}$$ for some given $w$ and $c_w$, to conclude that, for every $k\\geqslant m$, $$P(T_m=k)=\\sum_w\\frac{c_w}{w^{k+1-m}}$$ and $$P(T_m\\geqslant k)=\\sum_w\\frac{c_w}{w^{k-m}(w-1)}$$ – Did Apr 23 '18 at 10:04 If you toss a coin $n$ times, there are $2^n$ possible outcomes, each of which has probability $1/2^n$ of occurring. The number of favorable outcomes is found by subtracting those outcomes in which three consecutive heads do not occur from the total number of outcomes. Let $a_n$ be the number of sequences of length $n$ which do not contain three consecutive heads. Since it is not possible for three consecutive heads to occur until the coin has been tossed three times, $a_1 = 2$ and $a_2 = 4$. Since the only outcome in which three consecutive heads occurs when the coin is tossed three times is HHH, $a_3 = 2^3 - 1 = 7$. We write a recurrence relation for the number of sequences of length $n$ that do not contain three consecutive heads. We condition based on the most recent occurrence of tails. The most recent occurrence of tails must occur in one of the last three positions, otherwise we would have three consecutive heads. • An admissible sequence of length $n$ that ends in tails is formed by appending a T to an admissible sequence of length $n - 1$, of", "the sequence of scores is then: $\\color{red}{1}0123234$. Despite a tail on the sixth turn, he won specifically on the eighth turn and not earlier. Your final case is missing then the cases 1,6 | 2,6 | 3,6 | 4,6. You are correct in noting that 5,6 would fall under the first case. Correcting this then, there are a total of $14$ outcomes in the third case, for then $\\frac{16+16+14}{2^8}=\\frac{23}{128}$ giving the answer of $151$ as expected. • of course! Just an oversight. Very clear answer. Thank you. – Lucas May 31 '16 at 22:22 It boils down to counting the last ones. You'll have to have two $T$, and the arrival at $4$ net $H$ must occur exactly at toss $8$, which means it can't occur before. The way I saw this visually was to write down the times that had the winning toss on toss $7$. As soon as I got a $T$ on toss $7$ as I was writing these down, I knew I was double counting. So the two $T$s must be within the first six tosses, but cannot be in the fifth and sixth tosses (because I would have won on toss $4$). Here they are: $$TTHHHHHH$$ $$THTHHHHH$$ $$THHTHHHH$$ $$THHHTHHH$$ $$THHHHTHH$$ $$HTTHHHHH$$ $$HTHTHHHH$$ $$HTHHTHHH$$ $$HTHHHTHH$$ $$HHTTHHHH$$ $$HHTHTHHH$$ $$HHTHHTHH$$ $$HHHTTHHH$$ $$HHHTHTHH$$ Fourteen in all. \"Therefore,", "= a_{j}$ is defined as $P(X = a_{j}) = p_{j}$, for $j\\in\\{1,…n\\}$. As we already know from previous lessons, probabilities $p_{j}$ must satisfy 1. $0 \\leq p_{j} \\leq 1$, for each $j$ 2. $p_{1}+p_{2}+…+p_{n}=1$ Let’s use a coin flip as an example once more. Possible outcomes of flipping a coin are getting heads or tails. One possible random variable would be: $\\bullet$ $X=1$ if the flip of the coin is a head $\\bullet$ $X=0$ if the flip of the coin is a tail You may wonder, how did we come up with numbers $0$ and $1$? We chose them for no specific reason. We could write $\\bullet$ $X=35$ if the flip of the coin is a head $\\bullet$ $X=88$ if the flip of the coin is a tail and $X$ would still be a random variable representing the same event. However, the first example may be more intuitive. Lets flip a coin 4 times. Random variable $X$ could look like this: $\\bullet$ $X=0$ no heads $\\bullet$ $X=1$ if 1 head $\\bullet$ $X=2$ if 2 heads $\\bullet$ $X=3$ if 3 heads $\\bullet$ $X=4$ if 4 heads In conclusion, we gave numerical value to each outcome of an event. This set of values is a random variable $X=\\{0,1,2,3,4\\}$. #### Difference between traditional and random variables A variable is a symbol", "outcomes, all equally likely. Basically, $$S = \\set{(1, 1), \\ldots, (6, 6)}$$. Let $$A$$ represent the event of the sum being 5. Clearly, the combinations resulting in a total of 5 are $$(1, 4), (2, 3), (3, 2), (4, 1)$$. Since there are 4 outcomes, $$P(A) = \\frac{4}{36} = \\frac{1}{9}$$. If two indistinguishable dice were used, then we could no longer tell which die was which. Since the pairs in $$S$$ are no longer ordered, we remove the duplicates to obtain 21 possibilities. However, they now do not all have the same probability of occurring - $$(2, 2)$$ is only half as likely as $$(1, 2)$$ since $$(1, 2) = (2, 1)$$ when the dice are indistinguishable. ## Counting Techniques When there are $$n$$ outcomes that are all equally likely, then the probability of each outcome is $$\\frac{1}{n}$$. If job A can be done in P ways and job B can be done in Q ways, then we can do either job A or job B in $$P + Q$$ ways. Here, \"or\" becomes addition. For example, the probabiity of getting heads or tails is $$P(Heads) + P(Heads) = \\frac 1 2 + \\frac 1 2 = 1$$ If job A can be done in P ways, and for each of those P ways, job B can be", "\\bullet \\bullet \\bullet \\ 3 \\ \\bullet \\bullet \\bullet \\bullet \\ 4 \\ \\bullet \\bullet \\bullet \\bullet \\bullet \\ 5 \\ \\bullet \\bullet \\bullet \\bullet \\bullet \\bullet \\ 6 \\\\&= 6 \\ \\text{total outcomes}$ Step 2: Find the number of favorable outcomes. $\\text{favorable outcomes} &= \\bullet \\ 1 \\ \\bullet \\bullet \\ 2 \\ \\bullet \\bullet \\bullet \\ 3 \\ \\ {\\color{red}\\bullet \\bullet \\bullet \\bullet \\ 4} \\ \\bullet \\bullet \\bullet \\bullet \\bullet \\ 5 \\ \\bullet \\bullet \\bullet \\bullet \\bullet \\bullet \\ 6 \\\\&= 1 \\ \\text{favorable outcome}$ Step 3: Find the ratio of favorable outcomes to total outcomes. $\\text{Favorable}:\\text{Total}= 1:6$ While theoretical probability is based on the ideal, we can also figure out experimental probability. II. Recognize the Experimental Probability of an Event as the Ratio of Successful Outcomes to Trials Attempted Theoretical probability is based on an ideal situation. Since a flipped coin seems equally likely to land up or down, the theoretical probability of landing heads (or tails) is 1 out of 2. Whether or not the coin actually lands on heads (or tails) 1 out of every 2 flips in the real world is something you must determine with experimental probability. Experimental probability is probability based on doing actual experiments – flipping coins, spinning spinners, picking ping pong balls out of a jar,", "from each possible outcome to its probability. empiricaldist provides a class called Pmf that represents a probability mass function. To use Pmf you can import it like this: from empiricaldist import Pmf The following example makes a Pmf that represents the outcome of a coin toss. coin = Pmf() coin['tails'] = 1/2 coin probs tails 0.5 Pmf creates an empty Pmf with no outcomes. Then we can add new outcomes using the bracket operator. In this example, the two outcomes are represented with strings, and they have the same probability, 0.5. You can also make a Pmf from a sequence of possible outcomes. The following example uses Pmf.from_seq to make a Pmf that represents a six-sided die. die = Pmf.from_seq([1,2,3,4,5,6]) die probs 1 0.166667 2 0.166667 3 0.166667 4 0.166667 5 0.166667 6 0.166667 In this example, all outcomes in the sequence appear once, so they all have the same probability, $$1/6$$. More generally, outcomes can appear more than once, as in the following example: letters = Pmf.from_seq(list('Mississippi')) letters probs M 0.090909 i 0.363636 p 0.181818 s 0.363636 The letter M appears once out of 11 characters, so its probability is $$1/11$$. The letter i appears 4 times, so its probability is $$4/11$$. Since the letters in a string are not outcomes of a random process, I’ll use" ]

[ "flipping two coins, the number of possible outcomes are four. Let, H1 and T1 be the head and tail of the first coin and H2 and T2 be the head and tail of the second coin respectively and the sample space can be written as Sample Space, S = { (H1, H2), (H1, T2), (T1, H2), (T1, T2) } In general, if you have “n” coins, then the possible number of outcomes will be 2n. Example: If you toss 3 coins, “n” is taken as 3. Therefore, the possible number of outcomes will be 23 = 8 outcomes Sample space for tossing three coins is written as Sample space S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} ## A Die is Thrown When a single die is thrown, it has 6 outcomes since it has 6 faces. Therefore, the sample is given as S = { 1, 2, 3, 4, 5, 6} ## Two Dice are Thrown When two dice are thrown together, we will get 36 pairs of possible outcomes. Each face of the first die can fall with all the six faces of the second die. As there are 6 x 6 possible pairs, it becomes 36 outcomes. The 36 outcome pairs are written as: $\\begin{Bmatrix} (1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6) \\\\ (2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6) \\\\ (3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\\\ (4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6) \\\\", "I first tried to think about putting 2 unique people in a line of 6. Obviously the different amount of combinations would be $6\\cdot5=30$. Now if this were a circular permutation as in now 2 people sit in a circle of 6 chairs, each unique layout can be put into 6 different perspectives (because you can rotate), giving $\\frac{6\\cdot5}{6}$. Now each bullet is not unique unlike the persons in chairs, so I would do $\\frac{6\\cdot5}{6\\cdot2!} = 2.5$ which intuitively makes no sense... Anyways, I assumed there were 3 cases as outlined above that the bullets could be placed, and each would happen with the same probability so the probability of you living given you didn't spin is $$\\frac{2}{3}\\frac{1}{2} + \\frac{1}{3}\\frac{3}{4} = \\frac{7}{12}$$ I was wondering if this approach was reasonable and how exactly to mathematically deal with the different probabilities of arrangements of the bullets. Thanks - Your $3$ cases are not equally likely. – André Nicolas May 22 '12 at 21:18 Given the first shot is blank, if you don't spin then the probability that the next shot is a bullet is $\\frac{2}{5}$ with two bullets in five places. If you do it is $\\frac26=\\frac13$. So you should spin. The possible patterns are *BB*** **BB** ***BB* ****BB *B*B** *B**B* *B***B **B*B* **B**B ***B*B If the two bullets", "3 = $$\\frac{1}{12}$$. (ii) The sum 12 will occur, when (6, 6) appears, i.e., 6 on the coin as well as 6 on the die. ∴ No. of favourable cases = 1. ∴ Probability of getting the sum 12 is $$\\frac{1}{12}$$. Question 6. There are four men and six women on the city council, If one council member is selected for a committee at random, how likely is it that it is a woman? Solution: There are 10 members of the council. Any one may be selected for a committee. ∴ No. of exhaustive cases = 10. One woman out of 6 may be selected in 6 ways. ∴ No. of favourable cases = 6. ∴ Probability of selection of a woman = $$\\frac{6}{10}$$ = 0.6. Question 7. A fair coin is tossed four times and a person wins ₹ 1 for each head and lose ₹ 1.50 for each tail that turns up. From the sample space, calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts. Solution: There are five ways in which heads and tails appear. (i) No head and 4 tails, appear Money lost = ₹ 4 × 1.50 = ₹ 6.00 There is only 1 way when TTTT occurs. No. of", "\\right\\}$. Some simple sample spaces are discussed below. A coin is tossed When a coin is tossed, it has two possible outcomes: heads and tails. To be brief, heads is denoted by $H$ and tails is denoted by $T$. Thus the sample space consists of heads and tails. In set theory notation, we can write $S$ as: Two coins are tossed When two coins are tossed, there are four possible outcomes. Let ${H_1}$and ${T_1}$ denote the head and tail on the first coin and ${H_2}$ and ${T_2}$ denote the head and tail on the second coin. The sample space $S$ can be written as It may be noted that the sample space of the throw of two coins has $4$ possible points. A sample space of $3$ coins will have ${2^3} = 8$ possible points and for $n$ coins, the number of possible points will be${2^n}$. A die is thrown An ordinary die has six faces. These six faces contain $1,2,3,4,5,6$ dots on them. Thus for a single throw of a die, the sample space has $6$ possible outcomes, which are: Two dice are thrown A die has six faces. Each face of the first die can occur with all six faces of the second die. Thus there are $6 \\times 6 = 36$ possible pairs or points", "1, this results in a 1063 in 1064 chance of getting at least one heads. Neeraj from Wilson School developed a generalization for different numbers of possible outcomes: I noticed that when you add the probabilities together they make a whole. A quick way of figuring out how many times you get at least one head is, that it is always the no. Of possible outcomes minus one over the no. of possible outcomes So: if No of possible outcomes = n the equation would be: P= (n- 1)/n One student suggested how to calculate the number of desired outcomes: If the number of times flipped =p Then the number of outcomes that contain a head is$2^p-1$ So for flipping a coin $10$ times, the number of outcomes with at least one head is $2^{10}-1 = 1024 - 1 = 1023$ Luke from Maidstone Grammar School went further to investigate the next part of the question: When there are 4 green balls in the bag and there are 6 red balls the probability of randomly selecting a green ball is 0.4 ($\\frac{2}{5}$) and the probability selecting a red ball is 0.6 ($\\frac{3}{5}$). If a ball is selected and then replaced the probability of picking a red ball or a green ball is the same every time. When 3", "we would need $$6$$ pennies. Now we have $$4$$ coins for $$60$$ cents. Note that we need at least one quarter, since otherwise the maximum we could make is $$40$$ cents with $$4$$ dimes. One quarter leaves $$35$$ cents, which cannot be accomplished with $$3$$ coins ($$3$$ dimes would only achieve $$30$$ cents). This means that there are $$2$$ quarters, leaving $$10$$ cents. We can see that $$2$$ nickels can make this amount. Therefore, as only the first one was used, we must only have one dime. Thus, A is the correct answer. 21. Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is $$\\dfrac{1}{4}$$ $$\\dfrac{3}{8}$$ $$\\dfrac{1}{2}$$ $$\\dfrac{2}{3}$$ $$\\dfrac{3}{4}$$ ###### Solution(s): They can each either get one or zero heads. The probability that Keiko gets one head is $$\\dfrac{1}{2}.$$ The probability that Ephraim gets one head is $\\dfrac{1}{2} \\cdot \\dfrac{1}{2} + \\dfrac{1}{2} \\cdot \\dfrac{1}{2} = \\dfrac{1}{2}$(heads then tails or tails than heads). This means the probability that they both get one head is $\\dfrac{1}{2} \\cdot \\dfrac{1}{2} = \\dfrac{1}{4}.$ The probably that Keiko gets zero heads is $$\\dfrac{1}{2}.$$ The probability that Ephraim gets zero heads is if both flips are tails: $\\dfrac{1}{2} \\cdot \\dfrac{1}{2} = \\dfrac{1}{4}.$ Therefore, the probability that both flip zero heads is $\\dfrac{1}{2}", "that a fair gamble is a bet that has an expected value of zero, such as paying$0.50 to win $1 in a coin flip with heads or nothing if it lands tails. Fairly priced insurance is when the expected present value of the insurance premiums is equal to the expected loss from the disaster that the insurance protects against, such as the cost of rebuilding a home after a catastrophic fire. Which of the following statements is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Each person has$500 of initial wealth. A coin toss game is offered to each person at a casino where the player can win or lose $500. Each player can flip a coin and if they flip heads, they receive$500. If they flip tails then they will lose $500. Which of the following statements is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Each person has$256 of initial wealth. A coin toss game is offered to each person at a casino where the player can win or lose $256. Each player can flip a coin and if they flip heads, they receive$256. If they flip tails then they will lose $256. Which of the following statements is NOT correct? Mr", "fully reasonable. However, the probability that sometime or other there will be $3$ consecutive heads is $1$, so the answer is $7/64$. – André Nicolas Mar 21 '15 at 4:30", "get exactly $3$ heads is $$abc+acd+abd+bcd.$$ If we want the \"at least\" case then we have to add $abcd$. In the case of $10$ coins the number of possibilities for exactly $3$ heads is ${10 \\choose 3}=120$ and all have to be listed like above. In the \"at least\" case we have much more possibilities to be listed: $\\sum_{i=3}^{10} {10\\choose i}$ and all have to be depicted.", "all know that an unbiased coin has a head (H) & a tail (T) According to the question, a coin is tossed 4 times, possible outcomes are = 24=16$2^{4}=16$ Sample space = {TTTT, TTTH ,TTHT ,TTHH , THTT ,THTH ,THHT ,THHH ,HTTT ,HTTH , HTHT , HTHH , HHTT ,HHTH , HHHT ,HHHH } Question 4 Find the sample space of a coin and a die are tossed together. Sol: We all know that an unbiased coin has a head (H) & a tail (T) Possible outcomes when a die is thrown: 1, 2,3,4,5 and 6 Thus, sample space of a coin and a die when thrown together S = {T6 , T5 , T4 , T3 ,T2 , T1 , H6 , H5 , H4 ,H3 ,H2 ,H1 } Question 5 Find the sample space when a die is rolled and a coin is tossed only if in case there is head in the coin. Sol: We all know that a unbiased coin has a head (H) and a tail (T) Possible outcomes when a die is thrown: 1, 2,3,4,5 and 6 Thus, sample space of a coin and a die when thrown together S = { T , H6 , H5 , H4 ,H3 ,H2 ,H1 } Question 6 Room x is having 2 girls and", "is the expected number of turns…. if the outcome is two heads (H,H) then the prize money is \\$10, if heads, tales (H,T) then it is 7 dollars. Especially, here we mentioned all working Texting Simulator Codes. Parameters: r0: initial (first) seed a: scale factor , so that a*r0 give the first random number m: gives the max. C++ coin toss simulation program. I’ve seen several variations of these circuits that use the 555 to generate a clock signal which is then conditionally fed into a flip-flop when you push a button to “flip the coin”. You flip a coin 30 times and get heads 11 times, so the chance of getting heads is 11\\30. Works amazing and gives line of best fit for any data set. Sample Size Coin Toss Simulation. This resource is part of the Math at the Core: Middle School Collection. People also like. Dice Simulator: Dice are thought to be the oldest gambling device invented by man and have been around since before 2000 BC. Over a large number of tosses, though, the percentage of heads and tails will come to approximate the true probability of each outcome. Indeed, when I tried. 2 Unboxing Simulator Codes – How to Redeem? X is your damage, so if your damage is 2, you will earn", "– Neal Oct 11 '18 at 19:11 • You meant the total number of possibilities is $6!$, not the total probability. – N. F. Taussig Oct 11 '18 at 19:25 • @N.F.Taussig Could you be more specific? I cannot properly obtain where the issue is. – Mark Oct 11 '18 at 19:26 All the ways to seat all of the people $$6!$$ At least $$2$$ girls together. We attach 2 girls, and allocate $$5$$ objects. Making for $$5!$$ allocations $$5!$$ But we can swap the order of the girls. And we have 3 girls who we can make the \"odd-man-out\" $$5!\\cdot 6 = 6!$$ But we have double counted all of the cases where we have all 3 girls together. $$4!3!$$ $$6! - (6! - 4!3!) = 4!3! = 144$$ You don't state it, but I'm assuming the boys are indistinguishable and the girls are indistinguishable. Please clarify this point! Consider three girls. There must be at least one boy separating pairs: $$GBGBG$$. The only question then remains where the last boy can go: $$\\bullet G \\bullet B \\bullet G \\bullet B \\bullet G \\bullet$$. Note that some of these cases are identical, so you can eliminate them How many distinguishable places all told? • I dont know how to calculate that. And is my assumption wrong? –", "# Probability and Binomial distribution of Coin flip with two coins and two trails Consider two coins, $$A$$ and $$B$$. Let $$p_A = 0.6$$ be the probability that a flip of coin $$A$$ gives heads; let $$p_B = 0.4$$ be the corresponding probability for coin $$B$$. Consider the following experiment: $$\\bullet$$ First we pick a coin randomly (with probability $$1/2$$), then flip it. $$\\bullet$$ Then: – If the result of the first flip is heads, we then flip coin $$A$$ once. – Otherwise, we flip coin $$B$$ once. Thus in total we flip $$2$$ times. (The pick and the flips are jointly independent.) Here is a possible outcome of the experiment: $$\\bullet$$ 1. Pick $$A$$, flip $$A$$, get tails. $$\\bullet$$ 2. Flip $$B$$, get tails. Let $$X$$ be the number of heads we get; it is a random variable. (a) Compute the probability mass function of $$X$$, i.e., compute $$P(X = x)$$ for all possible values of $$X$$. Also find the cumulative distribution function $$P(X \\leq x)$$ for $$x \\in \\mathbb{R}$$. (b) Show that $$P(\\text{coin B is used in the second trial}) = 0.5$$. (c) Verify that $$X \\not∼ Binomial(2, 0.5)$$ by comparing the probability mass functions. (d) Consider the events $$C_1 = \\{\\text{first trial gives heads}\\}$$, $$C_2 = \\{\\text{second trial gives heads}\\}$$. Show that $$C_1$$ and $$C_2$$", "that now, in every case last turn will have only $$1$$ option, as it is same as the first one. Hence everytime i will be multiplying by 1 showing last turn has 1 option only. Case - 1: First 2 turns are same. $$4$$ ways.(every possible way will satisfy) Case-2 : First 2 turns are different and third turn is same as first. $$4 \\times 3 \\times 1 = 12$$ ways Case - 3: First 3 turns are different and fourth one is same as the first one : $$4 \\times 3 \\times 2 \\times 1 = 24$$ ways. Case-4 : First four turns are different and fifth turn is same as first turn : $$4 \\times 3 \\times 2 \\times 1 \\times 1 = 24$$ ways. Hence total number of favorable ways = $$64$$. Hence probability = $$\\frac{64}{196} = \\boxed{\\frac{16}{49}} .$$ 5 years ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they", "# Coin Flip Question I'm trying to do the maths on a coin toss game after 100 games but think I am failing. The rules are as follows. 2. we always bet that heads come up 3. minimum bet is 10 coins 4. maximum bet is 80 coins 5. if tails comes up, we lose our bet 6. if heads comes up, we win twice as much as we bet 7. the coin is fair and so the probability of heads is 50% 9. if we lose we double our bet 10. if we lose the maximum bet of 80 we start at a bet of 10 again 11. if we win we start at a bet of 10 again So in this set up, 50% of the time we would get our 10 coins back plus 10 additional coins. When we lose we need to lose four times in a row which I've figured out will be 6.25% which will cost 150 coins. 50% (10 coins) * 50% (20 coins) * 50% (40 coins) * (80 coins) 50% = 6.25% (150 coins total) Would I be right in thinking that this means that the chance of breaking even is 43.75%? So after 100 games we would have the win and loss as below and end up with", "\\bullet \\bullet \\bullet \\ 3 \\ \\bullet \\bullet \\bullet \\bullet \\ 4 \\ \\bullet \\bullet \\bullet \\bullet \\bullet \\ 5 \\ \\bullet \\bullet \\bullet \\bullet \\bullet \\bullet \\ 6 \\\\&= 6 \\ \\text{total outcomes}$ Step 2: Find the number of favorable outcomes. $\\text{favorable outcomes} &= \\bullet \\ 1 \\ \\bullet \\bullet \\ 2 \\ \\bullet \\bullet \\bullet \\ 3 \\ \\ {\\color{red}\\bullet \\bullet \\bullet \\bullet \\ 4} \\ \\bullet \\bullet \\bullet \\bullet \\bullet \\ 5 \\ \\bullet \\bullet \\bullet \\bullet \\bullet \\bullet \\ 6 \\\\&= 1 \\ \\text{favorable outcome}$ Step 3: Find the ratio of favorable outcomes to total outcomes. $\\text{Favorable}:\\text{Total}= 1:6$ While theoretical probability is based on the ideal, we can also figure out experimental probability. II. Recognize the Experimental Probability of an Event as the Ratio of Successful Outcomes to Trials Attempted Theoretical probability is based on an ideal situation. Since a flipped coin seems equally likely to land up or down, the theoretical probability of landing heads (or tails) is 1 out of 2. Whether or not the coin actually lands on heads (or tails) 1 out of every 2 flips in the real world is something you must determine with experimental probability. Experimental probability is probability based on doing actual experiments – flipping coins, spinning spinners, picking ping pong balls out of a jar,", "# Is it worth betting on this case? Let's imagine a coin-flip game, which uses an unbiased coin. Starting with X dollars, your total increases 50% every time you flip heads. But if the coin lands on tails, you lose 40% of your total. You can play this game N times. On each turn, you must bet the total amount you had on the last turn. Is it worth betting on this case? How can we formalize this in order to have this answer in the general case? Is it easier to formalize this for a specific value of X and N? Let us assume, for example X = 100 dollars and N = 100 turns. PS: This scenario appears in an article discussing some ideas of Ole Peters, a theoretical physicist who is claiming that Everything we know about modern economics is wrong. • It's more that everything Ole Peters claims to know about modern economics is wrong. – VARulle Dec 16 '20 at 1:55 The game has positive expected value, so you should play (assuming constant utility of each dollar, which tends to break down somewhat for very large sums of money). If you start with \\$X, after 1 round you have a 50% chance of having \\$1.5X, and a 50% chance of having \\$0.6X, for", "# Toss a fair coin a number of times. [closed] Toss a fair coin, number of times. You always bet on heads and gain one dollar if it is a head and lose one dollar if it is a tail. You stop whenever you gain 3 dollars or you lose 2 dollars. Then which of the following is the correct one? 1. you end up winning with $2\\over 3$ probability 2. you end up winning with $1\\over 3$ probability 3. you end up winning with $3\\over 4$ probability 4. you end up winning with $3\\over 5$ probability 5. None of the above is correct. If we don't know the number of trials, how can we do it? • Wining pobability: $$WWW$$ OR $$LWWWW$$ Feb 17 '16 at 8:14 • Then, how about LLWWWWW, LLLWWWWWW.... can have infinite trial... when should be stop? Feb 17 '16 at 8:18 • $$LL\\implies$$ \"you lose $2$ dollar, you are stopped\" Feb 17 '16 at 8:19 • How about LWLWLWLWLWLWWWW? Feb 17 '16 at 8:20 • Doe you stop when (a) you have won $3$ times in total or lost $2$ times in total (maximum trials would be $4$) or (b) when your net position is $+3$ or $-2$ or (c) you have won $3$ times consecutively or lost $2$ times consecutively? Feb", "compatible with a wealth growing over time. 4. Same anonymous. You wrote: \"These points are made very well in this Towers Watson article I mentioned in an earlier post.\" That article starts with a gross misconception of returns and I am sorry you did not catch that. A 10% return with 0.5 probability and -10% return with 0.5% probability has nothing to do with two consecutive returns of 10% and -10%. That is just a possible sequence like heads first and then tails. This is what they write in their footnote: \"We have been advised that some people, anchored in ensemble averages, will find this result hard to accept and so warrants further explanation. Consider betting $1 on the coin toss and playing for two rounds. Because it is a fair coin, one round will be heads and the other tails. If we get heads first, our$1 becomes $1.10 (a 10% gain). In the second round we get tails, a 10% loss. 10% of$1.10 is $0.11, so we end up with$0.99. Alternatively, if we get tails first, our $1 falls to$0.90. The subsequent heads wins us $0.09 and we, again, end up with$0.99. In the real world we can get lots of heads in a row, but we can also get lots of tails in a row. If", "even when rolled? #3: 3 numbers are selected at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction in lowest terms. #4: 9 fair coins are flipped. What is the probability that at least 4 are heads? #5: How many 3 digit numbers does not contain the digit 1 but have at least one digit that is 5? Answer: #1-$$\\dfrac {11} {16}$$ ; #2-$$\\dfrac {3} {4}$$ ; #3- $$\\dfrac {7} {8}$$ ; #4-$$\\dfrac {191} {256}$$ #5: 217" ]

[ "length and same breadth. Therefore, those two areas are equal. Hence all squares are not congruent. In mathematics, we say that two objects are similar if they have the same shape, but not necessarily the same size. Consider the rectangles shown below. They both have a perimeter of 12 units, but they are not the same triangle. are equal, then we have found two non-congruent triangles with equal perimeters and equal areas. \"IF TWO TRIANGLES HAVE THE SAME AREA THEN THEY ARE CONGRUENT\" Is this a true statement? In other words, if two figures A and B are congruent (see Fig.1) , then using a tracing paper, Fig-1. So, if two figures X and Y are congruent, they must have equal areas. (e) There is no AAA congruence criterion. 756/7 = 108 units2. Thus, a=d. It's very easy for two rectangles to have the same area and different perimeters,or the same perimeter and different areas. Yes, let's take two different rectangles:The first one is 4 inches by 5 inches.The second is 2 inches by 10 inches.Both of these have an area of 20 square inches, and they are not congruent. Dc ) = the area and perimeter of the two polygon have same! We have a/d = 1, we have found two non-congruent triangles equal... You have two", "it possible to fence a rectangular field using all of these pieces? #### #4 CONGRUENT RECTANGLES A rectangle $R$ has integer side lengths $m$ and $n$. Let $D(m,n)$ be the number of ways that $R$ can be cut into congruent rectangles with integer side lengths by means of cuts parallel to the sides of $R$ (with endpoints on opposite sides of $R$). Determine the perimeter $P$ of the rectangle for which $\\frac{D(m, n)}{P}$ is maximised. Puzzles contributed by Dr Michael Evans", "in order to enclose a 1200 square foot rectangular pig pen. The pig pen is adjacent to the barn so he only needs to form three sides of the rectangular area as shown below. What dimensions should the pen be? 5. A rectangle with perimeter 138 units is divided into 8 congruent rectangles as shown in the diagram below. Find the perimeter and area of one of the 8 congruent rectangles. Perimeter: The distance around a shape. Area: The space inside a shape. 2. (a) $x = \\pm 11$ (b) $x = \\pm 2 \\sqrt{5}$ (c) $x = 4,2$ 3. $4 + 4 + 7 + 7 = 22$ Aug 28, 2013 Jan 21, 2015", "areas are equal. Hence all squares are not congruent. In mathematics, we say that two objects are similar if they have the same shape, but not necessarily the same size. Consider the rectangles shown below. They both have a perimeter of 12 units, but they are not the same triangle. are equal, then we have found two non-congruent triangles with equal perimeters and equal areas. \"IF TWO TRIANGLES HAVE THE SAME AREA THEN THEY ARE CONGRUENT\" Is this a true statement? In other words, if two figures A and B are congruent (see Fig.1) , then using a tracing paper, Fig-1. So, if two figures X and Y are congruent, they must have equal areas. (e) There is no AAA congruence criterion. 756/7 = 108 units2. Thus, a=d. It's very easy for two rectangles to have the same area and different perimeters,or the same perimeter and different areas. Yes, let's take two different rectangles:The first one is 4 inches by 5 inches.The second is 2 inches by 10 inches.Both of these have an area of 20 square inches, and they are not congruent. Dc ) = the area and perimeter of the two polygon have same! We have a/d = 1, we have found two non-congruent triangles equal... You have two figures a and B are congruent (", "shorter sides congruent and the 2 longer sides also congruent, we'd have a rectangle.", "other two before, let ’ s center to the find the is... A rectangle and BDC is a polygon is within a circle is equal to 60 ( p! In calculus, it is perpendicular to that side a set of congruent isosceles...., either way this can give you a good estimate of the sides area area_red.", "necessarily the same size. Consider the rectangles shown below. They both have a perimeter of 12 units, but they are not the same triangle. are equal, then we have found two non-congruent triangles with equal perimeters and equal areas. \"IF TWO TRIANGLES HAVE THE SAME AREA THEN THEY ARE CONGRUENT\" Is this a true statement? In other words, if two figures A and B are congruent (see Fig.1) , then using a tracing paper, Fig-1. So, if two figures X and Y are congruent, they must have equal areas. (e) There is no AAA congruence criterion. 756/7 = 108 units2. Thus, a=d. It's very easy for two rectangles to have the same area and different perimeters,or the same perimeter and different areas. Yes, let's take two different rectangles:The first one is 4 inches by 5 inches.The second is 2 inches by 10 inches.Both of these have an area of 20 square inches, and they are not congruent. Dc ) = the area and perimeter of the two polygon have same! We have a/d = 1, we have found two non-congruent triangles equal... You have two figures a and B are congruent ( see Fig.1 ), then your two are. All parts equal one figure over the other completely the parallelogram lesson about congruent.. * ( DC ) =", "# Rectangles Squared Geometry Level 4 Two congruent rectangles of dimensions $$1 \\times x$$ are placed inside a square of side length 1. One of them is vertical, and the other is inclined. Find the side length $$x$$ to 3 decimal places. ×", "Mathematics # A rectangular closet has a perimeter of 10 feet. If the width of the closet is 3 feet what is the length of the closet? #### RosamondAfonso 3 years ago i think 30 i hope i helped a give me thanks", "# The perimeter of a rectangle is 68 inches. The perimeter equals twice the length of L inches, plus twice the width of 9 inches. What is the equation that expresses the perimeter of this rectangle? $2 L + 2 \\cdot 9 = 68$", "# Altered Square Geometry Level 1 The side length of a square was 26, but now two congruent quarter circles have been cut out, as seen above. What is the the perimeter of this new shape? ×", "17. Inscribed rectangle What is the perimeter of a rectangle that is inscribed in a circle whose diameter is 5 dm long? Answer: 14 dm", "# How Many Rectangles Get ready for a fresh new and free brain puzzle game. Draw your rectangles on Square-Inch Grid Paper and cut them out. how-many-rectangles. Discover (and save!) your own Pins on Pinterest. From there, we learned about a special type of quadrilateral whose opposite sides. W = 2L / (L - 2) Obviously, L has to be at least 3 for this to make sense. How many rectangles can be drawn with 1 8 cm as the perimeter? December 20, 2019 Bhavana Gan. Hi, Here we are with math Rectangles and square short trick. Lapghan: 36 x 48 inches = Starting chain of 38. More answers and solutions you can always find in our website. and number of rectangles. In this video, you will learn how to solve various type of rectangles and square easily. Opposite sides are parallel and of equal length (so it is a Parallelogram ). My son did -2-4-2-4 And one more 1-5-1-5 And one more 3-3-3-3 And I need one more rectangles number how you can make. Preview and details Files included (2) flp, 74 KB. A rectangle is a four-sided flat shape where every angle is a right angle (90°). The correct answer to the puzzle is 40 squares. Calculating it every time is cumbersome. For a 7x7 rubik", "side lengths of the triangle? The side lengths of the triangle = 9 centimeters. Explanation: Perimeter of Adriana’s shape = 72 centimeters Number of sides of the triangle shape = 8 The side lengths of the triangle = Perimeter of Adriana’s shape ÷ Number of sides of the triangle shape = 72 centimeters ÷ 8 = 9 centimeters. ### Eureka Math Grade 3 Module 7 Lesson 23 Homework Answer Key Question 1. Rosie draws a square with a perimeter of 36 inches. What are the side lengths of the square? The side lengths of the square = 9 inches. Explanation: Perimeter of the Square = 36 inches Number of sides in the square = 4 The side lengths of the square = Perimeter of the Square ÷ Number of sides in the square = 36 inches ÷ 4 = 9 inches. Question 2. Judith uses craft sticks to make two 24-inch by 12-inch rectangles. What is the total perimeter of the 2 rectangles? Total perimeter of the 2 rectangles = 144 inches. Explanation: Length of the rectangle = 24-inch Width of the rectangle = 12-inch Perimeter of the rectangle = 2 ( Length + Width) = 2 ( 24-inch + 12-inch ) = 2 × 36-inch = 72 inches. Total perimeter of the 2 rectangles = Perimeter of", "8. A rectangle and a semi-circle. The rectangle has a length of 9 inches and a width of 4.5 inches. The diameter of the circle matches the length. 9. A rectangle and a semi-circle. The rectangle has a length of 7 feet and a width of 4 feet. The diameter of the circle matches the length. 10. A rectangle and a semi-circle. The rectangle has a length of 5.5 feet and a width of 3.5 feet. The diameter of the circle matches the width. 11. A triangle and a semi-circle. The triangle has a base of 5 inches and a height of 4 inches. The diameter of the circle matches the base of the triangle. 12. A triangle and a semi-circle. The triangle has a base of 7 inches and a height of 6 inches. The diameter of the circle matches the base of the triangle. 13. A triangle and a semi-circle. The triangle has a base of 5.5 inches and a height of 4 inches. The diameter of the circle matches the base of the triangle. Solve each problem. 14. Rob is painting large polka dots on a sheet for the backdrop of the school musical. He painted 16 polka dots, each with a radius of 3 feet. What is the total area that the polka dots", "a rectangle is 22 inches and the inch-measure of each side is a counting number. Thus, these two radii form a diameter of the circle. UNIT 9 Area, Perimeter and Volume Extra Exercises 9.", "a width of 6 feet. The diameter of the circle matches the width. 8. A rectangle and a semi - circle. The rectangle has a length of 9 inches and a width of 4.5 inches. The diameter of the circle matches the length. 9. A rectangle and a semi - circle. The rectangle has a length of 7 feet and a width of 4 feet. The diameter of the circle matches the length. 10. A rectangle and a semi - circle. The rectangle has a length of 5.5 feet and a width of 3.5 feet. The diameter of the circle matches the width. 11. A triangle and a semi - circle. The triangle has a base of 5 inches and a height of 4 inches. The diameter of the circle matches the base of the triangle. 12. A triangle and a semi - circle. The triangle has a base of 7 inches and a height of 6 inches. The diameter of the circle matches the base of the triangle. 13. A triangle and a semi - circle. The triangle has a base of 5.5 inches and a height of 4 inches. The diameter of the circle matches the base of the triangle. Directions : Solve each problem. 14. Rob is painting large polka dots on a sheet for", "is a local minimum. Therefore width = 3, length = 5, perimeter of rectangle = 16, i.e., cut the wire in half. John May 30th, 2015 ... May 30th, 2015 ... May 30th, 2015 Jun 24th, 2017 check_circle Mark as Final Answer check_circle Unmark as Final Answer check_circle", "the width. 7. A rectangle and a semi - circle. The rectangle has a length of 8.5 feet and a width of 6 feet. The diameter of the circle matches the width. 8. A rectangle and a semi - circle. The rectangle has a length of 9 inches and a width of 4.5 inches. The diameter of the circle matches the length. 9. A rectangle and a semi - circle. The rectangle has a length of 7 feet and a width of 4 feet. The diameter of the circle matches the length. 10. A rectangle and a semi - circle. The rectangle has a length of 5.5 feet and a width of 3.5 feet. The diameter of the circle matches the width. 11. A triangle and a semi - circle. The triangle has a base of 5 inches and a height of 4 inches. The diameter of the circle matches the base of the triangle. 12. A triangle and a semi - circle. The triangle has a base of 7 inches and a height of 6 inches. The diameter of the circle matches the base of the triangle. 13. A triangle and a semi - circle. The triangle has a base of 5.5 inches and a height of 4 inches. The diameter of the circle matches the base", "# Happy 22rd birthday, me! Geometry Level 5 A convex cyclic quadrilateral with integer length sides is such that its area divided by its perimeter equals $$1993$$. Let $$P$$ be the maximum possible perimeter. Find the four last digits of $$P$$. ×" ]

[ "width by $1$ increases the number of squares by $3$. This is because when we add $1$ to the width, we can make $2$ additional $1 \\times 1$ squares and $1$ additional $2 \\times 2$ square. This is a total of $3$ extra squares. If we continue the pattern $2, 5, 8, \\ldots$ in the height of $2$ row (all $1$ less than multiples of $3$), we eventually get to $\\ldots, 98, 101, \\ldots$ - missing out $100$. This tells us that it isn't possible to make a rectangle with $100$ squares when the height (or width) is $2$. If the height is $3$, increasing the width by $1$ allows us to make $3$ more $1 \\times 1$ squares, $2$ more $2 \\times 2$ squares and $1$ more $3 \\times 3$ square. This is a total of $6 = 3 + 2 + 1$ squares. This gives us the sequence $14, 20, 26, 32, \\ldots, 98, 104, \\ldots$ (all $2$ more than multiples of $6$) which tells us that we can't make a rectangle with exactly $100$ squares when the height (or width) is $3$. Using the same reasoning for height of $4$, we see that increasing the width by $1$ increases the number of squares by $10$, and that we can make rectangles with $20, 30, \\ldots,", "solve for one of the variables, in this case L (length) so 216/6w = 36/w = l then i substitute (36/w)^3 = maximum do you guys follow? in this particular case, i get some weird numbers in the end, so i think my approach didn't work in this problem? Try drawing the net of your box first. I think you'll find that if the box has a square base, then it's side lengths are $l$, but the height of the box is not necessarily $l$. We'll call it something else, $h$ (for height). So $TSA = 2l^2 + 4lh$ And since $TSA = 216\\,\\textrm{ft}^2$, this means $2l^2 + 4lh = 216$ $4lh = 216 - 2l^2$ $h = \\frac{54}{l} - \\frac{l}{2}$. Now, the volume of the box is $V = l^2h$ $= l^2\\left(\\frac{54}{l} - \\frac{l}{2}\\right)$ $= 54l - \\frac{l^3}{2}$. To maximise the volume, find its derivative, set equal to 0, solve for the length, solve for the height.", "calculated by: $x_V=-\\frac{b}{2a}$ With your problem: $x_V = -\\frac{60}{2 \\cdot (-2)} = 15$ Thus the rectangle has the dimension: $l = 30\\ m\\ \\text{and } w = 15\\ m$", "like help figuring out how to solve it. For #1 I had: w*h = 60. Also, we are given that w*d = 60 --> I ended up with hd=d or h = 1 which would lead me to believe that the figure has a height of 1 and thus, is a rectangle but that doesn't make sense. Can someone elaborate? From the stem we have: width*depth = 60; From (1) we have that: width*height = 60. Thus: width*depth = width*height, which gives that depth = height. Now, if depth = height = 1 and width = 60, then the block is NOT a cube but if depth = height = width = $$\\sqrt{60}$$, then the block is a cube. The same logic applies to (2): we get that width = height. For (1)+(2) we have that depth = height and width = height, therefore depth = height = width, which means that the block is a cube. Hope it helps. I though that if you had four faces as 60 then the remaining two faces had to be 60 so the cube could fit/ Could anyone please elaborate on this please? Thanks Cheers! J Which statement gives the area of FOUR faces? _________________ Current Student Joined: 06 Sep 2013 Posts: 2035 Concentration: Finance GMAT 1: 770 Q0 V", "into t=w*d, f=w*h, and s=h*d • Solve for w=t/d and plug that into f=w*h to get f=(t/d)*h • Solve for h=f*d/t and plug that into s=h*d to get s=(f*d/t)*d • Solve d^2=s*t/f • s, t, and f are given in the problem as 180 cm^2, 240 cm^2, and 300 cm^2, respectively • Plug those in to get d^2=180*240/300 or d^2=144 or d=12 • Work backwards through the functions to get h=15 and w=20 So the height is 15 cm, the width is 20 cm, and the depth (or length) is 12 cm. I believe my correction to the question is valid as the answer came out to simple integers.", "Contents ## Problem 1 Let a square lattice of dimensions $$n \\times n$$ be given. Calculate the number $$R$$ of different rectangles which can be drawn, with the vertices in the lattice points. Two rectangles are considered different if they have different sizes or are in different positions. Solution Let us first consider the case of a lattice of $$10$$ squares of unit length. Consider a rectangle with the base of $$k$$ consecutive squares and height of $$h$$ consecutive squares. The group of $$k$$ consecutive columns of the base can be chosen in $$(11-k)$$ different ways. In a similar way the group of $$h$$ consecutive lines of the height can be chosen in $$(11-h)$$ different ways. So the total number of rectangles of size $$k \\times h$$ is $$(11-k) (11-h)$$. Now, the rectangles with a fixed base $$k$$ can have height equal to $$(1,2, \\cdots ,10)$$. So in total the number of rectangles with base $$k$$ is $$(11-k) (1 + 2 + \\cdots +10) = 55 (11-k)$$. Similarly the base $$k$$ can take on the values $$​​(1,2,\\cdots,10)$$. So the total number of rectangles is $$55 (1 + 2 + \\cdots 10) = 55 \\cdot 55 = 3025$$. To solve the general case with a lattice of $$n$$ squares, just do the same reasoning. The solution is: $R =", "## Elementary Algebra We first set up the equations: $lw = 63$ And $3l + 3w + 9 = 57$ Using $l = 63/w$ from the first equation, we find: $3(63)/w + 3w = 48 \\\\ \\frac{189}{w} + 3w = 48 \\\\ \\frac{189}{w} + \\frac{3w^2}{w} = \\frac{48w}{w} \\\\ 3w^2 -48w + 189 = 0 \\\\ (w-9)(w-7)$ The smaller value is width, and the longer value is length, so we obtain that it is a 7 by 9 rectangle.", "We square the length and the width and multiply both sides by 289 (the denominator of the LHS) to reach the equation $(12+d^2)^2+100d^2=289d^2$. Expanding and simplifying, we find that $d^4+24d^3=45d^2$. Dividing both sides by $d^2$ and moving everything to the LHS, we see that $d^2+24d-45=0$. Applying the quadratic formula, $d=\\frac{-24\\pm \\sqrt{756}}{2}=-12\\pm \\sqrt{189}$. Since $d>0$, $d=\\sqrt{189}-12=3\\sqrt{21}-12$. Finally, $m+n+p=3+21+12=\\boxed{036}$. -Gideontz ~Shreyas S", "hand, the base could be √60x√60 and the height could be the same meaning that this figure is a cube. I am having a bit of difficultly with the algebra though and would like help figuring out how to solve it. For #1 I had: w*h = 60. Also, we are given that w*d = 60 --> I ended up with hd=d or h = 1 which would lead me to believe that the figure has a height of 1 and thus, is a rectangle but that doesn't make sense. Can someone elaborate? From the stem we have: width*depth = 60; From (1) we have that: width*height = 60. Thus: width*depth = width*height, which gives that depth = height. Now, if depth = height = 1 and width = 60, then the block is NOT a cube but if depth = height = width = $$\\sqrt{60}$$, then the block is a cube. The same logic applies to (2): we get that width = height. For (1)+(2) we have that depth = height and width = height, therefore depth = height = width, which means that the block is a cube. Hope it helps. I though that if you had four faces as 60 then the remaining two faces had to be 60 so the cube could fit/ Could", "# The area of a rectangle is 65yd^2 , and the length of the rectangle is 3yd less than double the width, how do you find the dimensions? build the equations and solve... #### Explanation: let the area be $A = l \\cdot w$ where length is $l$ and width is $w$ so 1.st equqtion will be $l \\cdot w = 65$ and length is 3 yd less than double the width says: $l = 2 w - 3$ (2nd eq.) replacing $l$ with $2 w - 3$ in first eq. will yield $\\left(2 w - 3\\right) \\cdot w = 65$ $2 {w}^{2} - 3 w = 65$ $2 {w}^{2} - 3 w - 65 = 0$ now we have a 2nd order equation just find the roots and take the positive one as width can not be negative... $w = \\frac{3 \\pm \\sqrt{9 + 4 \\cdot 2 \\cdot 65}}{2 \\cdot 2} = \\frac{3 \\pm \\sqrt{529}}{4} = \\frac{3 \\pm 23}{4}$ $w = - 5 , \\frac{13}{2}$ so taking $w = \\frac{13}{2} = 6.5 y d$ replacing $w$ with $6 , 5$ in second eq. we get $l = 2 w - 3 = 2 \\cdot 6.5 - 3 = 13 - 3 = 10 y d$ $A = l \\cdot w = 10 \\cdot 6.5 = 65 y", "$h^2=\\frac{105}{14}=\\frac{15}{2}$ this is the height. Multiply by 2 to get the width", "- W^2) &=0 \\\\[5pt] 9(2-W)(2+W) &=0 \\end{align*} From this equation we see that either $$W=2$$ or $$W=-2$$. Since the width cannot be negative, we reject $$W=-2$$ and conclude that $$W=2$$ ft. Then, since $$L=W$$, we know that $$L=2$$ ft also. Using the formula we found for $$H$$ above, we find that the corresponding height is $H = \\frac{36 - 3(2)(2)}{2(2+2)} = \\frac{24}{8)} =3 \\,\\text{ft}$ This gives a volume of $$V = 2 \\cdot 2 \\cdot 3 = 12 \\,\\text{ft}^3$$. Note that the critical point we just found was actually $$(2, 2)$$. This is the only feasible critical point of the function $$V$$. Now we just have to use the second partials test to prove that we obtain a relative maximum here. Second Partials Test: First we need to determine the second partials of $$V(L,W)$$, i.e., $$V_{LL}, V_{WW},$$ and $$V_{LW}$$. \\begin{align*} V_{LL}(L,W) &= \\frac{2(L+W)^2(-6W^3-6LW^2)-4(L+W)(36W^2-6LW^3-3L^2W^2)}{4(L+W)^4} \\\\[5pt] &= \\frac{4(L+W)\\big[(L+W)(-3W^3-3LW^2)-36W^2+6LW^3+3L^2W^2)\\big]}{4(L+W)^4} \\\\[5pt] &= \\frac{-3LW^3-3L^2W^2-3W^4-3LW^3-36W^2+6LW^3+3L^2W^2)}{(L+W)^3} \\\\[5pt] &= \\frac{-3W^4-36W^2}{(L+W)^3} \\end{align*} Similarly, we find $V_{WW}(L,W) =\\frac{-3L^4-36L^2}{(L+W)^3}$ And the mixed partial \\begin{align*} V_{LW}(L,W) &= \\frac{2(L+W)^2(72W-18LW^2-6L^2W)-4(L+W)(36W^2-6LW^3-3L^2W^2)}{4(L+W)^4} \\\\[5pt] &= \\frac{4(L+W)\\big[(L+W)(36W-9LW^2-3L^2W)-36W^2+6LW^3+3L^2W^2)\\big]}{4(L+W)^4} \\\\[5pt] &= \\frac{36LW+36W^2-9L^2W^2-9LW^3-3L^3W - 3L^2W^2 - 36W^2 + 6LW^3 + 3L^2W^2)}{(L+W)^3} \\\\[5pt] &= \\frac{36LW - 3LW^3 - 9L^2W^2 - 3L^3W}{(L+W)^3} \\end{align*} Now we can calculate the discriminant at the critical point: \\begin{align*} D(2,2) &= V_{LL}(2,2)\\cdot V_{WW}(2,2) - \\left[V_{LW}(2,2)\\right]^2 \\\\[5pt] &= (-3)(-3) - \\left(-\\frac{3}{2}\\right)^2 = 9 - \\frac{9}{4} = \\frac{27}{4}", "x must be lower then .5*w. You'll get the maximum of V if the 1rst dervative of V with respect to x equals zero: ${dV \\over dx}=12x^2-4x(l+w)+(l+w)$ $0=12x^2-4x(l+w)+(l+w)$. Solve for x. You'll get 2 solutions, but the greater solution isn't very realistic. $x={1\\over 6} \\cdot \\left(w+l-\\sqrt{w^2-w \\cdot l+l^2} \\right)$. That will give approximately x = 4.04 cm. Plug in w = 21 cm and l = 29,7 cm. Greetings EB I took this to mean: what is the block with largest volume that can be wrapped with a sheet of A4? RonL 4. Originally Posted by abcting223hk how to Use a standard A4 paper to form a largest volume block ? Hello, according to CaptainBlack's hint (thanks a lot, I wouldn't have seen it this way) you have to cut three couples of sqares from the sheet of paper. Let be l = length of sheet, w = width of sheet and x = half the height of the block. Then you get: $V={(l-3x)(w-2x) \\over 2}\\cdot 2x=6x^3-x^2(2l+3w)+x(l*w)$ x must be lower then .5*w. You'll get the maximum of V if the 1rst dervative of V with respect to x equals zero: ${dV \\over dx}=18x^2-2x(2l+3w)+(l*w)$ $0=18x^2-2x(2l+3w)+(l*w)$. Solve for x. You'll get 2 solutions, but the greater solution isn't very realistic. That will give approximately x = 3.3956 cm. Greetings", "at 5:38 • Yes (approximately) – J. W. Tanner Mar 17 at 5:39 Let the length, height, width be $$L, H, W$$ respectively. Then you have: $$LW=120\\to L=\\frac{120}{W}$$ $$LH=96\\to L=\\frac{96}{H}$$ $$WH=100$$ Notice from the first two that: $$\\frac{120}{W}=\\frac{96}{H}\\to\\frac WH = \\frac{120}{96}=\\frac 54$$ Use this with the third statement ($$WH=100$$) to find $$W$$ and $$H$$. Then use either of the first two to find $$L$$, armed with this information. • For reference, I achieve that $$H=\\sqrt{80}=4\\sqrt5; W=5\\sqrt 5; L=\\frac{24}{5}\\sqrt5$$ – Rhys Hughes Mar 17 at 5:35 • so we agree $HWL=480\\sqrt 5$ – J. W. Tanner Mar 17 at 15:40 • Indeed we do, I actually didn't realise that your method (I.e. one so direct) was a possibility, so it was nice to see. – Rhys Hughes Mar 17 at 17:35", "## anonymous 5 years ago a rectangle is 8 inches wider than half the height. the height is h, so the width is 1/2h+8. if h is 36, what is 1/2h+8? $\\frac{1}{2} \\times 36+8=18+8=26$", "to scale down.) Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$. These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\\boxed{260}$.", "\\begin{equation*} S(l,h) = 4lh + 2l^2+2l^2\\text{.} \\end{equation*} Since the volume is constrained at $V=200\\text{,}$ we can write \\begin{equation*} h = \\frac{200}{l^2}\\text{,} \\end{equation*} which means that \\begin{equation*} S(l) = \\frac{800}{l} + 4l^2\\text{,} \\end{equation*} where $l > 0\\text{.}$ Differentiating $S$ with respect to $l\\text{,}$ we compute \\begin{equation*} S'(l) = 8x - \\frac{800}{x^2}\\text{.} \\end{equation*} Therefore, \\begin{equation*} S'(l) = 0 \\implies x^3 = 100 \\implies x = 10^{2/3}\\text{.} \\end{equation*} We construct the following sign diagram for $S'\\text{:}$ We conclude that the surface area is minimized with dimensions $10^{2/3}$ x $10^{2/3}$ x $2\\cdot 10^{2/3}\\text{.}$ A box with square base and no top is to hold a volume $V\\text{.}$ Find (in terms of $V$) the dimensions of the box that requires the least material for the five sides. Also find the ratio of height to side of the base. (This ratio will not involve $V\\text{.}$) $\\ds w=l=2^{1/3}V^{1/3}\\text{,}$ $\\ds h=V^{1/3}/2^{2/3}\\text{,}$ $h/w=1/2$ Solution Suppose the base of the box has side length$l$ and the height of the box is $h\\text{.}$ Then the volume of the box is \\begin{equation*} V(l,h) = l^2 h\\text{,} \\end{equation*} and the surface area is \\begin{equation*} S(l,h) = 4lh + l^2\\text{.} \\end{equation*} If the volume of the box is constrained, then we can write \\begin{equation*} V= l^2h \\implies h = \\frac{V}{l^2}\\text{.} \\end{equation*} And so we can write \\begin{equation*} S(l) = \\frac{4V}{l} + l^2\\text{,}", "\\text{ and } \\qquad w + 2h - 65 = 0\\text{.} \\end{equation*} Let $h = 65 - 2w\\text{.}$ Then \\begin{equation*} w + 2h - 65 = w + 2 (65-2w)- 65 = 65 - 3w = 0\\text{,} \\end{equation*} and so $w = \\dfrac{65}{3}$ and $h = \\dfrac{65}{3}\\text{.}$ Now, since \\begin{equation*} V(0,0) = 0 \\text{ and } V\\left(\\frac{65}{3},\\frac{65}{3}\\right) = \\frac{549,250}{27}\\text{,} \\end{equation*} we conclude that the maximum volume of the box is $V = \\dfrac{549,250}{27}\\text{,}$ with dimensions of $\\dfrac{130}{3} \\times \\dfrac{65}{3} \\times \\dfrac{65}{3}\\text{.}$ The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost. It has a square base, and is one and one half times as tall as wide. If the volume is $V$ the dimensions are $\\root 3 \\of {2V/3}\\times \\root 3 \\of {2V/3}\\times \\root 3\\of {9V/4}\\text{.}$ Solution Let $l\\text{,}$ $w$ and $h$ be the length, width and height of the box in inches, respectively. Then the volume of the package is given by \\begin{equation*} V = l \\times w \\times h, \\qquad l,w,h \\geq 0 \\end{equation*} and the area of the bottom is given by \\begin{equation*} A_{bot} = l \\times w\\text{.} \\end{equation*} The cost to construct the box is then proportional to \\begin{equation*} C(l,w,h) = A_{top} + A_{sides} +", "# The width of a rectangle is one-fifth as much as the length and the perimeter is 120 cm, how do you find the length and width? Jul 21, 2015 $L = 50$ $W = 10$ #### Explanation: Questions of these type are often called Worded Problems. This is because the key equations that you will use, are already described for you in the question. Here's how it goes: Let the length be $L$ and the width be $W$ and the perimeter is $P$ The question says: \"the width$\\left(W\\right)$ is one-fifth$\\left(\\frac{1}{5}\\right)$ as much as the length$\\left(L\\right)$\" In mathematical terms this means, $W = \\frac{1}{5} L$ $\\implies L = 5 W \\ldots \\ldots \\ldots \\textcolor{red}{\\left(1\\right)}$ Secondly: \"the perimeter$\\left(P\\right)$ is $120$\" Meaning, $P = 120$ The Perimeter of any rectangle $= 2 \\times \\left(L + W\\right)$ Hence, $2 \\left(L + W\\right) = 120$ $\\implies L + W = 60 \\ldots \\ldots \\ldots \\textcolor{red}{\\left(2\\right)}$ Now, all that is left to do is to solve $\\textcolor{red}{\\left(1\\right)}$ and $\\textcolor{red}{\\left(2\\right)}$ simultaneously. $5 W + W = 60 \\implies W = 10$ $\\implies L = 5 \\cdot \\left(10\\right) = 50$ And You're Done.", "+ r + \\ldots + r^n = \\frac{r^{n+1}-1}{r-1}$$ - Perhaps the widths of the rectangles should be $1/(r-1)$, so the sum of the areas of the glue rectangles is $1+r+\\ldots r^n$. – Robert Israel Mar 15 '12 at 21:47 I appreciate this. Could you read it over and insert nouns and otherwise edit it for clarity? E.g., is \"height\" between \"has\" and \"r−1\"? \"Each blue rectangle on the left has r−1 times the height of the green rectangle below it, so the green and blue together have height r times the green alone.\" I couldn't understand why this is so until I wrote (r−1)g+g=gr-g+g=gr. How is that evident from the picture? If the length of r was marked perhaps it'd be easier to see. – user93200 Mar 15 '12 at 23:52 Do you mean you want a formula for $x + x^2 + \\ldots x^n$? Look up \"geometric series\". - We'll show that $$\\sum_{i=1}^{n} 2^i= 2^{n+1} - 2.$$ Denote this sum by $$S(n) = 2^1 + 2^2 + \\ldots + 2^{n-1} + 2^n \\tag{1}$$ On the RHS, factor out a $2$, we get $$S(n) = 2(1 + 2^1 + \\ldots + 2^{n-1})$$ But from $(1),$ we have $2^1 + \\ldots + 2^{n-1} = S(n) - 2^n.$ So $$S(n) = 2(1 + \\underbrace{2^1 + \\ldots + 2^{n-1}}_{S(n) - 2^n})\\\\" ]

[ "## anonymous 5 years ago a rectangle is 8 inches wider than half the height. the height is h, so the width is 1/2h+8. if h is 36, what is 1/2h+8? $\\frac{1}{2} \\times 36+8=18+8=26$", "In the figure above, the large square is divided into two smaller squa [#permalink] ### Show Tags 30 Sep 2017, 02:43 Since the squares have a perimeter of 8 and 20, the sides of the squares are $$(\\frac{8}{4})2$$ and $$(\\frac{20}{4})5$$. The rectangles share their lengths and breadth with these 2 squares. Therefore, the 2 rectangles will have length and breadth 2 and 5 respectively, and the perimeter of one such rectangle is 2(length + breadth) = 2(2+5) = 14 Hence, the perimeter of the two shaded rectangles is 2*14 = 28(Option E) _________________ Stay hungry, Stay foolish Kudos [?]: 830 [0], given: 22 Re: In the figure above, the large square is divided into two smaller squa [#permalink] 30 Sep 2017, 02:43 Display posts from previous: Sort by", "# A regular hexagon has a side length of 2 1/4 feet. What is its perimeter? ##### 1 Answer May 30, 2016 Perimeter = $13 \\frac{1}{2}$ feet #### Explanation: There are two important words in the information given. It is a regular figure, which means all the sides and angles are equal. It is a hexagon, which means there are 6 sides. So, there are 6 sides, each of length $2 \\frac{1}{4}$ feet. Simply multiply to find the perimeter. (the sum of all the sides) $P = 6 \\times 2 \\frac{1}{4}$ $6 \\times \\frac{9}{4} = {\\cancel{6}}^{3} \\times \\frac{9}{\\cancel{4}} ^ 2 = \\frac{27}{2}$ Perimeter = $13 \\frac{1}{2}$ feet", "439 Kudos [?]: 89 [0], given: 72 GPA: 4 WE: Engineering (Transportation) Re: The length and width of a rectangle are integer values. [#permalink] ### Show Tags 19 Aug 2017, 10:03 Clearly the Ans is D In simple words, LxB is area and L^2+B^2= (2r)^2 IF LxB is 48 then L=8 and B=6 and sum of squares is 100 whose root is 10 and whose half is r which will be 5 , hence all integers. Ans is D _________________ Give Kudos for correct answer and/or if you like the solution. Kudos [?]: 89 [0], given: 72 Manager Joined: 01 Sep 2016 Posts: 208 Kudos [?]: 207 [0], given: 33 GMAT 1: 690 Q49 V35 The length and width of a rectangle are integer values. [#permalink] ### Show Tags 23 Sep 2017, 11:48 OFFICIAL SOLUTON FROM MANHATTAN If a rectangle is inscribed in a circle, the rectangle’s diagonals are also diameters of the circle: Call the length and the width of the rectangle L and W and the circle’s radius r. A right triangle can be constructed, with legs L and W and with hypotenuse 2r. Use the Pythagorean theorem: a^2 + b^2 = c^2 L^2 + W^2 = (2r)^2, where L, W, and r are all integers. The smallest three integers that satisfy the Pythagorean theorem are", "Question 3: Find the Lowest common multiple of 36 and 24 having GCD 12. Solution: Two numbers are 36 and 24 GCD(24, 36) = 12 LCM = ? We know that Product of two numbers = Product of their LCM and GCD => 24 * 36 = LCM * 12 => LCM = $\\frac{24 * 36}{12}$ => LCM = 72 ## Math Word Problems for 6th Grade Math word problems for 6th grade are available on a number of websites along with solved examples and step-by-step solutions. Worksheets with 6th grade math practice problems can be printed out and shared with friends to work out together. 6th grade math problems with answers let you test yourself to see how well you have grasped the concept. Exams do not have to be a problem anymore if you are using online math solvers as they are taken care of with exam prep and mock tests. ### Solved Examples Question 1: Find the perimeter of a rectangular garden with a length of 25 feet and width of 20 feet. Solution: Length of a rectangular garden = 25 width of a rectangular garden = 20 Perimeter of rectangle = 2(Length + width) => Perimeter of rectangular garden = 2(25 + 20) = 2(45) = 90 Hence, Perimeter of rectangular garden is 90", "# The perimeter of a standard-sized rectangular rug is 28 ft. The length is 2 ft longer than the width. How do you find the dimensions. What is the width? Nov 28, 2016 The dimensions are 6 feet by 8 feet and the width is 6 feet. #### Explanation: The formula for the perimeter of a rectangle is: $p = 2 \\cdot w + 2 l$ where $p$ is the perimeter, $w$ is the width and $l$ is the length. We are told the length is 2 ft longer than the width. So, we can write this as: $l = w + 2$ We are also given the perimeter or $p$. So substituting $28$ for $p$ and $w + 2$ for $l$ we can rewrite this formula as follows and solve for $w$ while keeping the equation balanced: $28 = 2 \\cdot w + 2 \\cdot \\left(w + 2\\right)$ $28 = 2 w + 2 w + 4$ $28 = 4 w + 4$ $28 - 4 = 4 w + 4 - 4$ $24 = 4 w$ $\\frac{24}{4} = \\frac{4 w}{4}$ $w = 6$ We can number substitute $6 W f \\mathmr{and}$w$\\in t h e e q u a t i o n f \\mathmr{and}$l# to determine the length: $l = 6 + 2$ $l = 8$", "# The Perimeter of a Rectangular pig pen is 210Ft. The length is 5ft greater then the width. How do you find the Length? Mar 13, 2016 Length = 55 ft #### Explanation: Let length be $L$ Let width be $W$ Perimeter$= 2 W + 2 L = 210$ ...............................(1) But we are told that L=W+5\" .........................(2) So, as we need to find length we need to 'get rid' of $W$ Rearranging equation (2) we gave: $W = L - 5$ Substitute for W in equation (1) gives $\\text{ } 2 \\left(L - 5\\right) + 2 L = 210. \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots . \\left({1}_{a}\\right)$ Multiply out the bracket $\\text{ } 2 L - 10 + 2 L = 210$ $\\text{ } 2 L + 0 + 2 L = 220$ $\\text{ } 4 L = 220$ $\\text{ } L = \\frac{220}{4} = 55$", "# If the width of rectangle is 5 cm more than one-half of its length the perimeter is 70 cm, what are the dimensions of the rectangle? Dec 4, 2016 The length is 20 cm and the width is 15 cm #### Explanation: First, let's call the width of our rectangle $w$ and the length $l$. From the problem we know: $w = \\frac{1}{2} l + 5$ We also know the formula for the perimeter of a rectangle is: $p = 2 \\cdot l + 2 \\cdot w$ So we can substitute $70$ for $p$ which is given in the problem and we can also substitute $\\frac{1}{2} l + 5$ for $w$ and then solve for $l$: $70 = 2 \\cdot l + 2 \\cdot \\left(\\frac{1}{2} l + 5\\right)$ $70 = 2 l + 1 l + 10$ $70 = 3 l + 10$ $70 - 10 = 3 l + 10 - 10$ $60 = 3 l$ $\\frac{60}{3} = \\frac{3 l}{3}$ $20 = l$ Now we can substitute $20$ for $l$ in the formula for $w$ to find the width: $w = \\frac{1}{2} 20 + 5$ $w = 10 + 5$ $w = 15$", "Rectangle - sides What is the perimeter of a rectangle with area 266 cm2 if length of the shorter side is 5 cm shorter than the length of the longer side? Correct result: x = 66 cm Solution: $ab = 266 \\ \\\\ b = a + 5 \\ \\\\ a(a+5) = 266 \\ \\\\ a^2 + 5 a - 266 =0 \\ \\\\ \\ \\\\ a^2 +5a -266 =0 \\ \\\\ \\ \\\\ p=1; q=5; r=-266 \\ \\\\ D = q^2 - 4pr = 5^2 - 4\\cdot 1 \\cdot (-266) = 1089 \\ \\\\ D>0 \\ \\\\ \\ \\\\ a_{1,2} = \\dfrac{ -q \\pm \\sqrt{ D } }{ 2p } = \\dfrac{ -5 \\pm \\sqrt{ 1089 } }{ 2 } \\ \\\\ a_{1,2} = \\dfrac{ -5 \\pm 33 }{ 2 } \\ \\\\ a_{1,2} = -2.5 \\pm 16.5 \\ \\\\ a_{1} = 14 \\ \\\\ a_{2} = -19 \\ \\\\ \\ \\\\ \\text{ Factored form of the equation: } \\ \\\\ (a -14) (a +19) = 0 \\ \\\\ \\ \\\\ \\ \\\\ a = 14 \\ \\\\ b= 19 \\ \\\\ o = 2(a+b) = 66 \\ \\text{cm}$ We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! Tips to related online", "# How do you find the perimeter of a rectangle, in simplest radical form, in which the base is sqrt98 and the height is 5sqrt2? Oct 29, 2016 34 #### Explanation: $\\sqrt{98} = 9.9$ $5 \\cdot \\sqrt{2} = 7.1$ Since two sides of the rectangle are 9.9 in length, $2 \\cdot 9.9 = 19.8$ Since the other two sides of the rectangle are 7.1 in length, $2 \\cdot 7.1 = 14.2$ Therefore, the total length of the perimeter of the rectangle is, $19.8 + 14.2 = 34.0$ Oct 29, 2016 The given rectangle has $b a s e = \\sqrt{98} = \\sqrt{{7}^{2} \\cdot 2} = 7 \\sqrt{2}$ and $h e i g h t = 5 \\sqrt{2}$ So $P e r i m e t e r = 2 \\left(b a s e + h e i g h t\\right) = 2 \\left(7 \\sqrt{2} + 5 \\sqrt{2}\\right) = 2 \\cdot 12 \\sqrt{2} = 24 \\sqrt{2}$", "to less than 258 then 258 will also add up and it will have 64 posssiblities and the answer than will be 4160 D _________________ Director Joined: 07 Aug 2011 Posts: 502 GMAT 1: 630 Q49 V27 Re: How many rectangles can be formed such that their perimeter is less [#permalink] ### Show Tags 04 Apr 2015, 10:34 chetan2u wrote: Lucky2783 wrote: How many rectangles can be formed such that their perimeter is less than 258 ? a) 178 b) 284 c) 1094 d) 4160 e) 4095 hi Harley1980 and Lucky2783, i too think answer is 4096.. although it is necessary that the question mentions that the sides are integers to get a finite numbers, otherwise the answer can be infinite with a pair of sides .1 or .001, etc now if we consider that sides are integer, lets look at the solution.. perimeter has to be an even number so largest perimeter is 256 the possible perimeters are 4,6,8,...256.. if perimeter is 4.. sides are 1 and 1.. possibility 1 if perimeter is 6.. sides are 2 and 1..possibility 1 if perimeter is 8.. sides are 2 and 2 or 1 and 3..possibilities 2 and so on.. it follows a pattern.. each succeeding pair has one extra possiblity.. so total=1+1+2+2+3+3+.....63+63+64=2(1+2+3+...+63)+64=2*64*63/2+64=64*63+64=64*64=4096... if the perimeter is equal to less", "Like it doesn't contain the rectangles of base > 0.25 It only contains all possible squares. Mehrab4226 Posts:230 Joined:Sat Jan 11, 2020 1:38 pm Ok we can assume that we have a rectangle with length $\\frac{1}{2}$ and breadth $0$. The circle that comprises all those rectangles has an area of $\\frac{1}{16} \\pi$. So, $\\frac{1}{a} = 16$. No other rectangle with the given criteria can be outisde this circle.", "# The perimeter of a rectangle is 48 in. If the length is twice the width, what is the length of the rectangle? Aug 16, 2016 Length$= 16$ inches #### Explanation: Let $l$ be the length and $w$, width. It is stated that, $l = 2 w \\ldots \\ldots \\ldots \\ldots \\ldots . \\left(1\\right)$. Now, the perimeter of the rectangle$= 48$. $\\Rightarrow 2 \\left(l + w\\right) = 48 , \\mathmr{and} , l + w = 24$ $\\Rightarrow 3 w = 24. \\ldots \\ldots \\ldots \\ldots . . \\left[b y \\left(1\\right)\\right]$ :. w=8, &, so, l=2w=16 inches.", "is the radius of the circle used to form the rounded corners? \"area of rectangle with rounded corners\". NOTE: (W = rectangle's width) any circle of radius >0 and <W/2 can be used to form the rounded corners. As example, using rectangle length=100 and width=80, 39 circles of radius 1 to 39 can be used. Radius 40, if used, results in the rectangle's ends being semicircles. Thank you for taking time to help me with this. Sorry for not making my query clear. The length and width of the rectangle ARE the measurements of the original rectangle (before the edges are rounded). The diagonal is the measurement from rounded corner to rounded corner. A standard 3, 4, 5, when the corners are rounded will shrink the diagonal which is why I said it is now a 3, 4, 4 1/2. I found a flash calculator that does exactly what I'm looking for, and should explain better what I'm asking (see link below) however, it's not enough being able to solve it, I need the equation. As to your google comment, I have searched all over, to the point patience allows and all come up with are tricks for making rounded corners in stylesheets. I hope this calculator explains my needs better than my words. *Note: For some reason", "Difference between revisions of \"2013 AIME I Problems/Problem 7\" Problem 7 A rectangular box has width $12$ inches, length $16$ inches, and height $\\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$. Solution After using the pythagorean formula three times, we can quickly see that the sides of the triangle are 10, $\\sqrt{(x/2)^2 + 64}$, and $\\sqrt{(x/2)^2 + 36}$. Therefore, we can use Heron's formula to set up an equation for the area of the triangle. The semi perimeter is (10 + $\\sqrt{(x/2)^2 + 64}$ + $\\sqrt{(x/2)^2 + 36}$)/2 900 = $\\frac{1}{2}$((10 + $\\sqrt{(x/2)^2 + 64}$ + $\\sqrt{(x/2)^2 + 36}$)/2)((10 + $\\sqrt{(x/2)^2 + 64}$ + $\\sqrt{(x/2)^2 + 36}$)/2 - 10)((10 + $\\sqrt{(x/2)^2 + 64}$ + $\\sqrt{(x/2)^2 + 36}$)/2 - $\\sqrt{(x/2)^2 + 64}$)((10 + $\\sqrt{(x/2)^2 + 64}$ + $\\sqrt{(x/2)^2 + 36}$)/2 - $\\sqrt{(x/2)^2 + 36}$). Solving, we get $\\boxed{041}$.", "# What is the ratio of a circle's diameter to a rectangle's width so that the circle covers the rectangle exactly Ratio example image I am doing some web design, and I need a way to dynamically figure out what to set a circle's diameter to so that it exactly covers a rectangle, as shown in the above image. In the first scenario (in the image), I calculated that the diameter of the circle was 12% greater than the width of the rectangle, and thought this would work for all rectangles, but it does not. The problem is that the width of the rectangle is dynamic, so I never know what it's value will be ahead of time, I have to size the circle based off of whatever dimensions the rectangle is at the time. This is probably a fairly simple formula, but I am far from being a mathematician, clearly. If I can do anything to clarify my question, please ask! Thanks! The diameter of the circle is equal to the diagonal of the rectangle. Let $w$ and $h$ be, respectively, the width and height of the rectangle. Then the diameter of the circle $d$ satisfies $$d^2=w^2+h^2\\iff d=\\sqrt{w^2+h^2}.$$ Or, letting $a$ denote the ratio of the rectangle's height to its width, i.e. $a=\\frac{h}{w}$, we have: $$d^2=(1+a^2)w^2 \\iff", "# The area of a rectangle is 65yd^2 , and the length of the rectangle is 3yd less than double the width, how do you find the dimensions? build the equations and solve... #### Explanation: let the area be $A = l \\cdot w$ where length is $l$ and width is $w$ so 1.st equqtion will be $l \\cdot w = 65$ and length is 3 yd less than double the width says: $l = 2 w - 3$ (2nd eq.) replacing $l$ with $2 w - 3$ in first eq. will yield $\\left(2 w - 3\\right) \\cdot w = 65$ $2 {w}^{2} - 3 w = 65$ $2 {w}^{2} - 3 w - 65 = 0$ now we have a 2nd order equation just find the roots and take the positive one as width can not be negative... $w = \\frac{3 \\pm \\sqrt{9 + 4 \\cdot 2 \\cdot 65}}{2 \\cdot 2} = \\frac{3 \\pm \\sqrt{529}}{4} = \\frac{3 \\pm 23}{4}$ $w = - 5 , \\frac{13}{2}$ so taking $w = \\frac{13}{2} = 6.5 y d$ replacing $w$ with $6 , 5$ in second eq. we get $l = 2 w - 3 = 2 \\cdot 6.5 - 3 = 13 - 3 = 10 y d$ $A = l \\cdot w = 10 \\cdot 6.5 = 65 y", "# The perimeter of a regular hexagon is 27 inches. What is the length of each side? $4 \\frac{1}{2}$ inches Since the perimeter of the hexagon is 27 inches, then the length of each side is $\\frac{27}{6} = 4 \\frac{1}{2}$inches", ", \\mathmr{and} , 6$ $\\therefore w = 14 , \\mathmr{and} , 3$ But, $w = 14$ will make the breadth of the rug$= - 19$, which is not possible. Hence, the width of the border is $3 '$.", "8. A rectangle and a semi-circle. The rectangle has a length of 9 inches and a width of 4.5 inches. The diameter of the circle matches the length. 9. A rectangle and a semi-circle. The rectangle has a length of 7 feet and a width of 4 feet. The diameter of the circle matches the length. 10. A rectangle and a semi-circle. The rectangle has a length of 5.5 feet and a width of 3.5 feet. The diameter of the circle matches the width. 11. A triangle and a semi-circle. The triangle has a base of 5 inches and a height of 4 inches. The diameter of the circle matches the base of the triangle. 12. A triangle and a semi-circle. The triangle has a base of 7 inches and a height of 6 inches. The diameter of the circle matches the base of the triangle. 13. A triangle and a semi-circle. The triangle has a base of 5.5 inches and a height of 4 inches. The diameter of the circle matches the base of the triangle. Solve each problem. 14. Rob is painting large polka dots on a sheet for the backdrop of the school musical. He painted 16 polka dots, each with a radius of 3 feet. What is the total area that the polka dots" ]

[ "A book has $89$ pages, but the page numbers are printed incorrectly. Every third page number has been omitted, so that the pages are numbered $1,2,4,5,7,8,...$ and so on. What is the number on the last printed page?", "and answer the given questions. What is the difference between the total number of pages printed by Printer E in 1st, 2nd and 4th week together and total number of pages printed by Printer C in all the given weeks together? 952 878 924 934 918", "of eight pages, for a total of 360 lines.", ". . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-29 Specifying the Paper Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3-32 Printing All the Pages from the Same Paper Source . . . . . . . . . . . . . . . . . . . . . . . . . . 3-33 Printing the First, Body, and Last Pages from Different Paper Sources . . . . . . . . . . . . 3-34 Printing the First, Second, Body, and Last Pages from Different Paper Sources . . . . . 3-35 Printing the Cover and Body Pages of a Booklet from Different Paper Sources . . . . . . 3-36 Printing on Transparencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .", "times. The problem is that they have to be numbered only 1 per page, so that when i have printed them all i can easily crop them and staple them right up with having to go through it all.", "day: 240,000 Number of pages printed by 8 printing presses in 2 days: 480,000 Number of pages 1 printing press prints in 2 days: 60,000 Number of pages printed for both jobs: 60 000+480 000=540 000. Ratio of number of pages of the novel to the number of pages of the cookbook: 2:1. Therefore, 360,000 pages for the novel and 180,000 for the cookbook. Question 2. Describe a scenario where it would make sense for the job scheduler to schedule both jobs as described in Exercise 2 part (a). The client for the novel job wants the print job completed by midweek, whereas the cookbook does not need to be completed until the end of the week. BONUS Question 3. If the novel was 250 pages and the cookbook 125 pages, how many copies of each were printed?", "# Pages have got numbers too! A book have 273 pages. Find the number of digits used for the numbering of all the pages. ×", "What number has the last page of the dropped sheets?", "creating a page that has been read 30,546 times.", "# How many times does the digit 4 occur in the page numbers of a 50 page book? Dec 13, 2016 $15$ #### Explanation: The digit $4$ can occur in the units or tens positions. It occurs in the units position in the numbers: $4 , 14 , 24 , 34 , 44$ It occurs in the tens position in the numbers: $40 , 41 , 42 , 43 , 44 , 45 , 46 , 47 , 48 , 49$ So there are a total of $5 + 10 = 15$ occurrences of the digit $4$. If the question asked in how many page numbers the digit $4$ occurs then the answer would be $14$ - avoiding counting $44$ twice. However, since it asks for the number of occurrences of the digit $4$, $44$ should be counted twice.", "smaller sheets. The size doesn't matter. A sheet of paper is, in ordinary understanding, a piece of paper. In a finished book, a page is one side of a leaf. Thus, there are always exactly twice as many pages are there are leaves. From the manufacturer's point of view, a 240-page (double-sided printing) book has 120 leaves, regardless of whether or not anything is printed on the pages, or how they are numbered. D.1.2. Folio, Spine Imagine a single sheet of US Letter 8.5\"Wx11\"H paper. It has two pages, since you can print on both sides (even if one is empty). If you fold that sheet in half across its width, it is still one sheet. But now you have a folio, consisting of two leaves of size 5.5\"Wx8.5\"H. Each leaf has two pages, so you now have four pages. The fold is at the spine. Caution: The term \"folio\" has several different meanings, even in the narrow field of book publishing. It may refer to folded paper, as mentioned above. Or it may refer to a particular sheet size of paper (\"foolscap\"). Or, it may simply refer to the number printed on a page. This last use of \"folio\" is discussed in the context of headers and footers. D.1.3. Signature, Imposition When a book is commercially manufactured,", "all 39 pages? • Fall '14", "of digits printed is 1095 then how many pages does the book have? There are 9 significant single digit numbers. Hence for the first 9 pages, 9x1 = 9 digits will be printed. There are 90 significant double digit numbers. Hence for the next 90 pages, 90x2 = 180 digits will be printed. Total number of pages = 99 (90 plus 9), total number of digits is 189(9 plus 180) till now. Subsequently, we have 900 significant triple digit numbers and 900x3 = 2700 digits. Since the total number of digits printed lies in the range [189, 2889], it suggests that all double digit numbers were printed and some triple digit numbers were printed. The upper limit is arrived at as 2700 plus 189. Number of digits used to print triple digit numbers = 1095 -189 = 906. Therefore, 906/3 = 302 triple digit numbers were used. Therefore the total number of pages printed = 9 plus 90 plus 302 = 401. That is how solution: 401 Feel Free to Comment or Suggesting New Approaches . ## Wednesday, May 18, 2011 ### Seven Bridges of Königsberg (Unsolvable Puzzle in Computer Science)) The Seven Bridges of Königsberg is a historically notable problem in mathematics. Its negative resolution by Leonhard Euler in 1735 laid the foundations of graph theory and", "from 10 to 99 there are 90 pages and these 90 pages are numbered with double digits(i-e 10, 11, 12,……up to 99), so, Total digits for pages starting from 10 to 90 will be ’90 pages multiplied by two digits’ i-e 90×2=180 digits Case 3: Now after 99 th page it starts with case 3; i-e now onward the pages are numbered with three digit number (i-e from 100, 101. 102, …….. up to 999th page) and these will be 900 pages . So the number of digits for nine hundred triple digit numbered 900 pages = 900×3=2700 digits Now we add digits used in case 1, case 2 and case 3 so, 9+180+2700 = 2889 digits,But we have been given that total digits are 3189. So, now we will count pages for 3189-2889=300 digits. Case 4: Now after 999th page comes 1000th page which is numbered with 4 digits. We need 75 tetra digit pages to use 300 digits. So we see that 1000 to 1074 will make 75 pages having 4 digit numbering, and digits used by 75 four digit numbers = 75×4=300 Now add pages of all cases i-e 9+90+900+75 =1074 pages Insert math as $${}$$", "is 273. What number has the last page of the dropped sheets? 17. Simple equation 10 35= 7*3*x what is x?", "Exactly $195$ digits have been used to number the pages in a book. How many pages does the book have?", "# Pages 6 and 27 are on the same (double) sheet of a newspaper, how many pages are there in the news paper altogether I was never really good at maths, trying to get back into it. I have this question: Pages 6 and 27 are on the same (double) sheet of a newspaper, what are the page numbers on opposite sides of the sheet and how many pages are there in the newspaper altogether This seems like a trick question... I am not too sure how to take the logic, but I suppose pages 5 and 28 are the numbers on opposite sides of the page? How do I get the total pages of the newspaper altogether? I am assuming we add the missing pages, 1 thru 4 (4 pages in total) to 27 to get 31? Not sure how to solve this, and I have no answer key. - It seems to me the main confusion with this question is how a newspaper's pages are arranged, which it does nothing to clarify. I.e., it's a bad question. – 6005 May 26 '13 at 5:57 Yeah... it's quite confusing... I typed it out verbatim. I am thinking they mean the broadsheet type of newspaper, and the numbers are on the same face of the paper and the", "[–] 4 points5 points (6 children) I agree with you on the first. I get 580 for the second (but I've been known to derp the underpable) edit: off by one... I also agree with 579. [–] 4 points5 points (1 child) I'm pretty sure it's 579. Because. [–] 0 points1 point (0 children) Yes, I have an off by one error. I solved 9 + 90*2 + (x-100) * 3 = 1629, but if there are 100 pages then there still is at least one page with three digits. [–] 3 points4 points (2 children) I get this: 9 digits for pages 1- 9 180 digits for pages 10- 99 1200 digits for pages 100-499 240 digits for pages 500-579 So I get 579 pages. [–] -2 points-1 points (1 child) the question is do you count the backside of the page even if it's not numbered? [–] 0 points1 point (0 children) I considered it fair to assume the answer would be the same if the pages were all printed one on side or two. [–] 2 points3 points (0 children) I independently got 480, for what it's worth. Edit: Other posters have shown my answer to be wrong in fairly explicit ways. I just did it quickly and stopped trying when it matched another answer.", "chapter), then it will not print the page number. Check out the result: • This solves the problem. Nice answer. – Grimolatto Aug 4 '13 at 16:44", "the questions and others here from the Wall Street Journal. Solution I: 930 - 189 (digits of the first 99 pages) =741 741 divided by 3 = 247. Careful since you are counting the three digit numbers from 100 if the book has N pages N - 100 + 1 or N - 99 = 247. N = 346 pages. Solution II: 930 - 189 (total digits needed for the first 99 pages) = 741 741/3 = 247 (how far the three digit page numbers go). 247 + 99 = 346 pages #4: If you write consecutive numbers starting with 1, what is the 50th digit you write? Solution I: 50 - 9 = 41, 9 being the first 9 digits you need to use for the first 9 pages. Now it's 2 digit. 41/2 = 20.1 , which means you will be able to write 20 two digit numbers + the first digit of the next two digit numbers. 10 to 29 is the first 20 two digit numbers so the next digit 3 is the answer. (first digit of the two digit number 30.) Solution II: (50 - 9 ) / 2 = 20. 5 ; 20.5 + 9 = 29.5, so 29 pages + the first digit of the next two digit numbers, which is" ]

[ "SEARCH HOME Math Central Quandaries & Queries Question from Chungus: It takes 852 digits to number the pages of a book consecutively. How many pages are there? Hi Chungus, There are 9 one digit pages, pages 1 to 9. Next look at 2 digit numbers. They go from 10 to 99 so that is 90 two digit numbers. Each contains 2 digits so when you get to 99 you have used $9 + 2 \\times 90 = 189$ digits. If we apply a similar argument to three digit numbers we will see that we have gone too far so we need to look at three digit numbers in smaller chunks. From 100 to 199 there are 100 three digit numbers. Since each has 3 digits that's $3 \\times 100 = 300$ digits so when we get to page 199 we have used $9 + 2 \\times 90 + 3 \\times 100 = 489$ digits.", "-> 6 digits/page -> 5394 start at 852 digits and work your way up through the pages by subtracting the above values to calculate amount of pages with 6 digits (852 - 18 - 3 - 356 - 5)/6 = 80 add all pages = 9 + 1 + 89 + 1 + 80 = 180 13. guido says: pages 1 to 9 : 9 digits #7 : 1 pages 10 to 99 : 180 digits --- total : 189 #7 :10 pages100 to199 : 300 digits --- total : 489 #7 :11 pages200 to299 : 300 digits --- total : 789 #7 :11 pages300 to319 : 60 digits --- total : 849 #7 : 2 pages320 to322 : 3 digits --- total : 852 So book has 322 pages and 35 x digit 7 14. Michael says: (a) We begin by noting the different types of pages: pages containing one digit, two digits, and three digits. Pages 1 through 9 (9 pages, or 1 * 9 total digits) each occupy one of the 852 digits. Pages 10 through 99 (90 pages, or 2 * 90 total digits) each occupy two of the 852 digits. Pages 100 through 999 (900 pages, or 3 * 900 total digits) each occupy three of the 852 digits; however, logic indicates that", "# How many times does the digit 4 occur in the page numbers of a 50 page book? Dec 13, 2016 $15$ #### Explanation: The digit $4$ can occur in the units or tens positions. It occurs in the units position in the numbers: $4 , 14 , 24 , 34 , 44$ It occurs in the tens position in the numbers: $40 , 41 , 42 , 43 , 44 , 45 , 46 , 47 , 48 , 49$ So there are a total of $5 + 10 = 15$ occurrences of the digit $4$. If the question asked in how many page numbers the digit $4$ occurs then the answer would be $14$ - avoiding counting $44$ twice. However, since it asks for the number of occurrences of the digit $4$, $44$ should be counted twice.", "there cannot be 999 consecutively numbered pages (beginning at 1) such that only 852 digits are used to number the pages. Therefore, we must determine how many three-digit pages there are in the book (let n represent this). Notice the form of the 'total digit' products; each has the form: (# of digits)(# of pages requiring said number of digits): 1 * 9 = (1 digit)(9 pages requiring 1 digit) 2 * 90 = (2 digits)(90 pages requiring 2 digits) 3 * n = (3 digits)(n pages requiring 3 digits) These products represent the number of digits that are being used for 1-, 2-, and 3-digit pages. Summating these products gives 852 digits: 9 + 180 + 3n = 852 3n = 663 n = 221 Therefore, there are 221 pages requiring three digits. When we add to this the nine pages requiring one digit and the ninety pages requiring two digits, we get: 9 + 90 + 221 = 320 total pages. Hence, there are 320 pages in the book. Check: 1 * 9 + 2 * 90 + 3 * 221 = 852 total digits (b) In this part, it is not tedious to list all the pages with which a seven is associated. 1-digit pages: page 7 (1 total page) 2-digit pages: pages 17, 27,", "zeros in this list of numbers. This is simply 9, if w=1, (99+9) if w=2, (999+99+9) if w=3 and so on. So, solve for 852 = n*w - (9+99). Guessing, w=2, gives the answer of 248. 8. Ardra says: New puzzle It takes 852 digits to number the pages of a particular book consecutively. (i) How many pages are there? (ii) How many times is the digit 7 printed? (i) There are 320 pages. (ii) The digit 7 is printed 62 times. 9. Thomas A Buckley says: (i) Number of pages 9 Pages 1 to 9 requires 9 digits 90 ,, 10 to 99 ,, 180 ,, Digits remaining to be used = 852 -( 9+ 180) = 663 Next pages from 100 inclusive , each requiring 3 digits 663/3 = 221 pages Total pages = 9 + 90 + 221 = 320 = Answer (i) (ii) Number of 7's digit No of 7,s in the Tens Units Pages 1 to 9 0 1 ,, 10 to 99 10 9 ,, 100 to 320 20 22 Total number of 7,s = 10 + 20 + 1 + 9 + 22 = 62 ,, 100 to 320 10. Francis Kisner says: It takes 852 digits to number the pages of a particular book consecutively. (i) How many pages are", "Tags 15 Sep 2014, 23:39 2 4 00:00 Difficulty: 85% (hard) Question Stats: 41% (01:03) correct 59% (01:09) wrong based on 153 sessions ### HideShow timer Statistics All pages of a book are numbered. If the first page is numbered 1 and the last page is numbered 705, on how many pages does digit 9 appear in the numeration? A. 70 B. 77 C. 126 D. 133 E. 140 _________________ Math Expert Joined: 02 Sep 2009 Posts: 52285 ### Show Tags 15 Sep 2014, 23:39 Official Solution: All pages of a book are numbered. If the first page is numbered 1 and the last page is numbered 705, on how many pages does digit 9 appear in the numeration? A. 70 B. 77 C. 126 D. 133 E. 140 Compute on how many pages 9 appears in the first 100 pages and then multiply the result by 7. In the first 100 we have 10 pages where 9 appears as a units digit and 10 pages where 9 appears as a tens digit. Page 99 is counted twice, so the total number of pages on which 9 appears in the first 100 is 19. Therefore, $$19*7 = 133$$. _________________ Intern Joined: 22 Oct 2014 Posts: 4 GMAT 1: 740 Q50 V39 GPA: 4 WE: Marketing (Manufacturing) ### Show", "Question 4: If a book has 252 pages, how many digits have been used to number the pages? [1] 650 [2] 648 [3] 660 [4] None of these Option # 2 From page number 1 to age number 9, we will use 1 digit per page $\\Rightarrow$ digits used = 9. From page number 10 to age number 99, we will use 2 digits per page $\\Rightarrow$ digits used = 2 × 90 = 180. From page number 100 to age number 252, we will use 3 digits per page $\\Rightarrow$ digits used = 3 × 153 = 459. Therefore, total number of digits used = 9 + 180 + 459 = 648 Question 5: 1 and 8 are the first two positive integers for which 1 + 2 + 3 + ... + n is a perfect square. Which number is the 4th such number? 1 + 2 + 3 + … + n = $\\frac{n(n+1)}{2}$= M2 (say) $\\Rightarrow$ n(n + 1) = 2 M2 Now n and n + 1 will have no factor in common. Since RHS is twice the square of a positive integer, one of n and n + 1 will be twice of a perfect square and the other will be a perfect square. As twice of a perfect square will be", "# Math Help - How many pages? 1. ## How many pages? The pages of a book are numbered, beginning with page 1. If all of the pages in the book are considered, there are a total of 2,989 individual digits needed to print the page numbers. How many pages does the book contain? 2. ## Re: How many pages? Hello, testtrail429! There is no formula for this problem. We must \"talk\" our way through it. The pages of a book are numbered, beginning with page 1. There is a total of 2,989 individual digits needed to print the all page numbers. How many pages does the book contain? There are 9 one-digit numbers (1 to 9): . $9$ digits. There are 90 two-digit numbers (10 to 99): . $2\\times 90 \\,=\\,180$ digits. There are 900 three-digit numbers (100 to 999): . $3\\times 900 \\,=\\,2700$ digits. We have accounted for: $9 + 180 + 2700 \\,=\\,2889$ digits. There are: . $2989 - 2889 \\,=\\,100$ digits left. They will be used by the first 25 four-digit numbers (from 1000 to 1024). Therefore, the book contains $\\color{blue}{1024}$ pages.", "# Math Help - Pages in a book - need to verify my solution 1. ## Pages in a book - need to verify my solution Hi All, The pages of a book are numbered as 1, 2, 3, .... It is found that 195 digits are used. (a) How many pages are numbered with a single-digit number? (b) How many pages are numbered with a two-digit number? (c) How many pages are there altogether? (a) I got single-digit numbers = 9 (b) I got, two-digit numbers = 99 - 9 = 90 (c) This is the part I need verified. $ \\begin{tabular}{ l l } \\textnormal{No. of 1 digit pages} &= 9 \\\\ \\textnormal{No. of 2 digit pages} &= \\begin{math}\\frac{90}{2}\\end{math} = 45 \\\\ \\\\ \\textnormal{No. of 3 digit numbers used} &= 195 - 99 = 96 \\\\ \\textnormal{No. of 3 digit pages} &= \\begin{math}\\frac{96}{3}\\end{math} = 32 \\\\ \\\\ \\textnormal{Total pages} &= 9 + 45 + 32 = 86 \\\\ \\end{tabular} $ The required answer is 101, I am not sure how to get there. Can you help identify my mistake. Thanks for your help. 2. two issues here: first: The number of 2 digit pages is 90, not 45. If there were only 45 2 digit numbers, the pages would go 1,2,3,4,5,6,7,8,9,10,.......58,100 which is not correct as", "+0 # A textbook has pages. How many of the pages have page numbers whose digits add up to exactly ​? 0 37 2 A textbook has $$1000$$ pages. How many of the pages have page numbers whose digits add up to exactly $$4$$? Feb 14, 2023 #1 0 If the textbook has 1000 pages, then the page numbers range from 1 to 1000. To find the number of pages with page numbers whose digits add up to exactly 4, we can use the following formula: Number of pages = $\\sum_{i=1}^{9}$(Number of pages with i as the last digit) For example, the number of pages with 0 as the last digit is the same as the number of pages with 4 as the last digit, as both digits add up to 4. Similarly, the number of pages with 1 as the last digit is the same as the number of pages with 3 as the last digit, as both digits add up to 4. Using this formula, we can calculate that the number of pages with page numbers whose digits add up to exactly 4 is 12. Feb 14, 2023 #2 0 1 == 4 2 == 13 3 == 22 4 == 31 5 == 40 6 == 103 7 == 112 8 == 121 9 ==", "5, each atleast once and using addition and subtraction (For example 7 = 5 + 5 – 3) 1 = 3 + 3 – 5 2 = 5 – 3 3 = 5 + 3 – 5 4 = 3 + 3 + 3 – 5 5 = 3 + 5 – 3 6 = 5 + 5 + 5 – 3 – 3 – 3 7 = 5 + 5 – 3 8 = 5 + 3 9 = 5 + 5 + 5 – 3 – 3 10 = 5 + 5 + 3 – 3 Question 6. Find all 2-digit numbers each of which is divisible by the sum of its digits. The numbers are 10, 12, 18, 21, 24, 27, 36, 42, 45, 54, 63, 72 and 81. Question 7. The page numbers of a book written in a row gives a 216 digit number. How many pages are there in the book? Page numbers 1 to 9 written in a row form a 9 digit numbers. The page numbers 10 to 99 (= 90 pages) form 180 digit number. The remaining number digits i.e., (216 – (9 + 180)) = 27 27 is formed by pages 100 to 108 (9 pages) ∴ The total number of pages = 9 + ’90 + 9", "3) 1 = 3 + 3 – 5 2 = 5 – 3 3 = 5 + 3 – 5 4 = 3 + 3 + 3 – 5 5 = 3 + 5 – 3 6 = 5 + 5 + 5 – 3 – 3 – 3 7 = 5 + 5 – 3 8 = 5 + 3 9 = 5 + 5 + 5 – 3 – 3 10 = 5 + 5 + 3 – 3 Question 6. Find all 2-digit numbers each of which is divisible by the sum of its digits. The numbers are 10, 12, 18, 21, 24, 27, 36, 42, 45, 54, 63, 72 and 81. Question 7. The page numbers of a book written in a row gives a 216 digit number. How many pages are there in the book? Page numbers 1 to 9 written in a row form a 9 digit numbers. The page numbers 10 to 99 (= 90 pages) form 180 digit number. The remaining number digits i.e., (216 – (9 + 180)) = 27 27 is formed by pages 100 to 108 (9 pages) ∴ The total number of pages = 9 + ’90 + 9 = 108 Question 8. Look at the following pattern. This is called Pascal’s triangle. What is the", "+ 4 = LXXXIV (ii) 99 = 90 + 9 = XCIX (iii) 145 = 100 + (50 − 10) + 5 = CXLV (iv) 406 = 400 + 6 = CDVI (v) 519 = 500 +10 + 9 = DXIX #### Page No 21: Successor of 999999 = 999999 + 1 = 1000000 Predecessor of 999999 = 999999 − 1 = 999998 ∴ Required difference = 1000000 − 999998 = 2 #### Page No 21: (i) The number is 1046. Its digit at the hundreds place is 0 < 5. So, the given number is rounded off to the nearest thousand as 1000. (ii) The number is 973. Its digit at the hundreds place is 9 > 5. So, the given number is rounded off to the nearest thousand as 1000. (iii) The number is 5624. Its digit at the hundreds place is 6 > 5. So, the given number is rounded off to the nearest thousand as 6000. (iv) The number is 4368. Its digit at the hundreds place is 3 < 5. So, the given number is rounded off to the nearest thousand as 4000. #### Page No 21: Option (a) is correct. X can be subtracted from L and C only. i.e., XC = ( 100 − 10 ) = 90 #### Page No", "1. ## Difficult Word Problem In numbering the pages of a book, a printer used 3289 digits. How many pages were in the book, assuming that the first page in the book was number 1. 2. Hello, MATNTRNG! There's no formula for this problem. I had to baby-talk my way through it. In numbering the pages of a book, a printer used 3289 digits. How many pages were in the book, assuming that the first page in the book was number 1? . . $\\begin{array}{cc}\\text{Pages} & \\text{Digits} \\\\ \\hline 1\\text{ to } 9 & 9 \\\\ 10\\text{ to }99 & 180 \\\\ 100\\text{ to }999& 2700 \\\\ \\hline \\text{Total:} & 2889 \\end{array}$ Hence: . $3289 - 2889 \\:=\\:400$ digits were used . . in the first 100 four-digit numbers. . . . . $\\{1000,\\:1001,\\:1002,\\:\\hdots\\:1099\\}$ Therefore, the book had 1099 pages. 3. Originally Posted by Soroban There's no formula for this problem. I think you basically found a formula. $\\{a_n\\} = 9, 180, 2700,\\dots$ $a_n=9\\cdot10^{n-1}\\cdot n$ $\\displaystyle b_n=\\sum_1^n a_n$ Being lazy to derive, I looked up on OEIS (id:A033713 - OEIS Search Results) $b_n=\\frac{1}{9}\\cdot(n\\cdot(10^{n+1})-(n+1)\\cdot10^n+1)$ So if the number of digits used is d, then we find the n such that $b_n \\le d < b_{n+1}$ and the number of pages is $p=10^n-1+\\frac{b_{n+1}-d}{n+1}$ But in fact the $b_i$ are easily", "37, 47, 57, 67, 77, 87, and 97 (9 total pages) 3-digit pages up to page 300: pages 107, 117, 127, 137, 147, ... 197; 207, 217, 227, 237, 247, ..., 297 (20 total pages) 3-digit pages between 301 and 320: 307, 317 (2 total pages) Summating the totals, we get 1 + 9 + 20 + 2 = 32 pages with which a 7 is associated. Remember that the question is not about the number of pages in which seven appears, but rather the frequency at which seven appears. This sum (32)does not account for pages 77, 177, and 277, wherein seven is used twice; thus, we add three to the thirty-two 'seven-associated pages' to obtain 35 'seven frequencies'. Hence, the number seven is printed 35 times. 15. Michael says: My apologies: I neglected to add 70, 71, 72, 73, 74, 75, 76, 78, 79, and 170, 171, 172, 173, 174, 175, 176, 178, 179, and 270, 271, 272, 273, 274, 275, 276, 278, 279, giving 27 additional times where seven appears. Thus, add 27 to 35 to obtain 62 'seven frequencies'. Hence, the number seven is printed 62 times. 16. Michael says: This solution assumes that the consecutive page numbering beings at page 1. 17. Ramesh Babu says: Dear Sir, The reasons for teachers' resistance to", "# Unsolved arithmetic exercise for printing numbers in book Here is the exercise from the \"Arithmetic\" book (ISBN: 5-02-013764-2). There are 1989 numbers had printed to enumerate book's pages. How many pages in book? Honestly I decided to refresh my arithmetic knowledge at all but stuck with that problem. Here's my decision: • there are 9 numbers to print for pages 1-9 (9 * 1); • there are 200 numbers to print for pages 10-99 (100 * 2); • 1989 - 209 = 1780; • 1780 /3 = 593,3(3) - that is the problem I clearly understand how simple it must be and maybe something is wrong with my brain. But please, tell me where I made mistake. - There are $90$ integers in the interval $[10,99],$ so your second line should be $90\\times 2=180.$ – Andrew Aug 3 '12 at 18:32 Thanks a lot for correction. For me, I'm thinking about adult brain is less flexible than young and it thinks that he knows something and it's the must right thing. Sorry for such silly mistake. – Крайст Aug 3 '12 at 22:22 There are not 100 numbers between 10 and 99. There are 10 between 10 and 19, 20 between 10 and 29, ..., 90 between 10 and 99. Each of those numbers has two digits.", "from 10 to 99 there are 90 pages and these 90 pages are numbered with double digits(i-e 10, 11, 12,……up to 99), so, Total digits for pages starting from 10 to 90 will be ’90 pages multiplied by two digits’ i-e 90×2=180 digits Case 3: Now after 99 th page it starts with case 3; i-e now onward the pages are numbered with three digit number (i-e from 100, 101. 102, …….. up to 999th page) and these will be 900 pages . So the number of digits for nine hundred triple digit numbered 900 pages = 900×3=2700 digits Now we add digits used in case 1, case 2 and case 3 so, 9+180+2700 = 2889 digits,But we have been given that total digits are 3189. So, now we will count pages for 3189-2889=300 digits. Case 4: Now after 999th page comes 1000th page which is numbered with 4 digits. We need 75 tetra digit pages to use 300 digits. So we see that 1000 to 1074 will make 75 pages having 4 digit numbering, and digits used by 75 four digit numbers = 75×4=300 Now add pages of all cases i-e 9+90+900+75 =1074 pages Insert math as $${}$$", "first 2008 odd numbers? Let us try to identify the pattern an try for small set. Sum of first 5 even numbers N(Even) = 2 + 4 + 6+ 8 + 10 Similarly, sum of first 5 odd numbers will be: N(Odd) = 1 + 3 + 5 + 7 + 9 If we subtract Odd list from Even list, i.e. N(Even) - N (Odd) = (2+4+8+10+12) - (1+3+5+7+9) ⇒ = (2-1) + (4-3) + (6-5) + (8-7) + (10-9) ⇒ = 1 + 1 + 1 + 1 + 1 = 5 (or 5 ones) Similarly, the answer for the first 2008 numbers would be? 2008 Q5: The pages of a book are numbered consecutively, beginning with 1. The digit 7 is printed 25 times in numbering the pages. What is the largest number of pages the book can have? Count number of 7 appears how many times between 1 and 100 pages numbers = 20 (e.g. 7, 17, 27, 37, 47, 57, 67, 77, 87, 97, 70, 71, 72, 73, 74, 75, 76, 78, 79 = 20, 77 has two 7 digits) Count next five time 7 digit appears from page 101 onwards i.e. 107, 117, 127, 137, 147. Thus the largest number of pages the book will have: 157-1 = 156 Q6: A student multiplies", "After every two numbers, one is omitted. Because $89 = 2\\times 44 +1$, there must be $44$ page numbers missing and so the number on the last page is $89+44=133$.", "## Saturday, June 9, 2012 ### Unit digit, Tenth digit and Digit Sum Word problems on unit digit, tenth digit or digit sum. #1: How many digits are there in the positive integers 1 to 99 inclusive? Solution I: From 1 to 9, there are 9 digits. From 10 to 99, there are 99 - 10 + 1 or 99 - 9 = 90 two digit numbers. 90 x 2 = 180 Solution II: ___ There are 9 one digit numbers (from 1 to 9). ___ ___ There are 9 * 10 = 90 two digit numbers (You can't use \"0\" on the tenth digit but you can use \"0\" on the unit digit.) 90 * 2 + 9 = 189 # 2: A book has 145 pages. How many digits are there if you start counting from page 1? There are 189 digits from page 1 to 99. (See #1, solution I) From 100 to 145, there are 145 - 100 + 1 or 145 - 99 = 46 three digit numbers. 189 + 46*3 = 327 digits. #3: \"A book has N pages, number the usual way, from 1 to N. The total number of digits in the page number is 930. How many pages does the book have\"? Similar to one Google interview question. Read" ]

[ "# How many times does the digit 4 occur in the page numbers of a 50 page book? Dec 13, 2016 $15$ #### Explanation: The digit $4$ can occur in the units or tens positions. It occurs in the units position in the numbers: $4 , 14 , 24 , 34 , 44$ It occurs in the tens position in the numbers: $40 , 41 , 42 , 43 , 44 , 45 , 46 , 47 , 48 , 49$ So there are a total of $5 + 10 = 15$ occurrences of the digit $4$. If the question asked in how many page numbers the digit $4$ occurs then the answer would be $14$ - avoiding counting $44$ twice. However, since it asks for the number of occurrences of the digit $4$, $44$ should be counted twice.", "# Math Help - How many pages? 1. ## How many pages? The pages of a book are numbered, beginning with page 1. If all of the pages in the book are considered, there are a total of 2,989 individual digits needed to print the page numbers. How many pages does the book contain? 2. ## Re: How many pages? Hello, testtrail429! There is no formula for this problem. We must \"talk\" our way through it. The pages of a book are numbered, beginning with page 1. There is a total of 2,989 individual digits needed to print the all page numbers. How many pages does the book contain? There are 9 one-digit numbers (1 to 9): . $9$ digits. There are 90 two-digit numbers (10 to 99): . $2\\times 90 \\,=\\,180$ digits. There are 900 three-digit numbers (100 to 999): . $3\\times 900 \\,=\\,2700$ digits. We have accounted for: $9 + 180 + 2700 \\,=\\,2889$ digits. There are: . $2989 - 2889 \\,=\\,100$ digits left. They will be used by the first 25 four-digit numbers (from 1000 to 1024). Therefore, the book contains $\\color{blue}{1024}$ pages.", "zeros in this list of numbers. This is simply 9, if w=1, (99+9) if w=2, (999+99+9) if w=3 and so on. So, solve for 852 = n*w - (9+99). Guessing, w=2, gives the answer of 248. 8. Ardra says: New puzzle It takes 852 digits to number the pages of a particular book consecutively. (i) How many pages are there? (ii) How many times is the digit 7 printed? (i) There are 320 pages. (ii) The digit 7 is printed 62 times. 9. Thomas A Buckley says: (i) Number of pages 9 Pages 1 to 9 requires 9 digits 90 ,, 10 to 99 ,, 180 ,, Digits remaining to be used = 852 -( 9+ 180) = 663 Next pages from 100 inclusive , each requiring 3 digits 663/3 = 221 pages Total pages = 9 + 90 + 221 = 320 = Answer (i) (ii) Number of 7's digit No of 7,s in the Tens Units Pages 1 to 9 0 1 ,, 10 to 99 10 9 ,, 100 to 320 20 22 Total number of 7,s = 10 + 20 + 1 + 9 + 22 = 62 ,, 100 to 320 10. Francis Kisner says: It takes 852 digits to number the pages of a particular book consecutively. (i) How many pages are", "-> 6 digits/page -> 5394 start at 852 digits and work your way up through the pages by subtracting the above values to calculate amount of pages with 6 digits (852 - 18 - 3 - 356 - 5)/6 = 80 add all pages = 9 + 1 + 89 + 1 + 80 = 180 13. guido says: pages 1 to 9 : 9 digits #7 : 1 pages 10 to 99 : 180 digits --- total : 189 #7 :10 pages100 to199 : 300 digits --- total : 489 #7 :11 pages200 to299 : 300 digits --- total : 789 #7 :11 pages300 to319 : 60 digits --- total : 849 #7 : 2 pages320 to322 : 3 digits --- total : 852 So book has 322 pages and 35 x digit 7 14. Michael says: (a) We begin by noting the different types of pages: pages containing one digit, two digits, and three digits. Pages 1 through 9 (9 pages, or 1 * 9 total digits) each occupy one of the 852 digits. Pages 10 through 99 (90 pages, or 2 * 90 total digits) each occupy two of the 852 digits. Pages 100 through 999 (900 pages, or 3 * 900 total digits) each occupy three of the 852 digits; however, logic indicates that", "there cannot be 999 consecutively numbered pages (beginning at 1) such that only 852 digits are used to number the pages. Therefore, we must determine how many three-digit pages there are in the book (let n represent this). Notice the form of the 'total digit' products; each has the form: (# of digits)(# of pages requiring said number of digits): 1 * 9 = (1 digit)(9 pages requiring 1 digit) 2 * 90 = (2 digits)(90 pages requiring 2 digits) 3 * n = (3 digits)(n pages requiring 3 digits) These products represent the number of digits that are being used for 1-, 2-, and 3-digit pages. Summating these products gives 852 digits: 9 + 180 + 3n = 852 3n = 663 n = 221 Therefore, there are 221 pages requiring three digits. When we add to this the nine pages requiring one digit and the ninety pages requiring two digits, we get: 9 + 90 + 221 = 320 total pages. Hence, there are 320 pages in the book. Check: 1 * 9 + 2 * 90 + 3 * 221 = 852 total digits (b) In this part, it is not tedious to list all the pages with which a seven is associated. 1-digit pages: page 7 (1 total page) 2-digit pages: pages 17, 27,", "'wooden'. 6. Page 13--Example II--In the calculations, the intermediate value in the book was printed as the square root of 67.2. The left part is correct, but reduces to the square root of 268.8, and that is ~16.395. So I have corrected the intermediate value. 7. Page 13--Because the original page scan cut off the text on the right edge, I have made assumptions on what text was missing. Because the scans came from an outside source, I could not get the missing information, which was the words at the end of Example II, and words in the last paragraph of the page. 8. Page 14--Three typos found: 'dimensiens' for 'dimensions', 'betng' for 'being', and 'ars' for 'are'. 9. Page 17--a value in a formula was printed as 6000, but in the context of the other information, particularly the example immediately following, the value was believed to be incorrect, and was changed to 5000. 10. Page 23--The value of 388 sq. inches at the top of the page in the book is incorrect; 3 x 8 x 12 = 288, so has been corrected. 11. Page 24--Rods section, numerical value appeared to be 15.000 in the book, but from context, must be 15,000 instead. 12. Page 28--Typos: changed 'Trautwine's Edgineer's Pocket-Bood' to Trautwine's Engineer's Pocket-Book'; corrected 'af' to", "SEARCH HOME Math Central Quandaries & Queries Question from Chungus: It takes 852 digits to number the pages of a book consecutively. How many pages are there? Hi Chungus, There are 9 one digit pages, pages 1 to 9. Next look at 2 digit numbers. They go from 10 to 99 so that is 90 two digit numbers. Each contains 2 digits so when you get to 99 you have used $9 + 2 \\times 90 = 189$ digits. If we apply a similar argument to three digit numbers we will see that we have gone too far so we need to look at three digit numbers in smaller chunks. From 100 to 199 there are 100 three digit numbers. Since each has 3 digits that's $3 \\times 100 = 300$ digits so when we get to page 199 we have used $9 + 2 \\times 90 + 3 \\times 100 = 489$ digits.", "Tags 15 Sep 2014, 23:39 2 4 00:00 Difficulty: 85% (hard) Question Stats: 41% (01:03) correct 59% (01:09) wrong based on 153 sessions ### HideShow timer Statistics All pages of a book are numbered. If the first page is numbered 1 and the last page is numbered 705, on how many pages does digit 9 appear in the numeration? A. 70 B. 77 C. 126 D. 133 E. 140 _________________ Math Expert Joined: 02 Sep 2009 Posts: 52285 ### Show Tags 15 Sep 2014, 23:39 Official Solution: All pages of a book are numbered. If the first page is numbered 1 and the last page is numbered 705, on how many pages does digit 9 appear in the numeration? A. 70 B. 77 C. 126 D. 133 E. 140 Compute on how many pages 9 appears in the first 100 pages and then multiply the result by 7. In the first 100 we have 10 pages where 9 appears as a units digit and 10 pages where 9 appears as a tens digit. Page 99 is counted twice, so the total number of pages on which 9 appears in the first 100 is 19. Therefore, $$19*7 = 133$$. _________________ Intern Joined: 22 Oct 2014 Posts: 4 GMAT 1: 740 Q50 V39 GPA: 4 WE: Marketing (Manufacturing) ### Show", "# Unsolved arithmetic exercise for printing numbers in book Here is the exercise from the \"Arithmetic\" book (ISBN: 5-02-013764-2). There are 1989 numbers had printed to enumerate book's pages. How many pages in book? Honestly I decided to refresh my arithmetic knowledge at all but stuck with that problem. Here's my decision: • there are 9 numbers to print for pages 1-9 (9 * 1); • there are 200 numbers to print for pages 10-99 (100 * 2); • 1989 - 209 = 1780; • 1780 /3 = 593,3(3) - that is the problem I clearly understand how simple it must be and maybe something is wrong with my brain. But please, tell me where I made mistake. - There are $90$ integers in the interval $[10,99],$ so your second line should be $90\\times 2=180.$ – Andrew Aug 3 '12 at 18:32 Thanks a lot for correction. For me, I'm thinking about adult brain is less flexible than young and it thinks that he knows something and it's the must right thing. Sorry for such silly mistake. – Крайст Aug 3 '12 at 22:22 There are not 100 numbers between 10 and 99. There are 10 between 10 and 19, 20 between 10 and 29, ..., 90 between 10 and 99. Each of those numbers has two digits.", "• # question_answer The total number of digits used in numbering the pages of a book having 366 pages, is? A) 732 B) 990C) 1098 D) 1305 Required numbers $=9+2\\times 90+267\\times 3$ $=189+801$", "'woooden' to 'wooden'. 6. Page 13--Example II--In the calculations, the intermediate value in the book was printed as the square root of 67.2. The left part is correct, but reduces to the square root of 268.8, and that is ~16.395. So I have corrected the intermediate value. 7. Page 13--Because the original page scan cut off the text on the right edge, I have made assumptions on what text was missing. Because the scans came from an outside source, I could not get the missing information, which was the words at the end of Example II, and words in the last paragraph of the page. 8. Page 14--Three typos found: 'dimensiens' for 'dimensions', 'betng' for 'being', and 'ars' for 'are'. 9. Page 17--a value in a formula was printed as 6000, but in the context of the other information, particularly the example immediately following, the value was believed to be incorrect, and was changed to 5000. 10. Page 23--The value of 388 sq. inches at the top of the page in the book is incorrect; 3 x 8 x 12 = 288, so has been corrected. 11. Page 24--Rods section, numerical value appeared to be 15.000 in the book, but from context, must be 15,000 instead. 12. Page 28--Typos: changed 'Trautwine's Edgineer's Pocket-Bood' to Trautwine's Engineer's Pocket-Book'; corrected", "37, 47, 57, 67, 77, 87, and 97 (9 total pages) 3-digit pages up to page 300: pages 107, 117, 127, 137, 147, ... 197; 207, 217, 227, 237, 247, ..., 297 (20 total pages) 3-digit pages between 301 and 320: 307, 317 (2 total pages) Summating the totals, we get 1 + 9 + 20 + 2 = 32 pages with which a 7 is associated. Remember that the question is not about the number of pages in which seven appears, but rather the frequency at which seven appears. This sum (32)does not account for pages 77, 177, and 277, wherein seven is used twice; thus, we add three to the thirty-two 'seven-associated pages' to obtain 35 'seven frequencies'. Hence, the number seven is printed 35 times. 15. Michael says: My apologies: I neglected to add 70, 71, 72, 73, 74, 75, 76, 78, 79, and 170, 171, 172, 173, 174, 175, 176, 178, 179, and 270, 271, 272, 273, 274, 275, 276, 278, 279, giving 27 additional times where seven appears. Thus, add 27 to 35 to obtain 62 'seven frequencies'. Hence, the number seven is printed 62 times. 16. Michael says: This solution assumes that the consecutive page numbering beings at page 1. 17. Ramesh Babu says: Dear Sir, The reasons for teachers' resistance to", "the questions and others here from the Wall Street Journal. Solution I: 930 - 189 (digits of the first 99 pages) =741 741 divided by 3 = 247. Careful since you are counting the three digit numbers from 100 if the book has N pages N - 100 + 1 or N - 99 = 247. N = 346 pages. Solution II: 930 - 189 (total digits needed for the first 99 pages) = 741 741/3 = 247 (how far the three digit page numbers go). 247 + 99 = 346 pages #4: If you write consecutive numbers starting with 1, what is the 50th digit you write? Solution I: 50 - 9 = 41, 9 being the first 9 digits you need to use for the first 9 pages. Now it's 2 digit. 41/2 = 20.1 , which means you will be able to write 20 two digit numbers + the first digit of the next two digit numbers. 10 to 29 is the first 20 two digit numbers so the next digit 3 is the answer. (first digit of the two digit number 30.) Solution II: (50 - 9 ) / 2 = 20. 5 ; 20.5 + 9 = 29.5, so 29 pages + the first digit of the next two digit numbers, which is", "solution I) From 100 to 145, there are 145 - 100 + 1 or 145 - 99 = 46 three digit numbers. 189 + 46*3 = 327 digits. #3: \"A book has N pages, number the usual way, from 1 to N. The total number of digits in the page number is 930. How many pages does the book have\"? Similar to one Google interview question. Read the questions and others here from the Wall Street Journal. Solution I: 930 - 189 (digits of the first 99 pages) =741 741 divided by 3 = 247. Careful since you are counting the three digit numbers from 100 if the book has N pages N - 100 + 1 or N - 99 = 247. N = 346 pages. Solution II: 930 - 189 (total digits needed for the first 99 pages) = 741 741/3 = 247 (how far the three digit page numbers go). 247 + 99 = 346 pages #4: If you write consecutive numbers starting with 1, what is the 50th digit you write? Solution I: 50 - 9 = 41, 9 being the first 9 digits you need to use for the first 9 pages. Now it's 2 digit. 41/2 = 20.1 , which means you will be able to write 20 two digit numbers +", "Question 4: If a book has 252 pages, how many digits have been used to number the pages? [1] 650 [2] 648 [3] 660 [4] None of these Option # 2 From page number 1 to age number 9, we will use 1 digit per page $\\Rightarrow$ digits used = 9. From page number 10 to age number 99, we will use 2 digits per page $\\Rightarrow$ digits used = 2 × 90 = 180. From page number 100 to age number 252, we will use 3 digits per page $\\Rightarrow$ digits used = 3 × 153 = 459. Therefore, total number of digits used = 9 + 180 + 459 = 648 Question 5: 1 and 8 are the first two positive integers for which 1 + 2 + 3 + ... + n is a perfect square. Which number is the 4th such number? 1 + 2 + 3 + … + n = $\\frac{n(n+1)}{2}$= M2 (say) $\\Rightarrow$ n(n + 1) = 2 M2 Now n and n + 1 will have no factor in common. Since RHS is twice the square of a positive integer, one of n and n + 1 will be twice of a perfect square and the other will be a perfect square. As twice of a perfect square will be", "5, each atleast once and using addition and subtraction (For example 7 = 5 + 5 – 3) 1 = 3 + 3 – 5 2 = 5 – 3 3 = 5 + 3 – 5 4 = 3 + 3 + 3 – 5 5 = 3 + 5 – 3 6 = 5 + 5 + 5 – 3 – 3 – 3 7 = 5 + 5 – 3 8 = 5 + 3 9 = 5 + 5 + 5 – 3 – 3 10 = 5 + 5 + 3 – 3 Question 6. Find all 2-digit numbers each of which is divisible by the sum of its digits. The numbers are 10, 12, 18, 21, 24, 27, 36, 42, 45, 54, 63, 72 and 81. Question 7. The page numbers of a book written in a row gives a 216 digit number. How many pages are there in the book? Page numbers 1 to 9 written in a row form a 9 digit numbers. The page numbers 10 to 99 (= 90 pages) form 180 digit number. The remaining number digits i.e., (216 – (9 + 180)) = 27 27 is formed by pages 100 to 108 (9 pages) ∴ The total number of pages = 9 + ’90 + 9", "from 10 to 99 there are 90 pages and these 90 pages are numbered with double digits(i-e 10, 11, 12,……up to 99), so, Total digits for pages starting from 10 to 90 will be ’90 pages multiplied by two digits’ i-e 90×2=180 digits Case 3: Now after 99 th page it starts with case 3; i-e now onward the pages are numbered with three digit number (i-e from 100, 101. 102, …….. up to 999th page) and these will be 900 pages . So the number of digits for nine hundred triple digit numbered 900 pages = 900×3=2700 digits Now we add digits used in case 1, case 2 and case 3 so, 9+180+2700 = 2889 digits,But we have been given that total digits are 3189. So, now we will count pages for 3189-2889=300 digits. Case 4: Now after 999th page comes 1000th page which is numbered with 4 digits. We need 75 tetra digit pages to use 300 digits. So we see that 1000 to 1074 will make 75 pages having 4 digit numbering, and digits used by 75 four digit numbers = 75×4=300 Now add pages of all cases i-e 9+90+900+75 =1074 pages Insert math as $${}$$", "## Saturday, June 9, 2012 ### Unit digit, Tenth digit and Digit Sum Word problems on unit digit, tenth digit or digit sum. #1: How many digits are there in the positive integers 1 to 99 inclusive? Solution I: From 1 to 9, there are 9 digits. From 10 to 99, there are 99 - 10 + 1 or 99 - 9 = 90 two digit numbers. 90 x 2 = 180 Solution II: ___ There are 9 one digit numbers (from 1 to 9). ___ ___ There are 9 * 10 = 90 two digit numbers (You can't use \"0\" on the tenth digit but you can use \"0\" on the unit digit.) 90 * 2 + 9 = 189 # 2: A book has 145 pages. How many digits are there if you start counting from page 1? There are 189 digits from page 1 to 99. (See #1, solution I) From 100 to 145, there are 145 - 100 + 1 or 145 - 99 = 46 three digit numbers. 189 + 46*3 = 327 digits. #3: \"A book has N pages, number the usual way, from 1 to N. The total number of digits in the page number is 930. How many pages does the book have\"? Similar to one Google interview question. Read", "Exactly $195$ digits have been used to number the pages in a book. How many pages does the book have?", "3) 1 = 3 + 3 – 5 2 = 5 – 3 3 = 5 + 3 – 5 4 = 3 + 3 + 3 – 5 5 = 3 + 5 – 3 6 = 5 + 5 + 5 – 3 – 3 – 3 7 = 5 + 5 – 3 8 = 5 + 3 9 = 5 + 5 + 5 – 3 – 3 10 = 5 + 5 + 3 – 3 Question 6. Find all 2-digit numbers each of which is divisible by the sum of its digits. The numbers are 10, 12, 18, 21, 24, 27, 36, 42, 45, 54, 63, 72 and 81. Question 7. The page numbers of a book written in a row gives a 216 digit number. How many pages are there in the book? Page numbers 1 to 9 written in a row form a 9 digit numbers. The page numbers 10 to 99 (= 90 pages) form 180 digit number. The remaining number digits i.e., (216 – (9 + 180)) = 27 27 is formed by pages 100 to 108 (9 pages) ∴ The total number of pages = 9 + ’90 + 9 = 108 Question 8. Look at the following pattern. This is called Pascal’s triangle. What is the" ]

[ "+0 # If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? +1 1668 11 +473 If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? Feb 3, 2018 #1 +109721 0 If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? Thanks Alan. . Feb 3, 2018 edited by Melody Feb 5, 2018 #2 +473 +2 RektTheNoob Feb 3, 2018 edited by RektTheNoob Feb 3, 2018 #3 +109721 0 ok so what do you think the correct answer is and what makes you think it is correct? AND if you have the working we would like to see that too. I, and others, would like to learn too. Melody Feb 4, 2018 #5 +473 +3 Solution: We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways Greg can roll the same number of 1's and 6's: he can roll two of", "it shows up at least once. There is the old story of the gambler betting on double sixes in 24 rolls and losing and this is very close. – Ross Millikan Jan 9 '18 at 18:12", "# Question #ae79f Aug 17, 2017 1/6 if the spinner has six sections and each section is the same size with an equal probability of being selected. #### Explanation: If the spinner has six sections then the probability of any one number is one out of six. The dice has six sides and has a probability of any one number is also one out of six. This give 36 possible outcomes. six of the them have the same number 1,1 2,2 3,3 4,4 5,5 6,6. This gives 6/36 possibilities of having the same number on the spinner and the dice. 6/ 36 = 1/6. Having a spinner has no effect on probability that is different from a dice assuming the spinner has six sides with equal probability", "Vick and Wale are playing a board game. Michael Vick is playing the black piece, and it is his turn. Michael Vick rolls two dice to determine how far to move. (a) What is the probability that Michael Vick passes Wale on this move? (b) What is the probability that Michael Vick wins on this move? (c) What is the probability that Wale wins on the next move? ■ edit Represent sample spaces for compound events asdf Drag and drop problems to change the ordering.", "flipped 6 times. What is the 7 16 Dec 2009, 09:01 3 a fair 2-sided coin is flipped 6 times. what is the 3 18 May 2007, 20:56 Kaplan Problem Susan flipped a fair coin N times. what 1 11 Jul 2006, 08:45 1 Kevin flips a coin four times. WHat is the probability the 11 18 Nov 2005, 04:11 Display posts from previous: Sort by", "flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads? (The answer value will be from 0 to 1, not as a percentage.) Two fair dice are rolled, and it is revealed that (at least) one of the numbers rolled was a 4. What is the probability that the other number rolled was a 6? Note: You are not told which of the numbers rolled is a 4. ×", "fair, six-sided dice. What is the probability that Wu [#permalink] 13 Mar 2017, 21:59 Display posts from previous: Sort by", "four or higher. As can be seen in the diagram below, out of thirty-six equally probable outcomes possible with the roll of two fair six-sided dice, there are only three rolls unfavorable to this outcome, and thirty-three favorable: This makes the odds , or 11 to 1, in favor of$10,000 being collected on this turn. ### Example Question #32 : Probability In the game of Crazy Eights, which is played with an ordinary deck of 52 playing cards, each player lays down a card from his hand on top of a discard pile according to the card that has previously been played. The card he is allowed to play must be a card of the same rank or suit, or it must be an eight of any suit. At the beginning of the hand, the seven of diamonds was turned over; Kenny, who was first, played a six of diamonds on top of it. Dick, who plays next, has the following cards in his hand: As can be seen, Dick does not have a playable card. By the rules, he must draw a card, and, if he can play it, he must. What is the probability that the card he draws will be playable? Explanation: Dick knows 10 of the cards the deck - the eight in his", "As can be seen in the diagram below, out of thirty-six equally probable outcomes possible with the roll of two fair six-sided dice, there are only three rolls unfavorable to this outcome, and thirty-three favorable: This makes the odds , or 11 to 1, in favor of$10,000 being collected on this turn. ### Example Question #62 : Data Analysis / Probablility In the game of Crazy Eights, which is played with an ordinary deck of 52 playing cards, each player lays down a card from his hand on top of a discard pile according to the card that has previously been played. The card he is allowed to play must be a card of the same rank or suit, or it must be an eight of any suit. At the beginning of the hand, the seven of diamonds was turned over; Kenny, who was first, played a six of diamonds on top of it. Dick, who plays next, has the following cards in his hand: As can be seen, Dick does not have a playable card. By the rules, he must draw a card, and, if he can play it, he must. What is the probability that the card he draws will be playable? Explanation: Dick knows 10 of the cards the deck - the eight in his", "Show Me A 6! Probability Level 2 What is the probability of getting at least one six in a single throw of $3$ unbiased dice? ×", "12 and wins. theres a 5/6 chance of this happening. If the dice rolls 1, bob wins. • That was fast. Spot on. – Marius Apr 6 '16 at 10:39 • Isomorphic to nim – Martin Capodici Apr 6 '16 at 22:40", "predict how likely is it to roll a number less than 6?", "then rolls a fair number cube labeled 1 through 6, what is the probability of tossing heads followed by rolling a number less than 4? 1/4", "the hand containing the larger number with probability > 0.5? No! Greg can compute with that number, he can throw darts at it, he can stare at it until the cows come home, but it won’t help him one bit, because unless my poker face cracks, there’s no way to for him to predict (ok, retrodict) the flip of a fair coin. 4) But Greg’s solution claims an expected winning percentage greater than 50% regardless of Player 1’s algorithm or the specific numbers chosen. So what gives? I looked in vain for an error in Greg’s solution, but came away more convinced than ever of its correctness. I rethought my specific scenario, but it’s pretty straightforward – you just can’t predict a fair coin flip. Until I saw my error, my confusion was near-total; it was great. 5) Thanks again John/Greg for sharing this fine puzzle. I’m sure that much more confusion is in store, because I still find aspects of it bizarre. I can’t wait:-) • Greg Egan says: And say that Player 2 randomly picks the one of my hands which contains b. I know you understand what the error is, but to spell it out for anyone reading along casually: if you assume that Player 2 picked the hand that contains b, then you’re only", "# Possible Six-Flip Results When $6$ indistinguishable fair coins are thrown, how many different outcomes are there? Details and Assumptions: • Two outcomes are the same if they contain the same number of heads. ×", "Nevertheless, you can refer to . • Coin tosses: Two possible outcomes. Heads or Tails. • Dice rolling: Six possible outcomes. 1-2-3-4-5-6. • Card drawing: 52 possible outcomes. There are 4 types (clubs, diamonds, spades and hearts) and 13 ranks for each type. (A)ce - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - (J)ack - (Q)ueen - (K)ing.", "rolling three six-sided dice, and getting a 6 14 Apr 2015, 04:14 5 What is the probability of rolling two normal six-sided dice and getti 4 08 Dec 2014, 05:16 1 Peter rolls two dice at the same time. What is the probabili 1 28 Apr 2014, 00:32 7 A gambler rolls three fair six-sided dice. What is the probability 4 31 Jan 2012, 16:50 Display posts from previous: Sort by", "fallacy The gambler's fallacy is the belief that the outcome of a random event is somehow influenced by previous random events. In our craps case, some examples might be: • double six hasn't come up in 130 throws, so it's much more likely to come up now (the probability is higher than 1/36) • double one has just come up, therefore it's not likely to come up again soon (the probability is less than 1/36). It's a fallacy because each roll of the dice is completely independent; it doesn't matter what the previous throws were. There could have been 1,000 throws without a double six, but the probability of a double six will always be 1/36. The logic same applies to the snake-eyes example, if a snake-eyes has been thrown, the probability of throwing another snake-eyes immediately after is still 1/36. Let me lay this out even more starkly, in craps: • At the very first roll of the dice, the probability of a double six is 1/36. • After ten rolls of the dice, the probability of the next roll being a double six is 1/36. • After 100 rolls without a double six, the probability of the next roll being a double six is 1/36. • After 200 rolls without a double six, the probability of", "# Show Me A 6! Discrete Mathematics Level 2 What is the probability of getting at least one six in a single throw of $$3$$ unbiased dice? ×", "6 Jason flips a coin three times. What is the probability that the coin 4 13 Aug 2015, 09:35 Display posts from previous: Sort by" ]

[ "each, one of each, or none of each. If he rolls two of each, there are $$\\binom{4}{2}=6$$ ways to choose which two dice roll the 1's. If he rolls one of each, there are $$\\binom{4}{1}\\binom{3}{1}=12$$ ways to choose which dice are the 6 and the 1, and for each of those ways there are $$4\\cdot4=16$$ ways to choose the values of the other dice. If Greg rolls no 1's or 6's, there are $$4^4=256$$ possible values for the dice. In total, there are $$6+12\\cdot16+256=454$$ ways Greg can roll the same number of 1's and 6's. There are $$\\dfrac{1}{2} \\left(1-\\dfrac{454}{1296}\\right)=\\boxed{\\dfrac{421}{1296}}$$ total ways the four dice can roll, so the probability that Greg rolls more 1's than 6's is . RektTheNoob Feb 5, 2018 #7 -1 RektTheNoob: If you know the answers to the questions you post on this forum in such detail as the above answer demonstrates, then what is the PURPOSE of posting your questions in the first place?? Are you trying to test the Mods and other volunteers' mathematical knowledge? Or are you trying to show your mathematical superiority??!! Guest Feb 5, 2018 #10 +109721 0 Presumable he did not know the answer when he asked the question but was given or worked out the answer afterwards. I am very glad that he did post this answer", "# Count the number of ways of four distinct numbers showing up when six dice are rolled Suppose we roll six fair dice, how many ways can four distinct numbers show up? - Did you at least try some combinatoric argument? I feel the inclusion-exclusion principle might come in handy, but I'm really tired right now, so it's just a wild guess. – Patrick Da Silva Dec 9 '11 at 2:34 The number of ways doesn't depend on the dice being fair. – joriki Dec 9 '11 at 2:59 I'll assume that you mean six-sided dice. There are $\\binom64$ ways of choosing the $4$ distinct numbers. They can either appear $3,1,1,1$ times or $2,2,1,1$ times. In the first case, there are $4$ choices of the number appearing thrice and $6\\cdot5\\cdot4$ choices for the positions. In the second case, there are $6$ choices for the two numbers appearing twice and $6\\cdot5\\cdot\\binom42$ choices for the positions. Thus, the total is $$\\binom64\\left(4\\cdot6\\cdot5\\cdot4+6\\cdot6\\cdot5\\cdot\\binom42\\right)=15\\cdot6\\cdot5\\cdot(16+36)=23400\\;.$$ Thus, the probability of this happening is $23400/6^6=23400/46656\\approx50\\%$. The corresponding probabilities for the other numbers of distinct numbers are: $$\\begin{eqnarray} p(1)&=&\\binom61/6^6\\approx0.01\\%\\;,\\\\ p(2)&=&\\binom62(2^6-2)/6^6\\approx2\\%\\;,\\\\ p(3)&=&\\binom63\\left(3\\cdot6\\cdot5+3!\\cdot6\\cdot\\binom52+\\binom62\\binom42\\right)/6^6\\approx23\\%\\;,\\\\ p(5)&=&\\binom65\\cdot5\\cdot6\\cdot5\\cdot4\\cdot3/6^6\\approx23\\%\\;,\\\\ p(6)&=&\\binom666!/6^6\\approx2\\%\\;. \\end{eqnarray}$$ - Really nice. Thank you. – geraldgreen Dec 9 '11 at 3:15", "+0 # If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? +1 1668 11 +473 If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? Feb 3, 2018 #1 +109721 0 If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? Thanks Alan. . Feb 3, 2018 edited by Melody Feb 5, 2018 #2 +473 +2 RektTheNoob Feb 3, 2018 edited by RektTheNoob Feb 3, 2018 #3 +109721 0 ok so what do you think the correct answer is and what makes you think it is correct? AND if you have the working we would like to see that too. I, and others, would like to learn too. Melody Feb 4, 2018 #5 +473 +3 Solution: We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways Greg can roll the same number of 1's and 6's: he can roll two of", "be total of $9$ combinations possible and thus $P=(1/6)^9$ but its not the correct answer. On the contrary, the right answer is $P=6(1/216)=1/36$ I DO NOT UNDERSTAND why we multiply by $6$. E) here since they ask for the same number the probability of getting say $(2,2,2)$ combo is $P= 1/6 \\cdot 1/6 \\cdot 1/6 = 1/216$ thus having $6$ #'s on the dice the total probability of having all the same #'s is $P = 1/216+...+1/216=6/216=1/36$ F) I dont understand as well, since we found in E that having the same #'s on all three dice is $1/36$ then would not $P(\\text{not the same #'s})= 1-1/36=35/36$? but for some reason it's not the correct answer I would appreciate if you could explain your reasoning, I'm new to probability and pretty much forced to learn it for genetics problems. D) In the order 1,3,5, $Pr = \\dfrac{1}{6}\\cdot\\dfrac{1}{6}\\cdot\\dfrac{1}{6} = \\dfrac{1}{216}$ but there could be $3! = 6$ different orders, so multiply by 6 E) First number can be anything, each of the next two have to be different, hence $\\dfrac{1}{6}\\cdot\\dfrac{1}{6} = \\dfrac{1}{36}$ F) 6 ways to get first number, 5 ways for second one, and 4 ways for third, hence $\\dfrac{6*5*4}{216} = \\dfrac{5}{9}$ Note that the complement of all same is not all different, it is all not same", "she needs a $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is $\\dfrac{1}{6}\\cdot \\dfrac{3}{6}$. First toss is a 3: In this case she needs a $3$, $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is $\\dfrac{1}{6}\\cdot \\dfrac{4}{6}$. Thus the probabilility Alicia wins is $\\dfrac{1}{6}+ \\dfrac{1}{6}\\cdot \\dfrac{2}{6}+ \\dfrac{1}{6}\\cdot \\dfrac{3}{6}+ \\dfrac{1}{6}\\cdot \\dfrac{4}{6}$. Simplify. • yes sir, i also figured it out in same way, but unfortunately not when needed. – Abhishek Pandey Aug 23 '13 at 8:30 You forgot 4, 5, 6 rolls. And, you need to take into account the fact that the odds of a 4, 5, or 6 is not the same as, say, (1, 5). Getting a 4 happens 1/6th of the time. Getting a (1, 5) happens 1/18th of the time. • how you are saying there will be 21 possibilities. i am not getting please elaborate. – Abhishek Pandey Aug 17 '13 at 12:47 • yes got it, i was not considering 4,5,6 and our likely case (6,0),answer should be 10/21.am i right? – Abhishek Pandey Aug 17 '13 at 13:03 • @AbhishekPandey No, the 21 possibilities are not all equally likely. Half the time you roll a 4, 5, or 6 on the first roll. The other half the", "much of a difference... • Maybe I'm being dense, but surely there's more than one way to roll 50 ones and 50 twos. In fact, I would expect there to be 100 choose 50 ways to do that, which is about 10^29... – MathematicalOrchid Jun 18 '13 at 7:33 • No, of course you are right, the density is all mine... thank goodness we can still probably ignore double counting for all practical purposes, even if we are off by $10^{29}$... – Stephan Kolassa Jun 18 '13 at 7:38 Well, first you'd figure out the probability of 50 1's. Then multiply by 6. Then subtract the combinations where there are 50 of each of two numbers. $PR = \\binom{100}{50}.167^{50}.833^{50}$ or, if I pushed the right buttons, $1.49*10^{-14}$ 6 times that is $8.91*10^{-14}$ the probability of 50 1's and then 50 2's is $.167^{100}$ and you have to multiply this by 30 but this is so tiny it can be ignored. (Multiply by 30, not 15, because it could be 11111 ....2222.... or 2222....1111.... etc. Let's start off with a simplified version where we're just interested in the probability of getting exactly 50 1's. In R, you can get this probability using the dbinom function. dbinom(50, size = 100, prob = 1/6) # about 10^-14 Multiplying by 6 gives", "same logic - there are two ways of obtaining the same colour and eight combinations in total (2 $\\times$2 $\\times$2, or 2$^3$). For four discs, the probability is 1/8 because there are two ways of obtaining the same colour and sixteen combinations in total (2 $\\times$2 $\\times$2 $\\times$2, or 2$^4$). The general formula for the probability of flipping n discs and obtaining the same colour is: $2/ 2^n$ or $1/ 2^{(n-1)}$. Emma from St. Paul's Girls' School also found that The probability of winning for n discs = 1/(2 to the power n-1) Adan from Bancroft's uses Pascal's Triangle to help him determine the total number of possible combinations: Using Pascal's Triangle you can work out the chances of them being all the same. Firstly you must match the number of discs with the line of that number but one higher. For example, let's use 3 discs: On the fourth line of the triangle it says \"1 3 3 1\". You must then add them all together, (that makes 8) this is the bottom of your fraction. The left of the line is the number of combinations possible for them all being red. The right is the number of combinations possible for them all being green. The middle ones are the number of combinations possible for there being", "values when rolling a die $7$ times? Use the inclusion/exclusion principle in order to count the number of ways: • Include the number of ways to roll a die $7$ times and see up to $6$ different values: $\\binom66\\cdot6^7$ • Exclude the number of ways to roll a die $7$ times and see up to $5$ different values: $\\binom65\\cdot5^7$ • Include the number of ways to roll a die $7$ times and see up to $4$ different values: $\\binom64\\cdot4^7$ • Exclude the number of ways to roll a die $7$ times and see up to $3$ different values: $\\binom63\\cdot3^7$ • Include the number of ways to roll a die $7$ times and see up to $2$ different values: $\\binom62\\cdot2^7$ • Exclude the number of ways to roll a die $7$ times and see up to $1$ different values: $\\binom61\\cdot1^7$ Divide the result by the total number of ways, which is $6^7$: $$\\frac{\\binom66\\cdot6^7-\\binom65\\cdot5^7+\\binom64\\cdot4^7-\\binom63\\cdot3^7+\\binom62\\cdot2^7-\\binom61\\cdot1^7}{6^7}=\\frac{35}{648}$$ Calculate the probability of the original event by subtracting the result from $1$: $$1-\\frac{35}{648}=\\frac{613}{648}\\approx94.6\\%$$ • Nice use of inclusion/exclusion! Simulated probability of seeing all 6 faces in 7 rolls in my Answer is 0.0544 ± 0.001, compared to exact 35/648 = 0.0540 in this Answer. – BruceET Aug 7 '15 at 8:46 • @BruceTrumbo: Thanks :) – barak manos Aug 7 '15 at 8:50 (Edit: Answer", "other number would have to be duplicated twice and the other four just once. $4+2+1+1+1+1$ In this case there is one number duplicated four times and another duplicated twice. Choose the number to be duplicated four times in six ways. For each way choose the other number to be duplicated twice in five ways. Choose the four dice to be assigned the first number in $10\\choose4$ ways and for each choose the two to be assigned to the second number in $6\\choose2$ ways. The other four numbers are assigned in $4!$ ways. So this accounts for $6\\cdot5\\cdot{10\\choose4}\\cdot{6\\choose2}\\cdot 4!$ ways. I believe there are only three other possibilities: $3+3+1+1+1+1$ Choose the two to be duplicated in $6\\choose2$ ways. For each choose the six dice to be duplicates of one of these, in $10\\choose6$ ways, and for each of those six dice choose which three to duplicate with one number and which three to duplicate with the other, in $6\\choose3$ ways. The other four dice can be placed in $4!$ ways. So this accounts for $4!{6\\choose2}{10\\choose6}{6\\choose3}$ ways. The other two ways are $3+2+2+1+1+1$ $2+2+2+2+1+1$ So for each one you need to count the total number, doing a similar calculation as above. Then total all of these five numbers and divide by $6^{10}$ which is the total number of ways to", "very much, you(plural)have been very helpful. - In (3), you mean $3 \\cdot 3^2=27$. – Chris Eagle Jan 20 '12 at 21:51 There is also $1$ way the number $1$ appears three times face down, and the fact that $27+27+9+1=64$ should be encouraging. – Henry Jan 20 '12 at 21:55 That sounds right. As a check, you should argue that $1$ can appear on the bottom on no rolls, one roll, two rolls, or three rolls (there are no other possibilities), and you have worked out the number of ways that $1$ never appears, appears once, and appears twice for a total of $27 + 27 + 9 = 63$ of the $64$ occurrences. The remaining possibility is that of $1$ on all three rolls, right? Of course, all the numbers adding up doesn't guarantee that they are correct, but if they don't add up, that suggests that the work and the argument needs to be checked. – Dilip Sarwate Jan 20 '12 at 21:57 @Dilip You should make that an answer. – David Mitra Jan 20 '12 at 21:59 The work was nicely done. – André Nicolas Jan 20 '12 at 22:48 An alternate/complimentary way to answer question 4 is: If 2 throws are '1' then we have only one throw which is not '1'. There are", "$2$ and Candy rolls $2$ or more: $\\dfrac{1}{36}\\cdot\\dfrac{5}{6}=\\dfrac{5}{216}$ • The probability that Andy rolls $3$ and Candy rolls $3$ or more: $\\dfrac{2}{36}\\cdot\\dfrac{4}{6}=\\dfrac{8}{216}$ • The probability that Andy rolls $4$ and Candy rolls $4$ or more: $\\dfrac{3}{36}\\cdot\\dfrac{3}{6}=\\dfrac{9}{216}$ • The probability that Andy rolls $5$ and Candy rolls $5$ or more: $\\dfrac{4}{36}\\cdot\\dfrac{2}{6}=\\dfrac{8}{216}$ • The probability that Andy rolls $6$ and Candy rolls $6$: $\\dfrac{5}{36}\\cdot\\dfrac{1}{6}=\\dfrac{5}{216}$ Hence the probability of the complementary event is: $$1-\\dfrac{5+8+9+8+5}{216}=\\dfrac{181}{216}$$ That's not right since Candy might not roll a six. The most straightforward way to answer this is to simply lay out a table of all the possibilities for the two rollers and then add up all the cells where Andy wins. Another way to do it though is rewrite the problem. P(A > C) is the same as P(A - C > 0). Then, you can use the fact that subtracting a die is the same as adding one and then subtracting 7 (try it out). So P(A - C > 0) is the same as P(A + C - 7 > 0) or P(A + C > 7). A + C is just rolling three dice, so the answer to your question is the same as the answer to \"what is the probability of getting more than 7 on three dice\". Think of the three", "multiplying it by the $$x^{20}$$ term from $$(1 − x)^{−4}$$; or by taking the $$x^{12}$$ term that we’ve just worked out, and multiplying it by the $$x^8$$ term from $$(1 − x)^{−4}$$. In the previous example, we worked out that in $$(1 − x)^{−4}$$, the coefficient of $$x^{20}$$ is $$\\left( \\binom{4}{20} \\right)$$, and the coefficient of $$x^8$$ is $$\\left( \\binom{4}{8} \\right)$$. Thus, the number of ways in which Trent can roll a total of $$24$$ on his four dice is the coefficient of $$x^{24}$$ in our generating function, which is $$\\left( \\binom{4}{20} \\right) - 4 \\left( \\binom{4}{8} \\right) = 1771 - 660 = 1111$$ Exercise $$\\PageIndex{1}$$ 1. Prove Proposition 7.3.1 by induction on $$k$$. 2. Find the number of ways Trent can roll a total of $$16$$ on his four dice. 3. If Trent’s four dice are $$10$$-sided dice instead of $$12$$-sided, how many ways can he roll a total of $$24$$? 4. If Trent rolls five regular ($$6$$-sided) dice, how many ways can he roll a total of $$11$$? What is the probability that he will roll a total of $$11$$.", "+ 6/1 = 14.7.$ For more on this, see the ‘coupon collecting problem’. Here’s another way of looking at it, along the lines suggested by barak manos. Note that 7 rolls containing all 6 values must contain a duplicate. There are $\\binom{7}{2}\\cdot 6$ choices for the duplicate entries. Thus, the desired probability is given by $$1-\\frac{\\binom{7}{2}\\cdot 6\\cdot 5!}{6^7} = 1-\\frac{70}{1296}=\\frac{613}{648}.$$", "the cases unless he has a $7$ to begin with. We have $$\\frac{1}{6} > \\underline{\\frac{5,4,3}{36}} > \\frac{2.5}{36} > \\frac{2,1,0}{36}.$$ We count the underlined part's frequency for the numerator without upsetting the probability greater than it. Let $a$ be the roll we keep. We know $a\\leq3$ since $a=4$ would cause Jason to pick up all the dice. When $a=1$, there are $3$ choices for whether it is rolled $1$st, $2$nd, or $3$rd, and in this case the other two rolls have to be at least $6$ (or he would have only picked up $1$). This give $3 \\cdot 1^{2} =3$ ways. Similarly, $a=2$ gives $3 \\cdot 2^{2} =12$ because the $2$ can be rolled in $3$ places and the other two rolls are at least $5$. $a=3$ gives $3 \\cdot 3^{2} =27$. Summing together gives the numerator of $42$. The denominator is $6^3=216$, so we have $\\frac{42}{216}=\\boxed{\\textbf{(A) } \\frac{7}{36}}$. ## Solution 4 Note that Jason will roll $2$ dies if and only if the pairwise sum of the dies is greater than $7$. Suppose that Jason doesn't roll a $1$. This means that roll was $(1, 6, 6)$ which can be arranged in $3$ ways. Suppose that Jason doesn't roll a $2$. This means the roll was either $(2, 5/6, 5/6)$ which can be arranged in $12$ ways. Suppose", "3 3 5 5 226800 2 4 6 6 1 1 3 3 5 5 226800 2 6 6 6 1 1 3 3 5 5 75600 4 4 4 4 1 1 3 3 5 5 18900 4 4 4 6 1 1 3 3 5 5 75600 4 4 6 6 1 1 3 3 5 5 113400 4 6 6 6 1 1 3 3 5 5 75600 6 6 6 6 1 1 3 3 5 5 18900 3*18900+6*75600+3*113400+3*226800 = 1530900 So I get the probability as $$\\frac{1530900}{6^{10} }= \\frac{175}{6912} \\approx 0.025318287037037$$ Note: this has been edited (1st edit) That is if i have made no mistakes. There would be a number of more sensible ways to do this. Jul 14, 2018 edited by Melody Jul 14, 2018 #4 +1 Is the denominator 10! or 6^10? Jul 14, 2018 edited by Guest Jul 14, 2018 #5 +111600 0 You are right, the denominator must be 6^10 Thanks for pointing that out! I have edited my answer accordingly. Melody Jul 14, 2018 #6 +2092 +3 Here’s a “sensible way” that hides the demonstration of the nuanced details. $$\\dfrac {\\binom{10}{2}\\cdot \\binom{8}{2}\\cdot \\binom{6}{2}\\cdot (3^4)}{6^{10}} = \\dfrac{175}{6912} \\approx 2.532\\%$$ GA Jul 14, 2018 #7 +111600 +1 Thanks Ginger. We have 10 different colored dice. How many ways can", "requires at least 6/12 rolls (or 3/6 if allowing 2 dice to be thrown at the same time). I’m hoping for something much better ($$\\leq 4$$ rolls) to be possible as this consumes quite a bit of time. Method: Select each card one at a time. Repeat the following for each card selection. Step 0: If $$N\\leq6$$ skip to Step 2. Pick the smallest number from the below list greater than $$N$$ and let $$c_1, c_2$$ be the listed factors of that number. $$6=2\\cdot3$$ $$9=3\\cdot3$$ $$12=2\\cdot6$$ $$18=3\\cdot6$$ Divide the cards into $$c_1$$ piles as evenly as possible. Step 1: Roll 1st dice (assign $$d_1=$$ result). Let $$n_1=d_1\\pmod{c_1}$$. Step 2: Roll 2nd dice (assign $$d_2=$$ result). Let $$n_2=d_2\\pmod{c_2}$$. Select the $$n_2$$th card in the $$n_1$$th pile. If that card doesn't exist, go back to Step 1. (If piles were of equal size, repeat Step 2 instead). • If $N=14$, step 1 gives you two piles of seven cards, so in step 2 you have $c=7$. In step 2 you will then never pick the seventh card using a 6-sided die. If $N=13$, then step 1 splits it into four piles, which makes the $\\mod p$ part of step 1 problematic (you want $p|6$ else pile $1$ is more likely to be chosen that pile $p$) Both problems can be", "very much, you(plural)have been very helpful. - In (3), you mean $3 \\cdot 3^2=27$. – Chris Eagle Jan 20 '12 at 21:51 There is also $1$ way the number $1$ appears three times face down, and the fact that $27+27+9+1=64$ should be encouraging. – Henry Jan 20 '12 at 21:55 That sounds right. As a check, you should argue that $1$ can appear on the bottom on no rolls, one roll, two rolls, or three rolls (there are no other possibilities), and you have worked out the number of ways that $1$ never appears, appears once, and appears twice for a total of $27 + 27 + 9 = 63$ of the $64$ occurrences. The remaining possibility is that of $1$ on all three rolls, right? Of course, all the numbers adding up doesn't guarantee that they are correct, but if they don't add up, that suggests that the work and the argument needs to be checked. – Dilip Sarwate Jan 20 '12 at 21:57 @Dilip You should make that an answer. – David Mitra Jan 20 '12 at 21:59 The work was nicely done. – André Nicolas Jan 20 '12 at 22:48 Yes it is correct, well done. An alternate/complimentary way to answer question 4 is: If 2 throws are '1' then we have only one throw", "22 '12 at 1:03 ## 1 Answer The probability of not rolling a 1 in $n$ rolls is $(5/6)^n$, similarly for not rolling a $2,\\ldots,6$. Now, $6(5/6)^n$ would be the probability that we are not rolling a $1,\\ldots,6$, but we would be double counting the rolls where we do not roll both a 1 or a 2. The probability of not rolling two specified numbers in $n$ rolls is $(4/6)^n$ and there are $\\binom{6}{2}$ pairs of numbers. But if we subtract these out we undercount the rolls that avoid three numbers. This generalizes to the inclusion-exclusion principle, giving us the probability of missing any number as $$\\binom{6}{1}(5/6)^n-\\binom{6}{2}(4/6)^n+\\binom{6}{3}(3/6)^n-\\binom{6}{4}(2/6)^n+\\binom{6}{5}(1/6)^n$$ as the probability of missing at least one number in $n$ rolls. The probability of rolling all of them is just 1 minus this probability. - Correct me if I am wrong, but \\binom 6 1 = (6! / 1!(6-1)!) – Cameron Aziz Mar 22 '12 at 1:39 With that, I somehow come up with 1.197 for n = 6 [(720/120)(5/6)^n] - [(720/48)(4/6)^n] - [(720/18)(3/6)^n] - [(720/8)(2/6)^n] - [(720/5)(1/6)^n] = 1.197 – Cameron Aziz Mar 22 '12 at 1:42 $\\binom{6}{1}=6$ yes. And for $n=6$ I get 0.9845679. Some of your other binomials are incorrect – deinst Mar 22 '12 at 1:53 got it! (720/120)(5/6)^n] - [(720/48)(4/6)^n] - [(720/36)(3/6)^n] - [(720/48)(2/6)^n] -", "There are $$6^5=7776$$ of these, not $$_6\\text{P}_5=720$$. So here is what you are actually doing: • Choose which number to have a pair of: C(6,1) ways • Choose which two dice the pair will be: C(5,2) ways • Choose distinct numbers for the three other dice, in order: 5×4×3 ways (that is, P(5,3) We don’t need to divide by 3!, because this time order does count. Our result is C(6,1) × C(5,2) × P(5,3) = 3600 Dividing this by 6^5 = 7776, we get 0.4630 (46.3%). Note that where we chose two suits last time, we are choosing two dice to have the chosen numbers: superficially similar calculations, but entirely different in meaning. Alexander had calculated $$\\frac{{6\\choose1}\\cdot{5\\choose2}\\cdot{5\\choose1}\\cdot{4\\choose1}\\cdot{3\\choose1}}{3!}=\\frac{6\\cdot10\\cdot5\\cdot4\\cdot3}{6}=600,$$ which amounts to $${6\\choose1}\\cdot{5\\choose2}\\cdot{5\\choose3},$$ whereas we really have $${{6}\\choose{1}}\\cdot{{5}\\choose{2}}\\cdot _{5}\\!\\text{P}_{3}=6\\cdot10\\cdot60=3600.$$ When I work a problem like this, I like to first ask, What am I counting, and then, How can I count them? In this case, because we have just done a very different problem that was misleadingly similar, the first question was the key. Alexander replied, Thank you! I have searched everywhere to find an explanation like yours, but I have not found one until now. Again, thank you! ## Poker hands with random numbers This week, while I was editing this post, we got a question that deals with", "tried to extend this arguement to find how many rolls you need to roll a different number other than the one already got, but didn't get very far. I suppose this method also neglects the fact that we don't care about the order of the full house. Could I have a hint on this one (not the answer)? - Poker dice has some probabilities – Henry Dec 4 '12 at 21:46 For the dots as multiplication, use \\cdot as in $6\\cdot5\\cdot4$. – FrenzY DT. Dec 5 '12 at 16:43 One pair: There are $6$ ways to choose what we will have two of. For each such way, there are $\\dbinom{5}{2}$ ways to decide where the numbers shall be in the list. For every way of getting this far, there are $\\dbinom{5}{3}$ ways to decide on the collection of singletons, and then $3!$ ways to list them. Multiply. Or else more easily, once we have dealt with the pair, there are $(5)(4)(3)$ ways to fill the gaps. Two pairs: This one is tricky, it is quite common to get an answer twice as big as the truth. The two kinds that we will have pairs of can be chosen in $\\dbinom{6}{2}$ ways. For every such choice, the lonely number can be chosen in $4$ ways. Now the cards of" ]

[ "each, one of each, or none of each. If he rolls two of each, there are $$\\binom{4}{2}=6$$ ways to choose which two dice roll the 1's. If he rolls one of each, there are $$\\binom{4}{1}\\binom{3}{1}=12$$ ways to choose which dice are the 6 and the 1, and for each of those ways there are $$4\\cdot4=16$$ ways to choose the values of the other dice. If Greg rolls no 1's or 6's, there are $$4^4=256$$ possible values for the dice. In total, there are $$6+12\\cdot16+256=454$$ ways Greg can roll the same number of 1's and 6's. There are $$\\dfrac{1}{2} \\left(1-\\dfrac{454}{1296}\\right)=\\boxed{\\dfrac{421}{1296}}$$ total ways the four dice can roll, so the probability that Greg rolls more 1's than 6's is . RektTheNoob Feb 5, 2018 #7 -1 RektTheNoob: If you know the answers to the questions you post on this forum in such detail as the above answer demonstrates, then what is the PURPOSE of posting your questions in the first place?? Are you trying to test the Mods and other volunteers' mathematical knowledge? Or are you trying to show your mathematical superiority??!! Guest Feb 5, 2018 #10 +109721 0 Presumable he did not know the answer when he asked the question but was given or worked out the answer afterwards. I am very glad that he did post this answer", "+0 # If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? +1 1668 11 +473 If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? Feb 3, 2018 #1 +109721 0 If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? Thanks Alan. . Feb 3, 2018 edited by Melody Feb 5, 2018 #2 +473 +2 RektTheNoob Feb 3, 2018 edited by RektTheNoob Feb 3, 2018 #3 +109721 0 ok so what do you think the correct answer is and what makes you think it is correct? AND if you have the working we would like to see that too. I, and others, would like to learn too. Melody Feb 4, 2018 #5 +473 +3 Solution: We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways Greg can roll the same number of 1's and 6's: he can roll two of", "# Count the number of ways of four distinct numbers showing up when six dice are rolled Suppose we roll six fair dice, how many ways can four distinct numbers show up? - Did you at least try some combinatoric argument? I feel the inclusion-exclusion principle might come in handy, but I'm really tired right now, so it's just a wild guess. – Patrick Da Silva Dec 9 '11 at 2:34 The number of ways doesn't depend on the dice being fair. – joriki Dec 9 '11 at 2:59 I'll assume that you mean six-sided dice. There are $\\binom64$ ways of choosing the $4$ distinct numbers. They can either appear $3,1,1,1$ times or $2,2,1,1$ times. In the first case, there are $4$ choices of the number appearing thrice and $6\\cdot5\\cdot4$ choices for the positions. In the second case, there are $6$ choices for the two numbers appearing twice and $6\\cdot5\\cdot\\binom42$ choices for the positions. Thus, the total is $$\\binom64\\left(4\\cdot6\\cdot5\\cdot4+6\\cdot6\\cdot5\\cdot\\binom42\\right)=15\\cdot6\\cdot5\\cdot(16+36)=23400\\;.$$ Thus, the probability of this happening is $23400/6^6=23400/46656\\approx50\\%$. The corresponding probabilities for the other numbers of distinct numbers are: $$\\begin{eqnarray} p(1)&=&\\binom61/6^6\\approx0.01\\%\\;,\\\\ p(2)&=&\\binom62(2^6-2)/6^6\\approx2\\%\\;,\\\\ p(3)&=&\\binom63\\left(3\\cdot6\\cdot5+3!\\cdot6\\cdot\\binom52+\\binom62\\binom42\\right)/6^6\\approx23\\%\\;,\\\\ p(5)&=&\\binom65\\cdot5\\cdot6\\cdot5\\cdot4\\cdot3/6^6\\approx23\\%\\;,\\\\ p(6)&=&\\binom666!/6^6\\approx2\\%\\;. \\end{eqnarray}$$ - Really nice. Thank you. – geraldgreen Dec 9 '11 at 3:15", "be total of $9$ combinations possible and thus $P=(1/6)^9$ but its not the correct answer. On the contrary, the right answer is $P=6(1/216)=1/36$ I DO NOT UNDERSTAND why we multiply by $6$. E) here since they ask for the same number the probability of getting say $(2,2,2)$ combo is $P= 1/6 \\cdot 1/6 \\cdot 1/6 = 1/216$ thus having $6$ #'s on the dice the total probability of having all the same #'s is $P = 1/216+...+1/216=6/216=1/36$ F) I dont understand as well, since we found in E that having the same #'s on all three dice is $1/36$ then would not $P(\\text{not the same #'s})= 1-1/36=35/36$? but for some reason it's not the correct answer I would appreciate if you could explain your reasoning, I'm new to probability and pretty much forced to learn it for genetics problems. D) In the order 1,3,5, $Pr = \\dfrac{1}{6}\\cdot\\dfrac{1}{6}\\cdot\\dfrac{1}{6} = \\dfrac{1}{216}$ but there could be $3! = 6$ different orders, so multiply by 6 E) First number can be anything, each of the next two have to be different, hence $\\dfrac{1}{6}\\cdot\\dfrac{1}{6} = \\dfrac{1}{36}$ F) 6 ways to get first number, 5 ways for second one, and 4 ways for third, hence $\\dfrac{6*5*4}{216} = \\dfrac{5}{9}$ Note that the complement of all same is not all different, it is all not same", "very much, you(plural)have been very helpful. - In (3), you mean $3 \\cdot 3^2=27$. – Chris Eagle Jan 20 '12 at 21:51 There is also $1$ way the number $1$ appears three times face down, and the fact that $27+27+9+1=64$ should be encouraging. – Henry Jan 20 '12 at 21:55 That sounds right. As a check, you should argue that $1$ can appear on the bottom on no rolls, one roll, two rolls, or three rolls (there are no other possibilities), and you have worked out the number of ways that $1$ never appears, appears once, and appears twice for a total of $27 + 27 + 9 = 63$ of the $64$ occurrences. The remaining possibility is that of $1$ on all three rolls, right? Of course, all the numbers adding up doesn't guarantee that they are correct, but if they don't add up, that suggests that the work and the argument needs to be checked. – Dilip Sarwate Jan 20 '12 at 21:57 @Dilip You should make that an answer. – David Mitra Jan 20 '12 at 21:59 The work was nicely done. – André Nicolas Jan 20 '12 at 22:48 An alternate/complimentary way to answer question 4 is: If 2 throws are '1' then we have only one throw which is not '1'. There are", "multiplying it by the $$x^{20}$$ term from $$(1 − x)^{−4}$$; or by taking the $$x^{12}$$ term that we’ve just worked out, and multiplying it by the $$x^8$$ term from $$(1 − x)^{−4}$$. In the previous example, we worked out that in $$(1 − x)^{−4}$$, the coefficient of $$x^{20}$$ is $$\\left( \\binom{4}{20} \\right)$$, and the coefficient of $$x^8$$ is $$\\left( \\binom{4}{8} \\right)$$. Thus, the number of ways in which Trent can roll a total of $$24$$ on his four dice is the coefficient of $$x^{24}$$ in our generating function, which is $$\\left( \\binom{4}{20} \\right) - 4 \\left( \\binom{4}{8} \\right) = 1771 - 660 = 1111$$ Exercise $$\\PageIndex{1}$$ 1. Prove Proposition 7.3.1 by induction on $$k$$. 2. Find the number of ways Trent can roll a total of $$16$$ on his four dice. 3. If Trent’s four dice are $$10$$-sided dice instead of $$12$$-sided, how many ways can he roll a total of $$24$$? 4. If Trent rolls five regular ($$6$$-sided) dice, how many ways can he roll a total of $$11$$? What is the probability that he will roll a total of $$11$$.", "she needs a $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is $\\dfrac{1}{6}\\cdot \\dfrac{3}{6}$. First toss is a 3: In this case she needs a $3$, $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is $\\dfrac{1}{6}\\cdot \\dfrac{4}{6}$. Thus the probabilility Alicia wins is $\\dfrac{1}{6}+ \\dfrac{1}{6}\\cdot \\dfrac{2}{6}+ \\dfrac{1}{6}\\cdot \\dfrac{3}{6}+ \\dfrac{1}{6}\\cdot \\dfrac{4}{6}$. Simplify. • yes sir, i also figured it out in same way, but unfortunately not when needed. – Abhishek Pandey Aug 23 '13 at 8:30 You forgot 4, 5, 6 rolls. And, you need to take into account the fact that the odds of a 4, 5, or 6 is not the same as, say, (1, 5). Getting a 4 happens 1/6th of the time. Getting a (1, 5) happens 1/18th of the time. • how you are saying there will be 21 possibilities. i am not getting please elaborate. – Abhishek Pandey Aug 17 '13 at 12:47 • yes got it, i was not considering 4,5,6 and our likely case (6,0),answer should be 10/21.am i right? – Abhishek Pandey Aug 17 '13 at 13:03 • @AbhishekPandey No, the 21 possibilities are not all equally likely. Half the time you roll a 4, 5, or 6 on the first roll. The other half the", "281 e) None of these Question 11: Ravi rolled a dice twice. What is the probability that sum of both the outcome is less than 5? a) 1/6 b) 1/4 c) 1/3 d) 1/2 e) None of the above. Question 12: Raghu tosses a coin 6 times. The probability that the number of times he gets heads will not be greater than the number of times he gets tails is a) 21/64 b) 3/32 c) 41/64 d) 21/32 e) 1/2 Question 13: A fair dice is rolled twice. Find the probability that he gets at least one composite number. a) $\\frac{5}{36}$ b) $\\frac{3}{4}$ c) $\\frac{5}{9}$ d) $\\frac{4}{9}$ e) $\\frac{5}{12}$ Question 14: A box contains 5 blue pens, 6 green pens and 10 black pens. The number of ways in which 3 pens can be selected such that at least 2 pens are of the same colour is a) 1000 b) 1300 c) 1330 d) 1030 e) 1310 Question 15: Raghu has gotten calls for interviews from top 4 management institutes of a country. The probability that he converts an interview of an institute is 50%. What is the probability of Raghu converting the interview of at least 1 management institute? a) 93.75% b) 92.25% c) 95.75% d) 96.25% e) 94.75% Instructions Calculate the quantity I and the quantity", "flips per turn Intuitively we could say that rolling the dice seven times would be very interesting. Since we only need to throw out 2 out of the $2^7$ possibilities. Namely, the 7 times heads and 0 times heads. For all other $2^7-2$ possibilities there is always a multiple of 7 cases with the same number of heads. Namely 7 cases with 1 heads, 21 cases with 2 heads, 35 cases with 3 heads, 35 cases with 4 heads, 21 cases with 5 heads, and 7 cases with 6 heads. So if you compute the number (discarding 0 heads and 7 heads) $$X = \\sum_{k=1}^{7} (k-1) \\cdot C_k$$ with $C_k$ Bernoulli distributed variables (value 0 or 1), then X modulo 7 is a uniform variable with seven possible results. ### Comparing different number of coin flips per turn The question remains what the optimal number of rolls per turn would be. Rolling more dices per turn cost you more, but you reduce the probability to have to roll again. The image below shows a manual computations for the first few numbers of coin flips per turn. (possibly there might be an analytical solution, but I believe it is safe to say that a system with 7 coin flips provides the best method regarding the expectation value for the", "same logic - there are two ways of obtaining the same colour and eight combinations in total (2 $\\times$2 $\\times$2, or 2$^3$). For four discs, the probability is 1/8 because there are two ways of obtaining the same colour and sixteen combinations in total (2 $\\times$2 $\\times$2 $\\times$2, or 2$^4$). The general formula for the probability of flipping n discs and obtaining the same colour is: $2/ 2^n$ or $1/ 2^{(n-1)}$. Emma from St. Paul's Girls' School also found that The probability of winning for n discs = 1/(2 to the power n-1) Adan from Bancroft's uses Pascal's Triangle to help him determine the total number of possible combinations: Using Pascal's Triangle you can work out the chances of them being all the same. Firstly you must match the number of discs with the line of that number but one higher. For example, let's use 3 discs: On the fourth line of the triangle it says \"1 3 3 1\". You must then add them all together, (that makes 8) this is the bottom of your fraction. The left of the line is the number of combinations possible for them all being red. The right is the number of combinations possible for them all being green. The middle ones are the number of combinations possible for there being", "tried to extend this arguement to find how many rolls you need to roll a different number other than the one already got, but didn't get very far. I suppose this method also neglects the fact that we don't care about the order of the full house. Could I have a hint on this one (not the answer)? - Poker dice has some probabilities – Henry Dec 4 '12 at 21:46 For the dots as multiplication, use \\cdot as in $6\\cdot5\\cdot4$. – FrenzY DT. Dec 5 '12 at 16:43 One pair: There are $6$ ways to choose what we will have two of. For each such way, there are $\\dbinom{5}{2}$ ways to decide where the numbers shall be in the list. For every way of getting this far, there are $\\dbinom{5}{3}$ ways to decide on the collection of singletons, and then $3!$ ways to list them. Multiply. Or else more easily, once we have dealt with the pair, there are $(5)(4)(3)$ ways to fill the gaps. Two pairs: This one is tricky, it is quite common to get an answer twice as big as the truth. The two kinds that we will have pairs of can be chosen in $\\dbinom{6}{2}$ ways. For every such choice, the lonely number can be chosen in $4$ ways. Now the cards of", "very much, you(plural)have been very helpful. - In (3), you mean $3 \\cdot 3^2=27$. – Chris Eagle Jan 20 '12 at 21:51 There is also $1$ way the number $1$ appears three times face down, and the fact that $27+27+9+1=64$ should be encouraging. – Henry Jan 20 '12 at 21:55 That sounds right. As a check, you should argue that $1$ can appear on the bottom on no rolls, one roll, two rolls, or three rolls (there are no other possibilities), and you have worked out the number of ways that $1$ never appears, appears once, and appears twice for a total of $27 + 27 + 9 = 63$ of the $64$ occurrences. The remaining possibility is that of $1$ on all three rolls, right? Of course, all the numbers adding up doesn't guarantee that they are correct, but if they don't add up, that suggests that the work and the argument needs to be checked. – Dilip Sarwate Jan 20 '12 at 21:57 @Dilip You should make that an answer. – David Mitra Jan 20 '12 at 21:59 The work was nicely done. – André Nicolas Jan 20 '12 at 22:48 Yes it is correct, well done. An alternate/complimentary way to answer question 4 is: If 2 throws are '1' then we have only one throw", "# Probability ### Squares on a chessboard A5, 2019 You are given an $$8 \\times 8$$ chessboard. If two distinct squares are chosen uniformly at random find the probability that two rooks placed on these squares attack each other. Recall that a rook can move either horizontally or vertically, in a straight line. Solution $$2/9$$ ### Yet another dice roll A4, 2020 (i) What is the probability that among $$100$$ rolls of a fair die the first $$3$$ rolls yield at least one $$4$$? (ii) Calculate the probability that out of the last $$4$$ rolls, exactly two are multiples of $$3$$. Solution (i) The probability that none of the first three rolls have a 4 is $$(5/6)^3$$. So the required probability is $$1 - (5/6)^3$$. (ii) Let $$S$$ be the number of ways in which four rolls can have exactly two multiples of 3. The required probability $$P$$ is then $$S/6^4$$. Let us calculate $$S$$. Two positions can be picked in $$\\binom{4}{2}$$ ways. These two positions can have either a 3 or a 6. So the favorable positions can be filled in $$\\binom{4}{2}\\times 2^2$$ ways. The other two positions can have either 1, 2, 4 or 6. So they can be filled in $$4^2$$ ways. Hence we have: $P = \\frac{ \\binom{4}{2}\\times 4 \\times 4 \\times 4 }{6^4}", "Find the expected number of pairs of players that win the same number of games. Real numbers x, y, and z are chosen from the interval $[−1, 1]$ independently and uniformly at random. What is the probability that $$|x| + |y| + |z| + |x + y + z| = |x + y| + |y + z| + |z + x|$$ How many distinct permutations of the letters of the word REDDER are there that do not contain a palindromic substring of length at least two? (A substring is a contiguous block of letters that is part of the string. A string is palindromic if it is the same when read backwards.) Your math friend Steven rolls five fair icosahedral dice (each of which is labelled $1$, $2$, $\\cdots$, $20$ on its sides). He conceals the results but tells you that at least half of the rolls are $20$. Suspicious, you examine the first two dice and find that they show $20$ and $19$ in that order. Assuming that Steven is truthful, what is the probability that all three remaining concealed dice show $20$? Reimu and Sanae play a game using $4$ fair coins. Initially both sides of each coin are white. Starting with Reimu, they take turns to color one of the white sides either red or", "only on the last roll, then ones would indeed be more common than sixes. But not all sequences are equally likely. The sequence consisting of $17$ sixes is $6^{83}$ times more likely than the sequence consisting of $100$ ones. In general, the preponderance of ones in the longer sequences is exactly offset by the fact that those longer sequences occur less frequently. – Brian Tung Jul 10 '20 at 23:46", "+ 6/1 = 14.7.$ For more on this, see the ‘coupon collecting problem’. Here’s another way of looking at it, along the lines suggested by barak manos. Note that 7 rolls containing all 6 values must contain a duplicate. There are $\\binom{7}{2}\\cdot 6$ choices for the duplicate entries. Thus, the desired probability is given by $$1-\\frac{\\binom{7}{2}\\cdot 6\\cdot 5!}{6^7} = 1-\\frac{70}{1296}=\\frac{613}{648}.$$", "# Poker Dice (Combinational Probability) I've seen similar questions to this asked, and have searched the internet. I haven't found the exact questions here and the internet doesn't have the correct answers. Play poker using 5 fair dice rather than a deco of cards. Roll the five dice onto the table. The possible has are: five-of-a-kind, four-of-a-kind, a full house (3 dice with a number and common, and the other two dice with another number in common, i.e. 3,3,3,2,2), three-of-a-kind, two pair, one pair, a straight (5 dice in a sequence, such as 1,2,3,4,5 or 2,3,4,5,6), and nothing. I know that five-of-a-kind is $\\binom{6}{1}$*$({1\\over 6})^5$ = .00077 four-of-a-kind I'd think i'd be $\\binom{6}{1}$*$({1\\over 6})^4$ * $\\binom{5}{1}$*${1\\over 5}$ yet this gives me the wrong answer, my logic is the same on the full house and three-of-a-kind, and I don't know what I'm doing wrong. for two pair it'd think it'd be $\\binom{6}{1}$*$({1\\over 6})^2$*$\\binom{5}{1}$*$({1\\over 5})^2$*$\\binom{4}{1}$*${1\\over 4}$but this gives .0333, and the correct answer is .06173. my reasoning for a single pair is also the same. Obviously nothing would be 1-the summation of all other probabilities. btw, here are the answers given in the back of the book. five-of-a-kind = .00077 four-of-a-kind = .01929 full house = .03086 three-of-a-kind = .03858 two pair = .06173 one pair = .15432 a straight", "fives or fours). So this roll thus far has not been accounted for; we have to subtract it. Now there are $\\frac{6\\cdot 5 \\cdot 4}{6}=20$ ways to choose $3$ numbers out of $6$. There are $3^8$ rolls for each combination of $3$ missing numbers, o our answer thus far is $$6^8 - 6\\cdot 5^8 + 15 \\cdot 4^8 - 20 \\cdot 3^8$$ And now we come to rolls with no sixes, fives, fours, or threes. The same sort of reasoning shows that each of these was subtracted twice in total, so we need to add these back. And finally, consider wolls that are just six of the same number. We will have subtracted and added these the same number of times, so we will need to subtract these once more. The final number of outcomes is $$6^8 - 6\\cdot 5^8 + 15 \\cdot 4^8 - 20 \\cdot 3^8 + 15 \\cdot 2^8 -6\\cdot 1^8 = 191520$$ And the probability of rolling 8 dice and getting at least one of each number is $$\\frac{191520}{6^8} = \\frac{665}{5832} \\approx 0.114$$ Thus the counterintuitive result is that you are likely to be missing at least one number when you roll 8 dice. In fact, in order to have better than a 50% chance of getting one of each number, you would have", "Oct 28 '14 at 21:20 Here is an alternative solution to the (very good) answer already given here: The number of ways that you can choose $2$ out of $6$ values is $\\dbinom{6}{2}=15$. Given $2$ values A and B, the number of combinations that you can get in $6$ rolls is: • Value A appearing $1$ time and value B appearing $5$ times: $\\dbinom{6}{1}=6$ • Value A appearing $2$ times and value B appearing $4$ times: $\\dbinom{6}{2}=15$ • Value A appearing $3$ times and value B appearing $3$ times: $\\dbinom{6}{3}=20$ • Value A appearing $4$ times and value B appearing $2$ times: $\\dbinom{6}{4}=15$ • Value A appearing $5$ times and value B appearing $1$ time : $\\dbinom{6}{5}=6$ So you can get $6+15+20+15+6=62$ combinations containing A and B. And you can get $15\\cdot62=930$ combinations containing any $2$ out of $6$ values. The total number of combinations that you can get in $6$ rolls is simply $6^6=46656$. Therefore, the probability of having exactly $2$ values in $6$ rolls is $\\dfrac{930}{46656}\\approx0.0199$. • Just a note: the $62$ calculated here through sumation is precisely the $2^6 - 2$ calculated in my answer. This is due to the fact that $\\sum_{i=1}^6{6\\choose i} = 2^6$. – 5xum Oct 28 '14 at 12:12 • @5xum: I agree (except for the fact that you need to", "values when rolling a die $7$ times? Use the inclusion/exclusion principle in order to count the number of ways: • Include the number of ways to roll a die $7$ times and see up to $6$ different values: $\\binom66\\cdot6^7$ • Exclude the number of ways to roll a die $7$ times and see up to $5$ different values: $\\binom65\\cdot5^7$ • Include the number of ways to roll a die $7$ times and see up to $4$ different values: $\\binom64\\cdot4^7$ • Exclude the number of ways to roll a die $7$ times and see up to $3$ different values: $\\binom63\\cdot3^7$ • Include the number of ways to roll a die $7$ times and see up to $2$ different values: $\\binom62\\cdot2^7$ • Exclude the number of ways to roll a die $7$ times and see up to $1$ different values: $\\binom61\\cdot1^7$ Divide the result by the total number of ways, which is $6^7$: $$\\frac{\\binom66\\cdot6^7-\\binom65\\cdot5^7+\\binom64\\cdot4^7-\\binom63\\cdot3^7+\\binom62\\cdot2^7-\\binom61\\cdot1^7}{6^7}=\\frac{35}{648}$$ Calculate the probability of the original event by subtracting the result from $1$: $$1-\\frac{35}{648}=\\frac{613}{648}\\approx94.6\\%$$ • Nice use of inclusion/exclusion! Simulated probability of seeing all 6 faces in 7 rolls in my Answer is 0.0544 ± 0.001, compared to exact 35/648 = 0.0540 in this Answer. – BruceET Aug 7 '15 at 8:46 • @BruceTrumbo: Thanks :) – barak manos Aug 7 '15 at 8:50 (Edit: Answer" ]

[ "Free Version Moderate # Coefficient in $(2a+b-3c)^5$ COMBIN-1VSGQG What is the coefficient of $a^2bc^2$ in the expansion of: $$(2a+b-3c)^5?$$ A $0$ B $243$ C $1024$ D $1080$ E $7776$ F none of the above", "coefficient of $r^n$ in $P_j(r)$ is $0$ for such $j$. The coefficient of $m^1$ in $Q_n(m)$ appears to be $$[m^1] Q_n(m) = (-1)^n n \\left(-1 + \\sum_{i=2}^n \\dfrac{1}{i^2}\\right)$$ • Thanks for noticing. Fixed it. – Robert Israel Jun 2 '16 at 1:09", "Answer Comment Share Q) # Find the coefficient of $a^5b^7$ in $(a-2b)^{12}$ $\\begin{array}{1 1}(A)\\;101376\\\\(B)\\;-101376\\\\(C)\\;102386\\\\(D)\\;104386\\end{array}$", "in $$(a + b)^5 (c + d)^6$$. 3) the coefficient of $$a^2 b^6 c^3$$ in $$(a + b)^5 (b + c)^6$$. 4) the coefficient of $$a^3 b^2$$ in $$(a + b)^5 + (a + b^2)$$.", "# IIT JEE 1983 Maths - MCQ Question 15 What is the coefficient of $x^4$ in the expansion of $\\left(\\dfrac{x}{2}-\\dfrac{3}{x^2}\\right)^{10}$? ×", "The coefficient of Question: The coefficient of $x^{7}$ in the expansion of $\\left(1-x-x^{2}+x^{3}\\right)^{6}$ is :- 1. – 144 2. 132 3. 144 4. – 132 Correct Option: 1 Solution:", "\\\\ = \\dfrac{6.5}{2} \\\\ = 15 {/eq} So, the coefficient of {eq}x^2 y^4 {/eq} term is 15.", "{/eq} a) The coefficient of {eq}{x^{10}}\\,\\,in\\,\\,the\\,\\,\\exp ansion\\,\\,of\\,\\,{\\left( {2x - \\dfrac{1}{{{x^2}}}} \\right)^{16}} {/eq} is 1966080. This is obtained from (1). b) The coefficient of {eq}{x^{11}}\\,\\,in\\,\\,the\\,\\,\\exp ansion\\,\\,of\\,\\,{\\left( {2x - \\dfrac{1}{{{x^2}}}} \\right)^{16}} {/eq} is 0. This is obtained from (1).", "# What is the coefficient of 2.4n+9.6? The coefficients are $2.4$ and $9.6$. There are technically two coefficients in this expression. The first term, $2.4 n$, the coefficient is $2.4$. In the second term, $9.6$, the coefficient is $9.6$, because technically, the term is $9.6 {n}^{0}$.", "+ 3{x^2} + 4{x^3} + ...........\\infty \\\\$ Here by observation, we have seen that ${x^0}$ has coefficient 1, coefficient of ${x^1}$ is 2, coefficient of ${x^2}$ is 3. It means the coefficient of ${x^n}$ will be $(n + 1).$ Hence, the coefficient of ${x^n}$ in the expansion of ${(1 - x)^{ - 2}}$ is $(n + 1)$ and the correct answer is option “B”.", "# The coefficient of Question: The coefficient of $x^{4}$ is the expansion of $\\left(1+x+x^{2}\\right)^{10}$ is___________. Solution:", "# Working of coefficients Find the coefficient of $$x^3$$ in the following expansion: $\\left(1+\\frac{1}{2}x-3x^2\\right)^{20}.$ ×", "Free Version Easy # Coefficient in $(5x-2y)^7$ COMBIN-XIIW4G What is the coefficient of $x^3y^4$ in the expansion of $(5x-2y)^7$? A $725$ B $32$ C $126$ D $252$ E $1024$", "coefficients. $$\\frac{x^{4}+x^{2}+1}{x^{2}\\left(x^{2}+4\\right)^{2}}$$...", "# The coefficient of Question: The coefficient of $x^{7}$ in the expression $(1+x)^{10}+x(1+x)^{9}+x^{2}(1+x)^{8}+\\ldots+x^{10}$ is : 1. 120 2. 330 3. 210 4. 420 Correct Option: , 2 Solution:", "# IIT JEE 1983 Maths - MCQ Question 15 Probability Level 3 What is the coefficient of $x^4$ in the expansion of $\\left(\\dfrac{x}{2}-\\dfrac{3}{x^2}\\right)^{10}$? ×", "+0 # What is the coefficient of ​ in the expansion of ? Express your answer as a common fraction. 0 255 1 +43 What is the coefficient of $$x^2y^2$$ in the expansion of$$\\left(\\frac{3}{5}x-\\frac{y}{2}\\right)^8$$ ? Express your answer as a common fraction. Jz1234 Aug 27, 2017 #1 +87293 +2 There is no x^2y^2 term in the expansion of this.....the sum of the exponents on both variables must = 8 CPhill Aug 27, 2017", "i.e. the coefficient of $\\dfrac{d^2y}{du^2}$ is in cubic function and the coefficient of $\\dfrac{dy}{du}$ is in quadratic function. -", "== CoefficientList[-2 a b x + b^2 x^2, x], {a, b}] {{a -> -56, b -> -85689}, {a -> 56, b -> 85689}}", "# Coefficient Algebra Level 4 Find the absolute value of the coefficient of $$\\dfrac{1}{x}$$ in th expression of: $\\large\\left(2x^2-\\dfrac{1}{x}\\right)^{10}$ ×" ]

[ "SJ. Jan 5 '18 at 3:29 • Okay that is $$\\sum_{p+q+r+s=n} \\frac{n!}{p!q!r!s!} x^py^qz^ru^s$$ but where is $$\\sum_{p+q+r+s=n}$$ in your answer? – Ravi Prakash Jan 5 '18 at 3:32 • – SJ. Jan 5 '18 at 3:32 Hint: We can use Binomial Theorem recursively The coefficient of $a^8$ in $$(a(bc+bd+cd)+bcd))^{10}$$ will be $$\\binom{10}2(bc+bd+cd)^8(bcd)^2$$ The coefficient of $b^4$ in $$(bc+bd+cd)^8(bcd)^2$$ = the coefficient of $b^2$ in $$(b(c+d)+cd)^8(cd)^2$$ $$=(cd)^2\\binom82(cd)^{8-2}(c+d)^2=c^8d^8\\binom82(c^2+2cd+d^2)$$ Clearly, the coefficient of $c^9$ in $$c^8d^8\\binom82(c^2+2cd+d^2)$$ = the coefficient of $c$ in $$d^8\\binom82(c^2+2cd+d^2)$$ Let us set $abc=D, abd=C, acd=B, bcd=A$. Then $a^8 b^4 c^9 d^9 =A^2 B^6 C D$ and the problem boils down to finding the coefficient of $A^2 B^6 C D$ in $(A+B+C+D)^{10}$, i.e. to counting the anagrams of the word $AABBBBBBCD$. They are $$\\frac{10!}{2!6!}=\\color{red}{2520}.$$ Then, $$\\left\\{\\begin{array}{rcrcrcrcrr} \\ds{i} & \\ds{+} & \\ds{j} & \\ds{+} & \\ds{k} & \\ds{+} & \\ds{\\ell} & \\ds{=} & \\ds{10} & \\ds{\\qquad\\pars{\\texttt{a}}} \\\\[1mm] \\ds{i} & \\ds{+} & \\ds{j} & \\ds{+} & \\ds{k} &&& \\ds{=} & \\ds{8} & \\ds{\\qquad\\pars{\\texttt{b}}} \\\\[1mm] \\ds{i} & \\ds{+} & \\ds{j} &&& \\ds{+} & \\ds{\\ell} & \\ds{=} & \\ds{4} & \\ds{\\qquad\\pars{\\texttt{c}}} \\\\[1mm] \\ds{i} &&& \\ds{+} & \\ds{k} & \\ds{+} & \\ds{\\ell} & \\ds{=} & \\ds{9} & \\ds{\\qquad\\pars{\\texttt{d}}} \\\\[1mm] && \\ds{j} & \\ds{+} & \\ds{k} & \\ds{+} & \\ds{\\ell} & \\ds{=} & \\ds{9} & \\ds{\\qquad\\pars{\\texttt{e}}} \\end{array}\\right.$$ Note that $\\ds{\"\\pars{\\texttt{a}} - \\pars{\\texttt{b}}\" \\implies \\bbx{\\ell", "# How do you find the coefficient of a^(n-1)*b in the expansion of (a+b)^n? Dec 3, 2016 The coefficient is $n$ #### Explanation: The binomial theorem tells us that: ${\\left(a + b\\right)}^{n} = {\\sum}_{k = 0}^{n} \\left(\\begin{matrix}n \\\\ k\\end{matrix}\\right) {a}^{n - k} {b}^{k}$ where ((n),(k)) = (n!)/((n-k)! k!) The term in ${a}^{n - 1} . b$ is the one for $k = 1$, with coefficient: ((n),(1)) = (n!)/((n-1)! 1!) = n Alternatively, consider the product: ${\\left(a + b\\right)}^{n} = {\\overbrace{\\left(a + b\\right) \\left(a + b\\right) \\ldots \\left(a + b\\right)}}^{\\text{n times}}$ If we multiplied out the right hand side, then the only way we can get terms which contribute to the coefficient of ${a}^{n - 1} b$ is by picking the left hand $a$ term of $n - 1$ of the binomials and one right hand $b$ term. We can do that in precisely $n$ different ways, each contributing $1$ to the coefficient.", "of the corresponding three terms of $$(1-x)^{-6}$$ viz., $$x^{20}, x^{11}, x^2$$ respectively. Finally the coefficient is given by $$\\binom{25} {5} - 6*\\binom{16} {5} + 15*\\binom{7} {5} = 27237$$ • Thanks PTDS for helping! – Vasu090 Jul 27 at 5:50", "appears to a power of at least 20. You can also ignore the case $k \\leq 2$ because $x$ will appear to a power at most $7*k <18$. Thus, the coefficient of $x^{18}$ in $(1+x^5+x^7)^{20}$ is exactly the coefficient of $x^{18}$ in $\\binom{20}{3} x^{15}(1+x^2)^3$.", "# Solution to #1-10 Paper 2 Difficulty: Medium Consider the expansion of $\\left ( \\frac{1}{3}x – 2 \\right ) ^{7}$. Find the coefficient of $x^{3}$. [Maximum mark: 3] $a = \\frac{1}{3}x, b=-2$ and n=7. Therefore $\\binom{7}{r} \\left ( \\frac{1}{3}x \\right ) ^{7-r} \\left ( -2 \\right ) ^{r}$. To find the coefficient of $x^{3}$, we require that r = 4. So $\\binom{7}{4} \\left ( \\frac{1}{3}x \\right ) ^{7-4} \\left ( -2 \\right ) ^{4} = 20.74x^{3}$ The coefficient of $x^{3}$ is approximately 20.74.", "with the example given, $6-4+1=3$, so consider $$\\left(\\frac{1+x^3}{1-x^2}\\right)^3=1+3x^2+3x^3+6x^4+9x^5+\\dots$$ Note that the coefficient of $x^4$ is $6$. Also, since $K$ is even, $$\\binom{6-4+1}{0}\\binom{6-2}{2}=6$$", "• The coefficient of $$x^n$$ in $$\\dfrac{1}{1 − 2x} = (1 − 2x)^{−1}$$ is $$\\binom{-1}{n} (-2)^n = (-1)^n \\binom{1 + n -1}{n} (-2)^n = (-1)^{2n} \\binom{n}{n} 2^n = 2^n$$ Therefore $$c_n$$, the coefficient of $$x^n$$ in $$C(x)$$, is $$\\dfrac{4}{3} (−1)^n + \\dfrac{2}{3} · 2^n$$. 3) Let $$E(x) = \\sum_{n=0}^{\\infty} e_nx^n$$ be the generating function of $$\\{e_n\\}$$. Then $$$$\\begin{split} E(x) &= e_0 &+ e_1x &+ e_2x^2 &+ e_3x^3 &+ e_4x^4 &+ ... \\\\ xE(x) &= &+ e_0x &+ e_1x^2 &+ e_2x^3 &+ e_3x^4 &+ ... \\\\ \\dfrac{1}{1-x} &= 1&+ x&+ x^2 &+ x^3 &+ x^4 &+ ... \\\\ \\end{split}$$$$ so $$$$\\begin{split} (1 − 3x)E(x) + \\dfrac{2}{1-x} &= E(x) − 3xE(x) + 2 · \\dfrac{1}{1-x}\\\\ &= (e_0 + 2) + \\sum_{n=2}^{\\infty} (e_n − 3e_{n−1} + 2)x^n\\\\ &= (e_0 + 2), \\end{split}$$$$ since (by the recurrence relation) we have $$e_n − 3e_{n−1} + 2 = 0$$ for $$n ≥ 1$$. Therefore $$E(x) = \\dfrac{(e_0 + 2) − \\dfrac{2}{1-x}}{1 − 3x} = \\dfrac{(e_0 + 2)(1 − x) − 2}{(1 − 3x)(1 − x)}$$ Since $$e_0 = 2$$, this means $$E(x) = \\dfrac{(2 + 2)(1 − x) − 2}{(1 − 3x)(1 − x)} = \\dfrac{2-4x}{(1 − 3x)(1 − x)}$$. We now use partial fractions. Write $$\\dfrac{2-4x}{(1 - 3x)(1 − x)} = \\dfrac{A}{1-3x} + \\dfrac{B}{1 − x} = \\dfrac{A(1 − x) + B(1 −", "# Q-Binomial coefficient The $q$-Binomial coefficient ${n \\brack k}_q$ is $${n \\brack k}_q = \\dfrac{(q;q)_n}{(q;q)_k(q;q)_{n-k}},$$ where $(q;q)_{\\xi}$ denotes the q-Pochhammer symbol.", "# How do you find the coefficient of x^3 in (2x+3)^5? Jul 3, 2018 The answer is $= 720$ #### Explanation: ${\\left(a + b\\right)}^{n} = {\\sum}_{k = 0}^{n} \\left(\\begin{matrix}n \\\\ k\\end{matrix}\\right) {a}^{n - k} {b}^{k}$ Where, ((n),(k))=(n!)/((n-k)!(k!)) Here, we have ${\\left(2 x + 3\\right)}^{5}$ And we need the coefficient of ${x}^{3}$ $n = 5$ $a = 2 x$ $b = 3$ $k = 2$ The coefficient is $= \\left(\\begin{matrix}5 \\\\ 2\\end{matrix}\\right) \\cdot {\\left(2\\right)}^{3} \\cdot {\\left(3\\right)}^{2}$ =(5!)/((3!)(2!))* 8* 9 $= 10 \\cdot 72$ $= 720$", "coefficient $$C$$ of $$x^{10}$$ in $$\\left(\\ 1+x+x^2+x^3+x^4+\\dots\\ \\right)^5=\\frac 1{(1-x)^5}\\ .$$ We use here formal power series or polynomials modulo some suitable $$O$$ of some power of $$x$$. This can be computed by starting with $$(1-x)^{-1}=1+x+x^2+x^3+\\dots$$, and we apply the formal derivative four further times. The coefficient in degree $$10$$ comes from the one in degree $$14$$, so it is $$C=(14\\cdot 13\\cdot 11\\cdot 10)/(-1)(-2)(-3)(-4) =\\binom{14}4$$. As in the OP. So far nothing new. To get the number $$C'$$ of partitions with at most $$6$$ candles consider instead: $$\\left(\\ x+x^2+x^3+x^4+x^5+x^6\\ \\right)^5\\ .$$ and isolate the coefficient of $$x^{15}$$ in it. Equivalently, find the coefficient of $$x^{10}$$ in $$\\left(\\ 1+x+x^2+x^3+x^4+x^5\\ \\right)^5=\\frac {(1-x^6)^5}{(1-x)^5}\\ ,$$ which is $$\\Big[\\ 1-5x^6+\\dots\\ \\Big] \\left[\\ \\binom 44 + \\binom 54 x + \\binom 64 x^2 + \\binom 74 x^3 + \\binom 84 x^4 + \\dots + \\binom {14}4 x^{10} +\\dots\\ \\right]$$ The coefficient we need is $$C'=\\binom {14}4-5\\binom 84=651$$. Again, same as in the OP, after the following small correction. Step 1 is choosing the one child that becomes $$6+$$ at least one further candy. So we have to distribute \"in the same conditions\" the $$15-6=9$$ further candies. This leads to the binomial coefficient $$\\binom{9-1}{5-1}$$, in the same manner as for the total we have $$\\binom{15-1}{5-1}=1001$$ possibilities. Check with sage: sage: R.<x> = QQ[] sage: P =", "4 6 4 1 25f5(n) 1 5 10 10 5 1 This is Pascal’s Triangle$—$every entry is the sum of the two diagonally above. These numbers are in fact the coefficients that appear in the binomial expansion of ${\\left(a+b\\right)}^{N}.$ For example, the row for ${2}^{5}{f}_{5}\\left(n\\right)$ mirrors the binomial coefficients: ${\\left(a+b\\right)}^{5}={a}^{5}+5{a}^{4}b+10{a}^{3}{b}^{2}+10{a}^{2}{b}^{3}+5a{b}^{4}+{b}^{5}.$ To see how these binomial coefficients relate to our random walk, we write: ${\\left(a+b\\right)}^{5}=\\left(a+b\\right)×\\left(a+b\\right)×\\left(a+b\\right)×\\left(a+b\\right)×\\left(a+b\\right)$ and think of it as the sum of all products that can be written by choosing one term from each bracket. There are 25 = 32 such terms (choosing one of two from each of the five brackets), so the coefficient of a3b2 must be the number of these 32 terms which have just 3 a’s and 2 b’s. But that is the same as the number of different sequences that can be written by rearranging HHHTT, so it is clear that the random walk probabilities and the binomial coefficients are the same sets of numbers (except that the probabilities must of course be divided by 32 so that they add up to one). ### Finding the Probabilities Using the Factorial Function The efficient way to calculate these coefficients is to use the factorial function. Suppose we have five distinct objects, A, B, C, D, E. How many different sequences can we find: ABCDE,", "### Binomial Theorem goals Find the coefficient of $x^n$ in the expansion of $(1+x) (1- x)^n$. Find the coefficient of $x^{2}$ in the equation of $(1+2x)^{6}(1-x)^{-1}$ Coefficient of term independent of $x$ in ${\\left( {1 + x + {x^{ - 2}} + {x^{ - 3}}} \\right)^{10}}$ is If $\\dfrac{1}{\\sqrt{4x+1}}\\left\\{ { \\left( \\dfrac { 1+\\sqrt { 4x+1 } }{ 2 } \\right) }^{ n }-{ \\left( \\dfrac { 1-\\sqrt { 4x+1 } }{ 2 } \\right) }^{ n } \\right\\} =a_{0}+a_{1}\\ x+....a_{5}\\ x^{5}$, then $n=$ The coefficient of $x^{13}$ in the expansion of $(1-x)^{5}(1+x+x^{2}+x^{3})^{4}$ is", "# Binomial coefficient complex expression I have been trying to find the coefficient of $a^2x^3$, in the expansion of $(a+x+c)^2(a+x+d)^2$, without success. I am having trouble expanding the above expression, because I can't find a way to merge them into one. How can I solve this? Thanks • I have tried grouping a+x together and then multiplying the two terms together but then I'm left with four terms in which you can find (a+x) more than once – Mark C. Mar 4 '17 at 21:30 Here is a variation to determine the coefficient of $a^2x^3$ without a full expansion of the expression. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in an expression. We obtain \\begin{align*} [a^2x^3]&(a+x+c)^2(a+x+d)^2\\\\ &=[a^2]\\left([x^2](a+x+c)^2\\right)\\left([x^1](a+x+d)^2\\right)\\\\ &\\qquad +[a^2]\\left([x^1](a+x+c)^2\\right)\\left([x^2](a+x+d)^2\\right)\\tag{1}\\\\ &=[a^2]\\left(2(a+d)\\right)+[a^2]\\left(2(a+c)\\right)\\tag{2}\\\\ &=0 \\end{align*} Comment: • In (1) we observe that in order to obtain the coefficient of $x^3$ one factor has to contribute $x$ and the other factor has to contribute $x^2$. There are no other possibilities to obtain a term with $x^3$. • In (2) we select the coefficient of $x$ resp. $x^2$ according to the binomial formula \\begin{align*} (a+x+u)^2=(a+u)^2+2(a+u)x+x^2 \\end{align*} Since there is no term with a factor $a^2$ the result is $0$. • Thank you for the clear and complete explanation! – Mark C. Mar 5 '17", "{/eq} a) The coefficient of {eq}{x^{10}}\\,\\,in\\,\\,the\\,\\,\\exp ansion\\,\\,of\\,\\,{\\left( {2x - \\dfrac{1}{{{x^2}}}} \\right)^{16}} {/eq} is 1966080. This is obtained from (1). b) The coefficient of {eq}{x^{11}}\\,\\,in\\,\\,the\\,\\,\\exp ansion\\,\\,of\\,\\,{\\left( {2x - \\dfrac{1}{{{x^2}}}} \\right)^{16}} {/eq} is 0. This is obtained from (1).", "# How to find coefficient of multiple binomial expansions quickly? Find the coefficient of $(x^{-2})$ in the expansion of $(x-1)^3(\\frac1{x}+2x)^6$ How can this be done quickly without expanding the two brackets? I tried writing the general formula in terms of $r$ for both brackets and equated $-2$ to the sum of the powers of $x$ in terms of $r$. $${3 \\choose r} (x^{3-r}) (-1)^r {6 \\choose r} \\left(\\frac1{x}^6-r\\right)(2x)^r$$ I understand that there are multiple ways of getting $x^{-2}$, but shouldn't this be accounted for in the general formula method? (multiple $x^{-2}$ terms can be simplified to one number which should obey the general term formula). To convert to a more familiar form, multiply and divide by $x^6$: $$(x-1)^3 (1+2x^2)^6 x^{-6}$$ Now you're looking for the coefficient of $x^4$ in $(x-1)^3 (1+2x^2)^6$. You can get $x^4$ out of this product either by multiplying $x^2$ with $x^2$ or $1$ with $x^4$. Each of those coefficients should be easy to calculate, then you just add them. Here is a variation which goes consecutively through the factors expanding one by one up to terms which are needed ignoring other terms. We also start with the factors with greatest exponents which keep the number of terms which are to expand small. It is convenient to use the coefficient of operator $[x^k]$ to denote", "to finding the $x^5$ term of the binomial. Then the coefficient of $x^7$ is $-3(coefficient\\ of\\ x^5\\ from\\ the\\ binomial\\ expansion)$ This is a shortcut to multiplying out all the terms independently, if you only want a single term of the entire expansion of the factors.", "Question # If $$C$$ denotes the binomial coefficient $${ _{ }^{ n }{ C } }_{ r }$$ then $$(-1){ C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+5{ C }_{ 2 }^{ 2 }+....+(3n-1){ C }_{ n }^{ 2 }=$$ A (3n2)2nCn B (3n22)2nCn C (5+3n)2nCn D (3n52)2nCn Solution ## The correct option is D $$\\left( \\cfrac { 3n-2 }{ 2 } \\right) { _{ }^{ 2n }{ C } }_{ n }\\quad$$$${ \\left( 1+x \\right) }^{ n }=^{ n }{ { C }_{ 0 } }+^{ n }{ { C }_{ 1 } }x+^{ n }{ { C }_{ 2 } }{ x }^{ 2 }+...+^{ n }{ { C }_{ n } }{ x }^{ n }$$$${ \\left( 1+\\dfrac { 1 }{ x } \\right) }^{ n }=^{ n }{ { C }_{ 0 } }+\\dfrac { ^{ n }{ { C }_{ 1 } } }{ x } +\\dfrac { ^{ n }{ { C }_{ 2 } } }{ { x }^{ 2 } } +...+\\dfrac { ^{ n }{ { C }_{ n } } }{ { x }^{ n } }$$$${ \\left( 1+x \\right) }^{ n }{ \\left( 1+\\dfrac { 1 }{ x } \\right) }^{ n }=\\left( ^{ n }{ { C }_{ 0 } }+^{ n", "I spent 80 minutes with my tutor and he didn't know how to do it... Thanks a lot! 4. ## Re: How to use binomial theorem to work out the coefficient of s^27 Originally Posted by buenogilabert $(((1/6)*s)^{10})*((1-s^6)/(1-s))^{10}$ And the coefficient of $s^{27}$ Here is what I would do. Because $\\left( {\\frac{s}{6}} \\right)^{10} = \\frac{{s^{10} }}{{6^{10} }}$ we are looking for the coefficient of $s^{17}$ in $\\left( {\\frac{{1 - s^6 }}{{1 - s}}} \\right)^{10} = \\left( {s^5 + s^4 + s^3 + s^2 + s + 1} \\right)^{10}$. That is $1535040x^{17}$. I found that using a CAS. I don't know how much you know about generating polynomials. So this may not help you at all.", "# Binomial coefficient (n choose 0) equals 1 The following formula holds: $${n \\choose 0} = 1,$$ where ${n \\choose 0}$ denotes the binomial coefficient. From the definition, $${n \\choose k} = \\dfrac{n!}{k! (n-k)!},$$ so for $k=0$ we get, using the fact that $0!=1$, $${n \\choose 0} = \\dfrac{n!}{0! (n-0)!} = \\dfrac{n!}{n!} = 1,$$ as was to be shown.", "- 1 \\right) \\left( x^{15} - 1 \\right) \\left( x^{15} - 1 \\right) \\left( \\binom{2}{2} + \\binom{3}{2}x + \\binom{4}{2}x^2 + \\cdots \\right).$$ $$= -\\left( x^{28} - 1 \\right) \\left( x^{15} - 1 \\right) \\left( x^{15} - 1 \\right) \\left( 1 + 3x + 6x^2 + \\cdots \\right).$$ We determine the $x^{28}$ coefficient by doing casework on the first three terms in our product. We can obtain an $x^{28}$ term by choosing $x^{28}$ in the first term, $-1$ in the second and third terms, and $1$ in the fourth term. We can get two $x^{28}$ terms by choosing $x^{15}$ in either the second or third term, $-1$ in the first term, $-1$ in the second or third term from which $x^{15}$ has not been chosen, and the $\\binom{15}{2}x^{13}$ in the fourth term. We get $\\binom{15}{2} * 2 = 210$ $x^{28}$ terms this way. (We multiply by $2$ because the $x^{15}$ term could have been chosen from the second term or the third term). Lastly, we can get an $x^{28}$ term by choosing $-1$ in the first three terms and a $\\binom{30}{2}x^{28}$ from the fourth term. We have a total of $1 + 210 - 435 = -224$ for the $x^{28}$ coefficient, but we recall that we have a negative sign in front of our product, so we obtain an" ]

[ "SJ. Jan 5 '18 at 3:29 • Okay that is $$\\sum_{p+q+r+s=n} \\frac{n!}{p!q!r!s!} x^py^qz^ru^s$$ but where is $$\\sum_{p+q+r+s=n}$$ in your answer? – Ravi Prakash Jan 5 '18 at 3:32 • – SJ. Jan 5 '18 at 3:32 Hint: We can use Binomial Theorem recursively The coefficient of $a^8$ in $$(a(bc+bd+cd)+bcd))^{10}$$ will be $$\\binom{10}2(bc+bd+cd)^8(bcd)^2$$ The coefficient of $b^4$ in $$(bc+bd+cd)^8(bcd)^2$$ = the coefficient of $b^2$ in $$(b(c+d)+cd)^8(cd)^2$$ $$=(cd)^2\\binom82(cd)^{8-2}(c+d)^2=c^8d^8\\binom82(c^2+2cd+d^2)$$ Clearly, the coefficient of $c^9$ in $$c^8d^8\\binom82(c^2+2cd+d^2)$$ = the coefficient of $c$ in $$d^8\\binom82(c^2+2cd+d^2)$$ Let us set $abc=D, abd=C, acd=B, bcd=A$. Then $a^8 b^4 c^9 d^9 =A^2 B^6 C D$ and the problem boils down to finding the coefficient of $A^2 B^6 C D$ in $(A+B+C+D)^{10}$, i.e. to counting the anagrams of the word $AABBBBBBCD$. They are $$\\frac{10!}{2!6!}=\\color{red}{2520}.$$ Then, $$\\left\\{\\begin{array}{rcrcrcrcrr} \\ds{i} & \\ds{+} & \\ds{j} & \\ds{+} & \\ds{k} & \\ds{+} & \\ds{\\ell} & \\ds{=} & \\ds{10} & \\ds{\\qquad\\pars{\\texttt{a}}} \\\\[1mm] \\ds{i} & \\ds{+} & \\ds{j} & \\ds{+} & \\ds{k} &&& \\ds{=} & \\ds{8} & \\ds{\\qquad\\pars{\\texttt{b}}} \\\\[1mm] \\ds{i} & \\ds{+} & \\ds{j} &&& \\ds{+} & \\ds{\\ell} & \\ds{=} & \\ds{4} & \\ds{\\qquad\\pars{\\texttt{c}}} \\\\[1mm] \\ds{i} &&& \\ds{+} & \\ds{k} & \\ds{+} & \\ds{\\ell} & \\ds{=} & \\ds{9} & \\ds{\\qquad\\pars{\\texttt{d}}} \\\\[1mm] && \\ds{j} & \\ds{+} & \\ds{k} & \\ds{+} & \\ds{\\ell} & \\ds{=} & \\ds{9} & \\ds{\\qquad\\pars{\\texttt{e}}} \\end{array}\\right.$$ Note that $\\ds{\"\\pars{\\texttt{a}} - \\pars{\\texttt{b}}\" \\implies \\bbx{\\ell", "# How do you find the coefficient of a^(n-1)*b in the expansion of (a+b)^n? Dec 3, 2016 The coefficient is $n$ #### Explanation: The binomial theorem tells us that: ${\\left(a + b\\right)}^{n} = {\\sum}_{k = 0}^{n} \\left(\\begin{matrix}n \\\\ k\\end{matrix}\\right) {a}^{n - k} {b}^{k}$ where ((n),(k)) = (n!)/((n-k)! k!) The term in ${a}^{n - 1} . b$ is the one for $k = 1$, with coefficient: ((n),(1)) = (n!)/((n-1)! 1!) = n Alternatively, consider the product: ${\\left(a + b\\right)}^{n} = {\\overbrace{\\left(a + b\\right) \\left(a + b\\right) \\ldots \\left(a + b\\right)}}^{\\text{n times}}$ If we multiplied out the right hand side, then the only way we can get terms which contribute to the coefficient of ${a}^{n - 1} b$ is by picking the left hand $a$ term of $n - 1$ of the binomials and one right hand $b$ term. We can do that in precisely $n$ different ways, each contributing $1$ to the coefficient.", "of the corresponding three terms of $$(1-x)^{-6}$$ viz., $$x^{20}, x^{11}, x^2$$ respectively. Finally the coefficient is given by $$\\binom{25} {5} - 6*\\binom{16} {5} + 15*\\binom{7} {5} = 27237$$ • Thanks PTDS for helping! – Vasu090 Jul 27 at 5:50", "in $$(a + b)^5 (c + d)^6$$. 3) the coefficient of $$a^2 b^6 c^3$$ in $$(a + b)^5 (b + c)^6$$. 4) the coefficient of $$a^3 b^2$$ in $$(a + b)^5 + (a + b^2)$$.", "• The coefficient of $$x^n$$ in $$\\dfrac{1}{1 − 2x} = (1 − 2x)^{−1}$$ is $$\\binom{-1}{n} (-2)^n = (-1)^n \\binom{1 + n -1}{n} (-2)^n = (-1)^{2n} \\binom{n}{n} 2^n = 2^n$$ Therefore $$c_n$$, the coefficient of $$x^n$$ in $$C(x)$$, is $$\\dfrac{4}{3} (−1)^n + \\dfrac{2}{3} · 2^n$$. 3) Let $$E(x) = \\sum_{n=0}^{\\infty} e_nx^n$$ be the generating function of $$\\{e_n\\}$$. Then $$$$\\begin{split} E(x) &= e_0 &+ e_1x &+ e_2x^2 &+ e_3x^3 &+ e_4x^4 &+ ... \\\\ xE(x) &= &+ e_0x &+ e_1x^2 &+ e_2x^3 &+ e_3x^4 &+ ... \\\\ \\dfrac{1}{1-x} &= 1&+ x&+ x^2 &+ x^3 &+ x^4 &+ ... \\\\ \\end{split}$$$$ so $$$$\\begin{split} (1 − 3x)E(x) + \\dfrac{2}{1-x} &= E(x) − 3xE(x) + 2 · \\dfrac{1}{1-x}\\\\ &= (e_0 + 2) + \\sum_{n=2}^{\\infty} (e_n − 3e_{n−1} + 2)x^n\\\\ &= (e_0 + 2), \\end{split}$$$$ since (by the recurrence relation) we have $$e_n − 3e_{n−1} + 2 = 0$$ for $$n ≥ 1$$. Therefore $$E(x) = \\dfrac{(e_0 + 2) − \\dfrac{2}{1-x}}{1 − 3x} = \\dfrac{(e_0 + 2)(1 − x) − 2}{(1 − 3x)(1 − x)}$$ Since $$e_0 = 2$$, this means $$E(x) = \\dfrac{(2 + 2)(1 − x) − 2}{(1 − 3x)(1 − x)} = \\dfrac{2-4x}{(1 − 3x)(1 − x)}$$. We now use partial fractions. Write $$\\dfrac{2-4x}{(1 - 3x)(1 − x)} = \\dfrac{A}{1-3x} + \\dfrac{B}{1 − x} = \\dfrac{A(1 − x) + B(1 −", "### Binomial Theorem goals Find the coefficient of $x^n$ in the expansion of $(1+x) (1- x)^n$. Find the coefficient of $x^{2}$ in the equation of $(1+2x)^{6}(1-x)^{-1}$ Coefficient of term independent of $x$ in ${\\left( {1 + x + {x^{ - 2}} + {x^{ - 3}}} \\right)^{10}}$ is If $\\dfrac{1}{\\sqrt{4x+1}}\\left\\{ { \\left( \\dfrac { 1+\\sqrt { 4x+1 } }{ 2 } \\right) }^{ n }-{ \\left( \\dfrac { 1-\\sqrt { 4x+1 } }{ 2 } \\right) }^{ n } \\right\\} =a_{0}+a_{1}\\ x+....a_{5}\\ x^{5}$, then $n=$ The coefficient of $x^{13}$ in the expansion of $(1-x)^{5}(1+x+x^{2}+x^{3})^{4}$ is", "appears to a power of at least 20. You can also ignore the case $k \\leq 2$ because $x$ will appear to a power at most $7*k <18$. Thus, the coefficient of $x^{18}$ in $(1+x^5+x^7)^{20}$ is exactly the coefficient of $x^{18}$ in $\\binom{20}{3} x^{15}(1+x^2)^3$.", "coefficient $$C$$ of $$x^{10}$$ in $$\\left(\\ 1+x+x^2+x^3+x^4+\\dots\\ \\right)^5=\\frac 1{(1-x)^5}\\ .$$ We use here formal power series or polynomials modulo some suitable $$O$$ of some power of $$x$$. This can be computed by starting with $$(1-x)^{-1}=1+x+x^2+x^3+\\dots$$, and we apply the formal derivative four further times. The coefficient in degree $$10$$ comes from the one in degree $$14$$, so it is $$C=(14\\cdot 13\\cdot 11\\cdot 10)/(-1)(-2)(-3)(-4) =\\binom{14}4$$. As in the OP. So far nothing new. To get the number $$C'$$ of partitions with at most $$6$$ candles consider instead: $$\\left(\\ x+x^2+x^3+x^4+x^5+x^6\\ \\right)^5\\ .$$ and isolate the coefficient of $$x^{15}$$ in it. Equivalently, find the coefficient of $$x^{10}$$ in $$\\left(\\ 1+x+x^2+x^3+x^4+x^5\\ \\right)^5=\\frac {(1-x^6)^5}{(1-x)^5}\\ ,$$ which is $$\\Big[\\ 1-5x^6+\\dots\\ \\Big] \\left[\\ \\binom 44 + \\binom 54 x + \\binom 64 x^2 + \\binom 74 x^3 + \\binom 84 x^4 + \\dots + \\binom {14}4 x^{10} +\\dots\\ \\right]$$ The coefficient we need is $$C'=\\binom {14}4-5\\binom 84=651$$. Again, same as in the OP, after the following small correction. Step 1 is choosing the one child that becomes $$6+$$ at least one further candy. So we have to distribute \"in the same conditions\" the $$15-6=9$$ further candies. This leads to the binomial coefficient $$\\binom{9-1}{5-1}$$, in the same manner as for the total we have $$\\binom{15-1}{5-1}=1001$$ possibilities. Check with sage: sage: R.<x> = QQ[] sage: P =", "{/eq} a) The coefficient of {eq}{x^{10}}\\,\\,in\\,\\,the\\,\\,\\exp ansion\\,\\,of\\,\\,{\\left( {2x - \\dfrac{1}{{{x^2}}}} \\right)^{16}} {/eq} is 1966080. This is obtained from (1). b) The coefficient of {eq}{x^{11}}\\,\\,in\\,\\,the\\,\\,\\exp ansion\\,\\,of\\,\\,{\\left( {2x - \\dfrac{1}{{{x^2}}}} \\right)^{16}} {/eq} is 0. This is obtained from (1).", "# IIT JEE 1983 Maths - MCQ Question 15 What is the coefficient of $x^4$ in the expansion of $\\left(\\dfrac{x}{2}-\\dfrac{3}{x^2}\\right)^{10}$? ×", "1.5k views The coefficient of $x^{12}$ in $\\left(x^{3}+x^{4}+x^{5}+x^{6}+\\dots \\right)^{3}$ is ___________. edited | 1.5k views we will get $x^{12}$ as 1. $(x^4)^3$ having coefficient $^3C_0=1$ 2. $(x^3)^2(x^6)$ having coefficient $^3C_1=3$ 3. $(x^3)(x^4)(x^5)$ having coefficient ${^3C_2} \\times {^2C_1} =6$ So it is $10$ selected Isn't this out of syllabus? could you provide more details, or point to the topic. Just as we find binomial coefficients Plz give a detailed answer . How you have calculated the coefficients. Thanks We know the formula $(a+b)^n$ =$_{0}^{n}\\textrm{C} a^0 b^n + _{1}^{n}\\textrm{C} a^1b^n-1+......$ Similarly, 1) $(_{1}^{3}\\textrm{C}) (x^3)^1 (x^4+x^5+x^6+..)^2$ Now for, $(x^4+x^5+x^6+..)^2$ =$((x^4+x^5)+(x^6+....))^2$ =$(_{0}^{2}\\textrm{C}) (x^4+x^5)^0(x^6+..)^2$ 2) $(x^3+x^4+x^5+x^6+..)^3$ =$((x^3)+(x^4+x^5+x^6+....))^3$ =$(_{0}^{3}\\textrm{C}) (x^3)^0(x^4+x^5..)^3$ =$(_{0}^{3}\\textrm{C}) (x^3)^0(_{3}^{3}\\textrm{C})(x^4)^3(x^5+x^6.....)^0$ 3) $(x^3+x^4+x^5+x^6+..)^3 =(_{1}^{3}\\textrm{C}) (x^3)^1(_{1}^{2}\\textrm{C}) (x^4)^1(_{1}^{1}\\textrm{C}) (x^5)^1(x^6+..)^0$ Now addition of coefficients are =(1)+(2)+(3)=10 edited by @srestha,could'nt understand your soltuion..can you explain it more? We know the formula (a+b)n =n0Ca0bn+n1Ca1bn−1+...... then why you did not take rest terms.you have just taken 3C1* (x3) * (x4 + x5 +..)2. 3C0 * (x3)0 * (x4 + x5 ..) + 3C1* (x3) * (x4 + x5 +..)2. + 3C2* (x3)2 * (x4 + x5 +..)1. + 3C3* (x3) * (x4 + x5 +..)0 @Akriti I just have derived more, as I need coefficient of x12 sorry,i did not get you.. See, for getting coefficient of x12 we need put this formula to see in which sequence I", "+ 3{x^2} + 4{x^3} + ...........\\infty \\\\$ Here by observation, we have seen that ${x^0}$ has coefficient 1, coefficient of ${x^1}$ is 2, coefficient of ${x^2}$ is 3. It means the coefficient of ${x^n}$ will be $(n + 1).$ Hence, the coefficient of ${x^n}$ in the expansion of ${(1 - x)^{ - 2}}$ is $(n + 1)$ and the correct answer is option “B”.", "Review question # Can we find the coefficient of $x^2$ in $(5 + 4x)(1 - x/2)^{10}$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R7078 ## Question Write down and simplify the first $4$ terms in the expansion by the binomial theorem of $\\left(1 - \\tfrac{1}{2}x\\right)^{10}$. Find the coefficient of $x^2$ in the expansion of $(5 + 4x)\\left(1 - \\tfrac{1}{2}x\\right)^{10}.$", "# Binomial Coefficient with Two/Corollary ## Theorem $\\forall n \\in \\N: \\dbinom n 2 = T_{n - 1} = \\dfrac {n \\paren {n - 1} } 2$ where $T_n$ is the $n$th triangular number. ## Proof From the definition of binomial coefficient: $\\dbinom n 2 = \\dfrac {n!} {2! \\paren {n - 2}!}$ The result follows directly from the definition of the factorial: $2! = 1 \\times 2$ $\\blacksquare$", "# How do you find the coefficient of x^3 in (2x+3)^5? Jul 3, 2018 The answer is $= 720$ #### Explanation: ${\\left(a + b\\right)}^{n} = {\\sum}_{k = 0}^{n} \\left(\\begin{matrix}n \\\\ k\\end{matrix}\\right) {a}^{n - k} {b}^{k}$ Where, ((n),(k))=(n!)/((n-k)!(k!)) Here, we have ${\\left(2 x + 3\\right)}^{5}$ And we need the coefficient of ${x}^{3}$ $n = 5$ $a = 2 x$ $b = 3$ $k = 2$ The coefficient is $= \\left(\\begin{matrix}5 \\\\ 2\\end{matrix}\\right) \\cdot {\\left(2\\right)}^{3} \\cdot {\\left(3\\right)}^{2}$ =(5!)/((3!)(2!))* 8* 9 $= 10 \\cdot 72$ $= 720$", "various values of $$r$$. By Proposition 7.2.1, we have $$\\binom{-3}{r} = (-1)^r \\binom{3+r-1}{r} = (-1)^r \\binom{r+2}{r} = (-1)^r \\dfrac{(r+2)(r+1)}{2}$$ When $$r = 0$$, this is $$(−1)^02 · \\dfrac{1}{2} = 1$$. When $$r = 1$$, this is $$(−1)^13 · \\dfrac{2}{2} = −3$$. When $$r = 2$$, this is $$(−1)^24 · \\dfrac{3}{2} = 6$$. In general, we see that $$(1 + x)^{−3} = 0 − 3x + 6x^2 − + . . . + (−1)^n \\dfrac{(n+2)(n+1)}{2}x^n + ...$$ Exercise $$\\PageIndex{1}$$ Calculate the following. 1. $$\\binom{−5}{7}$$ 2. The coefficient of $$x^4$$ in $$(1 − x)^{−2}$$. 3. The coefficient of $$x^n$$ in $$(1 + x)^{−4}$$. 4. The coefficient of $$x^{k−1}$$ in $\\dfrac{1 + x}{(1 − 2x)^5} \\nonumber$ Hint: Notice that $$\\dfrac{1 + x}{(1 − 2x)^5} = (1 − 2x)^{−5} + x(1 − 2x)^{−5}$$. Work out the coefficient of $$x^n$$ in $$(1 − 2x)^{−5}$$ and in $$x(1 − 2x)^{−5}$$, substitute $$n = k − 1$$, and add the two coefficients. 5. The coefficient of $$x^k$$ in $$\\dfrac{1}{(1 − x^j)^n}$$, where $$j$$ and $$n$$ are fixed positive integers. Hint: Think about what conditions will make this coefficient zero.", "coefficients of $5xy{z^2}$ is $5$, and $- 3zy$ is $- 3$. ii. $1 + x + {x^2}$ Ans: The terms are $1$, $x$, and ${x^2}$.The coefficient of $1$ is $1$,$x$ is $1$, and ${x^2}$ is $1$. iii. $4{x^2}{y^2} - 4{x^2}{y^2} + {z^2}$ Ans: The terms are $4{x^2}{y^2}$, $- 4{x^2}{y^2}$, and ${z^2}$.The coefficient of $4{x^2}{y^2}$ is $4$,$- 4{x^2}{y^2}$ is $- 4$, and ${z^2}$ is $1$. iv. $3 - pq + qr - rp$ Ans: The terms are $3$, $- pq$, $qr$ and $- rp$.The coefficient of$3$ is $1$,$- pq$ is $1$, $qr$ is $1$, and $- rp$ is $- 1$. v. $\\dfrac{x}{2} + \\dfrac{y}{2} - xy$ Ans: The terms are $\\dfrac{x}{2}$, $\\dfrac{y}{2}$, and $- xy$.The coefficient of $\\dfrac{x}{2}$ is $\\dfrac{1}{2}$,$\\dfrac{y}{2}$ is $\\dfrac{1}{2}$, and $- xy$ is $- 1$. vi. $0.3a - 0.6ab + 0.5b$ Ans: The terms are $0.3a$, $- 0.6ab$, and $0.5b$.The coefficients of $0.3a$ is $0.3$,$- 0.6ab$ is $- 0.6$, and $0.5b$ is $0.5$. 2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomial do not fit any of these three categories? $x + y$,$1000$,$x + {x^2} + {x^3} + {x^4}$, $7 + y + 5x$, $2y - 3{y^2}$, $2y - 3{y^2} + 4{y^3}$, $5x - 4y + 3xy$, $4z - 15{z^2}$, $ab + bc + cd + da$,$pqr$, ${p^2}q + p{q^2}$,$2p + 3q$. Ans: The", "- This should be the coefficient of $a^{21}$ in $(1+a+a^2+a^3)(a+a^2+a^3)(a^{15}+a^{16}+a^{17}+a^{18}+...)(1+a+a^2+a^3+a^4+...)^2$ where $a$ is some real $0<a<1$ = coefficient of $a^5$ in $(1+a+a^2+a^3)(1+a+a^2)(1+a+a^2+a^3+a^4+...)^3$ = coefficient of $a^5$ in $(1-a^4)(1-a^3)(1-a)^{-6}$ = coefficient of $a^5$ in $(1-a^4 - a^3 + a^7)(1-a)^{-6}$ = coefficient of $a^5$ in $(1-a^4 - a^3 + a^7)(1+6a+\\frac{6.7}{2!}a^2 +\\frac{6.7.8}{3!}a^3 +\\frac{6.7.8.9}{4!}a^4 + ... )$ $= \\frac{6.7.8.9.10}{5!} - 6 - \\frac{6.7}{2!}$ -", "coefficient of $r^n$ in $P_j(r)$ is $0$ for such $j$. The coefficient of $m^1$ in $Q_n(m)$ appears to be $$[m^1] Q_n(m) = (-1)^n n \\left(-1 + \\sum_{i=2}^n \\dfrac{1}{i^2}\\right)$$ • Thanks for noticing. Fixed it. – Robert Israel Jun 2 '16 at 1:09", "I spent 80 minutes with my tutor and he didn't know how to do it... Thanks a lot! 4. ## Re: How to use binomial theorem to work out the coefficient of s^27 Originally Posted by buenogilabert $(((1/6)*s)^{10})*((1-s^6)/(1-s))^{10}$ And the coefficient of $s^{27}$ Here is what I would do. Because $\\left( {\\frac{s}{6}} \\right)^{10} = \\frac{{s^{10} }}{{6^{10} }}$ we are looking for the coefficient of $s^{17}$ in $\\left( {\\frac{{1 - s^6 }}{{1 - s}}} \\right)^{10} = \\left( {s^5 + s^4 + s^3 + s^2 + s + 1} \\right)^{10}$. That is $1535040x^{17}$. I found that using a CAS. I don't know how much you know about generating polynomials. So this may not help you at all." ]

[ "+0 # Help. I have been stuck 0 114 4 Given that $$\\binom{15}{8}=6435$$$$\\binom{16}{9}=11440$$, and $$\\binom{16}{10}=8008$$, find $$\\binom{15}{10}$$. I know that I should be able to do this but I can't figure it out. Aug 10, 2019 #1 +23295 +3 Given that $$\\dbinom{15}{8}=6435$$, $$\\dbinom{16}{9}=11440$$ , and $$\\dbinom{16}{10}=8008$$, find $$\\dbinom{15}{10}$$. Formula: $$\\dbinom{n}{k} + \\dbinom{n}{k+1} = \\dbinom{n+1}{k+1}$$ $$\\begin{array}{|lrcll|} \\hline & \\mathbf{\\dbinom{n}{k} + \\dbinom{n}{k+1}} &=& \\mathbf{\\dbinom{n+1}{k+1}} \\\\\\\\ (1) & \\dbinom{15}{8} + \\dbinom{15}{9} &=& \\dbinom{16}{9} \\\\ (2) & \\dbinom{15}{9} + \\dbinom{15}{10} &=& \\dbinom{16}{10} \\\\ \\hline (1)-(2): & \\dbinom{15}{8} + \\dbinom{15}{9} -\\left( \\dbinom{15}{9} + \\dbinom{15}{10}\\right) &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & \\dbinom{15}{8} - \\dbinom{15}{10} &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & 6435 - \\dbinom{15}{10} &=& 11440-11440 \\\\ & 6435 - \\dbinom{15}{10} &=& 3432 \\\\ & \\dbinom{15}{10} &=& 6435 - 3432 \\\\ & \\mathbf{\\dbinom{15}{10}} &=& \\mathbf{3003} \\\\ \\hline \\end{array}$$ Aug 11, 2019 #2 0 Thank you so much!! I can't keep all the formulas in my head sometimes. Aug 11, 2019 #3 +1655 +3 There are really too many formulas... tommarvoloriddle Aug 12, 2019 #4 +105540 +1 You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from. There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level. Most times this does", "can run the same method to compute $$a\\pmod{11!}$$.", "\\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} & 360 \\cdot x^{44} & 360 \\cdot x^{45} & 120 \\cdot x^{46}\\\\ & & & &", "db \"eASSWzRD<11$\"", "(\\mathbf{A}\\mathbf{B})^{-1} \\dbinom12 = \\dbinom13 } \\\\ \\hline \\end{array}$$ Apr 17, 2020", "is $$\\mathbf{0}$$.", ")^{10} \\\\\\\\ && ( a + b + c )^{n} \\\\ &=& \\sum \\limits_{k=0}^{n} \\sum \\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} &", "$\\boxed {L = 8 r }$", "this case trivially):$2^3 = 8$The$2$nd of the$1$st pair of consecutive integers whose product is a primorial:$1 \\times 2 = 2 = 2 \\#$$3$rd Term The$3$rd integer$m$after$0$,$1$such that$m! + 1$(its factorial plus$1$) is prime:$2! + 1 = 2 + 1 = 3$The$3$rd integer after$0$,$1$such that its double factorial plus$1$is prime:$2!! + 1 = 3$The$3$rd integer after$0$,$1$which is (trivially) the sum of the increasing powers of its digits taken in order:$2^1 = 2$The$3$rd subfactorial after$0$,$1$:$2 = 3! \\paren {1 - \\dfrac 1 {1!} + \\dfrac 1 {2!} - \\dfrac 1 {3!} }$The$3$rd integer$n$after$-1$,$0$such that$\\dbinom n 0 + \\dbinom n 1 + \\dbinom n 2 + \\dbinom n 3 = m^2$for integer$m$:$\\dbinom 2 0 + \\dbinom 2 1 + \\dbinom 2 2 + \\dbinom 2 3 = 2^2$The$3$rd integer$m$after$0$,$1$such that$m^2 = \\dbinom n 0 + \\dbinom n 1 + \\dbinom n 2 + \\dbinom n 3$for integer$n$:$2^2 = \\dbinom 2 0 + \\dbinom 2 1 + \\dbinom 2 2 + \\dbinom 2 3$The$3$rd palindromic integer after$0$,$1$which is the index of a palindromic triangular number$T_2 = 3$The$3$rd integer$n$after$0$,$1$such that$2^n$contains no zero in its decimal representation:$2^2 = 4$The$3$rd integer$n$after$0$,$1$such that$5^n$contains no zero in its decimal representation:$5^2 = 25$The$3$rd integer$n$after$0$,$1$such that both$2^n$and$5^n$have no zeroes:$2^2 = 4, 5^2 = 25$The$3$rd integer after$0$,$1$which is palindromic in both decimal and ternary:$2_{10} = 2_3$The$3$rd Fibonacci number after$1$,$1$:$2 =", "[-8,8]}$.", "let $N_1=N-\\dbinom{x_1}k$.\" Did you mean $N_1=N-\\dbinom{x_k}{k}$? That would be consistent with what you wrote in your next paragraph: $N_1=\\dbinom{x_{k-1}}{k-1}+\\ldots+\\dbinom{x_1}{1}$ – tychicus Aug 6 '13 at 7:21 • @tychicus: Yes, I did; I’ve fixed it now. (The same mistake was in the original version of your hint, and I must have unconsciously copied that.) – Brian M. Scott Aug 6 '13 at 12:02", "## Intermediate Algebra for College Students (7th Edition) $x^{10}-5x^8+10x^6-10x^4+5x^2-1$ Apply Binomial Theorem. $(a+b)^n=\\sum_{r=0}^n\\dbinom{n}{r}(a)^{n-r}b^r$ $(x^2-1)^5=\\dbinom{5}{0}(x^2)^5(-1)^0+\\dbinom{5}{1}(x^2)^4(-1)^1+\\dbinom{5}{2}(x^2)^3(-1)^2+\\dbinom{5}{3}(x^2)^2(-1)^3+\\dbinom{5}{4}(x^2)^1(-1)^4+\\dbinom{5}{5}(x^2)^0(-1)^5$ or, $=x^{10}-5x^8+10x^6+10x^4(-1)+5x^2+(1)(1)(-1)$ or, $=x^{10}-5x^8+10x^6-10x^4+5x^2-1$", "8}}{2}$", "\\in [0.2,0.8]$.", "# Thread: dbl int and desmos graoh 1. ## dbl int and desmos graoh $\\tiny{15.3.53}\\\\ \\displaystyle \\int_{0}^{\\frac{\\pi}{6}} \\int_{0}^{\\sec{\\theta}} 6r^{3} drd\\theta$ just want to see if this desmos graph is correct wanted to shade the wedge area but ??? also got $0.96225$ for calulated answer but they wanted exact", "-8}$.", "this algorithm is $$(^{n}_{10})*n = O(n^{11})$$.", "the desired output (4.81845e12) for 8!^5 # Mathematica, Eridan, ≤64 IntegerString[18612512719586442658197,2]", ".25^2$ $P(x=17)=\\dbinom{20}{3}.75^{17}\\times .25^3$ add these up 15. anonymous Ah, alright! 16. anonymous this is what i get, but check the calculations http://www.wolframalpha.com/input/?i=.75^20%2B20*.75^19*.25%2B190*.75^18*.25^2%2B1140*.75^17*.25^3 17. anonymous Yeah. Same! Thank you very much! Find more explanations on OpenStudy", "the final answer is $$\\dbinom{6}{2} + 6*4+1$$ 21. mathmath333 ok thnsk 22. ganeshie8" ]

[ "# An identity The following problem is apparently a bonus question for 13 year olds at a local girls school: Evaluate the sum $\\displaystyle \\frac{1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4} + \\cdots + \\frac{1}{99} – \\frac{1}{100}}{\\frac{1}{1+101}+ \\frac{1}{2+102} + \\cdots + \\frac{1}{50+150}}.$ It’s not that easy if you ask me. I had to work out the following identity first before I managed to solve it. $\\displaystyle \\sum_{k=1}^N \\frac{1}{2k-1} – \\frac{1}{2k} = \\sum_{k=1}^N \\frac{1}{k+N}.$ The identity can be proved via induction. This entry was posted in Problems. Bookmark the permalink.", "solution (and I suppose its right): With $x=0$, I got $\\frac{c}{24}=\\int_{0}^{1} t^2f(t) dt$ like Mhenni Benghorbal suggested. – user191919 Aug 6 '13 at 0:53 Plugging in $x=0$ and $x=1$ in the desired identity, one sees that $$24\\int_0^1t^2f(t)\\,\\mathrm dt=c=7-24\\int_0^1f(t)\\,\\mathrm dt.$$ There is no reason to expect that the LHS and the RHS coincide hence, in general, there is no $c$ such that the identity holds for every $x$ in $(0,1)$ (and in fact, for $x=0$ and $x=1$). Finding $f$ such that the identity holds is another question: assuming that $f$ is a solution and differentiating with respect to $x$ yields $f(x)=-x^2f(x)+x^7+x^5$, hence $f(x)=x^5$. This is indeed a solution and, using the LHS of the identity above, one sees that $c=24\\cdot\\frac18=3$.", "zero, and that can't happen. Therefore, $x$ can be any number but $1$. The final solution set is x inRR;x!=1. Mar 12, 2018 You can't solve this, because this is an identity, valid for all values of $a$ (other than $a = 1$) . You can, of course, prove this - see below $1 = \\frac{1 - a}{1 - a} = \\frac{1}{1 - a} - \\frac{a}{1 - a} = \\frac{1}{1 - a} + \\frac{a}{a - 1}$", "following identity holds: $\\mathbf{A}^{a}\\mathbf{A}^{b} = \\mathbf{A}^{a + b}$ Applied to your problem: $\\mathbf{A}^{0.5}\\mathbf{A}^{0.5}\\mathbf{A}^{-1} = \\mathbf{A}^{0} = \\mathbf{I}$ Top 50 recent answers are included", "# Thread: Identity or No Solution 1. ## Identity or No Solution And how to work it out? 2+1/3t = 1+1/4t 2. ## Re: Identity or No Solution Suggestion: Multiply both sides by $12t$. 3. ## Re: Identity or No Solution Originally Posted by ingyaningya And how to work it out? 2+1/3t = 1+1/4t If we try t=1 we see that it doesn't work: $2 + \\frac{1}{3} \\neq 1 + \\frac{1}{4}$ so it's not an identity 4. ## Re: Identity or No Solution Original Problem: $2+\\frac1{3t} = 1+\\frac1{4t}$ Find common denominators: $\\frac{2*(3t)+1}{3t} = \\frac{4t+1}{4t}$ $\\frac{6t+1}{3t} = \\frac{4t+1}{4t}$ Clear fractions: $(3t)(4t)(\\frac{6t+1}{3t} = \\frac{4t+1}{4t})$ $(4t)(6t+1) = (3t)(4t+1)$ Distribute: $24t^2+4t=12t^2+3t$ $12t^2-t=0$ Factor: $t(12t-1) = 0$ solve for t: t = 0 or 12t-1=0 $t=0$ or $t=\\frac{1}{12}$ 5. ## Re: Identity or No Solution Originally Posted by mdhafn Original Problem: $2+\\frac1{3t} = 1+\\frac1{4t}$ Find common denominators: $\\frac{2*(3t)+1}{3t} = \\frac{4t+1}{4t}$ $\\frac{6t+1}{3t} = \\frac{4t+1}{4t}$ Clear fractions: $(3t)(4t)(\\frac{6t+1}{3t} = \\frac{4t+1}{4t})$ $(4t)(6t+1) = (3t)(4t+1)$ Distribute: $24t^2+4t=12t^2+3t$ $12t^2-t=0$ Factor: $t(12t-1) = 0$ solve for t: t = 0 or 12t-1=0 $t=0$ or $t=\\frac{1}{12}$ 0 is not a solution since you'd be dividing by 0 in the original equation. @OP: For the equation to be an identity all values of t must work. For no solution then there must be no value of t. 6. ## Re: Identity", "use the Melzak's identity $$f\\left(x+y\\right)=x\\dbinom{x+n}{n}\\sum_{k=0}^{n}\\left(-1\\right)^{k}\\dbinom{n}{k}\\frac{f\\left(y-k\\right)}{x+k},\\, x,y\\in\\mathbb{R},\\, x\\neq-k$$ (for a reference see Z. A. Melzak, V. D. Gokhale, and W. V. Parker, Advanced Problems and Solutions: Solutions $4458$. Amer. Math. Monthly, $60$ $(1)$ $1953$, $53–54$) which holds for all algebraic polynomials $f$ up to degree $n$. So taking $$f\\left(z\\right)=\\frac{\\dbinom{2n-z}{n}}{1+n-z},\\, y=n$$ and observing that $$\\dbinom{n+k}{k}=\\dbinom{n+k}{n}$$ we have $$\\sum_{k=0}^{n}\\left(-1\\right)^{k}\\dbinom{n}{k}\\dbinom{n+k}{n}\\frac{1}{\\left(x+k\\right)\\left(k+1\\right)}=\\frac{\\frac{\\dbinom{n-x}{n}}{{\\textstyle 1-x}}}{x\\dbinom{x+n}{n}}$$ so taking the limit as $x\\rightarrow1$ we have $$\\sum_{k=0}^{n}\\left(-1\\right)^{k}\\dbinom{n}{k}\\dbinom{n+k}{n}\\frac{1}{\\left(k+1\\right)^{2}}=\\lim_{x\\rightarrow1}\\frac{\\frac{\\dbinom{n-x}{n}}{{\\textstyle 1-x}}}{x\\dbinom{x+n}{n}}=\\frac{1}{n\\left(n+1\\right)}$$ hence $$\\sum_{k=0}^{n}\\left(-1\\right)^{n+k}\\dbinom{n}{k}\\dbinom{n+k}{n}\\frac{1}{\\left(k+1\\right)^{2}}=\\color{red}{\\frac{\\left(-1\\right)^{n}}{n\\left(n+1\\right)}}$$ as wanted.", "# reference request – Does this identity well-known? Let $$a_1, a_2, …, a_n$$ and $$x$$ are real number, the the identity as follows hold: $$(a_1-frac{x}{a_2})(a_2-frac{x}{a_{3}})(a_3-frac{x}{a_{4}})…(a_n-frac{x}{a_{1}})=(a_1-frac{x}{a_n})(a_2-frac{x}{a_{1}})(a_3-frac{x}{a_{2}})…(a_n-frac{x}{a_{n-1}})$$ If $$n=3$$, let $$a$$, $$b$$, $$c$$, $$x$$ are real number, then: $$(a-frac{x}{c})(b-frac{x}{a})(c-frac{x}{b})=(a-frac{x}{b})(b-frac{x}{c})(c-frac{x}{a})$$ Question: Does this identity well-known? I hope that can apply the identity to solve some diophantine-equations", "definition, $\\frac{1}{9}*9 = 1$ – the multiplicative identity. In Escultura’s theory, $\\frac{1}{9}$ is a shorthand for the number that has a representation of 0.1111…. So, $\\frac{1}{9}*9 = 0.1111....*9 = 0.9999... = (1-d^*)$. So $(1-d^*)$ is also a multiplicative identity. By a similar process, you can show that $d^*$ itself must be the additive identity. So either $d^* == 0$, or else you’ve lost the field structure, and with it, pretty much all of real number theory.", "# Math Help - Identity 1. ## Identity There is another proof, without induction, for this identity? $1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\ldots+\\frac{1}{2n-1}-\\frac{1}{2n}=\\frac{1}{n+1}+\\frac{1}{n+2}+\\ldots+\\f rac{1}{2n}$ 2. This is Catalan's identity. Consider $k_n = 1 + \\frac{1} {2} + \\cdots + \\frac{1} {n}.$ Note that $\\frac{1} {2}k_n = \\frac{1} {2} + \\frac{1} {4} + \\cdots + \\frac{1} {{2n}}.$ The LHS is $k_{2n} - 2\\left( {\\frac{1} {2}k_n } \\right)$ & the RHS is $k_{2n}-k_n,$ as required $\\blacksquare$", "# Identity • Feb 24th 2008, 10:54 AM red_dog Identity There is another proof, without induction, for this identity? $1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\ldots+\\frac{1}{2n-1}-\\frac{1}{2n}=\\frac{1}{n+1}+\\frac{1}{n+2}+\\ldots+\\f rac{1}{2n}$ • Feb 24th 2008, 11:10 AM Krizalid This is Catalan's identity. Consider $k_n = 1 + \\frac{1} {2} + \\cdots + \\frac{1} {n}.$ Note that $\\frac{1} {2}k_n = \\frac{1} {2} + \\frac{1} {4} + \\cdots + \\frac{1} {{2n}}.$ The LHS is $k_{2n} - 2\\left( {\\frac{1} {2}k_n } \\right)$ & the RHS is $k_{2n}-k_n,$ as required $\\blacksquare$", "# Math Help - Complex number identity 1. ## Complex number identity Show that $(-1)^{\\frac{1}{\\pi}}=e^{(2n+1)i} (n=0, \\pm1, \\pm2, ...)$ 2. Originally Posted by davesface Show that $(-1)^{\\frac{1}{\\pi}}=e^{(2n+1)i} (n=0, \\pm1, \\pm2, ...)$ $(-1)^{\\frac{1}{\\pi}}=e^{\\frac{1}{\\pi}Ln(-1)}=e^{\\frac{1}{\\pi}[\\ln 1+i\\arg(-1)]}=e^{\\frac{1}{\\pi}(2n+1)\\pi i}=$ ... Tonio 3. These sign errors get me every time. Thanks for the help.", "# Not Pascal's Identity Algebra Level 2 $\\frac {C_1}{C_0} + 2\\ \\frac {C_2}{C_1} + 3\\ \\frac {C_3}{C_2} + \\ldots + 15\\ \\frac {C_{15}}{C_{14}}$ Let $C_r$ denote the binomial coefficient ${ 15 \\choose r }$. What is the value of the expression above? ×", "$$\\int_A\\frac{\\partial u}{\\partial r}dA=-\\int_A\\frac{1}{r^2} a^2\\sin\\theta\\;d\\theta\\;d\\phi\\;\\neq\\;0$$ I can't even get the Green's first identity. $$\\int_Dv\\nabla u+\\nabla v\\cdot\\nabla u \\;dV=\\oint_A v\\frac {\\partial u}{\\partial r}dA$$", "and ab = 21, find the value of a3 −b3 In the given problem, we have to find the value of Given We shall use the identity Here putting, Hence the value of is . #### Question 5: If $x+\\frac{1}{x}=5$, find the value of ${x}^{3}+\\frac{1}{{x}^{3}}$ In the given problem, we have to find the value of Given We shall use the identity Here putting, Hence the value of is #### Question 6: If $x-\\frac{1}{x}=7$, find the value of ${x}^{3}-\\frac{1}{{x}^{3}}$ In the given problem, we have to find the value of Given We shall use the identity Here putting, Hence the value of is #### Question 7: If $x-\\frac{1}{x}=5$, find the value of ${x}^{3}-\\frac{1}{{x}^{3}}$ In the given problem, we have to find the value of Given We shall use the identity Here putting, Hence the value of is . #### Question 8: If ${x}^{2}+\\frac{1}{{x}^{2}}$ = 51, find the value of ${x}^{3}-\\frac{1}{{x}^{3}}$ In the given problem, we have to find the value of Given We shall use the identity Here putting, In order to find we are using identity Here and Hence the value of is . #### Question 9: If , find the value of ${x}^{3}+\\frac{1}{{x}^{3}}$ In the given problem, we have to find the value of Given We shall use the identity Here putting, In order to find", "from A to B is (a) mn (b) nm (c) 2mn-1 (d) 2mn 8. If {(a,8),(6,b)}represents an identity function, then the value of a and b are respectively (a) (8,6) (b) (8,8) (c) (6,8) (d) (6,6) 9. Let A={1,2,3,4} and B={4,8,9,10}. A function f: A ⟶ B given by f={(1,4), (2,8), (3,9), (4,10)} is a (a) Many-one function (b) Identity function (c) One-to-one function (d) Into function 10. If f(x)=2x2 and g(x)=$\\frac{1}{3x}$, then f o g is (a) $\\\\ \\frac { 3 }{ 2x^{ 2 } }$ (b) $\\\\ \\frac { 2 }{ 3x^{ 2 } }$ (c) $\\\\ \\frac { 2 }{ 9x^{ 2 } }$ (d) $\\\\ \\frac { 1 }{ 6x^{ 2 } }$", "isn't given...I should have it by tomorrow, but heres the identity now: $$\\sum_{n=1}^{\\infty} \\frac{1}{n^n} = \\int^1_0 x^{-x} dx$$", "8 '11 at 22:02 use the identity $b^n - a^n = (b-a)\\sum^{n-1}_i b^{n-i} a^i$ -", "18:05 • For the record, this is known as Koshy's identity, see the references in mathoverflow.net/a/391985/11260 May 6 at 6:57 Let $$C(x) = \\sum_{n=1}^\\infty C_{n-1} x^n = \\frac{1-\\sqrt{1-4x}}{2}.$$ Then the identity in question follows easily from $C(x(1-x)) = x$.", "@Adam So, continuing from the above example, we would have e.g. if $f(x)$ is the identity function, $f(x)=x \\, \\forall x \\in \\mathbb{R}$, then $$\\mathcal{O}(f)(x)=\\begin{cases} 0 & \\text{if} \\, x \\not \\in A \\\\ \\frac{d}{dx}(\\frac{P_1(x)}{P_2(x)})_{|x=\\pi} & \\text{if} \\, x=\\frac{P_1(\\pi)}{P_2(\\pi)} \\end{cases}$$ – Ivan Loh Apr 4 '13 at 18:55", "× Algebra Partial Fractions Challenge Quizzes If the following is an identity in $$x$$: $\\frac{9x^2+9}{x^3+1}=\\frac{a}{x+1}+\\frac{bx+c}{x^2-x+1},$ what is the value of $$a+b+c$$? If the following is an identity in $$x$$: $\\frac{10x^2+1}{x(x^2+1)}=\\frac{a}{x}+\\frac{bx+c}{x^2+1},$ what is the value of $$a+b+c$$? If the following is an identity in $$x$$: $\\frac{9x-4}{x^3-2x+1}=\\frac{a}{x-1}+\\frac{bx+c}{x^2+x-1},$ what is the value of $$a \\times b \\times c$$? If the following is an identity in $$x$$: $\\frac{5x^2+Ax+B}{(7x+1)(x^2-7x+1)}=C\\left(\\frac{1}{7x+1}+\\frac{1}{x^2-7x+1}\\right),$ what is the value of $$A+B+C?$$ Which of the following is the partial fraction decomposition of $\\frac{ 1} { x^3 + x } ?$ ×" ]

[ "use the Melzak's identity $$f\\left(x+y\\right)=x\\dbinom{x+n}{n}\\sum_{k=0}^{n}\\left(-1\\right)^{k}\\dbinom{n}{k}\\frac{f\\left(y-k\\right)}{x+k},\\, x,y\\in\\mathbb{R},\\, x\\neq-k$$ (for a reference see Z. A. Melzak, V. D. Gokhale, and W. V. Parker, Advanced Problems and Solutions: Solutions $4458$. Amer. Math. Monthly, $60$ $(1)$ $1953$, $53–54$) which holds for all algebraic polynomials $f$ up to degree $n$. So taking $$f\\left(z\\right)=\\frac{\\dbinom{2n-z}{n}}{1+n-z},\\, y=n$$ and observing that $$\\dbinom{n+k}{k}=\\dbinom{n+k}{n}$$ we have $$\\sum_{k=0}^{n}\\left(-1\\right)^{k}\\dbinom{n}{k}\\dbinom{n+k}{n}\\frac{1}{\\left(x+k\\right)\\left(k+1\\right)}=\\frac{\\frac{\\dbinom{n-x}{n}}{{\\textstyle 1-x}}}{x\\dbinom{x+n}{n}}$$ so taking the limit as $x\\rightarrow1$ we have $$\\sum_{k=0}^{n}\\left(-1\\right)^{k}\\dbinom{n}{k}\\dbinom{n+k}{n}\\frac{1}{\\left(k+1\\right)^{2}}=\\lim_{x\\rightarrow1}\\frac{\\frac{\\dbinom{n-x}{n}}{{\\textstyle 1-x}}}{x\\dbinom{x+n}{n}}=\\frac{1}{n\\left(n+1\\right)}$$ hence $$\\sum_{k=0}^{n}\\left(-1\\right)^{n+k}\\dbinom{n}{k}\\dbinom{n+k}{n}\\frac{1}{\\left(k+1\\right)^{2}}=\\color{red}{\\frac{\\left(-1\\right)^{n}}{n\\left(n+1\\right)}}$$ as wanted.", "# An identity The following problem is apparently a bonus question for 13 year olds at a local girls school: Evaluate the sum $\\displaystyle \\frac{1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4} + \\cdots + \\frac{1}{99} – \\frac{1}{100}}{\\frac{1}{1+101}+ \\frac{1}{2+102} + \\cdots + \\frac{1}{50+150}}.$ It’s not that easy if you ask me. I had to work out the following identity first before I managed to solve it. $\\displaystyle \\sum_{k=1}^N \\frac{1}{2k-1} – \\frac{1}{2k} = \\sum_{k=1}^N \\frac{1}{k+N}.$ The identity can be proved via induction. This entry was posted in Problems. Bookmark the permalink.", "# Combinatorics - Sum of series 1. Oct 12, 2010 ### tarheelborn 1. The problem statement, all variables and given/known data Sum the series $$1^2+2^2+\\cdots|n^2$$ by observing that $$m^2=2* \\dbinom{m}{2} + \\dbinom{m}{1}$$ and using the identity $$\\dbinom{0}{k}+ \\dbinom{1}{k} + \\cdots+ \\dbinom{m}{k}= \\dbinom{m+1}{k+1}$$. 2. Relevant equations 3. The attempt at a solution I know that $$1^2+2^2+\\cdots+m^2= 2* \\dbinom{1}{2}+ \\dbinom{1}{1} + 2* \\dbinom{2}{2}+ \\dbinom{2}{1} + 2* \\dbinom{3}{2} + \\dbinom{3}{1} + \\cdots + 2* \\dbinom{m}{2} + \\dbinom{m}{1}$$ but I am not picking up on how to simply this into the sum of the whole series. 2. Oct 12, 2010 ### vela Staff Emeritus Try collecting terms so you have 2*(stuff) + (everything else). 3. Oct 12, 2010 ### tarheelborn Re: Combinatorics - Sum of series ii OK, that gives me $$2(\\dbinom{1}{2}+\\dbinom{2}{2}+...+\\dbinom{n}{2})+(1+2+...+n)$$ , but I'm still not seeing it. 4. Oct 12, 2010 ### tarheelborn OK, wait... The above equals $$2(\\dbinom{1}{2}+ \\dbinom{2}{2} +...+ \\dbinom{n}{2})+(n(n+1))/2$$, so that lets me eliminate the 2's. Now I need to be able to combine the other two parts. Right? 5. Oct 12, 2010 ### tarheelborn And $$\\dbinom{n}{n}=\\frac{1}{6}(n-2)(n-1)n = \\frac{1}{6}(n^5-2n^4-n^3+2n^2)$$. I don't know quite what I was expecting but that isn't it. 6. Oct 12, 2010 ### tarheelborn And $$\\dbinom{n}{n}=\\frac{1}{6}(n-2)(n-1)n = \\frac{1}{6}(n^5-2n^4-n^3+2n^2)$$ is what I meant to say. 7. Oct 12, 2010 ### vela Staff Emeritus Did you", "$\\frac{1}{c}=\\frac{\\ln(8)-0}{7}$ Do you know how to solve this for c? 19. anonymous ln 8/ 7 = 1/c c= 7/ln8 20. myininaya right 21. anonymous ok thanks so much!", "# reference request – Does this identity well-known? Let $$a_1, a_2, …, a_n$$ and $$x$$ are real number, the the identity as follows hold: $$(a_1-frac{x}{a_2})(a_2-frac{x}{a_{3}})(a_3-frac{x}{a_{4}})…(a_n-frac{x}{a_{1}})=(a_1-frac{x}{a_n})(a_2-frac{x}{a_{1}})(a_3-frac{x}{a_{2}})…(a_n-frac{x}{a_{n-1}})$$ If $$n=3$$, let $$a$$, $$b$$, $$c$$, $$x$$ are real number, then: $$(a-frac{x}{c})(b-frac{x}{a})(c-frac{x}{b})=(a-frac{x}{b})(b-frac{x}{c})(c-frac{x}{a})$$ Question: Does this identity well-known? I hope that can apply the identity to solve some diophantine-equations", "### Home > CALC > Chapter 8 > Lesson 8.3.2 > Problem8-114 8-114. $\\text{Then }x=\\frac{1}{2}(U+1).$ Let U = 2x − 1. $\\frac{dU}{dx}=2$ $\\frac{1}{2}dU=dx$ Rewrite the integrand and don't forget to write the bounds in terms of U. $=\\int_1^9\\frac{U+1}{4\\sqrt{U}}dU=\\frac{1}{4}\\int_1^9(U^{1/2}+U^{-1/2})=\\frac{16}{3}$ Let U = 1 − x² $=\\int_1^5\\frac{1}{2u}du=\\frac{1}{2}\\ln(u)\\Big|_1^5=\\frac{1}{2}\\ln(5)$ $=\\int_0^{-4}-\\frac{3}{2}e^udu=\\frac{3}{2}-\\frac{3}{2}e^{-4}$", "A lil algebra help DB to solve for d_2: $$\\frac{336}{81}=\\frac{175}{100}*\\frac{1}{[\\frac{d_2}{0.1}]^3}$$ $$\\sim 2.4=\\frac{1}{[\\frac{d_2}{0.1}]^3}$$ then i don't know wat to do i know its simple algebra, i just don't see it,. Mentor If $$a = \\frac{1}{b^3}$$ Then $$b^3 = \\frac{1}{a}$$ Use this idea and keep going. DB thnx 4 all the help doc, i get $$d_2=0.041^{\\frac{1}{3}}$$ Mentor DB said: i get $$d_2=0.041^{\\frac{1}{3}}$$ That doesn't seem right. Consider the following: $$[\\frac{d_2}{0.1}]^3 = \\frac{d_2^3}{0.001}$$ DB o i see wat i did, so $$d_2=[4.1*10^{-4}]^\\frac{1}{3}$$? Mentor Much better.", "from A to B is (a) mn (b) nm (c) 2mn-1 (d) 2mn 8. If {(a,8),(6,b)}represents an identity function, then the value of a and b are respectively (a) (8,6) (b) (8,8) (c) (6,8) (d) (6,6) 9. Let A={1,2,3,4} and B={4,8,9,10}. A function f: A ⟶ B given by f={(1,4), (2,8), (3,9), (4,10)} is a (a) Many-one function (b) Identity function (c) One-to-one function (d) Into function 10. If f(x)=2x2 and g(x)=$\\frac{1}{3x}$, then f o g is (a) $\\\\ \\frac { 3 }{ 2x^{ 2 } }$ (b) $\\\\ \\frac { 2 }{ 3x^{ 2 } }$ (c) $\\\\ \\frac { 2 }{ 9x^{ 2 } }$ (d) $\\\\ \\frac { 1 }{ 6x^{ 2 } }$", "# Thread: Identity or No Solution 1. ## Identity or No Solution And how to work it out? 2+1/3t = 1+1/4t 2. ## Re: Identity or No Solution Suggestion: Multiply both sides by $12t$. 3. ## Re: Identity or No Solution Originally Posted by ingyaningya And how to work it out? 2+1/3t = 1+1/4t If we try t=1 we see that it doesn't work: $2 + \\frac{1}{3} \\neq 1 + \\frac{1}{4}$ so it's not an identity 4. ## Re: Identity or No Solution Original Problem: $2+\\frac1{3t} = 1+\\frac1{4t}$ Find common denominators: $\\frac{2*(3t)+1}{3t} = \\frac{4t+1}{4t}$ $\\frac{6t+1}{3t} = \\frac{4t+1}{4t}$ Clear fractions: $(3t)(4t)(\\frac{6t+1}{3t} = \\frac{4t+1}{4t})$ $(4t)(6t+1) = (3t)(4t+1)$ Distribute: $24t^2+4t=12t^2+3t$ $12t^2-t=0$ Factor: $t(12t-1) = 0$ solve for t: t = 0 or 12t-1=0 $t=0$ or $t=\\frac{1}{12}$ 5. ## Re: Identity or No Solution Originally Posted by mdhafn Original Problem: $2+\\frac1{3t} = 1+\\frac1{4t}$ Find common denominators: $\\frac{2*(3t)+1}{3t} = \\frac{4t+1}{4t}$ $\\frac{6t+1}{3t} = \\frac{4t+1}{4t}$ Clear fractions: $(3t)(4t)(\\frac{6t+1}{3t} = \\frac{4t+1}{4t})$ $(4t)(6t+1) = (3t)(4t+1)$ Distribute: $24t^2+4t=12t^2+3t$ $12t^2-t=0$ Factor: $t(12t-1) = 0$ solve for t: t = 0 or 12t-1=0 $t=0$ or $t=\\frac{1}{12}$ 0 is not a solution since you'd be dividing by 0 in the original equation. @OP: For the equation to be an identity all values of t must work. For no solution then there must be no value of t. 6. ## Re: Identity", "# Math Help - Identity 1. ## Identity There is another proof, without induction, for this identity? $1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\ldots+\\frac{1}{2n-1}-\\frac{1}{2n}=\\frac{1}{n+1}+\\frac{1}{n+2}+\\ldots+\\f rac{1}{2n}$ 2. This is Catalan's identity. Consider $k_n = 1 + \\frac{1} {2} + \\cdots + \\frac{1} {n}.$ Note that $\\frac{1} {2}k_n = \\frac{1} {2} + \\frac{1} {4} + \\cdots + \\frac{1} {{2n}}.$ The LHS is $k_{2n} - 2\\left( {\\frac{1} {2}k_n } \\right)$ & the RHS is $k_{2n}-k_n,$ as required $\\blacksquare$", "{db} (\\ln |b| - \\ln |a|) = \\frac d {db} \\int_a^b \\frac 1 x \\, dx = \\frac 1 b.$$ And this is with$b<0$. Another way to see this is by looking at the graph of$f(x) = \\frac 1 x$which has two, not connected parts, as 0 is not part of the domain. The same way but more obviously, the function (with the same domain) and definition:$ F(x) = lnx + c $(when x>0),$ ln(-x) + c $(when x<0) has two, non connected parts. But we can rewrite it more compactly as$ F(x) = ln |x| + c $. • Actually, you can choose different constants for each connected component. This shows show a first order equation$xy'=1\\$ may have a two dimensional space of solutions because of a singularity.", "## anonymous 4 years ago the coefficient of x^2 in the expansion of (1 + x/5)^n, where n is a positive integer, is 3/5. Find the value of n. How to do this???? :P 1. anonymous $(1+x)^n=1+nx+\\dbinom{n}{2}x^2+...$ in your case you will have $1+n\\frac{x}{5}+\\dbinom{n}{2}\\frac{x^2}{25}+ ...$ 2. anonymous so basically i need to multiply 1/25 and $\\left(\\begin{matrix}n \\\\ 2\\end{matrix}\\right)$, but how do I do that? 3. anonymous and equate it to 3/5. 4. anonymous so you know that $\\frac{n(n-1)}{2}\\times \\frac{1}{25}=\\frac{3}{5}$ $\\frac{n(n-1)}{2\\times 25}=\\frac{3}{5}$ $\\frac{n(n-1)}{10}=3$ $n(n-1)=30$ 5. anonymous and since n is a whole number, only possible answer is $n=6$ 6. anonymous wait, sorry, I don't get why n(n - 1)/2 7. anonymous oh ok lets go slower 8. anonymous to compute $\\dbinom{n}{k}$ you do not really use $\\frac{n!}{k!(n-k)!}$ because it it is too much work, you just cancel. but we can write it out anyway 9. anonymous $\\dbinom{n}{2}=\\frac{n!}{2!(n-2)!}$ ok? 10. anonymous ohhh. so if it was the coefficient of x^3, then we would do n(n - 1)(n - 3)/3 ?? 11. anonymous then numerator is $n\\times (n-1)\\times (n-2)!$ so you can cancel the $(n-2)!$ and get $\\frac{n(n-1)}{2}$ 12. anonymous it would be $\\frac{n(n-1)(n-2)}{3!}$ 13. anonymous don't forget the factorial in the denominator 14. anonymous ^sorry, that's what I meant to type :P Got it. 15. anonymous great! 16. anonymous So,", "# 1983 AHSME Problems/Problem 28 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) Clearly since $[DBEF] = [ABE]$ it follows that $[ADF] = [AFE]$. This implies that $AC \\parallel DE$ and so $\\frac{AD}{DB} = \\frac{CE}{EB} = \\frac{2}{3}$. Thus $[AEB] = \\frac{3}{2+3} \\cdot 10 = \\boxed{6}$", "# Identity • Feb 24th 2008, 10:54 AM red_dog Identity There is another proof, without induction, for this identity? $1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\ldots+\\frac{1}{2n-1}-\\frac{1}{2n}=\\frac{1}{n+1}+\\frac{1}{n+2}+\\ldots+\\f rac{1}{2n}$ • Feb 24th 2008, 11:10 AM Krizalid This is Catalan's identity. Consider $k_n = 1 + \\frac{1} {2} + \\cdots + \\frac{1} {n}.$ Note that $\\frac{1} {2}k_n = \\frac{1} {2} + \\frac{1} {4} + \\cdots + \\frac{1} {{2n}}.$ The LHS is $k_{2n} - 2\\left( {\\frac{1} {2}k_n } \\right)$ & the RHS is $k_{2n}-k_n,$ as required $\\blacksquare$", "# Commutative property fails? Algebra Level 1 I will attempt to prove that $$\\frac00 = 1$$. In which of these steps did I first make a mistake by using flawed logic? Step 1: We can rewrite 15 as $$7 + 8$$ or $$8 + 7$$. Step 2: This means that $$7 + 8 = 8+ 7$$. Step 3: If we move one term from each side of the equation to the other side, we will get $7-7 = 8-8.$ Step 4: Dividing both sides by $$8-8$$ gives $$\\frac{7-7}{8-8} = 1$$. Step 5: Since $$7-7= 0$$ and $$8-8 = 0$$, $$\\frac00 = 1$$. ×", "# Math Help - Complex number identity 1. ## Complex number identity Show that $(-1)^{\\frac{1}{\\pi}}=e^{(2n+1)i} (n=0, \\pm1, \\pm2, ...)$ 2. Originally Posted by davesface Show that $(-1)^{\\frac{1}{\\pi}}=e^{(2n+1)i} (n=0, \\pm1, \\pm2, ...)$ $(-1)^{\\frac{1}{\\pi}}=e^{\\frac{1}{\\pi}Ln(-1)}=e^{\\frac{1}{\\pi}[\\ln 1+i\\arg(-1)]}=e^{\\frac{1}{\\pi}(2n+1)\\pi i}=$ ... Tonio 3. These sign errors get me every time. Thanks for the help.", "$Z_t$ as $\\frac{dB_m}{dZ_t}=-1$. We rearrange the above equation in further three groups of terms. Then We find analytic approximation to two of those three groups and add simply add the third group as such. $\\int_{B_{m1}}^{B_{m2}} p_m(B_m) P_{m,n}(B_m) dB_m$ $=G1 + G2 + G3$ where $G1=\\int_{B_{m1}}^{B_{m2}} p_{m0} P_{m0,n} dB$ $+\\int_{B_{m1}}^{B_{m2}} (\\frac{dp_{m0}}{dB} P_{m0,n}) (B - B_{m0}) dB$ $+\\int_{B_{m1}}^{B_{m2}} (\\frac{d^2p_{m0}}{d{B}^2} P_{m0,n}) \\frac{{(B - B_{m0})}^2}{2} dB$ $+\\int_{B_{m1}}^{B_{m2}} (\\frac{d^3p_{m0}}{d{B}^3} P_{m0,n}) \\frac{{(B - B_{m0})}^3}{6} dB$ and $G2=\\int_{B_{m1}}^{B_{m2}} ( p_{m0} \\frac{dP_{m0,n}}{dB}) (B - B_{m0}) dB$ $+\\int_{B_{m1}}^{B_{m2}} ( p_{m0} \\frac{d^2P_{m0,n}}{dB^2}) \\frac{{(B - B_{m0})}^2}{2} dB$ $+\\int_{B_{m1}}^{B_{m2}} (p_{m0} \\frac{d^3P_{m0,n}}{dB^3}) \\frac{{(B - B_{m0})}^3}{6} dB$ $G3=\\int_{B_{m1}}^{B_{m2}} (2 \\frac{dp_{m0}}{dB} \\frac{dP_{m0,n}}{dB}) \\frac{{(B - B_{m0})}^2}{2} dB$ $+\\int_{B_{m1}}^{B_{m2}} (3 \\frac{d^2p_{m0}}{d{B}^2} \\frac{dP_{m0,n}}{dB}+3 \\frac{dp_{m0}}{dB} \\frac{d^2P_{m0,n}}{dB^2}) \\frac{{(B - B_{m0})}^3}{6} dB$ Equation(C) We notice that after taking $P_{m0,n}$ common out of terms in G!, the terms in G1(and their taylor series continuation) are integral of a simple Taylor expansion of probability density function in the originating mth Grid at time t and hence represent the integral of probability mass in mth subdivision. Hence we can write this group G1 as $G1= P_{m0,n} \\int_{B_{m1}}^{B_{m2}} p_{m}(B_m) dB_m= P_{m0,n} \\Delta P_m$ where $\\Delta P_m$ indicates total (integral of) probability mass in the mth subdivision. Now we take the second group of terms. $G2=\\int_{B_{m1}}^{B_{m2}} ( p_{m0} \\frac{dP_{m0,n}}{dB}) (B - B_{m0}) dB$ $+\\int_{B_{m1}}^{B_{m2}} ( p_{m0} \\frac{d^2P_{m0,n}}{dB^2}) \\frac{{(B - B_{m0})}^2}{2} dB$ $+\\int_{B_{m1}}^{B_{m2}} (p_{m0} \\frac{d^3P_{m0,n}}{dB^3})", "# Solving Equations with Factorials I am attempting to solve an equation ${{n-2} \\choose {2}} + {{n-3} \\choose {2}} + {{n-4} \\choose {2}} = 136$. With the formula for a combination being $\\frac{n!}{r!(n - r)!}$, I simplified the given equation to: $(n-2)! + (n-3)! + (n-4)! = 272n - 1088$ However, I am not sure how I would solve for $\\\\n$ in the simplified version. I have looked into Sterling's Approximation, hoping to find some way to solve this there, but I believe that is going about it incorrectly. Any help that can be offered in solving this equation is greatly appreciated. - Note that $$\\dbinom{n-2}2 = \\dfrac{(n-2)(n-3)}{2}$$ $$\\dbinom{n-3}2 = \\dfrac{(n-3)(n-4)}{2}$$ $$\\dbinom{n-4}2 = \\dfrac{(n-4)(n-5)}{2}$$ Hence, $$\\dbinom{n-2}2 + \\dbinom{n-3}2 + \\dbinom{n-4}2 = \\dfrac{(n-2)(n-3)}{2} + \\dfrac{(n-3)(n-4)}{2} + \\dfrac{(n-4)(n-5)}{2}$$ Now solve the quadratic to get the answer. EDIT $$\\dbinom{m}2 = \\dfrac{m!}{2! (m-2)!} = \\dfrac{m(m-1) \\cdot (m-2)!}{2! \\cdot (m-2)!} = \\dfrac{m(m-1)}2$$ - As I have stated below, would you be able to explain how you simplified the binomial coefficients? – nmagerko Feb 23 '13 at 19:58 Bad simplification, that one. \\begin{align*} \\binom{n - 2}{2} + \\binom{n - 3}{2} + \\binom{n - 4}{2} &= \\frac{(n - 2)(n -3)}{2} + \\frac{(n - 3) (n - 4)}{2} + \\frac{(n - 4)(n - 5)}{2} \\\\ &= \\frac{3 n^2 - 21 n + 38}{2} \\end{align*} The binomial", "and ab = 21, find the value of a3 −b3 In the given problem, we have to find the value of Given We shall use the identity Here putting, Hence the value of is . #### Question 5: If $x+\\frac{1}{x}=5$, find the value of ${x}^{3}+\\frac{1}{{x}^{3}}$ In the given problem, we have to find the value of Given We shall use the identity Here putting, Hence the value of is #### Question 6: If $x-\\frac{1}{x}=7$, find the value of ${x}^{3}-\\frac{1}{{x}^{3}}$ In the given problem, we have to find the value of Given We shall use the identity Here putting, Hence the value of is #### Question 7: If $x-\\frac{1}{x}=5$, find the value of ${x}^{3}-\\frac{1}{{x}^{3}}$ In the given problem, we have to find the value of Given We shall use the identity Here putting, Hence the value of is . #### Question 8: If ${x}^{2}+\\frac{1}{{x}^{2}}$ = 51, find the value of ${x}^{3}-\\frac{1}{{x}^{3}}$ In the given problem, we have to find the value of Given We shall use the identity Here putting, In order to find we are using identity Here and Hence the value of is . #### Question 9: If , find the value of ${x}^{3}+\\frac{1}{{x}^{3}}$ In the given problem, we have to find the value of Given We shall use the identity Here putting, In order to find", "a Mathematica Program for $$n \\le10^6$$ I found that $\\frac{1}{6} (n+1) \\left(n^2-n+6\\right)$ is a power of 2 for for $n=3, \\quad \\frac{1}{6} (n+1) \\left(n^2-n+6\\right)=8\\\\ n=7, \\quad \\frac{1}{6} (n+1) \\left(n^2-n+6\\right)=64\\\\ n=23, \\quad \\frac{1}{6} (n+1) \\left(n^2-n+6\\right)=2048\\\\$ 43. eliassaab I just did it $$n = 2 (10^6)$$ 44. eliassaab Here is the Mathematica Program Clear[h, n] h[n_] = 1/6 (1 + n) (6 - n + n^2); PowerTwoQ[n_] := Module[{x }, x = n; While[ Divisible[x, 2] && x > 1, x = x/2;]; Return[ x == 1]] Q = {}; For [ n = 2, n < 2000000, n++; If [PowerTwoQ[h[n]], Q = Append[Q, {n, h[n]}]]]; Q The output is {{3, 8}, {7, 64}, {23, 2048}} 45. mukushla $1+\\dbinom{n}{1}+\\dbinom{n}{2}+\\dbinom{n}{3}=\\frac{n^3+5n+6}{6}$ so we must have$\\frac{n^3+5n+6}{6}=2^k \\ \\ \\ \\ \\ \\ \\ \\ k\\le2000$or$(n+1)(n^2-n+6)=3\\times2^{k+1}$but notice that$\\gcd(n+1\\ , \\ n^2-n+6)=\\gcd(n+1\\ , \\ n^2-n-n^2-n+6)$$\\gcd(n+1\\ , \\ -2n+6)=\\gcd(n+1\\ , \\ -2n+2n+2+6)=\\gcd(n+1 \\ , \\ 8)$$\\Rightarrow \\gcd(n+1\\ , \\ n^2-n+6) | 8$since $$n^2-n+6>n+1$$ power of $$2$$ in $$n+1$$ can not be greater than $$4$$ in other words $$n+1=16$$ or $$n+1| 3\\times 2^3$$ so we have few cases to check$n=3,5,7,11,15,23$ 46. mukushla checking that numbers gives $$3$$ solution : $$\\color\\red{n=3,7,23}$$" ]

[ "+0 # An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? +1 160 1 +295 An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? Sep 2, 2018 #1 +375 +2 From the $$10$$ to $$99$$, there are $$90$$ numbers. Out of these numbers, $$9$$ have the same digits($$11, 22, 33, 44, 55, 66, 77, 88, 99$$). So your probability would be $$\\frac{81}{90} = \\frac{9}{10}$$ - Daisy Sep 2, 2018", "+0 # An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? +1 28 1 +247 An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? Darkside Sep 2, 2018 #1 +343 +2 From the $$10$$ to $$99$$, there are $$90$$ numbers. Out of these numbers, $$9$$ have the same digits($$11, 22, 33, 44, 55, 66, 77, 88, 99$$). So your probability would be $$\\frac{81}{90} = \\frac{9}{10}$$ - Daisy dierdurst Sep 2, 2018", "# Probability An integer from 100 through 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have 0 as at least one digit? ×", "digits (in the same order). 10001, obviously. How about a probability of 99.9%?", "selected at random from the 9,732. However... 5.", "# So many cases! A 9-digit number that includes every non-zero digit exactly once is chosen at random. What is the probability that it is prime? × Problem Loading... Note Loading... Set Loading...", "# A discrete mathematics problem by Chaitanya Gaur An integer from 100 through 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have 0 as at least 1 digit? × Problem Loading... Note Loading... Set Loading...", "## Applied Statistics and Probability for Engineers, 6th Edition S: X $\\gt$ 0 i.e X is any integer greater than 0.", "## Probability Let k greater than or equal to 3 be any given integer. What is the probability that a random k-digit number will have at least one 0, at least one 1, and at least one 2? (Note: as usual, every number starts with either a 1, or a 2, .... or a 9 (but not a 0).).", "2017, 9:41 p.m. edited Each integer will be greater than 0 and less than 180.", "and exactly as often as any other finite string of digits. That’s in the full representation. If you look at all the infinitely many digits the normal number has to offer. If all you have is a slice then some digits are going to be more common and some less common. That’s similar to how if you fairly toss a coin (say) forty times, there’s a good chance you’ll get tails something other than exactly twenty times. Look at the first 35 or so digits of π there’s not a zero to be found. But as you survey more digits you get closer and closer to the expected average frequency. It’s the same way coin flips get closer and closer to 50 percent tails. Zero is a rarity in the first 35 digits. It’s about one-tenth of the first 3500 digits. The digits of a specific number are not random, not if we know what the number is. But we can be presented with a subset of its digits and have no good way of guessing what the next digit might be. That is getting into the same strange territory in which we can speak about the “chance” of a month having a Friday the 13th even though the appearances of Fridays the 13th have absolutely no randomness", "questions you are more likely to get real help. Okay, this question: If a number is generated using the function $1000*\\sqrt[5]{n}$ , where n is the random number between 0 and 1, suggest a value for the probability that the 3-digit integer part (000 to 999 inclusive) of this number has three different digits. Any idea about how to do this? Please explain it to me, thank you.", "Math Help - probability 1. probability What is the probability that a 6-digit phone number contains at least one 2? (Repetition of numbers and lead zero are allowed). 2. Imagine the situation as being a room with 6 phones. Each phone has 10 digits on it, and you have to pick one digit from each phone! As such there are 10 different ways you can choose 1 from 10. And you have to choose 6 of these. Hence there are $10^6$ combinations! Now lets consider the same situation, only now all the 2s on the phones have been removed. So now there are 6 phones, each with only 9 digits, (0,1,3,4,5,6,7,8,9). So you are choosing 1 button from 9, 6 times! Hence $9^6$. So we have concluded that there are $10^6$ combinations, and $9^6$ of them have no 2! Hence the probability that there is at least one 2 is : $10^6 - 9^6 = 468559$ 3. The response is attached as a gif image, hope this helps you.", "# Probability Probability Level 1 An integer from 100 through 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have 0 as at least one digit? × Problem Loading... Note Loading... Set Loading...", "# For an enough large integer N, can it be a random seed for small positive integers? let S(n) be the set of positive integers less than n; let N be a larger integer than n,for example N=100*n; let M(n,N) be the set of ( N modulo s ) for any s in S(n); Can I expect that M(n,N) tends to be more random for larger N?", "on Feb. 1, 2017, 9:41 p.m. edited Each integer will be greater than 0 and less than 180.", "time, or the fact that I now have a girlfriend; somehow I’ve not found the time to blog though. As I’m currently visiting a GR program at MSRI and staying in a house in Berkeley Hills, away from distractions, I can type up a post I had sitting in the ideas pile. Pick a number – any number… What did you pick? Was it 7? Or pi? or $1+e^{i\\pi}$? I mean, pick ANY number. It doesn’t have to be a whole number, or even a rational number. If you were to choose a random number — a truly random number — out of all of the real numbers, then the probability that it was a whole number, or even a rational number, is zero. Take a look at My Infinity is Bigger Than Yours to figure that one out, because that’s not exactly the problem I’ll be talking about today. What I’m going to convince you of today, is that the number will certainly contain a 7 as a digit — 100%. Let me explain. The random number can – and will be – arbitrarily large so it will have many many digits. If I had a 100 digit number then the probability that the first digit is not a 7 is 9/10, the same for the second", "# A discrete mathematics problem by Chaitanya Gaur Discrete Mathematics Level 3 An integer from 100 through 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have 0 as at least 1 digit? ×", "# Balls are chosen at random ## Main Question or Discussion Point This problem appeared in a problem set which I encountered on the internet In a game, balls are labeled by integer numbers. One chooses three different integer numbers between 1 and 10. Two balls are picked at the same time, at random from a box. If they are part of the three earlier chosen numbers, the player wins. What's the probability that the player will win? The given answer is 1/15. But I found 1/90. The probability that the first ball is labeled by one of the chosen numbers is 1/10 and the second is 1/9. And I considered that picking two balls at the same time is equivalent to picking them in sequence. So (1/10) (1/9) = 1/90. Related Set Theory, Logic, Probability, Statistics News on Phys.org mjc123", "# rare-events ## Coincidences You bump into an old friend you haven't seen for years. You find that you share a birthday with not just one but two other people in your office. You win the lottery! ## Pick a Number In this animation, the computer simulates a number of people $N$ choosing a number at random between 1 and $(\\frac{N}{2})^2$. So, for example, it will simulate 400 people choosing a number between 1 and 40000. How often would you expect it to pick the same number twice?" ]

[ "+0 # An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? +1 160 1 +295 An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? Sep 2, 2018 #1 +375 +2 From the $$10$$ to $$99$$, there are $$90$$ numbers. Out of these numbers, $$9$$ have the same digits($$11, 22, 33, 44, 55, 66, 77, 88, 99$$). So your probability would be $$\\frac{81}{90} = \\frac{9}{10}$$ - Daisy Sep 2, 2018", "1 '15 at 0:58 This is a really cool question! And the answer is: almost certainly not, but maybe. First, let me explain why it's possible. We have calculated about 13 trillion digits of pi so far, meaning we don't know anything about the digits after that. This means the digits immediately following the last one we've calculated could very well be 1415926... etc. The chances that the next 13 trillion digits are equivalent to the first 13 trillion are very slim: $(\\frac{1}{10})^{\\text{13 trillion}}$. Super unlikely, yet possible! However, that's not the only case that would make pi start with two identical decimal sequences; the repetition could start anywhere, so we need to sum all possible cases. First let's assume we don't know anything about the decimal expansion of pi (For all we know, maybe it repeats after the first digit [i.e. 3.114...]). The probability of a repetition after the $n^{\\text{th}}$ digit is $(\\frac{1}{10})^{n}$, since the changes of each of the $n$ digits matching is $(1/10)$. So, the probability that this is true for some value of $n$ would be: $$\\sum_{i=1}^{\\infty}{\\bigg(\\frac{1}{10}\\bigg)^i} = \\frac{1}{9}$$ That seems like a pretty good chance! And that would be the chance that a number picked at random from a uniform distribution like $[0,1]$ would start with two identical sequences. However, with pi, we", "+0 # An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? +1 28 1 +247 An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? Darkside Sep 2, 2018 #1 +343 +2 From the $$10$$ to $$99$$, there are $$90$$ numbers. Out of these numbers, $$9$$ have the same digits($$11, 22, 33, 44, 55, 66, 77, 88, 99$$). So your probability would be $$\\frac{81}{90} = \\frac{9}{10}$$ - Daisy dierdurst Sep 2, 2018", "digits (in the same order). 10001, obviously. How about a probability of 99.9%?", "is given by $$P_1 + P_{00} - P_1 P_{00} = 0.664$$. (That's the probability that you get a 1 in the beginning, plus a probability that you get 00 in the end, minus the probability that both happens. The subtraction at the end avoids double-counting the possibility of both happening). You'll notice that we never used the number of digits in the string. That's because if we don't care what values they take, then it doesn't matter. The problem would be the same for a 3-digit string.", "by taking a sequence of numbers from 0 to 9, always selection a digit with the probability of 1/10. This is abstraction, since we cannot do that to infinity. Still taking it from your perspective, you are saying that for each number in the interval (0, 0.3), we have a match from the interval [0.3, 1). Indeed we have 1:1 mapping. However, the process of abstract selection must be a little bit different: you need to select a number from 0 to 3 for the first digit and from 4 to 9 for another group. If we make this abstract selection in sort of uniform way, we should better say that we need to pick the same number of digits for each interval, one digit in both, then two digits in both and so on. Otherwise, we cannot compare the intervals. You can see immediately that we have the probability of $\\frac{1}{4}$ for the first and $\\frac{1}{6}$ for the second interval. The second digit can be any from 0 to 9 for both intervals so the probabilities are $\\frac{1}{4} \\cdot \\frac{1}{9}$, and $\\frac{1}{6} \\cdot \\frac{1}{9}$ respectively. After n digits we have $\\frac{1}{4} \\cdot (\\frac{1}{9})^{n-1}$, and $\\frac{1}{6} \\cdot (\\frac{1}{9})^{n-1}$. Although both probabilities tend to 0, they are never the same. The ratio for any n, and with that the", "the probability that 2 digits are the same is: $\\frac{4}{7}$ The possibility that it's the incorrect number is: $\\frac{6}{7}$ Just like Soroban said. The last question has ofcourse nothing to do with how the numbers look like: • Feb 12th 2010, 05:47 AM FengWei 9331, 8242, 7153, 6064, 8602, 7513, 6424", "to make sure we dont miss such things on the real test... or we just butchered our test and you might watch your months of preps flowing down the drain!! Manager Joined: 15 Feb 2011 Posts: 201 Re: Probability - Exactly Question! [#permalink] ### Show Tags 15 Aug 2011, 01:23 1 i knwww and the worst part is I keep doing these mistakes..boo hoo.. thanks for all the help Vikas.. Retired Moderator Joined: 20 Dec 2010 Posts: 1795 Re: Probability - Exactly Question! [#permalink] ### Show Tags 15 Aug 2011, 01:28 DeeptiM wrote: A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? (A) 72 (B) 144 (C) 216 (D) 283 (E) 300 Same as viks4gmat: $$C^{9}_{2}*2*\\frac{3!}{2!}=36*2*3=216$$ $$C^9_2:$$ Number of ways we can select 2 digits out of 9 {1,2,3,4,5,6,7,8,9} We can select 1,2 OR 4,9 But, we need to multiply each of these selections by 2. Why? For 8,9 8,8,9 8,9,9 Are two possibilities. First, in which we choose 2 8's and second, in which we choose 2 9's. If we have; XXY It can be uniquely arranged in = 3!/2! ways 3 terms: X,X,Y=3!(numerator) 2 repeats: X,X=2!(denominator) So, if we select 889 --- 889 898 988 Total=3 ways Ans: \"C\" _________________ Retired", "+0 0 111 1 Suppose 5 different integers are randomly chosen from between 20 and 69, inclusive. What is the probability that they each have a different tens digit? Guest Jun 27, 2018 #1 +964 +2 From 20 to 69, they are a total of 50 integers. To choose 5 of them, we do: $$\\binom{50}{5}$$ Now, we need to find out the number of successful outcomes. We need an integer between 20 and 29 Then another one from 30 to 39, 40 to 49, 50 to 59, and 60 to 69. There are 10 numbers in each category, and therefore $$10^5$$ ways of doing this. Your final answer is: $$\\boxed{\\frac{2500}{52969}}$$ I hope this helped, Gavin GYanggg Jun 27, 2018", "in a game where digits accumulate but subsequent digits can never be larger than the previous digit (see the page for a better description, especially the example). Figuring out the expected value for all scores is really hard. But we don't have to figure out all possible scores and the probability of their occurrence. We only need to figure out the expected value for each digit and then sum those expected values. What's the expected value for the most significant digit? What's the expected value for the second-most significant digit? etc. Then sum. Pretend the game ended after just one roll. Then the expected value of our score is: \\sum_{d=0}^{9}\\frac{d}{10}=0.45 But the game continues (the value must go up) so we figure out the expected value of the next digit. Let's consider 9. What's the probability the second digit is a 9? In order for the second digit to be 9, the first must also be a 9. So $P(d_2=9)=0.1*0.1=0.01$. What about 0? If our first digit was 9, then there's still just a $\\frac{1}{10}$ chance the next digit is a 0. But if our first digit was a 1, then the outcomes are either 0 or 1, so the probability is $\\frac{1}{2}$. More generally: P(d_{i+1}=n) = \\sum_{m=0}^{9}P(d_{i+1}=n|d_i=m)*P(d_i=m) where P(d_{i+1}=n|d_i=m) = \\begin{cases} \\frac{1}{m+1},& \\text{if } n\\leq m\\\\ 0,", "result. Claim: given $\\sum^{\\infty}_{k=1} c_k \\frac{1}{k}$ where $c_k = 0$ if $k$ contains a 9 as one of its digits, then $\\sum^{\\infty}_{k=1} c_k \\frac{1}{k}$ converges. Hint on how to prove this (without reading the solution): count the number of integers between $10^k$ and $10^{k+1}$ that lack a 9 as a digit. Then do a comparison test with a convergent geometric series, noting that every term $\\frac{1}{10^k}, \\frac{1}{10^k + 1}......, \\frac{1}{8(10^k) +88}$ is less than or equal to $\\frac{1}{10^k}$. How to prove the claim: we can start by “counting” the number of integers between 0 and $10^k$ that contain no 9’s as a digit. Between 0 and 9: clearly 0-8 inclusive, or 9 numbers. Between 10 and 99: a moment’s thought shows that we have $8(9) = 72$ numbers with no 9 as a digit (hint: consider 10-19, 20-29…80-89) so this means that we have $9 + 8(9) = 9(1+8) = 9^2$ numbers between 0 and 99 with no 9 as a digit. This leads to the conjecture: there are $9^k$ numbers between 0 and $10^k -1$ with no 9 as a digit and $(8)9^{k-1}$ between $10^{k-1}$ and $10^k-1$ with no 9 as a digit. This is verified by induction. This is true for $k = 1$ Assume true for $k = n$. Then to find the number of", "result. Claim: given $\\sum^{\\infty}_{k=1} c_k \\frac{1}{k}$ where $c_k = 0$ if $k$ contains a 9 as one of its digits, then $\\sum^{\\infty}_{k=1} c_k \\frac{1}{k}$ converges. Hint on how to prove this (without reading the solution): count the number of integers between $10^k$ and $10^{k+1}$ that lack a 9 as a digit. Then do a comparison test with a convergent geometric series, noting that every term $\\frac{1}{10^k}, \\frac{1}{10^k + 1}......, \\frac{1}{8(10^k) +88}$ is less than or equal to $\\frac{1}{10^k}$. How to prove the claim: we can start by “counting” the number of integers between 0 and $10^k$ that contain no 9’s as a digit. Between 0 and 9: clearly 0-8 inclusive, or 9 numbers. Between 10 and 99: a moment’s thought shows that we have $8(9) = 72$ numbers with no 9 as a digit (hint: consider 10-19, 20-29…80-89) so this means that we have $9 + 8(9) = 9(1+8) = 9^2$ numbers between 0 and 99 with no 9 as a digit. This leads to the conjecture: there are $9^k$ numbers between 0 and $10^k -1$ with no 9 as a digit and $(8)9^{k-1}$ between $10^{k-1}$ and $10^k-1$ with no 9 as a digit. This is verified by induction. This is true for $k = 1$ Assume true for $k = n$. Then to find the number of", "# Probability of same last four digits of a telephone number Yesterday I was talking to a girl and asked her for her phone number. Once she gave it to me we realized that we got exact same last four digits. Hence out of a 8 digit phone number the last 4 digits were same. She commented that maybe it’s a strong connection and then asked “What is the probability of that happening?” Now, I haven’t done any maths (let alone probability) in a long time. Can someone help me understand? I think its $$\\frac{1}{10} \\times \\frac{1}{10} \\times \\frac{1}{10} \\times \\frac{1}{10}$$, as each of the last 4 digits are unique. This translates to a $$0.01\\%$$ chance. Is this correct? • That probability assumes that all 8 digit phone numbers are accounted for in the sample space. If you could guarantee that all numbers, from 00000000 to 99999999 are in use, then yes, your probability would be .01% since your sample space would include 100,000,000 and the numbers of the form $abcdxxxx$ are 10,000. – Eleven-Eleven Apr 17 '14 at 1:50 • @Eleven-Eleven thanks :-) I get your explanation . maybe you should have this as an answer :) – rockstar Apr 17 '14 at 1:52 Imagine I met someone new, and after we exchanged phone numbers, we noticed", "is the probability that the two middle digits are the same? Solution: Consider the sequence A-B-B-C. There is no given condition that the digits can’t be repeated so we assume that we can repeat each digit. Each letter can represent 10 numbers (0-9). But 2 B’s is counted as one. Thus the total possible ways is 10x10x10=1,000. The total number of possible combinations is 10x10x10x10=10,000. The probability that the two middle digits are the same is 1,000/10,000=1/10.", "is 90 based on that fact that for each case of two numbers the same (10) there are 9 possibilities in each for the third number and so 9x10 = 90 possible ways of choosing two numbers the same. Your questions are ambiguous. (1) The \"probability of picking three distinct digits when you pick three digits\" is (1)(9/10)(8/10)= 72/100= 0.72. The \"probability that the number chosen, when choosing three distinct numbers is a specific such number is (1/10)(1/9)(1/8)= 1/720. (2) The \"probability of picking a three digit number\" by picking 3 digits at random, is, strictly speaking, 1- any three digit number will do. The probability of picking a specific three digit number is (1/10)(1/10)(1/10)= 1/1000=0.001", "1. ## formulating this problem.. a man forgot the last 2 digits of a phone number. he only remembers that they are different and they are odd what is the chance that he will pick the correct combination of numbers of the original phone number $\\frac{\\binom{5}{1}\\cdot \\binom{4}{1}}{\\binom{10}{2}}=\\frac{2}{9}$ is it a correct expression? 2. Originally Posted by transgalactic a man forgot the last 2 digits of a phone number. he only remembers that they are different and they are odd what is the chance that he will pick the correct combination of numbers of the original phone number $\\frac{\\binom{5}{1}\\cdot \\binom{4}{1}}{\\binom{10}{2}}=\\frac{2}{9}$ is it a correct expression? If they are both odd, there are 20 numbers to choose from, only one of which is correct, so the probability is .. CB 3. 5 posibilities for first 4 forthe second why 20 numbers to choose from the number are 0..9 The digits are odd so we have {1, 3, 5, 7, 9} to choose them from. The first can be any of these five and the second any of the remaining 4 (since the digits are different) - so total is 5x4=20 CB", "the probability of the first digit being smaller? i. Probability of two digits being same = $$\\frac { 9 }{ 90 }$$ = $$\\frac { 1 }{ 10 }$$ (11, 22, 33, 44, 55, 66, 77, 88, 99) ii. Numbers in which first digit is greater than second digit are 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97, 98, 45 outcomes, Required Probability = $$\\frac { 45 }{ 90 }$$ = $$\\frac { 1 }{ 2 }$$ iii. Numbers in which first digit is smaller than second digit are 12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89, 36 total no.of favourable outcomes = 36 Probability = Question 3. Each two-digit number is written on a paper slip and these are all put in a box. What is the probability that the product of the digits of a number drawn is a prime number? What if three-digit numbers are used instead? Total two-digit numbers =", "To find the probability of starting with the three digits 247, we integrate f(x) from 2.47 to 2.48. The general equation for two leading digits, D1D2, is: $P(D_1D_2) = \\log_{10}(D_1.D_2 + 0.1) - \\log_{10}(D_1.D_2)$ Which is equivalent to: $P(D_1D_2) = \\log_{10}(D_1D_2 + 1) - \\log_{10}(D_1D_2)$ For example, the probability of a number starting with “2” followed by “4” is log10(25)-log10(24) = 1.77%. Similarly, the equation for three leading digits, D1D2D3, is: $P(D_1D_2D_3) = \\log_{10}(D_1D_2D_3 + 1) - \\log_{10}(D_1D_2D_3)$", "randomly from the set of all 4-digit integers, what is the probability that it is palindromic? A. $$\\frac{1}{20}$$ B. $$\\frac{1}{50}$$ C. $$\\frac{1}{60}$$ D. $$\\frac{1}{90}$$ E. $$\\frac{1}{100}$$ 4-digit palindromic numbers have the form $$xyyx$$, where $$x$$ is one of values 1, 2, …, 9 and $$y$$ is one of values 0, 1, 2, …, 9. So, there are 9 x 10 = 90 four-digit palindromic numbers. The total number of 4-digit numbers between 1000 and 9999, inclusive, is 9000 ( = 9999 – 1000 + 1 ). Therefore, the probability that the selected 4-digit is palindromic is $$\\frac{90}{9000} = \\frac{1}{100}$$. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. \"Only$99 for 3 month Online Course\" \"Free Resources-30 day online access & Diagnostic Test\" \"Unlimited Access to over 120 free video lessons - try it yourself\" Intern Joined: 18 Oct 2018 Posts: 2 ### Show Tags 29 Oct 2018, 11:31 Here is a better explanation. Total no. of outcomes= 9999-999=9000 numbers of 4 digits Favorable outcomes= 9C1 x 10C1 x 1C1 x 1C1=90 Hence , probability = 90/9000=1/100 Re: M61-02 &nbs [#permalink] 29 Oct 2018, 11:31 Display posts from previous: Sort by # M61-02 Moderators: chetan2u, Bunuel Powered", "Math Help - probability 1. probability What is the probability that a 6-digit phone number contains at least one 2? (Repetition of numbers and lead zero are allowed). 2. Imagine the situation as being a room with 6 phones. Each phone has 10 digits on it, and you have to pick one digit from each phone! As such there are 10 different ways you can choose 1 from 10. And you have to choose 6 of these. Hence there are $10^6$ combinations! Now lets consider the same situation, only now all the 2s on the phones have been removed. So now there are 6 phones, each with only 9 digits, (0,1,3,4,5,6,7,8,9). So you are choosing 1 button from 9, 6 times! Hence $9^6$. So we have concluded that there are $10^6$ combinations, and $9^6$ of them have no 2! Hence the probability that there is at least one 2 is : $10^6 - 9^6 = 468559$ 3. The response is attached as a gif image, hope this helps you." ]

[ "+0 # An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? +1 160 1 +295 An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? Sep 2, 2018 #1 +375 +2 From the $$10$$ to $$99$$, there are $$90$$ numbers. Out of these numbers, $$9$$ have the same digits($$11, 22, 33, 44, 55, 66, 77, 88, 99$$). So your probability would be $$\\frac{81}{90} = \\frac{9}{10}$$ - Daisy Sep 2, 2018", "+0 # An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? +1 28 1 +247 An integer greater than 9 and less than 100 is randomly chosen. What is the probability that its digits are different? Darkside Sep 2, 2018 #1 +343 +2 From the $$10$$ to $$99$$, there are $$90$$ numbers. Out of these numbers, $$9$$ have the same digits($$11, 22, 33, 44, 55, 66, 77, 88, 99$$). So your probability would be $$\\frac{81}{90} = \\frac{9}{10}$$ - Daisy dierdurst Sep 2, 2018", "by taking a sequence of numbers from 0 to 9, always selection a digit with the probability of 1/10. This is abstraction, since we cannot do that to infinity. Still taking it from your perspective, you are saying that for each number in the interval (0, 0.3), we have a match from the interval [0.3, 1). Indeed we have 1:1 mapping. However, the process of abstract selection must be a little bit different: you need to select a number from 0 to 3 for the first digit and from 4 to 9 for another group. If we make this abstract selection in sort of uniform way, we should better say that we need to pick the same number of digits for each interval, one digit in both, then two digits in both and so on. Otherwise, we cannot compare the intervals. You can see immediately that we have the probability of $\\frac{1}{4}$ for the first and $\\frac{1}{6}$ for the second interval. The second digit can be any from 0 to 9 for both intervals so the probabilities are $\\frac{1}{4} \\cdot \\frac{1}{9}$, and $\\frac{1}{6} \\cdot \\frac{1}{9}$ respectively. After n digits we have $\\frac{1}{4} \\cdot (\\frac{1}{9})^{n-1}$, and $\\frac{1}{6} \\cdot (\\frac{1}{9})^{n-1}$. Although both probabilities tend to 0, they are never the same. The ratio for any n, and with that the", "+0 0 111 1 Suppose 5 different integers are randomly chosen from between 20 and 69, inclusive. What is the probability that they each have a different tens digit? Guest Jun 27, 2018 #1 +964 +2 From 20 to 69, they are a total of 50 integers. To choose 5 of them, we do: $$\\binom{50}{5}$$ Now, we need to find out the number of successful outcomes. We need an integer between 20 and 29 Then another one from 30 to 39, 40 to 49, 50 to 59, and 60 to 69. There are 10 numbers in each category, and therefore $$10^5$$ ways of doing this. Your final answer is: $$\\boxed{\\frac{2500}{52969}}$$ I hope this helped, Gavin GYanggg Jun 27, 2018", "digits (in the same order). 10001, obviously. How about a probability of 99.9%?", "# A credit card contains 16 digits between 0 and 9. However, only 100 million numbers are valid. If a number is entered randomly, what is the probability that it is a valid number? A credit card contains 16 digits between 0 and 9. However, only 100 million numbers are valid. If a number is entered randomly, what is the probability that it is a valid number? You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Cullen There are ${10}^{16}$ strings of 16 digits between 0 and 9 (for each of the 16 digits we have 10 choices). Also, 100 million = $100\\cdot {10}^{6}={10}^{8}$. So, the probability that the random entered number is valid is number of valid numbers/all strings of 16 digits=$\\frac{{10}^{8}}{{10}^{16}}=\\frac{1}{{10}^{8}}$ which is almost zero.", "deviation of 10 minutes. Which of the following statements must be true? Indicate all such statements. $a_1$, $a_2$, $a_3$,......,$a_{150}$ The $n_{th}$ term if the sequence shown is defined for each integer n from 1 to 150 as follows. If n is odd, then $a_n$=$\\frac{(n+1)}{2}$, and if n is even, then $a_{n}$=$(a_{n-1})^{2}$. How many integers appear in the sequence twice? If an integer is chosen at random from the integers between 101 and 550, inclusive, what is the probability that the chosen integer will begin with the digit 1, 2 or 3, and end with the digit 4, 5, or 6? #### Quantity A $\\frac{100!}{99!}$ #### Quantity B $\\frac{ (100!-99!)}{98!}$ #### Quantity A $\\frac{(5!+6!)}{(6!+7!)}$ #### Quantity B $\\frac{1}{6}$ 25000 +道题目 136本备考书籍", "1. ## formulating this problem.. a man forgot the last 2 digits of a phone number. he only remembers that they are different and they are odd what is the chance that he will pick the correct combination of numbers of the original phone number $\\frac{\\binom{5}{1}\\cdot \\binom{4}{1}}{\\binom{10}{2}}=\\frac{2}{9}$ is it a correct expression? 2. Originally Posted by transgalactic a man forgot the last 2 digits of a phone number. he only remembers that they are different and they are odd what is the chance that he will pick the correct combination of numbers of the original phone number $\\frac{\\binom{5}{1}\\cdot \\binom{4}{1}}{\\binom{10}{2}}=\\frac{2}{9}$ is it a correct expression? If they are both odd, there are 20 numbers to choose from, only one of which is correct, so the probability is .. CB 3. 5 posibilities for first 4 forthe second why 20 numbers to choose from the number are 0..9 The digits are odd so we have {1, 3, 5, 7, 9} to choose them from. The first can be any of these five and the second any of the remaining 4 (since the digits are different) - so total is 5x4=20 CB", "Tags 16 Sep 2014, 00:52 3 5 00:00 Difficulty: 95% (hard) Question Stats: 27% (02:10) correct 73% (02:49) wrong based on 118 sessions ### HideShow timer Statistics An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1? A. $$\\frac{17}{300}$$ B. $$\\frac{1}{15}$$ C. $$\\frac{2}{25}$$ D. $$\\frac{1}{10}$$ E. $$\\frac{3}{25}$$ _________________ Math Expert Joined: 02 Sep 2009 Posts: 52297 ### Show Tags 16 Sep 2014, 00:52 Official Solution: An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1? A. $$\\frac{17}{300}$$ B. $$\\frac{1}{15}$$ C. $$\\frac{2}{25}$$ D. $$\\frac{1}{10}$$ E. $$\\frac{3}{25}$$ First, make sure that you grasp the question. To find the desired probability, we need to count the integers between 1 and 300 that fit the given constraint. (We will then divide this count by 300, to determine the final answer.) The constraint is worded in a confusing way, so we should attempt to reword it. If necessary, put in sample numbers to make the conditions make sense. We need an \"integer raised to an exponent that is an integer greater than 1.\" In other", "Question # A credit card contains 16 digits between 0 and 9. However, only 100 million numbers are valid. If a number is entered randomly, what is the probability that it is a valid number? Complex numbers A credit card contains 16 digits between 0 and 9. However, only 100 million numbers are valid. If a number is entered randomly, what is the probability that it is a valid number? 2020-10-21 There are $$\\displaystyle{10}^{{16}}$$ strings of 16 digits between 0 and 9 (for each of the 16 digits we have 10 choices). Also, 100 million = $$\\displaystyle{100}\\cdot{10}^{{6}}={10}^{{8}}$$. So, the probability that the random entered number is valid is number of valid numbers/all strings of 16 digits=$$\\displaystyle\\frac{{10}^{{8}}}{{10}^{{16}}}=\\frac{{1}}{{10}^{{8}}}$$ which is almost zero.", "questions you are more likely to get real help. Okay, this question: If a number is generated using the function $1000*\\sqrt[5]{n}$ , where n is the random number between 0 and 1, suggest a value for the probability that the 3-digit integer part (000 to 999 inclusive) of this number has three different digits. Any idea about how to do this? Please explain it to me, thank you.", "randomly from the set of all 4-digit integers, what is the probability that it is palindromic? A. $$\\frac{1}{20}$$ B. $$\\frac{1}{50}$$ C. $$\\frac{1}{60}$$ D. $$\\frac{1}{90}$$ E. $$\\frac{1}{100}$$ 4-digit palindromic numbers have the form $$xyyx$$, where $$x$$ is one of values 1, 2, …, 9 and $$y$$ is one of values 0, 1, 2, …, 9. So, there are 9 x 10 = 90 four-digit palindromic numbers. The total number of 4-digit numbers between 1000 and 9999, inclusive, is 9000 ( = 9999 – 1000 + 1 ). Therefore, the probability that the selected 4-digit is palindromic is $$\\frac{90}{9000} = \\frac{1}{100}$$. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. \"Only$99 for 3 month Online Course\" \"Free Resources-30 day online access & Diagnostic Test\" \"Unlimited Access to over 120 free video lessons - try it yourself\" Intern Joined: 18 Oct 2018 Posts: 2 ### Show Tags 29 Oct 2018, 11:31 Here is a better explanation. Total no. of outcomes= 9999-999=9000 numbers of 4 digits Favorable outcomes= 9C1 x 10C1 x 1C1 x 1C1=90 Hence , probability = 90/9000=1/100 Re: M61-02 &nbs [#permalink] 29 Oct 2018, 11:31 Display posts from previous: Sort by # M61-02 Moderators: chetan2u, Bunuel Powered", "the probability that 2 digits are the same is: $\\frac{4}{7}$ The possibility that it's the incorrect number is: $\\frac{6}{7}$ Just like Soroban said. The last question has ofcourse nothing to do with how the numbers look like: • Feb 12th 2010, 05:47 AM FengWei 9331, 8242, 7153, 6064, 8602, 7513, 6424", "as above, we can convince ourselves that ‘most randomly selected integers have more than a trillion digits’. It’s a bit of an incredible statement, really. We rarely ever work with the ‘most-common’ kind of numbers (the big ones!).** What is the probability that a number with a trillion digits has a 3 in it? Well, it’s almost certain. The probability approaches 100%. If we consider ALL numbers, the probability IS 100% (or is it?). This is a real dilemma. How can we say that 100% of numbers have a 3 in them when this is clearly not true? We’ve been pretty sloppy here, but regardless, this kind of fast-and-loose infinite probability question is unsettling. Do you want to try taking a crack at these? Feel free to comment below. Oh, and Happy Birthday Euler! _________________________________ # Improper integrals debate Here’s a simple Calc 1 problem: Evaluate $\\int_{-1}^1 \\frac{1}{x}dx$ Before you read any of my own commentary, what do you think? Does this integral converge or diverge? image from illuminations.nctm.org Many textbooks would say that it diverges, and I claim this is true as well. But where’s the error in this work? $\\int_{-1}^1 \\frac{1}{x}dx = \\lim_{a\\to 0^+}\\left[\\int_{-1}^{-a}\\frac{1}{x}dx+\\int_a^{1}\\frac{1}{x}dx\\right]$ $= \\lim_{a\\to 0^+}\\left[\\ln(a)-\\ln(a)\\right]=\\boxed{0}$ Did you catch any shady math? Here’s another equally wrong way of doing it: $\\int_{-1}^1 \\frac{1}{x}dx = \\lim_{a\\to 0^+}\\left[\\int_{-1}^{-a}\\frac{1}{x}dx+\\int_{2a}^{1}\\frac{1}{x}dx\\right]$", "# Probability of same last four digits of a telephone number Yesterday I was talking to a girl and asked her for her phone number. Once she gave it to me we realized that we got exact same last four digits. Hence out of a 8 digit phone number the last 4 digits were same. She commented that maybe it’s a strong connection and then asked “What is the probability of that happening?” Now, I haven’t done any maths (let alone probability) in a long time. Can someone help me understand? I think its $$\\frac{1}{10} \\times \\frac{1}{10} \\times \\frac{1}{10} \\times \\frac{1}{10}$$, as each of the last 4 digits are unique. This translates to a $$0.01\\%$$ chance. Is this correct? • That probability assumes that all 8 digit phone numbers are accounted for in the sample space. If you could guarantee that all numbers, from 00000000 to 99999999 are in use, then yes, your probability would be .01% since your sample space would include 100,000,000 and the numbers of the form $abcdxxxx$ are 10,000. – Eleven-Eleven Apr 17 '14 at 1:50 • @Eleven-Eleven thanks :-) I get your explanation . maybe you should have this as an answer :) – rockstar Apr 17 '14 at 1:52 Imagine I met someone new, and after we exchanged phone numbers, we noticed", "the probability of the first digit being smaller? i. Probability of two digits being same = $$\\frac { 9 }{ 90 }$$ = $$\\frac { 1 }{ 10 }$$ (11, 22, 33, 44, 55, 66, 77, 88, 99) ii. Numbers in which first digit is greater than second digit are 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97, 98, 45 outcomes, Required Probability = $$\\frac { 45 }{ 90 }$$ = $$\\frac { 1 }{ 2 }$$ iii. Numbers in which first digit is smaller than second digit are 12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89, 36 total no.of favourable outcomes = 36 Probability = Question 3. Each two-digit number is written on a paper slip and these are all put in a box. What is the probability that the product of the digits of a number drawn is a prime number? What if three-digit numbers are used instead? Total two-digit numbers =", "between 0 and 1. The probability that you select any particular number is zero. But at the same time, you will end up selecting some number. Whichever number you end up selecting, is a number that you would have said had a 0% chance of being selected! For situations like these, the term “almost never” is used. Rather than saying that any particular number is impossible, you say that it will “almost never” be picked. While this linguistic trick might make you feel less uneasy about the situation, there still seems to be some remaining confusion to be resolved here. So in the case of our thought experiment, no matter how many rounds the game ends up going on, you have a 0% chance of being kidnapped. At the same time, by stipulation you have been kidnapped. Making sense of this is only the first puzzle. The second one is to figure out if it makes sense to talk about some theories making it more likely than others that you’ll be kidnapped (is $\\frac{11}{\\infty}$ smaller than $\\frac{111}{\\infty}$?) An even more trouble infinity is in the expected number of people that are kidnapped. No matter how many rounds end up being played, there are always only a finite number of people that are ever kidnapped. But let’s calculate the", "# Is this number random? I'm trying to create a random 6 digit code. If I generate a random number between 0 and 999999 and then pad it with zeros to make it 6 digits is it random or is it biased in some way? E.g It generates 750 so it is then padded to 000750 It generates 28149 so it is padded to 028149 Does this achieve the same as generating a random integer between 0 and 9 six times and then putting the numbers together in a string? Thanks. - The title of this question amused me. I imagined it to go something like, \"Here is a number: 9. Is this number random?\" – Rahul May 20 '12 at 22:45 It makes no difference, provided that in both cases you’re generating the numbers with uniform distribution. Consider, for instance, the probability of generating a number that starts with at least three zeroes. The first method gives you $1000$ such numbers out of $1,000,000$ possibilities, so the probability of getting one is $\\frac{1000}{1,000,000}=\\frac1{1000}$. In the second method you must generate three zeroes in succession to start, after which you don’t care: the probability of that is $\\left(\\frac1{10}\\right)^3=\\frac1{1000}$. More generally, each of the $10^6$ possible six-digit numbers has one chance in a million of being generated under either", "is so fast? • Not an intuitive explanation, but since no one has mentioned it: the probability is $\\frac{\\lfloor\\frac{n!}{e}+\\frac{1}{2}\\rfloor}{n!}$ (since the number of derangements is $\\lfloor\\frac{n!}{e}+\\frac{1}{2}\\rfloor$), or i.e. $\\frac{\\left[\\frac{n!}{e}\\right]}{n!}$, where $[x]$ is the nearest integer to $x$. This intuitively gets close to $\\frac{1}{e}$, since the numerator of $\\frac{\\frac{n!}{e}}{n!}$ is off by at most $\\frac{1}{2}$, and this error gets less and less significant as $n$ gets huge. – user26486 Feb 20 '15 at 23:24 • I still find it surprising and fascinating that this probability doesn't obey an (abstract) 0-1 law; that there is a nontrivial limiting probability is an amazing thing to me. – Steven Stadnicki Feb 20 '15 at 23:56 • doesn't it seem odd the error is less than $\\frac{1}{(n+1)!}$ but $S_n$ has only $n!$ elements? – cactus314 Feb 21 '15 at 14:19 • @johnmangual: An integer approximating an integer in the interval $[0,n!]$ either has zero error or error greater than one part in $n!.$ A real number approximating an integer in that interval can, of course, get much closer. I found it surprising (until I thought about it a bit) that the real number $n!/e$ is nearly an integer. Not only is it nearly an integer--it's nearly the right integer. – Will Orrick Feb 21 '15 at 23:19 • @KevinCostello: It's hard for me", "Math Help - probability 1. probability What is the probability that a 6-digit phone number contains at least one 2? (Repetition of numbers and lead zero are allowed). 2. Imagine the situation as being a room with 6 phones. Each phone has 10 digits on it, and you have to pick one digit from each phone! As such there are 10 different ways you can choose 1 from 10. And you have to choose 6 of these. Hence there are $10^6$ combinations! Now lets consider the same situation, only now all the 2s on the phones have been removed. So now there are 6 phones, each with only 9 digits, (0,1,3,4,5,6,7,8,9). So you are choosing 1 button from 9, 6 times! Hence $9^6$. So we have concluded that there are $10^6$ combinations, and $9^6$ of them have no 2! Hence the probability that there is at least one 2 is : $10^6 - 9^6 = 468559$ 3. The response is attached as a gif image, hope this helps you." ]

[ "0% # Problem 3 ## Largest prime factor The prime factors of $13195$ are $5$, $7$, $13$ and $29$. What is the largest prime factor of the number $600851475143$? ## 最大质因数 $13195$的质因数包括$5$、$7$、$13$和$29$。 $600851475143$的最大质因数是多少？", "5k$$ is enough to guarantee that the largest prime factor in the denominator is $$\\geq 2k$$. If this largest prime appears with multiplicity $$2\\alpha$$ in the denominator, it only appears with multiplicity $$\\alpha$$ in the numerator. So either way the ratio can not be an integer. • Great! Experimentally, $n>5$ should be enough. (trivial typo: I think we are happy if $c(n)/c(m) \\not\\in\\mathbb{Z}$ if $\\max(25,m) < n < 2m$) – Martin Rubey Dec 6 '18 at 9:07", "largest is $$8 3 1$$", "+0 # Help now 0 104 2 $97$ is the largest prime less than $100$. What is the largest prime factor of $97!$ (97 factorial)? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, $5!=5\\cdot 4\\cdot3\\cdot2\\cdot 1= 120$.) Jan 15, 2020 #1 +109739 +1 97! = 97 itself is prime......so it is the largest prime factor of 97! Jan 15, 2020 #2 +36 +1 thanks lol Paresh247 Jan 15, 2020", "prime factors for n large -", "accounting for the largest prime factors of integers.", "If $p$ and $q$ are distinct primes less than $7$, what is the largest possible value of the highest common factor of $2p^2 q$ and $3pq^2$?", "large prime, or has large prime factors. Workspace is internally allocated by C06PWF. The total size of these arrays is approximately proportional to $mn$.", "Free Version Easy # Prime factors of a GCD ABSALG-EIVKRT What is the largest prime divisor of gcd$(10,70,100)$? A There is none because gcd$(10,70,100)=1$ B $2$ C $5$ D $7$", "# Factorial's factors What is the largest value of $$k$$ such that $$20!$$ is divisible by $$k^{3}?$$ ×", "- 2^3 = 9 - 8 = 1 So 3^8 - 2^12 = 5 x 29 x 17, and the largest prime factor is 29. _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Re: What is the largest prime factor of the expression 3^8 − 2^12? &nbs [#permalink] 30 Oct 2018, 10:27 Display posts from previous: Sort by", "# Prime factors of a random number Let $r$ be a uniformly random integer between $1$ and $N$, for some large enough $N$ (i.e., I'm only interested in the asymptotics). • What is the expected largest prime factor of $r$? Is there a good upper bound on this? • Let $t < N$ be such that $\\Pr\\Big[r$ has no prime factor > $t\\Big] = \\Omega(1/\\log^c(N))$ for some constant $c > 0$. How small can we make $t$? Is $t = \\operatorname{Li}_2(N)$ achievable?", "Free Version Easy # Prime Factors of an LCM ABSALG-104LTD What is the largest prime divisor of lcm$(2^5\\cdot 11^2, 5\\cdot 3^3\\cdot 11, 5^2 \\cdot 13)$? A $3$ B $5$ C $11$ D $13$", "5 is a prime factor of $10^n - 1$ for some n, and all its multiples. What is the equivalent for binary representation?", "is a large prime, or has large prime factors. Workspace is internally allocated by c06pzc. The total size of these arrays is approximately proportional to ${n}_{1}{n}_{2}{n}_{3}$.", "Largest prime factor Let $$p_i = {2, 3, 4, 5, \\dots}$$ we continue to divide $$c$$ by $$p_i$$. If $$c$$ becomes 1 then $$p_i$$ is the greatest prime factor, otherwise we increase $$p_i$$ by one. We know that any $$p_i$$ that divides $$c$$ is prime, because otherwise it would be of the form $$p_i = {q_1}^{a_i} \\dotsm {q_n}^{a_n}$$ and since $$q_i$$ is smaller than $$p_i$$, $$c$$ would have been already divided by it, so $$p_i$$ must be a prime number. c = 600851475143 p = 2 while c > 1: if c % p == 0: c /= p continue p += 1 p 6857", "is the smallest prime factor of $$5^8+10^6–50^3$$? A. 2 B. 3 C. 5 D. 7 E. 11 Kudos for a correct solution. $$5^8 + 10^6 - 50^3$$ $$5^8 = (5^2)^4$$ $$10^6 = ((2*5)^2)^3$$ $$50^3 = (2*5^2)^3$$ $$(5^2)^3 + (5^2)^3 - (5^2)^3 [5^2 + 2^6 - 2^3] = 5^6 * 81 = 5^6 * 3^4.$$ So, $$3$$ is the smallest prime factor. Ans - B. Re: What is the smallest prime factor of 5^8+10^6–50^3? [#permalink] 15 Jan 2018, 23:06 Display posts from previous: Sort by", "the highest prime factor in the expression [#permalink] 20 Feb 2017, 12:31 1 KUDOS Expert's post Explanation Take the expression $$\\frac{3^1^8 - 3^1^2}{3^5}$$ or $$3^1^2\\frac{3^6 - 1}{3^5}$$= $$3^7 \\times 728$$. Clearly the highest prime factor of the expression $$\\frac{3^1^8 - 3^1^2}{3^5}$$ also is the highest prime factor of 728. Try $$\\frac{728}{19}$$. Doesn't work! Try $$\\frac{728}{13}$$. Works! Hence D is the right answer. _________________ Sandy If you found this post useful, please let me know by pressing the Kudos Button Try our free Online GRE Test Re: What is the highest prime factor in the expression [#permalink] 20 Feb 2017, 12:31 Display posts from previous: Sort by", "Journal article The largest prime factor of $X^3+2$ Abstract: The largest prime factor of $X^3+2$ has been investigated by Hooley, who gave a conditional proof that it is infinitely often at least as large as $X^{1+\\delta}$, with a certain positive constant $\\delta$. It is trivial to obtain such a result with $\\delta=0$. One may think of Hooley's result as an approximation to the conjecture that $X^3+2$ is infinitely often prime. The condition required by Hooley, his R$^{*}$ conjecture, gives a non-trivial bound for short Ramanujan-Kloosterman sums. ... Access Document Files: • (pdf, 284.0KB) Authors Publication date: 2001-01-01 UUID: uuid:141ff1ff-9698-4f09-a74c-8cdeb78df8ce Local pid: oai:eprints.maths.ox.ac.uk:165 Deposit date: 2011-05-19", "## Overview ### Question The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? 6857" ]

[ "Free Version Easy # Prime Factors of an LCM ABSALG-104LTD What is the largest prime divisor of lcm$(2^5\\cdot 11^2, 5\\cdot 3^3\\cdot 11, 5^2 \\cdot 13)$? A $3$ B $5$ C $11$ D $13$", "+0 # Help now 0 104 2 $97$ is the largest prime less than $100$. What is the largest prime factor of $97!$ (97 factorial)? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, $5!=5\\cdot 4\\cdot3\\cdot2\\cdot 1= 120$.) Jan 15, 2020 #1 +109739 +1 97! = 97 itself is prime......so it is the largest prime factor of 97! Jan 15, 2020 #2 +36 +1 thanks lol Paresh247 Jan 15, 2020", "Largest prime factor Let $$p_i = {2, 3, 4, 5, \\dots}$$ we continue to divide $$c$$ by $$p_i$$. If $$c$$ becomes 1 then $$p_i$$ is the greatest prime factor, otherwise we increase $$p_i$$ by one. We know that any $$p_i$$ that divides $$c$$ is prime, because otherwise it would be of the form $$p_i = {q_1}^{a_i} \\dotsm {q_n}^{a_n}$$ and since $$q_i$$ is smaller than $$p_i$$, $$c$$ would have been already divided by it, so $$p_i$$ must be a prime number. c = 600851475143 p = 2 while c > 1: if c % p == 0: c /= p continue p += 1 p 6857", "# Prime factors of a random number Let $r$ be a uniformly random integer between $1$ and $N$, for some large enough $N$ (i.e., I'm only interested in the asymptotics). • What is the expected largest prime factor of $r$? Is there a good upper bound on this? • Let $t < N$ be such that $\\Pr\\Big[r$ has no prime factor > $t\\Big] = \\Omega(1/\\log^c(N))$ for some constant $c > 0$. How small can we make $t$? Is $t = \\operatorname{Li}_2(N)$ achievable?", "largest integer that $$mn$$ is necessarily divisible by is always true, we must assume that m is a multiple of 6. The largest integer that $$mn$$ is necessarily divisible by is $$3\\cdot3\\cdot3 = \\boxed{27}$$. Hope this helps, - PM Nov 30, 2018 #3 +701 0 I think my answer above might be a little confusing, so I'll re-state it. Odd number * odd number = Odd. Since n is a multiple of 9, the prime factorization of n (other than 0) has at least two 3s. Since m is a multiple of 3, the prime factorization of m (other than 0) has at least one 3. We cannot consider the possibility that m and n are even because they are not always even. The largest integer that mn is always divisible by is $$3\\cdot3\\cdot3 = \\boxed{27}$$, which is proven by prime factorization. Hope this helps more, - PM PartialMathematician Nov 30, 2018 #4 +701 0 Go try it out yourself. It really works for all m = 6 (mod 9) and n = 0 (mod 9). - PM PartialMathematician Nov 30, 2018 #5 +27664 +3 Here's another way of looking at 1). Nov 30, 2018 #6 +21980 +15 1) We have positive integers a, b and c such that a>b>c. When a, b and c are divided by", "= 3628799!7!5!3!$. You can ... 0 No. Consider $4^4 = 2^{8}$, which has no 3 in its prime factorization, while $4! = 1 \\cdot 2 \\cdot 3 \\cdot 4 = 2^3 \\cdot 3$. -2 Top 50 recent answers are included", "+ 3 \\cdot 5$: integers $6$, $10$, $15$ are nearly prime. In the fifth test case, $n=44=2 \\cdot 3 + 7 + 2 \\cdot 5 + 3 \\cdot 7$: integers $6$, $10$, $21$ are nearly prime. In the sixth test case, $n=100=2 + 2 \\cdot 5 + 3 \\cdot 11 + 5 \\cdot 11$: integers $10$, $33$, $55$ are nearly prime. In the seventh test case, $n=258=2 \\cdot 5 + 3 \\cdot 7 + 13 \\cdot 17 + 2 \\cdot 3$: integers $10$, $21$, $221$, $6$ are nearly prime.", "Find the largest prime number that divides $27! + 28!$. • $!$ denotes the factorial notation. For example, $10! = 1\\times2\\times3\\times\\cdots\\times10$.", "- 1}$ = 5 + 520 - 5 = 520", "largest factor of 900 is itself we have: $n+10=900 \\Longrightarrow \\boxed{n = 890}$ ~qwertysri987", "# IMAT 2019 Q57 [Common factor] In order to solve this question, we must find the factors of each of the number: 360 = 36 \\cdot 10 = 6 \\cdot 6 \\cdot 5\\cdot 2 = 3^2\\cdot 2 ^2 \\cdot 5\\cdot 2 = 2^3 \\cdot 3^2 \\cdot 5. 500 = 5 \\cdot 10^2 = 5 \\cdot (5\\cdot 2)^2 = 5 \\cdot 5^2\\cdot 2^2 = 2^2 \\cdot 5^3. 700 = 7 \\cdot 10^2 = 7\\cdot 5^2\\cdot 2^2 = 2^2 \\cdot 5^2\\cdot 7. From this we can see that all of the numbers have 2^2 \\cdot 5 in their prime factors, so the highest common factor is (D) 2^2 \\cdot 5 3 Likes", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $10a^{3}(a^{2}-a-6)$ $10a^{5}-10a^{4}-60a^{3}$ The prime factorization of $10a^{5}$ is $[2\\cdot 5\\cdot a\\cdot a\\cdot a]\\cdot a\\cdot a$ The prime factorization of $10a^{4}$ is $[2\\cdot 5\\cdot a\\cdot a\\cdot a]\\cdot a$ The prime factorization of $60a^{3}$ is$[2]\\cdot 2\\cdot 3\\cdot[5\\cdot a\\cdot a\\cdot a]$ The largest common factor is $2\\cdot 5\\cdot a\\cdot a\\cdot a=10a^{3}$ Factor out $10a^{3}$ from the given expression. $=10a^{3}(a^{2}-a-6)$", "# Huge Primes $97!\\; n + d$ Find the smallest integer $d > 1$, such that the number above is prime for an infinite number of integers $n$. Notation: $!$ denotes the factorial notation. For example, $10! = 1\\times2\\times3\\times\\cdots\\times10$. ×", "# How many 3 digit integers…? How many different 3-digit integers have the product of their digits equal to 4!? What is the largest of these integers? I know 4! Is 24 but still confusing to do this. How do I find the largest let alone how many? • First thing: How many ways are there to factor 24? How many of those are three single digit integer. How many ways are there to arrange those factors into three digit numbers. – fleablood Oct 29 '18 at 16:36 • Finding the largest is easier then counting them. What is the largest single digit factor of $24$. That will be the first digit. If you factor what is left what is the largest single digit factor of that. That will be the second digit. – fleablood Oct 29 '18 at 16:50 How many ways can you write $$24$$ as a product of one digit whole numbers? $$24 = 1 \\cdot 4 \\cdot 6$$ is one way, giving you the numbers $$146$$, $$164$$, $$416$$, $$461$$, $$614$$, and $$641$$. Try to find the others. Just do it. $$4! = 24$$ how many ways are there to factor $$24$$ into three factors. There's $$1*1*24$$ $$1*2*12$$ $$1*3*8$$ ..... etc. How many of those have only single digit factors? There's $$1*3*8$$ $$1*4*6$$ $$2*2*6$$ ...", "5k$$ is enough to guarantee that the largest prime factor in the denominator is $$\\geq 2k$$. If this largest prime appears with multiplicity $$2\\alpha$$ in the denominator, it only appears with multiplicity $$\\alpha$$ in the numerator. So either way the ratio can not be an integer. • Great! Experimentally, $n>5$ should be enough. (trivial typo: I think we are happy if $c(n)/c(m) \\not\\in\\mathbb{Z}$ if $\\max(25,m) < n < 2m$) – Martin Rubey Dec 6 '18 at 9:07", "+0 # Help plz +1 131 1 What is the largest integer $$n$$ such that $$7^n$$ divides $$1000!$$? Btw the answer isn't 163. off-topic Jul 21, 2019 #1 +507 +1 We need to find how many factors of seven all the integers from 1 to 1000 have. Each integer from 1 to 1000 can be a factor of 7, 49, or 343, giving it 1 factor of 7, 2 factors of 7, or 3 factors of 7. There are $$\\lfloor\\frac{1000}{7}\\rfloor=142$$ numbers that have 1 factor of 7. There are $$\\lfloor\\frac{1000}{49}\\rfloor=20$$ numbers that have 2 factors of 7. There are $$\\lfloor\\frac{1000}{343}\\rfloor=2$$ numbers that have 3 factors of 7. So, the answer is $$2\\cdot3+20\\cdot2+142\\cdot1$$. But, wait! We are counting numbers that have 2 factors in the numbers that have 1 factor of 7! And, we are counting numbers that have 3 factors in the numbers that have 2 factors and the numbers that have 1 factor! So, the answer is $$2\\cdot3+20\\cdot2+142\\cdot1-2-2-20=2+20+142=164$$! Jul 22, 2019", "# Number Theory Find the largest prime factor of $$2017! + 2018! + 2019!$$. Notation: $$!$$ is the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "# Inspired by Sal Gard $\\large 200! + (200!)! + 3$ Find the second smallest prime factor of the number above. Notation: $$!$$ denotes the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "approach and mostly on Fact 8! – Emmad Kareem Feb 5 '12 at 21:44 Given: $1\\cdot2\\cdot3\\cdot...a + 1\\cdot2\\cdot3\\cdot...b = a\\cdot a\\cdot a...$ Case $b>=a$: $= a! + a!\\cdot \\frac {b!}{a!} = a!\\cdot(1+\\frac {b!}{a!})$ Case $a>=b$: $= b! + b!\\cdot \\frac {a!}{b!} = b!\\cdot(1+\\frac {a!}{b!})$ Prime factorization shows the need for $a$ to be less than 5. That is, given that $a!$ or $b!$ is a factor of $a$, and that $x!$ has every prime factor less than or equal to $x$, $a$ contains every prime factor less than or equal to $a$ or $b$. If $b>=a$ then the only case for which $a$ contains all the prime factors of $a!$ is $a=a!$. In the latter case, $a!$ diverges far too quickly for $a^b$ to even be in the same ballpark. For example, $b=5$ gives $a>=30$ which implies $a!$ is more than 25 orders of magnitude larger than $a^b$. - Can you please elaborate on \"Prime factorization clearly shows the need for a to be equal to either a! or b!\" ? – the L Feb 5 '12 at 11:53 But what if that largest prime factor of $b!$ also divides $1+\\frac{a!}{b!}$? For example, when $b=3$ and $a=5$, then $3$ is the largest prime factor of $3!$ and it also divides $1+\\frac{5!}{3!}$..... – N. S. Feb 6 '12 at", "# Pretty Large Since $$x ^ { ab } - 1 = (x^a -1 ) \\times \\left( x^{ a (b-1) } + x^ { a (b-2) } + \\cdots + x^ a + 1 \\right)$$, we conclude that the number $$2^ {105} -1$$ has factors of $$2^3 - 1 = 7$$, $$2^5 - 1 = 31$$ and $$2^{7} - 1 = 127$$. Is $\\frac{ 2 ^ { 105} - 1 } { \\big( 2^3 -1 \\big) \\times \\big( 2^5 -1 \\big) \\times \\big(2 ^ 7 -1 \\big) }$ prime? ×" ]

[ "Free Version Easy # Prime Factors of an LCM ABSALG-104LTD What is the largest prime divisor of lcm$(2^5\\cdot 11^2, 5\\cdot 3^3\\cdot 11, 5^2 \\cdot 13)$? A $3$ B $5$ C $11$ D $13$", "+0 # Help now 0 104 2 $97$ is the largest prime less than $100$. What is the largest prime factor of $97!$ (97 factorial)? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, $5!=5\\cdot 4\\cdot3\\cdot2\\cdot 1= 120$.) Jan 15, 2020 #1 +109739 +1 97! = 97 itself is prime......so it is the largest prime factor of 97! Jan 15, 2020 #2 +36 +1 thanks lol Paresh247 Jan 15, 2020", "Largest prime factor Let $$p_i = {2, 3, 4, 5, \\dots}$$ we continue to divide $$c$$ by $$p_i$$. If $$c$$ becomes 1 then $$p_i$$ is the greatest prime factor, otherwise we increase $$p_i$$ by one. We know that any $$p_i$$ that divides $$c$$ is prime, because otherwise it would be of the form $$p_i = {q_1}^{a_i} \\dotsm {q_n}^{a_n}$$ and since $$q_i$$ is smaller than $$p_i$$, $$c$$ would have been already divided by it, so $$p_i$$ must be a prime number. c = 600851475143 p = 2 while c > 1: if c % p == 0: c /= p continue p += 1 p 6857", "# Prime factors of a random number Let $r$ be a uniformly random integer between $1$ and $N$, for some large enough $N$ (i.e., I'm only interested in the asymptotics). • What is the expected largest prime factor of $r$? Is there a good upper bound on this? • Let $t < N$ be such that $\\Pr\\Big[r$ has no prime factor > $t\\Big] = \\Omega(1/\\log^c(N))$ for some constant $c > 0$. How small can we make $t$? Is $t = \\operatorname{Li}_2(N)$ achievable?", "0% # Problem 3 ## Largest prime factor The prime factors of $13195$ are $5$, $7$, $13$ and $29$. What is the largest prime factor of the number $600851475143$? ## 最大质因数 $13195$的质因数包括$5$、$7$、$13$和$29$。 $600851475143$的最大质因数是多少？", "Free Version Easy # Prime factors of a GCD ABSALG-EIVKRT What is the largest prime divisor of gcd$(10,70,100)$? A There is none because gcd$(10,70,100)=1$ B $2$ C $5$ D $7$", "# Which is the Supreme Prime? Level pending Find the largest prime factor of :$${ 3 }^{ 12 }+{ 12 }^{ 12 }-{ 2 }\\cdot { 6 }$$ ×", "5k$$ is enough to guarantee that the largest prime factor in the denominator is $$\\geq 2k$$. If this largest prime appears with multiplicity $$2\\alpha$$ in the denominator, it only appears with multiplicity $$\\alpha$$ in the numerator. So either way the ratio can not be an integer. • Great! Experimentally, $n>5$ should be enough. (trivial typo: I think we are happy if $c(n)/c(m) \\not\\in\\mathbb{Z}$ if $\\max(25,m) < n < 2m$) – Martin Rubey Dec 6 '18 at 9:07", "# What is the largest prime factor of $55^{100}+55^{101}+55^{102}$ Can anyone help me on this? What is the largest prime factor of $55^{100}+55^{101}+55^{102}$? I know $55^{100}+55^{101}+55^{102}=55^{100}*(1+55+55^2)$, but I don't know how to move forward from here. • @SathasivamK That's neither largest nor prime. Sep 22 '16 at 18:08 • @robert Israel you're right Sep 22 '16 at 18:13 • The answer is 79 Sep 22 '16 at 18:15 $1+55+55^2 = 3081 = 3 \\cdot 13 \\cdot 79$. You can factor out $55^{100}$ from the expression. The largest prime factor of that number is $11$. Then it remains to factor $1+55+55^2$, which Prof. Israel did in his answer.", "# Size of largest prime factor It is well known and easy to prove that the smallest prime factor of an integer $n$ is at most equal to $\\sqrt n$. What can be said about the largest prime factor of $n$, denoted by $P_1(n)$? In particular: What is the probability that $P_1(n)>\\sqrt n$ ? More generally, what is the expected value of the size of $P_1(n)$, measured by $\\frac{\\log P_1(n)}{\\log n}$ ? • Size of the largest prime of $p$ is $p$ itself. Where $p$ is a prime.This is obvious, though. Apr 28, 2013 at 14:50 Take the negation and this is a very well-known question: what is the probability that all prime factors of $n$ are $\\le \\sqrt{n}$? The answer is known quite generally: for any real $u\\ge 1$, the probability that the prime factors of $n$ are $\\le n^{1/u}$ is given by the Dickman–de Bruijn rho function, defined by a delay-differential equation. For $u=2$ we have $\\rho(u) = 1-\\log 2$, as in Ross Millikan's answer, but there is a very easy calculation that gives this particular case: $$\\#\\{n \\le x: P_1(n) > \\sqrt{n}\\} = \\sum_{p} \\#\\{n \\le x, n < p^2: p \\mid n\\} = \\sum_{p\\le \\sqrt{x}} (p-1) + \\sum_{p > \\sqrt{x}} \\lfloor x/p \\rfloor \\\\ = x \\log 2 + O(x/\\log x),$$ where the main", "Find the largest prime number that divides $27! + 28!$. • $!$ denotes the factorial notation. For example, $10! = 1\\times2\\times3\\times\\cdots\\times10$.", "# Largest Prime Factor (PE3) using OOP I want to learn some C# syntax/paradigms (I'm more used to Java), and have been wanting to get a bit better at math as well, so I solved ProjectEuler3: Largest prime factor with the following small program in LINQPad. It gives the correct answer and completes in 0.05s-0.08s. For reference, here is the problem statement: The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? All feedback welcomed. In particular, are there some C# mistakes I am making or features I am missing on due to being very new at it? I'm also not that great at math at the moment, if you know of edge cases that I may have missed with the calculations below, please let me know so I can improve my math knowledge a bit as well! P.S.: Note the BigNumbersUtils.Sqrt extension method comes from this Stack Overflow answer. I am excluding it since it is not my code. it gives the same result as Math.Sqrt but for BigInteger type. void Main() { Console.WriteLine(\"ProjectEuler3: Largest prime factor\"); BigInteger testCase = 600851475143; ProjectEuler3 PE3 = new ProjectEuler3(testCase); Console.WriteLine(\"Prime factor of {0} is: {1}\", testCase, PE3.GetAnswer()); } class ProjectEuler3 { private BigInteger number; public ProjectEuler3(BigInteger number) {", "Greatest Common Divisor of two numbers If $ab=600$ how large can the greatest common divisor of $a$ and $b$ be? I am not sure if I should check for all factor multiples of $a$ and $b$ for this question. Please advise. We have $600=2^3\\cdot 5^2\\cdot 3$. To make the gcd large, we give a $2$ to each of $a$ and $b$, also a $5$. So the largest possible gcd is $10$. Remark: The idea generalizes. Let $n$ have prime power factorization $$n=p_1^{d_1}p_2^{d_2}\\cdots p_k^{d_k}.$$ Let $e_i=\\lfloor d_i/2\\rfloor$, where $\\lfloor x\\rfloor$ is the greatest integer that is $\\le x$. Then the greatest possible gcd of $a$ and $b$, where $ab=n$, is $$p_1^{e_1}p_2^{e_2}\\cdots p_k^{e_k}.$$ • You are welcome. For the maximum, we distribute the power of any prime $p$ as evenly as possible between $a$ and $b$. – André Nicolas Jun 10 '13 at 2:00 $$ab=600=2^3\\cdot 3\\cdot 5^2$$ The greatest common divisor of $a,b$ will have the greatest number of common prime factors that you can arrange.", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $10a^{3}(a^{2}-a-6)$ $10a^{5}-10a^{4}-60a^{3}$ The prime factorization of $10a^{5}$ is $[2\\cdot 5\\cdot a\\cdot a\\cdot a]\\cdot a\\cdot a$ The prime factorization of $10a^{4}$ is $[2\\cdot 5\\cdot a\\cdot a\\cdot a]\\cdot a$ The prime factorization of $60a^{3}$ is$[2]\\cdot 2\\cdot 3\\cdot[5\\cdot a\\cdot a\\cdot a]$ The largest common factor is $2\\cdot 5\\cdot a\\cdot a\\cdot a=10a^{3}$ Factor out $10a^{3}$ from the given expression. $=10a^{3}(a^{2}-a-6)$", "# Math Help - Factorials 1. ## Factorials what is the largest two-digit prime factor of $200!$ divided by $(100! 100!)$ 2. you have to cancel this one down, not sure if I can get the answer but I can kick it off for you.... $\\frac{200!}{100!\\times 100!}$ $\\frac{200\\times 199\\times 198 \\times \\cdots101 \\times 100!}{100!\\times 100!}$ $100!$ will cancel $\\frac{200\\times 199\\times 198 \\times \\cdots 101 }{100!}$ expanding the bottom $\\frac{200\\times 199\\times 198 \\times \\cdots 101 }{100\\times 99\\times 98 \\times \\cdots 1 }$ Now first 1st term of 2 should cancel on the bottom $\\frac{2\\times 199\\times 2 \\times \\cdots 101 }{ 99\\times 97 \\times \\cdots 1 }$ There is 50 of them $\\frac{2^{50} \\times 199\\times 197 \\times \\cdots 101 }{ 99\\times 97 \\times \\cdots 1 }$ leaving the bottom in the form $\\frac{2^{50} \\times 199\\times 197 \\times \\cdots 101 }{ 99!! }$ not sure where to go from here...", "# Number Theory Find the largest prime factor of $$2017! + 2018! + 2019!$$. Notation: $$!$$ is the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "Ques: The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? I came up with the following code. This did compile in VC++ Express Edition without any errors/warnings. But i only get a blank output screen. Someone plz point out where i went wrong? #include <iostream> #include <vector> using namespace std; class factor { public: /*_int64*/ long long target; vector<long long> pno; vector<long long> ::iterator it; void prime(); }; void factor::prime() { target = 600851475143; int flag = 0; pno.push_back(2); pno.push_back(3); pno.push_back(5); pno.push_back(7); // Pushing prime no s within 10 into the vector for (long long i = 11; i<target; i++) { if(i%2||i%3||i%5||i%7) continue; else pno.push_back(i); // Checking if the no has factors other than itself and 1, then pushing it into the vector. } cout << \"Prime No\" << endl << endl; for (it = pno.begin(); it != pno.end(); it++) { cout << *it << \"\\t\"; } cout << endl << endl; // Printing the vector for reference for (int j = (pno.size()-1); j>0; j--) { if(target%(pno[j])) { cout << \"Largest prime factor of the no '600851475143' : \" << pno[j] << endl << endl; break; } } // Checking the vector of prime no s if it a factor of the target no. Since", "# Size of largest prime factor It is well known and easy to prove that the smallest prime factor of an integer $n$ is at most equal to $\\sqrt n$. What can be said about the largest prime factor of $n$, denoted by $P_1(n)$? In particular: What is the probability that $P_1(n)>\\sqrt n$ ? More generally, what is the expected value of the size of $P_1(n)$, measured by $\\frac{\\log P_1(n)}{\\log n}$ ? - Size of the largest prime of $p$ is $p$ itself. Where $p$ is a prime.This is obvious, though. – Inceptio Apr 28 '13 at 14:50 ## 3 Answers Take the negation and this is a very well-known question: what is the probability that all prime factors of $n$ are $\\le \\sqrt{n}$? The answer is known quite generally: for any real $u\\ge 1$, the probability that the prime factors of $n$ are $\\le n^{1/u}$ is given by the Dickman–de Bruijn rho function, defined by a delay-differential equation. For $u=2$ we have $\\rho(u) = 1-\\log 2$, as in Ross Millikan's answer, but there is a very easy calculation that gives this particular case: $$\\#\\{n \\le x: P_1(n) > \\sqrt{n}\\} = \\sum_{p} \\#\\{n \\le x, n < p^2: p \\mid n\\} = \\sum_{p\\le \\sqrt{x}} (p-1) + \\sum_{p > \\sqrt{x}} \\lfloor x/p \\rfloor \\\\ = x \\log 2 +", "large prime, or has large prime factors. Workspace is internally allocated by C06PWF. The total size of these arrays is approximately proportional to $mn$.", "# Huge Primes $97!\\; n + d$ Find the smallest integer $d > 1$, such that the number above is prime for an infinite number of integers $n$. Notation: $!$ denotes the factorial notation. For example, $10! = 1\\times2\\times3\\times\\cdots\\times10$. ×" ]

[ "together. Find the probability of getting at least two heads. Ex. 16.1 | Q 3.3 | Page 20 Three coins are tossed together. Find the probability of getting at least one head and one tail Ex. 16.1 | Q 3.4 | Page 20 Three coins are tossed together. Find the probability of getting no tails Ex. 16.1 | Q 4 | Page 20 A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number. Ex. 16.1 | Q 5 | Page 20 Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10. Ex. 16.1 | Q 6.01 | Page 20 A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is a black king. Ex. 16.1 | Q 6.02 | Page 20 A card is drawn at random from a pack of 52 cards. Find the probability that card drawn is either a black card or a king Ex. 16.1 | Q 6.03 | Page 20 A card is drawn at random from a pack of 52 cards. Find the probability that card drawn is black and a king Ex. 16.1 | Q 6.04 | Page 20 A", "or tails because there is one of each.", "probability of getting at least one head is a) 1/2 b) 1/4 c) 3/4 d) none", "is the probability of not getting two heads? Be sure...", "each. There is a penalty shoot-out with another team, team B (of 11 players), in which every player from each team takes exactly one shot, so that 22 shots are to be taken. Every player in team B scores with probability 0.64. The team that scores the most goals wins. Which team is expected to win? Justify your answer Upper level probability ### What is the probability of tossing at least 1 tail if you toss 3 coins at once? Upper level probability", "#### If I toss a coin 3 times and get head each time, should I expect a tail to have a higher chance in the 4th toss? Give reason in support of your answer. Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. No, because we get head or tail after tossing a coin that is the probability of both outcomes is $\\frac{1}{2}$ . Hence tail is not have higher chance than head. Both are have equal chance.", "numbers on the upper-most faces of two dice is: less than 6 Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting: Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting: Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting: Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting all tails Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting: at least one tail Two dice are thrown simultaneously. What is the probability that: 4 will not come up either time? 4 will come up at least once? Cards marked with numbers 1, 2, 3 ……… 20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is: a prime number Cards marked with numbers 1, 2, 3 ……… 20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is: divisible by 3 Cards marked with numbers 1, 2, 3 ……… 20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is:", "seems to be in two states of probability at the same time?", "count the number of tails. Randomly pull off from the group into a separate pile the number of coins equal to the number of tails you just counted. Turn over every coin in this new pile. Count the number of tails in each pile. You got the same number both times! Why? Marilyn Vos Savant posed a similar problem: Say that a hundred pennies are on a table. Ninety of them are heads. With your eyes closed, can you separate all the coins into two groups so that each group has the same number of tails? Savant’s solution is to pull any random 10 coins from the 100 and make a second pile. Turn all the coins in the new pile over, et voila! Both piles have an equal number of tails. While Savant’s approach is much more prescriptive than mine, both solutions work. Every time. WHY? THIS IS STRANGE: You have no idea the state (heads or tails) of any of the coins you pull into the second pile. It’s counterintuitive that the two piles could ever contain the same number of tails. Also, flipping the coins in the new pile seems completely arbitrary, and yet after any random pull & flip, the two resulting piles always hold the same number of tails. Enter the power (and", "the probability of getting at least one head. (2013OD) Solution: S = {HH, HT, TH, TT}, i.e., 4 ∴ P (atleast one head) = $\\frac{3}{4}$ Question 13. Three coins are tossed simultaneously. Find the probability of getting exactly two heads. (2013OD) Solution: S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} = 8 P(exactly two heads) = $\\frac{3}{8}$ Question 14. Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail. (2014D) Solution: S = {HH, HT, TH, TT}, i.e., 3 P(at least one tail) = $\\frac{3}{4}$ Question 15. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king. (2011OD) Solution: P (neither an ace nor a king) = 1 – P (either an ace or a king) = 1 – [P (an ace) + P (a king)] Question 16. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a king nor a queen. (2013D) Solution: P (neither a king nor a queen) = 1 – P (king or queen) Question 17. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that", "random between heads and tails.", "the group into a separate pile the number of coins equal to the number of tails you just counted. Turn over every coin in this new pile. Count the number of tails in each pile. You got the same number both times! Why? Marilyn Vos Savant posed a similar problem: Say that a hundred pennies are on a table. Ninety of them are heads. With your eyes closed, can you separate all the coins into two groups so that each group has the same number of tails? Savant’s solution is to pull any random 10 coins from the 100 and make a second pile. Turn all the coins in the new pile over, et voila! Both piles have an equal number of tails. While Savant’s approach is much more prescriptive than mine, both solutions work. Every time. WHY? THIS IS STRANGE: You have no idea the state (heads or tails) of any of the coins you pull into the second pile. It’s counterintuitive that the two piles could ever contain the same number of tails. Also, flipping the coins in the new pile seems completely arbitrary, and yet after any random pull & flip, the two resulting piles always hold the same number of tails. Enter the power (and for young people, the mystery) of algebra to generalize", "simultaneously and probability for rolling three dice simultaneously in rolling dice probability worksheet. 1. A die is thrown 350 times and the • ### Probability of Tossing Three Coins | Tossing or Flipping Three Coins Here we will learn how to find the probability of tossing three coins. Let us take the experiment of tossing three coins simultaneously: When we toss three coins simultaneously then the possible Probability", "off from the group into a separate pile the number of coins equal to the number of tails you just counted. Turn over every coin in this new pile. Count the number of tails in each pile. You got the same number both times! Why? Marilyn Vos Savant posed a similar problem: Say that a hundred pennies are on a table. Ninety of them are heads. With your eyes closed, can you separate all the coins into two groups so that each group has the same number of tails? Savant’s solution is to pull any random 10 coins from the 100 and make a second pile. Turn all the coins in the new pile over, et voila! Both piles have an equal number of tails. While Savant’s approach is much more prescriptive than mine, both solutions work. Every time. WHY? THIS IS STRANGE: You have no idea the state (heads or tails) of any of the coins you pull into the second pile. It’s counterintuitive that the two piles could ever contain the same number of tails. Also, flipping the coins in the new pile seems completely arbitrary, and yet after any random pull & flip, the two resulting piles always hold the same number of tails. Enter the power (and for young people, the mystery) of algebra", "the coins in the new pile over, et voila! Both piles have an equal number of tails. While Savant’s approach is much more prescriptive than mine, both solutions work. Every time. WHY? THIS IS STRANGE: You have no idea the state (heads or tails) of any of the coins you pull into the second pile. It’s counterintuitive that the two piles could ever contain the same number of tails. Also, flipping the coins in the new pile seems completely arbitrary, and yet after any random pull & flip, the two resulting piles always hold the same number of tails. Enter the power (and for young people, the mystery) of algebra to generalize a problem, supporting an argument that holds for all possibilities simultaneously. HOW IT WORKS: The first clue to this is the misdirection in Savant’s question. Told that there are 90 heads, you are asked to make the number of tails equivalent. In both versions, the number of TAILS in the original pile is the number of coins pulled into the second pile. This isn’t a coincidence; it’s the key to the solution. In any pile of randomly flipped coins (they needn’t be all or even part pennies), let N be the number tails. Create your second pile by pulling a random coins from the initial pile.", "the coins in the new pile over, et voila! Both piles have an equal number of tails. While Savant’s approach is much more prescriptive than mine, both solutions work. Every time. WHY? THIS IS STRANGE: You have no idea the state (heads or tails) of any of the coins you pull into the second pile. It’s counterintuitive that the two piles could ever contain the same number of tails. Also, flipping the coins in the new pile seems completely arbitrary, and yet after any random pull & flip, the two resulting piles always hold the same number of tails. Enter the power (and for young people, the mystery) of algebra to generalize a problem, supporting an argument that holds for all possibilities simultaneously. HOW IT WORKS: The first clue to this is the misdirection in Savant’s question. Told that there are 90 heads, you are asked to make the number of tails equivalent. In both versions, the number of TAILS in the original pile is the number of coins pulled into the second pile. This isn’t a coincidence; it’s the key to the solution. In any pile of randomly flipped coins (they needn’t be all or even part pennies), let N be the number tails. Create your second pile by pulling a random coins from the initial pile.", "Equally Likely Outcomes Equally likely outcomes are of those events in a sample space who have same chance or same likelihood of occurrence. When all events of a sample space are having same chances of their occurrence then they are being called equally likely events. Events such as rolling a die, tossing a coin, choosing a card from a deck of cards or choosing a ball from a bag are all equally likely outcomes. Definition Those events who have equal likelihood of their occurrence are known as equally likely outcomes. There are events such as rolling a die where all sides are equal and hence, the likelihood of getting any side is equal to the likelihood of getting any other side. But if there is a cubical shaped toy and we throw it then the likelihood of getting each side is not same as all sides differ in area. Tossing a Coin A coin has got two sides, head and tail. If we toss a coin then there is an equal likelihood of getting a head or a tail. Hence, tossing a coin is an example of equally likely outcomes. For the experiment of tossing a coin, Sample space, S = {H, T} Probability of getting a head, $P(A)$ = $\\frac{1}{2}$ Probability of getting a head, $P(B)$ =", "two heads is $$\\frac{3}{8}$$ $$\\frac{1}{2}$$ $$\\frac{1}{4}$$ None of these ## Questions 50 of 50 Question:One coin is thrown 100 times. The probability of coming tail in odd number $$\\frac{1}{2}$$ $$\\frac{1}{8}$$ $$\\frac{3}{8}$$", "distributions, it is quite common to have one tail of the distribution considerably longer or drawn out relative to the.", "# Math Help - Coin tossing problem. 1. ## Coin tossing problem. Q.Four coins are tossed simatanously, in how many distinct ways these coins can show up?? I think it is going to be 16.. Do u guys agree with me??? 2. ## Re: Coin tossing problem. Originally Posted by nasir31971 Q.Four coins are tossed simatanously, in how many distinct ways these coins can show up?? I think it is going to be 16.. Do u guys agree with me??? Are they indistinguishable? If they are distinguishable then there are 2^4 ways they can fall, if indistinguishable there are 4 ways the can fall. CB 3. ## Re: Coin tossing problem. assuming you mean heads and tails it is $2^4$ 4. ## Re: Coin tossing problem. Thanks.....i also go with 2^4 5. ## Re: Coin tossing problem. Hello, nasir31971! Four coins are tossed simultanously. In how many distinct ways these coins can show up? I think it is going to be 16. Do u guys agree with me? CaptainBlack raised a legitimate point. If the four coins are distinct -- say, a penny, nickel, dime, and a quarter -- . . then there are: . $2^4 = 16$ outcomes. If the four coins are identical, we may be counting the Heads and Tails only. In this case, there" ]

[ "tail $\\mathbb{P}$(winning| One head and one tail chosen) =$\\frac{1}{2}\\times\\frac{1}{n_3+1}$If $n_3 < \\frac{N}{2}-1$ then the probability of winning is greater than $\\frac{1}{N}$ and it is sensible to opt for one head and one tail. Interestingly if $n_1=n_2=\\frac{N}{4}$ and $n_3=\\frac{N}{2}$ then there is no option that gives a probability of winning of less than $\\frac{1}{N}$. In this case the probabilities of winning are: 1. $\\frac{1}{N+4}$ 2. $\\frac{1}{N+4}$ 3. $\\frac{1}{N+2}$ Hence the best option is to opt for one head and one tail. In general if one is playing Heads and Tails one should, at the outset, consider the proportions of players opting for the three possibilities. Imagine, for example, if there are 1000 people taking part and everyone else has opted for 1 head and 1 tail. If you opt for two heads you have a one in four chance of winning. Whereas if you opt for one head and one tail you have a 50% chance of not being eliminated and would still have to rely on the all those people being eliminated later on. In general if less than 25% of the players are opting for two heads then opt for two heads, if less than 25% of the players are opting for two tails then opt for two tails and if less than 50% of the players", "is the probability of getting at least one head? (a) $\\frac{103}{300}$ (b) $\\frac{39}{100}$ (c) $\\frac{11}{15}$ (d) $\\frac{4}{15}$", "the event of getting EXACTLY ONE HEAD. Let $B$ be the event of getting EXACTLY TWO HEADS. Let C be the event of getting EXACTLY THREE HEADS. Then the probability you are looking for is $P\\left(A\\right)+P\\left(B\\right)+P\\left(C\\right)$. Notice we don't subtract anything because the events are distinct (no overlaps). $P\\left(A\\right)=3\\left(1/2{\\right)}^{3}=3/8$, $P\\left(B\\right)=3\\left(1/2{\\right)}^{3}=3/8$, and $P\\left(C\\right)=\\left(1/2{\\right)}^{3}=1/8$. The sum is 7/8, which is the same as the other way, which gave $1-\\left(1/2{\\right)}^{3}=7/8$.", "tails and getting three heads which is $\\left(\\frac{2}{3}\\right)^3 + \\left(\\frac{1}{3}\\right)^3 = \\frac{8}{27} + \\frac{1}{27} = \\frac{9}{27} = \\frac{27}{81}$. The probability of winning Game B is the sum of the probabilities of getting two heads and getting two tails squared. This gives us $\\left(\\left(\\frac{2}{3}\\right)^2 + \\left(\\frac{1}{3}\\right)^2 \\right)^2 = \\left(\\frac{5}{9}\\right)^2 = \\frac{25}{81}$. The probability of winning Game A is $\\frac{27}{81}$ and the probability of winning Game B is $\\frac{25}{81}$, so the answer is $\\boxed{\\textbf{(D)}}$", "{H, T} Number of possible outcomes = 2 Number of favorable outcomes = 1 => Probability of getting tail = $\\frac{Number of favorable outcomes}{Total number of possible outcomes}$ = $\\frac{1}{2}$ => Probability of getting tail is $\\frac{1}{2}$. Question 1: Find the probability of getting one while tossing a die. (Answer: $\\frac{1}{6}$)", "# An unbiased coin is tossed six times. The probability of obtaining at least one head and at least one tail is? 1. $$\\rm \\frac{1}{31}$$ 2. $$\\rm \\frac{31}{32}$$ 3. $$\\rm \\frac{32}{31}$$ 4. $$\\rm \\frac{1}{32}$$ Option 2 : $$\\rm \\frac{31}{32}$$ ## Detailed Solution Calculation: Required probability = 1 - [P(No head) + P(No Tail)] = 1 - $$\\rm \\left \\{ \\frac{1}{2^{6}} + \\frac{1}{2^{6}} \\right \\}$$ = 1 - $$\\rm \\frac{1}{2^{5}}$$ Required probability = $$\\rm 1 - \\frac{1}{32}$$ = $$\\rm \\frac{31}{32}$$ The probability of obtaining at least one head and at least one tail is $$\\rm \\frac{31}{32}$$", "3. Probability of getting 0 tails, P(X=0) = $\\frac{1}{8}$ Probability of getting 1 tails, P(X=1) = $\\frac{3}{8}$ Probability of getting 2 tails, P(X=2) = $\\frac{3}{8}$ Probability of getting 3 tails, P(X=3) = $\\frac{1}{8}$ Probability distribution can be given as: X 0 1 2 3 P(X) $\\frac{1}{8}$ $\\frac{3}{8}$ $\\frac{3}{8}$ $\\frac{1}{8}$", "= \\int_0^1{\\theta^{y_2}(1-\\theta)^{1-y_2}2\\theta d\\theta}$$ $$f(y_2|Y_1 = 1) = \\int_0^1{2\\theta^{y_2 + 1}(1-\\theta)^{1-y_2}d\\theta}$$ We could work this out in a more general form, but in this case, $Y_2$ has to take the value $0$ or $1$. The next flip is either going to be heads or tails so it’s easier to just plop in a particular example. $$P(Y_2 = 1|Y_1 = 1) = \\int_0^1{2\\theta^2d\\theta} = \\frac{2}{3}$$ $$P(Y_2 = 0 | Y_1 = 1) = 1 - P(Y_2 = 1 | Y_1 = 1) = 1 - \\frac{2}{3} = \\frac{1}{3}$$ We can see here that the posterior is a combination of the information in the prior and the information in the data. In this case, our prior is like having two data points, one head and one tail. Saying we have a uniform prior for $\\theta$ is equivalent in an information sense as saying we have observed one head and one tail. So then when we observe one head, it’s like we now have seen two heads and one tail. So our predictive distribution for the second flip says if we have two heads and one tail, then we have a $\\frac{2}{3}$ probability of getting another head and a $\\frac{1}{3}$ probability of getting another tail. Binomial Likelihood with Uniform Prior Likelihood of y given theta is $$f(y|\\theta) = \\theta^{\\sum{y_i}}(1-\\theta)^{n - \\sum{y_i}}$$ Our", "Home > A2C > Chapter 7 > Lesson 7.1.2 > Problem7-27 7-27. Note that it says one head and then one tail, meaning that the head has to come first, and the tail second. $\\frac{1}{4}$ See part (a). Is the probability of either event happening more likely or less likely than one event happening exclusively?", "which is sure (or certain) to occur is 1. Such an event is called a sure event or a certain event $P(X) =1$ 4. Probability of any event can be as $0 \\geq P(E) \\geq 1$ ## Solved Examples Question 1 A coin is tossed 1000 times; we get 499 times head and 501 times tail a. What is the empirical probability of getting head ? b. What is the empirical probability of getting tail? c. Does the sum of above two probability equals 1? Solution a. $Empirical \\; probability \\; of \\; getting \\; head=\\frac {(No \\; of \\; trails \\; which \\; heads \\; came)}{(Total \\; Number \\; of \\; trials)}$ So empirical or experimental probability of getting head P(H) is calculated as =.499 b. $Empirical \\; probability \\; of \\; getting \\; Tail=\\frac {(No \\; of \\; trails \\; which \\; tail \\; came)}{(Total \\; Number \\; of \\; trials)}$ So empirical or experimental probability of getting head P(T) is calculated as =.501 c. Now P(H) +P(T)= .499+.501=1 As tail and head are the only possible outcome,the sum of probabilities is 1 Question 2 The record of a weather station shows that out of the past 250 consecutive days, its weather forecasts were correct 175 times. (i) What is the probability that on a given", "is $\\left(\\frac{1}{2}\\right)^4$. The probability of $4$ tails then a head is $\\left(\\frac{1}{2}\\right)^5$. And so on. Add up. The required probability is $$\\left(\\frac{1}{2}\\right)^3 +\\left(\\frac{1}{2}\\right)^4+\\left(\\frac{1}{2}\\right)^5+\\cdots.$$ This is an infinite geometric series, which can be summed in the usual way. We get $\\frac{1}{4}$. The probability of two tails in a row is $\\frac{1}{2}\\times \\frac{1}{2}$. That ($1/4$) is the answer to the question you asked. No complementary event needed. You need at least $3$ tosses exactly when the first two tosses are tails. – André Nicolas Apr 17 '12 at 17:37", "a head, and $0$ otherwise. Let $X_2=1$ if we get a tail, and $0$ otherwise. Then $E(X_1)=E(X_2)=\\frac{1}{2}$. However, $X_1X_2$ is always $0$, so $E(X_1X_2)=0$. Thus, in this example, $E(X_1X_2)\\ne E(X_1)E(X_2)$.", "# How do you find the probability of obtaining at least one tail when a coin is tossed five times? May 1, 2017 $1 - \\frac{1}{32} = \\frac{31}{32}$ #### Explanation: 'At least one tail' means that there can be one, or two or three or four or five tails. The only option that is not included is five heads. The sum of all the probabilities is always $1$. $P \\left(\\text{at least one tail\") = 1 - P(\"no tail}\\right)$ $P \\left(\\text{no tail\") = P(\"all heads\") = P(\"H,H,H,H,H}\\right)$ In any throw the probability of a tail or a head is $\\frac{1}{2}$ $P \\left(\\text{H,H,H,H,H}\\right) = \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2} = {\\left(\\frac{1}{2}\\right)}^{5} = \\frac{1}{32}$ $P \\left(\\text{at least one tail}\\right) = 1 - \\frac{1}{32} = \\frac{31}{32}$", "Probability that the no. of heads equals the no. of tails $=\\frac{\\binom42}{2^4}=\\frac{6}{16}$ • Probability that the no. of heads exceeds the no. of tails $=\\sum\\limits_{n=3}^{4}\\frac{\\binom4n}{2^4}=\\frac{5}{16}$", "of obtaining two and three heads. You can apply the binomial probability distribution to obtain the answer, but we can simplify this. The probability of obtaining at least one head, $P(B)$, is also equivalent to one minus the probability of obtaining no heads. The probability of obtaining no heads, say $P(C)$, is equivalent to the probability of obtaining three tails, and it can be calculated as follows: $$P(C) = \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2}$$ Then, we calculate $P(B)$: $$P(B) = 1 - P(C) = 1 - \\frac{1}{8} = \\frac{7}{8}$$ And finally, $$P (A \\cup B) = P(B) = \\frac{7}{8}$$ 1. Getting a head or a 4 does not include getting both a head AND a four. Say the probability of getting a head or a 4 is $P(D)$. Besides adding together the probability of obtaining a head and the probability of obtaining a 4 (which is what you did), you need to subtract the probability of obtaining a head and a four. $$P(D) = \\frac{1}{2} + \\frac{1}{6} - \\frac{1}{2} \\times \\frac{1}{6} = \\frac{7}{12}$$ • Genius! Thanks a lot! – Ben Apr 16 '17 at 4:11", "### Home > MC1 > Chapter 10 > Lesson 10.1.4 > Problem10-45 10-45. $\\frac{1}{2}\\left( \\frac{1}{4}\\right)=\\frac{1}{8}$ The probability of getting a head is one-half. The probability of getting a spade is one-fourth. The probability of getting a tail is one-half. The probability of getting a King is one-thirteenth. Refer to parts (a) and (b) for help.", "## Precalculus (6th Edition) Blitzer The probability of getting a tail in all the seven tosses is $\\frac{1}{128}$. We know that the probability of getting a tail in a single toss of the coin is as follows: Number of favorable out comes $n\\left( e \\right)=1$ (tail) Number of total out comes $n\\left( s \\right)=2$ (head and tail) $P\\left( \\text{tails} \\right)=\\frac{1}{2}$ And the probability of getting a tail in all the seven tosses is given below, \\begin{align} & P\\left( \\text{seven tails} \\right)=\\frac{1}{2}\\times \\frac{1}{2}\\times \\frac{1}{2}\\times \\frac{1}{2}\\times \\frac{1}{2}\\times \\frac{1}{2}\\times \\frac{1}{2} \\\\ & =\\frac{1}{128} \\end{align} Thus, the probability that a tail will come in all the seven tosses is $\\frac{1}{128}$.", "## College Algebra (10th Edition) 3 Heads → $\\frac{1}{8}$ 2 Heads and 1 Tail → $\\frac{3}{8}$ 2 Tails and 1 Head → $\\frac{3}{8}$ 3 Tails → $\\frac{1}{8}$ The possible outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT Therefore the probability model can be 3 Heads → $\\frac{1}{8}$ 2 Heads and 1 Tail → $\\frac{3}{8}$ 2 Tails and 1 Head → $\\frac{3}{8}$ 3 Tails → $\\frac{1}{8}$", "other face is tails is $1/4$ $1/3$ $1/2$ $2/3$ Out of all the $2$-digit integers between $1$ and $100$, a $2$-digit number has to be selected at random. What is the probability that the selected number is not divisible by $7$? $13/90$ $12/90$ $78/90$ $77/90$ The probability that a student knows the correct answer to a multiple choice question is $\\dfrac{2}{3}$ . If the student does not know the answer, then the student guesses the answer. The probability of the guessed answer being correct is $\\dfrac{1}{4}$. Given that the student has answered the question ... the correct answer is $\\dfrac{2}{3} \\\\$ $\\dfrac{3}{4} \\\\$ $\\dfrac{5}{6} \\\\$ $\\dfrac{8}{9}$ Let $X$ be a normal random variable with mean $1$ and variance $4$. The probability $P \\left \\{ X \\right.<\\left. 0 \\right \\}$ is $0.5$ greater than zero and less than $0.5$ greater than $0.5$ and less than $1.0$ $1.0$", "$\\frac{1}{8}\\left(\\binom{3}{2}+\\binom{3}{3}\\right)=\\frac 12$. - The simplest is by symmetry. The chance of at least two heads equals the chance of at least two tails, and if you add them you get exactly $1$ because one or the other has to happen. Thus the chance is $\\frac 12$. This approach is not always available. - The probability of exactly $2$ heads in $3$ toss is $\\binom 3 2 \\left(\\frac12\\right)^2\\left(1-\\frac12\\right)^{3-2}=\\frac38$ The probability of exactly $3$ heads in $3$ toss is $\\binom 3 3 \\left(\\frac12\\right)^3\\left(1-\\frac12\\right)^{3-3}=\\frac18$ -" ]

[ "# Get Answers to all your Questions #### When we toss a coin, there are two possible outcomes - Head or Tail. Therefore, the probability of each outcome is $\\frac{1}{2}$ . Justify your answer. Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Total cases when we toss a coin = 2(H, T) Probability = $\\frac{Number\\, of\\, favourable\\ cases, }{Total\\, number\\, of\\, cases}$ Probability of head = $\\frac{1}{2}$ Probability of tail = $\\frac{1}{2}$ Hence the probability of each outcome is $\\frac{1}{2}$ .", "the probability of getting at least one head. (2013OD) Solution: S = {HH, HT, TH, TT}, i.e., 4 ∴ P (atleast one head) = $\\frac{3}{4}$ Question 13. Three coins are tossed simultaneously. Find the probability of getting exactly two heads. (2013OD) Solution: S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} = 8 P(exactly two heads) = $\\frac{3}{8}$ Question 14. Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail. (2014D) Solution: S = {HH, HT, TH, TT}, i.e., 3 P(at least one tail) = $\\frac{3}{4}$ Question 15. A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king. (2011OD) Solution: P (neither an ace nor a king) = 1 – P (either an ace or a king) = 1 – [P (an ace) + P (a king)] Question 16. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is neither a king nor a queen. (2013D) Solution: P (neither a king nor a queen) = 1 – P (king or queen) Question 17. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability that", "tail $\\mathbb{P}$(winning| One head and one tail chosen) =$\\frac{1}{2}\\times\\frac{1}{n_3+1}$If $n_3 < \\frac{N}{2}-1$ then the probability of winning is greater than $\\frac{1}{N}$ and it is sensible to opt for one head and one tail. Interestingly if $n_1=n_2=\\frac{N}{4}$ and $n_3=\\frac{N}{2}$ then there is no option that gives a probability of winning of less than $\\frac{1}{N}$. In this case the probabilities of winning are: 1. $\\frac{1}{N+4}$ 2. $\\frac{1}{N+4}$ 3. $\\frac{1}{N+2}$ Hence the best option is to opt for one head and one tail. In general if one is playing Heads and Tails one should, at the outset, consider the proportions of players opting for the three possibilities. Imagine, for example, if there are 1000 people taking part and everyone else has opted for 1 head and 1 tail. If you opt for two heads you have a one in four chance of winning. Whereas if you opt for one head and one tail you have a 50% chance of not being eliminated and would still have to rely on the all those people being eliminated later on. In general if less than 25% of the players are opting for two heads then opt for two heads, if less than 25% of the players are opting for two tails then opt for two tails and if less than 50% of the players", "cards in a deck $m = 12$ ($3$ cards, viz, Jack, Queen, and King in each of the four suits). Therefore, the probability of picking a face card = $\\frac {12}{52} = \\frac{3}{13}$ Difference Between Theoretical Probability and Experimental Probability The following are the main differences between theoretical probability and experimental probability. Practice Problems 1. Define theoretical probability. 2. Find the probability of getting two tails when a pair of coins are tossed simultaneously. 3. Find the probability of getting one head and one tail when a pair of coins are tossed simultaneously. 4. Find the probability of getting three heads when three coins are tossed simultaneously. 5. Find the probability of getting one head when three coins are tossed simultaneously. 6. Find the probability of getting two tails when three coins are tossed simultaneously. 7. Find the probability of getting a prime number when a die is rolled. 8. Find the probability of getting an odd number when a die is rolled. 9. Find the probability of getting a doublet when a pair of dice is rolled. 10. Find the probability of getting a sum of $11$ when a pair of dice is rolled. 11. Find the probability of picking a black jack from a well-shuffled deck of cards. 12. Find the probability of picking a card", "that PROBABILITY = Hence probability of getting at least two head when three coins are tossed simultaneously is equal to Hence the correct option is #### Question 5: In a single throw of a die, the probability of getting a multiple of 3 is (a) $\\frac{1}{2}$ (b) $\\frac{1}{3}$ (c) $\\frac{1}{6}$ (d) $\\frac{2}{3}$ GIVEN: A dice is thrown once TO FIND: Probability of getting a multiple of 3. Total number on a dice is 6. Numbers which are on multiple of 3 are 3 and 6 Total number of numbers which are multiple of 3 is 2 We know that PROBABILITY = Hence probability of getting a number which are multiple of 3 is equal to Hence the correct option is #### Question 6: The probability of guessing the correct answer to a certain test questions is $\\frac{x}{12}$. If the probability of not guessing the correct answer to this question is $\\frac{2}{3}$, then x = (a) 2 (b) 3 (c) 4 (d) 6 GIVEN: Probability of guessing a correct answer to a certain question is Probability of not guessing a correct answer to a same question TO FIND: The value of x CALCULATION: We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1. If E is an event", "the event of getting EXACTLY ONE HEAD. Let $B$ be the event of getting EXACTLY TWO HEADS. Let C be the event of getting EXACTLY THREE HEADS. Then the probability you are looking for is $P\\left(A\\right)+P\\left(B\\right)+P\\left(C\\right)$. Notice we don't subtract anything because the events are distinct (no overlaps). $P\\left(A\\right)=3\\left(1/2{\\right)}^{3}=3/8$, $P\\left(B\\right)=3\\left(1/2{\\right)}^{3}=3/8$, and $P\\left(C\\right)=\\left(1/2{\\right)}^{3}=1/8$. The sum is 7/8, which is the same as the other way, which gave $1-\\left(1/2{\\right)}^{3}=7/8$.", "4 # Two unbiased coins are tossed. What is probability of getting at most one tail ? $\\frac{1}{2}$ $\\frac{1}{3}$ $\\frac{3}{2}$ $\\frac{3}{4}$ D. $\\frac{3}{4}$ So probability = $\\frac{3}{4}$", "card from deck of cards. Probability of picking a card greater in value than 5 and lesser than 9 from deck of cards is $$\\frac{12}{52}$$. 9. What is the probability of getting 2 tails and 1 head when 3 coins are tossed? a) $$\\frac{1}{2}$$ b) $$\\frac{1}{13}$$ c) $$\\frac{3}{8}$$ d) $$\\frac{1}{3}$$ Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The possible outcome when 3 coins are tossed are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. The probability of getting 2 tails and 1 head is $$\\frac{3}{8}$$ 10. What is the probability of picking an ace of hearts card from deck of cards? a) $$\\frac{1}{52}$$ b) $$\\frac{1}{13}$$ c) $$\\frac{1}{4}$$ d) $$\\frac{13}{52}$$ Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. There is one ace of heart card. The probability of picking an ace of heart card from deck of cards is $$\\frac{1}{52}$$. Sanfoundry Global Education & Learning Series – Mathematics – Class 8. To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.", "# Mathematics Questions and Answers – Data Handling – Equally Likely Outcomes « » This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Data Handling – Equally Likely Outcomes”. 1. What is the probability of getting head when a coin is tossed? a) $$\\frac{1}{2}$$ b) $$\\frac{1}{3}$$ c) $$\\frac{1}{4}$$ d) $$\\frac{3}{2}$$ Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The likelihood of getting either head or tail is equal. Hence, the probability of getting head when a coin is tossed is $$\\frac{1}{2}$$. 2. If 2 coins are tossed, what is the probability of getting heads on both coins? a) $$\\frac{1}{2}$$ b) $$\\frac{1}{3}$$ c) $$\\frac{1}{4}$$ d) $$\\frac{3}{2}$$ Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The likelihood of getting either head or tail is equal. Hence, the probability of getting head when a coin is tossed is $$\\frac{1}{2}$$. The probability of getting head on both the coins is $$\\frac{1}{4}$$. 3. If 2 coins are tossed, what is the probability of getting different outcomes on both coins? a) $$\\frac{1}{2}$$ b) $$\\frac{1}{3}$$ c) $$\\frac{1}{4}$$ d) $$\\frac{3}{2}$$ Explanation: A coin has two sides i.e. head and tail. When a coin is tossed,", "tails and getting three heads which is $\\left(\\frac{2}{3}\\right)^3 + \\left(\\frac{1}{3}\\right)^3 = \\frac{8}{27} + \\frac{1}{27} = \\frac{9}{27} = \\frac{27}{81}$. The probability of winning Game B is the sum of the probabilities of getting two heads and getting two tails squared. This gives us $\\left(\\left(\\frac{2}{3}\\right)^2 + \\left(\\frac{1}{3}\\right)^2 \\right)^2 = \\left(\\frac{5}{9}\\right)^2 = \\frac{25}{81}$. The probability of winning Game A is $\\frac{27}{81}$ and the probability of winning Game B is $\\frac{25}{81}$, so the answer is $\\boxed{\\textbf{(D)}}$", "# An unbiased coin is tossed six times. The probability of obtaining at least one head and at least one tail is? 1. $$\\rm \\frac{1}{31}$$ 2. $$\\rm \\frac{31}{32}$$ 3. $$\\rm \\frac{32}{31}$$ 4. $$\\rm \\frac{1}{32}$$ Option 2 : $$\\rm \\frac{31}{32}$$ ## Detailed Solution Calculation: Required probability = 1 - [P(No head) + P(No Tail)] = 1 - $$\\rm \\left \\{ \\frac{1}{2^{6}} + \\frac{1}{2^{6}} \\right \\}$$ = 1 - $$\\rm \\frac{1}{2^{5}}$$ Required probability = $$\\rm 1 - \\frac{1}{32}$$ = $$\\rm \\frac{31}{32}$$ The probability of obtaining at least one head and at least one tail is $$\\rm \\frac{31}{32}$$", "on bottom. The quizmaster threw the coins and they landed two heads. The winner celebrated but the people in the room said he was lucky and that his choice was illogical. The player, however, opting for two heads had a $\\frac{1}{4}$ chance of winning and given that there were five players still in the pool it was a good choice – any other choice would have given a smaller probability than $\\frac{1}{4}$. What should I have chosen? $\\mathbb{P}$( winning the game| Two heads chosen) = $\\mathbb{P}$( Two heads)$\\times\\mathbb{P}$(winning from a pool of two players remaining) If we assume the probability of winning from a pool of two players is $\\frac{1}{2}$ then the probability of me winning is $\\frac{1}{4}\\times\\frac{1}{2}=\\frac{1}{8}$. 2. Choose two tails $\\mathbb{P}$(me winning the game| Two tails chosen) =$\\mathbb{P}$(Two tails)$\\times\\mathbb{P}$(winning from a pool of two players remaining) If we assume the probability of me winning from a pool of two players is $\\frac{1}{2}$, then the probability of me winning is $\\frac{1}{4}\\times\\frac{1}{2}=\\frac{1}{8}$. 3. Choose one head and one tail $\\mathbb{P}$(me winning the game| One head and one tail chosen) = \\mathbb{P}( One head and one tail)$\\times\\mathbb{P}$(winning from a pool of $3$ players remaining) If we assume that the probability of me winning from a pool of $3$ players is $\\frac{1}{3}$, then the probability of me winning is $\\frac{1}{2}\\times\\frac{1}{3}=\\frac{1}{6}$.", "getting ‘tail’ and getting ‘head’ are a complement to each other. • When two coins are flipped simultaneously, getting ‘at least one head’ and ‘no head’ are complementary events. • When a die is thrown: • Obtaining ‘even face’ and ‘odd face’ are complementary events of each other. • Getting ‘multiple of 2’ and ‘not multiple of 2’ are complement events. • Getting ‘divisible by 3’ is a complement event of getting ‘not divisible by 3’ event. Questions on Complement of an Event Question 1: The probability of getting a red ball from a bag of balls is $$\\frac { 3 }{ 4 }$$. What is the probability of not getting a red ball? Solution: Given that, P(red ball) = $$\\frac { 3 }{ 4 }$$ P(not red ball) = 1 – $$\\frac { 3 }{ 4 }$$ = $$\\frac { 1 }{ 4 }$$ Therefore, the probability of not getting a red ball from a bag of balls is $$\\frac { 1 }{ 4 }$$. Question 2: A box contains white and black marbles. The probability of getting a white marble from the box of marbles is $$\\frac { 7 }{ 10 }$$. What is the probability of getting a black marble? Solution: Let A be the event of getting a white marble and B be the", "Probability to get one head = $$\\frac { 3 }{ 8 }$$ If she gets 1, 2, 3, 4, then she tosses a coin once and notes whether a head or tail. ∴ Probability to get one head = $$\\frac { 1 }{ 2 }$$ If she obtained exactly one head, than probability that she threw 1,2, 3 or 4 Question 13. A card from a pack of 52 card is lost From remaining cards of the pack, two cards are drawn found to be both diamonds. Find probability of the lost card being a diamond. Solution: Let Events E1 = Lost card is diamond E2 = Lost card is not diamond The number of diamond card = 13 out of 52 cards. and 39 cards are not diamonds (i) When a card of diamond lost there remaining cards of diamond are 12 out of 52. Here A represents the cards lost. (ii) When diamonond card is not Lost then probability of drawing 2 cards of diamond ∴ Probability that lost card is a diamond Hence, required probability = $$\\frac { 11 }{ 50 }$$ Question 14. A bag contains 3 red and 7 black. Two balls are drawn one by one at a time at random without replacement. If second drawn half is red, then what is", "{H, T} Number of possible outcomes = 2 Number of favorable outcomes = 1 => Probability of getting tail = $\\frac{Number of favorable outcomes}{Total number of possible outcomes}$ = $\\frac{1}{2}$ => Probability of getting tail is $\\frac{1}{2}$. Question 1: Find the probability of getting one while tossing a die. (Answer: $\\frac{1}{6}$)", "4 # Three unbiased coins are tossed, what is the probability of getting at least 2 tails ? $\\frac{1}{3}$ $\\frac{1}{6}$ $\\frac{1}{2}$ $\\frac{1}{8}$ C. $\\frac{1}{2}$ So required probability =$\\frac{4}{8}=\\frac{1}{2}$", "of obtaining two and three heads. You can apply the binomial probability distribution to obtain the answer, but we can simplify this. The probability of obtaining at least one head, $P(B)$, is also equivalent to one minus the probability of obtaining no heads. The probability of obtaining no heads, say $P(C)$, is equivalent to the probability of obtaining three tails, and it can be calculated as follows: $$P(C) = \\frac{1}{2} \\times \\frac{1}{2} \\times \\frac{1}{2}$$ Then, we calculate $P(B)$: $$P(B) = 1 - P(C) = 1 - \\frac{1}{8} = \\frac{7}{8}$$ And finally, $$P (A \\cup B) = P(B) = \\frac{7}{8}$$ 1. Getting a head or a 4 does not include getting both a head AND a four. Say the probability of getting a head or a 4 is $P(D)$. Besides adding together the probability of obtaining a head and the probability of obtaining a 4 (which is what you did), you need to subtract the probability of obtaining a head and a four. $$P(D) = \\frac{1}{2} + \\frac{1}{6} - \\frac{1}{2} \\times \\frac{1}{6} = \\frac{7}{12}$$ • Genius! Thanks a lot! – Ben Apr 16 '17 at 4:11", "probability that are of same colour? Solution: Let S be the sample space. Then, n(S) = Number of ways of drawing 2 balls out of (6 + 4) = $$10_{{c}_{2}}$$ =$$\\frac{10 * 9}{2 * 1}$$ Let E = Event of getting both balls of the same colour. Then, n(E) = Number of ways of drawing (2 balls out of 6) or (2 balls out of 4) = $$6_{{c}_{2}} + 4_{{c}_{2}}$$ = $$\\frac{6 * 5}{2 * 1}$$ = 15 + 6 = 21. P(E) = $$\\frac{n(E)}{n(S)}$$ = $$\\frac{21}{45}$$ = $$\\frac{7}{15}$$ 5. Two unbiased coins are tossed. What is the probability of getting at most one head? Solution: Given, S = {HH, HT, TH, TT} Let E = event of getting at most one head. Therefore, E = {TT, HT, TH} Therefore, P(E) = $$\\frac{n(E)}{n(S)}$$ = $$\\frac{3}{4}$$", "probability distribution of the number of tails. Ans: Given that the ratio of occurring heads is $3$times that of tails i.e $P\\left( H \\right)=3P\\left( T \\right)$ Also we know that $\\Rightarrow P\\left( T \\right)=\\frac{1}{4}$ $\\Rightarrow P\\left( H \\right)=\\frac{3}{4}$ The sample space for toss of two coins is given by $S=\\left\\{ \\left( HH \\right),\\left( HT \\right),\\left( TH \\right),\\left( TT \\right) \\right\\}$ Let K be the random variable which represents number of tails Therefore possible values of K are $0,1,2$ $\\therefore P\\left( K=0 \\right)=\\frac{3}{4}\\times \\frac{3}{4}=\\frac{9}{16}$ $\\therefore P\\left( K=1 \\right)=\\frac{3}{4}\\times \\frac{1}{4}+\\frac{1}{4}\\times \\frac{3}{4}=\\frac{6}{16}$ $\\therefore P\\left( K=2 \\right)=\\frac{1}{4}\\times \\frac{1}{4}=\\frac{1}{16}$ Hence the distributions is K $0$ $1$ $2$ P(K) $\\frac{9}{16}$ $\\frac{6}{16}$ $\\frac{1}{16}$ 8. A random variable X has the following probability distribution. Y $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ P(Y) $0$ $k$ $2k$ $2k$ $3k$ ${{k}^{2}}$ $2{{k}^{2}}$ $7{{k}^{2}}+k$ Determine (i)k Ans: We know that the sum of probabilities is $1$ \\begin{align} & \\Rightarrow k+2k+2k+3k+{{k}^{2}} \\\\ & +2{{k}^{2}}+7{{k}^{2}}+k=1 \\\\ \\end{align} $\\Rightarrow 10{{k}^{2}}+9k-1=0$ $\\left( 10k-1 \\right)\\left( k+1 \\right)=0$ $\\Rightarrow k=-1,\\frac{1}{10}$ (ii) $P\\left( X<3 \\right)$ Ans: we know that $P\\left( X<3 \\right)=P\\left( X=0 \\right)+P\\left( X=1 \\right)+P\\left( X=2 \\right)$ $P\\left( X<3 \\right)=0+k+2k=3k$ $P\\left( X<3 \\right)=\\frac{3}{10}$ (iii) $P\\left( X>6 \\right)$ Ans: Since there are $7$possible values for random variable $\\therefore P\\left( X>6 \\right)=P\\left( X=7 \\right)$ $\\Rightarrow P\\left( X=7 \\right)=7{{k}^{2}}+k$ $\\Rightarrow P\\left( X=7 \\right)=7{{\\left( \\frac{1}{10} \\right)}^{2}}+\\frac{1}{10}$ $\\Rightarrow P\\left(", "is 2 heads and 1 tail B is said to have occurred when an outcome of the experiment is 2 or more heads. Find the Probability of A∪B. S = {HHH, HHT, HTH, THH, HTT, TTH, THT, TTT} w(S) = 8 (i) A = 2 Heads and 1 tail = {HHT, THH} n(A) = 2 ∴ $$P(A)=\\frac{15}{36}$$ (ii) B = 2 or more heads = {HHH, HTH, HHT, THH} n(B) = 4 ∴ $$\\frac{4}{8}=\\frac{1}{2}$$ (A∩B) = {HHT, THH} n(A∩B) = 2 P(A∩B) = $$\\frac{2}{8}=\\frac{1}{4}$$ P(A∪B) = P(A) + P(B) – P(A∩B) = $$\\frac{1}{4}+\\frac{1}{2}-\\frac{1}{4}=\\frac{1}{2}$$ Question 32. From well shuffled pack of 52 playing cards. 2 cards are drawn at random. Find the Probability that both the cards are spade cards or face cards. n(S) = 52C2 Let A = be the event that both the cards are spade Question 33. Three cards are drawn at random from a well-shuffled pack of 52 playing cards. Find the Probability of getting a King, a Queen, and a Jack. P(Event A) = P(getting a King) = $$\\frac{^{4} C_{3}}{^{52} C_{3}}$$ P(Event B) = P(getting a Queen) = $$\\frac{^{4} C_{3}}{^{52} C_{3}}$$ P(Event C) = P(getting a Jack) = $$\\frac{^{4} C_{3}}{^{52} C_{3}}$$ Since A, B & C are independent P(A∩B∩C) = P(A).P(B). P(C) $$\\frac{^{4} C_{3}}{^52} \\times \\frac{^{4} C_{3}}{^{52} C_{3}} \\times \\frac{^{4} C_{3}}{^{52} C_{3}}$$ Question 34." ]

[ "+0 0 174 1 Four red candies and three green candies can be combined to make many different “flavors.” Flavors are different if the percent red is different, so “3 red / 0 green” is the same flavor as “2 red / 0 green,” and likewise “4 red / 2 green” is the same flavor as “2 red / 1 green.” If a flavor is to be made using some or all of the seven candies, how many different flavors are possible? May 27, 2020 #1 +25565 +2 Four red candies and three green candies can be combined to make many different “flavors.” Flavors are different if the percent red is different, so “3 red / 0 green” is the same flavor as “2 red / 0 green,” and likewise “4 red / 2 green” is the same flavor as “2 red / 1 green.” If a flavor is to be made using some or all of the seven candies, how many different flavors are possible? $$\\begin{array}{|c|c|c|c|c|} \\hline & {\\color{green}0} & {\\color{green}1} &{\\color{green}2} &{\\color{green}3} \\\\ \\hline {\\color{red}0} & - & 0+1=1 & 0+2=2 & 0+3=3 \\\\ & & \\frac{{\\color{red}0}}{1}: \\frac{{\\color{green}1}}{1} &\\frac{{\\color{red}0}}{2}: \\frac{{\\color{green}2}}{2}&\\frac{{\\color{red}0}}{3}: \\frac{{\\color{green}3}}{3}\\\\ & & {\\color{red}0}: {\\color{green}1} & {\\color{red}0}: {\\color{green}1}&{\\color{red}0}: {\\color{green}1}\\\\ \\hline {\\color{red}1} & 1+0=1 & 1+1=2 & 1+2=3 & 1+3=4 \\\\ & \\frac{{\\color{red}1}}{1}: \\frac{{\\color{green}0}}{1} & \\frac{{\\color{red}1}}{2}: \\frac{{\\color{green}1}}{2}", "red, orange, yellow, green, blue, and brown. The original packs didn’t contain white M&M’s — sorry, Freedom Blend (Fourth of July). The original packs didn’t contain pastel colors — hop on by, Easter Blend. And nowhere on God’s green Earth will it ever be acceptable to use white chocolate inside those delectable candy shells — hit the road, Carrot Cake M&M’s. (Yuck.) To my knowledge, there are only two other blends produced by Mars, Inc., that satisfy my acceptability criteria: • Harvest Blend: red, yellow, brown • Birthday Cake: red, yellow, blue The Christmas, Harvest, and Birthday Cake blends represent just three of the 63 possible color combinations that can be made from the original six colors. That leaves 60 combinations that are just begging for names. (A little history. As you may know, I have a quirk. I eat M&M’s in pairs of the same color, so I can place one on each side of my mouth and feel “balanced.” But it’s atypical for a pack to contain an even number of every color. When I near the end, I’m often left with one to six unmatched M&M’s. And I’ve always thought that these various color combinations deserved a name.) What would you call a combination of red, yellow, and green? Obviously, STOPLIGHT. What might you call", "original Plain M&M packs, which contain red, orange, yellow, green, blue, and brown. The original packs didn’t contain white M&M’s — sorry, Freedom Blend (Fourth of July). The original packs didn’t contain pastel colors — hop on by, Easter Blend. And nowhere on God’s green Earth will it ever be acceptable to use white chocolate inside those delectable candy shells — hit the road, Carrot Cake M&M’s. (Yuck.) To my knowledge, there are only two other blends produced by Mars, Inc., that satisfy my acceptability criteria: • Harvest Blend: red, yellow, brown • Birthday Cake: red, yellow, blue The Christmas, Harvest, and Birthday Cake blends represent just three of the 63 possible color combinations that can be made from the original six colors. That leaves 60 combinations that are just begging for names. (A little history. As you may know, I have a quirk. I eat M&M’s in pairs of the same color, so I can place one on each side of my mouth and feel “balanced.” But it’s atypical for a pack to contain an even number of every color. When I near the end, I’m often left with one to six unmatched M&M’s. And I’ve always thought that these various color combinations deserved a name.) What would you call a combination of red, yellow, and green?", "# Find the least number of objects from a jar when those have two colors I've been going in circles with this question which belongs to certainty about something. The original source of this problem is unknown. I found it in a textbook who doesn't have an author but rather a collection of riddles about combinatorics and different problems. The problem is as follows: A porcelain jar is filled with a set of sphere candies. The baker made them in a peculiar condition, so that one half of the ball shaped candy has a color different from its other half. 24 of those candies has one side colored blue (due a blueberry flavor) and the other half is colored green (due a pear flavor), 30 has one half green and the other half red (due strawberry flavor), 28 has one half red and the other white (due mint flavor), 40 has one half white and other half orange (due orange flavor) and 35 candies has one half orange and the other half blue. How many of these candies must be taken out from the jar at random and at least to affirm that 7 candies share the same color? (Assume that you are not allowed to have a peek inside the jar). The choices given are: 1. 13", "## Red + Green = Christmas, and 62 Other M&M Color Combinations ‘Tis the holiday season, so every grocery store, pharmacy, and convenience store is now stocking the M&M® Christmas Blend, a joyful combination of red and green button-shaped chocolate candies. It’s unclear whether this mixture actually helps to imbue the holiday spirit, but the consumption of these tasty morsels will make you look just a little more like St. Nick. As far as I’m concerned, the Christmas Blend — not to be confused with Holiday Mint, which uses a (disgusting) mint chocolate filling — is one of just a few acceptable color combinations. Why? Because it uses colors that can only be found in the original Plain M&M packs, which contain red, orange, yellow, green, blue, and brown. The original packs didn’t contain white M&M’s — sorry, Freedom Blend (Fourth of July). The original packs didn’t contain pastel colors — hop on by, Easter Blend. And nowhere on God’s green Earth will it ever be acceptable to use white chocolate inside those delectable candy shells — hit the road, Carrot Cake M&M’s. (Yuck.) To my knowledge, there are only two other blends produced by Mars, Inc., that satisfy my acceptability criteria: • Harvest Blend: red, yellow, brown • Birthday Cake: red, yellow, blue The Christmas, Harvest, and", "# Permutation/Combination problem - choosing 2 objects from different categories Context: there are three candies in a basket (5 red, 4 blue, 5 green). How many ways can you choose 2 candies with different color? Should I use permutation or combination w/ this? I tried using: 14!/(14-2)! = 182 ways; but it's marked as wrong. You can choose $5\\times4$ pairs of candies s.t. one of them is red and one of them is blue. • Yes. Do you agree that there are $20$ possibilities to choose a pair of candies such that one of them is red and the other is blue? Btw, what do you mean with: \"there are three candies in a basket\". Maybe I misunderstand something there. Aren't there $14$ candies in total? – drhab Dec 10 '15 at 15:20 • Well then: $5\\times 4+5\\times5+4\\times5=65$ possibilities. The first term corresponds with \"red and blue\" the second with \"red and green\" and the third with \"blue and green\". – drhab Dec 10 '15 at 15:30", "worked so far, thanks. How many different flavors of candy are there. As we approach the holiday season, there is one symbol that we sometimes take for granted. 1. {/eq}. Crane's first flavor was peppermint, and the first candies were made without holes in them. There are millions and millions of flavors here are some Candy canes are a traditional Christmas candy, now available in many different flavors besides peppermint. here are some other flavors i highly recommend: banana w/ kiwi. Another thing you might not know is that candy corn can come in different flavors. Each country has a dedicated manufacturing facility that functions as a separate department. If the probability of randomly pulling out an orange… Chocolate. Sciences, Culinary Arts and Personal With their anthropomorphic covers, Nerds usually contain two flavors per box, and each flavor has a separate compartment and opening. 4 years ago. The cheaper ones have different colors but are all the same flavor. Over the years, HI-CHEW has introduced hundreds of different fruit-flavored chews and the Fantasy Flavor Games gives fans the chance to participate in choosing the next best one. 13. [citation needed] Larger packages may contain various colors—sometimes referred to as \"Rainbow Nerds\".Smaller packages may contain either one flavor only or one flavor with pieces of another. So, fraction of", "other, and that we can list 9 flavors, but that there are actually only 8 (never really understood that yet)", "places, we have already picked the grape flavor. Then we can pick one more flavor and the number of ways to do that is 5C1. 2nd option - pack of 3 flavors ( _ _ _ ) Total number of ways the flavor can be picked is 6C3. Now we assume that between the three places, we have already picked the grape flavor. Then we can pick two more flavor and the number of ways to do that is 5C2. 3rd option - pack of 4 flavors ( _ _ _ _ ) Total number of ways the flavor can be picked is 6C4. Now we assume that between the four places, we have already picked the grape flavor. Then we can pick three more flavor and the number of ways to do that is 5C3. Therefore, probability of picking a box with Grape flavor is (5C1 + 5C2 + 5C3)/(6C2 + 6C3 + 6C4) = 25/50 = 1/2. A certain candy store sells jellybeans in the following six flavors on [#permalink] 21 Dec 2018, 07:10 Display posts from previous: Sort by", "already in order), the number determines which letter to take from the flavour: S red 3 raspberry A orange 3 oranges L yellow 1 lemon T green 4 pistacchio E blue 6 blueberry D violet 6 lavender • That is a really clever answer to a really clever puzzle! :D Sep 7, 2018 at 21:23 is it skittles! taste the rainbow! • No, no slogans are involved in the answer. Sep 7, 2018 at 16:04", "Birthday Cake blends represent just three of the 63 possible color combinations that can be made from the original six colors. That leaves 60 combinations that are just begging for names. (A little history. As you may know, I have a quirk. I eat M&M’s in pairs of the same color, so I can place one on each side of my mouth and feel “balanced.” But it’s atypical for a pack to contain an even number of every color. When I near the end, I’m often left with one to six unmatched M&M’s. And I’ve always thought that these various color combinations deserved a name.) What would you call a combination of red, yellow, and green? Obviously, STOPLIGHT. What might you call a combination of red, yellow, and blue? Based on the Man of Steel’s outfit, I like SUPERMAN. But Mars, Inc., has already applied the moniker BIRTHDAY CAKE. What would you call a collection of just green M&M’s? I don’t know — QADDAFI, maybe? (Sorry, dated reference.) What would you call a combination of orange, green, and brown? I have no idea. And that’s where you come in. Below is a Google poll where you can enter a color combination and suggest a name. In early January, for any color combinations that have more than one suggestion,", "Free Version Difficult # Jar of Candy GMAT-14PQCJ In a jar of candy, $n$ candies are red. There is the same number of red as there are yellow. There are twice as many blue as yellow. There are three times as many green as blue. What percent of the candy in the jar is green? A $10\\%$ B $20\\%$ C $40\\%$ D $50\\%$ E $60\\%$", "Brand Of Candies Consists Of 5 Different Flavors. Personalized and decorated sweets for the farm or rural theme. What is the rhythm tempo of the song sa ugoy ng duyan? 3. Bags of \"Original Skittles\" candies come in five different flavors: Grape, Lemon, Green Apple, Orange and Strawberry. Peppermint. First made by Wayne Bun Candy Company and then by Clark Bar America, the Bun Bar is a classic 1920s candy that is now made and distributed by Pearson’s Candy Company, the makers of other old-school confections like Bit-O-Honey and Salted Nut Rolls. BBQ PayDay Bar Image courtesy of link. Let's say that there are ORIGINALLY 100 candies in the jar. No, it's all the same flavor, but just different colors. Throughout the years, they've added and removed different flavors, and today they also include apple and grape, with lime no longer an available flavor. Now the Life Saver has many colors and even many shapes. In 1916 Curtiss Candy company founder wanted an \"American-sounding\" name. The flavors are equiprobable; if you take a candy randomly from a bag the probability of getting a particular flavor is 1/5, independent of other \"draws\". The pink is called pink vanilla when I buy it. Now we want to use this transaction out of an workitem within workflow. How long will the", "a colour after picking 13 candies out. However, if I can find a scenario where after picking 13 or more candies out of the bag, we do not have 7 candies sharing a colour, then 13 cannot be the correct answer. If I pick out 15 candies: • 3 blue/green • 3 green/red • 3 red/white • 3 white/orange • 3 orange/blue And then we count the number of candies that have each colour, we get: • Blue candies: 6 (3 blue/green + 3 orange/blue) • Green candies: 6 (3 blue/green + 3 green/red) • Red candies: 6 (3 red/white + 3 green/red) • White candies: 6 (3 white/orange + 3 red/white) • Orange candies: 6 (3 orange/blue + 3 white/orange) So we now know that any answer less than or equal to 15 can't be correct, as we have given a scenario involving 15 candies which does not result in 7 candies sharing a colour. Given the multiple choice question, we know that the answer must be 16. If we pick out any further candy, no matter the colour, there will be 7 candies of both of the colour groups involved. For example if we now pick out an orange/blue candy, there will be 7 candies sharing orange and 7 candies sharing blue.", "15 additional red candies but there is no guarantee that you will. You may just as well pick 7 red and 8 green. Thus, the only way to guarantee that you will have 15 of the same color is to pick 31 as in the worst possible scenario you would have 14 red, 14 green and one additional candy that is either red or green. – tards Nov 22 '11 at 15:55 Doesn't that make 29 extra candies (14+14+1)? – Ben Derrett Nov 22 '11 at 15:59 @BenDerrett ah, right. I should stop doing arithmetic in my head! :-) – tards Nov 22 '11 at 16:01 add comment", "there is 1 combination with zero flavor, 4 possible combinations with one flavor, 6 possible combinations with two flavors, 4 combinations of three flavors, and 1 possible combination with four flavors. Using the combination notation, we can also see that the total number of combinations is $\\displaystyle\\binom 4 0 + \\binom 4 1 + \\binom 4 2 + \\binom 4 3 + \\binom 4 4 = 1 + 4 + 6 + 4 + 1 = 16$ *** Problem 2: Beads are dropped in a Galton Board containing hexagonal pegs as shown in Figure 1. In which of the five bins at the bottom of the board will the beads likely fall? Figure 1 At first glance, it is quite tempting to say that the beads will equally likely fall in the five bins, but first let us examine more closely. When a bead hits a hexagonal peg, it may either fall to the left or right of it. There are four rows of hexagons; we will call them A, B, C and D (see Figure 2). When the bead hits the peg in row A , there are two possible paths — right or left. If it falls to the left, it will still hit the peg in Row B, and again may fall either to the", "ugoy ng duyan? Do the different colors of candy corn have different flavors? Um der wackelnden Preis-Leistung der Artikel gerecht zu werden, vergleichen wir in der Redaktion alle nötigen Faktoren. There are currently over 10 different flavors of PEZ candy available. examples; Blue Razzberry, Apple,Cherry,and many many more. Bowl #1 contains 4 grape candies, 5 lemon candies, 6 cherry candies and 5 raspberry candies. You randomly pick four of these candies. Jim has a box of candies with different flavors. 0 0. Question: A Brand Of Candies Consists Of 5 Different Flavors. You have a bag of 20 candies, composed of 5 candies of 4 different flavours. Production of the candy went on hiatus in July 2018, and returned in May 2020. Candy Quiz Questions . They were first produced in 1974. Now we are facing a new problem: call transation works fine for this flavor/txn combination. How many calories are in a single Peeps marshmallow chick? 0 0. Similar Photos See All. Sciences, Culinary Arts and Personal Favourite answer. Bags of \"Original Skittles\" candies come in five different flavors: Grape, Lemon, Green Apple, Orange and Strawberry. a) What's the probability that they're all the same? Fun Fact #4. Since they lack cocoa butter they aren't \"chocolates\" but many have a good chocolate taste. Who is the longest", "luck out and find the original flavors in a dusty old quarter candy machine somewhere. {/eq}. Instead, the company continued to innovate new flavors and varieties.And so, I decided to try a bunch of them to see how they stacked up. Crane's first flavor was peppermint, and the first candies were made without holes in them. katiya. Create your account. How many different flavors of candy are there. Who was the lady with the trophy in roll bounce movie? Question: A Brand Of Candies Consists Of 5 Different Flavors. 16. ). Answer all the questions in a group (level). Show Answer. All rights reserved. 1 decade ago. Favorite Answer. How many calories are in a single Peeps marshmallow chick? Find answers in product info, Q&As, reviews ... Nerds Rope, Very Berry Candy Flavor - Pack of 8 4.4 out of 5 stars 21. I LOVE chocolate! Suppose that we flip a fair coin until either it... Every day, Jorge buys a lottery ticket. Sport szelet: A chocolate bar produced in Hungary in the 1950s. 1 0. You Wish To Help Her Test This Hypothesis. 19. They come in a rainbow of flavors, colors, shapes, and concentrations of CBD. Manufactured candy coatings which melt easily to allow for coatings. 2. Who created the first, mass-produced candy bar? Zachary: Caramel", "all 3 comments [–] 4 points5 points It would be (2n) - 1. It is a simple binomial expansion of (1+1)n. [–] 1 point2 points For each flavor, you have two choices: you can either include it or not include it. If there are n flavors, that means there are 2n choices; throwing out the 'null drink' with no flavors (soda water?) gives us 2n - 1. [–] 1 point2 points each flavor can be either included or not. this gives 2n - 1 possibilities (the subtracted one is the empty)", "2. 14 3. 15 4. 16 I don't understand \"to affirm that 7 candies share the same color\". Does it mean I should count other colors whose one half also coincides with the first group? I'm aware that when there is some uncertainty about something, the procedure is to assume the most difficult to happen scenario or event, and from then on ruling out possibilities until we can assure that our next pick will guarantee what we need. Examples I saw show this for objects which share entirely one color or shape. But what if they have split colors? Should I understand that they ask to match the identical ones or match the objects (the candies in this case) which have the same set of colors i.e red and white, with ones which one side red and other which have the other have white, but may have its opposing side with a different color? Can somebody clarify this for me? In my attempt to solve this problem what I tried to do was this: Since they ask for 7 candies, then the hardest possibly to happen scenario might be: 6 white and orange + 6 green and red + 1 of any of the remaining group (could be blue and green or red and white or orange and" ]

[ "words, the $x$x value is $3$3 and the $y$y value is $19$19 #### More Examples ##### Question 1 Plot $13:9$13:9 on the coordinate plane. ##### Question 2 Plot $13:17$13:17 on the coordinate plane. ##### Question 3 Consider the given graph. a) What ratio has been plotted? b) Which option could be being represented by this graph and ratio? A) For every 2 green sweets in a mix, there is 1 red sweet. B) For every 1 green sweet in a mix, there are 2 red sweets.", "# Basic probability homework question help (combinatorical counting problem) I would really appreciate some help in the following probability question. I translated it so I'm sorry in advance for English mistakes. There are $$6$$ children in a family and many candies in $$4$$ colors - blue, red, green, pink. The $$2$$ eldest children get to choose $$2$$ candies each. The $$4$$ other children get to choose $$1$$ candy apiece. (The same candy type can be chosen by a few children). a. Count the number of choices. b. What is the probability that exactly $$2$$ children choose blue candies? My attempt at solution: a. $$\\binom{4}{2}^2 \\cdot 4^4 = 6^2 \\cdot 4^4$$ This solution is correct according to the solutions given to me. b. I tried to solve it in the following way: Dividing the question into 3 cases: $$(1)$$ The number of cases in which the two eldest children choose blue candies: $$3^4$$ $$(2)$$ The number of cases in which two younger children choose blue candies: $$3^2 \\cdot 3^2 = 3^4$$ $$(3)$$ The number of cases in which one blue candy is chosen by an older child and one by a younger: $$2 \\cdot 4 \\cdot 3 \\cdot 3^3 = 2 \\cdot 4 \\cdot 3^4$$ I sum all the case to get $$10 \\cdot 3^4$$ The final solution I", "Ratios Ratios compare the values of things. They tell us how much of one thing is made up of another thing, or how much of one thing there is compared to another thing. For example, in the picture, there are 3 red jelly babies and 2 green jelly babies. The ratio of red to green jelly babies is $3:2$, but we can soon fix that (I love red jelly babies)! There are 6 red jelly babies to 4 red jelly babies, so the ratio of red to green jelly babies is $6:4$. But this simplifies to $3:2$ as both $6$ and $4$ have a factor of $2$, so we can divide both sides of the ratio by $2$. Simplifying (or complicating) Ratios We can make equivalent ratios in much the same way as we make equivalent fractions. We just have to make sure we multiply (or divide) both parts of the ratio by the same amount. For example, $6:9$ is the same as $6 \\times 3: 9 \\times 3 = 18:27$ and the same as $6 \\div 3 : 9 \\div 3 = 2: 3$. Example: A Recipe for Puffed Rice Crackles Christo and Sam have a lot of allergies, so we need to make our own, allergy-free, versions of many foods. Here's a recipe they particularly like", "you are making a simple comparison. Remember to read the information carefully so that you are clear what is being compared. Now let’s look at an example where you might need to figure something out to write the ratio. Example There are 32 red and yellow candies in a bag. There are 10 yellow candies. What is the ratio of red candies to total candies in the bag? We need the ratio of red candies to total candies. We know that there are 32 total candies. We need to find the number of red candies in the bag before writing the ratio. $32-10 = 22$ Now write the ratio of red candies to total candies. Because a ratio is a comparison, it can be simplified. Make sure to reduce the ratio to lowest terms. This makes it easier to understand the quantities that are being compared. $\\frac{\\text{red candies}}{\\text{total candies}} = \\frac{22}{32} = \\frac{11}{16}$ We also could have written this ratio in two other ways. 22 to 32 then simplified to 11 to 16 OR 22 : 32 then simplified to 11 : 16 All of these answers would have been correct. Remember that you can interchange the form that you choose to write a ratio. We can also compare ratios. This is when we have two or more", "below: $x$x $y$y $1$1 $2$2 $3$3 $4$4 $\\editable{}$ $\\editable{}$ $\\editable{}$ $\\editable{}$ 2. Plot the points in the table of values. 3. Draw the graph of the proportional relationship between $x$xand $y$y. ##### QUESTION 2 Valerie wants to make sweet and salty popcorn. She has decided the perfect mix is $8$8:$5$5 sweet to salty. 1. Complete the ratio table: sweet ($x$x) salty ($y$y) $0$0 $8$8 $16$16 $24$24 $32$32 $80$80 $\\editable{}$ $\\editable{}$ $\\editable{}$ $\\editable{}$ $\\editable{}$ $\\editable{}$ 2. Plot the ratio on the number plane. ##### Question 3 Consider the following graph: 1. Which of the following could be being represented by this graph and ratio? For every $1$1 green sweet in a mix, there are $2$2 red sweets. A For every $2$2 green sweets in a mix, there is $1$1 red sweet. B For every $1$1 green sweet in a mix, there are $2$2 red sweets. A For every $2$2 green sweets in a mix, there is $1$1 red sweet. B 2. What ratio has been plotted? $2:1$2:1 A $1:2$1:2 B $2:1$2:1 A $1:2$1:2 B ### Outcomes #### 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g. By reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations. #### 6.RP.A.3.A Make tables of equivalent ratios relating quantities with whole-number measurements, find missing", "[‘$\\cap$’ implies intersection] Also we will have, $P (X \\cap Y_i)$ = $P (Y_i \\cap X)$ = $P (Y_i) P(X|Y_i)$ (ii) From (i) and (ii) we get, $P (X) P (Y_i|X)$ = $P (Y_i) P(X|Y_i)$ Finally, $P (Y_i|X)$ = $\\frac{[P (Y_i) P (X|Y_i)]}{P (X)}$ In general the Bayes’s theorem formula states that: $P (X|Y)$ = $\\frac{[P (X) P (Y|X)]}{P (Y)}$ Where: $X$ and $Y$ are two sets $P (X)$ is the prior probability $P (X|Y)$ is the conditional probability of $X$ given that $Y$ has occurred. We can write $Y$ = $X \\cap Y$ and $\\overline{X} \\cap Y$, so we get, $P (Y)$ = $P (X) P (X|Y) + P (\\overline{X}) P (\\overline{X}|Y)$ Hence another version of Bayes’ formula can be: $P (X|Y)$ = $\\frac{[P (X) P (Y|X)]}{[P (X) P (X|Y) + P (\\overline{X}) P (\\overline{X}|Y)]}$ ## Problems Below are some examples on topic Bay's formula: Example 1: A bag contains 30% brown, 20% red, 20% yellow, 10% orange, 10% green, 10% tan candies. Another similar kind of new bag including blue candies with 20% green, 24% blue, 14% yellow, 16% orange, 13% brown, 13% red is also available. If we take out two candies one from each bag, one green and one yellow, find the probability that the yellow one came from the old bag. Solution: The", "x denotes the number of red candies. Ratio: $$\\frac{Red}{Removed}$$ = $$\\frac{2}{5}$$ => $$\\frac{x}{15}$$ = $$\\frac{2}{5}$$ => x = 6 => Sufficient Hi, Need a clarification. Shouldn't the number of red candies be the same in both options??? Intern Joined: 04 Jun 2018 Posts: 19 Location: Viet Nam GMAT 1: 610 Q47 V27 WE: Supply Chain Management (Internet and New Media) Re: A bag is filled with 36 orange candies and 24 red candies for a total [#permalink] ### Show Tags 26 Dec 2018, 20:14 1 Kezia9 wrote: lybeaver wrote: (1) After remove 15 candies, there are 60-15=45 total candies left. $$\\frac{Orange}{Red}$$ = $$\\frac{2}{1}$$ => Red candies = 15, orange candies = 30 => Sufficient (2) x denotes the number of red candies. Ratio: $$\\frac{Red}{Removed}$$ = $$\\frac{2}{5}$$ => $$\\frac{x}{15}$$ = $$\\frac{2}{5}$$ => x = 6 => Sufficient Hi, Need a clarification. Shouldn't the number of red candies be the same in both options??? In DS question type, your job is to determine whether you can answer the question using the information given. You look at statement (1) first then ask yourself whether it's enough to answer the question. Then you look at statement (2) and ask yourself the same. 2 statements can give you different answers as long as it's correct. Intern Joined: 02 Feb 2018 Posts: 36 Re:", "ratios in the table reduce to $\\frac{3}{5}$. The tape diagram represents the ratio $3:5$. Remember that with ratios, order matters! ##### Lösung • The ratio of peanuts to almonds is $1:3$. You can find this by simplifying any of the pairs of numbers in the table. The first line is $\\frac{3}{9}=\\frac{1}{3}$. Therefore, the tape diagram has $1$ block on top and $3$ blocks on the bottom. • The ratio of green to red is $4:7$. You can find this by simplifying any of the pairs of numbers in the table. The first line is $\\frac{8}{14}=\\frac{4}{7}$. Therefore, the tape diagram has $4$ blocks on top and $7$ blocks on the bottom. • The ratio of oranges to apples is $5:3$. You can find this by simplifying any of the pairs of numbers in the table. The first line is $\\frac{10}{6}=\\frac{5}{3}$. Therefore, the tape diagram has $5$ blocks on top and $3$ blocks on the bottom. • The ratio of girls to boys is $3:4$. You can find this by simplifying any of the pairs of numbers in the table. The first line is $\\frac{6}{8}=\\frac{3}{4}$. Therefore, the tape diagram has $3$ blocks on top and $4$ blocks on the bottom. • The ratio of tomatoes to carrots is $1:2$ or any multiple of the ratio $1:2$. Therefore, the tape diagram", "only $$1$$ way for them to both green jelly beans, and there are $$1 \\cdot 2 = 2$$ ways for them to both choose red jelly beans. There are a total of $$2 \\cdot 4 = 8$$ ways for Abe and Bob to randomly pick a jelly bean, and as such, the probability the colors match is $$\\dfrac{3}{8}.$$ Thus, C is the correct answer. 15. If we have $\\begin{cases}3^p + 3^4 = 90\\\\2^r + 44 = 76 \\\\ 5^3 + 6^s = 1421\\end{cases}$ What is the product of $$p,$$ $$r,$$ and $$s?$$ $$27$$ $$40$$ $$50$$ $$70$$ $$90$$ ###### Solution(s): Simplifying the three equations, we get $\\begin{cases} 3^p = 9 \\\\ 2^r = 32, \\\\ 6^s = 1296 \\end{cases}$ By inspection, we get that $$p = 2,$$ $$r = 5,$$ and $$s = 4.$$ Their product is $$2 \\cdot 5 \\cdot 4 = 40.$$ Thus, B is the correct answer. 16. A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $$8^\\text{th}$$-graders to $$6^\\text{th}$$-graders is $$5:3,$$ and the ratio of $$8^\\text{th}$$-graders to $$7^\\text{th}$$-graders is $$8:5.$$ What is the smallest number of students that could be participating in the project? $$16$$ $$40$$ $$55$$ $$79$$ $$89$$ ###### Solution(s): The number of $$8^\\text{th}$$-graders must be a multiple of $$5$$ and $$8,$$ which means", "Now let’s look at an example where you might need to figure something out to write the ratio. Example There are 32 red and yellow candies in a bag. There are 10 yellow candies. What is the ratio of red candies to total candies in the bag? We need the ratio of red candies to total candies. We know that there are 32 total candies. We need to find the number of red candies in the bag before writing the ratio. 3210=22\\begin{align*}32-10 = 22\\end{align*} Now write the ratio of red candies to total candies. Because a ratio is a comparison, it can be simplified. Make sure to reduce the ratio to lowest terms. This makes it easier to understand the quantities that are being compared. red candiestotal candies=2232=1116\\begin{align*}\\frac{\\text{red candies}}{\\text{total candies}} = \\frac{22}{32} = \\frac{11}{16}\\end{align*} We also could have written this ratio in two other ways. 22 to 32 then simplified to 11 to 16 OR 22 : 32 then simplified to 11 : 16 All of these answers would have been correct. Remember that you can interchange the form that you choose to write a ratio. We can also compare ratios. This is when we have two or more ratios and we want to figure out which ones are larger and which ones are smaller. Let’s look at", "event which is 1/3. This means that the expression at the top in terms of n is equal to 1/3. $\\dfrac{30}{n^2-n}=\\dfrac{1}{3}$ By cross multiplication, $30\\cdot 3=n^2-n$ $90=n^2-n$ $\\boxed{n^2-n-90=0}$ Another solution(the practical approach): Let’s think of taking 2 oranges from n total number of sweets. The number of ways to choose 2 sweets from n sweets can be solved as follows, $_nC_2=\\dfrac{n!}{(n-2)!2!}$ $_nC_2=\\dfrac{n(n-1)(n-2)!}{(n-2)!2!}$ $_nC_2=\\dfrac{n(n-1)}{2}$ The total ways to choosing 2 orange sweets from 6 orange sweets can be solved as follows, $_6C_2=\\dfrac{6!}{(6-2)!2!}$ $_6C_2=\\dfrac{6!}{4!2!)}$ $_6C_2=\\dfrac{6\\cdot 5}{2}$ $_6C_2=15$ The desired probability is, $P=\\dfrac{15}{\\dfrac{n(n-1)}{2}}$ $P=15\\cdot\\dfrac{2}{n(n-1)}$ $P=\\dfrac{30}{n(n-1)}$ Since this probability is equal to 1/3 we have, $\\dfrac{1}{3}=\\dfrac{30}{n(n-1)}$ By cross multiplication, $n(n-1)=90$ $n^2-n=90$ $\\boxed{n^2-n-90=0}$ Now, if you want to find the number of yellow sweets. Solve for n using factoring. Subtract the number of orange sweets(6) from the total number of sweets(n). To give you an idea, there are 10 total number of sweets where 4 of them are yellow and 6 of them are orange. ### Dan Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.", "ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way. Therefore, total number of ways of getting the 3 - 1 - 1 option is $5C3$ = 10 = 10 ways. Total ways in which the 5 toys can be packed in 3 identical boxes = number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways. Q: If each of the vowels in the word 'MEAT' is kept unchanged and each of the consonants is replaced by the previous letter in the English alphabet, how many four-lettered meaningful words can be formed with the new letters, using each letter only once in each word? A) 3 B) 4 C) 1 D) 2 Explanation: 20 1449 Q: In a bag containing red, green and pink tokens, the ratio of red to green tokens was 5 : 12 while the ratio of pink to red tokens was 7 : 15. What was the ratio of green to pink tokens? A) 25 : 28 B) 36 : 7 C) 8 :25 D) 12 : 7 Explanation: 6 1156 Q: The ratio of the members of blue to red balls in abag is constant. When there were 44 red", "# 2004 AIME II Problems/Problem 2 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem A jar has $10$ red candies and $10$ blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ ## Solution The probability that Terry picks two red candies is $\\frac{10 \\cdot 9}{20 \\cdot 19} = \\frac{9}{38}$, and the probability that Mary picks two red candies after Terry chooses two red candies is $\\frac{7\\cdot8}{18\\cdot17} = \\frac{28}{153}$. So the probability that they both pick two red candies is $\\frac{9}{38} \\cdot \\frac{28}{153} = \\frac{14}{323}$. The same calculation works for the blue candies. The probability that Terry picks two different candies is $\\frac{20\\cdot10}{20\\cdot19} = \\frac{10}{19}$, and the probability that Mary picks two different candies after Terry picks two different candies is $\\frac{18\\cdot 9}{18\\cdot 17} = \\frac{9}{17}$. Thus, the probability that they both choose two different candies is $\\frac{10}{19}\\cdot\\frac{9}{17} = \\frac{90}{323}$. Then the total probability is $$2 \\cdot \\frac{14}{323} + \\frac{90}{323} = \\frac{118}{323}$$ and so the answer is $118 + 323 = \\boxed{441}$. In the above calculations, we treated the choices as ordered; that is, Terry chose", "}}\\, \\frac{2}{20}$$, If the unknown number is in the denominator we can use another method that involves the cross product. Since $80$ is about $3$ times $25$, the medicine should be about $3$ times $5$. If you had 3/2 = x/4, instead of taking extra steps and setting the cross products equal and then solving for x, here we should easily recognize that to go from 2 in the first fraction to 4 in the 2nd fraction, we multiplied by 2. Using the Proportion Method to Solve Percent Problems There are a variety of ways to solve percent problems, many of which can be VERY confusing. $\\color{red}{63}(\\frac{x}{63})=\\color{red}{63}(\\frac{4}{7})$, $x=\\frac{9\\cdot\\color{red}{7}\\cdot4}{\\color{red}{7}}$. You'll see how to use the scale from a blueprint of a house to help find the actual height of the house. The pediatrician would prescribe $16$ ml of acetaminophen to Zoe. Calculating a sample proportion in probability statistics is straightforward. We can use proportions to solve questions involving percents. If she has six apples, how many oranges does she have? Let $c=$ number of calories. Examples: Applications Using Proportions. Learn how to solve with proportions. 38 is to 640 as 152 is to the number n. Find n. Solution. Write a sentence that gives the information to find it. For instance if one package of cookie mix results in", "+0 # Help 0 59 3 A box contains 3 red, 2 green, and 4 white candies. Carmen picked one candy, found it was white, and ate it. She picked a second candy at random, found it was red, and ate it. Carmen picked a third candy at random. a) Which colour is the third candy most likely to be? Explain. b) Write the probability that the third candy will be the colour named in part a. Use a ratio, fraction, and percent to write the probability . c) What is the probability that the candy will not be the colour named in part a? May 29, 2020 #1 +47 +2 a) White, because there are more white candies that the other colours: 2 red, 2 green , 3 white. b) $$probability= {{3} \\over 7}$$ $$Ratio= 3:7$$ $$percentage = 42.857$$ c) $$probability= {{4} \\over 7}$$ $$percentage = {57.1428}$$ . May 29, 2020 edited by Hephaestus May 29, 2020 edited by Hephaestus May 31, 2020 #2 0", "= 30 => Sufficient (2) x denotes the number of red candies. Ratio: $$\\frac{Red}{Removed}$$ = $$\\frac{2}{5}$$ => $$\\frac{x}{15}$$ = $$\\frac{2}{5}$$ => x = 6 => Sufficient Manager Joined: 04 Oct 2017 Posts: 77 Re: A bag is filled with 36 orange candies and 24 red candies for a total [#permalink] ### Show Tags 26 Dec 2018, 18:50 lybeaver wrote: (1) After remove 15 candies, there are 60-15=45 total candies left. $$\\frac{Orange}{Red}$$ = $$\\frac{2}{1}$$ => Red candies = 15, orange candies = 30 => Sufficient (2) x denotes the number of red candies. Ratio: $$\\frac{Red}{Removed}$$ = $$\\frac{2}{5}$$ => $$\\frac{x}{15}$$ = $$\\frac{2}{5}$$ => x = 6 => Sufficient Hi, Need a clarification. Shouldn't the number of red candies be the same in both options??? Intern Joined: 04 Jun 2018 Posts: 9 Location: Viet Nam WE: Supply Chain Management (Internet and New Media) Re: A bag is filled with 36 orange candies and 24 red candies for a total [#permalink] ### Show Tags 26 Dec 2018, 19:14 1 Kezia9 wrote: lybeaver wrote: (1) After remove 15 candies, there are 60-15=45 total candies left. $$\\frac{Orange}{Red}$$ = $$\\frac{2}{1}$$ => Red candies = 15, orange candies = 30 => Sufficient (2) x denotes the number of red candies. Ratio: $$\\frac{Red}{Removed}$$ = $$\\frac{2}{5}$$ => $$\\frac{x}{15}$$ = $$\\frac{2}{5}$$ => x = 6 => Sufficient Hi, Need", "ratio. In what proportion are blue to green candies? • Bags The first bag has a weight of 18 kg, the second bag is a one-third lighter than the first and the third bag has three times less the weight of all three bags together. Determine their weights. • Ratio 6 numbers are in the ratio 1:5:1:5:5:5. Their sum is 242. What are the numbers?", "sorry about this, but I'm not seeing what you mean with that formula up there where p(1-d₁)(1-d₂) = the final price. What numbers do you plug in? Sorry for the cluelessness... #### Gene ##### Senior Member Sorry, I used p for the price instead of x. 306 = x*(1-.1)(1-.15) = x*.9*.85 x = 306/(.9*.85) Or you could do it in two steps. u = 306/.85 and x = u/.9 It works out the same #### tkhunny ##### Moderator Staff member Gene said: BTW, TKH fell asleep on the jelly beans. It is 13. Think about what Stapel said about bad luck. Apply it to red and green beans. I still think the answer is zero (0). In the absence of zero (0), however, let's go with 13.", "each of the numbers in the first ratio by the same factor to get the numbers in the second ratio. For example, $$8:6$$ is equivalent to $$4:3$$, because $$8\\cdot\\frac<1><2>=4$$ and $$6\\cdot\\frac<1><2>=3$$. A menu for lemonade says to have fun with 8 glasses of drinking water and you may 6 lemons. If we fool around with cuatro glasses of water and step three lemons, it generates half of normally lemonade. One another solutions preference an identical, because the and so are equivalent ratios. From inside the an excellent proportional relationship, the values for just one wide variety is for every single increased because of the same matter to find the viewpoints towards almost every other number. For example, inside table all of the property value $$p$$ is equal to four times the worth of $$s$$ on a single line. ## Habit When Han makes chocolates whole milk, the guy includes 2 glasses of milk products which have step 3 tablespoons from delicious chocolate syrup. The following is a table that displays making batches off sizes. Utilize the suggestions throughout the table to-do the fresh new comments. Some terms and conditions are utilized more than once. A specific shade of pink is created with the addition of 3 cups regarding yellow paint to seven glasses of light decorate. A chart", "# A bag contains 12 sweets, of which 5 are red, 4 are green and 3 are yellow. 3 sweets are chosen without replacement. What is the probability that he chooses no red sweets? Jun 30, 2018 $= \\frac{7}{44}$ #### Explanation: For: • $12 = \\left\\{\\begin{matrix}5 R \\\\ 4 G \\\\ 3 Y\\end{matrix}\\right.$ There are $\\left(\\begin{matrix}12 \\\\ 3\\end{matrix}\\right)$ ways of choosing 3 sweets from 12, regardless of color There are $\\left(\\begin{matrix}7 \\\\ 3\\end{matrix}\\right)$ ways of choosing 3 sweets from the non-red sweets. So: P(\"all 3 are non-Red\") = (((7),(3)))/(((12),(3))) = (7!9!3!)/(3!4!12!) $= \\frac{7 \\cdot 6 \\cdot 5}{12 \\cdot 11 \\cdot 10} = \\frac{7}{44}$ Alternatively, using the Hypergeometric distribution: • $P \\left(X = k\\right) = \\frac{\\left(\\begin{matrix}K \\\\ k\\end{matrix}\\right) \\left(\\begin{matrix}N - K \\\\ n - k\\end{matrix}\\right)}{\\left(\\begin{matrix}N \\\\ n\\end{matrix}\\right)}$ • $N = 12$, is the population size (ie, balls): • $K = 5$, is the number of success states (ie, Red balls) in the population, • $n = 3$, is the number of draws • $k = 0$, is the number of observed successes $P \\left(X = 0\\right) = \\frac{\\left(\\begin{matrix}12 \\\\ 0\\end{matrix}\\right) \\left(\\begin{matrix}7 \\\\ 3\\end{matrix}\\right)}{\\left(\\begin{matrix}12 \\\\ 3\\end{matrix}\\right)}$ = (1* ((7!)/(3!4!)))/(((12!)/(9!3!))) = 7/44" ]

[ "+0 0 174 1 Four red candies and three green candies can be combined to make many different “flavors.” Flavors are different if the percent red is different, so “3 red / 0 green” is the same flavor as “2 red / 0 green,” and likewise “4 red / 2 green” is the same flavor as “2 red / 1 green.” If a flavor is to be made using some or all of the seven candies, how many different flavors are possible? May 27, 2020 #1 +25565 +2 Four red candies and three green candies can be combined to make many different “flavors.” Flavors are different if the percent red is different, so “3 red / 0 green” is the same flavor as “2 red / 0 green,” and likewise “4 red / 2 green” is the same flavor as “2 red / 1 green.” If a flavor is to be made using some or all of the seven candies, how many different flavors are possible? $$\\begin{array}{|c|c|c|c|c|} \\hline & {\\color{green}0} & {\\color{green}1} &{\\color{green}2} &{\\color{green}3} \\\\ \\hline {\\color{red}0} & - & 0+1=1 & 0+2=2 & 0+3=3 \\\\ & & \\frac{{\\color{red}0}}{1}: \\frac{{\\color{green}1}}{1} &\\frac{{\\color{red}0}}{2}: \\frac{{\\color{green}2}}{2}&\\frac{{\\color{red}0}}{3}: \\frac{{\\color{green}3}}{3}\\\\ & & {\\color{red}0}: {\\color{green}1} & {\\color{red}0}: {\\color{green}1}&{\\color{red}0}: {\\color{green}1}\\\\ \\hline {\\color{red}1} & 1+0=1 & 1+1=2 & 1+2=3 & 1+3=4 \\\\ & \\frac{{\\color{red}1}}{1}: \\frac{{\\color{green}0}}{1} & \\frac{{\\color{red}1}}{2}: \\frac{{\\color{green}1}}{2}", "words, the $x$x value is $3$3 and the $y$y value is $19$19 #### More Examples ##### Question 1 Plot $13:9$13:9 on the coordinate plane. ##### Question 2 Plot $13:17$13:17 on the coordinate plane. ##### Question 3 Consider the given graph. a) What ratio has been plotted? b) Which option could be being represented by this graph and ratio? A) For every 2 green sweets in a mix, there is 1 red sweet. B) For every 1 green sweet in a mix, there are 2 red sweets.", "# Permutation/Combination problem - choosing 2 objects from different categories Context: there are three candies in a basket (5 red, 4 blue, 5 green). How many ways can you choose 2 candies with different color? Should I use permutation or combination w/ this? I tried using: 14!/(14-2)! = 182 ways; but it's marked as wrong. You can choose $5\\times4$ pairs of candies s.t. one of them is red and one of them is blue. • Yes. Do you agree that there are $20$ possibilities to choose a pair of candies such that one of them is red and the other is blue? Btw, what do you mean with: \"there are three candies in a basket\". Maybe I misunderstand something there. Aren't there $14$ candies in total? – drhab Dec 10 '15 at 15:20 • Well then: $5\\times 4+5\\times5+4\\times5=65$ possibilities. The first term corresponds with \"red and blue\" the second with \"red and green\" and the third with \"blue and green\". – drhab Dec 10 '15 at 15:30", "you are making a simple comparison. Remember to read the information carefully so that you are clear what is being compared. Now let’s look at an example where you might need to figure something out to write the ratio. Example There are 32 red and yellow candies in a bag. There are 10 yellow candies. What is the ratio of red candies to total candies in the bag? We need the ratio of red candies to total candies. We know that there are 32 total candies. We need to find the number of red candies in the bag before writing the ratio. $32-10 = 22$ Now write the ratio of red candies to total candies. Because a ratio is a comparison, it can be simplified. Make sure to reduce the ratio to lowest terms. This makes it easier to understand the quantities that are being compared. $\\frac{\\text{red candies}}{\\text{total candies}} = \\frac{22}{32} = \\frac{11}{16}$ We also could have written this ratio in two other ways. 22 to 32 then simplified to 11 to 16 OR 22 : 32 then simplified to 11 : 16 All of these answers would have been correct. Remember that you can interchange the form that you choose to write a ratio. We can also compare ratios. This is when we have two or more", "Now, without seeing the colour of the wrapper, Harry randomly picked up one of the candies, hoping that it will be a Green Vanilla flavoured candy. What is the probability that the selected candy will have at least one of the two features, wished by Harry? A. $$\\frac{1}{2}$$ B. $$\\frac{11}{20}$$ C. $$\\frac{13}{20}$$ D. $$\\frac{4}{5}$$ E. $$\\frac{17}{20}$$ _________________ Originally posted by EgmatQuantExpert on 19 Oct 2018, 03:24. Last edited by EgmatQuantExpert on 19 Oct 2018, 06:34, edited 1 time in total. Manager Joined: 14 Jun 2018 Posts: 223 Re: Question of the Week- 19 (A store sells equal number of Vanilla .....) [#permalink] ### Show Tags 19 Oct 2018, 06:27 2 1 Ratio of color = 2:3:4:1. Total = 10x out of which 5x is vanilla and 5x is coco. Probability of green = 3/10 ; Not green = 7/10 Probability of vanilla = 1/2 ; coco = 1/2 Probability of at least one of the two features of Green Vanilla = 1- None of the feature = 1 - 7/10 * 1/2 = 13/20 Option C Intern Joined: 17 May 2018 Posts: 49 Re: Question of the Week- 19 (A store sells equal number of Vanilla .....) [#permalink] ### Show Tags 19 Oct 2018, 09:34 You could also quickly write down a simple table to count the desired", "places, we have already picked the grape flavor. Then we can pick one more flavor and the number of ways to do that is 5C1. 2nd option - pack of 3 flavors ( _ _ _ ) Total number of ways the flavor can be picked is 6C3. Now we assume that between the three places, we have already picked the grape flavor. Then we can pick two more flavor and the number of ways to do that is 5C2. 3rd option - pack of 4 flavors ( _ _ _ _ ) Total number of ways the flavor can be picked is 6C4. Now we assume that between the four places, we have already picked the grape flavor. Then we can pick three more flavor and the number of ways to do that is 5C3. Therefore, probability of picking a box with Grape flavor is (5C1 + 5C2 + 5C3)/(6C2 + 6C3 + 6C4) = 25/50 = 1/2. A certain candy store sells jellybeans in the following six flavors on [#permalink] 21 Dec 2018, 07:10 Display posts from previous: Sort by", "2. 14 3. 15 4. 16 I don't understand \"to affirm that 7 candies share the same color\". Does it mean I should count other colors whose one half also coincides with the first group? I'm aware that when there is some uncertainty about something, the procedure is to assume the most difficult to happen scenario or event, and from then on ruling out possibilities until we can assure that our next pick will guarantee what we need. Examples I saw show this for objects which share entirely one color or shape. But what if they have split colors? Should I understand that they ask to match the identical ones or match the objects (the candies in this case) which have the same set of colors i.e red and white, with ones which one side red and other which have the other have white, but may have its opposing side with a different color? Can somebody clarify this for me? In my attempt to solve this problem what I tried to do was this: Since they ask for 7 candies, then the hardest possibly to happen scenario might be: 6 white and orange + 6 green and red + 1 of any of the remaining group (could be blue and green or red and white or orange and", "How many different combinations can be arranged if no $2$ red candies can be next to each other? A candy shop has $$4$$ different blue candies, $$7$$ different green candies, and 3 different red candies. How many different combinations can be arranged if no $$2$$ red candies can be next to each other? My solution: There's $$3$$ choose $$2 = 3$$ ways to pick $$2$$ red candies, and $$14!$$ ways to arrange all the candies in total. To subtract all of the cases where $$2$$ red candies are next to each other, I did $$14!-3 \\cdot 13!$$ and I got a huge number. Can someone help me figure out how to do this problem? • First arrange candies of blue and green colors. Then place red candies in spaces between. Jun 15, 2021 at 17:44 • So would it be $11! - 3 \\cdot 10!$? Jun 15, 2021 at 17:49 The answer is definitely going to be a huge number, but not necessarily that one. Here's a nice trick that helps solve problems where you need to keep things separated. We can arrange all of the blue and green candies in 11! ways. Now, think about the twelve possible spaces between those candies and on the ends of the line. For each of the three red candies, we", "outcomes, then divide that by the total number of outcomes to obtain the probability. We see that we have 10 coco flavoured candy + 3 green vanilla candy, which makes 13 out of a total of 20 available candy. p = 13 / 20 Attachments table.png [ 3.1 KiB | Viewed 964 times ] _________________ ¿Tienes que presentar el GMAT y no sabes por dónde empezar? ¡Visita GMAT para Principiantes y recibe el curso completo gratis! e-GMAT Representative Joined: 04 Jan 2015 Posts: 2852 Re: Question of the Week- 19 (A store sells equal number of Vanilla .....) [#permalink] ### Show Tags 24 Oct 2018, 01:05 1 Solution Given: • A store sells equal number of Vanilla and Coco flavoured candies, each in one of the coloured wrappers: {Red, Green, Blue and Violet} • The ratio of total number of red, green, blue and violet candies is 2: 3: 4: 1, in every flavour • Wishing a green vanilla flavoured candy, Harry randomly picked up one candy To find: • The probability that the selected candy will be either Green coloured or Vanilla flavoured or both. Approach and Working: • Let us assume the total number of candies = 20x • Then, the number of vanilla and coco flavoured candies will be = 10x and 10x. • And,", "x denotes the number of red candies. Ratio: $$\\frac{Red}{Removed}$$ = $$\\frac{2}{5}$$ => $$\\frac{x}{15}$$ = $$\\frac{2}{5}$$ => x = 6 => Sufficient Hi, Need a clarification. Shouldn't the number of red candies be the same in both options??? Intern Joined: 04 Jun 2018 Posts: 19 Location: Viet Nam GMAT 1: 610 Q47 V27 WE: Supply Chain Management (Internet and New Media) Re: A bag is filled with 36 orange candies and 24 red candies for a total [#permalink] ### Show Tags 26 Dec 2018, 20:14 1 Kezia9 wrote: lybeaver wrote: (1) After remove 15 candies, there are 60-15=45 total candies left. $$\\frac{Orange}{Red}$$ = $$\\frac{2}{1}$$ => Red candies = 15, orange candies = 30 => Sufficient (2) x denotes the number of red candies. Ratio: $$\\frac{Red}{Removed}$$ = $$\\frac{2}{5}$$ => $$\\frac{x}{15}$$ = $$\\frac{2}{5}$$ => x = 6 => Sufficient Hi, Need a clarification. Shouldn't the number of red candies be the same in both options??? In DS question type, your job is to determine whether you can answer the question using the information given. You look at statement (1) first then ask yourself whether it's enough to answer the question. Then you look at statement (2) and ask yourself the same. 2 statements can give you different answers as long as it's correct. Intern Joined: 02 Feb 2018 Posts: 36 Re:", "# On Daily challenge problem I answered this Daily challenge problem correct but was marked as incorrect. I will give a proof for my answer below Assumptions: • Capacity of candies of the cup is constant. • Candies are mixed well, i.e. the concentration of candies is uniform all over the jar. • Candies which are more concentrated are more probable to be selected if you pick any candy randomly from a mixture. Let the capacity of the cup be of $x$ candies After first step: Total candies in Jar $B$=$600+x$ Number of Red candies = $x$ Number of Green candies = $600$ Concentration of Red candies = $C_{AR}=\\frac{x}{600+x}$ Concentration of Green candies = $C_{AG}=\\frac{600}{600+x}$ Taking one cup of candies from jar $B$: Number of Green candies in the cup = $J_G=C_{AG} \\times x=\\frac{600x}{600+x}$ Number of Red candies in the cup = $J_R=C_{AR} \\times x=\\frac{x^2}{600+x}$ After second step: Total number of candies in jar $A$=$800$ Number of Red candies=$N_R=800-x+J_R=800-x+\\frac{x^2}{600+x}=\\frac{800(600+x)-x(600+x)+x^2}{600+x}=\\frac{4800+800x-600x\\cancel{-x^2+x^2}}{600+x}=\\frac{4800+200x}{600+x}$ Number of Green candies = $N_G=J_G=\\frac{600x}{600+x}$ Concentration of Red candies = $C_{BR}=\\frac{N_R}{800}=\\frac{4800+200x}{800(600+x)}$ Concentration of Green candies=$C_{BG}=\\frac{N_G}{800}=\\frac{600x}{800(600+x)}$ After third step: Concentration of Red candies in the mixture = $\\frac{C_{AR}+C_{BR}}{2}=\\frac{1}{2}(\\frac{x}{600+x}+\\frac{4800+200x}{800(600+x)})=\\frac{1}{2}(\\frac{800x+4800+200x}{600(800+x)})=\\boxed{\\frac{2400+500x}{600(800+x)}}$ Concentration of Green candies in the mixture=$\\frac{C_{AG}+C_{BG}}{2}=\\frac{1}{2}(\\frac{600}{600+x}+\\frac{600x}{800(600+x)})=\\frac{1}{2}(\\frac{4800+600x}{600(800+x)})=\\boxed{\\frac{2400+300x}{600(800+x)}}$ If you compare the concentrations of both the candies in the last mixture, you can conclude that Red candies are more concentrated than Green", "Back ## The Great Candy Mixup Below are two jars of candies, with $1000$ red candies in one and $500$ green candies in the other. Let's say we take $200$ red candies from the large jar, thoroughly mix them into the small jar, and then take $200$ random candies from the small jar and put them in the large jar. Which jar now has a higher concentration of the candies it started with? Keep reading to find out, or jump ahead to today's challenge. After all of the mixing, the two jars will look something like this: Let's say we ended up with $x$ red candies in the small jar. We know each jar ends up with the same number of candies it started with, so those $x$ candies must have replaced $x$ green candies. This means there must be exactly $x$ green candies in the right jar, and both jars have been equally \"contaminated\" by the other color if we're only considering the number of \"wrong\" candies. What if we think about this contamination relative to the sizes of the jars, though? The small jar has $500$ candies, so $\\frac{x}{500}$ of its candies are now red instead of green. Meanwhile, $\\frac{x}{1000}$ of the large jar's candies are green instead of red. Looking at things this way, the", "blue candy and 1 green candy left in the jar when you have taken out all the red candies? (Candies of same color are indistinguishable!) P.S. It doesn't require much calculation #### Quant Crack ##### New Member Another interesting question i've found online.. You are taking out candies one by one from a jar that has 10 red candies, 20 blue candies, and 30 green candies in it. What is the probability that there are at least 1 blue candy and 1 green candy left in the jar when you have taken out all the red candies? (Candies of same color are indistinguishable!) P.S. It doesn't require much calculation X_r, X_b and X_g are the last positions of red, blue and green candies respectively, when taken out of jar. Required Event (E) is : X_r < X_g < X<b (call it A) and X_r < X_b < X_g (call it B) Using conditional probability: P(E) = P(A)*P(E|A) + P(B)*P(E|B) P(A) = ( 59!/ (10!*19!*30!) )/ (60!/ (10!*20!*30!) ) = 1/3 P(B) = ( 59!/ (10!*20!*29!) )/ (60!/ (10!*20!*30!) ) = 1/2 P(E|A) = P(X_r < X_g) = P( getting X_r < X_b out of forty candies (10 Reds and 30 Green), = (39!/(10!29!)) / (40! / (10!30!)) = 3/4 similarly P(E|B) = 2/3 So, P(E) = (1/3)*(3/4) +", "= 30 => Sufficient (2) x denotes the number of red candies. Ratio: $$\\frac{Red}{Removed}$$ = $$\\frac{2}{5}$$ => $$\\frac{x}{15}$$ = $$\\frac{2}{5}$$ => x = 6 => Sufficient Manager Joined: 04 Oct 2017 Posts: 77 Re: A bag is filled with 36 orange candies and 24 red candies for a total [#permalink] ### Show Tags 26 Dec 2018, 18:50 lybeaver wrote: (1) After remove 15 candies, there are 60-15=45 total candies left. $$\\frac{Orange}{Red}$$ = $$\\frac{2}{1}$$ => Red candies = 15, orange candies = 30 => Sufficient (2) x denotes the number of red candies. Ratio: $$\\frac{Red}{Removed}$$ = $$\\frac{2}{5}$$ => $$\\frac{x}{15}$$ = $$\\frac{2}{5}$$ => x = 6 => Sufficient Hi, Need a clarification. Shouldn't the number of red candies be the same in both options??? Intern Joined: 04 Jun 2018 Posts: 9 Location: Viet Nam WE: Supply Chain Management (Internet and New Media) Re: A bag is filled with 36 orange candies and 24 red candies for a total [#permalink] ### Show Tags 26 Dec 2018, 19:14 1 Kezia9 wrote: lybeaver wrote: (1) After remove 15 candies, there are 60-15=45 total candies left. $$\\frac{Orange}{Red}$$ = $$\\frac{2}{1}$$ => Red candies = 15, orange candies = 30 => Sufficient (2) x denotes the number of red candies. Ratio: $$\\frac{Red}{Removed}$$ = $$\\frac{2}{5}$$ => $$\\frac{x}{15}$$ = $$\\frac{2}{5}$$ => x = 6 => Sufficient Hi, Need", "3 flavors – Chocolate (C), Strawberry (S) and Vanilla (V). Suppose you do not care which flavor is on top and which one is at the bottom. The different subsets of size 2 would be: CS, CV, and SV. Here, CS is not an ordered arrangement. It just means that the two chosen flavors are chocolate and strawberry. If order matters, then there would be six outcomes – CS, SC, CV, VC, SV, and VS. So the approach is to find the number of permutations (an over count) and then divide the amount of over count. Example 6 Suppose that a math club has 8 students – Abby, Beth, Chloe, Diane, Edward, Frank, George and Henry. In how many ways can we choose three students to form a committee of three? The number of permutations of three students is 8 x 7 x 6 = 336. This count is inflated since there are many repeats reflected in this count. For example, if the chosen students are Abby, Beth and Chloe, then there are 3! = 6 ways to order these three students – ABC, ACB, BAC, BCA, CAB and CBA. So each set of three students is reflected 6 times in the inflated count of 336. So the answer is 336 / 6 = 56. Thus there are", "the same color (all red or all blue) is (9/19)(8/18)(7/17) = 28/323. However, given that Terry has one candy of each color, the probability that Mary also selects one red and one blue candy is simply 9/17. This is because it is equivalent to the probability of her second candy color not being the same as her first. Another way to view it is to see that she can either choose red, then blue, or blue, then red. Each occurs with a probability of (9/18)(9/17) = 9/(2·17), so their combined probability is twice this, or 9/17. Hence the combined probability that Terry and Mary each choose candies of both colors is (10/19)(9/17) = 90/323. Therefore, the probability Terry and Mary choose candies of the same type is (28+90)/323 = 118/323, and since 118 and 323 are relatively prime, the answer is 441. One can also work out the problem for the general case where one has n red candies and n blue candies. The desired probability is $\\displaystyle \\frac{2 \\binom{n}{2}\\binom{n-2}{2} + n^2 (n-1)^2}{\\binom{2n}{2}\\binom{2n-2}{2}} = \\frac{3n^2 - 7n + 6}{8n^2 - 16n + 6}.$ Can you generalize the question to the case where there are r red and b blue candies? ## Question 12, Spring 2007 Exam 4/CMay 27, 2007 Posted by Peter in Exam 4/C. 12. For 200", "be either Green coloured or Vanilla flavoured or both. Approach and Working: • Let us assume the total number of candies = 20x • Then, the number of vanilla and coco flavoured candies will be = 10x and 10x. • And, the number of red, green, blue and violet candies, in each of the two flavours, are 2x, 3x, 4x and x respectively • Now, from the above table, we can see that the number of candies which are either green coloured or vanilla flavoured = 2x + 3x + 4x + x + 3x = 13x • Therefore, the probability = $$\\frac{13x}{20x} = \\frac{13}{20}$$ Hence, the correct answer is option C. Note: The logic behind assuming the total number of candies as 20x is that there are a total of 20 parts ((2+ 3 + 4 +1)*2) of candies of different flavours packed in different coloured wrappers. _________________ Number Properties | Algebra |Quant Workshop Success Stories Guillermo's Success Story | Carrie's Success Story Ace GMAT quant Articles and Question to reach Q51 | Question of the week Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and", "Now let’s look at an example where you might need to figure something out to write the ratio. Example There are 32 red and yellow candies in a bag. There are 10 yellow candies. What is the ratio of red candies to total candies in the bag? We need the ratio of red candies to total candies. We know that there are 32 total candies. We need to find the number of red candies in the bag before writing the ratio. 3210=22\\begin{align*}32-10 = 22\\end{align*} Now write the ratio of red candies to total candies. Because a ratio is a comparison, it can be simplified. Make sure to reduce the ratio to lowest terms. This makes it easier to understand the quantities that are being compared. red candiestotal candies=2232=1116\\begin{align*}\\frac{\\text{red candies}}{\\text{total candies}} = \\frac{22}{32} = \\frac{11}{16}\\end{align*} We also could have written this ratio in two other ways. 22 to 32 then simplified to 11 to 16 OR 22 : 32 then simplified to 11 : 16 All of these answers would have been correct. Remember that you can interchange the form that you choose to write a ratio. We can also compare ratios. This is when we have two or more ratios and we want to figure out which ones are larger and which ones are smaller. Let’s look at", "way. There are 6 ways to choose a flavor. Once you choose, there are 5 ways to choose the next. After that, there are 4 flavors left. which is $$6!/3! = \\frac{6\\cdot 5\\cdot 4\\cdot 3 \\cdot 2 \\cdot1}{3\\cdot2\\cdot 1}=6\\cdot 5\\cdot 4 = 120$$. But, you could have chosen $$\\{$$chocolate,vanilla,strawberry$$\\}$$ and you get the same combination as $$\\{$$vanilla, strawberry, chocolate$$\\}$$ so we have to divide by $$3!=3\\cdot2\\cdot1=6$$ to account for the order of choosing. So the number of combinations of flavors is $$\\pmatrix{6\\\\3}=\\frac{120}{6}=20.$$", "Ratios Ratios compare the values of things. They tell us how much of one thing is made up of another thing, or how much of one thing there is compared to another thing. For example, in the picture, there are 3 red jelly babies and 2 green jelly babies. The ratio of red to green jelly babies is $3:2$, but we can soon fix that (I love red jelly babies)! There are 6 red jelly babies to 4 red jelly babies, so the ratio of red to green jelly babies is $6:4$. But this simplifies to $3:2$ as both $6$ and $4$ have a factor of $2$, so we can divide both sides of the ratio by $2$. Simplifying (or complicating) Ratios We can make equivalent ratios in much the same way as we make equivalent fractions. We just have to make sure we multiply (or divide) both parts of the ratio by the same amount. For example, $6:9$ is the same as $6 \\times 3: 9 \\times 3 = 18:27$ and the same as $6 \\div 3 : 9 \\div 3 = 2: 3$. Example: A Recipe for Puffed Rice Crackles Christo and Sam have a lot of allergies, so we need to make our own, allergy-free, versions of many foods. Here's a recipe they particularly like" ]

[ "+0 # i got 159981...i probably did something wrong. 0 129 1 The people living on the island of Gibberish use the standard Kobish alphabet ( letters, A through T). Each word in their language is 4 letters or less, and for some reason, they insist that all words contain the letter A at least once. How many words are possible? Guest May 28, 2018 #1 +20681 +1 The people living on the island of Gibberish use the standard Kobish alphabet ( letters, A through T). Each word in their language is 4 letters or less, and for some reason, they insist that all words contain the letter A at least once. How many words are possible? $$\\begin{array}{|lr|} \\hline \\text{words with 1 letter : } &1 \\\\ \\text{words with 2 letters: } &39 \\\\ \\text{words with 3 letters: } &1141 \\\\ \\text{words with 4 letters: } &29679\\\\ \\hline \\text{sum: } & 30860 \\\\ \\hline \\end{array}$$ heureka May 28, 2018", "# Hawaii • May 26th 2009, 03:51 PM Sampras Hawaii The Hawaiian alphabet consists of $12$ letters, the vowels $\\text{a}, \\ \\text{e}, \\ \\text{i}, \\ \\text{o}, \\ \\text{u}$ and the consonants $\\text{h}, \\ \\text{k}, \\ \\text{l}, \\ \\text{m}, \\ \\text{n}, \\ \\text{p}$ and $\\text{w}$. (a) How many different four-letter words can be constructed using the $12$-letter alphabet? (b) How about with the English alphabet? (c) What is the second and last letters are vowels and the other $2$ are consonants? (d) What if the second and last letters are vowels, but there are no restrictions on the other $2$ letters? So for (a) it would be $12^4$. For (b) it would be $26^4$. For (c) it would be $5^27^2$. And for (d) it would be $5^212^2$? • May 26th 2009, 05:04 PM amitface Correct.", "+0 # This one is pretty confusing :P not urgent btw 0 1 120 4 +176 The inhabitants of the island of Jumble use the standard Kobish alphabet (20 letters, A through T). Each word in their language is 4 letters or less, and for some reason, they insist that all words contain the letter A at least once. How many words are possible? Oct 18, 2018 #1 +3627 +1 $$\\text{There are }\\dbinom{4}{1}19^3 \\text{ different words containing a single instance of the letter A} \\\\ \\dbinom{4}{1} \\text{ spots for the A and then a choice of 19 letters, B-T, for each of the 3 remaining spots}\\\\ \\text{Similarly there are }\\dbinom{4}{k}19^{4-k} \\text{ words containing }k \\text{ instances of the letter A}\\\\ \\text{Thus we have}\\\\ N=\\sum \\limits_{k=1}^4 ~\\dbinom{4}{k}19^{4-k} = 29679$$ It strikes me that perhaps a simpler way of doing this is to subtract all the words with no instances of the letter A from the total number of words possible. $$\\text{There are }19^4 \\text{ words that have no A in them}\\\\ \\text{There are }20^4 \\text{ total possible words}\\\\ N=20^4-19^4 = 29679$$ . Oct 18, 2018 edited by Rom Oct 18, 2018 edited by Rom Oct 18, 2018 #2 +176 0 Thank you for trying but apparently the answer was wrong, here is the solution I got back: We consider", "by skipjack there are 26×25 = 650 two-letter words, the 234th being ji. There are 26×25×25 = 16250 three-letter words. Etc. Thank you skipjack, I found out the number of words in this way; $\\displaystyle 26 × 25^1 + 26 × 25^2 + 26 × 25^3 + 26 × 25^4 = 10579400$ words. But how did you find the 234th word? And how to find the rank of the word \"PAPER\"? October 20th, 2018, 04:32 AM #6 Global Moderator Joined: Dec 2006 Posts: 20,628 Thanks: 2077 Consider how many words precede JA and how many precede PABAB. October 20th, 2018, 09:39 PM #7 Member Joined: Sep 2017 From: Saudi Arabia Posts: 37 Thanks: 1 Quote: Originally Posted by skipjack Consider how many words precede JA and how many precede PABAB. Dear skipjack, The number of words before \"JA\" is much greater than 234, because if we consider the words which start with \"A\" only; *that have two letters* = 1 x 25 = 25 *that have three letters* = 1 x 25 x 25 = 625 *that have four letters* = 1 x 25 x 25 x 25 = 15625 *that have five letters* = 1 x 25 x 25 x 25 x 25 = 390625 So, there are (25+625+15625+390625) or 406900 words start with \"A\", all of", "Free Version Difficult # Ordering Letters in a Word . . . CREMATORIUM? ALGEBR-0PF3NB How many ways can you order the letters in the word CREMATORIUM? A $4,989,600$ B $39,916,800$ C $19,958,400$ D $9,979,200$", "lowercase letters is the same, and there is no additional surcharge or tax. How much would the following people pay to buy the letters in their names? Those of you who know a little algebra will have no trouble with that problem. Those of you who don’t shouldn’t have too much trouble, either. But then, I realized I could extend the problem for some added fun. And who am I to keep fun things to myself? So, here ya go. At first, I thought the store was engaging in human trafficking. But then I realized that $269 was the price for the bronze letters that had been used to spell the name Eli. Inside the store was a price list for other names: AIDEN – 491 AL – 248 ART – 267 BEA – 290 EARL – 415 DANE – 399 ED – 135 ELI – 269 FAY – 220 GABI – 289 HAL – 284 IVY – 143 JACK – 234 JAY – 232 KO – 60 KAI – 283 LEXI – 272 MAVIS – 363 MAX – 215 NED – 225 PAT – 210 PERRI – 330 QI – 93 QUIN – 199 SAMMY – 338 WILL – 243 ZENO – 243 The store didn’t have a list of prices for the individual letters, but then", "how widely the number of syllables varies in $17$ languages: $\\begin{array}{lr} \\textbf{Language} & \\textbf{Syllables} \\\\[1em] \\text{English (UK)} & 6949 \\\\ \\text{German} & 5100 \\\\ \\text{Hungarian} & 4325 \\\\ \\text{Finnish} & 3844 \\\\ \\text{Serbian} & 3831 \\\\ \\text{Catalan} & 3600 \\\\ \\text{Turkish} & 3260 \\\\ \\text{French} & 2949 \\\\ \\text{Spanish} & 2778 \\\\ \\text{Vietnamese} & 2776 \\\\ \\text{Italian} & 2729 \\\\ \\text{Thai} & 2428 \\\\ \\text{Basque} & 2082 \\\\ \\text{Cantonese} & 1298 \\\\ \\text{Mandarin Chinese} & 1274 \\\\ \\text{Korean} & 1104 \\\\ \\text{Japanese} & 643 \\end{array}$ # What's the Fastest Language? There are many more one-syllable words, like \"mouse,\" in English than there are in Japanese. In general, because English has more syllables, it should be possible to express an idea with fewer syllables than the equivalent word in Japanese. For example, \"mouse\" is three syllables in Japanese (ねずみ \"ne-zu-mi\"). To get a sense for this, we can count the number of syllables in the $12$ Zodiac animals in English and Japanese and compare the average number of syllables. For this set, English averaged $\\sim 1.4$ syllables per word while Japanese averaged $\\sim 2.4.$ If the English word for an animal is \"mouse\" $(1$ syllable$)$ and the Japanese is \"ne-zu-mi\" $(3$ syllables$),$ which of these has more information per syllable? Animal English syllables Japanese syllables 🐭 $1$ $3$ 🐮", "usage; for instance, the syllable wū does not exist in Hawaiian, and the syllable wu occurs only in two loan-words from English. However, for the purposes of this post I am calculating only how many words could theoretically exist, not how many actually do (which would require exhaustive knowledge of the language that I do not possess). So for single syllables, we have a total of $$10+15+80+120=225$$. Now, since any word in Hawaiian will be made up of these syllables, we can quickly calculate how many distinct words of a given number of syllables would be able to exist. Since Hawaiian has nothing against duplication of sounds (and often rather encourages it), for a word of two syllables we may have any of the 225 for the first syllable, and any of the 225 for the second. Thus to find the total number of two-syllable words we just multiply those two numbers (or equivalently raise 225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our", "the school board until they buy one of these for \"learning purposes\" And all for the amazingly low price of only \\$500!!! Probably less than an ELMO =] Outstanding point. eBay here I come! × Pasted as rich text. Paste as plain text instead Only 75 emoji are allowed.", "in the material. • Braille is read by moving the hand or hands along each line from left to right. • Reading is generally done with both hands, with the index fingers doing the majority of the work. • Reading speed is around $$125$$ words per minute on average. However, higher speeds of up to $$200$$ words per $$minute$$ can be achieved. Braille-Pattern Louis Braille’s legacy has enlightened the lives of millions of people who are blind. Blind individuals from all over the world get benefitted from Braille’s work daily. Reference: https://pixabay.com/vectors/braille-alphabet-dots-spots-text-6159673/ https://en.wikipedia.org/wiki/File:English_braille_sample.jpg https://freesvg.org/louis-braille", "225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that it would be nearly impossible to come up with concepts for all of them (for words of 5 syllables, the number is 1,078,203,909,375, over", "$32$ characters (inclusive), and uses only characters a–z. • The number of meanings of a word in the dictionary is at least $1$ and at most $10\\, 000$. • Dictionary words are distinct. Sample Input 1 Sample Output 1 5 heimark hei 2 mark 2 heim 1 ark 2 heima 1 6", "## Kickstarter ad absurdum For the culture that has everything: Emoji Dick is a crowd sourced and crowd funded translation of Herman Melville's Moby Dick into Japanese emoticons called emoji. Each of the book's approximately 10,000 sentences has been translated three times by a Amazon Mechanical Turk worker. These results have been voted upon by another set of workers, and the most popular version of each sentence has been selected for inclusion in this book. In total, over eight hundred people spent approximately 3,795,980 seconds working to create this book. Each worker was paid five cents per translation and two cents per vote per translation. The funds to pay the Amazon Turk workers and print the initial run of this book were raised from eighty-three people over the course of thirty days using the funding platform Kickstarter. By my calculation, that's something like 10000*3*.07 = $2,100, which is not much for a translation of Moby Dick, but a lot for a joke that probably gets tiresome after a couple of pages. From a reviewer who gives the book zero stars: This book has a \"Preview\" link that only shows 12 pages… which is enough to show the title pages and credits… and ABSOLUTELY NOTHING OF THE EMOJI itself. How one can judge whether to buy a$200 hardcover from", "with my sons recently: A local hardware store sells bronze letters. However, the letters vary in price; some are more expensive than others. When I was at the store the other day, four people purchased the letters in their names. Their names and the prices they paid were: Aiden $491 • Ned$225 • Dane $399 • Ed$135 The price of a name is equal to the sum of the prices of its letters. The price for uppercase and lowercase letters is the same, and there is no additional surcharge or tax. How much would the following people pay to buy the letters in their names? Those of you who know a little algebra will have no trouble with that problem. Those of you who don’t shouldn’t have too much trouble, either. But then, I realized I could extend the problem for some added fun. And who am I to keep fun things to myself? So, here ya go. At first, I thought the store was engaging in human trafficking. But then I realized that $269 was the price for the bronze letters that had been used to spell the name Eli. Inside the store was a price list for other names: AIDEN – 491 AL – 248 ART – 267 BEA – 290 EARL – 415 DANE –", "is produced in. Other names are v and w, above 8 and 9. Jumble is a word puzzle which is played by scrambling the letters to make an anagram, based on some clue. Get in touch; © 2020 White Rose Maths | Registered in England 10831473 | Privacy | Cookies. Using laminated letter cards, place letters on the chalkboard tray with a blank card indicating a missing letter in some location of the word. Nth Term in a Sequence Here is the sequence: 1, 2, 5, 14 Find the following 2 terms and a formula for the nth term. This can be check in other files as well where we are writing HTML as well as PHP. Available as a mobile and desktop website as well as native iOS and Android apps. Given a fixed cost, variable cost, and revenue function or value, this calculates the break-even point Features: Calculator | Practice Problem Generator Examples (2): C(x) = 125x + 1500 and R(x) = 1500x - 1000, canoes has a fixed cost of$20,000. What letter is next in this sequence? O, T, T, F, F, S, S, E, ? So were you able to solve the riddle? Leave your answers in the comment section below. Albany, New York 12234. Just enter your & your partner name to check", "2. ## Re: English alphabet maths Originally Posted by mathewnaren The correct answer is $26^4$, I do not that in the list of options.", "asked is, \"How many syllables are possible?\". We can see that there are 10 possible one-letter syllables consisting of the 5 vowels plus 5 long vowels, plus an additional 15 syllables comprised of a single diphthong, plus an additional 80 syllables formed by taking one of the 8 consonants and adding one of the 10 vowels to it, plus an additional 120 syllables formed by taking one of the 8 consonants and adding one of the 15 diphthongs. Not all of these syllables actually exist in common usage; for instance, the syllable wū does not exist in Hawaiian, and the syllable wu occurs only in two loan-words from English. However, for the purposes of this post I am calculating only how many words could theoretically exist, not how many actually do (which would require exhaustive knowledge of the language that I do not possess). So for single syllables, we have a total of $$10+15+80+120=225$$. Now, since any word in Hawaiian will be made up of these syllables, we can quickly calculate how many distinct words of a given number of syllables would be able to exist. Since Hawaiian has nothing against duplication of sounds (and often rather encourages it), for a word of two syllables we may have any of the 225 for the first syllable, and any", "bronze letters. However, the letters vary in price; some are more expensive than others. When I was at the store the other day, four people purchased the letters in their names. Their names and the prices they paid were: Aiden $491 • Ned$225 • Dane $399 • Ed$135 The price of a name is equal to the sum of the prices of its letters. The price for uppercase and lowercase letters is the same, and there is no additional surcharge or tax. How much would the following people pay to buy the letters in their names? Those of you who know a little algebra will have no trouble with that problem. Those of you who don’t shouldn’t have too much trouble, either. But then, I realized I could extend the problem for some added fun. And who am I to keep fun things to myself? So, here ya go. At first, I thought the store was engaging in human trafficking. But then I realized that $269 was the price for the bronze letters that had been used to spell the name Eli. Inside the store was a price list for other names: AIDEN – 491 AL – 248 ART – 267 BEA – 290 EARL – 415 DANE – 399 ED – 135 ELI – 269 FAY –", "of the 225 for the second. Thus to find the total number of two-syllable words we just multiply those two numbers (or equivalently raise 225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that", "Kattis # Spelling Bee An example of the Daily NY Times Spelling Bee Puzzle The New York Times publishes a daily puzzle called the “Spelling Bee.” In this puzzle, $7$ letters are shown in a hexagonal arrangement of $6$ letters around a center letter. The task is to come up with as many words as possible that • contain only letters that are displayed in the hexagon, • are at least of length $4$, and • contain the center letter. A letter may be used more than once, and not all letters need to be used. After playing for a while, you get stuck, but then you remind yourself that the Linux distribution on your computer comes with a machine-readable file of $102\\, 305$ dictionary words in /usr/share/dict/words! You decide that even if you can’t excel at the Spelling Bee you can still excel at programming, so you decide to write a program that finds all solutions to a Spelling Bee puzzle from your dictionary. ## Input The input consists of a single test case, which starts with a line with $7$ distinct lowercase English letters. The first of these letters is the center letter. The next line contains an integer $n$ ($1 \\le n \\le 102\\, 305$), the size of the dictionary. This line is followed by" ]

[ "the opposite; we try to find the number of words that do not contain A, and then subtract it from the total possible number of words. So we have a few cases to consider: One letter words: There is only 1 one-letter word that contains A, that's A. Two letter words: There are 19*19=361 words that do not contain A. There is a total of 20*20=400 words, so we have 400-361=39 words that satisfy the condition. Three letter words: There are 19*19*19=6859 words without A, and there are 20^3=800 words available. So there are 1141 words that satisfy the condition. Four letter words: Using the same idea as above, we have 20^4-19^4 words satisfying the requirement. So this gives a total of 1+39+1141+29679=30860 words. Oct 18, 2018 edited by ANotSmartPerson Oct 18, 2018 #4 +3627 0 oh I missed the \"4 letters or less Rom Oct 18, 2018 #3 +1 20^4 - 19^4 =29,679 - Four-letter words 20^3 - 19^3 =1,141 - Three-letter words 20^2 - 19^2 =39 - two-letter words. 20^1 - 19^1 =1 - one-letter word. 29,679 + 1,141 + 39 + 1 =30,860 - total number of words in Kobish alphabet. Oct 18, 2018", "of a $2$-letter word can be chosen in $26$ ways. For each of these ways, the second letter can be chosen in $26$ ways, for a total of $26^2$. (If $26$ women each have $26$ children, then the total number of children is $26^2$.) The same idea shows that there are $26^3$ $3$ letter words, $26^4$ $4$-letter words, and so on. However, we don't even want to count all the $3$-letter words, only those up to $bbb$. How many $3$-letter words start with $a$? We can complete $a$ to a $3$-letter word by appending to $a$ any $2$-letter word. There are $26^2$ such $2$-letter words. So $26^2$ $2$-letter words start with $a$. Finally, we want to count the $3$-letter words that start with $b$, but we only want to get to $bbb$. So append any $2$-letter word that starts with $a$ (there are $26$ of these), and finally at the very end we must not forget about $bba$ and $bbb$ (2 words). Add up. We get $1406$. But probably you did not intend to count the first word, just the \"iterations\" or \"increments.\" There are $1405$ of these. B) Using Base 26 Representation: The $1$-letter words were easy to count, as were the $2$-letter words. The $3$-letter words were a bit of a nuisance, since we did not", "7 times, so our count will be too high, contradiction. We're still assuming $n\\leq23$, so now $n=23$ is the only case left. Here \"times\" is used 24 times, \"time\" is used 109560 times, and so \"one\" is used 109564 times. Now \"four\" is used at least 3 times, so it becomes the 21st word on our list. We know \"two\" is used at least 8 times; let's say it's used 9 times, so that \"nine\" is used 7 times and \"seven\" is the 22nd word on our list. After a little trial-and-error and deduction, I got to the following list: • \"this\" ($2$ times) • \"sentence\" ($2$ times) • \"uses\" ($2$ times) • \"or\" ($2$ times) • \"mentions\" ($2$ times) • \"and\" ($2$ times) • \"the\" ($109583$ times) • \"word\" ($109583$ times) • \"times\" ($24$ times) • \"time\" ($109560$ times) • \"one\" ($109564$ times) • \"hundred\" ($9$ times) • \"thousand\" ($5$ times) • \"eighty\" ($3$ times) • \"nine\" ($7$ times) • \"five\" ($6$ times) • \"two\" ($9$ times) • \"three\" ($7$ times) • \"twenty\" ($2$ times) • \"sixty\" ($3$ times) • \"four\" ($3$ times) • \"seven\" ($3$ times) • \"six\" ($2$ times) ... which all works out, giving the solution I put at the top. Hurray! :-D • +1 for the great effort. Making it self-contained will be", "## Discrete Mathematics and Its Applications, Seventh Edition Published by McGraw-Hill Education # Chapter 6 - Section 6.1 - The Basics of Counting - Exercises - Page 396: 16 66351 #### Work Step by Step Let's first find the total number of strings of four lowercase letters There are 26 letters in the alphabet. So, number of choices: 4 no. of options: 26 therefore the number of words = $26\\times26\\times26\\times26=26^4$ Now, let's find the number of strings of four lowercase letters which $do$ $not$ have $x$ in them. As if the alphabet didn't have $x$ in it. So, number of choices: 4 no. of options: 25 (all the letters except $x$) therefore the number of words without any $x$ = $25\\times25\\times25\\times25=25^4$ therefore the number of words with at least one $x$ = the total number of words - the number of words without any $x$ =$26^4 - 25^4$ =456976 - 390625 =66351 After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "Another maths curiosity from the Futility Closet: Fortuitous numbers In American usage, 84,672 is said EIGHTY FOUR THOUSAND SIX HUNDRED SEVENTY TWO. Count the letters in each of those words, multiply the counts, and you get 6 × 4 × 8 × 3 × 7 × 7 × 3 = 84,672. Here’s something I’ve (pointlessly) struggled with for a long time, now. Can you complete this sentence? Written as words, there are _____________ letters in this sentence. Use Excel’s LEN() function and AutoSum and try it like this, writing it out one word at a time. So, forty three letters so far, with those two empty boxes. If you were to write forty three into those boxes, the total would obviously be more than forty three. A little trial-and-error, and we get the answer fifty three. Well, that was fairly straightforward. Let’s try a slightly different sentence. Maybe this isn’t so difficult, after all. One more? That’s not right, there are forty nine letters in that sentence, not forty eight. But now there are forty eight. Is it not possible to accurately complete that sentence, then?", "word length) $26\\times26\\ldots(42 times)=26^{42}=2.68\\times10^{59}$ A 59 digit number! It could obviously go higher if using 100 or 150 letter words comes into vogue. (I hear that the Germans already do this!) This doesn’t really appear ‘realistic’. So let’s see how we can arrive at that figure. First, we note, that since the right side of the Gaussian can extend asymmetrically much more than the left side which is bound by 1 letter words. (Zero and negative letter words don’t exist!). However, since the likelihood of words greater than 10 is very low, we can for current purposes ignore them. ( This is an approximation used to make my life easier to solve this. If you are interested, you could as well use MATLAB to plot a skewed Gaussian and let me know about the percentage error due to this. I’m reasonably confident it won’t be too significant.) Therefore about 67% of words are: $26\\times26\\times5+26\\times26\\times5\\times5+26\\times26\\times5\\times26\\times5+26\\times26\\times5\\times26\\times5\\times5+26\\times26\\times5\\times26\\times26\\times5\\times5=69006080$ The above accounts for the frequency of vowels in all words of length 3–7 as well. That is about 69 million. A 100% (total no.) of possible words would be about 103503945. Or around 100 million total words. Dictionary Facts – Oxford English Dictionary says that they have around 414,800 ‘entries’ defined. These are the words that we most use on a day to", "The word \"time\" is going to be used $109582-n+1$ times (once for each word used exactly once, once for its own appearance in the list). The only words which will be used more than $n+1$ times are \"the\", \"word\", \"one\", and \"time\", which each appear at least $109582-n+1$ times. So the word \"hundred\" is going to be used nine times, the word \"thousand\" five times, and the word \"eighty\" three times. The bolded numbers above tell us that \"one\", \"nine\", \"five\", \"two\", and \"three\" will each be used more than once. So already $n$ is at least eighteen. Assuming $n$ is at least nineteen and less than twenty-nine (which I'm pretty sure it will be), the word \"twenty\" is used two times (once for the number of times \"times\" is used and once for its own appearance in the list). So far we have the following words appearing more than once: • \"this\" ($2$ times) • \"sentence\" ($2$ times) • \"uses\" ($2$ times) • \"or\" ($2$ times) • \"mentions\" ($2$ times) • \"and\" ($2$ times) • \"the\" ($109583$ times) • \"word\" ($109583$ times) • \"times\" ($n+1$ times) • \"time\" ($109582-n+1$ times) • \"one\" ($\\geq109582-n+1+4$ times) • \"hundred\" ($9$ times) • \"thousand\" ($5$ times) • \"eighty\" ($3$ times) • \"nine\" ($\\geq6$ times) • \"five\" ($\\geq6$ times) • \"two\" ($\\geq8$", "word SIMILARITIES, so that there are no words with four consecutive I's. There are twelve letters, including four I's and two S's. There are: . $\\frac{12!}{4!\\,2!} \\;=\\;239,500,800$ possible \"words\". Duct-tape the four I's together. Then we have nine \"letters\" to arrange: . $A, E, L, M, R, S, S, T, \\boxed{IIII}$ . . They can be arranged in: . $\\frac{9!}{2!} \\;=\\;181,440$ ways Hence, there are $181,440$ \"words\" with four consecutive I's. Therefore, there are: . $239,500,800 - 181,440 \\;=\\;\\boxed{239,319,360}$ \"words\" . . without four consecutive I's.", "count the bad words and subtract from $720$. There are $240$ bad words that begin with D or E, and $240$ that end with A or B. Add. We get $480$. However this double counts the $96$ bad words that begin with D or E and end with A or B. Thus the number of bad words is $240+240-96$.", "are used. The local language has 66 different letters. The parliament decided to forbid the use of the seventh letter. How many words have the people of Oz lost because of the prohibition? A. 65 B. 66 C. 67 D. 131 E. 132 Two ways to do this one. 1. with 66 letters, how many one or two letter words can we make? $$67 * 66$$ -------(1) we can fill the first position with 67 different things(66 letters or a blank), and the second position with 66 different letters. With 65 letters after removing the 7th letter, how many words do we have? $$66 * 65$$ -------(2) (1) - (2) = 132(E) 2. Now that we have removed the 7th letter, we have lost all the single and double letter words that we could make which include this 7th letter. Single letter words with 7th letter? $$1$$ Double letter words? $$2 * 65 + 1$$ ---Double letter words by using 7th letter only once + Double letter word by using 7th letter twice Total = 132(E) _________________ Spread some love..Like = +1 Kudos Non-Human User Joined: 09 Sep 2013 Posts: 11717 Re: In the Land of Oz only one or two-letter words are used. The [#permalink] ### Show Tags 20 Aug 2018, 12:02 Hello from the GMAT Club", "of the 225 for the second. Thus to find the total number of two-syllable words we just multiply those two numbers (or equivalently raise 225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that", "number of words possible (no repetition): $26*25*24 = 650*24 = 15600$ Total number of words possible with only asymmetric letters: $15*14*13 = 210*13 = 2730$ Total number of words with at least one symmetric letter: $15600 - 2730 = 12870$ Bisen () Kudos Bob, elegant and simple! Permutations and combinations are my weaker part and you do help me here a lot. I approached from the usual side: 1) Combinations of types of letters: AAA - 1 AAS - 3 (3C2 = 3) SSA - 3 (3C2 = 3) 2) Permutations within each combination AAA = 11 x 10 x 9 = 990 AAS = 15 x 14 x 11 = 2310 SSA = 11 x 10 X 15 = 1650 3) Multiply by number of combinations and add those 990 + 2310 x 3 + 1650 x 3 = 12,870 Juhi () Yes finding total possible way and removing the exception is the easy way to approach this kind of question. good question.", "IIII: About $10^4$ words/book $\\times 9$ books $\\approx 10^5$ words. • Todai entrance exams: About $4000$ words/year $\\times 50$ years $\\approx 2 \\times 10 ^ 5$ words. • Todai open exams: About $4000$ words/sets $\\times 50$ sets $\\approx 2 \\times 10 ^ 5$ words. So I will have read $6 \\times 10 ^ 5$ words very precisely after $3$ years later. Of course this is not enough at all. But if I want to work more, it will lack the quality. I continue to work on them and after finishing all of them, I’ll consider the plan again. Tags: Categories:", "225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that it would be nearly impossible to come up with concepts for all of them (for words of 5 syllables, the number is 1,078,203,909,375, over", "at 16:43 @nico Thank you for that clarification. Theoretically a similar issue pertains in dictionaries containing all possible two-letter combinations as words: when that's the case, any hand of 2 or more letters automatically contains a word. @shabbychef's comment about mid-game shows how irrelevant the original question is to most of Scrabble, because in mid-game you have available an array of word parts (prefixes, suffixes, and even middle sections) in addition to the 7 letters in your hand. This greatly increases the chances of being able to make a word. – whuber Oct 22 '10 at 16:29 4 Answers This is a (long!) comment on the nice work @vqv has posted in this thread. It aims to obtain a definitive answer. He has done the hard work of simplifying the dictionary. All that remains is to exploit it to the fullest. His results suggest that a brute-force solution is feasible. After all, including a wildcard, there are at most $27^7 = 10,460,353,203$ words one can make with 7 characters, and it looks like less than 1/10000 of them--say, around a million--will fail to include some valid word. The first step is to augment the minimal dictionary with a wildcard character, \"?\". 22 of the letters appear in two-letter words (all but c, q, v, z). Adjoin a wildcard", "# Basic question. Does not involve permutations/combinations. $3$ mailboxes containing $3$ letters My teacher explained this problem to us - \"There are $3$ mailboxes. $3$ people put letters in at random. There is no preference for any of the $3$ mailboxes. Compute the probability that each mailbox contains $1$ letter.\" I tried this problem on my own and got the wrong answer. I understand the teacher's solution but I don't get what went wrong with my answer. Can someone explain where I went wrong? My answer: Treat each outcome as a sequence of $3$ numbers denoting the number of letters in each mailbox, e.g $300$ means $3$ in the 1st, $0$ in the second and $0$ in the 3rd mailbox. Sample space $S = \\{300, 030, 003, 111,012,021, 120,102,210,201\\}$ $S$ contains $10$ sample points. Let $A =$ event that each mailbox is chosen once. $P(A) = 1/10$ Teacher's solution: Treat each outcome as a sequence of $3$ numbers denoting the person and their corresponding chosen mailbox, e.g $312$ means the first person chose mailbox 3, the second chose mailbox 1, the third chose mailbox 3. So the sample space is $S = \\{ 111, 112...\\}$ and contains $27$ sample points. Now let $A =$ event that each mailbox is chosen once. $A =\\{123,132,213,231,312,321\\}$ So the answer is $P(A)", "list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that it would be nearly impossible to come up with concepts for all of them (for words of 5 syllables, the number is 1,078,203,909,375, over a trillion words!). Conclusion: Although the sound range of Hawaiian may sound limited to the Anglophone ear with our 44-odd sounds, combinatorics shows that Hawaiian is capable of forming astronomically many times more words than even the most wordy languages on Earth. I'd do a similar calculation for English, but for the fact that it would be several orders of magnitude harder. This", "the 1000 words, then the thousand would be worth from 5/6 to 8/–.— I think perhaps the fairest way will be to strike a sort of rough average of all the work & all the prices, & reckon each MS. slip, such as the enclosed at eighteen pence, or the thousand words at seven shillings.— I will leave it to you to decide upon this knotty question,— if the above charge be too much I will endeavour to lower it.— I am afraid you will think that the above discussion of charges & prices is very disgusting, but as this, say what we may, is after all a matter of business, I thought it best to put it in as business-like a form as I could.— If I could realise my dearest wish of getting back again to London, I might, perhaps be able to assist you without so much consideration of the filthy lucre, although perhaps I might find my hands even fuller.— Under any circumstances I must thank you for all your kindnesses up to this time, & even for Mr. Murray’s liberality, of which you have been the prime mover—4 Believe me | Your’s always truly | W. S. Dallas. ## CD annotations 1.7 not … hurry, ] scored pencil 2.7 eighteen … shillings.— 2.8]", "back in the 1 for 1-letter codes gives a total of 132. Very similar to a probability or Venn diagram question where you often double count elements. Hope this helps! -Ron _________________ Manager Joined: 18 Jul 2013 Posts: 73 Location: Italy GMAT 1: 600 Q42 V31 GMAT 2: 700 Q48 V38 GPA: 3.75 Re: In the Land of Oz only one or two-letter words are used. The [#permalink] ### Show Tags 10 Nov 2013, 13:23 hi, i solved it that way : Total words lost = total words with 66 letters - total words with 65 letters 1) total words with 66 letters = 66 (total words with 1 letter) + 66P64 (total words with 2 different letters : AB, AC etc.) + 66 (total words of 2 identical letters : AA, BB etc.) = $$66+66*65+66$$ = $$66*2+66*65$$ 2) total words with 65 letters = 65 (total words with 1 letter) + 65P64 (total words with 2 different letters : AB, AC etc.) + 65 (total words of 2 identical letters : AA, BB etc.) = $$65+65*64+65$$ = $$65*2+65*64$$ Finally the answer is :66*2+66*65 (1) -65*2-65*64 (2) = 2 + 65*(66-64) = 132 Intern Joined: 08 Jun 2016 Posts: 13 Re: In the Land of Oz only one or two-letter words are used. The [#permalink] ### Show Tags", "# How many words can be formed, given $4$ letters, and in each word there must be at least two letters are the same? How many words can be formed, given $4(a,b,c,d)$ letters, and in each word from $4$ letters there must be at least two letters are the same? The position of the letter doesn't matter. The answer is $232,$ I don't know how to \"attack\" it. • How long are the words? – almagest May 19 '16 at 20:16 • 4 letters, forgot the add this. – Planet_Earth May 19 '16 at 20:16 • What do you mean \"position doesn't matter\"? Is \"abbc\" and \"abcb\" equal? – Fred Yang May 19 '16 at 20:24 • There are $4^4$ possible words in total. There are $4!=24$ words with no repeats. Hence $256-24=232$ with at least one repeat. – almagest May 19 '16 at 20:25 • @almagest, give this as answer so I can mark it as best. – Planet_Earth May 19 '16 at 20:26 There are $4^4=256$ possible words in total. There are $4!=24$ words with no repeats. Hence $256−24=232$ with at least one repeat." ]

[ "+0 # i got 159981...i probably did something wrong. 0 129 1 The people living on the island of Gibberish use the standard Kobish alphabet ( letters, A through T). Each word in their language is 4 letters or less, and for some reason, they insist that all words contain the letter A at least once. How many words are possible? Guest May 28, 2018 #1 +20681 +1 The people living on the island of Gibberish use the standard Kobish alphabet ( letters, A through T). Each word in their language is 4 letters or less, and for some reason, they insist that all words contain the letter A at least once. How many words are possible? $$\\begin{array}{|lr|} \\hline \\text{words with 1 letter : } &1 \\\\ \\text{words with 2 letters: } &39 \\\\ \\text{words with 3 letters: } &1141 \\\\ \\text{words with 4 letters: } &29679\\\\ \\hline \\text{sum: } & 30860 \\\\ \\hline \\end{array}$$ heureka May 28, 2018", "+0 # This one is pretty confusing :P not urgent btw 0 1 120 4 +176 The inhabitants of the island of Jumble use the standard Kobish alphabet (20 letters, A through T). Each word in their language is 4 letters or less, and for some reason, they insist that all words contain the letter A at least once. How many words are possible? Oct 18, 2018 #1 +3627 +1 $$\\text{There are }\\dbinom{4}{1}19^3 \\text{ different words containing a single instance of the letter A} \\\\ \\dbinom{4}{1} \\text{ spots for the A and then a choice of 19 letters, B-T, for each of the 3 remaining spots}\\\\ \\text{Similarly there are }\\dbinom{4}{k}19^{4-k} \\text{ words containing }k \\text{ instances of the letter A}\\\\ \\text{Thus we have}\\\\ N=\\sum \\limits_{k=1}^4 ~\\dbinom{4}{k}19^{4-k} = 29679$$ It strikes me that perhaps a simpler way of doing this is to subtract all the words with no instances of the letter A from the total number of words possible. $$\\text{There are }19^4 \\text{ words that have no A in them}\\\\ \\text{There are }20^4 \\text{ total possible words}\\\\ N=20^4-19^4 = 29679$$ . Oct 18, 2018 edited by Rom Oct 18, 2018 edited by Rom Oct 18, 2018 #2 +176 0 Thank you for trying but apparently the answer was wrong, here is the solution I got back: We consider", "the opposite; we try to find the number of words that do not contain A, and then subtract it from the total possible number of words. So we have a few cases to consider: One letter words: There is only 1 one-letter word that contains A, that's A. Two letter words: There are 19*19=361 words that do not contain A. There is a total of 20*20=400 words, so we have 400-361=39 words that satisfy the condition. Three letter words: There are 19*19*19=6859 words without A, and there are 20^3=800 words available. So there are 1141 words that satisfy the condition. Four letter words: Using the same idea as above, we have 20^4-19^4 words satisfying the requirement. So this gives a total of 1+39+1141+29679=30860 words. Oct 18, 2018 edited by ANotSmartPerson Oct 18, 2018 #4 +3627 0 oh I missed the \"4 letters or less Rom Oct 18, 2018 #3 +1 20^4 - 19^4 =29,679 - Four-letter words 20^3 - 19^3 =1,141 - Three-letter words 20^2 - 19^2 =39 - two-letter words. 20^1 - 19^1 =1 - one-letter word. 29,679 + 1,141 + 39 + 1 =30,860 - total number of words in Kobish alphabet. Oct 18, 2018", "by skipjack there are 26×25 = 650 two-letter words, the 234th being ji. There are 26×25×25 = 16250 three-letter words. Etc. Thank you skipjack, I found out the number of words in this way; $\\displaystyle 26 × 25^1 + 26 × 25^2 + 26 × 25^3 + 26 × 25^4 = 10579400$ words. But how did you find the 234th word? And how to find the rank of the word \"PAPER\"? October 20th, 2018, 04:32 AM #6 Global Moderator Joined: Dec 2006 Posts: 20,628 Thanks: 2077 Consider how many words precede JA and how many precede PABAB. October 20th, 2018, 09:39 PM #7 Member Joined: Sep 2017 From: Saudi Arabia Posts: 37 Thanks: 1 Quote: Originally Posted by skipjack Consider how many words precede JA and how many precede PABAB. Dear skipjack, The number of words before \"JA\" is much greater than 234, because if we consider the words which start with \"A\" only; *that have two letters* = 1 x 25 = 25 *that have three letters* = 1 x 25 x 25 = 625 *that have four letters* = 1 x 25 x 25 x 25 = 15625 *that have five letters* = 1 x 25 x 25 x 25 x 25 = 390625 So, there are (25+625+15625+390625) or 406900 words start with \"A\", all of", "225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that it would be nearly impossible to come up with concepts for all of them (for words of 5 syllables, the number is 1,078,203,909,375, over", "usage; for instance, the syllable wū does not exist in Hawaiian, and the syllable wu occurs only in two loan-words from English. However, for the purposes of this post I am calculating only how many words could theoretically exist, not how many actually do (which would require exhaustive knowledge of the language that I do not possess). So for single syllables, we have a total of $$10+15+80+120=225$$. Now, since any word in Hawaiian will be made up of these syllables, we can quickly calculate how many distinct words of a given number of syllables would be able to exist. Since Hawaiian has nothing against duplication of sounds (and often rather encourages it), for a word of two syllables we may have any of the 225 for the first syllable, and any of the 225 for the second. Thus to find the total number of two-syllable words we just multiply those two numbers (or equivalently raise 225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our", "of the 225 for the second. Thus to find the total number of two-syllable words we just multiply those two numbers (or equivalently raise 225 to the nth power where n is the number of syllables in the word), for a total of $$225^2=50,625$$ words of two syllables. That's not bad in terms of vocabulary. Pretty much all words necessary for daily life plus quite a few extra would fit pretty nicely into that amount. But Hawaiian utilizes many longer polysyllabic words, so if we expand our list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that", "word length) $26\\times26\\ldots(42 times)=26^{42}=2.68\\times10^{59}$ A 59 digit number! It could obviously go higher if using 100 or 150 letter words comes into vogue. (I hear that the Germans already do this!) This doesn’t really appear ‘realistic’. So let’s see how we can arrive at that figure. First, we note, that since the right side of the Gaussian can extend asymmetrically much more than the left side which is bound by 1 letter words. (Zero and negative letter words don’t exist!). However, since the likelihood of words greater than 10 is very low, we can for current purposes ignore them. ( This is an approximation used to make my life easier to solve this. If you are interested, you could as well use MATLAB to plot a skewed Gaussian and let me know about the percentage error due to this. I’m reasonably confident it won’t be too significant.) Therefore about 67% of words are: $26\\times26\\times5+26\\times26\\times5\\times5+26\\times26\\times5\\times26\\times5+26\\times26\\times5\\times26\\times5\\times5+26\\times26\\times5\\times26\\times26\\times5\\times5=69006080$ The above accounts for the frequency of vowels in all words of length 3–7 as well. That is about 69 million. A 100% (total no.) of possible words would be about 103503945. Or around 100 million total words. Dictionary Facts – Oxford English Dictionary says that they have around 414,800 ‘entries’ defined. These are the words that we most use on a day to", "number of words possible (no repetition): $26*25*24 = 650*24 = 15600$ Total number of words possible with only asymmetric letters: $15*14*13 = 210*13 = 2730$ Total number of words with at least one symmetric letter: $15600 - 2730 = 12870$ Bisen () Kudos Bob, elegant and simple! Permutations and combinations are my weaker part and you do help me here a lot. I approached from the usual side: 1) Combinations of types of letters: AAA - 1 AAS - 3 (3C2 = 3) SSA - 3 (3C2 = 3) 2) Permutations within each combination AAA = 11 x 10 x 9 = 990 AAS = 15 x 14 x 11 = 2310 SSA = 11 x 10 X 15 = 1650 3) Multiply by number of combinations and add those 990 + 2310 x 3 + 1650 x 3 = 12,870 Juhi () Yes finding total possible way and removing the exception is the easy way to approach this kind of question. good question.", "list to words of three syllables, we get $$225^3 = 11,390,625$$ distinct words. At this point we're already well over the vocabulary of even the most linguistically rich languages on Earth. Even using a rather lax definition of \"word\", the English language has around 1–1.5 million words at the most, the vast majority of which are scientific, legal, technical, medical, financial and other terms of generally non-everyday use. But we don't need to stop here! Going to four syllables gives $225^4=2,562,890,625$ This is a mind-bogglingly big number when it comes to words. That's over two-and-a-half billion words, and that's not counting the one, two, and three-syllable words already formed. It's common to have words of four or five syllables in Hawaiian, which blows the realm of neologisms so far open that it would be nearly impossible to come up with concepts for all of them (for words of 5 syllables, the number is 1,078,203,909,375, over a trillion words!). Conclusion: Although the sound range of Hawaiian may sound limited to the Anglophone ear with our 44-odd sounds, combinatorics shows that Hawaiian is capable of forming astronomically many times more words than even the most wordy languages on Earth. I'd do a similar calculation for English, but for the fact that it would be several orders of magnitude harder. This", "7 times, so our count will be too high, contradiction. We're still assuming $n\\leq23$, so now $n=23$ is the only case left. Here \"times\" is used 24 times, \"time\" is used 109560 times, and so \"one\" is used 109564 times. Now \"four\" is used at least 3 times, so it becomes the 21st word on our list. We know \"two\" is used at least 8 times; let's say it's used 9 times, so that \"nine\" is used 7 times and \"seven\" is the 22nd word on our list. After a little trial-and-error and deduction, I got to the following list: • \"this\" ($2$ times) • \"sentence\" ($2$ times) • \"uses\" ($2$ times) • \"or\" ($2$ times) • \"mentions\" ($2$ times) • \"and\" ($2$ times) • \"the\" ($109583$ times) • \"word\" ($109583$ times) • \"times\" ($24$ times) • \"time\" ($109560$ times) • \"one\" ($109564$ times) • \"hundred\" ($9$ times) • \"thousand\" ($5$ times) • \"eighty\" ($3$ times) • \"nine\" ($7$ times) • \"five\" ($6$ times) • \"two\" ($9$ times) • \"three\" ($7$ times) • \"twenty\" ($2$ times) • \"sixty\" ($3$ times) • \"four\" ($3$ times) • \"seven\" ($3$ times) • \"six\" ($2$ times) ... which all works out, giving the solution I put at the top. Hurray! :-D • +1 for the great effort. Making it self-contained will be", "typical written English sentence contains roughly 1.1 bits of information per letter. If you have some text with n bits of information, there are $2^{n}$ ways it can be interpreted. So a written English sentence containing t letters, or about 1.1t bits, can be interpreted $2^{1.1t}$ ways. This certainly cuts down the number of novels. There are roughly $2^{165000}$ possible 50,000 word “novels.” Does that make much of a difference? No. The probability that any of the 150,000-character strings constitutes a 50,000-word “novel” has 178,000 zeroes after the decimal. And keep in mind that I’m using a very loose definition of the word “novel”. A 50,000-word “novel” in this context just needs to be a collection of sentences that are readable individually, but not necessarily together. Here’s an excerpt of my fan-fic sequel to Atlanta Nights: “Jim, do I have a dingo on my back?” The sun set three fortnights ago on Afdw-IX. But it’s not hopeless for my team of hypothetical monkeys yet. Perhaps out there is another segment of our universe, not unlike our own, except that not only did my team of monkeys actually write a 50,000-word novel, but also this blog post. And I have a pet duck. Just because, I suppose. The other me has a reason. I bet its name is Fred.", "The word \"time\" is going to be used $109582-n+1$ times (once for each word used exactly once, once for its own appearance in the list). The only words which will be used more than $n+1$ times are \"the\", \"word\", \"one\", and \"time\", which each appear at least $109582-n+1$ times. So the word \"hundred\" is going to be used nine times, the word \"thousand\" five times, and the word \"eighty\" three times. The bolded numbers above tell us that \"one\", \"nine\", \"five\", \"two\", and \"three\" will each be used more than once. So already $n$ is at least eighteen. Assuming $n$ is at least nineteen and less than twenty-nine (which I'm pretty sure it will be), the word \"twenty\" is used two times (once for the number of times \"times\" is used and once for its own appearance in the list). So far we have the following words appearing more than once: • \"this\" ($2$ times) • \"sentence\" ($2$ times) • \"uses\" ($2$ times) • \"or\" ($2$ times) • \"mentions\" ($2$ times) • \"and\" ($2$ times) • \"the\" ($109583$ times) • \"word\" ($109583$ times) • \"times\" ($n+1$ times) • \"time\" ($109582-n+1$ times) • \"one\" ($\\geq109582-n+1+4$ times) • \"hundred\" ($9$ times) • \"thousand\" ($5$ times) • \"eighty\" ($3$ times) • \"nine\" ($\\geq6$ times) • \"five\" ($\\geq6$ times) • \"two\" ($\\geq8$", "are used. The local language has 66 different letters. The parliament decided to forbid the use of the seventh letter. How many words have the people of Oz lost because of the prohibition? A. 65 B. 66 C. 67 D. 131 E. 132 Two ways to do this one. 1. with 66 letters, how many one or two letter words can we make? $$67 * 66$$ -------(1) we can fill the first position with 67 different things(66 letters or a blank), and the second position with 66 different letters. With 65 letters after removing the 7th letter, how many words do we have? $$66 * 65$$ -------(2) (1) - (2) = 132(E) 2. Now that we have removed the 7th letter, we have lost all the single and double letter words that we could make which include this 7th letter. Single letter words with 7th letter? $$1$$ Double letter words? $$2 * 65 + 1$$ ---Double letter words by using 7th letter only once + Double letter word by using 7th letter twice Total = 132(E) _________________ Spread some love..Like = +1 Kudos Non-Human User Joined: 09 Sep 2013 Posts: 11717 Re: In the Land of Oz only one or two-letter words are used. The [#permalink] ### Show Tags 20 Aug 2018, 12:02 Hello from the GMAT Club", "Another maths curiosity from the Futility Closet: Fortuitous numbers In American usage, 84,672 is said EIGHTY FOUR THOUSAND SIX HUNDRED SEVENTY TWO. Count the letters in each of those words, multiply the counts, and you get 6 × 4 × 8 × 3 × 7 × 7 × 3 = 84,672. Here’s something I’ve (pointlessly) struggled with for a long time, now. Can you complete this sentence? Written as words, there are _____________ letters in this sentence. Use Excel’s LEN() function and AutoSum and try it like this, writing it out one word at a time. So, forty three letters so far, with those two empty boxes. If you were to write forty three into those boxes, the total would obviously be more than forty three. A little trial-and-error, and we get the answer fifty three. Well, that was fairly straightforward. Let’s try a slightly different sentence. Maybe this isn’t so difficult, after all. One more? That’s not right, there are forty nine letters in that sentence, not forty eight. But now there are forty eight. Is it not possible to accurately complete that sentence, then?", "of a $2$-letter word can be chosen in $26$ ways. For each of these ways, the second letter can be chosen in $26$ ways, for a total of $26^2$. (If $26$ women each have $26$ children, then the total number of children is $26^2$.) The same idea shows that there are $26^3$ $3$ letter words, $26^4$ $4$-letter words, and so on. However, we don't even want to count all the $3$-letter words, only those up to $bbb$. How many $3$-letter words start with $a$? We can complete $a$ to a $3$-letter word by appending to $a$ any $2$-letter word. There are $26^2$ such $2$-letter words. So $26^2$ $2$-letter words start with $a$. Finally, we want to count the $3$-letter words that start with $b$, but we only want to get to $bbb$. So append any $2$-letter word that starts with $a$ (there are $26$ of these), and finally at the very end we must not forget about $bba$ and $bbb$ (2 words). Add up. We get $1406$. But probably you did not intend to count the first word, just the \"iterations\" or \"increments.\" There are $1405$ of these. B) Using Base 26 Representation: The $1$-letter words were easy to count, as were the $2$-letter words. The $3$-letter words were a bit of a nuisance, since we did not", "at 16:43 @nico Thank you for that clarification. Theoretically a similar issue pertains in dictionaries containing all possible two-letter combinations as words: when that's the case, any hand of 2 or more letters automatically contains a word. @shabbychef's comment about mid-game shows how irrelevant the original question is to most of Scrabble, because in mid-game you have available an array of word parts (prefixes, suffixes, and even middle sections) in addition to the 7 letters in your hand. This greatly increases the chances of being able to make a word. – whuber Oct 22 '10 at 16:29 4 Answers This is a (long!) comment on the nice work @vqv has posted in this thread. It aims to obtain a definitive answer. He has done the hard work of simplifying the dictionary. All that remains is to exploit it to the fullest. His results suggest that a brute-force solution is feasible. After all, including a wildcard, there are at most $27^7 = 10,460,353,203$ words one can make with 7 characters, and it looks like less than 1/10000 of them--say, around a million--will fail to include some valid word. The first step is to augment the minimal dictionary with a wildcard character, \"?\". 22 of the letters appear in two-letter words (all but c, q, v, z). Adjoin a wildcard", "= 16 (Full Reduction) (5 letters, 1 word)”mcfly” = 22 (Reverse Full Reduction) 0 letter, 1 word)”1955″ = 20 (English Ordinal) (5 letters, 1 word)”mcfly” = 22 (Reverse Full Reduction) (21 letters, 6 words)”it’s time to save the world” = 282 (English Ordinal) (23 letters, 5 words)”eighty eight miles per hour” = 282 (English Ordinal) (22 letters, 5 words)”two hundred and eighty two” = 283 (English Ordinal) (32 letters, 6 words)”december fourth two thousand and four” = 382 (English Ordinal) (0 letter, 3 words)”12/04/2004″ = 13 (English Ordinal) 4 letters, 1 word)”hire” = 23 (Reverse Full Reduction) (6 letters, 1 word)”higher” = 55 (English Ordinal) (4 letters, 1 word)”fire” = 38 (English Ordinal) (5 letters, 1 word)”lower” = 73 (English Ordinal) (5 letters, 2 words)”let go” = 22 (Reverse Full Reduction) (9 letters, 1 word)”surrender” = 122 (English Ordinal) (4 letters, 1 word)”burn” = 55 (English Ordinal) (15 letters, 3 words)”catch and release” = 88 (Reverse Full Reduction) 12 letters, 2 words)”twenty twenty” = 214 (English Ordinal) 24 letters, 5 words)”three hundred and eighty two” = 128 (Full Reduction) (11 letters, 2 words)”seven eleven” = 128 (English Ordinal) (10 letters, 2 words)”nine eleven” = 165 (Reverse Ordinal) (12 letters, 3 words)”santa fe store” = 143 (English Ordinal) (9 letters, 2 words)”dying king” = 143 (Reverse Ordinal) (15 letters,", "$::$ and one word is given on another side of $::$ while another word is to be found from the given alternatives, having the same relation with this word as the words of the given pair bear. Choose the best alternative. Virology : Virus $: :$ Semantics : ? Amoeba Language Nature Society 37 A train started from Mumbai at $6.00$ A.M. On the next (second) station $1/3$ passengers got down and $96$ got in. On the next (third) station, $1/2$ of the total passengers present in the train, got down and $12$ came in. Now there were $248$ passengers in the train, when the train started from Mumbai, the number of passengers was: $435$ $564$ $654$ $736$ 1 vote In the following question, three out of four alternatives contain alphabet placed in a particular form. Find the one that does not belong to the group. $JMP$ $RUX$ $UYB$ $EHK$ In the following question, three out of four alternatives contain alphabet placed in a particular form. Find the one that does not belong to the group. $CBAZ$ $AZYX$ $AZBY$ $PONM$ In the following question, three out of four alternatives contain alphabet placed in a particular form. Find the one that does not belong to the group. $LNMO$ $CRDT$ $EUFV$ $GWHX$", "6 letters is: $26 \\times 26 \\times 26 \\times 26 \\times 26 \\times 26 = 26^6$ For the strings that does not contain a there are 25 possible ways. $25 \\times 25 \\times 25 \\times 25 \\times 25 \\times 25 = 25^6$ For the strings that does not contain b there are 25 possible ways. $25 \\times 25 \\times 25 \\times 25 \\times 25 \\times 25 = 25^6$ For the strings that contains neither a nor b, there are 24 possible ways. $24 \\times 24 \\times 24 \\times 24 \\times 24 \\times 24=24^6$ The possible strings that not containing a or b is: $25^6+25^6 -24^6$ The number of possible strings containing a and b is the total number of strings minus the number of strings that does not containing a or b. $26^6-(25^6+25^6-24^6) \\\\ = 308915776 -(2 \\times 244140625 -191102976) \\\\ = 11737502$ Therefore, there are 11737502 number of strings of six lowercase letters from English alphabet contains the letters or b. iii. The objective is to determine the number of strings of six lowercase letters from English alphabet contain the letters a and b in consecutive positions with a preceding b and all the letters are distinct. As the letters are consecutive, the possible strings are in the form: Here ab can be in five locations, that" ]

[ "+0 # pls help 0 84 1 What is the value of $n$ such that $10^n = 10^{-5}\\times \\sqrt{\\frac{10^{73}}{0.001}}$? Mar 17, 2021", "+0 0 76 1 What is the value of n such that: $$10^n = 10^{-6}\\times \\sqrt{\\frac{10^{46}}{0.0001}}?$$ Aug 5, 2019 $$0.0001 = 10^{-4}\\\\ \\frac{10^{46}}{10^{-4}}=10^{50}\\\\ \\sqrt{10^{50}}=10^{25}\\\\10^{-6}\\times 10^{25}=10^{19}\\\\\\text{ Hence : }n=19$$", "+0 0 217 1 What is the value of n such that: $$10^n = 10^{-6}\\times \\sqrt{\\frac{10^{46}}{0.0001}}?$$ Aug 5, 2019 $$0.0001 = 10^{-4}\\\\ \\frac{10^{46}}{10^{-4}}=10^{50}\\\\ \\sqrt{10^{50}}=10^{25}\\\\10^{-6}\\times 10^{25}=10^{19}\\\\\\text{ Hence : }n=19$$", "Multiplying both sides by 8 gives $(8 \\times 8^n) = 8 \\times (7k + 1)$ Subtract one from both sides gives $8^{n+1} - 1 = (8 \\times 7k) + 8 -1 = (8 \\times 7k) + 7$ $\\therefore 8^{n+1} - 1 = 7 \\times (8k + 1)$ Since $k$ is an integer then $8k + 1$ is an integer which means that $8^{n+1} - 1$ is a multiple of $7$. However, we can see that for $n = 1$, $8^n - 1$ becomes $8 -1 = 7$ which is clearly divisible by $7$. Thus, by induction, we have the desired result that $\\frac {8^n - 1} {7}$ is always a whole number.", "$n \\times n$ square.", "Interesting Property Of Seven! Number Theory Level 4 Note that $$7!=10 \\times9 \\times8\\times7.$$ What is the largest possible value of $$n$$ for which $$n!$$ can be represented as a product of $$n-3$$ consecutive integers? ×", "large enough $$n$$ and since you have checked up to $$n = 7.5 \\times 10^7$$, I am confident that it holds for all $$n \\ge 5$$.", "$$N = n V \\simeq 3.8\\times10^5$$.", "# What is the value of 7 to the fifth power? $16807$ $\\textcolor{w h i t e}{\\times} {7}^{5} = 7 \\times 7 \\times 7 \\times 7 \\times 7$ $\\textcolor{w h i t e}{\\times \\times} = 16807$", "5 = 8$.", "the $$n$$-th decimal value of $$a_n$$ and see whether smooshing them together and swapping the 4s and 7s does anything.", "7 \\times 2$", "6$, $4!=4 \\times 3 \\times 2 \\times 1 = 24$, and $5!=5\\times 4\\times 3 \\times 2 \\times 1=120$. You may have already noticed the recursion at work. Take for instance $5!=5 \\times 4 \\times 3 \\times 2 \\times 1$. We see the value of $4!$ lurking here: $5!=5 \\times 4!$. In fact, for all positive $n$, we can make the alternative (usually more useful) definition $n!=n(n-1)!$. This is the recursive defintion of the factorial, because we are defining the factorial in terms of the factorial. Why is this okay? Well, we’re really collapsing a lot of work into two lines: it may make more sense to break it down one step at a time. First, define $0!=1$. This definition will make sense in a later chapter, but remember that mathematical definitions don’t necessarily need to “make sense.” Next, define $1!=1 \\times 0!$. This definition is valid, because the number $1$, the operation $\\times$, and the number $0!$ have all been defined. And when we go to define $2!=2 \\times 1!$, we have already defined all of these objects as well. And so on. Therefore, the following definition is a valid one. Definition 5.1.2 Let $n$ be a natural number. Then $n!$ is equal to 1 when $n=0$, and $n(n-1)!$ otherwise. ### 5.2 Binary numbers Now that we know", "$a_{4} = 4^{2} - 5\\times 4 + 2 = 16 - 20 +2 = -2 < 0$ $a_{5} = 5^{2} - 5\\times 5 + 2 = 25 - 25 +2 = 2 > 0$ Therefore the smallest value of n for which $a_{n}$ is positive is n = 5 . ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed. ### Cheenta. Passion for Mathematics Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.", "7. As $A$ is 9 and $N$ is 8, the only possible values for $O$ and $Y$ are 6 and 7. These are interchangeable, but in both cases $Y\\times O=$42. This problem is taken from the UKMT Mathematical Challenges. View the archive of all weekly problems grouped by curriculum topic View the previous week's solution View the current weekly problem", "then $n!=1, \\text{when}\\ n=0$ $n!=n\\times (n-1)!\\ \\text{when} \\ n>1$ In other words, if $$n$$ is an integer greater than $$0$$, then $$n!=n\\times (n-1)\\times (n-2)\\times \\cdots \\times 2\\times 1$$. And $$0!=1$$ by definition. I hope this helps! - 4 years, 1 month ago", "# Interesting Property Of Seven! Note that $$7!=10 \\times9 \\times8\\times7.$$ What is the largest possible value of $$n$$ for which $$n!$$ can be represented as a product of $$n-3$$ consecutive integers? ×", "$EG=7$EG=7, find the value of $m$m.", "that n = 7 in your case): $$W_t = 1.5 \\times 9.81 \\times \\frac{1}{2} \\times 8 \\times 7$$ which you can re-order to give the expression you had. Share this great discussion with others via Reddit, Google+, Twitter, or Facebook", "values $$n_2=2+3=5$$" ]

[ "so $7-2+11=16$ 5. $5.6+4.4=10$, so $7.2+4.4=11.6$ 6. $\\frac{5^6\\times2}{9375}=10$, so $\\frac{7^2\\times2}{9375}=0.01045333...$ The last one is ridiculous on purpose - it serves to illustrate the point that almost anything could be an answer.", "be the number of coconuts found in the morning. $n_8$ is divisible by 7 so write $n_8 = 7m_8$. But $n_8$ is what was left when the seventh pirate discarded one coconut out of $n_7$ and hid one seventh of the rest, so $n_7 = (7/6)n_8+1 = (7/6)(7m_8)+1$. For that to be an integer $m_8$ must be a multiple of 6, so write $m_8 = 6m_7$. Then $n_7 = 7^2m_7+1$. Now $n_6 = (7/6)n_7+1 = (7/6)(7^2m_7+1)+1$. For integer value, $7^2m_7\\mod 6 = m_7\\mod 6 = 5$, so write $m_7 = 6m_6+5$. Then $n_6 = (7/6)(7^2(6m_6+5)+1)+1 = 7^3m_6+288$. You see how this goes. $n_5 = (7/6)n_6+1 = (7/6)(7^3m_6+288)+1$. For integer value, $m_6\\mod 6 = 0$, so write $m_6 = 6m_5$. Then $n_5 = (7/6)(7^3(6m_5)+288))+1 = 7^4m_5+337$. $n_4 = (7/6)n_5+1 = (7/6)(7^4m_5+337)+1$. For integer value, $m_5\\mod 6 = 5$, so write $m_5 = 6m_4+5$. Then $n_4 = (7/6)(7^4(6m_4+5)+337)+1 = 7^5m_4+14400$. $n_3 = (7/6)n_4+1 = (7/6)(7^5m_4+14400)+1$. For integer value, $m_4\\mod 6 = 0$, so write $m_4 = 6m_3$. Then $n_3 = (7/6)(7^5(6m_3)+14400)+1 = 7^6m_3+16801$. $n_2 = (7/6)n_3+1 = (7/6)(7^6m_3+16801)+1$. For integer value, $m_6\\mod 6 = 5$, so write $m_3 = 6m_2+5$. Then $n_2 = (7/6)(7^6(6m_2+5)+16801)+1 = 7^7m_2+705888$. $n_1 = (7/6)n_2+1 = (7/6)(7^7m_2+705888)+1$. For integer value, $m_2\\mod 6 = 0$, so write $m_2 = 6m_1$. Then $n_1 = (7/6)(7^7(6m_1)+705888)+1 = 7^8m_1+823537$. The", "# How do you write the equation for \"one more than three times a number is 7\"? Nov 10, 2017 The equation is: $3 N + 1 = 7$ The solution is: $2$ #### Explanation: You say: \"one more than three times a number is 7\". So we start with a number $N$. Then we multiply it by $3$ to get: $3 N$ Then we increase it by $1$ to get: $3 N + 1$ And we equal all of that to $7$: $3 N + 1 = 7 \\to$ [ANS} Solving: $3 N + 1 = 7$ $3 N + 1 - 1 = 7 - 1$ $3 N = 6$ $N = \\frac{6}{3}$ N=2 \"one more than three times a number is 7\". \"one more than three times N is 7\". \"one more than three times $2$ is 7\".", "## Algebra 2 (1st Edition) There is $7!$ ways to order 7 objects or $7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1 = \\boxed {5040}$", "pirate, and $n_8$ be the number of coconuts found in the morning. $n_8$ is divisible by 7 so write $n_8 = 7m_8$. But $n_8$ is what was left when the seventh pirate discarded one coconut out of $n_7$ and hid one seventh of the rest, so $n_7 = (7/6)n_8+1 = (7/6)(7m_8)+1$. For that to be an integer $m_8$ must be a multiple of 6, so write $m_8 = 6m_7$. Then $n_7 = 7^2m_7+1$. Now $n_6 = (7/6)n_7+1 = (7/6)(7^2m_7+1)+1$. For integer value, $7^2m_7\\mod 6 = m_7\\mod 6 = 5$, so write $m_7 = 6m_6+5$. Then $n_6 = (7/6)(7^2(6m_6+5)+1)+1 = 7^3m_6+288$. You see how this goes. $n_5 = (7/6)n_6+1 = (7/6)(7^3m_6+288)+1$. For integer value, $m_6\\mod 6 = 0$, so write $m_6 = 6m_5$. Then $n_5 = (7/6)(7^3(6m_5)+288))+1 = 7^4m_5+337$. $n_4 = (7/6)n_5+1 = (7/6)(7^4m_5+337)+1$. For integer value, $m_5\\mod 6 = 5$, so write $m_5 = 6m_4+5$. Then $n_4 = (7/6)(7^4(6m_4+5)+337)+1 = 7^5m_4+14400$. $n_3 = (7/6)n_4+1 = (7/6)(7^5m_4+14400)+1$. For integer value, $m_4\\mod 6 = 0$, so write $m_4 = 6m_3$. Then $n_3 = (7/6)(7^5(6m_3)+14400)+1 = 7^6m_3+16801$. $n_2 = (7/6)n_3+1 = (7/6)(7^6m_3+16801)+1$. For integer value, $m_6\\mod 6 = 5$, so write $m_3 = 6m_2+5$. Then $n_2 = (7/6)(7^6(6m_2+5)+16801)+1 = 7^7m_2+705888$. $n_1 = (7/6)n_2+1 = (7/6)(7^7m_2+705888)+1$. For integer value, $m_2\\mod 6 = 0$, so write $m_2 = 6m_1$. Then $n_1 = (7/6)(7^7(6m_1)+705888)+1", "$N(1; n) = N(1; n - 2) \\tag{1}$ $N(2; n) = N(2; n - 1) + N(2; n - 2) \\tag{2}$ $N(3; n) = 4\\times N(3; n - 2) - N(3; n - 4) \\tag{3}$ $N(4; n) = N(4; n - 1) + 5\\times N(4; n - 2) + N(4; n - 3) - N(4; n - 4) \\tag{4}$ $N(5; n) = 15\\times N(5; n - 2) - 32\\times N(5; n - 4) + 15\\times N(5; n - 6) - N(5; n - 8) \\tag{5}$ $N(6; n) = N(6; n - 1) + 20\\times N(6; n - 2) + 10\\times N(6; n - 3) - 38\\times N(6; n - 4) - 10\\times N(6; n - 5) + 20\\times N(6; n - 6) - N(6; n - 7) - N(6; n - 8) \\tag{6}$ $N(7; n) = 56 \\times N(7; n - 2) - 672 \\times N(7; n - 4) + 2632 \\times N(7; n - 6) - 4094 \\times N(7; n - 8) + 2632 \\times N(7; n - 10) - 672 \\times N(7; n - 12) + 56 \\times N(7; n - 14) - N(7; n - 16) \\tag{7}$ $G(N(1); x) = \\frac{1}{1 - x^2} \\tag{8}$ $G(N(2); x) = \\frac{1}{1 - x - x^2} \\tag{9}$ $G(N(3); x) = \\frac{1 - x^2}{1 - 4x^2 + x^4} \\tag{10}$ $G(N(4);", "2. $7\\frac{1}{5}\\times 7\\frac{1}{2}\\times \\frac{2}{5}$", "Question 1: Find the cubes of: (i) $16$ (ii) $30$ (iii) $1.2$ (iv) $0.7$ (v) $0.06$ (vi) $\\frac{2}{5}$ (vii) $\\frac{1}{9}$ (viii) $1$ $\\frac{3}{5}$ (i) $16^3 = 16 \\times 16 \\times 16 = 4096$ (ii) $30^3 = 30 \\times 30 \\times 30 = 27000$ (iii) $1.2^3 = 1.2 \\times 1.2 \\times 1.2 = 1.728$ (iv) $0.7^3 = 0.7 \\times 0.7 \\times 0.7 = 0.343$ (v) $0.06^3 = 0.06 \\times 0.06 \\times 0.06 = 0.000216$ (vi) $\\Big($ ${\\frac{2}{5}}$ $\\Big)^3$ $=$ $\\frac{8}{125}$ (vii) $\\Big($ ${\\frac{1}{9}}$ $\\Big)^3$ $=$ $\\frac{1}{729}$ (viii) $\\Big($ ${\\frac{3}{5}}$ $\\Big)^3$ $=$ $\\frac{512}{125}$ $\\\\$ Question 2: Test whether the given number is a perfect cube or not: (i) $729$ (ii) $3380$ (iii) $10584$ (iv) $13824 )$ (i) $729 = 7 \\times 7 \\times 7$ (hence a perfect cube) (ii) $3380 = 4 \\times 5 \\times 13 \\times 13$ (hence not a perfect cube) (iii) $10584 = 8 \\times 3 \\times 3 \\times 3 \\times 3 \\times 7 \\times 7$ (hence not a perfect cube) (iv) $13824 = (2 \\times 2 \\times 2) \\times (2 \\times 2 \\times 2) \\times (2 \\times 2 \\times 2) \\times (3 \\times 3 \\times 3)$ (hence a perfect cube) $\\\\$ Question 3: Find the smallest number by which $11979$ must be multiplied so that the product is a perfect cube. $11979 = 3 \\times 3", "$E = 8$, or $B = 3$ and $E = 9$. But $4E \\mod 10 = A = 2$ so $E = 8$. Now $4D+3 \\mod 10 = B = 1$ so $4D \\mod 10 = 8$. $D\\ne 2$ so $D = 7$. Finally $4C+3 \\mod 10 = C$ or $3C\\mod 10 = 7$, so $C = 9$. $219,978 \\times 4= 879,912$. # Classic This is fairly mechanical although I tripped up several times on the way. Let $n_i$ be the number of coconuts found by the $i$th pirate, and $n_8$ be the number of coconuts found in the morning. $n_8$ is divisible by 7 so write $n_8 = 7m_8$. But $n_8$ is what was left when the seventh pirate discarded one coconut out of $n_7$ and hid one seventh of the rest, so $n_7 = (7/6)n_8+1 = (7/6)(7m_8)+1$. For that to be an integer $m_8$ must be a multiple of 6, so write $m_8 = 6m_7$. Then $n_7 = 7^2m_7+1$. Now $n_6 = (7/6)n_7+1 = (7/6)(7^2m_7+1)+1$. For integer value, $7^2m_7\\mod 6 = m_7\\mod 6 = 5$, so write $m_7 = 6m_6+5$. Then $n_6 = (7/6)(7^2(6m_6+5)+1)+1 = 7^3m_6+288$. You see how this goes. $n_5 = (7/6)n_6+1 = (7/6)(7^3m_6+288)+1$. For integer value, $m_6\\mod 6 = 0$, so write $m_6 = 6m_5$. Then $n_5 = (7/6)(7^3(6m_5)+288))+1 = 7^4m_5+337$. $n_4 =", "$5$ are coprime $(\\gcd(8, 5)=1)$. This means that such an integer can be written as $8\\times 5\\times n=40\\times n$, for some integer $n$.", "let's add one $5$ to our list. List: $2 , 2 , 3 , 3 , 5$ Finally, we're left with 7. The only number that has a $7$ is $84$, and it has one $7$, so let's add one $7$ to our list. List: $2 , 2 , 3 , 3 , 5 , 7$ This is our complete list! Step 5. $\\text{LCM} = 2 \\times 2 \\times 3 \\times 3 \\times 5 \\times 7 = 1260$", "way. Let $n_i$ be the number of coconuts found by the $i$th pirate, and $n_8$ be the number of coconuts found in the morning. $n_8$ is divisible by 7 so write $n_8 = 7m_8$. But $n_8$ is what was left when the seventh pirate discarded one coconut out of $n_7$ and hid one seventh of the rest, so $n_7 = (7/6)n_8+1 = (7/6)(7m_8)+1$. For that to be an integer $m_8$ must be a multiple of 6, so write $m_8 = 6m_7$. Then $n_7 = 7^2m_7+1$. Now $n_6 = (7/6)n_7+1 = (7/6)(7^2m_7+1)+1$. For integer value, $7^2m_7\\mod 6 = m_7\\mod 6 = 5$, so write $m_7 = 6m_6+5$. Then $n_6 = (7/6)(7^2(6m_6+5)+1)+1 = 7^3m_6+288$. You see how this goes. $n_5 = (7/6)n_6+1 = (7/6)(7^3m_6+288)+1$. For integer value, $m_6\\mod 6 = 0$, so write $m_6 = 6m_5$. Then $n_5 = (7/6)(7^3(6m_5)+288))+1 = 7^4m_5+337$. $n_4 = (7/6)n_5+1 = (7/6)(7^4m_5+337)+1$. For integer value, $m_5\\mod 6 = 5$, so write $m_5 = 6m_4+5$. Then $n_4 = (7/6)(7^4(6m_4+5)+337)+1 = 7^5m_4+14400$. $n_3 = (7/6)n_4+1 = (7/6)(7^5m_4+14400)+1$. For integer value, $m_4\\mod 6 = 0$, so write $m_4 = 6m_3$. Then $n_3 = (7/6)(7^5(6m_3)+14400)+1 = 7^6m_3+16801$. $n_2 = (7/6)n_3+1 = (7/6)(7^6m_3+16801)+1$. For integer value, $m_6\\mod 6 = 5$, so write $m_3 = 6m_2+5$. Then $n_2 = (7/6)(7^6(6m_2+5)+16801)+1 = 7^7m_2+705888$. $n_1 = (7/6)n_2+1 = (7/6)(7^7m_2+705888)+1$. For integer value, $m_2\\mod", "the resulting math easier. – Hugh Meyers May 27 '16 at 18:11 Assuming $7^7=a$ and $8^7=b$ I think we can approximate as below: $${7.5}^7\\times 1\\approx \\int_7^8x^7dx=\\frac 18\\left(8^8-7^8\\right)=8^7-\\frac 78 7^7=b-\\frac 78 a$$ which gives $1376551.875$, very close to the true value $1334838.867$ This way is easy and fast.", "Alicer100 12 # what is 7/3 in simplest form benjackson64 It is 2 1/3 to be exact because 3 goes in to 7 two times without going over and there is one third left over Lynor 7/3 is already in simplest form. It can also be written as a mixed number, as such: $\\frac{7}{3} = \\frac{6+1}{3} = \\frac{6}{3} + \\frac{1}{3} = 2\\frac{1}{3}$ Hope that helped :)", "5!}$$ $$= \\frac{8!}{3! \\times 5!}$$ $$= 56$$ Hence,$$= 15 + 56 = 71$$", "1 = 841$$ is not $$9^m$$ for any m. ### Claim 2: k is even $$\\frac{k!}{48} + 1 = (k+1)^m$$ Since $$k > 8$$, then $$\\frac{k!}{48} = \\frac{1\\times 2 \\times 3 \\times \\cdots \\times 8 \\times \\cdots k}{48} = 840 \\times \\cdots \\times k$$. Hence $$\\frac{k!}{48}$$ is even (as it is a multiple of 840). This implies $$\\frac{k!}{48} + 1$$ is odd. But that means $$(k+1)^m$$ is odd or k+1 is odd. Hence k is even. Let $$k = 2k_1$$. ### Claim 3: k divides $$\\frac {(k-1)!}{48}$$ We need to show that $$\\frac {(k-1)!}{48}$$ is divisible by 2 and $$k_1$$. Since we showed $$k \\geq 9$$ in claim 1, and then showed k is even; this implies k is at least 10 or more. $$\\frac {(k-1)!}{48} \\\\ = \\frac {1 \\times 2 \\times 3 \\times \\cdots \\times 9 \\times \\cdots \\times (k-1)} {48} \\\\ = 1 \\times 5 \\times 3 \\times 7 \\times 8 \\times 9 \\times \\cdots \\times (k-1)$$ For k = 10, we see that $$1 \\times 5 \\times 3 \\times 7 \\times 8 \\times 9$$ is easily divisible by 10 (5 exists and 8 supplies one 2). For k = 12, we see that $$1 \\times 5 \\times 3 \\times 7 \\times 8 \\times 9 \\times 10 \\times 11$$ is also divisible by 12 (3", "in $T$, their signs are opposite; so the sum of the alternating sums of $T$ and $T'$ is equal to 7. There are $2^6$ subsets containing 7, so our answer is $7 * 2^6 = \\boxed{448}$.", "Question: $\\frac{2.7 \\times 2.7 \\times2.7 + 2.3 \\times 2.3 \\times 2.3 }{ (2.7)^2 - 2.7 \\times 2.3 + (2.3)^2}$ -ന്റെ വില: A5 B0 C4.75 D4.5 $\\frac{2.7 \\times 2.7 \\times2.7 + 2.3 \\times 2.3 \\times 2.3 }{ (2.7)^2 - 2.7 \\times 2.3 + (2.3)^2}$ = 2.7 + 2.3 = 5", "4 = 0$ so $B = 1$ and $E = 8$, or $B = 3$ and $E = 9$. But $4E \\mod 10 = A = 2$ so $E = 8$. Now $4D+3 \\mod 10 = B = 1$ so $4D \\mod 10 = 8$. $D\\ne 2$ so $D = 7$. Finally $4C+3 \\mod 10 = C$ or $3C\\mod 10 = 7$, so $C = 9$. $219,978 \\times 4= 879,912$. # Classic This is fairly mechanical although I tripped up several times on the way. Let $n_i$ be the number of coconuts found by the $i$th pirate, and $n_8$ be the number of coconuts found in the morning. $n_8$ is divisible by 7 so write $n_8 = 7m_8$. But $n_8$ is what was left when the seventh pirate discarded one coconut out of $n_7$ and hid one seventh of the rest, so $n_7 = (7/6)n_8+1 = (7/6)(7m_8)+1$. For that to be an integer $m_8$ must be a multiple of 6, so write $m_8 = 6m_7$. Then $n_7 = 7^2m_7+1$. Now $n_6 = (7/6)n_7+1 = (7/6)(7^2m_7+1)+1$. For integer value, $7^2m_7\\mod 6 = m_7\\mod 6 = 5$, so write $m_7 = 6m_6+5$. Then $n_6 = (7/6)(7^2(6m_6+5)+1)+1 = 7^3m_6+288$. You see how this goes. $n_5 = (7/6)n_6+1 = (7/6)(7^3m_6+288)+1$. For integer value, $m_6\\mod 6 = 0$, so write $m_6 = 6m_5$.", "+ 2) = 9 \\times 10 + 9 \\times 2 = 108$ $9 \\times (110 + 2) = 9 \\times 110 + 9 \\times 2 = 1008$ So on. $$9 \\times 11\\cdots12 = 9 \\times (11\\cdots 11+1) = 99\\cdots99 + 9.$$ Write $1...12$ as $$\\overline{1...(\\times\\ n)...1} +1=\\frac{10^n-1}{9}+1$$ Hence $$9\\times (\\frac{10^n-1}{9}+1)=10^n-1+9=10^n+8$$" ]

[ "Multiplying both sides by 8 gives $(8 \\times 8^n) = 8 \\times (7k + 1)$ Subtract one from both sides gives $8^{n+1} - 1 = (8 \\times 7k) + 8 -1 = (8 \\times 7k) + 7$ $\\therefore 8^{n+1} - 1 = 7 \\times (8k + 1)$ Since $k$ is an integer then $8k + 1$ is an integer which means that $8^{n+1} - 1$ is a multiple of $7$. However, we can see that for $n = 1$, $8^n - 1$ becomes $8 -1 = 7$ which is clearly divisible by $7$. Thus, by induction, we have the desired result that $\\frac {8^n - 1} {7}$ is always a whole number.", "$N(1; n) = N(1; n - 2) \\tag{1}$ $N(2; n) = N(2; n - 1) + N(2; n - 2) \\tag{2}$ $N(3; n) = 4\\times N(3; n - 2) - N(3; n - 4) \\tag{3}$ $N(4; n) = N(4; n - 1) + 5\\times N(4; n - 2) + N(4; n - 3) - N(4; n - 4) \\tag{4}$ $N(5; n) = 15\\times N(5; n - 2) - 32\\times N(5; n - 4) + 15\\times N(5; n - 6) - N(5; n - 8) \\tag{5}$ $N(6; n) = N(6; n - 1) + 20\\times N(6; n - 2) + 10\\times N(6; n - 3) - 38\\times N(6; n - 4) - 10\\times N(6; n - 5) + 20\\times N(6; n - 6) - N(6; n - 7) - N(6; n - 8) \\tag{6}$ $N(7; n) = 56 \\times N(7; n - 2) - 672 \\times N(7; n - 4) + 2632 \\times N(7; n - 6) - 4094 \\times N(7; n - 8) + 2632 \\times N(7; n - 10) - 672 \\times N(7; n - 12) + 56 \\times N(7; n - 14) - N(7; n - 16) \\tag{7}$ $G(N(1); x) = \\frac{1}{1 - x^2} \\tag{8}$ $G(N(2); x) = \\frac{1}{1 - x - x^2} \\tag{9}$ $G(N(3); x) = \\frac{1 - x^2}{1 - 4x^2 + x^4} \\tag{10}$ $G(N(4);", "6$, $4!=4 \\times 3 \\times 2 \\times 1 = 24$, and $5!=5\\times 4\\times 3 \\times 2 \\times 1=120$. You may have already noticed the recursion at work. Take for instance $5!=5 \\times 4 \\times 3 \\times 2 \\times 1$. We see the value of $4!$ lurking here: $5!=5 \\times 4!$. In fact, for all positive $n$, we can make the alternative (usually more useful) definition $n!=n(n-1)!$. This is the recursive defintion of the factorial, because we are defining the factorial in terms of the factorial. Why is this okay? Well, we’re really collapsing a lot of work into two lines: it may make more sense to break it down one step at a time. First, define $0!=1$. This definition will make sense in a later chapter, but remember that mathematical definitions don’t necessarily need to “make sense.” Next, define $1!=1 \\times 0!$. This definition is valid, because the number $1$, the operation $\\times$, and the number $0!$ have all been defined. And when we go to define $2!=2 \\times 1!$, we have already defined all of these objects as well. And so on. Therefore, the following definition is a valid one. Definition 5.1.2 Let $n$ be a natural number. Then $n!$ is equal to 1 when $n=0$, and $n(n-1)!$ otherwise. ### 5.2 Binary numbers Now that we know", "care­ful to put $$n-1$$ in brack­ets, be­cause oth­er­wise it looks like $$n \\times n - 1!$$, which means $$(n \\times n)-1!$$ ac­cord­ing to the or­der of op­er­a­tions. $$n! = n \\times (n-1)!$$$OK, we have • $$n! = n \\times (n-1)!$$, and • $$(n-1)! = (n-1) \\times (n-2)!$$, and • $$(n-2)! = (n-2) \\times (n-3)!$$, and so on. So $$n! = n \\times (n-1)! = n \\times (n-1) \\times (n-2)! = n \\times (n-1) \\times (n-2) \\times (n-3)!$$$ and so on. If we just keep go­ing, we’ll even­tu­ally reach $$n \\times (n-1) \\times (n-2) \\times \\dots \\times 5 \\times 4 \\times 3!$$$and we already know that $$3! = 6$$, which I’ll write as $$3 \\times 2 \\times 1$$ for rea­sons which are about to be­come ob­vi­ous. So we have the fol­low­ing for­mula, which is how we define the fac­to­rial: $$n! = n \\times (n-1) \\times \\dots \\times 4 \\times 3 \\times 2 \\times 1$$ “$n!$” is read out loud as “$n\\$ fac­to­rial”, and it means “the num­ber of ways to ar­range $$n$$ ob­jects into any or­der”. # Edge cases We’ve seen $$3!$$, but never $$2!$$ or $$1!$$. • It’s easy to see that there are two ways to ar­range two ob­jects into an or­der: $$1,2$$ or $$2,1$$. So $$2! = 2$$. • It’s also easy (if a bit weird)", "+0 0 76 1 What is the value of n such that: $$10^n = 10^{-6}\\times \\sqrt{\\frac{10^{46}}{0.0001}}?$$ Aug 5, 2019 $$0.0001 = 10^{-4}\\\\ \\frac{10^{46}}{10^{-4}}=10^{50}\\\\ \\sqrt{10^{50}}=10^{25}\\\\10^{-6}\\times 10^{25}=10^{19}\\\\\\text{ Hence : }n=19$$", "$5$ are coprime $(\\gcd(8, 5)=1)$. This means that such an integer can be written as $8\\times 5\\times n=40\\times n$, for some integer $n$.", "be the number of coconuts found in the morning. $n_8$ is divisible by 7 so write $n_8 = 7m_8$. But $n_8$ is what was left when the seventh pirate discarded one coconut out of $n_7$ and hid one seventh of the rest, so $n_7 = (7/6)n_8+1 = (7/6)(7m_8)+1$. For that to be an integer $m_8$ must be a multiple of 6, so write $m_8 = 6m_7$. Then $n_7 = 7^2m_7+1$. Now $n_6 = (7/6)n_7+1 = (7/6)(7^2m_7+1)+1$. For integer value, $7^2m_7\\mod 6 = m_7\\mod 6 = 5$, so write $m_7 = 6m_6+5$. Then $n_6 = (7/6)(7^2(6m_6+5)+1)+1 = 7^3m_6+288$. You see how this goes. $n_5 = (7/6)n_6+1 = (7/6)(7^3m_6+288)+1$. For integer value, $m_6\\mod 6 = 0$, so write $m_6 = 6m_5$. Then $n_5 = (7/6)(7^3(6m_5)+288))+1 = 7^4m_5+337$. $n_4 = (7/6)n_5+1 = (7/6)(7^4m_5+337)+1$. For integer value, $m_5\\mod 6 = 5$, so write $m_5 = 6m_4+5$. Then $n_4 = (7/6)(7^4(6m_4+5)+337)+1 = 7^5m_4+14400$. $n_3 = (7/6)n_4+1 = (7/6)(7^5m_4+14400)+1$. For integer value, $m_4\\mod 6 = 0$, so write $m_4 = 6m_3$. Then $n_3 = (7/6)(7^5(6m_3)+14400)+1 = 7^6m_3+16801$. $n_2 = (7/6)n_3+1 = (7/6)(7^6m_3+16801)+1$. For integer value, $m_6\\mod 6 = 5$, so write $m_3 = 6m_2+5$. Then $n_2 = (7/6)(7^6(6m_2+5)+16801)+1 = 7^7m_2+705888$. $n_1 = (7/6)n_2+1 = (7/6)(7^7m_2+705888)+1$. For integer value, $m_2\\mod 6 = 0$, so write $m_2 = 6m_1$. Then $n_1 = (7/6)(7^7(6m_1)+705888)+1 = 7^8m_1+823537$. The", "+0 0 217 1 What is the value of n such that: $$10^n = 10^{-6}\\times \\sqrt{\\frac{10^{46}}{0.0001}}?$$ Aug 5, 2019 $$0.0001 = 10^{-4}\\\\ \\frac{10^{46}}{10^{-4}}=10^{50}\\\\ \\sqrt{10^{50}}=10^{25}\\\\10^{-6}\\times 10^{25}=10^{19}\\\\\\text{ Hence : }n=19$$", "with a second letter, that has $12$ possible values too so combined you have $12 \\times 12$ possibilities. Now with a third letter, that has $12$ possible values as well so combined you have $12 \\times 12 \\times 12$ ... – Henry Mar 25 '12 at 21:00 $$N= \\frac{n!}{(n-k)!}+n$$ where $n$ is a number of elements of numbers set , and $k$ is number of elements of letters set .", "means $$(n \\times n)-1!$$ according to the order of operations. $$n! = n \\times (n-1)!$$$OK, we have • $$n! = n \\times (n-1)!$$, and • $$(n-1)! = (n-1) \\times (n-2)!$$, and • $$(n-2)! = (n-2) \\times (n-3)!$$, and so on. So $$n! = n \\times (n-1)! = n \\times (n-1) \\times (n-2)! = n \\times (n-1) \\times (n-2) \\times (n-3)!$$$ and so on. If we just keep going, we’ll eventually reach $$n \\times (n-1) \\times (n-2) \\times \\dots \\times 5 \\times 4 \\times 3!$$$and we already know that $$3! = 6$$, which I’ll write as $$3 \\times 2 \\times 1$$ for reasons which are about to become obvious. So we have the following formula, which is how we define the factorial: $$n! = n \\times (n-1) \\times \\dots \\times 4 \\times 3 \\times 2 \\times 1$$ “$n!$” is read out loud as “$n\\$ factorial”, and it means “the number of ways to arrange $$n$$ objects into any order”. # Edge cases We’ve seen $$3!$$, but never $$2!$$ or $$1!$$. • It’s easy to see that there are two ways to arrange two objects into an order: $$1,2$$ or $$2,1$$. So $$2! = 2$$. • It’s also easy (if a bit weird) to see that there is just one way of arranging one object into an order: $$1$$ is", "pirate, and $n_8$ be the number of coconuts found in the morning. $n_8$ is divisible by 7 so write $n_8 = 7m_8$. But $n_8$ is what was left when the seventh pirate discarded one coconut out of $n_7$ and hid one seventh of the rest, so $n_7 = (7/6)n_8+1 = (7/6)(7m_8)+1$. For that to be an integer $m_8$ must be a multiple of 6, so write $m_8 = 6m_7$. Then $n_7 = 7^2m_7+1$. Now $n_6 = (7/6)n_7+1 = (7/6)(7^2m_7+1)+1$. For integer value, $7^2m_7\\mod 6 = m_7\\mod 6 = 5$, so write $m_7 = 6m_6+5$. Then $n_6 = (7/6)(7^2(6m_6+5)+1)+1 = 7^3m_6+288$. You see how this goes. $n_5 = (7/6)n_6+1 = (7/6)(7^3m_6+288)+1$. For integer value, $m_6\\mod 6 = 0$, so write $m_6 = 6m_5$. Then $n_5 = (7/6)(7^3(6m_5)+288))+1 = 7^4m_5+337$. $n_4 = (7/6)n_5+1 = (7/6)(7^4m_5+337)+1$. For integer value, $m_5\\mod 6 = 5$, so write $m_5 = 6m_4+5$. Then $n_4 = (7/6)(7^4(6m_4+5)+337)+1 = 7^5m_4+14400$. $n_3 = (7/6)n_4+1 = (7/6)(7^5m_4+14400)+1$. For integer value, $m_4\\mod 6 = 0$, so write $m_4 = 6m_3$. Then $n_3 = (7/6)(7^5(6m_3)+14400)+1 = 7^6m_3+16801$. $n_2 = (7/6)n_3+1 = (7/6)(7^6m_3+16801)+1$. For integer value, $m_6\\mod 6 = 5$, so write $m_3 = 6m_2+5$. Then $n_2 = (7/6)(7^6(6m_2+5)+16801)+1 = 7^7m_2+705888$. $n_1 = (7/6)n_2+1 = (7/6)(7^7m_2+705888)+1$. For integer value, $m_2\\mod 6 = 0$, so write $m_2 = 6m_1$. Then $n_1 = (7/6)(7^7(6m_1)+705888)+1", "× # 0!=1? Let n be any integer so,$$n!$$=$$1\\times2\\times3\\times\\dots\\times (n-1)\\times n$$ and, $$n-1!$$=$$1\\times2\\times3\\times\\dots\\times (n-1)$$ now, multiplying and dividing R.H.S. of $$n-1!$$ by n, we get $\\frac { 1\\times 2\\times 3\\times ....\\times n-1\\times n }{ n }$ which is actually $\\frac { n! }{ n }$ now, put $$n=1$$ in $$n-1!$$, we get $$1-1!$$=$$0!$$ on L.H.S. and $\\frac { 1! }{ 1 }$=$$1$$ in R.H.S. so, $$0!$$=$$\\boxed1$$ Please give some more solutions to this..Thank you! Note by Yoogottam Khandelwal 1 year, 11 months ago Sort by: If n! is defined as the product of all positive integers from 1 to n, then: 1! = 1*1 = 1 2! = 1*2 = 2 3! = 123 = 6 4! = 123*4 = 24 ... n! = 123...(n-2)(n-1)n and so on. Logically, n! can also be expressed n*(n-1)! . Therefore, at n=1, using n! = $$n*(n-1)!$$ 1! = 1*0! which simplifies to 1 = 0! · 1 year, 11 months ago The most comprehensible explanation by far...! Thank you :) · 1 year, 6 months ago You are welcome... · 1 year, 6 months ago Using words alone you can determine 0!=1. factorials are used to tell us how many ways there are to arrange n amount of objects. if there are 0 objects there is only 1 way to arrange", "# How do you write the equation for \"one more than three times a number is 7\"? Nov 10, 2017 The equation is: $3 N + 1 = 7$ The solution is: $2$ #### Explanation: You say: \"one more than three times a number is 7\". So we start with a number $N$. Then we multiply it by $3$ to get: $3 N$ Then we increase it by $1$ to get: $3 N + 1$ And we equal all of that to $7$: $3 N + 1 = 7 \\to$ [ANS} Solving: $3 N + 1 = 7$ $3 N + 1 - 1 = 7 - 1$ $3 N = 6$ $N = \\frac{6}{3}$ N=2 \"one more than three times a number is 7\". \"one more than three times N is 7\". \"one more than three times $2$ is 7\".", "\\frac{5}{3} \\times (n-2)$$ by $$f_n$$ and multiply to $$n \\times (n-1)$$: $$$$n \\times (n-1) \\times C_n = n \\times (n-1) \\times f_n + 6 \\times \\sum_{k=0}^{n-2} {(n-k-1) \\times C_k} \\tag{4}$$$$ Write formula (4) for $$n+1$$: $$$$n \\times (n+1) \\times C_{n+1} = n \\times (n+1) \\times f_n + 6 \\times \\sum_{k=0}^{n-1} {(n-k) \\times C_k} \\tag{5}$$$$ Subtract (4) from (5), we have: $$$$n \\times (n+1) \\times C_{n+1} - n \\times (n-1) \\times C_n = n \\times (n+1) \\times f_n - n \\times (n-1) \\times f_n + 6 \\times \\sum_{k=0}^{n-2} C_k + 6 \\times C_{n-1} \\tag{6}$$$$ Denote $$n \\times (n+1) \\times C_{n+1} - n \\times (n-1) \\times C_n$$ by $$X_n$$ and $$n \\times (n+1) \\times f_n - n \\times (n-1) \\times f_n$$ by $$F_n$$: $$$$X_n = F_n + 6 \\times \\sum_{k=0}^{n-2} C_k + 6 \\times C_{n-1} \\tag{7}$$$$ Write formula (7) for $$n+1$$: $$$$X_{n+1} = F_{n+1} + 6 \\times \\sum_{k=0}^{n-1} C_k + 6 \\times C_n \\tag{8}$$$$ Subtract (7) from (8), we have: $$$$X_{n+1} - X_n = F_{n+1} - F_n + 6 \\times C_n \\tag{9}$$$$ Resolving of $$F_{n+1} - F_n$$ gives: $$$$X_{n+1} - X_n = 2 + 10 \\times n + 6 \\times C_n \\tag{10}$$$$ The function $$X_n$$ is substituted into equation (10), which yields the following equation: $$$$(n+1) \\times (n+2) \\times C_{n+2} -2 \\times n(n+1) \\times C_{n+1} + \\left[n \\times (n-1) -", "## Algebra 2 (1st Edition) n-th term:$\\quad a_{n}=1.2+7.8n$, The next term is $\\quad a_{5}=40.2$ Note that $16.8-9=7.8,$ $24.6-16.8=7.8,$ $32.4-24.6=7.8$ .... and we can also rewrite the first term as $1.2+7.8$ (to fit the pattern) $n=1,\\quad 9=1.2+7.8$ $n=2,\\quad 16.8=9+7.8\\quad=1.2+2\\times 7.8$ $n=3,\\quad 24.6=16.8+7.8=1.2+3\\times 7.8$ $n=4,\\quad 32.4=24.6+7.8=1.2+4\\times 7.8\\quad...$ So, the nth term is $a_{n}=1.2+n\\times 7.8$, and the next term is $a_{5}=32.4+7.8=40.2$ (also, $1.2+5\\times 7.8=40.2$ )", "so $7-2+11=16$ 5. $5.6+4.4=10$, so $7.2+4.4=11.6$ 6. $\\frac{5^6\\times2}{9375}=10$, so $\\frac{7^2\\times2}{9375}=0.01045333...$ The last one is ridiculous on purpose - it serves to illustrate the point that almost anything could be an answer.", "then $n!=1, \\text{when}\\ n=0$ $n!=n\\times (n-1)!\\ \\text{when} \\ n>1$ In other words, if $$n$$ is an integer greater than $$0$$, then $$n!=n\\times (n-1)\\times (n-2)\\times \\cdots \\times 2\\times 1$$. And $$0!=1$$ by definition. I hope this helps! - 4 years, 1 month ago", "## Algebra 2 (1st Edition) There is $7!$ ways to order 7 objects or $7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1 = \\boxed {5040}$", "way. Let $n_i$ be the number of coconuts found by the $i$th pirate, and $n_8$ be the number of coconuts found in the morning. $n_8$ is divisible by 7 so write $n_8 = 7m_8$. But $n_8$ is what was left when the seventh pirate discarded one coconut out of $n_7$ and hid one seventh of the rest, so $n_7 = (7/6)n_8+1 = (7/6)(7m_8)+1$. For that to be an integer $m_8$ must be a multiple of 6, so write $m_8 = 6m_7$. Then $n_7 = 7^2m_7+1$. Now $n_6 = (7/6)n_7+1 = (7/6)(7^2m_7+1)+1$. For integer value, $7^2m_7\\mod 6 = m_7\\mod 6 = 5$, so write $m_7 = 6m_6+5$. Then $n_6 = (7/6)(7^2(6m_6+5)+1)+1 = 7^3m_6+288$. You see how this goes. $n_5 = (7/6)n_6+1 = (7/6)(7^3m_6+288)+1$. For integer value, $m_6\\mod 6 = 0$, so write $m_6 = 6m_5$. Then $n_5 = (7/6)(7^3(6m_5)+288))+1 = 7^4m_5+337$. $n_4 = (7/6)n_5+1 = (7/6)(7^4m_5+337)+1$. For integer value, $m_5\\mod 6 = 5$, so write $m_5 = 6m_4+5$. Then $n_4 = (7/6)(7^4(6m_4+5)+337)+1 = 7^5m_4+14400$. $n_3 = (7/6)n_4+1 = (7/6)(7^5m_4+14400)+1$. For integer value, $m_4\\mod 6 = 0$, so write $m_4 = 6m_3$. Then $n_3 = (7/6)(7^5(6m_3)+14400)+1 = 7^6m_3+16801$. $n_2 = (7/6)n_3+1 = (7/6)(7^6m_3+16801)+1$. For integer value, $m_6\\mod 6 = 5$, so write $m_3 = 6m_2+5$. Then $n_2 = (7/6)(7^6(6m_2+5)+16801)+1 = 7^7m_2+705888$. $n_1 = (7/6)n_2+1 = (7/6)(7^7m_2+705888)+1$. For integer value, $m_2\\mod", "large enough $$n$$ and since you have checked up to $$n = 7.5 \\times 10^7$$, I am confident that it holds for all $$n \\ge 5$$." ]

[ "yellow, blue, green, white so that each flag consisted of three different colors? • Cars plates How many different licence plates can country have, given that they use 3 letters followed by 3 digits?", "and positions 3, 4, and 5 can each hold 1 of 10. g he3llo9, i8 am ho9tti8e3. All of the combinations of the permitted characters (1 lette. A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. There are 26 letters in the alphabet so there are 26 26 26 = 17576 possible 3 letter \"words\". In the second space, you can place any of the digits from 0 to 9. Question 399402: 5. 1 A password to a computer account is to consist of 2 lower-case letters followed by four digits. (a) How many passwords are possible if there are no restrictions? 1757600 (b) How many passwords are possible if none of the. What is the probability of correctly guessing Arabella's password? asked by Blargh on April 24, 2016; stats. How many different license plates are possible if the first letter must be a vowel (a, e, i, o, u), and repetition of letters is not permitted, but repetition of digits is permitted?. Which of the following expressions represents the number of different outfits consisting of 1 pair of shorts, 1 shirt, and 1 pair of sandals? 2. Following is the syntax for isdigit() method −. The letter identifier spesifies further restrictions for the field. Complexity:", "650 unique 2 letter combinations which can be used as the first two letters of the license plate. Two combinations that differ only in ordering of their characters are the same combination. 50400 is the number of ways to arrange 10 letters (alphabets) word \"STATISTICS\" by using Permutations (nPr) formula. There are 95^8 possible combinations of 8 characters. Open a browser in PNG free application web site and go to the Merger tool. Believe it or not, the letter Z has not always been the last letter of the alphabet; in the Greek alphabet, it had a respectable place at number seven. Correct answer: for the letters there are 26 possibilities for each of the 3 tracks and for the numbers there are 10 possibilities for each of the 3 tracks. Unless indicated, rates listed are. b) Adding another digit increases the number of plates by. (Compare Problem 5, Topic 26. How many words of length 10 can you make from these, such that • all the A’s are used. One can also use the combination formula for this problem: n C r = n! / (n-r)! r! Therefore: 5 C 3 = 5! / 3! 2! = 10 (Note: an example of a counting problem in which order would matter is a lock or passcode situation. For", "# Math Help - Quick Question one Permutations/Combinations 1. ## Quick Question one Permutations/Combinations Hey all, I had a few questions regarding some probability theory which have me completely stumped, would someone mind pointing me in the right direction? Question #1- Assume that all license plates in Kentucky consist of three numbers followed by three letters. How many ways are there to get a Kentucky license plate with the number 41 appearing (without any digits between) and with the first two letters of your name appearing consecutively in order? (If the first two letters of your name are the same you should calculate this problem using any two distinct letters.) Question #2 - Suppose a conference has 12 basketball teams. At the end of the season the four best teams play for the championship. How many different sets of four teams could be in the championship? 2. ## Re: Quick Question one Permutations/Combinations Originally Posted by Danny09 Hey all, I had a few questions regarding some probability theory and have found myself completely stumped, would someone mind pointing me in the right direction? Question #1- Assume that all license plates in Texas consist of three numbers followed by three letters. How many ways are there to get a Kentucky license plate with the number 41 appearing (without any", "which do not meet the requirements for three-character assignments. They are keyed by the two-letter Post Office or supplemental abbreviation (listed below) of the state with which they are associated. The two-letter code appears in the first two, middle, or last two positions of the four-character code.", "to the nearest thousandth of a percent, the probability that a password chosen at random will contain the letters A, D, and T. asked • 08/08/19 A password consists of two letters of the alphabet followed by 4 digits chosen from 0 to 9. Calculate the new mean. Repetition is not permitted. You need not simplify your answers. if a license plate consists of two letters followed by three numbers, how many license plates are possible if the letters and numbers may be repeated? 2. Letter Codes: An identification code is to consist of 2 letters followed by 4 digits. B The probability is equal to 0. How many different plates can be made a. A password consists of 2 letters followed by 2 digits. Since any three digit number can go with any three digit \"word\" there. But digits. How many different ID numbers are possible with this system? (Q 2 (0) 000 A state has reoistered 8 million automobiles. How many passwords are possible if the letters must be uppercase? 3 letters: 26^3 ways 3 digits: 10^3 ways----# of passwords: 26^3*1000-----b. total no of ways in which u can select two letters is 24 * 24. A password must have at least eight characters. Lesson 19 - Raw Strings and Multiline Strings. A secret code consists", "x 26 x 25 x 24 x 23 = calculator. Decem51ber). For example, if you want to use the alphabets a,b,c and three. 2) combine the nibbles into a single binary number. if a license plate consists of two letters followed by three numbers, how many license plates are possible if the letters and numbers may be repeated? 2. Write a Java regular expression to match license plates that start with 4 digits and end with two uppercase letters. First and second number - 10 possibilities each. An encyclopedia has eight volumes. How many different passwords can be formed? Algebra -> Customizable Word Problem Solvers -> Age -> SOLUTION: A password consists of 2 letters followed by 2 digits. therefore total number of such codes = C 2 26 * 2! * C 2 9 * 2! = 26 * 25 2 * 2 * 9 * 8 2 * 2 = 26 * 25 * 72 = 46800 (b) 2 letters out of 26 letters can be chosen in C 2 26 w a y s and it can be arranged in 2! ways. The first set, which contains four characters, must consist of an alphanumeric character followed by two numeric characters followed by an alphanumeric character. The same with the second A from the left. (a)", "681412: A password consists of two letters of the alphabet followed by three digits chosen from 0 to 9. A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. If repetitions are allowed in the letters but not in the digits _____ ---- How many different computer passwords are possible? a. Example 2: Each user on a computer system has a password which must be six to eight characters long. We have 1,757,600 combinations available for license plates. In English language these include the 26 letters in the. The same with the second A from the left. How many such license plates are there? The number of possible license plates:. A password must have at least eight characters. You have said three different letters of the alphabet followed by four different digits. In this case you first are choosing one letter out of 26, followed by one lett. Suppose a password must consist of two letters followed by three digits. Repeats are allowed. This password generator tool runs locally on your Windows, Mac or Linux computer, as well as your iOS or Android device. A computer passcode consists of five letters followed by two digits which cannot be zero. A code word consists of three letters of", "figure out the number of ways to choose each and then multiply them. For the initial sign-on your Username and Password will be the same. A password consists of three digits, 0 through 9, followed by three letters from an alphabet having 26 letters. The password has 3 letters followed by 3 digits. How many different possible passwords are there?. Glad we were able to help! fn+F11 fixed mine- thank you! my keyboard instead of typing 'E' it 'pastes information,, my 'o' changes the scgreen to full screen and 'u' adds a h so it becomes 'hu' and now each time I type 'p' this sign. A license plate is to consist of three letters followed by two digits. Repeated characters are allowed. You have to multiply the possibilities - you need two letters and five numbers, so for each one you figure out the number of ways to choose each and then multiply them. In the first space, you can place any of the digits from 0 to 9. How many different words can you make from the one word \" mississippi\" , with out using two letters twice , and with out using the actual word? W hat do the numbers in my wi driver's license number mean? the letter is the 1st letter of my", "How many different passwords can be formed? - 3404286. a license plate has 2 letters followed by 4 digits. EXAMPLE: Student: John B. Solve the problem. Repeats are allowed. \" The parentheses set the boundaries for the group. d where each letter can represent 1, 2, or 3 digits, and the periods are required. Then find the last two digits of 2 raised to that power. (10 points) 2. Calculate the new mean. Example 2: Each user on a computer system has a password which must be six to eight characters long. Up to + total six digit passwords. There are 26 letters in the alphabet so there are 26 26 26 = 17576 possible 3 letter \"words\". The revised regular expression is: Regular expression: \\d{3}\\-\\d{2}\\-\\d{4} Matches: All social security numbers of the form 123-12-1234. 5 Certain automobile license plates consist of a sequence of three letters fol-lowed by three digits. Repetition is not permitted. * The letters will be the first two letters of your last name plus the first two letters of your first name making sure the first and third letters are capitalized. When letters or digits are allowed to repeat, we can write the same digits or letter again and again, so total number of ways or total numbers get increased but when repetition", "1, 3, 7 or 9 (because there are no even digits left to go around. · The % (modulo) operator. ecopoint 1,152 views. A combination of mandatory letters and characters, all uppercase. That's 5 places. Passwords are almost always case sensitive, for a given letter there are $26 \\times 2=52$ possibilities, and for the digits we have $10$ possibilities. Password strength is the measure of a password's efficiency to resist password cracking attacks. A password consists of two letters of the alphabet followed by three digits from CGNB 293 at Tenaga National University, Kajang. Alphanumeric characters refer to a set of characters consisting of both letters and numbers. In California, each automobile license plate consists of a single digit followed by three letters, followed by three digits. How many different possible passwords are there? how to slove. Second and third - 26 possibilities each. This is what I have so far:. a license plate has 2 letters followed by 4 digits. Page 1 of 2 Version 1 Appendix C - Valid Postcode Format Version 1 - Published 28 April 2017 Guidance on recording valid postcodes All postcodes on the ILR must be in upper case. How many different passwords can be formed? Answer by Fombitz(32378) (Show Source): You can put this solution on YOUR website! Digits - (0-9)", ", 8 C 5 , 5 · 6 · 7 . 11. In a certain state, license plates consist of 3 letters followed by 3 digits (0 to 9) followed by a letter. How many license plates can be formed if numbers and letters may not repeat? 12. There are 25 points marked on a circle. How many line segments can be drawn connecting any two points on the circle? 13. The language of the ancient “Immaculatans” contained the vowels a,e,i,o,u and the consonants c,l,m,n,s,t. a) How many 3 letter “words” are possible if each contained two different vowels and a consonant between them? b) How many 5 letter “words” with 2 different vowels at the beginning and the end and a group of 3 different consonants between them? 14. In how many ways can you randomly guess at the answers on a 10 question multiple choice quiz if each question has 5 possible responses from which to choose? 15. At Monmouth Park you bet the 1st place/2 nd place finish in 3 races. If the first race has 8 horses, the second race has 9 horses and the third race has 8 horses, what is the probability that you will choose the winning duo in all 3 races? Show work and give a solution. 16. Eight people", "cars, Code 3xx, registration office Erlangen Picture taken by myself on 2005-09-25... 1979 TI-99/4 with RF modulator, optional Speech Synthesizer, keyboard overlays, and a cartridge. ... The typeface used on most European number plates for cars distinguishes the two symbols partially in this manner (having a more rectangular or wider shape for the capital O than the digit 0), but in several countries a further distinction is made by slitting open the digit 0 on the upper right side (as in German plates). This typeface is called fälschungserschwerende Schrift (abbr.: FE Schrift) in Germany, meaning \"script which is harder to falsify\". Typefaces used on United Kingdom plates do not differentiate between the two as there can never be any ambiguity if the design is correctly spaced (the vehicle \"numbers\" are allocated in a manner that avoids any such confusion). The same applies to postal codes in the United Kingdom. Car redirects here. ... FE-Schrift or fälschungserschwerende Schrift (tamper-hindering font) is the typeface that has been used since November 2000 on vehicle registration plates in Germany. ... ## Etymology The word \"zero\" came via French zéro from Venetian zero, which (together with \"cipher\") came via Italian zefiro from Arabic صفر, şafira = \"it was empty\", şifr = \"zero\", \"nothing\", which was used to translate Sanskrit śūnya ( शून्य ),", "number of “mackerels,” the words that don’t share letters with only that state, and the number of unique characters in the state name. As expected, these exhibit a negative correlation, to the tune of -0.53, meaning the more unique characters a state has, the fewer of these words we expect. The states with the fewest number of “mackerels” both have 7 unique characters, Connecticut (9 “mackerels”) and Michigan (7 “mackerels”). The states with the most unique characters are North Dakota and New Mexico, each with 9 unique characters. As expected, both of these are in the bottom 6 in terms of number of “mackerels,” with 54 and 30 respectively. See the code for this project on my GitHub!", "c), (blue, s).\" Thus, we get 6 distinct objects (Fig 1). From the illustration given above we note that \\ \\ \\ \\ \\ \\ \\ \\ \\ color{blue } (A × B = {\"(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)\"}.) \"Again, consider the two sets :\" color{purple}( ✍️ A = {DL, MP, KA},) where DL, MP, KA represent Delhi, Madhya Pradesh and Karnataka, respectively and color{purple} (B = {01,02, 03}) representing codes for the licence plates of vehicles issued by DL, MP and KA . ● If the three states, Delhi, Madhya Pradesh and Karnataka were making codes for the licence plates of vehicles, with the restriction that the code begins with an element from set A, ● which are the pairs available from these sets and how many such pairs will there be (Fig 2)? ● The available pairs are: (DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03) and the product of set A and set B is given by color{purple}(A × B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03)}.) ● It can easily be seen that there will be 9 such pairs in the Cartesian product, since there are 3 elements in each of the sets A and B. This gives us 9 possible codes. \\color{red} \\ox \\color{red} \\mathbf(COMMON \\ CONFUSION) Also", "type of input mask, for example, to help users enter part numbers or other forms of inventory correctly. A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. JavaScript: Prompts and turning Strings in Numbers using parseInt() and parseFloat() Learning objectives: · Getting values into the program using prompt(). But digits. They also know that the second digit is either 4 or 7. A password consists of 2 letters followed by 2 digits. B){EE, EO, OO}. Repeats are allowed. Each group consists of zero or more digits, and may be followed by a comma. Round your answer to three decimal places. So 10 options. A book number with the literal text, mandatory first and last digits, and any combination of letters and characters between those digits. A password consists of three letters followed by a two-digit number. How many different plates can be made a. Imagine that you have four empty spaces. (a) What is the probability that a randomly chosen plate contains the number 9999? (Round your answer to six decimal places. 8 The exponent marker consists of a single letter followed by an optional sign and a sequence of digits, which are interpreted as the power of ten by which the number before the exponent", "number of distinct MSTs in a definitive way.", "x 8 x 7. The Nebraska license plate has 6 characters, each one filled with a number (0-9) or a letter. Although, I think a lot of things like this, it's always best to reason through than try to figure out if some formula applies to it. How many different \"words\" can be made from the letters in the word MISSISSIPPI? 5 How many permutations of 8 letters can be made from P A R A L L E L. where n is the number of letters and nr represents the number the number of times any letter is repeated. Excel > Combinations > Return all combinations. How many possible seating arrangements. Moe dul 19 964 on sLse 2. To see the answer, pass your mouse over the colored area. Question 122836: How many permutations of the leters ABCDEFG contain (a)the string BCD (b)the string CFGA (c)the string BA & GF (d)the string ABC & DE (e)the string. Repetition is not allowed. How many ways can 5 Scrabble tiles be placed in a row? Any time you have a number in the form N x (N-1) x (N-2) x. There are 3 possible ways a letter in the first position can be selected - either a, b or c. The group of all permutations of a set M", "1. ## Permutation trouble Hello, I am wondering whether I solved the following correctly. \"How many license plates containing seven characters can be created from letters and numbers if the first two characters must be uppercase letters and the last five characters must be digits from 0 to 9? Repetition of numbers is allowed. Repetition of letters is not allowed\". Here's my solution. 1) I assumed a universal set for the license plate as U = {uppercase letter, uppercase letter, number, number, number, number, number}. 2) For the first two characters I assumed a sample set A ={A,B,C...Z}, for the last five characters there's the sample set B = {0,1,2,3...9}. 3) Considering the repetition of letters is not allowed and that the order matters in set U, I determined 650 possible permutations in set A, following the formula n!/(n-r)!, which comes down to 26!/(26-2)! = 650 permutations. So set A has 650 possible permutations. 4) Considering the repetitions of numbers is allowed and that the order in set U matters, I determined 100.000 possible permutations in set B, following the formula n^r, which comes down to 10^5 = 100.000 permutations. So set B has 100.000 possible permutations. 5) Now set A and B are added together, giving what I think are 100.650 possible license plates. Are my thoughts", "a certain department, 18 drive foreign and 10 drive domestic cars. If five of these…" ]

[ "# Counting different kinds of permutations • What's the number of strings of digits and letters of length $7$ without repetition? • Solution 1. Pick $2$ digits, $4$ consonants, and $1$ vowel and then line them up: $\\binom{10}{2}\\binom{21}{4} \\cdot 5 \\cdot 7!$ • Solution 2. Pick $2$ positions in the string for the digits, $4$ places for the consonants, leaving $1$ for the vowel. Then fill the spots: $\\binom72\\binom54 \\cdot 10 \\cdot 9 \\cdot 21 \\cdot 20 \\cdot 19 \\cdot 18 \\cdot 5.$ I think the methods above also count the permutations of MISSISSIPPI. But what about the problem below: • License plates in some state consist of $3$ different digits followed by $3$ different letters. How many of these plates are there? We choose places first and for each of those we choose symbols: $\\binom63\\binom33P(10, 3)P(26, 3).$ But apparently that's wrong. The actual answer is $P(10, 3)P(26, 3).$ What makes these two problems different? Why does choosing places first in the second problem results in wrong answer? For the pick of digits you have exactly 10 choices for first pick and 9 for the second and then 8 for the thrid giving a total of $$10\\times9\\times8=P(10,3)$$ similarly for the letters you have 26 choices for the first and 25 for the second and 24 for the third", "### Home > A2C > Chapter 12 > Lesson 12.4.2 > Problem12-190 12-190. In the past, many states had license plates composed of three letters followed by three digits ($0\\text{ to }9$). Recently, many states have responded to the increased number of cars by adding one digit ($1\\text{ to }9$) ahead of the three letters. How many more license plates of the second type are possible? What is the probability of being randomly assigned a license plate containing $\\text{ALG }2$? Since repetition is allowed, the total number of license plates of the first type is $26\\cdot26\\cdot26\\cdot10\\cdot10\\cdot10 = 17,576,000$. The total number of license plates of the second type is $9\\cdot26\\cdot26\\cdot26\\cdot10\\cdot10\\cdot10 = 158,184,000$. Find the difference. Find the number of license plates containing $\\text{ALG }2$. $9\\cdot1\\cdot1\\cdot1\\cdot1\\cdot10\\cdot10 =$ $P(\\text{ALG }2)=\\frac{900}{158,184,000}=?$", "$\\text{Total number of licences} - 26\\cdot 25\\cdot 24\\cdot 23\\cdot 10\\cdot 9\\cdot 8\\cdot 7$. Where the total number of licences is $26^4\\cdot 10^4$ which is what you calculated. Assuming the letters and numbers can appear in any order: You can obtain this answer by multiplying the answer obtained above by $\\dfrac{8!}{4!4!}$ which is the number of permutations of $8$ things in which $4$ objects are of one kind and $4$ objects are of the other kind. This is the same as the answer given by Omnitic. Please note that as pointed out by Omnitic, the following does not restrict the number of letters and digits to $4$: Assuming the numbers or letters can be filled in any order (and are not restricted to be $4$): The licences have $8$ places which can be filled in with a number or a letter. Thus each place can have one of $26+10=36$ values. Now, number of licences with no letters or numbers repeated: $36\\cdot 35\\cdot 34\\cdot\\cdot\\cdot 30\\cdot29$. So, number of licences with at least one letter or number repeated: $\\text{Total number of licences} - 36\\cdot 35\\cdot 34\\cdot\\cdot\\cdot 30\\cdot29$ The total number is given by $36^8$. - If the licence plate is of the form $LLLLNNNN$ where L stands for letter and N for integer between 0 and 9 inclusive then no. Number", "# Combinations with Restrictions Automobile license plates for a state consist of four letters followed by a dash and two single digits. How many different license plate combinations are possible if exactly one letter is repeated exactly once, but digits cannot be repeated? We have $$\\binom{4}{2}$$ = 6 positions for the two identical letters to occupy And we have (25)(24) ways to choose the other two letters The total number of \"words\" = 6 * 26 *25*24 = 93600 Since the digits cannot be repeated = 10 * 9 = 90 The total possibilities = 93600 *90 =8,424,000 Is this correct? • I think it is right – Tojrah Apr 18 at 17:50 Let's say the repeated letter is $$A$$; you are correct that then number of ways we can put two $$A$$ on a plate is $$6$$. The are $$25 \\cdot 24$$ ways to pick the other two letters. But what if repeated letter is $$B$$? We get the same number of arrangements. So I think the correct number of arranging letters is $$26 \\cdot 25 \\cdot 24 \\cdot 6$$. Everything looks correct.", "second letters with first letter B, and so on. We add 26 to itself 26 times. Or quicker: there are $$26 \\cdot 26$$ choices for the first two letters. Now for each choice of the first two letters, we have 26 choices for the third letter. That is, 26 third letters for the first two letters AA, 26 choices for the third letter after starting AB, and so on. There are $$26 \\cdot 26$$ of these $$26$$ third letter choices, for a total of $$(26\\cdot26)\\cdot 26$$ choices for the first three letters. And for each of these $$26\\cdot26\\cdot26$$ choices of letters, we have a bunch of choices for the remaining digits. In fact, there are going to be exactly 1000 choices for the numbers. We can see this because there are 1000 three-digit numbers (000 through 999). This is 10 choices for the first digit, 10 for the second, and 10 for the third. The multiplicative principle says we multiply: $$10\\cdot 10 \\cdot 10 = 1000\\text{.}$$ All together, there were $$26^3$$ choices for the three letters, and $$10^3$$ choices for the numbers, so we have a total of $$26^3 \\cdot 10^3$$ choices of license plates. Careful: “and” doesn't mean “times.” For example, how many playing cards are both red and a face card? Not $$26 \\cdot 12\\text{.}$$ The", "= 80,000,000$$. For a veteran plate, there are 24 choices for the first letter, followed by 24 choices for the second letter. There are 10 choices for the first digit, and 10 choices for the second digit. Using the product rule, the total number of license plates in this category is $$242 \\cdot 102 = 57,600$$. Finally, for a motorcycle plate, there are 24 choices for the first letter, followed by 24 choices for the second letter. There are 10 choices for the first digit, 10 choices for the second digit, and 10 choices for the third digit. Using the product rule, the total number of license plates in this category is $$242 \\cdot 103 = 576,000$$. Using the sum rule, we see that the total number of license plates is $$13,824,000 + 80,000,000 + 57,600 + 576,000$$ which is $$94,457,600$$. It doesn’t always happen that the sum rule is applied first to break the problem down into cases, followed by the product rule within each case. In some problems, these might occur in the other order. Sometimes there may seem to be one “obvious” way to look at the problem, but often there is more than one equally effective analysis, and different analyses might begin with different rules. In Example 2.3.1, we could have begun by noticing", "of the $26$ letters ($A - Z$ is twenty six letters) can be placed in the sixth position. This is seen as: $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{A - Z}$ $\\underline{A - Z}$ $\\underline{A - Z}$ $\\underline{10}$ $\\underline{10}$ $\\underline{10}$ $\\underline{26}$ $\\underline{26}$ $\\underline{26}$ Number of different license plates = $10 * 10 * 10 * 26 * 26 * 26 = 17,576,000$ license plates (4.) The United States social security number (SSN) contains nine digits. How many different social security numbers are possible if: (a.) repetition of digits are allowed? (b.) repetition of digits are not allowed? (c.) repetition of digits are allowed and the first number cannot begin with a $0$? (d.) repetition of digits are not allowed and the first number cannot begin with a $0$? This is a case of the Fundamental Counting Principle There are ten digits from $0 - 9$ $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ (a.) repetition of digits are allowed. Any of the $10$ digits ($0 - 9$ is ten digits) can be placed in the any of the nine positions. $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{10}$ $\\underline{10}$ $\\underline{10}$ $\\underline{10}$ $\\underline{10}$ $\\underline{10}$", "letters and two digits. Motorcycle plates have two letters followed by three digits. Setting aside vanity plates and ignoring the fact that some three-letter words are avoided, how many license plates are available to individuals in Alberta for their cars or motorcycles? Solution To answer this question, there is a natural division into four cases: regular car plates with three digits; regular car plates with four digits; veteran plates; and motorcycle plates. For a regular car plate with three digits, there are 24 choices for the first letter, followed by 24 choices for the second letter, and 24 choices for the third letter. There are 10 choices for the first digit, 10 choices for the second digit, and 10 choices for the third digit. Using the product rule, the total number of license plates in this category is $$243 \\cdot 103 = 13,824,000$$. For a regular car plate with four digits, there are 20 choices for the first letter, followed by 20 choices for the second letter, and 20 choices for the third letter. There are 10 choices for the first digit, 10 choices for the second digit, 10 choices for the third digit, and 10 choices for the fourth digit. Using the product rule, the total number of license plates in this category is $$203 \\cdot 104", "letters in alphabetic order, and $\\binom{10}{4}$ counts the ways to pick four digits in descending order. If the first digit is allowed to be $0$, $\\binom{10}{4}$ is also the number of ways to pick four digits in ascending order. This assumption is likely to hold for license plates; otherwise, $10$ must be replaced by $9$. The solution that you already know tells you how to deal with distinct, but not ordered elements. If letters and digits may come in any order, for every choice of two distinct letters and four distinct digits there are $6!$ ways to arrange them. When the letters come before the digits, we reason similarly, but we separately count the ways to fill the letter positions and the ways to fill the digit positions. For example, if the letters only need to be distinct, while the digits must be in decreasing order, we get $\\binom{26}{2} \\cdot 2! \\cdot \\binom{10}{4}$. Do you see how to continue?", "# License plate consisting of 4 letters and 4 numbers While doing homework today, the following question popped into my head: Can you easily calculate the amount of unique license plates consisting of 4 letters and 4 numbers in any order? It doesn't seem to be easily possibly; $$26^3 * 10^3 * 8!$$ would include repeats (such as AAA123 and AAA123; As are in different positions) Thanks! • Could you clarify? AAA123 and AAA123 are identical, and neither has $4$ letters or $4$ numbers. – Brian M. Scott Mar 28 '13 at 6:52 If you want license plates consisting of four numbers and four letters in any possible arrangement, then you must first choose $4$ of the $8$ spots for your numbers (or symmetrically, your letters). Then there are $10^4$ choices for your numbers and $26^4$ choices for your letters. This means that in total you have $\\binom{8}{4}26^4\\cdot 10^4$ distinct license plates. If you make the restriction that the letters come before the numbers (or the numbers before the letters) then you must choose the first four spots for your letters and the last four spots for your numbers. This simply removes the binomial coefficient in the above, leaving $26^4\\cdot 10^4$ distinct license plates.", "choices. Example: Calculate the number of combinations of (69 choose 5) = 11 238 513, and multiply by (26 choose 1) = 26 for a total of 292 201 338 combinations. So there are more 6-character license plate combinations than there. For example 0! is a special case factorial. We can't really say that there are 4 choices for the first letter and proceed from there, since the number of choices for the second letter depend on which letter was chosen for the first. B In Utah, a license plate consists of 3 digits followed by 3 letters. A combination lock opens with the correct four-letter code. Below are Total 240 words made out of this word. ∴ By the multiplication (c ounting) rule, total number of solutions = 4×. McNeal & Urban safes used letters rather than numbers on the dial. (b) Calculate the number of different 5-digit numbers which can be formed using the digits 0,1,2,3,4 without repetition and assuming that a number cannot begin with 0. Algorithm : The main idea behind the solution to this problem is to find out the combination of 0,1 and 3. If each digit in a 3-digit lock contains the numbers 0 through 9, then each dial in the lock can be set to one of 10 options (0,", "permutations repetitions/no repetitions License plates consist of sequence of 3 letters followed by 3 digits. How can they be arranged if (i) no repetition of letters is permitted, how many possible license plates are there? should it be $26P3 \\cdot (10^3)$ or $3! \\cdot (10^3)$? Your first answer is correct. Since repetition of letters is not permitted, there are $26$ ways of choosing the first letter, $25$ ways of choosing the second letter, and $24$ ways of choosing the third letter. Therefore, there are $P(26, 3) = 26 \\cdot 25 \\cdot 24$ ways of selecting the three letters. Since repetition of digits is permitted, there are ten choices for each of the three digits. Thus, the number of license plates that can be formed is $$P(26, 3) \\cdot 10^3 = 26 \\cdot 25 \\cdot 24 \\cdot 10^3 = 15,600,000$$", "License plates are made using 3 letters followed by 2 digits. How many plates can be made if repetition of letters and digits is allowed? Jan 18, 2017 We have $1 , 757 , 600$ combinations available for license plates. Explanation: Number on license plates are of the form $L L L D D$, where $L$ represents a letter and $D$ represents a digit. As $L$ can be anything from $A$ to $Z$, there are $26$ combinations for that and as repetition is allowed, for second and third letters, we again have $26$ combinations available and thus $26 \\times 26 \\times 26 = 17576$ combinations for letters. But digits are from $0$ to $9$ i.e. $10$ combinations for each place and tolal $10 \\times 10 = 100$ combinations. Hence for $L L L D D$, we have $1 , 757 , 600$ combinations available for license plates.", "together, giving what I think are 100.650 possible license plates. The results are multiplied not added: $(650)\\cdot(10^5)$ 3. ## Re: Permutation trouble For the first letter you have 26 possible choices. For the 2nd letter you have 25. For each of the numbers you have 10 choices. There are 5 of these. so the total number of unique license plates is given by $N = 26 \\cdot 25 \\cdot 10^5 = 65,000,000$", "# Difference between revisions of \"2006 AMC 10A Problems/Problem 18\" ## Problem A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible? $\\mathrm{(A) \\ } 10^4\\times 26^2\\qquad\\mathrm{(B) \\ } 10^3\\times 26^3\\qquad\\mathrm{(C) \\ } 5\\times 10^4\\times 26^2\\qquad\\mathrm{(D) \\ } 10^2\\times 26^4\\qquad\\mathrm{(E) \\ } 5\\times 10^3\\times 26^3\\qquad$ ## Solution There are $10\\cdot10\\cdot10\\cdot10 = 10^4$ ways to choose 4 digits. There are $26 \\cdot 26 = 26^2$ ways to choose the 2 letters. For the letters to be next to each other, they can be the 1st and 2nd, 2nd and 3rd, 3rd and 4th, 4th and 5th, or the 5th and 6th characters. So, there are $5$ choices for position. Therefore the number of distinct license plates is $5\\times 10^4\\times 26^2 \\Rightarrow C$ Invalid username Login to AoPS", "the number of possible combinations is 36 6, which comes out to 2,176,782,336. (4 4 ) = 1. Introduction : In this C programming tutorial, we will learn how to print all combinations of three predefined numbers. P (10,3) = 720. How many license plates can be manufactured with three letters followed by three digits? No repetition 3. Suppose you have totally forgotten the combination to your locker. Find the number of different permutations of the letters of the word MASSACHUSETTS. The formulas definitely save time when we are asked to find the number of permutations of a larger set. The second letter spot can be any one of the remaining 25 letters. Therefore, group these vowels and consider it as a single letter. Generate A List Of All Possible 4 Digits Combinations; In some cases, we may need to generate a list of all possible 4 digits combinations of number 0 to 9, which means to generate a list of 0000, 0001, 0002…9999. Number of ways to arrange these $6$ letters $=6!2!=360$ All the $2$ vowels are different. How many combinations are there? Abe, Bob, Carol, Dee. For example, suppose we have a set of three letters: A, B, and C. If the questions have 4,3 and 2 solutionsvely, find the total number of solutions. Find the", "## Algebra and Trigonometry 10th Edition $138,240,000$ distinct license plate numbers are possible. There are 24 choices for each of the 3 letters. There are 10 choices for each of the 4 digits. Using the Fundamental Counting Principle: $24\\times24\\times24\\times10\\times10\\times10\\times10=138,240,000$", "# Difference between revisions of \"1999 AMC 8 Problems/Problem 15\" ## Problem Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}. When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters? $\\text{(A)}\\ 24 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 36 \\qquad \\text{(D)}\\ 40 \\qquad \\text{(E)}\\ 60$ ## Solution ### Solution 1 There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \\cdot 3 \\cdot 4 = 60$ license plates. Adding $2$ letters to the start gives $7\\cdot 3 \\cdot 4 = 84$ plates. Adding $2$ letters to the middle gives $5 \\cdot 5 \\cdot 4 = 100$ plates. Adding $2$ letters to the end gives $5 \\cdot 3 \\cdot 6 = 90$ plates. Adding a letter to the start and middle gives $6 \\cdot 4 \\cdot 4 = 96$ plates. Adding a letter to the start and end gives $6 \\cdot 3", "a license plate which has $2$ letters and $3$ digits, there are: $26 \\times 26 \\times 10 \\times 10 \\times 10 = 676 , 000$ possibilities. Hope this helps.", "four for the third. This means that $5\\cdot 3\\cdot 4=60$ plates could have been made. If two letters are added to the second set, then $5\\cdot 5\\cdot 4=100$ plates can be made. If one letter is added to each of the second and third sets, then $5\\cdot 4\\cdot 5=100$ plates can be made. None of the other four ways to place the two letters will create as many plates. So, $100-60=\\boxed{40}$ ADDITIONAL plates can be made.So the correct choice is $\\boxed{D}$" ]

[ "# Counting different kinds of permutations • What's the number of strings of digits and letters of length $7$ without repetition? • Solution 1. Pick $2$ digits, $4$ consonants, and $1$ vowel and then line them up: $\\binom{10}{2}\\binom{21}{4} \\cdot 5 \\cdot 7!$ • Solution 2. Pick $2$ positions in the string for the digits, $4$ places for the consonants, leaving $1$ for the vowel. Then fill the spots: $\\binom72\\binom54 \\cdot 10 \\cdot 9 \\cdot 21 \\cdot 20 \\cdot 19 \\cdot 18 \\cdot 5.$ I think the methods above also count the permutations of MISSISSIPPI. But what about the problem below: • License plates in some state consist of $3$ different digits followed by $3$ different letters. How many of these plates are there? We choose places first and for each of those we choose symbols: $\\binom63\\binom33P(10, 3)P(26, 3).$ But apparently that's wrong. The actual answer is $P(10, 3)P(26, 3).$ What makes these two problems different? Why does choosing places first in the second problem results in wrong answer? For the pick of digits you have exactly 10 choices for first pick and 9 for the second and then 8 for the thrid giving a total of $$10\\times9\\times8=P(10,3)$$ similarly for the letters you have 26 choices for the first and 25 for the second and 24 for the third", "the numbers. If repetitions are allowed in the letters but not in the digits _____ ---- How many different computer passwords are possible? a. So to find the total number of letter combinations we multiply the number of choices for the first digit by the number of choices for the next digit, in this case 26*26, for a total of 676. For example, if you want to use the alphabets a,b,c and three. To get the number of permutations here, you need to multiply the different values each of the characters can have: $26\\cdot10\\cdot10\\cdot10=26,000$ 2) A letter followed by four numbers. The Most Beautiful Equation in Math - Duration: 3:50. In this case you first are choosing one letter out of 26, followed by one lett. if a license plate consists of two letters followed by three numbers, how many license plates are possible if the letters and numbers may be repeated? 2. a 7-character password consists of two numbers (from 1 to 9) followed by three letters (from a to z) followed by two more numbers (from 1 to 9). An encyclopedia has eight volumes. For example, John Doe 123-45-6789 would be jdo56789. 4 million passwords each containing four digits, the DataGenetics Web site reported that the most commonly used four-digit sequence (representing 11 percent of. 5", "x 26 x 25 x 24 x 23 = calculator. Decem51ber). For example, if you want to use the alphabets a,b,c and three. 2) combine the nibbles into a single binary number. if a license plate consists of two letters followed by three numbers, how many license plates are possible if the letters and numbers may be repeated? 2. Write a Java regular expression to match license plates that start with 4 digits and end with two uppercase letters. First and second number - 10 possibilities each. An encyclopedia has eight volumes. How many different passwords can be formed? Algebra -> Customizable Word Problem Solvers -> Age -> SOLUTION: A password consists of 2 letters followed by 2 digits. therefore total number of such codes = C 2 26 * 2! * C 2 9 * 2! = 26 * 25 2 * 2 * 9 * 8 2 * 2 = 26 * 25 * 72 = 46800 (b) 2 letters out of 26 letters can be chosen in C 2 26 w a y s and it can be arranged in 2! ways. The first set, which contains four characters, must consist of an alphanumeric character followed by two numeric characters followed by an alphanumeric character. The same with the second A from the left. (a)", "yellow, blue, green, white so that each flag consisted of three different colors? • Cars plates How many different licence plates can country have, given that they use 3 letters followed by 3 digits?", "figure out the number of ways to choose each and then multiply them. For the initial sign-on your Username and Password will be the same. A password consists of three digits, 0 through 9, followed by three letters from an alphabet having 26 letters. The password has 3 letters followed by 3 digits. How many different possible passwords are there?. Glad we were able to help! fn+F11 fixed mine- thank you! my keyboard instead of typing 'E' it 'pastes information,, my 'o' changes the scgreen to full screen and 'u' adds a h so it becomes 'hu' and now each time I type 'p' this sign. A license plate is to consist of three letters followed by two digits. Repeated characters are allowed. You have to multiply the possibilities - you need two letters and five numbers, so for each one you figure out the number of ways to choose each and then multiply them. In the first space, you can place any of the digits from 0 to 9. How many different words can you make from the one word \" mississippi\" , with out using two letters twice , and with out using the actual word? W hat do the numbers in my wi driver's license number mean? the letter is the 1st letter of my", "and positions 3, 4, and 5 can each hold 1 of 10. g he3llo9, i8 am ho9tti8e3. All of the combinations of the permitted characters (1 lette. A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. There are 26 letters in the alphabet so there are 26 26 26 = 17576 possible 3 letter \"words\". In the second space, you can place any of the digits from 0 to 9. Question 399402: 5. 1 A password to a computer account is to consist of 2 lower-case letters followed by four digits. (a) How many passwords are possible if there are no restrictions? 1757600 (b) How many passwords are possible if none of the. What is the probability of correctly guessing Arabella's password? asked by Blargh on April 24, 2016; stats. How many different license plates are possible if the first letter must be a vowel (a, e, i, o, u), and repetition of letters is not permitted, but repetition of digits is permitted?. Which of the following expressions represents the number of different outfits consisting of 1 pair of shorts, 1 shirt, and 1 pair of sandals? 2. Following is the syntax for isdigit() method −. The letter identifier spesifies further restrictions for the field. Complexity:", "repetition is not allowed, we. To get the number of permutations here, you need to multiply the different values each of the characters can have: $26\\cdot10\\cdot10\\cdot10=26,000$ 2) A letter followed by four numbers. [letters], [numbers], [special characters (as above)]. We know that the last digit must be zero, so the only remaining option for the fifth digit is 5. There are 26 letters in the alphabet so there are 26 26 26 = 17576 possible 3 letter \"words\". Example: AA111, if repeats are allowed. Example: has 2,a,b,c means that an entry must have at least two of the letters a, b and c. Up to + total six digit passwords. JavaScript: Prompts and turning Strings in Numbers using parseInt() and parseFloat() Learning objectives: · Getting values into the program using prompt(). For us to allow you to reset your password, we need to first confirm your identity. if a license plate consists of two letters followed by three numbers, how many license plates are possible if the letters and numbers may be repeated? 2. Each old license plate consisted of a letter followed by four digits. There wouldn't be a case for more than two letters or 6 numbers. A company assigns an ID code to each of its employees that consists of 1, 2 or 3 uppercase", "How many different passwords can be formed? - 3404286. a license plate has 2 letters followed by 4 digits. EXAMPLE: Student: John B. Solve the problem. Repeats are allowed. \" The parentheses set the boundaries for the group. d where each letter can represent 1, 2, or 3 digits, and the periods are required. Then find the last two digits of 2 raised to that power. (10 points) 2. Calculate the new mean. Example 2: Each user on a computer system has a password which must be six to eight characters long. Up to + total six digit passwords. There are 26 letters in the alphabet so there are 26 26 26 = 17576 possible 3 letter \"words\". The revised regular expression is: Regular expression: \\d{3}\\-\\d{2}\\-\\d{4} Matches: All social security numbers of the form 123-12-1234. 5 Certain automobile license plates consist of a sequence of three letters fol-lowed by three digits. Repetition is not permitted. * The letters will be the first two letters of your last name plus the first two letters of your first name making sure the first and third letters are capitalized. When letters or digits are allowed to repeat, we can write the same digits or letter again and again, so total number of ways or total numbers get increased but when repetition", "$\\text{Total number of licences} - 26\\cdot 25\\cdot 24\\cdot 23\\cdot 10\\cdot 9\\cdot 8\\cdot 7$. Where the total number of licences is $26^4\\cdot 10^4$ which is what you calculated. Assuming the letters and numbers can appear in any order: You can obtain this answer by multiplying the answer obtained above by $\\dfrac{8!}{4!4!}$ which is the number of permutations of $8$ things in which $4$ objects are of one kind and $4$ objects are of the other kind. This is the same as the answer given by Omnitic. Please note that as pointed out by Omnitic, the following does not restrict the number of letters and digits to $4$: Assuming the numbers or letters can be filled in any order (and are not restricted to be $4$): The licences have $8$ places which can be filled in with a number or a letter. Thus each place can have one of $26+10=36$ values. Now, number of licences with no letters or numbers repeated: $36\\cdot 35\\cdot 34\\cdot\\cdot\\cdot 30\\cdot29$. So, number of licences with at least one letter or number repeated: $\\text{Total number of licences} - 36\\cdot 35\\cdot 34\\cdot\\cdot\\cdot 30\\cdot29$ The total number is given by $36^8$. - If the licence plate is of the form $LLLLNNNN$ where L stands for letter and N for integer between 0 and 9 inclusive then no. Number", "# Difference between revisions of \"2006 AMC 10A Problems/Problem 18\" ## Problem A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible? $\\mathrm{(A) \\ } 10^4\\times 26^2\\qquad\\mathrm{(B) \\ } 10^3\\times 26^3\\qquad\\mathrm{(C) \\ } 5\\times 10^4\\times 26^2\\qquad\\mathrm{(D) \\ } 10^2\\times 26^4\\qquad\\mathrm{(E) \\ } 5\\times 10^3\\times 26^3\\qquad$ ## Solution There are $10\\cdot10\\cdot10\\cdot10 = 10^4$ ways to choose 4 digits. There are $26 \\cdot 26 = 26^2$ ways to choose the 2 letters. For the letters to be next to each other, they can be the 1st and 2nd, 2nd and 3rd, 3rd and 4th, 4th and 5th, or the 5th and 6th characters. So, there are $5$ choices for position. Therefore the number of distinct license plates is $5\\times 10^4\\times 26^2 \\Rightarrow C$ Invalid username Login to AoPS", "1, 3, 7 or 9 (because there are no even digits left to go around. · The % (modulo) operator. ecopoint 1,152 views. A combination of mandatory letters and characters, all uppercase. That's 5 places. Passwords are almost always case sensitive, for a given letter there are $26 \\times 2=52$ possibilities, and for the digits we have $10$ possibilities. Password strength is the measure of a password's efficiency to resist password cracking attacks. A password consists of two letters of the alphabet followed by three digits from CGNB 293 at Tenaga National University, Kajang. Alphanumeric characters refer to a set of characters consisting of both letters and numbers. In California, each automobile license plate consists of a single digit followed by three letters, followed by three digits. How many different possible passwords are there? how to slove. Second and third - 26 possibilities each. This is what I have so far:. a license plate has 2 letters followed by 4 digits. Page 1 of 2 Version 1 Appendix C - Valid Postcode Format Version 1 - Published 28 April 2017 Guidance on recording valid postcodes All postcodes on the ILR must be in upper case. How many different passwords can be formed? Answer by Fombitz(32378) (Show Source): You can put this solution on YOUR website! Digits - (0-9)", "to the nearest thousandth of a percent, the probability that a password chosen at random will contain the letters A, D, and T. asked • 08/08/19 A password consists of two letters of the alphabet followed by 4 digits chosen from 0 to 9. Calculate the new mean. Repetition is not permitted. You need not simplify your answers. if a license plate consists of two letters followed by three numbers, how many license plates are possible if the letters and numbers may be repeated? 2. Letter Codes: An identification code is to consist of 2 letters followed by 4 digits. B The probability is equal to 0. How many different plates can be made a. A password consists of 2 letters followed by 2 digits. Since any three digit number can go with any three digit \"word\" there. But digits. How many different ID numbers are possible with this system? (Q 2 (0) 000 A state has reoistered 8 million automobiles. How many passwords are possible if the letters must be uppercase? 3 letters: 26^3 ways 3 digits: 10^3 ways----# of passwords: 26^3*1000-----b. total no of ways in which u can select two letters is 24 * 24. A password must have at least eight characters. Lesson 19 - Raw Strings and Multiline Strings. A secret code consists", "### Home > A2C > Chapter 12 > Lesson 12.4.2 > Problem12-190 12-190. In the past, many states had license plates composed of three letters followed by three digits ($0\\text{ to }9$). Recently, many states have responded to the increased number of cars by adding one digit ($1\\text{ to }9$) ahead of the three letters. How many more license plates of the second type are possible? What is the probability of being randomly assigned a license plate containing $\\text{ALG }2$? Since repetition is allowed, the total number of license plates of the first type is $26\\cdot26\\cdot26\\cdot10\\cdot10\\cdot10 = 17,576,000$. The total number of license plates of the second type is $9\\cdot26\\cdot26\\cdot26\\cdot10\\cdot10\\cdot10 = 158,184,000$. Find the difference. Find the number of license plates containing $\\text{ALG }2$. $9\\cdot1\\cdot1\\cdot1\\cdot1\\cdot10\\cdot10 =$ $P(\\text{ALG }2)=\\frac{900}{158,184,000}=?$", "(2) 0 is never the first of the 4 digits. How many possible license plates are there for the state? A. 5,189,184 B. 5,616,000 C. 5,630,625 D. 5,184,000 E. 5,183,424 Explanation: This is a combination problem whose answer can be calculated by multiplying the correct numbers. 1. We need to know how many different combinations of two letters are possible if the letters O and I are never used. This means there are 24 letters available, and so the number of combinations of letters is 24 x 24 = 576. 2. Next, we need to know how many 4 digit numbers there are if we disallow 0 from being the first digit. What this means is that we want to know how many numbers there are between 1000 and 9999, inclusive. This is determined by subtracting 999 from 9999. The result is 9000. 3. Finally, we multiply the results of 1 and 2: 576 × 9000 = 5,184,000. So D is the correct answer. Combinations Example 2) If a combination lock has 40 numbers on it, and each combination for unlocking the lock uses 3 of the numbers with repetitions allowed, then how many different combinations are possible? A. 24,000 B. 44,000 C. 82,000 D. 64,000 E. 36,000 Explanation: The order of the numbers in a combination is", "of the $26$ letters ($A - Z$ is twenty six letters) can be placed in the sixth position. This is seen as: $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{A - Z}$ $\\underline{A - Z}$ $\\underline{A - Z}$ $\\underline{10}$ $\\underline{10}$ $\\underline{10}$ $\\underline{26}$ $\\underline{26}$ $\\underline{26}$ Number of different license plates = $10 * 10 * 10 * 26 * 26 * 26 = 17,576,000$ license plates (4.) The United States social security number (SSN) contains nine digits. How many different social security numbers are possible if: (a.) repetition of digits are allowed? (b.) repetition of digits are not allowed? (c.) repetition of digits are allowed and the first number cannot begin with a $0$? (d.) repetition of digits are not allowed and the first number cannot begin with a $0$? This is a case of the Fundamental Counting Principle There are ten digits from $0 - 9$ $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ (a.) repetition of digits are allowed. Any of the $10$ digits ($0 - 9$ is ten digits) can be placed in the any of the nine positions. $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{0 - 9}$ $\\underline{10}$ $\\underline{10}$ $\\underline{10}$ $\\underline{10}$ $\\underline{10}$ $\\underline{10}$", "# Combinations with Restrictions Automobile license plates for a state consist of four letters followed by a dash and two single digits. How many different license plate combinations are possible if exactly one letter is repeated exactly once, but digits cannot be repeated? We have $$\\binom{4}{2}$$ = 6 positions for the two identical letters to occupy And we have (25)(24) ways to choose the other two letters The total number of \"words\" = 6 * 26 *25*24 = 93600 Since the digits cannot be repeated = 10 * 9 = 90 The total possibilities = 93600 *90 =8,424,000 Is this correct? • I think it is right – Tojrah Apr 18 at 17:50 Let's say the repeated letter is $$A$$; you are correct that then number of ways we can put two $$A$$ on a plate is $$6$$. The are $$25 \\cdot 24$$ ways to pick the other two letters. But what if repeated letter is $$B$$? We get the same number of arrangements. So I think the correct number of arranging letters is $$26 \\cdot 25 \\cdot 24 \\cdot 6$$. Everything looks correct.", "¢ ˘1287 5-cardhandsthatarealldiamonds,orallclubs,orallspades. It only checks for the proper number of digits or the code terminator #. The letter can be in any position in the password. In South Carolina, the license plate format is three letters followed by three digits. But you can not permute the available letters. For a position which contains a digit, there are 10 di erent ways to ll that position. I'm thinking about going back to Girlfriend 7. Since there are ( 6 2) pairs of positions, we must subtract ( 6 2) ⋅ 10 2 ⋅ 36 4 to get rid of the double-counting. The remaining four characters can be either a letter or a digit, so 36 possibilities exist for each of these. How many different teams could be formed? [5] 2. We can think of 3-digits codes as permutations of #10# digits chosen #3# digits at a time since no digits are repeated. How many times more 5-character passwords can you make using numbers and special characters in addition to upper- and lowercase letters than just with upper- and lowercase letters?. If the letter can be anywhere in the password, then the number of possible passwords is 786,240 * 6 = 4,717,440. For example, if you set a preamble of @@ and scanned bar code data of 12345, @@12345", "+0 # Plz help! 0 185 2 +120 License plates from different states follow different alpha-numeric formats, which dictate which characters of a plate must be letters and which must be numbers. Florida has license plates with an alpha-numeric format like the one pictured. North Dakota, on the other hand, has a different format, also pictured. Assuming all 10 digits are equally likely to appear in the numeric positions, and all 26 letters are equally likely to appear in the alpha positions, how many more license plates can Florida issue than North Dakota? Jul 6, 2020 #1 -1 Use multiplication principle: $$26^4 \\cdot 10^2 - 26^2 \\cdot 10^4 = 38937600$$ Jul 6, 2020 #2 +120 +1 Florida issues license plates in which the first three and last slots are filled with letters, and the fourth and fifth are filled with digits. Thus, there are $$26^4 \\cdot 10^2$$ Florida license plates possible. North Dakota, however, issues license plates in which the first three slots are filled with letters and the last three slots are filled with digits. There are thus $$26^3 \\cdot 10^3$$ possible North Dakota license plates. Multiplying these out and taking the difference yields an answer of $$\\boxed{28121600}$$. Jul 6, 2020", "the number of possible combinations is 36 6, which comes out to 2,176,782,336. (4 4 ) = 1. Introduction : In this C programming tutorial, we will learn how to print all combinations of three predefined numbers. P (10,3) = 720. How many license plates can be manufactured with three letters followed by three digits? No repetition 3. Suppose you have totally forgotten the combination to your locker. Find the number of different permutations of the letters of the word MASSACHUSETTS. The formulas definitely save time when we are asked to find the number of permutations of a larger set. The second letter spot can be any one of the remaining 25 letters. Therefore, group these vowels and consider it as a single letter. Generate A List Of All Possible 4 Digits Combinations; In some cases, we may need to generate a list of all possible 4 digits combinations of number 0 to 9, which means to generate a list of 0000, 0001, 0002…9999. Number of ways to arrange these $6$ letters $=6!2!=360$ All the $2$ vowels are different. How many combinations are there? Abe, Bob, Carol, Dee. For example, suppose we have a set of three letters: A, B, and C. If the questions have 4,3 and 2 solutionsvely, find the total number of solutions. Find the", "681412: A password consists of two letters of the alphabet followed by three digits chosen from 0 to 9. A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. If repetitions are allowed in the letters but not in the digits _____ ---- How many different computer passwords are possible? a. Example 2: Each user on a computer system has a password which must be six to eight characters long. We have 1,757,600 combinations available for license plates. In English language these include the 26 letters in the. The same with the second A from the left. How many such license plates are there? The number of possible license plates:. A password must have at least eight characters. You have said three different letters of the alphabet followed by four different digits. In this case you first are choosing one letter out of 26, followed by one lett. Suppose a password must consist of two letters followed by three digits. Repeats are allowed. This password generator tool runs locally on your Windows, Mac or Linux computer, as well as your iOS or Android device. A computer passcode consists of five letters followed by two digits which cannot be zero. A code word consists of three letters of" ]

[ "Talk:Perfect power Disambiguation This entry needs disambiguation from the Galvin Institute's Perfect Power, which is all about better electrical power. —Preceding unsigned comment added by 134.134.139.70 (talk) Wikipedia is not a dictionary attempting to list all possible meanings of a term. Wikipedia:Disambiguation#Deciding to disambiguate says: Disambiguation is required whenever, for a given word or phrase on which a reader might use the \"Go button\", there is more than one Wikipedia article to which that word or phrase might be expected to lead. There doesn't appear to be information about Galvin Institute's Perfect Power in Wikipedia, so no disambiguation is needed. PrimeHunter (talk) 21:52, 20 November 2009 (UTC) Computer Program for Perfect Powers Has anyone figured out a computer program for determining perfect powers in chronological order (A001597), excluding 1? Where n is the instance in the set and y is the value of that instance: n=1, y=4 n=2, y=8 n=3, y=9 n=4, y=16 n=5, y=25 n=6, y=27 n=7, y=32 n=8, y=36 What about a math equation? Jeanlovecomputers (talk) 01:19, 14 February 2010 (UTC) It's not too hard to come up with an algorithm for producing an ordered list of perfect powers: 1. Start with a set S containing 22 = 4, and an empty list L. 2. Find the smallest member of S - call it x. Then", "# How to quickly identify perfect powers In a test I'll take there may be a question such as the following: A perfect power is an integer that can be written as $a^b$, $a$ and $b$ being integers greater or equal to 2. One of the following numbers is not a perfect power, which one is it? • 125 • 216 • 1000 • 500 • 2500 The first three (125, 216, 1000) are simply primeNumber^something, but the last two aren't. Keeping in mind that I can't use a calculator and should spend on average only 1 minute per question, I was wondering what's the best method I can use to resolve such questions or more difficult ones (where there are less obvious wrong solutions, such as 1000). • $216=6^3$ so it is not a prime power. Nor is $1000=10^3$. But they are perfect powers. $2500=50^2$. Only $500$ fails. – alex.jordan Nov 3 '13 at 17:13 Once you are down to $500$ and $2500$, you can note that each has a factor of $10^2=2^25^2$ and no more factors of $2$. This tells you the power must be a square. Then you can just ask which of $5$ and $25$ is a square. Hint: If $n=p_1^{k_1} ...p_i^{k_i}$ then you can prove that $n$ is a perfect power if and", "125 is a perfect power (it can be expressed as rn, where r = 5 and n = 3 (both integers > 1)) x = 25 is a perfect power (it can be expressed as rn, where r = 5 and n = 2 (both integers > 1)) But x = 5 is not a perfect power (it cannot be expressed as rn, where r and p are integers > 1. The first entries of this unnamed set would be: 0, 1, 2, 3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 28 The missing integers from this set are: 4, 8, 9, 16, 25, 27... (the perfect powers). I am concluding that there is no official or agreed-upon name for this class of integers. Also, I cannot find that sequence in http://oeis.org. Maybe I'll add it. Thanks, -j Mark44 Mentor I would like to know if there is an official name for the class of integers that are (not) perfect powers. A perfect power is a number that can be expressed as xn, where x and n are both integers > 1. I have been calling these integers \"roots\" - since they do not have any integer roots of their own, and they are the roots of", "General Management (Energy and Utilities) Followers: 2 Kudos [?]: 74 [0], given: 387 Re: If x and y are distinct positive integers, what is the value [#permalink] 10 Mar 2015, 01:46 Bunuel wrote: Suvorov wrote: Bunuel wrote: only one such pair is possible $$x=4>y=2$$: $$x^y=4^2=16=2^4=y^x$$ --> $$x^4-y^4=240$$. Sufficient. How do you come to this conclusion? Just picking numbers/knowing or is there some mathematical rule for this? I understand that it revolves around the power of 2, but can't put my finger on it. I think it's worth remembering that $$4^2=16=2^4$$, I've seen several GMAT questions on number properties using this (another useful property $$8^2=4^3=2^6=64$$). But if you don't know this property: Given: $$x^y = y^x$$ and $$x>y$$, also $$x$$ and $$y$$ are distinct positive integers. Couple of things: $$x$$ and $$y$$ must be either distinct positive odd integers or distinct positive even integers (as odd in ANY positive integer power is odd and even in ANY positive integer power is even). After testing several options you'll see that $$x=4$$ and $$y=2$$ is the only possible scenario: because, when $$y\\geq{2}$$ and $$x>{4}$$, then $$x^y$$ (bigger value in smaller power) will be always less than $$y^x$$ (smaller value in bigger power): $$5^3<3^5$$, or $$6^2<2^6$$, or $$8^2<2^8$$, or $$10^2<2^{10}$$, .... Want to solve a GMAT like problem using this property 8^2=4^3=2^6=64", "# Find the n-th perfect power! A perfect power is a number of the form $$\\a^b\\$$, where $$\\a>0\\$$ and $$\\b>1\\$$. For example, $$\\125\\$$ is a perfect power because it can be expressed as $$\\5^3\\$$. # Goal Your task is to write a program/function that finds the $$\\n\\$$-th perfect power, given a positive integer $$\\n\\$$. # Specs • The first perfect power is $$\\1\\$$ (which is $$\\1^2\\$$). • Input/output in any reasonable format. • Built-ins are allowed. # Scoring This is . Shortest solution in bytes wins. # Testcases input output 1 1 2 4 3 8 4 9 5 16 6 25 7 27 8 32 9 36 10 49 • Uptil what number should this work? Infinity? – ghosts_in_the_code May 1 '16 at 9:59 • A reasonable amount. – Leaky Nun May 1 '16 at 10:00 • What about a language that uses only a data type of one bit? – ghosts_in_the_code May 1 '16 at 10:09 • @Agawa001 Yes it is a standard loophole which are no longer funny. – flawr May 1 '16 at 10:12 • – Martin Ender May 1 '16 at 10:15 # Jelly, 11 bytes µÆE;¬g/’µ#Ṫ ### Background Every positive integer $$\\k\\$$ can be factorized uniquely as the product of powers of the first $$\\m\\$$ primes, i.e., $$\\k=p_1^{\\alpha_1}\\cdots p_m^{\\alpha_m}\\$$, where $$\\\\alpha_m>0\\$$. We", "is -1 and power is even number. Then answer will always be 1. Now by equating base to -1, we get x=1,x=0. However, as per our condition if x=1, the power(x+2) becomes odd, thus only x=0 can be taken as a true value. So we have 1 value from this condition. 3) if power is zero(0) then, irrespective of the base, answer will be 1. Thus, by equating power(x+2) to 0, (x+2)=0, x=-2 thus we have totally 4 values(2,-1,0,-2) from all the three above conditions. Hope it clears your doubt. Power is not ignored, in fact, every possibility is considered! Cheers Tx Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8373 Location: Pune, India Suppose x is an integer such that (x^2−x−1)^(x+2)=1 [#permalink] ### Show Tags 21 Sep 2015, 22:49 1 anik1989 wrote: VeritasPrepKarishma wrote: anik1989 wrote: what will be the ans? a or d? There are 4 values x can take: -2, -1, 0, 2 (look at smyarga's solution above) thank you karishma. honestly i dod not get that ans. why power has been ignored in sloution?? $$a^b = 1$$ In which all cases can this happen? Case 1: a = 1, b can be anything e.g. $$1^4$$, $$1^{-3}$$, $$1^0$$ Case 2: a = -1, b must be an even integer e.g. $$(-1)^2$$, $$(-1)^{-4}$$ Case 3:", "Volume 24 1988 Issue 1 There are three items I wish to bring to the reader's attention. Issue 2 Suppose we have two quantities, $x$ and $y$, and we are given that $y$ may be expressed as an infinite power series in $x$: $$y = a_1 x + a_2x^2 + a_3x^3 + \\cdots$$ Issue 3 In probability theory and in everyday language we use the word 'random' and, for many of us, our first encounter with the word would have been in drawing a name at random from a hat.", "# Tag Archives: number-theory ## Sum of the first factorials a perfect power? Consider the following sum: $$F(n) = 1! + 2! + 3! + \\cdots + n!$$ Can this sum ever be a perfect power? A perfect power here is defined as $x^n$ for some natural number $x$ and some natural number $n\\geq 2$. Some observations immediately come to mind: $F(1) = 1$, and that’s trivially a […] ## Montreal Spring ’13 Conferences: Number Theory and Algebra This is a fairly recent picture of the McGill campus: However, soon the spell of unbearable heat will dawn on the city and there will be plenty of fun things to do outside. Despite the hot sun we shouldn’t neglect the indoor activities, such as the many awesome conferences and workshops that will be going […]", "reducing a random x-bits value to a power of two that is smaller (i.e. a smaller bit-size). This works since you have a random value whose max value is $2^x-1$ and so you have $2^x$ possible values in your range, and any power of two $2^y, y is naturally dividing a larger power of two: $\\frac{2^x}{2^y}=2^{x-y}$ without any remainder. This notably means that you can simply generate random bytes and then reduce them to a smaller bit-length without risking a modulo bias. Rejection sampling combined with modulo works well, as long as you are very careful with your bounds, because otherwise you might have the same problem as Cryptocat, where they were doing rejection sampling from 0 to 250 inclusive before computing the modulo reduction by 10. This means that they had 251 possible random values instead of 250, and the 251th value, $250$, was mapped to 0 instead of being rejected, and thus they had 2 chances out of 26 to get 0 and 1 chance out of 26 to get any other digit from 0 to 9. Okay, but I was mentioning ways to use the modulo in a safe manner, and actually it is worth noting here that when you have a number $p$ and a random value in the range $[0, 2^n[$, then the", "power... That for any given positive integer check box Superscript as shown below of., Jaroslaw … the following is a prime number ) 38 416 Please help, thanks are unblocked number as! Negative value a, its fourth roots of a number ( and its representation the!, Jaroslaw … the following is a perfect fourth power, what is the root! Half tones ) have our 1st ever SIGNED ALBUM largest perfect 4th powers are there integers that are zeros! Of k to 4 exponent has an even exponent then it is used times! Root, we have something to the Format Menu and choose the first 1,000 PEOPLE... Behind a web filter, Please make sure that the domains * and! Of the given number product 46400k is a perfect fourth powers ) 38 416 Please help, thanks ( steps... 67, written exclusively with the digits 6 and 7 the product 46400k is a perfect square written exclusively the! It, once it 's released 38 416 Please help, thanks then notice the! And so on until it is a perfect fourth power of 4 the. Fit the bill is 2^4, or the power of 4, if it is list! X ) and 4 is the square of some binomial which are perfect powers, 3 2 − 3. And 4 is the", "Demonstration, with Cuisenaire rods, of the perfect power nature of 4, 8, and 9 In mathematics, a perfect power is a natural number that is a product of equal natural factors, or, in other words, an integer that can be expressed as a square or a higher integer power of another integer greater than one. More formally, n is a perfect power if there exist natural numbers m > 1, and k > 1 such that mk = n. In this case, n may be called a perfect kth power. If k = 2 or k = 3, then n is called a perfect square or perfect cube, respectively. Sometimes 0 and 1 are also considered perfect powers (0k = 0 for any k > 0, 1k = 1 for any k). Examples and sums A sequence of perfect powers can be generated by iterating through the possible values for m and k. The first few ascending perfect powers in numerical order (showing duplicate powers) are (sequence A072103 in the OEIS): ${\\displaystyle 2^{2}=4,\\ 2^{3}=8,\\ 3^{2}=9,\\ 2^{4}=16,\\ 4^{2}=16,\\ 5^{2}=25,\\ 3^{3}=27,}$ ${\\displaystyle 2^{5}=32,\\ 6^{2}=36,\\ 7^{2}=49,\\ 2^{6}=64,\\ 4^{3}=64,\\ 8^{2}=64,\\dots }$ The sum of the reciprocals of the perfect powers (including duplicates such as 34 and 92, both of which equal 81) is 1: ${\\displaystyle \\sum _{m=2}^{\\infty }\\sum _{k=2}^{\\infty }{\\frac {1}{m^{k))}=1.}$ which", "will receive a summary of all topics I bump in your profile area as well as via email. _________________ Senior Manager Joined: 01 Nov 2013 Posts: 357 GMAT 1: 690 Q45 V39 WE: General Management (Energy and Utilities) Followers: 6 Kudos [?]: 169 [0], given: 403 Re: If x and y are distinct positive integers, what is the value [#permalink] ### Show Tags 10 Mar 2015, 01:46 Bunuel wrote: Suvorov wrote: Bunuel wrote: only one such pair is possible $$x=4>y=2$$: $$x^y=4^2=16=2^4=y^x$$ --> $$x^4-y^4=240$$. Sufficient. How do you come to this conclusion? Just picking numbers/knowing or is there some mathematical rule for this? I understand that it revolves around the power of 2, but can't put my finger on it. I think it's worth remembering that $$4^2=16=2^4$$, I've seen several GMAT questions on number properties using this (another useful property $$8^2=4^3=2^6=64$$). But if you don't know this property: Given: $$x^y = y^x$$ and $$x>y$$, also $$x$$ and $$y$$ are distinct positive integers. Couple of things: $$x$$ and $$y$$ must be either distinct positive odd integers or distinct positive even integers (as odd in ANY positive integer power is odd and even in ANY positive integer power is even). After testing several options you'll see that $$x=4$$ and $$y=2$$ is the only possible scenario: because, when $$y\\geq{2}$$ and $$x>{4}$$, then", "counting: (Highest - lowest) + 1. If unsure in the moment, test with small numbers. Ex.: How many integers from 1 to 4? List them: 1, 2, 3, 4. Count. There are four integers from 1 to 4. Then: (4-1)= 3. Not correct. ADD 1. (3+1) = 4 _________________ SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here. Instructions for living a life. Pay attention. Be astonished. Tell about it. -- Mary Oliver Retired Moderator Joined: 27 Oct 2017 Posts: 1260 Location: India GPA: 3.64 WE: Business Development (Energy and Utilities) Re: What is the probability that a randomly selected integer from a list [#permalink] ### Show Tags 26 Mar 2018, 08:32 total number = 500-101+1 = 400 favorable case = 100 ( from 300 to 399) required probability = 100/400 = 1/4 _________________ Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 8117 Location: United States (CA) Re: What is the probability that a randomly selected integer from a list [#permalink] ### Show Tags 27 Mar 2018, 11:33 Bunuel wrote: What is the probability that a randomly selected integer from a list of consecutive integers from 101 to 500, inclusive will have a hundreds digit of 3? (A) 1/5 (B)", "I recently stumbled over this tweet: Suppose you think you’re 80% likely to have left your laptop power adapter somewhere inside a case with 4 otherwise-identical compartments. You check 3 compartments without finding your adapter. What’s the probability that the adapter is inside the remaining compartment? Eliezer Yudkowski I thought I had trained my mind to update whenever new information arrives, but it still didn’t. My first thought was still “80% because nothing has changed about the bag”. But by looking through 3 compartments of the bag, you have gained information and should update your estimates! So I thought it through and came up with a quite general and useful rule: The odds of finding $x$ are updated by the fraction of space you haven’t looked at yet. In Eliezer’s example, the prior odds where 4:1 that your power adapter is somewhere inside the case. So far you have looked through 3/4 of the case, so the fraction of space you haven’t looked at is 1/4. The odds of finding your adapter in the remaining compartment are therefore 1:1 (or 50% probability). This works whenever you make a prediction about finding something in a region of space (or time!) and look through some part of it without finding anything. ## Why is that? Let’s derive this rule. The", "simply counting the proportion of values that equal or exceed $$x$$. Given that the six number formulae comprise the very much greater part of those that are admissible in numerical pursuit and that we should expect them to yield the largest results, we can reasonably approximate this test by considering them alone, as is done by deck 7 for results up to fifty thousand. Deck 7: The Logarithmic Probabilities Well it certainly looks rather close to a straight line to my eye, but unfortunately this is not a terribly sensitive test and to conclude that the results of arithmetical pursuit formulae governed by a power law distribution we shall need some weightier evidence. ### A Theoretical Justification Now it is most certainly the case that the results cannot be exactly power law distributed since their greatest value is $100 \\times 100 \\times 100 \\times 100 \\times 50 \\times 50 = 250,000,000,000$ beyond which the probabilities must trivially be zero. Nevertheless it may yet be the case that, beneath this value, they are approximately power law distributed. To figure whether it is or not, I suggest that we analyse a much simpler game; starting from a value of one, toss a coin and if it comes up heads we double the value and play again starting from the doubled", "of the reduced power will be the key to not over or under count. This is especially important for $a_i = 16, 64, 81$ where it is a composite integer power of some $a_j$ and also when the numerator and denominator of the exponent fraction has a greatest common divider that is greater than 1. The advantage is that it runs extremely fast even with large value of $a$ and $b$.", "probability definitely will go to 0 (even monotonically), so it does not portray an indeterminate form. A better example would perhaps be a basketball player that gets better and better with the number of throws. – Sam Nov 16 '10 at 17:16 The purpose of my answer was to give a simple intuitive explanation for why something close to $1$ raised to a power of large $N$ may result in an arbitrary value. – Shai Covo Nov 16 '10 at 17:36 I think this is a nice answer. It makes the point that it's not how close to $1$ you are in an absolute sense that matters, but rather how this compares to the size of the exponent. – Pete L. Clark May 4 '12 at 0:28 $$1^\\infty = 1^{\\frac{1}{0}} = 0\\sqrt{1}$$ so in limits $1^\\infty$ isn't necessarily $1\\times 1 \\times1 ....$", "bias since $401$ does not divide $65536$ evenly. Now if $x$ is a uniformly distributed random number between $0$ and $65535$, then it belongs to one of these bins. That is $x_i \\leq x < x_{i+1}$ for some $0 \\leq i \\leq 400$. Using the formula for $x_i$ you should be able to calculate the exact value of $i$. Once you know $i$ you should select the $i$th integer between $125$ and $525$, that is, the integer $125 + i$. If your generator is fast and you do not need many random numbers, you could just generate random numbers between 0 and 65535 and keep only the ones that are in the range you want.", "# CLSPWR - Editorial Author: Srikkanth R Editorialist: Aman Dwivedi Easy # PREREQUISITES: Maths, Precomputation # PROBLEM: An integer x is said to be a Perfect Power if there exists positive integers a and b (i.e a, and b should be \\geq 1) such that x = a^{b+1}. Given an integer n, find the closest Perfect Power to n. If there are multiple Perfect Powers equidistant from n and as close as possible to n, find the smallest one. More formally, find the smallest integer x such that x is a Perfect Power and |n - x| \\leq |n - y| over all Perfect Powers y. # EXPLANATION: Have a close look at the constraints of N, it’s just 1e18. Why just man, it’s so big? Remember we will be dealing with the powers and it will take few iterations to get to 1e18 even if we start will the lowest base possible i.e 2. It just takes 60 iterations to reach a value of 1e18. Now let’s look at how we can solve this problem. One of the basic ideas is to iterate for all bases and their powers. The problem with this approach is that the numbers that can act as bases are very large. A number as large as 1e9 can act as a base.", "wow, it's dividable by 81. the probability of the outcome 15313425691932 is unimaginably small. but that doesn't prove anything. openended_option: before drawing a specific random number, the probability of a random number to be dividable by 81, is about 1%. no big deal. I agree, no big deal. But you have not defined any specificity or meaning to the number that is chosen. Any number from 0 to 10^100000 is possible, and each of the possibilities is as probable and any other one. But big deal. This is all Shannon information theory. What is important is the specificity. If you want to specify that the number has to be wholly divisible by 81, that is good. You defined a probability target/meaning. But that is a pretty high probability, a little better than 1 in 10^2. But now lets make a probability of 1 in 10^150. You will still have quite a few hits within the range you defined. So now we have to correspond your range to something real, like time. 1 in 10^150 relates to every single atom in the known universe, the maximum rate per second at which transitions in physical states can occur (10^45), and billion times longer than the age of the universe. That is the probability ID proper measures against; if you cannot" ]

[ "words, we need an integer raised to the 2nd, 3rd, 4th, etc. power. In even other words, we need perfect squares, perfect cubes, etc. that are between 1 and 300, inclusive. Now we need to count these perfect squares, etc. in an efficient way. Let's start with the squares. What's the biggest perfect square less than 300? Test numbers if necessary. $$16^2 = 256$$ and $$17^2 = 289$$, but $$18^2 = 324$$. Thus, we have 12 through 172, for 17 perfect squares. Now, let's count the cubes. Leave out $$1^3$$, since we've already counted 1 as $$1^2$$. $$5^3 = 125$$ and $$6^3 = 216$$, but $$7^3 = 343$$. Thus, we seem to have 5 more integer powers ($$2^3$$ through $$6^3$$, inclusive) - but careful! $$4^3 = 64$$, which we've already counted as a square ($$8^2$$), so we only have 4 more integer powers. This gives us a cumulative total of 21. What about the fourth powers - $$2^4$$, $$3^4$$, etc.? We've already counted any of these that matter, because they are also perfect squares: $$2^4 = 42$$, $$3^4 = 9^2$$, etc. We can leave out any higher even powers for the same reason ($$2^6 = 8^2$$, $$2^8 = 16^2$$, etc.). However, we must consider additional odd powers, continuing to leave out 1 raised to any power. $$2^5 =", "Button Try our free Online GRE Test Director Joined: 20 Apr 2016 Posts: 815 WE: Engineering (Energy and Utilities) Followers: 9 Kudos [?]: 586 [0], given: 113 Re: GRE Math Challenge #81-One integer will be randomly selected [#permalink] 14 Dec 2017, 06:46 sandy wrote: One integer will be randomly selected from the integers 11 to 60, inclusive. What is the probability that the selected integer will be a perfect square or a perfect cube? (A) 0.1 (B) 0.125 (C) 0.16 (D) 0.5 (E) 0.9 Here Total number of integers from 11 to 60 = 50 perfect squares from 11 to 60 are = 16 , 25, 36 and 49 perfect cube = 27 Hence there are 5 integers that can be either a perfect square or a perfect cube Therefore probability = $$\\frac{5}{50}$$= 0.1 _________________ If you found this post useful, please let me know by pressing the Kudos Button Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html Re: GRE Math Challenge #81-One integer will be randomly selected [#permalink] 14 Dec 2017, 06:46 Display posts from previous: Sort by", "# How many perfect powers are there amoung the first 1000 positive integers A positive integer $$n$$ is called a perfect power if and only if there exist two positive integers $$p,q$$, both are greater than 1, such that $$n=p^q$$. For example $$8=2^3, 36=6^2$$ are perfect powers, but $$20=2^2*5, 11=11^1$$ are not. Let $$A=\\{1,2,\\dots,1000\\}$$. How many elements of $$A$$ is a perfect power? I hope there exist a beautiful way to solve this, perhaps without calculating the number of primes untill 1000 or considering all the cases. Is there a formula? Thanks in advance! I don't think there is anything better than counting up each power. There are $$\\lfloor \\sqrt {1000} \\rfloor=31$$ squares and a similar formula for each power. You are correct that you only need to consider primes as all the fourth powers are squares as well. You also don't need to count any higher than $$10$$ because $$2^{10} \\gt 1000$$ and you need to only count $$1$$ once. The sixth powers have been counted twice-once as squares and once as cubes, so need to be subtracted once. This would apply to any number the product of two primes, but six is the only one less than $$10$$.", "+0 0 164 1 An integer is randomly chosen from the integers 1 through 100, inclusive. What is the probability that the chosen integer is a perfect square or a perfect cube, but not both? Express your answer as a common fraction. So do you find all the cubes and squares from 1-100 and then eliminate the ones that are both cubes and squares and then put it in $$\\frac{result}{100}$$because there are 100 different total numbers? Jul 2, 2018 #1 +104 0 A number is a perfect square and a perfect cube if and only if it is a perfect sixth power. Note that 10^2=100 and 4^3<100<5^3, while 2^6<100<3^6=9^3. Hence, there are 10 squares and 4 cubes between 1 and 100, inclusive. However, there are also 2 sixth powers, so when we add 10+4 to count the number of squares and cubes, we count these sixth powers twice. However, we don't want to count these sixth powers at all, so we must subtract them twice. This gives us a total of 10+4-2*2=10 different numbers that are perfect squares or perfect cubes, but not both. Thus, our probability is 10/100=1/10. HOPE THIS HELPS! Mar 16, 2019", "$$17^2 = 289$$, but $$18^2 = 324$$. Thus, we have 12 through 172, for 17 perfect squares. Now, let's count the cubes. Leave out $$1^3$$, since we've already counted 1 as $$1^2$$. $$5^3 = 125$$ and $$6^3 = 216$$, but $$7^3 = 343$$. Thus, we seem to have 5 more integer powers ($$2^3$$ through $$6^3$$, inclusive) - but careful! $$4^3 = 64$$, which we've already counted as a square ($$8^2$$), so we only have 4 more integer powers. This gives us a cumulative total of 21. What about the fourth powers - $$2^4$$, $$3^4$$, etc.? We've already counted any of these that matter, because they are also perfect squares: $$2^4 = 42$$, $$3^4 = 9^2$$, etc. We can leave out any higher even powers for the same reason ($$2^6 = 8^2$$, $$2^8 = 16^2$$, etc.). However, we must consider additional odd powers, continuing to leave out 1 raised to any power. $$2^5 = 32$$, $$3^5 = 243$$, but $$4^5 =$$ something greater than 300, as we can see by considering that $$4^5 = 4^4 \\times 4 = 16^2 \\times 4 = 256 \\times 4$$. So we have two more integer powers ($$2^5$$ and $$3^5$$), for a cumulative total of 23. Seventh powers: $$2^7 = 128$$, but $$3^7 =$$ something much greater than 300 (since $$3^7 = 3^5 \\times", "## Perfect Squares and Perfect Cubes List all integers less than $20,\\!000$ that are both perfect squares and perfect cubes. Source: NCTM Mathematics Teacher, January 2006 Solution $9=3^2$ is a perfect square but not a perfect cube. On the other hand, $8=2^3$ is a perfect cube but not a perfect square. For an integer to be both perfect square and perfect cube, the exponent must be a multiple of $2$ and $3$, that is, the integer must be a sixth power. $0^6=0$ $1^6=1$ $2^6=64$ $3^6=729$ $4^6=4096$ $5^6=15625$ $6^6=46656>20000$ All integers less than $20,\\!000$ that are both perfect squares and perfect cubes $0,1,64,729,4096,15625$ Answer: $0,1,64,729,4096,15625$", "Show Tags 27 Jul 2015, 19:31 For me at least, the key to understanding this question was to carefully read what they were asking for. So, what are all of the unique integers found among perfect squares, cubes, etc. between 1 and 300? Perfect squares can be counted manually... 20^2 is 400 so it must be less than 20. 15^2 is 225 so plug and chug. You'll get 17^2, so that's 17 numbers for perfect squares. Perfect cubes (just look for the highest cube first and eliminate any repeated integers). 7^3 is 7*49= 343, so it's all cubed integers below 7. 2 is 8, 3 is 9 but repeats, 4 is 64 but repeats, 5 is 125, 6 is 216 Repeat for the other powers and count the unique integers. Kudos [?]: 44 [0], given: 40 An integer between 1 and 300, inclusive, is chosen at random [#permalink] 27 Jul 2015, 19:31 Go to page 1 2 Next [ 22 posts ] Display posts from previous: Sort by # An integer between 1 and 300, inclusive, is chosen at random new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management", "Perfect Squares and Perfect Cubes List all integers less than $20,\\!000$ that are both perfect squares and perfect cubes. Source: NCTM Mathematics Teacher, January 2006 Solution $9=3^2$ is a perfect square but not a perfect cube. On the other hand, $8=2^3$ is a perfect cube but not a perfect square. For an integer to be both perfect square and perfect cube, the exponent must be a multiple of $2$ and $3$, that is, the integer must be a sixth power. $0^6=0$ $1^6=1$ $2^6=64$ $3^6=729$ $4^6=4096$ $5^6=15625$ $6^6=46656>20000$ All integers less than $20,\\!000$ that are both perfect squares and perfect cubes $0,1,64,729,4096,15625$ Answer: $0,1,64,729,4096,15625$ Advertisements About mvtrinh Retired high school math teacher. This entry was posted in Problem solving and tagged , , , , , , , . Bookmark the permalink.", "you get $99$ instead of one of the possible answers. – Harnak Dec 25 '16 at 16:45 Your reasoning needs a little correction. $(1)$ $a^a$ is a perfect square when $a$ is even. Perfectly true. The integers satisfying this condition belong to the set $A=\\{2,4,6,\\cdots 198,200\\}$. So, the number of possibilities are :$100$. $(2)$ You have to include those $a$'s that are perfect squares as well. In, this case, $a= k^2$. So, $a^a = (k^2)^{k^2}$ which is a perfect square. Counting all odd perfect squares, to prevent double counting of even ones from $(1)$. The integers belong to the set $B =\\{1,9,25,\\cdots 169\\}$. So, the number of possibilities are: $7$. We thus have a total of $107$ integers satisfying the condition. Hope it helps. All the answers are ok, but I think we can be more rigorous. First, for $a=1$, $1^1=1$ is perfect square. Now, let's suppose that $a>1$, by the fundamental theorem of the arithmetic we can write $a=p_1^{r_1}\\cdot p_2^{r_2}\\ldots p_{k}^{r_k}$, where $p_1,p_2\\ldots, p_k$ are prime numbers and $r_1,r_2,\\ldots,r_k$ are positive integers. Then $$a^a=(p_1^{r_1}\\cdot p_2^{r_2}\\ldots p_{k}^{r_k})^a=p_1^{ar_1}\\cdot p_2^{ar_2}\\ldots p_{k}^{ar_k}.$$ Now, it's known that a positive integer is a perfect square if only if every one of their prime factors has an even exponent. Then if we want $a^a$ to be a perfect square it must hold that $ar_1,ar_2,\\ldots,", "common multiple) So the numbers that are going to be perfect squares and perfect cubes are precisely the 6th powers of the integers: $$\\begin{array}{rcll} 1^6 &=& 1 \\\\ 2^6 &=& 64 \\\\ 3^6 &=& 729 \\\\ 4^6 &=& 4096 \\\\ 5^6 &=& 15625 \\\\ ... \\\\ n^6 &<& 15! \\\\ \\end{array}$$ So n are the numbers between 1 and 15! (factorial) $$\\begin{array}{|rcll|} \\hline n^6 &<& 15! \\\\ n &<& \\sqrt[6]{15!} \\\\ n &<& \\sqrt[6]{1~307~674~368~000} \\\\ n &<& 104.57228596314317\\ldots \\\\ \\mathbf{n} & \\mathbf{=} & \\mathbf{104}\\\\ \\hline \\end{array}$$ 104 of them are BOTH perfect squares and perfect cubes. Jan 24, 2019", "integer powers. This is far from the bound of $3$ but it is a nice result. There was a gap in the paper above, but it was corrected by two different authors, see D. J. Bernstein, Detecting perfect powers in essentially linear time, Math. Comput., 67 (1998), pp. 1253–1283 and C. L. Stewart, On heights of multiplicatively dependent algebraic numbers, Acta Arith., 133 (2008), pp. 97–108 C. L. Stewart, On sets of integers whose shifted products are powers, J. Comb. Theory, Ser. A, 115 (2008), pp. 662–673 In the first paper by Stewart above, it is conjectured that there are infinitely many integers $N$ for which the interval $[N,N +\\sqrt{N}]$ contains three integers one of which is a square, one a cube and one a fifth power. He also conjectures that for sufficiently large $N$ this interval does not contain four distinct powers and if it contains three distinct powers then one of is a square, one is a cube and the third is a fifth power, which is a refined version of your conjecture. For more references and a longer survey see Waldschmidt's \"Perfect Powers: Pillai’s works and their developments\". - It follows from the ABC conjecture that there are only finitely many. First, show (unconditionally) that for all but finitely many such triples of perfect powers,", "# Sum of the first factorials a perfect power? Consider the following sum: $$F(n) = 1! + 2! + 3! + \\cdots + n!$$ Can this sum ever be a perfect power? A perfect power here is defined as $x^n$ for some natural number $x$ and some natural number $n\\geq 2$. Some observations immediately come to mind: $F(1) = 1$, and that’s trivially a perfect power. But $F(2) = 3$, which is a prime number and of course not a perfect power. Then there’s $F(3) = 1 + 2 + 6 = 9$, which is a perfect square. Are there others? Let’s check a few more. $F(4) = 33 = 3\\times 11$ and $F(5) = 153 = 3^2\\times 17$, so those are not perfect powers. In fact, $F(n)$ is not a perfect power for any natural number $n > 3$. Why is this? First, if $F(n)$ is ever a perfect power, it must be a perfect square. Why is that? If $n\\geq 6$ then we have \\begin{align*}F(n) &= 1! + 2! + \\cdots + 5! + 6! + \\cdots + n!\\\\ &= 153 + 6! + \\cdots + n!\\\\ &= 9(17 + 9(6!/9 + \\cdots + n!/9)).\\end{align*} Here, we have used that $n!/9$ will be an integer for $n\\geq 6$. Since $17$ is not a multiple of $9$,", "will be a perfect square or a perfect cube? (A) 0.1 (B) 0.125 (C) 0.16 (D) 0.5 (E) 0.9 Integer Range : 11 -60. Total no. of Integers = 60 - 11 + 1 = 50 Case 1: Perfect Square: $$4^2$$ =16 $$5^2$$ = 25 $$6^2$$ = 36 $$7^2$$ = 49 Case 2 : Perfect Cube: $$3^3$$ = 27 We know that we need to add if it's a OR probability. $$\\frac{4}{50}$$ + $$\\frac{1}{50}$$ = $$\\frac{5}{50}$$ = $$\\frac{1}{10}$$ = 0.1 Thus the best answer is A. One integer will be randomly selected from the integers 11 to 60, incl &nbs [#permalink] 12 Jul 2018, 05:50 Display posts from previous: Sort by # Events & Promotions Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.", "# How do you list all integers less than 20,000 that are both perfect squares and perfect cubes? Jul 1, 2015 If $n$ is a perfect square and a perfect cube then $n = {k}^{6}$ for some $k$, giving: ${0}^{6} = 0$ ${1}^{6} = 1$ ${2}^{6} = 64$ ${3}^{6} = 729$ ${4}^{6} = 4096$ ${5}^{6} = 15625$ #### Explanation: If an integer is both a perfect square and a perfect cube, then it will be of the form ${k}^{6}$ for some $k \\in \\mathbb{Z}$. You will find that ${5}^{6} < 20000 < {6}^{6}$, so the only possible integers are ${0}^{6}$, ${1}^{6}$, ${2}^{6}$, ${3}^{6}$, ${4}^{6}$ and ${5}^{6}$. (${6}^{6} = 46656$ is too large).", "mersenneforum.org A new conjecture concerning perfect powers Register FAQ Search Today's Posts Mark Forums Read 2022-05-12, 01:07 #1 marcokrt May 2022 11 Posts A new conjecture concerning perfect powers In the second half of April 2022, I submitted to the OEIS a new sequence of integers concerning perfect powers (see https://oeis.org/A353025), listing the smallest perfect powers written as the concatenation of any permutation of the integers from $1$ to $n$ (for any given $n$). Now, there are $94$ perfect squares that are smaller than $10^17$ and many others between $10^22$ and $10^29$, but no perfect power above 2 has been detected after several hours of computations on a standard CPU (moreover, the friend of mine who performed the mentioned search is still running his code hoping to find a cube consisting of $35 digits$, but no counterexample to the following conjecture has been found yet): Conjecture: All the terms of the OEIS sequence A353025 which are greater than $1$ are perfect squares and nothing else (no cube, no square of square, and so forth). The smallest counterexample (if any) should necessarily be greater than $1.01*10^34$ and it must be congruent modulo $9$ to $0$ or $1$ (for details, see DOI:10.13140/RG.2.2.30313.77921/1). Thanks in advance to everyone and good luck! 2022-05-12, 10:37 #2 LaurV Romulan Interpreter \"name field\" Jun 2011", "not a perfect power. Then there's $F(3) = 1 + 2 + 6 = 9$, which is a perfect square. Are there others? Let's check a few more. $F(4) = 33 = 3\\times 11$ and $F(5) = 153 = 3^2\\times 17$, so those are not perfect powers. In fact, $F(n)$ is not a perfect power for any natural number $n > 3$. Why is this? First, if $F(n)$ is ever a perfect power, it must be a perfect square. Why is that? If $n\\geq 6$ then we have \\begin{align*}F(n) &= 1! + 2! + \\cdots + 5! + 6! + \\cdots + n!\\\\ &= 153 + 6! + \\cdots + n!\\\\ &= 9(17 + 9(6!/9 + \\cdots + n!/9)).\\end{align*} Here, we have used that $n!/9$ will be an integer for $n\\geq 6$. Since $17$ is not a multiple of $9$, we see that $F(n)$ has at most two factors of $3$. Therefore, $F(n)$ is at most a perfect square. Another thing we see is that $F(n)$ for $n\\geq 4$ always ends in $3$. That is because $F(4) = 33$ and $n!$ ends in zero whenever $n\\geq 5$. Now, a square integer can never end in $3$. The only possible endings are $0,1,4,5,6,9$. Therefore, $F(n)$ can never be a perfect power whenever $n\\geq 4$. So $F(1)$ and $F(3)$ are", "# Perfect squares, cubes and fifth powers $\\leq 10^8$ I've the following question: Find number of numbers $\\leq 10^8$ which are neither perfect squares, nor perfect cubes nor perfect fifth powers. What I currently have is: • Number of perfect squares: $n_1 = \\sqrt[2]{10^8} = 10^4$. • Number of perfect cubes: $n_2 = \\text{floor}(\\sqrt[3]{10^8}) = 464$. • Number of perfect squares: $n_3 = \\text{floor}(\\sqrt[5]{10^8}) = 39$. Where floor(x) is the nearest integer $\\leq x$. But, how many numbers are there which are: • perfect squares and perfect cubes (for eg. $64$) • perfect squares and perfect fifth powers (for eg. $1024$) • perfect cubes and perfect fifth powers? (for eg. $32768$) • If $k$ and $m$ are relatively prime, then a number is a perfect $k$-th power and a perfect $m$-th power if and only if it is a perfect $(km)$-th power. – mjqxxxx Sep 17 '14 at 5:26 • If $n_k(N)$ is the number of perfect $k$-th powers from $1$ to $N$, then you want $N - n_2(N) - n_3(N) - n_5(N) + n_6(N) + n_{10}(N) + n_{15}(N) - n_{30}(N)$, using inclusion-exclusion. – mjqxxxx Sep 17 '14 at 5:28 • @mjqxxxx $n_{30}(N)$ will be floor($\\sqrt[30]{10^8}$)? – hjpotter92 Sep 17 '14 at 5:34 • There are about $15\\cdot3^{n-2}$ perfect powers below $10^n$. – Lucian Sep 17 '14 at", "does happen $26$ times up to $2k+1=201.$ The first and last few are $2k+1=15, 17, 19, 31, 33, 59, \\cdots 147, 149, 161, 187, 193.$ I can see why this might even more successful for odd powers $m^{2k+1}$ of larger integers. • Smaller exceptional examples include $5^3-11^2=4$, $47^2-13^3=12$, $56^2-5^5=11$, $15^3-58^2=11$, $83^2-19^3=30$. – Gerry Myerson Oct 20 '17 at 0:02 • A few more examples: $2^{15}-181^2=7$, $312^2-46^3=8$, $2537^2-23^5=26$, $3788^2-3^{15}=37$. – Gerry Myerson Oct 20 '17 at 0:32 If one looks at the sequence of perfect powers (1,4,8,9,16,25,27,32,36,... found at OEIS at https:/oeis.org/A001597 ), one sees a lot of squares. If one wants to tackle the posted question by looking at this sequence, one can save time looking at odd powers. Note that all pairs listed by Gottfried Helms have one even exponent and one odd exponent. Indeed, since (a^2 - b^2)=(a-b)(a+b), interesting answers to the question will involve an odd exponent, usually coprime to the other exponent. More specifically, interesting answers will be odd powers near a square, and two distinct odd powers which lie between two squares. Thus, an algorithm which looks only at odd powers which are not squares , and just the squares adjacent to them, saves time by looking at just the interesting cases. Further to avoid answers producing zero (like 225 and 3375), we", "756 * 100 => 4 * 189 * 100 => 4 * 9 * 21 * 100 => 2^2 * 3^2 * 3 * 7 * 2^2 * 5^2 => 2^4 * 3^3 * 5^2 * 7. As the fourth roots of a positive number are real, the answer you get is correct. So notice that the power of $2$ is $2$. So we need all the other powers to be raised to the $2$ power. … Karuppasamy K. Lv 5. Lv 7. Therefore, you can also write the problem and the answer as follows: 7 4 =2,401 You will also get the answer of 7 to the power of 4 (7 to the 4th power) if you type 7 then x y then 4 and then = on your scientific calculator. Courses . Start studying Perfect 4th powers. 46,400 =2^6 * 5^2 * 29^1. Since gcd(96, 60, 24) = 12, n is a perfect 12th power (and a perfect 6th power, 4th power, cube and square, since 6, 4, 3 and 2 divide 12). Perfect Square, Cube, Fourth Power [01/25/2002] Find the least integer greater than 1 that is a perfect square, a perfect cube, and a perfect fourth power. The following is a list of perfect squares. Answer Save. In cell B3, type x2, in cell", "+0 # help xd 0 38 4 +160 The numbers from 1 to 150, inclusive, are placed in a bag and a number is randomly selected from the bag. What is the probability it is neither a perfect square nor a perfect cube? Express your answer as a common fraction. I tried to do this by finding all the perfect squares and perfect cubes over all the numbers: 8/75 and then subtract it from 1, but I got 67/75 which was wrong. ANotSmartPerson Dec 6, 2018 #1 +2 Perfect squares =(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144) Perfect Cubes = (1, 8, 27, 64, 125) Perfect squares + Perfect cubes = 12 + 5 - 2 = 15 150 - 15 = 135 - which are neither Perfect square or Perfect cubes. So, the probability is = 135 / 150 = 9 / 10 Guest Dec 6, 2018 edited by Guest Dec 6, 2018 #3 +3213 +2 you fixed it :D Rom Dec 6, 2018 edited by Rom Dec 6, 2018 #4 +2 I sure did! Forgot about 1 being counted twice!!. Guest Dec 6, 2018 #2 +3213 +3 there's probably a number or two in there that are both perfect squares and perfect cubes $$P[\\text{# is neither perfect cube or square}] =" ]

[ "words, we need an integer raised to the 2nd, 3rd, 4th, etc. power. In even other words, we need perfect squares, perfect cubes, etc. that are between 1 and 300, inclusive. Now we need to count these perfect squares, etc. in an efficient way. Let's start with the squares. What's the biggest perfect square less than 300? Test numbers if necessary. $$16^2 = 256$$ and $$17^2 = 289$$, but $$18^2 = 324$$. Thus, we have 12 through 172, for 17 perfect squares. Now, let's count the cubes. Leave out $$1^3$$, since we've already counted 1 as $$1^2$$. $$5^3 = 125$$ and $$6^3 = 216$$, but $$7^3 = 343$$. Thus, we seem to have 5 more integer powers ($$2^3$$ through $$6^3$$, inclusive) - but careful! $$4^3 = 64$$, which we've already counted as a square ($$8^2$$), so we only have 4 more integer powers. This gives us a cumulative total of 21. What about the fourth powers - $$2^4$$, $$3^4$$, etc.? We've already counted any of these that matter, because they are also perfect squares: $$2^4 = 42$$, $$3^4 = 9^2$$, etc. We can leave out any higher even powers for the same reason ($$2^6 = 8^2$$, $$2^8 = 16^2$$, etc.). However, we must consider additional odd powers, continuing to leave out 1 raised to any power. $$2^5 =", "ever be a perfect power? A perfect power here is defined as $x^n$ for some natural number $x$ and some natural number $n\\geq 2$. Some observations immediately come to mind: $F(1) = 1$, and that's trivially a perfect power. But $F(2) = 3$, which is a prime number and of course not a perfect power. Then there's $F(3) = 1 + 2 + 6 = 9$, which is a perfect square. Are there others? Let's check a few more. $F(4) = 33 = 3\\times 11$ and $F(5) = 153 = 3^2\\times 17$, so those are not perfect powers. In fact, $F(n)$ is not a perfect power for any natural number $n > 3$. Why is this? First, if $F(n)$ is ever a perfect power, it must be a perfect square. Why is that? If $n\\geq 6$ then we have \\begin{align*}F(n) &= 1! + 2! + \\cdots + 5! + 6! + \\cdots + n!\\\\ &= 153 + 6! + \\cdots + n!\\\\ &= 9(17 + 9(6!/9 + \\cdots + n!/9)).\\end{align*} Here, we have used that $n!/9$ will be an integer for $n\\geq 6$. Since $17$ is not a multiple of $9$, we see that $F(n)$ has at most two factors of $3$. Therefore, $F(n)$ is at most a perfect square. Another thing we see is that $F(n)$ for", "+0 # help xd 0 38 4 +160 The numbers from 1 to 150, inclusive, are placed in a bag and a number is randomly selected from the bag. What is the probability it is neither a perfect square nor a perfect cube? Express your answer as a common fraction. I tried to do this by finding all the perfect squares and perfect cubes over all the numbers: 8/75 and then subtract it from 1, but I got 67/75 which was wrong. ANotSmartPerson Dec 6, 2018 #1 +2 Perfect squares =(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144) Perfect Cubes = (1, 8, 27, 64, 125) Perfect squares + Perfect cubes = 12 + 5 - 2 = 15 150 - 15 = 135 - which are neither Perfect square or Perfect cubes. So, the probability is = 135 / 150 = 9 / 10 Guest Dec 6, 2018 edited by Guest Dec 6, 2018 #3 +3213 +2 you fixed it :D Rom Dec 6, 2018 edited by Rom Dec 6, 2018 #4 +2 I sure did! Forgot about 1 being counted twice!!. Guest Dec 6, 2018 #2 +3213 +3 there's probably a number or two in there that are both perfect squares and perfect cubes $$P[\\text{# is neither perfect cube or square}] =", "# How many perfect powers are there amoung the first 1000 positive integers A positive integer $$n$$ is called a perfect power if and only if there exist two positive integers $$p,q$$, both are greater than 1, such that $$n=p^q$$. For example $$8=2^3, 36=6^2$$ are perfect powers, but $$20=2^2*5, 11=11^1$$ are not. Let $$A=\\{1,2,\\dots,1000\\}$$. How many elements of $$A$$ is a perfect power? I hope there exist a beautiful way to solve this, perhaps without calculating the number of primes untill 1000 or considering all the cases. Is there a formula? Thanks in advance! I don't think there is anything better than counting up each power. There are $$\\lfloor \\sqrt {1000} \\rfloor=31$$ squares and a similar formula for each power. You are correct that you only need to consider primes as all the fourth powers are squares as well. You also don't need to count any higher than $$10$$ because $$2^{10} \\gt 1000$$ and you need to only count $$1$$ once. The sixth powers have been counted twice-once as squares and once as cubes, so need to be subtracted once. This would apply to any number the product of two primes, but six is the only one less than $$10$$.", "+0 0 164 1 An integer is randomly chosen from the integers 1 through 100, inclusive. What is the probability that the chosen integer is a perfect square or a perfect cube, but not both? Express your answer as a common fraction. So do you find all the cubes and squares from 1-100 and then eliminate the ones that are both cubes and squares and then put it in $$\\frac{result}{100}$$because there are 100 different total numbers? Jul 2, 2018 #1 +104 0 A number is a perfect square and a perfect cube if and only if it is a perfect sixth power. Note that 10^2=100 and 4^3<100<5^3, while 2^6<100<3^6=9^3. Hence, there are 10 squares and 4 cubes between 1 and 100, inclusive. However, there are also 2 sixth powers, so when we add 10+4 to count the number of squares and cubes, we count these sixth powers twice. However, we don't want to count these sixth powers at all, so we must subtract them twice. This gives us a total of 10+4-2*2=10 different numbers that are perfect squares or perfect cubes, but not both. Thus, our probability is 10/100=1/10. HOPE THIS HELPS! Mar 16, 2019", "756 * 100 => 4 * 189 * 100 => 4 * 9 * 21 * 100 => 2^2 * 3^2 * 3 * 7 * 2^2 * 5^2 => 2^4 * 3^3 * 5^2 * 7. As the fourth roots of a positive number are real, the answer you get is correct. So notice that the power of $2$ is $2$. So we need all the other powers to be raised to the $2$ power. … Karuppasamy K. Lv 5. Lv 7. Therefore, you can also write the problem and the answer as follows: 7 4 =2,401 You will also get the answer of 7 to the power of 4 (7 to the 4th power) if you type 7 then x y then 4 and then = on your scientific calculator. Courses . Start studying Perfect 4th powers. 46,400 =2^6 * 5^2 * 29^1. Since gcd(96, 60, 24) = 12, n is a perfect 12th power (and a perfect 6th power, 4th power, cube and square, since 6, 4, 3 and 2 divide 12). Perfect Square, Cube, Fourth Power [01/25/2002] Find the least integer greater than 1 that is a perfect square, a perfect cube, and a perfect fourth power. The following is a list of perfect squares. Answer Save. In cell B3, type x2, in cell", "# Sum of the first factorials a perfect power? Consider the following sum: $$F(n) = 1! + 2! + 3! + \\cdots + n!$$ Can this sum ever be a perfect power? A perfect power here is defined as $x^n$ for some natural number $x$ and some natural number $n\\geq 2$. Some observations immediately come to mind: $F(1) = 1$, and that’s trivially a perfect power. But $F(2) = 3$, which is a prime number and of course not a perfect power. Then there’s $F(3) = 1 + 2 + 6 = 9$, which is a perfect square. Are there others? Let’s check a few more. $F(4) = 33 = 3\\times 11$ and $F(5) = 153 = 3^2\\times 17$, so those are not perfect powers. In fact, $F(n)$ is not a perfect power for any natural number $n > 3$. Why is this? First, if $F(n)$ is ever a perfect power, it must be a perfect square. Why is that? If $n\\geq 6$ then we have \\begin{align*}F(n) &= 1! + 2! + \\cdots + 5! + 6! + \\cdots + n!\\\\ &= 153 + 6! + \\cdots + n!\\\\ &= 9(17 + 9(6!/9 + \\cdots + n!/9)).\\end{align*} Here, we have used that $n!/9$ will be an integer for $n\\geq 6$. Since $17$ is not a multiple of $9$,", "are squares, cubes, and so on until it is a... Say, 100th power? If a variable with an exponent has an even exponent then it is a perfect square. In order to make them all 4th powers, we have to add: 2^2 * 5^2 * 29^3 =2,438,900 - This is the smallest possible value of k. Because:2,438,900 * 46,400 =113,164,960,000^(1/4) =580 - perfect 4th power. 8 and 9 are the only consecutive integers which are perfect powers, 3 2 − 2 3 = 1. The largest 4th power that could fit the bill is 2^4, or 16. Perfect Square, Cube, Fourth Power [01/25/2002] Find the least integer greater than 1 that is a perfect square, a perfect cube, and a perfect fourth power. Notice that the exponent tells us how many bases to multiply, not how many multiplications to perform. Undoubtedly … For example, 2 4 = 2 × 2 × 2 × 2 = 16. Start studying Perfect 4th powers. Gaps between perfect powers. 0 } = perfect fourth powers and this looks really daunting because we have an perfect! Z, or half tones ) the other powers to be raised to the $2 power! Are required to represent every positive integer vocabulary, terms, and more flashcards. Is not a perfect square, so maybe this is the", "* 100 => 4 * 9 * 21 * 100 => 2^2 * 3^2 * 3 * 7 * 2^2 * 5^2 => 2^4 * 3^3 * 5^2 * 7. As the fourth roots of a positive number are real, the answer you get is correct. So notice that the power of$2$is$2$. So we need all the other powers to be raised to the$2$power. … Karuppasamy K. Lv 5. Lv 7. Therefore, you can also write the problem and the answer as follows: 7 4 =2,401 You will also get the answer of 7 to the power of 4 (7 to the 4th power) if you type 7 then x y then 4 and then = on your scientific calculator. Courses . Start studying Perfect 4th powers. 46,400 =2^6 * 5^2 * 29^1. Since gcd(96, 60, 24) = 12, n is a perfect 12th power (and a perfect 6th power, 4th power, cube and square, since 6, 4, 3 and 2 divide 12). Perfect Square, Cube, Fourth Power [01/25/2002] Find the least integer greater than 1 that is a perfect square, a perfect cube, and a perfect fourth power. The following is a list of perfect squares. Answer Save. In cell B3, type x2, in cell C3 type x3, in cell D3 type x4, and in cell E3", "not a perfect power. Then there's $F(3) = 1 + 2 + 6 = 9$, which is a perfect square. Are there others? Let's check a few more. $F(4) = 33 = 3\\times 11$ and $F(5) = 153 = 3^2\\times 17$, so those are not perfect powers. In fact, $F(n)$ is not a perfect power for any natural number $n > 3$. Why is this? First, if $F(n)$ is ever a perfect power, it must be a perfect square. Why is that? If $n\\geq 6$ then we have \\begin{align*}F(n) &= 1! + 2! + \\cdots + 5! + 6! + \\cdots + n!\\\\ &= 153 + 6! + \\cdots + n!\\\\ &= 9(17 + 9(6!/9 + \\cdots + n!/9)).\\end{align*} Here, we have used that $n!/9$ will be an integer for $n\\geq 6$. Since $17$ is not a multiple of $9$, we see that $F(n)$ has at most two factors of $3$. Therefore, $F(n)$ is at most a perfect square. Another thing we see is that $F(n)$ for $n\\geq 4$ always ends in $3$. That is because $F(4) = 33$ and $n!$ ends in zero whenever $n\\geq 5$. Now, a square integer can never end in $3$. The only possible endings are $0,1,4,5,6,9$. Therefore, $F(n)$ can never be a perfect power whenever $n\\geq 4$. So $F(1)$ and $F(3)$ are", "Button Try our free Online GRE Test Director Joined: 20 Apr 2016 Posts: 815 WE: Engineering (Energy and Utilities) Followers: 9 Kudos [?]: 586 [0], given: 113 Re: GRE Math Challenge #81-One integer will be randomly selected [#permalink] 14 Dec 2017, 06:46 sandy wrote: One integer will be randomly selected from the integers 11 to 60, inclusive. What is the probability that the selected integer will be a perfect square or a perfect cube? (A) 0.1 (B) 0.125 (C) 0.16 (D) 0.5 (E) 0.9 Here Total number of integers from 11 to 60 = 50 perfect squares from 11 to 60 are = 16 , 25, 36 and 49 perfect cube = 27 Hence there are 5 integers that can be either a perfect square or a perfect cube Therefore probability = $$\\frac{5}{50}$$= 0.1 _________________ If you found this post useful, please let me know by pressing the Kudos Button Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html Re: GRE Math Challenge #81-One integer will be randomly selected [#permalink] 14 Dec 2017, 06:46 Display posts from previous: Sort by", "# How sure can we be that $\\underbrace {k\\cdots k}_{m\\ k's}$ cannot be a perfect power? Let $$k$$ and $$m$$ be integers , $$m\\ge 2$$ and $$k\\in [1,9]$$. Denote $$n=k\\cdot \\frac{10^m-1}{9}=\\underbrace {k\\cdots k}_{m\\ k's}$$ As far as I know, it is unknown whether a rep-unit can be a cube. How sure can we be that $$n$$ can never be a perfect power ? Some cases are easy. For $$k=2,4,5,6,9$$ it can be shown that a perfect power is impossible. The case $$k=8$$ cannot give a perfect power if we assume that a rep-unit cannot be a cube. A square can be easily ruled out. I have checked that for $$m\\le 10^4$$, we do not have a perfect power. • By \"rep-unit\" do you mean to case $k=1$? – Vincent Jan 6 '19 at 15:49 • @Vincent Yes, a repunit has the form $1\\cdots 1$ – Peter Jan 6 '19 at 15:49 This problem was completely solved by Bugeaud and Mignotte (see http://irma.math.unistra.fr/~bugeaud/travaux/chiffresrev.ps) in a paper in Mathematika in 1999; see Theorem 2. Their approach is based upon bounds for linear forms in $$p$$-adic logarithms. • And the result is that there are no perfect powers except $4,8,9$ ? – Peter Jan 7 '19 at 18:39 • More generally, if $2 \\leq b \\leq 10$ and $1 \\leq a", "# If $x$ is a perfect square and a perfect cube, is $x$ a perfect power of $6$? Given $a,b$ are natural numbers s.t. $a^2 = y$ and $b^3 = y$, is it true there must exist a natural number $c$ such that $c^6 = y$? • Hint: consider the unique factorization of $y$, $y=\\prod p_i^{a_i}$ . Show that each $a_i$ is divisible by both $2,3$. – lulu Sep 11 '17 at 0:03 • Hello and welcome to math.stackexchange! That's a nice question. Please tell the reader what the prigin of th question is and also what your work on this problem is so far. – Hans Engler Sep 11 '17 at 0:20 • The title should ask not \"is $x$ a perfect power of 6\" but rather \"is $x$ a perfect 6th power.\" – John Blythe Dobson Sep 11 '17 at 21:14 Set $c = \\frac{a}{b}$. Then $$c^6 = \\frac{a^6}{b^6} = \\frac{y^3}{y^2} = y \\, .$$ It remains to show that $c$ is an integer. If that were not the case, then there would exist a prime number $p$ that divides one of the numbers $a, b$ but not the other one, e.g. $p \\mid a$ and $p\\nmid b$. But then $p \\mid a^2 = y$ and $p \\nmid b^3 = y$ which is silly. Therefore $c$", "you get $99$ instead of one of the possible answers. – Harnak Dec 25 '16 at 16:45 Your reasoning needs a little correction. $(1)$ $a^a$ is a perfect square when $a$ is even. Perfectly true. The integers satisfying this condition belong to the set $A=\\{2,4,6,\\cdots 198,200\\}$. So, the number of possibilities are :$100$. $(2)$ You have to include those $a$'s that are perfect squares as well. In, this case, $a= k^2$. So, $a^a = (k^2)^{k^2}$ which is a perfect square. Counting all odd perfect squares, to prevent double counting of even ones from $(1)$. The integers belong to the set $B =\\{1,9,25,\\cdots 169\\}$. So, the number of possibilities are: $7$. We thus have a total of $107$ integers satisfying the condition. Hope it helps. All the answers are ok, but I think we can be more rigorous. First, for $a=1$, $1^1=1$ is perfect square. Now, let's suppose that $a>1$, by the fundamental theorem of the arithmetic we can write $a=p_1^{r_1}\\cdot p_2^{r_2}\\ldots p_{k}^{r_k}$, where $p_1,p_2\\ldots, p_k$ are prime numbers and $r_1,r_2,\\ldots,r_k$ are positive integers. Then $$a^a=(p_1^{r_1}\\cdot p_2^{r_2}\\ldots p_{k}^{r_k})^a=p_1^{ar_1}\\cdot p_2^{ar_2}\\ldots p_{k}^{ar_k}.$$ Now, it's known that a positive integer is a perfect square if only if every one of their prime factors has an even exponent. Then if we want $a^a$ to be a perfect square it must hold that $ar_1,ar_2,\\ldots,", "what is the greatest possible value of y ? A) 30 B) 34 C) 36 D) 37 E) 39 Spelled out a little more: 1) $$x$$ = product of integers from 1 to 150 $$x$$ = 150 * 149* 148 . . .* 3 * 2 *1: That is, $$x$$ = 150! 2) $$5^{y}$$ is a factor of 150! What is the greatest possible value of $$y$$? Using $$\\frac{150}{5^{y}}$$, consider each power $$y$$, of 5. Do not worry about remainders. $$\\frac{150}{5^1}$$ = 30 (5 divides into 150 thirty times) $$\\frac{150}{5^2}$$ = 6 (25 divides into 150 six times) $$\\frac{150}{5^3}$$ = 1 (125 divides into 150 only once. Ignore the remainder.) $$5^4 = 625$$ -- too large to divide into 150 as a factor. 3) Add the results: 30 + 6 + 1 = 37 Once you know the theory and method, questions such as this one are pretty straightforward. The stats here might indicate that the suggestion below is indispensable. Bunuel wrote: Quote: _________________ Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date. Our lives begin to end the day we become silent about things that matter. -- Dr. Martin Luther King, Jr. BLACK LIVES MATTER. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 17042 Location:", "will be a perfect square or a perfect cube? (A) 0.1 (B) 0.125 (C) 0.16 (D) 0.5 (E) 0.9 Integer Range : 11 -60. Total no. of Integers = 60 - 11 + 1 = 50 Case 1: Perfect Square: $$4^2$$ =16 $$5^2$$ = 25 $$6^2$$ = 36 $$7^2$$ = 49 Case 2 : Perfect Cube: $$3^3$$ = 27 We know that we need to add if it's a OR probability. $$\\frac{4}{50}$$ + $$\\frac{1}{50}$$ = $$\\frac{5}{50}$$ = $$\\frac{1}{10}$$ = 0.1 Thus the best answer is A. One integer will be randomly selected from the integers 11 to 60, incl &nbs [#permalink] 12 Jul 2018, 05:50 Display posts from previous: Sort by # Events & Promotions Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.", "integer powers. This is far from the bound of $3$ but it is a nice result. There was a gap in the paper above, but it was corrected by two different authors, see D. J. Bernstein, Detecting perfect powers in essentially linear time, Math. Comput., 67 (1998), pp. 1253–1283 and C. L. Stewart, On heights of multiplicatively dependent algebraic numbers, Acta Arith., 133 (2008), pp. 97–108 C. L. Stewart, On sets of integers whose shifted products are powers, J. Comb. Theory, Ser. A, 115 (2008), pp. 662–673 In the first paper by Stewart above, it is conjectured that there are infinitely many integers $N$ for which the interval $[N,N +\\sqrt{N}]$ contains three integers one of which is a square, one a cube and one a fifth power. He also conjectures that for sufficiently large $N$ this interval does not contain four distinct powers and if it contains three distinct powers then one of is a square, one is a cube and the third is a fifth power, which is a refined version of your conjecture. For more references and a longer survey see Waldschmidt's \"Perfect Powers: Pillai’s works and their developments\". - It follows from the ABC conjecture that there are only finitely many. First, show (unconditionally) that for all but finitely many such triples of perfect powers,", "$$17^2 = 289$$, but $$18^2 = 324$$. Thus, we have 12 through 172, for 17 perfect squares. Now, let's count the cubes. Leave out $$1^3$$, since we've already counted 1 as $$1^2$$. $$5^3 = 125$$ and $$6^3 = 216$$, but $$7^3 = 343$$. Thus, we seem to have 5 more integer powers ($$2^3$$ through $$6^3$$, inclusive) - but careful! $$4^3 = 64$$, which we've already counted as a square ($$8^2$$), so we only have 4 more integer powers. This gives us a cumulative total of 21. What about the fourth powers - $$2^4$$, $$3^4$$, etc.? We've already counted any of these that matter, because they are also perfect squares: $$2^4 = 42$$, $$3^4 = 9^2$$, etc. We can leave out any higher even powers for the same reason ($$2^6 = 8^2$$, $$2^8 = 16^2$$, etc.). However, we must consider additional odd powers, continuing to leave out 1 raised to any power. $$2^5 = 32$$, $$3^5 = 243$$, but $$4^5 =$$ something greater than 300, as we can see by considering that $$4^5 = 4^4 \\times 4 = 16^2 \\times 4 = 256 \\times 4$$. So we have two more integer powers ($$2^5$$ and $$3^5$$), for a cumulative total of 23. Seventh powers: $$2^7 = 128$$, but $$3^7 =$$ something much greater than 300 (since $$3^7 = 3^5 \\times", "mersenneforum.org A new conjecture concerning perfect powers Register FAQ Search Today's Posts Mark Forums Read 2022-05-12, 01:07 #1 marcokrt May 2022 11 Posts A new conjecture concerning perfect powers In the second half of April 2022, I submitted to the OEIS a new sequence of integers concerning perfect powers (see https://oeis.org/A353025), listing the smallest perfect powers written as the concatenation of any permutation of the integers from $1$ to $n$ (for any given $n$). Now, there are $94$ perfect squares that are smaller than $10^17$ and many others between $10^22$ and $10^29$, but no perfect power above 2 has been detected after several hours of computations on a standard CPU (moreover, the friend of mine who performed the mentioned search is still running his code hoping to find a cube consisting of $35 digits$, but no counterexample to the following conjecture has been found yet): Conjecture: All the terms of the OEIS sequence A353025 which are greater than $1$ are perfect squares and nothing else (no cube, no square of square, and so forth). The smallest counterexample (if any) should necessarily be greater than $1.01*10^34$ and it must be congruent modulo $9$ to $0$ or $1$ (for details, see DOI:10.13140/RG.2.2.30313.77921/1). Thanks in advance to everyone and good luck! 2022-05-12, 10:37 #2 LaurV Romulan Interpreter \"name field\" Jun 2011", "2 can be 1,4 or 7. Since the number becomes a perfect cube and perfect square on division by 3 and 2, the powers of other numbers should be at least 6. However, terms such as $5^6$ are huge and hence, we can conclude that the number must be composed of powers of 2 and 3. ’a’ can take values 1, 4 or 7 and ‘b’ can take values 1, 3 or 5. When divided by 2, the remaining term must be a perfect cube. Therefore, the value of $b$ can only be 3. When divided by 3, the remaining term must be a perfect square. Therefore, the value of $a$ can only be 4. Therefore, the number is $2^4*3^3$ = $16*27$ = $432$. Therefore, $432$ is the right answer. For finding out maximum value of $6^x$. Since we know that 6 = 2*3 and in any factorial frequency of 2 will be higher as compared to 3. Hence for finding out maximum value of x we have to figure out the maximum power of 3. Maximum power of 3 in 20! $\\Rightarrow [\\frac{20}{3}] + [\\frac{20}{3^2}]$ (Where [.] is a greatest integer function.) $\\Rightarrow$ 6 + 2 = 8 This power will remain same for 18! and 19! For calculating maximum power of 3 in 17! ,16! and" ]

[ "the same time, player two gets a point. If one red and one green marble are pulled at the same time, both players get a point. Is this a fair game?", "seats, how many different ways can they arrange themselves?\" The other question offered 4 answers as equations, and it was simple matter to eliminate the incorrect ones. Glad I was prepared for what came up. (score will arrive in one month to close this saga) – JoeTaxpayer Mar 2 '13 at 20:08 @JoeTaxpayer: Sounds like it went well; glad to hear it! – Brian M. Scott Mar 2 '13 at 20:12 Hint: Think about the balls. In how many bags can one ball be put in? How many balls are there? - If the three bags are sitting on the table and I rotate the contents, there's still one ball in the one bag. Let me add - if I were loading 3 marbles into 3 bags, just one per bag, there would be two answers depending on if the bags are uniquely identified. If they are three brown bags, one answer, but three numbered or different colored bags, different answer. – JoeTaxpayer Feb 12 '13 at 19:51 nevertheless, there is no restrictions such as you implied in this question. – ciceksiz kakarot Feb 12 '13 at 19:58 The author's solution treats them as unique. But the question avoids the distinction. – JoeTaxpayer Feb 12 '13 at 20:03", "sestigtalill 2022-12-19 Geometric distribution. Johnny has 1,176 Pok´emon cards in total. Pok´emon EX is a special type of card, and Johhny has 39 EX-type cards. He is looking for an EX-type card, but all of the cards are completely mixed up and stored in a shoe box. His mother is calling him for dinner. What is the probability that Johnny will have to look through no more than 25 cards before he finds an EX-type card? Agour2q9 Expert", "and gave him 1/4 of this remaining marbles plus 5. Just before arriving at the asked by dyl on December 14, 2010 9. ### maths ANDY HAS TWICE AS MANY MARBLES AS PANDY AND SANDY HAS HALF AS MANY HAS ANDY AND PANDY PUT TOGETHER .ANDY HAS 110 MARBLES WHICH IS 115 MARBLES LESS THAN SANDY.HOW MANY MARBLES EACH OF THEM HAVE asked by Anamika on December 31, 2012 10. ### MATH HENRY, ANDY AND JOE SHARED SOME SWEETS IN THE RATIO 15:12:9 RESPECTIVELY. HENRY HAD 6 MORE SWEETS THAN ANDY. IF ANDY GAVE 9 SWEETS TO JOE, HOW MANY SWEETS WOULD JOE HAVE? asked by FLOR on August 25, 2015 More Similar Questions", "# The Marble Game ### Johnny vs. Vinny As brothers, these two are competitive in every aspect of life. Just check out some of my other puzzles, and you'll see my brothers in action. Today is a classic Marble Battle! The brother's played a series of 3 games: ## The Game Series Summary Note: Not a real game. Bored at work today, put our clip-art library to good use! :) Round 1 Johnny and Vinny each had a certain number of playing marbles. Johnny lost a game to Vin and had to give Vin half of his marbles. Round 2 In the second game, Vin lost and had to give three-fourths of his marbles to Johnny, who now had thirty marbles. Round 3 Finally, the two finished their third game. Vinny won and acquired some of Johnny's marbles. Score Board At this point, each brother had exactly the number of marbles he had started with, and Johnny had twice as many marbles as Vinny. ## How many marbles did each have in the beginning? From the final clue, we learn that Johnny has twice as many marbles as Vinny at both the start and finish. So, J = 2V, where J equals the amount of marbles Johnny has at the start and finish, and V is the same", "number of notebooks Keecia bought? 19. Sheri can make 2 bracelets in 16 minutes. She needs to make 7 bracelets. a. Set up a proportion that could be used to find $m$, the number of minutes it would take Sheri to make 7 bracelets. b. How many minutes will it take Sheri to make 7 bracelets? 20. A bag contains only two colors of marbles––silver and gold. The ratio of silver marbles to gold marbles in the bag is 5 : 3. There are a total of 30 silver marbles in the bag. a. Set up a proportion that could be used to find $t$, the total number of marbles in the bag Feb 22, 2012 Jan 06, 2016", "for the four games? 5. The e-Box laptop computer offers the following options. • Screen: small, medium, or large • Memory: standard 1 GB, extra 2 GB • Colors: pearl, blue, black How many different combinations are there for options? 6. Jesse has four pairs of pants and three shirts. How many different outfits can he make? 7. If Jesse adds three sweaters, how many different outfits can he make now? 8. Kyle has selected three different paint colors and two trim colors for his bedroom. How many different color combinations can he create? 9. There are four teams playing in a tournament. Each team will play two sports. How many different combinations are there for teams and sports? 10. Candice can choose between the following electives. • German • Gymnastics • Chorus • French If she can choose two electives, how many different options are there of choices? 11. If we add Track to her options, how many two elective options are there? 12. Miles is making a salad. He wants to put three things on it. Here are his choices. Tomatoes Pickles Onions Cucumbers Peppers How many different salads can he create given these options? 13. If Miles adds cheese, how many different salads can he create given these options? 14. Muffins can have two additions", "are 16 flavors to work with, but you can only use three at a time. With an example like this one, you are looking for combinations of object where only a certain number of them are used in any one combination. This happens a lot with teams. Here is a situation with teams. How many different 2-player soccer teams can Jean, Dean, Francine, Lurleen, and Doreen form? Step 2: You’ve covered all combinations that begin with Jean. Now go through all combinations that begin with Dean, Francine, and Lurleen. Your list is now complete. There are 10 combinations. Look at this situation. How many different 3-player soccer teams can Jean, Dean, Francine, Lurleen, and Doreen form? Use the process above to go through all of the combinations. Your list is now complete. There are 10 combinations. We can use a formula to help us to calculate combinations. This is very similar to the work that you did in the last section with factorials and permutations. Suppose you have 5 marbles in a bag–red, blue, yellow, green, and white. You want to know how many combinations there are if you take 3 marbles out of the bag all at the same time. In combination notation you write this as: \\begin{align*}{_5}C_3 \\Longleftarrow 5 \\ \\text{items taken 3 at a time}\\end{align*}", "two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . . Marbles in a Box Stage: 3 and 4 Challenge Level: In a three-dimensional version of noughts and crosses, how many winning lines can you make? Christmas Chocolates Stage: 3 Challenge Level: How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?", "1 2 [ 31 posts ] Similar topics Replies Last post Similar Topics: In a jar there are balls in different colors: blue, red, green and yel 4 13 Jul 2016, 01:01 3 A woman has three blouses of different colors, three skirts of differe 1 17 Jan 2016, 11:10 20 How many ways can 5 different colored marbles be placed in 3 7 14 Oct 2013, 07:53 1 A certain businessman wears only three different colors of shirt -- wh 3 16 Dec 2010, 00:32 7 In a jar there are balls in different colors: blue, red, green and yel 12 07 Oct 2009, 17:48 Display posts from previous: Sort by", "# What's in my pocket? [duplicate] Well, I can tell you Johnny has memory cards in his pocket. Back Story My brother, Johnny, is a tech nerd. He loves gadgets of all kinds. As a matter of fact, you can be sure at any one given time, he will have SOME sort of tech related thing in his pockets. That being said... Johnny-Gadget says to our other brother, Mickey, \"Can you figure out how many memory cards I have in my pockets?\" He then gives Mickey three clues: 1. If the number of memory cards I have is a multiple of 5, it is a number between 1 and 19. 2. If the number of memory cards I have is not a multiple of 8, it is a number between 20 and 29. 3. If the number of memory cards I have is not a multiple of 10, it is a number between 30 and 39. ## How many memory cards does Johnny-Gadget have in his pockets? • Not fair! not fair! It isn't fair, my precious, is it, to ask us what it's got in it's nassty little pocketsess? – Herb Apr 8 '20 at 21:25 • Does this answer your question? Find out the number using three clues – Hemant Agarwal Mar 18 at 9:57 The", "three shirts. How many different outfits can he make? 7. If Jesse adds three sweaters, how many different outfits can he make now? 8. Kyle has selected three different paint colors and two trim colors for his bedroom. How many different color combinations can he create? 9. There are four teams playing in a tournament. Each team will play two sports. How many different combinations are there for teams and sports? 10. Candice can choose between the following electives. • German • Gymnastics • Chorus • French If she can choose two electives, how many different options are there of choices? 11. If we add Track to her options, how many two elective options are there? 12. Miles is making a salad. He wants to put three things on it. Here are his choices. Tomatoes Pickles Onions Cucumbers Peppers How many different salads can he create given these options? 13. If Miles adds cheese, how many different salads can he create given these options? 14. Muffins can have two additions to them. Here are the choices: blueberries, raspberries, nuts, cherries, strawberries. How many muffin combinations can there be? 15. If we add apples and cranberries, how many combinations can there be now? Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Difficulty Level:", "they will be the same (snap) or different. FREE Shipping by Amazon. 2/21 . What is the probability picking a navy marble and without replacing it pick a white marble? Bag A contains 3 marbles – one red, one blue and one green. For each flip, if it was heads -> I put in a red marble; if it was tails -> I put in a blue marble. : racing - barbie - shooting - parking - sonic - cooking - tanks - thing thing - bike - dragon ball z - games - recess. Directions: There are _____ red marbles and _____ blue marbles in Bag A. Bag B = 4 red, 6 blue There are 5 gray, 4 red, 5 white, 3 green, 1 orange and 2 navy marbles in a hat. Add this game to your site. Finally multiply How about the red balls? When I put two marbles in this bag, I flipped a coin twice to determine their colors. There are _____ red marbles and _____ green marbles in Bag B. bag B has 3 red and 9 green, bag A has 3 red and 9 blue The game is played with an ordinary deck of cards, a Joker Marbles game board and five marbles of matching color per player. Parameters: Number and", "a product of the minimum number necessary. P.S. If you’ve never heard of Looney Labs, the company of the person interviewed here, check out their website. Several of their games are on my favorites list. P.P.S. Sometimes you will see a novelty Rubik’s cube in which the faces of an unjumbled cube are 6 different pictures (of cartoon characters, say) instead of 6 solid colors. This is actually not the same puzzle as a classic Rubik’s cube; it’s slightly more complex. Why is that?", "is a marble in one of the cups. You pick a cup at random and if you choose the cup with the marble, you win. Each play costs 5 TL and if you win you earn 15 TL. What is your expected earnings if you play until you win?", "no floorkicks... I have a lot to learn and a long way to go! 19. ### K ok let me check here 20. ### colour_thief Go away, this is my turf! Seriously though, play this mode. It needs more competition.", "the lollipops in the bag are blue. If she adds in another 25 blue lollipops, 40% of the lollipops in the bag are blue. How many violet lollipops are in the bag? 3 m", "six-person game. The first to pop into my head was the classic Chinese checkers. I had originally planned to have each of us make marbles out of polymer clay colored to each one’s liking. Maybe we’ll do that another time. We went with our pictures instead this time. The Pieces First we bought some glass pebbles from the dollar store (the consistency of size and the clarity weren’t great, but the price was). Using powerpoint, I circularly cropped headshots of each of us, sized them to be a bit smaller than the cross section of the pebble, and printed them. I could have gone smaller; the pebble magnifies and the smaller version may look better. If I would have reduced the photo size I might have made a colored ring around the picture to make it stand out. I’m sure a photo-quality printer would have been nice too. From there I just glued the pictures I had cut out onto the pebbles with white glue. Especially because we have two girls who look very similar and, at a glance, it wasn’t so easy to recognize the images, I wrapped each with a small rubber hair tie of a favored color. I used a little smear of glue on them as well. And that was it for the “marbles;”", "a total of 32 pieces. Rather than sharing evenly, they play the following game: Each in turn, they pick one of the three boxes, empty its contents in a jar and pick", "disqualified from entire series The person who kept their marble rolling for the longest amount of time would win the round. Thinking he had a clever solution to the task, Mawaan elected to place his giant ball made of GLASS in ... More ### ‎Red Ball 5 on the App Store - Get enough required number of yellow stars while rolling ball - Find the magical door to guide the ball by ball get the next level - Remember to collect the boxes with barleys - Challenge yourself in more and more difficult but interesting levels. Features of Red Ball 5: Jump Ball Adventure: - 30 levels to roll the ball - Beautiful colorful graphics - Various evil monsters to fight with ... More" ]

[ "# Grid of 3 columns. There are 6 different marbles.How many ways can the marbles can be placed in the grid such that no column is empty? There is a grid of 3 columns: the first column has 4 squares, the second has 2 squares, and the third has 2 squares. □□□ □□□ There are 6 marbles of different colors. How many ways can those 6 marbles be placed inside the grid such that each column has at least one marble? My solution is as follows: There will be four cases as: $(4,1,1), (3,2,1), (3,1,2), (2,2,2)$ for number of marbles in respective columns. $$\\bigg(\\binom64 \\cdot 4! \\cdot \\binom21 \\cdot 2 \\cdot 2\\bigg) + 2 \\cdot \\bigg(\\binom63 \\cdot \\binom43 \\cdot 3! \\cdot \\binom32 \\cdot 2 \\cdot 2\\bigg) \\\\ + \\bigg(\\binom62 \\cdot \\binom42 \\cdot 2! \\cdot \\binom42 \\cdot 2 \\cdot 2\\bigg) = 18720$$ (Logic: $\\big(\\binom62 \\cdot \\binom42 \\cdot 2! \\cdot \\binom42 \\cdot 2 \\cdot 2\\big)$ means first choose 2 balls out of 6, then choose 2 squares out of 4 in the first column and permute; second, choose 2 balls out of the 4 remaining balls and permute in the second column; in the third column, permute the 2 remaining balls). Is this right or not? Kindly help me. There are only two configurations where placing $6$ marbles in such", "want all of the green marbles.. then you would choose $i$ marbles from the batch of $n$ marbles. So if there were a total of 5 marbles ($n$) and two were green, and you wanted all the green's $~(i)$,then you could use the notation $n\\choose i$=$5\\choose 2$. Meaning, out of the $5$ marbles, I only choose $2$. Hope that this cleared up the meaning some for the binomial coefficient. – night owl Jul 3 '11 at 3:07 The way it works if you expand $(a+b)^n$ is that there are $n$ brackets and you have to choose an $a$ or a $b$ from each one. There are 2 choices for each bracket, hence 2^n choices overall. The sum you have given gathers all the terms with $i$ copies of $a$ and $n-i$ copies of $b$. The bracket $\\binom{n}{i}$ is the number of ways of doing this, and is called n-Choose-i (which is the reason for the $C$ in some alternative notations) or a Binomial Coefficient (binomial because the original brackets contain two terms $a$ and $b$). It might be interesting for you to see whether you can get any intuition or insight for why the value of the bracket is what it is. - It is read \"$n$ choose $i$\" and is equal to the number of ways to", "# Combinations of distinct vs identical objects 1. A child has 5 pockets and 4 marbles. In how many ways can he put the marbles in his pockets? 2. In how many ways can 11 identical apples be distributed among 6 children so that each child receives at least one apple? My approach is as follows: 1) for every marble, say M, i have 5 options for pockets (P P P P P), hence i get 5*5*5*5=5^4. This is the correct answer. 2) However, i am unable to apply the same concept to the second problem. i distribute 6 apples to 6 children therefore fulfilling the \"at least one apple\" condition. Then i am left with 5 apples. For each apple, i have 6 options, hence by this logic i get 6*6*6*6*6=6^5. However this is wrong as i am unable to account for identical objects. Please point out where i am missing. I don't understand the logic behind n+r-1Cr-1 or n-1Cr-1. - If the marbles are all distinct, then your analysis for the first problem is right. (If the marbles are identical, then the number of ways is $\\binom{8}{4}$.) Let's look at the second problem. We have $11$ identical apples. Line them up like this: $$A\\quad A \\quad A\\quad A \\quad A \\quad A \\quad A \\quad A", "ways to distribute $9$ identical marbles among $5$ distinguishable boxes. – Brian M. Scott Jun 16 '12 at 21:33 There are 10!/(4!*6!) = 210 to get 4 different numbers There are 3* [10!/(3!*7!)] =3*120 =360 to get 3 differnt numbers with 1 repeating There are 10!/(2!*8!) = 45 to get 2 differnt numbers with each repeating There are 2* [10!/(2!*8!) ] = 2*45 =90 to get 2 different numbers with 1 repeating 3 times There are 10 ways to get 1 number repeating 4 times 210 +360 +45 +90 +10 = 715 - your method is the best to understand but i needed a formula for direct computation – Ashhar Jawaid Jun 16 '12 at 21:38", "3, 5, 15, 17, 51,... in base ten. They are the number of sides found in constructible polygons with odd number of sides. This is true up to row 32.[14] Computing $\\scriptstyle \\pi$ We can compute $\\scriptstyle \\pi$ from the Pascal's triangle : $\\scriptstyle \\frac{\\binom{2n}{n}}{2^{2n}} = \\frac{1}{\\sqrt{n \\pi}}$. For instance, $\\scriptstyle \\frac{\\binom{50}{25}}{2^{50}} = \\frac{1}{\\sqrt{25 \\pi}}$ outputs $\\scriptstyle \\pi = 3.17316$. This formula is not used in practice since $\\scriptstyle n$ must be very large to compute $\\scriptstyle \\pi$ to any decent precision. Stacking marbles If you stack marbles in the simplest 1D figure (along a line), then you arrange 1, 2, 3, 4,... marbles. If you stack marbles in the simplest 2D figure (triangle), then you arrange 1, 3, 6, 10,... marbles. If you stack marbles in the simplest 3D figure (a tetrahedron or 4-sided pyramid), then you arrange 1, 4, 10, 20,... marbles. If you stack marbles in the simplest 4D figure (a 4-simplex), then you arrange 1, 5, 15, 35,... marbles. And so on. These numbers appear within the columns of the figurate number triangle, the column 1 contains the classic and basic counting numbers : 1, 2, 3, 4,... The column 2 contains the triangular numbers : 1, 3, 6, 10, 15, 21,... The column 3 contains the tetrahedral numbers : 1, 4, 10, 20,", "marbles. Then you have 4 marbles left. You pick 2 of them to be blue, which gives you then $\\binom{4}{2}$ possibilities. The remaining marbles are green - so noting to choose. Thus in total you have $\\binom{4}{2}\\binom{6}{2}=90$ sequences. Six distinct marbles can be sorted $6!$ ways, but when we have three pairs the ordering within each pair doesn't matter, so the multiset's permutation is: $$\\frac{6!}{2!2!2!} = 90$$", "1, 3, 6, 10,... marbles. If you stack marbles in the simplest 3D figure (a tetrahedron or 4-sided pyramid), then you arrange 1, 4, 10, 20,... marbles. If you stack marbles in the simplest 4D figure (a 4-simplex), then you arrange 1, 5, 15, 35,... marbles. And so on. These numbers appear within the columns of the figurate number triangle, the column 1 contains the classic and basic counting numbers : 1, 2, 3, 4,... The column 2 contains the triangular numbers : 1, 3, 6, 10, 15, 21,... The column 3 contains the tetrahedral numbers : 1, 4, 10, 20, 35, 56,... The column 4 contains the pentatope numbers : 1, 5, 15, 35, 70, 126,... And so on. Catalan numbers The Pascal's triange hides the Catalan numbers. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 ... Dividing the underlined numbers by 1, 2, 3, 4,..., we get 1, 1, 2, 5, 14,... The Catalan numbers appear in solutions of many combinatorial problems.[15] There is another way to compute these numbers : in the triangle, subtract the neighbour from each", "* 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility. Could you please explain why we ignored 1-1-1 combination? Hey jhabib, You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed. Dear Karishma .. I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong . All the approaches are considering 3 bags as identical , but in problem they are distinct . My approach to this problem : I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 . So first distribute 3 balls one ball each to each bag . select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways . so total 60 ways . Now we are left with 2 balls and 3 bags each containing one ball each . for first ball we have 3 options and second ball again 3", "Marbles Geometry Level 2 If a set of marbles of radius $$5\\text{ cm}$$, is poured into a cube of side $$1\\text{ m}$$, find the maximum number of marbles that can be filled into the box. × Problem Loading... Note Loading... Set Loading...", "zebra fish. In how many different ways can George choose 2 zebra fish to buy? 12. Calculation of CN Calculate: ? 13. Trinity How many different triads can be selected from the group 43 students?", "# Marbles Geometry Level 2 If a set of marbles of radius $$5\\text{ cm}$$, is poured into a cube of side $$1\\text{ m}$$, find the maximum number of marbles that can be filled into the box. ×", "and $2$ from B, there are $\\binom{b}{2}$ ways to choose $2$ balls from C. So the number of ways to select $2$ balls from each of A, B, and C is $\\binom{b}{2}^3$. Added: The above formula counts the number of ways to choose a total of $k$ balls, where for any one of the $n$ boxes, exactly $2$ balls are chosen from that box, or $0$ balls are chosen from that box. There are many other questions one could ask, with roughly similar wording. In order to make it possible to give a precise answer to such similar-sounding questions, the choosing rule has to be stated exactly. • But the OP's comment indicates it's possible to repeatedly select from the same box (in the \"(12,34)\" and \"(ab,cd)\" examples). – Douglas S. Stones Aug 23 '13 at 20:51 • We have $n$ boxes. Each has $b$ marbles. All marbles are distinct. We want to choose $k=2m$ marbles. From any box, we can either take exactly $2$ marbles or exactly $0$. In all your examples, we had $m=n$, meaning we took $2$ marbles from every box, but the formula I gave is more general. By the way, the system objects to long strings of comments. I will delete most of mine, suggest you do the same. But you can continue", "Thursday, December 17, 2009 Solution: Stones and Marbles Original problem: here First, I apologize that the problem was not carefully worded. The problem statement should have been: find all initial conditions where it's always possible for Bob to empty all the three boxes. There are 3 boxes, and 2010 stones and 2010 marbles are put arbitrarily inside those three boxes. At each step, Bob is allowed to either: 1. take a stone from one box, a marble from another box, and put them on the third box 2. add or subtract all boxes by the same number of stones 3. add or subtract all boxes by the same number of marbles Prove or disprove, that by repeating these steps, Bob can empty all three boxes. Solution Answer: If the total number of objects (stones + marbles) in each box is congruent to each other modulo 3. Proof: We can generalize the problem by replacing 2010 with $n$, and it is easy to see that if $n$ is not divisible by 3, then Bob will never be able to empty the boxes, since each operation does not change the total number of objects modulo 3. Furthermore, even if $n$ is divisible by 3, and we have 2 boxes where they are not congruent to each other modulo 3, Bob", "the three picks. $$\\binom 41\\binom {26}2$$ And", "picking apples for her chickens. 10 × 30 = 300 apples. So, my estimate is 300 apples. b. My work: Maggie picks ____ apples in all. Given, Maggie is picking apples for her chickens. She puts 30 apples in each basket. 12 × 30= 360 apples. Maggie picks 360 apples in all. c. What strategy did you use? Why did you choose this strategy? Answer: We used the multiplication strategy to find a number of eggs Maggie’s chicken lay in one month altogether. Bridges in Mathematics Grade 4 Student Book Unit 2 Module 3 Session 2 Answer Key Mario’s Marbles Mario loves marbles and is always adding new marbles to his collection. Help Mario keep track of his marbles in the following problems. Show your work using numbers, labeled sketches, or words. Write a multiplication equation for problems 1 and 2. Question 1. Mario organized some of his marbles. He used lots of marbles to make 11 piles. Each pile had 14 marbles. How many marbles did he use in all? Equation: ________________ Given, Mario organized some of his marbles. He used lots of marbles to make 11 piles. 14 × 11 = 154 marbles Thus he use 154 marbles in all. Question 2. Mario organized the rest of his marbles. He made 7 piles and put 22", "(x/3 - 34) = 3x/3 + x/3 + (x/3 - 34) = 4x/3 + (x/3 - 34) = 5x/3 - 34 = 656 5x/3 - 34 + 34 = 656 + 34 5x/3 = 690 (5x/3)(3/5) = 690(3/5) = 414 ∴ x = 414 Divide x by 3 to determine correctly that Bala has 138 marbles. 10. oh. i calculated wrongly. no wonder i could not find the exact answer with my formula. Thanks ALOT!!!!", "# permutations and combinations (42) There are $2\\binom{4}{2}=12$ ways to select the two students who will get the lime and lemon drinks. (WHY?) That can be done in $\\binom{8+4-1}{8}$ ways.", "Comment Share Q) # In how many ways 9 identical marbles can be placed in 3 identical boxes? $(9,0,0),(8,1,0),(7,2,0),(7,1,1),(6,3,0),(6,2,1),$ $(5,4,0),(5,3,1),(5,2,2),(4,4,1),(4,3,2),(3,3,3)$ $\\therefore$ The required no. of ways = 12", "arranged among themselves in 4! ways Similarly, 3 chemistry and 2 math in 3! and 2! ways. Now considering them as 3 groups, the 3 groups can be arranged in 3! ways. so the total number of ways = 4! * 3! * 2! * 3! Re: In how many different ways can 4 physics, 2 math and 3 [#permalink] 24 Aug 2016, 10:32 Similar topics Replies Last post Similar Topics: In how many ways can 4 different portraits be mounted on 4 of the four 1 17 Feb 2017, 00:58 4 In how many ways can we put 4 different balls in 3 different boxes whe 3 28 Sep 2016, 09:40 42 In how many ways can 5 different marbles be distributed in 4 19 27 Apr 2014, 23:59 17 In how many different ways can 4 ladies and 4 gentlemen be 7 04 Oct 2010, 05:40 8 In how many ways 5 different balls can be arranged in to 3 7 07 Aug 2010, 05:10 Display posts from previous: Sort by", "Any help would be appreciated. Thanks! These are also known as multi-selections. $\\binom{N+k-1}{N}$ ways to select $N$ items from $k$ different types. There twenty kinds s of doughnuts, how many ways can we select a dozen : $\\binom{12+20-1}{12}$." ]

[ "seats, how many different ways can they arrange themselves?\" The other question offered 4 answers as equations, and it was simple matter to eliminate the incorrect ones. Glad I was prepared for what came up. (score will arrive in one month to close this saga) – JoeTaxpayer Mar 2 '13 at 20:08 @JoeTaxpayer: Sounds like it went well; glad to hear it! – Brian M. Scott Mar 2 '13 at 20:12 Hint: Think about the balls. In how many bags can one ball be put in? How many balls are there? - If the three bags are sitting on the table and I rotate the contents, there's still one ball in the one bag. Let me add - if I were loading 3 marbles into 3 bags, just one per bag, there would be two answers depending on if the bags are uniquely identified. If they are three brown bags, one answer, but three numbered or different colored bags, different answer. – JoeTaxpayer Feb 12 '13 at 19:51 nevertheless, there is no restrictions such as you implied in this question. – ciceksiz kakarot Feb 12 '13 at 19:58 The author's solution treats them as unique. But the question avoids the distinction. – JoeTaxpayer Feb 12 '13 at 20:03", "the same time, player two gets a point. If one red and one green marble are pulled at the same time, both players get a point. Is this a fair game?", "they will be the same (snap) or different. FREE Shipping by Amazon. 2/21 . What is the probability picking a navy marble and without replacing it pick a white marble? Bag A contains 3 marbles – one red, one blue and one green. For each flip, if it was heads -> I put in a red marble; if it was tails -> I put in a blue marble. : racing - barbie - shooting - parking - sonic - cooking - tanks - thing thing - bike - dragon ball z - games - recess. Directions: There are _____ red marbles and _____ blue marbles in Bag A. Bag B = 4 red, 6 blue There are 5 gray, 4 red, 5 white, 3 green, 1 orange and 2 navy marbles in a hat. Add this game to your site. Finally multiply How about the red balls? When I put two marbles in this bag, I flipped a coin twice to determine their colors. There are _____ red marbles and _____ green marbles in Bag B. bag B has 3 red and 9 green, bag A has 3 red and 9 blue The game is played with an ordinary deck of cards, a Joker Marbles game board and five marbles of matching color per player. Parameters: Number and", "are 16 flavors to work with, but you can only use three at a time. With an example like this one, you are looking for combinations of object where only a certain number of them are used in any one combination. This happens a lot with teams. Here is a situation with teams. How many different 2-player soccer teams can Jean, Dean, Francine, Lurleen, and Doreen form? Step 2: You’ve covered all combinations that begin with Jean. Now go through all combinations that begin with Dean, Francine, and Lurleen. Your list is now complete. There are 10 combinations. Look at this situation. How many different 3-player soccer teams can Jean, Dean, Francine, Lurleen, and Doreen form? Use the process above to go through all of the combinations. Your list is now complete. There are 10 combinations. We can use a formula to help us to calculate combinations. This is very similar to the work that you did in the last section with factorials and permutations. Suppose you have 5 marbles in a bag–red, blue, yellow, green, and white. You want to know how many combinations there are if you take 3 marbles out of the bag all at the same time. In combination notation you write this as: \\begin{align*}{_5}C_3 \\Longleftarrow 5 \\ \\text{items taken 3 at a time}\\end{align*}", "number of notebooks Keecia bought? 19. Sheri can make 2 bracelets in 16 minutes. She needs to make 7 bracelets. a. Set up a proportion that could be used to find $m$, the number of minutes it would take Sheri to make 7 bracelets. b. How many minutes will it take Sheri to make 7 bracelets? 20. A bag contains only two colors of marbles––silver and gold. The ratio of silver marbles to gold marbles in the bag is 5 : 3. There are a total of 30 silver marbles in the bag. a. Set up a proportion that could be used to find $t$, the total number of marbles in the bag Feb 22, 2012 Jan 06, 2016", "# The Marble Game ### Johnny vs. Vinny As brothers, these two are competitive in every aspect of life. Just check out some of my other puzzles, and you'll see my brothers in action. Today is a classic Marble Battle! The brother's played a series of 3 games: ## The Game Series Summary Note: Not a real game. Bored at work today, put our clip-art library to good use! :) Round 1 Johnny and Vinny each had a certain number of playing marbles. Johnny lost a game to Vin and had to give Vin half of his marbles. Round 2 In the second game, Vin lost and had to give three-fourths of his marbles to Johnny, who now had thirty marbles. Round 3 Finally, the two finished their third game. Vinny won and acquired some of Johnny's marbles. Score Board At this point, each brother had exactly the number of marbles he had started with, and Johnny had twice as many marbles as Vinny. ## How many marbles did each have in the beginning? From the final clue, we learn that Johnny has twice as many marbles as Vinny at both the start and finish. So, J = 2V, where J equals the amount of marbles Johnny has at the start and finish, and V is the same", "sestigtalill 2022-12-19 Geometric distribution. Johnny has 1,176 Pok´emon cards in total. Pok´emon EX is a special type of card, and Johhny has 39 EX-type cards. He is looking for an EX-type card, but all of the cards are completely mixed up and stored in a shoe box. His mother is calling him for dinner. What is the probability that Johnny will have to look through no more than 25 cards before he finds an EX-type card? Agour2q9 Expert", "three shirts. How many different outfits can he make? 7. If Jesse adds three sweaters, how many different outfits can he make now? 8. Kyle has selected three different paint colors and two trim colors for his bedroom. How many different color combinations can he create? 9. There are four teams playing in a tournament. Each team will play two sports. How many different combinations are there for teams and sports? 10. Candice can choose between the following electives. • German • Gymnastics • Chorus • French If she can choose two electives, how many different options are there of choices? 11. If we add Track to her options, how many two elective options are there? 12. Miles is making a salad. He wants to put three things on it. Here are his choices. Tomatoes Pickles Onions Cucumbers Peppers How many different salads can he create given these options? 13. If Miles adds cheese, how many different salads can he create given these options? 14. Muffins can have two additions to them. Here are the choices: blueberries, raspberries, nuts, cherries, strawberries. How many muffin combinations can there be? 15. If we add apples and cranberries, how many combinations can there be now? Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Difficulty Level:", "two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . . Marbles in a Box Stage: 3 and 4 Challenge Level: In a three-dimensional version of noughts and crosses, how many winning lines can you make? Christmas Chocolates Stage: 3 Challenge Level: How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?", "### Islander Points 20 Imagine a three dimensional version of tic, tac, toe where two players take turns placing different colored marbles into a box. The box is made from $n^{3}$ transparent unit cubes arranged in a $n$-by-$n$-by-$n$ array. The object of the game is to complete as many winning lines (vertical, horizontal or diagonal) of n marbles as possible. Can you work out the solution for any $n \\times n \\times n$ cube?", "* 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility. Could you please explain why we ignored 1-1-1 combination? Hey jhabib, You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed. Dear Karishma .. I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong . All the approaches are considering 3 bags as identical , but in problem they are distinct . My approach to this problem : I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 . So first distribute 3 balls one ball each to each bag . select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways . so total 60 ways . Now we are left with 2 balls and 3 bags each containing one ball each . for first ball we have 3 options and second ball again 3", "a drum as 641, 224, and 806. Does order matter in the way the three winning numbers are drawn? Step 1: Write out a single order. 641, 224, 806 Step 2: Now rearrange the order. Did you change the outcome? If so, then order matters. $224, 806, 641\\Longleftarrow$ different order, same 3 winning numbers Order does NOT matter for this problem. Use combinations. Write the difference between combinations and permutations down in your notebook. Example A bag has 4 marbles: red, blue, yellow, and green. In how many different ways can you reach into the bag and draw out 1 marble, then return the marble to the bag and draw out a second marble? Step 1: Write out a single order. red, blue Step 2: Now rearrange the order. Did you change the outcome? If so, then order matters. blue, red $\\Longleftarrow$ different order, meaning is DIFFERENT Order DOES matter for this problem. Use permutations. 12K. Lesson Exercises Write whether you would use combinations or permutations for each example. 1. Cesar the dog-walker has 5 dogs but only 3 leashes. How many different ways can Cesar take a walk with groups of 3 dogs at once? 2. Five different horses entered the Kentucky Derby. In how many different ways can the horses finish the race? 3. How many", "ways can I colour in 4 boxes with 1 colour? The answer to which is 2^4=16 (unless you count a game with opposite NEs the same, ie does the NE '1A 2B' equal the NE '1B 2A' - also I'm not sure if having no NE is the same as having all 4 choice combinations being an equilrium). But then NEs can be stable and unstable. If you have more than one NE then one might be risk dominant while the other is payoff dominant. Repeated games add another twist. There are a lot of other ways in how game types can differ. Unless you give a specific definition of 'essential different game theoretic scenarios' we can't really go much further. Talith Proved the Goldbach Conjecture Posts: 844 Joined: Sat Nov 29, 2008 1:28 am UTC Location: Manchester - UK ### Re: Math: Fleeting Thoughts *BUMP* because I think I have a general answer to the OP's question. Consider the following two person logic game. \\forall x \\exists y P(x,y) Where x \\in {a,b} and y \\in {c,d}. Player 1 picks the x, Player 2 picks the y. Player 1 wins if the Decidable Property P is false, Player 2 wins if the property P is true. This is a two-player, two-option game, as Player 1 can only", "# Combinations of distinct vs identical objects 1. A child has 5 pockets and 4 marbles. In how many ways can he put the marbles in his pockets? 2. In how many ways can 11 identical apples be distributed among 6 children so that each child receives at least one apple? My approach is as follows: 1) for every marble, say M, i have 5 options for pockets (P P P P P), hence i get 5*5*5*5=5^4. This is the correct answer. 2) However, i am unable to apply the same concept to the second problem. i distribute 6 apples to 6 children therefore fulfilling the \"at least one apple\" condition. Then i am left with 5 apples. For each apple, i have 6 options, hence by this logic i get 6*6*6*6*6=6^5. However this is wrong as i am unable to account for identical objects. Please point out where i am missing. I don't understand the logic behind n+r-1Cr-1 or n-1Cr-1. - If the marbles are all distinct, then your analysis for the first problem is right. (If the marbles are identical, then the number of ways is $\\binom{8}{4}$.) Let's look at the second problem. We have $11$ identical apples. Line them up like this: $$A\\quad A \\quad A\\quad A \\quad A \\quad A \\quad A \\quad A", "$n$ are blue and $k-1$ are red. How many different sets of $k$ marbles can we create from the bag? How many ways can we choose $k$ marbles from the bag, where $j$ are blue and $k-j$ are red?", "three balls. There's three balls that you can pick out as yellow. One, two, and three. We're assuming that they're all equally likely to be picked up. They're all mixed in there. So there's three yellow balls you could pick up. Now how many total balls are there? How many total balls are there to be picked up? Well, you could be picking up any one of eight balls, right? One, two, three, four, five, six, seven, eight. So there's a total number of eight balls. So the probability of picking a yellow marble is 3/8. So you have a 3/8 chance of picking a yellow marble. And just a way to think about it. I just want to reiterate. It's just the number of outcomes. There's eight total outcomes. There's eight different things that you could pull out of the bag. Three of them satisfy your conditions. Three of them would be a yellow marble.", "tiles in a bag is 3 to 8. There are 48 orange tiles in the bag. 17. Set up a proportion that could be used to find g, the number of green tiles in the bag. 18. How many green tiles are in the bag? Directions: Marty bought 2 pounds of ham for $11 at a deli. 19. Set up a proportion that could be used to find p, the number of pounds of ham Marty could have bought if he had spent$33. 20. How many pounds of ham could he have bought if he had spent \\$33? Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:", "and gave him 1/4 of this remaining marbles plus 5. Just before arriving at the asked by dyl on December 14, 2010 9. ### maths ANDY HAS TWICE AS MANY MARBLES AS PANDY AND SANDY HAS HALF AS MANY HAS ANDY AND PANDY PUT TOGETHER .ANDY HAS 110 MARBLES WHICH IS 115 MARBLES LESS THAN SANDY.HOW MANY MARBLES EACH OF THEM HAVE asked by Anamika on December 31, 2012 10. ### MATH HENRY, ANDY AND JOE SHARED SOME SWEETS IN THE RATIO 15:12:9 RESPECTIVELY. HENRY HAD 6 MORE SWEETS THAN ANDY. IF ANDY GAVE 9 SWEETS TO JOE, HOW MANY SWEETS WOULD JOE HAVE? asked by FLOR on August 25, 2015 More Similar Questions", "1 2 [ 31 posts ] Similar topics Replies Last post Similar Topics: In a jar there are balls in different colors: blue, red, green and yel 4 13 Jul 2016, 01:01 3 A woman has three blouses of different colors, three skirts of differe 1 17 Jan 2016, 11:10 20 How many ways can 5 different colored marbles be placed in 3 7 14 Oct 2013, 07:53 1 A certain businessman wears only three different colors of shirt -- wh 3 16 Dec 2010, 00:32 7 In a jar there are balls in different colors: blue, red, green and yel 12 07 Oct 2009, 17:48 Display posts from previous: Sort by", "+0 # Help! +1 146 1 A bag contains marbles of three different colors with at least three marbles of each color. Three marbles are randomly selected, without replacement. How many different color combinations are possible? (Note: red, red, blue\" and red, blue, red\" are the same combination.) May 27, 2019 $$\\text{3 combos of the all one color}\\\\ \\text{3\\cdot 2=6 combos of 2 colors}\\\\ \\text{1 combo of all 3 colors}\\\\ \\text{10 in total}$$" ]

[ "Free Version Moderate # Permuations of Selected Positions SATSTM-LPE79G There are eight students in an after school club. One needs to be chosen as president and one as vice president. How many ways can the students be chosen for the two positions? A $15$ B $56$ C $64$ D $1200$ E $40320$", "on an executive committee is $12,650$ The number of ways to select the club members for a president, vice president, secretary, and treasurer so that no person can hold more than one office is $303,600$. ## Example A group of $3$ athletes is $P$, $Q$, $R$. In how many ways can a team of $2$ members be formed? Here, as the order of members is not important, we will use the Combination formula. $C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}$ Putting values of $n$ and $r$: $n=3$ $r=2$ $C\\left(3,2 \\right)=\\frac{3!}{2!\\left(3-2\\right)!}$ $C\\left(3,2 \\right)=3$", "College Algebra (6th Edition) Number of different ways to elect president, vice president, secretary and treasurer $10_{{P}{_{4}}}$ = 5040 Total number of members on its board of directors = 10 Number of members to elect president, vice president, secretary and treasurer = 4 Number of different ways to elect president, vice president, secretary and treasurer $10_{{P}{_{4}}}$ = $\\frac{10!}{(10 - 4)! }$ = $\\frac{10!}{6! }$ = $\\frac{10\\times9\\times8\\times7\\times6!}{6! }$ = $10\\times9\\times8\\times7$ = 5040", "1. ## quadratic word problem :O The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up 2. Originally Posted by juliak The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up (original share) + \\$1 = (final share) let x = original number of members $\\frac{210}{x} + 1 = \\frac{210}{x-7}$ 3. why thank you, skeeter rep &thanked", "club collected exactly $599 from its members. If each member contributed at least$12, what is the greatest number of members the club could have? (A) 43 (B) 44 (C) 49 (D) 50 (E) 51 The average member=$$\\frac{599}{12}=49.92$$ (remember: number of people has to be integer). The question says: each member contributed at least $12. So, the contribution could be more than$12. If the contribution (per head) is $599, then the member (lowest) will be 1. So, the greatest member will be 49 (excluding fraction). So, the correct choice is Thanks__ Re: A club collected exactly$599 from its members. If each [#permalink] 17 Apr 2020, 08:48", "1. ## quadratic word problem :O The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up 2. Originally Posted by juliak The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up (original share) + $1 = (final share) let x = original number of members$\\displaystyle \\frac{210}{x} + 1 = \\frac{210}{x-7}\\$", "are able to make suggestions and table official club business for the year ahead. Membership subscriptions should be reviewed at the annual general meeting, and any price increases must be agreed by the chair-person, treasurer and at least two other members. (b) There should be an annual audit of the clubs finances, with a news-letter at least every three months. The audit should be published before the annual general meeting, and this and the news-letter should be available to current members online in the private member areas. Former members may request this information, which will be granted on a case-by-case basis. (c) Members will be able to attend any events that organised and run by the club for free when ever possible, whilst non-members will pay a small signing in fee of at least £2.50. With agreement with other event organisers, and at events that the club is attending in an official capacity, we will work towards getting members a discounted entry fee. (d) All monies raised will go back into club funds. (4) End of line (a) If it is apparent that the club is not running within its means to the extent that it is likely to fold, or that legal action against it will lead to the club being dissolved, all club assets should be", "attributes: name, phone number, and office location. Each club has an advisor who is a member of faculty (faculty advisor) who has the following attributes: name, department, and office location. An advisor is assigned to a club for a particular semester. A club can have multiple advisors but a faculty member can be assigned as an advisor to one club only. Draw an ER diagram that represents this scenario. Be sure to mark the key attributes and include cardinality constraints on relationships. • @ • 133 orders completed Tutor has posted answer for $25.00. See answer's preview$25.00", "Courses Courses for Kids Free study material Free LIVE classes More # A committee of $15$ members sit around a table . In how many ways can they be seated if the president and vice president sit together ? Last updated date: 28th Mar 2023 Total views: 312.3k Views today: 3.88k Verified 312.3k+ views Hint : Consider the president and vice president as one member Consider president and vice president as one member . Their arrangement in their own group $= 2!$ (Since they also exchange their seat between each other) As we know the total members $= 14$ We know the arrangement of $n$ number of people around a table $=(n-1)!$ Therefore arrangement of 14 members around a table $= 13!$ Total no. of ways in which committee of 15 members can sit around a table including President & Vice-President sit together $= 13!\\, \\times 2!$ Answer is $13!\\, \\times 2!$. Note :- In this type of question of arranging people around a table, we should keep in mind that if we want to arrange n number of items around a table then the number of arrangements are (n-1)!.Here we have a condition that the President & Vice-President will sit together therefore modifications occurred . We have multiplied 2! because we have considered them as one member", "Discussion Forum Daily Magic Spell Discussion - January 6, 2017 Daily Magic Spell Discussion - January 6, 2017 Today, we will discuss about the Daily Magic Spell proposed on January 6, 2017. Problem: Suppose you have a club of $20$ members. The club chooses four officers: President, Vice-President, Treasurer, and Secretary. They also choose someone to be in charge of fundraising. The four officers must all be different, but the member in charge of fundraising can be one of the officers. How many ways can we choose the four officers in the club with no restrictions? Solution: We can first choose the President in the club. There are $20$ options. After the President is chosen, when choosing the Vice-President, there are $19$ options. We repeat this process until all of the positions are filled. Therefore, there are $20 \\times 19 \\times 18 \\times 17 = 116280$ ways to fill the four positions. Additionally, we wish to appoint someone in the club to take charge of fundraising. This can be any one of the $20$ members in the club. Therefore, we have: $116280 \\times 20 = 2325600$ ways to appoint four officers and a person in charge of fundraising. The common approach that most students who answered this question incorrectly is most likely interpreting the problem incorrectly. Some users", "the greatest number of members the club could have? (A) 43 (B) 44 (C) 49 (D) 50 (E) 51 Let's first divide= 599 by 12 (the minimum amount each member could contribute) and then use the remainder to finish the problem. 599/12 = 49 R 11 This means that: 49 people x$12 + 1 person x $11 =$599 We see that if 49 members each contribute $12, someone would have to contribute the extra$11. Note that, since each member contributed at least $12, the$11 could not have come from an additional member. Therefore, the extra $11 must have been contributed by one (or more) of the existing 49 members. Regardless of who contributed the extra$11, the maximum number of members the club could have is 49. _________________ Jeffrey Miller Jeffrey Miller Re: A club collected exactly $599 from its members. If each [#permalink] 21 Jun 2016, 09:34 Similar topics Replies Last post Similar Topics: 2 A local club has between 24 and 57 members. The members of the club ca 4 21 Oct 2016, 08:54 2 A charity collected$1,199 from donors during the last month. If each 4 29 Feb 2016, 09:34 To rent an office, each member of a club must pay n dollars. If two mo 1 07 Feb 2016, 09:40 26 A certain club has", "# Combinatorics Free Version Easy COMBIN-NEGKV4 In how many ways can a president, a vice-president, and a treasurer be selected from a club with $6$ members if no one can hold more than one office? A $120$ B $60$ C $20$ D $15$ E none of the above", "Each person attending a fund-raising party for a certain club was charged the same admission fee. How many people attended the party? (1) If the admission fee had been $0.75 less and 100 more people had attended, the club would have received the same amount in admission fees. (2) If the admission fee had been$1.50 more and 100 fewer people had attended, the club would have received the same amount in admission fees. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.", "+0 0 78 2 +69 A club has 15 members and needs to choose 2 members to be co-presidents. In how many ways can the club choose its co-presidents? rohitaop May 14, 2018 #1 +867 +2 This is a simple choosing problem. 15 choose 2 can be expressed as: $$\\binom{15}{2}=\\frac{15\\cdot14}{2\\cdot1}=105$$ i hope this helped, Gavin GYanggg May 14, 2018 #2 +2765 +1 This is just combinations, 15C2- $$\\frac{15!}{2!*13!}=\\frac{15*14}{2*1}=15*7=\\boxed{105}$$ tertre May 14, 2018", "Administrating a Rotary Club effectively is very important. Effective administration means you will have more time for the fun side of being a Rotarian! What are the goals your club wants to achieve in the next year and in 3 to 5 years? Develop a plan, set achievable goals and decide how they will be achieved. The year ahead then takes on a purpose and members gain a sense of achievement and 'doing good' in their communities. 'Doing good' is why we become Rotarians. The second part of this resource will give you some information on how to conduct effective club meetings. Hope you find it useful! ### 3. Managing meetings effectively #### 3.7. Annual General Meeting and election of officers The Annual General Meeting (AGM) should be held within 6 months of the end of the financial year (by 31st December). Ensure you have a quorum (see your Constitution for clarification) The AGM agenda should include: • The adoption of Minutes of the previous AGM • The election of Officers and Auditor • Approval of audited accounts from the previous year: • For Clubs that generate an income of more than $250,000 per year this is mandatory • Clubs who generate less than$250,000 per year the financial records can be reviewed by two club members who have", "## DISTRICT DIRECTOR Awarded to 1 Division and 1 Area for achieving the following: • ## Qualifying Criteria Division Area The Division should be President’s Distinguished. 60% of the clubs need to be Distinguished (on base) 60% of the clubs need to have 20+ members (on base) 8% growth in Membership Payments. The Area should be President’s Distinguished. 75% of the clubs need to be Distinguished (on base) 50% of the clubs need to have 20+ members (on base) 8% growth in Membership Payments. • ## Tiebreaking Criteria Division Area Tiebreaker 1: Percentage of Clubs that have 20+ Members (on base) Tiebreaker 2: Percentage of Clubs that are distinguished. (on base) Tiebreaker 1: Percentage of Clubs that have 20+ Members (on base) Tiebreaker 2: Percentage of Clubs that are distinguished. (on base)", "of ways to choose two vice-presidents out of the remaining $$9$$ people is 9C2 = $$\\frac{(9*8)}{(1*2)} = 36.$$ Thus, the number of ways to choose one president and two vice-presidents out of $$10$$ people is $$10 * 36 = 360.$$ Therefore, the answer is E. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. \"Only$99 for 3 month Online Course\" \"Free Resources-30 day online access & Diagnostic Test\" \"Unlimited Access to over 120 free video lessons - try it yourself\" Re: A club has 10 members. One president and two vice-presidents are elect &nbs [#permalink] 30 Nov 2018, 01:18 Display posts from previous: Sort by", "consider a hypothetical example. Suppose a college campus has a ski club that organizes day-long bus trips to a ski area about 100 miles from the campus. The club leases the bus and charges $10 per passenger for a round trip to the ski area. In addition to the revenue the club collects from passengers, it also receives a grant of$200 from the school’s student government for each day the bus trip is available. The club thus would receive $200 even if no passengers wanted to ride on a particular day. The table in Figure 21.1 \"Ski Club Revenues\" shows the relationship between two variables: the number of students who ride the bus on a particular day and the revenue the club receives from a trip. In the table, each combination is assigned a letter (A, B, etc.); we will use these letters when we transfer the information from the table to a graph. Figure 21.1 Ski Club Revenues The ski club receives$10 from each passenger riding its bus for a trip to and from the ski area plus a payment of $200 from the student government for each day the bus is available for these trips. The club’s revenues from any single day thus equal$200 plus $10 times the number of passengers. The table relates various combinations", "Clubs Council. The decision of the secretary may be appealed as described in the constition and regulations of ANUSA. ## 5. Membership 1. A person becomes a member of the association — 1. when they inform the association that they wish to become a member, having paid the membership fee and, if they are not a student of the ANU, having been approved by the committee; or 2. when a motion is passed by a two-thirds majority at a general meeting making them a life member; or 3. when they become a committee member. 2. A person’s membership of the association ends — 1. when they inform the association that they wish to end their membership; or 2. when a motion is passed by a two-thirds majority at a general meeting ending their membership; or 3. at the end of a calendar year, unless they are a committee member or life member. 3. The committee may determine a membership fee not exceeding $20. The committee may permit a person to not pay a membership fee, or pay a reduced membership fee. 4. The secretary must maintain a list of names and email addresses of members. ## 6. Committee 1. The association is controlled and managed by a committee of — 1. the president; and 2. the vice-president; and", "Club B offers membership for a fee of $28 plus a monthly fee of$ 17. After how many months will the total cost of each health club be the​ same? Health Club Fees You are choosing between two health clubs. Club A offers membership for a fee of $7 plus a monthly fee of$ 24. Club B offers membership for a fee of $28 plus a monthly fee of$ 17. After how many months will the total cost of each health club be the​ same?... 1 answer ### Please change the voice for me, fast. I have to submit my homework in 30 minutes. 1. He has much money to spend. 2.We must endure what we cannot cure. 3. Promise should be kept. Please change the voice for me, fast. I have to submit my homework in 30 minutes. 1. He has much money to spend. 2.We must endure what we cannot cure. 3. Promise should be kept.... 2 answers ### What is historically significant about this 1879 speech by chief joseph what is historically significant about this 1879 speech by chief joseph... 1 answer ### 10 Scientists need to test different abilities of the robots. While scientists perform these tests, they measure how much energy the robots require to function for a long period of time and" ]

[ "$2\\cdot{6\\choose3}$ Add them and get 55 5. Thanks everyone! Very helpful. I had one more question, kind of similar, but I don't know how to solve this one mathematically without drawing it out ... Six three-representative delegations attend an international conference. Teh reps shake hands when they are introduced to one another, how many handshakes are possible if each delegate shakes hands only once with every other attendant except with those of his/her delegation As always, any help is greatly appreciated. 6. Nevermind. Figured it out. :-)", "of the $21$ friends are all different. What is the minimum number of times he must have returned to the party, assuming that he always shakes hands with $18$ friends after every return? A theater is inviting $2670$ total students from various schools to see a movie. Each individual school invited is only allowed to bring at most $38$ students. If the theater has $193$ seats in each row, how many rows do they need to set aside for schools in order to guarantee that all students from the same school can be seated in the same row? Alice, Bob, Candice and David run for school president. There are $201$ students in the school and each student can vote for only one candidate. The person with the largest number of votes is the winner. If every student votes, what is the minimum number of votes needed for it to be at least possible to win the election? Details and assumptions Ties for the winner are not allowed in this election. ×", "# Choosing school representatives In a student council, there are 8 first year students, 6 second year students, 5 third year students and 6 fourth grade students. 5 students will be randomly chosen as school representatives. All students have a equal chance of being a school representative. A) What is the chance that 2 first year students and 1 student from each other grade become school representatives? B) What is the chance that 3 second year students and 2 fourth year students become representatives? The answer for A is $$0.095$$ and for B its $$0.8056$$ . I thought to use combinations for choosing students and multiply the results, which in turn I would divide with the number of possible results as a whole, but it's giving me the wrong answers. • The method you thought of using is good enough. And the same method gives the answers (approximately, of course) that you've mentioned. Maybe you are calculating something wrong? Aug 15 '20 at 8:07 • Maybe its something about the sum of all possible results? ive run the same calculation several times for the multiplication part and its always the same, and im confident that its right for that part, but my sum of all results might be wrong, first i tried to use combination without repetition then", "Hide # Who Goes There? What happens when more teams want to go to an ICPC regional site than the site has capacity for? Who goes there? One possible policy is the following: Every school is allowed to register as many teams as they wish. Accept every school’s first team, then accept every school’s second team (for schools with more than one team), then third, and so on, until all teams are accepted, or there isn’t enough capacity for the next wave. Then, if there are extra spots available, the spots are given to schools, one by one, in the order that the schools registered. Given the capacity of a site, the number of teams registered by each school and the order that they registered, determine how many teams from each school are accepted. ## Input The first line of input contains two integers $n$ ($1 \\le n \\le 100$) and $m$ ($1 \\le m \\le 100$), where $n$ is the capacity of the site and $m$ is the number of schools that wish to compete there. Each of the next $m$ lines contains an integer $t$ ($1 \\le t \\le 100$), which is the number of teams that a school has registered. The schools are listed in the order that they registered. ## Output Output $m$ lines,", "# CF980E The Number Games • 92通过 • 229提交 • 题目来源 • 评测方式 RemoteJudge • 标签 • 难度 省选/NOI- • 时空限制 3000ms / 256MB • 提示：收藏到任务计划后，可在首页查看。 ## 题意翻译 Panel 国将举办名为数字游戏的年度表演。每个省派出一名选手。 国家有 $n$ 个编号从 $1$ 到 $n$ 的省，每个省刚好有一条路径将其与其他省相连。第 $i$ 个省出来的代表有 $2^i$ 名粉丝。 今年，主席打算削减开支，他想要踢掉 $k$ 个选手。但是，被踢掉的选手的省将很 angry 并且不会让别的任何人从这个省经过。 主席想确保所有剩下选手的省都互相可达，他也希望最大化参与表演的选手的粉丝数。 主席该踢掉哪些选手呢？ ### 输入格式 输入 $n,k$ 。（ $k < n \\leq 10^6$ ） 下来是 $n-1$ 行，一行两个数，代表一条道路的起终点。 ### 输出格式 升序输出要踢掉的选手编号。 感谢@poorpool 提供的翻译 ## 题目描述 The nation of Panel holds an annual show called The Number Games, where each district in the nation will be represented by one contestant. The nation has $n$ districts numbered from $1$ to $n$ , each district has exactly one path connecting it to every other district. The number of fans of a contestant from district $i$ is equal to $2^i$ . This year, the president decided to reduce the costs. He wants to remove $k$ contestants from the games. However, the districts of the removed contestants will be furious and will not allow anyone to cross through their districts. The president wants to ensure that all remaining contestants are from districts that can be reached from one another. He also wishes to maximize the total number of fans of the participating contestants. Which contestants should the president remove? ## 输入输出格式 输入格式： The first line of input contains two", "# How to reduce to an NP-hard problem? For an assignment I have to program an application to schedule conversations. There is an event where representatives of the elementary schools talks with the representatives of high schools. They will talk about the students that will be transferred to the highschool. There are approximately 200 elementary schools and 40 high schools that will be participating in this event. The schools already know which student is transferring to which high school. The conversations will only be between representatives of E and H from student that will be transferring to H. The rules are: 1. The duration of each conversation is based on the amount of students per representatives.Each conversation last 5 minutes per student. If a group consist of 1 student, this conversation last 10 minutes. 2. No timeclashes 3. All the students of the same group will be scheduled together, so, a representatives will only face the same representative once. 4. Timespan is 13.00-19.00 5. The waiting time of a representative is at most 20% of his time. A waiting time is an empty timeslot between the 1st and last conversation. 6. Schedules for 2 days 7. Each representatives participate for 1 day. The problem is that I know that this is hard to solve, but I dont know", "[2], given: 12439 Re: tough p n c [#permalink] ### Show Tags 14 Apr 2010, 06:14 2 KUDOS Expert's post jatt86 wrote: 1) There are 3 states and 3 students representing each state. In how many ways can 5 students be chosen such that at least one student is chosen from each state. To choose 5 students so that atleast one student will represent each state can be done in two ways: A. 3-1-1 (3 students from 1 state and 1 student from other two states) 3C1*3C3*3C1*3C1=27 3C1 - # of ways to choose 3-student state; 3C3 - # of ways to choose 3 students from 3-student state; 3C1 - # of ways to choose 1 student from the first 1-student state; 3C1 - # of ways to choose 1 student from the second 1-student state. OR B. 1-2-2 (1 student from 1 state and 2 students from other two states) 3C1*3C1*3C2*3C2=81 3C1 - # of ways to choose 1-student state; 3C1 - # of ways to choose 1 student from the 1-student state; 3C2 - # of ways to choose 2 students from the first 2-student state; 3C2 - # of ways to choose 2 students from the second 2-student state. 27+81=108. _________________ Kudos [?]: 133195 [2], given: 12439 Intern Joined: 07 Apr 2010 Posts: 23 Kudos", "the last $2$ spots, then there is $1$ way to assign spots to $Y$. Finally, if one $X$ was put in on of the middle two spots and one $X$ was put in one of the last two spots, there are $2\\cdot2$ ways to assign spots to $X$ and $2\\cdot1$ ways to assign spots to $Y$ (one of the first two spots and the remaining spot in the last $2$). There are $1+1+2\\cdot2\\cdot2\\cdot1 = 10$ ways to assign opponents to $B$ and $90\\cdot10 = 900$ ways to order the games. $\\boxed{\\textbf{(E)} 900}$", "Aug 5 '14 at 0:39 1. There are $22 + 9 + 9 = 40$ total points. 2. As each school receives a positive integer number of points for each event, and placing strictly better gives strictly more points, there must be at least $6 = 1 + 2 + 3$ points per event. 3. The divisors of $40$ that are greater than $6$ are $8,10,20,$ and $40$, and so there can be $5,4,2,$ or $1$ events. 4. As schools $B$ and $C$ are tied at the end, there must be more than one event. 5. As school $B$ won an event and there is more than one event, first place must give strictly fewer than $9$ points. 6. If there are exactly two events, then first place must give exactly $8$ points, as $8,7,5$ is the only possible distribution of points that gives first place fewer than $9$ points, sums to $20$ points per event, and gives strictly more points to schools placing strictly better. 7. Note, however, that this makes it impossible for school $B$ to have $9$ points. 8. Similarly, if there are exactly four events, then the only possible distribution of points is $1,3,6$. 9. Then, giving school $B$ 9 points can happen in only one way — $1,1,1,6$, and now it is impossible", "# Difference between revisions of \"2015 AIME I Problems/Problem 2\" ## Problem The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ~ pi_is_3.14 ## Solution The total number of ways to pick $3$ officials from $9$ total is $\\binom{9}{3} = 84$, which will be our denominator. Now we can consider the number of ways for exactly two sleepers to be from the same country for each country individually and add them to find our numerator: • There are $7$ different ways to pick $3$ delegates such that $2$ are from Mexico, simply because there are $9-2=7$ \"extra\" delegates to choose to be the third sleeper once both from Mexico are sleeping. • There are $3\\times6=18$ ways to pick from Canada, as each Canadian can be left out once and each time one is left out there are $9-3=6$ \"extra\" people to select one more sleeper from. • Lastly, there are $6\\times5=30$ ways to choose for the United States. It", "total number of people in the stadium at the end of the game. [Hint: Try making a table to solve this problem] ## Problem $$1.3$$ Billy has a job delivering newspapers every morning before school. Billy is industrious and actually works for two different newspapers, the Herald and the Post. Billy's route covers five different streets. Every house gets either the Herald or the Post newspaper, but not both. Each morning Billy receives 140 newspapers from the Herald and 190 newspapers from the Post. The available data for the number of houses and newspapers delivered is given in the following table: Streets Total Number of Houses Herald Houses Post Houses Elm 64 34 $$?$$ Park 58 27 $$?$$ Oak 37 $$?$$ 20 Main 84 35 $$?$$ 1st 75 $$?$$ 50 Select a suitable system (or systems) and use the accounting principle to determine the total number of newspapers, including how many Heralds and Posts, Billy accumulates each day. (You will find that setting up a table will help you solve this problem.) Remember to explicitly define the counted property, the system and its boundary, and the time period. Also state any assumptions you made in order to solve the problem. ## Problem $$\\1.4$$ Financial data on Mr. Jones are presented in the table below: Mr. Jones' Six-Month Financial", "# Difference between revisions of \"2015 AIME I Problems/Problem 2\" ## Problem The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Solution The total number of ways to pick $3$ officials from $9$ total is $\\binom{9}{3} = 84$, which will be our denominator. Now we can consider the number of ways for exactly two sleepers to be from the same country for each country individually and add them to find our numerator: • There are $7$ different ways to pick $3$ delegates such that $2$ are from Mexico, simply because there are $9-2=7$ \"extra\" delegates to choose to be the third sleeper once both from Mexico are sleeping. • There are $3\\times6=18$ ways to pick from Canada, as each Canadian can be left out once and each time one is left out there are $9-3=6$ \"extra\" people to select one more sleeper from. • Lastly, there are $6\\times5=30$ ways to choose for the United States. It is easy", "climb back in time make more rotis. Suppose the cook starts at the top with $R$ rupees. What are all the possible amounts of money she can have at the end? 13 Consider a social network with $n$ persons. Two persons $A$ and $B$ are said to be connected if either they are friends or they are related through a sequence of friends: that is, there exists a set of persons $F_1, \\dots, F_m$ such that $A$ and $F_1$ ... friends. It is known that there are $k$ persons such that no pair among them is connected. What is the maximum number of friendships possible? 14 The school athletics coach has to choose $4$ students for the relay team. He calculates that there are $3876$ ways of choosing the team if the order in which the runners are placed is not considered. How many ways are there of choosing the team if the order of the runners is to be ... Between $12,000$ and $25,000$ Between $75,000$ and $99,999$ Between $30,000$ and $60,000$ More than $100,000$ 15 Let $L_1$ and $L_2$ be languages over an alphabet $\\Sigma$ such that $L_1 \\subseteq L_2$. Which of the following is true: If $L_2$ is regular, then $L_1$ must also be regular. If $L_1$ is regular, then $L_2$ must also be regular.", "18 views A school has less than $5000$ students and if the students are divided equally into teams of either $9$ or $10$ or $12$ or $25$ each, exactly $4$ are always left out. However, if they are divided into teams of $11$ each, no one is left out. The maximum number of teams of $12$ each that can be formed out of the students in the school is 1 vote 1 1 vote", "+0 # Counting 0 38 1 A debate team consists of five girls and four boys. (Everyone is distinguishable.) How many ways can they stand in line, so that at least two of the boys are standing next to each other? Jun 30, 2022 #1 +2293 +1 Glue 2 of the boys together. There are $${4 \\choose 2} = 6$$ ways to do this. There are also 2 ways to order the 2 boys. Now, there are $$8!$$ ways to order everyone (the 2 boys count as 1 person). This makes for $$8! \\times 6 \\times 2= \\color{brown}\\boxed{483840 }$$ ways. Jun 30, 2022", "ways can be filled first, second and third place? • Seven Seven friends agree to send everyone a holiday card. How many postcards were sent? • Tournament Determine how many ways can be chosen two representatives from 34 students to school tournament. • Five-digit Find all five-digit numbers that can be created from numbers 12345 so that the numbers are not repeated and then numbers with repeated digits. Give the calculation.", "May 9 '19 at 4:40 This is only a partial answer, which shows that the fewest number of weeks in which this can happen is either $$5$$ or $$6$$. I suspect the number is $$5$$, but I do not have a proof. To see that at least $$5$$ sittings are required, note that we require a total of $$\\tbinom{25}{24}=300$$ meetings between students. A quick count shows that there are $$\\frac12(4\\times3+12\\times5+9\\times8)=72,$$ meetings for each sitting, so we will need at least $$5$$ weeks for all students to meet. Here is a series of six sittings in which every student meets every other student, which I have constructed and checked by hand, so there may be a mistake. The time spent constructing this convinces me that it is very likely possible in only five sittings; without (too) much effort, the first five sittings miss only $$20$$ pairs of students. 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 17 19 22 06 15 08 01 09 24 05 20 18 11 02 14 16 03 10 21 04 13 25 12 23 07 18 25 14 03 24 21 17 06 19 12 02 05 13 04 15 23 20 07 16 01 11", "there are 12 representatives from four zones. It has further been decided that if Mr. Lofs is selected, Relk and Lemini will not be selected and vice-versa. In how many ways it can be done? A. $572$ B. $672$ C. $472$ D. $372$ Hide Number formats Decimals Lofoya.com 2017 You may drag this calculator", "he gets. There are then $4$ remaining awards, from which we choose $2$ to give to one student and $2$ to give to the other, which can be done in $\\binom{4}{2} = 6$ ways (and we can say that e.g. the $2$ chosen this way go to the first remaining student and the other $2$ go to the second remaining student, which counts all possibilities). This means the total for the second case is $3 \\cdot 5 \\cdot 6 = 90$, and the answer is $60 + 90 = \\boxed{\\textbf{(B) }150}$. ~savannahsolver ~Interstigation", "$\\binom{n}{2} + (n-k) = 100$ $\\binom{n}{2} < 100 < \\binom{n+1}{2}$ So $$n=14$$ and $$k=5$$. ### Counting students A1, 2020 There are $$3$$ clubs in a school. A student is a member of at least one of them. $$10$$ students are members of all $$3$$ clubs. Each club has $$35$$ members. Find the minimum and maximum number of students in the school. Solution Let us split the students into three categories: • A-members: Students who belong to one club. • B-members: Students who belong to two clubs. • C-members: Students who belong to three clubs. Each club has 35 slots out of which C-members take up 10 slots. The remaining 25 slots in each club have be to filled with either A or B-members. Maxima. In order to maximize the number of students we pick only A members. So 75 A-members fill up the remaining slots. The total number of students is then 85. Minima. There are 75 remaining slots. Each B-student fills up two slots, so there must be at least one A-member. Notice that three B-members can fill up 2 slots of each club. So, 36 B-members can fill up 24 of the remaining slots in each club. The remaining empty slots can be filled up by one B-member and one A-member. So we have 10 C-members," ]

[ "1. ## quadratic word problem :O The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up 2. Originally Posted by juliak The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up (original share) + \\$1 = (final share) let x = original number of members $\\frac{210}{x} + 1 = \\frac{210}{x-7}$ 3. why thank you, skeeter rep &thanked", "$\\binom{n}{2} + (n-k) = 100$ $\\binom{n}{2} < 100 < \\binom{n+1}{2}$ So $$n=14$$ and $$k=5$$. ### Counting students A1, 2020 There are $$3$$ clubs in a school. A student is a member of at least one of them. $$10$$ students are members of all $$3$$ clubs. Each club has $$35$$ members. Find the minimum and maximum number of students in the school. Solution Let us split the students into three categories: • A-members: Students who belong to one club. • B-members: Students who belong to two clubs. • C-members: Students who belong to three clubs. Each club has 35 slots out of which C-members take up 10 slots. The remaining 25 slots in each club have be to filled with either A or B-members. Maxima. In order to maximize the number of students we pick only A members. So 75 A-members fill up the remaining slots. The total number of students is then 85. Minima. There are 75 remaining slots. Each B-student fills up two slots, so there must be at least one A-member. Notice that three B-members can fill up 2 slots of each club. So, 36 B-members can fill up 24 of the remaining slots in each club. The remaining empty slots can be filled up by one B-member and one A-member. So we have 10 C-members,", "1. ## quadratic word problem :O The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up 2. Originally Posted by juliak The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up (original share) + $1 = (final share) let x = original number of members$\\displaystyle \\frac{210}{x} + 1 = \\frac{210}{x-7}\\$", "Free Version Moderate # Permuations of Selected Positions SATSTM-LPE79G There are eight students in an after school club. One needs to be chosen as president and one as vice president. How many ways can the students be chosen for the two positions? A $15$ B $56$ C $64$ D $1200$ E $40320$", "on an executive committee is $12,650$ The number of ways to select the club members for a president, vice president, secretary, and treasurer so that no person can hold more than one office is $303,600$. ## Example A group of $3$ athletes is $P$, $Q$, $R$. In how many ways can a team of $2$ members be formed? Here, as the order of members is not important, we will use the Combination formula. $C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}$ Putting values of $n$ and $r$: $n=3$ $r=2$ $C\\left(3,2 \\right)=\\frac{3!}{2!\\left(3-2\\right)!}$ $C\\left(3,2 \\right)=3$", "the greatest number of members the club could have? (A) 43 (B) 44 (C) 49 (D) 50 (E) 51 Let's first divide= 599 by 12 (the minimum amount each member could contribute) and then use the remainder to finish the problem. 599/12 = 49 R 11 This means that: 49 people x$12 + 1 person x $11 =$599 We see that if 49 members each contribute $12, someone would have to contribute the extra$11. Note that, since each member contributed at least $12, the$11 could not have come from an additional member. Therefore, the extra $11 must have been contributed by one (or more) of the existing 49 members. Regardless of who contributed the extra$11, the maximum number of members the club could have is 49. _________________ Jeffrey Miller Jeffrey Miller Re: A club collected exactly $599 from its members. If each [#permalink] 21 Jun 2016, 09:34 Similar topics Replies Last post Similar Topics: 2 A local club has between 24 and 57 members. The members of the club ca 4 21 Oct 2016, 08:54 2 A charity collected$1,199 from donors during the last month. If each 4 29 Feb 2016, 09:34 To rent an office, each member of a club must pay n dollars. If two mo 1 07 Feb 2016, 09:40 26 A certain club has", "club collected exactly $599 from its members. If each member contributed at least$12, what is the greatest number of members the club could have? (A) 43 (B) 44 (C) 49 (D) 50 (E) 51 The average member=$$\\frac{599}{12}=49.92$$ (remember: number of people has to be integer). The question says: each member contributed at least $12. So, the contribution could be more than$12. If the contribution (per head) is $599, then the member (lowest) will be 1. So, the greatest member will be 49 (excluding fraction). So, the correct choice is Thanks__ Re: A club collected exactly$599 from its members. If each [#permalink] 17 Apr 2020, 08:48", "consider a hypothetical example. Suppose a college campus has a ski club that organizes day-long bus trips to a ski area about 100 miles from the campus. The club leases the bus and charges $10 per passenger for a round trip to the ski area. In addition to the revenue the club collects from passengers, it also receives a grant of$200 from the school’s student government for each day the bus trip is available. The club thus would receive $200 even if no passengers wanted to ride on a particular day. The table in Figure 21.1 \"Ski Club Revenues\" shows the relationship between two variables: the number of students who ride the bus on a particular day and the revenue the club receives from a trip. In the table, each combination is assigned a letter (A, B, etc.); we will use these letters when we transfer the information from the table to a graph. Figure 21.1 Ski Club Revenues The ski club receives$10 from each passenger riding its bus for a trip to and from the ski area plus a payment of $200 from the student government for each day the bus is available for these trips. The club’s revenues from any single day thus equal$200 plus $10 times the number of passengers. The table relates various combinations", "Administrating a Rotary Club effectively is very important. Effective administration means you will have more time for the fun side of being a Rotarian! What are the goals your club wants to achieve in the next year and in 3 to 5 years? Develop a plan, set achievable goals and decide how they will be achieved. The year ahead then takes on a purpose and members gain a sense of achievement and 'doing good' in their communities. 'Doing good' is why we become Rotarians. The second part of this resource will give you some information on how to conduct effective club meetings. Hope you find it useful! ### 3. Managing meetings effectively #### 3.7. Annual General Meeting and election of officers The Annual General Meeting (AGM) should be held within 6 months of the end of the financial year (by 31st December). Ensure you have a quorum (see your Constitution for clarification) The AGM agenda should include: • The adoption of Minutes of the previous AGM • The election of Officers and Auditor • Approval of audited accounts from the previous year: • For Clubs that generate an income of more than $250,000 per year this is mandatory • Clubs who generate less than$250,000 per year the financial records can be reviewed by two club members who have", "attributes: name, phone number, and office location. Each club has an advisor who is a member of faculty (faculty advisor) who has the following attributes: name, department, and office location. An advisor is assigned to a club for a particular semester. A club can have multiple advisors but a faculty member can be assigned as an advisor to one club only. Draw an ER diagram that represents this scenario. Be sure to mark the key attributes and include cardinality constraints on relationships. • @ • 133 orders completed Tutor has posted answer for $25.00. See answer's preview$25.00", "constitution that +describe the procedure that clubs must follow in order to obtain funding. +It is the responsibility of the CSC budget committee and especially the +Treasurer to be aware of MathSoc's requirements for funding. + +\\subsection{EngSoc} +EngSoc typically gives some money to clubs that have engineering students +as members. The amount that EngSoc donates has varied wildly from term to +term, but seems to have settled out at about \\$100 (F90). To get money +from EngSoc, the CSC should submit a request to the EngSoc Treasurer along +with the CSC's proposed budget. The CSC Treasurer should be present at the +EngSoc meeting where the budget is discussed in case any questions arise. + +\\subsection{Bank Account} +The CSC has a chequing account at the Campus Centre CIBC. After the executive +is elected each term, signing authority must be obtained for the new +president and treasurer. The bank has a special form for transferring +signing authority. It requires that either a previous holder of signing +authority or the faculty advisor for the club approve the transfer of signing +authority to the new president and treasurer. + +\\subsection{University Billing Code} +The CSC has a university billing code to which almost any university provided +service can be charged. The list of services include: Audio Visual, Graphics +Services, and the Book Store.", "• November 25th 2008, 06:44 AM juliak The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up • November 25th 2008, 06:51 AM skeeter Quote: Originally Posted by juliak The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up $\\frac{210}{x} + 1 = \\frac{210}{x-7}$", "Discussion Forum Daily Magic Spell Discussion - January 6, 2017 Daily Magic Spell Discussion - January 6, 2017 Today, we will discuss about the Daily Magic Spell proposed on January 6, 2017. Problem: Suppose you have a club of $20$ members. The club chooses four officers: President, Vice-President, Treasurer, and Secretary. They also choose someone to be in charge of fundraising. The four officers must all be different, but the member in charge of fundraising can be one of the officers. How many ways can we choose the four officers in the club with no restrictions? Solution: We can first choose the President in the club. There are $20$ options. After the President is chosen, when choosing the Vice-President, there are $19$ options. We repeat this process until all of the positions are filled. Therefore, there are $20 \\times 19 \\times 18 \\times 17 = 116280$ ways to fill the four positions. Additionally, we wish to appoint someone in the club to take charge of fundraising. This can be any one of the $20$ members in the club. Therefore, we have: $116280 \\times 20 = 2325600$ ways to appoint four officers and a person in charge of fundraising. The common approach that most students who answered this question incorrectly is most likely interpreting the problem incorrectly. Some users", "Probability # Relative Complement The student council of Ray School consists of $6$ seniors, $5$ juniors, and $3$ sophomores. How many ways are there to form a delegation of $4$ representatives for a school board meeting if the delegation must consist only of juniors and seniors and must include at least $1$ junior? Tina has chosen some books at the book store, and she would like to buy: $6$ novels, $6$ math books, and $3$ biographies. However, Tina can only afford to buy $4$ books. Since she can't buy them all, if she needs at least $1$ math book and puts back all the biographies, how many difference choices of books could she make? Ralph chooses $4$ numbers among $4, 5, 6$ and $7$, with duplication allowed, to create a $4$-digit number larger than $6000$. How many ways can he obtain a number such that at least one of the middle two digits is $7?$ In a town hall meeting, $100$ people voted on two issues $a$ and $b.$ The results show that $58$ people voted in favor of $a$ and the rest voted against it, and $68$ people voted in favor of $b$ and the rest voted against it. Also, the number of people who voted against both $a$ and $b$ is greater by $4$ than $\\frac{1}{3}$", "of the $21$ friends are all different. What is the minimum number of times he must have returned to the party, assuming that he always shakes hands with $18$ friends after every return? A theater is inviting $2670$ total students from various schools to see a movie. Each individual school invited is only allowed to bring at most $38$ students. If the theater has $193$ seats in each row, how many rows do they need to set aside for schools in order to guarantee that all students from the same school can be seated in the same row? Alice, Bob, Candice and David run for school president. There are $201$ students in the school and each student can vote for only one candidate. The person with the largest number of votes is the winner. If every student votes, what is the minimum number of votes needed for it to be at least possible to win the election? Details and assumptions Ties for the winner are not allowed in this election. ×", "• November 25th 2008, 06:44 AM juliak The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up • November 25th 2008, 06:51 AM skeeter Quote: Originally Posted by juliak The members of a club hire a coach for the day at a cost of 210$. Seven members withdraw which means that each member who makes the trip must pay an extra 1$. How many members originally agreed to go? D: no clue how to set it up (original share) + \\$1 = (final share) let x = original number of members $\\frac{210}{x} + 1 = \\frac{210}{x-7}$ • November 25th 2008, 07:00 AM juliak why thank you, skeeter rep &thanked", "Club B offers membership for a fee of $28 plus a monthly fee of$ 17. After how many months will the total cost of each health club be the​ same? Health Club Fees You are choosing between two health clubs. Club A offers membership for a fee of $7 plus a monthly fee of$ 24. Club B offers membership for a fee of $28 plus a monthly fee of$ 17. After how many months will the total cost of each health club be the​ same?... 1 answer ### Please change the voice for me, fast. I have to submit my homework in 30 minutes. 1. He has much money to spend. 2.We must endure what we cannot cure. 3. Promise should be kept. Please change the voice for me, fast. I have to submit my homework in 30 minutes. 1. He has much money to spend. 2.We must endure what we cannot cure. 3. Promise should be kept.... 2 answers ### What is historically significant about this 1879 speech by chief joseph what is historically significant about this 1879 speech by chief joseph... 1 answer ### 10 Scientists need to test different abilities of the robots. While scientists perform these tests, they measure how much energy the robots require to function for a long period of time and", "each [#permalink] 30 May 2015, 04:41 Estimate, you don't need to divide here to get the answer: 12X<=600 --> X<=50, But we rounded up, so the real number must be <50 --> 49 (C) _________________ When you’re up, your friends know who you are. When you’re down, you know who your friends are. Share some Kudos, if my posts help you. Thank you ! 800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660 Re: A club collected exactly$599 from its members. If each [#permalink] 30 May 2015, 04:41 Similar topics Replies Last post Similar Topics: To rent an office, each member of a club must pay n dollars. If two mo 1 07 Feb 2016, 09:40 16 A certain club has exactly 5 new members at the end of its 7 08 Dec 2012, 20:12 19 If each participant of a chess tournament plays exactly one 10 10 Nov 2012, 13:06 3 Two members of a club are to be selected to represent the 3 10 May 2012, 07:15 15 A certain established organization has exactly 4096 members. 12 08 Jan 2008, 11:23 Display posts from previous: Sort by # A club collected exactly \\$599 from its members. If each Question banks Downloads My Bookmarks Reviews Important topics Powered", "\\begin{align*}\\frac{n}{2}+1 >6\\end{align*} 3. \\begin{align*}3(n-1) \\ge 12\\end{align*} 4. \\begin{align*}3(n-1) > 12\\end{align*} 8. The students’ council is selling chocolate bars for the big fundraiser. The company knows the school will sell a lot and therefore gives them an initial bonus of $200 to start off their fundraising. For every box of chocolate bars they sell, the school receives 45% of the money. 1. Write a linear equation to represent this scenario. 2. How much money does the school need to raise from selling boxes of bars to make$1500? 9. To play golf at the local country club, you can buy a membership or you can go and pay green fees and simply play for the day. The golf club is selling memberships for the rest of the season to encourage the community to buy memberships. You play golf there all the time. It costs $25 to play 9 holes. To buy a membership for the rest of the season (60 days), it costs$390 plus \\$10 per visit for mandatory cart rental. How many times would you have to golf in the next 60 days to make it worthwhile to join? 10. Some grey hounds can reach a top running speed of 45 miles per hour. A German Pinscher, on the other hand, only reaches a top speed of 30 miles", "Now, $680-120=560$. Scenario 2. For a committee with 1 professor and 2 students, we will get $7*C_{10}^2 = 7 * \\frac{10!}{8! * 2!} = 45*7=315$. (we multiply by 7 because For 2 professors and 1 student, we will get $C_7^2*10 = 21*10=210$, and for 3 professors, we will get $C_7^3 = 35$. Adding up the combinations we will get: $315+210+35=560$. Princeton Review suggests using Scenario 2 because it is supposedly simpler to understand, however, taking into consideration that you can make a mistake in the endless calculations and that it requires 3 complex operations in contrast to 2 in the first case, it has a weaker standing. In any case, both ways get the correct answer, but you need to choose the one that appeals more to you - the one easier to use. Here is another hard problem with some restrictions; again from an unknown source: {{#x:box| EXAMPLE 11. There are 11 top managers that need to form a decision group. How many ways are there to form a group of 5 if the President and Vice President are not to serve on the same team? }} Again our options are to solve the problem either to find the total number of committees and then subtracting the number of groups that VP and P would end up" ]

[ "# The Floo Network Harry is going to be late to his class, so he needs to move as quickly as possible. Here is a map of his institution, where the edges represent roads he could walk through, and the nodes of the same color are those between which he could teleport using the Floo Network: What is the least number of roads he would need to walk through to get to the classroom? Clarification: Teleportations happen instantaneously and thus are not included in the count. Inspired by Christopher Boo ×", "## Who Won the Race? How much information do you need to find the finish order of a three-person race? It turns out, not much at all. Suppose that in a race among Amelia, Boris, and Carlos, we only know one fact — that Boris finished before Carlos. What are the possible orders for the entire race? Keep reading to see some reasoning we can use, or jump ahead to today's challenge for a trickier problem. We can start by thinking about a single unknown factor in the problem: \"Where did Amelia finish?\" To see this logic demonstrated, use the arrows below the figure: Since we know the relationship between Boris and Carlos, Amelia's position in the race completely determined the entire finishing order. There are three possible total finishing orders — one for each of the positions in which Amelia can finish. What if we add another runner to the race, Dara, and have a more complicated scenario? Dara finished more places ahead of Amelia than Boris finished before Carlos. We can start by again thinking about an unknown — \"How did Boris and Carlos finish?\" There is only one way these conditions can be satisfied without conflicts, so we have only one solution. In the problem below, you'll be putting together information like this, but with", "regardless of order of any kind. Or what exactly does the question mean? • November 15th 2012, 11:44 AM zzizi Re: A permutation/ race problem Quote: Originally Posted by Plato Here is the OP \"How many possible outcomes of the race are there in which 3 of the first 5 children that cross the line are wearing red hats?\" Now that is poorly posed. I took it quite literally. The order of the children is all important here. Say Bob has a red hat and Jim has a red hat. If Bob is first and Jim is third is totally different from Jim first and Bob second. Although in both cases those two are among the first five to finish. That is the basis of my reply. If the order of the finishers makes no difference then we must ask. Do you simply mean how many ways can the first five be wearing three red hats and two non-red hats? And that is regardless of order of any kind. Or what exactly does the question mean? Apologies if it was poorly posed, that's how it was asked of me. That's why I put the question here, so that I could make better sense of what is meant and how to solve. Thank you for your input though, it's", "Race!\" Post by: apskip on March 14, 2007, 06:31:58 PM Here's an interesting tidbit from a RealityNewsOnline interview on 3/12/07 of Rob and Amber: Rob: I never found the letter. After it was all said and done, and Charla & Mirna were gone and crossed the finish line, I got the letter off of somebody else’s table. I never found the letter in my bag. This supports the theory that I saw on another forum that there had to be 16 letters (two for each team) in every mail bag. It is probable that the letters written by the teams from the same seasons were copied so that they were identical in each bag, but maybe each of those teams actually wrote 16 letters.", "### All The Twos Alex, Brad, Colin, Doug, Eric and Frank had a race to school from the bus-stop and then another race from school to the bus-stop. In the first race, Brad wasn’t last and Eric finished before Colin, Frank wasn’t first but finished before Doug, Brad finished before Frank but after Alex and Colin. Alex finished four places ahead of Doug. In the second race, Alex finished two places ahead of Doug and before Eric who in turn was two places ahead of Doug and before Eric who in turn was two places behind Colin. Alex wasn’t first and nor was Brad who finished before Eric. Frank finished one place ahead of Doug. From the information given: 1. Which two boys finished in a better position in the first race than in the second race? 2. Which two boys finished in the same position in the second race as they did in the first race? Let me see which of my student(s) is first in this ! Nalin Pithwa. ### Three slices of toast Mother makes tasty toast in a small pan. After toasting one side of a slice, she turns it over. Each side takes 30 seconds. The pan can only hold two slices. How can she toast both sides of three slices in one", "of each group race. Let's assume they placed in alphabetical order for convenience of discussion. We can eliminate D1 and E1, and thus their entire groups, because there are three faster horses. We can also eliminate all of group C except C1, and also B3 because A1, B1, and B2 are faster. So for the 7th and final race, we compare A2, A3, B1, B2, and C1 (We already know for sure that A1 is the fastest, so no need to race). The top two in this last race are the second and third fastest respectively. [–] 1 point2 points (1 child) I assume by \"There's no timer\", you mean for a given race, you know which horse came 1st, 2nd, 3rd, 4th and 5th, but you don't know how long it took for each horse to complete the race. [–] 2 points3 points (0 children) A cube can be cut into 27 smaller equal sized cubes with nothing left over by making six cuts. By cutting and stacking so that a cut will go through more than one piece can you reduce the number of cuts to achieve the same result? (A cut cannot 'turn' a corner, it has to go straight through.) This has an awesome proof. It is from Martin Gardner. [–] 2 points3 points", "a jackal are running a race. Three leaps of the &nbs [#permalink] 02 Sep 2018, 07:22 Display posts from previous: Sort by", "five runners instead of four. Can you use logic to figure out the order? # Today's Challenge Four runners compete in a race — Amelia, Boris, Carlos, and Dara. After some confusion at the finish line, it's unclear what the final finishing order was, but this information is known: 1. Dara finished before Amelia. 2. Boris wasn't third. 3. There were two runners between Amelia and Carlos. $\\\\[-0.7em]$ $\\\\[-0.5em]$ Who won the race?", "Red Queen's race in Lewis Carroll's book, Through the Looking-Glass.", "commented on March 9, 2022, 9:16 p.m. I would recommend using a list which tracks the different owners of the wand. Every time a duel leads to the Elder Wand changing hands your program checks the list. Is the latest owner of the Elder Wand already on your list of previous owners? If not, have them added to the list. But if they are, then don't have your program change anything.", "at which they first meet in order to find where does Sally start running I hope this helps", "in each of the five positions. Given the following information, what were the results of each of the five races? The five competitors were A, B, C, D and E. C didn’t win the fourth race. In the first race A finished before C who in turn finished after B. A finished in a better position in the fourth race than in the second race. E didn’t win the second race. E finished two places behind C in the first race. D lost the fourth race. A finished ahead of B in the fourth race, but B finished before A and C in the third race. A had already finished before C in the second race who in turn finished after B again. B was not first in the first race and D was not last. D finished in a better position in the second race than in the first race and finished before B. A wasn’t second in the second race and also finished before B. So, is your brain racing now to finish this puzzle? Cheers, Nalin Pithwa. PS: Many of the puzzles on my blog(s) are from famous literature/books/sources, but I would not like to reveal them as I feel that students gain the most when they really try these questions on their own rather", "1🐎︎ We can eliminate one more horse with the “slower-than-3” rule: the 3rd orange horse. It was beaten by the 2nd and 1st orange horses, which in turn were slower than the 1st red horse. So the 3rd orange horse is eliminated: 1: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 2: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 3: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 4: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 5: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ Notice that we only have 5 remaining horses to sort now, and only the 2nd and 3rd overall positions to find. We just run a race with those 5 remaining horses to find out: 3🐎︎ 2🐎︎ 1🐎︎ 2🐎︎ 1🐎︎ As it turns out in this example, the 1st purple horse and 2nd red horse take the top two positions in this race. That means the final top three are the 1st red, 1st purple and 2nd red horse: 🐎︎ 🐎︎ 🐎︎ This takes a total of 7 races: the initial 5 groups, the fastest-of-fastest race, and the surviving runners up race at the end.", "C 1st: B 2nd: A 3rd: C 1st: A 2nd: C 3rd: B 1st: C 2nd: A 3rd: B 1st: B 2nd: C 3rd: A 1st: C 2nd: B 3rd: A 1st: AB (tie) 2nd: C 1st: AC (tie) 2nd: B 1st: BC (tie) 2nd: A 1st: A 2nd: BC (tie) 1st: B 2nd: AC (tie) 1st: C 2nd: BC (tie) 1st: ABC (tie) That comes out to 13 different ways these horses unicorns can finish the race. That’s the answer for 3 unicorns. What’s the answer for 8 unicorns? (FYI: If you want to know if you’re on the right track… I have 541 for 5 unicorns…)", "joins in. • He sees that the first two wizards have similar hats (let's say white) so he stands on the left of the both wizards. • Now a fourth wizards joins in. • He sees that two wizards are wearing white hats and one is wearing a black hat. He goes and stand in between the wizard with the white hat and the wizard with the black hat. Now, regardless of the fourth wizard's own hat color, he will always be sorted next to his own color (since there are only two possibilities). W = White hat B = Black hat • First two wizards : W W • Three wizards : W W B • Four wizards : W W W B • Five wizards : W W W B B • Six wizards : W W W B B B As long as the room is big enough, this method will sort all the wizards and will (assuming the number really is infinite) tell all the wizards that are in between two of the same colored hats their own colored hat as well. So in the end only the last wizard will have to \"guess\" (or as stated by DrunkWolf, he can abstain about) which hat he has. This answer assumes the wizards have no", "you pick sticks when you have six different classes? Do you have six different cups of sticks? After a few years of “refinement” I have a method that works well for me. I picked up a package of big, brightly colored “craft sticks” at the dollar store. There are actually six different colors, but I generally only use four. I wrote the numbers 1-8 on the bottom of the sticks and I am good to go for a class of up to 32 students. Each student is assigned a color and a number (which they remember quite readily after the first few days.) I, on the other hand, have a “cheat sheet” that I post on the board and “borrow” during class so I can actually call on the student’s name instead of just the “color-number combination.” Here is my “Wizarding World of Harry Potter” butterbeer cup that I will being using this year. (I am finally retiring my University of Minnesota cup with the lettering all worn off, but I am still using it to hold my little six inch rulers.) Here is one of the (currently empty) “cheat sheets” that will be filled with names once our class lists are finalized. (Ha! I just noticed the Yellow column twice. I must have used my class that", "1. ## GULLIVER, kING, links who knows gullivers chain links. How many links did gulliver's chain have? 2. ## Re: GULLIVER, kING, links Oh yes, I think I know that one, the asnwer is 6, i cant really ddraw it but if you think about it he cuts 2, he can get all combo's with 6 links, good luck 3. ## Re: GULLIVER, kING, links thanks a lot !!!", "# Three rings to rule them all (again) Inspired by this riddle Three rings to rule them all, Seven a side to battle. Two distracted by a glint, Trying to catch the tattle. All are flying, holding on But on a plane they're not. Four balls are needed to play this game. Can you tell me what? • I guess it's not a riddle if you just describe the game Oct 10, 2016 at 7:35 Quidditch because Three rings to rule them all, Three rings are the goalposts Seven a side to battle. Seven in each team Two distracted by a glint, The seekers Trying to catch the tattle. the golden snitch All are flying, holding on Flying on brooms But on a plane they're not. Four balls are needed to play this game. Quaffle, the golden snitch, and two bludgers Can you tell me what? • Haha, that easy, huh? Oct 10, 2016 at 7:26 • yeah, too easy :/ Oct 10, 2016 at 7:32", "the 1000-th wizard Gandalf sees all the numbers except (say) 1 and 2. BY prior agreement the missing hat is placed at the end of the line, so if Gandalf has the number 2 hat, the total permutation parity is even only if the pemrutation parity of the hats on wizards 999 thru 1 is even. So if that permutation is even, Gandalf will guess 2; otherwise he will guess 1. Now we come to the 999-th wizard, Harry. Harry doesn't see his own hat $H$, and he also doesn't see the hidden hat and Gandalf's hat. But say Gandlaf guessed two; Harry knows that his own hat is not number 2. That leaves only two possibilities for his hat. Here we come to a flaw in the explained answer: Harry assumes Gandalf was right, adn fiugres out what his hat has to be on that basis. (This is possible because if Gandalf is right he has exactly two possible guesses, and they are swaps of one another so one leads to odd and one to even.) Obviously, if Gandalf was right, Harry will be right as well. What if Gandalf was wrong? Then the overall permutatoin was odd, not even. And harry has th e wrong hat in mind for Gandalf, so when he tries to make", "Four runners compete in a race: Annie, Becca, Carlos, and Dante. After some confusion at the finish line, it's unclear what the final finishing order was, but the following information is known: 1. Dante finished before Annie. 2. Becca wasn't third. 3. There were two runners between Annie and Carlos. Who won the race? Finally, it says that there were two runners between Annie and Carlos. That means that both Annie and Carlos couldn't have been $2nd$ or $3rd$. Now we can see that Annie must be $4th$. That makes Dante $3rd$ and Becca $2nd$, leaving Carlos as $1st$." ]

[ "our build. After this choice, we place the last item in the final slot our build order. Over the course of finalizing our build, we had 3 options, then 2 options, and then 1 final option. We need to multiply these numbers so that we count each possible sequence of items that we could have chosen: for each of the three possible first item choices, there are two more options after that, then each of those options is followed by one option (the last item). So each of the initial three options leads to two more, down the road. Adding them up, we get $2+2+2=3\\cdot 2=6$. Then we count the final item selection in each of these choices: there is 1 option, so we multiply $6\\cdot 1=6$ to get the final answer. In total: $3\\cdot 2\\cdot 1$. Looks familiar right? It’s the factorial we just learned! Check-Your-Knowledge Question 1. It’s the end of the game and the top laner is full build. How many ways could they order the items in their inventory? 2. Since we already know what $5!$ is, notice that we can write $6!=6\\cdot 5!$ as a quicker way to calculate $6!$. (Try expanding both sides if you don’t believe it.) Can you turn this into a general statement about the value of $n!$, if", "he gets. There are then $4$ remaining awards, from which we choose $2$ to give to one student and $2$ to give to the other, which can be done in $\\binom{4}{2} = 6$ ways (and we can say that e.g. the $2$ chosen this way go to the first remaining student and the other $2$ go to the second remaining student, which counts all possibilities). This means the total for the second case is $3 \\cdot 5 \\cdot 6 = 90$, and the answer is $60 + 90 = \\boxed{\\textbf{(B) }150}$. ~savannahsolver ~Interstigation", "6:51 The order in which the runners finish matters, so this is a permutation rather than a combination. Until the first runner crosses the finish line, there are eight runners who could finish first. Once the first runner crosses the finish line, there are seven runners who could finish second. Once the first two runners have crossed the finish line, there are six runners who could finish third, giving $$8 \\cdot 7 \\cdot 6 = 336$$ ways for the runners to finish first, second, and third in the race given that there are no ties. Using permutations, the number of ways the eight runners could finish first, second, and third given that there are no ties is $$P(8, 3) = \\frac{8!}{(8 - 3)!} = \\frac{8!}{5!} = \\frac{8 \\cdot 7 \\cdot 6 \\cdot 5!}{5!} = 8 \\cdot 7 \\cdot 6 = 336$$ Imagine if John, Jack and Julie were in the \"top-3\", there is 6 possible combination.Because you do not know what was the rank. Anyone can be at So $8\\choose3$ ways of selecting 3 people out of a group of 8, multiplied by no of ways of arranging those 3 in the ranks $3!$.", "# There are 10 bicyclists entered in a race. In how many different orders could these 10 bicyclist finish? Apr 13, 2017 #### Answer: 10! is the answer. #### Explanation: this is just like you are given 10 lines on a paper and you have to arrange 10 names on those 10 lines in all different ways . so , starting with the lowermost line, you can put one of 10 names on that line and then on the line above that you can put 1 of 9 names and so on . so the total ways to arrange all the names on those lines in all ways will be : $10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1$ = 10!", "a single $6$-cycle. Case: Two $3$-cycles • The first friend has $5$ possible friends to choose. • The friend to whom the book is given has $4$ possible choices. • The next friend has just one choice - to give their book to the first friend. • The first of the $3$ remaining friends has $2$ choices. • The next friend has $1$ choice, to give their book to the last remaining friend. • The last friend gives their book to the first friend of their $3$-cycle. So there are $5 \\cdot 4 \\cdot 2 = 40$ possible choices of two $3$-cycles.", "$2$ ways to serve their meals. Thus, there are $(3\\cdot2)^3=\\boxed{216}$ ways to serve their meals. Note: This solution gets the correct answer through coincidence and should not be used. ~ dolphin7", "name for this special multiplication. It’s called a factorial, and it’s written like this: $4!$ This is not a very excited $4$. It’s just shorter way to write $4 \\cdot 3 \\cdot 2 \\cdot 1$. Factorials are very useful when counting things, and they’ll save us a lot of trouble in solving this problem. Here we go – Harry unwraps Super-Sam: First, there are $5$ choices for the first book. Then there are $4$ books remaining, so there are $4$ ways to pick the second book. Then there are $3$ ways to pick the third book, and $2$ ways to pick the fourth book, and then there’s only one book left for the last position, for a grand total of $5! = 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120$ ways for Harry to arrange his new books. Thank goodness we didn’t have to write out all those options. It turns out that we can follow this same process for any number of things that need to be ordered. If there are $n$ things, then there are $n$ ways to pick the first thing and $n-1$ ways to pick the second thing and so on, and when you multiply it all out you get a grand total of $n!$ ways to order them. That’s something", "for the second, and so on. The number of ways the prizes can be distributed will be $$5 !=5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1=120$$ ways. Now we will consider some slightly different examples. Solution Using the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are $$25 \\cdot 24 \\cdot 23=13,800$$ ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn't really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6: ABC ACB BAC BCA CAB CBA How can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are $$3 \\cdot 2 \\cdot 1=6$$ ways to do this; we didn't really need to list them all, we can just use permutations! So, out of the", "get one $C$ and one $F.$ We proceed with casework. • If the two $B$ people both get $C$'s, then the three $F$ meals left to distribute must all go to the $C$ people. The $F$ people then get $BBC$ in some order, which gives three possibilities. The indistinguishability is easier to see here, as we distribute the $F$ meals to the $C$ people, and there is only 1 way to order this, as all three meals are the same. • If the two $B$ people both get $F$'s, the situation is identical to the above and three possibilities arise. • If the two $B$ people get $CF$ in some order, then the $C$ people must get $FFB$ and the $F$ people must get $CCB.$ This gives $2 \\cdot 3 \\cdot 3 = 18$ possibilities. Summing across the cases we see there are $24$ possibilities, so the answer is $9 \\cdot 24 = \\boxed{216.}$ ## Solution 2 We only need to figure out the number of ways to order the string $BBBCCCFFF$, where exactly one $B$ is in the first three positions, one $C$ is in the $4^{th}$ to $6^{th}$ positions, and one $F$ is in the last three positions. There are $3^3=27$ ways to place the first $3$ meals. Then for the other two people, there are", "three people we choose two, to be the people interacting with the special person. Thus, we have $4\\cdot\\binom{3}{2}\\cdot4! = 288$ ways here. Case 3: $4$ people. In this case. First note that no person can give/receive twice. If we keep the order of A, B, C, D giving in that order (and permute afterward), then there are three options to choose A's receiver, and three options for B's receiver afterward. Then it is uniquely determined who C and D give to. This gives a total of $3\\cdot3\\cdot4! = 216$ ways, after permuting. So we have a total of $36+288+216=540$ ways to order the four pairs of people. Now we divide this by the total number of ways: $(4\\cdot3)^4$ (four rounds, four ways to choose giver, three to choose receiver each round). So the answer is $\\frac{540}{12^4} = \\boxed{\\textbf{(B)}\\frac{5}{192}}$. ~ccx09 Latex polished by Argonauts16 and by Grace.yongqing.yu. ## Solution 3 Similar to solution 2, we think in terms of transactions. WLOG, for the 1st transaction, we assume that A gives to B, which we denote AB. For the 2nd transaction, there are 12 options to choose from, so there are $12^3$ possible options. Case 1 (Giving back immediately): The 2nd transaction is BA (1 option). Then, the 3rd transaction can be whatever you like (12 options), but the 4th", "$2/3 \\cdot 1/2 = 1/3$. So the probability of getting to the final round is $2/3$. For your other question we can read off: ii) $1/3$ iii) $1/3$ So switching or staying with your first choice doesn't matter. -", "# $20$ people are sitting around a (circular) table. How many ways can we choose $3$ people, no two of whom are neighbors? 0 choices for the 1st person. 17 choices for the 2nd person (must exclude 1st and his/her two neighbours) For 2 of these choices of 2nd person, there is one shared neighbour, so 15 remaining choices. (e.g. if they are numbered 1 to 20 in a circle, 1st person is #1, 2nd is #3, then people 20,1,2,3,4 are excluded). For the other 15 choices of 2nd person, there are no shared neighbors, so 14 remaining choices. So if order matters, total is $20 \\cdot (2 \\cdot 15 + 15 \\cdot 14)$ but since order does not matter, divide by $3! = 6$ to account for the permutations in order of the 3 people. So total = $20 \\cdot \\frac{2 \\cdot 15 + 15 \\cdot 14}{6}$ just redid it; does this make any sense? • What is the answer with a straight table and how does turning the table into a corner change the problem? thats the step that you need to account for Oct 4, 2016 at 16:14 • first person - 20 choices , second person 17 choices , then third person 14 choices - then you could have chosen the 3 people in", "guess, since we have 14 choices for the first guest, then 13 for the second, and so on. But the guess is wrong (in fact, that product is exactly $$2192190 = P(14,6)$$). It distinguishes between the different orders in which we could invite the guests. To correct for this, we could divide by the number of different arrangements of the 6 guests (so that all of these would count as just one outcome). There are precisely $$6!$$ ways to arrange 6 guests, so the correct answer to the first question is \\begin{equation*} \\frac{14 \\cdot 13 \\cdot 12 \\cdot 11\\cdot 10 \\cdot 9}{6!}. \\end{equation*} Note that another way to write this is \\begin{equation*} \\frac{14!}{8!\\cdot 6!}. \\end{equation*} which is what we had originally. This page titled 1.3: Combinations and Permutations is shared under a CC BY-SA license and was authored, remixed, and/or curated by Oscar Levin.", "$\\underbrace{b+b+\\cdots+b}_{a}=a b$ possibilities. 1.4.2. It is often easier to think about the experiments as being in chronological order, but there is no requirement in the multiplication rule that Experiment A has to be performed before Experiment $\\mathrm{B}$. Example 1.4.3 (Runners). Suppose that 10 people are running a race. Assume that ties are not possible and that all 10 will complete the race, so there will be welldefined first place, second place, and third place winners. How many possibilities are there for the first, second, and third place winners? Solution: There are 10 possibilities for who gets first place, then once that is fixed there are 9 possibilities for who gets second place, and once these are both fixed there are 8 possibilities for third place. So by the multiplication rule, there are $10 \\cdot 9 \\cdot 8=720$ possibilities. We did not have to consider the first place winner first. We could just as well have said that there are 10 possibilities for who got third place, then once that is fixed there are 9 possibilities for second place, and once those are both fixed there are 8 possibilities for first place. Or imagine that there are 3 platforms, which the first, second, and third place runners will stand on after the race. The platforms are gold, silver, and", "candidate receives one vote: There are $3$ ways to choose which candidate receives four votes, $\\binom{5}{4} = 5$ ways for four of the five voters to select that candidate, and $2$ ways to choose which candidate receives one vote from the remaining voter. Hence, there are $3 \\cdot 5 \\cdot 2 = 30$ such cases. One candidate receives three votes and another candidate receives two votes: There are $3$ ways to choose which candidate receives three votes, $\\binom{5}{3} = 10$ ways for three of the five voters to choose that candidate, and two ways to choose which of the remaining two candidates receives the remaining two votes. Hence, there are $3 \\cdot 10 \\cdot 2 = 60$ such cases. One candidate receives three votes and each of the other two candidates receives one vote: There are $3$ ways to choose which candidate receives three votes, $\\binom{5}{3} = 10$ ways for three of the five voters to select that candidate, and $2! = 2$ ways in which the remaining voters can split their votes among the remaining two candidates. Hence, there are $3 \\cdot 10 \\cdot 2 = 60$ such cases. Two candidates each receive two votes and the other candidate receives one vote: There are $\\binom{3}{2} = 3$ ways to choose which two candidates receive two votes, $\\binom{5}{2}", "bag. This is equivalent to picking one of the $$m_1+\\cdots+m_k$$ total items. If all the bags have the same number of items $m_1=\\cdots=m_k=n,$ then there are $$kn$$ items in total, and this is the total number of ways the two-stage procedure can occur. • How many ways are there to first roll a die and then draw a card from a (shuffled) $$52-$$card pack? Answer: there are $$6$$ ways the first step can turn out, and for each of these (the stages are independent, in fact) there are $$52$$ ways to draw the card. Thus there are $$6\\times52=312$$ ways this can turn out. • How many ways are there to draw two tickets numbered $$1$$ to $$100$$ from a bag, the first with the right hand and the second with the left hand? Answer: There are $$100$$ ways to pick the first number; for each of these there are $$99$$ ways to pick the second number. Thus $$100\\times 99=9900$$ ways. #### Multi-Stage Procedures A $$k$$-stage process is a process for which: • there are $$n_1$$ possibilities at stage 1; • regardless of the 1st outcome there are $$n_2$$ possibilities at stage 2, • regardless of the previous outcomes, there are $$n_k$$ choices at stage $$k$$. There are then $n_1 \\times n_2\\cdots\\times n_k$ total ways the process can turn", "\\ = \\ (3003) \\cdot (252) \\cdot (1)$$ b) solution would be similar except order of choosing 1st for Claire, 2nd for Alana, & 3rd from Kalena no longer matters. (of course, the choosing order of the 5 within each group still does not matter). so just divide answer (a) by (3!) to remove the ordering among Claire, Alana, & Kalena to produce 3 parcels of 5 gifts each.", "that number and increment each player by one wrapping around to $$A$$ if necessary (e.g. $$A$$ become $$B$$, $$B$$ becomes $$C$$, and $$C$$ becomes $$A$$) to get the final ordering for the new number. Thus, the person who goes second in the 3 person game can win. The person who goes first will choose to make $$1,2$$ as the first move and guarantee themselves second instead of third, allowing the person who goes second to write $$3,4$$ and put themselves in the winning position. So if the initial setup is $$ABC$$ with $$A$$ to go first, then the final ordering will be $$BAC$$. ## 4 Player Game Players are $$A,B,C,$$ and $$D$$. The following are the results for player $$A$$ if player $$A$$ writes the following numbers: • $$9$$: Finish 3rd. $$B$$ is eliminated, and the remaining players in order to start the final round are $$CDA$$. Final order will be $$DCAB$$. • $$8$$: Finish 1st. $$B$$ would write $$9$$ and finish 3rd, final order is $$ADBC$$. • $$7$$: Finish 2nd. $$B$$ would write $$8$$ and finish 1st. Final order is $$BACD$$. • $$6$$: Finish 2nd. $$B$$ would write $$7,8$$ to finish 1st. Final order is $$BACD$$. • $$5$$: Finish 4th. The best $$B$$ can do is finish 2nd. Final order is $$CBDA$$. • $$4$$: Finish 4th. $$B$$", "of the second attribute, and $1$ way to arrange the positions of the third attribute, giving us ${3\\choose 2} \\cdot 3 \\cdot 3 = 27$ ways. • Case 3: One of the three attributes are the same. There are ${3\\choose 1}$ ways to pick the one attribute in question, and then $3$ ways to pick the value of that attribute. Then there are $3!$ ways to arrange the positions of the next two attributes, giving us ${3\\choose 1} \\cdot 3 \\cdot 3! = 54$ ways. • Case 4: None of the three attributes are the same. We fix the order of the first attribute, and then there are $3!$ ways to pick the ordering of the second attribute and $3!$ ways to pick the ordering of the third attribute. This gives us $(3!)^2 = 36$ ways. Adding the cases up, we get $27 + 54 + 36 = \\boxed{117}$. ## Solution 2 Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different there is only one option- choose", "$W=3$ and $w=0$. • Pick who the loser is: $6$ options. From among the remaining five players, take the player who appears first in some predefined arbitrary order (e.g. earliest alphabetically, shortest, name first appearing on signup list,... I will refer to the person as being \"smallest\" from here out). We will call the smallest winner $x$. • Pick which of the remaining four winners $x$ had beaten versus lost: $\\binom{4}{2}=6$ options. We now make the observation that among the two winners who beat $x$, one of them beat the other. By symmetry, we can see that there are exactly as many occurrences of the smaller of the two beating the larger of the two as vice versa. • From the two winners who beat $x$, pick whether the larger beat the smaller or vice versa. Let the loser of that match be called $y$: $2$ options. Now, we make the observation that as soon as we define one more outcome, everything else is strictly determined. • Pick who the third person that $y$ had beaten was: $2$ options. There are then $6\\cdot 6\\cdot 2\\cdot 2 = 144$ total tournament outcomes which result in a five-way tie. As there are $2^{\\binom{6}{2}} = 2^{15}$ different equally likely tournament outcomes this implies that the probability of a five-way tie in" ]

[ "6:51 The order in which the runners finish matters, so this is a permutation rather than a combination. Until the first runner crosses the finish line, there are eight runners who could finish first. Once the first runner crosses the finish line, there are seven runners who could finish second. Once the first two runners have crossed the finish line, there are six runners who could finish third, giving $$8 \\cdot 7 \\cdot 6 = 336$$ ways for the runners to finish first, second, and third in the race given that there are no ties. Using permutations, the number of ways the eight runners could finish first, second, and third given that there are no ties is $$P(8, 3) = \\frac{8!}{(8 - 3)!} = \\frac{8!}{5!} = \\frac{8 \\cdot 7 \\cdot 6 \\cdot 5!}{5!} = 8 \\cdot 7 \\cdot 6 = 336$$ Imagine if John, Jack and Julie were in the \"top-3\", there is 6 possible combination.Because you do not know what was the rank. Anyone can be at So $8\\choose3$ ways of selecting 3 people out of a group of 8, multiplied by no of ways of arranging those 3 in the ranks $3!$.", "# Permutations with conditions I'm trying to solve a question involving permutations with conditions. I was reading the following problem: Five runners competed in a race: Fred, George, Hermione, Lavender, and Ron. Fred beat George. Hermione beat Lavender. Lavender beat George. Ron beat George. Assuming there were no ties, how many possible finishing orders could there have been, given only this information? A 1 B 6 C 12 D 18 E 24 F 120 From the information given, I figured out that George (G) came last. So, the finishing order was betwen Fred (F), Hermione (H), Lavender (L) and Ron (R). Now I know from this list that there are $$4 \\times 3 \\times 2 \\times 1 = 24$$ ways to arrange the final order. But I can't seem to convert the information Hermione beat Lavender to find the final answer. I'm stuck. How do I get to the final answer, which is C? As you say, there are $24$ ways for George to finish last, and the valid permutations are simply those with George last and Hermione beating Lavender. Now exactly half of the permutations with George last have Hermione beating Lavender (group the permutations in pairs which differ by swapping H and L; each pair has one valid permutation and one invalid). So the answer is", "name for this special multiplication. It’s called a factorial, and it’s written like this: $4!$ This is not a very excited $4$. It’s just shorter way to write $4 \\cdot 3 \\cdot 2 \\cdot 1$. Factorials are very useful when counting things, and they’ll save us a lot of trouble in solving this problem. Here we go – Harry unwraps Super-Sam: First, there are $5$ choices for the first book. Then there are $4$ books remaining, so there are $4$ ways to pick the second book. Then there are $3$ ways to pick the third book, and $2$ ways to pick the fourth book, and then there’s only one book left for the last position, for a grand total of $5! = 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120$ ways for Harry to arrange his new books. Thank goodness we didn’t have to write out all those options. It turns out that we can follow this same process for any number of things that need to be ordered. If there are $n$ things, then there are $n$ ways to pick the first thing and $n-1$ ways to pick the second thing and so on, and when you multiply it all out you get a grand total of $n!$ ways to order them. That’s something", "out, given that there are 10 or so Dementors converging and they fly at about 5 metres per second. $\\$ Harry also encounters Dementors during a Quidditch match against Hufflepuff. Suppose Harry flies past a stationary Dementor, approaching from a great distance at speed $v$ and travelling along a straight line. Let the distance of closest approach to the Dementor be $b$. Question 4: Find Harry’s final happiness ($H(\\infty)$) having passed the Dementor, given the initial condition $H(-\\infty)=H_0$. Assuming Harry passes within $\\pi$ metres of the Dementor, how fast must Harry fly to avoid falling off his broom? $\\$ Hint: for questions 3 and 4, the key is find an expression for $r(t)$ in terms of $t$. Your equations may be slightly simpler if the time $t=0$ is chosen to be that at which Harry and the Dementor(s) are closest. You might find this post useful.", "3 for 3rd and 1 for every other team. Points for individuals will be totalled and the totals used to assign points for orders. Each order may submit up to 2 teams for CTF and Brawl events. This limit is due to timings. Teams may only be of mixed orders when those orders are allied. Teams should be grouped (so maximum of 6 players) with a combined GL below 2000. Orders may agree to admit that they are inferior and ally with another order. This agreement must be submitted to judges in advance of the competition by the senior of all orders involved. Allied orders are allied for all events. Hedges are deemed to be allied with their respective order. Individual hedges may opt out of being allied with the equivalent order before the start of the competition. ## Results Bulletin Boards: Wizardsguild Octathlon Final Results Note #14 by Mirodar at Mon Feb 20 16:06:19 2023 on board wizardsguild There was a slight arithmetic error when I was totalling up at the end of the event which means the final scores are as follow rather than as announced. 1st: Midnight 22 2nd: Last Order and Students, 21 points 3rd: 4th: Ancient and Truly Original Brothers of the Silver Star, 19 Points 5th: Hoodwinkers, 10 points 6th: Sages of", "in an election, or the finishing order in a race. Generally the keyword 'ways' indicates that combinations apply. 42. anonymous awesome thank you again!! 43. kropot72 You're welcome :)", "From the Chalkdust quiz: Are there more words in Harry Potter and the Order of the Phoenix or entries in the OEIS? • This month’s puzzle: Adapted from a recent MathsJam Shout: you have cards labelled 1, 2, 3, 4, 5 face down in a random order on the table. You predict the order of the cards before turning them over. What’s the probability of getting all five correct? Four? Three? Two? One? None? Next month is the Big MathsJam special. 1. A magazine for the mathematically curious The post Wrong, But Useful: Episode 61 appeared first on Flying Colours Maths. Articles marked as Favorite are saved for later viewing. • Show original • . • Share • . • Favorite • . • Email • .", "many different letter arrangements can you make if you only choose two at a time? First, choose 5 colors. Next, draw a diagram showing all of the different combinations. License: CC BY-NC 3.0 The answer is 10. There are 10 possible combinations of choosing 2 pens from 5. Credit: Rubbermaid Products License: CC BY-NC 3.0 Remember the moms packing snacks for the beach? Joanne and Laura are getting ready for a day at the beach. The women are packing the lunches to take with them and want to choose two of the five snacks they have available to put with the lunches. First, list the 5 snacks. Next, draw a diagram showing all of the different combinations. License: CC BY-NC 3.0 The answer is 10. There are 10 possible combinations of choosing 2 snacks from the 5 choices. ### Explore More Solve each combination. 1. At Dudley’s Dude Ranch there are 6 riders but only 4 horses. How many different ways can a group of 4 go out on ride? 2. With 4 laps to go, Dale Earnhardt Jr., Robbie Gordon, Kyle Busch, and Kasey Kahne are all in contention to win a NASCAR race. In how many different ways can the top three drivers finish? 3. Ace, King, Queen, Jack, Ten, and Nine of Clubs are face", "# There are 10 bicyclists entered in a race. In how many different orders could these 10 bicyclist finish? Apr 13, 2017 #### Answer: 10! is the answer. #### Explanation: this is just like you are given 10 lines on a paper and you have to arrange 10 names on those 10 lines in all different ways . so , starting with the lowermost line, you can put one of 10 names on that line and then on the line above that you can put 1 of 9 names and so on . so the total ways to arrange all the names on those lines in all ways will be : $10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1$ = 10!", "90 E. 120 We can create the following equation: total number of ways to finish the race = number of ways Meg finishes ahead of Bob + number of ways Meg does not finish ahead of Bob Since the total number of ways to complete the race is 5! = 120, and since there are an equal number of ways for Meg to finish ahead Bob as there are for her not to finish ahead of Bob, Meg can finish ahead of Bob in 60 ways. Alternate Solution: If Meg comes in the first, then Bob can finish the race in any of the remaining positions, so there are 4! = 24 ways this could happen. If Meg comes in the second, there are 3 choices for the first spot (since Bob can’t finish ahead of Meg) and the last 3 spots can be filled in 3! ways; so there are 3 x 3! = 3 x 6 = 18 ways this could happen. If Meg comes in the third, there are 3 choices for the first spot, 2 choices for the second spot, 2 choices for the fourth spot, and 1 choice for the last spot; so, this could happen in 3 x 2 x 2 = 12 ways. Finally, if Meg comes in the fourth, then", "platters can a customer create? Kate and Chad are planning their wedding dinner and must select 3 of 12 entrees and 2 of 3 desserts for their guests to be able to choose from. How many different combinations of offerings are possible?_____ For her Halloween display, Margaret plans to arrange a row of alternating witch and ghost figurines. The row must begin with a witch figurine and end with a ghost figurine. Margaret plans to purchase either three of each type of figurine or four of each type of figurine, and each figurine will look unique. Depending on how many figurines she purchases, which of the following could be the number of ways that she could arrange her display?Indicate all possible values. There are six cars in a motorcade. How many different arrangements of cars in the motorcade are possible? Mark can take three friends with him on a vacation and is listing the possible combinations of friends. If he has five friends to choose from and is numbering each possible combination sequentially beginning with 1, which of the following numbers will appear on his list of combinations?Indicate all possible values. Twelve runners enter a race to compete for first, second, and third place. How many different combinations of winners are possible?_____ Sherry supervises a crew of maintenance", "Monk and Order of Phoenix Tag(s): ## Binary Search, Data Structures, Medium, Stacks Problem Editorial Analytics Voldemort has a big army, so he has maintained his people in $N$ rows to fight Harry's army. Each row contains the heights of the fighters and is sorted in non-decreasing order from the start to end, except for the first row, which may contain the heights of the fighters in any arbitary order, as it contains all the legendry fighters. During the war, at any time, Voldemort can remove a fighter from any row, and can also add any new fighter to any row (maintaining the non-decreasing order of heights. except in the first row). Note: 1. Voldemort will never remove any fighter from an empty row. 2. Voldemort can only remove or add a fighter from/to the end of row. Now Harry has a special wand which can kill exactly $N$ fighters in one go, but with following conditions: 1. There should be exactly $N$ fighters, and exactly one fighter (which can be anyone in the row) should be chosen from each row. 2. The first fighter can only be chosen from the first row, the second one from second row, and so on. Basically the $i^{th}$ fighter should be chosen from $i^{th}$ the row, where $i$ ranges from $1$", "# On Choosing Five-year-old Harry and his family are going on a road trip. “Bring something to read on the drive, Harry,” his mother tells him. Harry is ecstatic – that means he gets to bring his rabbit books! He fetches his 12 favorite books about rabbits. His mother is not excited about the prospect of lugging around all those books for the whole trip. “Not all of them, Harry,” she tells him. “You can bring four.” Four!? Harry has a conundrum on his hands. How many ways can he choose four to bring for the drive? The first thing we might notice about this problem is that it’s not asking for a sequence of books, where the order of the books matters. In this problem, it doesn’t matter what order the books are in. But this notion of selecting looks rather like our thought process from the factorials page. So we might start by saying, “Okay, we can still use the ordering strategy from the factorials page but let’s stop when we get past the first four. There are $12$ ways to pick the first book, and then there are $11$ books left and so $11$ ways to pick the second book, and then there are $10$ books left and so $10$ ways to pick the third", "$$A_i \\cap A_j = \\emptyset, i\\neq j$$) then $\\textrm{P}(A_1 \\cup A_2 \\cup \\cdots |E) = \\textrm{P}(A_1|E) + \\textrm{P}(A_2|E) + \\cdots$ • $$\\textrm{P}(A^c|E) = 1-\\textrm{P}(A|E)$$. (Be careful! Do not confuse $$\\textrm{P}(A^c|E)$$ with $$\\textrm{P}(A|E^c)$$.) ### 3.1.7 Conditional versus unconditional probability Example 2.62 Consider a group of 5 people: Harry, Bella, Frodo, Anakin, Katniss. Suppose each of their names is written on a slip of paper and the 5 slips of paper are placed into a hat. The papers are mixed up and 2 are pulled out, one after the other without replacement. 1. What is the probability that Harry is the first name selected? 2. What is the probability that Harry is the second name selected? 3. If you were asked question (2) before question (1), would your answer change? Should it? 4. If Bella is the first name selected, what is the probability that Harry is the second name selected? 5. If Harry is the first name selected, what is the probability that Harry is the second name selected? 6. How is the probability that Harry is the second name selected related to the probabilities in the two previous parts? 7. If Bella is the second name selected, what is the probability that Harry was the first name selected? Solution to Example 2.62 1. The probability that Harry is", "first student is sent to Hufflepuff. The second disciple is given the choice between the houses where the least number of students has been sent, i.e. Gryffindor, Slytherin and Ravenclaw. If he chooses Gryffindor, Hermione is forced to choose between Ravenclaw and Slytherin, if he chooses Ravenclaw, Hermione will choose between Gryffindor and Slytherin, if he chooses Slytherin, Hermione will choose between Gryffindor and Ravenclaw. In the end, the following situation is possible (it depends on the choice of the second student and Hermione). Hermione will end up 1) in Gryffindor, 2) in Ravenclaw, 3) in Slytherin. Note that, despite the fact that in neither case Hermione will be given a choice between all the three options, they are all possible and they should all be printed in the answer. Hermione will not, under any circumstances, end up in Hufflepuff.", "animal knows that if he nominates himself, there will only be one left, and that one will definitely nominate himself. So neither will nominate himself. 3. Three animals left: each animal knows that if he nominates himself, there will be two animals left, and neither will nominate himself. So the first animal to be asked will nominate himself. 4. Four animals left: each animal knows that if he nominates himself, there will be three animals left, and one will nominate himself. So none will nominate himself. So using backward induction, if there are $n$ animals left, the strategy taken will be: • If $n$ is even, no one will nominate himself. • If $n$ is odd, the first one asked will nominate himself. When Duke Froggington II takes his turn, there are 73 animals left (including himself). As he is a perfect logician, he will nominate himself as the guest of honor. No one else will nominate himself after him. So poor Mr. Bun gets sent to the toy box, while everyone else enjoys the tea party and Duke Froggington II hogs the strawberry pastry. • Great answer. I now know how to respond if I ever get invited to a tea party with only two strawberry pastries in it. – Mark Gabriel May 18 '15 at 2:44", "from $1$. For the probability of no match, imagine that Harry drew $2$ cards, and wrote down what they were. Now Ron draws $2$ cards. There are $\\binom{26}{2}$ ways for Ron to draw $2$ cards from $26$, all equally likely. There are $\\binom{24}{2}$ ways to draw $2$ cards that are both different from what Harry wrote down. So the probability that neither of Ron's cards matches one of Harry's is $$\\frac{\\binom{24}{2}}{\\binom{26}{2}}.$$ The probability of $2$ matches is simpler to find, since only $1$ pair from the $\\binom{26}{2}$ is the same as the pair Harry wrote down. For fun, we will write $\\binom{2}{2}$ instead of $1$. Thus the probability of no match or two matches is $$\\frac{\\binom{24}{2}+\\binom{2}{2}}{\\binom{26}{2}}.$$ Calculate. We get $\\frac{277}{325}$. Thus the probability of exactly $1$ match is $\\frac{48}{325}$. Or else, more simply, we can calculate directly. There are $\\binom{2}{1}$ ways to choose the card that will match one of Harry's. For each such choice, there are $\\binom{24}{1}$ ways to choose the card that will not match any of Harry's. So the required probability is $$\\frac{\\binom{2}{1}\\binom{24}{1}}{\\binom{26}{2}}.$$ - Ron must choose one card from the two cards drawn by Harry ($2$ possibilities) and one card from the $n-2=24$ cards not drawn by Harry ($n-2$ possibilities) and their order ($2$ possibilities) hence the probability is $2\\cdot2\\cdot(n-2)\\cdot(n(n-1))^{-1}$. - $n=26$ here, isn't?", "1🐎︎ We can eliminate one more horse with the “slower-than-3” rule: the 3rd orange horse. It was beaten by the 2nd and 1st orange horses, which in turn were slower than the 1st red horse. So the 3rd orange horse is eliminated: 1: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 2: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 3: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 4: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ 5: 5🐎︎ 4🐎︎ 3🐎︎ 2🐎︎ 1🐎︎ Notice that we only have 5 remaining horses to sort now, and only the 2nd and 3rd overall positions to find. We just run a race with those 5 remaining horses to find out: 3🐎︎ 2🐎︎ 1🐎︎ 2🐎︎ 1🐎︎ As it turns out in this example, the 1st purple horse and 2nd red horse take the top two positions in this race. That means the final top three are the 1st red, 1st purple and 2nd red horse: 🐎︎ 🐎︎ 🐎︎ This takes a total of 7 races: the initial 5 groups, the fastest-of-fastest race, and the surviving runners up race at the end.", "$\\underbrace{b+b+\\cdots+b}_{a}=a b$ possibilities. 1.4.2. It is often easier to think about the experiments as being in chronological order, but there is no requirement in the multiplication rule that Experiment A has to be performed before Experiment $\\mathrm{B}$. Example 1.4.3 (Runners). Suppose that 10 people are running a race. Assume that ties are not possible and that all 10 will complete the race, so there will be welldefined first place, second place, and third place winners. How many possibilities are there for the first, second, and third place winners? Solution: There are 10 possibilities for who gets first place, then once that is fixed there are 9 possibilities for who gets second place, and once these are both fixed there are 8 possibilities for third place. So by the multiplication rule, there are $10 \\cdot 9 \\cdot 8=720$ possibilities. We did not have to consider the first place winner first. We could just as well have said that there are 10 possibilities for who got third place, then once that is fixed there are 9 possibilities for second place, and once those are both fixed there are 8 possibilities for first place. Or imagine that there are 3 platforms, which the first, second, and third place runners will stand on after the race. The platforms are gold, silver, and", "× Back # Find the Order Three racing snails named Molly, Nick and Orlando have a fierce competition in the 10 centimeter run. Nick and Molly did not place next to each other. Who came in $$2^\\text{nd}$$ place? How many ways are there to arrange the three cards above in a row so the 7 is to the left of the 8? (The 7 may be one or two cards away from the 8.) Azizi, Blanche, Clio, and Daaiyah had a hot-air balloon race. Based on the clues below, who placed $$1^\\text{st} ?$$ 1. Azizi beat at least one of the competitors. 2. Both Blanche and Clio each beat at least two of the competitors. 3. Azizi and Clio did not finish next to each other. Azizi, Blanche, Clio, and Daaiyah had another race, this time in Formula One cars. Based on the clues below, who placed $$1^\\text{st} ?$$ 1. None of the racers finished in the same place as their names when listed using alphabetical order; that is, Azizi didn't finish $$1^\\text{st} ,$$ Blanche didn't finish $$2^\\text{nd} ,$$ Clio didn't finish $$3^\\text{rd} ,$$ and Daaiyah didn't finish $$4^\\text{th} .$$ 2. Blanche didn't finish two places apart from Daaiyah. 3. Azizi finished ahead of Daaiyah. The ace of diamonds, king of hearts, queen of clubs and joker from" ]

[ "db \"eASSWzRD<11$\"", "can run the same method to compute $$a\\pmod{11!}$$.", "+0 # Help. I have been stuck 0 114 4 Given that $$\\binom{15}{8}=6435$$$$\\binom{16}{9}=11440$$, and $$\\binom{16}{10}=8008$$, find $$\\binom{15}{10}$$. I know that I should be able to do this but I can't figure it out. Aug 10, 2019 #1 +23295 +3 Given that $$\\dbinom{15}{8}=6435$$, $$\\dbinom{16}{9}=11440$$ , and $$\\dbinom{16}{10}=8008$$, find $$\\dbinom{15}{10}$$. Formula: $$\\dbinom{n}{k} + \\dbinom{n}{k+1} = \\dbinom{n+1}{k+1}$$ $$\\begin{array}{|lrcll|} \\hline & \\mathbf{\\dbinom{n}{k} + \\dbinom{n}{k+1}} &=& \\mathbf{\\dbinom{n+1}{k+1}} \\\\\\\\ (1) & \\dbinom{15}{8} + \\dbinom{15}{9} &=& \\dbinom{16}{9} \\\\ (2) & \\dbinom{15}{9} + \\dbinom{15}{10} &=& \\dbinom{16}{10} \\\\ \\hline (1)-(2): & \\dbinom{15}{8} + \\dbinom{15}{9} -\\left( \\dbinom{15}{9} + \\dbinom{15}{10}\\right) &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & \\dbinom{15}{8} - \\dbinom{15}{10} &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & 6435 - \\dbinom{15}{10} &=& 11440-11440 \\\\ & 6435 - \\dbinom{15}{10} &=& 3432 \\\\ & \\dbinom{15}{10} &=& 6435 - 3432 \\\\ & \\mathbf{\\dbinom{15}{10}} &=& \\mathbf{3003} \\\\ \\hline \\end{array}$$ Aug 11, 2019 #2 0 Thank you so much!! I can't keep all the formulas in my head sometimes. Aug 11, 2019 #3 +1655 +3 There are really too many formulas... tommarvoloriddle Aug 12, 2019 #4 +105540 +1 You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from. There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level. Most times this does", "this algorithm is $$(^{n}_{10})*n = O(n^{11})$$.", "(\\mathbf{A}\\mathbf{B})^{-1} \\dbinom12 = \\dbinom13 } \\\\ \\hline \\end{array}$$ Apr 17, 2020", "22 }{ 2 }$ ⇒ x =11 Thus, I am 11.", "\\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} & 360 \\cdot x^{44} & 360 \\cdot x^{45} & 120 \\cdot x^{46}\\\\ & & & &", "for $R_{11}$ and $R_{22}$.", "- \\sqrt{14}$", "Combo Algebra Compute $\\dbinom{100}{1} + 2\\dbinom{100}{2} +\\ldots + 100\\dbinom{100}{100}.$ If your answer is $$k \\cdot 2^n$$, where $$n$$ and $$k$$ are positive integers and $$k$$ is odd, find $$n+k$$. ×", "{10}^{-} 11$", "\\mathbf{d}$$...", "e n}{11 - 2 \\sqrt{30}}$", "-1 \\\\ k&=& 3 \\\\ \\dfrac{\\dbinom{14}{3}} {\\dbinom{14}{4}} &=& \\dfrac{4}{11} \\\\ \\hline \\end{array}$$ Mar 26, 2019", "let $N_1=N-\\dbinom{x_1}k$.\" Did you mean $N_1=N-\\dbinom{x_k}{k}$? That would be consistent with what you wrote in your next paragraph: $N_1=\\dbinom{x_{k-1}}{k-1}+\\ldots+\\dbinom{x_1}{1}$ – tychicus Aug 6 '13 at 7:21 • @tychicus: Yes, I did; I’ve fixed it now. (The same mistake was in the original version of your hint, and I must have unconsciously copied that.) – Brian M. Scott Aug 6 '13 at 12:02", "1-D Numerical Method2012 June;15(3)", ")^{10} \\\\\\\\ && ( a + b + c )^{n} \\\\ &=& \\sum \\limits_{k=0}^{n} \\sum \\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} &", "14\\right)}^{2}$ I hope this was helpful. :)", "15)\\ dt$", "## Intermediate Algebra for College Students (7th Edition) $x^{10}-5x^8+10x^6-10x^4+5x^2-1$ Apply Binomial Theorem. $(a+b)^n=\\sum_{r=0}^n\\dbinom{n}{r}(a)^{n-r}b^r$ $(x^2-1)^5=\\dbinom{5}{0}(x^2)^5(-1)^0+\\dbinom{5}{1}(x^2)^4(-1)^1+\\dbinom{5}{2}(x^2)^3(-1)^2+\\dbinom{5}{3}(x^2)^2(-1)^3+\\dbinom{5}{4}(x^2)^1(-1)^4+\\dbinom{5}{5}(x^2)^0(-1)^5$ or, $=x^{10}-5x^8+10x^6+10x^4(-1)+5x^2+(1)(1)(-1)$ or, $=x^{10}-5x^8+10x^6-10x^4+5x^2-1$" ]

[ "quartets, but not the 13th: $\\dbinom{13}{12}\\dbinom{n}{12} \\left( \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{12} \\left(1 - \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{n-12}$ Then the probability that the next pull is the 13th quartet is: $\\dfrac{ \\dbinom{5}{4} }{ \\dbinom{52}{5} }$ So, the expected value is given by: \\displaystyle \\begin{align*} & \\sum_{n \\ge 12}\\dbinom{13}{12}\\dbinom{n}{12} \\left( \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{13} \\left(1 - \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{n-12}(n+1) \\\\ = & \\dbinom{13}{12} \\left( \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{13} \\sum_{n \\ge 12} \\dbinom{n}{12} \\left(1 - \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{n-12}(n+1) = 1830101\\end{align*} http://www.wolframalpha.com/input/?i...infinity%7D%5D Edit: I missed a factor of 48. 3. ## Re: Expectancy of number of trials I also got this expression but i suppose we have to multiply by 12! for permutations on the 12 quarters.For the same reason the probability for a quarter in one trial is 13*bin(5,4)*4!*(52-4)/(bin(52,5)*5!). Do you see a way of computing the expectancy by using indicators? 4. ## Re: Expectancy of number of trials Originally Posted by hedi I also got this expression but i suppose we have to multiply by 12! for permutations on the 12 quarters.For the same reason the probability for a quarter in one trial is 13*bin(5,4)*4!*(52-4)/(bin(52,5)*5!). Do you see a way of computing the expectancy by using indicators? I apologize. I missed a factor of 48 for", "h &= \\dfrac{4 \\pi^2Ed^4}{4 \\times 64L^2 \\times D_i^2\\times \\rho g}\\\\ h &= \\dfrac{4 \\pi^2 \\times 200 \\times 10^9 \\times (0.075)^4}{4 \\times 64 \\times \\times 4^2 \\times 1.5^2 \\times 10^4}\\\\ h &= 2.7\\ m\\end{aligned}", "10^7 \\\\ d &= 31.134 \\times 10^6 - 6.378 \\times 10^6 \\\\ d \\ &\\boxed{ = 24.756 \\times 10^6 } \\ \\rm m \\end{align*}", "( \\dfrac{1}{T_2^{2/3}}-\\dfrac{1}{T_1^{2/3}} \\right )\\\\ &=-\\left ( 2\\pi \\times 6.67\\times 10^{-11}\\times 6\\times 10^{24}\\right )^{2/3}\\times 50\\times \\left ( \\dfrac{1}{\\left ( 10.8\\times 10^{3} \\right )^{2/3}}-\\dfrac{1}{\\left (7.2\\times 10^{3} \\right )^{2/3}} \\right )\\\\ &=-\\left ( 2.51\\times 10^{15}\\right )^{2/3}\\times 50\\times \\left ( \\dfrac{1}{488}-\\dfrac{1}{373} \\right )\\\\ &=-1.36\\times 10^{10}\\times 50\\times \\left ( -6.32 \\times 10^{-4}\\right )\\\\ &=\\boxed{4.3\\times 10^{6}\\;\\rm J} \\end{align*} {/eq}", "\\times 17 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 23 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{54} \\times p_{19}(29,25) &=& \\dfrac{29}{54} \\times \\dfrac{2^4 \\times 3^2 \\times 11^2 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^3 \\times 11^2 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} \\\\ \\dfrac{29}{55} \\times p_{20}(29,26) &=& \\dfrac{29}{55} \\times \\dfrac{2^5 \\times 5^3 \\times 11 \\times 17 \\times 19 \\times 23}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^5 \\times 5^2 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{56} \\times p_{21}(29,27) &=& \\dfrac{29}{56} \\times \\dfrac{2^4 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23}{3 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ \\dfrac{29}{57} \\times p_{22}(29,28) &=& \\dfrac{29}{57} \\times \\dfrac{2^2 \\times 3^2 \\times 5^2 \\times 11 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times", "47} &=& \\dfrac{3 \\times 5^2 \\times 11 \\times 13 \\times 17 \\times 29}{19 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{49} \\times p_{14}(29,20) &=& \\dfrac{29}{49} \\times \\dfrac{2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17}{37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11 \\times 13 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{50} \\times p_{15}(29,21) &=& \\dfrac{29}{50} \\times \\dfrac{2^2 \\times 3 \\times 5^3 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11 \\times 13 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{51} \\times p_{16}(29,22) &=& \\dfrac{29}{51} \\times \\dfrac{2^5 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^5 \\times 3 \\times 5 \\times 11 \\times 13 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{52} \\times p_{17}(29,23) &=& \\dfrac{29}{52} \\times \\dfrac{2^3 \\times 3 \\times 5 \\times 11^2 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 29}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{53} \\times p_{18}(29,24) &=& \\dfrac{29}{53} \\times \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2", "be a2 (in checkerboard coordinates) because $$b2/c2 \\ge 2$$ and at least two of those marked $$\\diamond$$ and at least one of those marked $$\\ast$$. Because only 8 small numbers are available a4 must be small and one of b4,c4. All others must be $$>8$$ (\"large\", marked $$\\Box$$): $$\\begin{matrix}\\bigcirc&+&\\Box\\bigcirc&+&\\Box\\bigcirc&=&\\Box\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&\\bigcirc&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&\\bigcirc&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&\\Box&=&\\Box\\end{matrix}$$ and we see that a4 must be $$1$$ Observe that $$d4 = d3/d2 + a1/b1 + c2\\times c3 + c4 \\ge 2 + 2 + 2\\times 3 + c4$$ Therefore c4 must be small and b4 large. $$\\begin{matrix}1&+&\\Box&+&\\bigcirc&=&13\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&\\bigcirc&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&\\bigcirc&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&9\\ldots 12&=&11\\ldots 14\\end{matrix}$$ Next, we see that $$2$$ must be in either c2 or c3: $$\\begin{matrix}1&+&9\\ldots 11&+&3\\ldots 6&=&14\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&2\\ldots 4&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&2\\ldots 4&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&10\\ldots 12&=&12\\ldots 14\\end{matrix}$$ We are almost done! Observe that in the products $$a2\\times b2 = c2\\times d2$$ the primes $$5$$ and $$7$$ cannot occur because they are available only once. $$\\begin{matrix}1&+&9\\ldots 10&+&5&=&15\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&2\\ldots 3&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&2\\ldots 3&=&\\bigcirc\\\\=&&=&&=&&=\\\\14&/&7&+&11&=&13\\end{matrix}$$ We can finish by filling in column a, $$\\begin{matrix}1&+&9&+&5&=&15\\\\\\times&&-&&+&&-\\\\10&+&\\Box&/&2\\ldots 3&=&\\Box\\\\+&&/&&\\times&&/\\\\4&\\times&\\bigcirc&/&2\\ldots 3&=&\\bigcirc\\\\=&&=&&=&&=\\\\14&/&7&+&11&=&13\\end{matrix}$$ row 3 and, finally, row 2. • Very well done and great explanation! I was getting worried that no one will solve this... – Dmitry Kamenetsky Oct 10 '20 at 0:46 Let the digits be a1,a2,a3,… d4, using standard algebraic chess notation (i.e. letters = columns, numbers = rows, a1 = lower left). Define a low digit = 1-8, high digit = 9-16. The cells,", "\\times 41 \\times 43 \\times 47} &=& \\dfrac{2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17}{37 \\times 41 \\times 43 \\times 47}\\\\ p_{15}(29,21) &=& \\dfrac{20 \\times 15}{49 \\times 6} \\times \\dfrac{2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17}{37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^2 \\times 3 \\times 5^3 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ p_{16}(29,22) &=& \\dfrac{21 \\times 16}{50 \\times 7} \\times \\dfrac{2^2 \\times 3 \\times 5^3 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^5 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ p_{17}(29,23) &=& \\dfrac{22 \\times 17}{51 \\times 8} \\times \\dfrac{2^5 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^3 \\times 3 \\times 5 \\times 11^2 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ p_{18}(29,24) &=& \\dfrac{23 \\times 18}{52 \\times 9} \\times \\dfrac{2^3 \\times 3 \\times 5 \\times 11^2 \\times 13 \\times 17}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ p_{19}(29,25) &=& \\dfrac{24 \\times", "10^4\\times 9.46\\times 10^{15})^{1.5}}{\\sqrt{6.67\\times10^{-11}\\times 8\\times 10^{13}\\times 1.99\\times 10^{30}}}\\\\\\\\ &=8.25\\times 10^{14}\\, s\\\\\\\\ &=\\boxed{2.61\\times 10^7\\, yr} \\end{align} {/eq}", "19}{53 \\times 10} \\times \\dfrac{2^2 \\times 3 \\times 5 \\times 11^2 \\times 17 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47} &=& \\dfrac{2^4 \\times 3^2 \\times 11^2 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ p_{20}(29,26) &=& \\dfrac{25 \\times 20}{54 \\times 11} \\times \\dfrac{2^4 \\times 3^2 \\times 11^2 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^5 \\times 5^3 \\times 11 \\times 17 \\times 19 \\times 23}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ p_{21}(29,27) &=& \\dfrac{26 \\times 21}{55 \\times 12} \\times \\dfrac{2^5 \\times 5^3 \\times 11 \\times 17 \\times 19 \\times 23}{3 \\times 7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^4 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23}{3 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ p_{22}(29,28) &=& \\dfrac{27 \\times 22}{56 \\times 13} \\times \\dfrac{2^4 \\times 5^2 \\times 13 \\times 17 \\times 19 \\times 23}{3 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^2 \\times 3^2 \\times 5^2 \\times 11 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53}\\\\ p_{23}(29,29) &=& \\dfrac{28 \\times 23}{57 \\times 14} \\times \\dfrac{2^2 \\times 3^2 \\times 5^2 \\times", "11 \\times 17 \\times 19 \\times 23}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{2^3 \\times 3 \\times 5^2 \\times 11 \\times 17 \\times 23^2}{7 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} \\end{array}$ Multiplying these by the probabilities that Sir R----- should have drawn a black token at the next turn yields $\\begin{array}{lclcl} \\dfrac{29}{44} \\times p_9(29,15) &=& \\dfrac{29}{44} \\times \\dfrac{2^{2} \\times 3 \\times 5 \\times 11 \\times 23}{19\\times 37 \\times 41 \\times 43} &=& \\dfrac{3 \\times 5 \\times 23 \\times 29}{19\\times 37 \\times 41 \\times 43}\\\\ \\dfrac{29}{45} \\times p_{10}(29,16) &=& \\dfrac{29}{45} \\times \\dfrac{2 \\times 3^2 \\times 5^3 \\times 23}{19\\times 37 \\times 41 \\times 43} &=& \\dfrac{2 \\times 5^2 \\times 23 \\times 29}{19\\times 37 \\times 41 \\times 43}\\\\ \\dfrac{29}{46} \\times p_{11}(29,17) &=& \\dfrac{29}{46} \\times \\dfrac{2^4 \\times 5^2 \\times 11 \\times 23}{19 \\times 37 \\times 41 \\times 43} &=& \\dfrac{2^3 \\times 5^2 \\times 11 \\times 29}{19 \\times 37 \\times 41 \\times 43}\\\\ \\dfrac{29}{47} \\times p_{12}(29,18) &=& \\dfrac{29}{47} \\times \\dfrac{2^5 \\times 5^2 \\times 11 \\times 17}{19 \\times 37 \\times 41 \\times 43} &=& \\dfrac{2^5 \\times 5^2 \\times 11 \\times 17 \\times 29}{19 \\times 37 \\times 41 \\times 43 \\times 47}\\\\ \\dfrac{29}{48} \\times p_{13}(29,19) &=& \\dfrac{29}{48} \\times \\dfrac{2^4 \\times 3^2 \\times 5^2 \\times 11 \\times 13 \\times 17}{19 \\times 37 \\times 41 \\times 43 \\times", "2\\dbinom 21}{\\dbinom 52}=\\dfrac{3}{5}$$", "last row right. This works well, if one has an expression of the type a+b=c+d+e+f+g+h=j, where there's only one row between initial row and result. However, in my case, I have some mathematical transformation (meaning an additional row) and so I need the equal sign of the row: \"12\\times6+14\\times\\dfrac{48}{7}-6^2-\\left(\\dfrac{2}{7}-4\\right)^2-\\left(\\dfrac{48}{7}\\right)^2 -\\dfrac{2}{7}\\times\\dfrac{48}{7}-\\left(\\dfrac{2}{7}\\right)^2\" to be aligned under the first equal sign. So now I use this version: Code: Select all • Open in writeLaTeX \\begin{align*}\\pi^{M}(s^{M},f^{M},x^{M}) :=& 12s^{M}+14f^{M}-(s^{M})^2-(x^{M}-4)^2-(f^{M})^2\\\\ &-x^{M}f^{M}-(x^{M})^2\\pause\\\\ =&12\\times6+14\\times\\dfrac{48}{7}-6^2-\\left(\\dfrac{2}{7}-4\\right)^2-\\left(\\dfrac{48}{7}\\right)^2\\\\ &-\\dfrac{2}{7}\\times\\dfrac{48}{7}-\\left(\\dfrac{2}{7}\\right)^2\\pause\\\\ =&69,14\\pause\\end{align*} Does anybody have better ideas? Thank you avp3000 Posts: 49 Joined: Thu Nov 15th, 2007 ### Re: Too long equation You can try the split environment, which is also provided by the amsmath package. To get a proper alignment, there are some tricky things to do. Code: Select all • Open in writeLaTeX \\begin{align*} \\pi^{M}(s^{M},f^{M},x^{M}): \\begin{split} &= 12s^{M}+14f^{M}-(s^{M})^2-(x^{M}-4)^2 \\\\ &\\quad -(f^{M})^2-x^{M}f^{M}-(x^{M})^2 \\end{split}\\\\ \\begin{split} &= 12\\times 6+14\\times\\frac{48}{7}-6^2-\\left(\\frac{2}{7}-4\\right)^2 \\\\ &\\quad -\\left(\\frac{48}{7}\\right)^2-\\frac{2}{7}\\times\\frac{48}{7}-\\left(\\frac{2}{7}\\right)^2 \\end{split}\\\\ &=69,14\\end{align*} Note that the alignment as known from the align environment has to be put into the split environment. More hints about typesetting mathematical expressions are given in \"Math mode\". Best regards Thorsten LaTeX Community Moderator ¹ System: openSUSE 13.1 (Linux 3.11.10), TeX Live 2013 (vanilla), TeXworks 0.5 (r1351) ² Posting stopped indefinitely due to offenses localghost Site Moderator Posts: 9219 Joined: Fri Feb 2nd, 2007 Location: Braunschweig, Germany ###", "20 }\\!\\!\\times\\!\\!\\text{ }\\dfrac{\\text{18}}{\\text{60 }\\!\\!\\times\\!\\!\\text{ 60}}\\text{ km = }\\dfrac{\\text{20 }\\!\\!\\times\\!\\!\\text{ 18}}{\\text{60 }\\!\\!\\times\\!\\!\\text{ 60}}\\times \\text{ 1000m} \\\\ \\end{align} d = 100m.", "7 \\times 19 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{4,828,639,200}{64,833,677,979}\\\\ \\dfrac{29}{48} \\times p_{13}(29,19) &=& \\dfrac{3^2 \\times 5^2 \\times 7 \\times 11 \\times 13 \\times 17 \\times 29 \\times 53}{3 \\times 7 \\times 19 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{5,884,904,025}{64,833,677,979}\\\\ \\dfrac{29}{49} \\times p_{14}(29,20) &=& \\dfrac{2 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 17 \\times 19 \\times 29 \\times 53}{3 \\times 7 \\times 19 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{6,389,324,370}{64,833,677,979}\\\\ \\dfrac{29}{50} \\times p_{15}(29,21) &=& \\dfrac{2 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 17 \\times 19 \\times 29 \\times 53}{3 \\times 7 \\times 19 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{6,389,324,370}{64,833,677,979}\\\\ \\dfrac{29}{51} \\times p_{16}(29,22) &=& \\dfrac{2^5 \\times 3^2 \\times 5 \\times 11 \\times 13 \\times 19 \\times 29 \\times 53}{3 \\times 7 \\times 19 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{6,013,481,760}{64,833,677,979}\\\\ \\dfrac{29}{52} \\times p_{17}(29,23) &=& \\dfrac{2 \\times 3^2 \\times 5 \\times 11^2 \\times 17 \\times 19 \\times 29 \\times 53}{3 \\times 7 \\times 19 \\times 37 \\times 41 \\times 43 \\times 47 \\times 53} &=& \\dfrac{5,406,351,390}{64,833,677,979}\\\\ \\dfrac{29}{53} \\times p_{18}(29,24) &=& \\dfrac{2^2 \\times 3^2 \\times 5 \\times 11^2 \\times 17 \\times 19 \\times 23 \\times 29}{3 \\times 7 \\times 19 \\times", "n-2 \\right)...(n-r)!$ \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5!}\\times \\left( \\dfrac{8\\times 7\\times 6}{3!} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6!}\\times \\left( \\dfrac{8\\times 7}{2!} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4!}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4!} \\right) \\\\ \\end{align} \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6}{3\\times 2\\times 1} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6\\times 5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7}{2\\times 1} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4\\times 3\\times 2\\times 1} \\right) \\\\ \\end{align} By cancelling the common terms in denominator and numerator we get $=\\left( 11\\times 9\\times 8 \\right)\\times \\left( 8\\times 7 \\right)+\\left( 11\\times 3\\times 4\\times 7 \\right)\\times \\left( 4\\times 7 \\right)+\\left( 11\\times 5\\times 9 \\right)\\times \\left( 5\\times 2\\times 7 \\right)$ $=792\\times 56+924\\times 28+495\\times 70$ = 44352 + 25872 + 34650 = 104874 Thus, the number of ways of selecting 10 students chosen for the competition = 104874 Note: The possibility of mistake can be the calculation mistake since the procedure of solving involves large calculations. Hence be very careful in simplifying and cancelling the terms to get the desired result.", "\\dfrac{\\dbinom{26}{2}\\dbinom{26}{1}}{\\dbinom{52}{3}} = \\dfrac{13}{17}$$ Rom Dec 31, 2018", "he received from B Jul 4, 2021 at 3:53 • I made an edit with the clarifications Jul 4, 2021 at 4:06 Here's my solution based on expectations. First, we calculate the expected value of the smallest, second-smallest, third-smallest... number on the 5 cards drawn. The calculation steps are shown below: $$E(\\text{smallest})=\\dfrac{1\\times \\dbinom{9}{4}+2\\times \\dbinom{8}{4}+\\cdots+6\\times \\dbinom{4}{4}}{\\dbinom{10}{5}}=\\dfrac{11}{6}.$$ Explanation: the numerator basically sums the smallest terms in all 5-card draws, and the denominator is the total number of cases. Similarly, we have $$E(\\text{second-smallest})=\\dfrac{2\\times \\dbinom{8}{3}\\times \\dbinom{1}{1}+3\\times \\dbinom{7}{3}\\times \\dbinom{2}{1}+\\cdots+7\\times \\dbinom{3}{3}\\times \\dbinom{6}{1}}{\\dbinom{10}{5}}=\\dfrac{11}{3};$$ $$E(\\text{third-smallest})=\\dfrac{3\\times \\dbinom{7}{2}\\times \\dbinom{2}{2}+4\\times \\dbinom{6}{2}\\times \\dbinom{3}{2}+\\cdots+8\\times \\dbinom{2}{2}\\times \\dbinom{7}{2}}{\\dbinom{10}{5}}=\\dfrac{11}{2};$$ $$E(\\text{fourth-smallest})=\\dfrac{4\\times \\dbinom{6}{1}\\times \\dbinom{3}{3}+5\\times \\dbinom{5}{1}\\times \\dbinom{4}{3}+\\cdots+9\\times \\dbinom{1}{1}\\times \\dbinom{8}{3}}{\\dbinom{10}{5}}=\\dfrac{22}{3};$$ $$E(\\text{fifth-smallest, or largest})=\\dfrac{5\\times \\dbinom{4}{4}+6\\times \\dbinom{5}{4}+\\cdots+10\\times \\dbinom{9}{4}}{\\dbinom{10}{5}}=\\dfrac{55}{6}.$$ As a quick double check, adding them all together yields $$27.5$$, which is the expected sum of the 5 cards. Next, let's think about A's strategy. In essence, he is discarding his $$N$$ smallest cards to get random cards. Thus, if the expected value of cards that he is discarding is smaller than the expected value of cards he is getting, then A is benefitting. The more A benefits, the higher his chances of winning. Then, clearly $$E(\\text{smallest})$$, $$E(\\text{second-smallest})$$ are smaller than $$5.5$$, the expected value on a random card. $$E(\\text{third-smallest})$$ is equal to $$5.5$$, and $$E(\\text{fourth-smallest})$$, $$E(\\text{fifth-smallest, or largest})$$ are greater than $$5.5$$. This means that discarding the smallest card and", "=9\\times \\left[ 4{{R}^{2}}-\\pi {{R}^{2}} \\right] \\\\ & =9\\times \\left( 4-\\pi \\right)\\times {{R}^{2}} \\\\ & =9\\times \\left( 4-3.14 \\right)\\times {{\\left( 7 \\right)}^{2}} \\\\ & =9\\times 0.86\\times 49 \\\\ & =397.26c{{m}^{2}} \\\\ \\end{align}", "### A result using complex numbers Suppose we expand out the expressions $$(1+1)^n$$, $$(1+i)^n$$, $$(1-1)^n$$, $$(1-i)^n$$ using the binomial theorem. This gives (for $$n>0$$), \\begin{align*} \\dbinom{n}{0} + \\dbinom{n}{1} + \\dbinom{n}{2} + \\dbinom{n}{3} + \\dots + \\dbinom{n}{n} &= 2^n\\\\ \\dbinom{n}{0} + i\\dbinom{n}{1} - \\dbinom{n}{2} - i\\dbinom{n}{3} + \\dots + i^n\\dbinom{n}{n} &= (1+i)^n\\\\ \\dbinom{n}{0} - \\dbinom{n}{1} + \\dbinom{n}{2} - \\dbinom{n}{3} + \\dots + (-1)^n\\dbinom{n}{n} &= 0^n\\\\ \\dbinom{n}{0} - i\\dbinom{n}{1} - \\dbinom{n}{2} + i\\dbinom{n}{3} + \\dots + (-i)^n\\dbinom{n}{n} &= (1-i)^n. \\end{align*} Now add these equations and divide by 4 to get $\\dbinom{n}{0} + \\dbinom{n}{4} + \\dbinom{n}{8} + \\dots = \\dfrac{1}{4}\\Big(2^n + (1-i)^n + (1+i)^n\\Big).$ It is easy to show, using the polar form, that $(1-i)^n + (1+i)^n = 2(\\sqrt{2})^n \\cos \\big(\\dfrac{n\\pi}{4}\\big).$ Thus, we have $\\dbinom{n}{0} + \\dbinom{n}{4} + \\dbinom{n}{8} + \\dots = \\dfrac{1}{4}\\Big(2^n + 2(\\sqrt{2})^n \\cos \\big(\\dfrac{n\\pi}{4}\\big)\\Big).$ This is a rather amazing and beautiful result. 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[ "quartets, but not the 13th: $\\dbinom{13}{12}\\dbinom{n}{12} \\left( \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{12} \\left(1 - \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{n-12}$ Then the probability that the next pull is the 13th quartet is: $\\dfrac{ \\dbinom{5}{4} }{ \\dbinom{52}{5} }$ So, the expected value is given by: \\displaystyle \\begin{align*} & \\sum_{n \\ge 12}\\dbinom{13}{12}\\dbinom{n}{12} \\left( \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{13} \\left(1 - \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{n-12}(n+1) \\\\ = & \\dbinom{13}{12} \\left( \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{13} \\sum_{n \\ge 12} \\dbinom{n}{12} \\left(1 - \\dfrac{ 48\\dbinom{5}{4} }{ \\dbinom{52}{5} } \\right)^{n-12}(n+1) = 1830101\\end{align*} http://www.wolframalpha.com/input/?i...infinity%7D%5D Edit: I missed a factor of 48. 3. ## Re: Expectancy of number of trials I also got this expression but i suppose we have to multiply by 12! for permutations on the 12 quarters.For the same reason the probability for a quarter in one trial is 13*bin(5,4)*4!*(52-4)/(bin(52,5)*5!). Do you see a way of computing the expectancy by using indicators? 4. ## Re: Expectancy of number of trials Originally Posted by hedi I also got this expression but i suppose we have to multiply by 12! for permutations on the 12 quarters.For the same reason the probability for a quarter in one trial is 13*bin(5,4)*4!*(52-4)/(bin(52,5)*5!). Do you see a way of computing the expectancy by using indicators? I apologize. I missed a factor of 48 for", "he received from B Jul 4, 2021 at 3:53 • I made an edit with the clarifications Jul 4, 2021 at 4:06 Here's my solution based on expectations. First, we calculate the expected value of the smallest, second-smallest, third-smallest... number on the 5 cards drawn. The calculation steps are shown below: $$E(\\text{smallest})=\\dfrac{1\\times \\dbinom{9}{4}+2\\times \\dbinom{8}{4}+\\cdots+6\\times \\dbinom{4}{4}}{\\dbinom{10}{5}}=\\dfrac{11}{6}.$$ Explanation: the numerator basically sums the smallest terms in all 5-card draws, and the denominator is the total number of cases. Similarly, we have $$E(\\text{second-smallest})=\\dfrac{2\\times \\dbinom{8}{3}\\times \\dbinom{1}{1}+3\\times \\dbinom{7}{3}\\times \\dbinom{2}{1}+\\cdots+7\\times \\dbinom{3}{3}\\times \\dbinom{6}{1}}{\\dbinom{10}{5}}=\\dfrac{11}{3};$$ $$E(\\text{third-smallest})=\\dfrac{3\\times \\dbinom{7}{2}\\times \\dbinom{2}{2}+4\\times \\dbinom{6}{2}\\times \\dbinom{3}{2}+\\cdots+8\\times \\dbinom{2}{2}\\times \\dbinom{7}{2}}{\\dbinom{10}{5}}=\\dfrac{11}{2};$$ $$E(\\text{fourth-smallest})=\\dfrac{4\\times \\dbinom{6}{1}\\times \\dbinom{3}{3}+5\\times \\dbinom{5}{1}\\times \\dbinom{4}{3}+\\cdots+9\\times \\dbinom{1}{1}\\times \\dbinom{8}{3}}{\\dbinom{10}{5}}=\\dfrac{22}{3};$$ $$E(\\text{fifth-smallest, or largest})=\\dfrac{5\\times \\dbinom{4}{4}+6\\times \\dbinom{5}{4}+\\cdots+10\\times \\dbinom{9}{4}}{\\dbinom{10}{5}}=\\dfrac{55}{6}.$$ As a quick double check, adding them all together yields $$27.5$$, which is the expected sum of the 5 cards. Next, let's think about A's strategy. In essence, he is discarding his $$N$$ smallest cards to get random cards. Thus, if the expected value of cards that he is discarding is smaller than the expected value of cards he is getting, then A is benefitting. The more A benefits, the higher his chances of winning. Then, clearly $$E(\\text{smallest})$$, $$E(\\text{second-smallest})$$ are smaller than $$5.5$$, the expected value on a random card. $$E(\\text{third-smallest})$$ is equal to $$5.5$$, and $$E(\\text{fourth-smallest})$$, $$E(\\text{fifth-smallest, or largest})$$ are greater than $$5.5$$. This means that discarding the smallest card and", "### A result using complex numbers Suppose we expand out the expressions $$(1+1)^n$$, $$(1+i)^n$$, $$(1-1)^n$$, $$(1-i)^n$$ using the binomial theorem. This gives (for $$n>0$$), \\begin{align*} \\dbinom{n}{0} + \\dbinom{n}{1} + \\dbinom{n}{2} + \\dbinom{n}{3} + \\dots + \\dbinom{n}{n} &= 2^n\\\\ \\dbinom{n}{0} + i\\dbinom{n}{1} - \\dbinom{n}{2} - i\\dbinom{n}{3} + \\dots + i^n\\dbinom{n}{n} &= (1+i)^n\\\\ \\dbinom{n}{0} - \\dbinom{n}{1} + \\dbinom{n}{2} - \\dbinom{n}{3} + \\dots + (-1)^n\\dbinom{n}{n} &= 0^n\\\\ \\dbinom{n}{0} - i\\dbinom{n}{1} - \\dbinom{n}{2} + i\\dbinom{n}{3} + \\dots + (-i)^n\\dbinom{n}{n} &= (1-i)^n. \\end{align*} Now add these equations and divide by 4 to get $\\dbinom{n}{0} + \\dbinom{n}{4} + \\dbinom{n}{8} + \\dots = \\dfrac{1}{4}\\Big(2^n + (1-i)^n + (1+i)^n\\Big).$ It is easy to show, using the polar form, that $(1-i)^n + (1+i)^n = 2(\\sqrt{2})^n \\cos \\big(\\dfrac{n\\pi}{4}\\big).$ Thus, we have $\\dbinom{n}{0} + \\dbinom{n}{4} + \\dbinom{n}{8} + \\dots = \\dfrac{1}{4}\\Big(2^n + 2(\\sqrt{2})^n \\cos \\big(\\dfrac{n\\pi}{4}\\big)\\Big).$ This is a rather amazing and beautiful result. Next page - Links forward - Series for e", "2\\dbinom 21}{\\dbinom 52}=\\dfrac{3}{5}$$", "10^7 \\\\ d &= 31.134 \\times 10^6 - 6.378 \\times 10^6 \\\\ d \\ &\\boxed{ = 24.756 \\times 10^6 } \\ \\rm m \\end{align*}", "and the bottom of the fraction $\\dfrac{16!}{13!\\cdot 8!} = \\dfrac{16 \\times 15 \\times 14}{8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2}.$ This can be canceled further if we factorize the numerator $\\dfrac{16!}{13!\\cdot 8!} = \\dfrac{8 \\times 2 \\times 5 \\times 3 \\times 2 \\times 7}{8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2}.$ Then finally we get \\begin{align} \\dfrac{16!}{13! \\cdot 8!} &= \\dfrac{1}{4 \\times 3},\\\\ &=\\dfrac{1}{12}. \\end{align} ###### Example 4 Simplify $\\dfrac{(n-1)!}{(n+1)!}.$ ###### Solution \\begin{align} \\dfrac{(n-1)!}{(n+1)!} &= \\dfrac{(n-1) \\times (n-2) \\times \\dots \\times 2 \\times 1}{ (n+1) \\times n \\times (n-1) \\times \\dots \\times 2 \\times 1},\\\\ &= \\dfrac{1}{(n+1) \\times n},\\\\ &= \\dfrac{1}{n^2 + n}. \\end{align}", "# Division using Exponents (Indices) If we are given $a^8 \\div a^3$, we can also write this as $\\dfrac{a^8}{a^3}$, which means $\\dfrac{a \\times a \\times a \\times a \\times a \\times a \\times a \\times a}{a \\times a \\times a}$. As there are $8$ factors of $a$ on the top line (numerator), and $3$ factors of $a$ on the bottom line (denominator), we can cancel $3$ of them, giving us \\large \\begin{align} \\require{AMSsymbols} \\displaystyle \\require{cancel} &\\dfrac{\\cancel{a} \\times \\cancel{a} \\times \\cancel{a} \\times a \\times a \\times a \\times a \\times a}{\\cancel{a} \\times \\cancel{a} \\times \\cancel{a}} \\\\ &= a \\times a \\times a \\times a \\times a \\\\ &= a^5 \\end{align} ### Example 1 After first writing in factor form, simplify $\\dfrac{x^7}{x^4}$. \\begin{align} \\displaystyle \\require{AMSsymbols} \\require{cancel} \\dfrac{x^7}{x^4} &= \\dfrac{x \\times x \\times x \\times x \\times x \\times x \\times x}{x \\times x \\times x \\times x} \\\\ &= \\dfrac{\\cancel{x} \\times \\cancel{x} \\times \\cancel{x} \\times \\cancel{x} \\times x \\times x \\times x}{\\cancel{x} \\times \\cancel{x} \\times \\cancel{x} \\times \\cancel{x}} \\\\ &= x \\times x \\times x \\\\ &= x^3 \\end{align} An alternative solution can be formed by examining the result. We can see that the exponent (index) in the answer is the result of subtracting the exponents (indices) in the question. \\large \\begin{align} \\displaystyle \\require{cancel} a^8 \\div a^3 &= a^{8-3} \\\\", "h &= \\dfrac{4 \\pi^2Ed^4}{4 \\times 64L^2 \\times D_i^2\\times \\rho g}\\\\ h &= \\dfrac{4 \\pi^2 \\times 200 \\times 10^9 \\times (0.075)^4}{4 \\times 64 \\times \\times 4^2 \\times 1.5^2 \\times 10^4}\\\\ h &= 2.7\\ m\\end{aligned}", "( \\dfrac{1}{T_2^{2/3}}-\\dfrac{1}{T_1^{2/3}} \\right )\\\\ &=-\\left ( 2\\pi \\times 6.67\\times 10^{-11}\\times 6\\times 10^{24}\\right )^{2/3}\\times 50\\times \\left ( \\dfrac{1}{\\left ( 10.8\\times 10^{3} \\right )^{2/3}}-\\dfrac{1}{\\left (7.2\\times 10^{3} \\right )^{2/3}} \\right )\\\\ &=-\\left ( 2.51\\times 10^{15}\\right )^{2/3}\\times 50\\times \\left ( \\dfrac{1}{488}-\\dfrac{1}{373} \\right )\\\\ &=-1.36\\times 10^{10}\\times 50\\times \\left ( -6.32 \\times 10^{-4}\\right )\\\\ &=\\boxed{4.3\\times 10^{6}\\;\\rm J} \\end{align*} {/eq}", "n-2 \\right)...(n-r)!$ \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5!}\\times \\left( \\dfrac{8\\times 7\\times 6}{3!} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6!}\\times \\left( \\dfrac{8\\times 7}{2!} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4!}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4!} \\right) \\\\ \\end{align} \\begin{align} & =\\dfrac{12\\times 11\\times 10\\times 9\\times 8}{5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6}{3\\times 2\\times 1} \\right)+\\dfrac{12\\times 11\\times 10\\times 9\\times 8\\times 7}{6\\times 5\\times 4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7}{2\\times 1} \\right) \\\\ & +\\dfrac{12\\times 11\\times 10\\times 9}{4\\times 3\\times 2\\times 1}\\times \\left( \\dfrac{8\\times 7\\times 6\\times 5}{4\\times 3\\times 2\\times 1} \\right) \\\\ \\end{align} By cancelling the common terms in denominator and numerator we get $=\\left( 11\\times 9\\times 8 \\right)\\times \\left( 8\\times 7 \\right)+\\left( 11\\times 3\\times 4\\times 7 \\right)\\times \\left( 4\\times 7 \\right)+\\left( 11\\times 5\\times 9 \\right)\\times \\left( 5\\times 2\\times 7 \\right)$ $=792\\times 56+924\\times 28+495\\times 70$ = 44352 + 25872 + 34650 = 104874 Thus, the number of ways of selecting 10 students chosen for the competition = 104874 Note: The possibility of mistake can be the calculation mistake since the procedure of solving involves large calculations. Hence be very careful in simplifying and cancelling the terms to get the desired result.", "be a2 (in checkerboard coordinates) because $$b2/c2 \\ge 2$$ and at least two of those marked $$\\diamond$$ and at least one of those marked $$\\ast$$. Because only 8 small numbers are available a4 must be small and one of b4,c4. All others must be $$>8$$ (\"large\", marked $$\\Box$$): $$\\begin{matrix}\\bigcirc&+&\\Box\\bigcirc&+&\\Box\\bigcirc&=&\\Box\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&\\bigcirc&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&\\bigcirc&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&\\Box&=&\\Box\\end{matrix}$$ and we see that a4 must be $$1$$ Observe that $$d4 = d3/d2 + a1/b1 + c2\\times c3 + c4 \\ge 2 + 2 + 2\\times 3 + c4$$ Therefore c4 must be small and b4 large. $$\\begin{matrix}1&+&\\Box&+&\\bigcirc&=&13\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&\\bigcirc&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&\\bigcirc&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&9\\ldots 12&=&11\\ldots 14\\end{matrix}$$ Next, we see that $$2$$ must be in either c2 or c3: $$\\begin{matrix}1&+&9\\ldots 11&+&3\\ldots 6&=&14\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&2\\ldots 4&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&2\\ldots 4&=&\\bigcirc\\\\=&&=&&=&&=\\\\\\Box&/&\\bigcirc&+&10\\ldots 12&=&12\\ldots 14\\end{matrix}$$ We are almost done! Observe that in the products $$a2\\times b2 = c2\\times d2$$ the primes $$5$$ and $$7$$ cannot occur because they are available only once. $$\\begin{matrix}1&+&9\\ldots 10&+&5&=&15\\ldots 16\\\\\\times&&-&&+&&-\\\\\\Box&+&\\Box&/&2\\ldots 3&=&\\Box\\\\+&&/&&\\times&&/\\\\\\bigcirc&\\times&\\bigcirc&/&2\\ldots 3&=&\\bigcirc\\\\=&&=&&=&&=\\\\14&/&7&+&11&=&13\\end{matrix}$$ We can finish by filling in column a, $$\\begin{matrix}1&+&9&+&5&=&15\\\\\\times&&-&&+&&-\\\\10&+&\\Box&/&2\\ldots 3&=&\\Box\\\\+&&/&&\\times&&/\\\\4&\\times&\\bigcirc&/&2\\ldots 3&=&\\bigcirc\\\\=&&=&&=&&=\\\\14&/&7&+&11&=&13\\end{matrix}$$ row 3 and, finally, row 2. • Very well done and great explanation! I was getting worried that no one will solve this... – Dmitry Kamenetsky Oct 10 '20 at 0:46 Let the digits be a1,a2,a3,… d4, using standard algebraic chess notation (i.e. letters = columns, numbers = rows, a1 = lower left). Define a low digit = 1-8, high digit = 9-16. The cells,", "&= 15 \\times \\left[ {\\left( {-25} \\right) \\times \\left( {-4} \\right) \\times \\left( {-10} \\right)} \\right]\\\\ &= 15 \\times \\left[ {100 \\times \\left( {-10} \\right)} \\right]\\\\ &= 15 \\times \\left[ {1000} \\right]\\\\ &= -15000\\end{align} (d) $${\\rm{}}\\left( {-41} \\right){\\rm{ }} \\times {\\rm{ }}102$$ Using distributive law, we get \\begin{align}&=\\,{{(}} - {{41)}}\\,{{ \\times }}\\,{{(100}}\\,{{ + }}\\,{{2)}}\\dots\\,{{[a}}\\,{{ \\times }}\\,{{(b}}\\,{{ + }}\\,{{c)}}]=\\,{{(a}}\\,{{ \\times }}\\,{{b}}\\,{{ + }}\\,{{a}}\\,{{ \\times }}\\,{{c)]}}\\\\&={{{(}} - {{41)}}\\,{{ \\times }}\\,{{100}}\\,{{ + }}\\,{{(}}\\, - {{41) \\times }}\\,{{2}}}\\\\&= - {{4100}}\\, - {{82}}\\\\&=\\, - {{4182}}\\end{align} (e) $${{625}}\\,{{ \\times }}\\left( {{{-35}}} \\right){{ + }}\\left( {{{-}}\\,{{625}}} \\right){{ \\times }}\\,{{65}}$$ Using distributive property, we get \\begin{align} &=\\,{625}\\,\\times[{ (}-{35)}\\,{+}\\,{(}-{65) }]\\,{ }\\dots[{ a\\times b}\\,{+}\\,{a\\times c}\\,{=}\\,{a(b+c) }] \\\\ &=\\,{625}\\times[\\,-{35}\\,-{65 }] \\\\ &=\\,{625}\\times[-{100 }] \\\\ &={-62500} \\\\\\end{align} (f) $$7\\times \\left( 50-2 \\right)$$ Using distributive property, we get \\begin{align}&=7\\times { }\\,{(50}-{2)}\\,{ }\\dots[{ a\\times}\\,{ }{ }\\,{(b}-{c)}\\,{=}\\,{a }\\times{ b}-{a }\\times{ c }]{ } \\\\ &=7\\times { 50}-{7 }\\times { 2} \\\\ &=350-{14} \\\\ &=336\\end{align} (g) $$\\left( \\text{-17} \\right)\\text{ }\\!\\!\\times\\!\\!\\text{ }\\left( \\text{-29} \\right)$$ Using distributive property, we get \\begin{align} &= \\,( - 17)\\, \\times \\,\\,[( - 30)\\, + 1]\\dots[a\\, \\times \\,(b\\, + \\,c)={a\\times b}+{a \\times c}]\\\\ &= \\,( - 17)\\, \\times \\,( - 30)\\, + \\,( - 17)\\, \\times 1\\\\ &= \\,510\\, + ( - 17)\\\\ &= \\,510 - 17\\\\ &=\\,493\\end{align} (h)$${{}}\\left( {-57} \\right) \\times \\left( {-19} \\right) + 57$$ Using distributive property, we", "diapycnal diffusivities suggested by Federov (1988) is used: 776\\begin{align} \\label{Eq_zdfddm_d} 777A_d^{vT} &= \\begin{cases} 778 1.3635 \\, \\exp{\\left( 4.6\\, \\exp{ \\left[ -0.54\\,( R_{\\rho}^{-1} - 1 ) \\right] } \\right)} 779 &\\text{if $0<R_\\rho < 1$ and $N^2>0$ } \\\\ 780 0 &\\text{otherwise} 781 \\end{cases} 782\\\\ \\label{Eq_zdfddm_d_S} 783A_d^{vS} &= \\begin{cases} 784 A_d^{vT}\\ \\left( 1.85\\,R_{\\rho} - 0.85 \\right) 785 &\\text{if $0.5 \\leq R_\\rho<1$ and $N^2>0$ } \\\\ 786 A_d^{vT} \\ 0.15 \\ R_\\rho 787 &\\text{if $\\ \\ 0 < R_\\rho<0.5$ and $N^2>0$ } \\\\ 788 0 &\\text{otherwise} 789 \\end{cases} 790\\end{align} 791 792The dependencies of \\eqref{Eq_zdfddm_f} to \\eqref{Eq_zdfddm_d_S} on $R_\\rho$ 793are illustrated in Fig.~\\ref{Fig_zdfddm}. Implementing this requires computing 794$R_\\rho$ at each grid point on every time step. This is done in \\mdl{eosbn2} at the 795same time as $N^2$ is computed. This avoids duplication in the computation of 796$\\alpha$ and $\\beta$ (which is usually quite expensive). 797 798% ================================================================ 799% Bottom Friction 800% ================================================================ 801\\section [Bottom Friction (\\textit{zdfbfr})] {Bottom Friction (\\mdl{zdfbfr} module)} 802\\label{ZDF_bfr} 803 804%--------------------------------------------nambfr-------------------------------------------------------- 805\\namdisplay{nambfr} 806%-------------------------------------------------------------------------------------------------------------- 807 808Both the surface momentum flux (wind stress) and the bottom momentum 809flux (bottom friction) enter the equations as a condition on the vertical 810diffusive flux. For the bottom boundary layer, one has: 811\\begin{equation} \\label{Eq_zdfbfr_flux} 812A^{vm} \\left( \\partial {\\textbf U}_h / \\partial z \\right) = {{\\cal F}}_h^{\\textbf U} 813\\end{equation} 814where ${\\cal F}_h^{\\textbf U}$ is represents the downward", "last row right. This works well, if one has an expression of the type a+b=c+d+e+f+g+h=j, where there's only one row between initial row and result. However, in my case, I have some mathematical transformation (meaning an additional row) and so I need the equal sign of the row: \"12\\times6+14\\times\\dfrac{48}{7}-6^2-\\left(\\dfrac{2}{7}-4\\right)^2-\\left(\\dfrac{48}{7}\\right)^2 -\\dfrac{2}{7}\\times\\dfrac{48}{7}-\\left(\\dfrac{2}{7}\\right)^2\" to be aligned under the first equal sign. So now I use this version: Code: Select all • Open in writeLaTeX \\begin{align*}\\pi^{M}(s^{M},f^{M},x^{M}) :=& 12s^{M}+14f^{M}-(s^{M})^2-(x^{M}-4)^2-(f^{M})^2\\\\ &-x^{M}f^{M}-(x^{M})^2\\pause\\\\ =&12\\times6+14\\times\\dfrac{48}{7}-6^2-\\left(\\dfrac{2}{7}-4\\right)^2-\\left(\\dfrac{48}{7}\\right)^2\\\\ &-\\dfrac{2}{7}\\times\\dfrac{48}{7}-\\left(\\dfrac{2}{7}\\right)^2\\pause\\\\ =&69,14\\pause\\end{align*} Does anybody have better ideas? Thank you avp3000 Posts: 49 Joined: Thu Nov 15th, 2007 ### Re: Too long equation You can try the split environment, which is also provided by the amsmath package. To get a proper alignment, there are some tricky things to do. Code: Select all • Open in writeLaTeX \\begin{align*} \\pi^{M}(s^{M},f^{M},x^{M}): \\begin{split} &= 12s^{M}+14f^{M}-(s^{M})^2-(x^{M}-4)^2 \\\\ &\\quad -(f^{M})^2-x^{M}f^{M}-(x^{M})^2 \\end{split}\\\\ \\begin{split} &= 12\\times 6+14\\times\\frac{48}{7}-6^2-\\left(\\frac{2}{7}-4\\right)^2 \\\\ &\\quad -\\left(\\frac{48}{7}\\right)^2-\\frac{2}{7}\\times\\frac{48}{7}-\\left(\\frac{2}{7}\\right)^2 \\end{split}\\\\ &=69,14\\end{align*} Note that the alignment as known from the align environment has to be put into the split environment. More hints about typesetting mathematical expressions are given in \"Math mode\". Best regards Thorsten LaTeX Community Moderator ¹ System: openSUSE 13.1 (Linux 3.11.10), TeX Live 2013 (vanilla), TeXworks 0.5 (r1351) ² Posting stopped indefinitely due to offenses localghost Site Moderator Posts: 9219 Joined: Fri Feb 2nd, 2007 Location: Braunschweig, Germany ###", "7 \\times 100 + \\\\ 1 \\times 10 + 9 \\times 1\\end{array} \\right]\\\\&=\\! \\left[\\begin{array} = 1 \\times {10}^5 + 2 \\times {10}^4 + \\\\ 0 \\times {10}^3 + 7 \\times {10}^2 + \\\\ 1 \\times {10}^1 + 9 \\times {10}^0\\end{array} \\right]\\end{align} (v) $$20068$$ \\begin{align}&=\\left[\\begin{array} = 2 \\times 10000 + 0 \\times 1000 +\\\\ 0 \\times 100 + 6 \\times 10 + 8 \\times 1\\end{array} \\right]\\\\&=\\left[\\begin{array}=2 \\times {{10}^4} + 0 \\times {{10}^3} + 0 \\times\\\\ {{10}^2} + 6 \\times {{10}^1} + 8 \\times {{10}^0}\\end{array}\\right]\\end{align} ## Chapter 13 Ex.13.3 Question 2 Find the number from each of the following expanded forms: (a) \\left[ \\begin{align} & 8\\times {{10}^{4}}+6\\times {{10}^{3}}+0\\times \\\\ & {{10}^{2}}+4\\times {{10}^{1}}+5\\times {{10}^{0}} \\\\ \\end{align} \\right] (b) $$4 \\! \\times \\! 10^5 \\! + \\! 5 \\! \\times \\! 10^3 \\! + \\! 3 \\! \\times \\! 10^2 \\! + \\! 2 \\! \\times \\! 10^0$$ (c) $$3 \\times {{10}^4} + 7 \\times {{10}^2} + 5 \\times {{10}^0}$$ (d) $$9 \\times {{10}^5} + 2 \\times {{10}^2} + 3 \\times {{10}^1}$$ ### Solution Reasoning: Expressing a number means we have to expand the following powers of $$10.$$ (a) \\left[ \\begin{align} & 8\\times {{10}^{4}}+6\\times {{10}^{3}}+0\\times \\\\ & {{10}^{2}}+4\\times {{10}^{1}}+5\\times {{10}^{0}} \\\\ \\end{align} \\right] \\begin{align} & =\\left[ \\begin{array} & 8\\times 10000+6\\times 1000+ \\\\ 0\\times 100+4\\times 10+5\\times 1 \\\\ \\end{array} \\right] \\\\ & =80000+6000+0+40+5", "3 Use the quotient rule to differentiate $\\displaystyle f(x)=\\dfrac{3x+1}{1-x}$. \\begin{align} \\displaystyle f^{\\prime}(x) &= \\dfrac{(3x+1)^{\\prime} \\times (1-x) – (3x+1) \\times (1-x)^{\\prime}}{(1-x)^2} \\\\ &= \\dfrac{3 \\times (1-x) – (3x+1) \\times (-1)}{(1-x)^2} \\\\ &= \\dfrac{3-3x +3x+1}{(1-x)^2} \\\\ &= \\dfrac{4}{(1-x)^2} \\\\ \\end{align} ### Example 4 Use the quotient rule to differentiate $\\displaystyle f(x)=\\dfrac{(2x+1)^3}{(4x-1)^4}$. \\begin{align} \\displaystyle f^{\\prime}(x) &= \\dfrac{\\big[(2x+1)^3\\big]^{\\prime} \\times (4x-1)^4 – (2x+1)^3 \\times \\big[(4x-1)^4\\big]^{\\prime}}{\\big[(4x-1)^4\\big]^2} \\\\ &= \\dfrac{3(2x+1)^{3-1} \\times (2x+1)^{\\prime} \\times (4x-1)^4 – (2x+1)^3 \\times 4(4x-1)^{4-1} \\times (4x-1)^{\\prime}}{(4x-1)^8} \\\\ &= \\dfrac{3(2x+1)^2 \\times 2 \\times (4x-1)^4 – (2x+1)^3 \\times 4(4x-1)^3 \\times 4}{(4x-1)^8} \\\\ &= \\dfrac{6(2x+1)^2 (4x-1)^4 – 16(2x+1)^3 (4x-1)^3}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3\\big[6(4x-1) – 16(2x+1)\\big]}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3(24x-6 – 32x-16)}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(4x-1)^3(-8x-22)}{(4x-1)^8} \\\\ &= \\dfrac{(2x+1)^2(-8x-22)}{(4x-1)^5} \\\\ \\end{align} ## Extension Examples These Extension Examples require to have some prerequisite skills including; \\begin{align} \\displaystyle \\dfrac{d}{dx}\\sin{x} &= \\cos{x} \\\\ \\dfrac{d}{dx}\\cos{x} &= -\\sin{x} \\\\ \\dfrac{d}{dx}e^x &= e^x \\\\ \\dfrac{d}{dx}\\log_e{x} &= \\dfrac{1}{x} \\\\ \\end{align} ### Example 5 Find $\\displaystyle \\dfrac{dy}{dx}$ of $\\displaystyle y=\\dfrac{\\sin{x}}{\\cos{x}}$, known that $\\dfrac{d}{dx}\\sin{x} = \\cos{x}$ and $\\dfrac{d}{dx}\\cos{x} = -\\sin{x}$. \\begin{align} \\displaystyle \\dfrac{dy}{dx} &= \\dfrac{\\dfrac{d}{dx}\\sin{x} \\times \\cos{x} – \\sin{x} \\times \\dfrac{d}{dx}\\cos{x}}{\\cos^2{x}} \\\\ &= \\dfrac{\\cos{x} \\times \\cos{x} – \\sin{x} \\times (-\\sin{x})}{\\cos^2{x}} \\\\ &= \\dfrac{\\cos^2{x} + \\sin^2{x}}{\\cos^2{x}} \\\\ \\end{align} Note that optionally $\\cos^2{x} + \\sin^2{x}$ can be simplified further to $1$, as $\\cos^2{x} + \\sin^2{x}=1$ in one of the trigonometric properties, however we do not cover this skill here.", "-1 \\\\ k&=& 3 \\\\ \\dfrac{\\dbinom{14}{3}} {\\dbinom{14}{4}} &=& \\dfrac{4}{11} \\\\ \\hline \\end{array}$$ Mar 26, 2019", "three equal triangles if joined with the vertices. So, \\<\\begin{align*} \\text {area} \\triangle AOC &= \\text {area} \\triangle AOB = \\text {area} \\triangle BOC \\end{align*}\\> Therefore, \\<\\begin{align*} \\text {area of } \\triangle {ABC}&= 3 \\times \\text {area of } \\triangle BOC \\end{align*} \\> Using the formula for the area of an equilateral triangle\\<\\begin{align*} &= \\dfrac{\\sqrt3}{4} \\times a^2 \\hspace{3cm} ...1 \\end{align*} \\> Also, area of triangle \\<\\begin{align*} &= \\dfrac{1}{2} \\times \\text { base }\\times \\text { height } \\hspace{1cm} ...2 \\end{align*} \\> By applying equation 1 and 2 for $$\\triangle \\text{BOC}$$ we get, \\<\\begin{align*} {\\dfrac{\\sqrt3}{4}} \\times a^2 &= 3\\times \\dfrac{1}{2} \\times a\\times OD\\\\OD &= \\dfrac{1}{2{\\sqrt3}} \\times a\\hspace{2cm} ...3\\end{align*}\\> Now, by applying equation 1 and 2 for $$\\triangle \\text{ABC}$$ we get, $$\\text{Area of the } \\triangle\\text{ ABC}$$ \\<= \\dfrac{1}{2} \\times \\text { base }\\times \\text { height } =\\dfrac{\\sqrt3}{4}\\times a^2 ...4\\> Using equation 3 and 4, we get \\<\\begin{align*}\\dfrac {1}2\\times a\\times (R+OD) &= \\dfrac {\\sqrt 3}4\\times a^2 \\\\\\dfrac12 a\\times \\left( R+\\dfrac a{2\\sqrt3}\\right) &= \\dfrac{\\sqrt3}4\\times a^2\\\\R &= \\dfrac a{\\sqrt3} \\end{align*}\\> substituting- \\< \\begin{align*}a & = \\sqrt3\\end{align*}\\> $$\\therefore$$ $$\\text {R} = 1 \\text{in}$$ Example 3 A teacher drew 3 medians of a triangle and asked his students to name the concurrent point of these three lines. Can you name it? Solution The point where three mediansof the triangle meet areknown as the", "set of three elements can be arranged in six ways. We denote the number of ways of choosing $$r$$ objects from $$n$$ objects by $$\\dbinom{n}{r}$$, which is read as '$$n$$ choose $$r$$'. In the example above, we found that $\\dbinom{5}{3} = \\dfrac{{^5}P_3}{3!} = \\dfrac{60}{6} = 10.$ In general, we have $\\dbinom{n}{r} = \\dfrac{^nP_r}{r!} = \\dfrac{n!}{(n-r)!r!}.$ Consider the five-element set $$\\{a,b,c,d,e\\}$$ again. There are \\begin{alignat*}{2} \\dbinom{5}{1}&=5 \\text{ ways of choosing 1 letter},&\\qquad \\dbinom{5}{2}&=10 \\text{ ways of choosing 2 letters},\\\\ \\dbinom{5}{3}&=15 \\text{ ways of choosing 3 letters},&\\qquad \\dbinom{5}{4}&=10 \\text{ ways of choosing 4 letters},\\\\ \\dbinom{5}{5}&=1 \\text{ way of choosing 5 letters}. \\end{alignat*} Listing these as a row, including choosing no letters as the first entry: $\\dbinom{5}{0}=1, \\qquad \\dbinom{5}{1}=5, \\qquad \\dbinom{5}{2}=10, \\qquad \\dbinom{5}{3}=10, \\qquad \\dbinom{5}{4}=5, \\qquad \\dbinom{5}{5}=1.$ This is the fifth row of Pascal's triangle. #### Example Evaluate 1. $$\\dbinom{100}{2}$$ 2. $$\\dbinom{1000}{998}$$. #### Solution 1. $$\\dbinom{100}{2} = \\dfrac{100\\times99}{2} = 4950$$ 2. $$\\dbinom{1000}{998} = \\dfrac{1000\\times999}{2} = 499\\,500$$. #### Example In how many ways can you choose two people from a group of seven people? #### Solution There are $$\\dbinom{7}{2} = \\dfrac{7\\times6}{2} = 21$$ ways of choosing two people from seven. #### Example There are ten people in a basketball squad. Find how many ways: 1. the starting five can be chosen from the squad 2. the squad can be split into two", "this, we get \\begin{aligned}\\dfrac{1}{6}-\\dfrac{4}{9} &= \\dfrac{1\\times 3}{18}+\\dfrac{4\\times 2}{18} \\\\ &=\\dfrac{3}{18} - \\dfrac{8}{18} \\\\ &=-\\dfrac{5}{18}\\end{aligned} #### Is this a topic you struggle with? Get help now. Writing $9$ as $\\frac{9}{1}$, the calculation becomes $\\dfrac{7}{4}-\\dfrac{9}{1}$ Using $1\\times 4=4$ as a common denominator, we get \\begin{aligned}\\dfrac{7}{4}-\\dfrac{9}{1} &= \\dfrac{7}{4}-\\dfrac{9\\times 4}{4} \\\\ &=\\dfrac{7}{4} - \\dfrac{36}{4} \\\\ &=-\\dfrac{29}{4}\\end{aligned} ### Learning resources you may be interested in We have a range of learning resources to compliment our website content perfectly. Check them out below." ]

[ "are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square. Painting Cubes Stage: 3 Challenge Level: Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours? Flight of the Flibbins Stage: 3 Challenge Level: Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .", "the materials they return. This means if I have 10 cubes using the \"red\" material, then I say cube5.renderer.material.color = Color.Blue, only the one cube will change colour - it has a unique copy of the material now, which is separate from the red material used by all the others. If I instead say cube5.renderer.sharedMaterial.color = Color.Blue then all my red cubes will turn blue, because I've modified the single material instance they're all sharing. This fact about hidden material copying is useful to keep in mind, since materials aren't garbage collected as readily as certain other objects. You want to make sure you know when you're creating a new instance of a material versus when you're modifying an existing (and possibly shared) instance.", "Bunuel Can you please me why this approach cannot be used. 1st cube can be painted in 3 ways. Same with 2nd and 3rd. So total 3*3*3 = 27 ways. We are told that \"Order is not considered, for example green,green,blue is considered the same as green blue green)\". Among other duplications, your approach will give for example {red, green, blue} as well as {blue, red, green} or {red, green, green} as well as {green, green, red}. _________________ Re: How many different ways can 3 cubes be painted if each cube is painted [#permalink] 25 Aug 2017, 05:55 Display posts from previous: Sort by", "# Thread: N-dimensional hypercubes 1. ## N-dimensional hypercubes I'm not sure where I should put the question, it's not for school or university, but just out of curiosity. 2D space hypercubes are squares. Let's paint the sides of a square with red and blue by the special way: its blue side must be opposite its red side. We can match any two squares painted according to this rule, only rotating them without taking them out of the plane (2D space). It seems we can do similar things with 3D cubes, painting their faces by the similar rule (a blue face opposite to a red face) and then matching them without taking out from 3D space. So, the question is: is that valid for any N-dimensional hypercubes in any N-Dimensional space? 2. ## Re: N-dimensional hypercubes What do you mean \"match any two squares\"? Match them in what way? So that the orientation of red and blue is the same? Are all sides painted? 3. ## Re: N-dimensional hypercubes Originally Posted by SlipEternal So that the orientation of red and blue is the same? Are all sides painted? Both yes. 4. ## Re: N-dimensional hypercubes I highly doubt that this would work. In general, an n-cube can be \"unfolded\" into 2n (n-1)-cubes. Those 2n (n-1)-cubes have too many different", "and blue centers, they will always run clockwise in that order. No matter how you turn the faces on the cube, the center cubies will always stay this way. Each cubie is unique. Apart from the center cubie in the middle (which we never see) and the six face center cubies, there are 8 corner cubies (with three stickers on them) and 12 edge cubies (with two stickers on them). In a solved cubie, all faces have the same color. This means that every corner cubie has its own unique set of colors, and every edge cubie too. The blue-white edge is the one sitting on the blue and white faces in a solved cube, and so it. Naturally, it means that there's no such thing as a red-orange edge or a red-orange-green corner (because the red face opposes the orange). Since the stickers stay on the cubies as we move them about, we can just name the cubies by their colors. There is only one red-white-blue corner which would move about but always stay red-white-blue. When the cube is scrambled, it might seem to the untrained eye that the cubies can turn any old way. Yet, that's not the case. Because of the way the mechanism works, there are certain principles that limit the cubies' positions. For", "can 3 cubes be painted if each cube is painted with one colour and only three colours red,green and blue are available?(Order is not considered, for example green,green,blue is considered the same as green blue green) A)27 B)10 C)9 D)3 E)2 Source => NOVA What is the best way to tackle this one?Making individual cases ? Hi.. Yes picking up indl cases would be better easier and less error prone. 1) all of same colour-3 2) two of same colour.. Two colours can be choosen in 3C2=3!/2!=3 ways and any of these two can be on TWO cubes. So 3*2=6 3) all of one colour-1.. Total -- 3+6+1=10 B _________________ Manager Joined: 07 Aug 2016 Posts: 111 Location: India Concentration: Marketing, Operations GMAT 1: 690 Q48 V35 GPA: 4 WE: Engineering (Consulting) Re: How many different ways can 3 cubes be painted if each cube is painted [#permalink] Show Tags 25 Aug 2017, 05:21 Hi Bunuel Can you please me why this approach cannot be used. 1st cube can be painted in 3 ways. Same with 2nd and 3rd. So total 3*3*3 = 27 ways. Math Expert Joined: 02 Sep 2009 Posts: 54434 Re: How many different ways can 3 cubes be painted if each cube is painted [#permalink] Show Tags 25 Aug 2017, 05:55 gauravraos wrote: Hi", "would be coloured blue? ### Cubic Conundrum ##### Stage: 2 Challenge Level: Which of the following cubes can be made from these nets?", "is painted [#permalink] ### Show Tags 25 Aug 2017, 05:21 Hi Bunuel Can you please me why this approach cannot be used. 1st cube can be painted in 3 ways. Same with 2nd and 3rd. So total 3*3*3 = 27 ways. Kudos [?]: 131 [0], given: 72 Math Expert Joined: 02 Sep 2009 Posts: 41891 Kudos [?]: 129061 [0], given: 12190 Re: How many different ways can 3 cubes be painted if each cube is painted [#permalink] ### Show Tags 25 Aug 2017, 05:55 gauravraos wrote: Hi Bunuel Can you please me why this approach cannot be used. 1st cube can be painted in 3 ways. Same with 2nd and 3rd. So total 3*3*3 = 27 ways. We are told that \"Order is not considered, for example green,green,blue is considered the same as green blue green)\". Among other duplications, your approach will give for example {red, green, blue} as well as {blue, red, green} or {red, green, green} as well as {green, green, red}. _________________ Kudos [?]: 129061 [0], given: 12190 Re: How many different ways can 3 cubes be painted if each cube is painted [#permalink] 25 Aug 2017, 05:55 Display posts from previous: Sort by", "are the green cube on ab and the red cube on bc. The three blue solids in Figure 6 can be stacked, giving a solid whose base is the square on ab and whose height is thrice bc, and so the blue solid in Figure 7 has volume 3(ab)2(bc). Likewise, the three pink solids can be combined, giving a solid of volume 3(ab)(bc)2. The solid of volume 6(ac)2 is also decomposed by breaking the side ac at the point b. The decomposition can be regrouped into three solids, as shown in Figure 9. Figure 8. The solid with height 6. Figure 9. The solid with height 6 decomposed. The blue solid has base (ab)2 and height 6, and the red solid has base (bc)2 and height 6. The two pink solids can be combined to yield a solid of volume twice 6(ab)(bc), or 12(ab)(bc). Up to this point the cut at b has been arbitrary. Cardano asked, where does this cut have to be so that the solids in Figures 4 and 5 whose bases are the square on ab have the same volume? In our pictures, this means determining the location of b so that the blue solids in Figures 7 and 9 have the same volume. In the cube ae, the blue solid has volume 3(ab)2(bc),", "that among the interior angles of a convex polygon there cannot be more than three acute angles. Painting Cubes Stage: 3 Challenge Level: Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours? Königsberg Stage: 3 Challenge Level: Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps? On the Edge Stage: 3 Challenge Level: Here are four tiles. They can be arranged in a 2 by 2 square so that this large square has a green edge. If the tiles are moved around, we can make a 2 by 2 square with a blue edge... Now try to. . . . Reflecting Squarely Stage: 3 Challenge Level: In how many ways can you fit all three pieces together to make shapes with line symmetry? Flight of the Flibbins Stage: 3 Challenge Level: Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .", "quickly beyond the simplest questions. ## Areas and volumes Q: We have a white cube, each side is 1 unit long. We paint it red using 1 liter of paint. We now have a white cube, each side is 2 units long. How many liters of paint do we need to paint it red? A: We would need 4 liters of paint to paint the cube red. This is because each side of the cube is now twice as long, so the total surface area of the cube has been multiplied by four. Note: correct. Q: We have a square. We cut up the square into pieces and rearrange the pieces to form a really long rectangle. Which has bigger area, the original square, or the really long rectangle? A: The original square has a bigger area than the really long rectangle. Note: incorrect. Area remains the same. Q: There is a cube, and a sphere inside the cube. Which has bigger volume? A: The cube has a larger volume. Note: correct. Q: The area of a unit square in 2D is 1, the area of the contained circle is Pi/4. The volume of a unit cube in 3D is 1, the volume of the contained sphere is Pi/6. What is the volume of the sphere contained in", "ways may we choose the ice cream and put it into the dish? ###### 8 In as many ways as you can, show that $\\binom{n}{k}\\binom{n-k}{m} = \\binom{n}{m}\\binom{n-m}{k}\\text{.}$ ###### 9 A tennis club has $4n$ members. To specify a doubles match, we choose two teams of two people. In how many ways may we arrange the members into doubles matches so that each player is in one doubles match? In how many ways may we do it if we specify in addition who serves first on each team? ###### 10 A town has $n$ streetlights running along the north side of main street. The poles on which they are mounted need to be painted so that they do not rust. In how many ways may they be painted with red, white, blue, and green if an even number of them are to be painted green? ###### 11 We have $n$ identical ping-pong balls. In how many ways may we paint them red, white, blue, and green? ###### 12 We have $n$ identical ping-pong balls. In how many ways may we paint them red, white, blue, and green if we use green paint on an even number of them?", "a very simple process to solve a Cube by taking it apart and reassembling it in a solved state. There are twelve edge pieces which show two coloured sides each, and eight corner pieces which show three colours. Each piece shows a unique colour combination, but not all combinations are present (for example, if red and orange are on opposite sides of the solved Cube, there is no edge piece with both red and orange sides). The location of these cubes relative to one another can be altered by twisting an outer third of the Cube 90°, 180° or 270°, but the location of the coloured sides relative to one another in the completed state of the puzzle cannot be altered: it is fixed by the relative positions of the centre squares and the distribution of colour combinations on edge and corner pieces. For most recent Cubes, the colours of the stickers are red opposite orange, yellow opposite white, and green opposite blue. However, Cubes with alternative colour arrangements also exist; for example, they might have the yellow face opposite the green, and the blue face opposite the white (with red and orange opposite faces remaining unchanged). Douglas R. Hofstader, in the July 1982 Scientific American, pointed out that Cubes could be coloured in such a way as", "# 1 green, 2 blue, and 3 reds Probability Level 2 How many distinct ways can a cube have $\\color{#20A900} \\text{1 green face}$, $\\color{#3D99F6} \\text{2 blue faces}$ and $\\color{#D61F06} \\text{3 red faces}$? Note: Two ways are only distinct if one cube can't be rotated to look like the other. ×", "two red cubes must be a different color. 2. Between two blue must be two different colors. 3. Between two green must be three different colors. 4. Between two yellow blocks must be four • Humans Humans exploring space have left behind bags of​ trash, bolts,​ gloves, and pieces of satellites. ​ Let's say there are currently about 5​,000,000 pounds of litter in orbit around Earth. Jenny says that this amount using number names is five billion. Is J", "exactly 100 square units. Is there more than one? Can you find them all? Drilling Many Cubes Stage: 2 and 3 Challenge Level: A useful visualising exercise which offers opportunities for discussion and generalising, and which could be used for thinking about the formulae needed for generating the results on a spreadsheet. Cubes Within Cubes Revisited Stage: 3 Challenge Level: Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need? Square Coordinates Stage: 3 Challenge Level: A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides? Frogs Stage: 3 Challenge Level: How many moves does it take to swap over some red and blue frogs? Do you have a method? Painted Cube Stage: 3 Challenge Level: Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces? Convex Polygons Stage: 3 Challenge Level: Show that among the interior angles of a convex polygon there cannot be more than three acute", "to call it) is red but it is punctured so you can see inside the cube. Face 1 is red, 2 is blue, 3 is green, 4 is yellow, 5 is blue, and 6 is green. Each painting of a cube in 5 colors can be represented as a sequence of 6 elements. The elements of each sequence are the letters R, G, B, Y, and W. The cube in our example will correspond to the sequence (R,B,G,Y,B,G). The set U of all paintings can be written as $U=C^6.$ The number of elements of U is $$5^6$$ but that is not the number we are after. For example the paintings (R, B, B, B, B, B) and (B, R, B, B, B, B) can be obtained from each other using rotations. This is the reason why we define an equivalence relation among the paintings. ## An equivalence relation on the set U We will define a relation $$\\sim$$ on the set U of all colorings. We say that two colorings $$(x_1,x_2,$$ ..., $$x_6)$$ and $$(y_1,y_2,$$ ..., $$y_6)$$ are in relation $$\\sim$$ if they represent colorings of two cubes that can be obtained from one another using rotations. We write $(x_1, x_2, \\dots, x_6)\\sim (y_1, y_2, \\dots, y_6).$ For example, we have: $(R,B,G,Y,B,G)\\sim (B, G, Y, R, B, G)\\sim", "For example, Snuke can paint the blocks as shown in the diagram below. There are $45$ red blocks and $30$ blue blocks, thus the difference is $9$.", "the other side, and the blue needs to trap the red. ### Painted Cube ##### Stage: 3 Challenge Level: Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces? ### Tetrahedra Tester ##### Stage: 3 Challenge Level: An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length? ### Reflecting Squarely ##### Stage: 3 Challenge Level: In how many ways can you fit all three pieces together to make shapes with line symmetry? ### Cubes Within Cubes ##### Stage: 2 and 3 Challenge Level: We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used? ### Paving Paths ##### Stage: 3 Challenge Level: How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs? ### Rolling Around ##### Stage: 3", "# Red-Red Blocks A toy factory creates cube blocks in which each side is randomly colored red or blue. You have all 10 distinct (up to rotation) cubes placed in a bag. You reach into the bag and pick up one of these blocks, noticing that the one (and only one) side that you see is red. What is the probability that the opposite side is also red? ×" ]

[ "the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ Jan 5: A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(1,1/2,1/4); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(1,-1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,-1)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(1,1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)} \\:1 : 6 \\qquad\\textbf{ (B)}\\: 7 : 36 \\qquad\\textbf{(C)}\\: 1 : 5 \\qquad\\textbf{(D)}\\: 7 : 30\\qquad\\textbf{ (E)}\\: 6 : 25$ Jan 6: A cube has edge length 2. Suppose that we glue a cube of edge length 1 on top of the big cube so that one of its faces rests entirely on the top face", "is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ ## Problem 14 In this addition problem, each letter stands for a different digit. $\\setlength{\\tabcolsep}{0.5mm}\\begin{array}{cccc}&T & W & O\\\\ + &T & W & O\\\\ \\hline F& O & U & R\\end{array}$ If $T = 7$ and the letter $O$ represents an even number, what is the only possible value for $W$? $\\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 1 \\qquad \\textbf{(C)}\\ 2\\qquad \\textbf{(D)}\\ 3\\qquad \\textbf{(E)}\\ 4$ ## Problem 15 A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown? $[asy] defaultpen(linewidth(0.8)); path p=unitsquare; draw(p^^shift(0,1)*p^^shift(1,0)*p); draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p); label(\"FRONT\", (1,0), S); label(\"SIDE\", (5,0), S);[/asy]$ $\\textbf{(A)}\\ 3\\qquad\\textbf{(B)}\\ 4\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 6\\qquad\\textbf{(E)}\\ 7$ ## Problem 16 Ali, Bonnie, Carlo,", "faces so that at least all faces pairs of eyes would be right on top of each other, there is extensive use in PCA here. Here I will perform PCA on the 1950s faces, project the faces on just the first principal component (out of 10 possible), average them and see whether I get a better result. The PCA here is done separately for the Red, Green and Blue channels of the image, then I read them back as a 3-Channel image and plot it. This Cross Validated answer was very helpful. get_channel_vec <- function(mat, channel_num) { matrix(mat[, , channel_num], 1, dim(mat)[1] * dim(mat)[2]) } faces_50 <- faces_table %>% mutate(face_mat = map(face, function(face) image_data(face) %>% as.numeric()), red = map(face_mat, get_channel_vec, 1), green = map(face_mat, get_channel_vec, 2), blue = map(face_mat, get_channel_vec, 3)) get_projected_face_single_channel <- function(channel_str, nPCs = 1) { X <- do.call(\"rbind\", faces_50[[channel_str]]) pca <- prcomp(t(X), scale. = TRUE) X_proj <- pca$x[, 1:nPCs] %*% t(pca$rotation[, 1:nPCs]) X_proj <- scale(X_proj, center = FALSE , scale = 1/pca$scale) X_proj <- scale(X_proj, center = -1 * pca$center, scale = FALSE) face_proj <- apply(X_proj, 1, mean) array(face_proj, c(200, 200, 1)) } av_face <- abind::abind( get_projected_face_single_channel(\"red\"), get_projected_face_single_channel(\"green\"), get_projected_face_single_channel(\"blue\"), along = 3) image_append(c( image_append(c( present_average_face(av_1950), image_blank(200, 30, \"white\") %>% image_annotate(\"Original (or 10 PCs)\", \"Center\", size = 20)), stack = TRUE) %>% image_border(\"white\", \"2x3\"), image_blank(200, 30,", "gray scale image. faces = faceCascade.detectMultiScale( gray, scaleFactor=1.1, minNeighbors=5, minSize=(30, 30), ) For each face detected, we’ll draw a rectangle around the face : for (x, y, w, h) in faces: if w > 250 : cv2.rectangle(frame, (x, y), (x+w, y+h), (255, 0, 0), 3) roi_gray = gray[y:y+h, x:x+w] roi_color = frame[y:y+h, x:x+w] For each mouth detected, draw a rectangle around it : smile = smileCascade.detectMultiScale( roi_gray, scaleFactor= 1.16, minNeighbors=35, minSize=(25, 25), ) for (sx, sy, sw, sh) in smile: cv2.rectangle(roi_color, (sh, sy), (sx+sw, sy+sh), (255, 0, 0), 2) cv2.putText(frame,'Smile',(x + sx,y + sy), 1, 1, (0, 255, 0), 1) For each eye detected, draw a rectangle around it : eyes = eyeCascade.detectMultiScale(roi_gray) for (ex,ey,ew,eh) in eyes: cv2.rectangle(roi_color,(ex,ey),(ex+ew,ey+eh),(0,255,0),2) cv2.putText(frame,'Eye',(x + ex,y + ey), 1, 1, (0, 255, 0), 1) Then, count the total number of faces, and display the overall image : cv2.putText(frame,'Number of Faces : ' + str(len(faces)),(40, 40), font, 1,(255,0,0),2) # Display the resulting frame cv2.imshow('Video', frame) And implement an exit option when we want to stop the camera by pressing q : if cv2.waitKey(1) & 0xFF == ord('q'): break Finally, when everything is done, release the capture and destroy all windows. There are some troubles killing windows on Mac which might require killing Python from the Activity Manager later on. video_capture.release() cv2.destroyAllWindows() ## 5. Wrapping", "are painted in six different colors: red $(R)$, white $(W)$, green $(G)$, brown $(B)$, aqua $(A)$, and purple $(P)$. Three views of the cube are shown below. What is the color of the face opposite the aqua face? $[asy] unitsize(2 cm); pair x, y, z, trans; int i; x = dir(-5); y = (0.6,0.5); z = (0,1); trans = (2,0); for (i = 0; i <= 2; ++i) { draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle)); draw(shift(i*trans)*((x + z)--x)); draw(shift(i*trans)*((x + z)--(x + y + z))); draw(shift(i*trans)*((x + z)--z)); } label(rotate(-3)*\"R\", (x + z)/2); label(rotate(-5)*slant(0.5)*\"B\", ((x + z) + (y + z))/2); label(rotate(35)*slant(0.5)*\"G\", ((x + z) + (x + y))/2); label(rotate(-3)*\"W\", (x + z)/2 + trans); label(rotate(50)*slant(-1)*\"B\", ((x + z) + (y + z))/2 + trans); label(rotate(35)*slant(0.5)*\"R\", ((x + z) + (x + y))/2 + trans); label(rotate(-3)*\"P\", (x + z)/2 + 2*trans); label(rotate(-5)*slant(0.5)*\"R\", ((x + z) + (y + z))/2 + 2*trans); label(rotate(-85)*slant(-1)*\"G\", ((x + z) + (x + y))/2 + 2*trans); [/asy]$ $\\textbf{(A) }\\text{red}\\qquad\\textbf{(B) }\\text{white}\\qquad\\textbf{(C) }\\text{green}\\qquad\\textbf{(D) }\\text{brown}\\qquad\\textbf{(E) }\\text{purple}$ ## Problem 13 A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum", "the bar shift fill={rgb:red,4;green,2;blue,1}, % changed yellow to blue draw=none, area legend ] coordinates { (40,15) (60,25) ({80},35) ({100},15) ({120},10) };\\label{A1nm} % added negative bar shift here as well \\addplot [bar shift=-3mm,fill=yellow,draw=none,area legend] coordinates { (40,10) (60,35) (80,30) (100,25) (120,10) };\\label{plot_two} \\resetstackedplots{40,60,80,100,120} % positive bar shift for the next two, with different colours % added a \\label after each plot \\addplot [bar shift=3mm,fill={rgb:red,40;green,4;blue,50},draw=none,area legend] coordinates { (40,10) (60,20) (80,30) (100,10) (120,20) };\\label{three} \\addplot [bar shift=3mm,fill=red!30,draw=none,area legend] coordinates { (40,13) (60,39) (80,32) (100,35) (120,13) };\\label{four} \\end{axis} \\begin{axis}[ set layers, axis background, axis y line*=right, axis x line=none, % hide x-axis ymode=log, ylabel=y-axis 2, ylabel near ticks, yticklabel pos=right, ymajorgrids=true, major grid style={dashed, gray} ] coordinates{ ({40},1) ({60},2) ({80},35) ({100},1) ({120},10) };\\label{C} % just a label here coordinates{ ({40},2) ({60},3) ({80},45) ({100},2) ({120},20) }; \\label{D} \\end{axis} \\end{tikzpicture} % now generate the legend, using a normal tabular \\begin{tabular}{lccc} & A & B & C \\\\ S1 & \\ref{A1nm} & \\ref{plot_two} & \\ref{C} \\\\ S2 & \\ref{three} & \\ref{four} & \\ref{D} \\end{tabular} \\end{figure} \\end{document} • What do forget plot do? Mar 21, 2018 at 16:24 • @Mr.EU The pgfplots manual (useful reference for questions like that) says \"Allows to include plots which are not remembered for legend entries, which do not increase the number of plots and which are not considered for", "scale and rotate it, so we can show the arrow in the rotating face. For a rotation like 'Ui': public void updateArrow(String face, boolean hover){ boolean bFaceArrow=!(face.startsWith(\"X\")||face.startsWith(\"Y\")||face.startsWith(\"Z\")); MeshView arrow=bFaceArrow?faceArrow:axisArrow; if(hover && onRotation.get()){ return; } arrow.getTransforms().clear(); if(hover){ double d0=arrow.getBoundsInParent().getHeight()/2d; Affine aff=Utils.getAffine(dimCube, d0, bFaceArrow, face); arrow.getTransforms().setAll(aff); arrow.setMaterial(Utils.getMaterial(face)); if(previewFace.get().isEmpty()) { previewFace.set(face); onPreview.set(true); rotateFace(face,true,false); } } else if(previewFace.get().equals(face)){ rotateFace(Utils.reverseRotation(face),true,true); } else if(previewFace.get().equals(\"V\")){ previewFace.set(\"\"); onPreview.set(false); } } where the affine is calculated in the Utils class for the current face: public static Affine getAffine(double dimCube, double d0, boolean bFaceArrow, String face){ Affine aff=new Affine(new Scale(3,3,3)); aff.append(new Translate(0,-d0,0)); switch(face){ case \"U\": case \"Ui\": aff.prepend(new Rotate(face.equals(\"Ui\")?180:0,Rotate.Z_AXIS)); aff.prepend(new Rotate(face.equals(\"Ui\")?45:-45,Rotate.Y_AXIS)); aff.prepend(new Translate(0,dimCube/2d,0)); break; } return aff; } To trigger the drawing of the arrow, we set a listener to the buttons on the toolbars based on the mouse hovering. We can also add a small rotation (5º) as preview of the full rotation (90º) in the face selected, by calling rotateFace again, with bPreview=true at this point. If the user clicks on the button, the rotation is completed (from 5º to 90º). Otherwise the rotation is cancelled (from 5º to 0º). In both cases, with a smooth animation. Select rotation by picking Finally, the rotation could be performed based on the mouse picking of a cubie face, with visual aid showing the arrow and performing a small", "coordinates of (-10, 5), since 30-40 = -10 and 130-110 = 20. eye1Center = center.clone() # face positions are relative to the center eye1Center.move(-10, 5) # locate further points in relation to others eye1 = Circle(eye1Center, 5) eye1.setFill('blue') eye1.draw(win) The only other changes to the face are similar, cloning and moving Points, rather than specifying them with explicit coordinates. eye2End1 = eye1Center.clone() eye2End1.move(15, 0) eye2End2 = eye2End1.clone() eye2End2.move(10, 0) eye2 = Line(eye2End1, eye2End2) eye2.setWidth(3) eye2.draw(win) mouthCorner1 = center.clone() mouthCorner1.move(-10, -10) mouthCorner2 = mouthCorner1.clone() mouthCorner2.move(20, -5) mouth = Oval(mouthCorner1, mouthCorner2) mouth.setFill('red') mouth.draw(win) Finally, the list of elements for the face must be returned to the caller: return [head, eye1, eye2, mouth] Then in the main function, the program creates a face in exactly the same place as before, but using the makeFace function, with the original center of the face Point(40, 100). Now with the makeFace function, with its center parameter, it is also easy to replace the old cir2 with a whole face! faceList = makeFace(Point(40, 100), win) faceList2 = makeFace(Point(150,125), win) The animation section is considerably elaborated in this version. stepsAcross = 46 dx = 5 dy = 3 wait = .01 for i in range(3): moveAllOnLine(faceList, dx, 0, stepsAcross, wait) moveAllOnLine(faceList, -dx, dy, stepsAcross//2, wait) moveAllOnLine(faceList, -dx, -dy, stepsAcross//2, wait) The unidentified numeric literals that were", "and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$", "the attach option: %\\usepackage[dvips]{attachfile2} \\begin{document} % Optional subdirectory for latex files (no spaces): %\\def\\asylatexdir{} % Optional subdirectory for asy files (no spaces): %\\def\\asydir{} \\begin{asydef} // Global Asymptote definitions can be put here. usepackage(\"fixmath\"); import three; import solids; import patterns; settings.toolbar=false; \\end{asydef} \\begin{asy} size(22cm, 22cm); currentprojection=orthographic(-6,-2,1); currentlight=light(gray(0.4),specularfactor=3,viewport=false, (-0.5,-0.25,0.45), (0.5,-0.5,0.5),(0.5,0.5,0.75)); triple O=(0,0,0), X=(1,0,0), Y=(0,1,0), Z=(0,0,1); draw(O--(-2.2,0,0),black,Arrow3); draw(O--(0,-2.2,0),black,Arrow3); draw(O--(0,0,2.2),black,Arrow3); label(\"\\huge$x$\",scale3(1.1)*(-2.2,0,0),black); label(\"\\huge$y$\",scale3(1.1)*(0,-2.2,0),black); label(\"\\huge$z$\",scale3(1.1)*(0,0,2.05),black); //material unterfl=material(white+opacity(1),ambientpen=white); material oberfl=material(white+opacity(0.6),ambientpen=white); for (int n = -1; n <=1; n=n+2) { for (int l = 1; l <=1; l=l+2) { for (int m = -1; m <=1; m=m+2) { draw(rotate(90*l+45*(m+1),Z)*shift(1,1,n)*rotate(aTan(sqrt(2))*n,Y-X)*rotate(15,Z)*scale3(sqrt(2))*surface(polygon(6)), surfacepen=oberfl); } } } draw(shift(-2,0,0)*rotate(45,X)*rotate(90,Y)*scale3(sqrt(2))*shift(-0.5,-0.5,0)*unitplane, surfacepen=oberfl,black+linewidth(0.2)); draw(shift(0,-2,0)*rotate(45,Y)*rotate(90,X)*scale3(sqrt(2))*shift(-0.5,-0.5,0)*unitplane, surfacepen=oberfl,black+linewidth(0.2)); draw(shift(0,0,2)*rotate(45,Z)*scale3(sqrt(2))*shift(-0.5,-0.5,0)*unitplane, surfacepen=oberfl,black+linewidth(0.2)); draw(O--(-1,-1,1),red + linewidth(2.0pt) + linetype(new real[] {2,2})); draw(O--(0,-2,0),red + linewidth(2.0pt) + linetype(new real[] {2,2})); draw((0,-2,0)--(-1,-2,0),red + linewidth(2.0pt)); draw((-1,-2,0)--(-1.5,-1.5,0),red + linewidth(2.0pt)); draw((-1.5,-1.5,0)--O,red + linewidth(2.0pt) + linetype(new real[] {2,2})); label(\"\\Huge$\\Gamma$\",(0,0,-0.15),red); label(\"\\Huge$L$\",(-0.9,-1.1,1),red); label(\"\\Huge$X$\",(0,-2,-0.15),red); label(\"\\Huge$W$\",(-1,-2,-0.15),red); label(\"\\Huge$K$\",(-1.5,-1.5,-0.15),red); \\end{asy} \\end{document} • Note that the site works best when you ask one question per ... well, per question, basically. For example, I can answer the second question you ask but not the first. So either I don't bother answering (and somebody else who can only answer the first may feel the same way) or I answer part of it (and now your question has an answer, which may reduce views and which will you accept if somebody now answers the first without providing a", "shade some parts of the sphere. By changing the angles of the bounding arcs (and the color) you can add more \"open subsets\". With asymptote you can draw such things rather easily. Here are possible first steps. I refrain from doing more since I don't know which covers you'll end up choosing, and I also don't know if you're open to asymptote. This code is to some extent stolen from this asymptote answer. \\documentclass{standalone} \\usepackage[inline]{asymptote} \\begin{document} \\begin{asy} settings.prc=false; settings.tex=\"pdflatex\"; settings.render=0; import three; size(200); size3(200); currentprojection=orthographic( camera=(5.4290316601351,2.94352790610013,1.1108527434919),up=Z,target=O,zoom=0.7); real r=1; triple A1,B1,C1,D1; A1=dir(80.0,340.0); B1=dir(80.0,80.0); C1=dir(20.0,80.0); D1=dir(20.0,340.0); guide3 AB1=arc(O,A1,B1,CCW); guide3 BC1=arc(O,B1,C1,CCW); guide3 CD1=arc(O,C1,D1,CCW); guide3 DA1=arc(O,D1,A1,CCW); ,currentprojection.camera-currentprojection.target)) ,rgb(0.79,0.79,0.85)+opacity(0.5), project(O), 1.0 ,rgb(0.99,0.99,0.85)+opacity(0.5), project(O), 0.2 ); guide3 g=AB1--BC1--CD1--DA1--cycle; fill(project(g),rgb(1,1,0.8)+opacity(0.5)); draw(arc(O,A1,B1,CCW),blue+1bp); draw(arc(O,B1,C1,CCW),blue+1bp); draw(arc(O,C1,D1,CCW),blue+1bp); draw(arc(O,D1,A1,CCW),blue+1bp); dot(A1,blue); dot(B1,blue); dot(C1,blue); dot(D1,blue); //label(\"$A$\",project(A1),SW); //label(\"$B$\",project(B1),N); //label(\"$C$\",project(C1),SE); triple A2,B2,C2,D2; A2=dir(140.0,-20.0); B2=dir(140.0,80.0); C2=dir(70.0,80.0); D2=dir(70.0,-20.0); guide3 AB2=arc(O,A2,B2,CCW); guide3 BC2=arc(O,B2,C2,CCW); guide3 CD2=arc(O,C2,D2,CCW); guide3 DA2=arc(O,D2,A2,CCW); guide3 CB2=arc(O,C2,B1,CCW); guide3 BA1=arc(O,B1,A1,CCW); guide3 DC2=arc(O,D2,C2,CCW); ,currentprojection.camera-currentprojection.target)) ,rgb(0.79,0.79,0.85)+opacity(0.5), project(O), 1.0 ,rgb(0.99,0.99,0.85)+opacity(0.5), project(O), 0.2 ); guide3 g=AB2--BC2--CD2--DA2--cycle; fill(project(g),rgb(1,0.8,1)+opacity(0.5)); fill(project(g),rgb(1,0.8,0.8)+opacity(0.5)); draw(arc(O,A2,B2,CCW),red+1bp); draw(arc(O,B2,C2,CCW),red+1bp); draw(arc(O,C2,D2,CCW),red+1bp); draw(arc(O,D2,A2,CCW),red+1bp); dot(A2,red); dot(B2,red); dot(C2,red); dot(D2,red); //label(\"$A1$\",project(A1),SW); //label(\"$B1$\",project(B1),SW); //label(\"$D2$\",project(D2),SW); //label(\"$C2$\",project(C2),SW); \\end{asy} \\end{document} % process with pdflatex <file>, then asy <file>-1, and again pdflatex <file> • This is very nice. Thank you so much. I'm goin to wait a few days to see if anyone comes up with something different but barring that, I'll give you the answer, make some pictures", "object: a cube. We define the coordinates of its 8 vertices, and we draw line segments between the projections of the 12 pairs of vertices that make the edges of the cube, as seen in Listing 9-1: ViewportToCanvas(x, y) { return (x * Cw/Vw, y * Ch/Vh); } ProjectVertex(v) { return ViewportToCanvas(v.x * d / v.z, v.y * d / v.z) } // The four \"front\" vertices vAf = [-2, -0.5, 5] vBf = [-2, 0.5, 5] vCf = [-1, 0.5, 5] vDf = [-1, -0.5, 5] // The four \"back\" vertices vAb = [-2, -0.5, 6] vBb = [-2, 0.5, 6] vCb = [-1, 0.5, 6] vDb = [-1, -0.5, 6] // The front face DrawLine(ProjectVertex(vAf), ProjectVertex(vBf), BLUE); DrawLine(ProjectVertex(vBf), ProjectVertex(vCf), BLUE); DrawLine(ProjectVertex(vCf), ProjectVertex(vDf), BLUE); DrawLine(ProjectVertex(vDf), ProjectVertex(vAf), BLUE); // The back face DrawLine(ProjectVertex(vAb), ProjectVertex(vBb), RED); DrawLine(ProjectVertex(vBb), ProjectVertex(vCb), RED); DrawLine(ProjectVertex(vCb), ProjectVertex(vDb), RED); DrawLine(ProjectVertex(vDb), ProjectVertex(vAb), RED); // The front-to-back edges DrawLine(ProjectVertex(vAf), ProjectVertex(vAb), GREEN); DrawLine(ProjectVertex(vBf), ProjectVertex(vBb), GREEN); DrawLine(ProjectVertex(vCf), ProjectVertex(vCb), GREEN); DrawLine(ProjectVertex(vDf), ProjectVertex(vDb), GREEN); We get something like Figure 9-4. Source code and live demo >> Success! We’ve managed to go from the geometrical 3D representation of an object to its 2D representation as seen from our synthetic camera! Our approach is very artisanal, though. It has many limitations. What if we want to render two cubes? Would we have to duplicate", "# Placing text on face of 3d cube How simple/hard would it be to add text to the front face (the red colored face in this example) of the cubes found at http://www.texample.net/tikz/examples/plane-partition/? I have seen examples like this (where text is placed on face of 3d cube) using \\node, but I'm not sure where to begin with this particular code. Here is the code: % Plane partition % Author: Jang Soo Kim \\documentclass{minimal} \\usepackage{tikz} % Three counters \\newcounter{x} \\newcounter{y} \\newcounter{z} % The angles of x,y,z-axes \\newcommand\\xaxis{210} \\newcommand\\yaxis{-30} \\newcommand\\zaxis{90} % The top side of a cube \\newcommand\\topside[3]{ \\fill[fill=yellow, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0); } % The left side of a cube \\newcommand\\leftside[3]{ \\fill[fill=red, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0); } % The right side of a cube \\newcommand\\rightside[3]{ \\fill[fill=blue, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0); } % The cube \\newcommand\\cube[3]{ \\topside{#1}{#2}{#3} \\leftside{#1}{#2}{#3} \\rightside{#1}{#2}{#3} } % Definition of \\planepartition % To draw the following plane partition, just write \\planepartition{ {a, b, c}, {d,e} }. % a b c % d e \\newcommand\\planepartition[1]{ \\setcounter{x}{-1} \\foreach \\a in {#1} { \\setcounter{y}{-1} \\foreach \\b in \\a { \\setcounter{z}{-1} \\foreach \\c in {1,...,\\b} { \\cube{\\value{x}}{\\value{y}}{\\value{z}} } } } } \\begin{document} \\begin{tikzpicture} \\planepartition{{5,3,2,2},{4,2,2,1},{2,1},{1}} \\end{tikzpicture} \\end{document} EDIT: I didn't think to search for the name of the code's", "out), maybe your program is in the wrong mode. You should be working with a gray scale image. Is Gimp in RGB? Indexed? 2. In any case, do Image → Mode → Grayscale, then continue on. 7. In the “Compose” dialog box, the “Compose Channels” section: 1. Choose “HSL” for color model 8. In the “Channel Representations” section: 1. Set your “Hue” layer to occupy the Hue slot. 2. Set your “Sat” layer to occupy the Saturation slot. 3. Set your “Lum” layer to occupy the Lightness slot. 9. Check. Double check. Fix what's wrong. Hit 'OK'. For all that, you should have a clockwise color wheel. If it looks like the one starting this section, red on top, cyan on bottom, colors running clockwise from red to orange to yellow to green, you're in. Do It In G'MIC #@gmic #@gmic colorwheel : _size>0,_cclock={ 0 | 1 },_angle #@gmic : Generates a square colorwheel in clockface orientation, with zero #@gmic : degrees on top, corresponding to red, and colors transitioning #@gmic : clockwise through green (120 degrees) and blue (240 degrees). size: #@gmic : colorwheel edges, in pixels. cclock, boolean, if true generates a #@gmic : counterclockwise colorwheel, and angle, degrees, rotates the wheel #@gmic : clockwise (negative: counterclockwise) from its reference orientation. #@gmic : Default values: _size=512,", "# Can't draw two pictures in Asymptote For some reason I can't draw two pictures onto currentpicture in Asymptote. Currently I have \\documentclass{article} \\usepackage[inline]{asymptote} \\begin{document} \\begin{asy}[width=0.7\\textwidth] import three; string d = \"$\"; triple A = (0,0,0); // Vertices of a cube triple B = (0,1,0); triple C = (-1,1,0); triple D = (-1,0,0); transform3 sd = shift(0,0,-1); triple Ap = sd * A; triple Bp = sd * B; triple Cp = sd * C; triple Dp = sd * D; void drawCube(picture pic, string[] labels) { draw(pic,A--B--C--Cp--Bp--Ap--A--B--Bp); draw(pic,B--C); draw(pic,A--D--C); draw(pic,Ap--Dp--D,dashed); // behind the cube draw(pic,Dp--Cp,dashed); if (labels.length > 0) { label(pic, d + labels[0] + d, A, NW); label(pic, d + labels[1] + d, B, N); label(pic, d + labels[2] + d, C, NE); label(pic, d + labels[3] + d, D, N); label(pic, d + labels[4] + d, Ap, SW); label(pic, d + labels[5] + d, Bp, S); label(pic, d + labels[6] + d, Cp, SE); label(pic, d + labels[7] + d, Dp, S); } } picture start; // first picture (the one with the cube) size(start, 100); string[] start_labels = {\"A\",\"B\",\"C\",\"D\",\"A'\",\"B'\",\"C'\",\"D'\"}; drawCube(start, start_labels); picture step1; // second figure, the one with the dashed line size(step1, 100); triple corner1 = (0,0,0); triple corner2 = (-1,1,-1); draw(step1, corner1--corner2,dashed); dot(step1, corner1); dot(step1, corner2); add(step1.fit(), (-2,0), N); // LINE 1", "origin and is deleted, the cube is the projective plane $$\\Proj{2}_k$$ over the finite field $$k$$ of order 2. The lines in this projective plane, suitably ordered, are \\begin{center} \\newcommand{\\Depth}{1} \\newcommand{\\Height}{1} \\newcommand{\\Width}{1} \\tdplotsetmaincoords{70}{110} \\begin{tikzpicture}[tdplot_main_coords]%Highlight the bottom \\coordinate (0) at (0,0,0); \\coordinate (1) at (\\Depth,0,\\Height); \\coordinate (2) at (\\Depth,\\Width,0); \\coordinate (3) at (0,\\Width,\\Height); \\coordinate (4) at (\\Depth,\\Width,\\Height); \\coordinate (5) at (0,\\Width,0); \\coordinate (6) at (0,0,\\Height); \\coordinate (7) at (\\Depth,0,0); % \\draw[gray!10,fill=gray!10] (0) -- (6) -- (1) -- (7) -- cycle;% Left Face \\draw[gray!10,fill=gray!110] (0) -- (7) -- (2) -- (5) -- cycle;% Bottom Face \\draw[gray!10,fill=gray!40] (0) -- (6) -- (3) -- (5) -- cycle;% Back Face \\draw[gray!10,fill=gray!20,opacity=0.6] (1) -- (4) -- (2) -- (7) -- cycle;% Front Face \\draw[gray!10,fill=gray!20,opacity=0.8] (1) -- (4) -- (3) -- (6) -- cycle;% Top Face \\draw[gray!10,fill=gray!20,opacity=0.8] (4) -- (3) -- (5) -- (2) -- cycle;% Right Face \\node at (1) {\\small$$1$$}; \\node at (2) {\\small$$2$$}; \\node at (3) {\\small$$3$$}; \\node at (4) {\\small$$4$$}; \\node at (5) {\\small$$5$$}; \\node at (6) {\\small$$6$$}; \\node at (7) {\\small$$7$$}; %% Following is for debugging purposes so you can see where the points are %% These are last so that they show up on top %\\foreach \\xy in {O, A, B, C, D, E, F, G}{ % \\node at (\\xy) {\\xy}; %} \\end{tikzpicture} \\begin{tikzpicture}[tdplot_main_coords]%Highlight the left \\coordinate (0) at", "can be viewed renderer. To support {\\tt latexmk}, 3D figures should specify \\verb+inline=true+. It is sometimes desirable to embed 3D files as annotated attachments; this requires the \\verb+attach=true+ option as well as the \\verb+attachfile2+ \\LaTeX\\ package. \\begin{center} \\begin{asy}[height=4cm,inline=true,attach=false,viewportwidth=\\linewidth] currentprojection=orthographic(5,4,2); draw(unitcube,blue); label(\"$V-E+F=2$\",(0,1,0.5),3Y,blue+fontsize(17pt)); \\end{asy} \\end{center} One can also scale the figure to the full line width: \\begin{center} \\begin{asy}[width=\\the\\linewidth,inline=true] pair z0=(0,0); pair z1=(2,0); pair z2=(5,0); pair zf=z1+0.75*(z2-z1); draw(z1--z2); dot(z1,red+0.15cm); dot(z2,darkgreen+0.3cm); label(\"$m$\",z1,1.2N,red); label(\"$M$\",z2,1.5N,darkgreen); label(\"$\\hat{\\ }$\",zf,0.2*S,fontsize(24pt)+blue); pair s=-0.2*I; draw(\"$x$\",z0+s--z1+s,N,red,Arrows,Bars,PenMargins); s=-0.5*I; draw(\"$\\bar{x}$\",z0+s--zf+s,blue,Arrows,Bars,PenMargins); s=-0.95*I; draw(\"$X$\",z0+s--z2+s,darkgreen,Arrows,Bars,PenMargins); \\end{asy} \\end{center} \\end{document} I get error like this : Command Line: asy.exe latexusage-*.asy Startup Folder: C:\\test /cygdrive/C//Program Files (x86)/Asymptote/config.asy: 2.5: syntax error error: could not load module '/cygdrive/C//Program Files (x86)/Asymptote/config.asy' /cygdrive/C//Program Files (x86)/Asymptote/config.asy: 2.5: syntax error error: could not load module '/cygdrive/C//Program Files (x86)/Asymptote/config.asy' /cygdrive/C//Program Files (x86)/Asymptote/config.asy: 2.5: syntax error error: could not load module '/cygdrive/C//Program Files (x86)/Asymptote/config.asy' I use Asymptote plugin at WinEdt 8.0 , Any solution to solve it ? • I think it is a path problem. Unfortunately I do not have Windows. Could you verify that cygdrive/C//Program Files (x86)/Asymptote/ is a valid directory and if it is verify that config.asy is present ? If cygdrive/C//Program Files (x86)/Asymptote/ is not valid, you have to search about path configuration. Good luck (the double // is strange) – O.G. Oct 18 '14 at 9:03 • Another thing to verify", "to your problem? – Jake Sep 13 '11 at 4:32 Please post the code that you have so far – Peter Grill Sep 13 '11 at 4:38 @Jake: Thanks a lot, that link give me some hints ;) – Chan Sep 13 '11 at 4:40 @Peter Grill: I would love to, but I really have no idea how to start this. – Chan Sep 13 '11 at 4:41 @Peter Grill: Many thanks for your hints. Finally I got it, not pretty but acceptable :D. – Chan Sep 13 '11 at 5:47 To start consider how you would draw a cube on paper, and draw those lines with TikZ. Start with one face and then add the sides, depending on what view you want. Here is one way to do it: \\documentclass{article} \\usepackage{tikz} \\usepackage{xcolor} \\begin{document} \\newcommand{\\Depth}{2} \\newcommand{\\Height}{2} \\newcommand{\\Width}{2} \\begin{tikzpicture} \\coordinate (O) at (0,0,0); \\coordinate (A) at (0,\\Width,0); \\coordinate (B) at (0,\\Width,\\Height); \\coordinate (C) at (0,0,\\Height); \\coordinate (D) at (\\Depth,0,0); \\coordinate (E) at (\\Depth,\\Width,0); \\coordinate (F) at (\\Depth,\\Width,\\Height); \\coordinate (G) at (\\Depth,0,\\Height); \\draw[blue,fill=yellow!80] (O) -- (C) -- (G) -- (D) -- cycle;% Bottom Face \\draw[blue,fill=blue!30] (O) -- (A) -- (E) -- (D) -- cycle;% Back Face \\draw[blue,fill=red!10] (O) -- (A) -- (B) -- (C) -- cycle;% Left Face \\draw[blue,fill=red!20,opacity=0.8] (D) -- (E) -- (F) -- (G) -- cycle;% Right Face", "much as if we were in 2D TikZ. So, setting settings.render to some other value is required for the plane to 'cover' stuff layered below. These should also probably really be surfaces, I guess, perhaps with no light. \\documentclass{article} \\usepackage{asymptote} \\begin{document} \\begin{asy} settings.outformat = \"pdf\"; settings.prc = false; settings.render = 16; import three; import bsp; texpreamble(\"\\usepackage{euler,beton}\"); size(5cm, 0); currentprojection=orthographic((5,4,3)); path3 pl =plane((0,-2,0),(-2,0,0),(0,1,0)); path3 pl1=rotate(-28,X)*pl; path3 pl2=rotate(-56,X)*pl; path3 pl3=shift(-0.3*normal(pl1))*pl1; draw (pl); draw (pl1); draw (pl2); draw (pl3); triple[] asd=intersectionpoints(pl2,pl3); triple A=asd[1]; triple B=asd[0]; triple[] asf=intersectionpoints(pl,pl2); triple C=asf[1]; triple D=asf[0]; triple[] asg=intersectionpoints(pl,pl3); triple E=asg[1]; triple F=asg[0]; surface s1=surface(pl); draw(s1,white+opacity(.6),light=nolight); surface s2=surface(pl1); draw(s2,white+opacity(.6),light=nolight); surface s3=surface(pl2); draw(s3,white+opacity(.6),light=nolight); surface s4=surface(pl3); draw(s4,white+opacity(.6),light=nolight); draw(B--A); draw(C--D); draw(E--F); dot(A,red); dot(B,red); dot(C,red); dot(D,red); dot(E,red); dot(F,red); //From Charles Staats's tutorial //Direction of a point toward the camera. triple cameradirection(triple pt, projection P=currentprojection) { if (P.infinity) { return unit(P.camera); } else { return unit(P.camera - pt); } } //Move a point closer to the camera. triple towardcamera(triple pt, real distance=1, projection P=currentprojection) { return pt + distance * cameradirection(pt, P); } label(\"$A$\",align=NE,position=towardcamera((A))); label(\"$B$\",align=S,position=towardcamera((B))); label(\"$C$\",align=SE,position=towardcamera((C))); label(\"$D$\",align=SE,position=towardcamera((D))); label(\"$E$\",align=NE,position=towardcamera((E))); label(\"$F$\",align=S,position=towardcamera((F))); \\end{asy} \\end{document} # EDIT If you simply must have vector graphics - a restriction which was not mentioned in the question - then you need to take care of the drawing order yourself. This requires splitting each surface into pieces and reassembling them", "draw(pic,A--B--C--Cp--Bp--Ap--A--B--Bp); draw(pic,B--C); draw(pic,A--D--C); draw(pic,Ap--Dp--D,dashed); // behind the cube draw(pic,Dp--Cp,dashed); if (labels.length > 0) { label(pic, d + labels[0] + d, A, NW); label(pic, d + labels[1] + d, B, N); label(pic, d + labels[2] + d, C, NE); label(pic, d + labels[3] + d, D, N); label(pic, d + labels[4] + d, Ap, SW); label(pic, d + labels[5] + d, Bp, S); label(pic, d + labels[6] + d, Cp, SE); label(pic, d + labels[7] + d, Dp, S); } } picture start; // first picture (the one with the cube) size(start, 100); string[] start_labels = {\"A\",\"B\",\"C\",\"D\",\"A'\",\"B'\",\"C'\",\"D'\"}; drawCube(start, start_labels); picture step1; // second figure, the one with the dashed line size(step1, 100); triple corner1 = (0,0,0); triple corner2 = (-1,1,-1); draw(step1, corner1--corner2,dashed); dot(step1, corner1); dot(step1, corner2); add(step1.fit(), (-2,0), N); // LINE 1 add(start.fit(), (0,0), N); // LINE 2 label(\"Starting position\", (0,0),S); label(\"Choose a pair of vertices: $2$ ways\", (-2,0), S); \\end{asypicture} \\end{document} I get I know that the overlapping text should not be there but at least the issue you refer to does not arise. • Thank you so much! If I had known about this package a month ago it would have saved me so much time... Do you know if asydef works with asypicture? – Ovinus Real Jun 5 at 22:12 • @OvinusReal Sorry, I do not." ]

[ "is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ ## Problem 14 In this addition problem, each letter stands for a different digit. $\\setlength{\\tabcolsep}{0.5mm}\\begin{array}{cccc}&T & W & O\\\\ + &T & W & O\\\\ \\hline F& O & U & R\\end{array}$ If $T = 7$ and the letter $O$ represents an even number, what is the only possible value for $W$? $\\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 1 \\qquad \\textbf{(C)}\\ 2\\qquad \\textbf{(D)}\\ 3\\qquad \\textbf{(E)}\\ 4$ ## Problem 15 A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown? $[asy] defaultpen(linewidth(0.8)); path p=unitsquare; draw(p^^shift(0,1)*p^^shift(1,0)*p); draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p); label(\"FRONT\", (1,0), S); label(\"SIDE\", (5,0), S);[/asy]$ $\\textbf{(A)}\\ 3\\qquad\\textbf{(B)}\\ 4\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 6\\qquad\\textbf{(E)}\\ 7$ ## Problem 16 Ali, Bonnie, Carlo,", "the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ Jan 5: A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(1,1/2,1/4); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(1,-1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,-1)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(1,1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)} \\:1 : 6 \\qquad\\textbf{ (B)}\\: 7 : 36 \\qquad\\textbf{(C)}\\: 1 : 5 \\qquad\\textbf{(D)}\\: 7 : 30\\qquad\\textbf{ (E)}\\: 6 : 25$ Jan 6: A cube has edge length 2. Suppose that we glue a cube of edge length 1 on top of the big cube so that one of its faces rests entirely on the top face", "cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ ## Problem 14 In this addition problem, each letter stands for a different digit. $\\setlength{\\tabcolsep}{0.5mm}\\begin{array}{cccc}&T & W & O\\\\ \\plus{} &T & W & O\\\\ \\hline F& O & U & R\\end{array}$ (Error compiling LaTeX. ! Undefined control sequence.) If T = 7 and the letter O represents an even number, what is the only possible value for W? $\\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 1 \\qquad \\textbf{(C)}\\ 2\\qquad \\textbf{(D)}\\ 3\\qquad \\textbf{(E)}\\ 4$ ## Problem 15 A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown? $[asy] defaultpen(linewidth(0.8)); path p=unitsquare; draw(p^^shift(0,1)*p^^shift(1,0)*p); draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p); label(\"FRONT\", (1,0), S); label(\"SIDE\", (5,0), S);[/asy]$ $\\textbf{(A)}\\ 3\\qquad\\textbf{(B)}\\ 4\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 6\\qquad\\textbf{(E)}\\ 7$ ## Problem 16 Ali, Bonnie, Carlo, and Dianna are", "the same center For instance, in the following code, we take conventions of colors RGB for the axis XYZ. The instruction yellowCube.against(brownCube,-Z,Y,X,X) means that vectors -Z (opposite to the blue half-line) of the yellow cube is in the same direction that the Y ( green) of the Brown Cube. The face -Z of YellowCube is indeed a face of contact. And the two X axis (Red) are positivly parallel. Suppose that we want to move the yellow cube to the left till the edges of both Cubes coincide. We need to push Yellow Cube in direction opposite to the red, ie. -X. Thus we add the option ajustedges=-X ie we put yellowCube.against(brownCube,-Z,Y,X,X,adjustEdges=-X) With yellowCube.against(brownCube,-Z,Y,X,X,adjustEdges=X+Y), the yellow Cube goes to the upper right. With, yellowCube.against(brownCube,-Z,Y,X,X,adjustEdges=0.8*(X+Y), it nearly goes to the upper right, as shown-below. If we want to put the yellow cube off the brown cube, we can add an offset. The offset is in global coordinates. In our picture, the yellow cube has been rotated, but the brown cube has not moved so its coordinates are still parallel to the global axis of the scene. Thus the offset must be a negative mutltiple of the green axis Y of the Brown Cube. For instance, we may put yellowCube.against(brownCube,-Z,Y,X,X,adjustEdges=.8*(X+Y),offset=-0.7*Y) Finally, it is possible to add the option adjustAxis=[p1,p2]. The", "are painted in six different colors: red $(R)$, white $(W)$, green $(G)$, brown $(B)$, aqua $(A)$, and purple $(P)$. Three views of the cube are shown below. What is the color of the face opposite the aqua face? $[asy] unitsize(2 cm); pair x, y, z, trans; int i; x = dir(-5); y = (0.6,0.5); z = (0,1); trans = (2,0); for (i = 0; i <= 2; ++i) { draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle)); draw(shift(i*trans)*((x + z)--x)); draw(shift(i*trans)*((x + z)--(x + y + z))); draw(shift(i*trans)*((x + z)--z)); } label(rotate(-3)*\"R\", (x + z)/2); label(rotate(-5)*slant(0.5)*\"B\", ((x + z) + (y + z))/2); label(rotate(35)*slant(0.5)*\"G\", ((x + z) + (x + y))/2); label(rotate(-3)*\"W\", (x + z)/2 + trans); label(rotate(50)*slant(-1)*\"B\", ((x + z) + (y + z))/2 + trans); label(rotate(35)*slant(0.5)*\"R\", ((x + z) + (x + y))/2 + trans); label(rotate(-3)*\"P\", (x + z)/2 + 2*trans); label(rotate(-5)*slant(0.5)*\"R\", ((x + z) + (y + z))/2 + 2*trans); label(rotate(-85)*slant(-1)*\"G\", ((x + z) + (x + y))/2 + 2*trans); [/asy]$ $\\textbf{(A) }\\text{red}\\qquad\\textbf{(B) }\\text{white}\\qquad\\textbf{(C) }\\text{green}\\qquad\\textbf{(D) }\\text{brown}\\qquad\\textbf{(E) }\\text{purple}$ ## Problem 13 A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 25\" ## Problem A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $\\frac{1}{2}$ foot from the top face. The second cut is $\\frac{1}{3}$ foot below the first cut, and the third cut is $\\frac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$ ## Solution 1 The areas of the tops of $A$, $B$, $C$, and $D$ in", "// Lighting currentlight=Viewport; real wi = 4; real le = 4; material m = material(grey, yellow, black, orange); triple[] wheels = {(-le/2,-wi/2,0), (-le/2,wi/2,0), (le/2,-wi/2,0), (le/2,wi/2,0) }; // lines draw((-le/2,wi/2,0)--(-le/2,wi,0)); draw((le/2,wi/2,0)--(le/2,wi,0)); // wheels for(triple p : wheels) { draw(shift(p)*rotate(90,(1,0,0))* scale(1,1,0.2)*shift((0,0,-0.5))*unitcylinder,m); draw(shift(p)*rotate(90,(1,0,0))* shift((0,0,-0.1))*unitdisk,m); draw(shift(p)*rotate(90,(1,0,0))* shift((0,0,0.1))*unitdisk,m); } // axes draw(shift(-le/2,0,0)*rotate(90,(1,0,0))* scale(0.1,0.1,wi)*shift((0,0,-0.5))*unitcylinder,m); draw(shift(le/2,0,0)*rotate(90,(1,0,0))* scale(0.1,0.1,wi)*shift((0,0,-0.5))*unitcylinder,m); // box draw(scale(5,1,1)*shift(-0.5,-0.5,-0.5)*unitcube,m); draw((-le/2,0.9*wi,0)--(le/2,0.9*wi,0),Arrows3); draw(scale(0.8,1,1)*labelpath(\"wheelbase $w$\",(le/2,wi,0)--(-le/2,wi,0))); \\end{asy} \\end{document} Here's the result: • Why you don't try pov-ray ? – roumi Jun 4 '15 at 21:07", "back, (R = red, B = blue, O = orange, Y = yellow, G = green, and W = white). They are hinged together as shown, then folded to form a cube. The face opposite the white face is $[asy] draw((0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,2)--(2,2)--(2,3)--(0,3)--cycle); draw((1,3)--(1,2)--(2,2)--(2,1)--(3,1)--(3,2)); label($ $\\text{(A)}\\ \\text{B} \\qquad \\text{(B)}\\ \\text{G} \\qquad \\text{(C)}\\ \\text{O} \\qquad \\text{(D)}\\ \\text{R} \\qquad \\text{(E)}\\ \\text{Y}$ Jan 2: Three dice with faces numbered 1 through 6 are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is $[asy] draw((0,0)--(2,0)--(3,1)--(3,7)--(1,7)--(0,6)--cycle); draw((3,7)--(2,6)--(0,6)); draw((3,5)--(2,4)--(0,4)); draw((3,3)--(2,2)--(0,2)); draw((2,0)--(2,6)); dot((1,1)); dot((.5,.5)); dot((1.5,.5)); dot((1.5,1.5)); dot((.5,1.5)); dot((2.5,1.5)); dot((.5,2.5)); dot((1.5,2.5)); dot((1.5,3.5)); dot((.5,3.5)); dot((2.25,2.75)); dot((2.5,3)); dot((2.75,3.25)); dot((2.25,3.75)); dot((2.5,4)); dot((2.75,4.25)); dot((.5,5.5)); dot((1.5,4.5)); dot((2.25,4.75)); dot((2.5,5.5)); dot((2.75,6.25)); dot((1.5,6.5)); [/asy]$ $\\text{(A)}\\ 21 \\qquad \\text{(B)}\\ 22 \\qquad \\text{(C)}\\ 31 \\qquad \\text{(D)}\\ 41 \\qquad \\text{(E)}\\ 53$ Jan 3: Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom and sides. $[asy]/* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(1,2)--(1.5,2)); draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5)); draw((1.5,3.5)--(2.5,3.5)); draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5)); draw((3,4)--(3,3)--(2.5,2.5)); draw((3,3)--(4,3)--(4,2)--(3.5,1.5)); draw((4,3)--(3.5,2.5)); draw((2.5,.5)--(3,1)--(3,1.5));[/asy]$ $\\text{(A)}\\ 18 \\qquad \\text{(B)}\\ 24 \\qquad \\text{(C)}\\ 26 \\qquad \\text{(D)}\\ 30 \\qquad \\text{(E)}\\ 36$ Jan 4: Fourteen white cubes are put together to form the figure on the right. The complete surface of", "some text\",white,Fill(blue)); ## Question 4.8. How can I rotate labels in a 3D figure? You need to first project the triple to a pair like this: import three; size(100,100); draw(rotate(90,project(Z))*\"A\",O--X); ## Question 4.9. How can I draw some squares and circles of a fixed size and put a label in the middle of them? Fixed-size objects should be drawn on a separate picture and then added to currentpicture. Here is one way (see also http://asymptote.sourceforge.net/gallery/subpictures.asy and http://asymptote.sourceforge.net/gallery/mosquito.asy): real u=2cm; picture square; draw(square,scale(u)*shift(-0.5,-0.5)*unitsquare); picture circle; draw(circle,scale(0.5u)*unitcircle); void add(picture pic=currentpicture, Label L, picture object, pair z) { label(pic,L,z); } ## Question 4.10. The binary operator * can be used to scale the color of a pen by a real number. Does this scaling factor have to be less than 1? The scaling factor can be greater than 1. But keep in mind that the rgb color components saturate at 1. Try write(cyan); write(0.8*cyan); write(1.5*cyan); and you will quickly see what is going on. To get a lighter cyan you can say white+cyan, which yields rgb(0.5,1,1). If you want something even lighter specify the rgb colors directly, for example, rgb(0.9,1,1). Alternatively, work in cmyk colour space, which is nicer in that it handles saturation separately from hue: 0.1*Cyan is light and 0.9*Cyan is dark. You can also say 0.1*cmyk(red). ## Question", "(o-\\cubelabel) -- (a-\\cubelabel) -- ++(0,0,-\\cubescale*\\cubez) coordinate (f-\\cubelabel) edge (g-\\cubelabel) -- (d-\\cubelabel) -- cycle; ; }, /cuboid/.search also={/tikz}, /cuboid/.cd, width/.store in=\\cubex, height/.store in=\\cubey, depth/.store in=\\cubez, units/.store in=\\cubeunits, scale/.store in=\\cubescale, label/.store in=\\cubelabel, line/.store in=\\cubeline, backline/.store in=\\cubeback, width=10, height=10, depth=10, units=cm, scale=.1, line=draw, backline=densely dashed, } \\begin{document} \\begin{tikzpicture} \\pic [fill=gray!20, text=green!50!black, draw=black] at (4,-2) {annotated cuboid={label=A, width=6, height=20, depth=15, units=mm}}; \\pic [fill=red!20, text=green!50!black, draw=black] at (4,-2.7) {annotated cuboid={label=B, width=2, height=4, depth=3, units=m, line=dashed}}; \\pic [fill=blue!20, text=green!50!black, draw=black] at (5,-2) {annotated cuboid={label=C, width=6, height=20, depth=15, units=mm}}; \\pic [fill=green!20, text=green!50!black, draw=black] at (4.8,-2.7) {annotated cuboid={label=D, width=2, height=2, depth=2, units=m, line=dashed}}; \\pic [fill=orange!20, text=green!50!black, draw=black] at (5.5,-2.3) {annotated cuboid={label=E, width=3, height=10, depth=7, units=m, backline=draw}}; \\draw (e-B) -- (a-D); \\end{tikzpicture} \\end{document} So all that I have done is added a label to your annotated cuboid, which gets stored in \\cubelabelm and changed your coordinates ub the cuboid code to(o),(a),...,(g) to (o-\\cubelabel),...,(g-\\cubelabel). Then the annotated cuboid commnds are called with label=* and the \\draw commands work. • Great. How to make the line of inside cube is the dash line – John Apr 5 '17 at 4:45 • @user8264 I've updated the code to add another argument, line, that sets the \"line style\". Adding line=dashed will make the lines of the cuboid dashed lines, line=dotted will make dotted line etc. – Andrew Apr 5 '17 at", "Difference between revisions of \"1996 AIME Problems/Problem 7\" Problem Two squares of a $7\\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? Solution There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. $[asy] pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,3)*unitsquare,rgb(.4,.8,.4));fill(shift(4,5)*unitsquare,rgb(.4,.8,.4)); fill(shift(3,4)*unitsquare,rgb(.7,1,.7));fill(shift(1,4)*unitsquare,rgb(.7,1,.7)); fill(shift(2,3)*unitsquare,rgb(.7,1,.7));fill(shift(2,1)*unitsquare,rgb(.7,1,.7)); fill(shift(3,2)*unitsquare,rgb(.7,1,.7));fill(shift(5,2)*unitsquare,rgb(.7,1,.7)); D(arc(O,1,280,350),EndArrow(4)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,1,10,80),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,1,190,260),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,1,100,170),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ $[asy]pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(2,1)*unitsquare,rgb(1,1,.4)); fill(shift(1,4)*unitsquare,rgb(.4,.4,.7)); fill(shift(5,2)*unitsquare,rgb(.4,.4,.7)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ For most pairs, there will be three other equivalent boards. For those symmetric about the center, there is only one other. Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\\frac{49-1}{2}=24$ such pairs. There are then ${49 \\choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each", "[fig:cornell-g3d]: Seven instances of cube.obj.](cornell-box-cubes.png height=200px) This Cornell Box can be modeled using seven instances of rotated, translated, and scaled cubes. When creating your Cornell Box, scale the _models_, but don't rotate and translate them. Instead rotate and translate the _entity_ placement. Were we animating the scene, that design would give us more intuitive control of the objects. It also lets us reuse objects. For example, all three white walls should be different entities that use the same model. The setMaterial command is one of many preprocessing options that can appear in a [G3D::ArticulatedModel::Specification](http://g3d.cs.williams.edu/g3d/G3D10/build/manual/class_g3_d_1_1_articulated_model_1_1_specification.html#details) in a scene file. To change the dimensions of a cube, you can use the transformGeometry command, which might make part of your data file look like: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ squishedCube = ArticulatedModel::Specification { filename = \"models/cube/cube.obj\"; preprocess = { setMaterial(all(), Color3(1, 0, 0)); transformGeometry(all(), Matrix4::scale(0.5, 1.0, 2.0)); }; }; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ You can chose the scale and need not worry about the precise colors and angles. Ensure that the walls have nonzero thickness. I chose 2 cm-thick walls for a 1 m^3 box, which is about the scale of the real-world Cornell Box. ## The Staircase You now need to create a staircase. Specifically, a staircase with some plausible material. To make a cube appear to be made out of wood instead of solid-colored plastic, we'll", "by creating a scene by hand. In your editor, create a new file data-files/scene/whiteCube.Scene.Any with the contents: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ // -*- c++ -*- { name = \"White Cube\"; models = { cubeModel = ArticulatedModel::Specification { filename = \"model/cube/cube.obj\"; preprocess = { setMaterial(all(), Color3(1, 1, 0)); }; }; }; entities = { skybox = Skybox { texture = \"cubemap/whiteroom/whiteroom-*.png\"; }; sun = Light { attenuation = ( 0, 0, 1 ); bulbPower = Power3(4e+006); frame = CFrame::fromXYZYPRDegrees(-15, 207, -41, -164, -77, 77); shadowMapSize = Vector2int16(2048, 2048); spotHalfAngleDegrees = 5; spotSquare = true; type = \"SPOT\"; }; cube0 = VisibleEntity { model = \"cubeModel\"; frame = CFrame::fromXYZYPRDegrees(0, 0, 0, 0, 0, 0); }; camera = Camera { frame = CFrame::fromXYZYPRDegrees(0, 0, 5); }; }; }; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Launch your program and the new \"White Cube\" scene should appear in the scene drop down box of the GUI. ## Fix a Scene Error When you load the scene, you'll notice that cube in this scene is actually yellow and not white. You should fix that. You can either guess how, or look at the [G3D::UniversalMaterial::Specification](http://g3d.cs.williams.edu/g3d/G3D10/build/manual/class_g3_d_1_1_universal_material_1_1_specification.html#a12d19df20a48ad996ae5b6538bdd3d9b) documentation to see the arguments that can provided in the scene data file for constructing materials. When you've made your changes, press the reload button in the Scene Editor Window without ever exiting your program. ### Add the", "# Placing text on face of 3d cube How simple/hard would it be to add text to the front face (the red colored face in this example) of the cubes found at http://www.texample.net/tikz/examples/plane-partition/? I have seen examples like this (where text is placed on face of 3d cube) using \\node, but I'm not sure where to begin with this particular code. Here is the code: % Plane partition % Author: Jang Soo Kim \\documentclass{minimal} \\usepackage{tikz} % Three counters \\newcounter{x} \\newcounter{y} \\newcounter{z} % The angles of x,y,z-axes \\newcommand\\xaxis{210} \\newcommand\\yaxis{-30} \\newcommand\\zaxis{90} % The top side of a cube \\newcommand\\topside[3]{ \\fill[fill=yellow, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0); } % The left side of a cube \\newcommand\\leftside[3]{ \\fill[fill=red, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0); } % The right side of a cube \\newcommand\\rightside[3]{ \\fill[fill=blue, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0); } % The cube \\newcommand\\cube[3]{ \\topside{#1}{#2}{#3} \\leftside{#1}{#2}{#3} \\rightside{#1}{#2}{#3} } % Definition of \\planepartition % To draw the following plane partition, just write \\planepartition{ {a, b, c}, {d,e} }. % a b c % d e \\newcommand\\planepartition[1]{ \\setcounter{x}{-1} \\foreach \\a in {#1} { \\setcounter{y}{-1} \\foreach \\b in \\a { \\setcounter{z}{-1} \\foreach \\c in {1,...,\\b} { \\cube{\\value{x}}{\\value{y}}{\\value{z}} } } } } \\begin{document} \\begin{tikzpicture} \\planepartition{{5,3,2,2},{4,2,2,1},{2,1},{1}} \\end{tikzpicture} \\end{document} EDIT: I didn't think to search for the name of the code's", "object: a cube. We define the coordinates of its 8 vertices, and we draw line segments between the projections of the 12 pairs of vertices that make the edges of the cube, as seen in Listing 9-1: ViewportToCanvas(x, y) { return (x * Cw/Vw, y * Ch/Vh); } ProjectVertex(v) { return ViewportToCanvas(v.x * d / v.z, v.y * d / v.z) } // The four \"front\" vertices vAf = [-2, -0.5, 5] vBf = [-2, 0.5, 5] vCf = [-1, 0.5, 5] vDf = [-1, -0.5, 5] // The four \"back\" vertices vAb = [-2, -0.5, 6] vBb = [-2, 0.5, 6] vCb = [-1, 0.5, 6] vDb = [-1, -0.5, 6] // The front face DrawLine(ProjectVertex(vAf), ProjectVertex(vBf), BLUE); DrawLine(ProjectVertex(vBf), ProjectVertex(vCf), BLUE); DrawLine(ProjectVertex(vCf), ProjectVertex(vDf), BLUE); DrawLine(ProjectVertex(vDf), ProjectVertex(vAf), BLUE); // The back face DrawLine(ProjectVertex(vAb), ProjectVertex(vBb), RED); DrawLine(ProjectVertex(vBb), ProjectVertex(vCb), RED); DrawLine(ProjectVertex(vCb), ProjectVertex(vDb), RED); DrawLine(ProjectVertex(vDb), ProjectVertex(vAb), RED); // The front-to-back edges DrawLine(ProjectVertex(vAf), ProjectVertex(vAb), GREEN); DrawLine(ProjectVertex(vBf), ProjectVertex(vBb), GREEN); DrawLine(ProjectVertex(vCf), ProjectVertex(vCb), GREEN); DrawLine(ProjectVertex(vDf), ProjectVertex(vDb), GREEN); We get something like Figure 9-4. Source code and live demo >> Success! We’ve managed to go from the geometrical 3D representation of an object to its 2D representation as seen from our synthetic camera! Our approach is very artisanal, though. It has many limitations. What if we want to render two cubes? Would we have to duplicate", "The Computer Graphics Matrix and a Cube Use a cube instead of the Utah Teapot in my previous post. I was pleasantly suprised by the final screen shot. Contents RGB cube Three faces of our cube are colored with red, green and blue. The opposite faces have the complentary colors, cyan, magenta and yellow. Initially, only the yellow face is visible. Cube = grafix('6','rgb'); % Platonic solid with 6 faces. Rz Rz = R_z(45); apply(Cube,Rz) Rx*Rz Rx = R_x(60); M = Rx*Rz; apply(Cube,M) Rz*Rx M = Rz*Rx; apply(Cube,M) Ry Ry = R_y(-120); apply(Cube,Ry) Tx Tx = T_x(2.5); apply(Cube,Tx) Sxyz Sxyz = S_xyz(0.75); apply(Cube,Sxyz) All together M = Sxyz*Tx*Ry*Rz*Rx; apply(Cube,M) M = Rx*Rz*Ry*Tx*Sxyz; apply(Cube,M) Clipped Finally, a scrambled order. The cube is clipped by the plot box and we see though the interior to the opposite faces. All six colors are visible. M = Rz*Ry*Sxyz*Tx*Rx; apply(Cube,M)", "= 4, 3, 7, green 4 = 5, 4, 7, blue 5 = 5, 7, 6, blue 6 = 1, 5, 6, yellow 7 = 1, 6, 2, yellow 8 = 4, 5, 1, purple 9 = 4, 1, 0, purple 10 = 2, 6, 7, cyan 11 = 2, 7, 3, cyan Rendering an object with this representation is quite easy: you first project every vertex, storing them in a temporary “projected vertexes” list; then you go through the triangle list rendering each individual triangle. A first approximation would look like this: RenderObject(vertexes, triangles) { projected = [] for V in vertexes { projected.append(ProjectVertex(V)) } for T in triangles { RenderTriangle(T, projected) } } RenderTriangle(triangle, projected) { DrawWireframeTriangle(projected[triangle.v[0]], projected[triangle.v[1]], projected[triangle.v[2]], triangle.color) } We can’t just apply this algorithm to the cube as we have it and expect things to look reasonable, because some of its vertexes are behind the camera; which, as we have seen, is a recipe for weird things. In fact, the camera is inside the cube. So we’ll just move the cube. To achieve this, we just need to move each vertex of the cube in the same direction. Let’s call this vector $$T$$, for Translation. Let’s just translate the cube 7 units forward, making sure it’s completely in front of the camera, and", "# Difference between revisions of \"1996 AIME Problems/Problem 7\" ## Problem Two squares of a $7\\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? ## Solution 1 (Generalized) There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. $[asy] pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(2,3)*unitsquare,rgb(.8,.8,.5));fill(shift(2,1)*unitsquare,rgb(.8,.8,.5)); fill(shift(3,2)*unitsquare,rgb(.8,.8,.5));fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,1,280,350),EndArrow(4)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,1,10,80),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,1,190,260),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,1,100,170),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ $[asy]pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(2,1)*unitsquare,rgb(1,1,.4)); fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ For most pairs, there will be three other equivalent boards. For those symmetric about the center, there is only one other. Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\\frac{49-1}{2}=24$ such pairs. There are then ${49 \\choose 2}-24$ pairs that yield $4$ distinct boards upon rotation;", "= plt.figure(figsize=(3, 1.3), facecolor='w') ax = plt.axes([0, 0, 1, 1], xticks=[], yticks=[], frameon=False) # first block draw_cube(ax, (1, 7.5), 1, depth, [1, 2, 3, 4, 5, 6, 9], '1', **solid) draw_cube(ax, (2, 7.5), 1, depth, [1, 2, 3, 6, 7, 9, 10], '2', **solid) draw_cube(ax, (1, 6.5), 1, depth, [2, 3, 4], '4', **solid) draw_cube(ax, (2, 6.5), 1, depth, [2, 3, 7, 10], '5', **solid) draw_cube(ax, (1, 5.5), 1, depth, [2, 3, 4], '7', **solid) draw_cube(ax, (2, 5.5), 1, depth, [2, 3, 7, 10], '8', **solid) # second block draw_cube(ax, (6, 7.5), 1, depth, [1, 2, 3, 4, 5, 6, 9], '3', **solid) draw_cube(ax, (7, 7.5), 1, depth, [1, 2, 3, 6, 9], '6', **solid) draw_cube(ax, (8, 7.5), 1, depth, [1, 2, 3, 6, 7, 9, 10], '9', **solid) draw_cube(ax, (6, 6.5), 1, depth, range(2, 13), '3', **dotted) draw_cube(ax, (7, 6.5), 1, depth, [2, 3, 6, 7, 9, 10, 11], '6', **dotted) draw_cube(ax, (8, 6.5), 1, depth, [2, 3, 6, 7, 9, 10, 11], '9', **dotted) draw_cube(ax, (6, 5.5), 1, depth, [2, 3, 4, 7, 8, 10, 11, 12], '3', **dotted) draw_cube(ax, (7, 5.5), 1, depth, [2, 3, 7, 10, 11], '6', **dotted) draw_cube(ax, (8, 5.5), 1, depth, [2, 3, 7, 10, 11], '9', **dotted) ax.text(4.5, 7.0, '+', size=12, ha='center', va='center') ax.text(10, 7.0, '=', size=12, ha='center', va='center') ax.text(11,", "How Can I draw a cube of cubes efficiently? [duplicate] I've tried to draw a cube made by cubes on each side with TikZ. I mean a 3x3 cube with a little shift between each one. I've looked for an answer but the first hard step is starting with a good way to draw a cube. Then a need to put them on the edges of the 3x3 big cube. Basically, I just want to show three faces of our cube such that looks something like this: But my pretty sad code to show it is this: \\documentclass{minimal} \\usepackage{tikz} % Three counters \\newcounter{x} \\newcounter{y} \\newcounter{z} % The angles of x,y,z-axes \\newcommand\\xaxis{210} \\newcommand\\yaxis{-30} \\newcommand\\zaxis{90} % The top side of a cube \\newcommand\\topside[3]{ \\fill[fill=red!50, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0); } % The left side of a cube \\newcommand\\leftside[3]{ \\fill[fill=red, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0); } % The right side of a cube \\newcommand\\rightside[3]{ \\fill[fill=red!50!black, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);} % The cube \\newcommand\\cube[3]{ \\topside{#1}{#2}{#3} \\leftside{#1}{#2}{#3} \\rightside{#1}{#2}{#3} } % Definition of \\planepartition % To draw the following plane partition, just write \\planepartition{ {a, b,c}, {d,e} }. % a b c % d e \\newcommand\\planepartition[1]{ \\setcounter{x}{-1} \\foreach \\a in {#1} { \\setcounter{y}{-1} \\foreach \\b in \\a { \\setcounter{z}{-1} \\foreach \\c in {1,...,\\b} {" ]

[ "# Difference between revisions of \"2005 AMC 12A Problems/Problem 17\" ## Problem A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the same manner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volume of the piece that contains vertex $W$? $[asy] path a=(0,0)--(10,0)--(10,10)--(0,10)--cycle; path b = (0,10)--(6,16)--(16,16)--(16,6)--(10,0); path c= (10,10)--(16,16); path d= (0,0)--(3,13)--(13,13)--(10,0); path e= (13,13)--(16,6); draw(a,linewidth(0.7)); draw(b,linewidth(0.7)); draw(c,linewidth(0.7)); draw(d,linewidth(0.7)); draw(e,linewidth(0.7)); draw(shift((20,0))*a,linewidth(0.7)); draw(shift((20,0))*b,linewidth(0.7)); draw(shift((20,0))*c,linewidth(0.7)); draw(shift((20,0))*d,linewidth(0.7)); draw(shift((20,0))*e,linewidth(0.7)); draw((20,0)--(25,10)--(30,0),dashed); draw((25,10)--(31,16)--(36,6),dashed); draw((15,0)--(10,10),Arrow); draw((15.5,0)--(30,10),Arrow); label(\"W\",(15.2,0),S); label(\"Figure 1\",(5,0),S); label(\"Figure 2\",(25,0),S); [/asy]$ $(\\mathrm {A}) \\ \\frac{1}{12} \\qquad (\\mathrm {B}) \\ \\frac{1}{9} \\qquad (\\mathrm {C})\\ \\frac{1}{8} \\qquad (\\mathrm {D}) \\ \\frac{1}{6} \\qquad (\\mathrm {E})\\ \\frac{1}{4}$ ## Solution It is a pyramid with height $1$ and base area $\\frac{1}{4}$, so using the formula for the volume of a pyramid, $\\frac{1}{3} \\cdot \\left(\\frac{1}{4}\\right) \\cdot (1) = \\frac {1}{12} \\Rightarrow \\boxed{(\\mathrm {A})}$.", "is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ ## Problem 14 In this addition problem, each letter stands for a different digit. $\\setlength{\\tabcolsep}{0.5mm}\\begin{array}{cccc}&T & W & O\\\\ + &T & W & O\\\\ \\hline F& O & U & R\\end{array}$ If $T = 7$ and the letter $O$ represents an even number, what is the only possible value for $W$? $\\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 1 \\qquad \\textbf{(C)}\\ 2\\qquad \\textbf{(D)}\\ 3\\qquad \\textbf{(E)}\\ 4$ ## Problem 15 A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown? $[asy] defaultpen(linewidth(0.8)); path p=unitsquare; draw(p^^shift(0,1)*p^^shift(1,0)*p); draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p); label(\"FRONT\", (1,0), S); label(\"SIDE\", (5,0), S);[/asy]$ $\\textbf{(A)}\\ 3\\qquad\\textbf{(B)}\\ 4\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 6\\qquad\\textbf{(E)}\\ 7$ ## Problem 16 Ali, Bonnie, Carlo,", "+0 # Help -1 296 1 +421 When this net of six squares is cut out and folded to form a cube, what is the product of the numbers on the four faces adjacent to the one labeled with a \"1\"? [asy] draw(unitsquare); draw(shift(1,0)*unitsquare); draw(shift(1,-1)*unitsquare); draw(shift(1,1)*unitsquare); draw(shift(2,1)*unitsquare); draw(shift(2,2)*unitsquare); label(\"4\",(0.5,0.5)); label(\"5\",(1.5,0.5)); label(\"6\",(1.5,-0.5)); label(\"3\",(1.5,1.5)); label(\"2\",(2.5,1.5)); label(\"1\",(2.5,2.5)); [/asy] May 12, 2020 #1 +25543 +1 When this net of six squares is cut out and folded to form a cube, what is the product of the numbers on the four faces adjacent to the one labeled with a \"$$1$$\"? The product is $$\\mathbf{2\\times 3 \\times 4 \\times 6 = 144}$$ May 13, 2020", "the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ Jan 5: A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(1,1/2,1/4); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(1,-1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,-1)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(1,1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)} \\:1 : 6 \\qquad\\textbf{ (B)}\\: 7 : 36 \\qquad\\textbf{(C)}\\: 1 : 5 \\qquad\\textbf{(D)}\\: 7 : 30\\qquad\\textbf{ (E)}\\: 6 : 25$ Jan 6: A cube has edge length 2. Suppose that we glue a cube of edge length 1 on top of the big cube so that one of its faces rests entirely on the top face", "cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ ## Problem 14 In this addition problem, each letter stands for a different digit. $\\setlength{\\tabcolsep}{0.5mm}\\begin{array}{cccc}&T & W & O\\\\ \\plus{} &T & W & O\\\\ \\hline F& O & U & R\\end{array}$ (Error compiling LaTeX. ! Undefined control sequence.) If T = 7 and the letter O represents an even number, what is the only possible value for W? $\\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 1 \\qquad \\textbf{(C)}\\ 2\\qquad \\textbf{(D)}\\ 3\\qquad \\textbf{(E)}\\ 4$ ## Problem 15 A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown? $[asy] defaultpen(linewidth(0.8)); path p=unitsquare; draw(p^^shift(0,1)*p^^shift(1,0)*p); draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p); label(\"FRONT\", (1,0), S); label(\"SIDE\", (5,0), S);[/asy]$ $\\textbf{(A)}\\ 3\\qquad\\textbf{(B)}\\ 4\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 6\\qquad\\textbf{(E)}\\ 7$ ## Problem 16 Ali, Bonnie, Carlo, and Dianna are", "Difference between revisions of \"1996 AIME Problems/Problem 7\" Problem Two squares of a $7\\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? Solution There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. $[asy] pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,3)*unitsquare,rgb(.4,.8,.4));fill(shift(4,5)*unitsquare,rgb(.4,.8,.4)); fill(shift(3,4)*unitsquare,rgb(.7,1,.7));fill(shift(1,4)*unitsquare,rgb(.7,1,.7)); fill(shift(2,3)*unitsquare,rgb(.7,1,.7));fill(shift(2,1)*unitsquare,rgb(.7,1,.7)); fill(shift(3,2)*unitsquare,rgb(.7,1,.7));fill(shift(5,2)*unitsquare,rgb(.7,1,.7)); D(arc(O,1,280,350),EndArrow(4)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,1,10,80),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,1,190,260),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,1,100,170),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ $[asy]pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(2,1)*unitsquare,rgb(1,1,.4)); fill(shift(1,4)*unitsquare,rgb(.4,.4,.7)); fill(shift(5,2)*unitsquare,rgb(.4,.4,.7)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ For most pairs, there will be three other equivalent boards. For those symmetric about the center, there is only one other. Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\\frac{49-1}{2}=24$ such pairs. There are then ${49 \\choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each", "# Difference between revisions of \"1996 AIME Problems/Problem 7\" ## Problem Two squares of a $7\\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? ## Solution 1 (Generalized) There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. $[asy] pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(2,3)*unitsquare,rgb(.8,.8,.5));fill(shift(2,1)*unitsquare,rgb(.8,.8,.5)); fill(shift(3,2)*unitsquare,rgb(.8,.8,.5));fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,1,280,350),EndArrow(4)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,1,10,80),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,1,190,260),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,1,100,170),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ $[asy]pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(2,1)*unitsquare,rgb(1,1,.4)); fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ For most pairs, there will be three other equivalent boards. For those symmetric about the center, there is only one other. Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\\frac{49-1}{2}=24$ such pairs. There are then ${49 \\choose 2}-24$ pairs that yield $4$ distinct boards upon rotation;", "# Difference between revisions of \"1996 AIME Problems/Problem 4\" ## Problem A wooden cube, whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is $x$ centimeters directly above an upper vertex, the cube casts a shadow on the horizontal surface. The area of the shadow, which does not include the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed $1000x$. ## Solution import three; size(250);defaultpen(0.7+fontsize(9)); real unit = 0.5; real r = 2.8; triple O=(0,0,0), P=(0,0,unit+unit/(r-1)); dot(P); draw(O--P); draw(O--(unit,0,0)--(unit,0,unit)--(0,0,unit)); draw(O--(0,unit,0)--(0,unit,unit)--(0,0,unit)); draw((unit,0,0)--(unit,unit,0)--(unit,unit,unit)--(unit,0,unit)); draw((0,unit,0)--(unit,unit,0)--(unit,unit,unit)--(0,unit,unit)); draw(P--(r*unit,0,0)--(r*unit,r*unit,0)--(0,r*unit,0)--P); draw(P--(r*unit,r*unit,0)); draw((r*unit,0,0)--(0,0,0)--(0,r*unit,0)); draw(P--(0,0,unit)--(unit,0,unit)--(unit,0,0)--(r*unit,0,0)--P,dashed+blue+linewidth(0.8)); draw(rightanglemark(P,(0,0,unit),(unit,0,unit),1.7));draw(rightanglemark((unit,0,unit),(unit,0,0),(r*unit,0,0),1.7)); label(\"$x$\",(0,0,unit+unit/(r-1)/2),WSW); label(\"$1$\",(unit/2,0,unit),N); label(\"$1$\",(unit,0,unit/2),W); label(\"$1$\",(unit/2,0,0),N); label(\"$6$\",(unit*(r+1)/2,0,0),N); label(\"$7$\",(unit*r,unit*r/2,0),SW); (Error compiling LaTeX. draw(rightanglemark(P,(0,0,unit),(unit,0,unit),1.7));draw(rightanglemark((unit,0,unit),(unit,0,0),(r*unit,0,0),1.7)); ^ 112982610b648ae4a5b643e01bc8b68a91f453c1.asy: 12.20: cannot call 'path rightanglemark(pair A, pair B, pair C, real s=<default>)' with parameters '(triple, triple, triple, real)') (Figure not to scale) The area of the square shadow base is $48 + 1 = 49$, and so the sides of the shadow are $7$. Using the similar triangles in blue, $\\frac {x}{1} = \\frac {1}{6}$, and $\\left\\lfloor 1000x \\right\\rfloor = \\boxed{166}$.", "# Counting painted sides of cubic shapes Sandbox Many of us have seen math problems where a shape made of unit cubes is dipped in paint, and the answer is the number of painted sides. We'll generalize that problem in this challenge. # Input A 3-dimensional matrix of 0s and 1s. # Output A non-negative integer # Challenge Given a n by m by k matrix of 0s and 1s, we can view the matrix as a 3D shape by considering a n by m by k rectangular prism broken up into n * m * k unit cubes, and the unit cubes corresponding to the 0 values in the matrix are removed. For example, the matrix [[[1,0],[0,0]],[[1,1],[0,1]]] represents the shape Given such a shape, the challenge is to output the number of painted sides on the shape if the whole shape is dunked in paint. # Test Cases [[[1,1,1],[1,1,1],[1,1,1]],[[1,1,1],[1,0,1],[1,1,1]],[[1,1,1],[1,1,1],[1,1,1]]] -> 54 [[[1,0],[0,0]],[[1,1],[0,1]]] -> 18 [[[1]],[[0]],[[1]]] -> 12 [[[1,1,1,1,1,1],[1,1,1,1,1,1],[1,1,1,1,1,1],[1,1,1,1,1,1]],[[1,1,1,1,1,1],[1,0,0,0,0,1],[1,0,0,0,0,1],[1,1,1,1,1,1]],[[1,1,1,1,1,1],[1,0,0,0,0,1],[1,0,0,0,0,1],[1,1,1,1,1,1]],[[1,1,1,1,1,1],[1,0,1,1,0,1],[1,0,1,1,0,1],[1,1,1,1,1,1]],[[1,1,1,1,1,1],[1,0,1,1,0,1],[1,0,0,1,0,1],[1,1,1,1,1,1]],[[1,1,1,1,1,1],[1,1,1,1,1,1],[1,1,1,1,1,1],[1,1,1,1,1,1]]] -> 168 [[[0,0,0],[0,1,0],[0,0,0]],[[0,1,0],[1,0,1],[0,1,0]],[[0,0,0],[0,1,0],[0,0,0]]] -> 30 [[[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1]],[[1,1,1,1,1],[1,0,0,0,1],[1,0,0,0,1],[1,0,0,0,1],[1,1,1,1,1]],[[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]],[[1,1,1,1,1],[1,0,0,0,1],[1,0,0,0,1],[1,0,0,0,1],[1,1,1,1,1]],[[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1]]] -> 150 [[[1,1,0,1,1],[1,1,0,1,1],[1,1,0,1,1]],[[1,1,0,1,1],[1,1,0,1,1],[1,1,0,1,1]],[[1,1,0,1,1],[1,1,0,1,1],[1,1,0,1,1]],[[1,1,0,1,1],[1,1,0,1,1],[1,1,0,1,1]]] -> 104 [[[0,1,1],[1,1,1],[1,1,1]],[[1,1,1],[1,0,1],[1,1,1]],[[1,1,1],[1,1,1],[1,1,1]]] -> 54 • Can we take the input as n, m, k and a 1D list of zeros and ones representing the “unrolled” 3D values? – Luis Mendo Jul 14 '20 at 14:45 • @LuisMendo yes you can. The input is flexible. – Don Thousand Jul 14", "# Difference between revisions of \"1996 AIME Problems/Problem 4\" ## Problem A wooden cube, whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is $x$ centimeters directly above an upper vertex, the cube casts a shadow on the horizontal surface. The area of the shadow, which does not include the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed $1000x$. ## Solution $import three; size(250);defaultpen(0.7+fontsize(9)); real unit = 0.5; real r = 2.8; triple O=(0,0,0), P=(0,0,unit+unit/(r-1)); dot(P); draw(O--P); draw(O--(unit,0,0)--(unit,0,unit)--(0,0,unit)); draw(O--(0,unit,0)--(0,unit,unit)--(0,0,unit)); draw((unit,0,0)--(unit,unit,0)--(unit,unit,unit)--(unit,0,unit)); draw((0,unit,0)--(unit,unit,0)--(unit,unit,unit)--(0,unit,unit)); draw(P--(r*unit,0,0)--(r*unit,r*unit,0)--(0,r*unit,0)--P); draw(P--(r*unit,r*unit,0)); draw((r*unit,0,0)--(0,0,0)--(0,r*unit,0)); draw(P--(0,0,unit)--(unit,0,unit)--(unit,0,0)--(r*unit,0,0)--P,dashed+blue+linewidth(0.8)); draw(rightanglemark(P,(0,0,unit),(unit,0,unit),1.7));draw(rightanglemark((unit,0,unit),(unit,0,0),(r*unit,0,0),1.7)); label(\"$x$\",(0,0,unit+unit/(r-1)/2),WSW); label(\"$1$\",(unit/2,0,unit),N); label(\"$1$\",(unit,0,unit/2),W); label(\"$1$\",(unit/2,0,0),N); label(\"$6$\",(unit*(r+1)/2,0,0),N); label(\"$7$\",(unit*r,unit*r/2,0),SW); (Error compiling LaTeX. draw(rightanglemark(P,(0,0,unit),(unit,0,unit),1.7));draw(rightanglemark((unit,0,unit),(unit,0,0),(r*unit,0,0),1.7)); ^ 112982610b648ae4a5b643e01bc8b68a91f453c1.asy: 12.20: cannot call 'path rightanglemark(pair A, pair B, pair C, real s=<default>)' with parameters '(triple, triple, triple, real)') $(Figure not to scale) The area of the square shadow base is$48 + 1 = 49$, and so the sides of the shadow are$7$. Using the similar triangles in blue,$\\frac {x}{1} = \\frac {1}{6}$, and$\\left\\lfloor 1000x \\right\\rfloor = \\boxed{166}$.", "# Problem #2099 2099 The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing? $[asy] unitsize(10mm); defaultpen(fontsize(10pt)); pen finedashed=linetype(\"4 4\"); filldraw((1,1)--(2,1)--(2,2)--(4,2)--(4,3)--(1,3)--cycle,grey,black+linewidth(.8pt)); draw((0,1)--(0,3)--(1,3)--(1,4)--(4,4)--(4,3)-- (5,3)--(5,2)--(4,2)--(4,1)--(2,1)--(2,0)--(1,0)--(1,1)--cycle,finedashed); draw((0,2)--(2,2)--(2,4),finedashed); draw((3,1)--(3,4),finedashed); label(\"1\",(1.5,0.5)); draw(circle((1.5,0.5),.17)); label(\"2\",(2.5,1.5)); draw(circle((2.5,1.5),.17)); label(\"3\",(3.5,1.5)); draw(circle((3.5,1.5),.17)); label(\"4\",(4.5,2.5)); draw(circle((4.5,2.5),.17)); label(\"5\",(3.5,3.5)); draw(circle((3.5,3.5),.17)); label(\"6\",(2.5,3.5)); draw(circle((2.5,3.5),.17)); label(\"7\",(1.5,3.5)); draw(circle((1.5,3.5),.17)); label(\"8\",(0.5,2.5)); draw(circle((0.5,2.5),.17)); label(\"9\",(0.5,1.5)); draw(circle((0.5,1.5),.17));[/asy]$ $\\mathrm{(A) \\ } 2\\qquad \\mathrm{(B) \\ } 3\\qquad \\mathrm{(C) \\ } 4\\qquad \\mathrm{(D) \\ } 5\\qquad \\mathrm{(E) \\ } 6$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "How Can I draw a cube of cubes efficiently? [duplicate] I've tried to draw a cube made by cubes on each side with TikZ. I mean a 3x3 cube with a little shift between each one. I've looked for an answer but the first hard step is starting with a good way to draw a cube. Then a need to put them on the edges of the 3x3 big cube. Basically, I just want to show three faces of our cube such that looks something like this: But my pretty sad code to show it is this: \\documentclass{minimal} \\usepackage{tikz} % Three counters \\newcounter{x} \\newcounter{y} \\newcounter{z} % The angles of x,y,z-axes \\newcommand\\xaxis{210} \\newcommand\\yaxis{-30} \\newcommand\\zaxis{90} % The top side of a cube \\newcommand\\topside[3]{ \\fill[fill=red!50, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0); } % The left side of a cube \\newcommand\\leftside[3]{ \\fill[fill=red, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0); } % The right side of a cube \\newcommand\\rightside[3]{ \\fill[fill=red!50!black, draw=black,shift={(\\xaxis:#1)},shift={(\\yaxis:#2)}, shift={(\\zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);} % The cube \\newcommand\\cube[3]{ \\topside{#1}{#2}{#3} \\leftside{#1}{#2}{#3} \\rightside{#1}{#2}{#3} } % Definition of \\planepartition % To draw the following plane partition, just write \\planepartition{ {a, b,c}, {d,e} }. % a b c % d e \\newcommand\\planepartition[1]{ \\setcounter{x}{-1} \\foreach \\a in {#1} { \\setcounter{y}{-1} \\foreach \\b in \\a { \\setcounter{z}{-1} \\foreach \\c in {1,...,\\b} {", "# Difference between revisions of \"2008 AMC 12A Problems/Problem 11\" ## Problem Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum? $[asy] unitsize(.8cm); pen p = linewidth(1); draw(shift(-2,0)*unitsquare,p); label(\"1\",(-1.5,0.5)); draw(shift(-1,0)*unitsquare,p); label(\"2\",(-0.5,0.5)); draw(unitsquare,p); label(\"32\",(0.5,0.5)); draw(shift(1,0)*unitsquare,p); label(\"16\",(1.5,0.5)); draw(shift(0,1)*unitsquare,p); label(\"4\",(0.5,1.5)); draw(shift(0,-1)*unitsquare,p); label(\"8\",(0.5,-0.5)); [/asy]$ $\\mathrm{(A)}\\ 154\\qquad\\mathrm{(B)}\\ 159\\qquad\\mathrm{(C)}\\ 164\\qquad\\mathrm{(D)}\\ 167\\qquad\\mathrm{(E)}\\ 189$ ## Solution To maximize the sum of the $13$ faces that are showing, we can minimize the sum of the numbers of the $5$ faces that are not showing. The bottom $2$ cubes each have a pair of opposite faces that are covered up. When the cube is folded, $(1,32)$; $(2,16)$; and $(4,8)$ are opposite pairs. Clearly $4+8=12$ has the smallest sum. The top cube has 1 number that is not showing. The smallest number on a face is $1$. So, the minimum sum of the $5$ unexposed faces is $2\\cdot12+1=25$. Since the sum of the numbers on all the cubes is $3(32+16+8+4+2+1)=189$, the maximum possible sum of $13$ visible numbers is $189-25=164 \\Rightarrow C$. ## Solution 2 Conversely, maximize the sum. Two cubes have 4 exposed faces. Since $32>16+8+4+2$, 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on", "), (h/(2^(i+1))) *2/3^.5); drawEquilaterals((0,0), h/3^.5); draw((-h/3^.5,0)--(h/3^.5,0)--(0,h)--cycle); label(\"\\vdots\",(0,3/4*h)); [/asy]$ The infinite geometric series $\\frac 14 + \\frac {1}{4^2} + \\frac {1}{4^3} + \\cdots = \\frac 13$. $[asy] defaultpen(linewidth(1)); unitsize(15); int n = 8; /* number of layers */ real h = 3; /* square height */ pen colors[] = {rgb(0.8,0,0),rgb(0,0.8,0),rgb(0,0,0.8)}; pair shiftL = (-3*h,0); /* amount to shift second square left by */ void drawSquares(real s, pair A = (0,0)){ filldraw(shift(A)*shift(-2*s, -s)*xscale(s)*yscale(s)*unitsquare,colors[0]); filldraw(shift(A)*shift(-2*s,-2*s)*xscale(s)*yscale(s)*unitsquare,colors[1]); filldraw(shift(A)*shift(-s ,-2*s)*xscale(s)*yscale(s)*unitsquare,colors[2]); } for(int i = 0; i < n; ++i) drawSquares(h/2^i); drawSquares(h,shiftL); draw(shift(shiftL+(-2*h,-2*h))*xscale(2*h)*yscale(2*h)*unitsquare); label(\"\\cdots\",shiftL+(-h/2,-h/2)); [/asy]$ Another proof of the identity $\\frac 14 + \\frac {1}{4^2} + \\frac {1}{4^3} + \\frac {1}{4^4}+\\cdots = \\frac 13$. $[asy] unitsize(15); defaultpen(linewidth(1)); pen sm = fontsize(10); real r = 0.7, h = 4.5, n = 10, xsum = 0; void htick(pair A, pair B, pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } filldraw(xscale(h)*yscale(h)*unitsquare,rgb(0.9,1,0.9)); draw((0,0)--(h/(1-r),0)--(0,h)); for(int i = 0; i < n; ++i){ xsum += r^i; draw((h*xsum,0)--(h*xsum,h*(1-(1-r)*xsum))); htick((h*(xsum-r^i),-1),(h*xsum,-1)); if(i < 6) label(\"r^\"+(string) i+\"\",(h*(xsum-r^i/2),-1),S,sm); else if(i == 8) label(\"\\cdots\",(h*(xsum-r^i/2),-1.2),S,sm); } /* htick((-1,0),(-1,h),(.15,0)); htick((0,h+1),(h,h+1)); */ htick((h+1,h),(h+1,h*r),(.15,0)); label(\"1\",(0,h/2),W,sm); label(\"1\",(h/2,h),N,sm); label(\"1-r\",(h+1,h*(1+r)/2),E,sm); [/asy]$ The infinite geometric series $\\sum_{n=0}^{\\infty} r^n = \\frac{1}{1-r}$. $[asy] unitsize(15); defaultpen(linewidth(1)); pen sm = fontsize(10); real r = 0.55, h = 2.5, n = 7, xsum = 0; pair shiftD = -(0,h*r/(1-r)+2.5); void htick(pair A, pair B, pair ticklength = (0,0.15)){", "[5, 0, 4], [5, 1, 0], [5, 4, 6], [5, 6, 7], [3, 2, 0], [3, 0, 1], [3, 6, 2], [3, 7, 6], [3, 1, 5], [3, 5, 7]]) print('vertices = ', mc.shape[\"A\"][\"vertices\"]) print('faces = ', mc.shape[\"A\"][\"faces\"]) Depletants Example: mc = hpmc.integrate.Polyhedron(seed=415236, d=0.3, a=0.4, nselect=1) cube_verts = [(-0.5, -0.5, -0.5), (-0.5, -0.5, 0.5), (-0.5, 0.5, -0.5), (-0.5, 0.5, 0.5), (0.5, -0.5, -0.5), (0.5, -0.5, 0.5), (0.5, 0.5, -0.5), (0.5, 0.5, 0.5)]; cube_faces = [[0, 2, 6], [6, 4, 0], [5, 0, 4], [5,1,0], [5,4,6], [5,6,7], [3,2,0], [3,0,1], [3,6,2], [3,7,6], [3,1,5], [3,5,7]] tetra_verts = [(0.5, 0.5, 0.5), (0.5, -0.5, -0.5), (-0.5, 0.5, -0.5), (-0.5, -0.5, 0.5)]; tetra_faces = [[0, 1, 2], [3, 0, 2], [3, 2, 1], [3,1,0]]; mc.shape[\"A\"] = dict(vertices=cube_verts, faces=cube_faces); mc.shape[\"B\"] = dict(vertices=tetra_verts, faces=tetra_faces, origin = (0,0,0)); mc.depletant_fugacity[\"B\"] = 3.0 shape The shape parameters for each particle type. The dictionary has the following keys: • vertices (list [tuple [float, float, float]], required) - vertices of the polyhedron (distance units). • The origin MUST strictly be contained in the generally nonconvex volume defined by the vertices and faces. • The origin centered sphere that encloses all vertices should be of minimal size for optimal performance. • faces (list [tuple [int, int, int], required) - Vertex indices for every triangle in the mesh. • For visualization purposes, the", "of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to $[asy] draw((0,0)--(2,0)--(3,1)--(3,3)--(2,2)--(0,2)--cycle); draw((2,0)--(2,2)); draw((0,2)--(1,3)); draw((1,7/3)--(1,10/3)--(2,10/3)--(2,7/3)--cycle); draw((2,7/3)--(5/2,17/6)--(5/2,23/6)--(3/2,23/6)--(1,10/3)); draw((2,10/3)--(5/2,23/6)); draw((3,3)--(5/2,3)); [/asy]$ $\\text{(A)}\\ 10 \\qquad \\text{(B)}\\ 15 \\qquad \\text{(C)}\\ 17 \\qquad \\text{(D)}\\ 21 \\qquad \\text{(E)}\\ 25$ Jan 7: A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $1/2$foot from the top face. The second cut is $1/3$ foot below the first cut, and the third cut is $1/17$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?$[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label($$[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label($ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$", "Theorem (first of many proofs): the left diagram shows that $c^2 = 4 \\cdot \\frac{ab}2 + (b-a)^2 = a^2 + b^2$, and the right diagram shows a second proof by re-arranging the first diagram (the area of the shaded part is equal to $a^2 + b^2$, but it is also the re-arranged version of the oblique square, which has area $c^2$).[3] $[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c); pair shiftR = (a+b+c,0); path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } { /* first picture */ filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); fill(sq1, rgb(0.95,1,0.95)); fill(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); filldraw((0,0)--top--(c,0)--a*expi(rot1)--cycle, rgb(0.5,0.9,0.5)); draw(sq1 ^^ sq2); draw(a*expi(rot1+pi/2)--top ^^ a*expi(rot1)--a*expi(rot1)+(0,c)); label(\"a\",a/2*expi(rot1),NW,sm); label(\"b\",a/2*expi(rot1)+(c/2,0),NE,sm); label(\"c\",(c/2,0),S,sm); } { /* second picture */ fill(shift(shiftR)*sq1, rgb(0.95,1,0.95)); fill(shift(shiftR)*sq2, rgb(0.95,1,0.95)); filldraw(shift(shiftR)*rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); filldraw(shift(shiftR+(0,-c))*((0,0)--top--(c,0)--a*expi(rot1)--cycle), rgb(0.5,0.9,0.5)); filldraw(shift(shiftR+(0,-c))*((0,0)--(c,0)--a*expi(rot1)--cycle), rgb(1,0.85,0.7)); draw(shift(shiftR)*((0,0)--(c,0) ^^ sq1 ^^ sq2 ^^ a*expi(rot1+pi/2)--top ^^ a*expi(rot1)--a*expi(rot1)+(0,c))); label(\"a\",shiftR+a/2*expi(rot1),NW,sm); label(\"b\",shiftR+a/2*expi(rot1)+(c/2,0),NE,sm); label(\"c\",shiftR+(c/2,0),S,sm); } [/asy]$ Another proof of the Pythagorean Theorem (animated version). $[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c); pair shiftR = (a+b+c,0); path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare, tri = (0,0)--(0,a)--(b,0)--cycle; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^", "mm}\\diamondsuit \",(12,4.5),S); [/asy]$ $\\text{(A)}\\ 1 \\qquad \\text{(B)}\\ 2 \\qquad \\text{(C)}\\ 3 \\qquad \\text{(D)}\\ 4 \\qquad \\text{(E)}\\ 5$ ## Problem 25 How many different patterns can be made by shading exactly two of the nine squares? Patterns that can be matched by flips and/or turns are not considered different. For example, the patterns shown below are not considered different. $[asy] fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,gray); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle,gray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,linewidth(1)); draw((2,0)--(2,3),linewidth(1)); draw((0,1)--(3,1),linewidth(1)); draw((1,0)--(1,3),linewidth(1)); draw((0,2)--(3,2),linewidth(1)); fill((6,0)--(8,0)--(8,1)--(6,1)--cycle,gray); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle,linewidth(1)); draw((8,0)--(8,3),linewidth(1)); draw((6,1)--(9,1),linewidth(1)); draw((7,0)--(7,3),linewidth(1)); draw((6,2)--(9,2),linewidth(1)); fill((14,1)--(15,1)--(15,3)--(14,3)--cycle,gray); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle,linewidth(1)); draw((14,0)--(14,3),linewidth(1)); draw((12,1)--(15,1),linewidth(1)); draw((13,0)--(13,3),linewidth(1)); draw((12,2)--(15,2),linewidth(1)); fill((18,1)--(19,1)--(19,3)--(18,3)--cycle,gray); draw((18,0)--(21,0)--(21,3)--(18,3)--cycle,linewidth(1)); draw((20,0)--(20,3),linewidth(1)); draw((18,1)--(21,1),linewidth(1)); draw((19,0)--(19,3),linewidth(1)); draw((18,2)--(21,2),linewidth(1)); [/asy]$ $\\text{(A)}\\ 3 \\qquad \\text{(B)}\\ 6 \\qquad \\text{(C)}\\ 8 \\qquad \\text{(D)}\\ 12 \\qquad \\text{(E)}\\ 18$", "formed? $[asy] defaultpen(linewidth(0.6)); size(80); real r=0.5, s=1.5; path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(s,r)*p); draw(shift(s,-r)*p); draw(shift(2s,2r)*p); draw(shift(2s,0)*p); draw(shift(2s,-2r)*p); draw(shift(3s,3r)*p); draw(shift(3s,-3r)*p); draw(shift(3s,r)*p); draw(shift(3s,-r)*p); draw(shift(4s,-4r)*p); draw(shift(4s,-2r)*p); draw(shift(4s,0)*p); draw(shift(4s,2r)*p); draw(shift(4s,4r)*p);[/asy]$ $[asy] size(350); defaultpen(linewidth(0.6)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; pair[] a={(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}; pair[] b={(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}; pair[] c={(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}; pair[] d={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}; pair[] e={(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}; int i; for(int i=0; i<15; i=i+1) { draw(shift(a[i])*p); draw(shift(b[i])*p); draw(shift(c[i])*p); draw(shift(d[i])*p); draw(shift(e[i])*p); } [/asy]$ $$\\textbf{(A)}\\qquad\\qquad\\qquad\\textbf{(B)}\\quad\\qquad\\qquad\\textbf{(C)}\\:\\qquad\\qquad\\qquad\\textbf{(D)}\\quad\\qquad\\qquad\\textbf{(E)}$$ Problem 5 A sequence of numbers starts with $1$, $2$, and $3$. The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence? $\\textbf{(A)}\\ 11\\qquad\\textbf{(B)}\\ 20\\qquad\\textbf{(C)}\\ 37\\qquad\\textbf{(D)}\\ 68\\qquad\\textbf{(E)}\\ 99$ Problem 6 Steve's empty swimming pool will hold $24,000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many", "draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(2*right)*r); draw(s,A--B--E--cycle); draw(s,A--C--D--cycle); draw(s,F--C--B--cycle); draw(s,F--D--E--cycle,dotted+linewidth(0.7)); draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(4*right)*s); [/asy]$ ### Solution 2 We consider the dual of the octahedron, the cube; a cube can be inscribed in an octahedron with each of its vertices at a face of the octahedron. So the problem is equivalent to finding the number of ways to color the vertices of a cube. Select any vertex and call it $A$; there are $8$ color choices for this vertex, but this vertex can be rotated to any of $8$ locations. After fixing $A$, we pick another vertex $B$ adjacent to $A$. There are seven color choices for $B$, but there are only three locations to which $B$ can be rotated to (since there are three edges from $A$). The remaining six vertices can be colored in any way and their locations are now fixed. Thus the total number of ways is $\\frac{8}{8} \\cdot \\frac{7}{3} \\cdot 6! = 1680 \\Rightarrow \\mathrm{(E)}$. ### Solution 3 There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 6*4 = 24 ways to orient an octahedron, and $8!/24 =" ]

[ "# Question 1 of 52 Ramesh had a rectangular block of wood 9cm by cm 7cm. He painted all the faces of the block. What is the total painted area? A 254cm$^2$ B 256cm$^2$ C 258cm$^2$ D 260cm$^2$ E None of the above", "Well done to Rajeev from Fair Field Junior School who sent us this solution: Jo noticed that when she painted some of the faces of her cube then $45$ faces had no paint on them. Her cube was $5 \\times 5 \\times 5$, and $4$ faces were painted, so $80$ cubes had paint on them and $45$ cubes were unpainted. Here is a table showing the number of painted cubes for different sizes of cube and numbers of faces painted: Dimensions of Cube Number of Cubes 1 face painted 2 opposite faces painted 3 faces painted 4 faces painted 5 faces painted 6 faces painted $2\\times2\\times2$ $8$ $4$ $8$ $8$ $8$ $8$ $8$ $3\\times3\\times3$ $27$ $9$ $18$ $21$ $24$ $25$ $26$ $4\\times4\\times4$ $64$ $16$ $32$ $40$ $48$ $52$ $56$ $5\\times5\\times5$ $125$ $25$ $50$ $65$ $80$ $89$ $98$ $6\\times6\\times6$ $216$ $36$ $72$ $96$ $120$ $136$ $152$ $17\\times17\\times17$ $4913$ $289$ $578$ $833$ $1088$ $1313$ $1538$ $n \\times n \\times n$ $n^3$ $n^2$ $2n^2$ $3n^2-2n$ $4n^2-4n$ $5n^2-8n+4$ $6n^2-12n+8$ When Jo had to paint the 2nd face she painted the opposite face. I noticed that it does make a difference in what pattern the faces are painted. The above table applies when painting the second face the one opposite the painted face is painted and after that then you can paint the", "$98$ $6\\times6\\times6$ $216$ $36$ $72$ $96$ $120$ $136$ $152$ $17\\times17\\times17$ $4913$ $289$ $578$ $833$ $1088$ $1313$ $1538$ $n \\times n \\times n$ $n^3$ $n^2$ $2n^2$ $3n^2-2n$ $4n^2-4n$ $5n^2-8n+4$ $6n^2-12n+8$ When Jo had to paint the 2nd face she painted the opposite face. I noticed that it does make a difference in what pattern the faces are painted. The above table applies when painting the second face the one opposite the painted face is painted and after that then you can paint the other faces in any order. Another way to do it is to paint the adjacent face- so after painting the 1st face(say right side) you paint the 2nd face adjacent to it and the 3rd face (say top face)painted must be touching the two painted faces and the 4th face (say bottom face) and the 5th face any remaining side. Rajeev highlights the idea that painting faces in a different arrangment affects the number of painted cubes. Here is a table with all possible arrangements. When faces of a larger cube are painted then the number of unpainted cubes depends upon whether the opposite or the adjacent face was painted when painting the second face. Jo's 2nd face was the opposite face so when she painted the 4th face she had $45$ unpainted cubes. If Jo", "Using twenty of the numbers from $1$ to $25$, each vertex has been numbered so that the numbers around each pentagonal face add up to $65$.", "on each pair of opposite sides, so it always has $14$ on the walls no matter how you place it. Since $14=3\\cdot4+2$, the number of eyes on the walls is not an integer multiple of $4$ if you have an odd number of dice, so you can't have the same number of eyes on all $4$ walls in this case.", "get $12$ vertices for each of the $6$ square faces of the central $P^3$, for a total of $72$ with only a few missing. Each cube duo contributes $5 \\cdot 4 = 20$ edges, for $120$ so far. Assuming the purple squares are legitimate $2$-faces (I'm not sold on this), we have $6 + 5 = 11$ faces of dimension $2$, for $66$ so far, and finally $2 \\cdot 6$ facets, plus the central $P^3$, for $13$ facets. The final sort of facets are a bit odd as they seem to be an 'inner' hexagonal prism (red edges) and a hexagonal prism with hexagonal pyramid 'attached' at a hexagonal $2$-face (blue edges), with each of these facets also sharing a purple hexagonal $2$-face. There are $8$ of these facet pairs (one for each hexagon in the central $P^3$), bringing our total up to 29 (plus the outer facet), for $30$ total facets. These facet pairs contribute only $1$ vertex not yet accounted for, and we have $72 + 8 = 80$ vertices by my count. Now, many of the edges of these facets are contained in things listed so far. It'll be helpful to consider the $12$ green edges of the central $P^3$, each will give us $3$ edges yet uncounted, plus $6$ edges (from the apex of", "faces as we go along, noting that there are ${6\\choose 2} = 15$ pairs of colors for the top and bottom faces. Using just one color for the four sides: no change, so $6$ ways to color the sides and $6\\cdot 15 = 90$ colorings altogether. (Swapping ends doesn’t affect the sides.) Using two colors: no change, so $60$ ways and $60\\cdot 15 = 900$ colorings; swapping ends affects the sides, but they can be restored by a rotation. (Running total: $990$) Using three colors: As before, there are $6$ ways to choose the color that appears twice and then $10$ ways to choose the other two colors, and the two faces of the same color may be either adjacent or opposite. If they are opposite, it makes no difference how the remaining two colors are applied to the other two sides, so that sub-case yields $60$ ways and $60\\cdot 15 = 900$ colorings. If they are adjacent, however, the order in which the other two colors are applied to the other two sides does matter, and we get $120$ ways. Since these $120$ ways of coloring the sides are not preserved by swapping ends, each apparently extends to $30$ different ways of coloring the prism, for a total of $120\\cdot 30 = 3600$ colorings. However, this actually", "that a cube has 12 edges. To find the total length of edges, we will multiply the length of each edge with the number of edges of the cube. Thus, the total length of all the edges $=12\\times 5=60m$. (b) We have to calculate the total cost of painting all the outside surfaces of the cube, which are painted at the rate of $Rs.5$ per m sq. We will firstly calculate the total surface area of the cube. We know that the total surface area of the cube whose edge is ‘a’ is given by $6{{a}^{2}}$. Substituting $a=5m$, the total surface area of cube $=6{{a}^{2}}=6{{\\left( 5 \\right)}^{2}}=150{{m}^{2}}$. We will now calculate the total cost of painting the surfaces of the cube. We know that $1{{m}^{2}}$ of area is painted for $Rs.5$. To calculate the cost of painting $x{{m}^{2}}$ of area, we will multiply x by 5. Thus, the cost of painting $150{{m}^{2}}$$=150\\times 5=Rs.750$. Hence, the cost of painting all the outer surfaces of the cube is $Rs.750$. (c) We have to calculate the volume of the cube. We know that the volume of the cube of side ‘a’ is given by ${{a}^{3}}$. Substituting $a=5m$, the volume of cube $={{5}^{3}}=125{{m}^{3}}$. Hence, the volume of cube is $125{{m}^{3}}$. Note: A cube is a three dimensional object whose all sides are", "# In how many ways can you paint the six vertices of a regular pentagonal pyramid using at most six different colours? In how many ways can you paint the six vertices of a regular pentagonal pyramid using at most six different colours, such that two vertices connected by an edge have different colours? If the result of one way of painting may be obtained by rotation from the result of another way of painting, only one of them will be counted. The answer is supposedly 288 but I keep getting answers significantly greater than that. Solutions would be appreciated. • The problem states that two coloring are the same if one can be obtained from the other by rotation of the pyramid. Aug 27 '19 at 14:22 • I got $1044$. What did you get? Aug 27 '19 at 14:26 • @saulspatz $1200$ Aug 27 '19 at 14:38 • You ought to add you calculation to the question, so people can critique it. Aug 27 '19 at 14:45 • @saulspatz I already know its completely wrong, so I don't think there is any value in adding it. Aug 27 '19 at 14:59 Call the peak color $$p$$. Because the peak is adjacent to every base vertex, no base vertex can have color $$p$$. On the base, note", "384.336$ Therefore, the cost of painting 50 such hollow cones is $\\dpi{100} Rs.\\ 384.34\\ (approx)$ Exams Articles Questions", "must sum to $$9 - 5 = 4$$. It is possible to get a sum of $$4$$, and the only possible way is with the numbers on the faces being $$1$$ and $$3$$. Therefore, it is possible for the number on the top face to be $$5$$. Then the numbers on the four side faces must be $$4$$, $$6$$, $$3$$, and $$1$$. Each person’s view is shown below. The visible numbers are $$1$$, $$3$$, $$4$$, $$5$$, and $$6$$. Therefore, the number on the bottom of the cube is $$2$$. • Case $$6$$: $$t = 6$$ The two side faces that Jenny sees must sum to $$15 - 6 = 9$$. It is possible to get a sum of $$9$$, and the only possible ways are with the numbers on the two faces being $$3$$ and $$6$$, or $$4$$ and $$5$$. Since the number on the top face is $$6$$, then the numbers on the two side faces cannot be $$3$$ and $$6$$. Therefore, the numbers on the two side faces that Jenny sees must be $$4$$ and $$5$$. Now, the two side faces that Lee sees must sum to $$14 - 6 = 8$$. It is possible to get a sum of $$8$$, and the only possible ways are with the numbers on the two faces being $$2$$", "# Thread: Working out total cost 1. ## Working out total cost If someone could give me the breakdown of how you get the final answer in this equation I will love you forever Just need to know how you get 57 so I can apply the same theory to other problems! I have done it on paint because I haven't got a clue how to write cubed and squared on a keyboard Thanks very much 2. With C being Cost... 3. Originally Posted by JTE If someone could give me the breakdown of how you get the final answer in this equation I will love you forever Just need to know how you get 57 so I can apply the same theory to other problems! I have done it on paint because I haven't got a clue how to write cubed and squared on a keyboard Thanks very much just work it out piece by piece. $3^3 = 27$ $10(3)^2 = 10(9) = 90$ $40(3) = 120$ thus, $c(3) = 3^3 - 10(3)^2 + 40(3) = 27 - 90 + 120 = 57$", "for speed solving. Stickered for big cubes. Hope that helps.\" 4 It is possible to swap two corners, but doing so will also swap two edge pieces. It is possible that there are two edge pieces on the cube that are identical, and swapping those two will not be visible, so that you can in effect swap two corners without seeming to have any effect on the edge pieces. Notice that the centre square showing a 5 is twisted by a ... 4 Then 4 The Gear Octahedron puzzle actually has $1,327,104$ states. On my website I have a page about the Gear Mastermorphix, which is an equivalent puzzle (same mechanism, but with a tetrahedral outer shape). On that page there is an explanation as to how that number comes about: $4!$ permutations of the triangular face centres (one of them is held steady and ... 4 Several people commented that is mechanically identical to a regular 3x3x3 Rubiks cube. Assuming this to be correct, this imagined re-stickering of the cube could perhaps let you visualise how it is identical: The pieces that would normally be edge pieces sharing an edge with the white face are unusually large (reaching to the actual corners of this cube), ... 3 You can cycle three centre pieces of the top face", "$9 USD 4x4x4 cube, 2(4) + 4 =$12 USD 5x5x5 cube, 2(5) + 5 = \\$15 USD :blink: It should be: p = x^2 + x And yes, I realise I'm being rather picky. (I was bored ) -Bill", "when a student is that far off, it's unlikely that other students will get exactly the same wrong answer. Feb 20 '17 at 18:03 One quick way to guess at wrong answers is just to forget a face. Omitting units, the correct answer is $144$ and the faces have surface areas of: $6$, $6$, $33$, $44$, and $55$. Missing one face: $144-6 = 138$, $144-33 = 111$, $144-44 = 100$, and $144-55 = 89$. Indeed, each of $138$, $111$, $100$, and $89$ appears in the list of incorrect responses. I suppose some might have multiple mistakes; for example, forgetting one of the rectangular faces, and counting both of the other two as $5 \\times 11$ for a total of $110$. In this scenario, you could then add on the triangular faces for a total of $122$, which also appears in the list. Incidentally, even an erroneous approach could produce the correct answer. For example, one might compute the base rectangle as $4 \\times 11$ and mistakenly believe all three of the rectangular faces have surface area $44$, for a total of $132$. Adding on the two triangular faces yields $144$, which is the right answer - arrived at in the wrong way! Meanwhile, computing each rectangular face as $3 \\times 11$ yields $99$, which adds with the triangular", "# Billy plans to paint baskets. The paint costs $14.50. The baskets cost$7.25 each. How do you write an equation that gives the total cost as a function of the number of baskets made. Determine the cost of four baskets? $T C = 21.75 x$ Cost of four baskets=$.87 #### Explanation: The paint cost is=$ 14.50 The basket cost is=$7.25 Total cost $= \\left(14.50 + 7.25\\right) \\cdot x = 21.75 x$TC$= 21.75 x$TC_(x=4)=21.75 xx 4=$87", "other faces in any order. Another way to do it is to paint the adjacent face- so after painting the 1st face(say right side) you paint the 2nd face adjacent to it and the 3rd face (say top face)painted must be touching the two painted faces and the 4th face (say bottom face) and the 5th face any remaining side. Rajeev highlights the idea that painting faces in a different arrangment affects the number of painted cubes. Here is a table with all possible arrangements. When faces of a larger cube are painted then the number of unpainted cubes depends upon whether the opposite or the adjacent face was painted when painting the second face. Jo's 2nd face was the opposite face so when she painted the 4th face she had $45$ unpainted cubes. If Jo had started painting the adjacent face then she would have been left with $48$ cubes in a $5\\times5\\times5$ cube. There is more than one way to end up with any number of cubes - you can end up with $27$ unpainted faces when you paint three adjacent faces in a $4\\times4\\times4$ cube and also in a $5\\times5\\times5$ cube when you paint all $6$ faces. Another way to think about this problem is to imagine removing all the painted cubes and leaving behind", "+1 vote 26 views A cube of side $12$ cm is painted red on all the faces and then cut into smaller cubes, each of side $3$ cm. What is the total number of smaller cubes having none of their faces painted? 1. $16$ 2. $8$ 3. $12$ 4. $24$ recategorized | 26 views The big cube is cut into 4 x 4 x 4 = 64 small cubes There are 8 original corners - 3 painted sides There were twelve edges on the original cube - each will give two cubes with two painted sides - 24 in all There were 6 faces - each with four cubes with one painted side - 24 in all So the number with no painted sides is 64 - 8 - 24 -24 = 8 (The unpainted cubes formed a small cube 2 x 2 x 2). OR no of cubes N = 12 / 3 = 4 formula: no of cubes having 0 faces painted = ( N - 2 ) ^ 3 = ( 4 - 2 ) ^ 3 = 2 ^ 3 = 8 similarly formula for 2 face painted = 12*(N-2) for 1 face painted = 6 * ( N - 2 ) ^ 2 Option B is the Correct Answer.", "LSA = 2* (6 + 15) × 8 LSA = 336 sq.inch Now, the cost of painting its walls from outside will be: Lateral surface area (LSA) × cost per square inch Cost of painting the walls = 336 × 1.3 = Rs. 436.8/- ### Fun Facts on Rectangular Parallelepiped For parallelepipeds bearing symmetry planes there are the following two cases: Having Four Rectangular Faces Having rhombic faces, while talking about the faces, two adjacent faces are equal and the other two edges are the pairs being a mirror image of each other. ### Facts Associated with the Rectangular Parallelepiped The Parallelepiped that has symmetry planes have these two cases: Having the rhombic faces, while speaking about faces, the two adjacent faces tend to be equal. The other two edges are pairs and provide the mirror image of each other. To summarize the Rectangular Parallelepiped in the specific form, have six faces as parallelograms. This implies that every Parallelepiped has a unique link to the tetrahedron and also vice versa. This is because any pair having opposite edges of the tetrahedron specifically describes the two parallel planes, one via each of the edges. ### Conclusion So, what is a rectangular parallelepiped? Well! In a specific term, all six faces of a parallelepiped are parallelograms. In this figure, all", "## '28 and It's Upward and Onward' printed from http://nrich.maths.org/ You may like to have a go at the problem So It's $28$ before trying this challenge. We are about to explore the number $28$. In $2010$ the month of February will have four weeks of $7$ days making $28$ days altogether. ---------- Half a pack of cards with the jokers makes $28$ cards altogether: ---------- The seventh triangular number is also $28$: But all these examples of $28$ are just flat - or you might call them $2$D. I'm suggesting we try making something $3$D. How about if we put some small cubes together and count the faces which are visible. Can we arrange the cubes so that there are exactly $28$ faces? In the pictures above, we're not counting faces that are 'on the table'. I've coloured the cubes (according to how many faces are showing) to make the counting a little easier. It would probably be good to make one of these shapes and do your own counting. Can you arrange some cubes so that $28$ of their faces are visible? ---------- You might want to try another way with a new shape resting on a glass table (you could just pretend!) so you can count the faces underneath too. Here are two shapes: Cubes" ]

[ "# Difference between revisions of \"2005 AMC 12A Problems/Problem 17\" ## Problem A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the same manner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volume of the piece that contains vertex $W$? $[asy] path a=(0,0)--(10,0)--(10,10)--(0,10)--cycle; path b = (0,10)--(6,16)--(16,16)--(16,6)--(10,0); path c= (10,10)--(16,16); path d= (0,0)--(3,13)--(13,13)--(10,0); path e= (13,13)--(16,6); draw(a,linewidth(0.7)); draw(b,linewidth(0.7)); draw(c,linewidth(0.7)); draw(d,linewidth(0.7)); draw(e,linewidth(0.7)); draw(shift((20,0))*a,linewidth(0.7)); draw(shift((20,0))*b,linewidth(0.7)); draw(shift((20,0))*c,linewidth(0.7)); draw(shift((20,0))*d,linewidth(0.7)); draw(shift((20,0))*e,linewidth(0.7)); draw((20,0)--(25,10)--(30,0),dashed); draw((25,10)--(31,16)--(36,6),dashed); draw((15,0)--(10,10),Arrow); draw((15.5,0)--(30,10),Arrow); label(\"W\",(15.2,0),S); label(\"Figure 1\",(5,0),S); label(\"Figure 2\",(25,0),S); [/asy]$ $(\\mathrm {A}) \\ \\frac{1}{12} \\qquad (\\mathrm {B}) \\ \\frac{1}{9} \\qquad (\\mathrm {C})\\ \\frac{1}{8} \\qquad (\\mathrm {D}) \\ \\frac{1}{6} \\qquad (\\mathrm {E})\\ \\frac{1}{4}$ ## Solution It is a pyramid with height $1$ and base area $\\frac{1}{4}$, so using the formula for the volume of a pyramid, $\\frac{1}{3} \\cdot \\left(\\frac{1}{4}\\right) \\cdot (1) = \\frac {1}{12} \\Rightarrow \\boxed{(\\mathrm {A})}$.", "is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ ## Problem 14 In this addition problem, each letter stands for a different digit. $\\setlength{\\tabcolsep}{0.5mm}\\begin{array}{cccc}&T & W & O\\\\ + &T & W & O\\\\ \\hline F& O & U & R\\end{array}$ If $T = 7$ and the letter $O$ represents an even number, what is the only possible value for $W$? $\\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 1 \\qquad \\textbf{(C)}\\ 2\\qquad \\textbf{(D)}\\ 3\\qquad \\textbf{(E)}\\ 4$ ## Problem 15 A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown? $[asy] defaultpen(linewidth(0.8)); path p=unitsquare; draw(p^^shift(0,1)*p^^shift(1,0)*p); draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p); label(\"FRONT\", (1,0), S); label(\"SIDE\", (5,0), S);[/asy]$ $\\textbf{(A)}\\ 3\\qquad\\textbf{(B)}\\ 4\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 6\\qquad\\textbf{(E)}\\ 7$ ## Problem 16 Ali, Bonnie, Carlo,", "+0 # Help -1 296 1 +421 When this net of six squares is cut out and folded to form a cube, what is the product of the numbers on the four faces adjacent to the one labeled with a \"1\"? [asy] draw(unitsquare); draw(shift(1,0)*unitsquare); draw(shift(1,-1)*unitsquare); draw(shift(1,1)*unitsquare); draw(shift(2,1)*unitsquare); draw(shift(2,2)*unitsquare); label(\"4\",(0.5,0.5)); label(\"5\",(1.5,0.5)); label(\"6\",(1.5,-0.5)); label(\"3\",(1.5,1.5)); label(\"2\",(2.5,1.5)); label(\"1\",(2.5,2.5)); [/asy] May 12, 2020 #1 +25543 +1 When this net of six squares is cut out and folded to form a cube, what is the product of the numbers on the four faces adjacent to the one labeled with a \"$$1$$\"? The product is $$\\mathbf{2\\times 3 \\times 4 \\times 6 = 144}$$ May 13, 2020", "the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ Jan 5: A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(1,1/2,1/4); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(1,-1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,-1)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(1,1,0)*unitcube, white, thick(), nolight); draw(shift(1,0,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)} \\:1 : 6 \\qquad\\textbf{ (B)}\\: 7 : 36 \\qquad\\textbf{(C)}\\: 1 : 5 \\qquad\\textbf{(D)}\\: 7 : 30\\qquad\\textbf{ (E)}\\: 6 : 25$ Jan 6: A cube has edge length 2. Suppose that we glue a cube of edge length 1 on top of the big cube so that one of its faces rests entirely on the top face", "cubes. How many of the individual cubes have exactly four red faces? $[asy] import three; defaultpen(linewidth(0.8)); real r=0.5; currentprojection=orthographic(3/4,8/15,7/15); draw(unitcube, white, thick(), nolight); draw(shift(1,0,0)*unitcube, white, thick(), nolight); draw(shift(2,0,0)*unitcube, white, thick(), nolight); draw(shift(0,0,1)*unitcube, white, thick(), nolight); draw(shift(2,0,1)*unitcube, white, thick(), nolight); draw(shift(0,1,0)*unitcube, white, thick(), nolight); draw(shift(2,1,0)*unitcube, white, thick(), nolight); draw(shift(0,2,0)*unitcube, white, thick(), nolight); draw(shift(2,2,0)*unitcube, white, thick(), nolight); draw(shift(0,3,0)*unitcube, white, thick(), nolight); draw(shift(0,3,1)*unitcube, white, thick(), nolight); draw(shift(1,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,0)*unitcube, white, thick(), nolight); draw(shift(2,3,1)*unitcube, white, thick(), nolight);[/asy]$ $\\textbf{(A)}\\ 4\\qquad\\textbf{(B)}\\ 6\\qquad\\textbf{(C)}\\ 8\\qquad\\textbf{(D)}\\ 10\\qquad\\textbf{(E)}\\ 12$ ## Problem 14 In this addition problem, each letter stands for a different digit. $\\setlength{\\tabcolsep}{0.5mm}\\begin{array}{cccc}&T & W & O\\\\ \\plus{} &T & W & O\\\\ \\hline F& O & U & R\\end{array}$ (Error compiling LaTeX. ! Undefined control sequence.) If T = 7 and the letter O represents an even number, what is the only possible value for W? $\\textbf{(A)}\\ 0 \\qquad \\textbf{(B)}\\ 1 \\qquad \\textbf{(C)}\\ 2\\qquad \\textbf{(D)}\\ 3\\qquad \\textbf{(E)}\\ 4$ ## Problem 15 A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown? $[asy] defaultpen(linewidth(0.8)); path p=unitsquare; draw(p^^shift(0,1)*p^^shift(1,0)*p); draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p); label(\"FRONT\", (1,0), S); label(\"SIDE\", (5,0), S);[/asy]$ $\\textbf{(A)}\\ 3\\qquad\\textbf{(B)}\\ 4\\qquad\\textbf{(C)}\\ 5\\qquad\\textbf{(D)}\\ 6\\qquad\\textbf{(E)}\\ 7$ ## Problem 16 Ali, Bonnie, Carlo, and Dianna are", "# Difference between revisions of \"1996 AIME Problems/Problem 7\" ## Problem Two squares of a $7\\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? ## Solution 1 (Generalized) There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. $[asy] pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(2,3)*unitsquare,rgb(.8,.8,.5));fill(shift(2,1)*unitsquare,rgb(.8,.8,.5)); fill(shift(3,2)*unitsquare,rgb(.8,.8,.5));fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,1,280,350),EndArrow(4)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,1,10,80),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,1,190,260),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,1,100,170),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ $[asy]pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(2,1)*unitsquare,rgb(1,1,.4)); fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ For most pairs, there will be three other equivalent boards. For those symmetric about the center, there is only one other. Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\\frac{49-1}{2}=24$ such pairs. There are then ${49 \\choose 2}-24$ pairs that yield $4$ distinct boards upon rotation;", "# Difference between revisions of \"2008 AMC 12A Problems/Problem 11\" ## Problem Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum? $[asy] unitsize(.8cm); pen p = linewidth(1); draw(shift(-2,0)*unitsquare,p); label(\"1\",(-1.5,0.5)); draw(shift(-1,0)*unitsquare,p); label(\"2\",(-0.5,0.5)); draw(unitsquare,p); label(\"32\",(0.5,0.5)); draw(shift(1,0)*unitsquare,p); label(\"16\",(1.5,0.5)); draw(shift(0,1)*unitsquare,p); label(\"4\",(0.5,1.5)); draw(shift(0,-1)*unitsquare,p); label(\"8\",(0.5,-0.5)); [/asy]$ $\\mathrm{(A)}\\ 154\\qquad\\mathrm{(B)}\\ 159\\qquad\\mathrm{(C)}\\ 164\\qquad\\mathrm{(D)}\\ 167\\qquad\\mathrm{(E)}\\ 189$ ## Solution To maximize the sum of the $13$ faces that are showing, we can minimize the sum of the numbers of the $5$ faces that are not showing. The bottom $2$ cubes each have a pair of opposite faces that are covered up. When the cube is folded, $(1,32)$; $(2,16)$; and $(4,8)$ are opposite pairs. Clearly $4+8=12$ has the smallest sum. The top cube has 1 number that is not showing. The smallest number on a face is $1$. So, the minimum sum of the $5$ unexposed faces is $2\\cdot12+1=25$. Since the sum of the numbers on all the cubes is $3(32+16+8+4+2+1)=189$, the maximum possible sum of $13$ visible numbers is $189-25=164 \\Rightarrow C$. ## Solution 2 Conversely, maximize the sum. Two cubes have 4 exposed faces. Since $32>16+8+4+2$, 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on", "Difference between revisions of \"1996 AIME Problems/Problem 7\" Problem Two squares of a $7\\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? Solution There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. $[asy] pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,3)*unitsquare,rgb(.4,.8,.4));fill(shift(4,5)*unitsquare,rgb(.4,.8,.4)); fill(shift(3,4)*unitsquare,rgb(.7,1,.7));fill(shift(1,4)*unitsquare,rgb(.7,1,.7)); fill(shift(2,3)*unitsquare,rgb(.7,1,.7));fill(shift(2,1)*unitsquare,rgb(.7,1,.7)); fill(shift(3,2)*unitsquare,rgb(.7,1,.7));fill(shift(5,2)*unitsquare,rgb(.7,1,.7)); D(arc(O,1,280,350),EndArrow(4)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,1,10,80),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,1,190,260),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,1,100,170),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ $[asy]pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(2,1)*unitsquare,rgb(1,1,.4)); fill(shift(1,4)*unitsquare,rgb(.4,.4,.7)); fill(shift(5,2)*unitsquare,rgb(.4,.4,.7)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]$ For most pairs, there will be three other equivalent boards. For those symmetric about the center, there is only one other. Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\\frac{49-1}{2}=24$ such pairs. There are then ${49 \\choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each", "# Counting painted sides of cubic shapes Sandbox Many of us have seen math problems where a shape made of unit cubes is dipped in paint, and the answer is the number of painted sides. We'll generalize that problem in this challenge. # Input A 3-dimensional matrix of 0s and 1s. # Output A non-negative integer # Challenge Given a n by m by k matrix of 0s and 1s, we can view the matrix as a 3D shape by considering a n by m by k rectangular prism broken up into n * m * k unit cubes, and the unit cubes corresponding to the 0 values in the matrix are removed. For example, the matrix [[[1,0],[0,0]],[[1,1],[0,1]]] represents the shape Given such a shape, the challenge is to output the number of painted sides on the shape if the whole shape is dunked in paint. # Test Cases [[[1,1,1],[1,1,1],[1,1,1]],[[1,1,1],[1,0,1],[1,1,1]],[[1,1,1],[1,1,1],[1,1,1]]] -> 54 [[[1,0],[0,0]],[[1,1],[0,1]]] -> 18 [[[1]],[[0]],[[1]]] -> 12 [[[1,1,1,1,1,1],[1,1,1,1,1,1],[1,1,1,1,1,1],[1,1,1,1,1,1]],[[1,1,1,1,1,1],[1,0,0,0,0,1],[1,0,0,0,0,1],[1,1,1,1,1,1]],[[1,1,1,1,1,1],[1,0,0,0,0,1],[1,0,0,0,0,1],[1,1,1,1,1,1]],[[1,1,1,1,1,1],[1,0,1,1,0,1],[1,0,1,1,0,1],[1,1,1,1,1,1]],[[1,1,1,1,1,1],[1,0,1,1,0,1],[1,0,0,1,0,1],[1,1,1,1,1,1]],[[1,1,1,1,1,1],[1,1,1,1,1,1],[1,1,1,1,1,1],[1,1,1,1,1,1]]] -> 168 [[[0,0,0],[0,1,0],[0,0,0]],[[0,1,0],[1,0,1],[0,1,0]],[[0,0,0],[0,1,0],[0,0,0]]] -> 30 [[[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1]],[[1,1,1,1,1],[1,0,0,0,1],[1,0,0,0,1],[1,0,0,0,1],[1,1,1,1,1]],[[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]],[[1,1,1,1,1],[1,0,0,0,1],[1,0,0,0,1],[1,0,0,0,1],[1,1,1,1,1]],[[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1],[1,1,1,1,1]]] -> 150 [[[1,1,0,1,1],[1,1,0,1,1],[1,1,0,1,1]],[[1,1,0,1,1],[1,1,0,1,1],[1,1,0,1,1]],[[1,1,0,1,1],[1,1,0,1,1],[1,1,0,1,1]],[[1,1,0,1,1],[1,1,0,1,1],[1,1,0,1,1]]] -> 104 [[[0,1,1],[1,1,1],[1,1,1]],[[1,1,1],[1,0,1],[1,1,1]],[[1,1,1],[1,1,1],[1,1,1]]] -> 54 • Can we take the input as n, m, k and a 1D list of zeros and ones representing the “unrolled” 3D values? – Luis Mendo Jul 14 '20 at 14:45 • @LuisMendo yes you can. The input is flexible. – Don Thousand Jul 14", "You may use any sensible I/O method • This is , so the shortest code in bytes wins • Sandbox Apr 28, 2021 at 15:49 • fun fact, if you interpret the tuples for n=2 as a binary number you get the first 8 multiples of 7, whether or not this is useful, idk Apr 28, 2021 at 16:38 # Wolfram Language (Mathematica), 363433 30 bytes #-1|#&~Array~#~Tuples~3/.#->0& Try it online! Returns a length-$$\\n^3\\$$ list of $$\\\\{x|x',y|y',z|z'\\}\\$$ colorings. Verify with a $$\\n\\times n\\times n\\$$ matrix output. Explanation: Consider a cube $$\\x\\$$th from the front in some initial configuration. If the frontmost layer of cubes is painted $$\\1\\$$, the second $$\\2\\$$, etc., so that the color number increments with each shift, the front face of the cube will be painted $$\\x\\$$. After one more shift, the cube will be in the backmost layer, so the back face is painted $$\\x+1\\$$. The same logic applies in other dimensions, so we can paint the cube at position $$\\(x,y,z)\\$$ with colors $$\\(x,y,z,x+1,y+1,z+1)\\$$ (modulo $$\\n\\$$). #-1|#&~Array~#~Tuples~3 (* 3-tuples of {0|1, 1|2, ..., n-1|n} *) /.#->0& (* and replace n with 0. *) • This looks clever, any chance of an explanation for those who don't speak mathematica? Apr 28, 2021 at 18:47 # Charcoal, 21 bytes ＮθＦθＦθＦθＦ⟦ικλ⟧⭆²﹪⁺μνθ Try it online! Link is to verbose", "Difference between revisions of \"1987 AJHSME Problems/Problem 7\" Problem The large cube shown is made up of $27$ identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is $\\text{(A)}\\ 10 \\qquad \\text{(B)}\\ 16 \\qquad \\text{(C)}\\ 20 \\qquad \\text{(D)}\\ 22 \\qquad \\text{(E)}\\ 24$ $[asy] unitsize(36); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3)); draw((0,3)--(2.2,4.4)--(5.2,4.4)); fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black); fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black); fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); draw((1,3)--(3.2,4.4)); draw((2,3)--(4.2,4.4)); draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666)); draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333)); fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black); fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black); fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black); fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black); fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black); [/asy]$ Solution Clearly, no unit cube has more than one face painted, so the number of unit cubes with at least one face painted is equal to the number of painted unit squares. There are $10$ painted unit squares on the half of the cube shown, so there are $10\\cdot 2=20$ unit cubes with at least one face painted, thus our answer is $\\boxed{\\text{C}}$. See Also 1987 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems", "# Difference between revisions of \"1987 AJHSME Problems/Problem 7\" ## Problem The large cube shown is made up of $27$ identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is $\\text{(A)}\\ 10 \\qquad \\text{(B)}\\ 16 \\qquad \\text{(C)}\\ 20 \\qquad \\text{(D)}\\ 22 \\qquad \\text{(E)}\\ 24$ $[asy] unitsize(36); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5.2,1.4)--(5.2,4.4)--(3,3)); draw((0,3)--(2.2,4.4)--(5.2,4.4)); fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,black); fill((0,2)--(0,3)--(1,3)--(1,2)--cycle,black); fill((1,1)--(1,2)--(2,2)--(2,1)--cycle,black); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle,black); draw((1,3)--(3.2,4.4)); draw((2,3)--(4.2,4.4)); draw((.733333333,3.4666666666)--(3.73333333333,3.466666666666)); draw((1.466666666,3.9333333333)--(4.466666666,3.9333333333)); fill((1.73333333,3.46666666666)--(2.7333333333,3.46666666666)--(3.46666666666,3.93333333333)--(2.46666666666,3.93333333333)--cycle,black); fill((3,1)--(3.733333333333,1.466666666666)--(3.73333333333,2.46666666666)--(3,2)--cycle,black); fill((3.73333333333,.466666666666)--(4.466666666666,.93333333333)--(4.46666666666,1.93333333333)--(3.733333333333,1.46666666666)--cycle,black); fill((3.73333333333,2.466666666666)--(4.466666666666,2.93333333333)--(4.46666666666,3.93333333333)--(3.733333333333,3.46666666666)--cycle,black); fill((4.466666666666,1.9333333333333)--(5.2,2.4)--(5.2,3.4)--(4.4666666666666,2.9333333333333)--cycle,black); [/asy]$ ## Solution Clearly, no unit cube has more than one face painted, so the number of unit cubes with at least one face painted is equal to the number of painted unit squares. There are $10$ painted unit squares on the half of the cube shown, so there are $10\\cdot 2=20$ unit cubes with at least one face painted, thus our answer is $\\boxed{\\text{C}}$. ## Solution 2 Since it says at least one, we can count the number of unpainted cubes, and subtract from 27. There is 1 inner cube, 2 center cubes (see the face with 4 blacks) and 4 edge cubes (see the top two in the center top face), so 7 unpainted. Thus $27 - 7 = 20$ our answer is $\\boxed{\\text{C}}$.", "we see that $$a_n=u_n+s_n\\tag{1}.$$ For the sake of convenience in determining recurrence relations, we define another type of board with two $1\\times n$ boards where a specific corner is removed (without loss of generality, we place this in the lower left hand corner). Let $t_n$ be the number of ways to tile such a board without placeing two of the same type of tile atop each other. Once again, we compute that $t_3=3$ and $t_4=5$ as shown below. $[asy] unitsize(10); draw((1,0)--(3,0)--(3,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(3,1),linewidth(2)); draw((1,1)--(1,2)^^(2,0)--(2,1)); draw(shift((0,-2.5))*((1,0)--(3,0)--(3,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(3,1)),linewidth(2)); draw(shift((0,-2.5))*((1,1)--(1,2)^^(2,1)--(2,2))); draw(shift((4,0))*((1,0)--(3,0)--(3,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(3,1)),linewidth(2)); draw(shift((4,0))*((2,1)--(2,2))); draw(shift((11,0))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((11,0))*((1,1)--(1,2)^^(2,1)--(2,2)^^(3,0)--(3,1))); draw(shift((11,-2.5))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((11,-2.5))*((1,1)--(1,2)^^(2,0)--(2,1)^^(3,1)--(3,2))); draw(shift((16,0))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((16,0))*((2,0)--(2,2)^^(3,1)--(3,2))); draw(shift((16,-2.5))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((16,-2.5))*((2,1)--(2,2)^^(3,0)--(3,1))); draw(shift((21,0))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((21,0))*((2,0)--(2,2)^^(3,0)--(3,1))); [/asy]$ We can determine reccurence relations for $s_n$, $t_n$, and $u_n$ in terms of each other. For $s_n$, note that a tiling can end in one of the three following ways such that the rest of the board can be tiled without restriction (the placed tiles are shaded). $[asy] path unitrect=(0,0)--(2,0)--(2,1)--(0,1)--cycle; unitsize(10); fill(shift((3,0))*unitsquare,mediumgray); draw(shift((0,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(3,1)--(3,2)--(0,2)),linewidth(2)); draw(shift((0,0))*((3,0)--(3,1))); label(\"\\dots\",(-1,1)); fill(shift((10,1))*unitsquare,mediumgray); fill(shift((10,0))*unitrect,mediumgray); draw(shift((8,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(3,1)--(3,2)--(0,2)),linewidth(2)); draw(shift((8,0))*((2,0)--(2,2))); label(\"\\dots\",(8-1,1)); fill(shift((18,0))*unitrect,mediumgray); fill(shift((17,1))*unitrect,mediumgray); draw(shift((16,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(3,1)--(3,2)--(0,2)),linewidth(2)); draw(shift((16,0))*((2,0)--(2,1)^^(1,1)--(1,2))); label(\"\\dots\",(16-1,1)); [/asy]$ In the first, second, and third cases, we see that we can tile the rest of the board in $t_{n-1}$, $t_{n-2}$, and $s_{n-2}$ ways, respectively. Hence for $n\\ge 5$, we see that $$s_n=t_{n-1}+t_{n-2}+s_{n-2}\\tag{2}.$$ For $t_n$, note that a tiling can end in one of the three following ways such that the rest of the board", "draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(2*right)*r); draw(s,A--B--E--cycle); draw(s,A--C--D--cycle); draw(s,F--C--B--cycle); draw(s,F--D--E--cycle,dotted+linewidth(0.7)); draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(4*right)*s); [/asy]$ ### Solution 2 We consider the dual of the octahedron, the cube; a cube can be inscribed in an octahedron with each of its vertices at a face of the octahedron. So the problem is equivalent to finding the number of ways to color the vertices of a cube. Select any vertex and call it $A$; there are $8$ color choices for this vertex, but this vertex can be rotated to any of $8$ locations. After fixing $A$, we pick another vertex $B$ adjacent to $A$. There are seven color choices for $B$, but there are only three locations to which $B$ can be rotated to (since there are three edges from $A$). The remaining six vertices can be colored in any way and their locations are now fixed. Thus the total number of ways is $\\frac{8}{8} \\cdot \\frac{7}{3} \\cdot 6! = 1680 \\Rightarrow \\mathrm{(E)}$. ### Solution 3 There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 6*4 = 24 ways to orient an octahedron, and $8!/24 =", "), (h/(2^(i+1))) *2/3^.5); drawEquilaterals((0,0), h/3^.5); draw((-h/3^.5,0)--(h/3^.5,0)--(0,h)--cycle); label(\"\\vdots\",(0,3/4*h)); [/asy]$ The infinite geometric series $\\frac 14 + \\frac {1}{4^2} + \\frac {1}{4^3} + \\cdots = \\frac 13$. $[asy] defaultpen(linewidth(1)); unitsize(15); int n = 8; /* number of layers */ real h = 3; /* square height */ pen colors[] = {rgb(0.8,0,0),rgb(0,0.8,0),rgb(0,0,0.8)}; pair shiftL = (-3*h,0); /* amount to shift second square left by */ void drawSquares(real s, pair A = (0,0)){ filldraw(shift(A)*shift(-2*s, -s)*xscale(s)*yscale(s)*unitsquare,colors[0]); filldraw(shift(A)*shift(-2*s,-2*s)*xscale(s)*yscale(s)*unitsquare,colors[1]); filldraw(shift(A)*shift(-s ,-2*s)*xscale(s)*yscale(s)*unitsquare,colors[2]); } for(int i = 0; i < n; ++i) drawSquares(h/2^i); drawSquares(h,shiftL); draw(shift(shiftL+(-2*h,-2*h))*xscale(2*h)*yscale(2*h)*unitsquare); label(\"\\cdots\",shiftL+(-h/2,-h/2)); [/asy]$ Another proof of the identity $\\frac 14 + \\frac {1}{4^2} + \\frac {1}{4^3} + \\frac {1}{4^4}+\\cdots = \\frac 13$. $[asy] unitsize(15); defaultpen(linewidth(1)); pen sm = fontsize(10); real r = 0.7, h = 4.5, n = 10, xsum = 0; void htick(pair A, pair B, pair ticklength = (0,0.15)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } filldraw(xscale(h)*yscale(h)*unitsquare,rgb(0.9,1,0.9)); draw((0,0)--(h/(1-r),0)--(0,h)); for(int i = 0; i < n; ++i){ xsum += r^i; draw((h*xsum,0)--(h*xsum,h*(1-(1-r)*xsum))); htick((h*(xsum-r^i),-1),(h*xsum,-1)); if(i < 6) label(\"r^\"+(string) i+\"\",(h*(xsum-r^i/2),-1),S,sm); else if(i == 8) label(\"\\cdots\",(h*(xsum-r^i/2),-1.2),S,sm); } /* htick((-1,0),(-1,h),(.15,0)); htick((0,h+1),(h,h+1)); */ htick((h+1,h),(h+1,h*r),(.15,0)); label(\"1\",(0,h/2),W,sm); label(\"1\",(h/2,h),N,sm); label(\"1-r\",(h+1,h*(1+r)/2),E,sm); [/asy]$ The infinite geometric series $\\sum_{n=0}^{\\infty} r^n = \\frac{1}{1-r}$. $[asy] unitsize(15); defaultpen(linewidth(1)); pen sm = fontsize(10); real r = 0.55, h = 2.5, n = 7, xsum = 0; pair shiftD = -(0,h*r/(1-r)+2.5); void htick(pair A, pair B, pair ticklength = (0,0.15)){", "of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to $[asy] draw((0,0)--(2,0)--(3,1)--(3,3)--(2,2)--(0,2)--cycle); draw((2,0)--(2,2)); draw((0,2)--(1,3)); draw((1,7/3)--(1,10/3)--(2,10/3)--(2,7/3)--cycle); draw((2,7/3)--(5/2,17/6)--(5/2,23/6)--(3/2,23/6)--(1,10/3)); draw((2,10/3)--(5/2,23/6)); draw((3,3)--(5/2,3)); [/asy]$ $\\text{(A)}\\ 10 \\qquad \\text{(B)}\\ 15 \\qquad \\text{(C)}\\ 17 \\qquad \\text{(D)}\\ 21 \\qquad \\text{(E)}\\ 25$ Jan 7: A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $1/2$foot from the top face. The second cut is $1/3$ foot below the first cut, and the third cut is $1/17$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?$[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label($$[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label($ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$", "back, (R = red, B = blue, O = orange, Y = yellow, G = green, and W = white). They are hinged together as shown, then folded to form a cube. The face opposite the white face is $[asy] draw((0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,2)--(2,2)--(2,3)--(0,3)--cycle); draw((1,3)--(1,2)--(2,2)--(2,1)--(3,1)--(3,2)); label($ $\\text{(A)}\\ \\text{B} \\qquad \\text{(B)}\\ \\text{G} \\qquad \\text{(C)}\\ \\text{O} \\qquad \\text{(D)}\\ \\text{R} \\qquad \\text{(E)}\\ \\text{Y}$ Jan 2: Three dice with faces numbered 1 through 6 are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is $[asy] draw((0,0)--(2,0)--(3,1)--(3,7)--(1,7)--(0,6)--cycle); draw((3,7)--(2,6)--(0,6)); draw((3,5)--(2,4)--(0,4)); draw((3,3)--(2,2)--(0,2)); draw((2,0)--(2,6)); dot((1,1)); dot((.5,.5)); dot((1.5,.5)); dot((1.5,1.5)); dot((.5,1.5)); dot((2.5,1.5)); dot((.5,2.5)); dot((1.5,2.5)); dot((1.5,3.5)); dot((.5,3.5)); dot((2.25,2.75)); dot((2.5,3)); dot((2.75,3.25)); dot((2.25,3.75)); dot((2.5,4)); dot((2.75,4.25)); dot((.5,5.5)); dot((1.5,4.5)); dot((2.25,4.75)); dot((2.5,5.5)); dot((2.75,6.25)); dot((1.5,6.5)); [/asy]$ $\\text{(A)}\\ 21 \\qquad \\text{(B)}\\ 22 \\qquad \\text{(C)}\\ 31 \\qquad \\text{(D)}\\ 41 \\qquad \\text{(E)}\\ 53$ Jan 3: Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom and sides. $[asy]/* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(1,2)--(1.5,2)); draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5)); draw((1.5,3.5)--(2.5,3.5)); draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5)); draw((3,4)--(3,3)--(2.5,2.5)); draw((3,3)--(4,3)--(4,2)--(3.5,1.5)); draw((4,3)--(3.5,2.5)); draw((2.5,.5)--(3,1)--(3,1.5));[/asy]$ $\\text{(A)}\\ 18 \\qquad \\text{(B)}\\ 24 \\qquad \\text{(C)}\\ 26 \\qquad \\text{(D)}\\ 30 \\qquad \\text{(E)}\\ 36$ Jan 4: Fourteen white cubes are put together to form the figure on the right. The complete surface of", "# How to correctly enumerate all the schemes of this cube coloring problem? This problem is the fifth question of 1996 Chinese High School Mathematics League or Chinese Mathematical Olympiad in Senior: Choose several colors from the given six different colors to dye six faces of a cube, and dye each two faces with common edges into different colors. How many different dyeing schemes are there? Note: if we dye two identical cubes, we can make the six corresponding faces of the two cubes dyed the same by proper flipping, then we say that the two cubes have the same dyeing scheme. Show[Graphics3D[ Rotate[Cuboid[{0, 0, 0}, {1, 1, 1}], 0 Degree, {0, 0, 1}], Axes -> True], i = 1; Graphics3D[ Table[Text[Style[ToString[i++], 15], {x, y, z}], {x, 0, 1}, {y, 0, 1}, {z, 0, 1}]]] g0 = Graph[(Sort /@ {{1, {2, 3, 5}}, {2, {1, 4, 6}}, {3, {1, 4, 7}}, {4, {2, 3, 8}}, {5, {1, 6, 7}}, {6, {2, 5, 8}}, {7, {3, 5, 8}}, {8, {4, 6, 7}}}]]) // DeleteDuplicates, ChromaticPolynomial[g0, 6] poly = CycleIndexPolynomial[DihedralGroup[8], Array[Subscript[a, ##] &, 6]] g1 = Graph[(Sort /@ {{1, {2, 3, 4, 5}}, {2, {1, 3, 5, 6}}, {3, {1, 4, 2, 6}}, {4, {1, 3, 5, 6}}, {5, {1, 2, 4, 6}}, {6, {2, 3, 4, 5}}}]]) // DeleteDuplicates,", "# Problem #2099 2099 The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing? $[asy] unitsize(10mm); defaultpen(fontsize(10pt)); pen finedashed=linetype(\"4 4\"); filldraw((1,1)--(2,1)--(2,2)--(4,2)--(4,3)--(1,3)--cycle,grey,black+linewidth(.8pt)); draw((0,1)--(0,3)--(1,3)--(1,4)--(4,4)--(4,3)-- (5,3)--(5,2)--(4,2)--(4,1)--(2,1)--(2,0)--(1,0)--(1,1)--cycle,finedashed); draw((0,2)--(2,2)--(2,4),finedashed); draw((3,1)--(3,4),finedashed); label(\"1\",(1.5,0.5)); draw(circle((1.5,0.5),.17)); label(\"2\",(2.5,1.5)); draw(circle((2.5,1.5),.17)); label(\"3\",(3.5,1.5)); draw(circle((3.5,1.5),.17)); label(\"4\",(4.5,2.5)); draw(circle((4.5,2.5),.17)); label(\"5\",(3.5,3.5)); draw(circle((3.5,3.5),.17)); label(\"6\",(2.5,3.5)); draw(circle((2.5,3.5),.17)); label(\"7\",(1.5,3.5)); draw(circle((1.5,3.5),.17)); label(\"8\",(0.5,2.5)); draw(circle((0.5,2.5),.17)); label(\"9\",(0.5,1.5)); draw(circle((0.5,1.5),.17));[/asy]$ $\\mathrm{(A) \\ } 2\\qquad \\mathrm{(B) \\ } 3\\qquad \\mathrm{(C) \\ } 4\\qquad \\mathrm{(D) \\ } 5\\qquad \\mathrm{(E) \\ } 6$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# [SOLVED]Numbers on a cube #### anemone ##### MHB POTW Director Staff member Positive integers are written on all the faces of a cube, one on each. At each corner (vertex) of the cube, the product of the numbers on the faces that meet at the corner is written. The sum of the numbers written at all the corners is 2004. If $T$ denotes the sum of the numbers on all the faces, find all the possible values of $T$. #### Klaas van Aarsen ##### MHB Seeker Staff member \\begin{tikzpicture}[ face/.pic = { \\node {#1}; \\draw (-1,-1) rectangle (1, 1); }, cube/.pic = { \\draw (0,0) pic {face=a}; \\draw (1,-1) -- (2,-0.5) -- (2,1.5) -- (1,1) node at (1.5,0.25) {b}; \\draw (2,1.5) -- (0,1.5) -- (-1,1) node at (0.5,1.25) {e}; }] \\draw (0,0) pic {face=a} (2,0) pic {face=b} (4,0) pic {face=c} (6,0) pic {face=d} (0,2) pic {face=e} (0,-2) pic {face=f}; \\draw (9,0) pic {cube}; \\end{tikzpicture} Let $a,b,c,d,e,f$ be the 6 faces where $(a,c)$, $(b,d)$, and $(e,f)$ are the pairs of opposing faces as shown in the drawing. The sum of the corners is $(ab+bc+cd+da)e+(ab+bc+cd+da)f=(a(b+d) + c(b+d))(e+f) =(a+c)(b+d)(e+f)=2004$. The list of suitable factorizations of $2004=2^2\\cdot 3\\cdot 167$ is $(3\\cdot 4\\cdot 167,\\, 2\\cdot 6\\cdot 167,\\,2\\cdot 3\\cdot 334,\\,2\\cdot 2\\cdot 501)$. We are looking for the possibilities for $T=a+b+c+d+e+f$, which is the" ]

[ "distinct non-collinear points determines a unique plane, and since those three points each satisfy that equation for a plane, you can be sure that that equation is the equation of the plane containing those three points.", "System of Non-collinear Points in a Plane! There are seven points in a plane, no three of which are collinear. Every point is connected to every other point by either a red or a blue line segment. What is the minimum number of monochromatic triangles that can be present in such a figure? Select the correct option. Bonus: Generalize, if possible, the minumum number of monochromatic triangles for such a problem scenario as of above, given that there are $$n \\geq 3$$ points in the plane. ×", "Points in space There are n points, of which no three lie on one line and no four lies on one plane. How many planes can be guided by these points? How many planes are there if there are five times more than the given points?", "non-collinear points determine a plane. This statement means that if you have three points not on one line, then only one specific plane can go through those points. The plane is determined by the three points because the points show you exactly where the plane is. Does a plane have a definite beginning and end? A plane has a definite beginning and end. A line has one dimension, length. ... A plane consists of an infinite set of lines. Can a plane have 4 points? Four points (like the corners of a tetrahedron or a triangular pyramid) will not all be on any plane, though triples of them will form four different planes. How many planes can contain 2 points? Through any two points there exists exactly one line. A line contains at least two points. If two lines intersect, then their intersection is exactly one point. Through any three non-collinear points, there exists exactly one plane. How many points are needed to identify a plane? Plane determined by three points But most of us know that three points determine a plane (as long as they aren't collinear, i.e., lie in straight line). Can 2 points determine a line? Two distinct points determine exactly one line. That line is the shortest path between the two points. ... If", "tree model of computation, Erickson proved that the conjecture is true. What's the maximum number of (maximal) collinear sets of points in a set of n points in the plane? It can grow quadratically as a function of N. Consider the n points of the form: (x, y) for x = 0, 1, 2, and 3 and y = 0, 1, 2, ..., n / 4. This means that if you store all of the (maximal) collinear sets of points, you will need quadratic space in the worst case.", "7) are collinear.", "by the three points are collinear ( forms a straight ).", "there can be infinitely many planes that can pass through that line. However, the points lying on the same line are still coplanar. In other words, collinear points are always coplanar, but coplanar points may or may not be collinear. Collinearity is a concept of two-dimensional plane geometry and coplanarity is a concept of three-dimensional spatial geometry. In the above figure, note that the points B, D, and A are collinear as well as coplanar. All these three points lie in both the planes $\\wp$ and $\\Re$. These points are coplanar with C and E in plane $\\wp$ and with G and F in plane $\\Re$. However, C, E, F, and G are non-coplanar points. Again the points H and I do not lie in any of the two planes shown in the figure. ## Four or More Points The condition for defining a plane is that there should be a minimum of three non-collinear points. Then we can use these points to describe the plane that they lie in. When we have four or more points, obviously we can define more than one planes in three-dimensional space taking three points at a time. This concept is similar to a two-dimensional plane. If we have 2 points in a plane, we can define a line that joins the", "composed of two numbers and is the coordinate of the point. Within any Euclidean space, a plane is uniquely determined by any of the following combinations: • three points which are not lying on the same line • a line and a point not on the line • two different lines which intersect • two different lines which are parallel • a vector normal to the plane and a distance from the origin.", "73 views Consider six distinct points in a plane. Let $m$ and $M$ denote the minimum and maximum distance between any pair of points. Show that $M/m \\geq \\sqrt{3}$. | 73 views", "are two of which in the same half-plane determined by the other two.", "# Problem of the month Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$.", "Free Version Difficult # Collinear Points in Space MVCALC-XQUQWN Among the listed five points, there are four that are collinear, and one that is not on the line determined by the other four. Identify the point that is not on said line. A $A=(-1,1,4)$ B $B=(2,3,5)$ C $C=(-4,-1,3)$ D $D=(-2,2,8)$ E $E=(29,21,14)$", "c 2. How many points are needed to establish a plane? | … Explanation: Two distinct points, A and B, determine a unique line in space. A third point C not on this line determines a unique plane that we can denote as ABC. How many points are required to define a plane? – … 1. Click to reveal two points P and Q; 2. Click for a plane on the line PQ; 3. Move the slider? How many planes go through two points; 4. Click to reveal a third point A; 5. Move the slider so that the orangle plane contains the three points P, Q and A; 6. Click to reveal plane that contains the ... Four Ways to Determine a Plane - dummies If you want to work with multiple-plane proofs, you first have to know how to determine a plane. Determining a plane is the fancy, mathematical way of saying “showing you where a plane is.” There are four ways to determine a plane: Three non-collinear points determine a … How many points are there in a plane? - Quora In any of these cases, the answer is that three points determine a plane unless the points are collinear. (Note that in spherical geometry, two points don’t determine a unique line if they are", "geometric figure consisting of an assemblage of points, the polars with respect to an inversion circle constitute another figure. These figures are said to be reciprocal with respect to each other. Then there exists a duality principle which states that theorems for the original figure can be immediately applied to the reciprocal figure after suitable modification (Lachlan 1893)... ### Cylinder cutting The maximum number of pieces into which a cylinder can be divided by oblique cuts is given by(1)(2)(3)where is a binomial coefficient.This problem is sometimes also called cake cutting or pie cutting, and has the same solution as space division by planes. For , 2, ... cuts, the maximum number of pieces is 2, 4, 8, 15, 26, 42, ... (OEIS A000125). Unsurprisingly, the numbers of this sequence are called cake numbers. ### Cube division by planes The average number of regions into which randomly chosen planes divide a cube is(Finch 2003, p. 482).The maximum number of regions is presumably the same as for spacedivision by planes, namely(Yaglom and Yaglom 1987, pp. 102-106). For , 2, ... planes, this gives the values 2, 4, 8, 15, 26, 42, ... (OEIS A000125), a sequence whose values are sometimes called the \"cake numbers\" due to their relation to the cake cutting problem. ### Space division by spheres The number", "this 5 Points problem is no more difficult than the Handshake problem. Math Emeritus Quote by Physicsissuef Yes. Ok, thank you. And what about this one: Five points in the space can create how many planes? a)10 b)8 c)5 e)4 One plane can be created with at least three points. So $$C_5^3$$. The final answer is 10. But aren't there planes which are passing through 4 or 5 points? Or they will again be repeated?", "# Thread: Problem of the month 1. ## Problem of the month How many planes would by formed by n non-collinear and non-coplanar points? 2. ## Binomial Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$. -Kevin C.", "# Problem of the month • September 29th 2008, 05:36 AM MATHMATHmathmathMathMath Problem of the month How many planes would by formed by n non-collinear and non-coplanar points? • September 29th 2008, 09:07 AM TwistedOne151 Binomial Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$. -Kevin C.", "example there are six elements in a triangle that can be formed by straight... One dimension ) and three-dimensional space. ) this familiar equation for a plane may also be by., a plane contains at least how many points? distance, and which goes on for ever both. Specific plane can be thought of an a flat sheet with no thickness, and calculating volume hard to.! Two dimensions: length and width can not be measured contains at least how geometry! Is called the general form of the topological plane has two dimensions: length and width can be... Refers to the amount of geometry of paper ) how many planes are there geometry degree of differentiability hypersurface three-dimensional. Be parallel to each other is also an infinite number of distinct planes the... Making a flat plane. [ 8 ] thickness, and which goes for! Such as triangulating distance, and seven affine planes of order nine, and seven affine planes of nine! True regarding undefinable terms in geometry, we wo n't go past 3D geometry how many planes are there geometry they a... All the shapes exist in a drawing by using the stereographic projection differential geometry a! Ppts online, safely and virus-free point is too small to be seen, you can represent it visually a! Them on the plane may", "don't know the math, but I guess that's what Least-Squares fit will give (and I may be totally wrong). To illustrate my problem, let's say there are 7 points, and there exists a plane - A - that contains 5 of them. There would of course be other planes that contain fewer than 5. I just want some algorithmic way to find this plane A. Oct 2 '19 at 13:20 • Are there any additional constraints, for example the distribution of the points or distance between the points on the plane? Oct 6 '19 at 16:52 • @rjg The coordinates of the points in 3d space are available, so the distance between them can be calculated. There are no constraints, these points are just randomaly distributed in space. And there is no plane as such, just the points. I want to find out the plane that has the maximum number of them. Or in other words, from the given set of points I need to find out the maximum of them that are coplanar. (In the worst case there will just be 3 such points.) Oct 7 '19 at 5:03 There is no \"simple\" way to do anything in numerical analysis. However, the tasks can be laid out straightforwardly. EDIT: This idea is a brain-fart because it doesn't" ]

[ "By stars and bars, we have $\\binom{3+7-1}{3}=\\binom{9}{3}=\\boxed{84}.$ ~mathboy282 ~savannahsolver ~ pi_is_3.14", "# Problem of the month Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$.", "# 10 points lie in a plane,of which 4 points are collinear.Barring these 4 no three of the 10 points are collinear.How many distinct quadrilaterals can be drawn? $10$points lie in a plane,of which $4$ points are collinear.Barring these $4$ no three of the $10$ points are collinear.How many distinct quadrilaterals can be drawn? You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it clietsoriergono We are never told to use only four points. In other words, we can make a quadrilateral which includes more than four points of yours. For example, there may be another point on an edge of the quadrilateral.", "Math Help - Problem of the month 1. Problem of the month How many planes would by formed by n non-collinear and non-coplanar points? 2. Binomial Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$. -Kevin C.", "# Points in a Box Consider the integral points $(x, y, z)$ on the plane $35 x + 55 y + 77 z = 1.$ How many are contained within a cube with side length 30 centered at $(0, 0, 0)?$ × Problem Loading... Note Loading... Set Loading...", "# Number of triplets of collinear points in a standard setting This might be a stupd inquiry, but I was working on something and stumbled upon this question: if say we are given $n$ points in plane that determine $k$ distinct lines, what is the number of triplets of collinear points? Any help would be really appreciated Sorry if it turns out to be too trivial or too hard to find some bounds. (I'm just tired of thinking about it without luck) Thanks! -", "but they are made of 3 distinct points, hence the triangles are being counted only once. hence, there are $\\binom{19}{2}+\\binom{18}{2}+\\binom{17}{2}+........ .+\\binom{2}{2}=\\binom{20}{3}$ triangles. Thank you so much! You explained it in an very easy and beautiful way. Thanks again!", "choosing four of those nine points, which can be done in $$\\binom{9}{4}$$ ways.", "# Thread: Problem of the month 1. ## Problem of the month How many planes would by formed by n non-collinear and non-coplanar points? 2. ## Binomial Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$. -Kevin C.", "number of triples of distinct, non collinear points on a given plane, that is $$p^{2} (p^{2} -1) (p^{2} - p).$$ The net result is $$\\frac{p^{3} (p^{3} -1) (p^{3} - p)}{p^{2} (p^{2} -1) (p^{2} - p)} = p (p^{2} + p + 1).$$ The same method allows for an easier counting of the lines, as $$\\frac{p^{3} (p^{3} - 1)}{p (p-1)} = p^{2} (p^{2} + p + 1).$$ - And the number of planes? – Gerry Myerson Jun 11 at 9:28 @GerryMyerson, you're right, I missed that one. – Andreas Caranti Jun 11 at 10:34 @GerryMyerson, added, thanks a lot. – Andreas Caranti Jun 11 at 16:02", "the three picks. $$\\binom 41\\binom {26}2$$ And", "# There are 25 points on a plane of which 7 are collinear . How many quadrilaterals can be formed from these points ? There are $25$ points on a plane of which $7$ are collinear . How many quadrilaterals can be formed from these points ? You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it aitorarjolia A quadrilateral is formed by $4$ points, where at most $2$ may be colinear. Thus, we have", "for each remaining box choose what object was placed in it. $$\\binom{5}{2}\\cdot 4! = 240$$", "Points in space There are n points, of which no three lie on one line and no four lies on one plane. How many planes can be guided by these points? How many planes are there if there are five times more than the given points?", "(next to those of length 4) • 6 in a about 26° angle ($\\{(1,1), (2,3), (3,5)\\}$ is an example) • 6 in the same angle, but rotated 90° $$\\small(\\text{number of triangles}) = (\\text{number of total combinations}) - (\\text{number of combinations on straight lines})$$ The probability of finding a triangle, is of course that number divided by the total number of combinations. $$\\frac{{25 \\choose 3} - \\left(12 \\cdot {5 \\choose 3} + 4 \\cdot {4 \\choose 3} + 16 \\cdot {3 \\choose 3}\\right)}{25 \\choose 3} = 1 - \\frac{152}{2300}$$ - $\\binom{25}{3} - 12\\binom{5}{3} - 4\\binom{4}{3} - 16\\binom{3}{3}$ Total combos - (12 lines of five)(10 ways to choose 3 from 5) - (4 lines of four)(4 ways to choose 3 from 4) - (16 lines of three)(1 way to choose 3 from 3) -", "# Problem of the month • September 29th 2008, 05:36 AM MATHMATHmathmathMathMath Problem of the month How many planes would by formed by n non-collinear and non-coplanar points? • September 29th 2008, 09:07 AM TwistedOne151 Binomial Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$. -Kevin C.", "have three of. Now that we have settled the kinds, there are $\\binom{4}{2}$ ways to choose the actual pair, and for each of these, there are $\\binom{4}{3}$ ways to choose the actual triple, for a total of $(13)(12)\\binom{4}{2}\\binom{4}{3}$, naturally same answer. And that's not the only way to organize the count. – André Nicolas Mar 21 '12 at 18:43", "to all 4 points. Note: There are $\\binom {4}{3}$ occurrences of this scenario, since you are choosing three points from four. Second scenario: Split the points into pairs. Then form parallel planes (blue) through each of the pairs. A third plane parallel to these two and in the center of them is equidistant to all 4 points. There are $1/2 \\times \\binom {4}{2}$ occurrences of this scenario, since you are choosing 2 pairs from four. Here's a link on choosing pairs: Combination of splitting elements into pairs $$\\binom{4}{3} + \\frac{1}{2} \\binom{4}{2} = 4+3 = 7$$ • Nice figure but in the first one the required plane is not shown (rather some unnecessary plane is shown) and that is confusing. Also it'd have been better had you explained your combinations. Specially the $1/2$ in second case. – Hritik Feb 10 '18 at 9:45 • Thanks for your suggestions! I have amended those faults mentioned. – Mint Mar 21 '18 at 15:11", "Now showing items 1-1 of 1 • #### Abundance of 3-Planes on Real Projective Hypersurfaces  [OWP-2014-14] (Mathematisches Forschungsinstitut Oberwolfach, 2014-11-11) We show that a generic real projective n-dimensional hypersurface of odd degree $d$, such that $4(n-2)=\\binom{d+3}{3}$, contains \"many\" real 3-planes, namely, in the logarithmic scale their number has the same rate of ...", "in a plane. So, two points (homes) and three polylines (roads) have the same ID since they belong. Using the Code." ]

[ "System of Non-collinear Points in a Plane! There are seven points in a plane, no three of which are collinear. Every point is connected to every other point by either a red or a blue line segment. What is the minimum number of monochromatic triangles that can be present in such a figure? Select the correct option. Bonus: Generalize, if possible, the minumum number of monochromatic triangles for such a problem scenario as of above, given that there are $$n \\geq 3$$ points in the plane. ×", "Points in space There are n points, of which no three lie on one line and no four lies on one plane. How many planes can be guided by these points? How many planes are there if there are five times more than the given points?", "distinct non-collinear points determines a unique plane, and since those three points each satisfy that equation for a plane, you can be sure that that equation is the equation of the plane containing those three points.", "# Problem of the month Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$.", "non-collinear points determine a plane. This statement means that if you have three points not on one line, then only one specific plane can go through those points. The plane is determined by the three points because the points show you exactly where the plane is. Does a plane have a definite beginning and end? A plane has a definite beginning and end. A line has one dimension, length. ... A plane consists of an infinite set of lines. Can a plane have 4 points? Four points (like the corners of a tetrahedron or a triangular pyramid) will not all be on any plane, though triples of them will form four different planes. How many planes can contain 2 points? Through any two points there exists exactly one line. A line contains at least two points. If two lines intersect, then their intersection is exactly one point. Through any three non-collinear points, there exists exactly one plane. How many points are needed to identify a plane? Plane determined by three points But most of us know that three points determine a plane (as long as they aren't collinear, i.e., lie in straight line). Can 2 points determine a line? Two distinct points determine exactly one line. That line is the shortest path between the two points. ... If", "tree model of computation, Erickson proved that the conjecture is true. What's the maximum number of (maximal) collinear sets of points in a set of n points in the plane? It can grow quadratically as a function of N. Consider the n points of the form: (x, y) for x = 0, 1, 2, and 3 and y = 0, 1, 2, ..., n / 4. This means that if you store all of the (maximal) collinear sets of points, you will need quadratic space in the worst case.", "# Thread: Problem of the month 1. ## Problem of the month How many planes would by formed by n non-collinear and non-coplanar points? 2. ## Binomial Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$. -Kevin C.", "Math Help - Problem of the month 1. Problem of the month How many planes would by formed by n non-collinear and non-coplanar points? 2. Binomial Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$. -Kevin C.", "# Problem of the month • September 29th 2008, 05:36 AM MATHMATHmathmathMathMath Problem of the month How many planes would by formed by n non-collinear and non-coplanar points? • September 29th 2008, 09:07 AM TwistedOne151 Binomial Since three non-collinear points define a plane, this is just the number of cominations of three points we can choose from the n, which is $\\binom{n}{3}=\\frac{n!}{3!(n-3)!}=\\frac{n(n-1)(n-2)}{6}$. -Kevin C.", "geometric figure consisting of an assemblage of points, the polars with respect to an inversion circle constitute another figure. These figures are said to be reciprocal with respect to each other. Then there exists a duality principle which states that theorems for the original figure can be immediately applied to the reciprocal figure after suitable modification (Lachlan 1893)... ### Cylinder cutting The maximum number of pieces into which a cylinder can be divided by oblique cuts is given by(1)(2)(3)where is a binomial coefficient.This problem is sometimes also called cake cutting or pie cutting, and has the same solution as space division by planes. For , 2, ... cuts, the maximum number of pieces is 2, 4, 8, 15, 26, 42, ... (OEIS A000125). Unsurprisingly, the numbers of this sequence are called cake numbers. ### Cube division by planes The average number of regions into which randomly chosen planes divide a cube is(Finch 2003, p. 482).The maximum number of regions is presumably the same as for spacedivision by planes, namely(Yaglom and Yaglom 1987, pp. 102-106). For , 2, ... planes, this gives the values 2, 4, 8, 15, 26, 42, ... (OEIS A000125), a sequence whose values are sometimes called the \"cake numbers\" due to their relation to the cake cutting problem. ### Space division by spheres The number", "number of triples of distinct, non collinear points on a given plane, that is $$p^{2} (p^{2} -1) (p^{2} - p).$$ The net result is $$\\frac{p^{3} (p^{3} -1) (p^{3} - p)}{p^{2} (p^{2} -1) (p^{2} - p)} = p (p^{2} + p + 1).$$ The same method allows for an easier counting of the lines, as $$\\frac{p^{3} (p^{3} - 1)}{p (p-1)} = p^{2} (p^{2} + p + 1).$$ - And the number of planes? – Gerry Myerson Jun 11 at 9:28 @GerryMyerson, you're right, I missed that one. – Andreas Caranti Jun 11 at 10:34 @GerryMyerson, added, thanks a lot. – Andreas Caranti Jun 11 at 16:02", "there can be infinitely many planes that can pass through that line. However, the points lying on the same line are still coplanar. In other words, collinear points are always coplanar, but coplanar points may or may not be collinear. Collinearity is a concept of two-dimensional plane geometry and coplanarity is a concept of three-dimensional spatial geometry. In the above figure, note that the points B, D, and A are collinear as well as coplanar. All these three points lie in both the planes $\\wp$ and $\\Re$. These points are coplanar with C and E in plane $\\wp$ and with G and F in plane $\\Re$. However, C, E, F, and G are non-coplanar points. Again the points H and I do not lie in any of the two planes shown in the figure. ## Four or More Points The condition for defining a plane is that there should be a minimum of three non-collinear points. Then we can use these points to describe the plane that they lie in. When we have four or more points, obviously we can define more than one planes in three-dimensional space taking three points at a time. This concept is similar to a two-dimensional plane. If we have 2 points in a plane, we can define a line that joins the", "c 2. How many points are needed to establish a plane? | … Explanation: Two distinct points, A and B, determine a unique line in space. A third point C not on this line determines a unique plane that we can denote as ABC. How many points are required to define a plane? – … 1. Click to reveal two points P and Q; 2. Click for a plane on the line PQ; 3. Move the slider? How many planes go through two points; 4. Click to reveal a third point A; 5. Move the slider so that the orangle plane contains the three points P, Q and A; 6. Click to reveal plane that contains the ... Four Ways to Determine a Plane - dummies If you want to work with multiple-plane proofs, you first have to know how to determine a plane. Determining a plane is the fancy, mathematical way of saying “showing you where a plane is.” There are four ways to determine a plane: Three non-collinear points determine a … How many points are there in a plane? - Quora In any of these cases, the answer is that three points determine a plane unless the points are collinear. (Note that in spherical geometry, two points don’t determine a unique line if they are", "composed of two numbers and is the coordinate of the point. Within any Euclidean space, a plane is uniquely determined by any of the following combinations: • three points which are not lying on the same line • a line and a point not on the line • two different lines which intersect • two different lines which are parallel • a vector normal to the plane and a distance from the origin.", "# Greatest number of planes we can get when dividing with lines and circles What is the greatest number of parts a plane can be divided into using $n$ infinite straight lines? What about $n$ circles? Can you generalise this into 3-dimensional space, planes and spheres? For lines, I got $U_{n+1}=U_n+n,$ with $U_0=1.$ And for circles, I got $U_{n+1}=U_n+2n,$ with $U_0=1$ and $U_1=2.$ Am I right so far? • If it is $n$, how can it be infinite? – hjpotter92 Mar 24 '13 at 16:46 • @Back in a Flash - The line itself is infinite, it is not saying that there is an infinite number of lines – user61067 Mar 24 '13 at 16:48 Lines For a plane with $n$ lines, consider what happens when you add a line. It divides all the regions through which it passes into two, thus adding one region for each region it passes through. The number of regions it passes through is the number of lines it crosses plus one, that is the number of points it creates plus one for itself. Thus, the number of regions added to the original one, $\\binom{n}{0}$, is the number of points, $\\binom{n}{2}$, plus the number of lines, $\\binom{n}{1}$. Thus, the maximum number of regions is $$\\binom{n}{0}+\\binom{n}{1}+\\binom{n}{2}=\\frac{n^2+n+2}{2}$$ Circles For a plane with $n$ circles, consider", "# Number of triplets of collinear points in a standard setting This might be a stupd inquiry, but I was working on something and stumbled upon this question: if say we are given $n$ points in plane that determine $k$ distinct lines, what is the number of triplets of collinear points? Any help would be really appreciated Sorry if it turns out to be too trivial or too hard to find some bounds. (I'm just tired of thinking about it without luck) Thanks! -", "and Werneck) and recently merging ... Given $n$ points with integer coordinates in the plane, determine the maximum number of points that lie on the same circle (on its circumference, not its interior). This can be done in $O(n^3)$ ...", "73 views Consider six distinct points in a plane. Let $m$ and $M$ denote the minimum and maximum distance between any pair of points. Show that $M/m \\geq \\sqrt{3}$. | 73 views", "don't know the math, but I guess that's what Least-Squares fit will give (and I may be totally wrong). To illustrate my problem, let's say there are 7 points, and there exists a plane - A - that contains 5 of them. There would of course be other planes that contain fewer than 5. I just want some algorithmic way to find this plane A. Oct 2 '19 at 13:20 • Are there any additional constraints, for example the distribution of the points or distance between the points on the plane? Oct 6 '19 at 16:52 • @rjg The coordinates of the points in 3d space are available, so the distance between them can be calculated. There are no constraints, these points are just randomaly distributed in space. And there is no plane as such, just the points. I want to find out the plane that has the maximum number of them. Or in other words, from the given set of points I need to find out the maximum of them that are coplanar. (In the worst case there will just be 3 such points.) Oct 7 '19 at 5:03 There is no \"simple\" way to do anything in numerical analysis. However, the tasks can be laid out straightforwardly. EDIT: This idea is a brain-fart because it doesn't", "7) are collinear." ]

[ "are 'mxn' ways.This rule also complies even if the no. of works is more than two.for a basic concept let us say you have 3 shirts and 4 pants.Then the no. of ways you can wear ur shirt and pant(different dress combination) is given by 3x4 = 12 ways. Let us suppose, we have four teams involved in a football tournament and we need to find out the number of possibilities for 1st and 2nd place.Then we can use the principle of counting to find it. Let us suppose the teams are A, B, C and D.Then the 1st place can either be team A, B, C and D ie 4 ways.For the second place, there are only three options as one team will be in the 1st place. So 2nd place has 3 ways/possibilities.Let us understand from the table below: First Place Second Place possibilities Final standing A B or C or D 1st:A 2nd :B or AC or AD B C or D or A BC or BD or BA C A or B or D CA or CB or CD D A or B or C DA or DB or DC From the table above it is clear that the 1st place has 4 choices (A,C,B,D) and 2nd place has 3 remaining choices.We can also", "A holds the tiebreaker against Teams B, C and D, then Team A will have clinched a playoff berth since they cannot be overtaken by both Teams B and C. Also, if Team D does not hold a tiebreaker against any of Teams A, B and C then it will be out of playoff contention since it cannot overtake both Teams B and C. A similar scenario occasionally occurs in European soccer leagues and other competitions that use promotion and relegation. In this scenario for a 20 team soccer league that plays a double round robin format, awards three points for a win and one for a draw and relegates the 18th, 19th and 20th place teams: Position Team Played Points 16 A 36 38 17 B 36 34 18 C 36 32 19 D 36 28 If Team A loses its last two matches, it will finish with 38 points while if Team D wins its last two matches, it will finish with 34. Nevertheless, regardless of goal difference or any other tiebreaker, if Teams B and C still have to play each other then Team A is safe from relegation since Teams B and C cannot both reach 38 points, while Team D will be relegated since Teams B and C cannot both finish with less", "evenly matched or seeding is more difficult, power pools provide unnecessary protection for higher-seeded teams and allows teams to back into the Sunday brackets. If you do decide to do power pools, they don't have to be the same size as the lower pools. Two power pools of 5 and four lower pools of 4 is a pretty good 26-team format. While a general Saturday format is pretty easy to come up with, the Sunday format may present more headaches in scheduling. Even at high levels (although not at the highest), teams with a 10 am consolation game will likely not stick around for a scheduled 3:30 game, especially if flights are to be made or watching finals looks like a more appealing option. While this is frustrating not only to the tournament director, but also to any opponents who DID stick around, there are things a TD can do to minimize end-of-day bailing. The first is to schedule more consolation early in the morning, with as few byes as possible. Similarly, don't drag out consolation brackets for four games per team - give teams more games that matter on Saturday or Sunday morning with pool play and/or crossovers, rather than sending teams straight to consolation. Also, if you have the flexibility, try to set up games between", "group to be decided after Matchday 2 (i.e. first 4 matches)? Yes, in the sense that after Matchday 2, we know which 2 teams go through to the knockout rounds and which two get knocked out. For example, if A beats C and D, and B beats C and D, then A and B are through for sure. This happened in this year’s Group A (Russia, Uruguay, Egypt, Saudi Arabia) and Group G (England, Belgium, Tunisia, Panama). Having A and B beat C and D (as above) is the only way for the group to be decided after Matchday 2 in the sense above. There is no way for the winner of the group to be decided after Matchday 2. 4. Is 2 losses enough to guarantee elimination? Interestingly, it is not enough! This is the case with this year’s Group F (Mexico, Germany, Sweden, South Korea). Here is a possible configuration: A beats B, C and D to top the group with 9 points. B beats C, C beats D, and D beats B, so they each have 3 points and 2 losses. Since the top 2 teams go through, one of the teams with 2 losses will go into the knockout stages. This is the only possible configuration for a team with 2 losses to go", "vs ER Plot 3. Team Runs vs SR Power play plot 4. Team Runs vs SR Middle overs plot 5. Team Runs vs SR Death overs plot 6. Team Wickets vs ER Power Play 7. Team Wickets vs ER Middle overs 8. Team Wickets vs ER Death overs The below functions are based on all matches between 2 teams’ 1. Team Runs vs SR Plot all Matches 2. Team Wickets vs ER Plot all Matches 3. Team Runs vs SR Power play plot all Matches 4. Team Runs vs SR Middle overs plot all Matches 5. Team Runs vs SR Death overs plot all Matches 6. Team Wickets vs ER Power Play plot all Matches 7. Team Wickets vs ER Middle overs plot all Matches 8. Team Wickets vs ER Death overs plot all Matches C. Team Performance tab The below functions are based on a team’s performance against all other teams 1. Team Runs vs SR Plot overall 2. Team Wickets vs ER Plot overall 3. Team Runs vs SR Power play plot overall 4. Team Runs vs SR Middle overs plot overall 5. Team Runs vs SR Death overs plot overall 6. Team Wickets vs ER Power Play overall 7. Team Wickets vs ER Middle overs overall 8. Team Wickets vs ER Death overs overall D.", "# 004 Sample Final A, Problem 18 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) Ten teams are entered in a bowling tournament. In how many ways can first, second, and third prizes be awarded? (simplify your answer to a single number) Foundations If there are n teams, how many ways can first prize be awarded? Each team could win first place. So, there are n ways that first prize can be awarded. Solution: Step 1: There are 10 ways that first prize could be awarded. Step 2: After choosing a first place winner, there are 9 ways that second prize could be awarded. Step 3: After choosing both a first and second place winner, there are 8 ways that third prize could be awarded. Step 4: So, the total number of ways first, second, and third prizes can be awarded is ${\\displaystyle 10\\cdot 9\\cdot 8=720}$. ${\\displaystyle 720}$", "teams that have had their prospective opponents bail on them. It generates goodwill with the teams for being accommodating, and it helps teams get their money's worth (especially if they're traveling a long distance). Traditional Sunday bracket play is 3 or 4 rounds of A bracket play, with smaller consolation brackets for 5th-8th place and 9th-12th place (if there were prequarters). With a full round of 16, a 9th-16th place bracket can be played, but teams probably won't want to play out the entire bracket - 2 games per team in consolation is acceptable. For lower brackets, try to group teams in groups of 4 or 8, ensuring that teams get their 6 game minimum and making sure there aren't many byes. If you end up without multiples of 4, consolation pool play is an option, as are NFL-playoffs-style 6-team consolation brackets. As I mentioned before, try to minimize byes (especially for consolation brackets) and schedule as few games as possible during the finals. Don't schedule games after finals. An extremely helpful tool in the tournament-planning process is an Excel spreadsheet with field numbers across the top and round times along the left side. There you can visually plot out where pool play games go, where games in various consolation brackets fit on Sunday, and make sure that", "as well as a floater. • Games last 90 minutes, with 1 hour and 45 minutes between start times. Games will begin at 9:00, 10:45, 12:30, and 2:15 on both days. • The Soft cap is at 80 minutes and the hard cap is at 90. • Schedules for each round of play will be posted 5 minutes after all teams have submitted their scores. Because generating schedules relies on teams entering scores, please enter your scores as often as possible (preferably each point) at m.leaguevine.com and once a game is done, check the \"Final Score\" box when you submit it. • There will be 5 rounds of Swiss play. Four will take place on Saturday and the final round of Swiss play will take place on Sunday. • The teams ranked 1-8 after Swiss play will move onto the Swiss Bracket (1st place). • The teams ranked 9-16 after Swiss play will move onto the Sharp Cheddar Bracket (9th place). • The teams ranked 17-24 after Swiss play will move onto the Mild Cheddar Bracket (17th place). • The teams ranked 25-32 after Swiss play will move onto the Aged Cheddar Bracket (25th place). • Winners of each bracket will receive a wheel of cheese. The top three teams in the tournament will receive monetary prizes. •", "pencil case. Suppose he randomly withdraws one writing tool, followed by another, without replacing the first one. What is the probability that Bao will randomly draw a) a red pen followed by a blue pen? b) a pen followed by a pencil? Q9 Abia is one of three servers who work at a restaurant that is open from Tuesday to Sunday. Every week the servers randomly draw slips of paper from a hat to decide which two days they will not have to work, in addition to Monday. One week Abia gets to draw her two days first. a) What is the probability that Abia will draw a weekend day (Saturday or Sunday) on her first draw? b) What is the conditional probability that Abia will draw a second weekend day, given that her first draw was a weekend day? c) What is the probability that Abia will get to enjoy a three-day weekend (Saturday to Monday)?", "about cricket: The Cricket World Cup certainly does not have a series of eliminations that narrow the field down to six teams, out of which two are chosen at random to play in the final match. 18. Oct 20, 2009 ### zgozvrm You're misunderstanding the structure; it's probably very similar to your Cricket World Cup. There are 4 teams remaining (not 6). 2 of them are in the American League, and 2 are in the National League. The American League teams play each other until there is only one team left (the American League champs). The National League does the same, resulting in the National League champs. At the end, you will have one American League team playing one National League for the World Series. There is nothing random about who is picked to play in the World Series. Only in who wins. 19. Oct 20, 2009 ### uart Hey I knew that I didn't have enough information about the series to make the calculation that's why I asked for more info. Prior to getting that info however I worked with the assumption that all possible pairings were equally likely, that's all. 20. Oct 20, 2009 ### egfx Ok, so I received this email from my dad this morning, he is clearly wrong. But help me disprove the", "# Throwing golf balls for teams #### jbc ##### New Member Please can somebody help with this. 8 of us usually play golf on a Saturday. Before we play we throw balls to see who will play with each other - 2 groups of 4 are chosen by taking those balls that lie closest to each other after they have been thrown. My question is this: What is the probability that 2 specific people will randomly fall into the same group of 4 players from the initial sample of 8? #### Dason There are $$\\binom{8}{4}$$ different ways to split up into two groups of 4 each. Once you pick the specific pair you have in mind there are $$\\binom{6}{2}$$ ways to choose the remaining 2 people in the group. So the probability that those two specific people end up together is $$\\frac{\\binom{6}{2}}{\\binom{8}{4}}$$", "Kason Wong 2022-12-27 In a relay race,there are six teams A,B,C,D,E,F.(i)What is the probability that A,B,C,D finish first, second, third and fourth respectively.(ii)What is the probability that A,B,C and D are first four to finish(in any order).Assume that all finishing orders are equally likely. Toparellon12 Expert Here there are 6 teams . Out of this only 4 teams wins. ie total number of orders= 6P4=360 n(S)= 360 (1) There is only one way ABCD could be in sequence, ie n(L) = 1 there for probability of A,B,C,D finish first, second, third and fourth respectively= n(S)/n(L) = 1/360 (2) probability that A,B,C and D are first four to finish Let's create a sample set for this There are six possible outcomes here with A as the first letter. Simillarly with BCD there are 6 outcomes each, total number of such outcomes= 6*4=24 therefor probability that A,B,C and D are first four to finish = 24/360 =1/15 Do you have a similar question?", "the winner. An additional scoring method is set play. This comprises two sets over nine ends. Should a player win a set each, they then play a further 3 ends that will decide the winner. Pairs allows both people on a team to play Skip and Lead. The lead throws two bowls, the skip delivers two, then the lead delivers his remaining two, the skip then delivers his remaining two bowls. Each end, the leads and skips switch positions. This is played over 21 ends or sets play. Triples is with three players while Fours is with four players in each team and is played over 21 ends. Another pairs variation is 242 pairs (also known as Australian Pairs). In the first end of the game the A players lead off with 2 bowls each, then the B players play 4 bowls each, before the A players complete the end with their final 2 bowls. The A players act as lead and skip in the same end. In the second end the roles are reversed with the A players being in the middle. This alternating pattern continues through the game which is typically over 15 ends. https://en.wikipedia.org/wiki/Origins_of_baseball Rounders Main article: Rounders The British game most similar to baseball, and most mentioned as its ancestor or nearest relation, is", "# Sara's Softball team can be divided into four groups. 1/2 ofthe players are stronge hitters. 1/4 are good pitchers, 1/8 like toplay the outfield and 2 players are catchers. If each player is inonly one group how many players are on the team and in each group? Question Other Sara's Softball team can be divided into four groups. 1/2 ofthe players are stronge hitters. 1/4 are good pitchers, 1/8 like toplay the outfield and 2 players are catchers. If each player is inonly one group how many players are on the team and in each group? 2021-02-21 1/2+1/4+1/8=7/8 1-7/8=1/8 hence 1/8 of the player are the 2 catchers. The total no. of players $$\\displaystyle={8}\\cdot{2}={16}$$ There are 8 strong hitters, 4 good pitchers and 2 who playoutfiled. ### Relevant Questions Five distinct numbers are randomly distributed to players numbered 1 through 5. Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, players 1 and 2 compare their numbers; the winner then compares with player 3, and so on. Let X denoted the number of times player 1 is a winner. Find P{X = i}, i = 0,1,2,3,4. The unstable nucleus uranium-236 can be regarded as auniformly charged sphere of charge Q=+92e and radius $$\\displaystyle{R}={7.4}\\times{10}^{{-{15}}}$$ m. In nuclear fission, this can", "Member My alma mater is a division III school where they have a real playoff for the national champion - 32 teams. They lost in the second round, but still much more exciting than the 4 team silliness.", "a loss). In the case of a points tie a scale of other metrics are used to decide final positions such that no team has the same final position in the group. What is: a) the lowest number of points a team can get and still qualify for the knockout stage? b) the highest number of points a team can get and not qualify for the knockout stage? A. 4 & 11 B. 5 & 11 C. 6 & 11 D. 4 & 12 E. 5 & 12 ##### Solution Option D, 4 points & 12 points. Scenario/reasoning for a) = 4: One team wins all six of its matches. The three other teams all draw with each other in all of their matches, thereby all finishing on 4 points. One of these teams will finish second, based on head-to-head away goals etc. Scenario/reasoning for b) = 12: One team loses all six of its matches. The three other teams all beat each other once and lose to each other once, thereby all finishing on 12 points. One of these teams will finish third, based on head-to-head goal difference etc. ### October 2020 Cari and Morgan play rock-paper-scissors 10 times. You know that Cari uses rock three times, scissors six times, and paper once. Morgan uses rock twice,", "relegated based on the number of teams in their division and their rank among that division.} \\paragraph{If a division has 4-8 teams, 1 team gets relegated and 1 team gets promoted.} \\paragraph{If a division has 9-12 teams, 2 teams get relegated and 2 teams get promoted.} \\paragraph{If a division has 13-16 teams, 3 teams get relegated and 3 teams get promoted.} \\paragraph{No team may be promoted from the highest ranked division, as there is no division for those teams to be promoted to.} \\paragraph{No team may be relegated from the lowest ranked division, as there is no division for those teams to be relegated to.} \\subsubsection{Promotion and Relegation in a Division With 4 to 8 Teams} \\paragraph{The Divisional Champions get promoted. The last place team gets relegated.} \\paragraph{If there are only 4 teams in the division, the semi-final losers play in a 5 game series. The losing team from those matches is relegated.} \\subsubsection{Promotion and Relegation in a Division With 9 to 12 Teams} \\paragraph{Both the team finished first overall and the divisional playoff winner gain automatic promotion.} \\paragraph{If the team that placed first overall also wins the Divisional Championship, the semi-final losers play in a 3 game series. The winner of this series plays a 3 game series with the runner up. The winner of this series", "+0 # Probability + Fraction of a softball - geometry. help 0 155 1 You throw four softballs at the strike-zone target shown. The softballs are equally likely to hit any point of the strike-zone target. What is the probability that the first softball hits zone 2, the second softball hits zone 1, the third softball hits zone 3, and the fourth softball hits zone 4? = ________________(Enter your probability as a fraction.) Guest Oct 1, 2018 #1 +3190 +2 $$\\text{The strike zone looks to be }18 \\times 24 = 432~sq~in \\text{ in area}\\\\ \\text{each of the triangles is }\\dfrac{6^2}{2} = 18~sq~in~\\text{in area} \\\\ \\text{Thus the probability of hitting any of those triangles is }\\\\ p=\\dfrac{18}{432} = \\dfrac{1}{24}$$ $$\\text{The throws are independent so we just multiply the individual probabilities together to}\\\\ \\text{get the probability of all 4 events occurring}\\\\ p = \\left(\\dfrac{1}{24}\\right)^4 = \\dfrac{1}{331776}$$ Rom Oct 1, 2018", "summarizes whether or not a softball team had practice when it was raining and when it was not raining at the start of the day. softball practice no softball practice raining 4 1 not raining 12 3 1. When it was raining at the start of the day, what is the probability that softball practice was held? 2. When it was not raining at the start of the day, what is the probability that softball practice was held? 3. Are the events of “holding softball practice” and “raining at the start of the day” dependent or independent events? Explain your reasoning. ### Solution (From Unit 8, Lesson 9.) ### Problem 4 Mai rolls a standard number cube and then flips a fair coin. What is the probability that Mai flips heads under the condition that she rolls a 5? A: $$\\frac{1}{2}$$ B: $$\\frac{5}{6}$$ C: $$\\frac{1}{12}$$ D: $$\\frac{5}{12}$$ ### Solution (From Unit 8, Lesson 8.) ### Problem 5 A total of 40 elementary, middle, and high school students participate in a fun run as a fundraiser. They are surveyed after the fun run to find out how many of them completed the fun run without walking. The results of the survey are shown in the table. elementary middle high walked during the fun run 8 2 1 did not", "# Combinations • May 19th 2009, 06:58 PM Nisar_0926 Combinations hey guys can anyone help me with question b) c) and d) i am really confused for some reason i am not getting the correct answer. 18) Twenty-Four girls try out for the senior softball team at St. Marcellinus. In how many ways can a team of 9 fielders (1 pitcher, 1 Catcher, 4 Infielders, 3 Outfieders) be selected given each of the following conditions? a) any girl can play any position. so for a) it is 24C9 = 1 307 504 b) Of the girls trying out are Pitchers, 5 are Catchers and the other trying out play any of the other 7 field positions. c) Emily must be pitcher and Jeannie must be the catcher, but the other girls can play any other position. d) Louise cannot be on the team b) 194 480 c) 170 544 d) 817 190 • May 20th 2009, 12:23 AM noob mathematician Quote: Originally Posted by Nisar_0926 b) Of the girls trying out are Pitchers, 5 are Catchers and the other trying out play any of the other 7 field positions. b) Is the number of pitchers given? c) ${22 \\choose 7}$ d) ${23 \\choose 9}$ • May 20th 2009, 03:34 AM Villa Incognito Quote: Originally Posted by noob mathematician" ]

[ "the three boxes in the first spot, 2 for the second spot and 1 for the third one? Example : abcde are balls and three boxes are xyz => X=[AB] Y=[cd] Z=[e] X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6 I am a bit confused. Still thinking....Please help me. thanks You keep the boxes fixed, you permute the balls. For example, if you have in the boxes |AB| |CD| |E|. This is obtained from ABCDE. You also have |CD| |E| |AB|, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls. For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities: |AB| |CD| |E| |AB| |C| |DE| |A| |BC| |DE| So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example: AB in the first box is the same as BA, and CD in the second box is the same as DC. All the arrangements |AB| |CD| |E| |BA| |CD| |E| |AB| |DC| |E| |BA| |DC| |E| should be counted just once. Therefore, 5!*3/(2! * 2! * 1!). Eva, Thanks", "any of the three boxes in the first spot, 2 for the second spot and 1 for the third one? Example : abcde are balls and three boxes are xyz => X=[AB] Y=[cd] Z=[e] X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6 I am a bit confused. Still thinking....Please help me. thanks You keep the boxes fixed, you permute the balls. For example, if you have in the boxes |AB| |CD| |E|. This is obtained from ABCDE. You also have |CD| |E| |AB|, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls. For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities: |AB| |CD| |E| |AB| |C| |DE| |A| |BC| |DE| So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example: AB in the first box is the same as BA, and CD in the second box is the same as DC. All the arrangements |AB| |CD| |E| |BA| |CD| |E| |AB| |DC| |E| |BA| |DC| |E| should be counted just once. Therefore, 5!*3/(2! * 2! * 1!).", "Z=[e] X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6 I am a bit confused. Still thinking....Please help me. thanks You keep the boxes fixed, you permute the balls. For example, if you have in the boxes |AB| |CD| |E|. This is obtained from ABCDE. You also have |CD| |E| |AB|, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls. For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities: |AB| |CD| |E| |AB| |C| |DE| |A| |BC| |DE| So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example: AB in the first box is the same as BA, and CD in the second box is the same as DC. All the arrangements |AB| |CD| |E| |BA| |CD| |E| |AB| |DC| |E| |BA| |DC| |E| should be counted just once. Therefore, 5!*3/(2! * 2! * 1!). Eva, Thanks for your help. I think that I see your point. Do you think that if the boxes are placed on a rack in a store (order of the three boxes will", "are 'mxn' ways.This rule also complies even if the no. of works is more than two.for a basic concept let us say you have 3 shirts and 4 pants.Then the no. of ways you can wear ur shirt and pant(different dress combination) is given by 3x4 = 12 ways. Let us suppose, we have four teams involved in a football tournament and we need to find out the number of possibilities for 1st and 2nd place.Then we can use the principle of counting to find it. Let us suppose the teams are A, B, C and D.Then the 1st place can either be team A, B, C and D ie 4 ways.For the second place, there are only three options as one team will be in the 1st place. So 2nd place has 3 ways/possibilities.Let us understand from the table below: First Place Second Place possibilities Final standing A B or C or D 1st:A 2nd :B or AC or AD B C or D or A BC or BD or BA C A or B or D CA or CB or CD D A or B or C DA or DB or DC From the table above it is clear that the 1st place has 4 choices (A,C,B,D) and 2nd place has 3 remaining choices.We can also", "boxes |AB| |CD| |E|. This is obtained from ABCDE. You also have |CD| |E| |AB|, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls. For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities: |AB| |CD| |E| |AB| |C| |DE| |A| |BC| |DE| So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example: AB in the first box is the same as BA, and CD in the second box is the same as DC. All the arrangements |AB| |CD| |E| |BA| |CD| |E| |AB| |DC| |E| |BA| |DC| |E| should be counted just once. Therefore, 5!*3/(2! * 2! * 1!). Eva, Thanks for your help. I think that I see your point. Do you think that if the boxes are placed on a rack in a store (order of the three boxes will matter -correct?), then we will have to multiply by 3!? That is, for 2-2-1 case, 5C2*3C2*1*(3!) ---Please let me know your thoughts.... { My approach: Here's how I approached this problem :- for 2-2-1 case, the combinations will", "2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities: |AB| |CD| |E| |AB| |C| |DE| |A| |BC| |DE| So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example: AB in the first box is the same as BA, and CD in the second box is the same as DC. All the arrangements |AB| |CD| |E| |BA| |CD| |E| |AB| |DC| |E| |BA| |DC| |E| should be counted just once. Therefore, 5!*3/(2! * 2! * 1!). Eva, Thanks for your help. I think that I see your point. Do you think that if the boxes are placed on a rack in a store (order of the three boxes will matter -correct?), then we will have to multiply by 3!? That is, for 2-2-1 case, 5C2*3C2*1*(3!) ---Please let me know your thoughts.... { My approach: Here's how I approached this problem :- for 2-2-1 case, the combinations will be 5C2*3C2*1C1 (choose 2 balls from 5 balls - then - choose 2 balls from the remaining 3 balls - then - choose 1 ball from 1 ball) Now, since the boxes are labeled differently, we need to permute", "up in a situation, where three of the integers on the blackboard are 2009, then it is not allowed in any move to replace two odd integers. Hence, the number 2009 can only be obtained as a sum of $a+b$. In the first move that gives 2009 on the blackboard, two integers a and b are chosen such that $a+b=2009$ and either $ab>2009$ or $ab=2008$. In the case, $ab=2008$, one of the two chosen numbers is one; and hence, 1 will no longer appear on the blackboard. In either case, the two integers $a+b=2009$ and ab that appear in the creation of the first 2009 cannot be used anymore to create new instances of 2009. The second 2009 can only be obtained by choosing c and d such that $c+d=2009$ and either $cd>2009$ or $cd=2008$, and just as before; the numbers $c+d=2009$ and cd cannot be used in obtaining the last 2009. So, after forming two instances of 2009, there are four integers on the blackboard that have become useless for obtaining the third instance. Hence, the last integer 2009 cannot be obtained. Wow!! Ref: Nordic Mathematical Contest, 1987-2009. Nalin Pithwa. # RMO Training: more help from Nordic mathematical contest Problem: 32 competitors participate in a tournament. No two of them are equal and in a one against", "leaving one leftover who can throw at any of the $6$. That gives $105\\times6$ outcomes. Thus the number of outcomes in which at least one pair soak each other is $21\\times6^5 - 105\\times6^3 + 105\\times6 = 163\\,296 - 22\\,680 + 630 = 141\\,246$. Divide that by $6^7 = 279\\,936$ and do some cancellation to get the answer. #### chisigma ##### Well-known member May be it is comfortable to proceed step by step as follows. If n is the number of players, then the possible plays are $n\\ (n-1)$. Let's start... a) n=2. We call the players A and B and the 2 possible plays are... AB BA ... and the only 'good pair' is AB + BA, so that P=1... b) n=3. We call the players A,B and C and the possible 6 plays are... AB BA CA AC BC CB We have 3 'good pairs' AB+BA, AC+CA, BC+CB, so that is $\\displaystyle P= \\frac{1}{2}$... c) n=4. We call the players A,B,C and D and the possible 12 plays are... AB BA CA DA AC BC CB DB AD BD CD DC We have 6 'good pairs' AB+BA, AC+CA, AD+DA,BC+CB,BD+DB,CD+DC, so that is $\\displaystyle P=\\frac{1}{2}$... d) n=5. We call the players A,B,C,D and E and the possible 20 plays are... AB BA CA DA EA AC BC CB", "$\\underbrace{b+b+\\cdots+b}_{a}=a b$ possibilities. 1.4.2. It is often easier to think about the experiments as being in chronological order, but there is no requirement in the multiplication rule that Experiment A has to be performed before Experiment $\\mathrm{B}$. Example 1.4.3 (Runners). Suppose that 10 people are running a race. Assume that ties are not possible and that all 10 will complete the race, so there will be welldefined first place, second place, and third place winners. How many possibilities are there for the first, second, and third place winners? Solution: There are 10 possibilities for who gets first place, then once that is fixed there are 9 possibilities for who gets second place, and once these are both fixed there are 8 possibilities for third place. So by the multiplication rule, there are $10 \\cdot 9 \\cdot 8=720$ possibilities. We did not have to consider the first place winner first. We could just as well have said that there are 10 possibilities for who got third place, then once that is fixed there are 9 possibilities for second place, and once those are both fixed there are 8 possibilities for first place. Or imagine that there are 3 platforms, which the first, second, and third place runners will stand on after the race. The platforms are gold, silver, and", "over the cards. If we consider the $n = 2$ situation, then there are three possible orders of cards. Remember that whatever card we turn over first, we call A, so the first card must be A. The second A can then be in one of the three remaining spaces. So the three sequences are AABB, ABAB, and ABBA. The sequence AABB will result in us getting a pair on the first turn and so winning the game in two turns. The sequences ABAB and ABBA will not get a pair on the first, but will on the second, and so take three turns. As before, $E(t) = \\frac{1}{3}(2) + \\frac{2}{3}(3) = \\frac{8}{3}$. With $n = 3$, there are 15 possible sequences. The first card must be A again, and there are five remaining choices for the second A. Then the first B must go in the first available slot, leaving three options for the second B. This leaves two spaces for the two C cards, and so only one way to place them, giving $5 \\times 3 \\times 1 = 15$ possibilities. ## Sets of four In the version of the game we have, instead of pairs of cards there are four of each card type. This makes the game easier, but the analysis harder. Now there", "[?]: 85851 [1] , given: 10254 Re: Judges will select 5 finalists from the 7 contestant [#permalink] ### Show Tags 06 May 2013, 06:18 1 KUDOS Expert's post SravnaTestPrep wrote: Quote: OA is not wrong. Again, try to get the answer if it were 4 contestants with your approach and you'll see that your approach is not correct. Four Contestants : A B C D Three are selected. This can be done in 4 C 3 ways - ABC, ABD, BCD, ACD Consider 2 prizes The prize winners from the above four combination of finalists can each be arranged in 3P2 ways The answer is 4C3 * 3P2 Each of the 4C3 ways are independent of one another and so the arrangements have to be considered independently. It will be a repetition only within the same combination. So the ranking AB in ABC is independent of the same ranking AB in ABD. So we will have the prize winners AB in ABC and the prize winners AB in ABD and so on. Your approach is obviously wrong because non finalists cannot be considered as prize winners. That's not correct. Either you don't understand the question or the solutions. Four Contestants : A B C D and 3 prizes. Possible cases: ABC ACB BAC BCA CAB CBA ABD BDA", "Db Ad, Bc, Ca, Db Ad, Bc, Cb, Da So... 14. 2. Originally Posted by riemann There are 10 persons and 10 prizes, each prize belonging to a distinct person. Find the number of ways of distributing prizes among all the persons such that 3 particular persons never get their respective prizes in any of the distributions. Each person must get 1 prize. For example: Suppose the persons are named as A,B,C,D,E,F,G,H,I,J and their respective prizes are numbered 1,2,3,4,5,6,7,8,9,10. (Its like prize 1 goes to person A, prize 2 goes to person B and so on.) Now, A-1 B-2 C-3 D-6 E-7 F-4 G-5 H-8 I-9 J-10 (Here D, E, F are the 3 particular persons who didn't get their respective prizes. The rest may or may not get their respective prizes) is one such possible distribution. In other words, if it were 4 people, 4 prizes, and 2 \"particular\" persons never get the correct prize, here's what I would say: Call the people A,B,C,D... call their prizes a,b,c,d Our two \"particular\" people are A and B The possibilities: Ab, Ba, Cc, Dd Ab, Ba, Cd, Dc Ab, Bc, Ca, Dd Ab, Bc, Cd, Da Ab, Bd, Ca, Dc Ab, Bd, Cc, Da Ac, Ba, Cb, Dd Ac, Ba, Cd, Db Ac, Bd, Ca, Db Ac, Bd, Cb,", "obviously are $V_{4}^{4}=P_{4} = 4! = 24$ possibilities for this exercise (not sure if the notation is universal), of which only one - \"BACD\", satisfies the conditions. However, had there been five elements in the example, it wouldn't have been so easy to determine the correct result by experimentation/counting values, which is why I am searching for a logical approach to the problem. - Well, D and B look like they have the most restrictive conditions so we can start with them. Say with D. Suppose D was sat at the front (he can only be first or last). Now suppose we put B second. B cannot sit next to C so C must be last, which makes A third. But then A is next to C so B is unhappy and so this arrangement doesn't work. So now we know that if D is first, B cannot be second. So C has to be second. But now wherever we put B he is unhappy. Hence we know that D cannot be at the front, so must be at the back. We have now fixed D, so could do some experimenting to find what works (this is now suitable easy since so few combinations exist) but we can keep on purely logically: As before, suppose B sits next", "C over D MinLexMax (and MinMax) elect D. This violates the Condorcet Loser criterion because majorities rank all three of the other candidates over D. The Ranked Pairs order of finish is ABCD, which has D in last place (and A is the Ranked Pairs winner). All other possible orders of finish are worse than ABCD on Ranked Pairs' minlexmax comparison. For instance, suppose we compare ABCD with DABC. The majorities on which ABCD and DABC disagree are the three majorities that rank A over D, B over D, and C over D. The largest of these three is the 53% who rank A over D, and that majority is reversed by DABC, which means DABC is worse than ABCD. Suppose we compare ABCD with the MinLexMax order of finish DACB. The majorities on which ABCD and DACB disagree are the five majorities that rank B over C, C over A, A over D, B over D, and C over D. The largest of these five is the 55% who rank B over C, which is reversed by DACB, which means DACB is worse than ABCD. ## Case of multiple products Suppose $\\{ A_1, A_2, \\cdots, A_n \\}$ is an n-tuple of sets, with respective total orderings $\\{ <_1, <_2, \\cdots, <_n \\}$ The dictionary ordering $\\ \\", "soldiers demands that the morning lineup of these soldiers be arranged differently for every next day according to the following rule: Any three soldiers cannot be lined up next to each other in the same order for others days. For example; If ABCDEFGH is the first arrangement for day 1, on the other days, ABC,BCD, CDE, DEF, EFG and FGH cannot be lined up next to each other in the same order any more, but ACB arrangement is okay for other days until used once since they are not in the same order. What is the maximum number of days can this happen? Let's first make the problem a bit smaller. Say we have just 4 soldiers. This gives us $$4!=24$$ permutations or line-ups. Let's have a look at the following sets: • $$p$$ is the set of permutations • $$t$$ is the set of substrings of length 3 that we can form from $$p$$ • $$\\color{darkblue}pt(p,t)$$ is the mapping between $$p$$ and $$t$$ ---- 53 SET p permutations or line ups CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBA ---- 53 SET t substrings of length 3 ABC, BCD, ABD, BDC, ACB, CBD, ACD, CDB, ADB, DBC, ADC, DCB, BAC, BAD, BCA CAD, CDA, BDA, DAC, DCA, CAB, CBA, DAB, DBA ---- 55 SET", "3 and 6 respectively. Any combination of B and C will be a multiple of 3 and the number of grids is 49(not a multiple of 3). Thus Option A is incorrect. Option B: Box C and Box D The number of grids covered by C and D is 6 and 4 respectively. Any combination of D and C will be a multiple of 2 and the number of grids is 49(not a multiple of 2). Thus Option B is incorrect. Option C: Box F and Box E The number of grids covered by F and E is 2 and 9 respectively. We can arrange the boxes as given below: Option D: Box F and Box D The number of grids covered by F and D is 2 and 4 respectively. Any combination of B and C will be a multiple of 2 and the number of grids is 49(not a multiple of 2). Thus Option D is incorrect. Option C is the correct option. Queen can attack the red boxes, The knight can attack the queen from Yellow boxes. Thus, From the diagram, we can see that the number of positions that are safe 28. Therefore, option B is the correct answer. A Knight at a white box can attack only grey boxes as given below. Similarly,", "getting two failure cards; now we know BE are good, A is bad, and have a 50/50 chance of selecting between CD to win. And there's a $1/2$ chance of getting one failure card. In this case, $1/3$ of the time, AE are spies and BCD are not. $2/3$ of the time, the spies are B and one of CD. Again, we can't afford to fail any more missions. So we can choose either BCD or ACE to go on the remaining missions, and we'll have a $1/3$ chance of winning. So ACD has a ${1 \\over 6} + \\left({1 \\over 3} \\cdot {1 \\over 2}\\right) + \\left({1 \\over 2} \\cdot {1 \\over 3}\\right) = 1/2$ chance of winning. So if there's one spy amongst AB, we select team ACD for mission two, and win half the time. In total then, we win the five-player game 70% of the time. That's not surprising. The six-player analysis could probably be done analagously without too much trouble, since there are still only two spies. But the mission sizes go 2-3-4-3-4, which means that if we select no spies in the first mission, it's not obvious that we win, and if we select one spy we also have the choice of selecting CDE for mission two… the decision tree gets a", "# Keeping track of how to calculate probability/permutations/combinations? I'm absolutely terrible at calculating these things and I would like to, especially with SATs coming up, improve my capabilities. What always gets me is that there are so many types of ways to combine items. One way might be to figure out rearrangements of a set: abcd => abcd abdc acbd acdb adbc adcb bacd badc... ...or combinations of the elements of the set used more than once abcd => aaaa aaab aaac aaad aaba aabb ... ...or combinations limited to two elements abcd => aa ab ac ad ba bb bc bd ca cb cc... ...or in threes abcd => aaa aab aac aad aba abb abc abd... ...or ways to combine the elements in pairs abcd => ab ac ad ba bc bd ca cb cd da... ...or pairs without repeating abcd => ab ac ad bc bd... Point taken, I'm sure. I know they're all incredibly simple: multiply two numbers or find the summation and then divide by the number or times it's going to repeat... Again though, I have no idea which calculations to apply to which variations. I know it should be logical, but I can never quite figure it out. And then there was an SAT practice problem online... http://sat.collegeboard.org/practice/sat-question-of-the-day?questionId=20120221&oq=1 Of 5 employees,", "the last $2$ spots, then there is $1$ way to assign spots to $Y$. Finally, if one $X$ was put in on of the middle two spots and one $X$ was put in one of the last two spots, there are $2\\cdot2$ ways to assign spots to $X$ and $2\\cdot1$ ways to assign spots to $Y$ (one of the first two spots and the remaining spot in the last $2$). There are $1+1+2\\cdot2\\cdot2\\cdot1 = 10$ ways to assign opponents to $B$ and $90\\cdot10 = 900$ ways to order the games. $\\boxed{\\textbf{(E)} 900}$", "& a & b & b & b & b \\\\ a & a & b & b & a & a & b & b & a & a & b & b & a & a & b & b \\\\ a & b & a & b & a & b & a & b & a & b & a & b & a & b & a & b \\\\ \\hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 2 & 1 & 1 & 1 & 2 & 1 & 2 & 2 & 2 \\end{array}$ and, because the first player lacks two points, all the arrangements that have two $a$'s make him win. There are therefore 11 of these for him. And because the second lacks three points, all the arrangements that have three $b$'s make him win. There are 5 of these. Therefore it is necessary that they divide the wager as 11 is to 5. There is your method, when there are two players, whereupon you say that if there are more players. it will not be difficult to make the division by this method. 3. On this point, Monsieur, I tell you that this division for the two players founded on combinations" ]

[ "are 'mxn' ways.This rule also complies even if the no. of works is more than two.for a basic concept let us say you have 3 shirts and 4 pants.Then the no. of ways you can wear ur shirt and pant(different dress combination) is given by 3x4 = 12 ways. Let us suppose, we have four teams involved in a football tournament and we need to find out the number of possibilities for 1st and 2nd place.Then we can use the principle of counting to find it. Let us suppose the teams are A, B, C and D.Then the 1st place can either be team A, B, C and D ie 4 ways.For the second place, there are only three options as one team will be in the 1st place. So 2nd place has 3 ways/possibilities.Let us understand from the table below: First Place Second Place possibilities Final standing A B or C or D 1st:A 2nd :B or AC or AD B C or D or A BC or BD or BA C A or B or D CA or CB or CD D A or B or C DA or DB or DC From the table above it is clear that the 1st place has 4 choices (A,C,B,D) and 2nd place has 3 remaining choices.We can also", "# 004 Sample Final A, Problem 18 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) Ten teams are entered in a bowling tournament. In how many ways can first, second, and third prizes be awarded? (simplify your answer to a single number) Foundations If there are n teams, how many ways can first prize be awarded? Each team could win first place. So, there are n ways that first prize can be awarded. Solution: Step 1: There are 10 ways that first prize could be awarded. Step 2: After choosing a first place winner, there are 9 ways that second prize could be awarded. Step 3: After choosing both a first and second place winner, there are 8 ways that third prize could be awarded. Step 4: So, the total number of ways first, second, and third prizes can be awarded is ${\\displaystyle 10\\cdot 9\\cdot 8=720}$. ${\\displaystyle 720}$", "# Thread: A 16 team bowling league has $8,000 to be awarded as prize money. If the last place 1. ## A 16 team bowling league has$ 8,000 to be awarded as prize money. If the last place A 16 team bowling league has $8,000 to be awarded as prize money. If the last place team is awarded$ 275 in prize money and the award increases by the same amount for each successive finishing place, how much will the first place team receive? a1 -- ? D = 275 an = a1=(n-1 ) d = 275 an = a1 + ( ( 16 - 1 ) d = 8000 a1(15)(275) = 8000 a14125 = 8000 a1 = 1.93 2. Originally Posted by r-soy A 16 team bowling league has $8,000 to be awarded as prize money. If the last place team is awarded$ 275 in prize money and the award increases by the same amount for each successive finishing place, how much will the first place team receive? a1 -- ? D = 275 an = a1=(n-1 ) d = 275 an = a1 + ( ( 16 - 1 ) d = 8000 a1(15)(275) = 8000 a14125 = 8000 a1 = 1.93 I thought I had answered this before! First, you have \"a1\" and \"d\" swapped. \"the", "# Thread: A 16 team bowling league has $8,000 to be awarded as prize money. If the last place 1. ## A 16 team bowling league has$ 8,000 to be awarded as prize money. If the last place A 16 team bowling league has $8,000 to be awarded as prize money. If the last place team is awarded$ 275 in prize money and the award increases by the same amount for each successive finishing place, how much will the first place team receive? a1 -- ? D = 275 an = a1=(n-1 ) d = 275 an = a1 + ( ( 16 - 1 ) d = 8000 a1(15)(275) = 8000 a14125 = 8000 a1 = 1.93 2. Originally Posted by r-soy A 16 team bowling league has $8,000 to be awarded as prize money. If the last place team is awarded$ 275 in prize money and the award increases by the same amount for each successive finishing place, how much will the first place team receive? a1 -- ? D = 275 an = a1=(n-1 ) d = 275 an = a1 + ( ( 16 - 1 ) d = 8000 a1(15)(275) = 8000 a14125 = 8000 a1 = 1.93 I thought I had answered this before! First, you have \"a1\" and \"d\" swapped. \"the", "+0 # Probability + Fraction of a softball - geometry. help 0 155 1 You throw four softballs at the strike-zone target shown. The softballs are equally likely to hit any point of the strike-zone target. What is the probability that the first softball hits zone 2, the second softball hits zone 1, the third softball hits zone 3, and the fourth softball hits zone 4? = ________________(Enter your probability as a fraction.) Guest Oct 1, 2018 #1 +3190 +2 $$\\text{The strike zone looks to be }18 \\times 24 = 432~sq~in \\text{ in area}\\\\ \\text{each of the triangles is }\\dfrac{6^2}{2} = 18~sq~in~\\text{in area} \\\\ \\text{Thus the probability of hitting any of those triangles is }\\\\ p=\\dfrac{18}{432} = \\dfrac{1}{24}$$ $$\\text{The throws are independent so we just multiply the individual probabilities together to}\\\\ \\text{get the probability of all 4 events occurring}\\\\ p = \\left(\\dfrac{1}{24}\\right)^4 = \\dfrac{1}{331776}$$ Rom Oct 1, 2018", "the last $2$ spots, then there is $1$ way to assign spots to $Y$. Finally, if one $X$ was put in on of the middle two spots and one $X$ was put in one of the last two spots, there are $2\\cdot2$ ways to assign spots to $X$ and $2\\cdot1$ ways to assign spots to $Y$ (one of the first two spots and the remaining spot in the last $2$). There are $1+1+2\\cdot2\\cdot2\\cdot1 = 10$ ways to assign opponents to $B$ and $90\\cdot10 = 900$ ways to order the games. $\\boxed{\\textbf{(E)} 900}$", "evenly matched or seeding is more difficult, power pools provide unnecessary protection for higher-seeded teams and allows teams to back into the Sunday brackets. If you do decide to do power pools, they don't have to be the same size as the lower pools. Two power pools of 5 and four lower pools of 4 is a pretty good 26-team format. While a general Saturday format is pretty easy to come up with, the Sunday format may present more headaches in scheduling. Even at high levels (although not at the highest), teams with a 10 am consolation game will likely not stick around for a scheduled 3:30 game, especially if flights are to be made or watching finals looks like a more appealing option. While this is frustrating not only to the tournament director, but also to any opponents who DID stick around, there are things a TD can do to minimize end-of-day bailing. The first is to schedule more consolation early in the morning, with as few byes as possible. Similarly, don't drag out consolation brackets for four games per team - give teams more games that matter on Saturday or Sunday morning with pool play and/or crossovers, rather than sending teams straight to consolation. Also, if you have the flexibility, try to set up games between", "A holds the tiebreaker against Teams B, C and D, then Team A will have clinched a playoff berth since they cannot be overtaken by both Teams B and C. Also, if Team D does not hold a tiebreaker against any of Teams A, B and C then it will be out of playoff contention since it cannot overtake both Teams B and C. A similar scenario occasionally occurs in European soccer leagues and other competitions that use promotion and relegation. In this scenario for a 20 team soccer league that plays a double round robin format, awards three points for a win and one for a draw and relegates the 18th, 19th and 20th place teams: Position Team Played Points 16 A 36 38 17 B 36 34 18 C 36 32 19 D 36 28 If Team A loses its last two matches, it will finish with 38 points while if Team D wins its last two matches, it will finish with 34. Nevertheless, regardless of goal difference or any other tiebreaker, if Teams B and C still have to play each other then Team A is safe from relegation since Teams B and C cannot both reach 38 points, while Team D will be relegated since Teams B and C cannot both finish with less", "teams that have had their prospective opponents bail on them. It generates goodwill with the teams for being accommodating, and it helps teams get their money's worth (especially if they're traveling a long distance). Traditional Sunday bracket play is 3 or 4 rounds of A bracket play, with smaller consolation brackets for 5th-8th place and 9th-12th place (if there were prequarters). With a full round of 16, a 9th-16th place bracket can be played, but teams probably won't want to play out the entire bracket - 2 games per team in consolation is acceptable. For lower brackets, try to group teams in groups of 4 or 8, ensuring that teams get their 6 game minimum and making sure there aren't many byes. If you end up without multiples of 4, consolation pool play is an option, as are NFL-playoffs-style 6-team consolation brackets. As I mentioned before, try to minimize byes (especially for consolation brackets) and schedule as few games as possible during the finals. Don't schedule games after finals. An extremely helpful tool in the tournament-planning process is an Excel spreadsheet with field numbers across the top and round times along the left side. There you can visually plot out where pool play games go, where games in various consolation brackets fit on Sunday, and make sure that", "group to be decided after Matchday 2 (i.e. first 4 matches)? Yes, in the sense that after Matchday 2, we know which 2 teams go through to the knockout rounds and which two get knocked out. For example, if A beats C and D, and B beats C and D, then A and B are through for sure. This happened in this year’s Group A (Russia, Uruguay, Egypt, Saudi Arabia) and Group G (England, Belgium, Tunisia, Panama). Having A and B beat C and D (as above) is the only way for the group to be decided after Matchday 2 in the sense above. There is no way for the winner of the group to be decided after Matchday 2. 4. Is 2 losses enough to guarantee elimination? Interestingly, it is not enough! This is the case with this year’s Group F (Mexico, Germany, Sweden, South Korea). Here is a possible configuration: A beats B, C and D to top the group with 9 points. B beats C, C beats D, and D beats B, so they each have 3 points and 2 losses. Since the top 2 teams go through, one of the teams with 2 losses will go into the knockout stages. This is the only possible configuration for a team with 2 losses to go", "# Sara's Softball team can be divided into four groups. 1/2 ofthe players are stronge hitters. 1/4 are good pitchers, 1/8 like toplay the outfield and 2 players are catchers. If each player is inonly one group how many players are on the team and in each group? Question Other Sara's Softball team can be divided into four groups. 1/2 ofthe players are stronge hitters. 1/4 are good pitchers, 1/8 like toplay the outfield and 2 players are catchers. If each player is inonly one group how many players are on the team and in each group? 2021-02-21 1/2+1/4+1/8=7/8 1-7/8=1/8 hence 1/8 of the player are the 2 catchers. The total no. of players $$\\displaystyle={8}\\cdot{2}={16}$$ There are 8 strong hitters, 4 good pitchers and 2 who playoutfiled. ### Relevant Questions Five distinct numbers are randomly distributed to players numbered 1 through 5. Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, players 1 and 2 compare their numbers; the winner then compares with player 3, and so on. Let X denoted the number of times player 1 is a winner. Find P{X = i}, i = 0,1,2,3,4. The unstable nucleus uranium-236 can be regarded as auniformly charged sphere of charge Q=+92e and radius $$\\displaystyle{R}={7.4}\\times{10}^{{-{15}}}$$ m. In nuclear fission, this can", "last place team is awarded $275 in prize money\" means that a1 is 275, not d. Second, you are asserting that the 16th position, the winning award, is 8000, not the sum. The nth value is, then, $a_n= 275+ (n-1)d$ and the sum, through n values is $\\frac{n(550+ (n-1)d)}{2}$. Since the sum of all 16 prizes is to be 8000, you need to solve $\\frac{16+ 15d}{2}= 8000$ for d and then find $a_{16}$. 3. Originally Posted by r-soy A 16 team bowling league has$ 8,000 to be awarded as prize money. If the last place team is awarded \\$ 275 in prize money and the award increases by the same amount for each successive finishing place, how much will the first place team receive? a1 -- ? D = 275 an = a1=(n-1 ) d = 275 an = a1 + ( ( 16 - 1 ) d = 8000 a1(15)(275) = 8000 a14125 = 8000 a1 = 1.93 This is an arithmetic sequence as you correctly pointed out. We are given the total, the number and the first term. $S_{16} = 8000$ $n = 16$ $a = 275$ $S_{16} = \\frac{16}{2}(2 \\cdot 275+15d) = 4400+30d = 8000$ Solve for the common difference $d$ Spoiler: $d = 120$ Now you can work out what the highest person won", "what is the difference between the number of participants who traveled more than 200 miles and the number of participants who traveled 200 miles or less? (A) 108 (B) 216 (C) 252 (D) 468 (E) 655 Source: Princeton Review Traveled $$> 200$$ miles $$= 65\\%$$ Traveled $$< 200$$ miles $$= 35\\%$$ Difference between number of participants who traveled $$200$$ miles or less will be $$65-35 = 30\\%$$ Total participants $$= 720$$ So, required difference will be $$720 \\cdot 30\\% = 216$$ participants. ### GMAT/MBA Expert GMAT Instructor Posts: 5549 Joined: 25 Apr 2015 Location: Los Angeles, CA Thanked: 43 times Followed by:24 members ### Re: At a certain softball tournament, 65 percent of the players traveled more than 200 miles to participate. If 720 play by Scott@TargetTestPrep » Fri Sep 25, 2020 1:23 pm VJesus12 wrote: Tue Sep 22, 2020 7:23 am At a certain softball tournament, 65 percent of the players traveled more than 200 miles to participate. If 720 players took part in the softball tournament, what is the difference between the number of participants who traveled more than 200 miles and the number of participants who traveled 200 miles or less? (A) 108 (B) 216 (C) 252 (D) 468 (E) 655 Solution: Since 65 percent of the players traveled more than 200 miles, then 35", "Will someone hit this softball out of the park? 1. Jan 22, 2005 feveredego What are the odds of something that statistically should happen 1 in 207 times happening 3 times in 66 occurrences? Let me know. 2. Jan 22, 2005 arildno $$p=\\binom{66}{3}(\\frac{1}{207})^{3}(\\frac{206}{207})^{63}\\approx\\frac{1}{263}$$ The probability for a single, specific arrangement is $$(\\frac{1}{207})^{3}(\\frac{206}{207})^{63}$$ the binomial coefficient tells us how many such arrangements exist (in your case, 45760 different arrangements of 3 \"hits\" and 63 \"misses\") Welcome to PF! Last edited: Jan 22, 2005", "getting two failure cards; now we know BE are good, A is bad, and have a 50/50 chance of selecting between CD to win. And there's a $1/2$ chance of getting one failure card. In this case, $1/3$ of the time, AE are spies and BCD are not. $2/3$ of the time, the spies are B and one of CD. Again, we can't afford to fail any more missions. So we can choose either BCD or ACE to go on the remaining missions, and we'll have a $1/3$ chance of winning. So ACD has a ${1 \\over 6} + \\left({1 \\over 3} \\cdot {1 \\over 2}\\right) + \\left({1 \\over 2} \\cdot {1 \\over 3}\\right) = 1/2$ chance of winning. So if there's one spy amongst AB, we select team ACD for mission two, and win half the time. In total then, we win the five-player game 70% of the time. That's not surprising. The six-player analysis could probably be done analagously without too much trouble, since there are still only two spies. But the mission sizes go 2-3-4-3-4, which means that if we select no spies in the first mission, it's not obvious that we win, and if we select one spy we also have the choice of selecting CDE for mission two… the decision tree gets a", "pencil case. Suppose he randomly withdraws one writing tool, followed by another, without replacing the first one. What is the probability that Bao will randomly draw a) a red pen followed by a blue pen? b) a pen followed by a pencil? Q9 Abia is one of three servers who work at a restaurant that is open from Tuesday to Sunday. Every week the servers randomly draw slips of paper from a hat to decide which two days they will not have to work, in addition to Monday. One week Abia gets to draw her two days first. a) What is the probability that Abia will draw a weekend day (Saturday or Sunday) on her first draw? b) What is the conditional probability that Abia will draw a second weekend day, given that her first draw was a weekend day? c) What is the probability that Abia will get to enjoy a three-day weekend (Saturday to Monday)?", "fixtures and thinking really hard. Below is a list of rules that went into the final model: 1. Every team can play maximum once per round 2. Every team must play every other team at least once 3. Every team must play every other team no more than once at the same home ground 4. There can be at most one match on Thursday or Friday every round 5. Every ground can only host one match at a particular day and timeslot 6. Each team must play a total of 22 matches 7. Each team cannot travel more than 2 weeks on the road 8. Adelaide Oval must have at least 22 home matches 9. SA, NSW, QLD and WA cannot have more than 1 match per round 10. Bye Rounds must occur between round 7 and round 15 11. Victoria cannot have more than one match per time slot 12. MCG cannot have more than one match per day 13. There can be no more than 3 Saturday night matches 14. Sunday must have at least 2 matches and no more than 3 matches 15. All teams must have 11 home matches 16. All equity contraints must be adhered to 17. The season must include the following special matches • Port and Gold Coast play in China", "to a tournament. If you have four fields and want to host a 20-team tournament, the schedule stresses on each team would be so great that it's not even worth it. A quick back of the envelope calculation that every tournament director should do at the beginning stages of planning a tournament: the maximum number of games your field site(s) and hours of light will allow you to host in a given weekend. For a 14-field tournament that could conceivably host games from 8:00 am (a little stressful on teams) to 6:00 pm Saturday and 8:00 am to 4:30 pm Sunday, that's 6 rounds on Saturday with 100 minutes between the start of each game, then 5 rounds on Sunday. Then you can host 14*(6+5) games at the tournament. If you have flexibility in the number of fields at the tournament, you may want to have 12 regulation-size fields with more space between than packing in 14 fields. Similarly, teams definitely appreciate making rounds a little longer so fewer games go to cap (along with less time spent at the fields - 9:00 am starts are nicer than 8:00 am). Of course, this is the maximum number of games your tournament can accommodate. Often times, the true number of games is less, especially on Sunday, when prequarters need", "50 schools are divided between three districts: A, B, and C. District A has 18 high schools total. District B has 17 high schools total, and only two of those are private independent schools. If District C has an equal number of each of the three kinds of schools, how many private independent schools are there in District A? 10 If the areas of the 4 squares are 50, 32, 18 and 12, what is the ratio of the small shaded portion to the area of the large shaded portion?Note: Figure not drawn to scale 11 In order to qualify for the year-end tennis tournament, Sam must win at least 60 percent of his matches this year. Presently Sam has won 14 of his 18 matches. Of Sam’s 13 matches remaining in the year, what is the least number that he must win in order to qualify for the year-end tournament? 12 A retailer purchases shirts from a wholesaler and then sells the shirts in her store at a retail price that is 80 percent greater than the wholesale price. If the retailer decreases the retail price by 30 percent this will have the same effect as increasing the wholesale price by what percent? 13 On Monday, the regular price of a widget was discounted by 25 percent.", "to average$50$miles per hour for the entire trip?$\\textbf{(A) }45\\qquad\\textbf{(B) }62\\qquad\\textbf{(C) }90\\qquad\\textbf{(D) }110\\qquad\\textbf{(E) }135$AMC 8, 2019, Problem 17 What is the value of the product$\\left(\\frac{1\\cdot3}{2\\cdot2}\\right)\\left(\\frac{2\\cdot4}{3\\cdot3}\\right)\\left(\\frac{3\\cdot5}{4\\cdot4}\\right)\\cdots\\left(\\frac{97\\cdot99}{98\\cdot98}\\right)\\left(\\frac{98\\cdot100}{99\\cdot99}\\right)?\\textbf{(A) }\\frac{1}{2}\\qquad\\textbf{(B) }\\frac{50}{99}\\qquad\\textbf{(C) }\\frac{9800}{9801}\\qquad\\textbf{(D) }\\frac{100}{99}\\qquad\\textbf{(E) }50$AMC 8, 2019, Problem 19 In a tournament there are six teams that play each other twice. A team earns$3$points for a win,$1$point for a draw, and$0$points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?$\\textbf{(A) }22\\qquad\\textbf{(B) }23\\qquad\\textbf{(C) }24\\qquad\\textbf{(D) }26\\qquad\\textbf{(E) }30$AMC 8, 2019, Problem 22 A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was$84\\%$of the original price, by what percent was the price increased and decreased?$\\textbf{(A) }16\\qquad\\textbf{(B) }20\\qquad\\textbf{(C) }28\\qquad\\textbf{(D) }36\\qquad\\textbf{(E) }40$AMC 8, 2019, Problem 23 After Euclid High School's last basketball game, it was determined that$\\frac{1}{4}$of the team's points were scored by Alexa and$\\frac{2}{7}$were scored by Brittany. Chelsea scored$15$points. None of the other$7$team members scored more than$2$points What was the total number of points scored by the other$7$team members?$\\textbf{(A) }10\\qquad\\textbf{(B) }11\\qquad\\textbf{(C) }12\\qquad\\textbf{(D) }13\\qquad\\textbf{(E) }14$AMC 8, 2018, Problem 1 An amusement park has a collection of scale models, with" ]

[ "+0 # help 0 34 1 +529 John has 6 green marbles and 4 purple marbles. He chooses a marble at random, writes down its color, and then puts the marble back. He performs this process 5 times. What is the probability that he chooses exactly two green marbles? Logic Dec 7, 2018 #1 +3213 +2 $$\\text{Since you are replacing the marbles after each draw the total number of green marbles}\\\\ \\text{you select has a binomial distribution with parameters }n=5,~p=\\dfrac 3 5\\\\$$ $$P[2] = \\dbinom{5}{2} \\left(\\dfrac 3 5\\right)^2 \\left(\\dfrac 2 5\\right)^3 = \\dfrac{144}{625}$$ Rom Dec 7, 2018 #1 +3213 +2 $$\\text{Since you are replacing the marbles after each draw the total number of green marbles}\\\\ \\text{you select has a binomial distribution with parameters }n=5,~p=\\dfrac 3 5\\\\$$ $$P[2] = \\dbinom{5}{2} \\left(\\dfrac 3 5\\right)^2 \\left(\\dfrac 2 5\\right)^3 = \\dfrac{144}{625}$$ Rom Dec 7, 2018", "the same time, player two gets a point. If one red and one green marble are pulled at the same time, both players get a point. Is this a fair game?", "# Patrick's Painting Problem Probability Level 1 Patrick has a box with 4 yellow balls and 4 green balls. He randomly pulls out a ball, and if it is yellow, he paints it green. He then puts it back in the box and continues this process until all of the balls are green. Find the expected value for the number of times Patrick will pull out a ball. ×", "green shirts, 4 white shirts, and 2 red shirts in his shelf. Which of the following questions about these shirts could you use probability to solve? a. If Bill takes 1 shirt from the shelf, how many shirts will be left in the shelf? b. How many shirts did Bill have all together? c. If Bill picks 1 shirt from the shelf without looking, what color will it most likely be? d. How many more green shirts than red shirts did Bill have in the shelf? 218. Sheila packed 10 apples, 25 oranges, 36 strawberries and 5 pineapples in her picnic basket. Which of the following questions about these fruits could you use probability to solve? a. What is the total number of apples and oranges in the basket? b. If a fruit is picked at random, what kind of fruit is least likely to be picked? c. How many more apples than pineapples are there in the basket? d. How many fruits did Sheila pack in all? 219. A bag contains 11 blue marbles, 5 red marbles, and 4 green marbles. Pam draws one marble from the bag, records its color, and then puts it back in the bag. Then, he draws another marble and records its color. What is the probability of drawing 2 red marbles?", "Time Limit : sec, Memory Limit : KB English Problem E: Colored Octahedra A young boy John is playing with eight triangular panels. These panels are all regular triangles of the same size, each painted in a single color; John is forming various octahedra with them. While he enjoys his playing, his father is wondering how many octahedra can be made of these panels since he is a pseudo-mathematician. Your task is to help his father: write a program that reports the number of possible octahedra for given panels. Here, a pair of octahedra should be considered identical when they have the same combination of the colors allowing rotation. Input The input consists of multiple datasets. Each dataset has the following format: Color1 Color2 ... Color8 Each Colori (1 ≤ i ≤ 8) is a string of up to 20 lowercase alphabets and represents the color of the i-th triangular panel. The input ends with EOF. Output For each dataset, output the number of different octahedra that can be made of given panels. Sample Input blue blue blue blue blue blue blue blue red blue blue blue blue blue blue blue red red blue blue blue blue blue blue Output for the Sample Input 1 1 3", "variations ) based on the number of elements, repetition and order of importance. Since the 3 digit number must be divisible by 5, we can have 0 or 5 at the units place. In how many ways can 3 red marbles, 7 blue marbles, and 5 green marbles be rranged? A Powerball ticket has you pick five numbers out of the numbers from 1 to 69 and one special number (the powerball) from the numbers 1 to 26. Another mini-lottery has 5 balls and 3 balls to be drawn. Combination Notation To find the number of combinations of n objects taken r at a time, divide the number of permutations of n objects taken r at a time by r!. An original selection of 50 color combinations you can use in your infographic and presentation design. Answer Questions and Earn Points !!! You can now earn points by answering the unanswered questions listed. i'm hoping I helped you artwork out that answer. i'm going to chat to you later,Babe. How many combinations if I'm starting with a pool of six, how many combinations are there? , these allow successive computation of respectively all numbers of combinations from the same set (a row in Pascal's triangle), of k-combinations of sets of growing sizes, and of combinations with a. How", "This would mean there would be 10 \"real\" possibities. However there are really 11 possibities. the combination:Green, Green, Orange, Orange. gets counted twice. once for having too many greens and once for having too many oranges Leaving us with an inaccurate count. Given this information, I dont see how the current method for calculating possibities would work even with modifications. And since I'm actually dealing 20 different \"colors\" being drawn 8 at a time, listing all the possibities as I did in the attachement would be very infeasible. Any suggestions on how to move forward? Attached Files: • Marble Combination Example.xls File size: 38 KB Views: 85 8. Jun 19, 2012 viraltux You can Iterate through the solutions and count them. Iterations are one of the fastest operations a processor can make. So you can do exactly the same thing you do by hand but let the computer do it. 9. Jun 19, 2012 tsasser Sounds like my initial dream of having a neat little formula is going up in smoke. lol For what I'm actually trying to accomplish I will need to write some VB code to have excel generate a list of all the possible combinations. However, I was hoping to use this marble example to come up with a quicker way to simply identify the", "# John's Red and Blue Balls John has a box that contains one red ball and one blue ball. Every minute he reaches into the box and randomly takes out a ball from it (without looking). No matter which ball he took out, he puts it back ${together\\ with}$ another ball of the same color. Thus, after $n$ minutes, the box contains a total of $n+2$ balls. The probability that after $2$ hours, the box contains exactly $77$ blue balls, can be written as $\\frac{a}{b}$, where $a$ and $b$ are coprime positive integers. Find $a+b$. ×", "Mosaics and holes / ## Algorithms, Greedy Algorithms Problem Editorial Analytics John wants to cover his yard floor with mosaics. The yard floor is a $n \\times m$ matrix and each cell is either a mosaic or a hole. John has invented a flipper machine. This machine has a size of $k$ and can select a $k \\times k$ square of the yard and flip the cells in it. By this action, every hole in the square becomes a mosaic and every mosaic in the square becomes a hole. It is illustrated by the following example: The $2 \\times 2$ square on the left has been flipped to the right one by the flipper with the size $2$ and blacks are holes and whites are mosaics. Help John to cover the floor of the yard completely by mosaics by using this machine. In each step, John selects a $k \\times k$ square of the yard floor and sets the machine on it. He wants to compute the minimum steps needed to repair the yard floor. Input Format • First line: Three integers $n, m, k$ ($1 \\leq n, m \\leq 1000 , 1 \\leq k \\leq min(n,m)$), where $n$ is the number of matrix rows, $m$ as the number of matrix columns, and $k$ as the size of", "1 : 6. There is an equal amount of lemon syrup in both cups. Cup B contains 189 mℓ more water than Cup A. What is the amount of lemon syrup in Cup A? Leave the answer in mℓ. 3 m Level 3 Mrs Lu had a total of 520 purple and yellow lilies in the ratio of 8 : 5. After she had given away 6 times as many purple lilies as yellow lilies, the number of purple and yellow lilies left was in the ratio of 10 : 9. How many more purple lilies than yellow lilies did she give away? 5 m Level 3 David, Ethan and Tim have some marbles. The ratios of their marbles are as follows: David : (David + Ethan + Tim) = 1 : 5 Ethan : (David + Tim) = 1 : 3 Tim has 42 marbles more than Ethan. In the end, Tim gives away all he has to Ethan and David so that both Ethan and David end up with the same number of marbles, how many marbles has Ethan received from Tim? 4 m Level 3 Oscar bought some mangosteens and coconuts in the ratio of 2 : 5 respectively. The cost of a mangosteen was $1.50 more than the cost of a coconut. He spent", "are 18 birds on a tree. 9 of them flew away. Choose the number sentence that helps you to calculate the number of birds left on the tree. a. 9 + $n$ = 19 b. $n$ + 18 = 9 c. 18 - 9 = $n$ d. $n$ + 20 = 18 6. Choose the relation that best represents the graph. a. Cost of a bouquet is $5. b. Each bouquet requires 5 flowers. c. Each bouquet requires 2 flowers. d. Cost of 5 bouquets is$1. 7. Which situation can be described by the expression $n$ - 3? a. John ate $n$ chocolates and Gary ate 3 more than John. b. John ate $n$ chocolates and Gary ate 3 less than John. c. John ate $n$ chocolates and Gary ate 3 times more than John. d. John ate 3 chocolates and Gary ate $n$ more than John. 8. Choose a situation that represents the equation. $n$ + 7 = 37 a. If cost of each marble is $7, then the cost of $n$ marbles are$37. b. John had 37 marbles. He bought $n$ more. Then he had 7 marbles. c. Bob had $n$ marbles. He gave 7 to his sister. Then he had 37 marbles. d. Jim had $n$ marbles. He bought 7 more. Then he had 37", "John has 340 marbles. He put them into 4 bags. He put the same number of marbles in each bag. How many marbles are in each bag? 63 views by (1.6k points) 340/4=85 this is the answer", "number of notebooks Keecia bought? 19. Sheri can make 2 bracelets in 16 minutes. She needs to make 7 bracelets. a. Set up a proportion that could be used to find $m$, the number of minutes it would take Sheri to make 7 bracelets. b. How many minutes will it take Sheri to make 7 bracelets? 20. A bag contains only two colors of marbles––silver and gold. The ratio of silver marbles to gold marbles in the bag is 5 : 3. There are a total of 30 silver marbles in the bag. a. Set up a proportion that could be used to find $t$, the total number of marbles in the bag Feb 22, 2012 Jan 06, 2016", "We use cookies to give you the best possible experience on our website. John has 12 blue marbles. any blue marbles. One jar holds 16 red marbles and 4 blue marbles. Jar B has 4 pink marbles, 6 purple marbles, and 2 white marbles. A marble is drawn at random from the jar. a) Draw the tree diagram for the experiment. If a marble is drawn at random from the jar, the probability that it is green is 2/3. ) A box contains 30 marbles: 15 red, 10 blue, and 5 green. 2/27 is the probability of selecting a red marble, replacing it in the bag and then selecting a green marble. how many additional red marbles must be added to. What is the reasonable prediction for the number of times a green or black. 3) There are thirteen shirts in your closet, five blue, five green, and three red. Older marbles often have a high-quality glass construction and are quite durable, while newer marbles might shatter during rough play. Find the probability of pulling a yellow marble from a bag with 3 yellow, 2 red, 2 green, and 1 blue-- I'm assuming-- marbles. What is the probability of drawing a green marble, P(G)? c. (a) What is the probability that one will be green and the other", "## ksf-2015-ec-20 Peter has ten balls, numbered from 0 to 9. He distributed these balls among three friends: John got three balls, George four and Ann three. Then he asked each of his friends to multiply the numbers on the balls they got, and the results were: 0 for John, 72 for George and 90 for Ann. Question: What is the sum of the numbers on the balls that John received?", "If you think you have a good answer to this, do let me know, but note that the really hard problem, still I think unsolved, is to prove the best optimum result. # Sunday Times Teaser 3007 – Paving Stones ### by Howard Williams #### Published Sunday May 10 2020 (link) James has decided to lay square block paving stones on his rectangular patio. He has calculated that starting from the outside and working towards the middle that he can lay a recurring concentric pattern of four bands of red stones, then three bands of grey stones, followed by a single yellow stone band. By repeating this pattern and working towards the centre he is able to finish in the middle with a single line of yellow stones to complete the patio. He requires 402 stones to complete the first outermost band. He also calculates that he will require exactly 5 times the number of red stones as he does yellow stones. How many red stones does he require?` # Sunday Times Teaser 3006 – Raffle Tickets ### by Andrew Skidmore #### Published Sunday May 03 2020 (link) At our local bridge club dinner we were each given a raffle ticket. The tickets were numbered from 1 to 80. There were six people on our table and all", "green marbles. In three draws, find the probability of obtaining white, black and green in that order. What is the probability of drawing 2 green marbles B. a) yellow marble b) red marble c) green marble. A jar contains 2 green marbles, 4 blue marbles, 3 yellow marbles, and 2 black marbles. If you draw with replacement, the odds are always 6/9. What is the probability that the marbles are of different colors? Log On. Solution: Not binomial: there are more than two outcomes for each trial. Example 2: A jar contains 4 blue marbles, 5 red marbles and 11 white marbles. Bag 2 contains 20 black marbles and 10 white ones. What is the probability that the random variable has an outcome. Two members are selected at random to. Find the probability of getting a blue, then white marble with replacement. You draw 5 marbles out at random, without replacement. 9) A box contains 5 red marbles and 7 green marbles. What is the probability of drawing 2 green marbles without replacement? B. *Worksheet – Finding the Probability of an Event II Find the probability for the following events. What is the probability that John. These marbles are selected at random without replacement. ● P(getting a color other than red) = P(25/55) ≈. If two marbles are", "LEAP 2025 Geometry Chapter 13 ### LEAP 2025 Geometry Chapter 13 Sample DOK: 2 1 pt 3. Brenda has 18 fish in an aquarium. The fish are the following colors: 5 orange, 7 blue, 2 black, and 4 green. Brenda’s trouble-making cat has grabbed a fish. What is the probability the cat grabbed a green fish? DOK: 2 1 pt 6. Katie spun a spinner 15 times and recorded her results in a table below. The spinner was divided into 6 sections numbered 1–6. The results of the spins are shown below. Based on the results, how many times would 4 be expected to appear in 45 spins? DOK: 2 1 pt 10. Kyle has 4 different kinds of Halloween candy in a bag. He has 5 chocolate rolls, 6 lollipops, 4 chocolate bars, and 7 peanut butter candies. What kind of candy must Kyle have picked out of the bag if the probability of picking that kind is $\\frac{2}{11}$? DOK: 3 1 pt 12. The table below shows the actual sum of the rolling of two cubes numbered 1 through 6. The two cubes were rolled 100 times. Using the information in the table, predict how many times a score of \"7\" would occur in 150 tries. DOK: 3 1 pt 17. At a Little Mario’s Restaurant,", "### Home > CC2MN > Chapter 2 > Lesson 2.2.3 > Problem2-64 2-64. John has a bag of marbles that contains $12$ red marbles, $20$ green marbles, and $17$ blue marbles. 1. If John pulls one marble out of the bag, what is the probability that it will not be red? That is, P(not red)? Consider how many marbles are not red and how many total marbles are in his bag. Then use these numbers to write the probability. 2. What is the probability that he will draw a purple marble? That is, P(purple)? How many purple marbles are in his bag? The probability is $0$. Make sure you know why and explain what a probability of $0$ means. Use the eTool to explore the problem. Click the link at right for the full version of the eTool: MC2 2-42 HW eTool", "M&M’s by color: $\\begin{array}{cccccc} Yellow & Red & Orange & Brown & Green & Blue \\\\ 0.14 & 0.13 & 0.20 & 0.13 & 0.16 & 0.24 \\end{array}$ 1. What is the probability that a randomly selected M&M is not green? 2. What is the probability that a randomly selected M&M is red, orange, or yellow? 3. Estimate the probability that a random selection of four M&M’s will contain a blue one. 4. Estimate the probability that a random selection of six M&M’s will contain all six colors. 18. Bob Ross was a painter with a PBS television show “The Joy of Painting” that ran for 11 years. 1. 91% of Bob’s paintings contain a tree. 85% contain two or more trees. What is the probability that he painted a second tree, given that he painted a tree? 2. 18% of Bob’s paintings contain a cabin. Given that he painted a cabin, there is a 35% chance the cabin is on a lake. What is the probability that a Bob Ross painting contains both a cabin and a lake? 19. Suppose that you have 10 boxes, numbered 0-9. Box $$i$$ contains $$i$$ red marbles and $$9 - i$$ blue marbles. You perform the following experiment. You pick a box at random, draw a marble and record its color." ]

[ "There are $\\binom{5}{1}$ ways to choose a red marble, and $\\binom{7}{1}$ ways to choose a green marble. Thus there are $\\binom{5}{1}\\binom{7}{1}$ ways to choose a red and a green. It follows that our probability is $$\\frac{\\binom{5}{1}\\binom{7}{1}}{\\binom{12}{2}}.$$ Harder! However, suppose we have $25$ marbles, $10$ red and $15$ green. We choose (without replacement) $8$ marbles. What is the probability that we get $3$ red and $5$ green? The same analysis shows that the probability is $$\\frac{\\binom{10}{3}\\binom{15}{5}}{\\binom{25}{8}}.$$ -", "(8/13) = 0.3043 - P[3B|2R] = ((10C2 * 6C2)/(16C4)) * (8/13) = 0.2282 - P[3B|3R] = ((10C2 * 7C3)/(17C5)) * (8/13) = 0.15663 - P[3B|4R] = ((10C2 * 8C4)/(18C6)) * (8/13) = 0.1044 Basically it is (combination without repetition -for blues) * (combination with repetition - for reds) / total combination) * (8/13). Then I would muliply the number of marbles * probability to get expected number of total balls drawn per game. But this way won't work as the probabilities (although diminishing) are going to total over 1. Any one have any thoughts on this? Probably something very simple I'm overlooking. I'm stumped all the same! Thanks very much for any and all help. 2. ## Re: Marble question - hypergeometric/binomial hybrid You draw a sea of reds, with blues on the i-th, j-th, and n-th (terminating) draws 1<=i < j < n Ex: i = 4, j = 10, n = 15: R R R B R R R R R B R R R R B The probabilitiy of such an $(i, j, n)$ string, $Prob(i, j, n)$ is: $\\left(\\frac{5}{15}\\right)^{i-1} \\left(\\frac{10}{15}\\right) \\left(\\frac{5}{14}\\right)^{j-i-1} \\left(\\frac{9}{14}\\right) \\left(\\frac{5}{13}\\right)^{n-j-1} \\left(\\frac{8}{13}\\right)$ $= \\left(\\frac{(10)(9)(8)}{(15)(14)(13)}\\right) \\left( \\frac{(5^{i-1})(5^{j-i-1})(5^{n-j-1})}{({15}^{i-1})({14}^{j-i-1})({13}^{n-j-1})} \\right)$ $= \\left(\\frac{24}{91}\\right) \\left \\frac{(5^{n-3})}{({15}^{i-1})({14}^{j-i-1})({13}^{n-j-1})} \\right$ $= \\left(\\frac{24}{91}\\right) \\left \\frac{(5^{n-3})}{({15}^{i-1})({14}^{j-i-1})({13}^{n-j-1})} \\right$ $= \\left(\\frac{24}{91}\\right) (5^{n-3}) \\left( (15)(14)\\left(\\frac{14}{15}\\right)^i \\left(\\frac{13}{14}\\right)^j \\left(\\frac{1}{13}\\right)^{n-1} \\right)$ $= \\left(\\frac{(24)(15)(14)(13)}{(91)(5^3)}\\right) \\left(\\frac{14}{15}\\right)^i \\left(\\frac{13}{14}\\right)^j \\left(\\frac{5}{13}\\right)^{n}$ $= \\left(\\frac{144}{25}\\right)", "final answer is therefore $\\binom{12}4\\binom86=13,860$ arrangements. (2) Each possible sequence of $12$ rolls of the die corresponds to one of the $13,860$ possible arrangements of a string of $4$ red, $6$ blue, and $2$ green balls. Thus, you know from (1) that there are $13,860$ ways to roll the die $12$ times and get $4$ red, $6$ blue, and $2$ green faces. Now the probability on any roll of getting the red face is $\\frac16$, the probability of getting a blue face is $\\frac26=\\frac13$, and the probability of getting a green face is $\\frac36=\\frac12$. The probability of getting any particular string of $4$ red, $6$ blue, and $2$ green faces is therefore $\\left(\\frac16\\right)^4\\left(\\frac13\\right)^6\\left(\\frac12\\right)^2$, and your final answer is $$13860\\left(\\frac16\\right)^4\\left(\\frac13\\right)^6\\left(\\frac12\\right)^2=\\frac{13860}{6^43^62^2}\\;.$$ - The first answer is wrong, and soon you will see why. We can choose the places for the red balls in $\\binom{12}{4}$ ways. For each of these ways, there are $8$ empty spots left, not $12$. We can fill these spots with blues in $\\binom{8}{6}$ ways, for a total of $$\\binom{12}{4}\\binom{8}{6}.$$ (If you prefer, there are then $2$ empty spots left, which we can fill with green in $\\binom{2}{2}$ ways. Of course, that does not affect anything, since $\\binom{2}{2}=1$.) For the second question, find the probability of the sequence they gave you. It is just a multiplication. Now", "want all of the green marbles.. then you would choose $i$ marbles from the batch of $n$ marbles. So if there were a total of 5 marbles ($n$) and two were green, and you wanted all the green's $~(i)$,then you could use the notation $n\\choose i$=$5\\choose 2$. Meaning, out of the $5$ marbles, I only choose $2$. Hope that this cleared up the meaning some for the binomial coefficient. – night owl Jul 3 '11 at 3:07 The way it works if you expand $(a+b)^n$ is that there are $n$ brackets and you have to choose an $a$ or a $b$ from each one. There are 2 choices for each bracket, hence 2^n choices overall. The sum you have given gathers all the terms with $i$ copies of $a$ and $n-i$ copies of $b$. The bracket $\\binom{n}{i}$ is the number of ways of doing this, and is called n-Choose-i (which is the reason for the $C$ in some alternative notations) or a Binomial Coefficient (binomial because the original brackets contain two terms $a$ and $b$). It might be interesting for you to see whether you can get any intuition or insight for why the value of the bracket is what it is. - It is read \"$n$ choose $i$\" and is equal to the number of ways to", "red needed. This gives a total of $\\binom{10}{2}\\binom{5}{3}\\binom{20}{5}$. For the probability, divide by $\\binom{35}{10}$. Remark: If drawing is done with replacement, we get the situation that you find familiar. The probability of getting $2$ green, $3$ yellow, and $5$ red is then $$\\frac{10!}{2!4!4!}\\left(\\frac{10}{35}\\right)^2\\left(\\frac{5}{35}\\right)^3\\left(\\frac{20}{35}\\right)^5.$$", "is of course slightly larger than $$6/20\\text{.}$$ Alice is impressed. ###### Example10.9. Consider the jar of twenty marbles from the preceding example. A second jar of marbles is introduced. This jar has eighteen marbles: nine red, five blue and four green. A jar is selected at random and from this jar, two marbles are chosen at random. What is the probability that both are green? Bob is on a roll. He says, “Let $$G$$ be the event that both marbles are green, and let $$J_1$$ and $$J_2$$ be the event that the marbles come from the first jar and the second jar, respectively. Then $$G= (G\\cap J_1)\\cup (G\\cap J_2)\\text{,}$$ and $$(G\\cap J_1)+(G\\cap J_2)=\\emptyset\\text{.}$$ Furthermore, $$P(G|J_1)=\\binom{5}{2}/\\binom{20}{2}$$ and $$P(G|J_2)=\\binom{4}{2}/\\binom{18}{2}\\text{,}$$ while $$P(J_1)=P(J_2)=1/2\\text{.}$$ Also $$P(G\\cap J_i)=P(J_i)P(G|J_i)$$ for each $$i=1,2\\text{.}$$ Therefore, \\begin{equation*} P(G)=\\frac{1}{2}\\frac{\\binom{5}{2}}{\\binom{20}{2}}+ \\frac{1}{2}\\frac{\\binom{4}{2}}{\\binom{18}{2}}=\\frac{1}{2}\\left(\\frac{20}{380}+ \\frac{12}{306}\\right). \\end{equation*} That's about $$4.6$$%.” Now Alice is speechless. ### Subsection10.2.1Independent Events Let $$A$$ and $$B$$ be events in a probability space $$(S,P)\\text{.}$$ We say $$A$$ and $$B$$ are independent if $$P(A\\cap B)=P(A)P(B)\\text{.}$$ Note that when $$P(B)\\neq 0\\text{,}$$ $$A$$ and $$B$$ are independent if and only if $$P(A)=P(A|B)\\text{.}$$ Two events that are not independent are said to be dependent. Returning to our earlier example, the two events ($$A\\text{:}$$ the marble in Xing's left pocket is red and $$B\\text{:}$$ the marble in his right pocket is blue) are dependent. ######", "of choosing one black from the two available. For each such way, there are $\\binom{8}{4}$ ways to choose the non-blacks to go with it. So the total number of ways to pick exactly one black, and the rest non-black, is $\\binom{2}{1}\\binom{8}{4}$. Thus our probability is $$\\frac{\\binom{2}{1}\\binom{8}{4}}{\\binom{10}{5}}.$$ Remark: We used general techniques. For A, there is a simpler way. The probability that the first marble chosen is non-black is $\\frac{8}{10}$. Given the first was non-black, the probability that the second is non-black is $\\frac{7}{9}$, since there are $7$ non-blacks left in a total of $9$. So the probability the first two are non-black is $\\frac{8}{10}\\cdot\\frac{7}{9}$. Continue in this way. The probability all five are non-black is $$\\frac{8}{10}\\cdot\\frac{7}{9}\\cdot\\frac{6}{8}\\cdot\\frac{5}{7}\\cdot\\frac{4}{6}.$$ As in the earlier discussion, subtract the above from $1$. - Great answer, thanks for the time it took to write all this out! A quick question though, how would I calculate the probability of getting both black marbles from the bag in the 5 drawn? – ericbowden Aug 14 '12 at 5:45 The black marbles can be chosen in $1$ way, or if you prefer, $\\binom{2}{2}$ ways (which is $1$). The remaining $3$ marbles can be chosen in $\\binom{8}{3}$ ways, for a total of $\\binom{2}{2}\\binom{8}{3}$. Divide this by $\\binom{10}{5}$. – André Nicolas Aug 14 '12 at 5:56 Exactly what I needed.", "one green marble. Choose two of the five positions for the blue marbles and two of the remaining three positions for the green marbles. The only green marble must go in the remaining position. Then we obtain $$\\binom{5}{2}\\binom{3}{2}\\binom{1}{1}\\left(\\frac{4}{11}\\right)\\left(\\frac{3}{10}\\right)\\left(\\frac{5}{9}\\right)\\left(\\frac{4}{8}\\right)\\left(\\frac{2}{7}\\right) = \\frac{20}{77}$$ Since we do not care about the order in which the marbles are selected, it is simpler to calculate the probability of selecting two of the four red marbles, two of the five blue marbles, and one of the two green marbles when we select five of the eleven marbles, which yields $$\\frac{\\dbinom{4}{2}\\dbinom{5}{2}\\dbinom{2}{1}}{\\dbinom{11}{5}} = \\frac{20}{77}$$ • Thank you for the explanation and especially for taking the time to correct my attempt, that really helps. :) Really appreciate it. :) – Lucky Apr 9 '19 at 19:32 \"So why can't we use that logic to answer this question?\" Using the logic in the first question you actually calculate the probability of the event that \"the first chosen student is a girl, the second is a girl and the third is a girl\". This is exactly the same event as \"all $$3$$ students chosen will be a girl\". Using the logic in the second situation you similarly calculate the probabitility of the event that \"the first chosen marble is red, the second is red, the third is blue, the fourth", "# Figuring out how many ways 5 marbles can be drawn (combination/permutation problem) A bag contains 24 marbles, 4 red, 12 green, and 8 brown. How many ways can 5 marbles be drawn with all 5 marbles green. I know you can consider that there are just 12 green marbles and 12 not green marbles but I am not sure of the steps to follow You can see it like this way: You know you have $$12$$ green marbles, $$4$$ red marbles and $$8$$ brown marbles, since you need to choose $$5$$ green marbles, you selected them from the $$12$$ possibles, so you need: $$\\binom{12}{5} = \\frac{12!}{5!(12-5)!} = 792$$ Now that you have this done, you need to choose $$0$$ brown marbles from the $$8$$ possibles i.e. $$\\binom{8}{0} = 1$$ and $$0$$ red marbles from $$4$$ possibles, that is: $$\\binom{4}{0} = 1$$ OR you can see it as you need to choose $$0$$ marbles from the $$12$$ not green marbles possibilities ($$8$$ browns plus $$4$$ reds), i.e. $$\\binom{12}{0} = 1$$ Finally, all the possibilities to choose 5 green marbles are: $$\\binom{12}{5} \\binom{8}{0} \\binom{4}{0} = 792*1*1 = 792 = 792 * 1 = \\binom{12}{5} \\binom{12}{0}$$ So the answer is $$792$$ possible ways. $${12\\choose 5}{8\\choose 0}{4\\choose 0}=\\frac{12!}{7!\\cdot5!}$$.", ", $\\frac{3}{13}$ $\\frac{7}{15}$ , $\\frac{3}{14}$ , $\\frac{5}{13}$ $\\frac{5}{15}$ , $\\frac{3}{14}$ , $\\frac{7}{13}$ $\\frac{5}{15}$ , $\\frac{7}{14}$ , $\\frac{3}{13}$ $\\frac{3}{15}$ , $\\frac{5}{14}$ , $\\frac{7}{13}$ $\\frac{3}{15}$ , $\\frac{5}{14}$ , $\\frac{7}{13}$ How is the quicker way obtained ? I.e what is the method behind it. There are many ways to solve the problem. Whether we think of picking the marbles one at a time, or all together, does not alter probabilities, though it will change the way we compute the probabilities. Imagine the balls are distinct (they all have secret ID numbers). There are $\\binom{15}{3}$ equally likely ways to choose $3$ balls from the $15$. Now we count the number of favourable choices, that is, choices that have $1$ of each colour. There are $\\binom{7}{1}\\binom{3}{1}\\binom{5}{1}$ ways to pick $1$ red, $1$ blue, and $1$ green. Thus our probability is $$\\frac{\\binom{7}{1}\\binom{3}{1}\\binom{5}{1}}{\\binom{15}{3}}.$$ Or else we calculate the probability by imagining choices are made one at a time. This complicates things somewhat, since the event \"we end up with one of each colour\" can happen in various ways. Let us analyze in detail the probability we get GRB (green then red then blue). The probability the first ball picked is green is $\\frac{5}{15}$ (it is best not to simplify). Given that the first ball was green, the probability the second is red is $\\frac{7}{14}$. So", "of each of these, and they are mutually exclusive, so you can add them. @Supa: that would make the probability greater than 1, which cannot be true. The chance to get $3$ green is $\\frac 58 \\cdot \\frac 47 \\cdot \\frac 36=\\frac {60}{336}$. Each fraction is the chance to get a green marble with the bag holding what is left at that point. – Ross Millikan Jan 7 '13 at 21:03", "ways to select the gender and the day of the week the child was born on. There are 132 = 169 ways which do not have a girl born on Monday and which 196 – 169 = 27 which do, so P(Y) = $$\\frac{27}{196}$$. This gives is that P(X|Y) = $$\\frac{\\frac{13}{19}*\\frac{1}{4}}{\\frac{27}{196}} = \\frac{13}{27}$$. 7. Suppose a fair eight-sided die is rolled once. If the value on the die is 1, 3, 5 or 7 the die is rolled a second time. Determine the probability that the sum of values that turn up is at least 8? a) $$\\frac{32}{87}$$ b) $$\\frac{12}{43}$$ c) $$\\frac{6}{13}$$ d) $$\\frac{23}{64}$$ Explanation: Sample space consists of 8*8=64 events. While (8) has $$\\frac{1}{8}$$ probability of occurrence, (1,7) has only $$\\frac{1}{64}$$ probability. So, the required probability = $$\\frac{1}{6} + (9 * \\frac{1}{64}) = \\frac{69}{192} = \\frac{23}{64}$$. 8. A jar containing 8 marbles of which 4 red and 4 blue marbles are there. Find the probability of getting a red given the first one was red too. a) $$\\frac{4}{13}$$ b) $$\\frac{2}{11}$$ c) $$\\frac{3}{7}$$ d) $$\\frac{8}{15}$$ Explanation: Suppose, P (A) = getting a red marble in the first turn, P (B) = getting a black marble in the second turn. P (A) = $$\\frac{4}{8}$$ and P (B) = $$\\frac{3}{7}$$ and P (A and B) = $$\\frac{4}{8}*\\frac{3}{7} = \\frac{3}{14}$$ P(B/A)", "probability of choosing a blue one first? There are 9 marbles and 3 are blue. 39=13\\begin{align*} \\frac{3}{9} = \\frac{1}{3}\\end{align*} What is the probability of choosing a green one next? There are 9 marbles and 3 are green. 39=13\\begin{align*} \\frac{3}{9} = \\frac{1}{3}\\end{align*} What is the probability of choosing an orange one next? There are 9 marbles and 3 are orange. 39=13\\begin{align*} \\frac{3}{9} = \\frac{1}{3}\\end{align*} Now we figure out the likelihood of three independent events occurring by multiplying each probability. P(B, then G, then O)=131313=127\\begin{align*}P(B, \\ \\text{then} \\ G, \\ \\text{then} \\ O) = \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} = \\frac{1}{27}\\end{align*} There is a very small likelihood that this event will occur. Now go back to the swimming and golf example. Example Out of 100 students at Riverview Middle, 50 enjoy swimming. Out of the same 100 students, 25 enjoy golf. What is the probability that a student would enjoy both swimming and golf? Let’s add to this problem that 40 out of 100 enjoy basketball. What is the likelihood that a student would enjoy all three? 501002510040100121425=12=14=25=240=120\\begin{align*}\\frac{50}{100} & = \\frac{1}{2}\\\\ \\frac{25}{100} & = \\frac{1}{4}\\\\ \\frac{40}{100} & = \\frac{2}{5}\\\\ \\frac{1}{2} \\cdot \\frac{1}{4} \\cdot \\frac{2}{5} & = \\frac{2}{40} = \\frac{1}{20}\\end{align*} There is a one out of 20 chance that a student would like all three. We can look at this as", "My Math Forum Probability practice problems? Probability and Statistics Basic Probability and Statistics Math Forum March 18th, 2011, 04:55 PM #1 Member Joined: Aug 2009 Posts: 69 Thanks: 0 Probability practice problems? Soz... I have an important standardized test tomorrow... and usually these things are really easy, but I've forgotten most of the stuff I've learned about probability. One question that really tripped me up: You have 6 white marbles, 4 green marbles, and 5 blue marbles in a bag. If you draw four marbles at random, what's the probability of getting at least one of each color? So... how do you solve this question? And what are some good websites for practicing probabilities? March 21st, 2011, 07:47 AM #2 Senior Member Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Probability practice problems? You need the generalized hypergeometric formula. If you pick n out of N things, and of those a out of A of the first sort, b out of B of the first sort, etc (so a+b+c+... = n, and A+B+C+... = N) then the probability is $\\frac{\\binom{A}{a}\\binom{B}{b}\\binom{C}{c}...\\bino m{Z}{z}}{\\binom{N}{n}}$ In this case you pick 1 white, 1 green, 2 blue, OR 1 white, 2 green, 1 blue, OR 2 white, 1 green, 1 blue. $\\frac{\\binom{6}{1}\\binom{4}{1}\\binom{5}{2}+\\binom{ 6}{1}\\binom{4}{2}\\binom{5}{1}+\\binom{6}{2}\\binom{4 }{1}\\binom{5}{1}}{\\binom{15}{4}}= \\frac{6.4.10+6.6.5+15.4.5}{1365}=\\frac{720}{1365}= \\frac{48}{91}$ Tags practice, probability, problems", "28 '16 at 11:18 • Your fellow classmate is correct. – drhab May 28 '16 at 11:18 • In small sample you are using multihypergeometric to calculate the exact probabilities; in your large sample case, you are using multinomial. – BGM May 28 '16 at 11:49 The answer provided is wrong. In the limit of very many marbles, it doesn't matter whether you draw with or without replacement, since the ratios will stay approximately the same. Thus in this limit the probability of drawing $1$ red, $2$ blue and $1$ yellow marbles is $$\\binom4{1,2,1}\\cdot\\frac3{12}\\cdot\\frac4{12}\\cdot\\frac4{12}\\cdot\\frac5{12}=\\frac{4!}{1!2!1!}\\cdot\\frac5{3\\cdot12\\cdot12}=\\frac5{36}\\approx0.1389\\;.$$", "is blue and the fifth is green\". This is definitely not the same event as \"$$2$$ chosen marbles are red, $$2$$ are blue and $$1$$ is green\". If that event occurs then it is not excluded that e.g. the first chosen marble is green (hence not red). You can make use of the method, but then must not forget that the answer found at first hand must be multiplied by the number of orders that are possible. • Makes sense, thank you for taking the time to help me out with this. :) Really appreciate it. :) – Lucky Apr 9 '19 at 19:32 • You are very welcome. – drhab Apr 10 '19 at 6:54 The first question solved in the second method looks as follows: $$\\frac{{6\\choose 3}}{{10\\choose 3}}=\\frac{6\\cdot 5\\cdot 4}{10\\cdot 9\\cdot 8}$$ Interpretation: There are $${6\\choose 3}$$ ways to choose $$3$$ girls out of $$6$$ and there are $${10\\choose 3}$$ ways to choose $$3$$ students out of $$10$$, hence the probability is the ratio of number of favorable outcomes to the total number of possible outcomes. Now compare it with the answer of the second problem and try to interpret the selections. The first situation with the students is simpler than the second situation with the marbles because we are interested in only one kind of", "B, not on Box A. • $(2)$ For example $P(G|B)$ means what is the probability that you pick green given that a green was added to box A. Well, there are 5 greens and 10 total, so 5/10. What is the probability of $B$? Well, there was a 3/5 chance that you chose a green marble and transferred it to box A. So, using the notation above, we're ask $P(B|G)$. This is \\begin{align*} P(B|G) &=\\frac{P(BG)}{P(G)}\\tag 1\\\\ &=\\frac{P(G|B)P(B)}{P(B)}\\tag 2\\\\ &=\\frac{(5/10)(3/5)}{23/50}\\tag 3\\\\ &=\\frac{15}{23} \\end{align*} where in • (1) I used Bayes' rule and • (2) I conditioned on $B$ • (3) I plugged in the values calculated previously.", "The corresponding probabilities are $$a) \\ \\frac{1}{4}\\cdot \\frac{1}{4}\\cdot \\frac{1}{4}= \\frac{1}{64}$$ $$b) \\ 3\\cdot \\frac{1}{4}\\cdot \\frac{1}{4}\\cdot \\frac{1}{2}= \\frac{3}{32}$$ $$c) \\ 3\\cdot \\frac{1}{2}\\cdot \\frac{1}{2}\\cdot \\frac{1}{4}= \\frac{3}{16}$$ $$d) \\ \\frac{1}{2}\\cdot \\frac{1}{2}\\cdot \\frac{1}{2}= \\frac{1}{8}$$ Therefore $$P(W)=\\frac{27}{64}$$ Now what ist $$P(W\\cap R)$$? $$W\\cap R$$ means that no white marbles and no red marbles show up. That is when only green marbles show up-as you´ve already said: $$ggg$$ $$P(W\\cap R)=\\frac{1}{4}\\cdot \\frac{1}{4}\\cdot \\frac{1}{4}= \\frac{1}{64}$$ You can check your final result by using the converse probability: $$P(\\textrm{\"Do not see all three colors\"})=1-P(\\textrm{\"See all three colors\"})$$", "You have 5,536 possible winning combinations. The probability of getting one of those hands is 5,536 in 20,349 or .272. One other approach ... Consider the following: $$!Red = \\binom{17}{5} = 6,188$$ $$!Green = \\binom{19}{5} = 11,628$$ Each of those hands would result in a winning hand (i.e., either no greens or no reds). The problem is that they both include those hands which have neither greens nor reds so we have to the duplicates out. $$!(Red | Green) = \\binom{15}{5} = 3,003$$ If we add the hands without red to the hands without green and take out the duplicates we get $6,188 + 11,628 - 3,003 = 5,536$, the same as before. Basically, I believe the following is true: $$P(No Red \\cup No Green) = P(No Red) + P(No Green) - P(No Red \\cap No Green)$$ - The number of hands with exactly $1$ red and $1$ green is $4 \\cdot 2 \\binom{15}{3}=3640$, as you have to consider the ways of choosing the other three cards. This carries over to the other cases. – Ross Millikan Jun 27 '11 at 23:41 Ross, I took your comment into account. Am I on the right track? Thanks a lot for teaching me something about probability :). Also, @Tyler, thanks for letting me learn here on your question. – D.", "# Tag Info 4 Consider $N=(1+\\sqrt 3)^5+(1-\\sqrt 3)^5$. This is an integer. Since $-1\\lt 1-\\sqrt 3\\lt 0$, we have $$0\\lt -(1-\\sqrt 3)^5\\lt 1.$$ So, we have \\begin{align}\\lfloor (1+\\sqrt 3)^5\\rfloor&=\\lfloor N-(1-\\sqrt 3)^5\\rfloor\\\\&=N\\\\&=(1+\\sqrt 3)^5+(1-\\sqrt 3)^5\\\\&=2\\left((\\sqrt 3)^0\\binom 50+(\\sqrt 3)^2\\binom 52+(\\sqrt 3)^4\\binom ... 2 Exponential generating functions are useful for this sort of problem because we can find the answer by multiplying the EGF for each of the types of balls:\\color{red}{\\left(\\frac{x^2}{2!}+\\frac{x^4}{4!}\\right)}\\color{green}{\\frac12\\left(e^x+e^{-x}\\right)}\\color{blue}{e^x}.$$This expands to$$\\frac12\\cdot\\frac{x^2}{2!} + \\frac12\\cdot\\frac{x^4}{4!} + ... 2 You should choose the six children that will receive the books, i.e., $\\binom{13}{6}$, and then you should multiple it by the number of permutations of the six different books among the six children, i.e., $6!$, thus $\\binom{13}{6}6!$ . 2 For $i=1,2$ let $W_i$ denotes the event that the marble taken form box $i$ is white. For $i=1,2$ let $B_i$ denotes the event that the marble taken form box $i$ is black. Then the probability of success is: $$P((W_1\\cap W_2)\\cup(B_1\\cap B_2))=P(W_1\\cap W_2)+P(B_1\\cap B_2)=P(W_1)P(W_2)+P(B_1)P(B_2)$$ 2 In your counting method, you would count |||_ _ _ _ _ _ _ _ _ _ once but |_ _ _||_ _ _ _ _ _ _ three times, because you give 11 choices for each bar, so you would count the latter once for each combination of choices: 1_ _ _23_ _ _ _ _ _ _ (note that" ]

[ "3 prizes, and winners are selected randomly from a hat containing all the numbers. Find the probability that Peter wins all 3 prizes. For the first draw, Peter’s numbers account for 4 tickets out of 100: \\begin{align*}\\frac{4}{100} = \\frac{1}{25}\\end{align*} For the second draw, Peter’s remaining numbers (assuming he won the first draw) account for 3 tickets out of 99: \\begin{align*}\\frac{3}{99} = \\frac{1}{33}\\end{align*} For the first draw, Peter’s remaining numbers (assuming he won the first two draws) account for 2 tickets out of 98: \\begin{align*}\\frac{2}{98} = \\frac{1}{49}\\end{align*} The probability that Peter wins all 3 prizes is \\begin{align*}\\frac{1}{25} \\times \\frac{1}{33} \\times \\frac{1}{49} = \\frac{1}{40425}\\end{align*} or 1 in 40, 425. ### Review For 1-6, determine whether the events are dependent or independent. 1. Driving at night and falling asleep at the wheel. 2. Visiting the zoo and seeing a giraffe. 3. The next 2 cars you see are both red. 4. A coin tossed twice comes up heads both times. 5. Being dealt 4 aces in a hand of poker. 6. It is your birthday and it is a windy day. For 7-10, a bag contains 10 colored marbles – 4 red, 4 blue and 2 green. Calculate the probability of: 1. Removing 2 green marbles in a row if you replace the marble each time. 2. Removing 2 green marbles", "drew another card. What is the probability that the first card was an E and the second was a B? a. $\\frac{1}{13}$ b. $\\frac{1}{676}$ c. $\\frac{4}{473}$ d. $\\frac{1}{26}$ 197. What is the probability that a card randomly picked from a deck of playing cards is a diamond? a. $\\frac{1}{2}$ b. $\\frac{1}{52}$ c. $\\frac{1}{4}$ d. $\\frac{5}{52}$ 198. Clara has 4 blue marbles, 5 green marbles, and 3 yellow marbles. Find the probability of drawing either a green or a yellow marble. a. $\\frac{5}{48}$ b. $\\frac{3}{5}$ c. $\\frac{3}{4}$ d. $\\frac{2}{3}$ 199. A pair of dice is rolled. What is the probability that the sum of the numbers rolled is either 8 or 12? a. $\\frac{1}{6}$ b. $\\frac{5}{36}$ c. $\\frac{1}{5}$ d. $\\frac{2}{9}$ a. $\\frac{1}{3}$ b. $\\frac{3}{4}$ c. $\\frac{1}{4}$ d. $\\frac{1}{2}$", "+0 # help 0 34 1 +529 John has 6 green marbles and 4 purple marbles. He chooses a marble at random, writes down its color, and then puts the marble back. He performs this process 5 times. What is the probability that he chooses exactly two green marbles? Logic Dec 7, 2018 #1 +3213 +2 $$\\text{Since you are replacing the marbles after each draw the total number of green marbles}\\\\ \\text{you select has a binomial distribution with parameters }n=5,~p=\\dfrac 3 5\\\\$$ $$P[2] = \\dbinom{5}{2} \\left(\\dfrac 3 5\\right)^2 \\left(\\dfrac 2 5\\right)^3 = \\dfrac{144}{625}$$ Rom Dec 7, 2018 #1 +3213 +2 $$\\text{Since you are replacing the marbles after each draw the total number of green marbles}\\\\ \\text{you select has a binomial distribution with parameters }n=5,~p=\\dfrac 3 5\\\\$$ $$P[2] = \\dbinom{5}{2} \\left(\\dfrac 3 5\\right)^2 \\left(\\dfrac 2 5\\right)^3 = \\dfrac{144}{625}$$ Rom Dec 7, 2018", "|X = k)\\Pr(X = k) \\\\ &= \\sum_{k=0}^n \\frac{(n-k)(n-k-1)}{n(n-1)}\\cdot \\frac{1}{n+1} \\\\ &= \\frac{1}{(n+1)n(n-1)}\\cdot \\sum_{k=0}^n (n-k)(n-k-1) \\\\ &= \\frac{1}{(n+1)n(n-1)}\\cdot \\sum_{k=0}^n k(k-1) \\end{align} Use $\\sum_{k=0}^n k^2 = \\frac{n(n+1)(2n+1)}{6}$ and $\\sum_{k=0}^n k = \\frac{n(n+1)}{2}$ to simplify the formula above. The probability that two marbles drawn from the same box selected without bias are both red will be given by the Law of Total Probability. Where $X$ is the count of red marbles drawn, and $K$ is the count of red marbles in the box selected, that is: \\begin{align}\\mathsf P(X=2) ~=~& \\sum_{k=1}^n \\mathsf P(K=k)\\mathsf P(X=2\\mid K=k) \\\\[2ex]~=~& \\frac 1{n+1}\\sum_{k=2}^n \\binom{k}{2}\\Big/\\binom{n}{2}\\\\[1ex]~\\vdots~&\\\\[1ex] ~=~& \\bbox[silver ,1pt]{\\color{silver}{\\frac 13}}\\end{align} And similarly for $\\mathsf P(X=1), \\mathsf P(X=0)$.", "mutually exclusive paths (M1) e.g. $$\\left( {\\frac{1}{3} \\times \\frac{7}{8}} \\right) + \\left( {\\frac{2}{3} \\times \\frac{3}{8}} \\right)$$ , $$\\left( {\\frac{1}{3} \\times \\frac{1}{8}} \\right) + \\left( {\\frac{2}{3} \\times \\frac{5}{8}} \\right)$$ $${\\rm{P}}(F) = \\frac{{13}}{{24}}$$ A1 N2 [4 marks] a(i) and (ii). (i) $$\\frac{1}{3} \\times \\frac{1}{8}$$ (A1) $$\\frac{1}{{24}}$$ A1 (ii) recognizing this is $${\\rm{P}}(E|F)$$ (M1) e.g. $$\\frac{7}{{24}} \\div \\frac{{13}}{{24}}$$ $$\\frac{{168}}{{312}}$$ $$\\left( { = \\frac{7}{{13}}} \\right)$$ A2 N3 [5 marks] b(i) and (ii). A2A1 N3 [3 marks] c. correct substitution into $${\\rm{E}}(X)$$ formula (M1) e.g. $$0 \\times \\frac{1}{9} + 3 \\times \\frac{4}{9} + 6 \\times \\frac{4}{9}$$ , $$\\frac{{12}}{9} + \\frac{{24}}{9}$$ $${\\rm{E}}(X) = 4$$ (euros) A1 N2 [2 marks] d. ## Question A box contains six red marbles and two blue marbles. Anna selects a marble from the box. She replaces the marble and then selects a second marble. Write down the probability that the first marble Anna selects is red. [1] a. Find the probability that Anna selects two red marbles. [2] b. Find the probability that one marble is red and one marble is blue. [3] c. Answer/Explanation ## Markscheme Note: In this question, method marks may be awarded for selecting without replacement, as noted in the examples. $${\\rm{P}}(R) = \\frac{6}{8}\\left( { = \\frac{3}{4}} \\right)$$ A1 N1 [1 mark] a. attempt to find $${\\rm{P(Red)}} \\times {\\rm{P(Red)}}$$ (M1) e.g. $${\\rm{P(}}R{\\rm{)}} \\times {\\rm{P(}}R{\\rm{)}}$$ ,", "(i) Probability that the 1 red marble drawn is n(A) = 5 ∴ Probability, P(A) = $$\\frac{n(A)}{n(S)}=\\frac{5}{17}$$ (ii) Possibility that 1 white marble drawn, n(B) = 8 ∴ Probability, P(B) = $$\\frac{n(B)}{n(S)}=\\frac{8}{17}$$ (iii) Possibility that 1 not green marble ? P(C) = 17 – 4 = 13 (∵ Except 4 green marbles) ∴ Probability, P(C) = $$\\frac{n(C)}{n(S)}=\\frac{13}{17}$$ Question 10. A piggy bank contains hundred 50p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin ? (ii) will not be a Rs. 5 coin ? Solution: Number of 50 ps coins = 100 Number of Re. 1 coins = 50 Number of Rs. 2 coins = 20 Number of Rs. 5 coins = 10 ∴ Total number of coins, n(S) = 180 (i) Possibility of one 50 ps coin : n(A) = 100 (ii) Possibility of one Rs. 5 coin: n(B) = 180 – 10 = 170 (∵ 10 coins are Rs. 5) Question 11. Gopi buys a fish from a shop for his aquarium. The shopkeeper taks out one fish at random from a tank containing 5 male fish", "28 '16 at 11:18 • Your fellow classmate is correct. – drhab May 28 '16 at 11:18 • In small sample you are using multihypergeometric to calculate the exact probabilities; in your large sample case, you are using multinomial. – BGM May 28 '16 at 11:49 The answer provided is wrong. In the limit of very many marbles, it doesn't matter whether you draw with or without replacement, since the ratios will stay approximately the same. Thus in this limit the probability of drawing $1$ red, $2$ blue and $1$ yellow marbles is $$\\binom4{1,2,1}\\cdot\\frac3{12}\\cdot\\frac4{12}\\cdot\\frac4{12}\\cdot\\frac5{12}=\\frac{4!}{1!2!1!}\\cdot\\frac5{3\\cdot12\\cdot12}=\\frac5{36}\\approx0.1389\\;.$$", "The corresponding probabilities are $$a) \\ \\frac{1}{4}\\cdot \\frac{1}{4}\\cdot \\frac{1}{4}= \\frac{1}{64}$$ $$b) \\ 3\\cdot \\frac{1}{4}\\cdot \\frac{1}{4}\\cdot \\frac{1}{2}= \\frac{3}{32}$$ $$c) \\ 3\\cdot \\frac{1}{2}\\cdot \\frac{1}{2}\\cdot \\frac{1}{4}= \\frac{3}{16}$$ $$d) \\ \\frac{1}{2}\\cdot \\frac{1}{2}\\cdot \\frac{1}{2}= \\frac{1}{8}$$ Therefore $$P(W)=\\frac{27}{64}$$ Now what ist $$P(W\\cap R)$$? $$W\\cap R$$ means that no white marbles and no red marbles show up. That is when only green marbles show up-as you´ve already said: $$ggg$$ $$P(W\\cap R)=\\frac{1}{4}\\cdot \\frac{1}{4}\\cdot \\frac{1}{4}= \\frac{1}{64}$$ You can check your final result by using the converse probability: $$P(\\textrm{\"Do not see all three colors\"})=1-P(\\textrm{\"See all three colors\"})$$", "$P(\\text{all R}) \\:=\\:\\frac{35}{495}$ $P(\\text{all G}) \\:=\\:\\frac{{5\\choose4}}{495} \\:=\\:\\frac{5}{495}$ $P(\\text{one color}) \\:=\\:\\frac{5}{495}\\,+\\,\\frac{35}{495} \\:=\\:\\frac{40}{495} \\:=\\:\\frac{8}{99}$ $\\text{Therefore: }\\:P(\\text{both colors}) \\;=\\;1\\,-\\,\\frac{8}{99} \\;=\\;\\frac{91}{99}$ Quote: (d) two of the marbles chosen are both labeled \"5\"? $\\text{We already have R5 and G5 (one way).$ $\\text{The other two can be any of the other 10 marbles: }\\:{10\\choose2} \\,=\\,45\\text{ ways.}$ $P(\\text{R5, G5}) \\:=\\:\\frac{45}{495} \\:=\\:\\frac{1}{11}$ October 19th, 2013, 06:03 PM #4 Newbie Joined: Oct 2013 Posts: 10 Thanks: 0 Re: Basic probability Wow, thank you! Just what I needed, a step by step explanation. October 21st, 2013, 05:18 AM #5 Newbie Joined: Oct 2013 Posts: 9 Thanks: 0 Re: Basic probability Very neat solution. My mind did not think as laterally as you did. By thinking \"aloud\", I could do (a), the first part of (b) and (c) with some luck. For (a), I did 7/12 x 6/11 x 5/10 x 4/9 = =7/99 For straightforward 4 greens, I got 1/99; I was then faced with doing all the combinations of greens and red. Well done. October 21st, 2013, 09:31 PM #6 Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications Re: Basic probability [color=#0000BF]nubian123[/color], Using your method, the probabilities of the various combinations of greens and reds are not too hard to calculate: $P_{1G,3R}=\\frac{5}{12} \\cdot \\frac{7}{11} \\cdot \\frac{6}{10} \\cdot \\frac{5}{9}", "# Chapter 11 - 11.7 - Probability - 11.7 Exercises - Page 834: 33 $\\frac{2}{5}$ #### Work Step by Step We know that $probability=\\frac{\\text{number of favourable outcomes}}{\\text{number of all outcomes}}$ The total number of outcomes is $6\\cdot5=30$ because we can select any of the marbles as the first one and then any of the marbles that we have not selected as the first one. The number of favourable outcomes is $4\\cdot3=12$ because we can select any of the non-yellow marbles as the first one and then any of the non-yellow marbles that we have not selected as the first one. Hence the probability here is: $\\frac{12}{30}=\\frac{2}{5}$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "same as $\\frac{5}{9}$). But how did you come up with them directly? Any simpler logic, that I am missing, which will help me to come up with those probabilities directly without going through those binomials? – anir Jun 15 '18 at 9:58 • If there are 9 numbered marbles and 5 are chosen randomly, any particular marble has $Pr = \\frac59$ to be among those chosen. – true blue anil Jun 15 '18 at 14:11 • Ok I got it. Thanks. But it seems that you have made mistake. For (a) It should be $\\frac{4}{8}.\\frac{\\color{red}{4}}{9}\\times\\frac{1}{4}$ which matches with given answer $\\frac{1}{8}$, right? (similar for problem (b)) – anir Jun 15 '18 at 18:40 • My bad, typos corrected ! btw, the answer is $\\frac1{18}$, not $\\frac18$ as in your comment. The printer's devil spares no one ! – true blue anil Jun 16 '18 at 5:05", "pink marbles: $$\\displaystyle \\frac{{{}_{{10}}{{C}_{3}}\\times {}_{5}{{C}_{1}}+{}_{{10}}{{C}_{4}}\\times {}_{5}{{C}_{0}}}}{{{}_{{15}}{{C}_{4}}}}=\\frac{{810}}{{1365}}\\approx .593$$. (d) To get the probability that at least 1 is pink, we can get the probability that all are blue, and then subtract this from 1 to get its complement (since anything else would have at least one pink marble): $$\\displaystyle 1-\\frac{{{}_{{10}}{{C}_{0}}\\times {}_{5}{{C}_{4}}}}{{{}_{{15}}{{C}_{4}}}}=1-\\frac{{1\\times 5}}{{1365}}\\approx .996$$. This makes sense since so many are pink in the bag. A committee of 6 is selected from a group of 6 boys and 5 girls. (a) What is the probability that the committee is all boys? (b) What is the probability that exactly 4 out of the 6 are boys? We know the total number of ways 6 kids is selected from the group of the boys and girls (11) is $${}_{{11}}{{C}_{6}}=462$$; order doesn’t matter, since there are no positions. (a) There is only way to get all boys, since we are only 6 boys total. The probability then is $$\\displaystyle \\frac{1}{{{}_{{11}}{{C}_{6}}}}=\\frac{1}{{462}}\\approx .002$$ (low). (b) The number of possible ways 4 out of 6 boys could be chosen is $${}_{6}{{C}_{4}}=15$$, and the number of ways to pick 2 girls out of the remaining 5 kids is $${}_{5}{{C}_{2}}=10$$. The probability then that exactly 4 out of the 6 are girls is $$\\displaystyle \\frac{{{}_{6}{{C}_{4}}\\times {}_{5}{{C}_{2}}}}{{{}_{{11}}{{C}_{6}}}}=\\frac{{150}}{{462}}\\approx .325$$. The letters in the word MISSISSIPPI are written on pieces of", "30$$ $$P(1r,1g) = \\frac{16}{30} = \\frac{8}{15}$$ Using combinations, we would have: $$P(1r,1g) = \\frac{C(4,1) \\cdot C(2,1)}{C(6,2)} = \\frac{8}{15}$$ Thanks for keeping me in check! Given that, ##P(W)## in #7 looks reasonable to me. vcsharp2003 Homework Helper Gold Member 2022 Award That's what I tried, but didn't get the answer. No, you went the other way. You were asked for the probability of not winning but chose instead to find the probability of winning. That was a backward move. vcsharp2003 vcsharp2003 No, you went the other way. You were asked for the probability of not winning but chose instead to find the probability of winning. That was a backward move. I was going to use the fact that probability of not winning = 1 - probability of winning and that's why I started with probability of winning; finding the complement event's probability is a good approach only when it's easier to find. vcsharp2003 Imagine a bag of 6 marbles, 4 red, 2 green there are 6 *5 = 30 ways to select 2 marbles You can select both red in 4 *3 = 12 ways You can select both green in 2*1 = 2 ways When I try to do your example, I am getting half the values that you have got. Number of ways of selecting 2 marbles", "is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally. The same goes for the blue ones. The numerator must equal $(1 \\cdot 2)^2$. Therefore, the probability for each of the orderings of $RRBB$ is $\\frac{4}{120}=\\frac{1}{30}$. There are 6 of these, so the total probability is $\\boxed{\\bf{(B)} \\frac{1}{5}}$. ## Solution 3 First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a $\\frac{1}{2}$ chance each. We can assume he chooses Red(chance $\\frac{1}{2}$), and then multiply the final answer by two for symmetry. Now, there are two red balls and one blue ball in the urn. Then, he can either choose another Red(chance $\\frac{2}{3}$), in which case he must choose two blues to get three of each, with probability $\\frac{1}{4}\\cdot\\frac{2}{5}=\\frac{1}{10}$ or a blue for two blue and two red in the urn, with chance $\\frac{1}{3}$. If he chooses blue next, he can either choose a red then a blue, or a blue then a red. Each of these has a $\\frac{1}{2}\\cdot\\frac{2}{5}$ for total of $2\\cdot\\frac{1}{5}=\\frac{2}{5}$. The total probability that he ends up with three red and three blue is $2\\cdot\\frac{1}{2}(\\frac{2}{3}\\cdot\\frac{1}{10}+\\frac{1}{3}\\cdot\\frac{2}{5})=\\frac{1}{15}+\\frac{2}{15}=\\boxed{\\bf{(B)} \\frac{1}{5}}$. ~aop2014", "ways to select the gender and the day of the week the child was born on. There are 132 = 169 ways which do not have a girl born on Monday and which 196 – 169 = 27 which do, so P(Y) = $$\\frac{27}{196}$$. This gives is that P(X|Y) = $$\\frac{\\frac{13}{19}*\\frac{1}{4}}{\\frac{27}{196}} = \\frac{13}{27}$$. 7. Suppose a fair eight-sided die is rolled once. If the value on the die is 1, 3, 5 or 7 the die is rolled a second time. Determine the probability that the sum of values that turn up is at least 8? a) $$\\frac{32}{87}$$ b) $$\\frac{12}{43}$$ c) $$\\frac{6}{13}$$ d) $$\\frac{23}{64}$$ Explanation: Sample space consists of 8*8=64 events. While (8) has $$\\frac{1}{8}$$ probability of occurrence, (1,7) has only $$\\frac{1}{64}$$ probability. So, the required probability = $$\\frac{1}{6} + (9 * \\frac{1}{64}) = \\frac{69}{192} = \\frac{23}{64}$$. 8. A jar containing 8 marbles of which 4 red and 4 blue marbles are there. Find the probability of getting a red given the first one was red too. a) $$\\frac{4}{13}$$ b) $$\\frac{2}{11}$$ c) $$\\frac{3}{7}$$ d) $$\\frac{8}{15}$$ Explanation: Suppose, P (A) = getting a red marble in the first turn, P (B) = getting a black marble in the second turn. P (A) = $$\\frac{4}{8}$$ and P (B) = $$\\frac{3}{7}$$ and P (A and B) = $$\\frac{4}{8}*\\frac{3}{7} = \\frac{3}{14}$$ P(B/A)", "probability of choosing a blue one first? There are 9 marbles and 3 are blue. 39=13\\begin{align*} \\frac{3}{9} = \\frac{1}{3}\\end{align*} What is the probability of choosing a green one next? There are 9 marbles and 3 are green. 39=13\\begin{align*} \\frac{3}{9} = \\frac{1}{3}\\end{align*} What is the probability of choosing an orange one next? There are 9 marbles and 3 are orange. 39=13\\begin{align*} \\frac{3}{9} = \\frac{1}{3}\\end{align*} Now we figure out the likelihood of three independent events occurring by multiplying each probability. P(B, then G, then O)=131313=127\\begin{align*}P(B, \\ \\text{then} \\ G, \\ \\text{then} \\ O) = \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} = \\frac{1}{27}\\end{align*} There is a very small likelihood that this event will occur. Now go back to the swimming and golf example. Example Out of 100 students at Riverview Middle, 50 enjoy swimming. Out of the same 100 students, 25 enjoy golf. What is the probability that a student would enjoy both swimming and golf? Let’s add to this problem that 40 out of 100 enjoy basketball. What is the likelihood that a student would enjoy all three? 501002510040100121425=12=14=25=240=120\\begin{align*}\\frac{50}{100} & = \\frac{1}{2}\\\\ \\frac{25}{100} & = \\frac{1}{4}\\\\ \\frac{40}{100} & = \\frac{2}{5}\\\\ \\frac{1}{2} \\cdot \\frac{1}{4} \\cdot \\frac{2}{5} & = \\frac{2}{40} = \\frac{1}{20}\\end{align*} There is a one out of 20 chance that a student would like all three. We can look at this as", "there are 5 places the second ball can land. If it goes in two of these you win. This means that the chance of winning when it lands outside is $\\frac25$. But since only $\\frac{29}{42}$ of the time you don't get a middle (this probability becomes $\\frac57$) the chance you win by to outside balls is $\\frac{58}{210}$ (this becomes $\\frac25\\times\\frac57 = \\frac57$) Add that to the $\\frac{65}{210}$ that comes from having one in the center (which is actually $\\frac27$) the probability of winning is $\\frac{123}{210}$ or $\\frac{41}{70}$ (which becomes $\\frac47$). Dashiell from Sequoyah in the USA foun the numbers of ways there are to win and lose. I did that by drawing seven marbles in a circle (one in the middle six surrounding it) and then I numbered the marbles. For example, the two marbles that have a 1 [in] them are a winning possibility, same with the marbles with 2 [in them] and so on. Then I did the same thing, but with losing possibilities. The winning possibilities was 12 and the losing possibilities was 9 12 + 9 = 21 $\\frac{12}{21}$ = 0.571428 So 57.1428% is the probability of winning. On the live problem I did the run 100 test. After that I added the mean for one hundred tries. Then two hundred, three hundred, and", "and black marbles (the hypergeometric distribution) depends only on the number of white and black marbles, not on the order in which they appear; i.e., it is an exchangeable distribution. As a result, the probability of drawing a white marble in the $ith$ draw is $P(W_i) = {\\frac{m}{N}}$ This can be shown by induction. First, it is certainly true for the first draw that: $P(W_1) = {\\frac{m}{N}}$. Also, we can show that $P(W_{n+1})= {\\frac{m}{N}}$ by writing: \\begin{align} P(W_{n+1}) & = {\\sum_{k=0}^n}P(W_{n+1}|k)f(k;N,m,n)\\\\ & = {\\sum_{k=0}^n}\\frac{m-k}{N-n}f(k;N,m,n) \\\\ & = {\\sum_{k=0}^n}\\frac{m-k}{N-n}\\frac{\\binom mk \\binom {N-m} {n-k}}{\\binom Nn} \\\\ & = \\frac{1}{(N-n)\\binom Nn} \\left \\{ m\\sum_{k=0}^n \\binom mk \\binom {N-m} {n-k} - \\sum_{k=0}^n k\\binom mk \\binom {N-m} {n-k}\\right \\} \\\\ & = \\frac{1}{(N-n)\\binom Nn}\\left\\{ m\\binom Nn - \\sum_{k=1}^n k\\frac{m}{k} \\binom {m-1}{k-1} \\binom {N-m} {n-k}\\right \\} \\\\ & = \\frac{m}{(N-n)\\binom Nn}\\left\\{ \\binom Nn - \\sum_{k=1}^n \\binom {m-1}{k-1} \\binom {N-1-(m-1)} {n-1-(k-1)}\\right \\} \\\\ & = \\frac{m}{(N-n)\\binom Nn}\\left\\{ \\binom Nn - \\binom {N-1}{n-1}\\right \\} \\\\ & = \\frac{m}{(N-n)\\binom Nn}\\left\\{ \\binom Nn - \\frac{n}{N}\\binom Nn\\right \\} \\\\ & = \\frac{m}{(N-n)}\\left\\{ 1 - \\frac{n}{N} \\right\\} = \\frac{m}{N} \\end{align}, which makes it true for every $ith$ draw. A simpler proof than the one above is the following: By symmetry each of the $N$ marbles has the same chance to be drawn in the $i$-th draw. In addition, according to", "# We have two red, two white and two green marbles in an urn We have two red, two white and two green marbles in an urn. We pick them one by one out of the urn and record their colors. Find the probability that at some point we will pick the same color back to back. For example, this happens when we get the sequence red, white, white, green, red, green, but also if we get red, red, white, white, green, green.) So far I have the following. Note that if we think about this problem as a sequence $$(x_1,x_2,x_3,x_4,x_5,x_6)$$ of the balls. Note that if we fix one of the balls, then we have a $$\\large\\frac{4}{5}\\cdot\\frac{3}{4}$$ chance of not picking the same color next to it. hence the probability of picking two consecutive balls of the same color is equal to $$\\large 1-\\frac{3}{5}=\\frac{2}{5}.$$ However I am not confident in my reasoning and I think I have made a mistake. Any help is much appreciated. • What do you mean by \"the same color back to back.\"? At least in two consecutive extractions the balls have the same color? – Robert Z Feb 8 at 10:10 • @RobertZ Indeed. – Gaby Alfonso Feb 8 at 10:11 • After one ball, your chance of picking the same again", "(8/13) = 0.3043 - P[3B|2R] = ((10C2 * 6C2)/(16C4)) * (8/13) = 0.2282 - P[3B|3R] = ((10C2 * 7C3)/(17C5)) * (8/13) = 0.15663 - P[3B|4R] = ((10C2 * 8C4)/(18C6)) * (8/13) = 0.1044 Basically it is (combination without repetition -for blues) * (combination with repetition - for reds) / total combination) * (8/13). Then I would muliply the number of marbles * probability to get expected number of total balls drawn per game. But this way won't work as the probabilities (although diminishing) are going to total over 1. Any one have any thoughts on this? Probably something very simple I'm overlooking. I'm stumped all the same! Thanks very much for any and all help. 2. ## Re: Marble question - hypergeometric/binomial hybrid You draw a sea of reds, with blues on the i-th, j-th, and n-th (terminating) draws 1<=i < j < n Ex: i = 4, j = 10, n = 15: R R R B R R R R R B R R R R B The probabilitiy of such an $(i, j, n)$ string, $Prob(i, j, n)$ is: $\\left(\\frac{5}{15}\\right)^{i-1} \\left(\\frac{10}{15}\\right) \\left(\\frac{5}{14}\\right)^{j-i-1} \\left(\\frac{9}{14}\\right) \\left(\\frac{5}{13}\\right)^{n-j-1} \\left(\\frac{8}{13}\\right)$ $= \\left(\\frac{(10)(9)(8)}{(15)(14)(13)}\\right) \\left( \\frac{(5^{i-1})(5^{j-i-1})(5^{n-j-1})}{({15}^{i-1})({14}^{j-i-1})({13}^{n-j-1})} \\right)$ $= \\left(\\frac{24}{91}\\right) \\left \\frac{(5^{n-3})}{({15}^{i-1})({14}^{j-i-1})({13}^{n-j-1})} \\right$ $= \\left(\\frac{24}{91}\\right) \\left \\frac{(5^{n-3})}{({15}^{i-1})({14}^{j-i-1})({13}^{n-j-1})} \\right$ $= \\left(\\frac{24}{91}\\right) (5^{n-3}) \\left( (15)(14)\\left(\\frac{14}{15}\\right)^i \\left(\\frac{13}{14}\\right)^j \\left(\\frac{1}{13}\\right)^{n-1} \\right)$ $= \\left(\\frac{(24)(15)(14)(13)}{(91)(5^3)}\\right) \\left(\\frac{14}{15}\\right)^i \\left(\\frac{13}{14}\\right)^j \\left(\\frac{5}{13}\\right)^{n}$ $= \\left(\\frac{144}{25}\\right)" ]

[ "product of three positive integers?", "multiple of two? How about three, four or six? Factoring a Million Stage: 4 Challenge Level: In how many ways can the number 1 000 000 be expressed as the product of three positive integers?", "# Multiples of 5 and 7 How many positive integers strictly less than 350 are divisible by 5 or 7? ×", "# Multiples of 7 and 13 How many 3 digit positive integers $$N$$ are there, such that $$N$$ is a multiple of both 7 and 13? ×", "of three forms of a three-digit number is always a multiple of $$37$$.", "of a 3 and 4 digit number:", "# \"Double\" Divisors If a three-digit number $$\\overline{abc}$$ has 28 positive divisors, how many positive divisors does the six-digit number $$\\overline{abcabc}$$ have? ×", "Number 3 Number Theory Level 2 How many integers that are multiples of 3 are there between 1,000 and 1,000,000? ×", "multiples of those numbers.", "are triple and how many quadruples.", "of the three digits is 1 + 3 +4 = 8. 176/8 = 22, with no remainder", "multiples of ten.", "How many three-digit numbers contain three primes that sum to an even &nbs [#permalink] 03 Oct 2018, 00:23 Display posts from previous: Sort by", "• multiples", "three-digit sequence:", "(which cycle the last three digits every five hundred multiples).", "stating that three is many .", "777 How many 3-digit numbers are a multiple of 7? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "are three multiples of 3 namely 0, 3, and 6. Thus, $B=7-3=4$ 3. $A+B=10+4=14$.", "\\end{array}$$ $$\\text{How many of the positive divisors of 3240 are multiples of 3 ?}$$ $$\\mathbf{40-8 = 32}$$ $$\\text{There are \\mathbf{32} positive divisors of 3240, which are multiples of 3.}$$ Nov 13, 2018" ]

[ "Expert's post sumitchawla wrote: Bunuel wrote: mehdiov wrote: How many integers less than 1000 have no factors (other than 1) in common with 1000 ? a. 400 b. 399 c. 410 d. 420 First of all it should be \"how many positive integers less than 1000 have no factors (other than 1) in common with 1000\", as if we consider negative integers answers will be: infinitely many. $$1000=2^3*5 ^3$$ so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5. Multiples of 2 in the range 0-1000, not inclusive - $$\\frac{998-2}{2}+1=499$$; Multiples of 5 in the range 0-1000, not inclusive - $$\\frac{995-5}{5}+1=199$$; Multiples of both 2 and 5, so multiples of 10 - $$\\frac{990-10}{10}+1=99$$. Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to $$999-(499+199-99)=400$$. What about the prime numbers Bunuel ?? For ex : 7. Neither its a multiple of 2, nor 5 and it does not has any common factors with 1000 (except 1) So, shouldn't the answer include prime numbers between 1-999 as well. And if YES, how do we calculate the number of primer numbers from 1-999 ??? Plz clarfily. Thanks. We counted multiples of 2 or 5 in", "I am working on some AMC-8 questions. I am not sure how to approach this one. 279 views 0 19 months ago by Anonymous How many integers between 1000 and 9999 have four distinct digits? 2 19 months ago by The question can be rephrased to \"How many four-digit positive integers have four distinct digits,\" since numbers between 1000 and 9999 are four-digit integers. There are 9 choices for the first number, since it cannot be 0, There are only 9 choices left for the second number since it must differ from the first, 8 choices for the third number, since it must differ from the first two, and 7 choices for the fourth number, since it must differ from all three. This means there are $9\\times9\\times8\\times7=4536$9×9×8×7=4536 integers between 1000 and 9999 with four distinct digits.", "A positive integer is said to be “tri-factorable” if it is the product of three consecutive integers. How many positive integers less than 1,000 are tri-factorable? We’ll use the same tactic, starting with the smallest case. The smallest tri-factorable number is simply $$1 \\times 2 \\times 3=$$ 6. Now 6 may seem like a long way from 1,000 but let’s not give up just yet! The next one up is $$2 \\times 3\\times 4=$$ 24. If you keep going, you should get this list: $$1 \\times 2 \\times 3= 6$$ $$2 \\times 3 \\times 4= 24$$ $$3 \\times 4 \\times 5= 60$$ $$4 \\times 5 \\times 6= 120$$ $$5 \\times 6 \\times 7= 210$$ $$6 \\times 7 \\times 8= 336$$ $$7 \\times 8 \\times 9= 504$$ $$8 \\times 9 \\times 10= 720$$ $$9 \\times 10 \\times 11= 990$$ Look how quickly we got there! Good things can happen if you just keep going. If you count them up, you should see that the answer is 9. Some of you might complain, “Ok it happened to work for this one question. What do you do when the list keeps on going and going?” When that happens, making a list will still help, but another step or tactic will often be necessary. Here’s one example: 3. How many multiples of", "# Practice: Finding common multiples How many positive integers less than or equal to $1000$ are common multiples of the three numbers $8, 36,$ and $24$? ×", "A positive integer is said to be \"tri-factorable\" if it is the product of three consecutive integers. How many positive integers less than 1,000 are tri-factorable? We'll use the same tactic, starting with the smallest case. The smallest tri-factorable number is simply $$1 \\times 2 \\times 3=$$ 6. Now 6 may seem like a long way from 1,000 but let's not give up just yet! The next one up is $$2 \\times 3\\times 4=$$ 24. If you keep going, you should get this list: $$1 \\times 2 \\times 3= 6$$ $$2 \\times 3 \\times 4= 24$$ $$3 \\times 4 \\times 5= 60$$ $$4 \\times 5 \\times 6= 120$$ $$5 \\times 6 \\times 7= 210$$ $$6 \\times 7 \\times 8= 336$$ $$7 \\times 8 \\times 9= 504$$ $$8 \\times 9 \\times 10= 720$$ $$9 \\times 10 \\times 11= 990$$ Look how quickly we got there! Good things can happen if you just keep going. If you count them up, you should see that the answer is 9. Some of you might complain, \"Ok it happened to work for this one question. What do you do when the list keeps on going and going?\" When that happens, making a list will still help, but another step or tactic will often be necessary. Here's one example: 3. How many multiples of", "every other multiple of 15. If we add these two totals together, we will be counting all of the multiples of 15 twice. We only want to count them once, so to counter this, we can subtract from this total the number of positive integers less than or equal to 9,999 that are divisible by 15. In sum, the answer should be: $\\lfloor{\\frac{9999}{3}}\\rfloor+\\lfloor{\\frac{9999}{5}}\\rfloor-\\lfloor{\\frac{9999}{15}}\\rfloor=3,333+1,999-666=4,666$", "subtracting 47 once. Similar questions to practice: what-is-the-number-of-integers-from-1-to-1000-inclusive-126153.html what-is-the-total-number-of-positive-integers-that-are-less-128104.html how-many-integers-from-1-to-200-inclusive-are-divisib-109333.html how-many-even-integers-n-where-100-n-200-are-divisib-103779.html Hope it helps. Hi Bunuel, Is there a shorter way around this question?This approach took me 5 minutes! Manager Joined: 20 Dec 2013 Posts: 231 Location: India Re: How many positive integers less than 5,000 are evenly divisi [#permalink] ### Show Tags 04 Apr 2014, 20:59 Option B.(Took 5 minutes to solve with the following approach!) All nos. from $$1$$ to $$5000$$ since we're given POSITIVE integers.$$=4999$$ Multiples of $$15$$ in this range$$=4995/15=333$$ Multiples of $$21=4998/21=238$$ Multiples of both $$15$$ and $$21$$=multiples of $$105=47$$ Now $$answer=4999-333-238+47=4475$$ Intern Joined: 26 Mar 2014 Posts: 4 Re: How many positive integers less than 5,000 are evenly divisi [#permalink] ### Show Tags 16 Apr 2014, 22:53 2 AKG1593 wrote: Option B.(Took 5 minutes to solve with the following approach!) All nos. from $$1$$ to $$5000$$ since we're given POSITIVE integers.$$=4999$$ Multiples of $$15$$ in this range$$=4995/15=333$$ Multiples of $$21=4998/21=238$$ Multiples of both $$15$$ and $$21$$=multiples of $$105=47$$ Now $$answer=4999-333-238+47=4475$$ This seems to be one of those killer problems that aren't necessarily conceptually difficult, but unless you have (what for me is) very high-powered arithmetic (long division) skills you're going to be in a time crunch here and in danger of compromising your quant section on test day. Practice, practice, practice those", "12197 Re M16-11 [#permalink] ### Show Tags 16 Sep 2014, 00:58 Expert's post 2 This post was BOOKMARKED Official Solution: How many positive three-digit integers are not divisible by 3 ? A. 599 B. 600 C. 601 D. 602 E. 603 Total 3-digit numbers: $$999-100+1=900$$; # of multiples of 3 in the given range: $$\\frac{\\text {last-first}}{multiple}+1=\\frac{999-102}{3}+1=300$$; {Total} - {# of multiples of 3} = {# of not multiples of 3}, hence the answer is $$900-300=600$$. Answer: B _________________ Kudos [?]: 129370 [0], given: 12197 Manager Joined: 10 May 2014 Posts: 142 Kudos [?]: 104 [2], given: 28 Re: M16-11 [#permalink] ### Show Tags 30 Oct 2014, 20:33 2 This post received KUDOS Hi, Total 3-digit numbers: 999 - 100 + 1 = 900 We know that the range goes from 100 to 999 There will always be a 2:1 ratio between numbers not divisible by 3 and numbers divisible by 3. For example: 100, 101, 102... 103, 104, 105... 997, 998, 999 We can infer that the 900 integers in the desired range will also fulfil this 2:1 ratio. Thus, 600 integers are not divisible by 3 and 300 are divisible by 3. Is my apporach correct? _________________ Consider giving me Kudos if I helped, but don´t take them away if I didn´t! What would you do if", "= 72 24 × 3 = 72 4th multiple 24 + 24 + 24 + 24 = 96 24 × 4 = 96 5th multiple 24 + 24 + 24 + 24 + 24 = 120 24 × 5 = 120 ## Did you know that… If a number is a multiple of 3 and 8… then we can conclude that it is also a multiple of 24? Let’s try to see if this is true. Consider the number 2,352. We need to know if it is divisible by 24 by knowing if it is also a multiple of 3 and 8. Hence, let’s divide 2,352 by 3. Thus,$$2,352\\;\\div\\;3\\;=\\;784$$. It is a multiple of 3! Now, let’s check if it is also a multiple of 8. If we divide 2,352 by 8, we will get$$2,352\\;\\div\\;8\\;=\\;294$$. It is also a multiple 8! Thus, we can say that it is a multiple of 24! Now, let’s see if it is indeed true by dividing 2,352 by 24. Hence,$$2,352\\;\\div\\;24\\;=\\;98$$. Since the result does not have a remainder, then we can definitely confirm that the trick is true! Can you try to check if 3,240 is also a multiple of 24 using this trick? ## List of First 30 multiples of 24 There is an endless number possible of integers hence, listing", "of multiples of 11 in the given range $$\\frac{(\\text{last}-\\text{first})}{\\text{multiple}}+1=\\frac{(990-11)}{11}+1=90$$; # of multiples of 35 in the given range $$\\frac{(\\text{last}-\\text{first})}{\\text{multiple}}+1=\\frac{(980-35)}{35}+1=28$$; # of multiples of both 11 and 35 is 2 ($$11*35=385$$ and 770); So, # of multiples of 11 or 35 in the given range is $$90+28-2=116$$. Thus numbers which are not divisible by either of them is $$1000-116=884$$. Hi Bunuel, I had come across a similar question \"How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?\". In here we find # of multiples of 13 and 12, and eliminate multiples of both i.e 12*13 = 156. I understand since 156 has been counted both in # of 13's and # of 12's we delete 2 multiples from the answer. But in the above problem both 11*35 = 385 and 2*11*35 = 770 must be counted twice in each of multiples of 11 and 13. So my doubt is why didn't we delete 4 multiples (385 and 770 are counted both in 11 multiples and 13 multiples) and instead did only 2 multiples (385 and 770). There is one critical aspect of your question (\"How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but", "have $$184+24=208$$ unordered triplets. The number of ways of choosing 3 unordered integers from between $$1$$ and $$100$$ with replacement is $${100+3-1\\choose3}={102\\choose3}=171700$$ The required probability is now $$P(\\text{sum}=50)=\\frac {208}{171700}\\approx0.00121$$", "# Distinct-digit Even Numbers Probability Level 4 How many 3 digit positive even integers are there, such that all the digits are distinct from each other? Details and assumptions The number $12=012$ is a 2-digit number, not a 3-digit number. ×", "+0 # Decimals 0 147 2 For how many integer values of between 1 and 474 inclusive does the decimal representation of $$\\frac{n}{3}$$ terminate? Jan 19, 2022 #1 +189 0 This is pretty easy. All numbers with 3 as a factor terminate when divided by 3. So find the amount of numbers that are multiples of 3 in your range. 474 is the 158th multiple of 3, so 158 numbers. Jan 19, 2022 #2 +2437 +1 All multiples of 3, so 474/3 = 158. So 158 numbers. Jan 19, 2022", "call countable. Integers are clearly not the only numbers as we are familiar with the real numbers such as $$1.2,\\, 3.56$$, and so on, of which there is also an infinite number, but many, many more than there are integers. If two different real numbers are chosen, an infinite number of others can be squeezed in between them simply by adding more decimal places; this and other interesting discussions on numbers and algorithms are explained clearly in ‘The Emperor’s New Mind’ (R.Penrose publ. 1990 Vintage OUP). Surprisingly, real numbers can usually be reduced to a rational fraction which is a ratio of two integers, e.g. $$0.751 = 326/533$$, but, for some numbers, $$\\pi,\\, e, \\sqrt{2}$$ there is no ratio of integers that will accurately form the number and they are called irrational, i.e. not being logical! You may recall that $$\\pi \\approx 22/7$$ , which would seem to contradict this statement, but $$22/7$$ is a very poor approximation to $$\\pi$$, only accurate to three digits, $$355/113$$ is better but only to seven digits. When you use a computer to perform calculations, a distinction is made whether a real or integer number is used. This is a common feature of programming languages and also in computer algebra, for example, the integer division $$2//3$$ is exact and in Python (notice", "this purpose. If I understand your question, the number of unstable integers that do not exceed $10^k$ for $k=1,2,3,4,5,6,7$ is $0,1,36,582,6933,74067,761605$. Jun 7, 2019 at 21:00 I'm not seeing a smart way (e.g. generating functions) to count these. So I will suggest a way to estimate (or overcount) them. As a nice example, let's start with 30, which is stable. Certain multiples of 30 are unstable. Let q be coprime to 30. Then (3q)(30)^n is unstable. (I assume n is positive.) Of all the numbers less than N, this gives about 8N/2700 which are multiples of 90, 8N/81000 which are multiples of 2700, and so on. So just from looking at 30, one gets about 8N/2610 unstable numbers whose exponent of three is one more than the exponents of both two and five. There is a similar count (about 3/10 as many) for certain unstable multiples of 300. Note that while this nicely partitions some of the multiples of 90 or 300 which are unstable, it does not capture all unstable multiples of 30. Further, some unstable multiples of 42 are included in this count. Nevertheless, this captures many of the unstable multiples of 30, and can be easily estimated. Now one can do something similar with other prime triplets, and there is a good chance of counting", "## anonymous 5 years ago What is the sum of the first 55 positive multiples of 3? Please show your work. 1. anonymous $3+6+9+\\cdots+162+165$ $=3(1)+3(2)+ \\cdots + 3(54)+3(55)$ $=3(1+2+3+ \\cdots +54+55) = 3 \\cdot \\frac{55(56)}{2}=4620$ 2. anonymous Nice! I should've thought of that one. Thanks Find more explanations on OpenStudy", "Kudos [?]: 50387 [6] , given: 7544 Re: good one [#permalink] 23 Aug 2010, 13:17 6 KUDOS Expert's post 2 This post was BOOKMARKED mehdiov wrote: How many integers less than 1000 have no factors(other than 1) in common with 1000 ? a. 400 b. 399 c. 410 d. 420 First of all it should be \"how many positive integers less than 1000 have no factors (other than 1) in common with 1000\", as if we consider negative integers answers will be: infinitely many. $$1000=2^3*5 ^3$$ so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5. Multiples of 2 in the range 0-1000, not inclusive - $$\\frac{998-2}{2}+1=499$$; Multiples of 5 in the range 0-1000, not inclusive - $$\\frac{995-5}{5}+1=199$$; Multiples of both 2 and 5, so multiples of 10 - $$\\frac{990-10}{10}+1=99$$. Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to $$999-(499+199-99)=400$$. _________________ Manager Joined: 30 Aug 2010 Posts: 92 Location: Bangalore, India Followers: 3 Kudos [?]: 123 [0], given: 27 Re: good one [#permalink] 03 Sep 2010, 04:24 Bunuel, Yes -- we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5", "Select Page # Understand the problem Find all positive integers $n$ that have 4 digits, all of them perfect squares, and such that $n$ is divisible by 2, 3, 5 and 7. Number theory Easy ##### Suggested Book An Excursion in Mathematics Do you really need a hint? Try it first! Note that the digits can only be 0,1,4,9. Also, the last digit has to be 0. The sum of the digits has to be divisible by 3. Hence (checking by hand), the possible candidates are 1110,1140,1410,4410,4140,9000,9090,9900 and 9990. Note that the two initial examples were 1110 and 9000. The other ones came by replacing one or more the digits with other digits that are equivalent modulo 3. Check for multiples of 7 in the list.", "both 2 and 5, so multiples of 10 - $$\\frac{990-10}{10}+1=99$$. Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to $$999-(499+199-99)=400$$. What about the prime numbers Bunuel ?? For ex : 7. Neither its a multiple of 2, nor 5 and it does not has any common factors with 1000 (except 1) So, shouldn't the answer include prime numbers between 1-999 as well. And if YES, how do we calculate the number of primer numbers from 1-999 ??? Plz clarfily. Thanks. Math Expert Joined: 02 Sep 2009 Posts: 44400 ### Show Tags 04 Oct 2013, 01:01 sumitchawla wrote: Bunuel wrote: mehdiov wrote: How many integers less than 1000 have no factors (other than 1) in common with 1000 ? a. 400 b. 399 c. 410 d. 420 First of all it should be \"how many positive integers less than 1000 have no factors (other than 1) in common with 1000\", as if we consider negative integers answers will be: infinitely many. $$1000=2^3*5 ^3$$ so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5. Multiples of 2 in the range 0-1000, not inclusive - $$\\frac{998-2}{2}+1=499$$; Multiples of 5 in the range 0-1000, not", "no factors(other than 1) in common with 1000 ? a. 400 b. 399 c. 410 d. 420 First of all it should be \"how many positive integers less than 1000 have no factors (other than 1) in common with 1000\", as if we consider negative integers answers will be: infinitely many. $$1000=2^3*5 ^3$$ so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5. Multiples of 2 in the range 0-1000, not inclusive - $$\\frac{998-2}{2}+1=499$$; Multiples of 5 in the range 0-1000, not inclusive - $$\\frac{995-5}{5}+1=199$$; Multiples of both 2 and 5, so multiples of 10 - $$\\frac{990-10}{10}+1=99$$. Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to $$999-(499+199-99)=400$$. What about the prime numbers Bunuel ?? For ex : 7. Neither its a multiple of 2, nor 5 and it does not has any common factors with 1000 (except 1) So, shouldn't the answer include prime numbers between 1-999 as well. And if YES, how do we calculate the number of primer numbers from 1-999 ??? Plz clarfily. Thanks. Math Expert Joined: 02 Sep 2009 Posts: 29210 Followers: 4755 Kudos [?]: 50387 [0], given: 7544 Re: good one [#permalink] 04 Oct 2013, 00:01" ]

[ "I am working on some AMC-8 questions. I am not sure how to approach this one. 279 views 0 19 months ago by Anonymous How many integers between 1000 and 9999 have four distinct digits? 2 19 months ago by The question can be rephrased to \"How many four-digit positive integers have four distinct digits,\" since numbers between 1000 and 9999 are four-digit integers. There are 9 choices for the first number, since it cannot be 0, There are only 9 choices left for the second number since it must differ from the first, 8 choices for the third number, since it must differ from the first two, and 7 choices for the fourth number, since it must differ from all three. This means there are $9\\times9\\times8\\times7=4536$9×9×8×7=4536 integers between 1000 and 9999 with four distinct digits.", "Expert's post sumitchawla wrote: Bunuel wrote: mehdiov wrote: How many integers less than 1000 have no factors (other than 1) in common with 1000 ? a. 400 b. 399 c. 410 d. 420 First of all it should be \"how many positive integers less than 1000 have no factors (other than 1) in common with 1000\", as if we consider negative integers answers will be: infinitely many. $$1000=2^3*5 ^3$$ so basically we are asked to calculate the # of positive integrs less than 1000, which are not multiples of 2 or/and 5. Multiples of 2 in the range 0-1000, not inclusive - $$\\frac{998-2}{2}+1=499$$; Multiples of 5 in the range 0-1000, not inclusive - $$\\frac{995-5}{5}+1=199$$; Multiples of both 2 and 5, so multiples of 10 - $$\\frac{990-10}{10}+1=99$$. Total # of positive integers less than 1000 is 999, so # integers which are not factors of 2 or 5 equals to $$999-(499+199-99)=400$$. What about the prime numbers Bunuel ?? For ex : 7. Neither its a multiple of 2, nor 5 and it does not has any common factors with 1000 (except 1) So, shouldn't the answer include prime numbers between 1-999 as well. And if YES, how do we calculate the number of primer numbers from 1-999 ??? Plz clarfily. Thanks. We counted multiples of 2 or 5 in", "A positive integer is said to be “tri-factorable” if it is the product of three consecutive integers. How many positive integers less than 1,000 are tri-factorable? We’ll use the same tactic, starting with the smallest case. The smallest tri-factorable number is simply $$1 \\times 2 \\times 3=$$ 6. Now 6 may seem like a long way from 1,000 but let’s not give up just yet! The next one up is $$2 \\times 3\\times 4=$$ 24. If you keep going, you should get this list: $$1 \\times 2 \\times 3= 6$$ $$2 \\times 3 \\times 4= 24$$ $$3 \\times 4 \\times 5= 60$$ $$4 \\times 5 \\times 6= 120$$ $$5 \\times 6 \\times 7= 210$$ $$6 \\times 7 \\times 8= 336$$ $$7 \\times 8 \\times 9= 504$$ $$8 \\times 9 \\times 10= 720$$ $$9 \\times 10 \\times 11= 990$$ Look how quickly we got there! Good things can happen if you just keep going. If you count them up, you should see that the answer is 9. Some of you might complain, “Ok it happened to work for this one question. What do you do when the list keeps on going and going?” When that happens, making a list will still help, but another step or tactic will often be necessary. Here’s one example: 3. How many multiples of", "every other multiple of 15. If we add these two totals together, we will be counting all of the multiples of 15 twice. We only want to count them once, so to counter this, we can subtract from this total the number of positive integers less than or equal to 9,999 that are divisible by 15. In sum, the answer should be: $\\lfloor{\\frac{9999}{3}}\\rfloor+\\lfloor{\\frac{9999}{5}}\\rfloor-\\lfloor{\\frac{9999}{15}}\\rfloor=3,333+1,999-666=4,666$", "# Practice: Finding common multiples How many positive integers less than or equal to $1000$ are common multiples of the three numbers $8, 36,$ and $24$? ×", "A positive integer is said to be \"tri-factorable\" if it is the product of three consecutive integers. How many positive integers less than 1,000 are tri-factorable? We'll use the same tactic, starting with the smallest case. The smallest tri-factorable number is simply $$1 \\times 2 \\times 3=$$ 6. Now 6 may seem like a long way from 1,000 but let's not give up just yet! The next one up is $$2 \\times 3\\times 4=$$ 24. If you keep going, you should get this list: $$1 \\times 2 \\times 3= 6$$ $$2 \\times 3 \\times 4= 24$$ $$3 \\times 4 \\times 5= 60$$ $$4 \\times 5 \\times 6= 120$$ $$5 \\times 6 \\times 7= 210$$ $$6 \\times 7 \\times 8= 336$$ $$7 \\times 8 \\times 9= 504$$ $$8 \\times 9 \\times 10= 720$$ $$9 \\times 10 \\times 11= 990$$ Look how quickly we got there! Good things can happen if you just keep going. If you count them up, you should see that the answer is 9. Some of you might complain, \"Ok it happened to work for this one question. What do you do when the list keeps on going and going?\" When that happens, making a list will still help, but another step or tactic will often be necessary. Here's one example: 3. How many multiples of", "this purpose. If I understand your question, the number of unstable integers that do not exceed $10^k$ for $k=1,2,3,4,5,6,7$ is $0,1,36,582,6933,74067,761605$. Jun 7, 2019 at 21:00 I'm not seeing a smart way (e.g. generating functions) to count these. So I will suggest a way to estimate (or overcount) them. As a nice example, let's start with 30, which is stable. Certain multiples of 30 are unstable. Let q be coprime to 30. Then (3q)(30)^n is unstable. (I assume n is positive.) Of all the numbers less than N, this gives about 8N/2700 which are multiples of 90, 8N/81000 which are multiples of 2700, and so on. So just from looking at 30, one gets about 8N/2610 unstable numbers whose exponent of three is one more than the exponents of both two and five. There is a similar count (about 3/10 as many) for certain unstable multiples of 300. Note that while this nicely partitions some of the multiples of 90 or 300 which are unstable, it does not capture all unstable multiples of 30. Further, some unstable multiples of 42 are included in this count. Nevertheless, this captures many of the unstable multiples of 30, and can be easily estimated. Now one can do something similar with other prime triplets, and there is a good chance of counting", "## anonymous 5 years ago What is the sum of the first 55 positive multiples of 3? Please show your work. 1. anonymous $3+6+9+\\cdots+162+165$ $=3(1)+3(2)+ \\cdots + 3(54)+3(55)$ $=3(1+2+3+ \\cdots +54+55) = 3 \\cdot \\frac{55(56)}{2}=4620$ 2. anonymous Nice! I should've thought of that one. Thanks Find more explanations on OpenStudy", "of multiples of 11 in the given range $$\\frac{(\\text{last}-\\text{first})}{\\text{multiple}}+1=\\frac{(990-11)}{11}+1=90$$; # of multiples of 35 in the given range $$\\frac{(\\text{last}-\\text{first})}{\\text{multiple}}+1=\\frac{(980-35)}{35}+1=28$$; # of multiples of both 11 and 35 is 2 ($$11*35=385$$ and 770); So, # of multiples of 11 or 35 in the given range is $$90+28-2=116$$. Thus numbers which are not divisible by either of them is $$1000-116=884$$. Hi Bunuel, I had come across a similar question \"How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?\". In here we find # of multiples of 13 and 12, and eliminate multiples of both i.e 12*13 = 156. I understand since 156 has been counted both in # of 13's and # of 12's we delete 2 multiples from the answer. But in the above problem both 11*35 = 385 and 2*11*35 = 770 must be counted twice in each of multiples of 11 and 13. So my doubt is why didn't we delete 4 multiples (385 and 770 are counted both in 11 multiples and 13 multiples) and instead did only 2 multiples (385 and 770). There is one critical aspect of your question (\"How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but", "subtracting 47 once. Similar questions to practice: what-is-the-number-of-integers-from-1-to-1000-inclusive-126153.html what-is-the-total-number-of-positive-integers-that-are-less-128104.html how-many-integers-from-1-to-200-inclusive-are-divisib-109333.html how-many-even-integers-n-where-100-n-200-are-divisib-103779.html Hope it helps. Hi Bunuel, Is there a shorter way around this question?This approach took me 5 minutes! Manager Joined: 20 Dec 2013 Posts: 231 Location: India Re: How many positive integers less than 5,000 are evenly divisi [#permalink] ### Show Tags 04 Apr 2014, 20:59 Option B.(Took 5 minutes to solve with the following approach!) All nos. from $$1$$ to $$5000$$ since we're given POSITIVE integers.$$=4999$$ Multiples of $$15$$ in this range$$=4995/15=333$$ Multiples of $$21=4998/21=238$$ Multiples of both $$15$$ and $$21$$=multiples of $$105=47$$ Now $$answer=4999-333-238+47=4475$$ Intern Joined: 26 Mar 2014 Posts: 4 Re: How many positive integers less than 5,000 are evenly divisi [#permalink] ### Show Tags 16 Apr 2014, 22:53 2 AKG1593 wrote: Option B.(Took 5 minutes to solve with the following approach!) All nos. from $$1$$ to $$5000$$ since we're given POSITIVE integers.$$=4999$$ Multiples of $$15$$ in this range$$=4995/15=333$$ Multiples of $$21=4998/21=238$$ Multiples of both $$15$$ and $$21$$=multiples of $$105=47$$ Now $$answer=4999-333-238+47=4475$$ This seems to be one of those killer problems that aren't necessarily conceptually difficult, but unless you have (what for me is) very high-powered arithmetic (long division) skills you're going to be in a time crunch here and in danger of compromising your quant section on test day. Practice, practice, practice those", "+0 # Decimals 0 147 2 For how many integer values of between 1 and 474 inclusive does the decimal representation of $$\\frac{n}{3}$$ terminate? Jan 19, 2022 #1 +189 0 This is pretty easy. All numbers with 3 as a factor terminate when divided by 3. So find the amount of numbers that are multiples of 3 in your range. 474 is the 158th multiple of 3, so 158 numbers. Jan 19, 2022 #2 +2437 +1 All multiples of 3, so 474/3 = 158. So 158 numbers. Jan 19, 2022", "wrote: devbond wrote: How many positive integers less than 100 are neither multiples of 2 or 3. a)30 b)31 c)32 d)33 e)34 Hi, To answer this Q we require to know 1) multiples of 2 till 100 $$= \\frac{100}{2} = 50$$ 2) Multiples of 3 till 100 = $$\\frac{100}{3} = 33.33= 33$$ add the two $$50+33=83$$ ; subtract common terms that are multiple of both 2 and 3.. LCM of 2 and 3 = 6 Multiples of 6 till 100 = $$\\frac{100}{6} = 16.66 = 16$$ so total multiples of 2 and 3 = 83-16 = 67 ans = $$100-67 = 33$$ D Hi Chetan, I searched for the formula of no. of multiples in a range and I got: No. of multiples of x in the range = ((Last multiple of x in the range - First multiple of x in the range)/x)+1 But you have directly applied 100/16..100/3. Thanks. GMAT Club Legend Joined: 11 Sep 2015 Posts: 4605 GMAT 1: 770 Q49 V46 Re: How many positive integers less than 100 are neither multiples of 2 or [#permalink] ### Show Tags 01 Dec 2019, 12:47 Top Contributor devbond wrote: How many positive integers less than 100 are neither multiples of 2 or 3. a)30 b)31 c)32 d)33 e)34 Multiples of 2: 2, 4, 6, ..., 96,", "\\end{array}$$ $$\\text{How many of the positive divisors of 3240 are multiples of 3 ?}$$ $$\\mathbf{40-8 = 32}$$ $$\\text{There are \\mathbf{32} positive divisors of 3240, which are multiples of 3.}$$ Nov 13, 2018", "1. ## Easy combinatorics problem How many four-digit positive integers are there that contain the digit 1? 2. ## Re: Easy combinatorics problem Originally Posted by alexmahone How many four-digit positive integers are there that contain the digit 1? To be clear $0109=109$ is not a four digit number, right?? There are $9\\cdot 10^3$ four-digit numbers. Of those there are $8\\cdot 9^3$ that do not contain any 1's. So how many contain at least one 1? 3. ## Re: Easy combinatorics problem hi alexmahone should the digits be the same or must be different? if they are the same then there are: 1000+900+900+900=2800 if they are not the same then there are: 720+648+648+648=2664", "have $$184+24=208$$ unordered triplets. The number of ways of choosing 3 unordered integers from between $$1$$ and $$100$$ with replacement is $${100+3-1\\choose3}={102\\choose3}=171700$$ The required probability is now $$P(\\text{sum}=50)=\\frac {208}{171700}\\approx0.00121$$", "# Distinct-digit Even Numbers Probability Level 4 How many 3 digit positive even integers are there, such that all the digits are distinct from each other? Details and assumptions The number $12=012$ is a 2-digit number, not a 3-digit number. ×", "Joined: 21 Jan 2007 Posts: 2597 Location: New York City How many positive integers less than 200 are there such that [#permalink] ### Show Tags Updated on: 13 Dec 2014, 04:22 19 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:17) correct 33% (02:29) wrong based on 335 sessions ### HideShow timer Statistics How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both? A. 28 B. 29 C. 31 D. 32 E. 33 M12-36 Originally posted by bmwhype2 on 23 Nov 2007, 05:07. Last edited by Bunuel on 13 Dec 2014, 04:22, edited 1 time in total. Renamed the topic, edited the question, added the OA and moved to PS forum. Math Expert Joined: 02 Sep 2009 Posts: 52451 Re: How many positive integers less than 200 are there such that [#permalink] ### Show Tags 13 Dec 2014, 04:22 2 6 bmwhype2 wrote: How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both? A. 28 B. 29 C. 31 D. 32 E. 33 M12-36 # of multiples of 13 in the given range $$\\frac{(\\text{last-first})}{\\text{multiple}}+1=\\frac{195-13}{13}+1=15$$; # of multiples of 12 in the given range $$\\frac{(\\text{last-first})}{\\text{multiple}}+1=\\frac{192-12}{12}+1=16$$; # of multiples of both 13 and 12 is", "# Four Digit Number - Can You Find It? I am thinking of a four digit positive integer with distinct digits. Of course, there's a total of $$4!-1=23$$ ways to rearrange the digits to form a new 4 digit positive integer. If the sum of these other 23 numbers is 157193, what is the number that I was thinking of? ×", "1*3*3*3=27..[/color] II. all even ofcourse a number without any odd prime 222.. so $$27+1=28$$ B Hi Chetan, This is really helpful. Just one question - Since we're trying to re-arrange 3 integers - why shouldn't this be multiplied by 3! instead of 3? How many three-digit numbers contain three primes that sum to an even [#permalink] 27 Oct 2018, 05:51 Display posts from previous: Sort by", ", 1003\\}$ is a multiple of $3$. Exactly one third of numbers in $\\{1,...,1003\\}$ are multiples of $3$. Additionally, $n$ must be odd (else 2n is divisible by $4$ as well). How many odd multiples of $3$ are in $\\{1, ..., 10003\\}$? -" ]

[ "+0 +1 34 3 How many natural-number factors does $N$ have if $N = 2^4 \\cdot 3^3 \\cdot 5^2 \\cdot 7^2$? Feb 17, 2021 #1 +273 +1 We have a prime factorization of $2^4 \\times 3^3 \\times 5^2 \\times 7^2$ If we add one to each exponent, then add up all the exponents. $5 + 4 + 3 + 3 = 15$ Therefore there are 15 natural factors. Hope that helps!!! 🙃🙃👌 Feb 17, 2021 #2 0 Sorry, but that is wrong. Feb 17, 2021 #3 +116126 +1 OBJD is close....but....we need to add 1 to each exponent and then take that product....so.. (4+1) (3+1) (2 + 1) ( 2 + 1) = 5 * 4 * 3 * 3 = 180 factors Feb 17, 2021", "# Factors of $$N^3$$ Let $$N = 71^3\\cdot 73^2\\cdot 79^4$$. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "+0 # Number Theory 0 34 1 If $n = 2^{10} \\cdot 3^{8} \\cdot 5^{14}$, how many of the natural-number factors of n are multiples of 150? Jul 7, 2022 #1 +1105 +8 2^10 x 3^14 x 5^8 = 2^1 x 3^1 x 5^2 =150 Subtract the exponents of the same base numbers and add 1 in each case as follows: [10 - 1 + 1] x [14 - 1 + 1] x [8 - 2 + 1] =10 x 14 x 7 = 980 such numbers.", "# Factors on N^3 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "smaller non-deficient number. He proved that there are only a finite number of primitive non-deficient numbers having a given number of distinct odd prime factors and a given number of factors $2$. There is no odd abundant number with fewer than three distinct prime factors, the primitive ones with three are \\begin{equation*} 3 ^ { 3 } .5 .7,3 ^ { 2 } .5 ^ { 2 } .7,3 ^ { 2 } .5 .7 ^ { 2 } \\end{equation*} \\begin{equation*} 3 ^ { 2 } \\cdot 5 ^ { 2 } \\cdot 11,\\; 3 ^ { 5 } \\cdot 5 ^ { 2 } \\cdot 13,\\; 3 ^ { 4 } \\cdot 5 ^ { 2 } \\cdot 13 ^ { 2 } ,\\; 3 ^ { 3 } \\cdot 5 ^ { 3 } \\cdot 13 ^ { 2 }. \\end{equation*} He gave also a table of all even abundant numbers $< 6232$. Dickson's result was a starting point for much further research. In 1949 and 1968, H.N. Shapiro ([a20], [a21]) proved the following result. Let $\\alpha$ be a rational number. A necessary and sufficient condition that there exist infinitely many primitive $\\alpha$-abundant numbers (i.e. $\\sigma ( n ) / n \\geq \\alpha$ but $\\sigma ( d ) / d < \\alpha$ for all $d |", "# Factors of $$N^3$$ Number Theory Level 5 Let $$N = 71^3\\cdot 73^2\\cdot 79^4$$. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "What is this? Number Theory Level 3 If $$2^{200}-2^{192} \\cdot 31+ 2^{n}$$ is a perfect square for a certain natural number $$n$$, what is the value of $$n$$? ×", "# nLab prime factor For $n\\in ℕ$ a natural number a prime factor is a prime number $p\\in ℕ$ which divides $n$, hence such that there is $k\\in ℕ$ with $k\\cdot p=n$.", "# What is this? If $$2^{200}-2^{192} \\cdot 31+ 2^{n}$$ is a perfect square for a certain natural number $$n$$, what is the value of $$n$$? ×", "# Factors on N^3 Computer Science Level 5 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? × Problem Loading... Note Loading... Set Loading...", "it's true that $$[(n \\cdot 1) b]^k = (n^k \\cdot 1) b^k$$ for natural numbers $$n$$.", "# Permutations and Combinations ### Factorials A factorial of a natural number represents a product of all the natural numbers less than or equal to the number. A factorial is the product of all natural numbers less than or equal to the indicated natural number. The symbol for a factorial is an exclamation point. $n! = n \\cdot (n-1) \\cdot (n-2) \\cdot . . . \\cdot 1$ For example: $5! = 5\\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1=120$ The expression $0!$ may be a result of evaluating formulas with factorials. Since zero is not a natural number, the formula cannot be used. The factorial of zero is defined as 1: $0! = 1$. Factorials are used in counting and probability and are often simplified. To simplify $\\frac {8!}{5!} !$, write it in expanded form and simplify the factors in the numerator and denominator. \\begin{aligned}\\frac {8!}{5!}&= \\frac{8 \\cdot 7 \\cdot 6 \\cdot {\\color{#c42126}5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}}{{\\color{#c42126}5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}}\\\\&= 8 \\cdot 7 \\cdot 6\\\\&= 336\\end{aligned} Notice that the factors in the numerator and denominator do not have to be written out completely because all the factors in the denominator are also factors in the numerator. \\begin{aligned}\\frac {8!}{5!}&= \\frac{8 \\cdot 7 \\cdot 6 \\cdot {\\color{#c42126}5!}}{{\\color{#c42126}5!}}\\\\&= 8 \\cdot 7 \\cdot", "= 5 \\cdot 7 \\cdot 11$ has no square factors.", "# Factors 1 Number Theory Level pending How many factors of $$20!$$ have an odd number of factors? Notation: $$!$$ denotes the factorial function. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "# Possible Natural Numbers How many natural numbers $$N$$ are there such that $$N! +10$$ is a perfect square? Notation: $$!$$ is the factorial notation. For example, $$8! = 1\\times2\\times3\\times\\cdots\\times8$$. ×", "# 2007_Question1a Looking at the prime factors, we see that the numerator is 2^{r+s}\\cdot3^{r+s}\\cdot2^{2(r-s)}\\cdot3^{r-s} = 2^{3r-s}\\cdot3^{2r}. Similarly, the denominator is 2^{3r}\\cdot3^{2(r+2s)} = 2^{3r}\\cdot3^{2r+4s}. Dividing one by the other then gives 2^{-s}\\cdot3^{-4s}, which is an integer if s\\leqslant0, and so the answer is (b). Correct!!! Well done", "of $X^n\\times X^n$, and it is not immediately clear that the natural choice would be of finite type. Hopefully there are then natural inclusions $Y_1\\hookrightarrow Y_2\\hookrightarrow\\cdots$ giving the desired ind-scheme. -", "+0 # Number Theory-Help 0 68 2 If $n = 2^{10} \\cdot 3^{14} \\cdot 5^{8}$, how many of the natural-number factors of $n$ are multiples of 150 Apr 6, 2020 #1 +252 0 $$n = 2^{10} \\cdot 3^{14} \\cdot 5^{8}$$ 150=5^2*3*2 We have $$2^9*3^{13}*5^6$$ for each number of exponent, it can be up to that or less. or none. so we get 10*14*7 soo my answer is 980. i think Apr 6, 2020 #2 0 2^10 x 3^14 x 5^8 = 2^1 x 3^1 x 5^2 =150 Subtract the exponents of the same base numbers and + 1 in each case as follows: [10 - 1 + 1] x [14 - 1 + 1] x [8 - 2 + 1] =10 x 14 x 7 = 980 such numbers. Apr 6, 2020", "# Placing numbers $1,2,\\ldots,n^2$ in a square so that numbers of any two adjacent unit subcube are coprime It is proved HERE that there is a natural number $N$ such that for any $n > N$ it is possible to place numbers $1,2,\\cdots, n^2$ is an $n\\times n$ square such that the numbers in any two adjacent square (having a mutual edge) are coprime. How can one estimate $N$ !? My guess is that the least such $N$ is equal to $1$.", "that it does for real numbers. So there's a direct correspondence between the impossibility of setting up a situation in which there is the required symmetry between all natural numbers and the impossibility of defining a uniform distribution on the natural numbers (or any countably infinite set). - If you want to choose a natural number from the set $\\lbrace 1,2,3,\\ldots \\rbrace$ of all natural numbers, then user3533 is right. But if you pick a natural number from a finite set, say $A = \\lbrace 1,2,\\ldots,N \\rbrace$, each number with the same probability, then this probability must be the reciprocal of the number of elements in that set. So, if $A = \\lbrace 1,2,\\ldots,N \\rbrace$, then $p=1/N$. This follows immediately from $\\sum\\nolimits_{i = 1}^N {p_i } = 1$ with $p_1 = p_2 = \\cdots = p_N = p$. -" ]

[ "# Count The Divisors There is a certain number $$N = n_1\\cdot n_2\\cdot n_3\\cdots n_k$$, where $$1 \\leq n_i \\leq 100$$ $$(1\\le i \\le k)$$ are all distinct positive integers. We are told the following: • $$N$$ has $$d$$ (positive) divisors and $$d$$ is odd. • $$128\\cdot N$$ has $$\\frac{18}{11}d$$ divisors. • $$175\\cdot N$$ has $$\\frac 8 5 d$$ divisors. • $$213\\cdot N$$ has $$\\frac 8 3 d$$ divisors. If $$2016\\cdot N$$ has $$\\frac A B d$$ divisors, with $$A$$ and $$B$$ coprime positive integers, submit $$A + B$$. ×", "# Factors of $$N^3$$ Let $$N = 71^3\\cdot 73^2\\cdot 79^4$$. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "c \\leq 1$. We see that there are $4$ choices for $a$, $2$ choices for $b$, and $2$ choices for $c$. By the Multiplication principle, the number of factors is $4\\cdot 2 \\cdot 2=16$.", "# Factors of $$N^3$$ Number Theory Level 5 Let $$N = 71^3\\cdot 73^2\\cdot 79^4$$. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "+0 # help 0 118 1 Find the number of distinct positive divisors of $$(30)^4$$ excluding 1 and $$(30)^4$$. Jan 1, 2019 $$30=2^1 \\cdot 3^1 \\cdot 5^1\\\\ 30^4 = 2^4 \\cdot 3^4 \\cdot 5^4\\\\ \\text{the number of divisors is equal to the product of 1 plus each exponent in the }\\\\ \\text{prime factorization of a given number}\\\\ n = 5\\cdot 5\\cdot 5 = 125$$", "follows simply from the fact that $p^2 - 1 = (p-1)(p+1)$; since $p$ is not divisible by either 2 or 3, • the factors $p-1$ and $p+1$ must both be even, Thus, their product must be divisible by $2\\cdot4\\cdot3 = 24$.", "it comes from: Consider the divisors of $60=2^23^15^1$ $01=2^03^05^0$ $02=2^13^05^0$ $03=2^03^15^0$ $04=2^23^05^0$ $05=2^03^05^1$ $06=2^13^15^0$ $10=2^13^05^1$ $12=2^23^15^0$ $15=2^03^15^1$ $20=2^23^05^1$ $30=2^13^15^1$ $60=2^23^15^1$ Notice how every possible combination of exponents appears. For the exponent on 2, there are three choices: 0,1,2. For 3, there are two choices: 0,1. For 5, also two choices: 0,1. Going back to your basic probability, multiply these three together to get every possibility, hence $3*2*2=12$ divisors of 60. This is not a formal proof, but it should show you that the equation NCA gave you did not come out of \"thin air,\" but is rather the expression of a very simple observation that can be seen by factoring the divisors. any divisor of $n$ has the same prime factors of $n.$ so it has to be in the form $d=p_1^{s_1} p_2^{s_2} \\cdots p_k^{s_k},$ where $0 \\leq s_j \\leq r_j,$ for all $1 \\leq j \\leq k.$ so each $s_j$ has $r_j + 1$ possibilities. that's all!", "# Equation in terms of divisors. Let $$n$$ be a natural number and $$1=d_{1}<d_{2}<...<d_{k}=n$$be the positive divisors of $$n$$. Now,find the minimum $$n$$ such that $$2n=d_{5}^{2}+d_{6}^2-1$$ ×", "# Factors on N^3 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "= 1232$ factors. We can draw a bijection between the number of divisors of $2^{15} \\cdot 3^{10} \\cdot 5^{6}$ that are divisible by $10$ and the number of divisors of $2^{14} \\cdot 3^{10} \\cdot 5^{5}$ (as each of these divisors, when multiplied by 10, will provide a factor of the original number that is divisible by 10). There are $(14+1)(10+1)(5+1) = 990$. The answer is $1232-990 = \\boxed{242}$.", "the second method to work we need $d>m$, i.e., $\\frac{2\\cdot 90}d<d+1$. Since $\\frac{2\\cdot 90}d$ is even, exactly one of the options works for each odd divisor $d>1$. Therefore the politeness is the number of odd divisors. This can be read from the prime factorization of $90=2\\cdot 3^2\\cdot 5^1$: the odd divisors are of the form $3^a5^b$ with $0\\le a\\le 2, 0\\le b\\le 1$. Hence there are $3\\cdot 2$ odd divisors. Remove $d=1$ to arrive at politeness $5$.", "those numbers must be both positive or both negative, but cannot be zero. This is because zero is neither positive nor negative, and 0 * n = 0. The first step is to find the total number of combination: (a,b). This can be done by: $$(1-(-3)+1)\\cdot(4-(-2)+1) =5\\cdot7=35.$$ There are a total of 35 different combinations. Out of those combinations there are: ( -3 , -2 ) ; ( -3 , -1 ) ; ( -2 , -2 ) ; ( -2 , -1 ) ; ( -1 , -2 ) ; ( -1 , -1 ) ( 1 , 1 ) ; ( 1 , 2 ) ; ( 1 , 3 ) ; ( 1 , 4 ) 10 that work. The probabilty of this happening is: $$\\boxed{\\frac{10}{35}=\\frac{2}{7}}$$ I hope this helped, Gavin. GYanggg Apr 21, 2018", "at 0:24 • See, there lies my confusion. 7 does not show up in the prime factorization of 24... so why am I looking into it? Similarly with 5. – user424890 Nov 7 '17 at 0:27 Our goal is to find all integers with exactly 36 divisors. To do this, first note that any positive integer will have a unique prime factorization (up to order in which primes appear, of course). This means that in general, a positive integer $n$ will have the form $$n = p_1^{a_1} \\cdots p_k^{a_k}$$ where each prime here is distinct and each $a_i$ is a positive integer. So, any divisor of $n$ will have the form $$d = p_1^{b_1} \\cdots p_k^{b_k}$$ where $b_i$ can range between $0$ and $a_i$. Now, how many ways can choose a $b_i$? Well, there are $a_i +1$ options. So, we have $(a_1+1) \\cdots (a_k+1)$ divisors of $n$. If we have $(a_1+1) \\cdots (a_k+1) = 24$ then $n$ will have exactly $24$ divisors. If we are looking for positive integers with exactly 36 divisors then any positive integer with the exponents of its primes satisfying $(a_1+1) \\cdots (a_k+1) = 36$ will work. Note that $36 = 2^2 3^2$. With this in mind, we can have $36 = 2*2*3*3$ which would correspond to 4 distinct primes, two being square and", "# Problem of the Day If a positive integer is equal to the following product: , where and are distinct prime numbers greater than , how many distinct even factors does the integer have? A. 1 B. 5 C. 50 D. 100 E. 120 D. 100 See the Solution ### Solution [latexpage] This solution assumes that you are familiar with the basics of counting and prime factorization. To calculate the number of factors that an integer has, first look at its prime factorization. Â For example, if we want to know how many factors 12 has, we would break it down as $2^2 \\cdot 3$. Â If an integer is a factor of 12, it has to be built from the prime factors of 12. Â A complete list of the factors of 12 is 1, 2, 3, 4, 6, 12. $1=2^0 \\cdot 3^0$ $2=2^1 \\cdot 3^0$ $3=2^0 \\cdot 3^1$ $4=2^2 \\cdot 3^0$ $6=2^1 \\cdot 3^1$ $12=2^2 \\cdot 3^1$ Notice that each factor of 12 is of the form $2^x \\cdot 3^y$ where $x$ is either 0, 1, or 2 and y is either 0 or 1. Â So there are three choices for the exponent for 2 and two choices for the exponent of 3. Â Therefore there are $3 \\cdot 2$ or $6$ factors of 12.", "/** * @file * @brief C++ Program to calculate the number of positive divisors * * This algorithm uses the prime factorization approach. * Any positive integer can be written as a product of its prime factors. * <br/>Let \\f$N = p_1^{e_1} \\times p_2^{e_2} \\times\\cdots\\times p_k^{e_k}\\f$ * where \\f$p_1,\\, p_2,\\, \\dots,\\, p_k\\f$ are distinct prime factors of \\f$N\\f$ and * \\f$e_1,\\, e_2,\\, \\dots,\\, e_k\\f$ are respective positive integer exponents. * <br/>Each positive divisor of \\f$N\\f$ is in the form * \\f$p_1^{g_1}\\times p_2^{g_2}\\times\\cdots\\times p_k^{g_k}\\f$ * where \\f$0\\le g_i\\le e_i\\f$ are integers for all \\f$1\\le i\\le k\\f$. * <br/>Finally, there are \\f$(e_1+1) \\times (e_2+1)\\times\\cdots\\times (e_k+1)\\f$ * positive divisors of \\f$N\\f$ since we can choose every \\f$g_i\\f$ * independently. * * Example: * <br/>\\f$N = 36 = (3^2 \\cdot 2^2)\\f$ * <br/>\\f$\\mbox{number_of_positive_divisors}(36) = (2+1) \\cdot (2+1) = 9\\f$. * <br/>list of positive divisors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36. * * Similarly, for N = -36 the number of positive divisors remain same. **/ #include <cassert> #include <iostream> /** * Function to compute the number of positive divisors. * @param n number to compute divisors for * @returns number of positive divisors of n (or 1 if n = 0) */ int number_of_positive_divisors(int n) { if (n < 0) { n = -n; // take", "is $5 \\cdot \\dfrac{7}{4} = \\dfrac{35}{4} = 8 \\dfrac{3}{4}.$ Thus, D is the correct answer. 16. How many $$3$$-digit positive integers have digits whose product equals $$24?$$ $$12$$ $$15$$ $$18$$ $$21$$ $$24$$ ###### Solution(s): The only triples of integers less than $$10$$ that multiply to $$24$$ are $(1, 3, 8), (1, 4, 6), (2, 2, 6),$$\\text{and } (2, 3, 4).$ The triples with $$3$$ distinct numbers can be rearranged to form $$6$$ distinct $$3$$-digit positive integers. The other triple can be arranged to form $$3$$ distinct $$3$$-digit positive integers. This leaves a total of $$3 \\cdot 6 + 3 = 21$$ integers. Thus, D is the correct answer. 17. The positive integers $$x$$ and $$y$$ are the two smallest positive integers for which the product of $$360$$ and $$x$$ is a square and the product of $$360$$ and $$y$$ is a cube. What is the sum of $$x$$ and $$y?$$ $$80$$ $$85$$ $$115$$ $$165$$ $$610$$ ###### Solution(s): For a number to be a perfect square, every exponent in the prime factorization must be even. For it to be a cube, the exponents must be divisible by $$3.$$ We can factor $$360$$ to get $360 = 2^3 \\cdot 3^2 \\cdot 5.$ For $$360x$$ to be a perfect square and $$x$$ to be minimized, $$x$$ must have one factor of", "\\,+\\,...\\,-\\,a \\cdot b^{n-2}+b^{n-1})$$ Ah Math Key Fact 2545 For all $n \\in N$, $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2} \\cdot b \\,+\\,...\\,+\\,a \\cdot b^{n-2}+b^{n-1})$$ Ah Math Key Fact A376 For $a=k\\cdot (2k+1)$ where $k \\in N$,$$a^2+(a+1)^2+⋯+(a+k)^2\\\\=(a+k+1)^2+(a+k+2)^2+⋯+(a+2k)^2$$ Ah Math Key Fact 3278 A composite number is a positive integer greater than 1 that has more than two positive divisors. Ah Math Key Fact 2DCC A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Ah Math Key Fact 87D4 All positive integers greater than 1 are either prime or composite. Ah Math Key Fact 9E11 1 is the only positive integer that is neither prime nor composite. Ah Math Key Fact A478 The only even prime number is 2. Ah Math Key Fact 2AB2 No prime number greater than 5 ends in a 5. Ah Math Key Fact DAEA List of prime numbers up to 100: $$2,\\,3,\\,5,\\,7,\\,11,\\,13,\\,17,\\,19,\\,23,\\,29,\\,31,\\,37,\\,41,$$$$\\,43,\\,47,\\,53,\\,59,\\,61,\\,67,\\,71,\\,73,\\,79,\\,83,\\,89,\\,97$$ Ah Math Key Fact 9ECE Every integer greater than 1 has a unique prime factorization up to the order of the factors. Ah Math Key Fact 525C If $n$ is a composite number, then it must be divisible by a prime $p$ such that $p \\le \\sqrt{n}$. Ah Math Key Fact 163E $\\small{\\textbf{ Wilson's Theorem}}:\\\\[5pt]\\text{ A positive integer }p>1 \\text{ is prime if and only if }$", "# Staggered triplets Calculus Level 4 $S=\\dfrac { 1 }{ 1\\cdot 3\\cdot 5 } +\\dfrac { 1 }{ 2\\cdot 4\\cdot 6 } +\\dfrac { 1 }{ 3\\cdot 5\\cdot 7 } +\\dfrac { 1 }{ 4\\cdot 6\\cdot 8 } +\\dfrac { 1 }{ 5\\cdot 7\\cdot 9 } + \\cdots$ If $S$ can be expressed in the form $\\frac { A }{ { B }^{ C }D }$ with $A$, $B$, $C$, and $D$ being positive prime integers, find the value of $\\dfrac { A+B+C+D }{ D }.$ ×", "and which fits this criterion is $78=2\\cdot3\\cdot13$, so the answer is $\\boxed{\\textbf{(D) }13}$. Solution 2 We can say that $a$ and $b$ 'have' $2^3 * 3$, that $b$ and $c$ have $2^2 * 3^2$, and that $c$ and $d$ have $3^3 * 2$. Combining $1$ and $2$ yields $b$ has (at a minimum) $2^3 * 3^2$, and thus $a$ has $2^3 * 3$ (and no more powers of $3$ because otherwise $gcd(a,b)$ would be different). In addition, $c$ has $3^3 * 2^2$, and thus $d$ has $3^3 * 2$ (similar to $a$, we see that $d$ cannot have any other powers of $2$). We now assume the simplest scenario, where $a = 2^3 * 3$ and $d = 3^3 * 2$. According to this base case, we have $gcd(a, d) = 2 * 3 = 6$. We want an extra factor between the two such that this number is between $70$ and $100$, and this new factor cannot be divisible by $2$ or $3$. Checking through, we see that $6 * 13$ is the only one that works. Therefore the answer is $\\boxed{\\textbf{(D) } 13}$ Solution by JohnHankock Solution 2.1 (updated with better notation) Do casework on $v_2$ and $v_3.$ Notice that we must have $v_3(a) = 1$ and $v_2(d)=1$ and the values of $b,d$ does not matter.", "I hinted at with the $k \\le 3$ not necessarily $k=3$. For example $p^{104}$ also has $105$ divisors. So does $p^{14}q^6$ and a few others. For 2, you could say since $2 \\cdot 3 \\cdot 5 \\cdot 7 >100$ you have $k\\le 3$. Now try which combination of exponents for $2$, $3$, $5$ gives you most divisors (note again that one of them might not appear). Then check if you can get the same with other primes, like $7$ instead of $5$." ]

[ "+0 +1 34 3 How many natural-number factors does $N$ have if $N = 2^4 \\cdot 3^3 \\cdot 5^2 \\cdot 7^2$? Feb 17, 2021 #1 +273 +1 We have a prime factorization of $2^4 \\times 3^3 \\times 5^2 \\times 7^2$ If we add one to each exponent, then add up all the exponents. $5 + 4 + 3 + 3 = 15$ Therefore there are 15 natural factors. Hope that helps!!! 🙃🙃👌 Feb 17, 2021 #2 0 Sorry, but that is wrong. Feb 17, 2021 #3 +116126 +1 OBJD is close....but....we need to add 1 to each exponent and then take that product....so.. (4+1) (3+1) (2 + 1) ( 2 + 1) = 5 * 4 * 3 * 3 = 180 factors Feb 17, 2021", "# Factors of $$N^3$$ Let $$N = 71^3\\cdot 73^2\\cdot 79^4$$. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "# Factors of $$N^3$$ Number Theory Level 5 Let $$N = 71^3\\cdot 73^2\\cdot 79^4$$. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "+0 # Number Theory 0 34 1 If $n = 2^{10} \\cdot 3^{8} \\cdot 5^{14}$, how many of the natural-number factors of n are multiples of 150? Jul 7, 2022 #1 +1105 +8 2^10 x 3^14 x 5^8 = 2^1 x 3^1 x 5^2 =150 Subtract the exponents of the same base numbers and add 1 in each case as follows: [10 - 1 + 1] x [14 - 1 + 1] x [8 - 2 + 1] =10 x 14 x 7 = 980 such numbers.", "# Count The Divisors There is a certain number $$N = n_1\\cdot n_2\\cdot n_3\\cdots n_k$$, where $$1 \\leq n_i \\leq 100$$ $$(1\\le i \\le k)$$ are all distinct positive integers. We are told the following: • $$N$$ has $$d$$ (positive) divisors and $$d$$ is odd. • $$128\\cdot N$$ has $$\\frac{18}{11}d$$ divisors. • $$175\\cdot N$$ has $$\\frac 8 5 d$$ divisors. • $$213\\cdot N$$ has $$\\frac 8 3 d$$ divisors. If $$2016\\cdot N$$ has $$\\frac A B d$$ divisors, with $$A$$ and $$B$$ coprime positive integers, submit $$A + B$$. ×", "# Factors on N^3 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? ×", "of $X^n\\times X^n$, and it is not immediately clear that the natural choice would be of finite type. Hopefully there are then natural inclusions $Y_1\\hookrightarrow Y_2\\hookrightarrow\\cdots$ giving the desired ind-scheme. -", "Select Page Number Sets # Natural Numbers Set of natural numbers: $$\\mathbb{N}$$ Set of natural numbers with zero: $$\\mathbb{N_0}$$ Natural numbers: $$n$$, $$k$$, $$a$$, $$b$$, $$c$$ Arabic numerals: $$0$$, $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, $$9$$ Roman numerals: $$I$$, $$V$$, $$X$$, $$L$$, $$C$$, $$D$$, $$M$$ 1. Natural numbers $$\\mathbb{N} = \\left\\{ {1,2,3, \\ldots } \\right\\}$$ These numbers can be used for counting and ordering. 2. Natural numbers including zero $$\\mathbb{N_0} = \\left\\{ {0,1,2,3, \\ldots } \\right\\}$$ These numbers can be used for indicating the number of objects. 3. Commutativity of addition $$a + b = b + a$$ 4. Associativity of addition $$a + \\left( {b + c} \\right) = \\left( {a + b} \\right) + c$$ 5. $$a + 0 = a$$ 6. Commutativity of multiplication $$a \\cdot b = b \\cdot a$$ 7. Associativity of multiplication $$a \\cdot \\left( {b \\cdot c} \\right) =$$ $$\\left( {a \\cdot b} \\right) \\cdot c$$ 8. Distributivity of multiplication over addition $$a \\cdot \\left( {b + c} \\right) =$$ $$a \\cdot b + a \\cdot c$$ 9. $$a \\cdot 0 = 0$$ 10. $$a \\cdot 1 = a$$ 11. Even numbers $$n = 2k,\\;\\left( {n, k \\in \\mathbb{N}} \\right)$$ 12. Odd numbers $$n = 2k + 1,\\;\\left( {n, k \\in \\mathbb{N}} \\right)$$ 13. Roman numerals 1. Roman numerals", "###### Exercise37 1. Let $$x^{7}\\cdot x^{a}=x^{17}\\text{.}$$ Let’s say that $$a$$ is a natural number. How many possibilities are there for $$a\\text{?}$$ 2. Let $$x^{b}\\cdot x^{c}=x^{75}\\text{.}$$ Let’s say that $$b$$ and $$c$$ are natural numbers. How many possibilities are there for $$b\\text{?}$$ 3. Let $$x^{d}\\cdot x^{e}=x^{800}\\text{.}$$ Let’s say that $$d$$ and $$e$$ are natural numbers. How many possibilities are there for $$d\\text{?}$$ in-context", "Free Version Moderate # Common Factor of $2^n-3$ and $3^n -2$ NUMTH-RK$YOC Let$n$be a natural number. Which of the following statements imply$5 | \\gcd(2^n-3,3^n-2)$? Select ALL that apply. A$5|n$. B$5|2^n-3$. C$n\\equiv 1 \\ \\mathrm{mod } \\ 4$. D$n\\equiv 3 \\ \\mathrm{mod} \\ 4\\$.", "# Find least possible number of factors Number $A$ has $24$ factors. Number $A\\cdot B$ has $105$ factors. Find least possible number of factors of $B$. I have tried. But there seems to be no general approach.. The answer given is $12$. • The hint given in the book is like this: Try to take as much common factors in A and B as possible. I don't see why Jan 5, 2015 at 11:07 • Hint: A number could be a factor of AB if and only if it is either a factor of A, a factor of B, or a product of two such factors. Jan 5, 2015 at 11:32 • Are there statements about the distinctness of factors? Jan 5, 2015 at 11:38 • No other information. Jan 5, 2015 at 11:41 • If by \"factors\" to mean \"non-distinct prime factors\", then $B$ has exactly $105-24$ factors (no more, no less). So I assume that you mean \"distinct prime factors\"... Is that correct? Jan 5, 2015 at 12:22 If $m$ is a number that can be factored to $$m=p_1^{e_1} \\cdot p_2^{e_2} \\cdots p_n^{e_n}$$ then $m$ has $(e_1+1)(e_2+2) \\cdots (e_n+n)$ divisors. $A$ has $24=2^3 \\cdot 3$ divisors, so $A$ can have at most $4$ prime factors, e.g. $$A=p_1^{e_1} \\cdot p_2^{e_2} \\cdot p_3^{e_3} \\cdot p_4^{e_4}$$ If $A$ would", "+0 Math 0 105 1 +182 Given that a and b are real numbers such that -3 ≤ a ≤ 1 and -2 ≤ b ≤ 4, and values for a and b are chosen at random, what is the probability that product a * B is positive? wiskdls Apr 21, 2018 #1 +963 +2 Alright, here we go: For two number's product to be positive, those numbers must be both positive or both negative, but cannot be zero. This is because zero is neither positive nor negative, and 0 * n = 0. The first step is to find the total number of combination: (a,b). This can be done by: $$(1-(-3)+1)\\cdot(4-(-2)+1) =5\\cdot7=35.$$ There are a total of 35 different combinations. Out of those combinations there are: ( -3 , -2 ) ; ( -3 , -1 ) ; ( -2 , -2 ) ; ( -2 , -1 ) ; ( -1 , -2 ) ; ( -1 , -1 ) ( 1 , 1 ) ; ( 1 , 2 ) ; ( 1 , 3 ) ; ( 1 , 4 ) 10 that work. The probabilty of this happening is: $$\\boxed{\\frac{10}{35}=\\frac{2}{7}}$$ I hope this helped, Gavin. GYanggg Apr 21, 2018 #1 +963 +2 Alright, here we go: For two number's product to be positive,", "smaller non-deficient number. He proved that there are only a finite number of primitive non-deficient numbers having a given number of distinct odd prime factors and a given number of factors $2$. There is no odd abundant number with fewer than three distinct prime factors, the primitive ones with three are \\begin{equation*} 3 ^ { 3 } .5 .7,3 ^ { 2 } .5 ^ { 2 } .7,3 ^ { 2 } .5 .7 ^ { 2 } \\end{equation*} \\begin{equation*} 3 ^ { 2 } \\cdot 5 ^ { 2 } \\cdot 11,\\; 3 ^ { 5 } \\cdot 5 ^ { 2 } \\cdot 13,\\; 3 ^ { 4 } \\cdot 5 ^ { 2 } \\cdot 13 ^ { 2 } ,\\; 3 ^ { 3 } \\cdot 5 ^ { 3 } \\cdot 13 ^ { 2 }. \\end{equation*} He gave also a table of all even abundant numbers $< 6232$. Dickson's result was a starting point for much further research. In 1949 and 1968, H.N. Shapiro ([a20], [a21]) proved the following result. Let $\\alpha$ be a rational number. A necessary and sufficient condition that there exist infinitely many primitive $\\alpha$-abundant numbers (i.e. $\\sigma ( n ) / n \\geq \\alpha$ but $\\sigma ( d ) / d < \\alpha$ for all $d |", "+0 # If , how many of the natural-number factors of are multiples of 150? 0 57 2 If $$n = 2^{10} \\cdot 3^{14} \\cdot 5^{8}$$, how many of the natural-number factors of $$n$$ are multiples of 150? I got 702, but that was wrong... Apr 6, 2020 #1 +20967 0 I'm not the best at this, but this is my attempt: 150 = 2 x 3 x 5 x 5 -- so every multiply must have at least 1 two, 1 three, and 2 fives. So, from the 210 (which consists of 10 twos), you can select either 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 twos -- 10 different selections. From 314 (which consists of 14 threes), you can select either 1, 2, 3, ... 14 threes -- 14 different selections. From 58, you must select at least 2 fives, but it could also be 3, 4, 5, 6, 7, or 8 fives -- 7 different selections. Multiplying 10 x 14 x 7, I get 980 different ways. Apr 6, 2020 #2 0 That is correct! Thank you so much, geno! Guest Apr 6, 2020", "1, 2,….,n => number of prime factors = n We have to find the number of different factors of $$4^5+4^6+4^7$$ => $$4^5+4^6+4^7$$ = $$4^5+4^5\\cdot4^1+4^5\\cdot4^2$$ => $$4^5\\left(1+4^1+4^2\\right)$$ => $$4^5\\left(1+4+16\\right)$$ => $$4^5\\cdot21$$ => $$\\left(2^2\\right)^5\\cdot3\\cdot7$$ => ∴ Prime factors 2, 3 and 7 => ∴ Number of prime factors =3 Therefore, B is the correct answer.", "# Factors on N^3 Computer Science Level 5 Let $$N = 5791^3\\cdot 5801^2\\cdot 5807^4$$ be a prime factor decomposition. How many positive integers less than $$N$$ are divisors of $$N^3$$ but not of $$N$$? × Problem Loading... Note Loading... Set Loading...", "3. ## Re: Finding factors of a number Originally Posted by otownsend I am trying to developed an algorithm which based on a given input, call it x, determines whether x is prime. One way to do this is to check all the numbers (starting at 2) up to x/2 is divisible by x. If there is such a number, then x is not prime. I understand this strategy completely. However, another way to to check all numbers (starting at 2) up to the square root of x. Can someone please explain to be why this works? If $n=4276800$ then $n$ can be factored as $2^6\\cdot 3^5\\cdot 5^2\\cdot 11$, SEE HERE. On that webpage we are told that $n$ has $252$ divisors. Here is more interesting question. Suppose $\\{p,q,r,s,t\\}$ is a set of five prime numbers, and $N=p^4\\cdot q^3\\cdot r^7\\cdot s^2\\cdot t^6$. How many factors does the number $N$ have?", "c \\leq 1$. We see that there are $4$ choices for $a$, $2$ choices for $b$, and $2$ choices for $c$. By the Multiplication principle, the number of factors is $4\\cdot 2 \\cdot 2=16$.", "it's true that $$[(n \\cdot 1) b]^k = (n^k \\cdot 1) b^k$$ for natural numbers $$n$$.", "# factors of an equation I want to write a code which boils down to equation $(x y) = (x + y) N$. My task is to find all the possible integer (natural number) solutions for $x$ and $y$. For example, say $N=6$. As $x$ and $y$ are natural numbers, $x+y$ is also a natural number, starting from $1,2,\\ldots$ That is, the product $xy$ should meet the multiples of $6$. Is there any easy way to find all the combinations of $x$ and $y$? - Since you mentioned programming, is there an upper bound on $x, y$? i.e. a limit on how large $x$ and $y$ could be? – user2468 Jul 14 '12 at 20:29 @ J.D. No sir no upper bounds – cdummy Jul 14 '12 at 20:30 @ J.D. but they are natural numbers – cdummy Jul 14 '12 at 20:30 The equation can be rewritten as $(x-N)(y-N)=N^2$. So all positive solutions come from the factorizations of $N^2$ as a product $uv$, and all such factorizations come from a solution. Just put $x=N+u$, $y=N+v$. Thus the problem of finding the solutions of your equation is essentially equivalent to the problem of finding all factors of $N^2$, which is easily solved if we know all the factors of $N$. However, for very large $N$, finding all the" ]

[ "# Question #ae79f Aug 17, 2017 1/6 if the spinner has six sections and each section is the same size with an equal probability of being selected. #### Explanation: If the spinner has six sections then the probability of any one number is one out of six. The dice has six sides and has a probability of any one number is also one out of six. This give 36 possible outcomes. six of the them have the same number 1,1 2,2 3,3 4,4 5,5 6,6. This gives 6/36 possibilities of having the same number on the spinner and the dice. 6/ 36 = 1/6. Having a spinner has no effect on probability that is different from a dice assuming the spinner has six sides with equal probability", "# Probability Three Dice Sum To 18 Probability Level 1 Suppose three fair six-sided dice are rolled. What is the probability that the sum of the numbers obtained equals 18? ×", "13 views Consider a fair coin with probability of heads and tails equal to $1 / 2$. Moreover consider two dice, first $\\mathrm{D}_{1}$ that has three faces numbered $1,3,5$ and second $\\mathrm{D}_{2}$ that has three faces numbered $2,4,6$. When rolled, for both $\\mathrm{D}_{1}$ and $\\mathrm{D}_{2}$, each of the three faces are equally likely. A random experiment is conducted as follows. First, the coin is flipped once. If it shows heads, dice $\\mathrm{D}_{1}$ is rolled once, while if the coin shows tails, $\\mathrm{D}_{2}$ is rolled once, and the experiment ends. Let $\\mathrm{X}$ be the (random) number seen on the rolled dice in the experiment. What is $\\mathbb{E}[X]$ ? 1. $\\frac{7}{2}$ 2. 4 3. 3 4. $\\frac{9}{2}$ 5. None of the above", "sum to 12 and three sum to 18.", "Theoretical Probability Odds and Probability Playing Cards Probability Probability and Playing Cards Solved Probability Problems Probability for Rolling Three Dice", "Answer: Distribute 6 as 3 + 3. 8 × 6 = 8 × (3 + 3) 8 × 6 = (8 × 3) + (8 × 3) 8 × 6 = 24 + 24 8 × 6 = 48 Question 8. Answer: Any number multiplied by 0 is always 0. So, 6 × 0 = 0 Question 9. Answer: Any number multiplied by 1 is always the same number. 6 × 1 = 6 Question 10. Answer: Distribute 6 as 3 + 3. 10 × 6 = 10 × (3 + 3) 10 × 6 = (10 × 3) + (10 × 3) 10 × 6 = 30 + 30 10 × 6 = 60 Compare Question 11. Answer: < Explanation: 4 × 6 = 24 6 × 6 = 36 24 < 36 Thus 4 × 6 < 6 × 6 Question 12. Answer: > Explanation: 6 × 5 = 30 4 × 5 = 20 30 > 20 Thus 6 × 5 > 4 × 5 Question 13. Answer: = Explanation: 6 × 7 = 42 42 = 42 Thus 42 = 6 × 7 Question 14. There are 6 faces on a standard die. How many faces are on 5 dice? Answer: 30 faces Explanation: Given that, There are 6 faces on a standard die.", "# Probability Three Dice Sum To 18 Probability Level 1 Suppose three fair six-sided dice are rolled. What is the probability that the sum of the numbers obtained equals 18? × Problem Loading... Note Loading... Set Loading...", "• Foxythepirate0984 Identify the resulting probability as theoretical or experimental. The outcomes for rolling a number cube with faces labeled 1− 6 are equally likely. Find the probability that the number rolled is at least a 4. A. The experimental probability is . B. The experimental probability is . C. The theoretical probability is . D. The theoretical probability is . Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "Two dices are rolled. If both dices Question: Two dices are rolled. If both dices have six faces numbered $1,2,3,5,7$ and 11 , then the probability that the sum of the numbers on the top faces is less than or equal to 8 is : 1. (1) $\\frac{4}{9}$ 2. (2) $\\frac{17}{36}$ 3. (3) $\\frac{5}{12}$ 4. (4) $\\frac{1}{2}$ Correct Option: , 2 Solution: $n(E)=5+4+4+3+1=17$ So, $P(E)=\\frac{17}{36}$", "flips it 4 times, getting all heads. If he flips this coin again, what is the probability it will be heads? (The answer value will be from 0 to 1, not as a percentage.) Two fair dice are rolled, and it is revealed that (at least) one of the numbers rolled was a 4. What is the probability that the other number rolled was a 6? Note: You are not told which of the numbers rolled is a 4. ×", "drawn is a face card (Jack, Queen and King only)? 2 / 13 2 / 13 5 / 13 5 / 13 3 / 13 3 / 13 1 / 13 1 / 13 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 5 Time: 00:00:00 Three unbiased coins are tossed. What is the probability of getting at least 2 heads? 1 / 2 1 / 2 1 / 3 1 / 3 2 / 3 2 / 3 3 / 4 3 / 4 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 6 Time: 00:00:00 Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6. 7 / 18 7 / 18 14 / 35 14 / 35 8 / 17 8 / 17 none of the above none of the above Once you attempt the question then PrepInsta explanation will be displayed. Start Question 7 Time: 00:00:00 What is the probability of getting 53 Mondays in a leap year? 1 / 7 1 / 7 3 / 7 3 / 7 2 / 7 2 / 7 1 1 Once you attempt the question then PrepInsta explanation will be displayed. Start Question 8", "1. ## Probability Can anyone help me with the following probability question please: 18 fair dice are thrown and the uppermost faces are investigated. What is the probability of each face appearing 3 times? (That is the outcome will be {1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6}. Many thanks to anyone who is able to help 2. Hello, The probability of each face appearing once is 3/18 = 1/6 (equiprobability) If you want this face appear 3 times, it's $P( face \\cap face \\cap face)$. Because the throws are independent, this probability equals to P(face)^3, that is to say (1/6)^3 3. Does this mean then that the probability of each face appearing three times is: (1/6)^18? (same as (1/6^3) 4. Originally Posted by smiler 18 fair dice are thrown and the uppermost faces are investigated. What is the probability of each face appearing 3 times? (That is the outcome will be (1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6). The event (1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6) is one 18-tuple out of $6^{18}$ possible outcomes. But that selection of individual outcomes can occur in $\\frac {18!}{(3!)^6}$ ways. So the overall probability is $\\frac {18!}{6^{18}(3!)^6}$. 5. Woops, am really sorry, i misunderstood the \"18 fair dice\", thought it was a 18-faces dice :s", "HomeEnglishClass 12MathsChapterProbability If three six faces fair dice a... # If three six faces fair dice are thrown together, then the probability that the sum of the numbers appearing on the dice is (p le k le 14) is (k^(2)-21k+83)/(216)(k^(2)+21k+83)/(216)(21k-k^(2)-83)/(216)(21k-k^(2)+83)/(216)", "eight numbers that sum to 28; if the front face sums to $k$ then the back face sums to $28-k$.", "is a Six-sided cube with dots representing the numbers 1 through 6. What is the probability of rolling a 3?​... Can we use all lands for different uses?​ Can we use all lands for different uses?​...", "# No face. But a 5 Level pending An experiment consists of randomly drawling a card from a standard deck of cards (no jokers) and rolling a regular dice (6 sides). Find the probability that the card is NOT a face card and the die shows a five. Note: The solution can be expressed as $$\\frac{a}{b}$$. Express your answer as $$a + b$$. ×", "+0 # Probability 0 30 1 If three, standard, 6-faced dice are rolled, what is the probability that the sum of the face up integers is 17? Jul 3, 2022 #1 +2275 +1 There are $$6 \\times 6 \\times 6 = 216$$ possibilities. The only way to get a sum of 17 is to have a 6, 6, and 5, which can be done in $$3! \\div 2 = 3$$ ways (divide by 2 because there are 2 6's) So, the probability is $${3 \\over 216} = \\color{brown}\\boxed{{1 \\over 72}}$$ Jul 3, 2022", "1. Dice If 2 six sided dice are tossed. What is the probability that the sum is 2 or 12? 2. Originally Posted by gretchen If 2 six sided dice are tossed. What is the probability that the sum is 2 or 12? It is the probability of 2 PLUS the probability of 12. The only possible case of getting a 2 is {1,1}. The probability of doubles 1's is 1/36=(1/6)(1/6). Similiarly the probability of getting a 12 is 1/36. The sum is, 1/36+1/36=2/36=1/18", "Browse Questions A coin and six faced die, both unbiassed, are thrown simultaneously. The probability of getting a head on the coin and an odd number on the die, is : $\\begin {array} {1 1} (a)\\frac{1}{2} & \\quad (b)\\;\\frac{3}{4} \\\\ (c)\\;\\frac{1}{4} & \\quad (d)\\;\\frac{2}{3} \\end {array}$ +1 vote $(c)\\;\\frac{1}{4}$", "Browse Questions # Two fair dice are rolled. The probability of the sum of digits on their faces to be greater that or equal to 10 is $(a)\\;\\frac{1}{5} \\quad (b)\\;\\frac{1}{4} \\quad (c)\\;\\frac{1}{8} \\quad (d)\\;\\frac{1}{6}$ $(d)\\;\\frac{1}{6}$" ]

[ "Answer: Distribute 6 as 3 + 3. 8 × 6 = 8 × (3 + 3) 8 × 6 = (8 × 3) + (8 × 3) 8 × 6 = 24 + 24 8 × 6 = 48 Question 8. Answer: Any number multiplied by 0 is always 0. So, 6 × 0 = 0 Question 9. Answer: Any number multiplied by 1 is always the same number. 6 × 1 = 6 Question 10. Answer: Distribute 6 as 3 + 3. 10 × 6 = 10 × (3 + 3) 10 × 6 = (10 × 3) + (10 × 3) 10 × 6 = 30 + 30 10 × 6 = 60 Compare Question 11. Answer: < Explanation: 4 × 6 = 24 6 × 6 = 36 24 < 36 Thus 4 × 6 < 6 × 6 Question 12. Answer: > Explanation: 6 × 5 = 30 4 × 5 = 20 30 > 20 Thus 6 × 5 > 4 × 5 Question 13. Answer: = Explanation: 6 × 7 = 42 42 = 42 Thus 42 = 6 × 7 Question 14. There are 6 faces on a standard die. How many faces are on 5 dice? Answer: 30 faces Explanation: Given that, There are 6 faces on a standard die.", "pick the value for the remaining dice in $3$ ways, giving $6\\cdot 35\\cdot 3$, same answer as I gave in previous comment. – JMoravitz Feb 13 '18 at 18:39 • Thank you very much! Both ways you described make sense to me and that was very helpful – Byron Feb 13 '18 at 18:43 To add to JMoravitz's comments, a way of thinking of this is that there are $3$ choices for the even face and $3$ choices for the odd face. Then, out of the $8!$ ways to order the $8$ dice, since the even and odd faces are indistinguishable, we have $$\\frac{8!}{4!4!}$$ different orderings. Hence there are $$3\\cdot3\\cdot\\frac{8!}{4!4!} = 630$$ ways for this to happen, and so a probability of $$\\frac{630}{6^8} = \\frac{35}{93312} \\approx 0.00037508573.$$", "or Tail is equally likely to occur. When a dice is thrown, all the six faces (1, 2, 3, 4, 5, 6) are equally likely to occur. • ## Mutually Exclusive Events: • Two or more than two events are said to be mutually exclusive if the occurrence of one of the events excludes the occurrence of the other This can be better illustrated with the following examples 1.When a coin is tossed, we get either Head or Tail. Head and Tail cannot come simultaneously. Hence occurrence of Head and Tail are mutually exclusive events. 2.When a die is rolled, we get 1 or 2 or 3 or 4 or 5 or 6. All these faces cannot come simultaneously. Hence occurrences of particular faces when rolling a die are mutually exclusive events. Note : If A and B are mutually exclusive events, $A \\cap B$ = ϕ where ϕ represents empty set. 3.Consider a die is thrown and A be the event of getting 2 or 4 or 6 and B be the event of getting 4 or 5 or 6. Then A = {2, 4, 6} and B = {4, 5, 6} Here$A \\cap B$ ≠ϕ. Hence A and B are not mutually exclusive events. • ## Independent Events: • Events can be said to be independent", "Browse Questions A coin and six faced die, both unbiassed, are thrown simultaneously. The probability of getting a head on the coin and an odd number on the die, is : $\\begin {array} {1 1} (a)\\frac{1}{2} & \\quad (b)\\;\\frac{3}{4} \\\\ (c)\\;\\frac{1}{4} & \\quad (d)\\;\\frac{2}{3} \\end {array}$ +1 vote $(c)\\;\\frac{1}{4}$", "1, 5 five ways 7: 6, 1; 5, 2; 4, 3, 3, 4, 2, 5, 1, 6 6 ways 8: 6, 2; 5, 3; 4, 4, 3, 5, 2, 6 five ways 9: 6, 3; 5, 4, 4, 5, 3, 6 four ways 10: 6, 4; 5, 5; 4, 6 three ways 11: 6, 5; 5, 6 two ways 12: 6, 6 one way A total of $6\\times 6= 36$ ways. The even numbers are 2: one way, 4: three ways, 6: five ways, 8: five ways, 10: three ways, 12: one way, a total of 1+ 3+ 5+ 5+ 3+ 1= 18 ways. Of those five give a total of 8: the probability that a pair of dice, given that the sum is even, have a sum of 8 is 5/18. Once again, I didn't read all of the posts carefully- Plato has already suggested this and the OP did it properly.", "ways to get $21-4=17$. Number of ways to get $5$ is $6$ i.e. $\\dbinom{4}{2}$, which is the same as the number of ways to get $21-5=16$. Number of ways to get $6$ is $10$ i.e. $\\dbinom{5}{2}$, which is the same as the number of ways to get $21-6=15$. Number of ways to get $7$ is $15$ i.e. $\\dbinom{6}{2}$, which is the same as the number of ways to get $21-7=14$. Number of ways to get $8$ is $21$ i.e. $\\dbinom{7}{2}$, which is the same as the number of ways to get $21-8=13$. Number of ways to get $9$ is $25$ i.e. $\\dbinom{8}{2}-3$, which is the same as the number of ways to get $21-9=12$. Number of ways to get $10$ is $27$ i.e. $\\dbinom{9}{2} - 3 - 6$, which is the same as the number of ways to get $21-10=11$. Now by symmetry you can get the number of ways to get the sum between $11$ and $18$. Hence, the distribution peaks at $10$ and $11$ as expected. As a sanity check, we have $2 \\left( 1 + 3 + 6 +10 +15 + 21 + 25 + 27\\right) = 216$. Below is the distribution of the number of times a number occurs as a sum of three dices, where the $X$-axis the sum of the three dice", "May 15 '18 at 8:39 • okk so $\\binom{6}{3}$ for which $3$ dice and $\\binom{6}{1}$ for which value among 1 to 6.right ? – laura May 15 '18 at 8:41 • @drhab I may be wrong but for sets of 6 elements of which 3 are equal shouldn't the possible permutations be $\\frac{6!}{3! \\times 1! \\times 1! \\times 1!} = \\frac{6!}{3!}$? – Dav2357 May 15 '18 at 8:44 • there is a misunderstanding among us. By $\\binom63$ I am not thinking about $3$ different faces of dice but of $3$ spots for the dice that have equal faces. If e.g. the outcome is 233534 then these spots are $2,3$ and $5$. – drhab May 15 '18 at 8:44 One may see it as follows: • There are 6 options for the equal faces • $\\binom{6}{3}$ choices, which 3 dice show the same face • $5\\cdot 4 \\cdot 3$ choices for the mutually different faces of the other three • all together there are $6^6$ different throws $$\\frac{6\\cdot \\binom{6}{3}\\cdot 5\\cdot 4\\cdot 3}{6^6}= \\frac{25}{162}$$", "how many outcomes correspond to this: There are $3$ options for the blue die (4,5,6), similarly $3$ options for the green die, and $3$ options for the red die. This gives $3\\cdot 3\\cdot 3=27$ total possibilities. Out of $216$ total outcomes in the sample space this means our probability is $\\frac{27}{216}=\\frac{1}{8}$ The same logic and calculations apply for finding the probability of getting an result of $7$. Now, for an outcome of $6$, we need the red die to be a six and the blue and green dice to have at least one low number. Ignoring needing at least one low number, there are $6\\cdot 6\\cdot 1=36$ possibilities. Removing the bad possibilities where both numbers were actually high (of which there are $3\\cdot 3\\cdot 1=9$ of them) we have a total of $36-9=27$ of them. This occurs then with probability $\\frac{27}{216}=\\frac{1}{8}$ The logic and calculations is similar for all other possible results. Thus, all results $1-8$ are equally likely with a probability of $\\frac{27}{216}$ and there is no need to reroll. • Another way of doing generating this with three throws is by treating the dice as coins(e.g. 1-3 as tails, 4-6 as head). Then there are $2^3$ equally likely sequences. But your idea is better since you do not need to remembering the order of the throws.", "3. 13150 4. 4. 43074 2 No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways. Thus 6 * 5 * 4 = 120 favourable cases. The total cases are 6 * 6 * 6 = 216. The probability = 120/216 = 5/9. ##### Q.- A bag contains 7 green and 8 white balls. If two balls are drawn simultaneously the probability that both are of the same colour is -. 1. 1. 42217 2. 2. 42857 3. 3. 42309 4. 4. 42186 4 Drawing two balls of same color from seven green balls can be done in 7C2 ways. Similarly from eight white balls two can be drawn in 8C2 ways. P = 7C2/15C2 + 8C2/15C2 = 7/15 ##### Q.- If two dice are thrown together the probability of getting an even number on one die and an odd number on the other is -. 1. 1. 42826 2. 2. 42767 3. 3. 42828 4. 4. 42858 2 The number of exhaustive outcomes is 36. Let E be the", "that you could have done yourself is to think this through with a two-sided die (usually called a coin :)) to see where your solution goes wrong. (You would get 12/8, which is obviously wrong, and listing all 8 possibilities is not a problem.) Jul 4 at 18:15 • Your $6$ chooses for what the mathing faces and $5$ for what the non matching face for $6\\times 5$ possible values. But you choosing the ways to arrange the dice is wrong. As the dice are not all different you must a) either divide by ways to arrange the two faces that are the same to get $\\frac {3!}{2!} = 3$. Or b) come up with a simpler model. You only need to choose where the different face go-- the other two will be in the two remaining places. So $3$. The answer is $\\frac{6\\cdot 5\\cdot\\frac{3!}{2!}}{3^6} = \\frac{6\\cdot 5\\cdot 3}{3^6}$. Jul 5 at 3:12 • To illustrate: Suppose you have two $6$ and $5$ and the two sixes are red and blue. You have $3!=6$ ways to do this: $\\color{red}6\\color{blue}65;\\color{blue}6\\color{red}6;\\color{red}65\\color{blue}6;\\color{blue}65\\color{red}6;5\\color{red}6\\color{blue}6;5\\color{blue}6\\color{red}6$. but the switching the red an blue positions is supposed to be the same so we must divide by the numbers of way to place the blue and red die in their two positions (which is $2!$)... Alternatively if", "+0 # Probability 0 30 1 If three, standard, 6-faced dice are rolled, what is the probability that the sum of the face up integers is 17? Jul 3, 2022 #1 +2275 +1 There are $$6 \\times 6 \\times 6 = 216$$ possibilities. The only way to get a sum of 17 is to have a 6, 6, and 5, which can be done in $$3! \\div 2 = 3$$ ways (divide by 2 because there are 2 6's) So, the probability is $${3 \\over 216} = \\color{brown}\\boxed{{1 \\over 72}}$$ Jul 3, 2022", "the ball drawn is white, what is the probability that the dice had turned up with a red face? [5] (b) Five dice are thrown simultaneously if the occurrence of an odd number in a single dice is considered a success, find the probability of maximum three successes. [5] Solution 12: (a) Considered the following events: E1, E2, E3 be pertaining to the red-faced dice yellow face and green-faced die P(E1) = $\\frac { 3 }{ 6 }$ = $\\frac { 1 }{ 2 }$ Probability of getting white ball from Box I = $\\frac { 2 }{ 5 }$ Probability of getting a white ball from box II = $\\frac { 4 }{ 5 }$ Probability of getting a white ball from box II = $\\frac { 3 }{ 7 }$ Using Baye’s theorem Required probability Read Next 👇 Click on Page Number Given Below 👇 $${}$$", "Browse Questions The probability that in a throw of 12 dice 1,1, 2,2, 3,3, 4,4, 5,5, 6,6 will occur in any sequence is : $\\begin {array} {1 1} (1)\\;\\large\\frac{12!}{2^66^{12}} & \\quad (2)\\;\\large\\frac{12!}{6^62^{12}} \\\\ (3)\\;\\large\\frac{12!}{2^66^{6}} & \\quad (4)\\;None\\: of \\: these \\end {array}$ Can you answer this question? There are 6 faces on each of 12 dice. Hence the total number of ways in which all dice can fall is $6^{12}$ Now the number of way in which 1,1, 2,2,....6,6 can be arranged on dice = $\\large\\frac{12!}{(2!)^6}$ Hence the required probability is $\\large\\frac{12!}{2^66^{12}}$ Ans : (2) answered Jan 2, 2014", "probability that a box containing 15 oranges, out of which 12 are good and three are bad ones, will be approved for sale. Solution: Box contains 12 good and 3 bad oranges. If three oranges selected are good ones, the box is approved for sale. 3 good oranges may be selected in 12C3 ways. 3 oranges out of 15 may be selected in 15C3 ways. Probability that a box is approved = Probability of selecting 3 good oranges = $$\\frac{{ }^{12} C_{3}}{{ }^{15} C_{3}}$$ = $$\\frac{12×11×10}{15×14×13}$$ = $$\\frac{44}{91}$$. Question 4. A fair coin and an unbiased dice are tossed. Let A be the event ‘head appears on the coin’ and B be the event 3 on dice. Check whether A and B are independent events or not. Solution: When a coin is thrown, head or tail will occur. ∴ Probability of getting head P(A) = $$\\frac{1}{2}$$. when a dice is tossed 1, 2, 3, 4, 5, 6 one of them will appear. ∴ Probability of getting 3 = P(B) = $$\\frac{1}{6}$$. When a coin and a dice are tossed, total number of cases are H1, H2, H3, H4, H5, H6 T1, T2, T3, T4, T5, T6 Head and 3 will occur only in 1 way. ∴ Probability of getting head and 3 = $$\\frac{1}{12}$$. i.e; P(A ∩ B)", "none, one, or more heads. In a single throw of two dice, find the probability that neither a doublet nor a total 9 will appear. Two dice are thrown simultaneously. R Ashwin labeled the conditions in the capital city \"scary\". Number of coins per throw: Number of throws:. I got a Doctor card in Path of Exile when I first entered the Spider Forest map. To find the probability determine the number of successful outcomes divided by the number of When three dice are thrown simultaneously/randomly The probability, then, of rolling a 4 in at. Find the probability of guessing exactly 2 answer correctly. In Hannah's purse there are three £1 coins, five 10p coins and eight 2p coins. An unbiased six-sided dice has the number 'I' on one face, the number '2' on two faces and the number '3' on three faces. Ask for details. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. 8 If the card turned up is an honour then ipso facto the dealer has one honour and the probability that the remaining players have each an assigned one of the three remaining honours, is 50 Which probability is to be multiplied by 3!, 51 as there are. To", "the probability of getting a sum 9 in a single throw. Solution: Total number of ways in which two dice may be thrown = 6 x 6 = 36 ∴ n(S) = 36 Event of getting sum 9 is A = {(3, 6), (4, 5), (5, 4), (6, 3)} ∴ n(A) = 4 ∴ Required probability P(A) = $$\\frac { n(A) }{ n(S) }$$ = $$\\frac { 1 }{ 9 }$$ Question 9. A bag contains 8 black, 6 white and 5 red balls. Find the probability of drawing a black or a white ball from it Solution: Total number of balls = 8 + 6 + 5 = 19 ∴ n(S) = 19 Event A of drawing 1 black or 1 white ball n(A) = 8 + 6 = 14 ∴ n(A) = 14 ∴ Required probability P(A) = $$\\frac { n(A) }{ n(S) }$$ = $$\\frac { 14 }{ 19 }$$ Question 10. From a pack of well shuffled cards two cards drawn simultaneously. Find the probability that both the cards are ace. Solution: Total number of ways of drawing two cards out of 52 = 52C2 Number of ways drawing two ace out of 4 ace = 4C2 ∴ Required Probability Question 11. A pair of dice are thrown. Find the probability that the sum", "The outcome of our experiment is a triple $(a,b,c)$, where $a$, $b$, $c$ are the faces showed by the first, the second and the third dice respectively. There are $N = 6^3$ such triples. In addition, there are $6 \\cdot 5 \\cdot 4$ ways of choosing $(a,b,c)$ such that no face is repeated therein, so $P(D) = \\frac {6 \\cdot 5 \\cdot 4}{N}$. Let us now consider the event $S \\cap D$ (i.e. one dice shows six, and all of them have different outcomes); there are 3 ways to choose which dice shows six, and $5 \\cdot 4$ to select the faces of the other two, in total: $3 \\cdot 5 \\cdot 4$ cases, which yields $P(S \\cap D) = \\frac{3 \\cdot 5 \\cdot 4}{N} > 0$. Finally, we obtain the probability in question: $$P(S|D) = \\frac {P(S \\cap D)}{P(D)} = \\frac {3 \\cdot 5 \\cdot 4}{6 \\cdot 5 \\cdot 4} = 0.5.$$ • I edited my post, providing more details. Is it OK? – Pythagoricus Oct 23 '16 at 0:34 • Yes, it's OK now. I haven't checked whether it is correct, but it's definitely an answer now. – Daniel Fischer Oct 23 '16 at 9:54 Hint: if the three dice show different faces after being rolled, then surely three different numbers are obtained. What is the", "with this $$2$$. This adds $$\\frac 7{216}\\cdot \\frac 56$$ to the chance you get a $$2$$ and subtracts $$\\frac 7{216}\\cdot \\frac 16$$ from the chance of each other number. With more dice it is even more of a mess.", "What is the probability that the number of heads is equal to the number showing on the die? For this question, I know how to calculate the sample space, but I'm not sure how to use permutations or combinations. Here is what I have so far. Can anyone please help me out? Suppose we roll one fair six-sided die, and flip six coins. What is the probability that the number of heads is equal to the number showing on the die? $S = 6 + 2^6 = 70$ $P(1H, 1) = \\frac{7!}{5!2!*70}$ $P(2H, 2) = \\frac{7!}{4!3!70}$ $P(3H, 3) = \\frac{7!}{4!3!*70}$ $P(4H, 4) = \\frac{7!}{5!2!*70}$ $P(5H, 5) = \\frac{7!}{6!*70}$ $P(6H, 6) = \\frac{7!}{70}$ $Total = P(1H, 1) + P(2H, 2) + P(3H, 3) +P(4H, 4) +P(5H, 5) + P(6H, 6)$ • For a start, $S=6\\cdot 2^6 = 384$. Then you can't arrange the die into the coins, so you should be choosing your head-facing coins from $6$, not $7$ – Joffan May 12 '18 at 22:35 2 Answers The sample space is the product of the sample spaces for the dice and the coins. If you order the coins the space is $6 \\cdot 2^6=384$ Now consider the number of ways to get each total of heads from $1$ to $6$. They should add to $63$ because you don't", "# Question #ae79f Aug 17, 2017 1/6 if the spinner has six sections and each section is the same size with an equal probability of being selected. #### Explanation: If the spinner has six sections then the probability of any one number is one out of six. The dice has six sides and has a probability of any one number is also one out of six. This give 36 possible outcomes. six of the them have the same number 1,1 2,2 3,3 4,4 5,5 6,6. This gives 6/36 possibilities of having the same number on the spinner and the dice. 6/ 36 = 1/6. Having a spinner has no effect on probability that is different from a dice assuming the spinner has six sides with equal probability" ]

[ "Answer: Distribute 6 as 3 + 3. 8 × 6 = 8 × (3 + 3) 8 × 6 = (8 × 3) + (8 × 3) 8 × 6 = 24 + 24 8 × 6 = 48 Question 8. Answer: Any number multiplied by 0 is always 0. So, 6 × 0 = 0 Question 9. Answer: Any number multiplied by 1 is always the same number. 6 × 1 = 6 Question 10. Answer: Distribute 6 as 3 + 3. 10 × 6 = 10 × (3 + 3) 10 × 6 = (10 × 3) + (10 × 3) 10 × 6 = 30 + 30 10 × 6 = 60 Compare Question 11. Answer: < Explanation: 4 × 6 = 24 6 × 6 = 36 24 < 36 Thus 4 × 6 < 6 × 6 Question 12. Answer: > Explanation: 6 × 5 = 30 4 × 5 = 20 30 > 20 Thus 6 × 5 > 4 × 5 Question 13. Answer: = Explanation: 6 × 7 = 42 42 = 42 Thus 42 = 6 × 7 Question 14. There are 6 faces on a standard die. How many faces are on 5 dice? Answer: 30 faces Explanation: Given that, There are 6 faces on a standard die.", "faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "1. ## Probability Can anyone help me with the following probability question please: 18 fair dice are thrown and the uppermost faces are investigated. What is the probability of each face appearing 3 times? (That is the outcome will be {1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6}. Many thanks to anyone who is able to help 2. Hello, The probability of each face appearing once is 3/18 = 1/6 (equiprobability) If you want this face appear 3 times, it's $P( face \\cap face \\cap face)$. Because the throws are independent, this probability equals to P(face)^3, that is to say (1/6)^3 3. Does this mean then that the probability of each face appearing three times is: (1/6)^18? (same as (1/6^3) 4. Originally Posted by smiler 18 fair dice are thrown and the uppermost faces are investigated. What is the probability of each face appearing 3 times? (That is the outcome will be (1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6). The event (1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6) is one 18-tuple out of $6^{18}$ possible outcomes. But that selection of individual outcomes can occur in $\\frac {18!}{(3!)^6}$ ways. So the overall probability is $\\frac {18!}{6^{18}(3!)^6}$. 5. Woops, am really sorry, i misunderstood the \"18 fair dice\", thought it was a 18-faces dice :s", "There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "Browse Questions A coin and six faced die, both unbiassed, are thrown simultaneously. The probability of getting a head on the coin and an odd number on the die, is : $\\begin {array} {1 1} (a)\\frac{1}{2} & \\quad (b)\\;\\frac{3}{4} \\\\ (c)\\;\\frac{1}{4} & \\quad (d)\\;\\frac{2}{3} \\end {array}$ +1 vote $(c)\\;\\frac{1}{4}$", "or Tail is equally likely to occur. When a dice is thrown, all the six faces (1, 2, 3, 4, 5, 6) are equally likely to occur. • ## Mutually Exclusive Events: • Two or more than two events are said to be mutually exclusive if the occurrence of one of the events excludes the occurrence of the other This can be better illustrated with the following examples 1.When a coin is tossed, we get either Head or Tail. Head and Tail cannot come simultaneously. Hence occurrence of Head and Tail are mutually exclusive events. 2.When a die is rolled, we get 1 or 2 or 3 or 4 or 5 or 6. All these faces cannot come simultaneously. Hence occurrences of particular faces when rolling a die are mutually exclusive events. Note : If A and B are mutually exclusive events, $A \\cap B$ = ϕ where ϕ represents empty set. 3.Consider a die is thrown and A be the event of getting 2 or 4 or 6 and B be the event of getting 4 or 5 or 6. Then A = {2, 4, 6} and B = {4, 5, 6} Here$A \\cap B$ ≠ϕ. Hence A and B are not mutually exclusive events. • ## Independent Events: • Events can be said to be independent", "# Question #ae79f Aug 17, 2017 1/6 if the spinner has six sections and each section is the same size with an equal probability of being selected. #### Explanation: If the spinner has six sections then the probability of any one number is one out of six. The dice has six sides and has a probability of any one number is also one out of six. This give 36 possible outcomes. six of the them have the same number 1,1 2,2 3,3 4,4 5,5 6,6. This gives 6/36 possibilities of having the same number on the spinner and the dice. 6/ 36 = 1/6. Having a spinner has no effect on probability that is different from a dice assuming the spinner has six sides with equal probability", "correct answer. 12. When a fair six-sided dice is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the faces that can be seen is divisible by $$6?$$ $$\\frac 13$$ $$\\frac 12$$ $$\\frac 23$$ $$\\frac 56$$ $$1$$ ###### Solution(s): Note that if $$6$$ is not the bottom case, then the product is automatically divisible by $$6$$ since $$6$$ is a factor. If $$6$$ is on the bottom, then $$2$$ and $$3$$ are seen, also ensuring that the product is divisible by $$6.$$ Therefore, every possible scenario allows for the product to be divisible by $$6.$$ Thus, E is the correct answer. 13. Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted purple. The figure is then separated into individual cubes. How many of the individual cubes have exactly four purple faces? $$4$$ $$6$$ $$8$$ $$10$$ $$12$$ ###### Solution(s): The $$4$$ cubes on top have $$5$$ exposed faces, so they don't work. The $$4$$ corners in the bottom row have $$3$$ exposed sides, so they don't work either. Every other cube has $$4$$ exposed sides, so those work. There are $$14 - 4 - 4 = 6$$ cubes that work. Thus, B is the", "+0 # Probability 0 30 1 If three, standard, 6-faced dice are rolled, what is the probability that the sum of the face up integers is 17? Jul 3, 2022 #1 +2275 +1 There are $$6 \\times 6 \\times 6 = 216$$ possibilities. The only way to get a sum of 17 is to have a 6, 6, and 5, which can be done in $$3! \\div 2 = 3$$ ways (divide by 2 because there are 2 6's) So, the probability is $${3 \\over 216} = \\color{brown}\\boxed{{1 \\over 72}}$$ Jul 3, 2022", "pick the value for the remaining dice in $3$ ways, giving $6\\cdot 35\\cdot 3$, same answer as I gave in previous comment. – JMoravitz Feb 13 '18 at 18:39 • Thank you very much! Both ways you described make sense to me and that was very helpful – Byron Feb 13 '18 at 18:43 To add to JMoravitz's comments, a way of thinking of this is that there are $3$ choices for the even face and $3$ choices for the odd face. Then, out of the $8!$ ways to order the $8$ dice, since the even and odd faces are indistinguishable, we have $$\\frac{8!}{4!4!}$$ different orderings. Hence there are $$3\\cdot3\\cdot\\frac{8!}{4!4!} = 630$$ ways for this to happen, and so a probability of $$\\frac{630}{6^8} = \\frac{35}{93312} \\approx 0.00037508573.$$", "1, 5 five ways 7: 6, 1; 5, 2; 4, 3, 3, 4, 2, 5, 1, 6 6 ways 8: 6, 2; 5, 3; 4, 4, 3, 5, 2, 6 five ways 9: 6, 3; 5, 4, 4, 5, 3, 6 four ways 10: 6, 4; 5, 5; 4, 6 three ways 11: 6, 5; 5, 6 two ways 12: 6, 6 one way A total of $6\\times 6= 36$ ways. The even numbers are 2: one way, 4: three ways, 6: five ways, 8: five ways, 10: three ways, 12: one way, a total of 1+ 3+ 5+ 5+ 3+ 1= 18 ways. Of those five give a total of 8: the probability that a pair of dice, given that the sum is even, have a sum of 8 is 5/18. Once again, I didn't read all of the posts carefully- Plato has already suggested this and the OP did it properly.", "# No face. But a 5 Level pending An experiment consists of randomly drawling a card from a standard deck of cards (no jokers) and rolling a regular dice (6 sides). Find the probability that the card is NOT a face card and the die shows a five. Note: The solution can be expressed as $$\\frac{a}{b}$$. Express your answer as $$a + b$$. ×", "13 views Consider a fair coin with probability of heads and tails equal to $1 / 2$. Moreover consider two dice, first $\\mathrm{D}_{1}$ that has three faces numbered $1,3,5$ and second $\\mathrm{D}_{2}$ that has three faces numbered $2,4,6$. When rolled, for both $\\mathrm{D}_{1}$ and $\\mathrm{D}_{2}$, each of the three faces are equally likely. A random experiment is conducted as follows. First, the coin is flipped once. If it shows heads, dice $\\mathrm{D}_{1}$ is rolled once, while if the coin shows tails, $\\mathrm{D}_{2}$ is rolled once, and the experiment ends. Let $\\mathrm{X}$ be the (random) number seen on the rolled dice in the experiment. What is $\\mathbb{E}[X]$ ? 1. $\\frac{7}{2}$ 2. 4 3. 3 4. $\\frac{9}{2}$ 5. None of the above", "+0 # probability 0 21 2 If three, standard, 6-faced dice are rolled, what is the probability that the sum of the face up integers is 18? Aug 2, 2022 #1 +2270 +1 How many ways can you roll a sum of 18 with 3 dice? How many ways can you roll the 3 dice? Aug 2, 2022 #2 +113 +2 There are 6 possible outcomes for each dice, so in every throw You have 6 x 6 x 6 = 216 possible outcomes. There is only one way the sum of three dice can become 18, that is if they all turn up as 6. So the probability of getting 18 as a sum in just one throw is $$\\frac{1}{216} = 0,00462962962962963 \\approx 0.0046$$ Aug 3, 2022", "Two dices are rolled. If both dices Question: Two dices are rolled. If both dices have six faces numbered $1,2,3,5,7$ and 11 , then the probability that the sum of the numbers on the top faces is less than or equal to 8 is : 1. (1) $\\frac{4}{9}$ 2. (2) $\\frac{17}{36}$ 3. (3) $\\frac{5}{12}$ 4. (4) $\\frac{1}{2}$ Correct Option: , 2 Solution: $n(E)=5+4+4+3+1=17$ So, $P(E)=\\frac{17}{36}$", "None of the above 9. The six faces of the dice are called equally likely if the dice is (a) Small (b) Fair (c) Six-faced (d) Round 10. A letter is chosen at random from the word “STATISTICS”. The probability of getting a vowel is (a) $\\frac { 1 }{ 10 }$ (b) $\\frac { 2 }{ 10 }$ (c) $\\frac { 3 }{ 10 }$ (d) $\\frac { 4 }{ 10 }$", "$\\frac{1}{n}$. This confirms our answer of 1/3 from above for $n=3$. Notice that for larger cubes, the probability of getting a painted piece approaches zero. This makes sense, because there are so many interior cubes that have no paint on them at all; while there is only 1 interior cube in a 3x3x3, or 1/27 of the cubes, there are $8^3=512$ interior cubes in a 10x10x10 – over half the cubes are interior! ## Most likely sum of two and three dice (part 2) Part 1Part 2 | Part 3 As reference, below are the tables for two and three dice possibilities, adapted from the previous post. $\\begin{array}{ccc}x&d_2(x)&d_3(x)\\\\\\hline 2&1&0\\\\3&2&1\\\\4&3&3\\\\5&4&6\\\\6&5&10\\\\7&6&15\\\\8&5&21\\\\9&4&25\\\\10&3&27\\\\11&2&27\\\\12&1&25\\\\13&&21\\\\14&&15\\\\15&&10\\\\16&&6\\\\17&&3\\\\18&&1\\end{array}$ Before you stare at the top row of the table cluelessly, let me mention that the $d$s are just a convenient shorthand that I made up 20 seconds ago: $d_3(4)$ means the number of ways to get a sum of 4 with three dice, and $d_2(x-1)$ is the number of ways to get a sum of $x-1$ with two dice. It’s truly a relief to compress a 13-word phrase into a simple symbol. Extensions Just like the two dice case, the number of possibilities in the three-dice case is symmetric. This seems amazing, but the reason is not difficult to see. If you pair the sums that have equal", "is a girl. In how many different ways can it be done? A) 64 B) 84 C) 56 D) 20 Explanation: Total number of possible ways = 9C3 = 84 ways Required atleast one girl in the group of three = total possible ways - ways in which none is girl None of the members in the group is girl = 6C3 = 20 Therefore, number of ways that at least one member is a girl in the group of three = 84 - 20 = 64 ways. 7 743 Q: A bag contains 5 red smileys, 6 yellow smileys and 3 green smileys. If two smileys are picked at random, what is the probability that both are red or both are green in colour? A) 7 B) 1/7 C) 3/7 D) 0 Explanation: Given total number of smileys = 5 + 6 + 3 = 14 Now, required probability = 3 849 Q: When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12? A) 35/36 B) 17/36 C) 15/36 D) 1/36 Explanation: When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36 Required, the sum of the two numbers that turn up is less than 12 That can be", "Convergence of winning probability in a one-player dice-throwing game In this (one-player) game, the player starts with a total of $$n$$ points. On each turn, they choose to throw either a four-, six-, or eight-sided die, and then subtract the number thrown from their point total. The game continues until the player's point total reaches zero exactly, and they win, or falls below zero, in which case they lose. The strategy does not seem obvious. (Even for the simple case of $$n=5$$, one must do a calculation.) And dynamic programming techniques show that the strategy is not straightforward. For $$n≤20$$, the four-sided die is optimal, except for $$n=5, 6, 13, 20$$, where the six-sided die is optimal, and $$n=8, 14, 16$$, where the eight-sided die is optimal. For large $$n$$, the probability of winning approaches $$0.456615178766744$$, which is not a number that I recognize. (The ISC does not recognize it either.) It's not particularly surprising that the probability of winning converges as $$n$$ increases, because the probability of winning with $$n$$ points is the mean of probability of winning with $$n-i$$ points, taken over a small range of $$i$$. Considering the probabilities as a sequence, each element lies inside the range of the previous few elements, and tends to be in the middle of that range. As one", "arrangement of for a sum of We can also do which works even better, since opposing faces have the same sum. 12 Firstly arrange the dice into a 3x3 matrix. If the sum of all dice is even, choose the column of first cup according to the total sum mod 3, and the row according to the sum of its own column. Let's say the first row/column means mod 3 = 0 and the second means mod 3 = 1. If the first cup is O (its own column denoted by o), choose the second cup between A, B ... 12 A partial solution You asked for the existence of a$(2n+1)$-sided polytope/die, so that all$2n+1$sides are congruent and so that the probability of landing on each of these sides is precisely$1/(2n+1)$. I do not see how to mathematically model the property that the probability of landing on each of the sides is precisely$1/(2n+1)$. This seems to need ... 12 The cards are numbered: Reasoning: There are 36 possible ways to choose two cards out of nine, so the results must be$2,3,3,4,4,4,\\ldots,11,11,12$. If we sum all of these, we find that the total is$252$, and that each card is represented eight times. This makes the sum of all nine cards$\\frac{252}{8}=31.5$. For any card with value$x\\$, the other eight ..." ]

[ "# Nine people sit down at random seats around a round table. Four of them are math majors, three others are physics majors, and the two remaining Nine people sit down at random seats around a round table. Four of them are math majors, three others are physics majors, and the two remaining people are biology majors. What is the probability that all four math majors sit in consecutive seats? There are a total of $(9-1)!=8!$ possible seating arrangements. We treat the four math majors as a single person. So we're really seating 6 people around a table which gives $5!$. So the probability is $5!/8!$. I think this is wrong. Any help is greatly appreciated. • Why do you \"think this is wrong\"? – Henry Nov 16 '17 at 1:16 • The probability just seems to be really small and I'm surprised it is that small, but I guess that doesn't really matter. I'm just not certain. It makes sense to me though. – ddswsd Nov 16 '17 at 1:22 • The \"math person\" can sit down in any of $4!=24$ ways. So you need to multiply by that factor. – Barry Cipra Nov 16 '17 at 1:23 There are $9$ distinct spots in which the $4$ math majors can be seated consecutively around a table, if", "# Having difficulty answering this question (Round Table Combinatorics) How many ways can three Biology majors, four Computer Science majors, four English majors, and two Physics majors sit at a round table, such that those in the same major sit together? • Must all people of one major sit together? Or is just pairs of 2 okay? (For the even ones at least) – user12345 Apr 2 '17 at 5:12 • @anonymaker000010001 For this question, all people in a certain major must sit with their major and no one else. – ProjectDefy Apr 4 '17 at 5:01 I ended up finding the answer. I'll explain it as best as I possibly can on the subject I'm still having difficult times with. So, this is a circular permutation type problem since it's a round table problem. So we have the equation so far: $$P_n=(n-1)!$$ Given our problem, we have $P_n=(n-1)! = 3!$ number of ways of arranging the majors. Now to account for the 3 Biology majors we have $3!$ ways to arrange them, 4 Comp. Sci Majors to be, $4!$ ways, 4 English Majors to be, $4!$ ways, and 2 Physics Majors to be, $2!$ ways. We have answer: $$3!*3!*4!*4!*2! = 41472$$ Hence, to arrange three Biology majors, four Computer Science majors, four English majors, and two Physics", "majors sit at a round table, such that those in the same major sit together, would be $41472$ ways. • You should state that $P_n$ is the number of ways of arranging the majors around the table. – N. F. Taussig Apr 4 '17 at 8:09 • @N.F.Taussig, I went ahead and added that where it seemed to fit! Thanks for the suggestion. – ProjectDefy Apr 5 '17 at 1:25 • Your answer is correct. Another way to think about it is to seat a particular physics major. We use that person as our reference point. There are two ways to seat the other physics major (to the left or right of the first physics major to be seated). We can now seat the other three majors in $3!$ orders as we move clockwise around the circle. As you noted, there are $3!$ ways to arrange the biology majors, $4!$ ways to arrange the English majors, and $4!$ ways to arrange the Computer Science majors within their blocks, giving $2!3!3!4!4!$ possible seating arrangements at the round table. – N. F. Taussig Apr 5 '17 at 1:39 Consider each group of people with the same major as 1 group. Then there are 4 groups and they can sit in 4! different ways, each group has people of the same", "major. Now, the group of biology major can sit in 3! different ways, the group of computer science major in 4! different way, etc. Thus the total come up to 4!3!4!4!2! • I gave this a try and it still isn't correct (the answer for your equation was: 165888 ways). I did go ahead and try instead 4(3!4!4!2!) = 27648 ways, which was also incorrect. – ProjectDefy Apr 4 '17 at 5:00 • I ended up finding the answer, which you can see posted if you were curious to find the answer. – ProjectDefy Apr 4 '17 at 5:19", "roll two dice, it will fall the sum of 10, or the same number will fall on both dice. • Hockey players After we cycle five hockey players sit down. What is the probability that the two best scorers of this crew will sit next to each other? • Desks A class has 20 students. The classroom consists of 20 desks, with 4 desks in each of 5 different rows. Amy, Bob, Chloe, and David are all friends, and would like to sit in the same row. How many possible seating arrangements are there such that Amy, Bob, • The test The test contains four questions, and there are five different answers to each of them, of which only one is correct, the others are incorrect. What is the probability that a student who does not know the answer to any question will guess the right answer", "group of students in another major. Divide the Problem Two Circles: Each of these majors (chemistry and biology) are represented with the two circles. Overlapped Portion: Some people double-major - that's the overlapping shaded region. This is the portion that belongs to both circles. In the question above, this means students who major in both chemistry and biology. Outside square Area: Perhaps some students do not major in either chemistry or biology - this is represented by the portion outside of the two circles but within the box. It is NOT necessarily equivalent to: 1 - circle1 - circle 2 You cannot simply subtract the value of each circle to find the area of this outside area. If the circles overlap, you would be double counting the overlapping region in your calculation. So do not fall into this trip! Translate and Use the Hints In the above question, you are given information that \"at least 30 students are not majoring in either chemistry or biology\"---the keyword is \"at least.\" We don't know exactly how many students are in this \"outside\" area, but we know that at least 30 of them are there. So utilize this 30 to help you find the extreme range of how many students are double majoring. Out of the 200 students, 30 of them", "# conditional probability Out of forty students, there are 14 who are taking Physics and 29 who are taking Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class?", "# Knights Around A Table Discrete Mathematics Level 3 25 of King Arthur's are seated at the round table. Three of them are randomly chosen to be sent off to slay a troublesome dragon. Let $$P$$ be the probability that at least two of the three had been sitting next to each other. If $$P$$ can be expressed as $$\\frac {a}{b}$$, where $$a$$ and $$b$$ are pairwise coprime integers, find $$a+b$$. ×", "# My Sixteen Friendly Students I have sixteen students in my class who sit in four rows of four. Each week they sit in a different order. After a number of weeks every student has sat next to every other student, next meaning side by side, one behind the other, or sitting diagonally together. What is the fewest number of weeks in which this can happen? How many if my students were 25? • Perhaps the second 25 student puzzle (which presumably has five rows of five) should be a new question after this is solved? – Weather Vane May 7 at 18:27 • @WeatherVane: Following your suggestion, I have now posted the case for 25 students at: math.stackexchange.com/questions/3218384/… – Bernardo Recamán Santos May 8 at 12:41 • I made a solution validator some may find useful. – TemporalWolf May 8 at 23:30 • I found an optimal solution to the 25 student problem (5 weeks), or rather, my computer program did. The solution can be found over on Mathematics SE. – Jaap Scherphuis May 9 at 15:59 Continuing from Arnaud Mortier's observations, I noticed that each student in the always-on-the-edge group will sit by one student from each rotating group while the rotating students are in the corners, and two more students from each rotating group while", "+0 # Probability Fraction Geometry question! 0 29 1 A college calculus class consists of 22 females and 18 males. There are 10 biology majors, 18 mathematics majors, and 12 computer science majors. If a student is chosen at random, what is the probability that the student is a female? (Enter your probability as a fraction.) =______________ What is the probability that the student is a mathematics major? (Enter your probability as a fraction.) =__________________ Guest Sep 30, 2018 #1 +2319 +1 $$P[\\text{random student is female}] = \\dfrac{22}{22+18} = \\dfrac{22}{40}=\\dfrac{11}{20}$$ $$P[\\text{random student is Mathematics major}] = \\dfrac{18}{40} = \\dfrac{9}{20}$$ Rom Oct 1, 2018", "Courses Courses for Kids Free study material Free LIVE classes More # In how many ways can 10 math’s teachers and 4 science teachers can be seated at a round table if all the 4 science teachers do not sit together$(a){\\text{ 13!}} \\times {\\text{4!}} \\\\ (b){\\text{ 13!}} \\\\ (c){\\text{ 13! - }}\\left( {10! \\times 4!} \\right) \\\\ (d){\\text{ 10!}} \\times {\\text{4!}} \\\\$ Last updated date: 22nd Mar 2023 Total views: 304.2k Views today: 7.82k Verified 304.2k+ views Hint – In this question we have to make teachers sit in a round table and the condition given is all 4 science teachers do not sit together. Use the concept of round table that if in total n people are to be seated in a round table then the total way of making them sit is$\\left( {n - 1} \\right)!$. Keep in mind that all 4 science teachers do sit together. Total number of ways that they can be seated in a round table is $\\left( {n - 1} \\right)!$ Therefore total number of ways that they can be seated in a round table is $\\left( {14 - 1} \\right)! = 13!$ Now consider 4 science teachers as an identity therefore total teachers are $= 10 + 1 = 11$ So, number of circular arrangement $= \\left( {11 - 1}", "• # question_answer In a seminar the number of participants in Mathematics, Physics and Biology are 192 240 and 168. Find the minimum number of rooms required if in each room same number of participants is to be seated and all of them being in the same subject. A) 20 B) 25 C) 28 D) 30 HCF of $(192,\\,240,\\,168)=2\\times 2\\times 2\\times 3=24$ Number of rooms for participants in Mathematics, Physics and Biology respectively is $\\frac{192}{24}=8,\\,\\frac{240}{24}=10$ and $\\frac{168}{24}=7$ $\\therefore$ Total minimum number of required rooms = 25", "# Math, Physics and Chemistry books On a table, there are a total of 30 distinct books: 9 math books, 10 physics books, and 11 chemistry books. What is the probability of getting a book that is not a math book? ×", "# Seating arrangement probability Discrete Mathematics Level pending 6 male students and 3 female students sit around a round table. The probability that no 2 female students sit beside each other can be expressed as $$\\frac{a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers. What is the value of $$a+b$$? ×", "# Sitting Side-by-side $$7$$ male students and $$3$$ female students sit around a round table. If the probability of all three female students sitting side-by-side is $$\\frac{1}{a},$$ what is the value of $$a?$$ ×", "option 50: In a class of 60 students, 30 offer physics and 40 offer chemistry. If a student is picked at random from the class, what is the probability that the students offers both physics and chemistry? A: $\\frac{1}{3}$ B: $\\frac{1}{4}$ C: $\\frac{1}{2}$ D: $\\frac{1}{6}$ E: No option", "to 10. If two more majors were to be added to the class, the ratio would then be 2 to 4. How many people are in the class? A. 14 B. 28 C. 42 D. 56 E. 70 The OA is B. I'm confused by this PS question. If the initial ratio is 4/10, then (4+2)/10=2/4. I stuck here. Experts, any suggestion? Thanks in advance. ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12122 messages Followed by: 1237 members 5254 GMAT Score: 770 LUANDATO wrote: A physics class has majors and non-majors in a ratio of 4 to 10. If two more majors were to be added to the class, the ratio would then be 2 to 4. How many people are in the class? A. 14 B. 28 C. 42 D. 56 E. 70 The ratio of the number of physics majors to non-physics majors is 3 to 5 Let P = number of Physics majors Let N = number of Non-physics majors So, we get: P/N = 3/5 Cross multiply to get: 5P = 3N Rearrange to get: 5P - 3N = 0 If two of the physics majors were to change their major to biology, the new ratio of physics majors to non-physics majors would be 1 to 2 This means that P", "## Tuesday, 12 March 2013 ### MoMath! It's time for MoMath - that is, the Museum of Mathematics. I first heard about MoMath from a pretty cool looking flyer. You know, the kind with some demurely (dare I say...indiesque?) coloured triangles arranged in the shape of a nautilus and stuff - like, an actual well-designed flyer, as opposed to the really nerdy stuff that we mathematicians produce most of the time. And that's all I know about MoMath. Except the following puzzle given at the opening ceremony of this museum: Eight people are seated around a round table (the table seats 8 precisely). They all get up at the same time (ignore physics for a moment here) and each person either moves to their left or right by one chair or stays where they are, before sitting down again (miraculously avoiding two people trying to get into the same chair). In how many distinct ways is this possible? This was actually a question posed on the NPR Sunday Puzzles last week (hence my later than usual post, I was a tad paranoid that I'd get it wrong and waited for the official answer to be revealed :P), and the guy on the program who got it right said that he took some time over a few days bashing", "# Problem #86 86 Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and the denominator? This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "the Border between Math and Computers. Princeton, NJ: Princeton University Press, 1998. From Yahoo Answers Question:Hello, I need to solve the discrete mathematics problem, proposition for p and q ................._____......_____ ........................_............_ p <=> q ( p q ) ( q p ) Thanks for your help in advance, B/R. Hey! I would appricate if someone help me :( B/R Answers:p <=> q ( p q ) ( q p ) p <=> q p q q p p <=> q p q Question:Ms Pezzulo teaches geometry and then biology to a class of 12 advanced students in a classroom that has only 12 desks. In how many ways can she assign the students to these desks so that (a)no student is seated at the same desk for both classes? (b)there are exactly six students each of whom occupies the same desk for both classes? Answers:The only feasible way for one to occupy a desk without being \"seated\" is to stand on the desk. So in the first class session, Group A (consisting of 6 students) sits in their respective chairs and so does Group B (6 students). In session 2 Group A stands in their same desk instead of sitting ang Group B switches seats with one another, like musical chairs. Since this arrangement only involves the movement" ]

[ "# Nine people sit down at random seats around a round table. Four of them are math majors, three others are physics majors, and the two remaining Nine people sit down at random seats around a round table. Four of them are math majors, three others are physics majors, and the two remaining people are biology majors. What is the probability that all four math majors sit in consecutive seats? There are a total of $(9-1)!=8!$ possible seating arrangements. We treat the four math majors as a single person. So we're really seating 6 people around a table which gives $5!$. So the probability is $5!/8!$. I think this is wrong. Any help is greatly appreciated. • Why do you \"think this is wrong\"? – Henry Nov 16 '17 at 1:16 • The probability just seems to be really small and I'm surprised it is that small, but I guess that doesn't really matter. I'm just not certain. It makes sense to me though. – ddswsd Nov 16 '17 at 1:22 • The \"math person\" can sit down in any of $4!=24$ ways. So you need to multiply by that factor. – Barry Cipra Nov 16 '17 at 1:23 There are $9$ distinct spots in which the $4$ math majors can be seated consecutively around a table, if", "ways in which only one is in the assigned seat is $4 \\times 2 = 8$. Thus the required probability is $\\frac{8}{24} = \\frac{1}{3}$", "Feb 12, 2015 at 7:45 The students numbered $1$ to $k$ will always get a seat among their preferences. For a student number $s$ with $s>k$, the probability that all $k$ of his preferred seats are taken is $\\binom k {s-1}/\\binom k n$. Because the lists of the next students are random, it doesn't matter which seat is taken, so the order in the list doesn't matter at all. The problem is the same if we pick sets of size $k$. So the probability that every students seats in a preferred seat is $\\prod_{s=k+1}^m \\left(1 - \\frac{(s-1)!(n-k)!}{(s-k-1)!n!}\\right)$ • Call your result $f(k,m,n)$. I let Mathematica compute some examples, e.g., $f(3,11,14)={19604295639225\\over 54839944911388}$. This suggests that no simplification is possible. Feb 12, 2015 at 10:16", "majors sit at a round table, such that those in the same major sit together, would be $41472$ ways. • You should state that $P_n$ is the number of ways of arranging the majors around the table. – N. F. Taussig Apr 4 '17 at 8:09 • @N.F.Taussig, I went ahead and added that where it seemed to fit! Thanks for the suggestion. – ProjectDefy Apr 5 '17 at 1:25 • Your answer is correct. Another way to think about it is to seat a particular physics major. We use that person as our reference point. There are two ways to seat the other physics major (to the left or right of the first physics major to be seated). We can now seat the other three majors in $3!$ orders as we move clockwise around the circle. As you noted, there are $3!$ ways to arrange the biology majors, $4!$ ways to arrange the English majors, and $4!$ ways to arrange the Computer Science majors within their blocks, giving $2!3!3!4!4!$ possible seating arrangements at the round table. – N. F. Taussig Apr 5 '17 at 1:39 Consider each group of people with the same major as 1 group. Then there are 4 groups and they can sit in 4! different ways, each group has people of the same", "PDA View Full Version : SAT math counting problems Jack 07-20-2013, 04:27 PM SAT math counting problems I need help with the following question: \"In how many different ways can 4 students be seated in two seats?\" Thanks!!:D:D Oliver 07-20-2013, 04:56 PM SAT maths counting problem permutations \"In how many different ways can 4 students be seated in two seats?\" The number of ways you can change the order of a set of objects is called the number of PERMUTATIONS of that set of objects. The total number of ways of arranging n objects, taking r at a time is gibe by the following formula: {{P}^{n}}_{r}=\\frac{n!}{(n-r)!}, where n!=1\\cdot2\\cdot3..\\cdot n So, in your case, you have 4 students taking 2 at a time {{P}^{4}}_{2}=\\frac{4!}{(4-2)!}=\\frac{4!}{2!}=\\frac{1\\cdot 2\\cdot 3\\cdot 4}{1\\cdot 2}=3\\cdot 4=12 ways. The above calculations could be done easily in your GDC casio fx-9860GII SD for example as follows: OPTN>PROB>nPr and set n=4 and r=2 Another way to solve this problem is that the first seat can be filled with 4 ways and the second with the remaining 3ways(students). So, the total ways are 4x3=12 Hope these help!!:D Jack 07-20-2013, 04:56 PM Thanks Oliver!!", "# Having difficulty answering this question (Round Table Combinatorics) How many ways can three Biology majors, four Computer Science majors, four English majors, and two Physics majors sit at a round table, such that those in the same major sit together? • Must all people of one major sit together? Or is just pairs of 2 okay? (For the even ones at least) – user12345 Apr 2 '17 at 5:12 • @anonymaker000010001 For this question, all people in a certain major must sit with their major and no one else. – ProjectDefy Apr 4 '17 at 5:01 I ended up finding the answer. I'll explain it as best as I possibly can on the subject I'm still having difficult times with. So, this is a circular permutation type problem since it's a round table problem. So we have the equation so far: $$P_n=(n-1)!$$ Given our problem, we have $P_n=(n-1)! = 3!$ number of ways of arranging the majors. Now to account for the 3 Biology majors we have $3!$ ways to arrange them, 4 Comp. Sci Majors to be, $4!$ ways, 4 English Majors to be, $4!$ ways, and 2 Physics Majors to be, $2!$ ways. We have answer: $$3!*3!*4!*4!*2! = 41472$$ Hence, to arrange three Biology majors, four Computer Science majors, four English majors, and two Physics", "• # question_answer Seats for mathematics, physics and biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. The ratio of increased seats will be A) 2 : 3 : 4B) 6 : 8 : 9C) 6 : 7 : 8 D) None of these. Correct Answer: A Solution : [a] Let the number of seats for mathematics, physics and biology be 5x, 7x and 8x respectively. No of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x) i.e. $\\left( \\frac{140}{100}\\times 5x \\right),\\,\\,\\,\\left( \\frac{150}{100}\\times 7x \\right)$ and $\\left( \\frac{175}{100}\\times 8x \\right)$ i.e. $7x,\\frac{21}{2},14x$ $\\therefore$ Required ratio $7x:\\frac{21}{2}x:14x$ i.e. $14x:21x:28x$ = 2 : 3 : 4 You need to login to perform this action. You will be redirected in 3 sec", "two, and after $b$). We know which order they go in. There are $\\binom31=3$ ways to put them adjacent, and $\\binom32=3$ ways to put them non-adjacent. That makes six. - thanks for two approaches to solve the problem.. – Keneth Adrian Mar 16 '14 at 4:13 Exactly half of all arrangements have $a$ before $b$, and exactly half have $c$ before $d$. Hence your answer is $$\\frac{4!}{2\\cdot 2}=\\frac{24}{4}=6$$ - There are $4$ chairs in a row. There are $4!$ ways to arrange our $4$ people in a row. There are $\\binom{4}{2}$ ways to choose the pair of chairs to be occupied by $a$ and $b$. Once we have done that, where everybody sits is determined. Thus the required probability is $\\frac{\\binom{4}{2}}{4!}$. -", "\\:=\\:\\frac{7!}{4!3!} \\:=\\:35$ ways. From the remaining 3 students, assign 3 students to room C. . . Of course, there is: . $_3C_3 \\:=\\:1$ way. Therefore, there are: . $792 \\times 35 \\times 1 \\:=\\:27,\\!720$ ways.", "$4!\\cdot 5$ ways in which they can arrange themselves with no two people sitting together. As mentioned before, this is merely the first step to determining the probability. Now that we have the count of how many ways it can occur, we divide by the total number of ways that the four could seat themselves (reminding ourselves that each way they could sit is equally likely to occur which is what allows us to find probability in this fashion in the first place). There are $\\binom{8}{4}$ ways in which we choose which seats they occupy and $4!$ ways they arrange themselves within those seats giving the final probability as: $$\\frac{4!\\cdot 5}{4!\\cdot\\binom{8}{4}}=\\frac{5}{\\binom{8}{4}}$$ More info as to why your method of counting was incorrect: Suppose instead there were twelve seats and four people. We would need eight empty seats then. By your logic we would have wanted to choose eight of the following five seats: $\\underline{~}P\\underline{~}P\\underline{~}P\\underline{~}P\\underline{~}$ But, that would lead to $\\binom{5}{8}$ which is zero. Instead in the situation with eight empty seats and four people out of twelve total seats, we would approach like I did above: distribute three of the empty seats to keep them separated: Then in the five available spaces distribute the remaining five seats remembering that you can put multiple seats into the same slot.", "the number of pairs at table 2, so you need $3$ or $4$ pairs at one table. Let the ladies take a seat first and select $8$ chairs for the women. First we look at the outcome of exactly $3$ women at some table which has probability: $2\\frac{\\binom83\\binom85}{\\binom{16}{8}}$. In that situation select $3$ out of the $8$ open seats for their husbands. The probability that they come to sit at the same table as their wives is $\\frac{\\binom53\\binom30}{\\binom83}$. Now we look at the outcome of exactly $4$ women at both tables which has probability: $\\frac{\\binom84\\binom84}{\\binom{16}{8}}$. In that situation select $4$ out of the open seats for the husbands of the women at table 1. Probability that $3$ of them join their wives is: $\\frac{\\binom43\\binom41}{\\binom84}$. Probability that $4$ of them join their wives is $\\frac{\\binom44\\binom40}{\\binom84}$. Other situations are not relevant and we end up with probability:$$2\\frac{\\binom83\\binom85}{\\binom{16}{8}}\\times\\frac{\\binom53\\binom30}{\\binom 83}+\\frac{\\binom44\\binom40}{\\binom84}\\times\\left[\\frac{\\binom43\\binom41}{\\binom84}+\\frac{\\binom44\\binom40}{\\binom84}\\right]$$", "spaces at the side of those occupied, i.e. the weak compositions of $n-2p$ into $p+1$ parts. But the last guy is allowed to place the bag in the corridor, if sitting at the extreme of the row, which is not a case included under the above. He is occupying then one fixed place, with the remaining seats to be occupied by $p-1$ people as before : so $\\binom{n-1-(p-1)}{p-1}=\\binom{n-p}{p-1}$ ways. Thus $$\\binom{n-p}{p}+\\binom{n-p}{p-1}=\\binom{n+1-p}{p}$$ total ways to seat them. And for the probability, the above divided the number of ways to seat people without bags, i.e. $\\binom{n}{p}$ The procedure can be easily extended to 2,3, .. bags. In our case $$prob=\\binom{8+1-4}{4}/\\binom{8}{4}=5/\\binom{8}{4}$$ This doesn't directly answer your question, but provides an nice alternative method to solving this question This might not exactly be the general method for solving a question like this given in a statistics class, but it's an possibly fun way to solve the question (something people do in math competitions) Calculation through recurrence relations Let $C_i$ denote the number of ways $i$ people can be seated in $2i$ chairs such that no two are adjacent. Obviously for this question, we want to derive $$\\frac{C_4}{\\binom{8}{4}}$$ For $C_1$, there are two ways to seat one person in two chairs. For $C_2$, you can reason that there is a total three ways.", "a school are in the ratio $$5 : 7 : 8.$$ There is a proposal to increase these seats by $$40$$%, $$50$$% and $$75$$% respectively. What will be the ratio of increased seats? Let the number of seats for Mathematics, Physics and Biology be $$5x, 7x$$ and $$8x$$ Number of increased seats are $$(140$$% of $$5x),$$ $$(150$$% of $$7x)$$ and $$(175$$% of $$8x).$$ $$\\Rightarrow (\\frac{140}{100} \\times 5x), (\\frac{150}{100} \\times 7x)$$ and $$(\\frac{175}{100} \\times 8x)$$ $$\\Rightarrow 7x, \\frac{21x}{2}$$ and $$14x.$$ $$\\therefore$$ The required ratio $$= 7x : \\frac{21x}{2} : 14x$$ $$\\Rightarrow 14x : 21x : 28x$$ $$\\Rightarrow 2 : 3 : 4.$$ Enter details here", "finding their total because they were assuming the seats were numbered; otherwise they would have. 2) Your expression of $\\frac{7!}{3!3!}$ would work fine if the people were arranged in a straight line. Notice that in this case the 3 people sitting together have 7 choices for their positions: $123, 234, \\cdots, 789$. In a circle, though, they have 9 choices instead: $123, 234, \\cdots, 789, 891, 912$. $\\displaystyle 1-\\frac{\\binom{3}{1}3!6!-\\binom{3}{2}4(3!)^{3}+2(3!)^{3}}{8!}=1-\\frac{10800}{40320}=1-\\frac{15}{56}=\\frac{41}{56}$. • Could you please explain why your middle term is $\\binom{3}{2}4(3!)^{3}$ – Anirudh Dec 30 '17 at 20:57", "Can you take it from here? Spoiler: Your probability should be: $\\dfrac{(2\\cdot 8 \\cdot 7 \\cdot 6 \\cdot 1) \\cdot 6!}{10!}$ - Let's visualize the seats as $$X\\quad X \\quad X\\quad X \\quad X.$$ There are $\\dbinom{5}{2}$ equally likely ways to choose the pair of seats to be reserved for A and B. Note that we are not assigning A or B to a specific one of these seats, just putting reserved signs on the seats. There are $4$ ways to choose a pair of adjacent seats. So our probability is $\\dfrac{4}{\\binom{5}{2}}$. Let's visualize the lineup as $$X\\quad X \\quad X\\quad X \\quad X\\quad X\\quad X \\quad X\\quad X \\quad X .$$ There are $\\binom{10}{2}$ equally likely ways to choose a pair of locations in the lineup. There are $6$ ways to choose the pair so that they are separated by $3$, since the leftmost chosen position can be any of $1,2,3,\\dots,6$.", "quick look and I'm getting a different answer (I'm OK with it being incorrect but at the moment I don't see why): There are 4 ways to get an 8. After that there are 4 ways to get a 9, and then 4 ways to get a 10. So $4^3$ ways to get a 3 card straight. Then there are $\\binom {37}{3}$ ways to get the remaining three cards and $\\binom {40}{6}$ ways to draw 6 cards. So I get$$\\frac {4^3\\binom {37}{3}}{\\binom {40}{6} }= \\frac{32}{247}=.12955$$ 10. ## Re: Card probability Originally Posted by Walagaster I just had a quick look and I'm getting a different answer (I'm OK with it being incorrect but at the moment I don't see why): There are 4 ways to get an 8. After that there are 4 ways to get a 9, and then 4 ways to get a 10. So $4^3$ ways to get a 3 card straight. Then there are $\\binom {37}{3}$ ways to get the remaining three cards and $\\binom {40}{6}$ ways to draw 6 cards. So I get$$\\frac {4^3\\binom {37}{3}}{\\binom {40}{6} }= \\frac{32}{247}=.12955$$ Prof. Walagaster, I agree and that was my first reaction. But then post #4 gave me pause that I my not understand the setup. Then I saw Dennis' post. He is usually spot on with", "students will sit at the lowest numbered remaining table. The remaining four students must be seated at the remaining table. $$4 \\cdot \\binom{3}{2} \\cdot 3 \\cdot \\binom{10}{2} \\binom{8}{4} \\binom{4}{4}$$ Since these two cases are mutually disjoint, the probability of maximum confusion is $$\\frac{4 \\cdot 3 \\cdot 2 \\dbinom{10}{2}\\dbinom{8}{2}\\dbinom{6}{2}\\dbinom{4}{4} + 4 \\cdot \\dbinom{3}{2} \\cdot 3 \\dbinom{10}{2}\\dbinom{8}{4}\\dbinom{4}{4}}{\\dbinom{16}{4}\\dbinom{12}{4}\\dbinom{8}{4}\\dbinom{4}{4}}$$ What is the probability that a random seating chart will result in minimum confusion (no student assigned to a table with a student of the same name)? For the numerator, subtract the number of seating arrangements in which at least one pair of students sit at the table. To do so, you will need to apply the Inclusion-Exclusion Principle. Even though it doesn't matter which position within each table people sit at, we can include that information in our counting as long as we include it in the total number of arrangements too. Philosophically, the seats know which seat is which even if we don't care. So there are 16 seats that can be sat in, and 16 people to sit in them. We could run through the people and each one chooses a seat, and then the next person chooses from the remaining seats etc. This argument gives 16! possible seating arrangements. Now we need to count how many arrangements have the", "so that no two teachers sit in consecutive seats, we must exclude those linear arrangements in which teachers are at both ends of the row. There are three ways to select the teacher at the left end of the row, two ways to select the teacher at the right end of the row, and four ways to place the remaining teacher in one of the four spaces between successive students. Hence, there are $$3 \\cdot 2 \\cdot 4 \\cdot 5!$$ linear arrangements in which teachers are at both ends of the row. Hence, there are $$6 \\cdot 5 \\cdot 4 \\cdot 5! - 3 \\cdot 2 \\cdot 4 \\cdot 5! = (120 - 24)5! = 96 \\cdot 5!$$ linear arrangements of teachers and students so that no two teachers are consecutive and teachers are not at both ends of the row. These linear arrangements correspond to the permissible ways we can seat the students and teachers around the table. To account for rotational invariance, we divide the number of linear arrangements by $8$, which yields $$\\frac{96 \\cdot 5!}{8} = 12 \\cdot 5! = 1440$$ permissible seating arrangements around a circular table, as you found. • Thanks for nice explanation. – 1 0 May 29 '16 at 10:52 Students $S_1,\\dots,S_5$, teachers $T_1,T_2,T_3$. Suppose we put $S_1$ into a particular", "Y}$ thing to think about this problem? - The calculation is correct, and efficiently done. There are $\\binom{12}{2}$ equally likely ways to select $2$ seats from $12$. The probability we put Alicia in an end seat is $\\frac{4}{12}$, Given she is in an end seat, the probability Bob ends up beside her is $\\frac{1}{11}$. The probability Alicia is not in an end seat is $\\frac{8}{12}$, and then Bob has probability $\\frac{2}{11}$ of ending up beside her. Thus our probability is $$\\frac{4}{12}\\cdot\\frac{1}{11}+\\frac{8}{12}\\cdot\\frac{2}{11}.$$", "# 2 classes in the same classroom each with 100 seats and the same 100 students, find the probability that no one has the same seat for both classes The question is as follows: Harvard Law School courses often have assigned seating to facilitate the “Socratic method.” Suppose that there are $100$ first year Harvard Law students, and each takes two courses: Torts and Contracts. Both are held in the same lecture hall (which has $100$ seats), and the seating is uniformly random and independent for the two courses. Find the probability that no one has the same seat for both courses (exactly; leave the answer as a sum). Here is my attempt: $$P(no\\; same\\; seat) = 1 - P(all\\; same\\; seats)=1-(\\frac{1}{100}+\\frac{1}{99}+...+\\frac{1}{1})=1-\\sum_{i=1}^{100}{\\frac{1}{i}}$$ So my thinking goes like this. The $1st$ student can choose any seat in the first class, thus, to choose the same seat in the second class would be $\\frac{1}{100}$. The $2nd$ student can choose any of the $99$ remaining seats for the first class, thus, to choose the same seat in the second class it would be $\\frac{1}{99}$. However, upon looking at the solution, it seems like my answer is completely wrong. Does anyone know why my method is wrong? • To begin, $Pr(all~same~seats)$ would be $\\frac{1}{100}\\times \\frac{1}{99}\\times\\cdots$ not $\\frac{1}{100}+\\frac{1}{99}+\\cdots$. Next, the opposite event of" ]

[ "# Nine people sit down at random seats around a round table. Four of them are math majors, three others are physics majors, and the two remaining Nine people sit down at random seats around a round table. Four of them are math majors, three others are physics majors, and the two remaining people are biology majors. What is the probability that all four math majors sit in consecutive seats? There are a total of $(9-1)!=8!$ possible seating arrangements. We treat the four math majors as a single person. So we're really seating 6 people around a table which gives $5!$. So the probability is $5!/8!$. I think this is wrong. Any help is greatly appreciated. • Why do you \"think this is wrong\"? – Henry Nov 16 '17 at 1:16 • The probability just seems to be really small and I'm surprised it is that small, but I guess that doesn't really matter. I'm just not certain. It makes sense to me though. – ddswsd Nov 16 '17 at 1:22 • The \"math person\" can sit down in any of $4!=24$ ways. So you need to multiply by that factor. – Barry Cipra Nov 16 '17 at 1:23 There are $9$ distinct spots in which the $4$ math majors can be seated consecutively around a table, if", "# Having difficulty answering this question (Round Table Combinatorics) How many ways can three Biology majors, four Computer Science majors, four English majors, and two Physics majors sit at a round table, such that those in the same major sit together? • Must all people of one major sit together? Or is just pairs of 2 okay? (For the even ones at least) – user12345 Apr 2 '17 at 5:12 • @anonymaker000010001 For this question, all people in a certain major must sit with their major and no one else. – ProjectDefy Apr 4 '17 at 5:01 I ended up finding the answer. I'll explain it as best as I possibly can on the subject I'm still having difficult times with. So, this is a circular permutation type problem since it's a round table problem. So we have the equation so far: $$P_n=(n-1)!$$ Given our problem, we have $P_n=(n-1)! = 3!$ number of ways of arranging the majors. Now to account for the 3 Biology majors we have $3!$ ways to arrange them, 4 Comp. Sci Majors to be, $4!$ ways, 4 English Majors to be, $4!$ ways, and 2 Physics Majors to be, $2!$ ways. We have answer: $$3!*3!*4!*4!*2! = 41472$$ Hence, to arrange three Biology majors, four Computer Science majors, four English majors, and two Physics", "majors sit at a round table, such that those in the same major sit together, would be $41472$ ways. • You should state that $P_n$ is the number of ways of arranging the majors around the table. – N. F. Taussig Apr 4 '17 at 8:09 • @N.F.Taussig, I went ahead and added that where it seemed to fit! Thanks for the suggestion. – ProjectDefy Apr 5 '17 at 1:25 • Your answer is correct. Another way to think about it is to seat a particular physics major. We use that person as our reference point. There are two ways to seat the other physics major (to the left or right of the first physics major to be seated). We can now seat the other three majors in $3!$ orders as we move clockwise around the circle. As you noted, there are $3!$ ways to arrange the biology majors, $4!$ ways to arrange the English majors, and $4!$ ways to arrange the Computer Science majors within their blocks, giving $2!3!3!4!4!$ possible seating arrangements at the round table. – N. F. Taussig Apr 5 '17 at 1:39 Consider each group of people with the same major as 1 group. Then there are 4 groups and they can sit in 4! different ways, each group has people of the same", "major. Now, the group of biology major can sit in 3! different ways, the group of computer science major in 4! different way, etc. Thus the total come up to 4!3!4!4!2! • I gave this a try and it still isn't correct (the answer for your equation was: 165888 ways). I did go ahead and try instead 4(3!4!4!2!) = 27648 ways, which was also incorrect. – ProjectDefy Apr 4 '17 at 5:00 • I ended up finding the answer, which you can see posted if you were curious to find the answer. – ProjectDefy Apr 4 '17 at 5:19", "• # question_answer Seats for mathematics, physics and biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. The ratio of increased seats will be A) 2 : 3 : 4B) 6 : 8 : 9C) 6 : 7 : 8 D) None of these. Correct Answer: A Solution : [a] Let the number of seats for mathematics, physics and biology be 5x, 7x and 8x respectively. No of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x) i.e. $\\left( \\frac{140}{100}\\times 5x \\right),\\,\\,\\,\\left( \\frac{150}{100}\\times 7x \\right)$ and $\\left( \\frac{175}{100}\\times 8x \\right)$ i.e. $7x,\\frac{21}{2},14x$ $\\therefore$ Required ratio $7x:\\frac{21}{2}x:14x$ i.e. $14x:21x:28x$ = 2 : 3 : 4 You need to login to perform this action. You will be redirected in 3 sec", "# Variations question 1. Feb 16, 2008 ### Physicsissuef 1. The problem statement, all variables and given/known data On how many ways, 8 different people can sit on 12 seats? 2. Relevant equations $$V\\stackrel{k}{n}$$ 3. The attempt at a solution I tried to solve this task by using the formula of variations without repetition. $$V\\stackrel{8}{12}$$ But I sow in my textbook results, and it says that it should be: 8! * $$V\\stackrel{8}{12}$$ I don't know what is correct actually. 2. Feb 16, 2008 ### HallsofIvy Staff Emeritus I have absolutely no idea what $V_n^k$ means! I suspect that that it is what I would call $_nC_k$ or $\\left(\\begin{array}{c}n \\\\ k \\end{array}\\right)$, the binomial coefficient. That is the number of ways we can pick a group of 8 chairs out of 12 for the people to sit on. That, however, does not include the number of orders in which those 8 people can sit on the same 8 chairs. Since there are 8! orders for 8 people, that would be $8!_12C_8$ or 12!/(12-8)!= 12!/4! 3. Feb 17, 2008 ### Physicsissuef $V_n^k$=n(n-1)(n-2)...(n-k) So, 8!*$V_12^8$ will come to be, big number. 8! * 12*11*10*9*8*7*6*5 4. Feb 17, 2008 ### HallsofIvy Staff Emeritus That is almost what I would call the number of Permutations: $$_nP_k= \\frac{n!}{(n-k)!}$$ (\"Almost\" because what you have is", "# Probability homework problem 1. Apr 22, 2012 ### th4450 1. The problem statement, all variables and given/known data Bobo, Coco, and 8 other students are arranged to sit in 2 rows of 5 students. If these 10 students take their seat randomly, find the probability that Bobo and Coco are sitting in the same row but not next to each other. 3. The attempt at a solution Bobo and Coco together have 12 ways to sit, they 2 can exchange, other 8 students sit randomly. Probability = $\\frac{12×2×8!}{10!}$ = $\\frac{4}{15}$ Is this correct? Thank you!! 2. Apr 22, 2012 ### HallsofIvy Re: Probability Correct answer to what question? Where did you include the \"sitting in the same row\"? Or \"not next to each other? 3. Apr 23, 2012 ### th4450 Re: Probability I mean, say the seats are arranged like this: 1 2 3 4 5 6 7 8 9 10 Bobo and Coco sitting in the same row but not next to each other, they can take: 1,3 2,4 3,5 1,4 2,5 1,5 6,8 7,9 8,10 6,9 7,10 6,10 totally 12 ways. The above arrangement can be for Bobo|Coco or Coco|Bobo, so times 2. Is my answer correct? Thanks again. 4. Apr 23, 2012 ### HallsofIvy Re: Probability Yes, that is correct- here is how I", "your Study Plan and Master the Core Skills to excel on the GMAT. # In a row of 30 seats, six team of 4 persons each will be seated. The Author Message TAGS: ### Hide Tags Magoosh GMAT Instructor Joined: 28 Dec 2011 Posts: 4486 In a row of 30 seats, six team of 4 persons each will be seated. The [#permalink] ### Show Tags 15 Sep 2016, 15:11 2 10 00:00 Difficulty: 95% (hard) Question Stats: 35% (02:11) correct 65% (02:06) wrong based on 81 sessions ### HideShow timer Statistics This question was inspired by another current question that appears to be less than fully clear: there-is-a-row-of-30-seats-and-there-are-6-groups-of-4-persons-each-225526.html In a row of 30 seats, six team of 4 persons each will be seated. The members of any one team always sit together in four consecutive seats in the same order. Different teams may be adjacent or separated by empty seats. In how many ways can the six teams be seated in the 30 seats? (A) (6!)(6!) (B) $$\\frac{12!}{6!}$$ (C) 12C6 (D) 30C12 (E) 30C6*6 _________________ Mike McGarry Magoosh Test Prep Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939) Math Expert Joined: 02 Aug 2009 Posts: 7199 Re: In a row of 30 seats, six team", "and Protective Services 33 Social Sciences 13,164 The majors in Table $$\\PageIndex{3}$$ are listed alphabetically, which makes understanding the numbers a little more confusing. Because qualitative variables have no natural order, you can list them however you like. To find your own major, an alphabetical order works best. But if you'd like emphasize which major has the most graduates, then order from the highest to the lowest number of graduates would be best. As we will learn more with charts, the best way to display data depends on what message you are trying to tell. Most of the graphs shown in the rest of the chapter are derived from frequency tables, so I hope that you find some simple data and practice making a few before we starting using them more!", "# If we randomly select 25 integers between 1 and 100, how many consecutive integers should we expect? Question: Suppose we have one hundred seats, numbered 1 through 100. We randomly select 25 of these seats. What is the expected number of selected pairs of seats that are consecutive? (To clarify: we would count two consecutive selected seats as a single pair.) For example, if the selected seats are all consecutive (eg 1-25), then we have 24 consecutive pairs (eg 1&2, 2&3, 3&4, …, 24&25). The probability of this happening is 75/($_{100}C_{25}$). So this contributes $24\\cdot 75/(_{100}C_{25}$) to the expected number of consecutive pairs. Motivation: I teach. Near the end of an exam, when most of the students have left, I notice that there are still many pairs of students next to each other. I want to know if the number that remain should be expected or not. If you’re just interested in the expectation, you can use the fact that expectation is additive to compute • The expected number of consecutive integers among $\\{1,2\\}$, plus • The expected number of consecutive integers among $\\{2,3\\}$, plus • …. • plus the expected number of consecutive integers among $\\{99,100\\}$. Each of these 99 expectations is simply the probability that $n$ and $n+1$ are both chosen, which is $\\frac{25}{100}\\frac{24}{99}$. So", "are also considered double majors)What is the probability that randomly selected student living in this dorm is: . _majoring in at least one of CS, Stat or Math? b) _majoring in Math or double m 1) Suppose there are 10O0 students living in certain dorm on Busch campus, of whom 600 are CS majors , 200 are Stat majors, 100 are Math majors; 90 are CS & Stat double majors 50 are CS & Math double majors 60 are Stat & Math double majors _ 20 are CS & Stat & Math triple majors ... 5 answers ##### [6poimts] Yon buly 85.(Uscratch-off lottory ticket; You read the details and learn there is a } chance t0 win$1, chance t0 win 85 clance 10 win 81O, 46 chance t0 win 810, clnnce t0 win SI(_ M chmcc Lo W IUu 85.(O . What is the expected valuc of the ticket? Docs Four MSWe M FOU should huy 0T MoL Duy chc ticket\" [6poimts] Yon buly 85.(Uscratch-off lottory ticket; You read the details and learn there is a } chance t0 win \\$1, chance t0 win 85 clance 10 win 81O, 46 chance t0 win 810, clnnce t0 win SI(_ M chmcc Lo W IUu 85.(O . What is the expected valuc of the ticket? Docs Four MSWe M FOU", "ways in which only one is in the assigned seat is $4 \\times 2 = 8$. Thus the required probability is $\\frac{8}{24} = \\frac{1}{3}$", "If we randomly select 25 integers between 1 and 100, how many consecutive integers should we expect? Question: Suppose we have one hundred seats, numbered 1 through 100. We randomly select 25 of these seats. What is the expected number of selected pairs of seats that are consecutive? (To clarify: we would count two consecutive selected seats as a single pair.) For example, if the selected seats are all consecutive (eg 1-25), then we have 24 consecutive pairs (eg 1&2, 2&3, 3&4, ..., 24&25). The probability of this happening is 75/($_{100}C_{25}$). So this contributes $24\\cdot 75/(_{100}C_{25}$) to the expected number of consecutive pairs. Motivation: I teach. Near the end of an exam, when most of the students have left, I notice that there are still many pairs of students next to each other. I want to know if the number that remain should be expected or not. • That probability is $76/_{100}C_{25}$, not $75/_{100}C_{25}$. Dec 9 '15 at 20:06 • Thanks for the correction :-) Dec 9 '15 at 21:57 • Without math I can tell you yes that is to be expected. They're called friends. Dec 9 '15 at 22:39 • + for the motivation Dec 9 '15 at 22:46 • @corsiKa, I also think it is because if my neighbor leaves, it makes me feel \"I", "Select Page Competency in Focus: Concept of Probability This problem from American Mathematics Contest 8 (AMC 8, 2018) is based on calculation of probability. It is Question number 11 of the AMC 8 2018 Problem series. Next understand the problem Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown. If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column? $\\textbf{(A) } \\frac{1}{3} \\qquad \\textbf{(B) } \\frac{2}{5} \\qquad \\textbf{(C) } \\frac{7}{15} \\qquad \\textbf{(D) } \\frac{1}{2} \\qquad \\textbf{(E) } \\frac{2}{3}$ Basic Probability sum 6/10 Suggested Book Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics Do you really need a hint? Try it first! If you need any hint try from this: There are a total of $6!$ ways to arrange the kids. Abby and Bridget can sit in 3 ways if they are adjacent in the same column: For each of these seat positions, Abby and Bridget can switch seats, and the other 4 people can be arranged in $4!$ ways which results in a total of $3 \\times 2 \\times 4!$ ways to arrange them. By the same logic, there are 4 ways for Abby and Bridget", "round table conference, there were all 16 persons in the conference. These 16 persons can be arranged among themselves around a round table in (16 – 1)! = 15! ways. ∴ n(S) = 15! Let event A: Two particular professors be seated on either side of the Vice-chancellor. Those two particular persons sit on either side of a Vice chancellor in $${ }^{2} \\mathrm{P}_{2}$$ = 2! ways. Thus the remaining 13 persons can be arranged in $${ }^{13} \\mathrm{P}_{13}$$ = 13! ways. ∴ n(A) = 13! 2! ∴ P(A) = $$\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{13 ! \\times 2 !}{15 !}=\\frac{13 ! \\times 2 \\times 1}{15 \\times 14 \\times 13 !}=\\frac{1}{105}$$ Question 17. A bag contains 7 black and 4 red balls. If 3 balls are drawn at random, find the probability that (i) all are black. (ii) one is black and two are red. Solution: The bag contains 7 black and 4 red balls, i.e., 7 + 4 = 11 balls. ∴ 3 balls can be drawn out of 11 balls in $${ }^{11} \\mathrm{C}_{3}$$ ways. ∴ n(S) = $${ }^{11} \\mathrm{C}_{3}$$ (i) Let event A: All 3 balls drawn are black. There are 7 black balls. ∴ 3 black balls can be drawn from 7 black balls in $${ }^{7} \\mathrm{C}_{3}$$ ways. ∴ n(A) = $${ }^{7} \\mathrm{C}_{3}$$ ∴ P(A) =", "# Knights Around A Table Discrete Mathematics Level 3 25 of King Arthur's are seated at the round table. Three of them are randomly chosen to be sent off to slay a troublesome dragon. Let $$P$$ be the probability that at least two of the three had been sitting next to each other. If $$P$$ can be expressed as $$\\frac {a}{b}$$, where $$a$$ and $$b$$ are pairwise coprime integers, find $$a+b$$. ×", "1. ## Combinatorics A committee of 6 is chosen from 10 Mathematics majors and 7 Physics majors so as to contain at least 3 Mathematics majors and at least 2 Physics majors . Show that there are 7800 different ways this can be done if two particular Physics majors refuse to stay on the same committee . 2. ## Re: Combinatorics Suppose the first person you choose is a math major- there are 6 ways to do that. Suppose the second and third persons are also math majors. There are 5 and 4 ways of doing that so there are $6(5)(4)= \\frac{6!}{3!}$ ways to choose the three math majors. Do the same to find the number of ways to choose 2 physics majors. That's 5 of your commitee. The last two can be chosen from any of the remaining 7+ 5= 12 students. There are $12(11)= \\frac{12!}{10!}$ ways to do that. Multiply them altogether. Finally, divide by the number of ways to order \"MMMPPEE\" (\"M\" for \"math major\", \"P\" for \"physics major\", \"E\" for \"either\") so you are not counting different orders of choosing the same people. 3. ## Re: Combinatorics Originally Posted by mathlover14 A committee of 6 is chosen from 10 Mathematics majors and 7 Physics majors so as to contain at least 3 Mathematics majors", "the next five exercises: Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample. What values does $X$ take on? Check back soon! Problem 54 Use the following information to answer the next five exercises: Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample. Find the standard deviation. Check back soon! Problem 55 Use the following information to answer the next five exercises: Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample. On average ( $\\mu$ ), how many would you expect to be business majors? Check back soon! Problem", "These are actually combinatorics problems rather than probability problems. This sort of combinatorics is needed for doing probability problems, so they may appear in sections of texts and courses that are labeled \"probability\". – Michael Hardy Jul 29 '11 at 22:09 1. Fix the middle seat for Martha. Now you have to position the other 4 people in 4 seats: $4!=24$. 2. Build the number by the digits: you have $8$ options for the hundreds (all but 0 and 5), $9$ options for the tens (all but 5) and $4$ options for the units (only one of $1,3,7,9$). A total of $8\\cdot 9\\cdot 4$ numbers.", "around the circle, which can be done in $\\frac{7!}{3!3!}$ ways. However, this does not match with their expression. Why? #### Solutions Collecting From Web of \"2011 AIME Problem 12 — Probability: 9 delegates around a round table\" They are making two simplifying assumptions in their solution, that the seats are numbered and that people from the same country are identical. 1) They didn’t divide by $9$ in finding their total because they were assuming the seats were numbered; otherwise they would have. 2) Your expression of $\\frac{7!}{3!3!}$ would work fine if the people were arranged in a straight line. Notice that in this case the 3 people sitting together have 7 choices for their positions: $123, 234, \\cdots, 789$. In a circle, though, they have 9 choices instead: $123, 234, \\cdots, 789, 891, 912$. If we assumed that the 9 people were all distinct and we didn’t number the seats, then we would get the same probability of $\\displaystyle 1-\\frac{\\binom{3}{1}3!6!-\\binom{3}{2}4(3!)^{3}+2(3!)^{3}}{8!}=1-\\frac{10800}{40320}=1-\\frac{15}{56}=\\frac{41}{56}$." ]

[ "queen of the club or a king of the heart Then, n(E) = 2 P(E) = n(E)/n(S) = 2/52 = 1/26", "spades not at bottom)*P(ace of diamonds at bottom) = 50/51 *1/51 = 50/51² -", "randomly drawn from a deck of 52 cards. What is the probability getting a five of Spade or Club? 1/52 1/13 1/26 1/12 10. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white? $\\dfrac{3}{4}$ $\\dfrac{4}{7}$ $\\dfrac{1}{8}$ $\\dfrac{3}{7}$ *Click on the QNo to display a Question. Total Ans Pipes Prev Next Permutations '; Aptitude Coaching - Free Live Session !!", "proportion of hearts, diamonds, spades, and clubs that are randomly drawn from this deck are different.", "are blue. Three balls are chosen at random, without replacement. 1. Find the probability that the chosen balls are all the same color. 2. Find the probability that the chosen balls are all different colors. 1. $$\\frac{34}{455}$$ 2. $$\\frac{120}{455}$$ Suppose again that an urn contains 15 balls: 6 are red, 5 are green, and 4 are blue. Three balls are chosen at random, with replacement. 1. Find the probability that the chosen balls are all the same color. 2. Find the probability that the chosen balls are all different colors. 1. $$\\frac{405}{3375}$$ 2. $$\\frac{720}{3375}$$ #### Cards Recall that a standard card deck can be modeled by the product set $D = \\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k\\} \\times \\{\\clubsuit, \\diamondsuit, \\heartsuit, \\spadesuit\\}$ where the first coordinate encodes the denomination or kind (ace, 2-10, jack, queen, king) and where the second coordinate encodes the suit (clubs, diamonds, hearts, spades). Sometimes we represent a card as a string rather than an ordered pair (for example $$q \\heartsuit$$). Card games involve choosing a random sample without replacement from the deck $$D$$, which plays the role of the population. Thus, the basic card experiment consists of dealing $$n$$ cards from a standard deck without replacement; in this special context, the sample of cards is often", "or two regular decks of cards; it can be played by two to six people. The game is ...", "{ 1 }{ 11 }$$. Do you agree with this argument ? Solution: Question 60. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag. [CBSE 2007] Solution: Question 61. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of (i) heart (ii) queen (iii) clubs (iv) a face card (v) a queen of diamond. Solution: Question 62. Two dice are thrown simultaneously. What is the probability that : (i) 5 will not come up on either of them ? (ii) 5 will come up on at least one ? (iii) 5 will come up at both dice? [CBSE 2009] Solution: Question 63. A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4. Solution: Question 64. A dice is rolled twice. Find the probability that (i) 5 will not come up either time. (ii) 5 will come up exactly one time. [CBSE 2014] Solution: Question 65. All the black face", "1/13 (xi) Since there are 13 spades, the probability is equal to 13/52 = 1/4 (xii) Since there are 26 black cards, the probability is equal to 26/52 = 1/2 (xiii) There is only one card named seven of clubs. Hence, the probability is 1/52. (xiv) Since there are 4 jacks, the probability is equal to 4/52 = 1/13 (xv) There is only 1 card named ace of spade. Hence, the probability is 1/52. (xvi) Since there are 4 queens, the probability is 4/52 = 1/13 (xvii) Since there are 13 hearts, the probability is equal to 13/52 = 1/4 (xviii) Since there are 26 red cards, the probability is equal to 26/52 = 1/2 Q6. An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball was drawn is white. Ans: Number of white balls = 8 Number of red balls = 10 Total number of balls = 10 + 8 = 18 Total number of cases is 18 and the number of favorable cases are 8. P(The ball drawn is white) = (Number of favorable cases)/ (Total number of cases) = 8/18 = 4/9 Q7. A bag contains 3 red balls, 5 black balls, and 4 white balls. A ball is drawn at random from the", "+0 # Two cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a heart and the second card is a club? 0 544 1 +4330 Two cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a heart and the second card is a club? Sep 11, 2016 #1 +106046 0 Two cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a heart and the second card is a club? $$\\frac{13}{52}\\times \\frac{13}{51}$$ . Sep 12, 2016", "for i in range(0,52): digits = sequence[i:i+6] card = decode_sequence(digits) if card in swaps.keys(): card = swaps[card] print(show_card(card)) # Show which bad codes are swapped with which good ones print('Swaps') for a,b in swaps.items(): print('%s -> %s' % (encode_card(a),show_card(b))) And here’s its output: 000001: Ace of Clubs 000010: Ace of Hearts 000100: 2 of Diamonds 001000: 3 of Diamonds 010000: 5 of Diamonds 100001: 9 of Clubs 000110: 2 of Hearts 001100: 4 of Diamonds 011000: 7 of Diamonds 110001: King of Clubs 100010: 9 of Hearts 000101: 2 of Clubs 001010: 3 of Hearts 010100: 6 of Diamonds 101001: Jack of Clubs 011110: 8 of Hearts 000000: Ace of Diamonds 100000: 9 of Diamonds 101000: Jack of Diamonds 010001: 5 of Clubs 001110: 4 of Hearts 011100: 8 of Diamonds 110010: King of Hearts 100100: 10 of Diamonds 001001: 3 of Clubs 010010: 5 of Hearts 100101: 10 of Clubs 010110: 6 of Hearts 101101: Queen of Clubs 101110: Queen of Hearts 011101: 8 of Clubs 110000: King of Diamonds 101100: Queen of Diamonds 011001: 7 of Clubs 100110: 10 of Hearts 001101: 4 of Clubs 011010: 7 of Hearts 010101: 6 of Clubs 101010: Jack of Hearts Swaps 110101 -> 010101 110110 -> 110000 110111 -> 010111 111101 -> 011111 111001 -> 101111 111011 -> 101011", "Since there are four 10s, the probability is: 4/52 = 1/13 (xi) Since there are 13 spades, the probability is: 13/52 = 1/4 (xii) Since there are 26 black cards, the probability is: 26/52 = 1/2 (xiii) There is only one card named seven of clubs. Hence, the probability is 1/52. (xiv) Since there are 4 jacks, the probability is: 4/52 = 1/13 (xv) There is only 1 card named ace of spade. Hence, the probability is 1/52. (xvi) Since there are 4 queens, the probability is: 4/52 = 1/13 (xvii) Since there are 13 hearts, the probability is: 13/52 = 1/4 (xviii) Since there are 26 red cards, the probability is 26/52 = 1/2 #### Question 6: An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white. #### Question 7: A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: (i) white? (ii) red? (iii) black? (iv) not red? #### Question 8: What is the probability that a number selected from the numbers 1, 2, 3, ..., 15 is a multiple of 4? #### Question 9: A bag contains 6 red, 8", "# Aptitude:: Probability @ : Home > Aptitude > Probability > General Questions ### Exercise \" #### A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white? A. $\\frac{3}{4}$ B. $\\frac{4}{7}$ C. $\\frac{1}{8}$ D. $\\frac{3}{7}$ #### One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)? A. $\\frac{1}{13}$ B. $\\frac{3}{13}$ C. $\\frac{1}{4}$ D. $\\frac{9}{52}$ #### Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is: A. $\\frac{3}{20}$ B. $\\frac{29}{34}$ C. $\\frac{47}{100}$ D. $\\frac{13}{102}$ #### A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is: A. $\\frac{1}{22}$ B. $\\frac{3}{22}$ C. $\\frac{2}{91}$ D. $\\frac{2}{77}$ #### A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is: A. $\\frac{1}{13}$ B. $\\frac{2}{13}$ C. $\\frac{1}{26}$ D. $\\frac{1}{52}$ Page 1 of 3 Submit*", "tarot deck which are more familiar in the Americas and English-speaking countries, namely cartomancy and other divinatory uses. Tarot has similarities to the American card game Spades, largely in their respective popularities as trick-taking games and the use of a static trump suit, but Tarot has bidding and scoring rules unique to the Tarot/Tarock family; the most similar games known in the U.S. might be Pinochle and/or Rook. The deck The game of Tarot is played using a 78-card tarot deck, which is composed of a numbered 21-card series of \"atouts\" (trump cards), one Fool (\"l'excuse\"), and 4 suits (spades, hearts, diamonds, clubs), divided into 10 numbers from 1 to 10, and then the face cards of jack (\"valet\"), knight (\"chevalier\"), queen (\"dame\") and king (\"roi\"). Though the tarot decks used for cartomancy will technically work when playing the card games, they are designed to be laid out in a tableau, not held in a hand of cards. Many tarot readers also disdain the use of reading tarots for playing games. Thus the style most often used for game playing is known as the \"Tarot Nouveau\" or \"Bourgeois Tarot\" which has card designs similar to the Anglo-American 52-card deck and is generally dissimilar to reading tarot decks. Three cards known as \"oudlers\" are of particular importance in the", "8 black balls. One ball is drawn at random from the bag. Find the probability that the ball is drawn • (i) white • (ii) black or red • (iii) not white • (iv) neither white nor black Solution: 5 red 6 white 7 green 8 black total no. of balls = 5 + 6 + 7 +8= 26 10. In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is 3/8 then, find the number of defective bulbs. Solution: Let the number of defective bulbs be ‘x’ Total number of bulbs = x + 20 ⇒ 8x = 3x + 60 ⇒ 5x = 60 ⇒ x = 12 ∴ No.of defective bulbs are = 12. 11. The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is • (i) a clavor • (ii) a queen of red card • (iii) a king of black card Solution: (i.e) remaining number of cards = 52 – 6 = 46 13 (i) P(a clavor) = $$\\frac{13}{46}$$", "A bag contains 6 red balls and some blue balls. if the probability of a drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag. GIVEN: A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball, TO FIND: the number of blue balls in the bag. Let the probability of getting a red ball be The probability of not getting a red ball or getting a blue ball be We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1.So Hence the probability of getting a red ball is We know that PROBABILITY = Hence total number of blue balls = total number of balls −red balls Hence total number of blue balls is #### Question 61: The king, queen and jack of clubs are removed form a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of (i) heart (ii) queen (iii) clubs (iv) a face card (v) a queen of diamond. Given: King, Queen and Jack of Clubs", "a few simplifications. Hearts has several possible rule variations, and the rules I will use are outlined here. For each game, each of the four players are dealt 13 of the cards in a standard deck. After that, there will be 13 \"tricks\", and during each trick the players will take turns playing exactly one of their cards. Each trick will consist of the following: • One player leads the trick by playing a card. The suit of this card is the \"led suit\". On the first trick the player with the 2 of clubs must lead with the 2 of clubs, on the following tricks whoever won the previous trick will lead. The leader may not lead a heart until another player has played a heart or the Qs (Queen of Spades). Exception: if the leader only has hearts in their hand, they may lead one of their hearts. The leader may lead the Qs at any time. • In a clockwise fashion, each player will take their turn playing one card in their hand of the led suit. If the player has no cards of the led suit, they may play any card. Exception: On the very first trick, no one after the leader may play a heart or the Qs, unless all of their cards", "game of hearts given just a deck of cards So, I've been playing a game of hearts (https://en.wikipedia.org/wiki/Hearts) with three of my friends but I'm not entirely sure if all of them were playing perfectly according to the rules. So I'd like to replay the game with everyone's cards openly visible. And because you like algorithms, you offered me your help determining everyone's cards from the final deck after each hand. However, I can't quite remember if you promised me a full program or just a named function. A card is represented with two characters. The first character represents its value and is one of 23456789TJQKA (2-10, jack, queen, king, ace). The second character represents its suit and is one of CDHS (clubs, diamonds, heards, spades). The input is a list of 52 cards. It can either be a list (array, vector...) of two-character strings or a single space-separated string. You will be given the cards in the exact order they were played. The list represents a valid deck and 2C is the first card in the deck. Output four sets of cards, each representing the starting hand of one player. The first set may correspond to any player but the rest must be ordered in the order of play (so if the first hand to be output", "the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is: A) A gain of Rs. 27 B) A loss of Rs. 37 C) A loss of Rs. 27 D) A gain of Rs. 37 Explanation: As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order). The overall resultant will remain same. So final amount with the person will be (in all cases): 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27 Hence the final result is: 64 − 27 37 A loss of Rs.37 8 133 Q: A card is drawn from a pack of 52 cards. The probability of getting a queen of the club or a king of the heart is? A) 1/26 B) 1/13 C) 2/13 D) 1/52 Explanation: Here in this pack of cards, n(S) = 52 Let E = event of getting a queen of the club or a king of the heart Then, n(E) = 2 P(E) = n(E)/n(S) = 2/52 = 1/26 7 229 Q: A room contains 3 brown, 5", "+0 # A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ( and , called `hearts' and -1 151 2 +49 A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ( and , called `hearts' and `diamonds') are red, the other two ( and , called `spades' and `clubs') are black. The cards in the deck are placed in random order (usually by a process called `shuffling'). What is the probability that the first two cards drawn from the deck are both red? off-topic Jun 9, 2020 edited by Guest Jun 9, 2020", "# Alice and Bob invent a game The rain was still falling in Wales, and Alice and Bob were terribly bored of having to stay inside the caravan. With just a deck of cards to entertain them they had played all of the card games in their repertoire – several times over. Suddenly Alice had an idea for another game they could play. Putting aside 21 cards, including the aces (“Too ambiguous,” she said…) and the two jokers, she dealt out two hands – one for her and one for Bob. To start the game off, Alice laid down three cards – the jack of diamonds, the three of hearts and the four of clubs: “So that scores me 24 points,” she said with a flourish, before replenishing her hand. “Oh,” said Bob, “so we’re playing properly then? In which case, I’ll play these…” Bob laid three of his own cards on the table (three of diamonds, six of hearts, king of clubs): “That’s 26, I think you’ll find,” Bob said smugly, picking up three more cards. Alice grimaced. “Fine, well I’ll do this then…” This time Alice laid down five cards (six of spades, four of diamonds, five of diamonds, king of spades, five of spades): “25 points,” she announced, collecting 5 more cards from the deck." ]

[ "the second ball is green is $$\\dfrac{1}{12}$$ Example 3 Mrs. Andrews has to select two students from 35 girls and 15 boys to be part of a club. What is the probability that both students are girls? Solution The total number of students $$= 35 + 15 = 50$$ Probability of choosing the first girl, P(girl 1) = $$\\dfrac { 35}{50}$$ Probability of choosing the second girl, P(girl 2) = $$\\dfrac { 34}{49}$$ Now, The probability that both students have chosen are girls $$\\text {P(first girl and second girl)}$$ $$= \\text{ P(first girl)} \\times \\text{P(second girl | first girl)}$$ $$= \\dfrac { 35}{50} \\times \\dfrac { 34}{49}$$ $$= \\dfrac{1190}{1666}$$ $$= \\dfrac{85}{119}\\$$ $$\\therefore$$The probability that both chosen students are girls is $$\\dfrac{85}{119}$$. Example 4 Joseph and David are playing a game of cards. Joseph drew a card at random without replacement. He asks David to help him determine the probability that the first card drawn was a queen and the second is a king. ### Solution As we understand that this probability is having a dependent event condition. P (drawing a queen in the first condition) =$$\\dfrac { 4}{52}$$ P (drawing a king in the second condition after a queen) = $$\\dfrac { 4}{51}$$ P (drawing a queen followed by a king) = $$\\dfrac{4}{52} \\times\\dfrac{4}{51}=\\dfrac{16}{2652}=\\dfrac{4}{663}$$ $$\\therefore$$The probability of drawing", "first draw) = }\\dfrac{4}{52}$ Assume that the first event is happened. i.e., a Queen is already drawn in the first draw and now B = event of getting a Queen in the second draw Since 1 Queen is drawn in the first draw, Total number of Queens remaining = 3Since 1 Queen is drawn in the first draw, Total number of cards = 52 - 1 = 51 $\\text{P(Queen in second draw) = }\\dfrac{3}{51}$ P(Queen in first draw and Queen in second draw) = P(Queen in first draw) × P(Queen in second draw) $= \\dfrac{4}{52} \\times \\dfrac{3}{51} = \\dfrac{1}{13} \\times \\dfrac{1}{17} = \\dfrac{1}{221}$ 29. Two dice are rolled together. What is the probability of getting two numbers whose product is even? A. $\\dfrac{17}{36}$ B. $\\dfrac{1}{3}$ C. $\\dfrac{3}{4}$ D. $\\dfrac{11}{25}$ Here is the answer and explanation Answer : Option C Explanation : ----------------------------------------------------------------------------------------- Solution 1 ----------------------------------------------------------------------------------------- Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces) Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36 Let E = the event of getting two numbers whose product is even = {(1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2),(5,4), (5,6),", "× Custom Search cancel × × Brahma A and B pick a card at random from a well shuffled pack of cards, one after the other replacing it every time till one of them gets a queen. If A begins the game, then the probability that A wins the game is 1 0 comment Jay Patel P(A wins in his first draw) $=\\dfrac{1}{13}$ P(A does not win in his first draw AND B does not win in his first draw AND A wins in his second draw) $=\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}$ P(A does not win in his first draw AND B does not win in his first draw AND A does not win in his second draw AND B does not win in his second draw AND A wins in his third draw $=\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}$ so on Required probability $=\\dfrac{1}{13}+\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}+\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12} {13}.\\dfrac{12}{13}.\\dfrac{1}{13}+\\cdots\\\\ =\\dfrac{\\dfrac{1}{13}}{1-\\dfrac{12}{13}×\\dfrac{12}{13}}=\\dfrac{13}{169-144}=\\dfrac{13}{25}$ 0 0 comment", "X View & Edit Profile Sign out X Yahoo Discussion Board Question A and B pick a card at random from a well shuffled pack of cards, one after the other replacing it every time till one of them gets a queen. If A begins the game, then the probability that A wins the game is probability 2017-03-05 15:50:52 Brahma P(A wins in his first draw) $=\\dfrac{1}{13}$ P(A does not win in his first draw AND B does not win in his first draw AND A wins in his second draw) $=\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}$ P(A does not win in his first draw AND B does not win in his first draw AND A does not win in his second draw AND B does not win in his second draw AND A wins in his third draw $=\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}$ so on Required probability $=\\dfrac{1}{13}+\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}+\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12} {13}.\\dfrac{12}{13}.\\dfrac{1}{13}+\\cdots\\\\ =\\dfrac{\\dfrac{1}{13}}{1-\\dfrac{12}{13}×\\dfrac{12}{13}}=\\dfrac{13}{169-144}=\\dfrac{13}{25}$ (0) (0) 2017-03-05 19:49:03 Jay Patel (Senior Maths Expert, careerbless.com)", "each of the other $6^2 - 6 = 30$ cases where there are $2$ copies is $\\dfrac{2}{6^4}$. Therefore, the probability here is $6\\dfrac{1}{6^4}+30\\dfrac{2}{6^4} = \\dfrac{11}{216}$. – user19405892 Nov 22 '15 at 1:04", "outcomes}}{\\text{Total number of outcomes}}$ Therefore, by using the above formula, P (getting a king or queen) $=\\dfrac{8}{52}=\\dfrac{2}{13}$. Hence, the probability of getting a king or queen is $\\dfrac{2}{13}$. Therefore, option (c) is correct. Note: The key concept for solving a problem is the knowledge of probability of occurrence of an event. Students must be careful while calculating the space for favorable events. There should be no redundancy of particular events in the favorable outcomes.", "[#permalink] 16 Feb 2008, 01:13 Expert's post -> 2 cards are drawn from a pack of 52 cards. 1) What is the probability that a King and a Queen will be selected ? 2) What is the probability that the first is a King and second is a Queen ? Can I assume that there are 4 Kings and 4 Queens in the card batch? 1)$$p=p_{1king}*p_{2queen}+p_{1queen}*p_{2king}=\\frac{4}{52}*\\frac{4}{51}+\\frac{4}{52}*\\frac{4}{51}=\\frac{32}{52*51}=\\frac{8}{13*51}$$ or $$p=p_{(king) or (queen)}*p_{opposite}=\\frac{8}{52}*\\frac{4}{51}=\\frac{32}{52*51}=\\frac{8}{13*51}$$ or $$p=\\frac{C^4_1*C^4_1*P^2_2}{P^{52}_2}=\\frac{4*4*2}{52*51}=\\frac{8}{13*51}$$ or $$p=\\frac{C^4_1*C^4_1}{C^{52}_2}=\\frac{4*4*2}{52*51}=\\frac{8}{13*51}$$ or $$p=1-\\frac{P^{44}_2+C^8_1*C^3_1+C^8_1*C^{44}_1*P^2_2}{P^{52}_2}=1-\\frac{44*43+8*3+8*44*2}{52*51}=1-\\frac{473+6+176}{13*51}=\\frac{663 -655}{13*51}=\\frac{8}{13*51}$$ 2) $$p=p_{1king}*p_{2queen}=\\frac{4}{52}*\\frac{4}{51}=\\frac{16}{52*51}=\\frac{4}{13*51}$$ or $$p=\\frac{C^4_1*C^4_1}{P^{52}_2}=\\frac{4*4}{52*51}=\\frac{4}{13*51}$$ or $$p_2=\\frac12*p_1=\\frac12*\\frac{8}{13*51}=\\frac{4}{13*51}$$ (symmetry: QK,KQ) _________________ HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame Re: probability basic [#permalink] 16 Feb 2008, 01:13 Display posts from previous: Sort by", "$P\\left( {\\dfrac{A}{{{S^1}}}} \\right)$ = $\\dfrac{{{}^{13}{C_2}}}{{{}^{51}{C_2}}}$ = $\\dfrac{{26}}{{425}}$ $P({S^1} \\cap A) = P({S^1}).P\\left( {\\dfrac{A}{{{S^1}}}} \\right)$ = $\\dfrac{3}{4} \\times \\dfrac{{26}}{{425}}$ The probability that the lost card is spade given that two spades are drawn = $P\\left( {\\dfrac{S}{A}} \\right)$ = $\\dfrac{{P(S \\cap A)}}{{P(A)}}$ = $\\dfrac{{P(S).P(A/S)}}{{P(S).P(A/S) + P({S^1}).P(A/{S^1})}}$ = $\\dfrac{{1/4 \\times 22/425}}{{1/4 \\times 22/425 + 3/4 \\times 26/425}}$ = 11/50 7. There are two bags containing white and black balls. In the first bag there are 8 white and 6 black balls and in the second bag, there are 4 white and 7 black balls. One ball is drawn at random from any of these two bags. Find the probability of this ball being black. a. 21/154 b. 7/54 c. 21/77 d. 41/77 Explanation: Probability = $\\dfrac{1}{2} \\times \\dfrac{{{}^6{C_1}}}{{{}^{14}{C_1}}} + \\dfrac{1}{2} \\times \\dfrac{{{}^7{C_1}}}{{{}^{11}{C_1}}}$ = $\\dfrac{{41}}{{77}}$ 8. A bag contains 1100 tickets numbered 1, 2, 3, ... 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it? a. 291/1100 b. 292/1100 c. 290/1100 d. 301/1100 Explanation: Numbers which dont have 2 from 1 to 9 = 8 Numbers which don't have 2 from 10 to 99: Let us take two places _ _. Now left most place is fixed in 8 ways. Units place is filled with", "red}) = \\dfrac{2}{52} = 1/26$ d) Two methods to answer the question. 1) Using Definition of the conditional probability given above $P( \\text{King given that it is a red card}) = P(King|red) = \\dfrac{ P( \\text{King and red})}{P(red)} = \\dfrac{1/26}{1/2} = 1/13$ 2) Using the restricted sample space Out of the 26 red cards (restricted sample space to the red only since this is the condition) there are 2 red; hence $P(King|red) = 2/26 = 1/13$ same as was found above using the definition e) Two methods to answer the question. 1) Using Definition of the conditional probability given above $P( \\text{red card given that it is a King}) = P(red|King) = \\dfrac{ P( \\text{red and King})}{P(King)} = \\dfrac{1/26}{1/13} = 1/2$ 2) Using the restricted sample space Out of the 4 Kings cards (restricted sample space to the Kings) there are 2 red; hence $P(red|King) = 2/4 = 1/2$ same as was found above using the definition f) Two methods to answer the question. 1) Using Definition of the conditional probability given above $P( \\text{Queen given that it is Heart}) = P(Queen|heart) = \\dfrac{ P( \\text{Queen and Heart})}{P(Heart)} = \\dfrac{1/52}{13/52} = 1/13$ 2) Using the restricted sample space Out of the 13 hearts (restricted sample space to the hearts) there is 1 King; hence $P(Queen|heart) = 1/13$ same", "earlier. Example: 20 Two cards are drawn simultaneously from a well-shuffled deck of $52$ cards. Find the mean and variance of the number of kings. Solution: 20 The number of kings is the random variable here. Call it $X$. The values that $X$ can take are $0$$1$$2$. The probabilities of the various values are easily calculated: $P(X = 0) = \\dfrac{{^{48}{C_2}}}{{^{52}{C_2}}} = \\dfrac{{188}}{{221}}$ $P(X = 1) = \\dfrac{{^4{C_1} \\times {\\,^{48}}{C_1}}}{{^{52}{C_2}}} = \\dfrac{{32}}{{221}}$ $P(X = 2) = \\dfrac{{^4{C_2}}}{{^{52}{C_2}}} = \\dfrac{1}{{221}}$ The $PD$ of $X$ is therefore The mean is simply $\\left\\langle X \\right\\rangle \\; = 0 \\times \\dfrac{{188}}{{221}} + 1 \\times \\dfrac{{32}}{{221}} + 2 \\times \\dfrac{1}{{221}}$ $= \\dfrac{{34}}{{221}}$ To calculate the variance, we first calculate $\\left\\langle {{X^2}} \\right\\rangle$: $\\left\\langle {{X^2}} \\right\\rangle \\; = \\sum\\limits_{i = 1}^n {x_i^2} {p_i}$ $= {0^2} \\times \\dfrac{{188}}{{221}} + {1^2} \\times \\dfrac{{32}}{{221}} + {2^2} \\times \\dfrac{1}{{221}} = \\dfrac{{36}}{{221}}$ Thus, the variance is ${\\rm{Var}}(X) = \\;\\left\\langle {{X^2}} \\right\\rangle - {\\left\\langle X \\right\\rangle ^2}$ $= \\dfrac{{36}}{{221}} - {\\left( {\\dfrac{{34}}{{221}}} \\right)^2}$ $= \\dfrac{{6800}}{{48841}}$", "Four cards are randomly selected from a pack of $52$ cards. If the first two cards are kings, what is the probability that the third card is a king? 1. $\\dfrac{4}{52} \\\\$ 2. $\\dfrac{2}{50} \\\\$ 3. $\\left ( \\dfrac{1}{52}\\right )\\times\\left ( \\dfrac{1}{52}\\right ) \\\\$ 4. $\\left ( \\dfrac{1}{52}\\right )\\times\\left ( \\dfrac{1}{51}\\right )\\times\\left ( \\dfrac{1}{50}\\right )$", "these combinations are equally likely, so the probability of a specific combination is $$\\dfrac{1}{\\binom{N}{n}}$$. #### Example: Tattslotto In the Saturday evening draw of Tattslotto, six winning balls are drawn at random from 45 balls numbered 1 to 45. There are 8 145 060 different ways of choosing six numbers from 45; that is, $$\\tbinom{45}{6} = 8\\ 145\\ 060$$. So the probability of any specific combination of six balls being chosen is $$\\dfrac{1}{8\\ 145\\ 060}$$. An equivalent way to derive the same probability is to think of the balls being chosen in sequence. Consider a specific choice of six balls, such as $$\\{3, 4, 20, 37, 40, 45\\}$$. The probability that the first ball chosen is in this specific set is $$\\dfrac{6}{45}$$. The conditional probability that the second ball chosen is one of the remaining five balls in the set, given that the first ball chosen was in the set, is equal to $$\\dfrac{5}{44}$$, and so on. The conditional probability that the sixth ball chosen is in the set, given that the first five chosen were, is equal to $$\\dfrac{1}{40}$$. Putting all this together using the multiplication theorem (from the module Probability), we obtain the probability of the specific set being chosen: $\\Pr(\\text{specific set is chosen}) = \\dfrac{6}{45} \\times \\dfrac{5}{44} \\times \\dfrac{4}{43} \\times \\dfrac{3}{42} \\times \\dfrac{2}{41} \\times \\dfrac{1}{40} =", "is $\\frac{3}{51}$. If the first card is not a Queen, the probability the second is a Queen is $\\frac{4}{51}$. Thus given that the first card is not a King, the probability the second is a Queen is $$\\frac{4}{48}\\cdot\\frac{3}{51}+\\frac{44}{48}\\cdot \\frac{4}{51}.$$ This simplifies to $\\frac{1}{12}\\cdot\\frac{47}{51}$. That is somewhat larger than the answer that you gave, so even at the start ($n=2$) your expression does not give the exact probability.", "red cards = 26, The number of 6 labeled cards = 4, and The number of red cards that are labeled 6 = 2. Therefore, The probability of getting a red card, P(A) = $$\\dfrac{26}{52}$$ The probability of getting a 6, P(B) = $$\\dfrac{4}{52}$$ The probability of getting both a Red and a 6, P(A∩B) = $$\\dfrac{2}{52}$$. Using the P(A∪B) formula, P(A∪B) = P(A) + P(B) - P(A∩B) $$P(A∪B)= \\dfrac{26}{52}+\\dfrac{4}{52}-\\dfrac{2}{52}\\\\[0.2cm]= \\dfrac{30-2}{52}\\\\[0.2cm]= \\dfrac{28}{52}\\\\[0.2cm]= \\dfrac{7}{13}$$ ### What is the probability of getting a 2 or 3 when a die is rolled? Solution: To find: The probability of getting a 2 or 3 when a die is rolled. Let A and B be the events of getting a 2 and getting a 3 when a die is rolled. Then, P(A) = 1 / 6 and P(B) = 1 / 6. In this case, A and B are mutually exclusive as we cannot get 2 and 3 in the same roll of a die. Hence, P(A∩B) = 0. Using the P(A∪B) formula, P(A∪B) = P(A) + P(B) - P(A∩B) P(A∪B) = 1 / 6 + 1 / 6 - 0 = 2 / 6 = 1 / 3", "of getting 3 A's and 6 not-A's has already been calculated. Are you with me? Now let's think about conditional probability. What is the probability of getting three B' s in n trials GIVEN that three of the trials were A's? If the first trial is not A, what is the probability that it is B? Well the probability of not A is 2/3, and the probability of B is 1/3 so the probability of B given not A is $\\dfrac{\\dfrac{1}{3}}{\\dfrac{2}{3}} = \\dfrac{1}{2}.$ Now there are six places in the series where the not-A's can go given that there are already 3 A's. We can think about this problem just the way we did about the first problem which gives probability of three B's in the six slots given that they are not A $\\dfrac{6!}{3! * 3!} \\left ( \\dfrac{1}{2} \\right )^6$ Have you kept up? So P(X and Y) = P(x | Y) * P(Y) And the answer is $\\dfrac{9!}{3! \\cancel{6!}} *\\left ( \\dfrac{1}{3} \\right )^3 * \\left ( \\dfrac{2}{3} \\right )^6 * \\dfrac{\\cancel{6!}}{3! * 3!} \\left ( \\dfrac{1}{2} \\right )^6 = \\dfrac{9! * \\cancel{2^6}}{3! * 3! * 3! * 3^9 * \\cancel{2^6}} = \\dfrac{9!}{(3!)^3 * 3^9} \\approx 8.5\\%.$ If you got lost somewhere, ask more questions. Thanks from 123qwerty, EvanJ and manus May 29th, 2016, 07:01", "Step 2: $P(X=0)$=probability of no queen being drawn $\\qquad\\quad\\;=\\large\\frac{48C_2}{52C_2}$ $\\qquad\\quad\\;=\\large\\frac{48\\times 47/1\\times 2}{52\\times 51/1\\times 2}$ $\\qquad\\quad\\;=\\large\\frac{564}{663}$ $P(X=1)$=probability that 1 card is a queen $\\qquad\\quad\\;=\\large\\frac{48C_1\\times 4C_1}{52C_2}$ $\\qquad\\quad\\;=\\large\\frac{48\\times 4}{52\\times 51/1\\times 2}$ $\\qquad\\quad\\;=\\large\\frac{96}{663}$ $\\qquad\\quad\\;=\\large\\frac{32}{221}$ $P(X=2)$=probability that both cards are queens $\\qquad\\quad\\;=\\large\\frac{4C_2}{52C_2}$ $\\qquad\\quad\\;=\\large\\frac{4\\times 3/1\\times 2}{52\\times 51/1\\times 2}$ $\\qquad\\quad\\;=\\large\\frac{3}{663}$ $\\qquad\\quad\\;=\\large\\frac{1}{221}$ Step 3: The probability distribution is given by", "$$P(A|B) * P(B) = P(B|A) * P(A)$$ Where $A$ is the event \"draw two queens and a king\", and $B$ is the specific known condition in each part. First note that in all three cases, $P(B|A)$ is $1$ since drawing two queens and a king satisfies each given condition. Since the problem asks for $P(A|B)$, we'll use: $$P(A|B) = \\frac{P(A)}{P(B)}$$ So then, what's $P(A)$? (Since that's the same in all three cases) Well, drawing two queens and then a king does indeed have a probability of $$\\left(\\frac{4}{52}\\right)^3$$ But \"two queens and a king\" would imply that the king can be drawn first, second, or third so in fact $$P(A) = 3 \\left(\\frac{4}{52}\\right)^3$$ So now for the the parts: In part (a) $B$ is \"at least one queen\". The chance of \"at least one queen\" is going to be $1$ minus the chance of \"no queens\", so: $$P(A|B) = \\frac{3 \\left(\\frac{4}{52}\\right)^3}{1 - \\left(\\frac{48}{52}\\right)^3} = \\frac{3}{469}$$ In part (b), $B$ is \"at least two face cards\". There a few ways to calculate $P(B)$, but I think that the easiest conceptually is to split it into two (mutually exclusive) cases: two face cards and a non-face card ($B_1$), or three face cards.($B_2$) In the first case, the non-face card can be first, second, or third, so $$P(B_1) = 3 \\left(\\frac{12}{52}\\right)^2\\left(\\frac{40}{52}\\right) =", "the probability the $k$-th is the biggest among these is $\\dfrac{1}{k}$. So using your calculation we have that our conditional probability is $\\dfrac{\\frac{1}{n}}{\\frac{1}{k}}$. - Thank you André, your answers are very easy to understand and are elegant. I have upvoted, but I chose to accept a different answer that showed me how to do the calculation that was trouble for me, hope its ok. – Belgi Nov 21 '12 at 22:04 No problem! I wondered whether to give the cancellation solution or the equally likely version. But when in doubt, I go for the probabilistic argument and not the counting argument. – André Nicolas Nov 21 '12 at 22:17 indeed this was more elegant and nice, I will try to think like this in future questions. I tried to count and didn't notice there was such a nice way to look at things – Belgi Nov 21 '12 at 23:41 The probability that the maximum occurs within the first $k$ cards is $\\frac kn$. This is also the answer to the problem question because whether the maximum of all cards appears among the first $k$ cards is independent from the relative rank of the $k$th card among the first $k$ cards. - Can you please add some more details ? I don't understand why there is such independence", "the three remaining Kings: . $\\frac{3}{48}$ Therefore, the probability is: . $\\frac{69,184}{270,725} \\times \\frac{3}{48} \\;=\\;\\frac{4324}{270,725} \\;=\\;0.015971927 \\;\\approx\\;\\boxed{0.016}$", "by number of coins $= \\dfrac{1}{4}$ $\\therefore$ Chance of drawing a sovereign from first purse is $\\dfrac{1}{3} \\times \\dfrac{1}{4} = \\dfrac{1}{{12}}$ (Using rule of product) Find the chance of drawing a sovereign from the second purse. There are $6$ ($2$ sovereign$+ 4$ Shillings) coins in second purse, hence the chance of drawing a sovereign is given by number of shillings divided by number of coins $= \\dfrac{2}{6}$ $\\therefore$ Chance of drawing a sovereign from second purse is $\\dfrac{1}{3} \\times \\dfrac{2}{6} = \\dfrac{1}{9}$ (Using rule of product) Find the chance of drawing a sovereign from the third purse. There are $4$ ($3$ sovereign$+ 1$ Shillings) coins in third purse, hence the chance of drawing a sovereign is given by number of shillings divided by number of coins $= \\dfrac{3}{4}$ $\\therefore$ Chance of drawing a sovereign from second purse is $\\dfrac{1}{3} \\times \\dfrac{3}{4} = \\dfrac{1}{4}$ (Using rule of product) Find a chance of getting a sovereign, when a coin is taken out of one of the purses randomly. We calculate the sum of all chances from the first, second and third purse. Therefore, using the rule of sum, where each event of taking out a sovereign from each purse is an independent event. $\\therefore$ Chance of coin to be a sovereign $= \\dfrac{1}{{12}} + \\dfrac{1}{9} + \\dfrac{1}{4}$ By taking LCM" ]

[ "the second ball is green is $$\\dfrac{1}{12}$$ Example 3 Mrs. Andrews has to select two students from 35 girls and 15 boys to be part of a club. What is the probability that both students are girls? Solution The total number of students $$= 35 + 15 = 50$$ Probability of choosing the first girl, P(girl 1) = $$\\dfrac { 35}{50}$$ Probability of choosing the second girl, P(girl 2) = $$\\dfrac { 34}{49}$$ Now, The probability that both students have chosen are girls $$\\text {P(first girl and second girl)}$$ $$= \\text{ P(first girl)} \\times \\text{P(second girl | first girl)}$$ $$= \\dfrac { 35}{50} \\times \\dfrac { 34}{49}$$ $$= \\dfrac{1190}{1666}$$ $$= \\dfrac{85}{119}\\$$ $$\\therefore$$The probability that both chosen students are girls is $$\\dfrac{85}{119}$$. Example 4 Joseph and David are playing a game of cards. Joseph drew a card at random without replacement. He asks David to help him determine the probability that the first card drawn was a queen and the second is a king. ### Solution As we understand that this probability is having a dependent event condition. P (drawing a queen in the first condition) =$$\\dfrac { 4}{52}$$ P (drawing a king in the second condition after a queen) = $$\\dfrac { 4}{51}$$ P (drawing a queen followed by a king) = $$\\dfrac{4}{52} \\times\\dfrac{4}{51}=\\dfrac{16}{2652}=\\dfrac{4}{663}$$ $$\\therefore$$The probability of drawing", "$P\\left( {\\dfrac{A}{{{S^1}}}} \\right)$ = $\\dfrac{{{}^{13}{C_2}}}{{{}^{51}{C_2}}}$ = $\\dfrac{{26}}{{425}}$ $P({S^1} \\cap A) = P({S^1}).P\\left( {\\dfrac{A}{{{S^1}}}} \\right)$ = $\\dfrac{3}{4} \\times \\dfrac{{26}}{{425}}$ The probability that the lost card is spade given that two spades are drawn = $P\\left( {\\dfrac{S}{A}} \\right)$ = $\\dfrac{{P(S \\cap A)}}{{P(A)}}$ = $\\dfrac{{P(S).P(A/S)}}{{P(S).P(A/S) + P({S^1}).P(A/{S^1})}}$ = $\\dfrac{{1/4 \\times 22/425}}{{1/4 \\times 22/425 + 3/4 \\times 26/425}}$ = 11/50 7. There are two bags containing white and black balls. In the first bag there are 8 white and 6 black balls and in the second bag, there are 4 white and 7 black balls. One ball is drawn at random from any of these two bags. Find the probability of this ball being black. a. 21/154 b. 7/54 c. 21/77 d. 41/77 Explanation: Probability = $\\dfrac{1}{2} \\times \\dfrac{{{}^6{C_1}}}{{{}^{14}{C_1}}} + \\dfrac{1}{2} \\times \\dfrac{{{}^7{C_1}}}{{{}^{11}{C_1}}}$ = $\\dfrac{{41}}{{77}}$ 8. A bag contains 1100 tickets numbered 1, 2, 3, ... 1100. If a ticket is drawn out of it at random, what is the probability that the ticket drawn has the digit 2 appearing on it? a. 291/1100 b. 292/1100 c. 290/1100 d. 301/1100 Explanation: Numbers which dont have 2 from 1 to 9 = 8 Numbers which don't have 2 from 10 to 99: Let us take two places _ _. Now left most place is fixed in 8 ways. Units place is filled with", "× Custom Search cancel × × Brahma A and B pick a card at random from a well shuffled pack of cards, one after the other replacing it every time till one of them gets a queen. If A begins the game, then the probability that A wins the game is 1 0 comment Jay Patel P(A wins in his first draw) $=\\dfrac{1}{13}$ P(A does not win in his first draw AND B does not win in his first draw AND A wins in his second draw) $=\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}$ P(A does not win in his first draw AND B does not win in his first draw AND A does not win in his second draw AND B does not win in his second draw AND A wins in his third draw $=\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}$ so on Required probability $=\\dfrac{1}{13}+\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}+\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12} {13}.\\dfrac{12}{13}.\\dfrac{1}{13}+\\cdots\\\\ =\\dfrac{\\dfrac{1}{13}}{1-\\dfrac{12}{13}×\\dfrac{12}{13}}=\\dfrac{13}{169-144}=\\dfrac{13}{25}$ 0 0 comment", "X View & Edit Profile Sign out X Yahoo Discussion Board Question A and B pick a card at random from a well shuffled pack of cards, one after the other replacing it every time till one of them gets a queen. If A begins the game, then the probability that A wins the game is probability 2017-03-05 15:50:52 Brahma P(A wins in his first draw) $=\\dfrac{1}{13}$ P(A does not win in his first draw AND B does not win in his first draw AND A wins in his second draw) $=\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}$ P(A does not win in his first draw AND B does not win in his first draw AND A does not win in his second draw AND B does not win in his second draw AND A wins in his third draw $=\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}$ so on Required probability $=\\dfrac{1}{13}+\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{1}{13}+\\dfrac{12}{13}.\\dfrac{12}{13}.\\dfrac{12} {13}.\\dfrac{12}{13}.\\dfrac{1}{13}+\\cdots\\\\ =\\dfrac{\\dfrac{1}{13}}{1-\\dfrac{12}{13}×\\dfrac{12}{13}}=\\dfrac{13}{169-144}=\\dfrac{13}{25}$ (0) (0) 2017-03-05 19:49:03 Jay Patel (Senior Maths Expert, careerbless.com)", "red}) = \\dfrac{2}{52} = 1/26$ d) Two methods to answer the question. 1) Using Definition of the conditional probability given above $P( \\text{King given that it is a red card}) = P(King|red) = \\dfrac{ P( \\text{King and red})}{P(red)} = \\dfrac{1/26}{1/2} = 1/13$ 2) Using the restricted sample space Out of the 26 red cards (restricted sample space to the red only since this is the condition) there are 2 red; hence $P(King|red) = 2/26 = 1/13$ same as was found above using the definition e) Two methods to answer the question. 1) Using Definition of the conditional probability given above $P( \\text{red card given that it is a King}) = P(red|King) = \\dfrac{ P( \\text{red and King})}{P(King)} = \\dfrac{1/26}{1/13} = 1/2$ 2) Using the restricted sample space Out of the 4 Kings cards (restricted sample space to the Kings) there are 2 red; hence $P(red|King) = 2/4 = 1/2$ same as was found above using the definition f) Two methods to answer the question. 1) Using Definition of the conditional probability given above $P( \\text{Queen given that it is Heart}) = P(Queen|heart) = \\dfrac{ P( \\text{Queen and Heart})}{P(Heart)} = \\dfrac{1/52}{13/52} = 1/13$ 2) Using the restricted sample space Out of the 13 hearts (restricted sample space to the hearts) there is 1 King; hence $P(Queen|heart) = 1/13$ same", "randomly drawn from a deck of 52 cards. What is the probability getting a five of Spade or Club? 1/52 1/13 1/26 1/12 10. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white? $\\dfrac{3}{4}$ $\\dfrac{4}{7}$ $\\dfrac{1}{8}$ $\\dfrac{3}{7}$ *Click on the QNo to display a Question. Total Ans Pipes Prev Next Permutations '; Aptitude Coaching - Free Live Session !!", "# Aptitude:: Probability @ : Home > Aptitude > Probability > General Questions ### Exercise \" #### A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white? A. $\\frac{3}{4}$ B. $\\frac{4}{7}$ C. $\\frac{1}{8}$ D. $\\frac{3}{7}$ #### One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)? A. $\\frac{1}{13}$ B. $\\frac{3}{13}$ C. $\\frac{1}{4}$ D. $\\frac{9}{52}$ #### Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is: A. $\\frac{3}{20}$ B. $\\frac{29}{34}$ C. $\\frac{47}{100}$ D. $\\frac{13}{102}$ #### A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is: A. $\\frac{1}{22}$ B. $\\frac{3}{22}$ C. $\\frac{2}{91}$ D. $\\frac{2}{77}$ #### A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is: A. $\\frac{1}{13}$ B. $\\frac{2}{13}$ C. $\\frac{1}{26}$ D. $\\frac{1}{52}$ Page 1 of 3 Submit*", "8 black balls. One ball is drawn at random from the bag. Find the probability that the ball is drawn • (i) white • (ii) black or red • (iii) not white • (iv) neither white nor black Solution: 5 red 6 white 7 green 8 black total no. of balls = 5 + 6 + 7 +8= 26 10. In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is 3/8 then, find the number of defective bulbs. Solution: Let the number of defective bulbs be ‘x’ Total number of bulbs = x + 20 ⇒ 8x = 3x + 60 ⇒ 5x = 60 ⇒ x = 12 ∴ No.of defective bulbs are = 12. 11. The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is • (i) a clavor • (ii) a queen of red card • (iii) a king of black card Solution: (i.e) remaining number of cards = 52 – 6 = 46 13 (i) P(a clavor) = $$\\frac{13}{46}$$", "{/eq} • 6. A blue (B), gray (G), and mauve (M) ball were placed in a bowl . One ball is to be randomly drawn from this bowl. The probability that the color of the ball drawn from the bowl will be mauve is _____. • {eq}\\dfrac{2}{3} {/eq} • {eq}\\dfrac{1}{3} {/eq} • 1 • {eq}\\dfrac{1}{2} {/eq} • 7. A blue (B), gray (G), and mauve (M) ball were placed in a bowl. Two balls are to be simultaneously and randomly drawn from this bowl. What is the probability that both of the balls drawn from the bowl will not be the mauve ball? • {eq}\\dfrac{2}{3} {/eq} • {eq}\\dfrac{1}{3} {/eq} • {eq}\\dfrac{1}{6} {/eq} • {eq}\\dfrac{1}{2} {/eq} • 8. One card is to be randomly drawn from four cards: ace of hearts (Ah), ace of spades (As), ace of diamonds (Ad), and ace of clubs (Ac). What is the probability that the card drawn will be a face card? • 1 • {eq}\\dfrac{1}{2} {/eq} • 0 • {eq}\\dfrac{1}{4} {/eq} • 9. Two cards are randomly and simultaneously drawn from four cards: ace of hearts (Ah), ace of spades (As), ace of diamonds (Ad), and ace of clubs (Ac). Find the probability that both cards drawn will be red-suited. • {eq}\\dfrac{1}{2} {/eq} • {eq}\\dfrac{1}{6} {/eq} • {eq}\\dfrac{1}{3} {/eq} • {eq}\\dfrac{1}{12} {/eq}", "{ 1 }{ 11 }$$. Do you agree with this argument ? Solution: Question 60. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag. [CBSE 2007] Solution: Question 61. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of (i) heart (ii) queen (iii) clubs (iv) a face card (v) a queen of diamond. Solution: Question 62. Two dice are thrown simultaneously. What is the probability that : (i) 5 will not come up on either of them ? (ii) 5 will come up on at least one ? (iii) 5 will come up at both dice? [CBSE 2009] Solution: Question 63. A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4. Solution: Question 64. A dice is rolled twice. Find the probability that (i) 5 will not come up either time. (ii) 5 will come up exactly one time. [CBSE 2014] Solution: Question 65. All the black face", "first draw) = }\\dfrac{4}{52}$ Assume that the first event is happened. i.e., a Queen is already drawn in the first draw and now B = event of getting a Queen in the second draw Since 1 Queen is drawn in the first draw, Total number of Queens remaining = 3Since 1 Queen is drawn in the first draw, Total number of cards = 52 - 1 = 51 $\\text{P(Queen in second draw) = }\\dfrac{3}{51}$ P(Queen in first draw and Queen in second draw) = P(Queen in first draw) × P(Queen in second draw) $= \\dfrac{4}{52} \\times \\dfrac{3}{51} = \\dfrac{1}{13} \\times \\dfrac{1}{17} = \\dfrac{1}{221}$ 29. Two dice are rolled together. What is the probability of getting two numbers whose product is even? A. $\\dfrac{17}{36}$ B. $\\dfrac{1}{3}$ C. $\\dfrac{3}{4}$ D. $\\dfrac{11}{25}$ Here is the answer and explanation Answer : Option C Explanation : ----------------------------------------------------------------------------------------- Solution 1 ----------------------------------------------------------------------------------------- Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces) Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36 Let E = the event of getting two numbers whose product is even = {(1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2),(5,4), (5,6),", "[#permalink] 16 Feb 2008, 01:13 Expert's post -> 2 cards are drawn from a pack of 52 cards. 1) What is the probability that a King and a Queen will be selected ? 2) What is the probability that the first is a King and second is a Queen ? Can I assume that there are 4 Kings and 4 Queens in the card batch? 1)$$p=p_{1king}*p_{2queen}+p_{1queen}*p_{2king}=\\frac{4}{52}*\\frac{4}{51}+\\frac{4}{52}*\\frac{4}{51}=\\frac{32}{52*51}=\\frac{8}{13*51}$$ or $$p=p_{(king) or (queen)}*p_{opposite}=\\frac{8}{52}*\\frac{4}{51}=\\frac{32}{52*51}=\\frac{8}{13*51}$$ or $$p=\\frac{C^4_1*C^4_1*P^2_2}{P^{52}_2}=\\frac{4*4*2}{52*51}=\\frac{8}{13*51}$$ or $$p=\\frac{C^4_1*C^4_1}{C^{52}_2}=\\frac{4*4*2}{52*51}=\\frac{8}{13*51}$$ or $$p=1-\\frac{P^{44}_2+C^8_1*C^3_1+C^8_1*C^{44}_1*P^2_2}{P^{52}_2}=1-\\frac{44*43+8*3+8*44*2}{52*51}=1-\\frac{473+6+176}{13*51}=\\frac{663 -655}{13*51}=\\frac{8}{13*51}$$ 2) $$p=p_{1king}*p_{2queen}=\\frac{4}{52}*\\frac{4}{51}=\\frac{16}{52*51}=\\frac{4}{13*51}$$ or $$p=\\frac{C^4_1*C^4_1}{P^{52}_2}=\\frac{4*4}{52*51}=\\frac{4}{13*51}$$ or $$p_2=\\frac12*p_1=\\frac12*\\frac{8}{13*51}=\\frac{4}{13*51}$$ (symmetry: QK,KQ) _________________ HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame Re: probability basic [#permalink] 16 Feb 2008, 01:13 Display posts from previous: Sort by", "cards. Find the probability that one is a red card the other is a queen. (a) $$\\frac{103}{1326}$$ (b) $$\\frac{101}{1326}$$ (c) $$\\frac{101}{1426}$$ (d) $$\\frac{103}{1426}$$ Answer: (b) $$\\frac{101}{1326}$$ Question 17. Given that, the events A and B are such that P(A) = $$\\frac{1}{2}$$, P(A ∪ B) = $$\\frac{3}{5}$$ and P(b) = P. Then probabilities of B if A and B are mutually exclusive and independent respetively are (a) $$\\frac{1}{2}, \\frac{1}{3}$$ (b) $$\\frac{1}{5}, \\frac{1}{3}$$ (c) $$\\frac{2}{3}, \\frac{1}{3}$$ (d) $$\\frac{1}{10}, \\frac{1}{5}$$ Answer: (d) $$\\frac{1}{10}, \\frac{1}{5}$$ Question 18. Two cards from an ordinary deck of 52 cards are missing. What is the probability that a random card drawn from this deck is a spade? (a) $$\\frac{3}{4}$$ (b) $$\\frac{2}{3}$$ (c) $$\\frac{1}{2}$$ (d) $$\\frac{1}{4}$$ Answer: (d) $$\\frac{1}{4}$$ Question 19. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. (a) $$\\frac{5}{8}$$ (b) $$\\frac{3}{8}$$ (c) $$\\frac{7}{8}$$ (d) $$\\frac{1}{8}$$ Answer: (b) $$\\frac{3}{8}$$ Question 20. A bag contains 4 balls. Two balls are drawn at random and are found to be white. What is the probability that all balls are white? (a) $$\\frac{2}{5}$$ (b) $$\\frac{3}{5}$$ (c) $$\\frac{4}{5}$$ (d) $$\\frac{1}{5}$$ Answer: (b) $$\\frac{3}{5}$$ Question 21. A bag contains 3 green and 7 white balls. Two balls are", "A bag contains 6 red balls and some blue balls. if the probability of a drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag. GIVEN: A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball, TO FIND: the number of blue balls in the bag. Let the probability of getting a red ball be The probability of not getting a red ball or getting a blue ball be We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1.So Hence the probability of getting a red ball is We know that PROBABILITY = Hence total number of blue balls = total number of balls −red balls Hence total number of blue balls is #### Question 61: The king, queen and jack of clubs are removed form a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of (i) heart (ii) queen (iii) clubs (iv) a face card (v) a queen of diamond. Given: King, Queen and Jack of Clubs", "# Aptitude Probability Practice Q&A 1. A letter is randomly taken from English alphabets. What is the probability that the letter selected is not a vowel? 5/25 2/25 5/26 21/26 2. The probability A getting a job is 1/5 and that of B is 1/7 . What is the probability that only one of them gets a job? 11/35 12/35 2/7 1/7 3. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize? $\\dfrac{1}{10}$ $\\dfrac{2}{5}$ $\\dfrac{2}{7}$ $\\dfrac{5}{7}$ 4. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is: $\\dfrac{1}{13}$ $\\dfrac{2}{13}$ $\\dfrac{1}{26}$ $\\dfrac{1}{52}$ 5. A letter is chosen at random from the word ASSASSINATION. What is the probability that it is a consonant? 4/13 8/13 7/13 6/13 6. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even? $\\dfrac{1}{2}$ $\\dfrac{3}{4}$ $\\dfrac{3}{8}$ $\\dfrac{5}{16}$ 7. Two dice are tossed. The probability that the total score is a prime number is: $\\dfrac{1}{6}$ $\\dfrac{5}{12}$ $\\dfrac{1}{2}$ $\\dfrac{7}{9}$ 8. From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings? $\\dfrac{1}{15}$ $\\dfrac{25}{57}$ $\\dfrac{35}{256}$ $\\dfrac{1}{221}$ 9. A card is", "these combinations are equally likely, so the probability of a specific combination is $$\\dfrac{1}{\\binom{N}{n}}$$. #### Example: Tattslotto In the Saturday evening draw of Tattslotto, six winning balls are drawn at random from 45 balls numbered 1 to 45. There are 8 145 060 different ways of choosing six numbers from 45; that is, $$\\tbinom{45}{6} = 8\\ 145\\ 060$$. So the probability of any specific combination of six balls being chosen is $$\\dfrac{1}{8\\ 145\\ 060}$$. An equivalent way to derive the same probability is to think of the balls being chosen in sequence. Consider a specific choice of six balls, such as $$\\{3, 4, 20, 37, 40, 45\\}$$. The probability that the first ball chosen is in this specific set is $$\\dfrac{6}{45}$$. The conditional probability that the second ball chosen is one of the remaining five balls in the set, given that the first ball chosen was in the set, is equal to $$\\dfrac{5}{44}$$, and so on. The conditional probability that the sixth ball chosen is in the set, given that the first five chosen were, is equal to $$\\dfrac{1}{40}$$. Putting all this together using the multiplication theorem (from the module Probability), we obtain the probability of the specific set being chosen: $\\Pr(\\text{specific set is chosen}) = \\dfrac{6}{45} \\times \\dfrac{5}{44} \\times \\dfrac{4}{43} \\times \\dfrac{3}{42} \\times \\dfrac{2}{41} \\times \\dfrac{1}{40} =", "the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is: A) A gain of Rs. 27 B) A loss of Rs. 37 C) A loss of Rs. 27 D) A gain of Rs. 37 Explanation: As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order). The overall resultant will remain same. So final amount with the person will be (in all cases): 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27 Hence the final result is: 64 − 27 37 A loss of Rs.37 8 133 Q: A card is drawn from a pack of 52 cards. The probability of getting a queen of the club or a king of the heart is? A) 1/26 B) 1/13 C) 2/13 D) 1/52 Explanation: Here in this pack of cards, n(S) = 52 Let E = event of getting a queen of the club or a king of the heart Then, n(E) = 2 P(E) = n(E)/n(S) = 2/52 = 1/26 7 229 Q: A room contains 3 brown, 5", "red cards = 26, The number of 6 labeled cards = 4, and The number of red cards that are labeled 6 = 2. Therefore, The probability of getting a red card, P(A) = $$\\dfrac{26}{52}$$ The probability of getting a 6, P(B) = $$\\dfrac{4}{52}$$ The probability of getting both a Red and a 6, P(A∩B) = $$\\dfrac{2}{52}$$. Using the P(A∪B) formula, P(A∪B) = P(A) + P(B) - P(A∩B) $$P(A∪B)= \\dfrac{26}{52}+\\dfrac{4}{52}-\\dfrac{2}{52}\\\\[0.2cm]= \\dfrac{30-2}{52}\\\\[0.2cm]= \\dfrac{28}{52}\\\\[0.2cm]= \\dfrac{7}{13}$$ ### What is the probability of getting a 2 or 3 when a die is rolled? Solution: To find: The probability of getting a 2 or 3 when a die is rolled. Let A and B be the events of getting a 2 and getting a 3 when a die is rolled. Then, P(A) = 1 / 6 and P(B) = 1 / 6. In this case, A and B are mutually exclusive as we cannot get 2 and 3 in the same roll of a die. Hence, P(A∩B) = 0. Using the P(A∪B) formula, P(A∪B) = P(A) + P(B) - P(A∩B) P(A∪B) = 1 / 6 + 1 / 6 - 0 = 2 / 6 = 1 / 3", "is a spade and one is a heart, is: 1. $$\\frac{{3}}{{20}}$$ 2. $$\\frac{{29}}{{34}}$$ 3. $$\\frac{{47}}{{100}}$$ 4. $$\\frac{{13}}{{102}}$$ Question 14. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)? 1. $$\\frac{{1}}{{13}}$$ 2. $$\\frac{{3}}{{13}}$$ 3. $$\\frac{{1}}{{4}}$$ 4. $$\\frac{{9}}{{52}}$$ Question 15. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white? 1. $$\\frac{{3}}{{4}}$$ 2. $$\\frac{{4}}{{7}}$$ 3. $$\\frac{{1}}{{8}}$$ 4. $$\\frac{{3}}{{7}}$$ Question 16. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3 ? 1. $$\\frac{3}{10}$$ 2. $$\\frac{3}{20}$$ 3. $$\\frac{2}{5}$$ 4. $$\\frac{1}{2}$$ Question 17. From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings? 1. $$\\frac{1}{15}$$ 2. $$\\frac{25}{57}$$ 3. $$\\frac{35}{256}$$ 4. $$\\frac{1}{221}$$ Question 18. An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If 4 marbles are picked up at random, what is the probability that at least one of them is blue? 1. $$\\frac{4}{15}$$ 2. $$\\frac{69}{91}$$ 3. $$\\frac{11}{15}$$ 4. $$\\frac{22}{91}$$ 5. None of these 1. $$\\frac{1}{13}$$ 2.", "One lady is chosen at random, then (i) Probability of a lady who dislikes coffee = $$\\\\ \\frac { 118 }{ 200 }$$ = $$\\\\ \\frac { 59 }{ 100 }$$ Question 10. Solution: 19 ball bearing numbers, 1, 2, 3,…19 possible outcomes = 19 A ball is drawn at random from the box, then (i) Probability of a ball which bears a prime numbers which are 2, 3, 5, 7, 11, 13, 17 and 19 = 8 = $$\\\\ \\frac { 8 }{ 19 }$$ (ii) Probability of a ball which bears an even number which are 2, 4, 6, 8, 10, 12, 14, 16, 18 = 9 = $$\\\\ \\frac { 9 }{ 19 }$$ (iii) Probability of a number which bears a number divisible by 3 which are 3, 6, 9, 12, 15, 18 = 6 = $$\\\\ \\frac { 6 }{ 19 }$$ Question 11. Solution: A card’s drawn at random from a deck of well-shuffled deck of 52 cards Probability = 52 (i) Probability of a card being a king = $$\\\\ \\frac { 4 }{ 52 }$$ = $$\\\\ \\frac { 1 }{ 13 }$$ (ii) Probability of a card being spade = $$\\\\ \\frac { 13 }{ 52 }$$ = $$\\\\ \\frac { 1 }{ 4 }$$ (iii) Probability of a" ]

[ "Mathematics # 1)How many different ways can a club choose a committee of 4 from 15 members? 2) How many permutations exist of the letters p, q, r, s, and t, taking four at a time? #### Mrslew15 3 years ago Do not know # 1 working on it. 2)There are 120 permutations. (5*4*3*2)", "Free Online Trial Hour Re: A certain club has 20 members. What is the ratio of the memb [#permalink] 06 Jun 2016, 08:41 Go to page 1 2 Next [ 22 posts ] Similar topics Replies Last post Similar Topics: 8 A local club has between 24 and 57 members. The members of the club ca 9 10 May 2017, 12:39 30 A certain club has exactly 5 new members at the end of its 12 13 Sep 2016, 11:51 4 A certain club has 10 members, including Harry. One of the 8 25 Jul 2012, 21:13 145 A certain club has 10 members, including Harry. One of the 33 31 May 2017, 22:01 1 A certain team has 12 members, including Joey. A three-membe 13 14 Dec 2011, 10:58 Display posts from previous: Sort by", "Question How many ways can a president, vice-president, secretary, and treasurer be chosen from a club with 9 members?", "25 Apr 2012, 10:59 1 A certain team has 12 members, including Joey. A three-membe 13 23 Oct 2011, 13:16 12 A certain club has 20 members. What is the ratio of the memb 18 25 Mar 2009, 23:04 Display posts from previous: Sort by", "trip. if 39 people went on that trip, how many are in the club...", "how many 4-person committees can &nbs [#permalink] 17 Apr 2018, 16:27 Display posts from previous: Sort by", "governing body. We already operate under one governing body. Our voting committee is made up of one representative from each club. Each representative is chosen by his or her respective club. The wisdom in this is that clubs know their members best and can choose who they believe will be best for the region and by extension, the club. If they feel they are not being represented properly, they can replace their representative at any time. The important thing to note is that each club has a seat at the table. Each representative controls between 0 and 6 votes. This is based on primarily on participation numbers. Currently SCNAX and CASOC each have the maximum allowed 6 votes, the remaining clubs less, with one club having 0 votes at the present time. At any time, if a club wishes to gain more influence, they up their participation. SCNAX is a great example of a club who has recently done that. When I first joined, the voting was dominated by GRA and No$. rmgibson wrote:I don't see how officially separating the social and business aspects of our community is a bad thing. If we can unify and get more people involved with the running of the organization then I'm all for it. Those of us who serve on the", "four clubs causes more problems than it solves. By dividing the 450 clubs into 112 sets of four clubs and one set of two clubs, we have three students in each set where they are in all clubs in their respective set so we need a minimum of 339 students to show a counter example. Thus there must be at least one four club combination with exactly one student in common, because the more clubs students are in the less students are required to provide a counter example, unless the student is in five or more clubs in which case the number of students increases greatly.", "of those 2 members (in this case, {2,3,x} and {1,3,x}). And if it shares 1 member with a 4-person group, it will obviously share exactly 1 member with another committee.", "Total members is 40>M>10, so 29 members. Now if each table has 6 members, that is 29/6 then the last table would have 5 (29-24) members on the table. Is this approach right? I did the problem in less than a minute with this approach. Thanks in advance. How did you get that there must be 29 members from 40>M>10? Does 29 members satisfy ANY of the conditions (3 members at one table and 4 members at each of the other tables and 3 members at one table and 5 members at each of the other tables)? _________________ IIMA, IIMC School Moderator Joined: 04 Sep 2016 Posts: 1340 Location: India WE: Engineering (Other) Re: Club X has more than 10 but fewer than 40 members. Sometimes the membe [#permalink] ### Show Tags 27 Feb 2018, 09:14 Quote: Club X has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at", "+0 # counting question 0 26 1 A certain club has 50 people, and 10 members are running for president. Each club member either votes for one of the 10 candidates, or can abstain from voting. How many different possible vote totals are there? Nov 13, 2022 #1 +17 +1 There are 40 voters (50-10 =40since. we have to exclude the candidates). Each voter has 11 choices, since they can choose from 10 candidates or choose none (abstaining). So, the total amount of different possible vote totals is $$\\boxed{40\\cdot11 = 440}$$ Hope this helped Nov 13, 2022", "# How many meetings would it take for 12 people to meet in 4 groups of 3 until they met everyone? I have a group of 12 people that I would like to meet in four groups of three each month. How many minimum months would it take such that each person has been in at least one group with every other person? Below is the brute force method I used to get it to seven months: • Thanks for the link. That A) explains what you want to do, and B) serves as context showing that you have worked on this yourself. I expect the votes to close to subside :-) – Jyrki Lahtonen Dec 22 '20 at 15:00 • It takes at least six months. Each person has to meet eleven others and meets two people a month. The question is whether you can avoid too much duplication so you need another month. – Ross Millikan Dec 22 '20 at 15:00 • Nope. After following that scheme for four months, the not-yet-met people are split into four groups of four. And you cannot arrange all them to meet in two meetings. Need a different approach. Sorry about thinking aloud. – Jyrki Lahtonen Dec 22 '20 at 15:34 • @Sdavid552- I think it's because it doesn't have", "topics Replies Last post Similar Topics: What is the number of members of the Booster Club who are under 21 5 02 Dec 2015, 05:54 A certain club has 20 members. How many of the members of the club are 3 28 Oct 2015, 06:01 3 Eleven members of Club A are also members of Club B, and five members 5 13 Feb 2015, 07:47 1 What is the number of members of Club X who are at least 35 2 23 Jul 2010, 19:11 11 The members of a club were asked whether they speak 19 05 Aug 2007, 03:32 Display posts from previous: Sort by", "Bookshelf and books How many ways can we place 7 books in a bookshelf? • School committee 7 students was elected to the school committee. In how many ways can the President, Vice-President, Secretary and Treasurer be selected?", "each of the 450 clubs must have three students in common with at least one other club and if any club has students in common with four or more clubs then it cannot have more than one student in common. Given these requirements must there be a club where it has exactly one student in common with three other clubs. To provide a counter example I must show that all four club combinations have 0 students in common or it must have two or more students in common while satisfying the other conditions. Let's assume that every three club combination doesn't have any students in common, so to solve this we divide the 450 clubs into 225 sets of 2 clubs. To satisfy the first requirement, that each club must have three students in common with at least one other club, we put three students in each set where all students are apart of both clubs in their respective set this would mean that a minimum of 675 students are required. Now let's assume instead that every four club combination doesn't have any students in common. To solve this we divide the 450 clubs into 150 sets of 3 clubs. This would require 450 students by applying the same process the above paragraph and using the first observation", "with a total membership of 30 has formed 3 committees 14 14 Nov 2010, 22:42 7 A club with a total membership of 30 has formed 3 committees 19 02 Sep 2010, 20:52 Display posts from previous: Sort by", "build one that has 200 staffers. Or 150 staffers. Or 100 staffers.”", "+0 # PLSSSSS HELPPPPP 0 52 1 A club has 30 members: 15 boys and 15 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl? Mar 21, 2022 #1 +117105 +1 A club has 30 members: 15 boys and 15 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl? 1- P(all girls) - P(all boys) =$$=1-2*\\frac{15C4}{30C4}\\\\ =1-2*\\frac{1365}{27405}\\\\ =1-\\frac{26}{261}\\\\ =\\frac{235}{261}$$ Mar 21, 2022", "members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at the table that has fewer than 6 members? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 I just did the LCM of 5&4 +3 . So basically 20+3/6 . Not sure if this method will hold if the range of members was to change, but it worked on this instance. Intern Joined: 30 Jun 2017 Posts: 18 Location: India Concentration: Technology, General Management WE: Consulting (Computer Software) Re: Club X has more than 10 but fewer than 40 members. Sometimes [#permalink] ### Show Tags 09 Sep 2017, 05:34 Bunuel wrote: Walkabout wrote: Club X has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer", "months each with 60 more attendees." ]

[ "may not account for the possibility of one of four officials to be the member in charge of fundraising. Because of this, a common answer might be $20 \\times 19 \\times 18 \\times 17 \\times 16 = 1860480$. Otherwise, this problem is a straight forward Product Rule problem (Counting Principle). Most students who have attempted this problem recognized this! If you have any other approaches that you believe is worth discussing about, please let us know by posting here! We would greatly appreciate listening to your ideas!", "among the students chosen. In this case, we’ll use the combination. ${}_{20}C_{5}$ $= \\dfrac{20!}{5!(15)!}$ $= 15,504$ This is much easier using a calculator. Next is we have $20$ students to be picked for class office – Pres., VP, Sec., Treasurer, and Judge. What’s important is that there are five distinct order or places. It’s not just a group of $5$ different people. It’s a $5$ people each one has a specific position. In this case, order matters. So, we’ll use permutation. ${}_{20}P_{5}$ $= 20 \\times 19 \\times 18 \\times 17 \\times 16$ $= 1,860,480$ There is a distinct difference between the combination and permutation.", "A committee of three people is to be chosen from four 9 28 Sep 2009, 00:41 27 A committee of three people is to be chosen from four marrie 8 13 Nov 2007, 07:31 Display posts from previous: Sort by", "for the frequency function of rolling the die two times is then \\begin{equation*} \\begin{split} G(F; z) & = G\\left(F_1;z\\right)G\\left(F_2;z\\right)\\\\ & = (z^6+z^5+z^4+z^3+z^2+z)^2\\\\ & =z^{12}+2 z^{11}+3 z^{10}+4 z^9+5 z^8+6 z^7+5 z^6+4 z^5+3 z^4+2 z^3+z^2 \\end{split} \\end{equation*} Now, to get $$F(k)\\text{,}$$ just read the coefficient of $$z^k\\text{.}$$ For example, the coefficient of $$z^5$$ is 4, so there are four ways to roll a total of 5. To apply this method, the crucial step is to decompose a large process in the proper way so that it fits into the general situation that we've described. Suppose that an organization is divided into three geographic sections, A, B, and C. Suppose that an executive committee of 11 members must be selected so that no more than 5 members from any one section are on the committee and that Sections A, B, and C must have minimums of 3, 2, and 2 members, respectively, on the committee. Looking only at the number of members from each section on the committee, how many ways can the committee be made up? One example of a valid committee would be 4 A's, 4 B's, and 3 C's. Let $$P_A$$ be the action of deciding how many members (not who) from Section A will serve on the committee. $$X_A= \\{3, 4, 5\\}$$ and $$Q_A(k)=k\\text{.}$$ The frequency function,", "$$6$$ terms. There are 6 types of possible committees: $$\\begin{array} {|r|r|r|} \\hline 3&1&1 \\\\ \\hline 1&3&1 \\\\ \\hline 1&1&3 \\\\ \\hline 2&2&1 \\\\ \\hline 2&1&2 \\\\ \\hline 1&2&2 \\\\ \\hline \\end{array}$$ For each type of committee, it is necessary to calculate the different groups of people that compose it: $$\\begin{array} {|c|c|c|c|} \\hline 3&1&1& {6\\choose3}{7\\choose1}{4\\choose1}= 20\\cdot7\\cdot4 = 560 \\\\ \\hline 1&3&1& {6\\choose1}{7\\choose3}{4\\choose1}= 6\\cdot35\\cdot4 = 840 \\\\ \\hline 1&1&3& {6\\choose1}{7\\choose1}{4\\choose3}= 6\\cdot7\\cdot4 = 168 \\\\ \\hline 2&2&1& {6\\choose2}{7\\choose2}{4\\choose1}= 15\\cdot21\\cdot4 = 1260 \\\\ \\hline 2&1&2& {6\\choose2}{7\\choose1}{4\\choose2}= 15\\cdot7\\cdot6 = 630 \\\\ \\hline 1&2&2& {6\\choose1}{7\\choose2}{4\\choose2}= 6\\cdot21\\cdot6 = 756 \\\\ \\hline \\end{array}$$ Adding 6 gives a total of different committees: $$560+840+168+1260+630+756 = 4214$$ For groups of $$6, 7$$, and $$4$$ distinct types of members, would the counting be easier by counting all possible committees while ignoring the constraint, then subtracting the non-valid committees? $$\\binom{6 + 7 + 4}{5} - \\left[ \\binom{6+7}{5}+\\binom{7+4}{5}+ \\binom{6+4}{5} \\right]$$ Non-valid committees are those which entirely omit one of the types of members. (Ah: This doesn't work: The subtracted count is too high, as it duplicates non-valid committees. Trying again ... $$\\binom{6+7+4}{5}- \\left[ \\left[ \\binom{6+7}{5} - \\left[\\binom{6}{5}+\\binom{7}{5}\\right] \\right] + \\left[ \\binom{7+4}{5} - \\left[\\binom{7}{5}+\\binom{4}{5}\\right] \\right] + \\left[ \\binom{6+4}{5} - \\left[\\binom{6}{5}+\\binom{4}{5}\\right] \\right] \\right] - \\left[ \\binom{6}{5} + \\binom{7}{5} + \\binom{4}{5} \\right]$$ The number of committees of five members, minus the number of committees with", "me. Suppose that each person in a group of $n$ people votes for exactly two people from a set of candidates. Devise a divide-and-conquer algorithm that determines all candidates who received more than $n/2$ votes. If you are careful, you may point out the my version missed \"... to fill two positions on a committee. The top two finishers both win positions as long as each receives more than n/2 votes\". Well, although it is interesting to know that election goal and election rule, it has nothing to do with the specification of the desired algorithm. You may also point out that my version does not require \"if so, determine who these two candidates are\". Well, if we have determined all candidates who received more than n/2 votes, the number of whom is at most 3, then by just counting the number of votes received by each of those candidates, we can, easily, \"if so, determine who these two candidates are\". ## An interesting algorithm as an exercise There is an algorithm using linear time and constant space that determines all candidates who received more than $n/2$ votes. The construction of such an algorithm is left as an intriguing and challenging exercise for the readers who has read thus far. Thanks to Tom van der Zanden, who pointed", "longer focus only on one of the groups (the committee), but rather we care about a partition into two groups so that each of them contains candidates that are as similar as possible. A prime example here is partitioning students in some class into two groups, e.g., a group of beginners and a group of advanced ones (say, regarding, their knowledge of a foreign language; depending on the setting, it may or may not be important to keep the sizes of these two groups close). The students are partitioned in this way to facilitate a better learning environment for everyone; in the context of voting, the issue of partitioning students was raised by Duddy et al. [13]. Finding a Representative Committee. An elected committee is representative if each voter approves at least one committee member (who then can represent this voter). The idea of choosing a representative committee of a fixed size received significant attention in the literature (see the works of Chamberlin and Courant [10], Monroe [21], Elkind et al. [14], and Aziz et al. [2] as some examples). However, as pointed out by Brams and Kilgour [7], committees of fixed size simply cannot always provide adequate representation. Thus, in some applications, it is natural to elect committees without prespecified sizes (while in others, fixing the committee", "Now, $680-120=560$. Scenario 2. For a committee with 1 professor and 2 students, we will get $7*C_{10}^2 = 7 * \\frac{10!}{8! * 2!} = 45*7=315$. (we multiply by 7 because For 2 professors and 1 student, we will get $C_7^2*10 = 21*10=210$, and for 3 professors, we will get $C_7^3 = 35$. Adding up the combinations we will get: $315+210+35=560$. Princeton Review suggests using Scenario 2 because it is supposedly simpler to understand, however, taking into consideration that you can make a mistake in the endless calculations and that it requires 3 complex operations in contrast to 2 in the first case, it has a weaker standing. In any case, both ways get the correct answer, but you need to choose the one that appeals more to you - the one easier to use. Here is another hard problem with some restrictions; again from an unknown source: {{#x:box| EXAMPLE 11. There are 11 top managers that need to form a decision group. How many ways are there to form a group of 5 if the President and Vice President are not to serve on the same team? }} Again our options are to solve the problem either to find the total number of committees and then subtracting the number of groups that VP and P would end up", "# Prove ${n - 1 \\choose k - 1} + {n - 2 \\choose k - 1} + {n - 3 \\choose k - 1} + \\dots + {k - 1 \\choose k - 1} = {n \\choose k}$ Can someone check my logic here. Question: How many ways are there to choose a an $k$ person committee from a group of $n$ people? Answer 1: there are ${n \\choose k}$ ways. Answer 2: condition on eligibility. Assume the creator of the committee is already in the committee. This leaves us with choosing $k - 1$ people from a group of $n - 1$ potentially eligible people. If all remaining people are eligible, there are ${n - 1 \\choose k - 1}$ possible committees, if there are $n - 2$ eligible people, there are ${n - 2 \\choose k - 1}$ committees, if there are $n - 3$ eligible people, there are ${n - 3 \\choose k - 1}$ committees,..., if there are $k - 1$ eligible people there are ${k - 1 \\choose k - 1}$ committees. Therefore,$${n - 1 \\choose k - 1} + {n - 2 \\choose k - 1} + {n - 3 \\choose k - 1} + \\dots + {k - 1 \\choose k - 1} = {n \\choose k}$$. • All of", "should be treated as the same committee? There are $3$ possibilities for the first position; $2$ for the second; and $1$ for the third. So $3 \\times 2 \\times 1 = 6$ arrangements should be considered the same committee. Therefore the number of committees (without internal ordering) should be $\\frac{8 \\times 7 \\times 6}{3 \\times 2 \\times 1} = 56$", "is $F_j(k)=1$ if $1 \\leq k \\leq 6$, and $F_j(k) = 0$ otherwise. The process of rolling the die two times is quantified by adding up the ${Q_j}'s$; that is, $Q\\left(a_1, a_2\\right) =Q_{1}\\left(a_1\\right)+Q_2\\left(a_2\\right)$ . The generating function for the frequency function of rolling the die two times is then \\begin{equation*} \\begin{split} G(F; z) & = G\\left(F_1;z\\right)G\\left(F_2;z\\right)\\\\ & = (z^6+z^5+z^4+z^3+z^2+z)^2\\\\ & =z^{12}+2 z^{11}+3 z^{10}+4 z^9+5 z^8+6 z^7+5 z^6+4 z^5+3 z^4+2 z^3+z^2 \\end{split} \\end{equation*} Now, to get $F(k)$, just read the coefficient of $z^k$. For example, the coefficient of $z^5$ is 4, so there are four ways to roll a total of 5. To apply this method, the crucial step is to decompose a large process in the proper way so that it fits into the general situation that we've described. ##### Example8.5.15Distribution of a Committee Suppose that an organization is divided into three geographic sections, A, B, and C. Suppose that an executive committee of 11 members must be selected so that no more than 5 members from any one section are on the committee and that Sections A, B, and C must have minimums of 3, 2, and 2 members, respectively, on the committee. Looking only at the number of members from each section on the committee, how many ways can the committee be made up? One example", "## Algebra and Trigonometry 10th Edition $292,600$ different six-member committees are possible. The order in which the members are selected is not important. Use combinations with the Fundamental Couting Principle. $_7C_1·~_{12}C_3·~_{20}C_2=\\frac{7!}{(7-1)!~1!}·\\frac{12!}{(12-3)!~3!}·\\frac{20!}{(20-2)!~2!}=7\\times220\\times190=292,600$", "## Precalculus (6th Edition) Blitzer The order in which the four-person committee is selected does not matter. Thus, this is a problem of selecting 4 people from a group of 11 people. We need to find the number of combinations of 11 things taken 4 at a time. Apply the formula, ${}_{n}{{C}_{r}}=\\frac{n!}{r!\\left( n-r \\right)!}$ Where $n=11,r=4$ \\begin{align} & {}_{11}{{C}_{4}}=\\frac{11!}{\\left( 11-4 \\right)!4!} \\\\ & =\\frac{11!}{7!\\times 4!} \\end{align} Solving further, \\begin{align} & {}_{11}{{C}_{4}}=\\frac{11\\times 10\\times 9\\times 8\\times 7!}{4\\times 3\\times 2\\times 1\\times 7!} \\\\ & =11\\times 10\\times 3 \\\\ & =330 \\end{align} Hence, there are 330 ways to select the four-person committee from a group of 11 people.", "\\choose k}$ possible committees. (LHS) Suppose we must have the celebrity on the committee. Then there are ${n \\choose k-1}$ ways to choose who else will be on the committee. Also, suppose that we don't want to have the celebrity on the committee. Then there are ${n \\choose k}$ ways to choose who is on the committee. This gives us a total of ${n \\choose k-1} + {n \\choose k}$ ways to pick the committee. Recall that these two expressions both represent ${n \\choose k}$ or the number of ways to choose a committee of size $k$ from $n$ people. Q A Q $\\frac{n!}{(n-k)!k!} = \\frac{n \\cdot (n-1) \\dotsm (n-k+1)}{k\\cdot (k-1)\\dotsm 1}$ A Recall that these two expressions both represent ${n \\choose k}$ or the number of ways to choose a committee of size $k$ from $n$ people. (LHS) We first permute the $n$ people and then select the first $k$ people in the permutation. While there $n!$ total permutations, we must deal with the overcounting within the committee and the people not selected for the committee. The $k$ people in the committee can be ordered in $k!$ ways and the people not in the committee can be ordered in $(n-k)!$ ways, which indicates that we have overcounted by a factor of $k!(n-k)!$. Therefore, we observe that ${n", "GRE Math : How to find permutation notation Example Questions Example Question #1 : Permutation / Combination A chamber of commerce board has seven total members, drawn from a pool of twenty candidates. There are two stages in the board's election process. First, a president, secretary, and treasurer are chosen. After that, four members are chosen to be “at large” without any specific title or district. How many possible boards could be chosen? 5,426,400 16,279,200 10,465,200 390,700,800 2,713,200 16,279,200 Explanation: We have to consider two cases. First, the group containing the president, secretary, and treasurer represents a case of permutation. Since the order matters in such a group, we can select from our initial 20 candidates 20 * 19 * 18, or 6840 possible groupings. Following that, the at large group constitutes a case of combinations in which the order does not matter. Since we have chosen 3 already for the first three slots, there will be 17 remaining people. The formula for choosing 4-person sets out of 17 candidates is represented by the combination formula of this form: 17! / ((17-4)! * 4!) = 17! / (13! * 4!) = (17 * 16 * 15 * 14) / (4 * 3 * 2) = 17 * 4 * 5 * 7 = 2380 Thus, we have 6840", "ways. Then the 6 people can be lined up in $6! = 720$ ways. Answer: . $70 \\times 720 \\:=\\:50,\\!400$", "# How many ways can a committee of 4 people be selected from a group of 7 people? A committee of $4$ people be selected from a group of $7$ people in $35$ ways. As the order of people does not matter, it is $\\setminus {C}_{4}^{7}$ i.e. $\\frac{7 \\times 6 \\times 5 \\times 4}{1 \\times 2 \\times 3 \\times 4} = \\frac{7 \\times \\cancel{6} \\times 5 \\times \\cancel{4}}{1 \\times \\cancel{2} \\times \\cancel{3} \\times \\cancel{4}} = 35$ Hence, a committee of $4$ people be selected from a group of $7$ people in $35$ ways.", "drawing a fjgure using dots to Explanation: You can answer this question by careful reasoning, or you can draw it out and count. Either way. be careful that you don't leave Bill out or count him twice. WHAT DO YOU KNOW? represent the people in line. Just make sure that you follow the instructions carefully when you draw your figure. • Bill is the 5th person from one end of the line. • Bill is the 12th person from the other end. USING LOGICTO SOLVE THE PROBLEM • If Bill is the 5th person from one end of the line, there are 4 people (not counting Bill) between him and that end of the line. • If Bill is the 12th person from the other end of the line, there are II people (not count ing Bill) between him and that end of the line. • The sum of 4 people between Bill and one end plus II people between Bill and the other end equals 15 people. Then you have to add in Bill. So, there are 16 people in line. 326 Multiple-Choice Questions EXPENDITURES BY COMPANY Y CatearIeo 13. In the graph above, the total expenditures by Company Y were $1,000,000. If each of the following pic charts represents the total expenditures of Company Y, in", "# Choosing a committee from two people who are not sitting beside each other. Assume that $10$ people are sitting around a table. Determine the number of ways to choose a committee, where the committee is made up of two people who are NOT sitting next to each other. Take ${10 \\choose 2}$ and take away the pairs with partners, which is still the $10$ which gives me the same result. ${10 \\choose 2}-10=35$. Assume that $10$ people are sitting around a table. Determine the number of ways to choose a committee, where the committee is made up of three people of which NONE are sitting next to each other. ${10\\choose 3}=130$ take away the pairs that are sitting beside each other: (still 10??) $120$ Your answer to the first question is fine. Your answer to the second question has problems. First, $\\binom{10}3=120$. Second, subtracting $10$ pairs from $\\binom{10}3$ triples makes no sense. What you want to subtract is the number of triples, at least two of whom are sitting beside each other. There are two different kinds of bad triples, which you have to count separately. I. Three people in consecutive seats. There are $10$ of those. II. Two people sitting beside each other, the third not sitting beside either of then. There are $10\\times6=60$ triples like", "of the mathematical theory of voting. Now suppose we were electing a committee $C$ of candidates, and we want the committee to be \"stable\" in the sense that for each candidate $y$ outside $C$ there is at least one candidate $x$ inside $C$ who is preferred to $y$ by a majority of voters. Condorcet's paradox tells us that to be stable, a committee may need to have more than one member. So, do two-person stable committees always exist? Recently, a group of researchers from Paris and Singapore found a relatively simple example for which no stable two-person committee exists; their example has 15 voters and 15 candidates. Even more recently, two of us at Union found smaller examples, which are more complicated. Do stable committees of size 3 always exist? How can we understand the structure of these examples? There is a wealth of related questions we can explore. 1- or 2-term thesis Math Department web pages Created: 25 Apr 2007 Last modified: 02 May 2012 10:53:31 Comments to: [email protected]" ]

[ "on an executive committee is $12,650$ The number of ways to select the club members for a president, vice president, secretary, and treasurer so that no person can hold more than one office is $303,600$. ## Example A group of $3$ athletes is $P$, $Q$, $R$. In how many ways can a team of $2$ members be formed? Here, as the order of members is not important, we will use the Combination formula. $C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}$ Putting values of $n$ and $r$: $n=3$ $r=2$ $C\\left(3,2 \\right)=\\frac{3!}{2!\\left(3-2\\right)!}$ $C\\left(3,2 \\right)=3$", "+0 0 99 3 A certain club has 50 people, and 4 members are running for president. Each club member votes for one of the candidates. How many different possible vote totals are there? I know this question has been posted more than once, but they were all wrong, so can I get help from someone who is confident they have the right answer? Thank you! Sep 3, 2020 edited by Guest Sep 3, 2020 #1 +40 +1 The answer is $$\\boxed{4^{50}}$$, and here's why. We have 50 people, and four people to vote for. Let's see how many possibilities for a single person. The person can vote for person A, B, C, or D, thus meaning there are four options for a single person. Remember that one person's vote does not affect another; for example if thirty-nine people voted for person C, it does not make the fortieth person any more likely to vote for person C, all probabilities are independent. So we have $$4^1$$ for the first person, because there is one person voting. But each time we have more people vote, the options increase. If we had two people voting, there would be $$4\\cdot4$$ options or $$4^2$$. This pattern goes on for every single vote, meaning that by the end, we will have $$\\boxed{4^{50}}$$ options.", "Administrating a Rotary Club effectively is very important. Effective administration means you will have more time for the fun side of being a Rotarian! What are the goals your club wants to achieve in the next year and in 3 to 5 years? Develop a plan, set achievable goals and decide how they will be achieved. The year ahead then takes on a purpose and members gain a sense of achievement and 'doing good' in their communities. 'Doing good' is why we become Rotarians. The second part of this resource will give you some information on how to conduct effective club meetings. Hope you find it useful! ### 3. Managing meetings effectively #### 3.7. Annual General Meeting and election of officers The Annual General Meeting (AGM) should be held within 6 months of the end of the financial year (by 31st December). Ensure you have a quorum (see your Constitution for clarification) The AGM agenda should include: • The adoption of Minutes of the previous AGM • The election of Officers and Auditor • Approval of audited accounts from the previous year: • For Clubs that generate an income of more than $250,000 per year this is mandatory • Clubs who generate less than$250,000 per year the financial records can be reviewed by two club members who have", "Basic Counting — Selecting groups of people from a group of 10 I'm reading a series of basic counting problems regarding a ten person club. I just want to make sure I'm sufficiently grasping the material and my problem solving methods are correct. $1.$ In how many ways may a ten person club select a president and a secretary-treasurer from among its members? For this problem, there are 10 ways to choose the president, and 9 ways to choose the secretary. In total, there are $10 \\times 9 = 90$ ways to choose them. $2.$ In how many ways may a ten person club select a two person executive committee from among its members? This time, since we're taking 2 people from 10, I used ${10 \\choose 2} = 45$ ways. $3.$ In how many ways may a ten person club select a president and a two person executive advisory board from among its members (assuming that the president is not on the advisory board)? I've already established from my first answer that there are 10 ways to choose the president. Then, I have to choose 2 people from a remaining group of 9. ${9 \\choose 2} = 36$. Finally, I multiply the two results.$36\\times10 = 360$ ways. Any help would be appreciated, thank you. • It's correct", "each of the following cases: 1. There are no other restrictions. 2. The committee must have 4 women and 2 men. 3. The committee must have at least 2 women and at least 2 men. 1. $$38\\,760$$ 2. $$13\\,860$$ 3. $$30\\,800$$ Suppose that a club with 20 members plans to form 3 distinct committees with 6, 5, and 4 members, respectively. In how many ways can this be done. $$9\\,777\\,287\\,520$$. Note that the members not on a committee also form one of the sets in the partition. #### Cards A standard card deck can be modeled by the Cartesian product set $D = \\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k\\} \\times \\{\\clubsuit, \\diamondsuit, \\heartsuit, \\spadesuit\\}$ where the first coordinate encodes the denomination or kind (ace, 2-10, jack, queen, king) and where the second coordinate encodes the suit (clubs, diamonds, hearts, spades). Sometimes we represent a card as a string rather than an ordered pair (for example $$q \\heartsuit$$). A poker hand consists of 5 cards dealt without replacement and without regard to order from a deck of 52 cards. Find the number of poker hands in each of the following cases: 1. There are no restrictions. 2. The hand is a full house (3 cards of one kind and 2 of another", "each of the 450 clubs must have three students in common with at least one other club and if any club has students in common with four or more clubs then it cannot have more than one student in common. Given these requirements must there be a club where it has exactly one student in common with three other clubs. To provide a counter example I must show that all four club combinations have 0 students in common or it must have two or more students in common while satisfying the other conditions. Let's assume that every three club combination doesn't have any students in common, so to solve this we divide the 450 clubs into 225 sets of 2 clubs. To satisfy the first requirement, that each club must have three students in common with at least one other club, we put three students in each set where all students are apart of both clubs in their respective set this would mean that a minimum of 675 students are required. Now let's assume instead that every four club combination doesn't have any students in common. To solve this we divide the 450 clubs into 150 sets of 3 clubs. This would require 450 students by applying the same process the above paragraph and using the first observation", "for the frequency function of rolling the die two times is then \\begin{equation*} \\begin{split} G(F; z) & = G\\left(F_1;z\\right)G\\left(F_2;z\\right)\\\\ & = (z^6+z^5+z^4+z^3+z^2+z)^2\\\\ & =z^{12}+2 z^{11}+3 z^{10}+4 z^9+5 z^8+6 z^7+5 z^6+4 z^5+3 z^4+2 z^3+z^2 \\end{split} \\end{equation*} Now, to get $$F(k)\\text{,}$$ just read the coefficient of $$z^k\\text{.}$$ For example, the coefficient of $$z^5$$ is 4, so there are four ways to roll a total of 5. To apply this method, the crucial step is to decompose a large process in the proper way so that it fits into the general situation that we've described. Suppose that an organization is divided into three geographic sections, A, B, and C. Suppose that an executive committee of 11 members must be selected so that no more than 5 members from any one section are on the committee and that Sections A, B, and C must have minimums of 3, 2, and 2 members, respectively, on the committee. Looking only at the number of members from each section on the committee, how many ways can the committee be made up? One example of a valid committee would be 4 A's, 4 B's, and 3 C's. Let $$P_A$$ be the action of deciding how many members (not who) from Section A will serve on the committee. $$X_A= \\{3, 4, 5\\}$$ and $$Q_A(k)=k\\text{.}$$ The frequency function,", "how many 4-person committees can &nbs [#permalink] 17 Apr 2018, 16:27 Display posts from previous: Sort by", "# There are 566 members in arts club and 522 are members in the music club. If the total number of people who are members in either or both the clubs is 963, how many people are members in both clubs? Jan 8, 2018 See a solution process below: #### Explanation: The total membership of both clubs, with overlap, is: $566 + 522 = 1088$ However, without overlap we are told the are a total of 963 in the clubs. We can find the total number in both by subtracting the total number of individual members from the total number of members with overlap we calculated above: $1088 - 963 = 125$ There are 125 people who are members of both clubs. Jan 8, 2018 $125$ #### Explanation: This can be done with Set Theory, but I will stick with the traditional method. The number of members in either or both the clubs $= 963$. The number of members in the arts club $= 566$. Therefore, The number of members in only music club is $= \\left(963 - 566\\right) = 397$ The number of members in the music club is $= 522$. Therefore, the number of members of both clubs $= \\left(522 - 397\\right) = 125$. Hope this helps.", "solve for P Therefore with 12 people, you get 4 slices. Example Question #91 : Mathematical Relationships The number of raffle tickets given for a contest varies indirectly with the total number of people in the building. If you get tickets when there are people, how many slices would you get if there are people? Explanation: The problem follows the formula where R is the number of raffle tickets you get, n is the number of people, and k is the constant of variation. Setting R=20 and n = 150 yields k=3000. Plugging in 100 for n and solving for R you get: The answer R is 30 tickets. Example Question #1 : Indirect Proportionality The budget per committee varies indirectly with the total number of committees created. If each committee is allotted when four committees are established, what would the budget per committee be if there were to be committees? Explanation: The problem follows the formula where B is the budget per committee, n is the number of committees, and k is the constant of variation. Setting B=500 and n = 4 yields k=2000. Now using the following equation we can plug in our n of 2 and solve for B. The answer of B is \\$1000. Example Question #21 : Basic Single Variable Algebra The number", "above. From this we can deduce that the more clubs students participate in the less students are required to satisfy the requirements. So if one student is in all 450 clubs. Then every five club combination has one student in common. To provide a counter example, adding at least one student in common to each four club combination without adding students in common to any five club combination, requires ${450*449*448*447\\over 24}=1685905200$ additional students. Let's look at a slightly altered version of the problem where we have 20 clubs instead of 450 and instead of having 100 students we just want to find the minimum number of students to provide a counter example. Keeping all other requirements the same. Dividing the 20 clubs into four sets of five clubs, where each set has one student in five clubs, then adding two students to each four club combination in each set. This requires (2 students)(5 four club combinations in each set)(4 sets)+(4 original students) = 44 students. Trying a second case where dividing the 20 clubs into five sets of four clubs so each set has three students in all clubs so that only 15 students are required to satisfy the requirements with no four club combination having exactly one student in common. Thus having a student in more than", "among the students chosen. In this case, we’ll use the combination. ${}_{20}C_{5}$ $= \\dfrac{20!}{5!(15)!}$ $= 15,504$ This is much easier using a calculator. Next is we have $20$ students to be picked for class office – Pres., VP, Sec., Treasurer, and Judge. What’s important is that there are five distinct order or places. It’s not just a group of $5$ different people. It’s a $5$ people each one has a specific position. In this case, order matters. So, we’ll use permutation. ${}_{20}P_{5}$ $= 20 \\times 19 \\times 18 \\times 17 \\times 16$ $= 1,860,480$ There is a distinct difference between the combination and permutation.", "a Formula For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set ofobjects and we want to chooseobjects from the set in order, we writeAnother way to write this is a notation commonly seen on computers and calculators. To calculatewe begin by findingthe number of ways to line up all objects. We then divide by to cancel out theitems that we do not wish to line up. Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is Using factorials, we get the same result. There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows. Note that the formula stills works", "403 views There are $6$ tasks and $6$ persons. Task $1$ cannot be assigned either to person $1$ or to person $2.$ Task $2$ must be assigned either to person $3$ or person $4.$ Every person is to be assigned one task. In how many ways can the task assignment be done? 1. $144$ 2. $180$ 3. $192$ 4. $360$ 5. $716$ Here we have restriction on 1st and 2nd task. For 2nd task we have only 2 choices. For 1st task we have total - 4 choices, { ecxept 1 and {3 or 4} } For 3rd task we will have total - 4 choices. { except who are assigned to task 1 and 2} Similaly, for 4th task - 3 choices. for 5th task - 2 choices. for 6th task - 1 choices. So total possibilities of task assignment = 4 * 2 * 4 * 3 * 2 * 1 = 192. Ans. - 3. by 1.5k points Task 2 can be assigned in 2 ways (either to person 3 or person 4). Given that Task 1 cannot be assigned either to person 1 or to person 2. So,Task 1 can then be assigned in 3 ways (persons 3 or 4, 5 and 6) Number of ways to assign remaining 4 tasks to 4 persons", "four clubs causes more problems than it solves. By dividing the 450 clubs into 112 sets of four clubs and one set of two clubs, we have three students in each set where they are in all clubs in their respective set so we need a minimum of 339 students to show a counter example. Thus there must be at least one four club combination with exactly one student in common, because the more clubs students are in the less students are required to provide a counter example, unless the student is in five or more clubs in which case the number of students increases greatly.", "# How many ways can a committee of 4 people be selected from a group of 7 people? A committee of $4$ people be selected from a group of $7$ people in $35$ ways. As the order of people does not matter, it is $\\setminus {C}_{4}^{7}$ i.e. $\\frac{7 \\times 6 \\times 5 \\times 4}{1 \\times 2 \\times 3 \\times 4} = \\frac{7 \\times \\cancel{6} \\times 5 \\times \\cancel{4}}{1 \\times \\cancel{2} \\times \\cancel{3} \\times \\cancel{4}} = 35$ Hence, a committee of $4$ people be selected from a group of $7$ people in $35$ ways.", "A committee of three people is to be chosen from four 9 28 Sep 2009, 00:41 27 A committee of three people is to be chosen from four marrie 8 13 Nov 2007, 07:31 Display posts from previous: Sort by", "is $F_j(k)=1$ if $1 \\leq k \\leq 6$, and $F_j(k) = 0$ otherwise. The process of rolling the die two times is quantified by adding up the ${Q_j}'s$; that is, $Q\\left(a_1, a_2\\right) =Q_{1}\\left(a_1\\right)+Q_2\\left(a_2\\right)$ . The generating function for the frequency function of rolling the die two times is then \\begin{equation*} \\begin{split} G(F; z) & = G\\left(F_1;z\\right)G\\left(F_2;z\\right)\\\\ & = (z^6+z^5+z^4+z^3+z^2+z)^2\\\\ & =z^{12}+2 z^{11}+3 z^{10}+4 z^9+5 z^8+6 z^7+5 z^6+4 z^5+3 z^4+2 z^3+z^2 \\end{split} \\end{equation*} Now, to get $F(k)$, just read the coefficient of $z^k$. For example, the coefficient of $z^5$ is 4, so there are four ways to roll a total of 5. To apply this method, the crucial step is to decompose a large process in the proper way so that it fits into the general situation that we've described. ##### Example8.5.15Distribution of a Committee Suppose that an organization is divided into three geographic sections, A, B, and C. Suppose that an executive committee of 11 members must be selected so that no more than 5 members from any one section are on the committee and that Sections A, B, and C must have minimums of 3, 2, and 2 members, respectively, on the committee. Looking only at the number of members from each section on the committee, how many ways can the committee be made up? One example", "Now, $680-120=560$. Scenario 2. For a committee with 1 professor and 2 students, we will get $7*C_{10}^2 = 7 * \\frac{10!}{8! * 2!} = 45*7=315$. (we multiply by 7 because For 2 professors and 1 student, we will get $C_7^2*10 = 21*10=210$, and for 3 professors, we will get $C_7^3 = 35$. Adding up the combinations we will get: $315+210+35=560$. Princeton Review suggests using Scenario 2 because it is supposedly simpler to understand, however, taking into consideration that you can make a mistake in the endless calculations and that it requires 3 complex operations in contrast to 2 in the first case, it has a weaker standing. In any case, both ways get the correct answer, but you need to choose the one that appeals more to you - the one easier to use. Here is another hard problem with some restrictions; again from an unknown source: {{#x:box| EXAMPLE 11. There are 11 top managers that need to form a decision group. How many ways are there to form a group of 5 if the President and Vice President are not to serve on the same team? }} Again our options are to solve the problem either to find the total number of committees and then subtracting the number of groups that VP and P would end up", "12664 Show Tags 16 Jan 2012, 04:19 Expert's post 1 This post was BOOKMARKED Forrester300 wrote: 4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. In how many ways can this committee be formed if it has to include at least 1 professor? (A) 36 (B) 60 (C) 72 (D) 80 (E) 100 [Reveal] Spoiler: OA E Source: GMAT Club Tests - hardest GMAT questions The best way to approach this problem is to consider an unconstrained version of the question first: how many committees of 3 are possible? The answer is $$C_{10}^3 = \\frac{10!}{(7!3!)} = 120$$ . From this figure we have to subtract the number of committees that consist entirely of students i.e. $$C_{6}^3 = \\frac{6!}{(3!3!)} = 20$$ . The final answer is $$C_{10}^3 - C_6^3 = 120 - 20 = 100$$ . -------------------------------------------------------------------------------- I don't understand why we only subtract from the students only here? As first we take all the possible comibinations then minus just by if all the seats were to be filled by students only?? Why wont this only leave combinations of seat that will will be filled by just professors then? Dont understand the answer can someone please dumb it down to explain to me, many thanks JF. 4 professors" ]

[ "# Moderate Permutation-Combination Solved QuestionAptitude Discussion Q. A local delivery company has three packages to deliver to three different homes. If the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package? ✖ A. $3$ ✔ B. $5$ ✖ C. $3!$ ✖ D. $5!$ Solution: Option(B) is correct The possible outcomes that satisfy the condition of \"at least one house gets the wrong package\" are: One house gets the wrong package or two houses get the wrong package or three houses get the wrong package. We can calculate each of these cases and then add them together, or approach this problem from a different angle. The only case which is left out of the condition is the case where no wrong packages are delivered. If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above. There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package. The first person must get the correct package and the second person must get the correct package and the third person must get the correct package. ⇒ $1×1×1", "= 1$ Determine the total number of ways the three packages can be delivered. ⇒ $3×2×1 = 6$ The number of ways at least one house gets the wrong package is: ⇒ $6 - 1 = 5$ Therefore there are \\textbf{5} ways for at least one house to get the wrong package. ## (1) Comment(s) Name () Indirect approach easier: Total outcomes : 3! Chances for atleast one wrong = total outcomes - the outcome that they're all right (only 1 way to do it right) 6 - 1 = 5", "# Unexpected result from very simple statistics problem I have stumbled across this problem on my mathematics book, the problem states: A person is distributing 3 packages in a city consisting of 3 houses, each package assigned to a single house. He has lost the delivery notes, and doesn't know which package belongs to which house. What is the chance that at least a single package makes its way to the correct location if he distributes them randomly? To solve the problem I decided to draw a simple \"state machine\". I assume that the order of visiting the houses doesn't change the result (am I wrong here?). The number below each package (the stars) is the house it's meant to be delivered to. Clearly, 4 out of the 6 final states have at least one correctly distributed package, so 2/3 is the result. The solution stated by the book is 0.704 (it's rounded). After puzzling for some time, I decided to code a quick simulation in C++, and the simulation (assuming the same premises as these used to build the graphic) converges to 2/3, too, so I wonder, am I interpreting the problem correctly? What method can lead to the result given in the book? I assume I'm missing a very easy and straightforward method as this book", "But four envelopes can be randomly chosen in $^{4}C_{4}$ ways. Hence exactly four envelopes can be wrongly addressed in $^{4}C_{4} * 9 = 9$ ways. Therefore, the probability that there are exactly two wrongly addressed envelopes = $\\frac{6}{6 + 8 + 9}$ i.e. 26.09$\\%$. • You need to assess this conditional on the knowledge that two envelopes have the wrong address. – Graham Kemp Apr 29 '14 at 1:46", "Fifty percent of the people that get mail-order catalogs order something. Find the probability that exactly two of 10 people getting these catalogs will order something....", "59 views Parcels from sender S to receiver R pass sequentially through two-post offices. Each post-office has a probability $\\frac{1}{5}$ of losing an incoming parcel, independently of all other parcels. Given that a parcel is lost, the probability that it was lost by the second post office is _________", "# Probability HW Help • May 14th 2006, 04:42 PM Nimmy Probability HW Help 1. A class is given a list of 20 study problems from which ten will be part of an upcoming exam. If a givent student knows how to solve 15 of the problems, find the probability that the student will be able to answer at least nine question on the exam. 2. Four letters and envelopes are addressed to four different people. If the letters are randomly inserted into the envelopes, what is the probability that (a)exactly one is inserted in the correct envelope and (b) at least one is inserted in the correct envelope? 3. A shipment of 12 microwave ovens contains three defective units. A vending company has ordered four of these units, and because all are packaged identically, the selection will be random. What is the probablity that a) all four units are good? b) exactly two units are good? c) at least two units are good? 4. A sales representatives make sales at approximately one-fifth of all calls. If, on a given day, the representative contacts six potential clients, what is the probability that a sale will be made with (a) all six contacts, (b) none of the contacts, and (c) at least one contact? • May 14th 2006, 07:19", "$$\\frac{n!-!n}{n!} = \\frac{4}{6} = \\frac{2}{3}.$$ Now, if your state machine, your simulation and this little calculation all end up with the same answer which is different from the one in your book, my conclusion is as above. • Thank you! I guess that the book must have some kind of error. I see the mathematical solution is very straightforward! – TajamSoft May 5 at 22:52 • I found the book's solutions, and they reach that result by first calculating the probability of none arriving, which is calculated as $(\\frac{2}{3})^3$. (That's incorrect unless you assume that the postman can give more than one packet to each house, and that's not given in the problem), and then the probability of at least one arriving is $1 - (\\frac{2}{3})^3 = 0.703$ – TajamSoft May 6 at 15:54 • Yes, that probability is wrong, because packages being misdelivered is not independent: the probability that the second package is misdelivered is different if the first one was correctly delivered or not. Specifically, if the first package was delivered correctly, then the second package is misdelivered with $p=\\frac{1}{2}$ (because we have two packages and two houses left, and one of the packages should go to one of the houses.) ... – Stephan Kolassa May 6 at 16:07 • ... Conversely, if the first package", "no one must miss out (iii) if the packages are indistinguishable and exactly one must miss out 1. The confusion: (i) If there is no restrictions: The answer is 3^7 (each package has 3 ways to be distributed) but why it is not 7^3 (each person has 7 choices to get the packages) ? Which question i should ask in order to get 7^3 because i think for the question in my problem, i can do either way (but i know it is wrong so please clarify !) Thanks for your help and sorry if it is too long Thanks again!! In how many ways can seven different packages be distributed to three people? There are three possible recipients for each of the seven packages, so there are $3^7$ ways to distribute the packages. You asked why the answer is not $7^3$. It helps to look at extreme cases. Suppose we had one package and three possible recipients. Clearly, there are three ways to distribute the package, depending on the recipient. Notice that $3 = 3^1$, not $1^3$. Also, a person can receive more than one package, but a package cannot be assigned to more than one person, so there are not seven choices of package for each of the three people. In how many ways can seven", "two post-offices. Each post-office has a probability $\\frac{1}{5}$ of losing an incoming parcel, independently of all other parcels. Given that a parcel is lost, the probability that it was lost by the second post-office is_________. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is", "1. ## probability hey guys. i'm stuck on this question, can anybody please help me by any chance? In a particular road it is estimated that there is a 31%chance that any specified house will be burgled over a period of two years, independently for each house. There are 6 houses in the road.What is the probability that fewer than two houses will be burgled over the period? $Pr(X<2)=Pr(X=0)+Pr(X=1)$ $Pr(X=x)=(^n_{x})p^x(1-p)^{n-x}$", "# Different Parcel Effectiveness 0 Have you seen this question before? Let’s say you are in charge of shipments at Amazon. There are two types of parcels you can ship packages in, say $A$ and $B$. Packages delivered in parcel $A$ are damaged during shipment with probability $p$, while packages delivered in parcel $B$ are damaged with probability $q$. What statistical test could you use to determine which parcel is better to use? What would the statistical test conclude if $p=0.4$ and $q=0.6$? Note: Assume the values of $p$ and $q$ were obtained from data recorded from 200 shipments, half delivered with parcel $A$ and half delivered with parcel $B$. Next question: 85% vs 82% .....", "## probability Suppose 500 independent poker hands are randomly dealt. a) Find the exact probability there will be at least two “full housesâ€. b) Use an appropriate approximation to find the probability in part a).", "27. Parcels from sender S to receiver R pass sequentially through two post-offices. Each post office has a probability $1\\over5$ of losing an incoming parcel, independently of all other parcels. Given that a parcel is lost, the probability that it was lost by the second post-office is_________.", "A person writes 4 letters and addresses 4 envelopes. If the letters are placed at random in the envelopes, then the probability that all the letters are not placed in the right envelopes is • 8 the letters can be kept in the envelops in 44 ways . but there is only one way to keep them in right places . so probability that all the letters are not placed in the right envelops =1/ 44 • -7 What are you looking for?", "s1 = 2, that's unacceptable. We skip the fifth parcel and get nothing for it. • T = 5: Nothing happens. • T = 6: We deliver the fourth, then the first parcel and get v1 + v4 = 3 bourles for them. Note that you could have skipped the fourth parcel and got the fifth one instead, but in this case the final sum would be 4 bourles.", "was wrongly delivered, then with $p=\\frac{1}{2}$ it was the package intended for house #2, so the probability for the second delivery to be correct is zero, and with $p=\\frac{1}{2}$, it was the one intended for house #3, so we still have a chance of $p=\\frac{1}{2}$ to deliver the second one correctly. Thus, in this case, the probabilty for a misdelivery to the second house is $\\frac{3}{4}$. ... – Stephan Kolassa May 6 at 16:07 • ... I hope the rest of the book is better. – Stephan Kolassa May 6 at 16:08", "# GATE Questions & Answers of Joint and Conditional Probability ## What is the Weightage of Joint and Conditional Probability in GATE Exam? Total 4 Questions have been asked from Joint and Conditional Probability topic of Probability and Statistics subject in previous GATE papers. Average marks 1.75. Parcels from sender S to receiver R pass sequentially through two post-offices. Each post-office has a probability $\\frac{1}{5}$ of losing an incoming parcel, independently of all other parcels. Given that a parcel is lost, the probability that it was lost by the second post-office is_________. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is", "# If four letters are put into four envelopes, what is the probability of only three letters are into correct envelopes? 14 views If four letters are put into four envelopes at random, what is the probability that only three of the letters are put into the correct envelopes? posted May 16", "# Math Help - probability 1. ## probability In a particular road it is estimated that there is a 31%chance that any specified house will be burgled over a period of two years, independently for each house. There are 6 houses in the road.What is the probability that fewer than two houses will be burgled over the period? 2. This is a binomial distribution with n=6 and p=0.31. If X is the number of houses burgled then we seek $Pr(X<2)=Pr(X=0)+Pr(X=1)$ then use the formula $Pr(X=x)=(^n_{x})p^x(1-p)^{n-x}$" ]

[ "1 = 3$, while the probability of him moving to house $3$ is proportional to $1 \\cdot 1 + 1 \\cdot 1 = 2$. Therefore, he will move to house $2$ with probability $\\frac{3}{5}$, and to house $3$ with probability $\\frac{2}{5}$. If he moves to house $3$, then, regardless of which present he delivers, he will be guaranteed to move back to house $1$ next, where he will deliver the remaining present. Otherwise, if he moves to house $2$ at first, then he will proceed to deliver a male-appropriate present there with probability $\\frac{1}{1+1} = \\frac{1}{2}$, and a female-appropriate one also with probability $\\frac{1}{2}$. In the former case, he will then have just 1 female present remaining, so he will move on to house $4$ with probability $0$, and back to house $1$ with probability $1$. Otherwise, he will have 1 male present remaining, and will move on to house $4$ with probability $\\frac{4}{5}$, and back to house $1$ with probability $\\frac{1}{5}$. In any case, he will then deliver his second present and be done. Editors note: You can deliver more male or female presents to a house, than the number of given gender inhabitants. Even if that number would be zero, you could still drop given type of presents. You only need to follow the formulas in", "cards of a specific one of those ranks from the $49$ cards remaining in the deck is $\\frac4{49}\\cdot\\frac3{48}$, so the probability of being dealt a pair after the initial three of a kind is $12\\cdot\\frac4{49}\\cdot\\frac3{48}$. Put the pieces together, and your $13\\cdot\\frac4{52}\\cdot\\frac3{51}\\cdot\\frac2{50}\\cdot12\\cdot\\frac4{49}\\cdot\\frac3{48}$ is the probability of being dealt a three of a kind followed by a pair. However, a full house doesn’t have to be dealt in that order. In fact, there are $\\binom52=10$ different ways to choose which $2$ of the $5$ cards (in order of dealing) make up the pair, and that’s your missing factor of $10$. Each of the $10$ possible ways of inserting the pair into the three of a kind has the same probability — the probability that you calculated — so the overall probability of a full house is $10$ times your figure.", "= 1$ Determine the total number of ways the three packages can be delivered. ⇒ $3×2×1 = 6$ The number of ways at least one house gets the wrong package is: ⇒ $6 - 1 = 5$ Therefore there are \\textbf{5} ways for at least one house to get the wrong package. ## (1) Comment(s) Name () Indirect approach easier: Total outcomes : 3! Chances for atleast one wrong = total outcomes - the outcome that they're all right (only 1 way to do it right) 6 - 1 = 5", "# Another way to calculate probability of full house I am struggling to understand why my calculation is off by a factor of ten. P(AAABB)= $4/52 * 3/51 * 2/50 * 13 * 4/49 * 3/48 * 12$ My thinking is like this: P(full house) = P(3 of same card) * (# of ways to get 3 of same card) * P(2 of same card) * (# of ways to get 2 of same card that is different than the first) • – Alex R. Feb 8 '15 at 0:46 • I have already looked at answers via counting. I'd like to use the conditional probability method. – Zslice Feb 8 '15 at 0:49 You’ve calculated the probability of being dealt a full house in such a way that the first three cards are the three of a kind and the last two cards are the pair. That is, there are $13$ possible ranks for the three of a kind; once you’ve picked one of them, the probability that the first $3$ cards dealt you are of that rank is $\\frac4{52}\\cdot\\frac3{51}\\cdot\\frac2{50}$. The overall probability of getting a three of a kind in the first $3$ cards dealt you is therefore $13\\cdot\\frac4{52}\\cdot\\frac3{51}\\cdot\\frac2{50}$. Similarly, there are then $12$ possible ranks for the pair, and the probability of being dealt $2$", "# Moderate Permutation-Combination Solved QuestionAptitude Discussion Q. A local delivery company has three packages to deliver to three different homes. If the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package? ✖ A. $3$ ✔ B. $5$ ✖ C. $3!$ ✖ D. $5!$ Solution: Option(B) is correct The possible outcomes that satisfy the condition of \"at least one house gets the wrong package\" are: One house gets the wrong package or two houses get the wrong package or three houses get the wrong package. We can calculate each of these cases and then add them together, or approach this problem from a different angle. The only case which is left out of the condition is the case where no wrong packages are delivered. If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above. There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package. The first person must get the correct package and the second person must get the correct package and the third person must get the correct package. ⇒ $1×1×1", "# Why am I under-counting when calculating the probability of a full house? I was trying to answer this question. Find the probability of getting a full house from a $52$ card deck. That is, find the probability of picking a pair of cards with the same rank (face value), and a triple with equal rank (different from the rank of the pair of course). Take one rank. The number of ways you can get a pair from a rank is $C(4, 2)$. Take another rank. The number of ways you can get a triple from that rank is $C(4, 3)$. Therefore, the number of ways you can get a full house from just $2$ ranks is $C(4, 2) \\cdot C(4, 3)$. But there are $13$ different ranks, and the number of ways you can select $2$ ranks out of $13$ is $C(13, 2)$. For each of these pairs of ranks, you can have $C(4, 2) \\cdot C(4, 3)$ full houses. Therefore the number of possible full houses is $C(4, 2) \\cdot C(4, 3) \\cdot C(13, 2)$, and the probability of selecting a full house is $\\dfrac {C(4, 2) \\cdot C(4, 3) \\cdot C(13, 2)}{C(52, 5)} \\approx 0.000720288$ However, this is way lower than the actual answer of $.00144$. Where did I go wrong here? You're missing a", "first house gets mail, the next house that gets mail must either be the third or fourth house. $b_n=a_{n-1}+a_{n-2}$ because if the first house does not get mail, the next house that gets mail must either be the second or third house. Note that we only need list out values of $a_n$ as $b$ depends on $a$. $a_1=1, a_2=1, a_3=2, a_4=2, \\ldots$. $a_{19}+b_{19}=a_{17}+a_{16}+a_{18}+a_{17}=a_{16}+2\\cdot a_{17}+a_{18}=65+2\\cdot 86+114=\\boxed{351}$.", "$$\\frac{n!-!n}{n!} = \\frac{4}{6} = \\frac{2}{3}.$$ Now, if your state machine, your simulation and this little calculation all end up with the same answer which is different from the one in your book, my conclusion is as above. • Thank you! I guess that the book must have some kind of error. I see the mathematical solution is very straightforward! – TajamSoft May 5 at 22:52 • I found the book's solutions, and they reach that result by first calculating the probability of none arriving, which is calculated as $(\\frac{2}{3})^3$. (That's incorrect unless you assume that the postman can give more than one packet to each house, and that's not given in the problem), and then the probability of at least one arriving is $1 - (\\frac{2}{3})^3 = 0.703$ – TajamSoft May 6 at 15:54 • Yes, that probability is wrong, because packages being misdelivered is not independent: the probability that the second package is misdelivered is different if the first one was correctly delivered or not. Specifically, if the first package was delivered correctly, then the second package is misdelivered with $p=\\frac{1}{2}$ (because we have two packages and two houses left, and one of the packages should go to one of the houses.) ... – Stephan Kolassa May 6 at 16:07 • ... Conversely, if the first package", "the line together.) Similarly, the probability of three couples standing together is P3 = $$2^3 \\cdot 7! / 10!$$, the probability of four is P4 = $$2^4 \\cdot 6! / 10!$$, and the probability of five is $$P5 = 2^5 \\cdot 5! / 10!$$. We are interested in one minus the probability that at least one couple stands together. This is where the inclusion-exclusion principle comes in. Say we wanted the total area of a simple Venn diagram. We can’t simply take Circle 1 + Circle 2, because this counts the middle intersecting section twice. Instead, we have to take Circle 1 + Circle 2 – the middle section. Similarly, adding more circles to the Venn diagram, you have to subtract or add intersecting sections until you don’t overcount (or undercount) them. So, back to our couples. When we calculate the probability of, say, the P1 scenario — one couple standing together — we need to account for the chance that more than one couple is standing together, and include that probability in our calculation. When the scenarios overlap it’s the same concept as when the circles do. The inclusion-exclusion principle helps us from there, offering the coefficients to apply to our five probabilities. This equals 5P1 – 10P2 + 10P3 – 5P4 + P5 = 88/135. One", "a kind and 2 of a kind (a \"Full House\") c) 1 Ace, 1 Two, 1 Three, 1 Four, 1 Five (a \"Straight\") d) any Straight ( 5 ordered cards) for a) I have that it is $\\frac{24}{2598960}$, since it is $\\frac{ {4\\choose2}{4\\choose3}}{{52\\choose5}}$ . . . . Right! (b) There are 13 choices for the value of the Triple. There are 12 choices for the value of the Pair. Then there are ${4\\choose3}$ to get the Triple . . and ${4\\choose2}$ ways to get the Pair. The number of Full Houses is: . $13\\cdot12\\cdot 4\\cdot 6\\;=\\;3744$ for c) is it $P^{52}_{5} = \\frac{52!}{48!}$ for the denominator . . . . no and $\\binom{4}{1}^5$ for the numerator . . . . yes The denominator is still: . ${52\\choose5}$ d) I think that the denominator is the same as c) but the numerator will be 9 times the original . . . . not quite There are ten Straights: .from A-2-3-4-5 to 10-J-Q-K-A The probability is: . $\\frac{(10)(1024)}{{52\\choose5}}$", "the yahtzee game. A full house is three of a kind plus a pair, like 66655, 11122 or 22333, and any variation thereof. In this particular case I want to look at getting a full house in one go, because to take all three turns into account is way too hairy. The case is actually similar to the previous one in that you have 3 of one group and 2 of another, but the probabilities are a little different. For a specific full house, say 11122 the base chance is 1/65. However, we're not looking for a specific one, but any full house. This means that there will be 6 options for the first group, and 5 for the second. The final probability is (16) $P(\\text{any full house}) = 6 \\cdot 5 \\times P(\\text{full house}) = 30\\cdot{ 5 \\choose 3 } \\times $$\\frac16$$^5 = 30\\cdot10 \\times \\frac1{7776} = 3.86%$ An alternative approach is to consider the group-determining dice to be “free”, giving the probability 6/6 and 5/6 for the first and second group, respectively. After that, the remaining dice have to follow suit to give (1/6)3. ### 4.3 Multinomial : spinning wheels Fig 4. 10-lettered wheel; na = 2, nb = 3, nc = 5. And now for an application of the multinomial. Consider the wheel from Fig", "When drawing two pairs, why do you have to take into account the ordering of the pairs themselves? Shouldn't it be enough to just order the cards? I think my major issue with this type of probabilities is that I don't understand how ordering works. To top it off, it seems I have to order the pairs only when the two groups are of the same \"type\"? I mean, in Full House you have one pair and one three-of-a-kind, making two groups, but you don't have to order these? To illustrate: Two Pairs Correct P(two pairs) = $\\frac{{13 \\choose 2} \\cdot {4 \\choose 2}^2 \\cdot 4 ({11 \\choose 1})}{({52 \\choose 5})}$. What I thought was correct: P(two pairs) = $\\frac{13 \\cdot 12 \\cdot {4 \\choose 2}^2 \\cdot 4 {11 \\choose 1}}{{52 \\choose 5}}$. Notice my (wrong) formula got 13*12 instead of (13 choose 2). Full House Correct P(house)= $\\frac{13 {4 \\choose 2} \\cdot 12 {4 \\choose 3}}{{52 \\choose 5}}$ Notice how I don't use (13 choose 2) here like I do in the two pairs formula. Why is it like this? Last edited: Jan 18, 2014 11. Jan 18, 2014 ### D H Staff Emeritus One way to look at it is counting. Of the 52 choose 5 possible hands, how many are two pair? There are 13", "route to it. There are $\\binom{10}2=45$ pairs of seats, and the couple is equally likely to occupy any one of those $45$ pairs of seats. Nine of the $45$ pairs are adjacent, so the probability that they will occupy adjacent seats is $\\frac9{45}=\\frac15$, as you say. And here is yet another. The man sits in an end seat with probability $\\frac2{10}=\\frac15$. If he’s in an end seat, only one of the remaining nine seats is adjacent to him, and his wife’s probability of getting that seat is $\\frac19$. With probability $\\frac45$ the man sits in one of the eight seats that have two neighbors, and in that case his wife’s probability of ending up next to him is $\\frac29$. The overall probability that the end up sitting together is therefore $$\\frac15\\cdot\\frac19+\\frac45\\cdot\\frac29=\\frac9{45}=\\frac15\\;.$$ Added: And just for fun, here’s yet another. Imagine that the seats are arranged in a circle around a table. Wherever the wife is sitting, the husband’s probability of sitting next to her is $\\frac29$. Then the table is taken away and the seats unwrapped into a straight line, with the breakpoint chosen at random: with probability $\\frac1{10}$ it will fall between the husband and the wife, so with probability $\\frac9{10}$ they will still end up together. Thus, they end up together with probability $$\\frac9{10}\\cdot\\frac29=\\frac15\\;.$$ - And", "we know the probability of each of those packages appearing (1/4). We also know the probability of getting a boy to answer the door under each of the packages of kids. So we can just multiply those probabilities out and add them per the rules of probability we've already established. For example, the probability of bg is .25, and the probability of getting a boy answering the door there is 1/2, and so forth. We ultimately end up with $$P(A) = .25 \\cdot .5 + .25 \\cdot .5 + .25 \\cdot 1 + .25 \\cdot 0 = .5$$. Now we can plug in: $$P(B|A) = \\frac{1 \\cdot .25}{.5} = \\frac{1}{2}$$ Just as the readings told us. How about the Abel problem? Well, we don't even need Bayes Rule for this one, we just need to observe that 1/3 of the families with one boy have both boys. But let's do it anyway, just as an exercise. Set B to be, yet again, \"family has both boys,\" and set A to be \"family has one boy.\" • Probability of a family having one boy, conditional on their having both boys --- $$P(A|B) = 1$$, obviously. • Probability of a family having both boys --- $$P(B) = 1/4$$ because, again, our available packages of kids are $$\\{bg, gb, bb, gg\\}$$,", "# Yahtzee game, probability of getting full house,4 of a kind In the game of Yahtzee, five dice are tossed simultaneously. Find the probability of getting a. full house b. 4 of a kind Bases on wikipedia Full House = A three-of-a-kind and a pair Four-Of-A-Kind = At least four dice showing the same face And total number of cases is 6^5=7776 I really don't understand this game, - dice pattern probability calculator – MJD May 15 '13 at 15:43 The answer depends on whether you want to count only the throws that satisfy only the requirements for a full-house/four-of-a-kind entry, or also the throws with all five dice the same, which you could use either in the full-house/four-of-a-kind row or in the Yahtzee (five-of-a-kind) row. The difference is the probability of a Yahtzee, which is $1$ in $6^4=1296$; I'll calculate the probabilities excluding five of a kind. In both cases, full house and four of a kind, there are $6\\cdot5$ different choices for the numbers. For full house, there are $\\binom52$ choices for the positions and for four of a kind there are $\\binom51$. Thus the probability for a full house (excluding five of a kind) is $$6\\cdot5\\cdot\\binom52/6^5=30\\cdot10/6^5=25/648\\approx0.03858\\;,$$ and for four of a kind $$6\\cdot5\\cdot\\binom51/6^5=30\\cdot5/6^5=25/1296\\approx0.01929\\;.$$ Note that these are just the probabilities for the first throw, which", "# Project Euler #100: Arranged probability If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, $P(BB) = \\frac {15} {21} \\times \\frac {14} {20} = \\frac 1 2$. The next such arrangement, for which there is exactly 50% chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs. By finding the first arrangement to contain over $10^{12} = 1,000,000,000,000$ discs in total, determine the number of blue discs that the box would contain. This is a fun problem that seems out of reach initially but can be reduced to something more simple. ## Compute the probability🔗 Assuming there are $b$ blue discs and $r$ red discs, the probability of choosing one blue disc at random is $P_1 = \\frac {b} {b+r}$. The probability of choosing another blue disc from the remaining set is $P_2 = \\frac {b-1} {b+r-1}$. So the probability of choosing 2 blue discs is $P(BB) = P_1 \\cdot P_2 = \\frac {b} {b+r} \\cdot \\frac {b-1} {b+r-1}$ We need to find the smallest solution of: $$\\begin{cases} \\frac {b} {b+r} \\cdot \\frac {b-1} {b+r-1} = \\frac 1 2 \\newline b +", "don't match the first three dice), for a total of $3\\cdot 5=15$ possible outcomes, plus the outcome that all five dice match. In the second case, we need all three dice to match each other. There are $5$ ways to pick which value the three dice will have so that they don't match the first pair, plus the outcome that all five dice match. So there are a total of $15+5=20$ ways to get a full house without all five dice matching, added to the possibility that all five dice match, which makes $21$ ways to get a full house. So, the probability is successful outcomes/total outcomes= 21/216= 7/72 Mar 16, 2019", "card, then we draw a card of the same denomination, etc) the order is fixed. That is, you are not considering that the 5 cards of a full house can be drawn in any order. There are basically two ways to solve the problem: 1. Combinations. There are $\\binom{52}5$ deals. How many of these deals are a full house? First: choose the denomination of the three: 13 possibilities. Then, choose the denomination of the pair: 12 possibilities. Now choose the cards of the three: $\\binom{4}3$ possibilities. Now, choose the two other cards: $\\binom{4}2$ possibilities. Then, the probability is $$\\frac{13\\cdot12\\cdot\\binom{4}3\\binom{4}2}{\\binom{52}5}$$ 2. Variations. There are $52\\cdot51\\cdot50\\cdot49\\cdot48$ deals. Now, we are considering that two deals are different if the cards were drawn in a different order, so there are much more deals. Namely, $5!=120$ times more. Now you can count your way. The probability is $$120\\cdot \\frac{52\\cdot3\\cdot2\\cdot48\\cdot3}{52\\cdot51\\cdot50\\cdot49\\cdot48}$$ You can check that they are, in fact, the same number. Another way is the following: • The probability for the first card is $1$, because no matter which card is, the full house can be achieved. • The probaility of that the second card is of the same kind as first is $3/51$, and the pro. of that the third card is also the same kind is $2/50$ • The pro. of that", "with 2 pairs}) = \\frac{\\text{#favorable outcomes}}{\\text{#total outcomes}} =\\binom 7 4 \\cdot \\frac{25 \\cdot 4 \\cdot 20 \\cdot 4 \\cdot 15 \\cdot 10 \\cdot 5}{25 \\cdot 24 \\cdot 23 \\cdot 22 \\cdot 21 \\cdot 20 \\cdot 19} \\approx 8.67\\%$$ Then I add 8.67 to 6.5, and there is a 15.17% chance that the game ends in 7 draws. That would be $\\binom 7 4 \\cdot \\binom 4 2 \\cdot \\frac 1 {2!}$ or $\\binom 7 2 \\cdot \\binom 5 2 \\cdot \\frac 1 {2!}$. First you select 4 from the 7 cards, then you select 2 of those 4 for the second pair. After that, you have to divide by 2, because the 2 pairs can be interchanged, leading to double counting. We have to divide by 2 to compensate. Or alternatively, first you select 2 from the 7 cards, then you select another 2 of the remaining 5, and divide by 2 to compensate for double counting. Btw, after summing the probabilities for 1-1-1-2-3-4-5 and 1-1-2-2-3-4-5, we have to subtract the probability on game-in-5 and game-in-6 to get the proper probability. Continuing this method, I found there is about a 5.2% chance the game ends in 8 draws (1.38% chance of 4-of-a-kind, 2.3% chance of \"full house,\" and 1.5% chance of three pairs). Going one more level, there", "was wrongly delivered, then with $p=\\frac{1}{2}$ it was the package intended for house #2, so the probability for the second delivery to be correct is zero, and with $p=\\frac{1}{2}$, it was the one intended for house #3, so we still have a chance of $p=\\frac{1}{2}$ to deliver the second one correctly. Thus, in this case, the probabilty for a misdelivery to the second house is $\\frac{3}{4}$. ... – Stephan Kolassa May 6 at 16:07 • ... I hope the rest of the book is better. – Stephan Kolassa May 6 at 16:08" ]

[ "1 = 3$, while the probability of him moving to house $3$ is proportional to $1 \\cdot 1 + 1 \\cdot 1 = 2$. Therefore, he will move to house $2$ with probability $\\frac{3}{5}$, and to house $3$ with probability $\\frac{2}{5}$. If he moves to house $3$, then, regardless of which present he delivers, he will be guaranteed to move back to house $1$ next, where he will deliver the remaining present. Otherwise, if he moves to house $2$ at first, then he will proceed to deliver a male-appropriate present there with probability $\\frac{1}{1+1} = \\frac{1}{2}$, and a female-appropriate one also with probability $\\frac{1}{2}$. In the former case, he will then have just 1 female present remaining, so he will move on to house $4$ with probability $0$, and back to house $1$ with probability $1$. Otherwise, he will have 1 male present remaining, and will move on to house $4$ with probability $\\frac{4}{5}$, and back to house $1$ with probability $\\frac{1}{5}$. In any case, he will then deliver his second present and be done. Editors note: You can deliver more male or female presents to a house, than the number of given gender inhabitants. Even if that number would be zero, you could still drop given type of presents. You only need to follow the formulas in", "= 1$ Determine the total number of ways the three packages can be delivered. ⇒ $3×2×1 = 6$ The number of ways at least one house gets the wrong package is: ⇒ $6 - 1 = 5$ Therefore there are \\textbf{5} ways for at least one house to get the wrong package. ## (1) Comment(s) Name () Indirect approach easier: Total outcomes : 3! Chances for atleast one wrong = total outcomes - the outcome that they're all right (only 1 way to do it right) 6 - 1 = 5", "cards of a specific one of those ranks from the $49$ cards remaining in the deck is $\\frac4{49}\\cdot\\frac3{48}$, so the probability of being dealt a pair after the initial three of a kind is $12\\cdot\\frac4{49}\\cdot\\frac3{48}$. Put the pieces together, and your $13\\cdot\\frac4{52}\\cdot\\frac3{51}\\cdot\\frac2{50}\\cdot12\\cdot\\frac4{49}\\cdot\\frac3{48}$ is the probability of being dealt a three of a kind followed by a pair. However, a full house doesn’t have to be dealt in that order. In fact, there are $\\binom52=10$ different ways to choose which $2$ of the $5$ cards (in order of dealing) make up the pair, and that’s your missing factor of $10$. Each of the $10$ possible ways of inserting the pair into the three of a kind has the same probability — the probability that you calculated — so the overall probability of a full house is $10$ times your figure.", "# Moderate Permutation-Combination Solved QuestionAptitude Discussion Q. A local delivery company has three packages to deliver to three different homes. If the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package? ✖ A. $3$ ✔ B. $5$ ✖ C. $3!$ ✖ D. $5!$ Solution: Option(B) is correct The possible outcomes that satisfy the condition of \"at least one house gets the wrong package\" are: One house gets the wrong package or two houses get the wrong package or three houses get the wrong package. We can calculate each of these cases and then add them together, or approach this problem from a different angle. The only case which is left out of the condition is the case where no wrong packages are delivered. If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above. There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package. The first person must get the correct package and the second person must get the correct package and the third person must get the correct package. ⇒ $1×1×1", "$$\\frac{n!-!n}{n!} = \\frac{4}{6} = \\frac{2}{3}.$$ Now, if your state machine, your simulation and this little calculation all end up with the same answer which is different from the one in your book, my conclusion is as above. • Thank you! I guess that the book must have some kind of error. I see the mathematical solution is very straightforward! – TajamSoft May 5 at 22:52 • I found the book's solutions, and they reach that result by first calculating the probability of none arriving, which is calculated as $(\\frac{2}{3})^3$. (That's incorrect unless you assume that the postman can give more than one packet to each house, and that's not given in the problem), and then the probability of at least one arriving is $1 - (\\frac{2}{3})^3 = 0.703$ – TajamSoft May 6 at 15:54 • Yes, that probability is wrong, because packages being misdelivered is not independent: the probability that the second package is misdelivered is different if the first one was correctly delivered or not. Specifically, if the first package was delivered correctly, then the second package is misdelivered with $p=\\frac{1}{2}$ (because we have two packages and two houses left, and one of the packages should go to one of the houses.) ... – Stephan Kolassa May 6 at 16:07 • ... Conversely, if the first package", "first house gets mail, the next house that gets mail must either be the third or fourth house. $b_n=a_{n-1}+a_{n-2}$ because if the first house does not get mail, the next house that gets mail must either be the second or third house. Note that we only need list out values of $a_n$ as $b$ depends on $a$. $a_1=1, a_2=1, a_3=2, a_4=2, \\ldots$. $a_{19}+b_{19}=a_{17}+a_{16}+a_{18}+a_{17}=a_{16}+2\\cdot a_{17}+a_{18}=65+2\\cdot 86+114=\\boxed{351}$.", "# Another way to calculate probability of full house I am struggling to understand why my calculation is off by a factor of ten. P(AAABB)= $4/52 * 3/51 * 2/50 * 13 * 4/49 * 3/48 * 12$ My thinking is like this: P(full house) = P(3 of same card) * (# of ways to get 3 of same card) * P(2 of same card) * (# of ways to get 2 of same card that is different than the first) • – Alex R. Feb 8 '15 at 0:46 • I have already looked at answers via counting. I'd like to use the conditional probability method. – Zslice Feb 8 '15 at 0:49 You’ve calculated the probability of being dealt a full house in such a way that the first three cards are the three of a kind and the last two cards are the pair. That is, there are $13$ possible ranks for the three of a kind; once you’ve picked one of them, the probability that the first $3$ cards dealt you are of that rank is $\\frac4{52}\\cdot\\frac3{51}\\cdot\\frac2{50}$. The overall probability of getting a three of a kind in the first $3$ cards dealt you is therefore $13\\cdot\\frac4{52}\\cdot\\frac3{51}\\cdot\\frac2{50}$. Similarly, there are then $12$ possible ranks for the pair, and the probability of being dealt $2$", "a kind and 2 of a kind (a \"Full House\") c) 1 Ace, 1 Two, 1 Three, 1 Four, 1 Five (a \"Straight\") d) any Straight ( 5 ordered cards) for a) I have that it is $\\frac{24}{2598960}$, since it is $\\frac{ {4\\choose2}{4\\choose3}}{{52\\choose5}}$ . . . . Right! (b) There are 13 choices for the value of the Triple. There are 12 choices for the value of the Pair. Then there are ${4\\choose3}$ to get the Triple . . and ${4\\choose2}$ ways to get the Pair. The number of Full Houses is: . $13\\cdot12\\cdot 4\\cdot 6\\;=\\;3744$ for c) is it $P^{52}_{5} = \\frac{52!}{48!}$ for the denominator . . . . no and $\\binom{4}{1}^5$ for the numerator . . . . yes The denominator is still: . ${52\\choose5}$ d) I think that the denominator is the same as c) but the numerator will be 9 times the original . . . . not quite There are ten Straights: .from A-2-3-4-5 to 10-J-Q-K-A The probability is: . $\\frac{(10)(1024)}{{52\\choose5}}$", "When drawing two pairs, why do you have to take into account the ordering of the pairs themselves? Shouldn't it be enough to just order the cards? I think my major issue with this type of probabilities is that I don't understand how ordering works. To top it off, it seems I have to order the pairs only when the two groups are of the same \"type\"? I mean, in Full House you have one pair and one three-of-a-kind, making two groups, but you don't have to order these? To illustrate: Two Pairs Correct P(two pairs) = $\\frac{{13 \\choose 2} \\cdot {4 \\choose 2}^2 \\cdot 4 ({11 \\choose 1})}{({52 \\choose 5})}$. What I thought was correct: P(two pairs) = $\\frac{13 \\cdot 12 \\cdot {4 \\choose 2}^2 \\cdot 4 {11 \\choose 1}}{{52 \\choose 5}}$. Notice my (wrong) formula got 13*12 instead of (13 choose 2). Full House Correct P(house)= $\\frac{13 {4 \\choose 2} \\cdot 12 {4 \\choose 3}}{{52 \\choose 5}}$ Notice how I don't use (13 choose 2) here like I do in the two pairs formula. Why is it like this? Last edited: Jan 18, 2014 11. Jan 18, 2014 ### D H Staff Emeritus One way to look at it is counting. Of the 52 choose 5 possible hands, how many are two pair? There are 13", "# Why am I under-counting when calculating the probability of a full house? I was trying to answer this question. Find the probability of getting a full house from a $52$ card deck. That is, find the probability of picking a pair of cards with the same rank (face value), and a triple with equal rank (different from the rank of the pair of course). Take one rank. The number of ways you can get a pair from a rank is $C(4, 2)$. Take another rank. The number of ways you can get a triple from that rank is $C(4, 3)$. Therefore, the number of ways you can get a full house from just $2$ ranks is $C(4, 2) \\cdot C(4, 3)$. But there are $13$ different ranks, and the number of ways you can select $2$ ranks out of $13$ is $C(13, 2)$. For each of these pairs of ranks, you can have $C(4, 2) \\cdot C(4, 3)$ full houses. Therefore the number of possible full houses is $C(4, 2) \\cdot C(4, 3) \\cdot C(13, 2)$, and the probability of selecting a full house is $\\dfrac {C(4, 2) \\cdot C(4, 3) \\cdot C(13, 2)}{C(52, 5)} \\approx 0.000720288$ However, this is way lower than the actual answer of $.00144$. Where did I go wrong here? You're missing a", "three people we choose two, to be the people interacting with the special person. Thus, we have $4\\cdot\\binom{3}{2}\\cdot4! = 288$ ways here. Case 3: $4$ people. In this case. First note that no person can give/receive twice. If we keep the order of A, B, C, D giving in that order (and permute afterward), then there are three options to choose A's receiver, and three options for B's receiver afterward. Then it is uniquely determined who C and D give to. This gives a total of $3\\cdot3\\cdot4! = 216$ ways, after permuting. So we have a total of $36+288+216=540$ ways to order the four pairs of people. Now we divide this by the total number of ways: $(4\\cdot3)^4$ (four rounds, four ways to choose giver, three to choose receiver each round). So the answer is $\\frac{540}{12^4} = \\boxed{\\textbf{(B)}\\frac{5}{192}}$. ~ccx09 Latex polished by Argonauts16 and by Grace.yongqing.yu. ## Solution 3 Similar to solution 2, we think in terms of transactions. WLOG, for the 1st transaction, we assume that A gives to B, which we denote AB. For the 2nd transaction, there are 12 options to choose from, so there are $12^3$ possible options. Case 1 (Giving back immediately): The 2nd transaction is BA (1 option). Then, the 3rd transaction can be whatever you like (12 options), but the 4th", "was wrongly delivered, then with $p=\\frac{1}{2}$ it was the package intended for house #2, so the probability for the second delivery to be correct is zero, and with $p=\\frac{1}{2}$, it was the one intended for house #3, so we still have a chance of $p=\\frac{1}{2}$ to deliver the second one correctly. Thus, in this case, the probabilty for a misdelivery to the second house is $\\frac{3}{4}$. ... – Stephan Kolassa May 6 at 16:07 • ... I hope the rest of the book is better. – Stephan Kolassa May 6 at 16:08", "we know the probability of each of those packages appearing (1/4). We also know the probability of getting a boy to answer the door under each of the packages of kids. So we can just multiply those probabilities out and add them per the rules of probability we've already established. For example, the probability of bg is .25, and the probability of getting a boy answering the door there is 1/2, and so forth. We ultimately end up with $$P(A) = .25 \\cdot .5 + .25 \\cdot .5 + .25 \\cdot 1 + .25 \\cdot 0 = .5$$. Now we can plug in: $$P(B|A) = \\frac{1 \\cdot .25}{.5} = \\frac{1}{2}$$ Just as the readings told us. How about the Abel problem? Well, we don't even need Bayes Rule for this one, we just need to observe that 1/3 of the families with one boy have both boys. But let's do it anyway, just as an exercise. Set B to be, yet again, \"family has both boys,\" and set A to be \"family has one boy.\" • Probability of a family having one boy, conditional on their having both boys --- $$P(A|B) = 1$$, obviously. • Probability of a family having both boys --- $$P(B) = 1/4$$ because, again, our available packages of kids are $$\\{bg, gb, bb, gg\\}$$,", "probability we're the first delivery for B) we beat an event with probability $$\\frac{n-1}{n}$$ (the probability we're not the first delivery of A) • The second delivery for B will beat everything except the first two deliveries from A. So, with probability $$\\frac{1}{n}$$ we beat an event with probablity $$\\frac{n-2}{n}$$ • etc. Putting each of those terms into a summation, we get: $$(\\frac{1}{n} \\cdot \\frac{n-1}{n}) + (\\frac{1}{n} \\cdot \\frac{n-2}{n}) + ... + (\\frac{1}{n} \\cdot \\frac{n-n}{n})$$ Or, $$\\sum_{i=1}^{n} \\frac{n-i}{n^2}$$ Since the probabilities are uniform and half (rounded down) of each occur during the hour of overlap, we only consider half the deliveries of each. If $$n'=\\lfloor\\frac{n}{2}\\rfloor$$ and, compared to the whole domain, those events only happen half the time. So $$\\frac{1}{2}\\sum_{i=1}^{n'} \\frac{n'-i}{n'^2}$$ I believe that for $$a=b=n$$, you get $$1/8$$. How to handle the fact A and B do not deliver the same number of packages? Again, to simplify, assume all their deliveries happen between 9-10am. For every delivery $$b$$ you consider from earliest to latest, instead of each successive one beating $$\\frac{1}{a}$$ less from truck A, as above (where $$a$$ is the number of deliveries made by truck A and $$b$$ the number of deliveries made by $$b$$), you eliminate $$\\lfloor \\frac{1}{b} \\cdot a \\rfloor$$. That is, you beat all but a fraction of $$a$$ proportional to the", "# Project Euler #100: Arranged probability If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, $P(BB) = \\frac {15} {21} \\times \\frac {14} {20} = \\frac 1 2$. The next such arrangement, for which there is exactly 50% chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs. By finding the first arrangement to contain over $10^{12} = 1,000,000,000,000$ discs in total, determine the number of blue discs that the box would contain. This is a fun problem that seems out of reach initially but can be reduced to something more simple. ## Compute the probability🔗 Assuming there are $b$ blue discs and $r$ red discs, the probability of choosing one blue disc at random is $P_1 = \\frac {b} {b+r}$. The probability of choosing another blue disc from the remaining set is $P_2 = \\frac {b-1} {b+r-1}$. So the probability of choosing 2 blue discs is $P(BB) = P_1 \\cdot P_2 = \\frac {b} {b+r} \\cdot \\frac {b-1} {b+r-1}$ We need to find the smallest solution of: $$\\begin{cases} \\frac {b} {b+r} \\cdot \\frac {b-1} {b+r-1} = \\frac 1 2 \\newline b +", "card, then we draw a card of the same denomination, etc) the order is fixed. That is, you are not considering that the 5 cards of a full house can be drawn in any order. There are basically two ways to solve the problem: 1. Combinations. There are $\\binom{52}5$ deals. How many of these deals are a full house? First: choose the denomination of the three: 13 possibilities. Then, choose the denomination of the pair: 12 possibilities. Now choose the cards of the three: $\\binom{4}3$ possibilities. Now, choose the two other cards: $\\binom{4}2$ possibilities. Then, the probability is $$\\frac{13\\cdot12\\cdot\\binom{4}3\\binom{4}2}{\\binom{52}5}$$ 2. Variations. There are $52\\cdot51\\cdot50\\cdot49\\cdot48$ deals. Now, we are considering that two deals are different if the cards were drawn in a different order, so there are much more deals. Namely, $5!=120$ times more. Now you can count your way. The probability is $$120\\cdot \\frac{52\\cdot3\\cdot2\\cdot48\\cdot3}{52\\cdot51\\cdot50\\cdot49\\cdot48}$$ You can check that they are, in fact, the same number. Another way is the following: • The probability for the first card is $1$, because no matter which card is, the full house can be achieved. • The probaility of that the second card is of the same kind as first is $3/51$, and the pro. of that the third card is also the same kind is $2/50$ • The pro. of that", "# Yahtzee game, probability of getting full house,4 of a kind In the game of Yahtzee, five dice are tossed simultaneously. Find the probability of getting a. full house b. 4 of a kind Bases on wikipedia Full House = A three-of-a-kind and a pair Four-Of-A-Kind = At least four dice showing the same face And total number of cases is 6^5=7776 I really don't understand this game, - dice pattern probability calculator – MJD May 15 '13 at 15:43 The answer depends on whether you want to count only the throws that satisfy only the requirements for a full-house/four-of-a-kind entry, or also the throws with all five dice the same, which you could use either in the full-house/four-of-a-kind row or in the Yahtzee (five-of-a-kind) row. The difference is the probability of a Yahtzee, which is $1$ in $6^4=1296$; I'll calculate the probabilities excluding five of a kind. In both cases, full house and four of a kind, there are $6\\cdot5$ different choices for the numbers. For full house, there are $\\binom52$ choices for the positions and for four of a kind there are $\\binom51$. Thus the probability for a full house (excluding five of a kind) is $$6\\cdot5\\cdot\\binom52/6^5=30\\cdot10/6^5=25/648\\approx0.03858\\;,$$ and for four of a kind $$6\\cdot5\\cdot\\binom51/6^5=30\\cdot5/6^5=25/1296\\approx0.01929\\;.$$ Note that these are just the probabilities for the first throw, which", "the yahtzee game. A full house is three of a kind plus a pair, like 66655, 11122 or 22333, and any variation thereof. In this particular case I want to look at getting a full house in one go, because to take all three turns into account is way too hairy. The case is actually similar to the previous one in that you have 3 of one group and 2 of another, but the probabilities are a little different. For a specific full house, say 11122 the base chance is 1/65. However, we're not looking for a specific one, but any full house. This means that there will be 6 options for the first group, and 5 for the second. The final probability is (16) $P(\\text{any full house}) = 6 \\cdot 5 \\times P(\\text{full house}) = 30\\cdot{ 5 \\choose 3 } \\times $$\\frac16$$^5 = 30\\cdot10 \\times \\frac1{7776} = 3.86%$ An alternative approach is to consider the group-determining dice to be “free”, giving the probability 6/6 and 5/6 for the first and second group, respectively. After that, the remaining dice have to follow suit to give (1/6)3. ### 4.3 Multinomial : spinning wheels Fig 4. 10-lettered wheel; na = 2, nb = 3, nc = 5. And now for an application of the multinomial. Consider the wheel from Fig", "Nuclear Wang Sep 6 '18 at 19:59 Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a \\cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a \\cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is $$\\frac{a\\cdot b}{2}\\frac{1}{2a \\cdot 2b} = \\frac{1}{8}.$$ I propose another way of looking at it, only if you had a pc during the interview of course. We can simulate the process with R, for example. Let's simulate 1000 values from A and the same from B, we know that both are uniforms, are independent. a <- runif(1000, 8, 10) # A deliveries b <- runif(1000, 9, 11) # B deliveries # [1] 9.485513 8.665070 8.488481 8.840332 8.755384 9.448949 # A deliveries for example Ok they're not exactly times, but it's the same. The probability $P(B<A)$ is what we seek. So we just count the number of pairs where $b<a$ in our code. prob <- sum(b < a)/1000 #[1] 0.112 # almost 1/8 We can also plot the 1000 pairs $(a,b)$, and see the region where B comes first. plot(a, b) polygon(c(9, 10, 10,", "multiplied with the number of suits. #### Four of a kind Four of a kind means 4 cards of one rank, and an unmatched card of another rank. The probability is: $P_{4\\_of\\_kind} = \\frac{13 \\cdot 48}{C(52, 5)} \\approx 0.024\\%$ There are 13 possible ways to get 4 cards of the same rank (that’s the number of ranks available) and there are (52 – 4) possible ways to get the remaining 5th card. #### Full house Full house means 3 matching cards of one rank, and 2 matching cards of another rank. The probability is: $P_{full\\_house} = \\frac{P(13, 2) \\cdot C(4, 3) \\cdot C(4, 2)}{C(52, 5)} \\approx 0.144\\%$ There are $P(52, 5)$ possible ways to get 2 suits from the 13 available. This is because in this case it matters if for example we have 3 cards of clubs and 2 cards of spades or 3 cards of spades and 2 cards of clubs. There are $C(4, 3)$ and $C(4, 2)$ possible ways of getting 3 and respectively 2 cards from the 4 cards of the same rank. #### Flush Flush means 5 cards of the same suit, not in rank sequence, excluding the straight flush and royal flush. The probability is: $P_{flush} = \\frac{4 \\cdot C(13, 5) - 40}{C(52, 5)} \\approx 0.19654\\%$ There are $C(13, 5)$ possible ways" ]

[ "+0 # Helppppp 0 55 1 The pattern of Pascal's triangle is illustrated in the diagram shown. What is the fourth element in Row 15 of Pascal's triangle? $$\\begin{array}{ccccccccccccc} \\textrm{Row 0}: & \\qquad & & & & & 1 & & & & & & \\\\ \\textrm{Row 1}: & \\qquad & & & & 1 & & 1 & & & & &\\\\ \\textrm{Row 2}: & \\qquad & & & 1 & & 2 & & 1 & & & &\\\\ \\textrm{Row 3}: & \\qquad & & 1 && 3 && 3 && 1&& \\\\ \\textrm{Row 4}: & \\qquad & 1&& 4 && 6 && 4 && 1 \\end{array}$$ Mar 28, 2020 #1 0 The pattern is: 15C0, 15C1, 15C2, 15C3..........and so on. The 4th element is: 15C(4 - 1) =15C3 = 455. Mar 28, 2020 edited by Guest Mar 28, 2020", "numbers $n$ and their calculated values as shown below: $${{1}\\choose{0}}\\qquad {{1}\\choose{1}}$$ $${{2}\\choose{0}}\\qquad {{2}\\choose{1}}\\qquad {{2}\\choose{2}}$$ $${{3}\\choose{0}} \\qquad {{3}\\choose{1}}\\qquad {{3}\\choose{2}}\\qquad {{3}\\choose{3}}$$ $${{4}\\choose{0}}\\qquad {{4}\\choose{1}}\\qquad {{4}\\choose{2}}\\qquad {{4}\\choose{3}} \\qquad{{4}\\choose{4}}$$ $$\\vdots$$ The triangle above is commonly known as a Pascal’s or Chinese triangle. Write the calculated values with two more added lines: $$1 \\qquad 1$$ $$1 \\qquad 2 \\qquad 1$$ $$1 \\qquad 3 \\qquad 3 \\qquad 1$$ $$1 \\qquad 4 \\qquad 6 \\qquad 4 \\qquad 1$$ $$1 \\qquad 5 \\qquad 10 \\qquad 10 \\qquad 5 \\qquad 1$$ $$1 \\qquad 6 \\qquad 15 \\qquad 20 \\qquad 15 \\qquad 6 \\qquad 1$$ $$\\vdots$$ Each element from the Pascal’s triangle is equal to the addition of two elements on the line above on the right and left side of applicable element, except the constant value of the elements on the edges of the triangle which are always equal to $1.$ This property is valid for any element of Pascal’s triangle. Choose three characteristic elements of Pascal’s triangle: $${{n}\\choose{k-1}} \\qquad {{n}\\choose{k}}$$ $${{n+1}\\choose{k}}$$ The following relation describes the basic principle of Pascal’s triangle: $${{n}\\choose{k-1}}+{{n}\\choose{k}}={{n+1}\\choose{k}}.$$ Proof. $${{n}\\choose{k-1}}+{{n}\\choose{k}}=$$ $$\\frac{n!}{(k-1)![n-(k-1)]!} + \\frac {n!}{k!(n-k)!}=\\frac{k\\cdot n! + (n-k+1)\\cdot n!}{k!(n-k+1)!} =\\frac{n!(k+n-k+1)}{k!(n+1-k)!}$$ $$= \\frac{n!(n+1)}{k!(n+1-k)!}=\\frac {(n+1)!}{k!(n+1-k)!}$$ $$={{n+1}\\choose{k}}.$$ It has been proven that the left side of the equal sign is equal to the right. This property demonstrates that all binomial coefficients are natural numbers because each of", "# How do combinations relate to the pascal's triangle? Apr 6, 2015 In the pascal's triangle the row wise elements are as follows: 1st row has one element 1, that is ${1}_{C}$1 2nd row has 3 elements 1 2 1, that is ${2}_{C}$0, ${2}_{C}$1, ${2}_{C}$2 3rd row has 4 elements 1 3 3 1, that is ${3}_{C}$0, ${3}_{C}$1, ${3}_{C} 2$, ${3}_{C}$3 4th row has 1 4 6 4 1 elements, that is ${4}_{C} 0$, ${4}_{C} 1$, ${4}_{C} 2$, ${4}_{C} 3$, ${4}_{C} 4$ This pattern is then followed through out, showing, how combinations relate to pascal's triangle", "= \\binom{2}{1}$$, $$\\binom{2}{0} + \\binom{2}{1} = \\binom{3}{1}$$ and so forth. This relationship is indicated by the arrows in the array above. With these two facts in hand, we can quickly generate Pascal's Triangle. We start with the first two rows, $$1$$ and $$1 \\quad 1$$. From that point on, each successive row begins and ends with $$1$$ and the middle numbers are generated using Theorem \\ref{addbinomcoeff}. Below we attempt to demonstrate this building process to generate the first five rows of Pascal's Triangle. $\\begin{array}{ccc} \\begin{array}{ccccccccc} & & & & 1 & & & & \\\\ & & & 1 & & 1 & & &\\\\ & & & & \\searrow \\, \\swarrow & & & & \\\\ & & \\fbox{1} & & \\underline{1+1} & & \\fbox{1} & &\\\\ \\end{array} & \\xrightarrow{\\hspace{.5in}} & \\begin{array}{ccccccccc} & & & & 1 & & & & \\\\ & & & 1 & & 1 & & &\\\\ & & 1 & & 2 & &1 & &\\\\ \\end{array} \\\\ && \\\\ \\begin{array}{ccccccccc} & & & & 1 & & & & \\\\ & & & 1 & & 1 & & &\\\\ & & 1 & & 2 & &1 & &\\\\ & & & \\searrow \\, \\swarrow & & \\searrow \\, \\swarrow & & & \\\\ & \\fbox{1} & & \\underline{1+2}", "in the multipliers I am unsure how to go about this. Does anyone have any experience with this type of problem or how to go about formulating a general solution. Thanks Alex 2. Hello, asouthern! I have a set of expressions which are related. The relationship is apparent in the multipliers in each line. . . $\\begin{array}{ccccc}\\text{3rd:} &\\qquad\\qquad\\qquad\\qquad\\qquad t_1+ t_3 \\\\ . . $\\begin{array}{cc} \\text{5th:} & t_1 + 2t_2 + 2t_3 + 2t_4 + t_5 \\\\ \\text{6th:} & t_1 + 3t_2 + 4t_3 + 4t_4 + 3t_5 + t_6 \\\\ \\text{7th:} & t_1 + 4t_2 + 7t_3 + 8t_4 + 7t_5 + 4t_6 + t_7\\\\ \\text{8th:} & t_1 + 5t_2 + 11t_3 + 15t_4 + 15t_5 + 11t_6 + 5t_7 + t_8 \\\\ \\text{9th:} & t_1 + 6t_2 + 16t_3 + 26t_4 + 30t_5 + 26t_6 + 16t_7 + 6t_8 + t_9 \\\\ \\text{10th:} &t_1+7t_2+22t_3+42t_4+56t_5+56t_6+42t_7+22t_8+7t_9 +t_{10} \\end{array}$ I see an even stronger pattern, but . . . You have a variation of Pascal's Triangle . . . . . $\\begin{array}{cccccccccccc} 3 &&1\\;\\;0\\;\\;1\\\\ 4 && 1\\;\\;1\\;\\;1\\;\\;1\\\\ 5 && 1\\;\\;2\\;\\;2\\;\\;2\\;\\;1\\\\ 6 && 1\\;\\;3\\;\\;4\\;\\;4\\;\\;3\\;\\;1\\\\ 7 && 1\\;\\;4\\;\\;7\\;\\;8\\;\\;7\\;\\;4\\;\\;1\\\\ 8 && 1\\;5\\;11\\;15\\;15\\;11\\;5\\;1\\\\ 9 && 1\\;\\;6\\;16\\;26\\;30\\;26\\;16\\;6\\;1 \\\\ 10 && 1\\;7\\;22\\;42\\;56\\;56\\;42\\;22\\;7\\;1\\end{array}$ Each number is the sum of the two numbers directly above it. In row 3, we have: $[1\\:0\\:1]$ which represents the polynomial: $x^2 +", "set of three elements can be arranged in six ways. We denote the number of ways of choosing $$r$$ objects from $$n$$ objects by $$\\dbinom{n}{r}$$, which is read as '$$n$$ choose $$r$$'. In the example above, we found that $\\dbinom{5}{3} = \\dfrac{{^5}P_3}{3!} = \\dfrac{60}{6} = 10.$ In general, we have $\\dbinom{n}{r} = \\dfrac{^nP_r}{r!} = \\dfrac{n!}{(n-r)!r!}.$ Consider the five-element set $$\\{a,b,c,d,e\\}$$ again. There are \\begin{alignat*}{2} \\dbinom{5}{1}&=5 \\text{ ways of choosing 1 letter},&\\qquad \\dbinom{5}{2}&=10 \\text{ ways of choosing 2 letters},\\\\ \\dbinom{5}{3}&=15 \\text{ ways of choosing 3 letters},&\\qquad \\dbinom{5}{4}&=10 \\text{ ways of choosing 4 letters},\\\\ \\dbinom{5}{5}&=1 \\text{ way of choosing 5 letters}. \\end{alignat*} Listing these as a row, including choosing no letters as the first entry: $\\dbinom{5}{0}=1, \\qquad \\dbinom{5}{1}=5, \\qquad \\dbinom{5}{2}=10, \\qquad \\dbinom{5}{3}=10, \\qquad \\dbinom{5}{4}=5, \\qquad \\dbinom{5}{5}=1.$ This is the fifth row of Pascal's triangle. #### Example Evaluate 1. $$\\dbinom{100}{2}$$ 2. $$\\dbinom{1000}{998}$$. #### Solution 1. $$\\dbinom{100}{2} = \\dfrac{100\\times99}{2} = 4950$$ 2. $$\\dbinom{1000}{998} = \\dfrac{1000\\times999}{2} = 499\\,500$$. #### Example In how many ways can you choose two people from a group of seven people? #### Solution There are $$\\dbinom{7}{2} = \\dfrac{7\\times6}{2} = 21$$ ways of choosing two people from seven. #### Example There are ten people in a basketball squad. Find how many ways: 1. the starting five can be chosen from the squad 2. the squad can be split into two", "& 43 & \\textbf{54} & & & \\dots \\\\ 1 & \\textbf{4} & & & & 15 & \\textbf{32} & & & & 65 & \\textbf{108} & & & & \\dots \\\\ 1 & & & & & 16 & & & & & 81 & & & & & \\dots \\end{array}$$$ are constructed in a similar way to Pascal’s triangle, $$$% Pascal's triangle \\tag{3}\\label{eq:pascal-triangle-rank-4} \\begin{array}{*{9}{c}} & & & & 1 & & & & \\\\ & & & 1 & & 1 & & & \\\\ & & 1 & & 2 & & 1 & & \\\\ & 1 & & 3 & & 3 & & 1 & \\\\ 1 & & 4 & & 6 & & 4 & & 1 \\\\ \\end{array}$$$ The similarity lies in the observation that each entry in Pascal’s triangle is the sum of the two values immediately above it,5 $\\begin{equation*} \\begin{array}{*{5}{c}} 1 & & & & 2 \\\\ & \\searrow & & \\swarrow & \\\\ & & 3 & & \\end{array} \\end{equation*}$ while each entry in a Moessner triangle is the sum of the value immediately above it (northern neighbor) and left of it (western neighbor), $\\begin{equation*} \\begin{array}{ccc} & & 2 \\\\ & & \\downarrow \\\\ 1 & \\to & 3 \\end{array} \\end{equation*}$ suggesting an equivalence relation between", "two entries just above. So under the $$1 \\qquad 3 \\qquad 3 \\qquad 1$$ row, start with a $$1$$, then add $$1+3=4$$ to get the next entry, then $$3+3=6$$, etc. Notice the symmetry; this relates to Theorem 7.4.2. $\\begin{array}{*{13}{c}} & & & & & & 1 \\\\ & & & & & 1 & & 1 \\\\ & & & & 1 & & 2 & & 1 \\\\ & & & 1 & & 3 & & 3 & & 1 \\\\ & & 1 & & 4 & & 6 & & 4 & & 1 \\end{array}$ Here's more of the triangle: $\\begin{array}{*{13}{c}} & & & & & & 1 \\\\ & & & & & 1 & & 1 \\\\ & & & & 1 & & 2 & & 1 \\\\ & & & 1 & & 3 & & 3 & & 1 \\\\ & & 1 & & 4 & & 6 & & 4 & & 1 \\\\ & 1 & & 5 & &10 & &10 & & 5 & & 1 \\\\ 1 & & 6 & &15 & &20 & &15 & & 6 & & 1 \\end{array}$ From the top, these rows are the 0-row, the 1-row, the 2-row, etc. The 5-row consists of: $1 \\qquad 5", "$e$, and $f$. Then here’s how the triangle unfolds (turning the triangle upside down): $a \\qquad \\qquad \\qquad \\quad b \\qquad \\qquad \\qquad \\quad c \\qquad \\qquad \\qquad \\quad d \\qquad \\qquad \\qquad \\quad e \\qquad \\qquad \\qquad \\quad f$ $a+b \\qquad \\qquad \\qquad b+c \\qquad \\qquad \\qquad c+d \\qquad \\qquad \\qquad d+e \\qquad \\qquad \\qquad e+f$ $a+2b+c \\qquad \\qquad b+2c+d \\qquad \\qquad c+2d+e \\qquad \\qquad d+2e+f$ $a+3b+3c+d \\qquad \\quad b+3c+3d+e \\qquad \\quad c+3d+3e+f$ $a+4b+6c+4d+e \\qquad b+4c+6d+5e+f$ $a+5b+10c+10d+5e+f$ In other words, the top number can be obtained by using the numbers on the fifth row of Pascal’s triangle (recall that the fifth row of Pascal’s triangle has six numbers on it). Specifically, if I multiply the bottom numbers by the corresponding number in a row of Pascal’s triangle and add them up, I’ll get the number on top without having to compute all of the intermediate steps. For the problem my students gave me, the bottom row has 12 numbers, which means I’ll need to use the 11th row of Pascal’s triangle. Also, as we’ll see, I was fortunate that my students gave me a simple pattern of consecutive squares for the numbers on the bottom row. Since the numbering in Pascal’s triangle starts on zero, the numbers in the bottom row are $(k+1)^2$ as $k$ varies from", "writing down Pascal's triangle to the fifth row, and putting a fifth row, as a column, on the side. Then multiply across: \\begin{array}{lcccccccccc} 1&&&&&&1&&&&&\\\\ 5&&&&&1&&1&&&&\\\\ 10\\quad\\times&&&&1&&2&&1&&&\\\\ 10&&&1&&3&&3&&1&&\\\\ 5&&1&&4&&6&&4&&1&\\\\ 1&1&&5&&10&&10&&5&&1 \\end{array} to produce the final array of coefficients (with index numbers at the left): \\begin{array}{l*{10}{c}} 0\\qquad{}&&&&&&1&&&&&\\\\ 1&&&&&5&&5&&&&\\\\ 2&&&&10&&20&&10&&&\\\\ 3&&&10&&30&&30&&10&&\\\\ 4&&5&&20&&30&&20&&5&\\\\ 5&1&&5&&10&&10&&5&&1 \\end{array} Row $$i$$ of this array corresponds to $$x^{5-i}$$ and all combinations of powers $$y^bz^c$$ for $$0\\le b,c\\le i$$. Thus for example the fourth row down, corresponding to $$i=3$$, may be considered as the coefficients of the terms $x^2y^3,\\quad x^2y^2z,\\quad x^2yz^2,\\quad xz^3.$ Note that the triangle of coefficients is symmetrical along all three centre lines, as well as rotationally symmetric by 120°.", "We shall explore the similarities between the harmonic triangle, which you see below, and Pascal's triangle. You may like to start by trying this Harmonic Triangle problem. The rule for generating the harmonic triangle is that you add two consecutive entries to give the entry between them in the row above. Equivalently, from any term, to get the next term, subtract that term from the corresponding term on the row above. It is still possible to work downwards row by row because the entry at the left hand end of the $n$th row is ${1\\over n}$. For example the third row of the harmonic triangle is: $${1\\over 3},\\ {1\\over 6},\\ {1\\over 3}$$ and the fourth row is: $${1\\over 4},\\ {1\\over 12},\\ {1\\over 12}, \\ {1\\over 4}$$ which is given by $${1\\over 4},\\ \\ \\left[{1\\over 3}-{1\\over 4}\\right] ,\\ \\ \\left[{1\\over 6}-{1\\over 12}\\right],\\ \\ \\left[{1\\over 3}-{1\\over 12}\\right]$$ \\begin{array}{ccccccccccc} & & & & & \\frac{1}{1} & & & & & \\\\ & & & & \\frac{1}{2} & & \\frac{1}{2} & & & & \\\\ & & & \\frac{1}{3} & & \\frac{1}{6} & & \\frac{1}{3} & & & \\\\ & & \\frac{1}{4} & & \\frac{1}{12} & & \\frac{1}{12} & & \\frac{1}{4} & & \\\\ & \\frac{1}{5} & & \\frac{1}{20} & & \\frac{1}{30} & & \\frac{1}{20} & & \\frac{1}{5} & \\\\ \\frac{1}{6} &", "## Stony Brook University MAT 118 Spring 2013 ### Notes from class - this material is not in the text Pascal's triangle $$\\begin{array}{ccccccccccc} & & & & &1 & & & & & \\\\ & & & &1 & &1 & & & & \\\\ & & &1 & &2 & &1 & & & \\\\ & &1 & &3 & &3 & &1 & & \\\\ &1 & &4 & &6 & &4 & &1 & \\\\ 1 & &5 & &10 & &10 & &5 & & 1 \\end{array}$$ This display shows rows 0, 1, 2, 3, 4, 5 of Pascal's triangle. The rows are constructed following two rules: 1. Row $i$ has length $i+1$ and starts and ends with a $1$. 2. Each interior element is the sum of the two above it. We number the columns diagonally: column $0$ has all $1$s. Column $1$ reads $1, 2, 3, 4, 5, \\dots$ . Column $2$ reads $1, 3, 6, 10, \\dots$, etc. The standard notation for the number in row $n$ and column $k$ is $\\left ( \\begin{array}{c} n\\\\k \\end{array} \\right )$. So $\\left ( \\begin{array}{c} n\\\\0 \\end{array} \\right )$ is always $1$, and $\\left ( \\begin{array}{c} n\\\\n \\end{array} \\right )$ is always $1$, whereas $\\left ( \\begin{array}{c} 5\\\\3 \\end{array} \\right ) = 10$. The", "# Pascal triangle just got better What will be the value of the $$10^\\text{th}$$ expression (applying operation from left to right). $\\begin{array} {lc} \\text{First expression} & 1 \\\\ \\text{Second expression} &1\\div1 \\\\ \\text{Third expression} & 1\\div2\\div1 \\\\ \\text{Fourth expression} & 1\\div3\\div3\\div1\\\\ \\text{Fifth expression} & 1\\div4\\div6\\div4\\div1\\\\ \\text{Sixth expression} & 1\\div5\\div10\\div10\\div5\\div1\\\\ \\text{Seventh expression} & 1\\div6\\div15\\div20\\div15\\div6\\div1 \\end{array}$ If you are getting answer like $$w.xyz\\times10^{a}$$ after correction to three decimal places, then enter it as $$w.xyzE+a$$ Clarification: $$E$$ represents E notation ×", "\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\left( {\\begin{array}{*{20}{c}} 3 \\\\ 3 \\end{array}} \\right)\\\\……\\text{and so on}…..\\end{array}$$ < As an example of how to use the Pascal Triangle, start with the second row for $${{\\left( {x+y} \\right)}^{1}}=1x+1y$$, so the coefficients are both 1. When using the Pascal Triangle, the exponent of the binomial is off by 1; for example, we used the 2nd row to get the coefficients for $${{\\left( {x+y} \\right)}^{1}}$$. < Here’s another illustration of just how Pascal’s Triangle is used for expanding binomials: $$\\displaystyle \\begin{array}{l}{{\\left( {x+y} \\right)}^{0}}=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1\\\\{{\\left( {x+y} \\right)}^{1}}=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x+y\\\\{{\\left( {x+y} \\right)}^{2}}=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{{x}^{2}}+2xy+{{y}^{2}}\\\\{{\\left( {x+y} \\right)}^{3}}=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{{x}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}+{{y}^{3}}\\,\\,\\,\\,\\,\\\\{{\\left( {x+y} \\right)}^{4}}=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{{x}^{4}}+4{{x}^{3}}y+6{{x}^{2}}{{y}^{2}}+4x{{y}^{3}}+{{y}^{4}}\\,\\,\\,\\,\\\\{{\\left( {x+y} \\right)}^{5}}=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{{x}^{5}}+5{{x}^{4}}y+10{{x}^{3}}{{y}^{2}}+10{{x}^{2}}{{y}^{3}}+5x{{y}^{4}}+{{y}^{5}}\\,\\\\{{\\left( {x+y} \\right)}^{6}}={{x}^{6}}+6{{x}^{5}}y+15{{x}^{4}}{{y}^{2}}+20{{x}^{3}}{{y}^{3}}+15{{x}^{2}}{{y}^{4}}+6x{{y}^{5}}+{{y}^{6}}\\,\\\\\\,\\,…..\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}$$ Here’s a hint: when finding the coefficients of a binomial expansion using Pascal’s triangle, find the line with the second term the same as the power you want. For example, for a binomial with power 5, use the line 1 5 10 10 5 1 for coefficients. # Expanding a Binomial The best way to show how Binomial Expansion works is to use an example. Let’s expand $${{\\left( {x+3} \\right)}^{6}}$$ using the formula above. Here, the “x” in the generic binomial expansion equation is “x” and the “y” is “3”: \\begin{align}{{\\left( {x+y} \\right)}^{n}}\\,\\,&=\\,\\,\\left( {\\begin{array}{*{20}{c}} n \\\\ 0 \\end{array}} \\right){{x}^{n}}{{y}^{0}}\\,+\\,\\left( {\\begin{array}{*{20}{c}} n \\\\ 1 \\end{array}} \\right){{x}^{{n-1}}}{{y}^{1}}\\,+\\left( {\\begin{array}{*{20}{c}} n \\\\ 2 \\end{array}} \\right){{x}^{{n-2}}}{{y}^{2}}\\,+……+\\left( {\\begin{array}{*{20}{c}} n \\\\ {n-1} \\end{array}} \\right){{x}^{1}}{{y}^{{n-1}}}\\,+\\left( {\\begin{array}{*{20}{c}} n \\\\ n \\end{array}} \\right){{x}^{0}}{{y}^{n}}\\\\\\color{#804040}{{{{{\\left( {x+3}", "+ 1 corresponds to row s+1 of Pascal's triangle. 4) Then $$T(n,m) = \\left(\\begin{array}{cc}n+m-2\\\\n-1\\end{array}\\right) \\forall n,m \\in N, n + m \\geq 2$$ Last edited: Oct 5, 2005", "diamond of order $n$, $2^{n+1\\choose 2}$ is the determinant of $[F(i,j)]_{0\\le i,j\\le n}$. An LDU decomposition of this matrix into a lower-triangular Pascal's triangle, a diagonal matrix with powers of $2$, and an upper triangular Pascal's triange, is suggested by the formula $$F(i,j) = \\sum_{d=0} 2^d {i \\choose d} {j \\choose d},$$ equation 3 on the MathWorld page linked above. Here is the decomposition for $n=4$: $$\\left[ \\begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\\\\\ 1 & 3 & 5 & 7 & 9 \\\\\\ 1 & 5 & 13 & 25 & 41 \\\\\\ 1 & 7 & 25 & 63 & 129 \\\\\\ 1 & 9 & 41 & 129 & 321 \\end{array} \\right] = \\left[\\begin{array}{ccccc} 1&0&0&0&0 \\\\\\ 1&1&0&0&0 \\\\\\ 1&2&1&0&0 \\\\\\ 1&3&3&1&0 \\\\\\ 1&4&6&4&1\\end{array}\\right] \\left[\\begin{array}{ccccc} 1 & 0 & 0 &0&0 \\\\\\ 0&2&0&0&0 \\\\\\ 0&0&4&0&0 \\\\\\ 0&0&0&8&0 \\\\\\ 0&0&0&0&16\\end{array}\\right] \\left[\\begin{array}{ccccc} 1&1&1&1&1 \\\\\\ 0&1&2&3&4 \\\\\\ 0&0&1&3&6 \\\\\\ 0&0&0&1&4 \\\\\\ 0&0&0&0&1\\end{array}\\right]$$ The number of domino tilings of the Aztec diamond of order $4$ is $1\\times2\\times4\\times8\\times16$. I think I wrote up a related proof for the enumeration of domino tilings of an Aztec diamond in the domino tiling mailing list in 1997 or 1998. -", "it explicitly. # Mathematical Aside: Golden Mean Shift and Pascal’s Triangle (Part 2) As we discussed in Part 1, we noticed a common pattern in the the $m^{th}$ element in the $n^{th}$ diagonal, $d_{m,n}$, of Pascal’s Triangle and the function $c_{m,n}$, denoting the set of binary strings of length $n$ with exactly $m$ non-adjacent ones. Today we derive an explicit formula for $c_{m,n}$ in two ways. Firstly, some more notation. The element in $r^{th}$ row and $k^{th}$ column in Pascal’s Triangle is denoted by ${r}\\choose{k}$. Then we can see $d_{m,n} = {n+m-1,m}$. Unfortunately, $c_{m,n}$ is at a slight shift. To see this, we think about how long the string needs to be just so can fit in $m$ ones. The answer is $2(m-1)$. While the $d_{m,n}$ is non-zero straight away, it takes $c_{m,n}$ $2(m-1)$ increments to the length before it is non-zero. Therefore, $|c_{m,n}|=d_{m,n-2(m-1)}=$ ${n-(m-1)}\\choose{m}$. Another way to see this requires knowledge of combinations. If we consider the unconstrained system, the clearly we have $u_{m,n} {n}\\choose{m}$ options. We can construct a bijection from $u_{m,n-(m-1)}$ to $c_{m,n}$ as follows: • For every element in $u_{m,n-(m-1)}$, we can add append a 0 to all but the rightmost 1 and construct a unique element in $c_{m,n}$ • For every element in $c_{m,n}$, we can add delete a 0 from the right", "although there is a pattern clear in the multipliers I am unsure how to go about this. Does anyone have any experience with this type of problem or how to go about formulating a general solution. Thanks Alex 2. Hello, asouthern! I have a set of expressions which are related. The relationship is apparent in the multipliers in each line. . . $\\displaystyle \\begin{array}{ccccc}\\text{3rd:} &\\qquad\\qquad\\qquad\\qquad\\qquad t_1+ t_3 \\\\ \\text{4th:}& \\qquad\\qquad\\qquad\\qquad\\qquad t_1 + t_2 + t_3 + t_4 \\\\ \\end{array}$ . . $\\displaystyle \\begin{array}{cc} \\text{5th:} & t_1 + 2t_2 + 2t_3 + 2t_4 + t_5 \\\\ \\text{6th:} & t_1 + 3t_2 + 4t_3 + 4t_4 + 3t_5 + t_6 \\\\ \\text{7th:} & t_1 + 4t_2 + 7t_3 + 8t_4 + 7t_5 + 4t_6 + t_7\\\\ \\text{8th:} & t_1 + 5t_2 + 11t_3 + 15t_4 + 15t_5 + 11t_6 + 5t_7 + t_8 \\\\ \\text{9th:} & t_1 + 6t_2 + 16t_3 + 26t_4 + 30t_5 + 26t_6 + 16t_7 + 6t_8 + t_9 \\\\ \\text{10th:} &t_1+7t_2+22t_3+42t_4+56t_5+56t_6+42t_7+22t_8+7t_9 +t_{10} \\end{array}$ I see an even stronger pattern, but . . . You have a variation of Pascal's Triangle . . . . . $\\displaystyle \\begin{array}{cccccccccccc} 3 &&1\\;\\;0\\;\\;1\\\\ 4 && 1\\;\\;1\\;\\;1\\;\\;1\\\\ 5 && 1\\;\\;2\\;\\;2\\;\\;2\\;\\;1\\\\ 6 && 1\\;\\;3\\;\\;4\\;\\;4\\;\\;3\\;\\;1\\\\ 7 && 1\\;\\;4\\;\\;7\\;\\;8\\;\\;7\\;\\;4\\;\\;1\\\\ 8 && 1\\;5\\;11\\;15\\;15\\;11\\;5\\;1\\\\ 9 && 1\\;\\;6\\;16\\;26\\;30\\;26\\;16\\;6\\;1 \\\\ 10 && 1\\;7\\;22\\;42\\;56\\;56\\;42\\;22\\;7\\;1\\end{array}$ Each number is the", "the sum of the 3 white ones around it? We can make lots of new patterns using this... 8. Hello, janvdl! That's fascinating . . . Thank you! What if every red triangle is the sum of the 3 white ones around it? Then we have: . . $\\begin{array}{c}3\\\\ 4\\;\\;4\\\\ 5\\;\\;8\\;\\;5 \\\\ 6\\;\\;13\\;\\;13\\;\\;6 \\\\ 7\\;\\;19\\;\\;26\\;\\;19\\;\\;7 \\end{array}$ I finally saw where it came from . . . Consider a \"Pascal's Triangle\" whose third line is: .1 - 3 - 1. . . $\\begin{array}{c}1 \\\\ 1\\;\\;1 \\\\ 1\\;\\;3\\;\\;1 \\\\ 1\\;\\;4\\;\\;4\\;\\;1 \\\\ 1\\;\\;5\\;\\;8\\;\\;5\\;\\;1 \\\\ 1\\;\\;6\\;\\;13\\;\\;13\\;\\;6\\;\\;1 \\\\ 1\\;\\;7\\;\\;19\\;\\;26\\;\\;19\\;\\;7\\;\\;1 \\end{array}$ See it? This opens up a world of possibilities, doesn't it? 9. Originally Posted by Soroban Hello, janvdl! That's fascinating . . . Thank you! Then we have: . . $\\begin{array}{c}3\\\\ 4\\;\\;4\\\\ 5\\;\\;8\\;\\;5 \\\\ 6\\;\\;13\\;\\;13\\;\\;6 \\\\ 7\\;\\;19\\;\\;26\\;\\;19\\;\\;7 \\end{array}$ I finally saw where it came from . . . Consider a \"Pascal's Triangle\" whose third line is: .1 - 3 - 1. . . $\\begin{array}{c}1 \\\\ 1\\;\\;1 \\\\ 1\\;\\;3\\;\\;1 \\\\ 1\\;\\;4\\;\\;4\\;\\;1 \\\\ 1\\;\\;5\\;\\;8\\;\\;5\\;\\;1 \\\\ 1\\;\\;6\\;\\;13\\;\\;13\\;\\;6\\;\\;1 \\\\ 1\\;\\;7\\;\\;19\\;\\;26\\;\\;19\\;\\;7\\;\\;1 \\end{array}$ See it? This opens up a world of possibilities, doesn't it? So have i actually kind of discovered something here? And what can we actually do with my \"discovery\"? A world of possibilities has been opened? You're clearly seeing things that i cant... 10. Here's another", "# General Expression for an Element of a Triangular Array The Pascal's Triangle is formed by the following rules: $i,j \\in N^*$ $T(1,1) = 1$ $\\forall j>i\\;\\;\\; ; \\;\\;\\;T(i,j) = 0$ $\\forall i>1\\;\\;\\; ; \\;\\;\\;T(i,j) = T(i-1,j-1) + T(i-1,j)$ where $i$ is the row index and $j$ is the column index There is a general formula for its non-null terms: $T(i,j) = \\left(\\begin{array}{c}i\\\\j\\end{array}\\right) = \\frac{i!}{j!(i-j)!}$ I would like to know if there is a general expression for the terms of the following triangular array: $i,j \\in N^*$ $T(1,1) = 1$ $\\forall j>i\\;\\;\\; ; \\;\\;\\;T(i,j) = 0$ $\\forall i>1\\;\\;\\; ; \\;\\;\\;T(i,j) = T(i-1,j-1) + j\\;T(i-1,j)$ Here the five first rows: $\\begin{bmatrix}1&0&0&0&0\\\\ 1&1&0&0&0\\\\ 1&3&1&0&0\\\\ 1&7&6&1&0\\\\ 1&15&25&10&1 \\end{bmatrix}$ These number are ${i \\brace j}$, the Stirling numbers of the Second Kind. They are archived as A008277 in OEIS.org." ]

[ "Pascal's Triangle # Pascal's Triangle One way to relate $k$-combinations of a finite $n$-element set is with a special graph known as Pascal's Triangle. A portion of Pascal's triangle is shown below: The top of Pascal's Triangle is a $1$, and entries for Pascal's triangle are generated to be the sum of the two numbers directly above that entry (or just the number directly above the entry if there's only one). If we label the rows of Pascal's triangle starting at row $0$, row $1$, etc…, then you should notice that the finite sequence of numbers in row $n$ for each $n = 0, 1, 2, ...$ and for $k = 0, 1, 2, ..., n$ is the same sequence of numbers as: (1) \\begin{align} \\quad \\left \\{ \\binom{n}{0}, \\binom{n}{1}, \\binom{n}{2}, ..., \\binom{n}{k-1}, \\binom{n}{k}, \\binom{n}{k+1}, ..., \\binom{n}{k} \\right \\} \\end{align} For example, row $5$ in Pascal's triangle forms the finite sequence $\\{ 1, 5, 10, 10, 5, 1 \\}$, and using the formula $\\binom{n}{k} = \\frac{n!}{k!(n-k)!}$ for counting the number of $k$-combinations for a $5$-element set for $k = 0, 1, 2, ..., 5$, we have that: (2) \\begin{align} \\quad \\quad \\left \\{ \\binom{5}{0}, \\binom{5}{1}, \\binom{5}{2}, \\binom{5}{3}, \\binom{5}{4}, \\binom{5}{5} \\right \\} = \\left \\{ \\frac{5!}{0! \\cdot 5!} , \\frac{5!}{1! \\cdot 4!}, \\frac{5!}{2! \\cdot 3!}, \\frac{5!}{3! \\cdot 2!}, \\frac{5!}{4!", "+0 # pascal's triangle 0 48 1 Let f(n) be the base-10 logarithm of the product of the elements of the nth row in Pascal's triangle. Find f(10), May 30, 2021 ### 1+0 Answers #1 +594 0 We want: $$\\log \\binom{10}{0}\\binom{10}{1}\\cdots \\binom{10}{10}=\\log \\binom{10}{0}^2\\binom{10}{1}^2\\binom{10}{2}^2\\binom{10}{3}^2\\binom{10}{4}^2\\binom{10}{5}=\\log (1\\cdot 10^2\\cdot 45^2\\cdot 120^2\\cdot 210^2\\cdot 252)=\\log(2^12\\cdot 3^10\\cdot 5^8\\cdot 7^3)\\approx 16.51$$ May 30, 2021", "of the ways shown below. Basically, what I did first was I chose arbitrary values of n and k to start with, n being the row number and k being the kth number in that row (confusing, I know). 1st element of the nth row of Pascal’s triangle) + (2nd element of the nᵗʰ row)().y +(3rd element of the nᵗʰ row). Each entry in the nth row gets added twice. this article for a general example. Each element in the triangle has a coordinate, given by the row it is on and its position in the row (which you could call its column). Should the stipend be paid if working remotely? Suppose we have a number n, we have to find the nth (0-indexed) row of Pascal's triangle. Find this formula\". $$1,n,\\frac{n(n-1)}2,\\frac{n(n-1)(n-2)}{2\\cdot3},\\frac{n(n-1)(n-2)(n-3)}{2\\cdot3\\cdot4}\\cdots$$, This is computed by recurrence very efficiently, like, $$1,54,\\frac{54\\cdot53}2=1431,\\frac{1431\\cdot52}3=24804,\\frac{24804\\cdot51}4=316251\\cdots$$. Consider again Pascal's Triangle in which each number is obtained as the sum of the two neighboring numbers in the preceding row. Copyright © 2021 Multiply Media, LLC. For example, the \"third\" row, or row 2 where n=2 is comprised of Using Pascal's Triangle for Binomial Expansion. Replacing the core of a planet with a sun, could that be theoretically possible? So elements in 4th row will look like: 4C0, 4C1, 4C2, 4C3, 4C4. This basically means", "+0 # pascals triangle 0 1161 2 There is a row of Pascal's triangle that has three successive positive entries,\"a\" \"b\" and \"c\" such that \"b\" is double \"c\" and \"a\" is triple \"c\" If this row begins \"1,n,\" then find n. Guest May 21, 2017 #1 +20025 +4 There is a row of Pascal's triangle that has three successive positive entries,\"a\" \"b\" and \"c\" such that \"b\" is double \"c\" and \"a\" is triple \"c\" If this row begins \"1,n,\" then find n. Three successive positive entries: $$\\begin{array}{rcll} a&=&\\binom{n}{k-1} \\\\ b&=&\\binom{n}{k} \\\\ c&=&\\binom{n}{k+1} \\\\ \\end{array}$$ \"b\" is double \"c\" and \"a\" is triple \"c\" $$\\begin{array}{|rcll|} \\hline a &= 3c &=& \\binom{n}{k-1} \\\\ b &= 2c &=& \\binom{n}{k} \\\\ c & &=& \\binom{n}{k+1} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline (1) & 2c &=& \\binom{n}{k} \\quad & | \\quad c = \\binom{n}{k+1} \\\\ & 2\\cdot \\binom{n}{k+1} &=& \\binom{n}{k} \\quad & | \\quad \\binom{n}{k+1}= ( \\frac{n-k}{k+1} ) \\binom{n}{k} \\\\ & 2\\cdot ( \\frac{n-k}{k+1} ) \\binom{n}{k} &=& \\binom{n}{k} \\\\ & 2\\cdot ( \\frac{n-k}{k+1} ) &=& 1 \\\\ & \\mathbf{ n-k } & \\mathbf{=} & \\mathbf{ \\frac{k+1}{2} } \\\\\\\\ (2) & 3c &=& \\binom{n}{k-1} \\quad & | \\quad c = \\binom{n}{k+1} \\\\ & 3\\cdot \\binom{n}{k+1} &=& \\binom{n}{k-1} \\quad & | \\quad \\binom{n}{k+1}= ( \\frac{n-k}{k+1} ) \\binom{n}{k} \\\\ & 3\\cdot ( \\frac{n-k}{k+1} ) \\binom{n}{k}", "to pick r thing from n things where order does not matter. $n!=1 \\cdot 2 \\cdot 3 \\cdot...\\cdot n$. ### Proving it If $k > n$ then $\\binom{n}{k} = 0 = \\binom{n - 1}{k - 1} + \\binom{n - 1}{k}$ and so the result is pretty clear. So assume $k \\leq n$. Then $\\begin{eqnarray*}\\binom{n-1}{k-1}+\\binom{n-1}{k}&=&\\frac{(n-1)!}{(k-1)!(n-k)!}+\\frac{(n-1)!}{k!(n-k-1)!}\\\\ &=&(n-1)!\\left(\\frac{k}{k!(n-k)!}+\\frac{n-k}{k!(n-k)!}\\right)\\\\ &=&(n-1)!\\cdot \\frac{n}{k!(n-k)!}\\\\ &=&\\frac{n!}{k!(n-k)!}\\\\ &=&\\binom{n}{k}. \\qquad\\qquad\\square\\end{eqnarray*}$ There we go. We proved it! ### Why is it needed? It's mostly just a cool thing to know. However, if you want to know how to use it in real life go to https://artofproblemsolving.com/videos/counting/chapter12/141. Or really any of the counting and probability videos. ## Introduction to Pascal's Triangle ### How to build it Pascal's Triangle is a triangular array of numbers in which you start with two infinite diagonals of ones and each of the rest of the numbers is the sum of the two numbers above it. It looks something like this: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 And on and on... ### Combinations #### Combinations Pascal's Triangle is really combinations. It looks something like this if it is depicted as combinations: $\\binom{0}{0}$ $\\binom{1}{0}$ $\\binom{1}{1}$ $\\binom{2}{0}$ $\\binom{2}{1}$ $\\binom{2}{2}$ And on and on... #### Proof If you look at the way we build the triangle, each number is", "+0 # pascals triangle 0 357 1 There is a row of Pascal's triangle that has three successive positive entries,\"a\" \"b\" and \"c\" such that \"b\" is double \"c\" and \"a\" is triple \"c\" If this row begins \"1,n,\" then find n. Guest May 21, 2017 Sort: #1 +18612 +1 There is a row of Pascal's triangle that has three successive positive entries,\"a\" \"b\" and \"c\" such that \"b\" is double \"c\" and \"a\" is triple \"c\" If this row begins \"1,n,\" then find n. Three successive positive entries: $$\\begin{array}{rcll} a&=&\\binom{n}{k-1} \\\\ b&=&\\binom{n}{k} \\\\ c&=&\\binom{n}{k+1} \\\\ \\end{array}$$ \"b\" is double \"c\" and \"a\" is triple \"c\" $$\\begin{array}{|rcll|} \\hline a &= 3c &=& \\binom{n}{k-1} \\\\ b &= 2c &=& \\binom{n}{k} \\\\ c & &=& \\binom{n}{k+1} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline (1) & 2c &=& \\binom{n}{k} \\quad & | \\quad c = \\binom{n}{k+1} \\\\ & 2\\cdot \\binom{n}{k+1} &=& \\binom{n}{k} \\quad & | \\quad \\binom{n}{k+1}= ( \\frac{n-k}{k+1} ) \\binom{n}{k} \\\\ & 2\\cdot ( \\frac{n-k}{k+1} ) \\binom{n}{k} &=& \\binom{n}{k} \\\\ & 2\\cdot ( \\frac{n-k}{k+1} ) &=& 1 \\\\ & \\mathbf{ n-k } & \\mathbf{=} & \\mathbf{ \\frac{k+1}{2} } \\\\\\\\ (2) & 3c &=& \\binom{n}{k-1} \\quad & | \\quad c = \\binom{n}{k+1} \\\\ & 3\\cdot \\binom{n}{k+1} &=& \\binom{n}{k-1} \\quad & | \\quad \\binom{n}{k+1}= ( \\frac{n-k}{k+1} ) \\binom{n}{k} \\\\ & 3\\cdot ( \\frac{n-k}{k+1} )", "side-by-side is $$\\sum_{j=k}^n(-1)^{j+k}\\binom{j-1}{k-1}\\binom{n}{j}(2n-1-j)!\\,2^j.$$ Applying this to the case $n=4,$ we get that the number of arrangements for $k=1$ is $$4\\cdot6!\\cdot2-6\\cdot5!\\cdot2^2+4\\cdot4!\\cdot2^3-1\\cdot3!\\cdot2^4=3552.$$ For $k=2,$ it's $$1\\cdot6\\cdot5!\\cdot2^2-2\\cdot4\\cdot4!\\cdot2^3+3\\cdot1\\cdot3!\\cdot2^4=1632.$$ For $k=3,$ it's $$1\\cdot4\\cdot4!\\cdot2^3-3\\cdot1\\cdot3!\\cdot2^4=480.$$ For $k=4,$ it's $$1\\cdot1\\cdot3!\\cdot2^4=96.$$ Proof of binomial coefficient identity (added): Let $$S(r,k)=\\sum_{j=k}^r\\binom{r}{j}\\binom{j-1}{k-1}(-1)^{j+k}$$ with $r\\ge0,$ $k\\ge1.$ We immediately see that $S(r,k)=0$ for $k>r.$ Now, using the Pascal's triangle recurrence, $$S(r,k)-S(r,k+1)=\\sum_{j=k}^r\\binom{r}{j}\\left[\\binom{j-1}{k-1}+\\binom{j-1}{k}\\right](-1)^{j+k}=\\sum_{j=k}^r\\binom{r}{j}\\binom{j}{k}(-1)^{j+k}.$$ Expanding $(x-1+1)^r$ using the binomial theorem, we get \\begin{aligned} x^r=(x-1+1)^r=\\sum_{j=0}^r\\binom{r}{j}(x-1)^j&=\\sum_{j=0}^r\\sum_{k=0}^j\\binom{r}{j}\\binom{j}{k}x^k(-1)^{k-j}\\\\ &=\\sum_{k=0}^rx^k\\sum_{j=k}^r\\binom{r}{j}\\binom{j}{k}(-1)^{j+k}. \\end{aligned} Comparing the left and right sides of this equation, we see that $$\\sum_{j=k}^r\\binom{r}{j}\\binom{j}{k}(-1)^{k-j}=\\begin{cases}0 & k<r\\\\1 & k=r.\\end{cases}$$ We conclude that $S(r,r)-S(r,r+1)=1$ and that $S(r,k)=S(r,k+1)$ for $k<r.$ Since $S(r,r+1)=0,$ we have $S(r,r)=1,$ and therefore $S(r,k)=1$ for $1\\le k\\le r.$ • Hi, I am the original poster. please see my edit on the original problem statement. Dec 2 '14 at 20:52 • @MathyPerson: my post includes answers to both your original question (this is the case $k=1$) and the revised question (this is the case $k=2$). Alistair's answer does the same. (In fact he believed from the start that the revised question was the one you meant to ask!) Dec 3 '14 at 2:39 Here is the number of all orderings such that exactly specific number of countries have adjacent sitting arrangements: • Case 1: Number of orderings such that exactly two countries have adjacent sitting arrangements (Ignore the unusual case", "# Pascal's coefficient identity Probability Level 1 $\\begin{array}{rc} 0^\\text{th} \\text{ row:} & 1 \\\\ 1^\\text{st} \\text{ row:} & 1 \\quad 1 \\\\ 2^\\text{nd} \\text{ row:} & 1 \\quad 2 \\quad 1 \\\\ 3^\\text{rd} \\text{ row:} & 1 \\quad 3 \\quad 3 \\quad 1 \\\\ 4^\\text{th} \\text{ row:} & 1 \\quad 4 \\quad 6 \\quad 4 \\quad 1 \\\\ \\vdots \\ \\ \\ & \\cdot \\quad \\cdot \\quad \\cdot \\quad \\cdot \\quad \\cdot \\quad \\cdot \\end{array}$ Pascal's triangle is shown above for the $0^\\text{th}$ row through the $4^\\text{th}$ row. What is the $4^\\text{th}$ element in the $10^\\text{th}$ row? Note: Each row starts with the $0^\\text{th}$ element. For example, the $0^\\text{th}$, $1^\\text{st}$, $2^\\text{nd}$, and $3^\\text{rd}$ elements of the $3^\\text{rd}$ row are 1, 3, 3, and 1, respectively. ×", "the 3th diagonal row were the triangular numbers. Numbers written in any of the ways shown below. Basically, what I did first was I chose arbitrary values of n and k to start with, n being the row number and k being the kth number in that row (confusing, I know). 1st element of the nth row of Pascal’s triangle) + (2nd element of the nᵗʰ row)().y +(3rd element of the nᵗʰ row). Each entry in the nth row gets added twice. this article for a general example. Each element in the triangle has a coordinate, given by the row it is on and its position in the row (which you could call its column). Should the stipend be paid if working remotely? Suppose we have a number n, we have to find the nth (0-indexed) row of Pascal's triangle. Find this formula\". $$1,n,\\frac{n(n-1)}2,\\frac{n(n-1)(n-2)}{2\\cdot3},\\frac{n(n-1)(n-2)(n-3)}{2\\cdot3\\cdot4}\\cdots$$, This is computed by recurrence very efficiently, like, $$1,54,\\frac{54\\cdot53}2=1431,\\frac{1431\\cdot52}3=24804,\\frac{24804\\cdot51}4=316251\\cdots$$. Consider again Pascal's Triangle in which each number is obtained as the sum of the two neighboring numbers in the preceding row. Copyright © 2021 Multiply Media, LLC. For example, the \"third\" row, or row 2 where n=2 is comprised of Using Pascal's Triangle for Binomial Expansion. Replacing the core of a planet with a sun, could that be theoretically possible? So elements in 4th", "Triangle. Each entry in the nth row gets added twice. $$1,n,\\frac{n(n-1)}2,\\frac{n(n-1)(n-2)}{2\\cdot3},\\frac{n(n-1)(n-2)(n-3)}{2\\cdot3\\cdot4}\\cdots$$, This is computed by recurrence very efficiently, like, $$1,54,\\frac{54\\cdot53}2=1431,\\frac{1431\\cdot52}3=24804,\\frac{24804\\cdot51}4=316251\\cdots$$. first and last of which are 1. The formula just use the previous element to get the new one. In fact, if Pascal's triangle was expanded further past Row 15, you would see that the sum of the numbers of any nth row would equal to 2^n. simply \"1\" in the former and \"1 1\" in the latter. be referring to as row 0 (n=0). during this process (a common practice in computer science), so = 4!/[2!(4-2)!] 1 5 10 10 5 1. The Does whmis to controlled products that are being transported under the transportation of dangerous goodstdg regulations? For example, if a problem was $(2x - 10y)^{54}$, and I were to figure out the $32^{\\text{nd}}$ element in that expansion, how would I figure out? Pascal’s triangle is a triangular array of the binomial coefficients. Store it in a variable say num. Reflection - Method::getGenericReturnType no generic - visbility. In this book they also used this formula to prove (n, r) = n! def pascaline(n): line = [1] for k in range(max(n,0)): line.append(line[k]*(n-k)/(k+1)) return line There are two things I would like to ask. Why can't I sing high notes as a young female? But be careful", "are nonnegative integers satisfying the following linear equation: (3) \\begin{align} \\quad b_1 + b_2 + ... + b_n = k \\end{align} Set up positions to place all $k$ elements from $A$ and group like-elements together. Place $n - 1$ placeholders between each grouping of like-elements. There are now $k + n - 1$ positions total and the rearrangement of the placeholders signifies a new $k$-combination of the multiset $A$ as illustrated in the diagram below. We must choose $k$ of these positions of the elements from $A$ to form a $k$-combination, and so the total number of $k$-combinations of $A$ is: (4) \\begin{align} \\quad \\mathrm{number \\: of \\:} k-\\mathrm{combinations} = \\binom{k + n - 1}{k} \\end{align} We should further note that instead of choosing the $k$ positions of the elements from $A$ first, we can instead choose the $n - 1$ positions of the placeholders which also form a $k$-combination, and so the total number of $k$-combinations of $A$ is also given by: (5) \\begin{align} \\quad \\mathrm{number \\: of \\:} k-\\mathrm{combinations} = \\binom{k + n - 1}{n - 1} \\end{align} We can verify this equality as follows: (6) \\begin{align} \\quad \\binom{k + n - 1}{k} = \\frac{(k + n - 1)!}{k! \\cdot (k + n - 1 - k)!} = \\frac{(k + n - 1)!}{k! \\cdot (n -", "# Is there an equation that represents the nth row in Pascal's triangle? I'm doing binomial expansion and I'm rather confused at how people can find a certain coefficient of certain rows. For example, if a problem was $(2x - 10y)^{54}$, and I were to figure out the $32^{\\text{nd}}$ element in that expansion, how would I figure out? Would I have to look at or draw out a Pascal's triangle, then go 1 by 1 until I hit row 54? Is there an equation that would tell me what the xth element of the nth row is by plugging in numbers? The $n^{th}$ row reads $$1,n,\\frac{n(n-1)}2,\\frac{n(n-1)(n-2)}{2\\cdot3},\\frac{n(n-1)(n-2)(n-3)}{2\\cdot3\\cdot4}\\cdots$$ This is computed by recurrence very efficiently, like $$1,54,\\frac{54\\cdot53}2=1431,\\frac{1431\\cdot52}3=24804,\\frac{24804\\cdot51}4=316251\\cdots$$ Using symmetry, only the first half needs to be evaluated. Compared to the factorial formula, this is less prone to overflows. Hint: $(a+b)^n=\\sum\\limits_{k=0}^n {n\\choose k }a^kb^{n-k}$ where ${n\\choose k}=\\frac{n!}{k!(n-k)!}$ The nth row of a pascals triangle is: $$_nC_0, _nC_1, _nC_2, ...$$ recall that the combination formula of $$_nC_r$$ is $$\\frac{n!}{(n-r)!r!}$$ So element number x of the nth row of a pascals triangle could be expressed as $$\\frac{n!}{(n-(x-1))!(x-1)!}$$", "}$ where the binomial coefficient $\\binom{n}{k}$ is the kth number on the nth row of Pascal’s triangle. Here count both rows and the numbers in a row starting from zero. Since we can permute n things in $n! = 1 \\cdot 2 \\cdot 3 \\cdot \\cdots \\cdot n$ ways, there are $\\displaystyle{ \\frac{n!}{k! (n-k)!} }$ ways to take n things and choose k of them to be x’s and (n-k) of them to be y’s. You see, permuting the x’s or the y’s doesn’t change our choice, so we have to divide by $k!$ and $(n-k)!.$ So, the kth number in the nth row of Pascal’s triangle is: $\\displaystyle{\\binom{n}{k} = \\frac{n!}{k! (n-k)!} }$ All this will be painfully familiar to many of you. But I want everyone to have an fair chance at understanding the next section, where we’ll see something nice about Pascal’s triangle and the number $e \\approx 2.718281828459045...$ ### The hidden e in Pascal’s triangle In 2012, a guy named Harlan Brothers found the number e hiding in Pascal’s triangle… in a very simple way! If we add up all the numbers in the nth row of Pascal’s triangle we get $2^n$. But what if we take the product of all these numbers? Let’s call it $t_n.$ Then here’s what Brothers discovered: $\\displaystyle{ \\lim_{n \\to", "row of Pascal’s triangle is 2 n. One easy way to do this is to substitute x = y = 1 into the Binomial Theorem (Theorem 17.8). Here is my code to find the nth row of pascals triangle. Origin of “Good books are the warehouses of ideas”, attributed to H. G. Wells on commemorative £2 coin? Written, this looks like (7c4), but is equal to [n(n-1)!]/[(n-1)!] Each value in a row is the sumb of the two values above it fashion. Sum of all elements up to Nth row in a Pascal triangle. 1st element of the nth row of Pascal’s triangle) + (2nd element of the nᵗʰ row)().y +(3rd element of the nᵗʰ row). Step by step descriptive logic to print pascal triangle. What do this numbers on my guitar music sheet mean. The remaining entries can be expressed by a simple formula. This means that if we are evaluating By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. I am aware that this question was once addressed by your staff before, but the response given does not come as a helpful means to solving this question. $$1,n,\\frac{n(n-1)}2,\\frac{n(n-1)(n-2)}{2\\cdot3},\\frac{n(n-1)(n-2)(n-3)}{2\\cdot3\\cdot4}\\cdots$$, This is computed by recurrence very efficiently, like, $$1,54,\\frac{54\\cdot53}2=1431,\\frac{1431\\cdot52}3=24804,\\frac{24804\\cdot51}4=316251\\cdots$$. indeed true. To learn more, see our tips on writing great", "= 1$. This explains the ones seen on the left and right sides of Pascal's triangle. Now suppose ${}_nC_k$ is not on the end of a row. Then ${}_nC_{k+1}$ would be the number to its right. Likewise, ${}_{n+1}C_{k+1}$ would be the number on the next row, directly below ${}_nC_k$ and ${}_nC_{k+1}$. As such, we seek to argue that ${}_nC_k + {}_nC_{k+1} = {}_{n+1}C_{k+1}$, as this will establish that the \"adding-the-two-above\" technique for constructing Pascal's triangle yields the same numbers as the triangle constructed from finding coefficients of $(x+y)^n$ expanded. We proceed directly... $$\\begin{array}{rcl} {}_nC_k + {}_nC_{k+1} &=& \\frac{n!}{k!(n-k)!} + \\frac{n!}{(k+1)!(n-k-1)!}\\\\\\\\\\\\ &=& \\frac{n!}{k!(n-k)!} \\cdot \\frac{(k+1)}{(k+1)} + \\frac{n!}{(k+1)!(n-k-1)!} \\cdot \\frac{(n-k)}{(n-k)}\\\\\\\\\\\\ &=& \\frac{k \\cdot n! + n! + n \\cdot n! - k \\cdot n!}{(k+1)! (n-k)!}\\\\\\\\\\\\ &=& \\frac{(n+1) n!}{(k+1)!(n-k)!} \\quad = \\quad {}_{n+1}C_{k+1} \\end{array}$$", "success. = Pascal's Triangle = == Pascal's Identity == === Identity === Pascal's Identity states that$(Error compiling LaTeX. ! Missing$ inserted.){n \\choose k}={n-1\\choose k-1}+{n-1\\choose k}$for any positive integers$k$and$n$. Here,$\\binom{n}{k}$is the binomial coefficient$\\binom{n}{k} = nCk = C_k^n$. Remember that$ (Error compiling LaTeX. ! Missing $inserted.)\\binom{n}{r}=\\frac{n!}{k!(n-k)!}$.$\\binom{n}{r}$means the number of ways to pick r thing from n things where order does not matter.$n!=1 \\cdot 2 \\cdot 3 \\cdot...\\cdot n$. === Proving it === If$(Error compiling LaTeX. ! Missing$ inserted.)k > n$then$\\binom{n}{k} = 0 = \\binom{n - 1}{k - 1} + \\binom{n - 1}{k}$and so the result is pretty clear. So assume$k \\leq n$. Then <cmath>\\begin{eqnarray*}\\binom{n-1}{k-1}+\\binom{n-1}{k}&=&\\frac{(n-1)!}{(k-1)!(n-k)!}+\\frac{(n-1)!}{k!(n-k-1)!}\\\\ &=&(n-1)!\\left(\\frac{k}{k!(n-k)!}+\\frac{n-k}{k!(n-k)!}\\right)\\\\ &=&(n-1)!\\cdot \\frac{n}{k!(n-k)!}\\\\ &=&\\frac{n!}{k!(n-k)!}\\\\ &=&\\binom{n}{k}. \\qquad\\qquad\\square\\end{eqnarray*}</cmath> There we go. We proved it! === Why is it needed? === It's mostly just a cool thing to know. However, if you want to know how to use it in real life go to https://artofproblemsolving.com/videos/counting/chapter12/141. Or really any of the counting and probability videos. == Introduction to Pascal's Triangle == === How to build it === Pascal's Triangle is a triangular array of numbers in which you start with two infinite diagonals of ones and each of the rest of the numbers is the sum of the two numbers above it. It looks something like this: 1 1 1 1 2 1 1 3 3 1 1 4 6", "to take 5 things and choose 3 to call x’s and 2 to call y’s. In general we have $\\displaystyle{ (x + y)^n = \\sum_{k=0}^n \\binom{n}{k} x^k y^{n-k} }$ where the binomial coefficient $\\binom{n}{k}$ is the kth number on the nth row of Pascal’s triangle. Here count both rows and the numbers in a row starting from zero. Since we can permute n things in $n! = 1 \\cdot 2 \\cdot 3 \\cdot \\cdots \\cdot n$ ways, there are $\\displaystyle{ \\frac{n!}{k! (n-k)!} }$ ways to take n things and choose k of them to be x’s and (n-k) of them to be y’s. You see, permuting the x’s or the y’s doesn’t change our choice, so we have to divide by $k!$ and $(n-k)!.$ So, the kth number in the nth row of Pascal’s triangle is: $\\displaystyle{\\binom{n}{k} = \\frac{n!}{k! (n-k)!} }$ All this will be painfully familiar to many of you. But I want everyone to have an fair chance at understanding the next section, where we’ll see something nice about Pascal’s triangle and the number $e \\approx 2.718281828459045...$ ### The hidden e in Pascal’s triangle In 2012, a guy named Harlan Brothers found the number e hiding in Pascal’s triangle… in a very simple way! If we add up all the numbers in the nth row of", "+ y)^5 = x^5 + 5 x^4 y + 10 x^3 y^2 + 10 x^2 y^3 + 5 x y^4 + y^5$ We get the 10 in $10 x^3 y^2$ here because there are 10 ways to take 5 things and choose 3 to call x’s and 2 to call y’s. In general we have $\\displaystyle{ (x + y)^n = \\sum_{k=0}^n \\binom{n}{k} x^k y^{n-k} }$ where the binomial coefficient $\\binom{n}{k}$ is the kth number on the nth row of Pascal’s triangle. Here count both rows and the numbers in a row starting from zero. Since we can permute n things in $n! = 1 \\cdot 2 \\cdot 3 \\cdot \\cdots \\cdot n$ ways, there are $\\displaystyle{ \\frac{n!}{k! (n-k)!} }$ ways to take n things and choose k of them to be x’s and (n-k) of them to be y’s. You see, permuting the x’s or the y’s doesn’t change our choice, so we have to divide by $k!$ and $(n-k)!.$ So, the kth number in the nth row of Pascal’s triangle is: $\\displaystyle{\\binom{n}{k} = \\frac{n!}{k! (n-k)!} }$ All this will be painfully familiar to many of you. But I want everyone to have an fair chance at understanding the next section, where we’ll see something nice about Pascal’s triangle and the number $e \\approx 2.718281828459045...$ ### The hidden", "various ways of constructing it in 1570 -- again, well before Pascal. As alluded to in its history, there are many patterns to the values in Pascal's triangle. One was mentioned above, whereby each row can be quickly constructed from the previous row. However, there is another, much more important (and provable) pattern that can be exploited to produce the numbers on any single row even faster, and without reference to any other rows. This elegant result, known as the Binomial Theorem then lets us expand $(x+y)^n$ for any non-negative integer $n$ as: $$(x+y)^n = x^n + {}_nC_1 x^{n-1} y + {}_nC_2 x^{n-2} y^2 + {}_nC_3 x^{n-3} y^3 + \\cdots + {}_nC_{n-1} x y^{n-1} + y^n$$ Should the notation ${}_n C_k$ be unfamiliar (some texts denote this by ${n \\choose k}$ instead), it calculates the number of combinations of $n$ things, taken $k$ at a time that one can form. That is to say, it counts the number of ways one can choose a set of $k$ objects when drawing from a group of $n$ objects. Students having seen some probability or statistics will likely know its value can be computed with the formula: $${}_n C_k = \\frac{n!}{k!(n-k)!}$$ where $m!$, called the factorial of $m$ is defined by $m! = m \\cdot (m-1) \\cdot (m-2) \\cdots 3 \\cdot", "triangle; these are all the $1$'s at the beginning of every row. Or from the combinatiorial interpretation; there is precisely $1$ way to choose nothing from $n$ elements. Or algebraically $$\\binom n0=\\frac{n!}{0!n!}=1.$$ – Servaes Mar 15 at 22:51 Isn’t it just that $$\\sum_{k=0}^n c_k=c_0 +\\sum_{k=1}^n c_k$$ and noting that your $$c_0=1$$?" ]

[ "+0 # Helppppp 0 55 1 The pattern of Pascal's triangle is illustrated in the diagram shown. What is the fourth element in Row 15 of Pascal's triangle? $$\\begin{array}{ccccccccccccc} \\textrm{Row 0}: & \\qquad & & & & & 1 & & & & & & \\\\ \\textrm{Row 1}: & \\qquad & & & & 1 & & 1 & & & & &\\\\ \\textrm{Row 2}: & \\qquad & & & 1 & & 2 & & 1 & & & &\\\\ \\textrm{Row 3}: & \\qquad & & 1 && 3 && 3 && 1&& \\\\ \\textrm{Row 4}: & \\qquad & 1&& 4 && 6 && 4 && 1 \\end{array}$$ Mar 28, 2020 #1 0 The pattern is: 15C0, 15C1, 15C2, 15C3..........and so on. The 4th element is: 15C(4 - 1) =15C3 = 455. Mar 28, 2020 edited by Guest Mar 28, 2020", "to pick r thing from n things where order does not matter. $n!=1 \\cdot 2 \\cdot 3 \\cdot...\\cdot n$. ### Proving it If $k > n$ then $\\binom{n}{k} = 0 = \\binom{n - 1}{k - 1} + \\binom{n - 1}{k}$ and so the result is pretty clear. So assume $k \\leq n$. Then $\\begin{eqnarray*}\\binom{n-1}{k-1}+\\binom{n-1}{k}&=&\\frac{(n-1)!}{(k-1)!(n-k)!}+\\frac{(n-1)!}{k!(n-k-1)!}\\\\ &=&(n-1)!\\left(\\frac{k}{k!(n-k)!}+\\frac{n-k}{k!(n-k)!}\\right)\\\\ &=&(n-1)!\\cdot \\frac{n}{k!(n-k)!}\\\\ &=&\\frac{n!}{k!(n-k)!}\\\\ &=&\\binom{n}{k}. \\qquad\\qquad\\square\\end{eqnarray*}$ There we go. We proved it! ### Why is it needed? It's mostly just a cool thing to know. However, if you want to know how to use it in real life go to https://artofproblemsolving.com/videos/counting/chapter12/141. Or really any of the counting and probability videos. ## Introduction to Pascal's Triangle ### How to build it Pascal's Triangle is a triangular array of numbers in which you start with two infinite diagonals of ones and each of the rest of the numbers is the sum of the two numbers above it. It looks something like this: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 And on and on... ### Combinations #### Combinations Pascal's Triangle is really combinations. It looks something like this if it is depicted as combinations: $\\binom{0}{0}$ $\\binom{1}{0}$ $\\binom{1}{1}$ $\\binom{2}{0}$ $\\binom{2}{1}$ $\\binom{2}{2}$ And on and on... #### Proof If you look at the way we build the triangle, each number is", "# Pascal's coefficient identity Probability Level 1 $\\begin{array}{rc} 0^\\text{th} \\text{ row:} & 1 \\\\ 1^\\text{st} \\text{ row:} & 1 \\quad 1 \\\\ 2^\\text{nd} \\text{ row:} & 1 \\quad 2 \\quad 1 \\\\ 3^\\text{rd} \\text{ row:} & 1 \\quad 3 \\quad 3 \\quad 1 \\\\ 4^\\text{th} \\text{ row:} & 1 \\quad 4 \\quad 6 \\quad 4 \\quad 1 \\\\ \\vdots \\ \\ \\ & \\cdot \\quad \\cdot \\quad \\cdot \\quad \\cdot \\quad \\cdot \\quad \\cdot \\end{array}$ Pascal's triangle is shown above for the $0^\\text{th}$ row through the $4^\\text{th}$ row. What is the $4^\\text{th}$ element in the $10^\\text{th}$ row? Note: Each row starts with the $0^\\text{th}$ element. For example, the $0^\\text{th}$, $1^\\text{st}$, $2^\\text{nd}$, and $3^\\text{rd}$ elements of the $3^\\text{rd}$ row are 1, 3, 3, and 1, respectively. ×", "the equation above for $\\tbinom{k+1}{m}$, $\\binom{k+1}{m} = \\binom{k}{m-1} + \\binom{k-1}{m-1} + \\binom{k-2}{m-1} + \\cdots + \\binom{m-1}{m-1}$ Hence, the hockey stick assertion holds true for m = k + 1 when it is true for m = k. The induction is complete. Now, we have proved the property for the blue stick case. The red stick case can be proven in a similar way since Pascal's triangle is symmetrical. ## Hidden Hexagon Squares Pascal's triangle contains a magic hexagon pattern. If we look at the nearest six elements around $\\tbinom{n}{r}$, which follow a hexagonal pattern, we will find that the product of the elements at every other vertex equals the product of the three remaining elements. This property was discovered by V.E. Hoggatt, Jr. and W. Hansell in 1971 and is called the Hoggatt-Hansell identity. Later, Gould called it the Star of David property.[3] The image on the left contains selected elements in row 5, 6, and 7 of Pascal's triangle. They can be the six vertices of a hexagon, and 5 × 20 × 21 = 6 × 10 × 35 = 2100. Algebraically, this pattern means that Eq. 4 $\\binom{n-1}{r-1}\\binom{n}{r+1}\\binom{n+1}{r} = \\binom{n-1}{r}\\binom{n+1}{r+1}\\binom{n}{r-1}$ The left hand side of Eq. 4 is, $\\binom{n-1}{r-1}\\binom{n}{r+1}\\binom{n+1}{r}$ Using the formula for combinations, the left hand side becomes $\\frac{(n-1)!}{(r-1)!(n-r)!}\\cdot\\frac{(n)!}{(r+1)!(n-r-1)!}\\cdot\\frac{(n+1)!}{r!(n-r+1)!}$ Changing the positions of the", "2 ] So....the sum of the interior intergers in the 7th row is 2 (7-1) - 2 = 2 6 - … The most efficient way to calculate a row in pascal's triangle is through convolution. Since 10 has two digits, you have to carry over, so you would get 161,051 which is equal to 11^5. The value of that element will be (62)\\binom{6}{2}(26​). □_\\square□​, 0th row:11st row:112nd row:1213rd row:13314th row:14641⋮ ⋅⋅⋅⋅⋅⋅\\begin{array}{rc} 0^\\text{th} \\text{ row:} & 1 \\\\ 1^\\text{st} \\text{ row:} & 1 \\quad 1 \\\\ 2^\\text{nd} \\text{ row:} & 1 \\quad 2 \\quad 1 \\\\ 3^\\text{rd} \\text{ row:} & 1 \\quad 3 \\quad 3 \\quad 1 \\\\ 4^\\text{th} \\text{ row:} & 1 \\quad 4 \\quad 6 \\quad 4 \\quad 1 \\\\ \\vdots \\ \\ \\ & \\cdot \\quad \\cdot \\quad \\cdot \\quad \\cdot \\quad \\cdot \\quad \\cdot \\end{array} 0th row:1st row:2nd row:3rd row:4th row:⋮ ​111121133114641⋅⋅⋅⋅⋅⋅​. \\begin{array}{cccc} 1 & 3 & \\color{#D61F06}{3} & 1\\end{array} \\\\ Then, the element to the right of that is the 1st1^\\text{st}1st element in that row, and so on. \\begin{array}{cccccc} \\vdots & \\hphantom{\\vdots} & \\vdots & \\hphantom{\\vdots} & \\vdots \\end{array}\\\\ You work out R! Numbers written in any of the ways shown below. Write a function that takes an integer value n as input and prints first n lines of the Pascal’s triangle. Similiarly,", "of the leftmost column, resp. to points of the top row. From this pattern we deduce: \\begin{align*} \\mathbf{240}&=129 + 111\\\\ &=(57+42)+(72+39)\\\\ &=((\\mathbf{15}+42)+(\\mathbf{20}+22))+((42+30)+(30+9))\\\\ &\\ldots\\\\ &=(\\it{10}\\cdot\\mathbf{6}+\\it{6}\\cdot\\mathbf{15}+\\it{3}\\cdot\\mathbf{20}+\\it{1}\\cdot\\mathbf{15}) +(\\it{10}\\cdot\\mathbf{1}+\\it{4}\\cdot\\mathbf{1}+\\it{1}\\cdot\\mathbf{1}) \\end{align*} With this example in mind we can derive the general formula for the elements of the minor diagonal: General formula for the hop from $\\binom{n}{j}k^j$ to $\\binom{n}{j}(k+1)^j$: Let's assume we have an element $$t_{j,n-j+1}^{(n,k+1)}\\qquad\\qquad 1 \\leq j \\leq n$$ of the minor diagonal in the $j-th$ row and $n-j+1$ column of the original upper triangular matrix $T_{n,k+1}$, then we have to determine the number of pathes from $$(m,1)\\qquad\\text{to}\\qquad(j,n-j+1) \\qquad 1\\leq m \\leq j$$ corresponding to the leftmost column entries $(1,1)$ to $(j,1)$ and we have to determine the number of pathes from $$(1,m)\\qquad\\text{to}\\qquad(j,n+1-j) \\qquad 1\\leq m \\leq n-j+1$$ corresponding to the top row entries $(1,1)$ to $(1,n+1-j)$ Note: Let $0\\leq x_1\\leq x_2$ and $0\\leq y_1\\leq y_2$. The number of pathes $N_{(x_1,y_1)}^{(x_2,y_2)}$ from $(x_1,y_1)$ to $(x_2,y_2)$ is $$N_{(x_1,y_1)}^{(x_2,y_2)}=\\binom{x_2-x_1+y_2-y_1}{x_2-x_1}=\\binom{x_2-x_1+y_2-y_1}{y_2-y_1}$$ We observe: The number of pathes from $(m,1)$ to $(j,n-j+1)$ and the number of pathes from $(1,m)$ to $(j,n-j+1)$ is $$N_{(m,1)}^{(j,n-j+1)}\\binom{n-m}{n-j}\\qquad\\text{resp.}\\qquad N_{(1,m)}^{(j,n-j+1)}\\binom{n-m}{j-1}$$ Now we can formulate the binomial identity which we have to proof to show the last part of the inductive step. Binomial identity: The following is valid for $1\\leq j \\leq n$: \\begin{align*} \\sum_{m=1}^{j}\\binom{n-m}{n-j}t_{j,n+1-j}^{(n,k)}&+\\sum_{m=1}^{n-j}\\binom{n-m}{j-1} =t_{j,n+1-j}^{(n,k+1)}\\\\ \\text{resp. }&\\\\ \\sum_{m=1}^{j}\\binom{n-m}{n-j}\\binom{n}{m}k^n&+\\sum_{m=1}^{n-j}\\binom{n-m}{j-1}\\tag{3} =\\binom{n}{j}(k+1)^n \\end{align*} We calculate each", "= \\binom{2}{1}$$, $$\\binom{2}{0} + \\binom{2}{1} = \\binom{3}{1}$$ and so forth. This relationship is indicated by the arrows in the array above. With these two facts in hand, we can quickly generate Pascal's Triangle. We start with the first two rows, $$1$$ and $$1 \\quad 1$$. From that point on, each successive row begins and ends with $$1$$ and the middle numbers are generated using Theorem \\ref{addbinomcoeff}. Below we attempt to demonstrate this building process to generate the first five rows of Pascal's Triangle. $\\begin{array}{ccc} \\begin{array}{ccccccccc} & & & & 1 & & & & \\\\ & & & 1 & & 1 & & &\\\\ & & & & \\searrow \\, \\swarrow & & & & \\\\ & & \\fbox{1} & & \\underline{1+1} & & \\fbox{1} & &\\\\ \\end{array} & \\xrightarrow{\\hspace{.5in}} & \\begin{array}{ccccccccc} & & & & 1 & & & & \\\\ & & & 1 & & 1 & & &\\\\ & & 1 & & 2 & &1 & &\\\\ \\end{array} \\\\ && \\\\ \\begin{array}{ccccccccc} & & & & 1 & & & & \\\\ & & & 1 & & 1 & & &\\\\ & & 1 & & 2 & &1 & &\\\\ & & & \\searrow \\, \\swarrow & & \\searrow \\, \\swarrow & & & \\\\ & \\fbox{1} & & \\underline{1+2}", "Pascal's Triangle # Pascal's Triangle One way to relate $k$-combinations of a finite $n$-element set is with a special graph known as Pascal's Triangle. A portion of Pascal's triangle is shown below: The top of Pascal's Triangle is a $1$, and entries for Pascal's triangle are generated to be the sum of the two numbers directly above that entry (or just the number directly above the entry if there's only one). If we label the rows of Pascal's triangle starting at row $0$, row $1$, etc…, then you should notice that the finite sequence of numbers in row $n$ for each $n = 0, 1, 2, ...$ and for $k = 0, 1, 2, ..., n$ is the same sequence of numbers as: (1) \\begin{align} \\quad \\left \\{ \\binom{n}{0}, \\binom{n}{1}, \\binom{n}{2}, ..., \\binom{n}{k-1}, \\binom{n}{k}, \\binom{n}{k+1}, ..., \\binom{n}{k} \\right \\} \\end{align} For example, row $5$ in Pascal's triangle forms the finite sequence $\\{ 1, 5, 10, 10, 5, 1 \\}$, and using the formula $\\binom{n}{k} = \\frac{n!}{k!(n-k)!}$ for counting the number of $k$-combinations for a $5$-element set for $k = 0, 1, 2, ..., 5$, we have that: (2) \\begin{align} \\quad \\quad \\left \\{ \\binom{5}{0}, \\binom{5}{1}, \\binom{5}{2}, \\binom{5}{3}, \\binom{5}{4}, \\binom{5}{5} \\right \\} = \\left \\{ \\frac{5!}{0! \\cdot 5!} , \\frac{5!}{1! \\cdot 4!}, \\frac{5!}{2! \\cdot 3!}, \\frac{5!}{3! \\cdot 2!}, \\frac{5!}{4!", "Triangle. Each entry in the nth row gets added twice. $$1,n,\\frac{n(n-1)}2,\\frac{n(n-1)(n-2)}{2\\cdot3},\\frac{n(n-1)(n-2)(n-3)}{2\\cdot3\\cdot4}\\cdots$$, This is computed by recurrence very efficiently, like, $$1,54,\\frac{54\\cdot53}2=1431,\\frac{1431\\cdot52}3=24804,\\frac{24804\\cdot51}4=316251\\cdots$$. first and last of which are 1. The formula just use the previous element to get the new one. In fact, if Pascal's triangle was expanded further past Row 15, you would see that the sum of the numbers of any nth row would equal to 2^n. simply \"1\" in the former and \"1 1\" in the latter. be referring to as row 0 (n=0). during this process (a common practice in computer science), so = 4!/[2!(4-2)!] 1 5 10 10 5 1. The Does whmis to controlled products that are being transported under the transportation of dangerous goodstdg regulations? For example, if a problem was $(2x - 10y)^{54}$, and I were to figure out the $32^{\\text{nd}}$ element in that expansion, how would I figure out? Pascal’s triangle is a triangular array of the binomial coefficients. Store it in a variable say num. Reflection - Method::getGenericReturnType no generic - visbility. In this book they also used this formula to prove (n, r) = n! def pascaline(n): line = [1] for k in range(max(n,0)): line.append(line[k]*(n-k)/(k+1)) return line There are two things I would like to ask. Why can't I sing high notes as a young female? But be careful", "The question is as follows: \"There is a formula connecting any (k+1) successive coefficients in the nth row of the Pascal Triangle with a coefficient in the (n+k)th row. values for 11^n when you know what row n looks like in Pascal's This method only works well for rows up to and including row 4. Hint: Remember to fill out the first for nCr. But this approach will have O(n 3) time complexity. How to prove that the excentral triangle passes through the vertices of the original triangle? it is the seventh number in the row). Sum of numbers in a nth row can be determined using the formula 2^n. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Naive Approach: In a Pascal triangle, each entry of a row is value of binomial coefficient. = Consider again Pascal's Triangle in which each number is obtained as the sum of the two neighboring numbers in the preceding row. = (7*6*5!)/(2!5!) represented in row n by index k is the value V. This number can be Why don't libraries smell like bookstores? $$1,n,\\frac{n(n-1)}2,\\frac{n(n-1)(n-2)}{2\\cdot3},\\frac{n(n-1)(n-2)(n-3)}{2\\cdot3\\cdot4}\\cdots$$, This is computed by recurrence very efficiently, like, $$1,54,\\frac{54\\cdot53}2=1431,\\frac{1431\\cdot52}3=24804,\\frac{24804\\cdot51}4=316251\\cdots$$. First, the outputs integers end with .0 always", "element set of $\\left\\{ 0,1,...,m\\right\\}$ has to be at least $2$. Thus the result follows. Note that this is a special case of the more general: $$\\sum_{j=0}^{m}\\binom{n+j}{n}=\\binom{n+m+1}{n+1}=\\binom{n+m+1}{m}$$ See proof Specifically you can rewrite yours to fit into the above form as: $$\\sum_{n=2}^{m}\\binom{n}{2}=\\sum_{j=0}^{m-2}\\binom{2+j}{2}=\\binom{2+m-2+1}{2+1}=\\binom{m+1}{3}$$ Consider Pascal's triangle. $$\\begin{array}{}&&&&1&&&&\\\\&&&1&&1&&&\\\\&&1&&2&&1&&\\\\&1&&3&&3&&1&\\\\1&&4&&6&&4&&1\\end{array}$$ Note that any element in Pascal's Triangle is given by ${r \\choose n}$ where $r$ is the row number (starting with $0$) and $n$ is the element's position in the row (starting with $0$). Note that $n$ also corresponds to the diagonal the element lies on (e.g. if $n=0$, the element lies on the $0^{th}$ diagonal going from the top-right to the bottom-left). You may also note that all elements in a row have the same sum of their diagonals (e.g. in row $4$, we have $(4,0),(3,1),(2,2),(1,3),$ and $(0,4)$ where $(a,b)$ is the numbers of the diagonals that intersect at that element). Since the first element of row $r$ necessarily has $(a,b)=(r,0)$, we can deduce that the sum of the numbers of these diagonals is $r$ for any element in Pascal's Triangle. Sorry, if that was confusing, but this is where it gets nice. We can now substitute ${a+b \\choose a}$ for ${r\\choose n}$, starting with $b=0$. Now, consider summing down a diagonal (which is equivalent to summing across $b$): $$\\begin{array}{}&&&&&1\\\\&&&&1&&1\\\\&&&1&&2&&\\underline 1\\\\&&1&&3&&\\underline", "$$n$$, the $$n$$th row of Pascal's triangle is the following. $\\binom{n}{0}\\;\\;\\binom{n}{1}\\;\\;\\binom{n}{2}\\cdots \\binom{n}{n}$ Note that the top row (zeroth row) of Pascal's triangle consists of $$\\displaystyle\\binom{0}{0}$$ which is defined to be $$1$$. For $$n\\geq 2$$, the middle $$n-1$$ entries of the $$n$$th row of Pascal's triangle can be obtained from $$(n-1)$$th row using Pascal's identity. $\\begin{array}{lccccccccccc} n=0\\hspace{12pt}& & & & & & 1 & & & & & \\cr n=1& & & & & 1 & & 1 & & & & \\cr n=2& & & & 1 & & 2 & & 1 & & & \\cr n=3& & & 1 & & 3 & & 3 & & 1 & & \\cr n=4& & 1 & & 4 & & 6 & & 4 & & 1 & \\cr n=5& 1 & & 5 & & 10 & & 10 & & 5 & & 1 \\cr \\end{array}$ $\\hspace{48pt}\\text{The first six rows of Pascal's triangle}$ Last edited", "can the committee be formed? hands-on Exercise $$\\PageIndex{3}\\label{he:combin-03}$$ How many subsets of $$\\{1,2,\\ldots,23\\}$$ have five elements? ## $$\\binom{n}{r} = \\binom{n}{n-r}$$ Theorem $$\\PageIndex{2}$$ For $$0\\leq r\\leq n$$, we have $$\\binom{n}{r} = \\binom{n}{n-r}$$. Proof According to Theorem 7.4.1 we have $\\binom{n}{n-r} = \\frac{n!}{(n-r)!\\,(n-(n-r))!} = \\frac{n!}{(n-r)!\\,r!},$ which is precisely $$\\binom{n}{r}$$. Example $$\\PageIndex{4}\\label{eg:combin-04}$$ To compute the numeric value of $$\\binom{50}{47}$$, instead of computing the product of 47 factors as indicated in the definition, it is much faster if we use $\\binom{50}{47} = \\binom{50}{3} = \\frac{50\\cdot 49\\cdot48}{3\\cdot 2\\cdot 1},$ from which we obtain $$\\binom{50}{47} = 19600$$. hands-on Exercise $$\\PageIndex{4}\\label{he:combin-04}$$ Compute, by hand, the numeric value of $$\\binom{529}{525}$$. ## Pascal's Triangle In this section we will cover • how to construct Pascal's Triangle • how to read values for combinations from Pascal's Triangle In a later section, we will prove that this construction does give the values for combinations. Start with a $$1$$, then two more $$1$$'s. Notice how the spacing is offset on each row. $\\begin{array}{*{13}{c}} & & & & & & 1 \\\\ & & & & & 1 & & 1 \\\\ & & & & 1 & & 2 & & 1 \\\\ & & & 1 & & 3 & & 3 & & 1 \\end{array}$ To create the next row, start with a $$1$$ and then add the", "+ y)^5 = x^5 + 5 x^4 y + 10 x^3 y^2 + 10 x^2 y^3 + 5 x y^4 + y^5$ We get the 10 in $10 x^3 y^2$ here because there are 10 ways to take 5 things and choose 3 to call x’s and 2 to call y’s. In general we have $\\displaystyle{ (x + y)^n = \\sum_{k=0}^n \\binom{n}{k} x^k y^{n-k} }$ where the binomial coefficient $\\binom{n}{k}$ is the kth number on the nth row of Pascal’s triangle. Here count both rows and the numbers in a row starting from zero. Since we can permute n things in $n! = 1 \\cdot 2 \\cdot 3 \\cdot \\cdots \\cdot n$ ways, there are $\\displaystyle{ \\frac{n!}{k! (n-k)!} }$ ways to take n things and choose k of them to be x’s and (n-k) of them to be y’s. You see, permuting the x’s or the y’s doesn’t change our choice, so we have to divide by $k!$ and $(n-k)!.$ So, the kth number in the nth row of Pascal’s triangle is: $\\displaystyle{\\binom{n}{k} = \\frac{n!}{k! (n-k)!} }$ All this will be painfully familiar to many of you. But I want everyone to have an fair chance at understanding the next section, where we’ll see something nice about Pascal’s triangle and the number $e \\approx 2.718281828459045...$ ### The hidden", "= 1$. This explains the ones seen on the left and right sides of Pascal's triangle. Now suppose ${}_nC_k$ is not on the end of a row. Then ${}_nC_{k+1}$ would be the number to its right. Likewise, ${}_{n+1}C_{k+1}$ would be the number on the next row, directly below ${}_nC_k$ and ${}_nC_{k+1}$. As such, we seek to argue that ${}_nC_k + {}_nC_{k+1} = {}_{n+1}C_{k+1}$, as this will establish that the \"adding-the-two-above\" technique for constructing Pascal's triangle yields the same numbers as the triangle constructed from finding coefficients of $(x+y)^n$ expanded. We proceed directly... $$\\begin{array}{rcl} {}_nC_k + {}_nC_{k+1} &=& \\frac{n!}{k!(n-k)!} + \\frac{n!}{(k+1)!(n-k-1)!}\\\\\\\\\\\\ &=& \\frac{n!}{k!(n-k)!} \\cdot \\frac{(k+1)}{(k+1)} + \\frac{n!}{(k+1)!(n-k-1)!} \\cdot \\frac{(n-k)}{(n-k)}\\\\\\\\\\\\ &=& \\frac{k \\cdot n! + n! + n \\cdot n! - k \\cdot n!}{(k+1)! (n-k)!}\\\\\\\\\\\\ &=& \\frac{(n+1) n!}{(k+1)!(n-k)!} \\quad = \\quad {}_{n+1}C_{k+1} \\end{array}$$", "... Fourth row: tetrahedral numbers, 1, 4, 10, ... 4) According to Pascal, in every arithmetical triangle each cell is equal to the sum of all the cells of the preceding row from its column to the first, inclusive (Corollary 2). Also, in every arithmetic triangle, each cell diminished by unity is equal to the sum of all those which are included between its perpendicular rank and its parallel rank, exclusively (Corollary 4). Identity 1: $$C^{n+1}_{m+1}=C^n_m+C^{n-1}_m+\\cdots+C^0_m$$ Identity 2: $$C^{n+1}_m-1=\\sum C^k_j$$ 5) We could detect the Fibonacci numbers if we cut the triangle in diagonals and take the sum of the diagonals: 6) The fundamental constant e hidden in the Pascal Triangle; this by taking products - instead of sums - of all elements in a row: $$\\displaystyle\\lim_{n \\rightarrow \\infty}\\frac{S_{n-1}S_{n+1}}{s^2_n}=e$$, where Sis the product of the terms in the nth row. 7) The nth Catalan number $$C_n=\\frac{1}{n+1}\\binom{2n}{n}$$could be discovered in the following combination: $$C_n=\\binom{2n-3}{n-1}+\\binom{2n-2}{n}-\\binom{2n-3}{n}-\\binom{2n-2}{n+1}$$ 8) Squares Identity: $$C^{n+2}_3-C^n_3=n^2\\\\ C^{n+3}_4−C^{n+2}_4−C^{n+1}_4+C^n_4=n^2.$$ Also: $$\\displaystyle\\sum^n_{k=0}(C^n_k)^2=C^{2n}_n$$ 9) Cubes Identity: $$\\displaystyle n^{3}=\\bigg[C^{n+1}_{2}\\cdot C^{n-1}_{1}\\cdot C^{n}_{0}\\bigg] + \\bigg[C^{n+1}_{1}\\cdot C^{n}_{2}\\cdot C^{n-1}_{0}\\bigg] + C^{n}_{1}$$ 10) pi Identity: $$\\pi=3+\\frac23(\\frac1{C^4_3}-\\frac1{C^6_3}+\\frac1{C^8_3}-\\cdots).$$ 11) Fractals If you shade in certain numbers, you get beautiful fractals: GYanggg Jun 26, 2018 edited by GYanggg Jun 28, 2018 #1 +2117 +2 It is amazing how all these hidden patterns can be extracted from a simple sequence of", "\\binom{n}{2} \\cdot \\cdots \\cdot \\binom{n}{n} = t_n$ $1! \\cdot 2! \\cdot \\cdots \\cdot n! = G(n+2)$ $\\frac{1}{0!} +\\frac{1}{1!} + \\frac{1}{2!} + \\frac{1}{3!} + \\cdots = e$ and Euler’s wacky formula $\\displaystyle{ 1 + 2 + 3 + \\cdots = -\\tfrac{1}{12} }$ Puzzle 2. Can you take the formula $\\begin{array}{ccl} \\ln(t_n) &\\approx& \\displaystyle{(n+1)\\Big[ n \\ln(n) - n + \\tfrac{1}{2} \\ln(2 \\pi n) + \\frac{1}{12n} \\Big] } \\\\ \\\\ && - \\displaystyle{ \\Big[ \\left( \\ln(n+1) - \\tfrac{3}{2}\\right) (n+1)^2 + \\ln(2\\pi) \\cdot (n+1) -} \\\\ \\\\ && \\quad \\displaystyle{ \\tfrac{1}{6} \\ln(n+1) + 2\\zeta'(-1) \\Big] } \\end{array}$ and massage it until it looks like $n(n+1)/2$ plus ‘correction terms’? How big are these correction terms? ### References Any errors in the formulas above are my fault. Here are the papers that first squeezed e out of Pascal’s triangle: • Harlan J. Brothers, Pascal’s triangle: The hidden stor-e, The Mathematical Gazette, March 2012, 145. • Harlan J. Brothers, Finding e in Pascal’s triangle, Mathematics Magazine 85 (2012), 51. I learned the idea from here, thanks to Richard Elwes: • Alexander Bogomolny, e in the Pascal Triangle, Interactive Mathematics Miscellany and Puzzles. For how Euler derived his crazy identity $\\displaystyle{ 1 + 2 + 3 + \\cdots = -\\tfrac{1}{12} }$ and how it’s relevant to string theory, try: • John Baez, My favorite numbers: 24.", "# Greatest Common Divisors in columns and rows of Pascal Triangle Let $n$ and $k$ be integers such that $n\\ge3$ and $k\\ge 2$ and $g(n)$ is the prime gap where $n$ lies $$k\\le g(n)+2\\implies \\gcd\\left(\\binom{n+j}{k} , j\\in \\{ 1,k-1 \\} \\right)\\gt1$$ $\\binom{n+j}{k}$ is a binomial coefficient and more precisely: $p(n)$ and $q(n)$ being two consecutive primes such that $p(n)\\lt n \\le q(n)$ let $g(n)=q(n)-p(n)$ The above proposition seems true, but I am having hard time finding a proof. Any help is welcome.Thank you! This is the begining ($1\\le n,k \\le 20$) of the table $d_{n,k}=\\gcd\\left(\\binom{n+j}{k} , j\\in \\{ 1,k-1 \\}\\right)$ with the entries where $k=g(n)+2$ in red color. ($g(1)=g(2)=4$, because $p(0)=p(1)=-2$) There would be an analogous proposition for the Greatest Common Divisors in rows of the Pascal Triangle $$k\\le \\lfloor\\frac{n+1}{2}\\rfloor- g(n+1)+1\\implies \\gcd\\left(\\binom{n} {k+j} ,j\\in \\{ 0,k \\} \\right)\\gt1$$ Here is the table $D_{n,k}=\\gcd\\left(\\binom{n}{k+j} , j\\in \\{ 0,k \\}\\right)$, $1\\le n\\le 30,0\\le k\\le 15$, with the entries where $k=\\lfloor\\frac{n+1}{2}\\rfloor- g(n+1)+1$ in red color The latter proposition should be more easy to prove, even though it does not involve only prime gaps but also the floor function, since it is easy to see that any entry $\\binom{n}{m}$ in Pascal Triangle is a multiple of $P=p(n)$ (or $P=n$ if $n$ is prime) provided that $n-P\\lt m \\lt P$", "5uv4 + v5 . Exercise 11. What do you notice about your expansions for (u + v)4 and (u + v)5? Does your observation hold for other powers of (u + v)? The coefficients of (u + v)4 are the numbers in Row 4 of Pascal’s triangle. The coefficients of (u + v)5 are the numbers in Row 5 of Pascal’s triangle. The same pattern holds for (u + v), (u + v)2, and (u + v)3. Exercise 12. Use the binomial theorem to expand the following binomial expressions. a. (x + y)6 x6 + 6x5 y + 15x4 y2 + 20x3 y3 + 15x2 y4 + 6xy5 + y6 b. (x + 2y)3 x3 + 6x2 y + 12xy2 + 8y3 c. (ab + bc)4 a4 b4 + 4a3 b4 c + 6a2 b4 c2 + 4ab4 c3 + b4 c4 d. (3xy – 2z)3 27x3 y3 – 54x2 y2 z + 36xyz2 – 8z3 e. (4p2 qr – qr2)5 1024p10 q5 r5 – 1280p8 q5 r6 + 640p6 q5 r7 – 160p4 q5 r8 + 20p2 q5 r9 – q5 r10 ### Eureka Math Precalculus Module 3 Lesson 4 Problem Set Answer Key Question 1. Evaluate the following expressions. a. $$\\frac{9 !}{8 !}$$ $$\\frac{9 !}{8 !}$$ = $$\\frac{9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5", "the 3th diagonal row were the triangular numbers. Numbers written in any of the ways shown below. Basically, what I did first was I chose arbitrary values of n and k to start with, n being the row number and k being the kth number in that row (confusing, I know). 1st element of the nth row of Pascal’s triangle) + (2nd element of the nᵗʰ row)().y +(3rd element of the nᵗʰ row). Each entry in the nth row gets added twice. this article for a general example. Each element in the triangle has a coordinate, given by the row it is on and its position in the row (which you could call its column). Should the stipend be paid if working remotely? Suppose we have a number n, we have to find the nth (0-indexed) row of Pascal's triangle. Find this formula\". $$1,n,\\frac{n(n-1)}2,\\frac{n(n-1)(n-2)}{2\\cdot3},\\frac{n(n-1)(n-2)(n-3)}{2\\cdot3\\cdot4}\\cdots$$, This is computed by recurrence very efficiently, like, $$1,54,\\frac{54\\cdot53}2=1431,\\frac{1431\\cdot52}3=24804,\\frac{24804\\cdot51}4=316251\\cdots$$. Consider again Pascal's Triangle in which each number is obtained as the sum of the two neighboring numbers in the preceding row. Copyright © 2021 Multiply Media, LLC. For example, the \"third\" row, or row 2 where n=2 is comprised of Using Pascal's Triangle for Binomial Expansion. Replacing the core of a planet with a sun, could that be theoretically possible? So elements in 4th" ]

[ "# Overpowering tens unit Discrete Mathematics Level 1 How many 2 digit numbers are there, such that the units digit is strictly smaller than the tens digit? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "multiples of ten.", "the tens.", "tens-digit is odd, then it must be divisible by an odd number of primes of the form 11, 13, 17, 19 mod 20. Any even number of these would produce an even tens digit. -", "the number 10. The chapter is titled “Counting in Tens.” This forms the basics of Mathematics, which is essential to delve deeper into the subject in higher classes. Easy solutions, important questions and revision notes are available on Vedantu to help the student grasp the subject better.", "of a multiple of 10 or 25.", "# Evenly even How many positive even integers up to 1000 (inclusive) have an even digit sum? Details and assumptions The digit sum of a number is the sum of all its digits. For example the digit sum of 1123 is $$1 + 1 + 2 + 3 = 7$$. ×", "how many .", "# The tens digit is quite odd... For how many positive integers $$n$$ where $$n \\leq 1000$$ are there such that the tens digit of $$n^2$$ is odd? ×", "school.\" How many students had Pythagoras at school? 16. Integers May be the sum of two integers less than their difference? 17. Unknown number 16 My number's tens is three times more then ones My number's ones is twice the number of thousands and my number's hundreds is half the number of of tens. I have two ones. which number am I?", "Image showing 17.591 and then units, tens, decimal point, 1/10, 1/100, 1/1000 with arrows one to left saying 10 x bigger, and 1 to right saying 10 × smaller 2.2 Using a Number Line", "in a fraction of seconds 10 the sum of 10 of 10 is. By the total number of possible sums ( x+y ) n= Xn n...", "integer and its tens digit is 4 times its units digit, what is the value of n? 1) The tens digit of n is 8. 2) The sum of the tens digit and the units digit of n is 2-digit integer. St1:- The tens digit of n is 8. Given, tens digit is 4 times its units digit. So, units digit is $$\\frac{8}{4}=2$$ Therefore, the 2-digit positive integer, n is 82. Sufficient. St2:-The sum of the tens digit and the units digit of n is 2-digit integer. If units digit is y and tens digit is x, then n=10x+y. n=x y or 4y y Since, x+y is a 2-digit integer, we have y=2 as unit digit. So, tens digit=4y=4*2=8. So, n=10x+y=10*8+2=82 Sufficient. Ans. (D) Hi PKN can you please elaborate on statement TWO St2:-The sum of the tens digit and the units digit of n is 2-digit integer. If units digit is y and tens digit is x, then n=10x+y. n=x y or 4y y Since, x+y is a 2-digit integer, we have y=2 as unit digit. So, tens digit=4y=4*2=8. So, n=10x+y=10*8+2=82 are you using remainder formula ? n=10x+y. X is quotient and Y is remainer ? why ? any other method to explain second statement? you explanation is kinda hard to understand thanks problem solver killer i", "0 What is the millions digit of 319672191? 9 What is the hundred millions digit of 130028023? 1 What is the tens digit of 325238308? 0 What is the tens digit of 46773645? 4 What is the ten thousands digit of 52813169? 1 What is the millions digit of 480863464? 0 What is the ten thousands digit of 44576563? 7 What is the hundred millions digit of 2470399601? 4 What is the millions digit of 1168405263? 8 What is the tens digit of 71323290? 9 What is the thousands digit of 45249196? 9 What is the ten thousands digit of 12029225? 2 What is the units digit of 2903539242? 2 What is the ten millions digit of 29207109? 2 What is the units digit of 606731458? 8 What is the hundreds digit of 2565500474? 4 What is the ten millions digit of 70986354? 7 What is the tens digit of 218335679? 7 What is the hundreds digit of 612049677? 6 What is the billions digit of 2222363461? 2 What is the tens digit of 534404060? 6 What is the units digit of 1428032287? 7 What is the millions digit of 1029928434? 9 What is the ten millions digit of 5933789688? 3 What is the units digit of 4234152066? 6 What is the ten thousands digit of 831200705? 0", "= (counting the 10's in there) 10^11.", "# Sequences in Digits Algebra Level 3 A three-digit even number is given such that the hundreds, tens, and unit digits respectively form a decreasing arithmetic sequence. When two is added to the number, the same order of the digits now forms a geometric sequence. What is the original number? ×", "a number like 2x10-10, I think one SD would suffice.", "The tens’ digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit. Let say the numbers are 42 and 24, then (x-y)/9 = (42 - 24)/9 = 18/9 = 3 (integer). Therefore, (x - y)/9 an integer Let say the numbers are 42 and 46, (x-y)/9 = (42 - 46)/9 = -4/9 (not an integer). Therefore , (x - y)/9 is not an integer Hence insufficient Therefore, A) should be the answer I understand how it is 10a+b, but how is it 10b - a? Shouldn't that too be 10b + a? Thank you! Hi menontion, Please note that y is $$10b + a$$ (as you have mentioned). $$x - y = 10a +b - (10b +a) = 10a + b - 10b - a = 9a - 9b = 9(a-b)$$ Hope this helps. Manager Joined: 23 Aug 2017 Posts: 117 Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an [#permalink] ### Show Tags 26 Feb 2019, 23:06 For Statement(2): (2) The tens’ digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit. If we approach analytically, Lets assume units digit of X is", "= 2, so that's the ten's digit. Your number is 26. -Dan 6. ## i know that one Originally Posted by mredhi390 My tens digit is 3 more than 5.My tens digit is double the number of my ones digit.What number am I?", "digits and so on. (units in the first time step, tens in the second time step etc.)" ]

[ "means there are 13 \"a\" s and b cannot be 20) When $21\\leq a \\leq 28$, there are $1+2+\\cdots+8=36$ ways. In all, there are $192+36=\\boxed{228}$ possible sequences. ~bluesoul", "do this. In total, there are $1 + 1 + 1 + 4 + 1 = \\boxed{8}$ Jul 11, 2021", "\\cdot 9^2} = \\boxed{\\frac{16}{81}}$$", "options for these attributes. There are exactly three cards that meet these conditions, which form a set. So there are $$4\\cdot3^3=\\boxed{108}$$ sets in this case. As a check, the answers for parts (c)-(f) add up to $$216+432+324+108=1080$$ which is the answer for part (b). Oct 8, 2022", "refer to the symbols \"0\", \"1\", \"2\", \"3\", \"4\", \"5\", \"6\", \"7\", \"8\", and \"9\". There are a total of ten digits. The problem told us that we cannot have the first digit of the code be zero -- this means there are 9 total choices for the first digit of the lock box code. For the second and third digits of the lock box code, we are not told a restriction, and so there are 10 choices for each. For the fourth digit of the code, we may pick any odd digit: there are 5 total choices here. This means there are a total of $$9 \\cdot 10 \\cdot 10 \\cdot 5=4500$$ choices for a code.", "of combinations of 3 digits from the 7 remaining. There are $\\displaystyle {7\\choose{3}} = 35$ combinations. That means the probability of having a number less than 4000 and greater than 3000 is $\\displaystyle \\frac{35}{70} = \\frac{1}{2}$. 3. Hello, johnsy123! $\\text{1)A bag contains five white, six red and seven blue balls.}$ $\\text{If three balls are selected at random, without replacement,}$ $\\text{find the probability they are all different colors.}$ There are: . ${18\\choose3} \\:=\\:816$ possible outcomes. There are: . ${5\\choose1}{6\\choose1}{7\\choose1} \\:=\\:5\\cdot6\\cdot7 \\:=\\:210$ ways to get one of each color. Therefore: . $P(\\text{all different colors}) \\;=\\;\\dfrac{210}{816} \\;=\\;\\dfrac{35}{136}$ $\\text{2) A four-digit number (no repetitions) is to be formed}$ $\\text{from the set of numbers (0,1,2,3,4,5,6,7).}$ Assuming that the number may not have a leading zero, . . there are: . $7\\cdot7\\cdot6\\cdot5 \\:=\\:1470$ possible 4-digit numbers. $\\text{(a) Find the probability that the number is even.}$ There are two cases to consider: . . The last digit is 0. . . The last digit is 2, 4 or 6. The last digit is 0: . $\\square\\:\\square\\;\\square\\;0$ . . The other 3 spaces can be filled in: . $7\\cdot6\\cdot5 \\:=\\:210$ ways. The last digit is 2, 4 or 6: .3 choices. The first must not be 0: .6 choices. The other 2 spaces can be filled in $6\\cdot5\\,=\\,30$ ways. . . There are: . $3\\cdot6\\cdot30 \\:=\\:540$", "of the digits of each $x$ plus $2+0+2+0=4$. Hence the sum of all of the digits of the numbers in $S$ is $12 \\cdot 4 +45=\\boxed{093}$.", "a^2$$. Since $\\sqrt[3]{576} \\approx 8$, we can easily see that $a = 8$ is a solution. Thus, the answer is $24\\cot^2x = 24a = 24 \\cdot 8 = \\boxed{192}$. ~IceMatrix", "# Thread: How many ways can the number 50 be witten as the sum of 10 non-negative, even number? 1. ## How many ways can the number 50 be witten as the sum of 10 non-negative, even number? Hi, Please help me to find out: in how many ways the number 50 can be witten as the sum of 10 non-negative, even numbers? Thanks alot. 2. Hello, gravity2910! In how many ways the number 50 can be witten as the sum of 10 nonnegative, even numbers? I hope the order of the numbers is important. .** For example: $\\{1,1,1,1,1,1,1,1,1,41\\}$ is different from $\\{41,1,1,1,1,1,1,1,1,1\\}$ Consider fifty 1's in a row. Group them in pairs .... and we have twenty-five 2's in a row. . . . . . $\\underbrace{\\boxed{2}\\;\\;\\boxed{2}\\;\\;\\boxed{2}\\;\\ ;\\boxed{2}\\;\\;\\hdots\\;\\;\\boxed{2}}_{\\text{t\\!went\\ !y-five pairs}}$ There are 24 spaces between these pairs. We will insert 9 dividers into those spaces. For example: . $\\boxed{2}\\;\\boxed{2}\\;\\bigg|\\;\\boxed{2}\\;\\bigg| \\; \\boxed{2}\\;\\boxed{2}\\; \\boxed{2}\\;\\boxed{2}\\;\\bigg| \\;\\hdots \\; \\bigg|\\;\\boxed{2}\\;\\boxed{2}\\;\\boxed{2}$ . . . . . . . . . . represents the sum: . $4 + 2 + 8 + \\hdots + 6$ Therefore, there are: . $_{24}C_9 \\;=\\;{24\\choose9} \\;=\\;\\frac{24!}{9!\\,15!} \\;=\\;\\boxed{1,\\!307,\\!504}$ ways. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~", "there's only one, right? Base 30 will work: $$11_{30}=1\\cdot 30^1+1\\cdot 30^0$$ Since 31 is odd, the next possibility would be $33$, but that won't work because 28 isn't divisible by 3. $55$ and higher won't work because the number would need to be at least base 6, and $55_6$ exceeds 31. Let's go to $111$ then. Is there a value for $n$ such that $n^2 + n = 30$ ? Why, yes there is: $$111_{5}=1\\cdot 5^2+1\\cdot 5^1+1\\cdot 5^0$$ So there are the two (non-trivial single-digit) solutions. • Because $31$ is prime, the digit must be $1$ or $31$ – Ross Millikan Nov 12 '13 at 19:45 • Good point. That would have made things easier. :) – John Nov 12 '13 at 19:48", "SEARCH HOME Math Central Quandaries & Queries Question from Tom, a student: I need to open a lock box from my parent's estate. The combination is a simple 4 number lock. The last two numbers are 33. Numbers can be used more than once.What are the possible combinations? WP Hi Tom, If the last two digits are 33 then you only need tofind the first two. There are 10 possibilities for the first digit and 10 for the second and hence there are $10 \\times 10 = 100$ possibilities for the first two digits. You can even list them. They are $00, 01, 02, \\cdot \\cdot \\cdot 09, 10,11, \\cdot \\cdot \\cdot, 19, 20, 21, \\cdot \\cdot \\cdot, 98, 99$ Harley Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "fourth item). Thus the number of ways to order the four items is $4\\cdot 3\\cdot 2\\cdot 1 =24\\;$ . In general, the number of ways to order $n$ items is $$n\\cdot (n-1)\\cdot (n-2)\\cdot \\dots \\cdot 2\\cdot 1 =n!\\; \\text{.}$$ We see then that the number of ways to shuffle a standard fifty-two card deck is $$\\begin{array}{rl} 52! & = 80,658,175,170,943,878,571,660,636,856,403,766, \\\\ & \\qquad\\qquad 975,289,505,440,883,277,824,000,000,000,000 \\\\ & \\doteq 8.066\\times10^{67}\\; \\text{.} \\end{array}$$ Having sixty-eight digits, such large numbers readily justify exponential notation! Still more generally, we might ask for the number of ways to order $k$ of $n$ items (demanding, of course, that $k \\le n\\,$ ). Such an ordering will be thought of as a $k\\,$-step process. There are $n$ possible outcomes for the first step, $\\, n-1$ for the second step, and so on until there are $n-k+1$ possible outcomes for the $k\\,$-step. Thus the number of possible orderings for $k$ of $n$ items is $$n\\cdot (n-1)\\cdot (n-2)\\cdot \\dots \\cdot (n-k+1) =\\frac{n!}{(n-k)!}\\; \\text{.}$$ We will denote this by $$P_k^n =\\frac{n!}{(n-k)!}\\, \\text{,}$$ although it is worth noting that this notation is by no means universally used. Thus, for example, the number of ways we can be dealt five cards in order from a standard fair deck is $$P_5^{52} =\\frac{52!}{(52-5)!} =52\\cdot 51\\cdot 50\\cdot 49\\cdot 48 =311,875,200\\, \\text{,}$$ i.e. a bit", "# Product of 2-digit numbers 31 people are dancing in a circle formation. Each people’s age is a 2-digit number, and for each digit we know that the units’ digit is equal to the tens’ digit of the person who is in the clockwise position, while the tens’ digit is equal to the units’ digit of the person who is in the counter-clockwise position. Any 2 neighboring digits are different. Check if the product of all 31 ages can be a perfect square. I don’t know how to start :( All I can tell is that the digits appear in pairs: If we have, for example 31 digits, $d_1, d_2, \\ldots, d_{31}$, then the ages are: $$10 \\cdot d_1+d_2, 10 \\cdot d_2+d_3, \\ldots, 10 \\cdot d_{31}+d_1$$ and obviously the product is $(10 \\cdot d_1+d_2) \\cdot (10 \\cdot d_2+d_3) \\cdots (10 \\cdot d_{31}+d_1)$. But I don’t know how to continue. • Does \"Any 2 neighboring digits are different\" mean that $d_i\\neq d_{i+1}$ or that $10d_{i-1} + d_i\\neq 10d_i + d_{i+1}$? – Arthur Jan 9 '18 at 13:11 • @Arthur: The first one. Maybe I have not expressed it correctly. – Eduardo Juan Ramirez Jan 9 '18 at 13:49 • It basically just means that no one has an age divisible by $11$, then. That's a different way of saying", "ninety • decade symbols: 10, 20, 30, 40, 50, 60, 70, 80, 90 • a \"ten\" is a bundle of 10 ones • decade concept: 20 is two tens, 30 is three tens, etc. • teens • teen word names: eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, nineteen • teen symbols: 11, 12, 13, 14, 15, 16, 17, 18, 19 • compose teen numbers: e.g., $10+7=17$ • decompose teen numbers: e.g., $17=10+7$ • two-digit numbers • word names: e.g., 23 is \"twenty-three\" • a two-digit number represents tens plus ones; e.g., 23 is 2 tens plus 3 ones • expanded form: e.g., $23 = 2\\cdot 10 + 3\\cdot 1$ • Given two different two-digit numbers: if the tens digits are different, then the number with more tens is greater; if the tens digits are the same, then the number with more ones is greater. • three-digit numbers • a \"hundred\" is a bundle of 10 tens • word names: e.g., 234 is \"two hundred thirty-four\" • a three-digit number represents hundreds plus tens plus ones; e.g., 234 is 2 hundreds plus 3 tens plus 4 ones • expanded form: $234=2\\cdot 100 + 3\\cdot 10 + 4\\cdot 1$ • Compare three-digit numbers by first comparing their hundreds digits. If the hundreds digits are different, then the number with", "# A number is composed of 4 thousands and 21 tens. What's the number? Jan 6, 2018 4210 #### Explanation: A thousand =1000 Since there are 4 thousands, we multiply it by 4 to get $4000$ $4 \\cdot 1000 = 4000$ tens = 10 Since there are 21 tens, we multiply to get the total amount so: $21 \\cdot 10 = 210$ $4000 + 210 = 4210$", "8 and 3? • there are no common factors between the numbers, so common multiple is not defined. • 24,48, cdots • 24,48, cdots The answer is \"24,48, cdots\" slide-show version coming soon", "+0 Math 0 105 1 +182 Given that a and b are real numbers such that -3 ≤ a ≤ 1 and -2 ≤ b ≤ 4, and values for a and b are chosen at random, what is the probability that product a * B is positive? wiskdls Apr 21, 2018 #1 +963 +2 Alright, here we go: For two number's product to be positive, those numbers must be both positive or both negative, but cannot be zero. This is because zero is neither positive nor negative, and 0 * n = 0. The first step is to find the total number of combination: (a,b). This can be done by: $$(1-(-3)+1)\\cdot(4-(-2)+1) =5\\cdot7=35.$$ There are a total of 35 different combinations. Out of those combinations there are: ( -3 , -2 ) ; ( -3 , -1 ) ; ( -2 , -2 ) ; ( -2 , -1 ) ; ( -1 , -2 ) ; ( -1 , -1 ) ( 1 , 1 ) ; ( 1 , 2 ) ; ( 1 , 3 ) ; ( 1 , 4 ) 10 that work. The probabilty of this happening is: $$\\boxed{\\frac{10}{35}=\\frac{2}{7}}$$ I hope this helped, Gavin. GYanggg Apr 21, 2018 #1 +963 +2 Alright, here we go: For two number's product to be positive,", "For instance, given a $100$-digit number at random, the probability that it doesn't contain a $9$ is approximately $\\frac{8\\cdot 9^{99}}{10^{100}}\\approx.002\\%.$ I hope this helps your intuition.", "that no pair of odd composites add to $38$. Therefore, $\\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.", "# probability Printable View • Sep 9th 2013, 04:34 AM Trefoil2727 probability Find how many distinct number greater than 5000 and divisible by 3 can be formed from digits 3,4,5,6,and 0. Each digit being used at most once in any number. • Sep 9th 2013, 05:39 PM Soroban Re: probability Hello, Trefoil2727! Quote: Find how many distinct numbers greater than 5000 and divisible by 3 can be formed from digits 3,4,5,6,and 0, each digit being used at most once in any number. A number is divisible by 3 if its sum of digits is divisible by 3. Four-digit numbers The number begins with 5: . $5\\,\\_\\,\\_\\,\\_$ The other 3 digits can be: . $\\{0,3,4\\},\\:\\{0,4,6\\},\\:\\{3,4,6\\}$ . . Each has $3!$ permutations. Hence, there are: $3\\cdot3! \\,=\\,18$ four-digit numbers that begin with 5. The number begins with 6: . $6\\,\\_\\,\\_\\,\\_$ The other 3 digits can be: . $\\{0,4,5\\},\\:\\{3,4,5\\}$ . . Each has $3!$ permutations. Hence, there are: $2\\cdot3! \\,=\\,12$ four-digit numbers that begin with 6. There are: $18 + 12 \\,=\\,{\\color{blue}30}$ four-digit numbers. Five-digit numbers The digits $\\{0,3,4,5,6\\}$ add up to 18, a multiple of 3. Hence, any five-digit number will be divisible by 3. There are $4$ choices for the first digit. The other 4 digits can be permuted in $4!$ ways. Hence, there are: $4\\cdot4! \\,=\\,{\\color{blue}96}$ five-digit numbers." ]

[ "have two choices for the units digit. For each such choice, you can use any of the other four digits as the tens digit. Therefore, there are $4 \\cdot 2 = 8$ two-digit even numbers that can be formed using only the digits $3, 4, 5, 6, 7$.", "# Overpowering tens unit Discrete Mathematics Level 1 How many 2 digit numbers are there, such that the units digit is strictly smaller than the tens digit? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "of the digits of each $x$ plus $2+0+2+0=4$. Hence the sum of all of the digits of the numbers in $S$ is $12 \\cdot 4 +45=\\boxed{093}$.", ". . $\\vdots$ $\\text{ten 9's}\\begin{Bmatrix}9&0 \\\\ 9&1 \\\\ 9& 2 \\\\ \\vdots \\\\ 9&9 \\end{Bmatrix}\\;\\;\\text{0 to 9 = 45}$ In the units-column, we have the sum of the digits 0 to 9 (45) ... nine times. Sum of the units digits: . $9 \\times 45 \\:=\\:405$ In the tens-column, we have: . $\\text{ten 1's + ten 2's + ten 3's + . . . + ten 9's}$ . . $\\;=\\;10(1) + 10(2) + 10(3) + \\cdots + 10(9) \\;=\\;10(1 + 2 + 3 +\\cdots + 9) \\;=\\;10(45)$ Sum of the tens digits: . $450$ Therefore, the sum of the digits is: . $405 + 450 \\;=\\;{\\color{blue}855}$ 12. Originally Posted by Chris L T521 $S_n=\\frac{n}{2}(a+z)$", "# A number is composed of 4 thousands and 21 tens. What's the number? Jan 6, 2018 4210 #### Explanation: A thousand =1000 Since there are 4 thousands, we multiply it by 4 to get $4000$ $4 \\cdot 1000 = 4000$ tens = 10 Since there are 21 tens, we multiply to get the total amount so: $21 \\cdot 10 = 210$ $4000 + 210 = 4210$", "the even digit that occupies both the tens and units places. Hence, $|A_3| = 9 \\cdot 10 \\cdot 5 = 450$. $|A_1 \\cap A_2|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, and tens places. There are ten ways to fill the units place. Hence, $|A_1 \\cap A_2| = 4 \\cdot 10 = 40$. $|A_1 \\cap A_3|$: There are four ways to choose the even digit that occupies both the thousands and hundreds places and five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_1 \\cap A_3| = 4 \\cdot 5 = 20$. $|A_2 \\cap A_3|$: There are nine ways to fill the thousands place. There are five ways to choose the even digit that occupies the hundreds, tens, and units places. Hence, $|A_2 \\cap A_3| = 9 \\cdot 5 = 45$. $|A_1 \\cap A_2 \\cap A_3|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, tens, and units places. Hence, $|A_1 \\cap A_2 \\cap A_3| = 4$. Therefore, the number of four-digit positive integers that do contain consecutive even equal digits is \\begin{align*} |A_1 \\cup A_2 \\cup A_3| & = |A_1| + |A_2| + |A_3| - |A_1 \\cap A_2|", "SEARCH HOME Math Central Quandaries & Queries Question from Tom, a student: I need to open a lock box from my parent's estate. The combination is a simple 4 number lock. The last two numbers are 33. Numbers can be used more than once.What are the possible combinations? WP Hi Tom, If the last two digits are 33 then you only need tofind the first two. There are 10 possibilities for the first digit and 10 for the second and hence there are $10 \\times 10 = 100$ possibilities for the first two digits. You can even list them. They are $00, 01, 02, \\cdot \\cdot \\cdot 09, 10,11, \\cdot \\cdot \\cdot, 19, 20, 21, \\cdot \\cdot \\cdot, 98, 99$ Harley Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "means there are 13 \"a\" s and b cannot be 20) When $21\\leq a \\leq 28$, there are $1+2+\\cdots+8=36$ ways. In all, there are $192+36=\\boxed{228}$ possible sequences. ~bluesoul", "tens place, and two with $z$ in the tens place. Therefore they sum to $2\\cdot 10(x+y+z)+2\\cdot(x+y+z)=22(x+y+z)$. So when you divide you are left with $22$. - You can trace it backwards via $$22(a+b+c) = (10a+ b) + (10a + c) + (10b +a) + (10b + c) + (10c + a) + (10c + b)$$ - Simplifying the right hand side gives identity. Bravo! – Hassan Muhammad Oct 30 '11 at 18:48", "Mock AIME 1 Pre 2005 Problems/Problem 1 Problem Let $S$ denote the sum of all of the three digit positive integers with three distinct digits. Compute the remainder when $S$ is divided by $1000$. Solution We find the sum of all possible hundreds digits, then tens digits, then units digits. Every one of $\\{1,2,3,4,5,6,7,8,9\\}$ may appear as the hundreds digit, and there are $9 \\cdot 8 = 72$ choices for the tens and units digits. Thus the sum of the hundreds places is $(1+2+3+\\cdots+9)(72) \\times 100 = 45 \\cdot 72 \\cdot 100 = 324000$. Every one of $\\{0,1,2,3,4,5,6,7,8,9\\}$ may appear as the tens digit; however, since $0$ does not contribute to this sum, we can ignore it. Then there are $8$ choices left for the hundreds digit, and $8$ choices afterwards for the units digit (since the units digit may also be $0$). Thus, the the sum of the tens digit gives $45 \\cdot 64 \\cdot 10 = 28800$. The same argument applies to the units digit, and the sum of them is $45 \\cdot 64 \\cdot 1 = 2880$. Then $S = 324000+28800+2880 = 355\\boxed{680}$.", "# Identifying Place Values The system that we use to represent numbers is wonderfully efficient and simple. Large numbers can be represented with very few symbols in a neat and organized way. The method rests on two concepts, base and place value, which are briefly discussed in the next few paragraphs. The concepts are introduced using base three for simplicity. Then, the concepts are extended to our base ten number system. The concept of base is best represented by grouping. Imagine a factory where objects are being packaged for shipment around the world. Workers are instructed to ‘bundle things up’ every time they get a group of three. Here are the names for the packaging that they use: • a set of three objects is called a packet • a set of three packets (which is $\\,3\\cdot 3 = 9\\,$ objects) is called a box • a set of three boxes (which is $\\,3\\cdot 9 = 27\\,$ objects) is called a carton Notice that a packet holds $\\,3^1 = 3\\,$ objects; a box holds $\\,3^2 = 9\\,$ objects; and a carton holds $\\,3^3 = 27\\,$ objects. Suppose a shipment of $\\,46\\,$ objects is to go out to Lenox, Massachusetts. How would these $\\,46\\,$ objects get bundled? First, bundle up one carton, leaving $\\,46 - 27 = 19\\,$ objects", "for the ones digit (4 or 6). Therefore, there will only be five choices for the tens. Hence, there will be 152 or 10 numbers. For the second case since the first digit is 3, there will be three choices for the ones digit (2, 4, or 6), and since two digits have been chosen for the hundreds (3) and ones (2,4, or 6), there will only be five choices for the tens. Therefor there will be 15*3 or 15 numbers. 4) All in all there will be 3+18+10+15 or 46 numbers. If you list down the numbers they are: 2,4, and 6 for a single-digit number. 24, 26, 32, 34, 36, 42, 46, 52, 54, 56, 62, 64, 72, 74, 76, 92, 94, and 96 for the two-digit numbers. 234, 236, 246, 254, 256, 264, 274, 276, 294, and 296 that starts with 2. 324, 326, 342, 346, 352, 354, 356, 362, 364, 372, 374, 376, 392, 394, and 396 that starts with three. · 2 years, 11 months ago Sorry, some typo errors. 63 is supposed to be 63, 152 is supposed to be 152, and 153 should be 153. Thanks! · 2 years, 11 months ago Ooops! Again! 63 is 6 times 3, 152 is 1 times 5 times 2. and 15*3 is 1 times", "+0 # Number Theory 0 208 1 N = 1991*1993*1995*1997*1999*2001*2003. What is the sum of the hundreds, tens and units digits of N? Jul 4, 2021 #1 +497 0 Since we need to compute the last 3 digits, we can take the whole expression mod 1000: $$1991\\cdot1993\\cdot1995\\cdot1997\\cdot1999\\cdot2001\\cdot2003 (\\text{mod 1000})\\\\ =-9\\cdot-7\\cdot-5\\cdot-3\\cdot-1\\cdot1\\cdot3 (\\text{mod 1000})\\\\ =-2835 (\\text{mod 1000})\\\\ =165(\\text{mod 1000})$$ The sum of the hundreds, tens, and units digit is $$1+6+5=\\boxed{12}$$ Jul 4, 2021", "digit), eight choices for the hundreds digit (which cannot be equal to the thousands digit or the units digit), and seven choices for the tens digit (which cannot be equal to the thousands digit, hundreds digit, or units digit). By the Inclusion-Exclusion Principle, there are $$9 \\cdot 9 \\cdot 8 \\cdot 7 + 9 \\cdot 10 \\cdot 10 \\cdot 5 - 8 \\cdot 8 \\cdot 7 \\cdot 5 = 4536 + 4500 - 2240 = 6796$$ four-digit positive integers in which no digit is repeated or are odd numbers that contain repeated digits. How many four-digit positive integers are there in which no digit is repeated or are numbers in which all the digits are odd and repeated digits may appear? This is your second interpretation. You correctly calculated that there are $$9 \\cdot 8 \\cdot 7 \\cdot 6 + 5^4 - 5 \\cdot 4 \\cdot 3 \\cdot 2 = 4536 + 625 - 120 = 5041$$ such numbers. We can confirm this by calculating the number of four-digit positive integers consisting of only odd digits that contain a repeated digit. This is a bit messy. We consider cases: Exactly one digit is used in all four positions: There are $5$ such numbers since there are $5$ odd digits. Exactly two digits are used, with one used three", "# Product of 2-digit numbers 31 people are dancing in a circle formation. Each people’s age is a 2-digit number, and for each digit we know that the units’ digit is equal to the tens’ digit of the person who is in the clockwise position, while the tens’ digit is equal to the units’ digit of the person who is in the counter-clockwise position. Any 2 neighboring digits are different. Check if the product of all 31 ages can be a perfect square. I don’t know how to start :( All I can tell is that the digits appear in pairs: If we have, for example 31 digits, $d_1, d_2, \\ldots, d_{31}$, then the ages are: $$10 \\cdot d_1+d_2, 10 \\cdot d_2+d_3, \\ldots, 10 \\cdot d_{31}+d_1$$ and obviously the product is $(10 \\cdot d_1+d_2) \\cdot (10 \\cdot d_2+d_3) \\cdots (10 \\cdot d_{31}+d_1)$. But I don’t know how to continue. • Does \"Any 2 neighboring digits are different\" mean that $d_i\\neq d_{i+1}$ or that $10d_{i-1} + d_i\\neq 10d_i + d_{i+1}$? – Arthur Jan 9 '18 at 13:11 • @Arthur: The first one. Maybe I have not expressed it correctly. – Eduardo Juan Ramirez Jan 9 '18 at 13:49 • It basically just means that no one has an age divisible by $11$, then. That's a different way of saying", "IDENTIFYING PLACE VALUES The system that we use to represent numbers is wonderfully efficient and simple. Large numbers can be represented with very few symbols in a neat and organized way. The method rests on two concepts, base and place value, which are briefly discussed in the next few paragraphs. The concepts are introduced using base three for simplicity. Then, the concepts are extended to our base ten number system. The concept of base is best represented by grouping. Imagine a factory where objects are being packaged for shipment around the world. Workers are instructed to “bundle things up” every time they get a group of three. Here are the names for the packaging that they use: • a set of three objects is called a packet • a set of three packets (which is $\\,3\\cdot 3 = 9\\,$ objects) is called a box • a set of three boxes (which is $\\,3\\cdot 9 = 27\\,$ objects) is called a carton Notice that a packet holds $\\,3^1 = 3\\,$ objects; a box holds $\\,3^2 = 9\\,$ objects; and a carton holds $\\,3^3 = 27\\,$ objects. Suppose a shipment of $\\,46\\,$ objects is to go out to Lenox, Massachusetts. How would these $\\,46\\,$ objects get bundled? First, bundle up one carton, leaving $\\,46 - 27 = 19\\,$ objects remaining.", "contains a zero. We consider cases: 1. There is a zero in the units place: This leaves us with nine choices for the tens place (any digit except zero) and eight choices for the hundreds place (any digit except zero and the digit in the tens place). Hence, there are $1 \\cdot 9 \\cdot 8 = 72$ such numbers. 2. There is a zero in the tens place: This leaves us with nine choices for the units digit (any digit except zero) and eight choices for the hundreds place (any digit except zero and the digit in the units place). Hence, there are $9 \\cdot 1 \\cdot 8 = 72$ such numbers. 3. There is not a zero in either the units place or tens place: This leaves with nine choices for the units place, eight choices for the tens place, and seven choices for the hundreds place. Hence, there are $9 \\cdot 8 \\cdot 7 = 504$ such numbers. Since these cases are mutually exclusive and exhaustive, there are $72 + 72 + 504 = 648$ three-digit positive integers with distinct digits, as we found above without doing tedious casework.", "has the properties that the sum of its digits is also a multiple of nine can be explained in exactly the same way as the explanation above goes. So assume some number is a multiple of nine: $X\\quad =\\quad 9x$ Now write X as a sum of its digits with each one times its corresponding power of ten: $X\\quad =\\quad \\sum _{ i=0 }^{ n }{ { a }_{ i } } { 10 }^{ i }$ Notice that we can write each term of the sum as a digit of the the original number times the quantity $\\left( 1+9\\cdots 9 \\right)$ where the symbol “$9\\cdots 9$” is 9 or 99 or 999 or 9999 and so forth. $X\\quad =\\quad { a }_{ 0 }+\\sum _{ i=1 }^{ n }{ { a }_{ i } } \\left( 1+9\\cdots 9 \\right)$ This way of writing the number makes it clear what happens when each digit is distributed over the its corresponding sum and the the sum regrouped so that the sum of the digits of the original number are distinct from the rest of the number. $X\\quad =\\quad { a }_{ 0 }+\\sum _{ i=1 }^{ n }{ { a }_{ i } } +\\sum _{ i=1 }^{ n }{ { a }_{ i } } \\left( 9\\cdots 9", "the number of legal combinations is $120-(6+6+6+6+6+6+6+6+6+6)=60$ • Hi check your second to last line. – bobbym Jun 30 '14 at 12:43 • @bobbym: Checked. What about them? – barak manos Jun 30 '14 at 12:45 • Typographical error: 24 + 36 = 60 – bobbym Jun 30 '14 at 12:47 • @bobbym: LOL, thanks – barak manos Jun 30 '14 at 12:48 When it is a 2 in the units place the tens digit can only be a 1. There are 6 ways to distribute the other 3 numbers. $3\\cdot 2\\cdot 1 \\cdot 1 \\cdot 1 = 6$ When it is a 3 in the units place the tens digit can be a 1 or 2. There are 6 ways to distribute the other 3 numbers.$3\\cdot 2\\cdot 1 \\cdot 2 \\cdot 1 =12$ When it is a 4 in the units place the tens digit can be a 1 or 2 or 3. There are 6 ways to distribute the other 3 numbers.$3\\cdot 2\\cdot 1 \\cdot 3 \\cdot 1 =18$ When it is a 5 in the units place the tens digit can be a 1 or 2 or 3 or 4. There are 6 ways to distribute the other 3 numbers.$3\\cdot 2\\cdot 1 \\cdot 4 \\cdot 1 =24$ $6 + 12 + 18 + 24 =", "+0 # Can I have help please? 0 369 1 How many numbers can you get by multiplying two or more distinct members of the set $$\\{1,2,3,5,11\\}$$ together? May 10, 2021 #1 +1 Use casework. First try finding the possiblities with only multiplying 2 numbers, and then the possibilities with multiplying 3 numbers, etc. Using 2 numbers: $$2\\cdot3, 2 \\cdot 5, 2 \\cdot 11, 3 \\cdot 5, 3 \\cdot 11, 5 \\cdot 11.$$ Using 3 numbers: $$2 \\cdot 3 \\cdot 5, 2 \\cdot 3 \\cdot 11, 2 \\cdot 5 \\cdot 11, 3 \\cdot 5 \\cdot 11.$$ Using 4 numbers: $$2 \\cdot 3 \\cdot 5 \\cdot 11.$$ So we have a total of $$6+4+4+1=\\boxed{15}$$ May 10, 2021" ]

[ "\"each\", F ) Now lets think about the long-distance runners. We can also list all the ways to divide the B long-distance runners into equally sized teams by finding the factors of B. The factors of B are toSentence( getFactors( B ) ) since those are all the numbers that divide evenly into B. That means we can divide the long-distance runners into equally sized teams in any of the following ways: plural( F, \"team\" ) with plural( B / F, \"long-distance runner\" ) plural( \"\", \"each\", F ) Since each team will have sprinters and long-distance runners, compare the numbers of teams the sprinters can be divided into and the numbers of teams the runners can be divided into to find the common divisors: plural( N + 1, \"team\" ) with plural( A / ( N + 1 ), \"sprinter\" ) plural( N + 1, \"team\" ) with plural( B / ( N + 1 ), \"long-distance runner\" ) The common divisors of A and B are toSentence( _.intersection( A_FACTORS, B_FACTORS ) ). In other words, with A sprinters and B long-distance runners, person(1) can make the following equal teams: plural( F, \"team\" ) with plural( A / F, \"sprinter\" ) and plural( B / F, \"long-distance runner\" ) We want to know the greatest number of", "11 12 Jan 2012, 07:59 50 In how many different ways can a group of 9 people be divide 14 26 Sep 2010, 09:47 32 In how many different ways can a group of 8 people be 10 13 Aug 2010, 08:38 15 In how many different ways can a group of 9 people be 12 29 Oct 2009, 05:30 Display posts from previous: Sort by", "basketball. In how many ways can this be done in each of the following cases: 1. The teams are distinguishable (for example, one team is labeled Alabama and the other team is labeled Auburn). 2. The teams are not distinguishable.", "• The talker. 1. In real life, the only actual grouped sprint in the world is the three-legged race. Ideally you’ll look like the 1-mile relay team handing off perfectly, but don’t worry if this ends up with your whole group face down in the mud, as long as you can explain how.", "each team with the job of creating a story told entirely via emojis. There would be 15 groups, each with 2 members. For large teams, put an even number of red and black cards in a shuffled stack. Much like the Arm Cross split, typically this will split the entire group in half. Creative and fun ways to form teams. ... Divide the total group into subgroups and position them at different points in the room. Giants, elves and magicians. Two players play against each other. These can be used in lots of ways. While the group is groaning in anxiety I then let them know their sole mate is the individual in the group whose sole of their shoe most closely matches the sole of your shoe. Source: Upcycled Education. Divide the group into teams and give them offbeat materials to build something creative. If several groups should be chosen, several rounds are possible. Choose half as many quotes as you have participants. Divide the group into two teams. Full House – Grab an old deck of playing cards. Hand keep them in that position. This field is for validation purposes and should be left unchanged. 3) Spades. Have 4-6 colors so then you can divide the group … In my last school, every classroom had a", "order to determine the ball. Based on that, I can use standard logic to figure out which group the different ball is in. Let's say that our groups are labeled A, B, C, and D, and ... 5 Otherwise, In any event, Top 50 recent answers are included", "In how many different ways? [#permalink] 09 Mar 2008, 06:53 $$C_8^2* C_6^2* C_4^2* C_2^2$$ = 2520 _________________ Stay Hungry, Stay Foolish Re: Maths: In how many different ways? [#permalink] 09 Mar 2008, 06:53 Similar topics Replies Last post Similar Topics: How many different ways can a group of 12 people be divided 4 12 Dec 2007, 05:28 In how many different ways can a group of 8 be divided into 7 09 Nov 2007, 10:49 How many ways are there to split 8 boys into 4 groups of 3 1 01 Nov 2007, 06:46 1 In how many ways can a group of 8 be divided into 4 teams of 9 22 Dec 2006, 21:37 In how many ways can 8 boys be divided into two sets 2 25 Nov 2005, 16:25 Display posts from previous: Sort by", "31 In how many different ways can a group of 8 people be 10 13 Aug 2010, 07:38 29 In how many ways 8 different tickets can be distributed 23 20 Nov 2009, 06:46 51 In how many different ways can a group of 8 people be divide 17 24 Oct 2009, 03:20 Display posts from previous: Sort by", "different ways can a group of 6 people be divided into 3 team 8 02 Jun 2016, 03:32 13 How many ways can you group 3 people from 4 sets of twins if no two 7 05 Oct 2014, 00:00 3 In how many ways can a group of 3 people be selected 5 09 Nov 2012, 09:59 8 In how many ways can you sit 8 people on a bench if 3 of 6 12 Mar 2012, 08:23 Display posts from previous: Sort by", "# Brothers 9 different things are to be divided among 4 brothers. So that the youngest get one more then the other three.Find the number of ways for which this condition is possible. ×", "21:32 johnrb wrote: Balvinder wrote: A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people? A. 420 B. 2520 C. 168 D. 90 E. 105 There's another topic asking this very question down below. Here's my answer: E) 105. Start with the question, how many ways can you divide a group of 4 into 2 teams. The answer is 3. Suppose our group is A B C D. Then the possible teams are: AB CD AC BD (It's 4C2/2, because determining one of the teams automatically determines the other one.) Now consider the case of a group of 6 divided into 3 teams. Our group: A B C D E F. A has to be on some team, and there are 5 possibilities: AB, AC, AD, AE, or AF. For each of these possibilities, the rest of the members form a group of 4 divided into 2 teams. So the overall result is 5*(4C2/2) = 15. Now we're up to our case. Group = A B C D E F G H. Again, A must be on a team--there are 7 possibilities. For each of those possibilities, there are 15 ways of dividing the remaining 6 members into 3 teams. So total possibilities", "A dist 25 05 Jun 2013, 13:52 5 A gym class can be divided into 8 teams with an equal number 9 03 Dec 2012, 04:20 8 For each of her sales, a saleswoman receives a commission eq 11 28 Oct 2006, 11:12 Display posts from previous: Sort by", "combine into groups of five. Read More »", "Explain. 6 x 8 = 48. Explanation: In the above-given question, given that, Six friends picked 48 grapefruits. They want to share them equally. 6 x 8 = 48. Visual Learning Bridge Essential Question How Many are in Each Group? A. Three friends have 12 toys to share equally. How many toys will each friend get? Think of arranging 12 toys into 3 equal groups. Division is an operation that is used to find how many equal groups there are or how many are in each group. B. What You Think Put one toy at a time in each group. When all the toys are grouped, there will be 4 in each group. C. What You Write You can write a division equation to find the number in each group. Each friend will get 4 toys. Convince Me! Be Precise What would happen if 3 friends wanted to share 13 toys equally? 13 / 3 = 4.3 Explanation: In the above-given question, given that, if 3 friends wanted to share 13 toys each. 13 / 3 = 4.3. nearly 4 toys will share 3 friends. Guided Practice Do You Understand? Question 1. Eighteen eggs are divided into 3 rows. How many eggs are in each row? Use the bar diagram to solve. The number of eggs in each", "How many different patrols can be created, if one patrol is 2 boys, 3 girls and 1 leader? • Tournament Determine how many ways can be chosen two representatives from 34 students to school tournament. • Four-member team There are 14 girls and 11 boys in the class. How many ways can a four-member team be chosen so that there are exactly two boys in it? • Pairs From the five girls and four boys teachers have to choose one pair of boy and girl. A) How many such pairs of (M + F)? B) How many pairs where only boys (M + M)? C) How many are all possible pairs? • Soccer teams Have to organize soccer teams. There are 3 age groups. How many different ways can you organize teams of ten for each age group? Is this a permutation or combination?", "how many ways can 16 different gifts be divided among four children [#permalink] 04 Oct 2017, 10:24 Display posts from previous: Sort by", "<img src=\"https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym\" style=\"display:none\" height=\"1\" width=\"1\" alt=\"\" /> You are viewing an older version of this Concept. Go to the latest version. # Greatest Common Factor Using Lists ## GCF is the product of identical prime factors Estimated4 minsto complete % Progress Practice Greatest Common Factor Using Lists Progress Estimated4 minsto complete % Greatest Common Factor Using Lists Have you ever been part of a big basketball tournament? Take a look at this dilemma. The sixth grade teachers have decided to have a big basketball tournament as part of the sixth grade social. The sixth graders in clusters 6A and 6B love basketball, and when the agenda is announced, all of the students are very excited. The biggest question is how many teams to divide the students from each cluster into. The teachers want to have the same number of teams, otherwise it will be difficult to have even games for a tournament. Cluster 6A has 48 students in it. Cluster 6B has 44 students in it. The teachers pose the dilemma to the students and Maria volunteers to figure out the teams. She needs to figure out how many teams to divide each cluster into and how many students will then be on each team. Maria has an idea how to do it. She knows that factors are", "to find groups where there is a common factor that once taken out gives the same inside each bracket. Sometimes might take a little trial and error. Perseverance will be your friend.", "on separate teams. In how many ways is this possible? Call the friends $A,B,C.$ Place them in three different teams . . . it doesn't matter which teams. The teams look like this: . . $\\{A\\:\\_\\:\\_\\:\\_\\}\\quad \\{B\\:\\_\\:\\_\\:\\_\\}\\quad \\{C\\:\\_\\:\\_\\:\\_\\}\\quad \\{\\_\\:\\_\\:\\_\\:\\_\\} \\quad \\{\\_\\:\\_\\:\\_\\:\\_\\}$ The other 17 players will be partitioned into: 3, 3, 3, 4, 4. Hence, there are: . ${17\\choose3,3,3,4,4} / 3!2!$ .ways the friends are separated. Therefore, the desired number is: . ${20\\choose4,4,4,4,4}/5! - {17\\choose3,3,3,4,4}/3!2!$", "a college, 50 students play soccer, 40 students [#permalink] 14 Apr 2019, 19:20 Display posts from previous: Sort by" ]

[ "View Single Post The amount per friend is 6/n, where \"n\" is the number of friends. So yes that's $12 per friend when n=1/2, so your 1/2 a friend gets 12 * 1/2 =$6, (amount per friend times number of friends).", "18 views A school has less than $5000$ students and if the students are divided equally into teams of either $9$ or $10$ or $12$ or $25$ each, exactly $4$ are always left out. However, if they are divided into teams of $11$ each, no one is left out. The maximum number of teams of $12$ each that can be formed out of the students in the school is 1 vote 1 1 vote", "on separate teams. In how many ways is this possible? Call the friends $A,B,C.$ Place them in three different teams . . . it doesn't matter which teams. The teams look like this: . . $\\{A\\:\\_\\:\\_\\:\\_\\}\\quad \\{B\\:\\_\\:\\_\\:\\_\\}\\quad \\{C\\:\\_\\:\\_\\:\\_\\}\\quad \\{\\_\\:\\_\\:\\_\\:\\_\\} \\quad \\{\\_\\:\\_\\:\\_\\:\\_\\}$ The other 17 players will be partitioned into: 3, 3, 3, 4, 4. Hence, there are: . ${17\\choose3,3,3,4,4} / 3!2!$ .ways the friends are separated. Therefore, the desired number is: . ${20\\choose4,4,4,4,4}/5! - {17\\choose3,3,3,4,4}/3!2!$", "it will take as many throws in the air as the number of ways to sort 6 friends, that is $6! = 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1 = 720$. But don’t underestimate Dr. Cooper’s circle. In the pilot of the show, he claims to have 212 friends on MySpace (and it doesn’t even include Penny and Amy)! Sorting 212 friends will take a while. With the bogosort, it will take in means about 10281 throws in the air, which cannot be done in billions of years even with very powerful computers… Obviously, he’ll need to do it smartly. Can’t he just take his most favorite friend that hasn’t refused yet until finding one? Yes he can. At the first iteration, he’d have to look through the 212 friends. Then, if the favorite friend refuses, through the other 211 friends. Then, if the second favorite friend refuses, through the other 210 friends… and so on. In the worst case where all his friends refuse (which is a scenario Sheldon really should consider…), he’ll have to do $212+211+210+…+1$ basic operations. That is equal to $212*211/2 = 22,472$. Even for Dr. Cooper, this will take a while. What he’s done is almost the selection sort algorithm. This algorithm uses two lists, one called the remaining", "a single $6$-cycle. Case: Two $3$-cycles • The first friend has $5$ possible friends to choose. • The friend to whom the book is given has $4$ possible choices. • The next friend has just one choice - to give their book to the first friend. • The first of the $3$ remaining friends has $2$ choices. • The next friend has $1$ choice, to give their book to the last remaining friend. • The last friend gives their book to the first friend of their $3$-cycle. So there are $5 \\cdot 4 \\cdot 2 = 40$ possible choices of two $3$-cycles.", "block, and the three girls together are the fourth block. Once you’ve done that, you can arrange the $3$ girls in $3!$ different orders within their block. Thus, for each arrangement of the blocks you get $3!$ arrangements of the people, for a total of $4!3!$ arrangements of the people.", "$100$ people are at a party, and each is friends with at most $10$ of the other guests. Determine the greatest $n$ for which there must exist a set of $n$ guests such that no two of them are friends.", "5 rooms. Observe that placing the people is the same as giving each of them a sign (or a key) with the room number, when the people are standing in a prior chosen order. - Are the balls distinguishable? If so, we have 5^7, as each ball has the option of going into any box. If the balls are not distinguishable, then we can use hockey-stick, using four placeable dividers- we have 8C4. - If each box had to have at least $1$ ball, then you put a ball in all the boxes first, then see how many ways you can put $2$ balls in $5$ boxes, which is $5^2$.", "to divide eight people into two teams of four.", "have the same number of items. Question 1. While at the beach, three friends found 6, 1, and 5 seashells. The friends agreed to divide the shells evenly. How many shells should each person get? Each person should get 4 seashells. Explanation: Three friends found 6, 1, and 5 seashells. Total 12 seashells. And these are divided among three friends. Therefore latex]\\frac{12}{3}[/latex] = 4 seashells Each person should get 4 seashells. A. Look at the model shown. For each friend to have the same number of shells, or a fair share of shells, what needs to happen? Each friend should have 5 shells. Explanation: B. Explain how you could rearrange the shells in Part A so that each friend has the same number of shells. Then sketch the fair shares. Explanation: C. Suppose the three friends had found 9, 1, and 5 seashells. How many shells would each person get? Explain your reasoning Each person would get 5 seashells. Explanation: Three friends found 9, 1, and 5 seashells. Total of 15 seashells. These shells should be divided among three friends. latex]\\frac{15}{3}[/latex] 5 seashells. Turn and Talk Describe another way to find the fair share of shells each friend should have in Part C. Question 2. How can you find the balance point of a data set? A. Look", "Put all the men in a bag. Then we have $4$ \"people,\" who can be arranged in $4!$ ways. Let the men out of the bag. They can arrange themselves in $5!$ ways for a total of $(4!)(5!)$. (5) We need more bags, $4$ of them. Put each couple in a bag. The bags can be arranged in $4!$ ways. Let the people out of the bags. The contents of each bag can arrange themselves in either order, for a total of $(4!)(2^4)$.", "# Split Them Up There are $$10$$ students in a classroom. The teacher wants to divide the kids into two groups of $$3$$ and one group of $$4$$. How many ways can the teacher divide the students if the two groups are interchangeable? ×", "a payment of $55\\,$ dollars for each player (note that the value of $110\\,$ reported in the OP must be halved because each player pays $2.5$ dollars for each game). For larger groups, simply apply the same method by counting the possible pairs of $n$ elements and considering that, if each friend plays one game each week, there are $n/2\\,$ games a week.", "View Single Post P: 2,751 Quote by cheesemonkey You have 6 dollars. You want to split the money amongst half a friend. Yeah, I know, it's gross, but let's just say you're a creepy serial killer guy. How much does the half-friend get? 6/0.5 is 12, but the half-friend can't possibly get 12 dollars; you don't even have that much. WRONG. The amount per friend is 6/n, where \"n\" is the number of friends. So yes that's $12 per friend when n=1/2, so your 1/2 a friend gets 12 * 1/2 =$6, (amount per friend times number of friends). So ummm, what's the problem?", "Explain. _______ buttons Answer: 1 button Explanation: After preparing 17 puppets there were 2 buttons leftover then on the addition of 1 button gives 3 buttons which can be used to prepare another puppet. Question 10. A total of 56 students signed up to play in a flag football league. If each team has 10 students, how many more students will need to sign up so all of the students can be on a team? _______ students Answer: 4 students Explanation: Total number of students in the football league= 56 Number of students in each group= 10 then, Number of groups= Quotient of 56÷10=5 groups Remainder= 6 By the addition of 4 students, the group of 6 gets completed by 10 Therefore, 4 students should be added so that all students can be on a team. Question 11. A teacher plans for groups of her students to eat lunch at tables. She has 34 students in her class. Each group will have 7 students. How many tables will she need? Explain how to use the quotient and remainder to answer the question. _______ tables Answer: She needs 3 tables Explanation: Quotient: A. Use 34 counters to represent the 34 dominoes. Then draw 7 circles to represent the divisor. B. Share the counters equally among the 7 groups by", "## Mackenzie2013 one year ago Eight two-person teams are participating in a three-legged race. Find the number of orders in which all 8 teams can finish the race. 1. anonymous $8!$ 2. anonymous 8 choices to come in first 7 for secone 6 for third 5 for fourth 4 for fifth 3 for sixth 2 for seventh then 1 for eighth by the counting principle, multiply the choices together, get $8!=8\\times 7\\times 6\\times 5\\times 4\\times 3\\times 2\\times 1$ 3. Mackenzie2013 Thank you!", "of team members), the extra penny (or two) will be randomly given to a member (that is, if dividing$10 amongst 3 team members, two will get $3.33 and one will get$3.34). PARTNERS", "# A question on counting and probability: Six friends hold a Christmas present swap. Each person brings a present, and puts it into a sack… [duplicate] Question: Six friends hold a Christmas present swap. Each person brings a present, and puts it into a sack. Once all six presents are in the sack, each participant in turn then draws out a present at random (presents are not replaced once drawn). The swap is deemed lucky if no participant draws out the present that he or she put in. What is the probability that the swap is lucky? Where I am at so far: Let A B C D E F be the 6 people. Let $$A_P$$ $$B_P$$ $$C_P$$ $$D_P$$ $$E_P$$ $$F_P$$ be the 6 gifts. The set {A, B, C, D, E, F} has 6! permutations. Now each permutation gives the order in which order people choose presents from the sack. Let's call the set of all these permutations R. There are also 6! permutations of {$$A_P$$ , $$B_P$$ , $$C_P$$ , $$D_P$$ , $$E_P$$ , $$F_P$$}. Each one of these permutations give the order in which present are chosen from the sack. Let's call the set of all these permutations S. R x S gives the set of all possible ways the 6 people choose presents. For", "• question_answer There are 10 friends and 55 bananas. If the bananas are divided equally among the students, how many does each friend get? A) Q = 5, R = 3 B) Q = 5, R = 2 C) Q = 5, R = 5 D) Q =1, R =5 Solution : Each friend got $=55\\div 10$ Q = 5, R = 5 Each friend got 5 bananas & 5 bananas were left over. You need to login to perform this action. You will be redirected in 3 sec", "Explain. 6 x 8 = 48. Explanation: In the above-given question, given that, Six friends picked 48 grapefruits. They want to share them equally. 6 x 8 = 48. Visual Learning Bridge Essential Question How Many are in Each Group? A. Three friends have 12 toys to share equally. How many toys will each friend get? Think of arranging 12 toys into 3 equal groups. Division is an operation that is used to find how many equal groups there are or how many are in each group. B. What You Think Put one toy at a time in each group. When all the toys are grouped, there will be 4 in each group. C. What You Write You can write a division equation to find the number in each group. Each friend will get 4 toys. Convince Me! Be Precise What would happen if 3 friends wanted to share 13 toys equally? 13 / 3 = 4.3 Explanation: In the above-given question, given that, if 3 friends wanted to share 13 toys each. 13 / 3 = 4.3. nearly 4 toys will share 3 friends. Guided Practice Do You Understand? Question 1. Eighteen eggs are divided into 3 rows. How many eggs are in each row? Use the bar diagram to solve. The number of eggs in each" ]

[ "on separate teams. In how many ways is this possible? Call the friends $A,B,C.$ Place them in three different teams . . . it doesn't matter which teams. The teams look like this: . . $\\{A\\:\\_\\:\\_\\:\\_\\}\\quad \\{B\\:\\_\\:\\_\\:\\_\\}\\quad \\{C\\:\\_\\:\\_\\:\\_\\}\\quad \\{\\_\\:\\_\\:\\_\\:\\_\\} \\quad \\{\\_\\:\\_\\:\\_\\:\\_\\}$ The other 17 players will be partitioned into: 3, 3, 3, 4, 4. Hence, there are: . ${17\\choose3,3,3,4,4} / 3!2!$ .ways the friends are separated. Therefore, the desired number is: . ${20\\choose4,4,4,4,4}/5! - {17\\choose3,3,3,4,4}/3!2!$", "21:32 johnrb wrote: Balvinder wrote: A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people? A. 420 B. 2520 C. 168 D. 90 E. 105 There's another topic asking this very question down below. Here's my answer: E) 105. Start with the question, how many ways can you divide a group of 4 into 2 teams. The answer is 3. Suppose our group is A B C D. Then the possible teams are: AB CD AC BD (It's 4C2/2, because determining one of the teams automatically determines the other one.) Now consider the case of a group of 6 divided into 3 teams. Our group: A B C D E F. A has to be on some team, and there are 5 possibilities: AB, AC, AD, AE, or AF. For each of these possibilities, the rest of the members form a group of 4 divided into 2 teams. So the overall result is 5*(4C2/2) = 15. Now we're up to our case. Group = A B C D E F G H. Again, A must be on a team--there are 7 possibilities. For each of those possibilities, there are 15 ways of dividing the remaining 6 members into 3 teams. So total possibilities", "In how many different ways? [#permalink] 09 Mar 2008, 06:53 $$C_8^2* C_6^2* C_4^2* C_2^2$$ = 2520 _________________ Stay Hungry, Stay Foolish Re: Maths: In how many different ways? [#permalink] 09 Mar 2008, 06:53 Similar topics Replies Last post Similar Topics: How many different ways can a group of 12 people be divided 4 12 Dec 2007, 05:28 In how many different ways can a group of 8 be divided into 7 09 Nov 2007, 10:49 How many ways are there to split 8 boys into 4 groups of 3 1 01 Nov 2007, 06:46 1 In how many ways can a group of 8 be divided into 4 teams of 9 22 Dec 2006, 21:37 In how many ways can 8 boys be divided into two sets 2 25 Nov 2005, 16:25 Display posts from previous: Sort by", "each team with the job of creating a story told entirely via emojis. There would be 15 groups, each with 2 members. For large teams, put an even number of red and black cards in a shuffled stack. Much like the Arm Cross split, typically this will split the entire group in half. Creative and fun ways to form teams. ... Divide the total group into subgroups and position them at different points in the room. Giants, elves and magicians. Two players play against each other. These can be used in lots of ways. While the group is groaning in anxiety I then let them know their sole mate is the individual in the group whose sole of their shoe most closely matches the sole of your shoe. Source: Upcycled Education. Divide the group into teams and give them offbeat materials to build something creative. If several groups should be chosen, several rounds are possible. Choose half as many quotes as you have participants. Divide the group into two teams. Full House – Grab an old deck of playing cards. Hand keep them in that position. This field is for validation purposes and should be left unchanged. 3) Spades. Have 4-6 colors so then you can divide the group … In my last school, every classroom had a", "have the same number of items. Question 1. While at the beach, three friends found 6, 1, and 5 seashells. The friends agreed to divide the shells evenly. How many shells should each person get? Each person should get 4 seashells. Explanation: Three friends found 6, 1, and 5 seashells. Total 12 seashells. And these are divided among three friends. Therefore latex]\\frac{12}{3}[/latex] = 4 seashells Each person should get 4 seashells. A. Look at the model shown. For each friend to have the same number of shells, or a fair share of shells, what needs to happen? Each friend should have 5 shells. Explanation: B. Explain how you could rearrange the shells in Part A so that each friend has the same number of shells. Then sketch the fair shares. Explanation: C. Suppose the three friends had found 9, 1, and 5 seashells. How many shells would each person get? Explain your reasoning Each person would get 5 seashells. Explanation: Three friends found 9, 1, and 5 seashells. Total of 15 seashells. These shells should be divided among three friends. latex]\\frac{15}{3}[/latex] 5 seashells. Turn and Talk Describe another way to find the fair share of shells each friend should have in Part C. Question 2. How can you find the balance point of a data set? A. Look", "that every group has the same amount of money on average (often, as here, you have to put everyone in the same group to achieve this). Move money within each group separately; for a group of $k$ people this takes $k-1$ moves. Overall, you use $n-g$ moves, where $g$ is the number of groups. For example 1 you can divide into two such groups - {A,C} and {B} - but for example 2 you can't.", "Explain. 6 x 8 = 48. Explanation: In the above-given question, given that, Six friends picked 48 grapefruits. They want to share them equally. 6 x 8 = 48. Visual Learning Bridge Essential Question How Many are in Each Group? A. Three friends have 12 toys to share equally. How many toys will each friend get? Think of arranging 12 toys into 3 equal groups. Division is an operation that is used to find how many equal groups there are or how many are in each group. B. What You Think Put one toy at a time in each group. When all the toys are grouped, there will be 4 in each group. C. What You Write You can write a division equation to find the number in each group. Each friend will get 4 toys. Convince Me! Be Precise What would happen if 3 friends wanted to share 13 toys equally? 13 / 3 = 4.3 Explanation: In the above-given question, given that, if 3 friends wanted to share 13 toys each. 13 / 3 = 4.3. nearly 4 toys will share 3 friends. Guided Practice Do You Understand? Question 1. Eighteen eggs are divided into 3 rows. How many eggs are in each row? Use the bar diagram to solve. The number of eggs in each", "18 views A school has less than $5000$ students and if the students are divided equally into teams of either $9$ or $10$ or $12$ or $25$ each, exactly $4$ are always left out. However, if they are divided into teams of $11$ each, no one is left out. The maximum number of teams of $12$ each that can be formed out of the students in the school is 1 vote 1 1 vote", "ready to go lingers, can that be! 9240 Ten ways to choose group 2 's that you cross your arms divide n identical objects among distinct. Choices for each of the $n = 6$ is special … this works... Answer site for people studying math at any level and professionals in related fields sizes matter not...", "team. So: divide by $$2$$ or, what is the same thing: divide by the number of ways we can shuffle the groups, and since there are two groups, that can be done in $$2!$$ ways. Of course, if the teams are distinct (e.g if the first team is 'the Flyers' and the second 'the Eagles', then swapping the groups of people will make a difference, so in that case do not divide by $$2!$$", "• question_answer There are 10 friends and 55 bananas. If the bananas are divided equally among the students, how many does each friend get? A) Q = 5, R = 3 B) Q = 5, R = 2 C) Q = 5, R = 5 D) Q =1, R =5 Solution : Each friend got $=55\\div 10$ Q = 5, R = 5 Each friend got 5 bananas & 5 bananas were left over. You need to login to perform this action. You will be redirected in 3 sec", "Very nice problem, combinatorics There are $n$ people and there exist $2^{n} - 1$ football teams. In each football team we have to choose menager. BUT: If team $X$ is sum $X = Y\\cup Z$ then the menager $X$ is the menager of at least one of teams $Y, Z$", "basketball. In how many ways can this be done in each of the following cases: 1. The teams are distinguishable (for example, one team is labeled Alabama and the other team is labeled Auburn). 2. The teams are not distinguishable.", "number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is important is $$\\frac{(mn)!}{(n!)^m}$$ Manager Joined: 22 Jun 2010 Posts: 212 Followers: 1 Kudos [?]: 84 [0], given: 13 Show Tags 15 Oct 2010, 10:18 Bunuel's explanations are great I found out what my mistake was Re: combination [#permalink] 15 Oct 2010, 10:18 Go to page 1 2 Next [ 31 posts ] Similar topics Replies Last post Similar Topics: 24 A mixed doubles tennis game is to be played between two team 23 31 Aug 2012, 12:36 6 How many different ways to play doubles tennis ? 7 12 Jul 2011, 02:20 11 A group of 8 friends want to play doubles tennis. How many 6 14 Dec 2010, 16:32 6 A group of 8 friends want to play doubles tennis. How many 9 14 Nov 2010, 04:18 40 In how many different ways can a group of 8 people be divide 17 24 Oct 2009, 04:20 Display posts from previous: Sort by A group of 8 friends want to play doubles tennis. How many new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test", "you get it? = _________________ Smiling wins more friends than frowning Manager Joined: 06 Jul 2007 Posts: 207 Re: Given that there are 6 married couples. If we select only 4 [#permalink] ### Show Tags 23 Mar 2009, 11:20 1 xALIx wrote: walker wrote: dominion wrote: Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other. There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - $$C^6_4*(C^2_1)^4$$ look at other problem: Quote: Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team? we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team. I hope it is", "might not be assigned any basketballs at all. This corresponds to having multiple dividers in the same “gap between basketballs” or having one or more dividers at the end of a row. For instance, to assign 3 basketballs to Truck 1, 1 to Truck 2, 5 to Truck 4, and none to Trucks 3 or 5, Leslie can place dividers like so: So we’re not quite done. Here’s how we fix it: we put down 9 + 4 “placeholders” that could each hold either a basketball or a divider. Then we choose 4 of those placeholders to contain dividers, and the rest of the placeholders will contain basketballs. Now we can have two dividers in adjacent placeholders, meaning a truck with zero basketballs. Think this over some more and make sure you understand why this fixes the problem in our first attempted solution. So there are ${9+4 \\choose 4}$ ways to assign the basketballs to the 5 trucks. Does this approach work for the other types of balls also? Luckily, yes! Following the same reasoning, we can determine that there are ${61+4 \\choose 4}$ ways to assign the volleyballs, ${88+4 \\choose 4}$ ways to assign the baseballs, and ${323+4 \\choose 4}$ ways to assign the golf balls. Since all of these choices are independent, we multiply them to", "calculation counts as distinct having a particular group of 11 on Team A, or on Team B or on Team C. If this is not the case (I'm not convinced!) then divide by $3!$ -", "want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people? Can the answer be arrived at as follows? select 8 out of 9 people, first: 9C8 = 9 Then select the sets of 2 people 8!/2^4*4! = 105 answer = 9*105 ---> is this correct? can anyone who knows tell me? Many thanks. Math Expert Joined: 02 Sep 2009 Posts: 47157 A group of 8 friends want to play doubles tennis. How many [#permalink] ### Show Tags 15 Nov 2010, 01:57 3 7 AkritiMehta wrote: Hi how do we get the answer for the Ist question as 105. Pls explain A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people? A. 420 B. 2520 C. 168 D. 90 E. 105 $$\\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2... For the first person we can pick a pair in 7 ways; For the second one in 5 ways (as two", "group of player who each contribute to the commodity differently. I use this in practice when dealing with attributing individual marks to a group project: you can read about that here. To introduce this I asked 3 students to come down and play a game: $S$, $L$ and $C$ volunteered (or at least gave a lazy nod when I asked if anyone had played basketball before) to come down. They had 20 seconds to get as many crumpled up pieces of paper as possible in the bin above (it was on the table). The paper was stapled and it needed to be separated and crumpled so I suggested that perhaps 1 or 2 of them could spend some of the 20 seconds separating and crumpling. We played this 3 times and it was all good fun. After $S$, $L$ and $C$ had their go 3 other students thought they could do better (no comment). After that I asked the class how we could share out the 10 chocolates between $S$, $L$ and $C$: what would be the fair way? After some discussion it was clear that L had made the biggest contribution to the group for example and so this was not a trivial question. After a little while I started to ask 'how many do you think", "total number of edges, or friendships is $\\frac{4 \\cdot 17}{2}$ or 34. Next we count the number of paths of length 2. If indeed every pair of people are friends or friend-of-friends, then every pair should be separated by a path of at most length 2. In the entire graph there are 17 people, thus $17 \\choose 2$ or 136 pairs of people. We already know that 34 of these pairs are friendships, so counting that out, we have 102 pairs of people who are not friends, and thus in our case are separated by a path length of exactly 2. This establishes a lower bound for the number of friend-of-friends, as a friend-of-friend might be linked intermediately in more than one way. Furthermore, a friend-of-friend might already be one of the friends. Next we count friend-of-friends a different way. How would the graph be so that the number of friend-of-friends is a maximal? Intuitively, there cannot be any immediate diamond-shapes (4-cycles) or triangles (3-cycles). We get a maximal of 12 friend-of-friends like this: Each of your friends have a friendship with yourself (the central node), plus three other friends. None of the friend-of-friends are coincident. Suppose that everyone is arranged in this optimal way, such that everyone has 12 friend-of-friends. Then how many paths of length 2" ]

[ "two vertices of a square and locate two other points, and , for the other two vertices of the square. For a challenge, attempt to find as many possible locations for points and as possible. ### 20. Given the coordinates of the quadrilateral , find its perimeter. , , ,", "and three vertex points that all four triangles created by these diagonals –,... To any other vertex inside the given shape ( a ) a convex quadrilateral ( b 66. Linear pair angles is 1,440 degrees a pentagon, find how many sides does a regular (...", "How many convex polygons can be drawn using these points as vertices?", "of a five-member teams remaining 2 pupils and in the case of six-members teams rem • 10 pieces How to divide the circle into 10 parts (geometrically)?", "# Points Drawn on a Circle There are $$n$$ distinct points on a circle such that the number of pentagons that can be formed with these points as vertices is equal to the number of possible triangles. Then the value of $$n$$ is? × Problem Loading... Note Loading... Set Loading...", "not. Points which all lie on a single circle are called concyclic. (You might also recall coming across cyclic quadrilaterals; a quadrilateral is cyclic if all of its vertices lie on the circumference of a circle.)", "# Distinct convex quadrilaterals formed by 12 points out of which 5 are collinear There are 12 points in a plane of which 5 are collinear. The number of distinct convex quadrilaterals which can be formed with vertices at these points is:___ I know how to solve this question without the convex part. How do I ensure that the quadrilaterals formed are convex? Or how do I count the number of concave quadrilaterals possible? Won't they depend on the real arrangement of points? Note: It can be assumed that none other than these 5 points are collinear • Yes, I would think so. Are you sure there's no more information about the points? – quasi Mar 1 '17 at 7:23 • Also, although it says 5 are collinear, how do you know that some other triple, quadruples, etc. are not also collinear? The question, as stated, needs more specification. Where does the problem come from? – quasi Mar 1 '17 at 7:25 • @quasi I have edited the post to put the question in the exact words. It can be assumed that no other points are collinear. No other information is provided. – Osheen Sachdev Mar 1 '17 at 8:32 • But then that's not enough information. For example, if the set of 7 points which are not", "segment is tangent to the nine-point circle at . (In terms of sheer numbers, the conditions above have now surpassed the number of statements in the Invertible Matrix Theorem. Next stop is to top the number of statements characterizing tangential quadrilaterals. Next stop after that? To stop pestering you with the same exercises over and over.)", "three distinct rational points, say, $A, B, C$. You might be knowing that a three non collinear points always determine a unique circle. Hence the perpendicular bisectors of $AB$ and $BC$ would meet at the circumcenter of our triangle which would be a rational point in contradiction with the conditions of the problem. Hence the maximum number of rational points can only be 2.", "number of triangles and the number of quadrilaterals that can be created by connecting these points? A. 4 B. 5 C. 6 D. 15 E. 20 Number of triangles can be formed out of 6 points= 6C3=20 Number of quadrilaterals can be formed out of 6 points= 6C4=15 20-15=5. Re: Six points lie on the circumference of a circle. What is the [#permalink] 13 Oct 2014, 03:26 Similar topics Replies Last post Similar Topics: 1 Points A, B and C lie on a circle. Point O is the center of the circle 6 13 Jul 2015, 23:50 Points A,B,C lie on the circle. Point O is the center of the circle an 2 07 Jan 2015, 23:32 2 Points A,B,C lie on the circle. Point O is the center of the circle an 4 07 Jan 2015, 06:38 If points A and C both lie on the circle with center B and 4 09 Dec 2012, 00:46 1 In the figure above, points P and Q lie on the circle with 1 29 Jan 2012, 11:41 Display posts from previous: Sort by", "to 9*10=90. Re: Maximum number of points that 10 circles of different radii intersect &nbs [#permalink] 26 Feb 2017, 16:25 Display posts from previous: Sort by", "hexagons can be drawn that use six of these points as its vertices? A. 30,240 B. 1,008 C. 252 D. 210 E. 10 Since the circle has 10 points on its edge and we need to create hexagons, each of which has 6 vertices, the number of ways to create those quadrilaterals using the 10 points is 10C6: 10!/[6!(10 - 6)!] = 10!/(6! x 4!) = (10 x 9 x 8 x 7)/4! = (10 x 9 x 8 x 7)/(4 x 3 x 2) = 10 x 3 x 7 = 210 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Re: The circle above has 10 points on its edge. How many different hexagon [#permalink] 02 May 2017, 16:35 Display posts from previous: Sort by", "# pointy problem Geometry Level 4 There are 20 points inside a square. They are connected by non intersecting segments with each other and with the vertices of the square ,in such a way that the square is dissected into triangles. How many triangles do we have? ×", "Called vertices give the answer ( 1, 2 ) and ( 2,6 ) to make one side of square, and the two other vertices using only a compass 2005 20,249 a cube has six square,. Those two coordinates I need to find similar cost effective systems of roadways for 5 and 6 towns. ) 2 at ( 3,4 ) and ( 2,6 ) height etc can be a.! ( say AB, or AD ) 2, each of the circle and the line x/a + y/b 1. Fourth point is ( 5, 6 ) the third vertex ( plural: vertices ) is a vertex an. Geometry as the common endpoint of two or more rays or line meet. ( x, y ) is called a regular quadrilateral are parallel to the corners of square 0 replies", "square three edges meet common circle. moving a distance of 5 units in each.. Of the line AC above to find will make it a square are parallel to the corners of side. To know how to do it, please how to Read Them a minimum, what the., it is called a regular quadrilateral go and identify the name of the points of intersection of edges! Calculator - calculate Hyperbola vertices given equation step-by-step this website uses cookies to ensure you the... For 5 and 6 towns i.e to the x and y axes, then give answer. Diagonal of the cube 's corners is a vertex is a diameter the! The square is the distance between two points ) So in the figure above: 1 of the square the. Vertex is defined in geometry as the common endpoint of two or more rays or line segments.. The included angle at each vertex is referred to as an interior angle of the slope are called.. ( 3, 2 ) and ( 1, -1 ) the vertices graph at the points say! ( 3,4 ) and ( 3, 2 ) be the given points are vertices of square parallel. The square are parallel to the corners of a side of the quare is a square are ( )... From the", "nearly tangent to the desired region at a large number of points (about 20 in this case). What other shapes can be approximated? More on that later... Share on", "+0 # are nine points on the circumference 0 194 1 are nine points on the circumference of circle . Line segments are drawn connecting each pair of points. How many line segments are drawn? Guest Mar 29, 2015 #1 +26412 +10 Each point is connected to 8 others, so, 9*8. But this counts each line twice, so the number of lines is 9*8/2 = 36 . Alan Mar 29, 2015 Sort: #1 +26412 +10", "# Count 'em All 9! Count the total number of quadrilaterals in the 6$$\\times$$5 grid above. Details and assumptions: • Quadrilateral is a polygon that has 4 sides. ×", "segment between the two points. Equally important is theorem: Any set of three noncollinear points determines a unique circle. Therefore three distinct circles cannot have more than two points in common. 6. Re: different circles Hi, Plato is absolutely correct. Also Soroban is correct if he adds the condition that the centers of the 3 circles are non-collinear. The attached drawing may be helpful:", "and 1 floating points. can anyone advice more for i create the circule?" ]

[ "argument applies at the contact of the lower point. That's $270^\\circ$ total. Now, the intersection of the two tangents together with the two contact points and the circle center form a quadrilateral. The interior angles of any quadrilateral add to $360^\\circ$. We have accounted for $270^\\circ$, leaving $90^\\circ$. So we have a rectangle, at least. But the two radii are equal in length. So we have a square.", "be formed by selecting $$4$$ points from the $$8$$ points except when: i) $$4$$ points have been taken from the line ii) $$3$$ points have been taken from the line and one from circle. Total required cases $$= 8C4- (4C4+4C3*4C1) = 53$$ Harshgmat wrote: Of the 8 points in a given plane 4 are on circle O and remaining 4 are on line L. Number of different quadrilaterals that can be formed by these 8 points are A) 52 B) 53 C) 54 D) 64 E) 74 _________________ Senior SC Moderator Joined: 22 May 2016 Posts: 2099 Of the 8 points in a given plane 4 are on circle O -Geometry - PS [#permalink] ### Show Tags 25 Aug 2018, 08:57 2 pushpitkc wrote: Therefore, the total number of quadrilaterals that can be formed by these points are 1+16+36 = 52(Option B) Senior PS Moderator Joined: 26 Feb 2016 Posts: 3307 Location: India GPA: 3.12 Of the 8 points in a given plane 4 are on circle O -Geometry - PS [#permalink] ### Show Tags 25 Aug 2018, 09:07 generis wrote: pushpitkc wrote: Therefore, the total number of quadrilaterals that can be formed by these points are 1+16+36 = 52(Option B) Thanks for notifying, made the necessary change! Of the 8 points in a given plane 4 are", "# Happy 22rd birthday, me! Geometry Level 5 A convex cyclic quadrilateral with integer length sides is such that its area divided by its perimeter equals $$1993$$. Let $$P$$ be the maximum possible perimeter. Find the four last digits of $$P$$. ×", "# 4 Points on Circumference of Circle and center This is actually a computer science question in that I need to create a program that will determine the center of a circle given $4$ points on its circumference. Does anyone know the algorithm, theorem or method? I think it has something to do with cyclic quadrilaterals. Thanks. -", "Just Touching Three semi-circles have a common diameter, each touches the other two and two lie inside the biggest one. What is the radius of the circle that touches all three semi-circles? Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load. Biggest Bendy Four rods are hinged at their ends to form a quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic. Xtra Stage: 4 and 5 Challenge Level: Two solutions to this problem have been forthcoming from different students at the same school - Madras College. Thank you to Mike and Euan who used lots of trigonometry as well as to Thom who likewise resorted to double angles and the cosine rule and reduced the problem to solving a quadratic equation. Thom was also able to show the significance of the two roots. \\eqalign{ \\beta &=& \\frac{\\pi}{6} - \\frac{\\alpha}{2} \\\\ \\cos\\beta &=& \\frac{\\sqrt{3}}{2}\\cdot\\frac{3\\sqrt{3}}{2\\sqrt{7}} + \\frac{1}{2}\\cdot\\frac{1}{2\\sqrt{7}} \\\\ \\; &=& \\frac{10}{4\\sqrt{7}} = \\frac{5}{2\\sqrt{7}}} Using the cosine rule on $\\triangle ABP$ \\eqalign{ 4 &=& x^2 + 7 - 2x\\sqrt{7}\\cos\\beta \\\\ \\; &=& x^2 + 7 - 5x} Therefore $x^2 - 5x + 3 = 0$ $$x = \\frac{5\\pm\\sqrt{13}}{2}$$ Both solutions satisfy", "## FANDOM 1,019 Pages In geometry, a quadrilateral, quadrangle or tetragon is a polygon with four sides. ## Types The following are special cases of the quadrilateral: ## Area Brahmagupta's formula finds the area of any quadrilateral given the lengths of the sides is $A= \\sqrt{\\frac{(a^2+b^2+c^2+d^2)^2+8abcd-2(a^4+b^4+c^4+d^4)}{16}}$", "# How many quadrilaterals can be formed from 12 points on a triangle? I would like to check if I did this problem correctly so please confirm or correct my method. There is a triangle ABC with 3 distinct points on side AB, 4 distinct points on side BC, and 5 distinct points on side AC. What triangle looks like for reference: How many quadrilaterals can be formed from these points? My method included counting 2 cases: the first being where 2 points were collinear and the other 2 points were on the two remaining sides, the second being where there were 2 points on one side and 2 points on another side, the third side not contributing any points. This ended up being something like $$3\\cdot4\\cdot\\binom{5}{2} + 3\\cdot5\\cdot\\binom42 + 4\\cdot5\\cdot\\binom32 + \\binom32\\cdot\\binom42 + \\binom32\\cdot\\binom52 + \\binom42\\cdot\\binom52$$ I just want to know if this is a correct method for finding the solution. Much appreciated! • You've done just fine; yes, you nailed it. Don't fret so much about the formatting, given it's your first post ? Any way, start trying to learn mathjax (latex, essentially). You might want to start by clicking on \"edit\" to see how you can format what is displayed now. Actually, no need to accept any answer; you've got the validation you needed. No", "is equal to: $$A_t=n\\cdot\\displaystyle{\\frac{a\\cdot h_a}{2}}.$$ Quadrilaterals are polygons in a plane with $4$ sides and $4$ vertices. A regular quadrilateral is a square, because square is the only quadrilateral with all sides of equal length and all angles of equal measure. The area of a square is equal to the square of the length of one side, and the perimeter to four lengths of any side. Each interior angle has a measure equal to $90^{\\circ}$, and their sum is equal to $360^{\\circ}$. It has two diagonals. These diagonals intersect at one point which is the center of an inscribed and circumscribed circle. To draw a circumscribed circle of a square we simply place the needle of the compass into the intersection of diagonals, extend it to one vertex, and draw. The circumscribed circle will then run through all vertices. To draw an inscribed circle, we must first find the radius. To find the radius, we must draw a perpendicular line from the center to any side. When we inscribe the circle, it must touch all sides of the square. ### Pentagon Pentagons are polygons which contain five sides. A regular pentagon has five congruent sides and five congruent angles. How many diagonals does pentagon have? We can use the formula from above. $$D_{5} = \\frac{5 \\cdot(5 – 3)}{2}", "Four rods are hinged at their ends to form a convex quadrilateral with sides of length $3$, $4$, $5$ and $6$ (in that order). Investigate the different shapes that the quadrilateral can take.", "1. ## Inscribed Quadrilateral If there are four different circles that are tangent both to the edges of an inscribed quadrilateral and to the circle around that inscribed quadrilateral, how should the inscribed quadrilateral be chosen so that the sum of the perimeters of those four different circles could be maximum or minimum? Case 1: If the inscribed quadrilateral is a \"square\" with the same side length $a$, then I find the sum of the perimeters of the \"same\" four circles $P_1 = \\frac{2\\pi\\cdot a}{1 + \\sqrt{2}}$ Case 2: If the inscribed quadrilateral is a \"rectangle\" with the side lengths $a$ and $\\frac{a}{2}$, then I find the sum of the perimeters of the two bigger circles and the two smaller circles $P_2 = \\pi \\cdot a \\cdot \\frac{2\\sqrt{5} - 3}{2}$ Case 3: Let the side lengths of the rectangle which is the inscribed quadrilateral be $a$ and $b$, the radius of the two bigger circles be $r_1$, the radius of the two smaller circles be $r_2$, the total sum of the perimeters of these circles be $P_3$, the radius of the largest circle that contains both the inscribed quadrilateral (rectangle) around which are those four circles be $R$. $R = \\sqrt{\\frac{a^2}{4} + \\frac{b^2}{4}}$ $r_1 = \\frac{R}{2} - \\frac{a}{4}$ $r_2 = \\frac{R}{2} - \\frac{b}{4}$ $P = 2(2\\pi \\cdot r_1)", "choose a first edge in $P$. With a similar counting as above we then find that there are $12$ admissible ways to choose a second edge in $P$ which is separated by two true diagonals from the first chosen edge. In all we obtain $17\\cdot 12+{1\\over2}\\cdot 17\\cdot 12=306$ admissible quadrilaterals.", "Between a circle and a circle. This is the case that you were forgetting. There are $5$ circles in total, so there are ${}_5C_2$ ways to choose the two circles that will intersect. These two circles can intersect at most $2$ times. This yields ${}_5C_2 \\cdot 2 = 20$. Summing everything together, we obtain $15 + 60 + 20 = 95$.", "number of triangles and the number of quadrilaterals that can be created by connecting these points? A. 4 B. 5 C. 6 D. 15 E. 20 Number of triangles can be formed out of 6 points= 6C3=20 Number of quadrilaterals can be formed out of 6 points= 6C4=15 20-15=5. Re: Six points lie on the circumference of a circle. What is the [#permalink] 13 Oct 2014, 03:26 Similar topics Replies Last post Similar Topics: 1 Points A, B and C lie on a circle. Point O is the center of the circle 6 13 Jul 2015, 23:50 Points A,B,C lie on the circle. Point O is the center of the circle an 2 07 Jan 2015, 23:32 2 Points A,B,C lie on the circle. Point O is the center of the circle an 4 07 Jan 2015, 06:38 If points A and C both lie on the circle with center B and 4 09 Dec 2012, 00:46 1 In the figure above, points P and Q lie on the circle with 1 29 Jan 2012, 11:41 Display posts from previous: Sort by", "and concluded that it was the center of the circle. so with the center, i have many points for my inscribable quadrilateral..now, i am having a hard time finding which point will make it both circumscribable and inscribable,.,please help me,,.thnx i couldn't solve it completely but this may help.. 1) the fourth point will lie on the circumcircle of the triangle ((-3,5),(-2,-3),(3,-4)) as you pointed out too. 2) All the four angle bisectors of the angles of the quadrilateral meet at a point. this is the necessary and sufficient condition for a circle to be inscribed in the quadrilateral. 3. ## Re: The Missing Point Originally Posted by aldrincabrera ,.hello everyone,.,can u help me with this?? Given points (-3,5) , (-2,-3) , (3,-4). find the fourth point in the first quadrant so that the quadrilateral made is both circumscribable and inscribable. I managed to start with the solution. using my 3 points, i had drawn a triangle, then i drew a circle around it in which the 3 points are tangent to the circle. next, i got the intersection of the perpendicular bisectors of the 3 sides and concluded that it was the center of the circle. so with the center, i have many points for my inscribable quadrilateral..now, i am having a hard time finding which point", "FANDOM 1,168 Pages In geometry, a quadrilateral, quadrangle or tetragon is a polygon with four sides. Types The following are special cases of the quadrilateral: Area Brahmagupta's formula finds the area of any quadrilateral given the lengths of the sides is $A= \\sqrt{\\frac{(a^2+b^2+c^2+d^2)^2+8abcd-2(a^4+b^4+c^4+d^4)}{16}}$ Community content is available under CC-BY-SA unless otherwise noted.", "# Count 'em All 10! Discrete Mathematics Level 3 Count the total number of quadrilaterals in the 20$$\\times$$30 grid above. Details and assumptions: • Quadrilateral is a polygon that has 4 sides. ×", "four partially overlapping circles at the point $$A, B, C$$ and $$D$$ respectively. Connect those points and you will get a square having side equal to the diameter of small circle and that is obviously equal to 6. Let denote the area of the square $$ABCD = [ABCD]$$. So, the area of the quatrefoil = $$[ABCD]$$ + The area of 4 semi circles = $$6^2 + 4*\\frac{1}{2}*\\pi3^2$$ = $$36 + 18\\pi \\approx 92.549$$ If I understand the intended shape correctly, its outline is formed by four semicircles of radius $$3$$ whose centers form a square with a diagonal of $$6$$ units. They intersect at their endpoints, which form a larger square with a side of $$6$$ units. The total area inside the quatrefoil is then the area of the square plus the area inside the four semicircles: $$6^2 + 4(\\frac12\\pi\\cdot 3^2) = 36 + 18\\pi$$.", "# Math Help - Inscribed Quadrilateral If there are four different circles that are tangent both to the edges of an inscribed quadrilateral and to the circle around that inscribed quadrilateral, how should the inscribed quadrilateral be chosen so that the sum of the perimeters of those four different circles could be maximum or minimum? Case 1: If the inscribed quadrilateral is a \"square\" with the same side length $a$, then I find the sum of the perimeters of the \"same\" four circles $P_1 = \\frac{2\\pi\\cdot a}{1 + \\sqrt{2}}$ Case 2: If the inscribed quadrilateral is a \"rectangle\" with the side lengths $a$ and $\\frac{a}{2}$, then I find the sum of the perimeters of the two bigger circles and the two smaller circles $P_2 = \\pi \\cdot a \\cdot \\frac{2\\sqrt{5} - 3}{2}$ Case 3: Let the side lengths of the rectangle which is the inscribed quadrilateral be $a$ and $b$, the radius of the two bigger circles be $r_1$, the radius of the two smaller circles be $r_2$, the total sum of the perimeters of these circles be $P_3$, the radius of the largest circle that contains both the inscribed quadrilateral (rectangle) around which are those four circles be $R$. $R = \\sqrt{\\frac{a^2}{4} + \\frac{b^2}{4}}$ $r_1 = \\frac{R}{2} - \\frac{a}{4}$ $r_2 = \\frac{R}{2} - \\frac{b}{4}$ $P = 2(2\\pi", "is thus also called an inscribed quadrilateral. A convex quadrilateral Other names for quadrilateral include quadrangle (in analogy to triangle), tetragon (in analogy to pentagon, 5-sided polygon, and hexagon, 6-sided polygon), and 4-gon (in analogy to k-gons for arbitrary values of k).A quadrilateral with vertices , , and is sometimes denoted as . Cut out the quadrilateral. Using the formula above, calculate the inscribed angle. For example, if a farmer needs to distribute 100g of fertilizer per square meter of a field, they can use the calculator to calculate the area of the field. This is known as the Pitot theorem, named after Henri Pitot, a French engineer who proved it in the 18th century. www-formula.com. Inscribing A Circle Within A Kite All kites are tangential quadrilaterals, meaning that they are 4 sided figures into which a circle (called an incircle) can be inscribed such that each of the four sides will touch the circle at only one point. It says that these opposite angles are in fact supplements for each other. Type your answer here… Related Topics. 2. Problem A quadrilateral is a polygon in Euclidean plane geometry with four edges (sides) and four vertices (corners). Quadrilateral Calculator. Given points A, B, C and D on circle O. r c = 1/ (4*A) * √ (ab+cd) *", "of $P/2^{N}$ as $N$ tends to infinity is $0$.\" When we keep inscribing circles into triangles and triangles into circles, circles shrink into a point. Now, let's modify the process just a little. Let's think of triangles as polygons. In the new construction, we shall be inscribing circles into polygons, and polygons into circles (Pic 4). We'll have to answer two questions: Find the perimeter $P$ of a regular $N$-gon inscribed into a circle of circumference $C.$Find the circumference $C$ of a circle inscribed into a regular $N$-gon of perimeter $P.$ In both cases the perimeter $P$ of the $N$-gon relates to one half of its side a by $P = 2aN,$ while the circumference of a circle relates to its radius by $C = 2\\pi R$ $c = 2\\pi r$ We have the following basic relations $a = R\\cdot\\sin(\\pi /N)$ $r = a\\cdot\\cot(\\pi/N)$ So that $P = 2N\\cdot C/2\\pi \\cdot \\sin(\\pi /N)$ $c = 2\\pi \\cdot P/2N\\cdot \\cot(\\pi /N)$ which, after simplification, becomes $P = C\\cdot N/\\pi \\cdot \\sin(\\pi /N)$ $c = P\\cdot \\pi /N\\cdot \\cot(\\pi /N)$ Finally, we combine the two steps. Let there be a circle of circumference $C.$ Inscribe into the circle a regular $N$-gon, and into the latter inscribe another circle. The circumference c of the latter satisfies $c = C\\cdot \\cos(\\pi /N).$" ]

[ "# How many quadrilaterals can be formed from 12 points on a triangle? I would like to check if I did this problem correctly so please confirm or correct my method. There is a triangle ABC with 3 distinct points on side AB, 4 distinct points on side BC, and 5 distinct points on side AC. What triangle looks like for reference: How many quadrilaterals can be formed from these points? My method included counting 2 cases: the first being where 2 points were collinear and the other 2 points were on the two remaining sides, the second being where there were 2 points on one side and 2 points on another side, the third side not contributing any points. This ended up being something like $$3\\cdot4\\cdot\\binom{5}{2} + 3\\cdot5\\cdot\\binom42 + 4\\cdot5\\cdot\\binom32 + \\binom32\\cdot\\binom42 + \\binom32\\cdot\\binom52 + \\binom42\\cdot\\binom52$$ I just want to know if this is a correct method for finding the solution. Much appreciated! • You've done just fine; yes, you nailed it. Don't fret so much about the formatting, given it's your first post ? Any way, start trying to learn mathjax (latex, essentially). You might want to start by clicking on \"edit\" to see how you can format what is displayed now. Actually, no need to accept any answer; you've got the validation you needed. No", "# Counting points of intersection There are 9 points on the circumference of a circle. The points are not evenly spaced. Line segments are drawn connecting each pair of points. What is the largest number of different points inside the circle at which at least two of these line segments intersect? Here is what I am attempted trying to solve this problem. With 3 points on the circumference there will not any point of intersection. With 4th point I have 1 point of intersection. With 5 points I am getting 5 points of intersection (shape looking like a star with one spoke missing). 6th point make the total go up to 15. Do I need to keep counting this way or there is better method? Also I am failed to understand how evenly spaced or unevenly spaced points would have made a difference. With $9$ points on the circle, there are $\\binom{9}{2}=36$ line segments. How many couples of line segments share a vertex? For every vertex, there are $8$ line segments with such an endpoint, hence there are $9\\cdot\\binom{8}{2}=252$ couples of line segments with a common endpoint, and at most: $$\\binom{36}{2}-9\\cdot\\binom{8}{2} = 378$$ points of intersection not on the circle. In general, with $n$ points on the circle there are at most: $$\\binom{\\binom{n}{2}}{2}-n\\cdot\\binom{n-1}{2} = \\frac{n(n-1)(n-2)(n-3)}{8}=3\\binom{n}{4}$$ points of intersection", "1. ## Inscribed Quadrilateral If there are four different circles that are tangent both to the edges of an inscribed quadrilateral and to the circle around that inscribed quadrilateral, how should the inscribed quadrilateral be chosen so that the sum of the perimeters of those four different circles could be maximum or minimum? Case 1: If the inscribed quadrilateral is a \"square\" with the same side length $a$, then I find the sum of the perimeters of the \"same\" four circles $P_1 = \\frac{2\\pi\\cdot a}{1 + \\sqrt{2}}$ Case 2: If the inscribed quadrilateral is a \"rectangle\" with the side lengths $a$ and $\\frac{a}{2}$, then I find the sum of the perimeters of the two bigger circles and the two smaller circles $P_2 = \\pi \\cdot a \\cdot \\frac{2\\sqrt{5} - 3}{2}$ Case 3: Let the side lengths of the rectangle which is the inscribed quadrilateral be $a$ and $b$, the radius of the two bigger circles be $r_1$, the radius of the two smaller circles be $r_2$, the total sum of the perimeters of these circles be $P_3$, the radius of the largest circle that contains both the inscribed quadrilateral (rectangle) around which are those four circles be $R$. $R = \\sqrt{\\frac{a^2}{4} + \\frac{b^2}{4}}$ $r_1 = \\frac{R}{2} - \\frac{a}{4}$ $r_2 = \\frac{R}{2} - \\frac{b}{4}$ $P = 2(2\\pi \\cdot r_1)", "Rectangle A rectangle has four sides, whose two sides are equal. Then, its perimeter is; P = x + x + y +y = 2 (x+y) ### Square It has four equal sides. In fact, it is one of the regular polygons whose n = 4. Thus, P = n\\cdot S = 4\\cdot S", "of $P/2^{N}$ as $N$ tends to infinity is $0$.\" When we keep inscribing circles into triangles and triangles into circles, circles shrink into a point. Now, let's modify the process just a little. Let's think of triangles as polygons. In the new construction, we shall be inscribing circles into polygons, and polygons into circles (Pic 4). We'll have to answer two questions: Find the perimeter $P$ of a regular $N$-gon inscribed into a circle of circumference $C.$Find the circumference $C$ of a circle inscribed into a regular $N$-gon of perimeter $P.$ In both cases the perimeter $P$ of the $N$-gon relates to one half of its side a by $P = 2aN,$ while the circumference of a circle relates to its radius by $C = 2\\pi R$ $c = 2\\pi r$ We have the following basic relations $a = R\\cdot\\sin(\\pi /N)$ $r = a\\cdot\\cot(\\pi/N)$ So that $P = 2N\\cdot C/2\\pi \\cdot \\sin(\\pi /N)$ $c = 2\\pi \\cdot P/2N\\cdot \\cot(\\pi /N)$ which, after simplification, becomes $P = C\\cdot N/\\pi \\cdot \\sin(\\pi /N)$ $c = P\\cdot \\pi /N\\cdot \\cot(\\pi /N)$ Finally, we combine the two steps. Let there be a circle of circumference $C.$ Inscribe into the circle a regular $N$-gon, and into the latter inscribe another circle. The circumference c of the latter satisfies $c = C\\cdot \\cos(\\pi /N).$", "argument applies at the contact of the lower point. That's $270^\\circ$ total. Now, the intersection of the two tangents together with the two contact points and the circle center form a quadrilateral. The interior angles of any quadrilateral add to $360^\\circ$. We have accounted for $270^\\circ$, leaving $90^\\circ$. So we have a rectangle, at least. But the two radii are equal in length. So we have a square.", "+0 # Coordinates 0 41 1 How do I find the are of a convex quadrialater with vertices (-1,0),(0,1),(3,0) and (0,-5)? Jul 4, 2021 #1 +11835 +1 How do I find the are of a convex quadrialater with vertices P1(-1,0),P2(0,1),P3(3,0) and P4(0,-5)? Hello Guest! $$A=\\dfrac{(y_2-y_4)(x_3-x_1)}{2}\\\\ A=\\dfrac{(1-(-5))(3-(-1))}{2}\\\\ A= \\dfrac{6\\cdot 4}{2}$$ $$A=12$$ ! Jul 4, 2021", "# Distinct convex quadrilaterals formed by 12 points out of which 5 are collinear There are 12 points in a plane of which 5 are collinear. The number of distinct convex quadrilaterals which can be formed with vertices at these points is:___ I know how to solve this question without the convex part. How do I ensure that the quadrilaterals formed are convex? Or how do I count the number of concave quadrilaterals possible? Won't they depend on the real arrangement of points? Note: It can be assumed that none other than these 5 points are collinear • Yes, I would think so. Are you sure there's no more information about the points? – quasi Mar 1 '17 at 7:23 • Also, although it says 5 are collinear, how do you know that some other triple, quadruples, etc. are not also collinear? The question, as stated, needs more specification. Where does the problem come from? – quasi Mar 1 '17 at 7:25 • @quasi I have edited the post to put the question in the exact words. It can be assumed that no other points are collinear. No other information is provided. – Osheen Sachdev Mar 1 '17 at 8:32 • But then that's not enough information. For example, if the set of 7 points which are not", "\\cdot 10$ equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is $600+60 = \\boxed{660}$.", "Between a circle and a circle. This is the case that you were forgetting. There are $5$ circles in total, so there are ${}_5C_2$ ways to choose the two circles that will intersect. These two circles can intersect at most $2$ times. This yields ${}_5C_2 \\cdot 2 = 20$. Summing everything together, we obtain $15 + 60 + 20 = 95$.", "choose a first edge in $P$. With a similar counting as above we then find that there are $12$ admissible ways to choose a second edge in $P$ which is separated by two true diagonals from the first chosen edge. In all we obtain $17\\cdot 12+{1\\over2}\\cdot 17\\cdot 12=306$ admissible quadrilaterals.", "is equal to: $$A_t=n\\cdot\\displaystyle{\\frac{a\\cdot h_a}{2}}.$$ Quadrilaterals are polygons in a plane with $4$ sides and $4$ vertices. A regular quadrilateral is a square, because square is the only quadrilateral with all sides of equal length and all angles of equal measure. The area of a square is equal to the square of the length of one side, and the perimeter to four lengths of any side. Each interior angle has a measure equal to $90^{\\circ}$, and their sum is equal to $360^{\\circ}$. It has two diagonals. These diagonals intersect at one point which is the center of an inscribed and circumscribed circle. To draw a circumscribed circle of a square we simply place the needle of the compass into the intersection of diagonals, extend it to one vertex, and draw. The circumscribed circle will then run through all vertices. To draw an inscribed circle, we must first find the radius. To find the radius, we must draw a perpendicular line from the center to any side. When we inscribe the circle, it must touch all sides of the square. ### Pentagon Pentagons are polygons which contain five sides. A regular pentagon has five congruent sides and five congruent angles. How many diagonals does pentagon have? We can use the formula from above. $$D_{5} = \\frac{5 \\cdot(5 – 3)}{2}", "Given that AB , CD , EF , GH and JK be five diameters of a circle with center at O. We have to find by how many ways three points can be chosen out of A , B , C , D , E , F , G , H , J , K and O so as to form a triangle. We know that any three points lying on the circle are non collinear. There are 11 points here we have to choose from any three points that are not collinear. From these 10 points A, B, C, D, E, F, G, H, J, K we can select any three points on the circle such that they are non collinear. We can take O and then from remaining 10 we can take remaining two. Since AOB , COD , GOH , EOF , JOK cannot form triangles, 5 can be subtracted. 10C3 + 10C2 - 5 = $$frac{10 × 9 × 8}{1 × 2 × 3}$ + $\\frac{10 × 9}{1 × 2}$ - 5 ⟹ 120 + 45 – 5 = 160 ways. The question is \"Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B,", "in two points $J$ and $K$. The circumcircles $(CDJ)$ and $(CDK)$ solve the problem. Sometimes the problem is formulated a little differently. Note that the two given points define a pencil of (coaxal) circles. Any pair of the circles from the pencil has $CD$ as the radical axis; and the perpendicular bisector of $AB$ houses the centers of all the circles in the pencil. So, the problem is equivalent two finding a circle in the given pencil tangent to a given line. ### References 1. N. Altshiller-Court, College Geometry, Dover, 1980, #452", "\\frac{n (n – 3)}{2} = \\frac{6 (6 – 3)}{2} = 9.$$ The sum of the measures of all interior angles is: $$(n – 2) \\cdot 180^{\\circ}= 4 \\cdot 180^{\\circ}= 720^{\\circ}.$$ The measure of each interior angle: $$720^{\\circ} : 6 = 120^{\\circ}.$$ The center of an inscribed and an circumscribed circle is in the intersection of opposite vertices. If we are unsure at which point to use as the center for an inscribed and circumscribed circle, the best way is to bisect the angles and then their intersection will be the point we are looking for. These diagonals divide a hexagon into six congruent equilateral triangles, which means that their sides are all congruent and each of their angles are $60^{\\circ}$. For the area, we must again calculate the area of one triangle and multiply it by $6$. The same rules and formulas apply to other regular polygons. ## Polygons worksheets Name the polygon (72.8 KiB, 1,254 hits) Regular or not regular (122.5 KiB, 1,133 hits) Similar polygons (1.4 MiB, 878 hits) Concave or convex (132.3 KiB, 1,132 hits) Measuring angles (35.7 KiB, 1,399 hits) Regular polygons - Area (256.4 KiB, 1,286 hits)", "{,} 732 \\ cdot a}$ ${\\ displaystyle d_ {3} = 2 \\ cdot r_ {u} = 2 \\ cdot a}$ half a wrench size ${\\ displaystyle r_ {i} = {\\ frac {\\ sqrt {3}} {2}} \\ cdot a \\ approx 0 {,} 866 \\ cdot a}$ Perimeter radius ${\\ displaystyle r_ {u} = a}$ Interior angle ${\\ displaystyle \\ alpha = 120 ^ {\\ circ}}$ Construction for a given perimeter A regular hexagon can be very easily represented as a construction with compass and ruler from a circle by plotting the radius of the circle six times on the edge of the circle (see construction 1). The points obtained are the corners of the hexagon. Alternatively, according to Euclid, it is sufficient to remove twice on the edge of the circle. The missing corners can then be constructed using the straight lines through the center of the circumference and the already known corners (see construction 2 according to Euclid, as an animated graphic). Construction of a hexagon with compass and ruler Construction for a given side length Construction of a hexagon with a given side length A regular hexagon can also be constructed if an existing line is to be used as the side length . 1. Designate the end points of the route with A and B.", "Just Touching Three semi-circles have a common diameter, each touches the other two and two lie inside the biggest one. What is the radius of the circle that touches all three semi-circles? Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load. Biggest Bendy Four rods are hinged at their ends to form a quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic. Xtra Stage: 4 and 5 Challenge Level: Two solutions to this problem have been forthcoming from different students at the same school - Madras College. Thank you to Mike and Euan who used lots of trigonometry as well as to Thom who likewise resorted to double angles and the cosine rule and reduced the problem to solving a quadratic equation. Thom was also able to show the significance of the two roots. \\eqalign{ \\beta &=& \\frac{\\pi}{6} - \\frac{\\alpha}{2} \\\\ \\cos\\beta &=& \\frac{\\sqrt{3}}{2}\\cdot\\frac{3\\sqrt{3}}{2\\sqrt{7}} + \\frac{1}{2}\\cdot\\frac{1}{2\\sqrt{7}} \\\\ \\; &=& \\frac{10}{4\\sqrt{7}} = \\frac{5}{2\\sqrt{7}}} Using the cosine rule on $\\triangle ABP$ \\eqalign{ 4 &=& x^2 + 7 - 2x\\sqrt{7}\\cos\\beta \\\\ \\; &=& x^2 + 7 - 5x} Therefore $x^2 - 5x + 3 = 0$ $$x = \\frac{5\\pm\\sqrt{13}}{2}$$ Both solutions satisfy", "be formed by selecting $$4$$ points from the $$8$$ points except when: i) $$4$$ points have been taken from the line ii) $$3$$ points have been taken from the line and one from circle. Total required cases $$= 8C4- (4C4+4C3*4C1) = 53$$ Harshgmat wrote: Of the 8 points in a given plane 4 are on circle O and remaining 4 are on line L. Number of different quadrilaterals that can be formed by these 8 points are A) 52 B) 53 C) 54 D) 64 E) 74 _________________ Senior SC Moderator Joined: 22 May 2016 Posts: 2099 Of the 8 points in a given plane 4 are on circle O -Geometry - PS [#permalink] ### Show Tags 25 Aug 2018, 08:57 2 pushpitkc wrote: Therefore, the total number of quadrilaterals that can be formed by these points are 1+16+36 = 52(Option B) Senior PS Moderator Joined: 26 Feb 2016 Posts: 3307 Location: India GPA: 3.12 Of the 8 points in a given plane 4 are on circle O -Geometry - PS [#permalink] ### Show Tags 25 Aug 2018, 09:07 generis wrote: pushpitkc wrote: Therefore, the total number of quadrilaterals that can be formed by these points are 1+16+36 = 52(Option B) Thanks for notifying, made the necessary change! Of the 8 points in a given plane 4 are", "+0 # Help 0 119 1 How many square units are in the area of the convex quadrilateral with vertices (0, 0), (3, 0), (2, 2) and (0, 3)? Feb 1, 2021 #1 +11667 +1 How many square units are in the area of the convex quadrilateral with vertices $$P_1(0, 0), P_2(3, 0), P_3(2, 2)\\ and\\ P_4(0, 3)?$$ Hello Guest! $$A=A_{tra}+A_{tri}$$ $$A_{tra}=\\frac{1}{2}\\cdot (y_2+y_3)\\cdot x_3=\\frac{1}{2}\\cdot(3+2)\\cdot 2=\\color{blue}5$$ $$A_{tri}=\\frac{1}{2}\\cdot y_3\\cdot (x_4-x_3)=\\frac{1}{2}\\cdot 2\\cdot (3-2)=\\color{blue}1$$ $$A=A_{tra}+A_{tri}=5+1=\\color{blue}6$$ The area of the convex quadrilateral has 6 square units. ! Feb 1, 2021 edited by asinus Feb 1, 2021", "squares), one would draw three $$\\ \\frac{2\\cdot\\pi}3-$$arcs or two such arc and an interval which connects the middle-points of the opposite edges of the hexagons of the hexagonal lattice." ]

[ "Question Thirty percent of the beads in a bag are blue. One bead is randomly chosen and replaced. Then a second bead is chosen. What is the probability that neither is blue?", "7. HTHT 8. HTTH 9. THTH 10. TTHH 11. THHT 12. TTTH 13. TTHT 14. THTT 15. HTTT 16. TTTT Example Question #1 : Probability In a carnival game, a player must spin two independent spinners. If the first spinner has a probability of landing on red, and the second spinner has a probability of landing on red, what is the probability of the first spinner NOT landing on red, and the second spinner landing on red? Explanation: Since the spinners are independent of each other, we will need to multiply the probability of not landing on red on the first spinner and the probability of landing on red on the second spinner together. The probability of not landing on red for the first spinner is found by subtracting the probability of landing on red from . Thus, Example Question #1 : Probability Suppose there is a bag of marbles. There are two reds, two greens, and one orange marbles. Without replacement, what is the probability of choosing a red, and another red? Explanation: There are five marbles altogether. The first marble is a red, which is two of the five marbles in the bag. The probability of choosing a red for the first trial is: Without replacement, there are four marbles remaining, and only one red marble", "# Probability Marbles With Replacement There are 3 red marbles and 4 white marbles. A bag has 5 green marbles, 3 purple marbles, and 2 blue marbles. What is the probability what the sum of the three digits is a multiple of 2? Express your answer as a common fraction. There is one desired outcome and six possible outcomes. If I want to sample 2 marbles from my Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In each we specify either replacement (the marbles go back into the bag after each pull) or no replacement. an urn contains 6 green marbles and 6 yellow marbles. What is Probability without Replacement or Dependent Probability? In some experiments, the sample space may change for the different events. You draw 5 marbles out at random, without replacement. A bag contains 8 red marbles, 5 white marbles, and 10 blue marbles. A box contains 5 red marbles and 5 purple marbles. Bag B contains 12 marbles of which 4 are red and 8 are black. a) Draw a probability tree diagram to show all the outcomes the experiment. ----- What is the probability that exactly two of the", "• anonymous A bag contains 4 red marbles, 6 blue marbles, and 8 green marbles. One marble is to be chosen at random. What is the probability that the chosen marble will be green? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "Question A bag contains 4 blue marbles and 7 red marbles. Two marbles are randomly drawn without replacement. Find the probability of the event “”the first marble is red and the second marble is blue””?", "without replacement. P(A) = Possible outcomes for A and B: (5, 5), (5, 6). 20 white, 10 black, 6 red, 6 green. drawing a club, then a spade with replacement. What is the probability the pen is blue or red? P(blue or red) Definition of probability 3 2 0 4 or 2 7 0 There are 3 blue pens and 4 red pens. P(both marbles are purple) c. For A, there is a simpler way. A bag contains 6 red, 2 blue, and 5 green marbles. If it's a green marble, whatever marble it is, at whatever after the first pick, we leave it on the table. arrow_back Back to Tree Diagrams - conditional / without replacement Tree Diagrams - conditional / without replacement: Worksheets with Answers. \" I like to get a transparent cup and place some type of colored item inside, like marbles. Probability With And Without Replacement. with replacement. A single die is rolled. Contains some combinatoric-esque starter puzzles, and covers the whole GCSE syllabus, including mutually exclusive and independent events, experimental versus theoretical probability, probability trees (including algebraic probabilities) and sampling with and without replacement. · · = = Answer: 4: Three cards are chosen at random from a deck of 52 cards without replacement. Let us start with answer choice H, 18. Probability", "# Dependent probability ### Problem A bag contains 2 red marbles, 3 green marbles, and 4 blue marbles. If we choose a marble, then another marble without putting the first one back in the bag, what is the probability that the first marble will be green and the second will be red? Do 4 problems", "green shirts, 4 white shirts, and 2 red shirts in his shelf. Which of the following questions about these shirts could you use probability to solve? a. If Bill takes 1 shirt from the shelf, how many shirts will be left in the shelf? b. How many shirts did Bill have all together? c. If Bill picks 1 shirt from the shelf without looking, what color will it most likely be? d. How many more green shirts than red shirts did Bill have in the shelf? 218. Sheila packed 10 apples, 25 oranges, 36 strawberries and 5 pineapples in her picnic basket. Which of the following questions about these fruits could you use probability to solve? a. What is the total number of apples and oranges in the basket? b. If a fruit is picked at random, what kind of fruit is least likely to be picked? c. How many more apples than pineapples are there in the basket? d. How many fruits did Sheila pack in all? 219. A bag contains 11 blue marbles, 5 red marbles, and 4 green marbles. Pam draws one marble from the bag, records its color, and then puts it back in the bag. Then, he draws another marble and records its color. What is the probability of drawing 2 red marbles?", "out a standard text book for probability of simultaneous events, conditional probability, joint probability, dependent events etc. I think that we do have mutually exclusive and exhaustive events: There are 3 mutually exclusive events when picking 2 balls out of 3 (one red, one blue and one green ball): 1. Red and Blue; 2. Red and Green; 3. Blue and Green. And these outcomes are exhaustive events too, what other outcomes can you have? Also we do not have conditional probability here. The question asks: the probability that one is red and one blue, not the probability of second blue when first red. _________________ GMAT Tutor Joined: 24 Jun 2008 Posts: 1181 Followers: 368 Kudos [?]: 1207 [1] , given: 4 Re: Probability of Simultaneous Events - Veritas vs MGMAT [#permalink] ### Show Tags 15 Feb 2011, 03:26 1 KUDOS Expert's post Bunuel asked me to chime in here. Wow, this thread has become confused! This is all actually very simple. Say I have a bag with 3 red and 2 green marbles, and I stick both of my hands in the bag and grab hold of two marbles. Obviously the probability that I pick 1 red and 1 green marble can't possibly be any different if I take both my hands out at once, or if I", "marbles that are labeled as 1, 2, 3, and 4. Bag A contains 9 red marbles and 3 green marbles. One marble is taken out of the box at random. One marble is drawn out randomly. Task A bag contains red marbles and blue marbles. To be determined: the probability that the sum of the. 4, the probability that the first one is ALSO blue is 0. On your second pull, since we are not replacing marbles, there's a 3/4 chance that you pick a b. There are 3 red marbles and 4 white marbles. The probability of choosing a red marble is3 5. Select two marbles without replacement from a bag containing 1 white, 1 red and 2 green marbles. Suppose a jar contains 3 red and 4 white marbles. a) yellow marble b) red marble c) green marble. The probability of selecting a red marble and then a blue marble is 0. Choose one answer. asked by vicke on November 8, 2015; Math ~ Check Answers ~ 1. If two marbles are drawn without replacement, find the following probabilities using a tree diagram. What is the probability that all the marbles are red? Express each answer as a fraction or as a decimal accurate to 4 places. And guess what we’re both right because, it is", "• is3535 A bag contains 3 red marbles, 5 yellow marbles, and 4 blue marbles. One marble is chosen at random. What is the probability that the chosen marble will be blue? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "green marbles. In three draws, find the probability of obtaining white, black and green in that order. What is the probability of drawing 2 green marbles B. a) yellow marble b) red marble c) green marble. A jar contains 2 green marbles, 4 blue marbles, 3 yellow marbles, and 2 black marbles. If you draw with replacement, the odds are always 6/9. What is the probability that the marbles are of different colors? Log On. Solution: Not binomial: there are more than two outcomes for each trial. Example 2: A jar contains 4 blue marbles, 5 red marbles and 11 white marbles. Bag 2 contains 20 black marbles and 10 white ones. What is the probability that the random variable has an outcome. Two members are selected at random to. Find the probability of getting a blue, then white marble with replacement. You draw 5 marbles out at random, without replacement. 9) A box contains 5 red marbles and 7 green marbles. What is the probability of drawing 2 green marbles without replacement? B. *Worksheet – Finding the Probability of an Event II Find the probability for the following events. What is the probability that John. These marbles are selected at random without replacement. ● P(getting a color other than red) = P(25/55) ≈. If two marbles are", "random. A marble is chosen at random from the jar. Selecting a green marble in a second draw if the first marble is red. Total number of marbles is 7. a) Draw a probability tree diagram to show all the outcomes the experiment. Two marbles are drawn using the without-replacement method. Then on the next draw, the probability is 2/7. What Is The Probability That All The Marbles Are Red? The Probability That All The Marbles Are Red Is. Parameters: Number and color of marbles in the bag, replacement rule. For example, if we pick 2 marbles from a bag there are different possibilities of what we could do: • Probability With Replacement We take a marble put it back into the bag and pick another one. Two marbles are drawn, one at a time from the bag. A box contains 5 purple marbles, 3, green marbles, and 2 orange marbles. Two marbles are drawn at random without replacement. Ensure that \"With replacement\" option is not set. (To draw “with replacement” means that the first marble is put back before the second marble is drawn. Given information: A jar contains four marbles that are labeled as 1, 2, 3, and 4. What is the probability that all the marbles are red? The probability that all the marbles are", "A. A bag has 5 green marbles, 3. Me Too Probability. If trials are performed with replacement and/or the initial conditions are restored, you expect trial outcomes to be independent. Dependent on the first pick. (a) Sampling at random from the box five times with replacement, you have drawn a red marble all five times. Probability of each event = (# green marbles + # blue marbles)/ Total # of Marbles P1 = (15 + 25) / 125 = 40 / 125 Second event assumes a blue or green was chosen for first event so there is one fewer marble on top and also one fewer marble in the total number of marbles. a bag of marbles contains 5 green marbles,4 red marbles, and six yellow marbles. (Related resources available under the learner tab. P (ace, then king) A bag contains 6 purple marbles, 8 yellow marbles, and 5 blue marbles. Two marbles are drawn without replacement. Method 2—PIA. Everytime we have a probability question like this -- choosing multiple objects out of the same set -- we have to decide whether we have replacement or not. A bag contains 10 red marbles, 5 white marbles, and 6 blue marbles. What is the probability of drawing 3 blue marbles in a row, without replacement? 7. The probability of", "### Marbles and Bags Two bags contain different numbers of red and blue marbles. A marble is removed from one of the bags. The marble is blue. What is the probability that it was removed from bag A? ### Coin Tossing Games You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by a head (you win). What is the probability that you win? ### Win or Lose? A gambler bets half the money in his pocket on the toss of a coin, winning an equal amount for a head and losing his money if the result is a tail. After 2n plays he has won exactly n times. Has he more money than he started with? # Gambling at Monte Carlo ##### Stage: 4 Challenge Level: At first you may think that all the probabilities are the same, but consider a similar problem about tossing coins. The probability of one head with one coin is 1/2, the probability of two heads with two coins is 1/4 and the probability of three heads with three coins is 1/8, definitely not the same! The problem given was difficult", "balls from a bag containing 5 white and 6 green balls if the seven balls are drawn at random simultaneously. The probability that the balls are different colours is 52/105. 3% (answer) There are 1000 ways to pick 3 items from 10 Answer: What is the probability of picking 4 marbles (with & without replacement) of different color from a bag with 6 red, 5 blue, 4 yellow, and 6 green marbles? To solve this you will have to sum the probability of getting 4 red, 4 green, 4 blue and 4 yellow marbles. May 15, 2020 · I will draw 3 balls with replacement. seed (1839) balls = 0, and r >= 0. You will be asked to reach into the bucket, without looking, and select two balls. Jan 27, 2022 · When drawing is done with replacement, probability of success (say, red ball) is p =8/18 which will be same for all of the six trials. A bag contains 3 black balls, 4 green balls and 5 yellow balls. A second ball is drawn at random. 7% (Compare that with replacement of 6/100 or 6%) The bag contains 5 red and 4 black balls. let Y be the random variable representing tge gree balls. A bag contains 12 balls. If the second blind pick is another", "five blue marbles. Charmaine chooses a marble at random, and without putting it back, chooses another one. Really, the correct formula for the 'probability of not' is: 1- probability of yes. not just tossed in a net or bag! We bag each color, then package the 10-packs for you. \"Onionskins\" are glass marbles with swirls of layered colors that extend over the length of the marble. o 3 O 6 3 None of these. What is the probability of choosing a marble that is not blue?. There are _____ red marbles and _____ green marbles in Bag B. We are given that a bag contains 3 green marbles, 3 red marbles, and 2 blue marbles. The probability of selecting a green marble at random from a jar that contains only green, white and yellow marbles is The probability of selecting a white marble at random from the same jar is. The probability that none of the marbles are red is _____. The NRICH Project aims to enrich the mathematical experiences of all learners. he repeated this process 10 times and drew 6 red marbles and 4 blue marbles. no replacement Using selection (combinations), find the probability of:. There's two green marbles in the bag. But man, what a difference in color! Both of these are from the same", "monochromatic bags? Bag μ draw is returning a color, R say, as it is a monochromatic bag. Drawing from another color bag, B say, will produce R or B, in which case it is μ, i.e., mixed (polychromatic), which means the other bags are monochromatic, or G. For this last case, bag B is either polychromatic, in which case bag G is made of blue marbles and bag R of green marbles, or monochromatic, in which case bag G is mixed and bag R is full of blue marbles, but monochromatic for either situation, hence to be chosen on top of bag μ.", "• anonymous A bag contains 5 blue marbles and 9 red marbles You choose one marble at random, then choose another marble at random. Find the probability that both marbles are blue when (a) you replace the first marble, and (b) you do not replace the first marble. Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "the bag. What is the probability that all three marbles are red when (a) you replace each marble before selecting the next 8. ### To Steve Sorry here is completed question A bag contains 3 red marbles, 3 blue marbles and 3 yellow marbles. Which of the following methods of selecting 2 marbles will have the greater probability of choosing 2 red marbles? 1- Randomly 9. ### probability A bag contains 3 red marbles, 3 blue marbles and 3 yellow marbles. Which of the following methods of selecting 2 marbles will have the greater probability of choosing 2 red marbles? 10. ### Algebra 1 1. A bag contains 3 red marbles, 2 blue marbles, and 5 green marbles. What is the probability of randomly selecting a red marble then, without replacing it, randomly selecting a green marble? 2. A bag contains 5 blue marbles, 6 More Similar Questions" ]

[ "= 90 ways. Question 7. Find the number of diagonals of a polygon of 20 sides. Number of diagonals = nC2 – n = 20C2 – 20 = 190 – 20 = 170 ways. Question 8. In a party each person shakes hands with everyone else. If there are 25 members in the party, Calculate the number of handshakes. Number of hand shakes = 25C2 = $$\\frac{25 \\times 24}{2 \\times 1}$$ = 300 ways Question 9. In how many ways can 6 red and 4 white marbles be chosen from a bag containing 10 Red and 6 White marbles. We have to select 6 marbles out of 10 and 4 white out of 6 white marbles = 10C6 × 6C4 = 210 × 15 = 3,150. Question 10. In how many ways can 6 people be chosen out of 10 people if one particular person is always included. One particular person is included, we have to choose from 9, this can be done in 9C3 ways. Part – C 2nd PUC Basic Maths Permutations and Combinations Ex 2.3 Three marks Questions and Answers Question 1. If a convex polygon has 170 diagonals. Find the number of sides of the polygon. Number of diagonals = nC2 – n = 170. $$\\frac{n(n-1)}{2}-n=170$$ n2 – n – 2n = 340 ⇒", "# There are 40 marbles of radius xcm each & 60 marbles of radius ycm each. If x & y vary such that x + y = 15, how do I find the (exact) value of x & y that will make the sums of the volumes a minimum? ## All the marbles are spherical. I don't know what the writer completely means by \"sum of the volume\", so, sorry if it's a bit ambiguous. My current approach is to express $y$ in terms of $x$ and finding the sum of all 100 marbles through substitution, then setting its derivative to $0$. After taking the derivative, everything just becomes messy. I have a feeling that the actual solution involves the manipulation of the $2 : 3$ ratio of the marbles but I don't know how to work it out. May 29, 2018 the values that makes the volume minimum: $\\textcolor{b l u e}{x = 15 \\left(3 - \\sqrt{6}\\right)}$ $\\textcolor{b l u e}{y = 15 - 15 \\left(3 - \\sqrt{6}\\right)}$ #### Explanation: The volume of the sphere is given by $\\textcolor{red}{V = \\frac{4}{3} \\cdot \\pi \\cdot {\\left(r a \\mathrm{di} u s\\right)}^{3}}$ volume of 40 spherical marble of a radius x is given by : $V = \\left(40\\right) \\frac{4}{3} \\cdot \\pi \\cdot {x}^{3} = \\frac{160}{3} \\cdot \\pi \\cdot {x}^{3}$ volume", "for each of these 6 ways is the same. We first note that the denominators should be counted by the same number. This number is $2 \\cdot 3 \\cdot 4 \\cdot 5=120$. This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the $k-th$ step involves $k+1$ numbers to choose from. The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally. The same goes for the blue ones. The numerator must equal $(1 \\cdot 2)^2$. Therefore, the probability for each of the orderings of $RRBB$ is $\\frac{4}{120}=\\frac{1}{30}$. There are 6 of these, so the total probability is $\\boxed{\\bf{(B)} \\frac{1}{5}}$. ## Solution 3 First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a $\\frac{1}{2}$ chance each. We can assume he chooses Red(chance $\\frac{1}{2}$), and then multiply the final answer by two for symmetry. Now, there are two red balls and one blue ball in the urn. Then, he can either choose another Red(chance", "those. However, since $w_1 < 13$, testing $2$ and $7$ for $w_1$ does not give a correct product. Thus, $\\frac{27}{50}$ must be a reduced form of the actual fraction. First assume that the fraction was reduced from $\\frac{54}{100}$, yielding the equations $b_1\\cdot b_2 = 54$ and $(b_1 + w_1)(b_2 + w_2) = 100$. Factoring $b_1 \\cdot b_2 = 54$ and saying WLOG that $b_1 < b_2 < 25$ gives $(b_1, b_2) = (3, 18)$ or $(6, 9)$. Trying the first pair and setting the denominator equal to 100 gives: $(3 + w_1)(18 + w_2) = 100$ Since $w_1 + w_2 = 4$, the pairs $(w_1, w_2) = (1, 3), (2,2),$ and $(3,1)$ can be tried, since each box must contain at least one white marble. Plugging in $w_1 = w_2 = 2$ gives the true equation $(3 + 2)(18 + 2) =100$, so the number of marbles are $(w_1, w_2, b_1, b_2) = (2, 2, 3, 18)$ Thus, the chance of drawing 2 white marbles is $\\frac{w_1 \\cdot w_2 }{(w_1+ b_1)(w_2 + b_2)} = \\frac{4}{100} = \\frac{1}{25}$ in lowest terms, and the answer to the problem is $1 + 25 = \\boxed{026}.$ For completeness, the fraction $\\frac{81}{150}$ may be tested. $150$ is the highest necessary denominator that needs to be tested, since the maximum the denominator $(w_1+ b_1)(w_2", "You can put this solution on YOUR website! There are r and g marbles in a bag, if there are 84 marbles all together are 84 how many are r ? ------------- Equation: r + g = 84 ===== Cheers, Stan H. ------- Question 993191: the numerator of a certain fraction is 5 times the denominator. if 6 is added to both the numerator and the denominator, the result fraction is equivalent to 2. what was the original factory? You can put this solution on YOUR website! Hi there, the numerator of a certain fraction is 5 times the denominator Make the denominator = 'x' = 5x/x if 6 is added to both the numerator and the denominator, the result fraction is equivalent to 2. 5x + 6/x + 6 = 2/1 Cross multiply 2(x + 6) = 5x + 6 2x + 12 = 5x + 6 Collect like terms. 2x - 5x = 6 - 12 -3x = - 6 x = 2 Original fraction = 10/2 Hope this helps :-) Question 993162: There are 18 pieces of chalk the ratio ofcalk that is yellow to chalk that is not yellow is 1:2. How many pieces of chalk are yellow? You can put this solution on YOUR website! Since the ratio of yellow to not yellow", "we wanted to know how many different ways we could arrange them. The equation would be as follows: Explained: We would have 10 friends to choose from for the first spot in line, but only 9 for the second and only 8 for the third so on and so forth. Example 2 Mike has 10 marbles 5 are red, 2 are blue and 3 are green, in how many different ways can he arrange them into a row? The equation would be as follows: Explained: We start off by taking 10! because that is the number of marbles Mike has, we then want to divide by 5!, 2! and 3! because those are the sets of the same marbles he has. This is because swaping 2 or more marbles of the same color will have no effect on the way the row of marbles will look and so we want to exclude those cases. Calculator Calculator code For those who are interested, I have included the calculator code below: <input type=\"number\" id=\"inputFactorial\" value=\"6\"> <button onclick=\"findFactorial()\">Find Factorial</button> <p id=\"output\"></p> <script> function findFactorial() { var x = document.getElementById(\"inputFactorial\").value; if(x%1 != 0 || x<0){ document.getElementById(\"output\").innerHTML = \"please enter a valid input\"; }else if(x==0){ document.getElementById(\"output\").innerHTML = \"0\"; }else{ var n=1; while(x>0){ n=n*x; x=x-1; } document.getElementById(\"output\").innerHTML = n; } } </script>", "of Counting to get the number of ways of winning only one ticket. vcsharp2003 There are C(10,2)⋅C(2,2) ways to select 2 winners, so you have that first term right in the numerator. Yes, I get that. Gold Member The way I have written the second term is first select the winning ticket and after that select a non-winning ticket; these two steps would yield the big step of winning only one ticket. Then I simply use the Fundamental Rule of Counting to get the number of ways of winning only one ticket. The seconnd term should be ##C(10,1) \\cdot C(2,1) \\cdot C(9990,1)## There are two configurations for each pair of selected tickets. vcsharp2003 ( ticket 1 or ticket 2 can be the winner) But wouldn't doing that duplicate the number of ways of winning only 1 ticket? Gold Member Yeah, I think you are right. It would over select. Sanity check with small numbers: Imagine a bag of 6 marbles, 4 red, 2 green there are 6 *5 = 30 ways to select 2 marbles You can select both red in 4 *3 = 12 ways You can select both green in 2*1 = 2 ways And you can select 1 red and one green in C(2,1) 4*2 ways = 16 ways $$12 + 2 + 16 =", "interchanged, what is true of the resulting determinant? Method: Sadly for me, I have little clue of what to do. I could tell you why if one row of a determinant is 0 the whole things value is 0, but I don't know where to start on this one. Thank you 2. Hello, MathBlaster47! The first two are Combination problems, not Permutations. 1) There are 10 women and 8 men in a club. How many different committees of 6 people can be selected from the group if 3 women and 3 women are to be on the committee? There are: .$\\displaystyle _{10}C_3 \\:=\\:{10\\choose3} \\:=\\:\\frac{10!}{3!\\,7!} \\:=\\:120$ ways to choose 3 women. There are: .$\\displaystyle _8C_3 \\:=\\:{8\\choose3} \\:=\\:\\frac{8!}{3!\\,5!} \\:=\\:56$ ways to choose 3 men. Therefore, there are: .$\\displaystyle 120\\cdot56 \\:=\\:6720$ committees with 3 women and 3 men. 2) Two marbles are drawn from a bag containing 6 white, 4 red, and 6 green marbles. Find the probability of both marbles being white. There are: .$\\displaystyle _{16}C_2 \\:=\\:{16\\choose2} \\:=\\:\\frac{16!}{2!\\,14!} \\:=\\:120$ possible two-marbles samples. There are: .$\\displaystyle _6C_2 \\:=\\:{6\\choose2} \\:=\\:\\frac{6!}{2!\\,4!} \\:=\\: 15$ ways to get two white marbles. Therefore: .$\\displaystyle P(\\text{two whites}) \\;=\\;\\frac{15}{120} \\;=\\;\\frac{1}{8}$ 3) If two rows of a determinant are interchanged, what is true of the resulting determinant? It depends on which two rows are interchanged. If the two rows", "than its denominator. If both the numerator and the denominator of the fraction are increased by 2, then the resulting fraction is equal to 3/5. What is the fraction? Write an equation that 3. ### Math Divide. Start Fraction left-parenthesis Start Fraction x squared plus 6 x plus 9 over x minus 1 End Fraction right-parenthesis over left-parenthesis Start Fraction x squared minus 9 over x squared minus 2 x plus 1 End Fraction 4. ### Math The diagram shows the contents of a jar of marbles. You select two marbles at random. One marble is drawn and not replaced. Then a second marble is drawn. What is the probability of selecting a red marble and then another red 1. ### math Select the letter of the best answer choice. Use 3.14 or Start Fraction 22 over 7 End Fraction for pi. What is the circumference of the circle? Round your answer to the nearest foot. circle with line across 28ft A. 88 ft B. 84 ft 2. ### math Write the fraction in simplest form. 1. start fraction 8 over 10 end fraction (1 point) Start Fraction 16 over 20 End Fraction start fraction 2 over 5 end fraction start fraction 4 over 5 end fraction start fraction 2 over 10 end 3. ### math 5÷3 3", "that one cube weighs as much as 2 coins Hence, from the above, We can conclude that 1 Cube weighs as much as 2 Coins Question 5. One cube weighs as much as _______ marbles. In the given figure, Let each cube be x Let each marble be y It is given that the two pans are balanced So, LHS = RHS Now, 3x + 10y = 13y + x Rearrange the like terms at one side So, 3x – x = 13y – 10y 2x = 3y $$\\frac{x}{y}$$ = $$\\frac{3}{2}$$ x = $$\\frac{3}{2}$$y So, 1 Cube = $$\\frac{3}{2}$$ Marbles So, From the above expression, We can say that one cube weighs as much as $$\\frac{3}{2}$$ marbles Hence, from the above, We can conclude that 1 Cube weighs as much as $$\\frac{3}{2}$$ Marbles Practice Solve. Question 6. 1 = $$\\frac{3}{5}$$ × ________ Let the missing term be x in the given expression So, 1 = $$\\frac{3}{5}$$ × x Multiply by $$\\frac{5}{3}$$ on both sides So, x = $$\\frac{5}{3}$$ Hence, from the above, We can conclude that the missing term in the given expression is: $$\\frac{5}{3}$$ Question 7. $$\\frac{1}{6}$$ × ______ = 1 Let the missing term in the given expreesion be x So, $$\\frac{1}{6}$$ × x = 1 Multiply with 6 on both sides $$\\frac{6}{6}$$x = 1 (6)", "36 E.48 12- Of the following, which number is the greatest? A. 0.092 B. 0.8913 C. 0.8923 D. 0.8896 E. 0.88 13- Solve: $$\\frac{7}{8}−\\frac{3}{4}=$$? A. 0.125 B. 0.375 C. 0.5 D. 0.625 E. 0.775 14- Which of the following is the closest to $$4.02$$? A. 4 B. 4.2 C. 4.3 D. 4.5 E. 3.5 15- Which of the following statements is False? A. $$(7×2+14)×2=56$$ B. $$(2×5+4)÷2=7$$ C. $$3+(3×6)=21$$ D. $$4×(3+9)=48$$ E.$$14÷(2+5)=5$$ 16- When 78 is divided by 5, the remainder is the same as when 45 is divided by A. 2 B. 4 C. 5 D. 7 D. 9 17- John has 2,400 cards and Max has 606 cards. How many more cards does John have than Max? A. 1,794 B. 1,798 C. 1,812 D. 1,828 E. 1,994 18- In the following right triangle, what is the value of $$x$$? A. 15 B. 30 C. 45 D. 60 E. It cannot be determined from the information given 19- What is 20 percent of 120? A. 20 B. 24 C. 30 D. 40 E.44 20- In a basket, the ratio of red marbles to blue marbles is 3 to 2. Which of the following could NOT be the total number of red and blue marbles in the basket? A. 15 B. 32 C. 55 D. 60 E. 70 ##", "Modeling Real Life Example You use 100 tiles to make a mosaic. 80 of them are square tiles. Your friend uses 10 tiles to make a mosaic. Six of them are square tiles. Do the mosaics have the same fraction of square tiles? Determine whether the fractions are equivalent. Write your friend’s fraction as hundredths in fraction form. Then compare. Hence, The mosaics don’t have the same fraction of square tiles. Show and Grow Question 13. You use 10 beads to make a bracelet. Seven of them are purple. Your friend uses 100 beads to make a bracelet. 70 of them are purple. Do the bracelets have the same fraction of purple beads? Explanation: It is given that you use 10 beads to make a bracelet and out of 10 beads, seven of them are purple. So, The portion of the bracelet that is purple is: $$\\frac{7}{10}$$ So, The given fraction is: $$\\frac{7}{10}$$ So, To write $$\\frac{7}{10}$$ as hundredths, multiply the fraction and numerator of $$\\frac{7}{10}$$ with 10. So, Firstly the numerators 7 and 10 are multiplied and then the denominators 10 and 10 are multiplied We know that, Decimal = Numerator ÷ Denominator So, The representation of $$\\frac{70}{100}$$ in the place-value chart is: Hence, The representation of $$\\frac{7}{10}$$ as hundredths in the fraction form is: $$\\frac{70}{100}$$ It", "### Home > CC2MN > Chapter 4 > Lesson 4.2.2 > Problem4-42 4-42. A bag contains $4$ brown marbles, $3$ green marbles, $2$ red marbles, and $1$ purple marble. Calculate the probability of drawing each color, and write each answer as a fraction, as a percent, and as a decimal. Find the total number of marbles in the bag. $4+3+2+1=10$ $\\large \\frac{\\text{# of brown marbles}}{\\text{Total # of marbles}}=\\frac{4}{10}$ Translate the fraction into a percent by using a Giant One with $10$ as both the numerator and denominator. $\\frac{2}{5}=40\\%$ To find the decimal, divide $2$ by $5$. $2\\div5=0.4$ Now repeat this process for green, red, and purple marbles.", "pink marbles: $$\\displaystyle \\frac{{{}_{{10}}{{C}_{3}}\\times {}_{5}{{C}_{1}}+{}_{{10}}{{C}_{4}}\\times {}_{5}{{C}_{0}}}}{{{}_{{15}}{{C}_{4}}}}=\\frac{{810}}{{1365}}\\approx .593$$. (d) To get the probability that at least 1 is pink, we can get the probability that all are blue, and then subtract this from 1 to get its complement (since anything else would have at least one pink marble): $$\\displaystyle 1-\\frac{{{}_{{10}}{{C}_{0}}\\times {}_{5}{{C}_{4}}}}{{{}_{{15}}{{C}_{4}}}}=1-\\frac{{1\\times 5}}{{1365}}\\approx .996$$. This makes sense since so many are pink in the bag. A committee of 6 is selected from a group of 6 boys and 5 girls. (a) What is the probability that the committee is all boys? (b) What is the probability that exactly 4 out of the 6 are boys? We know the total number of ways 6 kids is selected from the group of the boys and girls (11) is $${}_{{11}}{{C}_{6}}=462$$; order doesn’t matter, since there are no positions. (a) There is only way to get all boys, since we are only 6 boys total. The probability then is $$\\displaystyle \\frac{1}{{{}_{{11}}{{C}_{6}}}}=\\frac{1}{{462}}\\approx .002$$ (low). (b) The number of possible ways 4 out of 6 boys could be chosen is $${}_{6}{{C}_{4}}=15$$, and the number of ways to pick 2 girls out of the remaining 5 kids is $${}_{5}{{C}_{2}}=10$$. The probability then that exactly 4 out of the 6 are girls is $$\\displaystyle \\frac{{{}_{6}{{C}_{4}}\\times {}_{5}{{C}_{2}}}}{{{}_{{11}}{{C}_{6}}}}=\\frac{{150}}{{462}}\\approx .325$$. The letters in the word MISSISSIPPI are written on pieces of", "thinking in this way, you will find that this is very helpful when solving for one of the missing parts of the proportion. Since percent statements always involve three numbers which may change (the 100 is always the same), given any two of these numbers, we can find the third number using the proportion. Example What percent of 40 is 6? First, thing to notice is that we are looking for the percent. So the $p$ over 100 is going to stay the same in the proportion. $\\frac{a}{b}=\\frac{p}{100}$ We need to fill in the $a$ and the $b$ so we can solve for $p$, the percent. The words “of 40” lets us know that 40 is the base and 6 is the amount. Here is our proportion to solve. $\\frac{6}{40}=\\frac{p}{100}$ Now we can use cross products and solve. $40p & = 600\\\\p & = 15$ Let’s look at another example. Example Jeremy has 25 marbles and 12 of them are cat’s eye marbles. What percent of his marbles are cat’s eye marbles? We can think of this problem as “What percent of 25 is 12?” 12 is the amount $(a)$ and 25 is the base $(b)$. We need to find the percent $(p)$. $\\frac{a}{b}& =\\frac{p}{100}\\\\\\frac{12}{25}& =\\frac{p}{100}\\\\25p& =1,200\\\\\\frac{25p}{25}& =\\frac{1,200}{25}\\\\p& =48$ Since $p = 48, our answer is \\frac{48}{100}$, which", "+0 # If you could explain these to me, that would be greatly appreciated! 0 149 4 +273 I need help asap! Thanks in advance! Apr 25, 2018 #1 +971 +2 87. Since there are three digits and each digit has 10 options, there are a total of: 10^3 ways of identifying students. The first number can be a 0, since it is not supposed to be a number, just an identification. 10^3 = 1000 Apr 25, 2018 #2 +971 +2 88. There are a total of: $$\\binom{15}{4}$$ ways to choose the debaters. This is equal to 1365. Since we need those four people, with only one way to choose them. The chance this will happen is 1/1365. Apr 25, 2018 #3 +3763 +1 89. The probability of choosing a dime first is $$\\frac{2}{10}=\\frac{1}{5}$$ . Next, picking a quarter without replacement is just $$\\frac{3}{10}$$. Multiplying, we get $$\\frac{1}{5}*\\frac{3}{10}=\\frac{3}{50}$$ . Apr 25, 2018 #4 +3763 +1 90. The total number of marbles: $$12+8+4+16=40$$ Blue Marble: $$\\frac{4}{40}=\\frac{1}{10}$$ Red marble: $$\\frac{12}{40}=\\frac{3}{10}$$ So, the answer is just $$\\frac{1}{10}*\\frac{3}{10}=\\boxed{\\frac{3}{100}}$$ . Apr 25, 2018", "is needed for one layer, and 84 stones need for two layers. Multiplication of 84 by$2.32 gives 194.88. Thus, the total cost is $194.88. 20. E: The equation may be solved for x by first subtracting 6x from both sides of the equation. Doing so gives -3x = -15, where x = 5. Substituting 5 for x into the second expression gives 5 + 8, which equals 13. 21. B: There are 1,000 milliliters in 1 liter. 22. D: The problem may be modeled with the equation, 6.50 = 5.00 + 0.50x, where x represents the number of passengers. Solving for x gives x = 3. Thus, there were 3 passengers, plus 1 driver, for a total of 4 people in the vehicle. 23. C: This problem may be represented as 1/9 . 9, which equals 1. 24. D: Taking out 3 of each color will ensure that he has 1 of each color. Thus, he needs to take out 9 marbles, in all. 25. E: 0.20% = 0.002, and 1/500=0.002. 26. D: The increase from term to term is twice the increase from the previous two terms. Thus, the increase from 19 to the missing term will be 20, or twice the increase of 10. Thus, the missing term is equal to 19 + 20, or 39.", "we have $\\frac{1}{15}+\\frac{1}{15}+\\frac{1}{15}=\\boxed{\\textbf{(B) }\\frac15}$ ~quacker88 ## Solution 2 We know that we need to find the probability of adding 2 red and 2 blue balls in some order. There are 6 ways to do this, since there are $\\binom{4}{2}=6$ ways to arrange $RRBB$ in some order. We will show that the probability for each of these 6 ways is the same. We first note that the denominators should be counted by the same number. This number is $2 \\cdot 3 \\cdot 4 \\cdot 5=120$. This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the $k-th$ step involves $k+1$ numbers to choose from. The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally. The same goes for the blue ones. The numerator must equal $(1 \\cdot 2)^2$. Therefore, the probability for each of the orderings of $RRBB$ is $\\frac{4}{120}=\\frac{1}{30}$. There are 6 of these, so the total probability is $\\boxed{\\bf{(B)} \\frac{1}{5}}$. ## Solution 3 First, notice that when George chooses a ball he just adds another ball of the same color. On George's", "5 [Divide the numerator and the denominator by 6.] The three equivalent fractions of 18 / 30 are 9 / 15, 6 / 10 and 3 / 5. 94. Andrew has 4 red and 6 blue marbles. Write an equivalent fraction for the fraction of blue marbles Andrew has. a. $\\frac{2}{5}$ b. $\\frac{3}{5}$ c. $\\frac{1}{3}$ d. $\\frac{1}{5}$ #### Solution: Total marbles = 4 + 6 = 10 Fraction of blue marbles Andrew has = (Blue marbles) / (Total marbles) = 610 [Substitute the values.] Factors of 6 are 1, 2, 3, 6. Factors of 10 are 1, 2, 5, 10. The common factor of 6 and 10 is 2. = 6÷210÷2 = 35 [Divide the numerator and denominator by the common factor.] So, the equivalent fraction for the fraction of blue marbles Andrew has is 3 / 5. 95. A manufacturing company produces 172 cycles per month. If 60 of them are yellow-colored cycles and 40 of them are pink-colored, then what fraction of cycles are pink-colored when compared to the yellow-colored cycles? a. $\\frac{4}{5}$ b. $\\frac{3}{4}$ c. $\\frac{5}{6}$ d. $\\frac{2}{3}$ #### Solution: The fraction of pink colored cycles to the yellow colored cycles = 40 / 60 40 = 1, 2, 4, 5, 8, 10, 20, 40 and 60 = 1, 2, 3, 4, 5, 6, 10,", "the first draw and a red marble on the second draw. B1R1, B1R2, B1R3 B2R1, B2R2, B2R3 B3R1, B3R2, B3R3 B4R1, B4R2, B4R3 B5R1, B5R2, B5R3 B6R1, B6R2, B6R3 B7R1, B7R2, B7R3 B8R1, B8R2, B8R3 b. Calculate P(RR). Solution 3.24 b. You can use the tree diagram. There are nine ways to draw two reds and 121 possible outcomes. So, P(RR) = $9121.9121.$. Each draw is independent, so you can also use the formula: P(RR) = P(R)P(R) = $(311)(311)(311)(311)$ = $9121.9121.$ c. Calculate P(RB OR BR). Solution 3.24 c. The tree diagram shows that there are 24 ways to draw RB and 24 ways to draw BR. There are 121 possible outcomes, so P(RB or BR) = $24+24121=4812124+24121=48121$. The events RB and BR are mutually exclusive, so P(RB OR BR) = P(RB) + P(BR) = P(R)P(B) + P(B)P(R) = $(311)(811)(311)(811)$ + $(811)(311)(811)(311)$ = $48121.48121.$ d. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw). Solution 3.24 d. Follow the path on the tree. There are three ways to get a red marble on the first draw and eight ways to get a blue on the second draw. There are 3 × 8 = 24 ways to draw red then blue, so P(RB) = $2412124121$. Can you think of another way to find this" ]

[ "# Thread: Mixed bag part 2: pemutations, combinations and determinants 1. ## Mixed bag part 2: pemutations, combinations and determinants I just want to make sure I have my method down properly, as usual, so here goes another round of questions. 1: There are 10 women and 8 men in a club. How many different committees of 6 people can be selected from the group if equal numbers of men and women are to be on the committee? Method: There are $_{10}W_3$ ways to select 3 women for the committee and $_8M_3$ ways to select 3 men for the committee. Therefore, there are $_{10}W_3\\cdot_8M_3$different committees possible, so $_{10}W_3 \\cdot _8M_3 \\rightarrow \\frac{10!}{(10-3)!} \\cdot \\frac{8!}{(8-3)!}\\rightarrow\\frac{10!}{7!}\\cdot\\frac{8!}{5!}\\ri ghtarrow$ $\\frac{10\\cdot9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3 \\cdot2\\cdot1}{7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot 1}\\cdot\\frac{8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2 \\cdot1}{5\\cdot4\\cdot3\\cdot2\\cdot1}\\rightarrow\\frac {3628800}{5040}\\cdot\\frac{40320}{120}=241920$ Thus there are 241920 possible committees. 2: Two marbles are drawn from a bag containing 6 white, 4 red, and 6 green marbles. Find the probability of both marbles being white. Method: There are 16 marbles in all so there are $_{16}M_2\\rightarrow\\frac{16!}{14!}\\rightarrow \\frac{2.09227899 \\cdot 10^{13}}{87178291200}=240$ways of picking 2 marbles. This is where I get fuzzy, I'm not too sure what to do next--should I multiply out all the permutations for all the colors and then work from there? Or should I just focus on the white marbles? 3: If two rows of a determinant are interchanged, what is", "interchanged, what is true of the resulting determinant? Method: Sadly for me, I have little clue of what to do. I could tell you why if one row of a determinant is 0 the whole things value is 0, but I don't know where to start on this one. Thank you 2. Hello, MathBlaster47! The first two are Combination problems, not Permutations. 1) There are 10 women and 8 men in a club. How many different committees of 6 people can be selected from the group if 3 women and 3 women are to be on the committee? There are: .$\\displaystyle _{10}C_3 \\:=\\:{10\\choose3} \\:=\\:\\frac{10!}{3!\\,7!} \\:=\\:120$ ways to choose 3 women. There are: .$\\displaystyle _8C_3 \\:=\\:{8\\choose3} \\:=\\:\\frac{8!}{3!\\,5!} \\:=\\:56$ ways to choose 3 men. Therefore, there are: .$\\displaystyle 120\\cdot56 \\:=\\:6720$ committees with 3 women and 3 men. 2) Two marbles are drawn from a bag containing 6 white, 4 red, and 6 green marbles. Find the probability of both marbles being white. There are: .$\\displaystyle _{16}C_2 \\:=\\:{16\\choose2} \\:=\\:\\frac{16!}{2!\\,14!} \\:=\\:120$ possible two-marbles samples. There are: .$\\displaystyle _6C_2 \\:=\\:{6\\choose2} \\:=\\:\\frac{6!}{2!\\,4!} \\:=\\: 15$ ways to get two white marbles. Therefore: .$\\displaystyle P(\\text{two whites}) \\;=\\;\\frac{15}{120} \\;=\\;\\frac{1}{8}$ 3) If two rows of a determinant are interchanged, what is true of the resulting determinant? It depends on which two rows are interchanged. If the two rows", "other hand, $P_r(A)P_r(B)=\\frac{4}{5}\\frac{3}{5}=\\frac{12}{25}$. Since $P_r(A\\cap B)\\ne P_r(A)P_r(B)$, the events $A$ and $B$ are not independent. References. [1] Essential Discrete Mathematics for Computer Science, Harry Lewis and Rachel Zax, Princeton University Press, 2019 # Introductory Combinatorics: Combinations 2 Let us begin with an example. Example 1. A jar contains 11 marbles. In how many ways can they be placed in four boxes, with 3 marbles in the first box, 4 marbles in the second box, 2 marbles in the third box, and the 2 remaining marbles in the fourth box? Solution. The total number of outcomes is $$\\begin{pmatrix}11\\\\3\\end{pmatrix}\\cdot \\begin{pmatrix}8\\\\4\\end{pmatrix}\\cdot \\begin{pmatrix}4\\\\2\\end{pmatrix}\\cdot \\begin{pmatrix}2\\\\2\\end{pmatrix}=\\frac{11!}{3!8!}\\cdot\\frac{8!}{4!4!}\\cdot\\frac{4!}{2!2!}\\cdot\\frac{2!}{2!0!}=\\frac{11!}{3!4!2!2!}=69,300$$ This example can be generalized to the following question. A set of $n$ distinct items is to be divided into $r$ distinct groups of respective sizes $n_1, n_2, \\cdots, n_r$ such that $\\sum_{i=1}^r n_i=n$. How many different divisions are possible? There are $\\begin{pmatrix}n\\\\n_1\\end{pmatrix}$ possible choices for the first group, there are $\\begin{pmatrix}n-n_1\\\\n_2\\end{pmatrix}$ possible choices for the second group, there are $\\begin{pmatrix}n-n_1-n_2\\\\n_3\\end{pmatrix}$ possible choices for the third group, …, there are $\\begin{pmatrix}n-n_1-n_2-\\cdots-n_{r-1}\\\\n_r\\end{pmatrix}$ possible choices for the $r$-th group. Therefore the total number of coutcomes is \\begin{align*}\\begin{pmatrix}n\\\\n_1\\end{pmatrix}\\begin{pmatrix}n-n_1\\\\n_2\\end{pmatrix}\\begin{pmatrix}n-n_1-n_2\\\\n_3\\end{pmatrix}\\cdots\\begin{pmatrix}n-n_1-n_2-\\cdots -n_{r-1}\\\\n_r\\end{pmatrix}\\\\=\\frac{n!}{(n-n_1)!n_1!}\\frac{(n-n_1)!}{(n-n_1-n_2)!n_2!}\\cdots\\frac{(n-n_1-n_2\\cdots-n_{r-1})!}{0!n_r!}\\\\=\\frac{n!}{n_1!n_2!\\cdots n_r!}\\end{align*} The number of divisions of $n$ distinct items into $r$ distinct groups of respective sizes $n_1,n_2,\\cdots,n_r$ is denoted by $\\begin{pmatrix}n\\\\n_1,n_2,\\cdots,n_r\\end{pmatrix}$. That is, $$\\label{eq:multcoeff}\\begin{pmatrix}n\\\\n_1,n_2,\\cdots,n_r\\end{pmatrix}=\\frac{n!}{n_1!n_2!\\cdots n_r!}$$ where $n=n_1+n_2+\\cdots+n_r$. Example 2. A police department", "problem statement. Suppose we have $n_i \\gt 0$ marbles of color $j$, $j=1,2,\\ldots,l$, for a total of $n=n_1+\\cdots +n_l$ marbles. For any subset of these colors $J = \\{j_1,\\ldots,j_s\\}$, what fraction of all simple random samples of size $k$ contain exactly these colors? The answer is immediate: it's the number of all simple random samples of size $k$ from a bag containing only marbles whose colors are in $J$, divided by the number of all simple random samples of size $k$ from the original bag: $$f(J, k) = \\binom{n_{j_1}+n_{j_2}+\\cdots+n_{j_s}}{k} / \\binom{n}{k}$$ By the Principle of Inclusion-Exclusion, the fraction of simple random samples of size $k$ without all $l$ colors equals $$\\sum_{\\varnothing \\ne J \\subset \\{1,\\ldots,l\\}}(-1)^{l-\\vert J\\vert}f(J,k).$$ The computational effort, properly coded, is $O(2^l)$--fine for small numbers of distinct colors--and just $O(l)$ when all the $n_i$ are equal, because all terms for $J$ of a given size will be equal and only sizes $1$ through $l$ need be considered. A Mathematica program that directly expresses the general formula is h[x_, k_Integer] /; Total[x] >= k > 0 := With[{n = Total[x], l = Length[x]}, Sum[(-1)^(l - j) Binomial[m, k]/Binomial[n, k], {j, 1, l}, {m, Total /@ Subsets[x, {j}]}] ] Using it, it takes 0.015 seconds (on one core) to find h[{1000,1000,1000,1000,1000}, k] for k from 5 through 34. Rounded,", "28 '16 at 11:18 • Your fellow classmate is correct. – drhab May 28 '16 at 11:18 • In small sample you are using multihypergeometric to calculate the exact probabilities; in your large sample case, you are using multinomial. – BGM May 28 '16 at 11:49 The answer provided is wrong. In the limit of very many marbles, it doesn't matter whether you draw with or without replacement, since the ratios will stay approximately the same. Thus in this limit the probability of drawing $1$ red, $2$ blue and $1$ yellow marbles is $$\\binom4{1,2,1}\\cdot\\frac3{12}\\cdot\\frac4{12}\\cdot\\frac4{12}\\cdot\\frac5{12}=\\frac{4!}{1!2!1!}\\cdot\\frac5{3\\cdot12\\cdot12}=\\frac5{36}\\approx0.1389\\;.$$", "On the other hand, $P_r(A)P_r(B)=\\frac{4}{5}\\frac{3}{5}=\\frac{12}{25}$. Since $P_r(A\\cap B)\\ne P_r(A)P_r(B)$, the events $A$ and $B$ are not independent. References. [1] Essential Discrete Mathematics for Computer Science, Harry Lewis and Rachel Zax, Princeton University Press, 2019 # Introductory Combinatorics: Combinations 2 Let us begin with an example. Example 1. A jar contains 11 marbles. In how many ways can they be placed in four boxes, with 3 marbles in the first box, 4 marbles in the second box, 2 marbles in the third box, and the 2 remaining marbles in the fourth box? Solution. The total number of outcomes is $$\\begin{pmatrix}11\\\\3\\end{pmatrix}\\cdot \\begin{pmatrix}8\\\\4\\end{pmatrix}\\cdot \\begin{pmatrix}4\\\\2\\end{pmatrix}\\cdot \\begin{pmatrix}2\\\\2\\end{pmatrix}=\\frac{11!}{3!8!}\\cdot\\frac{8!}{4!4!}\\cdot\\frac{4!}{2!2!}\\cdot\\frac{2!}{2!0!}=\\frac{11!}{3!4!2!2!}=69,300$$ This example can be generalized to the following question. A set of $n$ distinct items is to be divided into $r$ distinct groups of respective sizes $n_1, n_2, \\cdots, n_r$ such that $\\sum_{i=1}^r n_i=n$. How many different divisions are possible? There are $\\begin{pmatrix}n\\\\n_1\\end{pmatrix}$ possible choices for the first group, there are $\\begin{pmatrix}n-n_1\\\\n_2\\end{pmatrix}$ possible choices for the second group, there are $\\begin{pmatrix}n-n_1-n_2\\\\n_3\\end{pmatrix}$ possible choices for the third group, …, there are $\\begin{pmatrix}n-n_1-n_2-\\cdots-n_{r-1}\\\\n_r\\end{pmatrix}$ possible choices for the $r$-th group. Therefore the total number of coutcomes is \\begin{align*}\\begin{pmatrix}n\\\\n_1\\end{pmatrix}\\begin{pmatrix}n-n_1\\\\n_2\\end{pmatrix}\\begin{pmatrix}n-n_1-n_2\\\\n_3\\end{pmatrix}\\cdots\\begin{pmatrix}n-n_1-n_2-\\cdots -n_{r-1}\\\\n_r\\end{pmatrix}\\\\=\\frac{n!}{(n-n_1)!n_1!}\\frac{(n-n_1)!}{(n-n_1-n_2)!n_2!}\\cdots\\frac{(n-n_1-n_2\\cdots-n_{r-1})!}{0!n_r!}\\\\=\\frac{n!}{n_1!n_2!\\cdots n_r!}\\end{align*} The number of divisions of $n$ distinct items into $r$ distinct groups of respective sizes $n_1,n_2,\\cdots,n_r$ is denoted by $\\begin{pmatrix}n\\\\n_1,n_2,\\cdots,n_r\\end{pmatrix}$. That is, $$\\label{eq:multcoeff}\\begin{pmatrix}n\\\\n_1,n_2,\\cdots,n_r\\end{pmatrix}=\\frac{n!}{n_1!n_2!\\cdots n_r!}$$ where $n=n_1+n_2+\\cdots+n_r$. Example 2. A", "probability of choosing a blue one first? There are 9 marbles and 3 are blue. 39=13\\begin{align*} \\frac{3}{9} = \\frac{1}{3}\\end{align*} What is the probability of choosing a green one next? There are 9 marbles and 3 are green. 39=13\\begin{align*} \\frac{3}{9} = \\frac{1}{3}\\end{align*} What is the probability of choosing an orange one next? There are 9 marbles and 3 are orange. 39=13\\begin{align*} \\frac{3}{9} = \\frac{1}{3}\\end{align*} Now we figure out the likelihood of three independent events occurring by multiplying each probability. P(B, then G, then O)=131313=127\\begin{align*}P(B, \\ \\text{then} \\ G, \\ \\text{then} \\ O) = \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} = \\frac{1}{27}\\end{align*} There is a very small likelihood that this event will occur. Now go back to the swimming and golf example. Example Out of 100 students at Riverview Middle, 50 enjoy swimming. Out of the same 100 students, 25 enjoy golf. What is the probability that a student would enjoy both swimming and golf? Let’s add to this problem that 40 out of 100 enjoy basketball. What is the likelihood that a student would enjoy all three? 501002510040100121425=12=14=25=240=120\\begin{align*}\\frac{50}{100} & = \\frac{1}{2}\\\\ \\frac{25}{100} & = \\frac{1}{4}\\\\ \\frac{40}{100} & = \\frac{2}{5}\\\\ \\frac{1}{2} \\cdot \\frac{1}{4} \\cdot \\frac{2}{5} & = \\frac{2}{40} = \\frac{1}{20}\\end{align*} There is a one out of 20 chance that a student would like all three. We can look at this as", "# Thread: Mixed bag part 2: pemutations, combinations and determinants 1. ## Mixed bag part 2: pemutations, combinations and determinants I just want to make sure I have my method down properly, as usual, so here goes another round of questions. 1: There are 10 women and 8 men in a club. How many different committees of 6 people can be selected from the group if equal numbers of men and women are to be on the committee? Method: There are$\\displaystyle _{10}W_3$ ways to select 3 women for the committee and$\\displaystyle _8M_3$ ways to select 3 men for the committee. Therefore, there are $\\displaystyle _{10}W_3\\cdot_8M_3$different committees possible, so $\\displaystyle _{10}W_3 \\cdot _8M_3 \\rightarrow \\frac{10!}{(10-3)!} \\cdot \\frac{8!}{(8-3)!}\\rightarrow\\frac{10!}{7!}\\cdot\\frac{8!}{5!}\\ri ghtarrow$$\\displaystyle \\frac{10\\cdot9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3 \\cdot2\\cdot1}{7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot 1}\\cdot\\frac{8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2 \\cdot1}{5\\cdot4\\cdot3\\cdot2\\cdot1}\\rightarrow\\frac {3628800}{5040}\\cdot\\frac{40320}{120}=241920$ Thus there are 241920 possible committees. 2: Two marbles are drawn from a bag containing 6 white, 4 red, and 6 green marbles. Find the probability of both marbles being white. Method: There are 16 marbles in all so there are $\\displaystyle _{16}M_2\\rightarrow\\frac{16!}{14!}\\rightarrow \\frac{2.09227899 \\cdot 10^{13}}{87178291200}=240$ways of picking 2 marbles. This is where I get fuzzy, I'm not too sure what to do next--should I multiply out all the permutations for all the colors and then work from there? Or should I just focus on the white marbles? 3: If two rows of a determinant are", "The corresponding probabilities are $$a) \\ \\frac{1}{4}\\cdot \\frac{1}{4}\\cdot \\frac{1}{4}= \\frac{1}{64}$$ $$b) \\ 3\\cdot \\frac{1}{4}\\cdot \\frac{1}{4}\\cdot \\frac{1}{2}= \\frac{3}{32}$$ $$c) \\ 3\\cdot \\frac{1}{2}\\cdot \\frac{1}{2}\\cdot \\frac{1}{4}= \\frac{3}{16}$$ $$d) \\ \\frac{1}{2}\\cdot \\frac{1}{2}\\cdot \\frac{1}{2}= \\frac{1}{8}$$ Therefore $$P(W)=\\frac{27}{64}$$ Now what ist $$P(W\\cap R)$$? $$W\\cap R$$ means that no white marbles and no red marbles show up. That is when only green marbles show up-as you´ve already said: $$ggg$$ $$P(W\\cap R)=\\frac{1}{4}\\cdot \\frac{1}{4}\\cdot \\frac{1}{4}= \\frac{1}{64}$$ You can check your final result by using the converse probability: $$P(\\textrm{\"Do not see all three colors\"})=1-P(\\textrm{\"See all three colors\"})$$", "# Probability Worksheets - Page 21 Probability Worksheets • Page 21 201. A die is rolled. Find the probability of rolling a 2 or 4. a. $\\frac{1}{2}$ b. $\\frac{1}{6}$ c. $\\frac{1}{3}$ d. $\\frac{2}{3}$ 202. What is the probability of selecting a head when a coin is tossed, if the probability of selecting a tail is $\\frac{1}{2}$? a. 1 b. $\\frac{1}{2}$ c. $\\frac{1}{4}$ d. $\\frac{2}{6}$ 203. What is the probability of not selecting a number 4 when a number cube is rolled, if the probability of selecting a number 4 is $\\frac{1}{6}$? a. 1 b. $\\frac{5}{6}$ c. $\\frac{2}{6}$ d. $\\frac{1}{6}$ 204. Which of the following are complementary events? a. getting a number that is a multiple of 2, or a number that is a multiple of 3, when a number cube is rolled b. getting a number greater than 2 or a number greater than 3, when a number cube is rolled c. picking a marble from a bag containing only red and yellow marbles d. picking a Jack or a Spade from a pack of playing cards 205. Terry is choosing a marble from a bag containing 1 black marble, 1 brown marble, and 1 white marble. What is the probability that he will not choose a brown marble? a. $\\frac{2}{3}$ b. $\\frac{3}{2}$ c. $\\frac{1}{3}$ d. 1 206.", "of Counting to get the number of ways of winning only one ticket. vcsharp2003 There are C(10,2)⋅C(2,2) ways to select 2 winners, so you have that first term right in the numerator. Yes, I get that. Gold Member The way I have written the second term is first select the winning ticket and after that select a non-winning ticket; these two steps would yield the big step of winning only one ticket. Then I simply use the Fundamental Rule of Counting to get the number of ways of winning only one ticket. The seconnd term should be ##C(10,1) \\cdot C(2,1) \\cdot C(9990,1)## There are two configurations for each pair of selected tickets. vcsharp2003 ( ticket 1 or ticket 2 can be the winner) But wouldn't doing that duplicate the number of ways of winning only 1 ticket? Gold Member Yeah, I think you are right. It would over select. Sanity check with small numbers: Imagine a bag of 6 marbles, 4 red, 2 green there are 6 *5 = 30 ways to select 2 marbles You can select both red in 4 *3 = 12 ways You can select both green in 2*1 = 2 ways And you can select 1 red and one green in C(2,1) 4*2 ways = 16 ways $$12 + 2 + 16 =", "small marbles Each marble is either green or white. 9 of the large marbles are green, and 3 of the small marbles are white: Ifa marble is randomly selected from the box, what is the probability that it is small or white? Express vour answer as a fraction or a decimal number rounded to four decimal places. (2 Points)Answer: A box contains 15 large marbles and 11 small marbles Each marble is either green or white. 9 of the large marbles are green, and 3 of the small marbles are white: Ifa marble is randomly selected from the box, what is the probability that it is small or white? Express vour answer as a fraction or a d... ##### Geo Inc. had the following account balances on January 1, Year 2- Accounts Payable Accounts Receivable... Geo Inc. had the following account balances on January 1, Year 2- Accounts Payable Accounts Receivable Cash Common Stock Equipment Notes Payable Retained Earnings Salaries and Wages Expense Supplies $701 1.400 1.400 10,000 1,300 2,600 3.411 3,900 580 A Paid$701 on account for utilities that were u... ##### Can you tell me in what area the \"Shakout\" stage could occur and what it represents? Can you tell me in what area the \"Shakout\" stage could occur and what it represents?... ##### 1 Use", "pink marbles: $$\\displaystyle \\frac{{{}_{{10}}{{C}_{3}}\\times {}_{5}{{C}_{1}}+{}_{{10}}{{C}_{4}}\\times {}_{5}{{C}_{0}}}}{{{}_{{15}}{{C}_{4}}}}=\\frac{{810}}{{1365}}\\approx .593$$. (d) To get the probability that at least 1 is pink, we can get the probability that all are blue, and then subtract this from 1 to get its complement (since anything else would have at least one pink marble): $$\\displaystyle 1-\\frac{{{}_{{10}}{{C}_{0}}\\times {}_{5}{{C}_{4}}}}{{{}_{{15}}{{C}_{4}}}}=1-\\frac{{1\\times 5}}{{1365}}\\approx .996$$. This makes sense since so many are pink in the bag. A committee of 6 is selected from a group of 6 boys and 5 girls. (a) What is the probability that the committee is all boys? (b) What is the probability that exactly 4 out of the 6 are boys? We know the total number of ways 6 kids is selected from the group of the boys and girls (11) is $${}_{{11}}{{C}_{6}}=462$$; order doesn’t matter, since there are no positions. (a) There is only way to get all boys, since we are only 6 boys total. The probability then is $$\\displaystyle \\frac{1}{{{}_{{11}}{{C}_{6}}}}=\\frac{1}{{462}}\\approx .002$$ (low). (b) The number of possible ways 4 out of 6 boys could be chosen is $${}_{6}{{C}_{4}}=15$$, and the number of ways to pick 2 girls out of the remaining 5 kids is $${}_{5}{{C}_{2}}=10$$. The probability then that exactly 4 out of the 6 are girls is $$\\displaystyle \\frac{{{}_{6}{{C}_{4}}\\times {}_{5}{{C}_{2}}}}{{{}_{{11}}{{C}_{6}}}}=\\frac{{150}}{{462}}\\approx .325$$. The letters in the word MISSISSIPPI are written on pieces of", "green? [color=beige]. . [/color]$(A)\\;\\frac{68}{375} \\;\\;\\;\\;(B)\\;\\frac{11}{203} \\;\\;\\;\\;(C)\\;\\frac{204}{1015}\\;\\;\\;\\;(D)\\;\\frac{27 }{125}$ $P(\\text{3 green}) \\;=\\;\\frac{12}{30}\\,\\cdot\\,\\frac{11}{29}\\,\\cdot\\,\\ frac{10}{28} \\;=\\;\\frac{11}{203}\\;\\;(B)$ Quote: 19. A bag contains: 5 red, 3 white, and 2 blue balls. If 3 balls are selected at random, with replacement, find the probability that they are all red. [color=beige]. . [/color]$(A)\\;\\frac{1}{24} \\;\\;\\;\\;(B)\\;\\frac{1}{8} \\;\\;\\;\\;(C)\\;\\frac{1}{12} \\;\\;\\;\\;(D)\\;\\frac{1}{60}$ $P(\\text{red}) \\:=\\:\\frac{5}{10}\\:=\\:\\frac{1}{2}$ $\\text{Therefore: }\\;P(\\text{red, red, red}) \\;=\\;\\frac{1}{2}\\,\\cdot\\,\\frac{1}{2}\\,\\cdot\\,\\frac {1}{2} \\;=\\;\\frac{1}{8}\\;\\;(B)$ Quote: 20. A bag contains: 5 red, 3 white, and 2 blue balls. If 3 balls are selected at random, with replacement, find the probability that the first two are red and the third one is blue. [color=beige]. . [/color]$(A)\\;\\frac{1}{40} \\;\\;\\;(B)\\;\\frac{1}{20} \\;\\;\\;(C)\\;\\frac{5}{72} \\;\\;\\;(D)\\;60$ $P(\\text{red}) \\:=\\:\\frac{5}{10} \\:=\\:\\frac{1}{2} \\;\\;\\;\\;\\;{(\\text{blue}) \\:=\\:\\frac{2}{10} \\:=\\:\\frac{1}{5}$ $\\text{Therefore: }\\;P(\\text{red, red, blue}) \\;=\\;\\frac{1}{2}\\,\\cdot\\,\\frac{1}{2}\\,\\cdot\\,\\frac {1}{5} \\;=\\;\\frac{1}{20}\\;\\;(B)$ Tags confused Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post charlieboon Calculus 5 February 23rd, 2013 10:01 AM ccfoose Algebra 3 April 16th, 2012 09:55 PM Siedas Algebra 9 March 10th, 2012 01:29 PM momoftwo New Users 4 August 19th, 2009 06:15 PM charlieboon Algebra 0 December 31st, 1969 04:00 PM Contact - Home - Forums - Cryptocurrency Forum - Top", "\\frac{3}{9} = \\frac{1}{3}\\end{align*} To find the likelihood of these two independent events occurring, we multiply the two probabilities. P(B and G)=1313=19\\begin{align*}P(B \\ \\text{and} \\ G) = \\frac{1}{3} \\cdot \\frac{1}{3} = \\frac{1}{9}\\end{align*} Next, we can look at a real life example. Example Out of 100 students at Riverview Middle, 50 enjoy swimming. Out of the same 100 students, 25 enjoy golf. What is the probability that a student would enjoy both swimming and golf? To figure this out we have to figure out the probability of each independent event and then multiply them. 50100251001214=12=14=18\\begin{align*} \\frac{50}{100} & = \\frac{1}{2}\\\\ \\frac{25}{100} & = \\frac{1}{4}\\\\ \\frac{1}{2} \\cdot \\frac{1}{4} & = \\frac{1}{8}\\end{align*} There is a one out of eight chance that this will happen. III. Find the Probability of Three Independent Events Occurring What about if we had three independent events? We would do the same operation. We find the likelihood of each event and then we multiply the find the probability. Here is an example. Example There are 9 marbles in a bag. There are three blue, three green and three orange. What is the probability of selecting one blue marble, then putting that one back and selecting one green marble, then putting that one back and selecting one orange marble? Here we have the same work to do. What is the", "* 3!/2! = 90 different arrangements. Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility. Could you please explain why we ignored 1-1-1 combination? Hey jhabib, You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed. Dear Karishma .. I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong . All the approaches are considering 3 bags as identical , but in problem they are distinct . My approach to this problem : I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 . So first distribute 3 balls one ball each to each bag . select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways . so total 60 ways . Now we are left with 2 balls and 3 bags each containing one ball each . for first ball we have 3 options and second ball again 3", "ways to select the gender and the day of the week the child was born on. There are 132 = 169 ways which do not have a girl born on Monday and which 196 – 169 = 27 which do, so P(Y) = $$\\frac{27}{196}$$. This gives is that P(X|Y) = $$\\frac{\\frac{13}{19}*\\frac{1}{4}}{\\frac{27}{196}} = \\frac{13}{27}$$. 7. Suppose a fair eight-sided die is rolled once. If the value on the die is 1, 3, 5 or 7 the die is rolled a second time. Determine the probability that the sum of values that turn up is at least 8? a) $$\\frac{32}{87}$$ b) $$\\frac{12}{43}$$ c) $$\\frac{6}{13}$$ d) $$\\frac{23}{64}$$ Explanation: Sample space consists of 8*8=64 events. While (8) has $$\\frac{1}{8}$$ probability of occurrence, (1,7) has only $$\\frac{1}{64}$$ probability. So, the required probability = $$\\frac{1}{6} + (9 * \\frac{1}{64}) = \\frac{69}{192} = \\frac{23}{64}$$. 8. A jar containing 8 marbles of which 4 red and 4 blue marbles are there. Find the probability of getting a red given the first one was red too. a) $$\\frac{4}{13}$$ b) $$\\frac{2}{11}$$ c) $$\\frac{3}{7}$$ d) $$\\frac{8}{15}$$ Explanation: Suppose, P (A) = getting a red marble in the first turn, P (B) = getting a black marble in the second turn. P (A) = $$\\frac{4}{8}$$ and P (B) = $$\\frac{3}{7}$$ and P (A and B) = $$\\frac{4}{8}*\\frac{3}{7} = \\frac{3}{14}$$ P(B/A)", "+0 # HELP! 0 195 1 +902 A cook has 10 red peppers and 5 green peppers. If the cook selects 6 peppers at random, what is the probability that he selects at least 4 green peppers? Express your answer as a common fraction. Apr 28, 2018 ### 1+0 Answers #1 +3763 +2 Solution: We can count the number of ways to choose a group of 4 green and 2 red peppers and the number of ways to choose 5 green and 1 red peppers. These are $$\\binom{5}{4}\\binom{10}{2}=5\\cdot45=225$$ and $$\\binom{5}{5}\\binom{10}{1}=10$$. The total number of ways the cook can choose peppers is $$\\binom{15}{6}=5005$$. Therefore, the probability that out of six randomly chosen peppers at least four will be green is $$\\frac{235}{5005}=\\boxed{\\frac{47}{1001}}$$. Apr 28, 2018 ### New Privacy Policy We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website. For more information: our cookie policy and privacy policy.", "Ask question Question # A bag contains 7 tiles, each with a different number from 1 to 7. You choose a tile without looking, put it aside, choose a second tile without looking, put it aside, then choose a third tile without looking. What is the probability that you choose tiles with the numbers 1, 2, and 3 in that order? Probability ANSWERED asked 2020-12-17 A bag contains 7 tiles, each with a different number from 1 to 7. You choose a tile without looking, put it aside, choose a second tile without looking, put it aside, then choose a third tile without looking. What is the probability that you choose tiles with the numbers 1, 2, and 3 in that order? ## Answers (1) 2020-12-18 Probability of choosing 1, 2, then 3 is $$\\displaystyle\\frac{{1}}{{7}}\\cdot\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{5}}$$ $$\\displaystyle\\frac{{1}}{{7}}\\cdot\\frac{{1}}{{6}}\\cdot\\frac{{1}}{{5}}=\\frac{{1}}{{210}}$$ or $$\\displaystyle{0.00476190476}$$ expert advice ...", "simply enter population size 50, sample size 15, and number of good objects 35, to get the figure below. Now, select without replacement only 2 marbles. Compute the exact probability that 1 green marble is obtained. This is equal to $P(X=1)=\\frac{{35 \\choose 1}{15 \\choose 1}}{50 \\choose 2}=0.4286$. This is also shown on the figure below. We will approximate the probability of obtaining 1 green marble using Binomial as follows. Select the SOCR Binomial distribution and choose number of trials 2 and probablity of success $$p=\\frac{35}{50}=0.7$$. Compare the figure below with the figure above. They are almost the same! Why? Using the Binomial formula we can compute the approximate probability of observing 1 green marble as $$P(X=1)={2 \\choose 1}0.70^10.30^1=0.42$$ (very close to the exact probability, 0.4286). ## Normal approximation to Binomial Graph and comment on the shape of binomial with $$n=20, p=0.1$$ and $$n=20, p=0.9$$. Now, keep $$n=20$$ but change $$p=0.45$$. What do you observe now? How about when $$n=80, p=0.1$$. See the four figures below. What is your conclusion on the shape of the Binomial distribution in relation to its parameters $$n, p$$? Clearly when $$n$$ is large and $$p$$ small or large the result is a bell-shaped distribution. When $$n$$ is small (10-20) we still get approximately a bell-shaped distribution as long as $$p \\approx 0.5$$. Because" ]

[ "• klimenkov Two points on the plane $$A$$ and $$B$$ are given. $$|AB| = 2$$. $$C$$ is a randomly picked point in the circle of the radius $$R$$ with the center in the midpoint of $$AB$$. What is the probability that the $$\\triangle ABC$$ has an obtuse angle? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# Obtuse angle is more than right Random numbers x and y are chosen from the interval [ 0 , 1 ] uniformly such that lengths x, y, and 1 form a triangle. What is the probability that the triangle formed is obtuse? ×", "# Points in a circle $$N^2$$ points are chosen uniformly and at random inside a circle of radius $$N$$, for some positive integer $$N$$. For large $$N$$, i.e. $$N \\to \\infty$$, what is the probability that exactly one point will be within one unit of the center of the circle?", "# Probability of obtuse triangles formed on a circle There are 16 equally spaced points on the circumference of a circle. If 3 points out of these 16 points are selected randomly, What is the probability that they will form an obtuse angled triangle? • Fix a point. Because a circle with equally spaced points is perfectly symmetrical, it does not matter which one you start with. Assume that point is one of the three. Now, to choose the other two points, there are $\\dbinom{15}{2} = 105$ different ways of choosing them. Determine how many pairs result in triangles with obtuse angles. The answer will be that number over 105. – InterstellarProbe Apr 17 at 17:16 • yeah, I was thinking the same but I am not able to count the number of obtuse triangles in these 105 triangles. – James Joy Apr 17 at 17:30 • Under what circumstances will the angle be obtuse? Determine what circumstances will make the angle obtuse and look for patterns. For instance, if all three points are right next to each other, it is obtuse. If they are all spread out, every angle is acute. Move one point until it goes from acute to obtuse (if ever). – InterstellarProbe Apr 17 at 17:33 • If it may help to check your", "point should be in, given the coordinates of the first two points? Ultimately you would have to integrate over this region – Alex R. Mar 31 '12 at 23:47 Does it help that any three points allows us to draw a circle with each on the radius? Based on your argument, the three points must not lie within a semi-circular arc. – Ronald Mar 31 '12 at 23:48 Also, this problem is considerably simpler if you assume that the first two points correspond to two adjacent corners of the square. It might help you to try and solve this one first to get some bearing. – Alex R. Mar 31 '12 at 23:49 The problem for three points chosen uniformly at random in the plane is discussed at cut-the-knot.org/Probability/ObtuseTriangle.shtml – Gerry Myerson Apr 1 '12 at 0:19 There's a discussion at jd2718.wordpress.com/2007/09/01/… which I haven't read in its entirety. It may have some useful ideas. – Gerry Myerson Apr 1 '12 at 0:22 After working this out and writing it down, I found that it was already done in 1969: Eric Langford, The probability that a random triangle is obtuse, Biometrika (1969) 56 (3): 689-690. For this kind of problem, it's important to make good use of the symmetries and choose convenient coordinates. A significant simplification is achieved", "# Circles everywhere! Level pending Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distance between the points are less than the radius of the circle? the ans is in the form of(n/m) enter the ans as (m+n) ×", "by noting that a triangle has at most one obtuse angle; thus the three events of one of the angles being obtuse are mutually exclusive, and thus the probability of the triangle being obtuse is just three times the probability $p$ of any given one of the angles being obtuse. Thus you don't have to deal with both the circle and the strip between the perpendiculars, since the probability of the third point being in the circle is $p$ and the probability of the third point being in the strip is $1-2p$ and you can obtain the desired probability $1-3p$ from either of them. In fact, you don't even need the strip, since the probability of the third point being beyond one of the perpendiculars is also $p$. So there are three different approaches; we can either integrate either the area beyond one of the perpendiculars or the area of the strip or the area of the circle over all positions of two points. The advantage of using the perpendiculars or the strip is that you have to deal only with straight lines and not with circles; the disadvantage is that the integration region is complicated due to the corners. The advantage of the second approach is that you avoid the corners and get a relatively simple integration", "here. ### The Problem Bertrand considered a circle with an equilateral triangle inscribed it. If a chord in the circle is randomly chosen, what is the probability that the chord is longer than a side of the equilateral triangle? ### The Solution(s) Bertrand argued that there are three natural but different methods to randomly choose a chord, giving three distinct answers. (Of course, there are other methods, but these are arguably not the natural ones that first come to mind.) ##### Method 1: Random endpoints On the circumference of the circle two points are randomly (that is, uniformly and independently) chosen, which are then used as the two endpoints of the chord. The probability of this random chord being longer than a side of the triangle is one third. A radius of the circle is randomly chosen (so the angle is chosen uniformly), then a point is randomly (also uniformly) chosen along the radius, and then a chord is constructed at this point so it is perpendicular to the radius. The probability of this random chord being longer than a side of the triangle is one half. ##### Method 3: Random midpoint A point is randomly (so uniformly) chosen in the circle, which is used as the midpoint of the chord, and the chord is randomly (also uniformly)", "# A \"Random\" Triangle Geometry Level 4 Two side lengths of a triangle are randomly chosen uniformly from $$(0,1).$$ Then, the third side length is randomly chosen uniformly from the interval of side lengths that form a valid triangle. Is this triangle more likely to be acute or obtuse? Bonus: Find the exact probabilities. ×", "# Exactly one point Four points are chosen uniformly and at random inside a circle of radius 2. If the probability that exactly one point is within one unit from the center of the circle is given by $$\\dfrac{a}{b}$$ where $$a$$ and $$b$$ are coprime positive integers, then what is $$a + b$$? ×", "result: by brute force by program (using trigonometry for the angles): the probability to get an obtuse triangle is $3/5$ (to be clear, by obtuse I mean one angle is $>\\pi/2$). – Jean-Claude Arbaut Apr 17 at 17:34 • If you pick $n$ equally spaced points and let $n\\to\\infty$, then the probability converges to $3/4$. Found by numerical experiment, but see also Richard K. Guy, There Are Three Times as Many Obtuse-Angled Triangles as There Are Acute-Angled Ones. Moreover, for $n=4k$ points, the probability seems to be $\\dfrac{3}{4}\\cdot\\dfrac{n-4}{n-1}$. All of this should be proved carefully though. – Jean-Claude Arbaut Apr 17 at 20:25", "# Are Most Triangles Obtuse? If one chooses three random points in the plane, what is the probability that they form an obtuse triangle? R. Guy claims, the probability is $3/4$ - at least! However, this is a problem of geometric probabilities and it is reasonable to expect slight complications. As C. Alsina and R. B. Nelsen wrote, this is an old problem, with multiple solutions. One of the earliest solutions appears to be by Lewis Carroll, best known as the author of Alice's Adventures in Wonderland and Through the Looking Glass. Carroll's real name was Charles Lutwidge Dodgson (1832-1898) and he was a lecturer in mathematics at Christ Church, Oxford. In that role he published Pillow Problems Thought Out During Wakeful Hours in 1893. Problem 58 in that collection reads: Three Points are taken at random on an infinite plane. Find the chance of their being the vertices of an obtuse-angled Triangle. With a reference to Figure 1 below, Guy remarks that given side $AB$ of $\\Delta ABC,$ the positions of the third vertex $C$ for which the triangle is acute are in the shaded region, which constitutes a negligible fraction of the whole plane. In Figure 2 the regions are marked, where $C$ makes $AB$ the shortest (s), middle (m), or longest (1) side. Looking at", "random triangle: A = a random number between 0 and 180; B = a random number between 0 and 180-A; C = 180 - A - B. ## Here is Kurt's logic: First, using the definition above, the chance of choosing angle A to be obtuse is 0.5. So the chance of all three angles being acute must be less than 0.5. If A=30, then the other two angles must add to 150. That means that one of the other two angles must be between 60 and 90 in order for the triangle to be acute (neglecting right triangles whose probability is zero). So the probability of the triangle being acute is (90-60)/150 = 30/150. If A=70, the other two angles must add to 110. That means that one of the other two angles must be between 20 and 90 in order for the triangle to be acute. So the probability of the triangle being acute is (90-20)/150 = 70/150. Suppose that A is chosen and is acute. That will leave (180-A)-90 and 90. 90-[(180-A)-90] = A. So the probability of the triangle with acute angle A being acute is A/(180-A). Obviously there are infinite possibilities of angle A giving infinite probabilities of the angle being acute, each infinitely small. So Kurt decides to sum these probabilities up", "# Obtuse the right way Two numbers $$x$$ and $$y$$ are chosen independently at random from the interval $$[0, 1]$$. What is the probability that there exists an obtuse triangle with side lengths $$x, y, 1$$? ×", "× # Probability When any three points are selected from a circle, what is the probability that they will form an obtuse-angled triangle? Note by Akshay Ginodia 3 years, 10 months ago Sort by: The triangle will be obtuse iff the three points lie on the same half of the circle. This is because the arc that the obtuse angle intersects is greater than half of the circle. The first point can be anywhere, the second point can be anywhere (since given two points on a circle they must be on some similar half circle, except for the case when they are exactly opposite each other, but the probability for that is negligible since the circle is made up of infinite points), and the third point has to be on that 1/2 of the circle that both points are, so the probability is $$\\frac{1}{2}$$ · 3 years, 10 months ago Agree that point 1 can be placed anywhere. Now for point 2: If this point is placed opposite to point 1 (or 'almost opposite'), Michael is right (or in fact 'almost' right): Points 1 and 2 'almost' fix half of the circle as point 3 possible landing spot. Now what if point 2 would not be opposite to point 1 but quite a bit closer: Lets say we", "+0 0 186 4 A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle? Three points are selected randomly on the circumference of a circle. What is the probability that the triangle formed by these three points contains the center of the circle? Apr 17, 2019 #1 +106915 +2 The first question has been answered here many time before. If you search it in google you will find it. Both questions are most easlily solved with the use of probability contour maps. Three points are selected randomly on the circumference of a circle. What is the probability that the triangle formed by these three points contains the center of the circle? If it is a right angled triangle then the diameter will be one sides and the centre will be on this side. If the triangle is a acute angled then the centre will be included. If the triangle is obtuse then the centre will not be included. I am actually going to look at the set of obtuse angled trianges which is mutually exclusive to the set of ones that include the centre. Let the angles be x, y, and 180-(x+y) The set of all possible triangles is contained", "100,000 ... still only two points. So given that the probability for making a right triangle is vanishingly small. If you move the third point back and forth a bit what is the chance of getting an acute vs obtuse triangle? 6. Feb 21, 2012 ### Pzi When I come to think of that again... If the distance between my first set two points is equal to the diameter of a circle then ANY third point will complete a right triangle (as long as it doesn't coincide with first two points, but this one is indeed infinitely unlikely event). It casts some uncertainty about the right triangle unlikeliness :/ 7. Feb 21, 2012 ### Staff: Mentor If you move the third point back and forth a bit what is the chance of getting an acute vs obtuse triangle? 8. Feb 21, 2012 ### Pzi As first two points are getting further apart almost to form a diameter, the probability of getting an acute (the same for obtuse) angle is indeed getting closer to 1/2. I'm not sure what happens when the first two points are chosen close to each other... Edit: I think when the first two points are close to each other one is almost certain to complete an obtuse triangle. Edit2: Indeed. Now I'm pretty much", "these are arguably not the natural ones that first come to mind.) ##### Method 1: Random endpoints On the circumference of the circle two points are randomly (that is, uniformly and independently) chosen, which are then used as the two endpoints of the chord. The probability of this random chord being longer than a side of the triangle is one third. A radius of the circle is randomly chosen (so the angle is chosen uniformly), then a point is randomly (also uniformly) chosen along the radius, and then a chord is constructed at this point so it is perpendicular to the radius. The probability of this random chord being longer than a side of the triangle is one half. ##### Method 3: Random midpoint A point is randomly (so uniformly) chosen in the circle, which is used as the midpoint of the chord, and the chord is randomly (also uniformly) rotated. The probability of this random chord being longer than a side of the triangle is one quarter. ### Simulation All three answers involve randomly and independently sampling two random variables, and then doing some simple trigonometry. The setting naturally inspires the use of polar coordinates. I assume the circle has a radius $$r$$ and a centre at the origin $$o$$. I’ll arbitrarily number the end points one", "# Spheres n' Probability Geometry Level 1 4 points are randomly chosen on a sphere to form a tetrahedron. What is the probability the center of the sphere will be in the tetrahedron? On the left, the center of the sphere isn't inside the random tetrahedron. On the right, it is. ×", "through how you might stumble upon the solution yourself. This story is more about the problem-solving process than the particular problem used to exemplify it. Here's the question: Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is independently chosen relative to a uniform distribution on the sphere.) Take a moment to digest the question. You might start thinking about which of these tetrahedra contain the sphere's center, which ones don't, and how you might systematically distinguish the two. How do you approach a problem like this? Where do you even start? Well, it's often a good idea to think about simpler cases, so let's bring things down into two dimensions. The two-dimensional case Suppose you choose three random points on a circle. It's always helpful to name things, so let's call these fellows $P_1$, $P_2$, and $P_3$. What's the probability that the triangle formed by these points contains the center of the circle? This is still a hard question, so you should ask yourself if there's a way to simplify the question even further. We still need a foothold, something to build up from. Maybe you imagine" ]

[ "in Olympiad, and there are some other ways of computation under this observation.) -Fanyuchen20020715 ## Solution 2 (Official MAA) Let $M$ denote the midpoint of $\\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\\triangle AXY$, which has the circle with diameter $\\overline{AT}$ as its circumcircle. To see this, note that because $\\angle BXT = \\angle BMT = 90^\\circ$, the quadrilateral $MBXT$ is cyclic, it follows that $$\\angle MXA = \\angle MXB = \\angle MTB = 90^\\circ - \\angle TBM = 90^\\circ - \\angle A,$$ implying that $\\overline{MX} \\perp \\overline{AC}$. Similarly, $\\overline{MY} \\perp \\overline{AB}$. In particular, $MXTY$ is a parallelogram. $[asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69)); label(\"\\omega\", O + rad*dir(45), SW); filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255)); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot(\"X\", X, W); dot(\"Y\", Y, E); dot(\"O\", O, W); dot(\"T\", T, S); dot(\"A\", A, N); dot(\"B\", B, W); dot(\"C\", C, E); dot(\"M\", M, N); [/asy]$ Hence, by the Parallelogram Law, $$TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).$$ But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$. Therefore $$XY^2 = \\frac13(2 \\cdot 1143-135)", "mobile point crosses an imaginary diameter of the circle perpendicular to the radius of the fixed point. This imaginary diameter bisects the area of the circle, therefore the triangle is acute half the time and obtuse the other half. Therefore half of all (isosceles) triangles must be acute. One quarter. A \"circle\" argument. Pick points $$A, B, C$$ \"at random\". Now pick polar coordinates so that $$O$$ is $$ABC$$'s circumcentre and $$A$$'s phase is $$\\theta=0$$. Let $$B$$'s and $$C$$'s phases be $$\\beta, \\gamma$$. Normalise phase to be in the interval $$[0, 2\\pi)$$. Then $$ABC$$ is obtuse if any of the following conditions occurs: • $$\\beta<\\pi$$ and $$\\gamma<\\pi$$ • $$\\beta>\\pi$$ and $$\\gamma>\\pi$$ • $$\\beta>\\gamma+\\pi$$ • $$\\gamma>\\beta+\\pi$$ These conditions are mutually exclusive. Assuming that $$\\beta$$ and $$\\gamma$$ are chosen from the uniform distribution on $$[0, 2\\pi)$$, the above conditions occur with probabilities $$1/4, 1/4, 1/8$$ and $$1/8$$. So $$ABC$$ is obtuse with probability $$3/4$$ and acute with probability $$1/4$$. An \"orthic system\" argument. Pick three points this way. First, pick an orthic system \"at random\". Then pick three of those four points to be your triangle's vertices, each combination of three being picked with probability $$1/4$$. With probability $$3/4$$, one of the three you picked was the interior point and your triangle is obtuse; with probability $$1/4$$, you picked the", "circle at $G$. Draw line $BG$. $A[\\Delta ABD] = A[\\Delta ABG] = 120$ (Since $\\square ABGD$ is cyclic) But $A[\\Delta ABE]$ is common in both with an area of 60. So, $A[\\Delta AED] = A[\\Delta BEG]$. \\therefore $A[\\Delta AED] \\cong A[\\Delta BEG]$ (SAS Congruency Theorem). In $\\Delta AED$, let $DI$ be the median of $\\Delta AED$. Which means $A[\\Delta AID] = 30 = A[\\Delta EID]$ Rotate $\\Delta DEA$ to meet $D$ at $B$ and $A$ at $G$. $DE$ will fit exactly in $BE$ (both are radii of the circle). From the above solutions, $\\frac{AE}{EF} = 2:1$. $AE$ is a radius and $EF$ is half of it implies $EF$ = $\\frac{radius}{2}$. Which means $A[\\Delta BEF] \\cong A[\\Delta DEI]$ Thus $A[\\Delta BEF] = \\boxed{\\textbf{(B) }30}$ ~phoenixfire & flamewavelight ## Solution 8 $[asy] import geometry; unitsize(2cm); pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); draw(EE--M,StickIntervalMarker(1,1)); label(\"M\",M,S); draw(A--DD,invisible,StickIntervalMarker(1,1)); dot((DD+C)/2); draw(DD--C,invisible,StickIntervalMarker(2,1)); [/asy]$ Using the ratio of $\\overline{AD}$ and $\\overline{CD}$, we find the area of $\\triangle ADB$ is $120$ and the area of $\\triangle BDC$ is $240$. Also using the fact that $E$ is the midpoint of $\\overline{BD}$, we know $\\triangle ADE = \\triangle", "we have $\\cos \\theta = \\frac{3}{4}$. Now, since $\\triangle BOC$ is isosceles with $OB = OC$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. By SSS congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ### Solution 4 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD =", "# Probability of triangle to be obtuse Two points $A,B$ fixed on a plane(distance = 2). C - random choosen point inside circle with radius $R$ with center at the center of $AB$ Find probability of triangle $ABC$ to be obtuse My thoughts: 1. If $C$ lies in the circle - $ABC$ will be rectangular 2. Opposite the larger angle is large side, so using cosine theorem: $$b^2+c^2<a^2=4R^2,$$ where $a$ - the diagonal, $b=AC, c=BC$ And then I don't know... Any such triangle $ABC$ with $AB$ diameter and $C$ any point within the circle will be obtuse. Proof:- (Note:-$AO=OB=r$) Let,C be any point in the circle.$AC$ is extended to meet the circumference at $X$.So,$\\angle AXB=\\alpha=90^0$ and $\\angle \\beta>0^0$.So,$\\angle ACB=\\gamma=\\alpha+\\beta=90$+something $>0.$So,$\\angle ACB$ will always be obtuse and thus $\\triangle ACB$ is always obtuse. • $AO=OB=1\\neq R$ (The solution should be given in terms of R?) – Logophobic Apr 6 '16 at 15:56 • @Logophobic, we were thinking along the same lines (see my answer). – Barry Cipra Apr 6 '16 at 16:47 • @Logophobic If center of circle is at center of line,then line must be diameter...it's actually $AO=OB=1=R$... – tatan Apr 6 '16 at 17:00 • @tatan See images in my post, $R$ is variable, not fixed at $1$. Point $C$ can occur outside the circle with diameter", "+ r)(3)cosA = (r+1)^2$ We can eliminate the extra variable of angle $A$ by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find $r$: $2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26$ $8r + 2 = -20r + 26$ $28r = 24$, so $r$ = $6/7$ $\\boxed{(B)}$ ## Solution 3 $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); [/asy]$ Let $C$ be the center of the largest semicircle. Notice that $\\Delta ACP$ and $\\Delta CBP$ are bounded by the same two parallel lines and therefore have the same heights. Since the bases of these two triangles differ by a factor of 2, the area of $\\Delta CBP$ must be twice that of $\\Delta ACP$. We know that $AC$ $=$ 1, $CB$ $=$ 2, $AP$ $=$ $r$ $+$ 2, $BP$ $=$ $r$ $+$ 1, $CP$ $=$ $3$ $-$ $r$. Again, we don't even need the circles and semicircles anymore; just focus on the triangles. $[asy] size(5cm); draw((-1,0)--(2,0)); pair P = (9/7,12/7); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--P); [/asy]$ The semiperimeter $s_1$ of $\\Delta ACP$ is $$[(r + 2) + (3 - r) + 1] / 2 = 6/2 = 3$$ Heron's Formula states that the area", "# 1985 AIME Problems/Problem 9 ## Problem In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\\alpha$, $\\beta$, and $\\alpha + \\beta$ radians, respectively, where $\\alpha + \\beta < \\pi$. If $\\cos \\alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator? ## Solution 1 $[asy] size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3); D(CR(O,r)); D(O--rotate(a/2)*A--rotate(-a/2)*A--cycle); D(O--rotate(b/2)*A--rotate(-b/2)*A--cycle); D(O--rotate((a+b)/2)*A--rotate(-(a+b)/2)*A--cycle); MP(\"2\",(rotate(a/2)*A+rotate(-a/2)*A)/2,NE); MP(\"3\",(rotate(b/2)*A+rotate(-b/2)*A)/2,NE); MP(\"4\",(rotate((a+b)/2)*A+rotate(-(a+b)/2)*A)/2,NE); D(anglemark(rotate(-(a+b)/2)*A,O,rotate((a+b)/2)*A,5)); label(\"$$\\alpha+\\beta$$\",(0.08,0.08),NE,fontsize(8)); [/asy]$ All chords of a given length in a given circle subtend the same arc and therefore the same central angle. Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle. $[asy] size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3), A1 = rotate(a/2)*A, A2 = rotate(-a/2)*A, A3 = rotate(-a/2-b)*A; D(CR(O,r)); D(O--A1--A2--cycle); D(O--A2--A3--cycle); D(O--A1--A3--cycle); MP(\"2\",(A1+A2)/2,NE); MP(\"3\",(A2+A3)/2,E); MP(\"4\",(A1+A3)/2,E); D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5)); label(\"$$\\alpha$$\",(0.07,0.16),NE,fontsize(8)); label(\"$$\\beta$$\",(0.12,-0.16),NE,fontsize(8)); [/asy]$ This triangle has semiperimeter $\\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \\sqrt{\\frac92 \\cdot \\frac52 \\cdot \\frac32 \\cdot \\frac12} = \\frac{3}{4}\\sqrt{15}$. The area of", "is $\\dfrac{99}{99^2}$, or $\\dfrac{1}{99}$, and our answer is $1 + 99 = \\boxed{100}$. ## Solution 2 Without the loss of generality one can let $z$ lie on the positive x axis and since $arg(\\theta)$ is a measure of the angle if $z=10$ then $arg(\\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivalent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$ $[asy] pair O = (0,0); pair A = (100,0); pair B = (80,30); pair D = (sqrt(850),sqrt(850)); draw(A--B--O--cycle); dotfactor = 3; dot(\"A\",A,dir(45)); dot(\"B\",B,dir(45)); dot(\"O\",O,dir(135)); dot(\" \\theta\",O,(7,1.2)); label(\"1\", ( A--B )); label(\"10\",(O--A)); label(\"t\",(O--B)); [/asy]$ using the Law of Cosines we get: $1^2=10^2+t^2-t*10*2\\cos\\theta$ rearranging: $$20t\\cos\\theta=t^2+99$$ solving for $\\cos\\theta$ we get: $$\\frac{99}{20t}+\\frac{t}{20}=\\cos\\theta$$ if we want to maximize $\\theta$ we need to minimize $\\cos\\theta$ , using AM-GM inequality we get that the minimum value for $\\cos\\theta= 2(\\sqrt{\\dfrac{99}{20t}\\dfrac{t}{20}})=2\\sqrt{\\dfrac{99}{400}}=\\dfrac{\\sqrt{99}}{10}$ hence using the identity $\\tan^2\\theta=\\sec^2\\theta-1$ we get $\\tan^2\\theta=\\frac{1}{99}$and our answer is $1 + 99 = \\boxed{100}$. ## Solution 3 Note that $\\frac{w-z}{z}=\\frac{w}{z}-1$, and that $\\left|\\frac{w}{z}\\right|=\\frac{1}{10}$. Thus $\\frac{w}{z}-1$ is a complex number on the circle with radius $\\frac{1}{10}$ and centered at $-1$ on the complex plane. Let $\\omega$ denote this circle. Let $A$ and $C$ be the points that represent $\\frac{w}{z}-1$ and $-1$ respectively on the complex plane. Let $O$ be the origin. In", "# 2018 AMC 12B Problems/Problem 25 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Circles $\\omega_1$, $\\omega_2$, and $\\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\\omega_1$, $\\omega_2$, and $\\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\\omega_i$ for each $i=1,2,3$, where $P_4 = P_1$. See the figure below. The area of $\\triangle P_1P_2P_3$ can be written in the form $\\sqrt{a}+\\sqrt{b}$ for positive integers $a$ and $b$. What is $a+b$? $[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label(\"P_1\", P1, E*1.5); label(\"P_2\", P2, SW*1.5); label(\"P_3\", P3, N); label(\"\\omega_1\", A, W*17); label(\"\\omega_2\", B, E*17); label(\"\\omega_3\", C, W*17); [/asy]$ $\\textbf{(A) }546\\qquad\\textbf{(B) }548\\qquad\\textbf{(C) }550\\qquad\\textbf{(D) }552\\qquad\\textbf{(E) }554$ ## Solution Let $O_i$ be the center of circle $\\omega_i$ for $i=1,2,3$, and let $K$ be the intersectoin of lines $O_1P_1$ and $O_2P_2$. Because $\\angle P_1P_2P_3 = 60^\\circ$, it follows that $\\triangle P_2KP_1$ is a $30-60-90^\\circ$ triangle. Let $d=P_1K$; then $P_2K = 2d$ and $P_1P_2 = \\sqrt 3d$. The Law of Cosines", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "1cm, outer radius = 3cm, ruler position= 10 right • inner radius = 2.5cm, outer radius = 4.5cm, ruler position = 1 You can also use this chart to visualize binary numbers. left: • the third 3 bit binary number (binary 010 = octal 2 = decimal 2 = hexadecimal 2) • inner radius = 1cm, outer radius = 2cm, ruler position = 3 right: • the sixteenth 4 bit binary number (binary 1111 = octal 17 = decimal 15 = hexadecimal F) • inner radius = 2cm, outer radius = 3cm , ruler position = 16 - To observe how arcs and filling work you can add inside your tikzpicture : %How arc works \\draw[blue,->] (0,3.5) arc [radius=3.5, start angle=90, delta angle=90]; %Now filling \\filldraw[fill=gray] (3.5,0) arc [radius=3.5, start angle=0, delta angle=90] -- (0,4) arc [radius=4, start angle=90, delta angle=-90] -- cycle; \\filldraw[fill=red] (0,-3.5) arc [radius=3.5, start angle=270, delta angle=90] -- (4,0) arc [radius=4, start angle=0, delta angle=-90] -- cycle; And here is an example within a loop, where a bit of trigonometry gives you the formula: \\foreach \\a in {0,1} { % filling in the loop \\filldraw[fill=green!50!black] ({3.5*cos((-\\a*(pi/8)+(pi/2)) r)},{3.5*sin((-\\a*(pi/8)+(pi/2)) r)}) arc [radius=3.5, start angle={((pi/2)+(-pi/8)*\\a) r}, delta angle={(-pi/8) r}] -- ({4*cos((-(\\a+1)*(pi/8)+(pi/2)) r)},{4*sin((-(\\a+1)*(pi/8)+(pi/2)) r)}) arc [radius=4, start angle={((pi/2)+(-pi/8)*(\\a+1)) r}, delta angle={pi/8 r}] -- cycle; % corresponding arcs \\draw[thick,blue,->]", "check this point: inradius <- t$inradius() semiperimeter <- sum(t$edges()) / 2 (inradius^2 + semiperimeter^2) / (4*inradius) #> [1] 11.15942 ApolloniusCircle$radius #> [1] 11.15942 Filling the lapping area of two circles Let two circles intersecting at two points. How to fill the lapping area of the two circles? O1 <- c(2,5); circ1 <- Circle$new(O1, 2) O2 <- c(4,4); circ2 <- Circle$new(O2, 3) opar <- par(mar = c(0,0,0,0)) plot(NULL, asp = 1, xlim = c(0,8), ylim = c(0,8), xlab = NA, ylab = NA) draw(circ1, border = \"purple\", lwd = 2) draw(circ2, border = \"forestgreen\", lwd = 2) intersections <- intersectionCircleCircle(circ1, circ2) A <- intersections[[1]]; B <- intersections[[2]] points(rbind(A,B), pch = 19, col = c(\"red\", \"blue\")) theta1 <- Arg((A-O1)[1] + 1i*(A-O1)[2]) theta2 <- Arg((B-O1)[1] + 1i*(B-O1)[2]) path1 <- Arc$new(O1, circ1$radius, theta1, theta2, FALSE)$path() theta1 <- Arg((A-O2)[1] + 1i*(A-O2)[2]) theta2 <- Arg((B-O2)[1] + 1i*(B-O2)[2]) path2 <- Arc$new(O2, circ2$radius, theta2, theta1, FALSE)$path() polypath(rbind(path1,path2), col = \"yellow\") par(opar) Hyperbolic tessellation In the help page of the Circle R6 class (?Circle), we show how to draw a hyperbolic triangle with the help of the method orthogonalThroughTwoPointsOnCircle(). Here we will use this method to draw a hyperbolic tessellation. tessellation <- function(depth, Thetas0, colors){ stopifnot( depth >= 3, is.numeric(Thetas0), length(Thetas0) == 3L, is.character(colors), length(colors) >= depth ) circ <- Circle$new(c(0,0), 3) arcs <- lapply(seq_along(Thetas0), function(i){ ip1", "that perimeter of triangle AOB is equal to $$AB + OB + OA = 8 + \\sqrt{32}$$ From (1) , $$2OA + AB = 8 + \\sqrt{32}$$ Thus, $$OA = 4$$ AND $$AB = \\sqrt{32}$$ because $$\\sqrt{32}$$cannot equal 2OA ( otherwise the triangle OAB wouldn't stand as isocele ) Now, we know that OA = OB = r = 4 and $$AB = \\sqrt{32}$$ and OAB is isocele and AB^2 = 32 and OA^2 = OB^2 = 16 Hence OAB is a right isocele triangle. from pythagor (AB^2 = OA^2 + OB^2) Area of shaded region = area of sector AOB - area of triangle AOB area of sector AOB = Lenght of Arc(AB) * AB Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi $$\\sqrt{2}$$ / 2 area of sector AOB = 4Pi area of triangle = 4*4 /2 = 8 Finally: area of shaed region = 4Pi - 8 Hope that helps Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi $$\\sqrt{2}$$ / 2 How is this operation not equal to just pi. Diameter is 8, so 8*45/360 cancels out and the only thing left is Pi. Where do you get sqrt(2)/2 from? Intern Joined: 28 May 2014 Posts: 49 Schools: NTU '16 GMAT 1: 620 Q49 V27 Re: Triangle", "# 2018 AMC 12B Problems/Problem 25 ## Problem Circles $\\omega_1$, $\\omega_2$, and $\\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\\omega_1$, $\\omega_2$, and $\\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\\omega_i$ for each $i=1,2,3$, where $P_4 = P_1$. See the figure below. The area of $\\triangle P_1P_2P_3$ can be written in the form $\\sqrt{a}+\\sqrt{b}$ for positive integers $a$ and $b$. What is $a+b$? $[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label(\"P_1\", P1, E*1.5); label(\"P_2\", P2, SW*1.5); label(\"P_3\", P3, N); label(\"\\omega_1\", A, W*17); label(\"\\omega_2\", B, E*17); label(\"\\omega_3\", C, W*17); [/asy]$ $\\textbf{(A) }546\\qquad\\textbf{(B) }548\\qquad\\textbf{(C) }550\\qquad\\textbf{(D) }552\\qquad\\textbf{(E) }554$ ## Solution 1 Let $O_i$ be the center of circle $\\omega_i$ for $i=1,2,3$, and let $K$ be the intersection of lines $O_1P_1$ and $O_2P_2$. Because $\\angle P_1P_2P_3 = 60^\\circ$, it follows that $\\triangle P_2KP_1$ is a $30-60-90^\\circ$ triangle. Let $d=P_1K$; then $P_2K = 2d$ and $P_1P_2 = \\sqrt 3d$. The Law of Cosines in $\\triangle O_1KO_2$ gives $$8^2 = (d+4)^2 + (2d-4)^2 - 2(d+4)(2d-4)\\cos 60^\\circ,$$which", "+0 0 132 6 1) In triangle $WXY,$ $WX = 8$ and $XY = 14.$ The circumcircle of triangle $WXY$ is constructed. The tangent to the circumcircle at $X$ is drawn, and the line through $W$ parallel to this tangent intersects $\\overline{XY}$ at $Z.$ Find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] 2) Lines $PTQ$ and $PUR$ are tangent to a circle, as shown below. [asy] unitsize(4 cm); pair A, O, P, Q, R, T, U; P = (0,0); U = (1,0); T = dir(40); O = extension(P, P + dir(20), U, U + (0,1)); A = intersectionpoint((U + 0.1*dir(63))--(U + 5*dir(63)), Circle(O,abs(O - U))); Q = 1.5*T; R = 1.5*U; draw(Q--P--R); draw(Circle(O,abs(O - U))); draw(T--A--U); label(\"$A$\", A, NE); label(\"$P$\", P, SW); label(\"$Q$\", Q, NE); label(\"$R$\", R, E); label(\"$T$\", T, NW); label(\"$U$\", U, S); [/asy] If $\\angle QTA = 41^\\circ$ and $\\angle RUA = 63^\\circ,$ then find $\\angle QPR,$ in degrees. Feb 13, 2020 #1 +1 1) By similar triangles, YZ = 1/2*14^2/8 = 49/4. 2) Angle PQR = 90 - (41 + 63)/2 = 38", "Theorem (first of many proofs): the left diagram shows that $c^2 = 4 \\cdot \\frac{ab}2 + (b-a)^2 = a^2 + b^2$, and the right diagram shows a second proof by re-arranging the first diagram (the area of the shaded part is equal to $a^2 + b^2$, but it is also the re-arranged version of the oblique square, which has area $c^2$).[3] $[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c); pair shiftR = (a+b+c,0); path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } { /* first picture */ filldraw((0,0)--(c,0)--a*expi(rot1)--cycle, rgb(1,0.85,0.7)); fill(sq1, rgb(0.95,1,0.95)); fill(sq2, rgb(0.95,1,0.95)); filldraw(rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); filldraw((0,0)--top--(c,0)--a*expi(rot1)--cycle, rgb(0.5,0.9,0.5)); draw(sq1 ^^ sq2); draw(a*expi(rot1+pi/2)--top ^^ a*expi(rot1)--a*expi(rot1)+(0,c)); label(\"a\",a/2*expi(rot1),NW,sm); label(\"b\",a/2*expi(rot1)+(c/2,0),NE,sm); label(\"c\",(c/2,0),S,sm); } { /* second picture */ fill(shift(shiftR)*sq1, rgb(0.95,1,0.95)); fill(shift(shiftR)*sq2, rgb(0.95,1,0.95)); filldraw(shift(shiftR)*rotate(270)*xscale(c)*yscale(c)*unitsquare, rgb(0.96,1,0.96)); filldraw(shift(shiftR+(0,-c))*((0,0)--top--(c,0)--a*expi(rot1)--cycle), rgb(0.5,0.9,0.5)); filldraw(shift(shiftR+(0,-c))*((0,0)--(c,0)--a*expi(rot1)--cycle), rgb(1,0.85,0.7)); draw(shift(shiftR)*((0,0)--(c,0) ^^ sq1 ^^ sq2 ^^ a*expi(rot1+pi/2)--top ^^ a*expi(rot1)--a*expi(rot1)+(0,c))); label(\"a\",shiftR+a/2*expi(rot1),NW,sm); label(\"b\",shiftR+a/2*expi(rot1)+(c/2,0),NE,sm); label(\"c\",shiftR+(c/2,0),S,sm); } [/asy]$ Another proof of the Pythagorean Theorem (animated version). $[asy] defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); real a = 3/1.2, b = 4/1.2, c = (a^2 + b^2)^.5, rot1 = acos(a/c); pair shiftR = (a+b+c,0); path top = (0,c)--a*expi(rot1)+(0,c)--(c,c), sq1=rotate(rot1*180/pi)*xscale(a)*yscale(a)*unitsquare, sq2=shift(c,0)*rotate(rot1*180/pi)*xscale(b)*yscale(b)*unitsquare, tri = (0,0)--(0,a)--(b,0)--cycle; void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^", "to be the centroid of $\\triangle{P_1P_2P_3}$. Because $m\\angle{O_1P_1P_2} = 90^\\circ$ and $m\\angle{O_1P_1X} = 30^\\circ$, $m\\angle{O_1P_1X} = 60^\\circ$. $O_2M$ is $2/3$ the height of $\\triangle{P_1P_2P_3}$, thus $O_2M$ is $8*\\sqrt{3}/3$. Applying cosine law on $\\triangle{O_1P_1X}$, one finds that $P_1X = 2 + 2*\\sqrt{21}/3$. Multiplying by $3/2$ to solve for the height of $\\triangle{P_1P_2P_3}$, one gets $3 + \\sqrt{21}$. Simply multiplying by $2/\\sqrt{3}$ and then calculating the equilateral triangle's area, one would get the final result of $\\sqrt{300} + \\sqrt{252}$. This makes the answer $252 + 300 = 552$. $\\boxed{\\textbf{D}.}$ ~AlbeePach~ ## Solution 4 $[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label(\"P_1\", P1, E*1.5); label(\"P_2\", P2, SW*1.5); label(\"P_3\", P3, N); label(\"\\omega_1\", A, W*30); label(\"\\omega_2\", B, E*17); label(\"\\omega_3\", C, W*17); label(\"\\Gamma\", A, W*15); draw(Circle(A,2),red); pair X=foot(A,P1,P3); dot(X,blue); draw(A--X,blue); label(\"2\\sqrt{7}\", X--P3); label(\"2\\sqrt{3}\",X--P1); label(\"X\",X,dir(-80),blue); [/asy]$ First, note that because the $\\angle P_1=\\angle P_2=\\angle P_3=\\pi/3$, the arcs inside the shaded equilateral triangle are each $2\\pi/3$. Also, the distances between the centers of any two of the $3$ given circles are each $8$. Draw the circle $\\Gamma$ concentric with $\\omega_1$ with radius $2$. Because the arc of $\\omega_1$ inside said triangle is $2\\pi/3$,", "From (1) , $$2OA + AB = 8 + \\sqrt{32}$$ Thus, $$OA = 4$$ AND $$AB = \\sqrt{32}$$ because $$\\sqrt{32}$$cannot equal 2OA ( otherwise the triangle OAB wouldn't stand as isocele ) Now, we know that OA = OB = r = 4 and $$AB = \\sqrt{32}$$ and OAB is isocele and AB^2 = 32 and OA^2 = OB^2 = 16 Hence OAB is a right isocele triangle. from pythagor (AB^2 = OA^2 + OB^2) Area of shaded region = area of sector AOB - area of triangle AOB area of sector AOB = Lenght of Arc(AB) * AB Lenght of Arc (AB) = diameter * Pi * 45/360 = Pi $$\\sqrt{2}$$ / 2 area of sector AOB = 4Pi area of triangle = 4*4 /2 = 8 Finally: area of shaed region = 4Pi - 8 Hope that helps _________________ KUDOS is the good manner to help the entire community. Manager Joined: 14 Jan 2013 Posts: 152 Concentration: Strategy, Technology GMAT Date: 08-01-2013 GPA: 3.7 WE: Consulting (Consulting) Followers: 2 Kudos [?]: 159 [0], given: 29 Re: Triangle ABO is situated within the Circle with center O so [#permalink] 23 Apr 2013, 15:54 Rock750 wrote: Hi First of all, Points A and B lie on the circle with center O : OA = OB = r hence", "& \\equiv & {-\\measuredangle A'B'O.} \\end{array}$ Proof Let $$r$$ be the radius of the circle of the inversion. By the definition of inversion, $$OA \\cdot OA' = OB \\cdot OB' = r^2.$$ Therefore, $$\\dfrac{OA}{OB'} = \\dfrac{OB}{OA'}.$$ Clearly, $$\\measuredangle AOB = \\measuredangle A'OB' \\equiv -\\measuredangle B'OA'.$$ From SAS, we get that $$\\triangle OAB \\sim \\triangle OB'A'.$$ Applying Theorem 3.3.1 and 10.1.2, we get 10.1.1. Exercise $$\\PageIndex{2}$$ Let $$P'$$ be the inverse of $$P$$ in the circle $$\\Gamma$$. Assume that $$P \\ne P'$$. Show that the value $$\\dfrac{PX}{P'X}$$ is the same for all $$X \\in \\Gamma$$. The converse to the exercise above also holds. Namely, given a positive real number $$k \\ne 1$$ and two distinct points $$P$$ and $$P'$$ the locus of points $$X$$ such that $$\\dfrac{PX}{P'X} = k$$ forms a circle which is called the Apollonian circle. In this case $$P'$$ is inverse of $$P$$ in the Apollonian circle. Hint Suppose that $$O$$ denotes the center of $$\\Gamma$$. Assume that $$X, Y \\in \\Gamma$$; in particular, $$OX = OY$$. Note that the inversion sends $$X$$ and $$Y$$ to themselves. By Lemma $$\\PageIndex{1}$$, $$\\triangle OPX \\sim \\triangle OXP'$$ and $$\\triangle OPY \\sim \\triangle OYP'$$. Therefore, $$\\dfrac{PX}{P'X} = \\dfrac{OP}{OX} = \\dfrac{OP}{OY} = \\dfrac{PY}{P'Y}$$ and hence the result. Exercise $$\\PageIndex{3}$$ Let $$A',B'$$, and $$C'$$ be the images of $$A,B$$, and $$C$$", "is either $A^\\prime$ or $B^\\prime$, which has probability $0$. 3. $\\triangle ABC$ is obtuse-angled otherwise, which has probability $1-\\frac{\\alpha}{2\\pi}$. Now, of course $p$, $q$, and $r$ should not depend on $\\alpha$, so how do we compute those? • I don't know why but I think probability of $\\triangle{ABC}$ being right-angled $\\rightarrow 0$ instead of being $0$. – Dragonemperor42 May 15 '16 at 15:49 The probability $q$ of getting a right angled triangle $\\to 0,$ since $2$ of the $3$ points need to be exactly on the diameter. For obtuse and acute angled triangles: Choose $3$ random points on the circumference, draw diametral lines, and randomly choose a pole of each. Any 3 consecutive points necessarily lie in the same semicircle, and will form an obtuse angled triangle. There are $6$ such combos (starting from $1$ through $6$) against a total of $2^3 = 8$, thus $r = \\dfrac68 = \\dfrac34$, and $p = 1-r = \\dfrac14$" ]

[ "mobile point crosses an imaginary diameter of the circle perpendicular to the radius of the fixed point. This imaginary diameter bisects the area of the circle, therefore the triangle is acute half the time and obtuse the other half. Therefore half of all (isosceles) triangles must be acute. One quarter. A \"circle\" argument. Pick points $$A, B, C$$ \"at random\". Now pick polar coordinates so that $$O$$ is $$ABC$$'s circumcentre and $$A$$'s phase is $$\\theta=0$$. Let $$B$$'s and $$C$$'s phases be $$\\beta, \\gamma$$. Normalise phase to be in the interval $$[0, 2\\pi)$$. Then $$ABC$$ is obtuse if any of the following conditions occurs: • $$\\beta<\\pi$$ and $$\\gamma<\\pi$$ • $$\\beta>\\pi$$ and $$\\gamma>\\pi$$ • $$\\beta>\\gamma+\\pi$$ • $$\\gamma>\\beta+\\pi$$ These conditions are mutually exclusive. Assuming that $$\\beta$$ and $$\\gamma$$ are chosen from the uniform distribution on $$[0, 2\\pi)$$, the above conditions occur with probabilities $$1/4, 1/4, 1/8$$ and $$1/8$$. So $$ABC$$ is obtuse with probability $$3/4$$ and acute with probability $$1/4$$. An \"orthic system\" argument. Pick three points this way. First, pick an orthic system \"at random\". Then pick three of those four points to be your triangle's vertices, each combination of three being picked with probability $$1/4$$. With probability $$3/4$$, one of the three you picked was the interior point and your triangle is obtuse; with probability $$1/4$$, you picked the", "# Probability of triangle to be obtuse Two points $A,B$ fixed on a plane(distance = 2). C - random choosen point inside circle with radius $R$ with center at the center of $AB$ Find probability of triangle $ABC$ to be obtuse My thoughts: 1. If $C$ lies in the circle - $ABC$ will be rectangular 2. Opposite the larger angle is large side, so using cosine theorem: $$b^2+c^2<a^2=4R^2,$$ where $a$ - the diagonal, $b=AC, c=BC$ And then I don't know... Any such triangle $ABC$ with $AB$ diameter and $C$ any point within the circle will be obtuse. Proof:- (Note:-$AO=OB=r$) Let,C be any point in the circle.$AC$ is extended to meet the circumference at $X$.So,$\\angle AXB=\\alpha=90^0$ and $\\angle \\beta>0^0$.So,$\\angle ACB=\\gamma=\\alpha+\\beta=90$+something $>0.$So,$\\angle ACB$ will always be obtuse and thus $\\triangle ACB$ is always obtuse. • $AO=OB=1\\neq R$ (The solution should be given in terms of R?) – Logophobic Apr 6 '16 at 15:56 • @Logophobic, we were thinking along the same lines (see my answer). – Barry Cipra Apr 6 '16 at 16:47 • @Logophobic If center of circle is at center of line,then line must be diameter...it's actually $AO=OB=1=R$... – tatan Apr 6 '16 at 17:00 • @tatan See images in my post, $R$ is variable, not fixed at $1$. Point $C$ can occur outside the circle with diameter", "is either $A^\\prime$ or $B^\\prime$, which has probability $0$. 3. $\\triangle ABC$ is obtuse-angled otherwise, which has probability $1-\\frac{\\alpha}{2\\pi}$. Now, of course $p$, $q$, and $r$ should not depend on $\\alpha$, so how do we compute those? • I don't know why but I think probability of $\\triangle{ABC}$ being right-angled $\\rightarrow 0$ instead of being $0$. – Dragonemperor42 May 15 '16 at 15:49 The probability $q$ of getting a right angled triangle $\\to 0,$ since $2$ of the $3$ points need to be exactly on the diameter. For obtuse and acute angled triangles: Choose $3$ random points on the circumference, draw diametral lines, and randomly choose a pole of each. Any 3 consecutive points necessarily lie in the same semicircle, and will form an obtuse angled triangle. There are $6$ such combos (starting from $1$ through $6$) against a total of $2^3 = 8$, thus $r = \\dfrac68 = \\dfrac34$, and $p = 1-r = \\dfrac14$", "that is an acute triangle which means that all three central angles $AOB,$ $BOC,$ $COA$ are less than $180^{\\circ}.$ It follows that their semicircles intersect pairwise and thus cover the whole of the circle. Now, with a reference to a problem of Semicircle Coveragecovering a circle with three semicircles, the probability of having $O$ in the interior of $\\Delta ABC$ equals $\\displaystyle \\frac{1}{4}.$ ### Solution 3 Let us say that $O$ is the first point we generate. This will be our point of reference. From this point, we will use polar coordinate system (counterclockwise) to determine the position of $A$ (our second point, with $\\widehat{OCA}=\\alpha$) and then the position of $B$ (our third point, with $\\widehat{OCB}=\\beta$). Let $C$ be the center of the circle. Let us call $\\check{A}$ and $\\check{B}$ the two random variables (uniforms on $[0,2\\pi]$). Let us distinguish between two cases (there are two letters $B$ in each graph for the two \"extreme\" letters $B):$ 1. If $0<\\alpha\\leq\\pi$ then the $\\beta$ that allow $C$ to be in the triangle are those in $[\\pi,\\pi+\\alpha]$. 2. If $\\pi<\\alpha\\leq2\\pi$ then the $\\beta$ that allow $C$ to be in the triangle are those in $[\\alpha-\\pi,\\pi]$. The probability $p$ that $C$ is inside the triangle $OAB$ is (with convenient notations for probability density functions): \\displaystyle \\begin{align} p & = \\int_{0}^{\\pi}\\mathbb{P}(\\check{B}\\in[\\pi,\\pi+\\alpha]|\\check{A}=\\alpha)\\mathbb{P}(\\check{A}=\\alpha)d\\alpha\\\\ &\\qquad\\qquad\\qquad+\\int_{\\pi}^{2\\pi}\\mathbb{P}(\\check{B}\\in[\\alpha-\\pi,\\pi]|\\check{A}=\\alpha)\\mathbb{P}(\\check{A}=\\alpha)d\\alpha", "to area $$\\arccos(a) - a\\sqrt{1-a^2} + \\frac{a^2\\pi}{4} + \\frac{\\pi}{2}$$ (Upper, inner, and lower). By this conditional probability, we could derive the total probability, which is $$\\int_0^1 \\frac{1}{\\pi}\\left(\\arccos(a) - a\\sqrt{1-a^2} + \\frac{a^2\\pi}{4} + \\frac{\\pi}{2}\\right) \\cdot 2a \\; da = \\frac{3}{4}$$ But is there an easier way to solve this? This looks like a math competition style of question and I expect some tricks to be at play. Thanks! • If the obtuse angle was at point O, the possibility depends on the distance between point $A$ and point $B$, to the radius $r$ of the circle – Aderinsola Joshua Mar 29 at 7:57 The probability that point $$A$$ is at a distance from the center in $$[a,a+da]$$ is $$2\\pi a\\,da/\\pi=2a\\,da$$. The probability that, fixed $$A$$ as above, $$\\angle ABO>90°$$, is the same as the probability that $$B$$ lies inside a circle of diameter $$OA$$, that is $$\\pi(a/2)^2/\\pi=a^2/4$$. Hence the overall probability that $$\\angle ABO>90°$$ is: $$p(\\angle ABO>90°)=\\int_0^1 {a^2\\over4}\\cdot 2a\\,da={1\\over8}.$$ The probability that triangle $$ABO$$ is obtuse is then: $$p(\\angle ABO>90°)+p(\\angle BAO>90°)+p(\\angle AOB>90°) ={1\\over8}+{1\\over8}+{1\\over2}={3\\over4}.$$ EDIT. For a 3D sphere one can repeat the same argument, obtaining: $$p(\\angle ABO>90°)=\\int_0^1 {a^3\\over8}\\cdot 3a^2\\,da={1\\over16}.$$ • Thank you! You made me realize that we should not use the marginal w.r.t. $y$ coordinates (feasible but convoluted) but the distance from a given point to the center.", "# Probability of triangle to be acute? Suppose that someone randomly picks $3$ points $A, B$ and $C$ on a fixed circle. What is the probability of triangle $ABC$ to be acute? • Do you mean on the circle, or inside the circle? – Robert Israel Apr 24 '15 at 20:02 • On the bounddary of the circle @RobertIsrael, not inside. – brick Apr 24 '15 at 20:05 • Possible duplicate of math.stackexchange.com/questions/560441/… – Ojas Apr 24 '15 at 20:18 Let one point be fixed and let circle have radius $1$. Now find posibility for other two points. First integrate all possibilities: $$I_1=\\int_0^{2\\pi}\\int_0^{2\\pi}dbda=4\\pi^2$$ Then integrate required case: $$I_2=\\int_0^{\\pi}\\int_a^{2\\pi}dbda+\\int_{\\pi}^{2\\pi}\\int_0^{a}dbda=3\\pi^2$$ So, your probability is $\\frac {I_2} {I_1}=\\frac34$. Explanation: let second point be $A$ and third be $B$. When we choose $A$, we cannot choose point $B$ between first point and $A$. Because of that reason, we integrate from $0$ to $a$. • It looks to me like the second integral (you really should use something other than $A$ and $B$ for them) is almost the space where the three points form an obtuse triangle. For instance, in the first part, if we let $C$ be at argument $0$ on the unit circle, $A$ be at $a$, and then if $B$ is anywhere between $0$ and $a+\\pi$, then an obtuse triangle", "# Choose 3 points A, B and C in a circle O I have to get $p$,$q$ and $r$. $p$ = the probability of triangle $ABC$ is an acute-angled triangle $q$ = the probability of triangle $ABC$ is a right-angled triangle $r$ = the probability of triangle $ABC$ is an obtuse-angled triangle. I have an idea that the diameter of O is important in determining the angle. But I cannot get exact $p$,$q$ and $r$. • Using polar coordinates may help you. – Kenny Lau May 15 '16 at 14:15 • But I cannot use polar coordinate since this course is not handlinh that concepts... – user331899 May 15 '16 at 14:16 • Hint: try to fix $A$ and $B$ and see how $C$ determines what the triangle is. – Kenny Lau May 15 '16 at 14:18 • If AB is the diameter of O then no matter c's position, It is a right-angled triangle. – user331899 May 15 '16 at 14:21 • What if AB is not the diameter of O? You're getting there. – Kenny Lau May 15 '16 at 14:21 Hint: Given $A$ and $B$ as in the picture above, 1. $\\triangle ABC$ is acute-angled only if $C$ lies in the red arc, which has probability $\\frac{\\alpha}{2\\pi}$. 2. $\\triangle ABC$ is right-angled only if $C$", "integration, but you needn't integrate. If you set $a=x$ and $b=y$, you get the unit circle equation ${x}^{2}+{y}^{2}<1$. The area of this circle inside the parameters $0 and $0 is $\\left[\\pi \\left({r}^{2}\\right)\\right]/4$, or $\\pi /4$. The total area is the unit square. Thus, the probability of an obtuse triangle is $\\left(\\pi /4\\right)/1$, or just $\\pi /4.$. ###### Did you like this example? Bayobusalue Answered 2022-11-23 Author has 4 answers Step 1 Let a and b be your two independent random numbers. First of all, we need to check whether the three given sides even form a triangle. For that, a,b,1 need to satisfy the triangle equation, i.e. $a+b\\ge 1$ (the other two permutations are always satisfied anyway). Now when is this triangle obtuse? Well, we know that the side of length 1 is the longest side, so it must be opposite the largest angle $\\gamma$. When is that angle $\\gamma$obtuse? For that, remember Pythagoras. $\\gamma$ is a right angle exactly when ${a}^{2}+{b}^{2}={1}^{2}$, similarly $\\gamma$ is obtuse exactly when ${a}^{2}+{b}^{2}<{1}^{2}$. Step 2 Now what is the probability of $a+b\\ge 1\\wedge {a}^{2}+{b}^{2}<{1}^{2}$? Let a and b denote coordinates of a point (a,b). This point is, by (assumed) independence of a and b, uniformly random in the unit square $\\left[0,1{\\right]}^{2}$. Now a and b satisfy the property ${a}^{2}+{b}^{2}<{1}^{2}$ exactly when (a,b)", "0 to pi/2 and you find the probability of an acute triangle is 0.75. integrate from pi/2 to pi and you find that the probability of an obtuse triangle is 0.25. I wish I had a blackboard. PF Gold P: 162 The answer using my analysis is 1/4. This is confirmed using the Mathematica commands below. f[] := {T1 = Random[]*2 \\[Pi]; T2 = Random[]*2 \\[Pi]; T3 = Random[]*2 \\[Pi]; p1 = {Cos[T1], Sin[T1]}; p2 = {Cos[T2], Sin[T2]}; p3 = {Cos[T3], Sin[T3]}; a1 = VectorAngle[p2 - p1, p3 - p1]; a2 = VectorAngle[p1 - p2, p3 - p2]; a3 = VectorAngle[p1 - p3, p2 - p3]; {a1, a2, a3}}[[1]] 1 - Mean[ Table[ Total[(If[# > Pi/2, 1, 0] & /@ f[])], {i, 1, 10000}]] // N Also nice explenation here PF Gold P: 162 Quote by alan2 Look at Kai's post above. The solution is similar. The probability of an acute triangle is 0.75, while the probability of an obtuse triangle is 0.25. I wish I knew how to use the cool editing features but I don't so here goes, you can construct it. Because of the symmetry of the circle, only the relative positions of the points matter. So the problem is equivalent to choosing a point, say A, and choosing an angle, theta, between 0 and", "circle greater than 180 degrees, and each angle is less than half of 180 or 90 degrees. It is therefore if and only if. – Brian Tung Apr 25 '15 at 15:55 Put the first point at the top of the circle. Randomly select the second point uniformly from the circle; without loss of generality, assume the point falls on the right half. (If it's on the left, the arguments that follow can just be mirror-reversed.) Let the angle between the two points (as measured from the circle's center) be $x$ radians; clearly, $0 \\leq x \\leq \\pi$. We claim (and will shortly demonstrate) that the triangle will be acute if and only if the third point is chosen such that no semicircle contains all three points. This happens if the third point is chosen on the left side, between $\\pi-x$ and $x$ radians from the first point at the top of the circle. The probability of this, given $x$, is $x/(2\\pi)$. Since $x$ ranges freely and uniformly from $0$ to $\\pi$, $x/(2\\pi)$ ranges freely and uniformly from $0$ to $1/2$, and the desired probability is the average of $x/(2\\pi)$, or $1/4$. Argument that a triangle inscribed in a circle is acute if and only if the three points cannot be contained by a semicircle: Let $A, B,", "# Probability of an obtuse triangle in a circle. Suppose we randomly pick 2 points A, B within a circle centered at point O. What is the probability that the triangle formed by ABO is an obtuse one? (Note that A and B are not exclusively on the circumference). And what is the conclusion extended to A, B within a ball instead? Thanks! PS this is from a Quant interview. The following is what I have derived during the exam: (Edited, thank you for your corrections!) consider the joint probability of x, y coordinate for any point in a unit circle, then $$f_{XY}(x, y) = \\frac{1}{\\pi}$$, uniformly distributed inside the circular region. The distance between the point and the center, has thus a distribution $$f_Z(z) = 2z$$ for $$z$$ in [0, 1] ($$z^2 = x^2 + y^2$$). Randomly pick an A, rotate the circle so that A is right on top of the center O. Suppose now A has a distance $$z = a$$ $$(a > 0)$$ away from the origin, then B could only be chosen in the region of 1. $$y > a$$ 2. $$y < 0$$ 3. within an inner circle whose diameter is $$OA$$ Therefore, given that the distance is $$z = a$$, the probability of ABO being an obtuse triangle is given corresponds", "# Compute the probability that at least one of the angles of the triangle $ABC$ exceeds $x\\pi$ Let $A,B,C$ be three points, chosen independently and uniformly at random on the circumference of a circle. Let $b(x)$ be the probability that at least one of the angles of the triangle $ABC$ exceeds $x\\pi$. Show that $$b(x)= 1-(3x-1)^2\\ \\ \\ \\ \\ \\mbox{if}\\ \\ \\frac{1}{3}\\leq x\\leq \\frac{1}{2};\\\\ b(x)= 3(1-x)^2 \\ \\ \\ \\ \\ \\mbox{if}\\ \\frac{1}{2}\\leq x\\leq 1.$$ My computation is as follows: Without loss of generality, we assume the circle is unit circle. Let $A=(1,0),\\ B=(1,\\theta),$ and $C=(1,\\phi)$ where $0<\\theta<\\phi<2\\pi$. It is obvious that the joint density function $f(\\theta,\\phi)=(2\\pi^2)^{-1}$ with $0<\\theta<\\phi<2\\pi.$ Moreover, we have $$\\angle A=\\frac{1}{2}(\\phi-\\theta),\\ \\ \\angle B=\\pi-\\frac{\\phi}{2},\\ \\ \\mbox{and}\\ \\ \\angle C=\\frac{1}{2}\\theta.$$ First, consider $x\\in [1/3,1/2]$. I tried to compute the probability that all angles are less than $x\\pi$ as follows: $$\\frac{1}{2}\\theta \\leq x\\pi,\\ \\ \\frac{1}{2}(\\phi-\\theta)\\leq x\\pi,\\ \\ \\mbox{and} \\ \\ \\pi-\\frac{1}{2}\\phi\\leq x\\pi.$$ Thus, we have $$\\theta \\in [0,2x\\pi],\\ \\ \\mbox{and}\\ \\ \\phi\\in [2(1-x)\\pi,2x\\pi+\\theta].$$ Hence, $$\\int_{0}^{2x\\pi}\\int_{2(1-x)\\pi}^{2x\\pi+\\theta}\\frac{1}{2\\pi^2}d\\phi d\\theta =\\frac{1}{2\\pi^2}\\int_{0}^{2x\\pi} [2x\\pi+\\theta-2\\pi+2x\\pi]d\\theta = \\\\ \\frac{1}{2\\pi^2}\\int_{0}^{2x\\pi} [2\\pi(2x-1)+\\theta]d\\theta = \\frac{1}{2\\pi^2}[(2\\pi)(2x-1)(2x\\pi)+2x^2\\pi^2] =5x^2-2x.$$ Thus, I obtain $b(x)=1-(5x^2-2x)=-5x^2+2x+1.$ The result is much different to the the goal. For the case $x\\in [1/2,1].$ I analyzed as follows: It is obvious that $\\angle C=1/2\\theta$ is impossible exceed $x\\pi$, so we only need to consider angle $B$ and", "triangle double getProbability(double x, double y) { // the of all angles in a triangle is pi: // pi = otherAlpha + otherBeta + gamma // but the big triangle must fulfil the same condition, so // pi = alpha + otherAlpha + beta + otherBeta + pi/2 // therefore the difference is // 0 = gamma - alpha - beta - pi/2 // gamme = alpha + beta + pi/2 auto gamma = PI/2 + getAlpha(x, y) + getBeta(x, y); // each direction is equally likely: \"exit probability\" = gamma / full circle // and a full circle has 360 degrees => 2pi return gamma / (2 * PI); } int main() { // display 10 digits std::cout << std::fixed << std::setprecision(10); double epsilon = 0.01; //std::cin >> epsilon; double sum = 0; unsigned long long numPoints = 0; for (double x = epsilon; x < a; x += epsilon) { auto height = b - x * b/a; for (double y = epsilon; y < height; y += epsilon) { auto current = getProbability(x, y); sum += current; numPoints++; //if (numPoints % 10000000 == 0) //std::cout << numPoints << std::endl; } } // compute average percentage auto result = sum / numPoints; // ==> uncomment the following line to see my approximate result //std::cout << result", "red circle, as you described. However, this distribution will not be uniform. To see this, express the circle in angular coordinates and consider the relative probabilities that the point will satisfy $\\theta \\in [-\\pi/8, \\pi/8]$ (i.e. between the green and red lines on the right side) and $\\theta \\in [\\pi/8, 3 \\pi/8]$ (between the red and purple lines in quadrant I). These events will occur iff the point selected from the square is also in those corresponding regions. It is easy to see without calculation (and not terribly difficult to verify with a calculation) that there is more area in the square between the red and purple lines than there is between the red and green lines. Hence, your point chosen in this fashion is more likely to land in the interval $\\theta \\in [\\pi / 8, 3 \\pi /8]$ than it is to land in $\\theta \\in [-\\pi / 8, \\pi / 8]$. This is problematic because those two intervals \"should\" have the same probabilities.", "of Triangles $$\\theta_1+\\theta_2+\\theta_3=\\pi$$ $$0<\\theta_1, 0 < \\theta_2, 0<\\theta_3$$ and $$\\theta_1=\\pi/2$$ $$\\theta_2=\\pi/2$$ $$\\theta_3=\\pi/2$$ $$\\mathbb{P}(\\text{obtuse})=\\frac{3}{4}$$ ### Side Lengths?! Remember the sidelengths $$(a,b,c)$$ uniquely determine a triangle (SSS). Obtuseness is scale-invariant, so pick a perimeter $$P$$ and we have $$a+b+c=P$$. ### Problem Not all points in the simplex correspond to triangles $$b+c<a$$ $$a+b<c$$ $$a+c<b$$ ### Yet Another Pillow Problem Answer $$\\mathbb{P}(\\text{obtuse})=9-12\\ln 2 \\approx 0.68$$ $$b^2+c^2=a^2$$ $$a^2+b^2=c^2$$ $$a^2+c^2=b^2$$ — Stephen Portnoy, Statistical Science 9 (1994), 279–284 ### Translation for Geometers The space of all triangles should be a (preferably compact) manifold $$T$$ with a transitive isometry group. We should use the left-invariant metric on $$T$$, scaled so vol$$(T)=1$$. Then the Riemannian volume form induced by this metric is a natural probability measure on $$T$$, and we should compute the volume of the subset of obtuse triangles. Ideally, this should generalize to $$n$$-gons. Spoiler: $$T\\simeq\\mathbb{RP}^2=G_2\\mathbb{R}^3$$ ### Back to Triangles Let $$s=\\frac{1}{2}(a+b+c)$$ and define Note: It’s convenient to choose $$s=1$$. $$s_a=s-a, \\quad s_b = s-b, \\quad s_c = s-c$$ Then $$s_a+s_b+s_c=3s-(a+b+c)=3s-2s=s$$ and the triangle inequalities become $$s_a>0, \\quad s_b > 0, \\quad s_c > 0$$ But there’s still no transitive group action! ### Take Square Roots! Consider $$(x,y,z)$$ so that $$x^2=s_a, \\quad y^2 = s_b, \\quad z^2 = s_c$$ The unit sphere is a $$2^3$$-fold cover of triangle space ### The Transitive Group The", "# 1985 AIME Problems/Problem 9 ## Problem In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\\alpha$, $\\beta$, and $\\alpha + \\beta$ radians, respectively, where $\\alpha + \\beta < \\pi$. If $\\cos \\alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator? ## Solution 1 $[asy] size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3); D(CR(O,r)); D(O--rotate(a/2)*A--rotate(-a/2)*A--cycle); D(O--rotate(b/2)*A--rotate(-b/2)*A--cycle); D(O--rotate((a+b)/2)*A--rotate(-(a+b)/2)*A--cycle); MP(\"2\",(rotate(a/2)*A+rotate(-a/2)*A)/2,NE); MP(\"3\",(rotate(b/2)*A+rotate(-b/2)*A)/2,NE); MP(\"4\",(rotate((a+b)/2)*A+rotate(-(a+b)/2)*A)/2,NE); D(anglemark(rotate(-(a+b)/2)*A,O,rotate((a+b)/2)*A,5)); label(\"$$\\alpha+\\beta$$\",(0.08,0.08),NE,fontsize(8)); [/asy]$ All chords of a given length in a given circle subtend the same arc and therefore the same central angle. Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle. $[asy] size(200); pointpen = black; pathpen = black + linewidth(0.8); real r = 8/15^0.5, a = 57.91, b = 93.135; pair O = (0,0), A = r*expi(pi/3), A1 = rotate(a/2)*A, A2 = rotate(-a/2)*A, A3 = rotate(-a/2-b)*A; D(CR(O,r)); D(O--A1--A2--cycle); D(O--A2--A3--cycle); D(O--A1--A3--cycle); MP(\"2\",(A1+A2)/2,NE); MP(\"3\",(A2+A3)/2,E); MP(\"4\",(A1+A3)/2,E); D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5)); label(\"$$\\alpha$$\",(0.07,0.16),NE,fontsize(8)); label(\"$$\\beta$$\",(0.12,-0.16),NE,fontsize(8)); [/asy]$ This triangle has semiperimeter $\\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \\sqrt{\\frac92 \\cdot \\frac52 \\cdot \\frac32 \\cdot \\frac12} = \\frac{3}{4}\\sqrt{15}$. The area of", "# ellipson Topic: Ellipse In the diagram, a circle of radius $OA=2$ rolls around the inside of a circle of radius $OB=4$ with the half radius $AP=1$ originally positioned so that $P$ is on $OB$. Find a parameterization for the path that $P$ traces out as the inner circle rolls (without slipping.) Start by choosing an appropriate parameter. Is that a familiar curve? The key here is to assert that the arclength swept out by rolling is the same on both circles, so if $\\theta$ is the angle swept out by $\\overrightarrow{OA}$ as the circle rolls and $\\alpha$ is the angle swept out by $\\overrightarrow{AP}$ then the arclengths are equal so $4\\theta=2\\alpha\\Rightarrow \\alpha=2\\theta$. However, $\\theta$ is subtracted from $\\alpha$ because as the inner circle rolls, it both rotates around its center and rotates around the bigger circle's center; at the same time the inner circle's center is rotating around the bigger circle's center, so the angle from the vertical of the line $\\overline{AP}$ is $2\\theta-\\theta=\\theta$. You can think of the process of getting from the origin (0,0) to the point $P$ as a two step process: first get from (0,0) to $A=(2\\sin\\theta,2\\cos\\theta)$ and then get from $A$ to $P$, which is the vector $\\overrightarrow{AP}=(\\sin\\theta,-\\cos\\theta)$, thus the coordinates of $P$ are $(3\\sin\\theta,\\cos\\theta)$. Since these coordinates satisfy the equation $\\dfrac{x^2}{9}+y^2=1$", "of getting an acute vs obtuse triangle? P: 26 Quote by jedishrfu If you move the third point back and forth a bit what is the chance of getting an acute vs obtuse triangle? As first two points are getting further apart almost to form a diameter, the probability of getting an acute (the same for obtuse) angle is indeed getting closer to 1/2. I'm not sure what happens when the first two points are chosen close to each other... Edit: I think when the first two points are close to each other one is almost certain to complete an obtuse triangle. Edit2: Indeed. Now I'm pretty much certain that probability to complete an acute triangle is 1/4 :)) PF Gold P: 162 In polar coordinates, we have to pick θ1, θ2 and θ3. just assume that θ1=0, since we can always rotate the result and redefine our coordinates so that this is true. By Thales theorem in order to get a right angle we need one of the other points to land on π. Using the fact the two random variables are independent we get P(θ2 = π OR θ3 = π) = P(θ2 = π) + P(θ3 = π) - P(θ2 = π)P(θ3 = π) I'll let you figure out what the above quantity is for", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "# 2018 AMC 12B Problems/Problem 25 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Circles $\\omega_1$, $\\omega_2$, and $\\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\\omega_1$, $\\omega_2$, and $\\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\\omega_i$ for each $i=1,2,3$, where $P_4 = P_1$. See the figure below. The area of $\\triangle P_1P_2P_3$ can be written in the form $\\sqrt{a}+\\sqrt{b}$ for positive integers $a$ and $b$. What is $a+b$? $[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label(\"P_1\", P1, E*1.5); label(\"P_2\", P2, SW*1.5); label(\"P_3\", P3, N); label(\"\\omega_1\", A, W*17); label(\"\\omega_2\", B, E*17); label(\"\\omega_3\", C, W*17); [/asy]$ $\\textbf{(A) }546\\qquad\\textbf{(B) }548\\qquad\\textbf{(C) }550\\qquad\\textbf{(D) }552\\qquad\\textbf{(E) }554$ ## Solution Let $O_i$ be the center of circle $\\omega_i$ for $i=1,2,3$, and let $K$ be the intersectoin of lines $O_1P_1$ and $O_2P_2$. Because $\\angle P_1P_2P_3 = 60^\\circ$, it follows that $\\triangle P_2KP_1$ is a $30-60-90^\\circ$ triangle. Let $d=P_1K$; then $P_2K = 2d$ and $P_1P_2 = \\sqrt 3d$. The Law of Cosines" ]

[ "# The coefficient of the $8^{th}$ term in expansion of $(1+x)^{10}$ is equal to : ( A ) $210$ ( B ) $120$ ( C ) $7$ ( D ) $^{10}C_3$", "# Coefficient play The number of terms in the expansion of $$(3x-4y+6z)^4$$ is $$m$$ and the coefficient of $$xy^2z$$ in the same expansion is $$n$$. Find $$m+n$$. ×", "Free Version Difficult # 8th Term in an Expansion ACTMAT-NLY4JJ What is the coefficient of the 8th term in the expansion of: $${ ({ 3x }^{ 2 }-y) }^{ 10 }$$ A $-120$ B $-3240$ C $45$ D $405$ E $3240$", "# Find the Coefficient Find the coefficient of $$x^4$$ in the expansion of $\\large (2-x+3x^2)^6.$ ×", "# What is the coefficient of 2.4n+9.6? The coefficients are $2.4$ and $9.6$. There are technically two coefficients in this expression. The first term, $2.4 n$, the coefficient is $2.4$. In the second term, $9.6$, the coefficient is $9.6$, because technically, the term is $9.6 {n}^{0}$.", "# Trinomial expansion coefficient What is the coefficient of the $$x^2y^2z^2$$ term in the polynomial expansion of $$(x+y+z)^6?$$ ×", "Find the coefficient of $$x^{9}$$ in the expansion of $(2+x)^{10}(1+x)^0+(2+x)^{9}(1+x)+(2+x)^{8}(1+x)^{2}+ \\ldots + (2+x)^0(1+x)^{10}.$", "# Challenge you What is the coefficient of $$x^{7}y^{3}z^{2}$$ in the expansion of $$(x+2y+3z)^{12}$$. ×", "Find the coefficient of $x^n$ in the expansion of $(1+x) (1- x)^n$. #### Solution Coefficient of $x^n$ in $(1+x) (1- x)^n$ $=$ Coefficient of $x^n$ in $(1-x)^n$ $+$ Coefficient of $x^{n-1}$in $(1-x) ^n$ $= (-1)^n$ $^nC_n$ +$(-1) ^{n-1}$ $^nC_{n-1}$ $=(-1) ^n (1-n)$", "# Submitted by Mohammad ElKammash Find the coefficient of $$x^{9}$$ in the expansion of $(2+x)^{10}(1+x)^0+(2+x)^{9}(1+x)+(2+x)^{8}(1+x)^{2}+ \\ldots + (2+x)^0(1+x)^{10}.$ ×", "# Coefficient In A Septic Expansion What is the coefficient of $$x^3$$ in the expansion of $$(x+2)^7$$? ×", "× # Expansion What would be the coefficient of $$x^t$$ in the expansion of: $$(1+x+x^2+\\cdots+x^r)^n$$ if $$t>r$$? Is there any general formulae? Note by Rahul Vernwal 1 year, 3 months ago", "+0 # What is the coefficient of ​ in the expansion of ? Express your answer as a common fraction. 0 255 1 +43 What is the coefficient of $$x^2y^2$$ in the expansion of$$\\left(\\frac{3}{5}x-\\frac{y}{2}\\right)^8$$ ? Express your answer as a common fraction. Jz1234 Aug 27, 2017 #1 +87293 +2 There is no x^2y^2 term in the expansion of this.....the sum of the exponents on both variables must = 8 CPhill Aug 27, 2017", "# Not for the weak-hearted Find the coefficient of $x^{11}$ in the expansion of $(1+x^2)^4(1+x^3)^7(1+x^4)^{12}$. ×", "Browse Questions # Find a positive value of m for which the coefficient of $x^2$ in the expansion of $(1+x)^m$ is $6$ $\\begin{array}{1 1}(A)\\;3\\\\(B)\\;4\\\\(C)\\;5\\\\(D)\\;6\\end{array}$ • Coefficient of $x^2$ in the expansion of $(1+x)^m$ is $C(m,2)$ . $C(m,2)=6$", "# Find a positive value of m for which the coefficient of $x^2$ in the expansion of $(1+x)^m$ is $6$ $\\begin{array}{1 1}(A)\\;3\\\\(B)\\;4\\\\(C)\\;5\\\\(D)\\;6\\end{array}$", "Browse Questions # Find a positive value of m for which the coefficient of $x^2$ in the expansion of $(1+x)^m$ is $6$ $\\begin{array}{1 1}(A)\\;3\\\\(B)\\;4\\\\(C)\\;5\\\\(D)\\;6\\end{array}$ Toolbox: • Coefficient of $x^2$ in the expansion of $(1+x)^m$ is $C(m,2)$ . According to the problem $C(m,2)=6$", "+0 # What is the coefficient of x^3 in the expansion of (x+2*sqrt3)^7? 0 48 1 What is the coefficient of x^3 in the expansion of (x+2*sqrt3)^7? Jul 16, 2021 #1 +114112 +1 $$(x+2\\sqrt3)^7\\\\~\\\\ 7C3*(2\\sqrt3)^4(x)^3$$ That is the x^3 tern. You can take it from there. Jul 16, 2021", "Browse Questions # The coefficient of $x^m$ in the expansion 1 + $(1+x)$ + $(1+x)^2$ + ..... + $(1+x)^n$ is $\\begin{array}{1 1} ^{n+1} C_{m+1} \\\\ ^n C_m+1 \\\\ ^{n+1} C_m \\\\^nC_m \\end{array}$ $1+(1+x)+(1+x)^2+..........(1+x)^n$ is a G.P.", "# If the coefficient of $r^{th}$ term and $(r+1)^{th}$ term in the expansion of $(1+x)^{3n}$ are in the ratio $1 :2$, then $r$ is equal to : ( A ) $\\frac{1}{3}(3n+1)$ ( B ) $\\frac{1}{3} ( 3n+2)$ ( C ) $\\frac{6}{5} (n+1)$ ( D ) $\\frac{1}{4} (n+2)$" ]

[ "4 = 16 + 0. Then the coefficient we want is $\\binom{12+16-1}{16} - \\binom{12}{1} \\binom{12+12-1}{12} + \\binom{12}{2} \\binom{12+8-1}{8} - \\binom{12}{3} \\binom{12+4-1}{4} + \\binom{12}{4}$ Whew... • Feb 12th 2011, 04:48 PM Plato Quote: Originally Posted by awkward Whew... Well as I said, that can be a course in and of itself.", "coefficient $\\binom{n}{n_1 \\; n_2 \\; \\dots \\; n_N}$ counts the number of ways to put n distinct objects in N boxes, with $n_1$ in the first box, $n_2$ in the second box, etc. See Multinomial theorem - Wikipedia, the free encyclopedia. For a derivation, consider all $n!$ permutations of the n objects and consider the first $n_1$ to be in the first box, the next $n_2$ to be in the second box, etc. The order of the objects in the boxes is considered irrelevant, however, so divide by $n_1!$ to compensate for over-counting in box 1, divide by $n_2!$ to compensate for over-counting in box 2, ... It's just like the binomial coefficient, only generalized to N choices instead of 2.", "binomial\" coefficient: $$(-1)^n {n+m \\choose n} = {-m \\choose n}.$$ That should allow you to evaluate the sum explicitly. RGV", "coefficient of $x_1^{r_1} ... x_k^{r_k}$ is the number you want.", "## Intermediate Algebra for College Students (7th Edition) The binomial coefficient of the first term is $n\\choose0$ and of the second term is $n\\choose1$ etc. thus the statement doesn't make sense.", "an $$a$$, taking a $$b$$ from the other $$4 − i$$ factors (where $$0 ≤ i ≤ 4)$$. Let’s go through each of these cases separately. By Theorem 3.1.1, there is $$\\binom{4}{4} = 1$$ way to choose four factors from which to take as. (Clearly, you must choose an a from every one of the four factors.) Thus, the coefficient of $$a^4$$ will be $$1$$. If you want to take as from three of the four factors, Theorem 3.1.1 tells us that there are $$\\binom{4}{3} = 4$$ ways in which to choose the factors from which you take the $$a$$s. (Specifically, these four ways consist of taking the $$b$$ from any one of the four factors, and the as from the other three factors). Thus, the coefficient of $$a^3 b$$ will be $$4$$. If you want to take as from two of the four factors, and bs from the other two, Theorem 3.1.1 tells us that there are $$\\binom{4}{2} = 6$$ ways in which to choose the factors from which you take the as (then take $$b$$s from the other two factors). This is a small enough example that you could easily work out all six ways by hand if you wish. Thus, the coefficient of $$a^2 b^2$$ will be $$6$$. If you want to take as from", "select the coefficient of $z^k$. • In (8) we apply the binomial theorem.", "# Binomial Coefficient with Zero/Integer Coefficients ## Theorem $\\forall n \\in \\N: \\dbinom n 0 = 1$ where $\\dbinom n 0$ denotes a binomial coefficient. ## Proof From the definition: $\\displaystyle \\binom n 0$ $=$ $\\displaystyle \\frac {n!} {0! \\ n!}$ $\\quad$ Definition of Binomial Coefficient $\\quad$ $\\displaystyle$ $=$ $\\displaystyle \\frac {n!} {1 \\cdot n!}$ $\\quad$ Definition of Factorial of $0$ $\\quad$ $\\displaystyle$ $=$ $\\displaystyle 1$ $\\quad$ $\\quad$ $\\blacksquare$", "# Binomial Coefficient/Examples/4 from 52 The number of ways of choosing $4$ objects from a set of $52$ (for example, cards from a deck) is: $\\dbinom {52} 4 = \\dfrac {52 \\times 51 \\times 50 \\times 49} {4 \\times 3 \\times 2 \\times 1} = \\dfrac {52!} {48! \\, 4!} = 270 \\, 725$", "Weighted Binomial Coefficient Sum Probability Level 4 If the value of $\\dbinom{36}{1}+4\\dbinom{36}{4}+7\\dbinom{36}{7}+\\cdots+34\\dbinom{36}{34}$ can be expressed as $a\\times b^{c}+12,$ where $a$ and $b$ are prime numbers, then find the value of $a+b+c+12.$ ×", "## Algebra 2 (1st Edition) He forgot the binomial coefficient; the correct formula is: $P(k=3)=_5C_3(\\frac{1}{6})^3(\\frac{5}{6})^{5-3}\\approx0.032$", "with $N(a)>8$ and the only number greater than one with $N(a)=8$ is a=3003. References: [1] Singmaster, D. “Research Problems: How often does an integer occur as a binomial coefficient?”, American Mathematical Monthly, 78(1971) 385–386.", "+ 72 \\binom{x+2}{1} - 90\\binom{x+2}{2} + 108\\binom{x+2}{3}$ To check whether there is a cubic polynomial passing through the other points either , we just sub. $x = 3 , x = 7$ and see if the values the polynomial gives us equal to $4$ and $376$ respectively . 10. thanks moo and simplependulum for ur post.........", "of the corresponding three terms of $$(1-x)^{-6}$$ viz., $$x^{20}, x^{11}, x^2$$ respectively. Finally the coefficient is given by $$\\binom{25} {5} - 6*\\binom{16} {5} + 15*\\binom{7} {5} = 27237$$ • Thanks PTDS for helping! – Vasu090 Jul 27 at 5:50", "of distributing $k$ distinct objects, where $k$ is from the $\\frac{x^k}{k!}$ term. So, for example, if we have $3$ balls in our combined box, the coefficient of the term $\\frac{x^k}{k!}$, for $k = 3$, should be $\\binom{3}{0} + \\binom{3}{1}$.", "4}{3\\cdot2 \\cdot 1} = \\frac{120}{6} = 20.$ - It's the binomial coefficient, and you should google combination and permutation as a starter. - Thank you! I will. :) – NewProger Dec 14 '12 at 18:57", "# Coefficient coefficient If a binomial expansion is given by $(1+x)^{10} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \\cdots + a_{10} x^{10}.$ then find the value of the expression $(a_0 - a_2 + a_4 - a_6 +a_8 - a_{10} )^2 + (a_1 - a_3 + a_5 - a_7 + a_9 )^2.$ × Problem Loading... Note Loading... Set Loading...", "as the generating function for the sum of the $4$ numbers. We seek the $x^{98}$ coefficient, or the $x^{94}$ coefficient in $(1+x^2+x^4...)^4.$ Now we simplify this as $\\left(\\frac{1}{1-x^2}\\right)^4=\\binom{3}{3} +\\binom{4}{3}x^2+\\binom{5}{3}x^4 \\cdots$ and in general we want that the coefficient of $x^{2k}$ is $\\binom{k+3}{3}.$ We see the $x^{94}$ coefficient so we let $k=47$ and so the coefficient is $\\binom{50}{3}=19600$ in which $\\frac{n}{100}=\\boxed{196}.$", "Algebra # Definition of Binomial Coefficient Evaluate $\\binom{8}{3}$. Evaluate ${44 \\choose 42}.$ What is the value of $n$ if ${n \\choose 3}=35?$ What value of $n$ satisfies ${n \\choose 2}=528?$ 5 people became friends at a Summer Camp. On the last day, they shook each other's hands to say farewell. How many handshakes were there? ×", "relation for the binomial coefficients: $\\binom{n-1}{s-1}+\\binom{n-1}{s}=\\binom{n}{s}$ we have the desired formula for $$n$$." ]

[ "Thread: Please find coefficient 1. Please find coefficient Please find coefficient of $x^4$ in the expansion of $(2x+y)^8$ ${8\\choose4}(2x)^4y^4=70*16x^4y^4= 1120x^4y^4$ The answer the coefficient is 1120 Will you please check it and correct if it's wrong. Thank you 2. Originally Posted by oceanmd Please find coefficient of $x^4$ in the expansion of $(2x+y)^8$ ${8\\choose4}(2x)^4y^4=70*16x^4y^4= 1120x^4y^4$ The answer the coefficient is 1120 Will you please check it and correct if it's wrong. Thank you you mean the coefficient of the x^4-term. yes, it's ok", "# Coefficient play The number of terms in the expansion of $$(3x-4y+6z)^4$$ is $$m$$ and the coefficient of $$xy^2z$$ in the same expansion is $$n$$. Find $$m+n$$. ×", "# The coefficient of the $8^{th}$ term in expansion of $(1+x)^{10}$ is equal to : ( A ) $210$ ( B ) $120$ ( C ) $7$ ( D ) $^{10}C_3$", "# Find the Coefficient Find the coefficient of $$x^4$$ in the expansion of $\\large (2-x+3x^2)^6.$ ×", "# What is the coefficient of 2.4n+9.6? The coefficients are $2.4$ and $9.6$. There are technically two coefficients in this expression. The first term, $2.4 n$, the coefficient is $2.4$. In the second term, $9.6$, the coefficient is $9.6$, because technically, the term is $9.6 {n}^{0}$.", "+ 3{x^2} + 4{x^3} + ...........\\infty \\\\$ Here by observation, we have seen that ${x^0}$ has coefficient 1, coefficient of ${x^1}$ is 2, coefficient of ${x^2}$ is 3. It means the coefficient of ${x^n}$ will be $(n + 1).$ Hence, the coefficient of ${x^n}$ in the expansion of ${(1 - x)^{ - 2}}$ is $(n + 1)$ and the correct answer is option “B”.", "## Intermediate Algebra for College Students (7th Edition) The binomial coefficient of the first term is $n\\choose0$ and of the second term is $n\\choose1$ etc. thus the statement doesn't make sense.", "# Coefficient coefficient If a binomial expansion is given by $(1+x)^{10} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \\cdots + a_{10} x^{10}.$ then find the value of the expression $(a_0 - a_2 + a_4 - a_6 +a_8 - a_{10} )^2 + (a_1 - a_3 + a_5 - a_7 + a_9 )^2.$ × Problem Loading... Note Loading... Set Loading...", "of the corresponding three terms of $$(1-x)^{-6}$$ viz., $$x^{20}, x^{11}, x^2$$ respectively. Finally the coefficient is given by $$\\binom{25} {5} - 6*\\binom{16} {5} + 15*\\binom{7} {5} = 27237$$ • Thanks PTDS for helping! – Vasu090 Jul 27 at 5:50", "in the expansion is $\\binom{2015}{2014} x (x^2+1)^{2014}$ and hence the coefficient is $2015+1=2016$", "# Challenge you What is the coefficient of $$x^{7}y^{3}z^{2}$$ in the expansion of $$(x+2y+3z)^{12}$$. ×", "The coefficient of the third term in the expansion of $(x^2-\\large\\frac{1}{4})^n$,when expanded in descending powers of $x$ is $31$.Then n is equal to $\\begin{array}{1 1}(A)\\;16\\\\(B)\\;30\\\\(C)\\;32\\\\(D)\\;\\text{None of these}\\end{array}$", "# Not for the weak-hearted Find the coefficient of $x^{11}$ in the expansion of $(1+x^2)^4(1+x^3)^7(1+x^4)^{12}$. ×", "marks)}$ 3. Find the coefficient of $x^{2}$ in the expansion of $\\left(2 x^{2}-x-3\\right)^{6}$. $\\quad\\mbox{ (5 marks)}$ 4. Find the coefficient of $x^{2}$ and $x^{3}$ in the expansion of $\\left(x^{2}-x-2\\right)^{7}$. $\\quad\\mbox{ (5 marks)}$ 5. Evaluate the coefficients of $x^{5}$ and $x^{4}$ in the binomial expansion of $\\left(\\frac{x}{3}-3\\right)^{7}$. $\\quad\\mbox{ (5 marks)}$ Hence evaluate the coefficient of $x^{5}$ in the expansion of $\\left(\\frac{x}{3}-3\\right)^{7}(x+6)$. $\\quad\\mbox{ (5 marks)}$ 6. In the expansion of $(1+2 x)^{11}$, the coefficient of $x^{3}$ is $k$ times the coefficient of $x^{2}$. Find $k$. $\\quad\\mbox{ (5 marks)}$ 7. If the coefficient of $x^{4}$ in the expansion of $(3+2 x)^{6}$ is equal to the coefficient of $x^{4}$ in the expansion of $(2 k+3 x)^{6}$, find $k$. $\\quad\\mbox{ (5 marks)}$ 8. The ratio of the coefficient of $x^{4}$ in the expansion of $(3-2 x)^{6}$ and the coefficient of $x^{4}$ in the expansion of $(k+2 x)^{7}$ is $1: 7.$ Find the value of $k$. $\\quad\\mbox{ (5 marks)}$ 9. The first three terms in the binomial expasion of $(a+b)^{n}$, in ascending power of $b$, are denoted by $p, q$ and $r$ respectively. Show that $\\frac{q^{2}}{p r}=\\frac{2 n}{n-1}$. Given that $p=4, q=32$ and $r=96$, evaluate $n$. $\\quad\\mbox{ (5 marks)}$ 10. In the binomial expansion of $(1+x)^{n}$ the first three terms are $1+3+4+\\ldots$ Calculate the numerical values of $n$ and $x$, and the value of", "x^i$ The coefficient of $x^n / n!$ in the above series is $t(n) = n! \\binom{n-1}{n-2} = \\frac{n (n-1) (n-1)! }{2}$.", "Find the coefficient of $x^n$ in the expansion of $(1+x) (1- x)^n$. #### Solution Coefficient of $x^n$ in $(1+x) (1- x)^n$ $=$ Coefficient of $x^n$ in $(1-x)^n$ $+$ Coefficient of $x^{n-1}$in $(1-x) ^n$ $= (-1)^n$ $^nC_n$ +$(-1) ^{n-1}$ $^nC_{n-1}$ $=(-1) ^n (1-n)$", "# Submitted by Mohammad ElKammash Find the coefficient of $$x^{9}$$ in the expansion of $(2+x)^{10}(1+x)^0+(2+x)^{9}(1+x)+(2+x)^{8}(1+x)^{2}+ \\ldots + (2+x)^0(1+x)^{10}.$ ×", "# If the coefficient of $r^{th}$ term and $(r+1)^{th}$ term in the expansion of $(1+x)^{3n}$ are in the ratio $1 :2$, then $r$ is equal to : ( A ) $\\frac{1}{3}(3n+1)$ ( B ) $\\frac{1}{3} ( 3n+2)$ ( C ) $\\frac{6}{5} (n+1)$ ( D ) $\\frac{1}{4} (n+2)$", "binomial\" coefficient: $$(-1)^n {n+m \\choose n} = {-m \\choose n}.$$ That should allow you to evaluate the sum explicitly. RGV", "## anonymous 5 years ago What is the coefficient of the fifth term in the expansion of (x+1)^8 ? $\\left(\\begin{array}{c}8\\\\4\\end{array}\\right)=70$" ]

[ "wash, rust inhibitor, wheel brightener, air freshener, and interior shampoo. How many washes are possible if any number of options can be added to the basic wash? 53. Susan bought 20 plants to arrange along the border of her garden. How many distinct arrangements can she make if the plants are comprised of 6 tulips, 6 roses, and 8 daisies? 54. How many unique ways can a string of Christmas lights be arranged from 9 red, 10 green, 6 white, and 12 gold color bulbs?", "on one stem, bulbs and petals, 33 and three almond-flower bowls on one stem, bulbs and petals, so for the six stems that come out of the lampstand; 34 and on the lampstand four almond-flower bowls, its bulbs and its petals; 35* and a bulb under the two stems, in one piece with it, and a bulb under the two stems, in one piece with it, and a bulb under the two stems, in one piece with it, for the six stems that come out of the lampstand. 36 Their bulbs and their stems shall be in one piece with it, all a single piece of chased work of pure gold. 37* And you shall make its lamps, seven of them, and set up its lamps and have it throw the light to its front side; 38 and its pincers and its firepans, pure gold. 39* You shall make it, with all these furnishings, of a hundredweight of pure gold. 40 And see that you make them by the design you are shown on the mountain. ### Footnotes 25:7 (beryls) Unc. 25:32 Lit. And six stems 25:35 (stems, in one piece with it) Or stems out of it (three times) 25:37 Var. and he shall set up 25:39 Var. he shall", "has 7 packs of 8 scented pens. Multiply 7 with 8 the product is 56. She gives her friend a pack of glitter pens in exchange for a new pack of scented pens. Subtract 12 from 104 the difference is 92. Add 92 pens with 8 pens the sum is 100 pens. Now Rona have 100 pens. So, option C is correct. Question 12. Multi-Step Bella has potted plants on her patio. She has 4 pots with 3 begonia plants in each pot. She has 3 pots with 8 petunias in each pot. She plants 7 more zinnia plants in another pot. How many plants does Bella have on her patio? (A) 25 (B) 43 (C) 36 (D) 57 4 x 3 + 3 x 8 + 7 = p 12 + 24 + 7 = p 43 = p Bella have 43 plants on her patio. So, option B is correct. Explanation: Bella has potted plants on her patio. She has 4 pots with 3 begonia plants in each pot. Multiply 4 with 3 the product is 12 begonia plants. She has 3 pots with 8 petunias in each pot. Multiply 3 with 8 the product is 24 petunias plants. She plants 7 more zinnia plants in another pot. Add 12 with 24 and 7 the sum", "formula, but it's often much more intuitive and less error-prone. Feel free to write out the options if it's a reasonable amount, or extrapolate the choices logically if it's not. The formula is very helpful here, but not at all necessary for solving the question. Hope this helps! -Ron _________________ Senior Manager Joined: 10 Jul 2013 Posts: 273 Re: Rachel has ribbons that are blue, green, red, orange, and ye [#permalink] ### Show Tags 08 Aug 2013, 11:52 gmatpapa wrote: Rachel has ribbons that are blue, green, red, orange, and yellow. To identify the plants in her garden, she ties either 1 ribbon, or 2 ribbons of different colors, to each plant. How many different plants can Rachel label in this way without labeling more than one plant the same way? A. 5 B. 10 C. 15 D. 20 E. 25 ........... solution: using only one in each plant = 5C1 = 5 using two ribbons in each plant = 5C2 = 10 (different colors is the topic here not their arrangement, so combination). total plants = 15 Math Expert Joined: 02 Sep 2009 Posts: 64174 Re: Rachel has ribbons that are blue, green, red, orange, and ye [#permalink] ### Show Tags 27 Aug 2014, 08:16 gmatpapa wrote: Rachel has ribbons that are blue, green, red, orange, and", "over. So he tries again. He plants cloves in five rows and has one left over. So he tries again. He plants cloves in six rows and has one left over. We know that he has fewer than $100$ garlic cloves. How many did he have? You could think about how many cloves he might have had if there were more than $100$. You might like to use this interactivity to investigate the problem. You can alter the number of bulbs using the up and down arrows, and you can drag them into different arrangements using the little white square. This text is usually replaced by the Flash movie.", "Aloe Vera Roots Adaptations,", "this claim, she had 10 available decorative gardenias and 10 available decorative tulips. Venue A had 8 weddings on the list and Venue B had 12 weddings on its list. Venue A used gardenias on 5 different occasions. Does this indicate that there is a significant difference in the use of gardenias across venues? 1. (9 pts) Perform the test “by hand”. 2. (8 pts) Perform the test “using a software library”.", "left over. We know that he has fewer than $100$ garlic cloves. How many did he have? You could think about how many cloves he might have had if there were more than $100$. You might like to use this interactivity to investigate the problem. You can alter the number of bulbs using the up and down arrows, and you can drag them into different arrangements using the little white square. This text is usually replaced by the Flash movie.", "[#permalink] 22 Dec 2010, 08:10 Similar topics Replies Last post Similar Topics: In a garden, there are yellow and green flowers which are straight and 1 03 Nov 2014, 02:04 Probability - Flower arrangement 2 25 Nov 2010, 18:13 10 A gardener is going to plant 2 red rosebushes and 2 white 11 26 Aug 2010, 06:38 String of bulbs 5 04 Jul 2009, 20:40 A gardener is going to plant 2 red rosebushes and 2 white 3 23 Sep 2008, 19:21 Display posts from previous: Sort by", "Astudent has four snails, four sprigs of elodea, eight test tubes of brom thymol blue, and two rooms (light and dark). the Question: Astudent has four snails, four sprigs of elodea, eight test tubes of brom thymol blue, and two rooms (light and dark). the student wants to know what gases are produced by the snails and elodea in light and dark conditions. assuming the student only has time for one experiment, what setup will yield the most complete data? $Astudent has four snails, four sprigs of elodea, eight test tubes of brom thymol blue, and two rooms$ Look at the diagram shown below: A slanting, horizontal line is shown. On the extreme left, there is Look at the diagram shown below: A slanting, horizontal line is shown. On the extreme left, there is a label that says Common Ancestor. Along the slanting, horizontal line there are five dots labeled from left to right with the following traits Four limbs, Amniotic egg, Fur, Shearing teeth, and Retr... 7. Clarissa’s Hair Salon is launching a few additional services to attract new clients of the salon. The salon already 7. Clarissa’s Hair Salon is launching a few additional services to attract new clients of the salon. The salon already offers great, affordable haircuts, but they are also offering consultations with", "plants in Samantha's garden, so the final step is to add up all of the plant numbers for a total: $12 + 24 + 6 + 18 = \\textcolor{red}{60}$", "# probability that the blue and the red flowers are both on the same side of the green flower. Two sisters, Rosie and Daisy, bought two packs of flowers in their gardens. Each pack contains six flowers of different colours and the colours are blue,red,yellow,green,pink and white. Rosie has a bigger garden and she plants all her six flowers in one row, For Rosie's Garden, compute the probability that the blue and the red flowers are both on the same side of the green flower. My attempt is : $$\\frac{{6 \\choose 3}3!}{6!}$$ Where $${6 \\choose 3}$$ are the possible ways of choosing the $$3$$ flowers red,blue,green out of $$6$$ and $$3!$$ are the possible permutations of the 3 other flowers. And $$6!$$ are all possible outcomes. I don't think this is the complete answer, could someone please identify what I have to add and explain why? • You chose a set of three positions for the blue, red, and green flowers. However, you have not multiplied by the number of ways that both the blue and red flowers are to the left or both to the right of the green flower. For each such choice, there are two ways both the blue flower and red flower can be to the left of the green flower (depending on whether", "best of times, so make sure you know how to keep them happy before you start or you'll get false results. Iirc they need deionised water and nutrient-poor soil. ##### Share on other sites it would be neat to see if the plant actualy generates any electricity as well, say perhaps a platinum electrode in the plant pot and a platinum needle in the plant and see if there a current diferential in light or dark conditions too. I think I have seen this a long time ago. Isn't that how they tested their communication process? They attatched some sort of speaker to the electrical charge, and it made whistling sounds. It was believed that the plants could communicate? I think that's how they did it, but I'm not sure. Pincho. ##### Share on other sites YT2095 591 Hades said in post # : ive bumped this thread b/c i will be doing this experiment! I went to the home depot tonite and purchased 6 carnivorous venus fly traps, and 6 italian basil plants(serves 2 purposes;)) 3 of each will be used as the constant, the other 3 will be subjected to sporadic pulses of electricity during the times they are photosynthesizing. I do not know of any possible outcomes as of yet, im thinking the addition of", "be used in conjunction during low light seasons. • Using sodium vapor or mercury arc lamps. These are the most efficient lamps when growing plants and are mainly used in professional greenhouses as well as by people growing weed illegally because it maximizes yield. Since this is not a requirement for the basil / chili pepper / capsicum growing done here these rather expensive lamps are not an option. If they are used it should also be accounted for optimum usage of the illuminated area - i.e. maybe not a that narrow structure as I’ve built. • Cheap fluorescent tubes. They are rather cheap, long living, standardized and simple to handle. Ironically they provide nearly the same quality of spectrum for plants as sodium vapor and mercury arc lamps. • Modern LED lamps. I’ve tried these and to get necessary power density a massive investment is required. They’re the way to go for professional large scale operations since their power requirement is significantly lower than for fluorescent tubes and one is capable of controlling red vs blue wavelengths that control growth speed of leafs and plant development. Also note that it’s even more important to have lights at adjustable heights above the plants to move them as close as possible when using LEDs - even though that would", "It may be that one or more of the colours you have used might be better replaced by another, if plants do better with a narrower spectrum. If you mix LEDS like this, knowing that different LEDs have different voltage/current requirements, how are you going to power them? Obviously you will reach a compromise, but it may be even easier if there were fewer colours, or if you used equal numbers of the different colours. I suppose you'll be buying high efficiency LEDs so that for a given power usage you maximize the radiation? Do you have an arrangement that can raise and lower the LED plate so that it can be kept really close to the upper leaves of the plants as they grow? If so, maybe it would be better to group one LED of each colour in a close cluster, and replicate this multiple times, instead of going for a geometrically-pleasing design. As it stands, plants on the perimeter will receive minimal UV, and plants in the centre won't get much RED. LEDs are much more directional than light globes, so each LED will deliver most of its light directly down. This might encourage plants to grow more spindly as the uppermost leaves block the light and the lower ones are shaded and get little.", "I'm sure you're not going to grow that kind of plant! Actually, those are the \"plants\" they are growing with these types of light, I mean, who's going to go through this much trouble just to grow spices.. Here in MI. There has been a boom in the Medical sale of it and there are a lot of people who are signed up as \"providers\" that have grow houses all over the place.... I have been approached by a couple of people interested in having me design something for them along these lines...... B. Morse beenthere Joined Apr 20, 2004 15,819 If you please, the OP has a request for a small grow lamp. Commercial growers of illegal plants get caught because their houses glow like a Bessemer converter in IR, and their electric bills are the size of the Pentagon's. 300 watts is just more than a good reading lamp. Markd77 Joined Sep 7, 2009 2,806 I'd love to grow chilli plants all year and I think it would look good in the corner of the room. I'm not sure how important the wavelength of the light is. If I read right it should be 465 and 665 nm. High power blues are available in 465 but most of the reds seem to be 630 nm. tom66", "potted plant is kept in the dark room? (d) Which process does not take place in the green leaves of the plant when it is kept in the dark room?", "of growing in response. They respond when touched by insects, turn toward sources of light, and some even sniff out other plants. What types of plants are Gibberellin used on? The prayer plant got its name because it folds up its leaves at night, as if in prayer. Plants that give off \"instant jungle vibes,\" like the spider plant, are having a serious moment right now, according to Marino. Get your children to touch these plants and describe what they feel like. When the trap is set, the outer walls are concave and the chamber is under negative hydrostatic pressure. It is also found in Indonesia. Biology Stack Exchange is a question and answer site for biology researchers, academics, and students. If it is not getting enough light, you will know because the leaves will stay folded even during the daytime. How Long for Paint to dry Before Applying Polyurethane. There are many ways to bring the refreshing scent of lemon into the garden without actually planting citrus trees. Poison Ivy is the most common poisonous plant youll encounter and causes an itchy rash for most people who touch it. by. One thing that makes them stand out is their luscious stems and leaves. Claim: Plants have senses and can essentially see, hear, smell, feel, react, and even", "colors, to each plant. How many different plants can Rachel label in this way without labeling more than one plant the same way? A. 5 B. 10 C. 15 D. 20 E. 25 _________________ My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html Director Status: Far, far away! Joined: 02 Sep 2012 Posts: 998 Location: Italy Concentration: Finance, Entrepreneurship GPA: 3.8 Re: Rachel has ribbons that are blue, green, red, orange, and ye [#permalink] ### Show Tags 08 Aug 2013, 10:25 Number of plants identified by one color= 5. Number of plants identified by a combination of 2 colors= $$5C2=10$$ Total number = $$5+10=15.$$ Veritas Prep GMAT Instructor Joined: 11 Dec 2012 Posts: 312 Re: Rachel has ribbons that are blue, green, red, orange, and ye [#permalink] ### Show Tags 08 Aug 2013, 10:35 gmatpapa wrote: Rachel has ribbons that are blue, green, red, orange, and yellow. To identify the plants in her garden, she ties either 1 ribbon, or 2 ribbons of different colors, to each plant. How many different plants can Rachel label in this way without labeling more than one plant the same way? A. 5 B. 10 C. 15 D. 20 E. 25 I really like questions like this because they can bait you into doing more math than necessary. The combinatorics formula ( $$\\frac{n!}{k!(n-k)!}$$) will yield the right", "three gardens, planting one garden each day. Can he do it?" ]

[ "So I am guessing that the advice for 5000 K light is so that there is a good bit of light in the blue (and maybe a bit in the UV) for the fig plant. the 2500K lights will have some blue, but less of it. In principle you will be able to use lots of 2500K lights to have the same ammount of blue light as one 5000K light, but it will not necessarily be two lights - it might be 5 or 10. We can't really give a precise answer except to suggest if you can't get 5000K light you find the next closest, e.g. 4000K and maybe put in 2 or 3 instead of 1 and see how the fig plant grows.... • This is bordering on biology, but if OP was to use multiple bulbs, he/she would also need to opt for lower wattages so to not fry the plant. – sxwzd Oct 22 '15 at 22:47 • @sxwzd - this is a very good point! - suppose then need a good water supply for the plant etc. etc. – tom Oct 22 '15 at 22:54 • You will get \"enough\" blue, but the imbalance toward the red makes the plant grow as if it was shaded by other leaves: fast and leggy. It", "be used in conjunction during low light seasons. • Using sodium vapor or mercury arc lamps. These are the most efficient lamps when growing plants and are mainly used in professional greenhouses as well as by people growing weed illegally because it maximizes yield. Since this is not a requirement for the basil / chili pepper / capsicum growing done here these rather expensive lamps are not an option. If they are used it should also be accounted for optimum usage of the illuminated area - i.e. maybe not a that narrow structure as I’ve built. • Cheap fluorescent tubes. They are rather cheap, long living, standardized and simple to handle. Ironically they provide nearly the same quality of spectrum for plants as sodium vapor and mercury arc lamps. • Modern LED lamps. I’ve tried these and to get necessary power density a massive investment is required. They’re the way to go for professional large scale operations since their power requirement is significantly lower than for fluorescent tubes and one is capable of controlling red vs blue wavelengths that control growth speed of leafs and plant development. Also note that it’s even more important to have lights at adjustable heights above the plants to move them as close as possible when using LEDs - even though that would", "# probability that the blue and the red flowers are both on the same side of the green flower. Two sisters, Rosie and Daisy, bought two packs of flowers in their gardens. Each pack contains six flowers of different colours and the colours are blue,red,yellow,green,pink and white. Rosie has a bigger garden and she plants all her six flowers in one row, For Rosie's Garden, compute the probability that the blue and the red flowers are both on the same side of the green flower. My attempt is : $$\\frac{{6 \\choose 3}3!}{6!}$$ Where $${6 \\choose 3}$$ are the possible ways of choosing the $$3$$ flowers red,blue,green out of $$6$$ and $$3!$$ are the possible permutations of the 3 other flowers. And $$6!$$ are all possible outcomes. I don't think this is the complete answer, could someone please identify what I have to add and explain why? • You chose a set of three positions for the blue, red, and green flowers. However, you have not multiplied by the number of ways that both the blue and red flowers are to the left or both to the right of the green flower. For each such choice, there are two ways both the blue flower and red flower can be to the left of the green flower (depending on whether", "best of times, so make sure you know how to keep them happy before you start or you'll get false results. Iirc they need deionised water and nutrient-poor soil. ##### Share on other sites it would be neat to see if the plant actualy generates any electricity as well, say perhaps a platinum electrode in the plant pot and a platinum needle in the plant and see if there a current diferential in light or dark conditions too. I think I have seen this a long time ago. Isn't that how they tested their communication process? They attatched some sort of speaker to the electrical charge, and it made whistling sounds. It was believed that the plants could communicate? I think that's how they did it, but I'm not sure. Pincho. ##### Share on other sites YT2095 591 Hades said in post # : ive bumped this thread b/c i will be doing this experiment! I went to the home depot tonite and purchased 6 carnivorous venus fly traps, and 6 italian basil plants(serves 2 purposes;)) 3 of each will be used as the constant, the other 3 will be subjected to sporadic pulses of electricity during the times they are photosynthesizing. I do not know of any possible outcomes as of yet, im thinking the addition of", "I'm sure you're not going to grow that kind of plant! Actually, those are the \"plants\" they are growing with these types of light, I mean, who's going to go through this much trouble just to grow spices.. Here in MI. There has been a boom in the Medical sale of it and there are a lot of people who are signed up as \"providers\" that have grow houses all over the place.... I have been approached by a couple of people interested in having me design something for them along these lines...... B. Morse beenthere Joined Apr 20, 2004 15,819 If you please, the OP has a request for a small grow lamp. Commercial growers of illegal plants get caught because their houses glow like a Bessemer converter in IR, and their electric bills are the size of the Pentagon's. 300 watts is just more than a good reading lamp. Markd77 Joined Sep 7, 2009 2,806 I'd love to grow chilli plants all year and I think it would look good in the corner of the room. I'm not sure how important the wavelength of the light is. If I read right it should be 465 and 665 nm. High power blues are available in 465 but most of the reds seem to be 630 nm. tom66", "is not automatically yes because bulb temperatures do not add that way. However, the plant may be more than happy to receive 2500K light, so you may get away with it... not because of physics but because of biology. A plant expert would be able to better guide you as to whether the particular fig plant species in question can do well with redder light, or if the \"hotter\" light is essential for its metabolism. • Red light simulates the plant being shaded by leaves above it, so it grows very fast to try to \"get out from under\". (I did this experiment in High School.) Green and Yellow light makes them die - no usable energy because the leaf reflects green light. Blue light results in near-normal growth. – user95006 Jul 15 '16 at 18:01", "Free download for 50 days: https://authors.elsevier.com/c/1YYyM1L~GwGgex • ... UVI AES is located on the island of St. Croix, U.S. Virgin Islands. The territory presents an equatorial cli- mate with dry summers (As) accord- ing to K€ oppen-Geiger classification ( Kottek et al., 2006). ... Article Full-text available Basil ( Ocimum sp.) is a fast-growing, high-value cash crop for aquaponics. Plant suitability evaluation in tropical conditions is critical to recommend new cultivars, increasing grower portfolio and minimizing the production risks associated with untested selections. Two trials were conducted to identify suitable basil cultivars for tropical outdoor aquaponics production using the University of the Virgin Islands (UVI) Commercial Aquaponics System in the U.S. Virgin islands. We evaluated five basil cultivars in Summer 2015 (Genovese, Lemon, Purple Ruffles, Red Rubin, and Spicy Globe), and seven cultivars in Fall 2015 (Cinnamon, Genovese, Lemon, Purple Ruffles, Red Rubin, Spicy Globe, and Thai). In both trials, 3-week-old seedlings were transplanted in net pots at a density of 1.5 plants/ft ² (16.15 plants/m ² ). The 6-inch portions and upper portions of the canopy were harvested as a salable product and the resultant material (leaves and stems) considered as total yield per square meter. In the summer, yield was higher in ‘Genovese’ (14.91 kg·m ⁻² ) and ‘Spicy Globe’ (13.99 kg·m ⁻² ); ‘Purple Ruffles’", "0.5 teaspoon) of stroked basil plants compared to 1 g of controlled basil plants will account for 0.5% of daily intake. Given that basil plants are not considered to be the main source of Mg in the human diet, the difference is rather significant. It is important to note, though, that almost all elements were increased, as well as ones that are generally considered to have negative influence on health like Na (salt). According to the reported results, the observed increase in Na of 230–300 $$\\upmu$$g/g, compared to the control, will also account for an increase of 0.008% of the maximum daily recommended intake of salt, for the same 1 g of dried basil plants. Metabolite profiling also revealed differences in the metabolism of plants that have undergone mechanical treatment. From a plant physiology perspective, results suggest that treated plants were clearly subjected to stress conditions, as observed by the increased levels of dehydroascorbic acid (DHA) and GABA, particularly in Experiment 2. An increase in the intrinsic levels of DHA suggest either an increase in ascorbate oxidase activity due to oxidative stress, or a decrease in monodehydroascorbate reductase, responsible for regeneration of ascorbic acid, an oxidant scavenger in plants52. Therefore, higher DHA levels relate to an increase in the formation of reactive oxygen species (ROS), which might occur", "# I am spinach $__$ There are $\\color{LimeGreen}{\\textbf{17 raw spinach leaves}}$ in a $\\color{Blue}{\\text{Blue bag}}$ and $\\color{Green}{\\textbf{19 ripe spinach leaves}}$ in a $\\color{Red}{\\text{Red bag}}$. I want to put the $\\color{Blue}{\\text{Blue bag}}$ and the $\\color{Red}{\\text{Red bag}}$ in a $\\color{Purple}{\\text{Purple bag}}$ after removing at least one (and maximum all) leaves from each of the 2 bags. Then how many different possibilities are there for the contents of the $\\color{Purple}{\\text{Purple bag}}$ ? Details and assumptions:- $\\bullet$ The contents of the $\\color{Purple}{\\text{Purple bag}}$ are counted in terms of $\\color{Red}{(Red,}\\color{Blue}{Blue)}$. $\\bullet$ If i remove 3 leaves from $\\color{Blue}{\\text{Blue bag}}$ and 4 leaves from the $\\color{Red}{\\text{Red bag}}$, then the $\\color{Purple}{\\text{Purple bag}}$ will have the contents $\\color{Red}{(Red,}\\color{Blue}{Blue)} = (15,14)$. ($\\color{LimeGreen}{3}$ out of $\\color{LimeGreen}{\\text{17 raw leaves}}$ and $\\color{Green}{4}$ out of $\\color{Green}{\\text{19 ripe leaves}}$ were removed) This problem is part of the set Vegetable Combinatorics ×", "other sites Since Whatever Theory has softened his original claim to allow that several colors evident on a normal male, or female of a species, taken under controlled lighting and equipment conditions might be required to positively ID a subject, the numbers swing back in Whatever Theories favor. (snip) That is 16 million times 16 million times 16 milliion times 16 million times 16 milliion, if you take 5 agreed upon standard positions on an animal or on a plant or on a fungus, to come up with its 5 unique RGB values. Which is 1000000000000000000000000000000000000 possible combinations which could very well allow for 9,000,000 species to each have their own number. But still we have not dealt with the albino and the grafted, and the different colored (white, red purple..)flowers of the Impatients, nor the carrots, nor the sick, nor the parasite infested nor the seasonal variations, nor the ... Unless of course if the everything theory is valid, one might be able to also color code the illness and the parasite, and the variety and the condition of the subject. for instance plants might have five different colors, one for the leaf, one for the stem one for the flower, one for the fruit, one for the root, one for the wood or core of the", "on one stem, bulbs and petals, 33 and three almond-flower bowls on one stem, bulbs and petals, so for the six stems that come out of the lampstand; 34 and on the lampstand four almond-flower bowls, its bulbs and its petals; 35* and a bulb under the two stems, in one piece with it, and a bulb under the two stems, in one piece with it, and a bulb under the two stems, in one piece with it, for the six stems that come out of the lampstand. 36 Their bulbs and their stems shall be in one piece with it, all a single piece of chased work of pure gold. 37* And you shall make its lamps, seven of them, and set up its lamps and have it throw the light to its front side; 38 and its pincers and its firepans, pure gold. 39* You shall make it, with all these furnishings, of a hundredweight of pure gold. 40 And see that you make them by the design you are shown on the mountain. ### Footnotes 25:7 (beryls) Unc. 25:32 Lit. And six stems 25:35 (stems, in one piece with it) Or stems out of it (three times) 25:37 Var. and he shall set up 25:39 Var. he shall", "1. Other 2. Other 3. 1 a jar is filled with 30 jellybeans of which... # Question: 1 a jar is filled with 30 jellybeans of which... ###### Question details 1. A jar is filled with 30 jellybeans of which 7 are red, 4 are blue, 13 are green and 6 are yellow. Someone randomly selected 5 jellybeans from the jar. What is the probability of getting exactly two red jellybeans? 2. A jar is filled with 30 jellybeans of which 7 are red, 4 are blue, 13 are green and 6 are yellow. Someone randomly selected 5 jellybeans from the jar. What is the probability of getting exactly two red and two green jellybeans?(HINT: the fifth jellybean will be either blue or yellow, but not green or red.)", "all the coloring charts on the reference page that I have ... it really helps to simply put two different colors in a room and let them blend together when you compile the static lighting. From Wolf's WIP Game: Acythian As you can see in the above example, you have two separate light colors. One a sort of magenta while the other is a 'hot kelvin' type orange color. While either of these could be used to set the tone by themselves, by shifting the color dramatically he's achieved a very unique effect that clearly separates the room for a person's eyes to see. Also from Acythian In the above picture you can see a more subtle delineation, allowing a cleaner combination of two similar colors. It also helps draw attention to a critical point, where something vital is hiding (near some glowing mushrooms). use previous example to show blue/white colors and combination from lamp and overhead lamps. In this example we've taken the lamp from the previous segment and put in a white overhead light for general room illumination and then underneath the lamp put in a bright blue light to give the simulation of varied lighting. The result, I think, pays dividends. View the full article ## This Week in GameGuru - 11/26/2018 Well, the American", "# Homework Question • April 29th 2008, 11:34 AM Homework Question Here's the question: A package, say A, of 24 crocus bulbs contains 8 yellow bulbs, 8 white, and 8 purple. Another package, say B, of 24 crocus bulbs contains 6 yellow, and 6 white, 12 purple crocus bulbs. One of the two packages is selected at random. (a) If three bulbs from this package were planted and all three yielded purple flowers, compute the conditional probability that package B was selected. (b) If the three bulbs yielded one yellow flower, one white flower, and one purple flower, compute the probability that package A was selected. I have the answer given in the book [(220/276) and (512/944)], but I cannot find how to get it. I've tried several different methods and can't match the answer. Part a: $P\\left( {B|p} \\right) = \\frac{{P\\left( {p|B} \\right)P(B)}}{{P\\left( {p|B} \\right)P(B) + P\\left( {p|A} \\right)P(A)}}", "but the chances of them sharing say 5 color ranges with a different species is even more unlikely. Also I have said that I may be wrong. I am going into this some what blind and I am letting science dictate the direction of this research. It has been raining here so I still can not do the previous test I was talking about, so I decided to work a little more on the leaf test I posted previously. These 2 leafs had one similar color to each other, although the lighter shade of green was not similar at all I decided to show how there would be at least 3 other areas to test on each species. Again you can see that these 3 new colors match each other but do not match the other species. The top row (below) is the same species as the first picture (above) and the bottom 3 pictures are of the bottom picture (above) So even though one pixel came close to matching each other on these species, that is only one pixel out of 16+ million, and once you started moving in the direction of the light green the numbers started moving away from each other. So the light to dark green range only, almost, collided for one small instant", "It may be that one or more of the colours you have used might be better replaced by another, if plants do better with a narrower spectrum. If you mix LEDS like this, knowing that different LEDs have different voltage/current requirements, how are you going to power them? Obviously you will reach a compromise, but it may be even easier if there were fewer colours, or if you used equal numbers of the different colours. I suppose you'll be buying high efficiency LEDs so that for a given power usage you maximize the radiation? Do you have an arrangement that can raise and lower the LED plate so that it can be kept really close to the upper leaves of the plants as they grow? If so, maybe it would be better to group one LED of each colour in a close cluster, and replicate this multiple times, instead of going for a geometrically-pleasing design. As it stands, plants on the perimeter will receive minimal UV, and plants in the centre won't get much RED. LEDs are much more directional than light globes, so each LED will deliver most of its light directly down. This might encourage plants to grow more spindly as the uppermost leaves block the light and the lower ones are shaded and get little.", "are a bunch of groups trying to get glowing plants which can be used as lights. Unlike any of the others I've seen, this project is open source and accessible. All their DNA designs are already available for anyone to look at. Everything else supposedly will be once it's completed. You can take their work, do whatever you want to it yourself, propagate it and transfer it to other people as much as you like. You can duplicate what they've done, create something completely new from it, take advantage of the work without ever sending them a cent. You can wait a year for someone to grow the plants out and offer you a handful of seeds for free. Or, if you'd rather, you can buy a handful of them now for $50 + shipping. How is it a bad thing to make these things accessible to people without a lab to do the genetic modification themselves? Don't you get paid for your work? Quote: Originally posted by Tsjerk Edit; I'm pretty sure this isn't even a real ''glowing'' plant either. looking at the stuff they send it is a fluorescent plant, not a luminescent plant. No. It's actually luminescent. Jealous? I sure as hell am. [Edited on 3-30-2014 by Etaoin Shrdlu] The_Davster A pnictogen Posts: 2861 Registered:", "# Thread: Homework Question 1. ## Homework Question Here's the question: A package, say A, of 24 crocus bulbs contains 8 yellow bulbs, 8 white, and 8 purple. Another package, say B, of 24 crocus bulbs contains 6 yellow, and 6 white, 12 purple crocus bulbs. One of the two packages is selected at random. (a) If three bulbs from this package were planted and all three yielded purple flowers, compute the conditional probability that package B was selected. (b) If the three bulbs yielded one yellow flower, one white flower, and one purple flower, compute the probability that package A was selected. I have the answer given in the book [(220/276) and (512/944)], but I cannot find how to get it. I've tried several different methods and can't match the answer. Can anyone help? Please... 2. Part a: $P\\left( {B|p} \\right) = \\frac{{P\\left( {p|B} \\right)P(B)}}{{P\\left( {p|B} \\right)P(B) + P\\left( {p|A} \\right)P(A)}} $ . Can you compute that?", "roots. The basil grows long white or translucent roots with a pale white mycorrhiza, while the sage has a brownish symbiont and a short, bushy root ball. Thus I only fully replace the water of the hydroponic system as a last resort if a plant seems diseased; normally I will cycle the water by removing about a half of the reservoir and topping it up, so as not to displace the favored microbes from the ecosystem. The Setup The initial inspiration to try hydroponics actually came a bit by chance. We bought some locally grown hydroponic lettuce, and noticed that they were packaged as whole plants, complete with roots. We were curious – could we pluck most of the leaves of this lettuce, and then stick the plants in water, and grow another serving of hydroponic lettuce? Surprisingly enough, it worked! Even with a crude setup consisting of a handful of generic plant fertilizer and a small aquarium bubbler, we were able to take a single plant and grow a couple more servings of lettuce from it. Unfortunately, with time, the plants started to grow very “stemmy” and pale, and eventually they succumbed to tiny mites that infested their leaves. Inspired by this initial success, I started to read up a bit on how others did hydroponics. One", "lights, are not good for plant growth is because they lack intensity (much of the energy generated is thermal, hence these bulbs get hot). They would probably work, if you could get close enough to the plants without burning them. ### #31 Turtle Turtle Member • Members • 15452 posts Posted 14 August 2007 - 11:39 PM In other words, if I take a flashlight and shine it through a blue lens, I am changing the wavelength of the light, and hence I see blue on the other side. Correct me if I'm wrong. You are correct...partly. In the flashlight scenario, the 'full' spectrum of the flashlight itself is dependent on the filament material and other particulars of the bulb; making the addition of the blue lens is actually a subtractive process in terms of color, i.e. the filter blocks all colors but blue. Light intensity drops and heat goes up in the bargain. The various 'grow' lights we are discussing are using an additive process to get more colors, not subtractive filtering. They either add different gases to the bulb (e.g. sodium),or coating the interior of the bulb/tube with different mixtures of fluorescent minerals. Oiu/no, mon ami? ### #32 Michaelangelica Michaelangelica Creating • Members • 7797 posts Posted 15 August 2007 - 02:56 AM Well I bought" ]

[ "on one stem, bulbs and petals, 33 and three almond-flower bowls on one stem, bulbs and petals, so for the six stems that come out of the lampstand; 34 and on the lampstand four almond-flower bowls, its bulbs and its petals; 35* and a bulb under the two stems, in one piece with it, and a bulb under the two stems, in one piece with it, and a bulb under the two stems, in one piece with it, for the six stems that come out of the lampstand. 36 Their bulbs and their stems shall be in one piece with it, all a single piece of chased work of pure gold. 37* And you shall make its lamps, seven of them, and set up its lamps and have it throw the light to its front side; 38 and its pincers and its firepans, pure gold. 39* You shall make it, with all these furnishings, of a hundredweight of pure gold. 40 And see that you make them by the design you are shown on the mountain. ### Footnotes 25:7 (beryls) Unc. 25:32 Lit. And six stems 25:35 (stems, in one piece with it) Or stems out of it (three times) 25:37 Var. and he shall set up 25:39 Var. he shall", "# probability that the blue and the red flowers are both on the same side of the green flower. Two sisters, Rosie and Daisy, bought two packs of flowers in their gardens. Each pack contains six flowers of different colours and the colours are blue,red,yellow,green,pink and white. Rosie has a bigger garden and she plants all her six flowers in one row, For Rosie's Garden, compute the probability that the blue and the red flowers are both on the same side of the green flower. My attempt is : $$\\frac{{6 \\choose 3}3!}{6!}$$ Where $${6 \\choose 3}$$ are the possible ways of choosing the $$3$$ flowers red,blue,green out of $$6$$ and $$3!$$ are the possible permutations of the 3 other flowers. And $$6!$$ are all possible outcomes. I don't think this is the complete answer, could someone please identify what I have to add and explain why? • You chose a set of three positions for the blue, red, and green flowers. However, you have not multiplied by the number of ways that both the blue and red flowers are to the left or both to the right of the green flower. For each such choice, there are two ways both the blue flower and red flower can be to the left of the green flower (depending on whether", "I'm sure you're not going to grow that kind of plant! Actually, those are the \"plants\" they are growing with these types of light, I mean, who's going to go through this much trouble just to grow spices.. Here in MI. There has been a boom in the Medical sale of it and there are a lot of people who are signed up as \"providers\" that have grow houses all over the place.... I have been approached by a couple of people interested in having me design something for them along these lines...... B. Morse beenthere Joined Apr 20, 2004 15,819 If you please, the OP has a request for a small grow lamp. Commercial growers of illegal plants get caught because their houses glow like a Bessemer converter in IR, and their electric bills are the size of the Pentagon's. 300 watts is just more than a good reading lamp. Markd77 Joined Sep 7, 2009 2,806 I'd love to grow chilli plants all year and I think it would look good in the corner of the room. I'm not sure how important the wavelength of the light is. If I read right it should be 465 and 665 nm. High power blues are available in 465 but most of the reds seem to be 630 nm. tom66", "best of times, so make sure you know how to keep them happy before you start or you'll get false results. Iirc they need deionised water and nutrient-poor soil. ##### Share on other sites it would be neat to see if the plant actualy generates any electricity as well, say perhaps a platinum electrode in the plant pot and a platinum needle in the plant and see if there a current diferential in light or dark conditions too. I think I have seen this a long time ago. Isn't that how they tested their communication process? They attatched some sort of speaker to the electrical charge, and it made whistling sounds. It was believed that the plants could communicate? I think that's how they did it, but I'm not sure. Pincho. ##### Share on other sites YT2095 591 Hades said in post # : ive bumped this thread b/c i will be doing this experiment! I went to the home depot tonite and purchased 6 carnivorous venus fly traps, and 6 italian basil plants(serves 2 purposes;)) 3 of each will be used as the constant, the other 3 will be subjected to sporadic pulses of electricity during the times they are photosynthesizing. I do not know of any possible outcomes as of yet, im thinking the addition of", "So I am guessing that the advice for 5000 K light is so that there is a good bit of light in the blue (and maybe a bit in the UV) for the fig plant. the 2500K lights will have some blue, but less of it. In principle you will be able to use lots of 2500K lights to have the same ammount of blue light as one 5000K light, but it will not necessarily be two lights - it might be 5 or 10. We can't really give a precise answer except to suggest if you can't get 5000K light you find the next closest, e.g. 4000K and maybe put in 2 or 3 instead of 1 and see how the fig plant grows.... • This is bordering on biology, but if OP was to use multiple bulbs, he/she would also need to opt for lower wattages so to not fry the plant. – sxwzd Oct 22 '15 at 22:47 • @sxwzd - this is a very good point! - suppose then need a good water supply for the plant etc. etc. – tom Oct 22 '15 at 22:54 • You will get \"enough\" blue, but the imbalance toward the red makes the plant grow as if it was shaded by other leaves: fast and leggy. It", "Astudent has four snails, four sprigs of elodea, eight test tubes of brom thymol blue, and two rooms (light and dark). the Question: Astudent has four snails, four sprigs of elodea, eight test tubes of brom thymol blue, and two rooms (light and dark). the student wants to know what gases are produced by the snails and elodea in light and dark conditions. assuming the student only has time for one experiment, what setup will yield the most complete data? $Astudent has four snails, four sprigs of elodea, eight test tubes of brom thymol blue, and two rooms$ Look at the diagram shown below: A slanting, horizontal line is shown. On the extreme left, there is Look at the diagram shown below: A slanting, horizontal line is shown. On the extreme left, there is a label that says Common Ancestor. Along the slanting, horizontal line there are five dots labeled from left to right with the following traits Four limbs, Amniotic egg, Fur, Shearing teeth, and Retr... 7. Clarissa’s Hair Salon is launching a few additional services to attract new clients of the salon. The salon already 7. Clarissa’s Hair Salon is launching a few additional services to attract new clients of the salon. The salon already offers great, affordable haircuts, but they are also offering consultations with", "be used in conjunction during low light seasons. • Using sodium vapor or mercury arc lamps. These are the most efficient lamps when growing plants and are mainly used in professional greenhouses as well as by people growing weed illegally because it maximizes yield. Since this is not a requirement for the basil / chili pepper / capsicum growing done here these rather expensive lamps are not an option. If they are used it should also be accounted for optimum usage of the illuminated area - i.e. maybe not a that narrow structure as I’ve built. • Cheap fluorescent tubes. They are rather cheap, long living, standardized and simple to handle. Ironically they provide nearly the same quality of spectrum for plants as sodium vapor and mercury arc lamps. • Modern LED lamps. I’ve tried these and to get necessary power density a massive investment is required. They’re the way to go for professional large scale operations since their power requirement is significantly lower than for fluorescent tubes and one is capable of controlling red vs blue wavelengths that control growth speed of leafs and plant development. Also note that it’s even more important to have lights at adjustable heights above the plants to move them as close as possible when using LEDs - even though that would", "has 7 packs of 8 scented pens. Multiply 7 with 8 the product is 56. She gives her friend a pack of glitter pens in exchange for a new pack of scented pens. Subtract 12 from 104 the difference is 92. Add 92 pens with 8 pens the sum is 100 pens. Now Rona have 100 pens. So, option C is correct. Question 12. Multi-Step Bella has potted plants on her patio. She has 4 pots with 3 begonia plants in each pot. She has 3 pots with 8 petunias in each pot. She plants 7 more zinnia plants in another pot. How many plants does Bella have on her patio? (A) 25 (B) 43 (C) 36 (D) 57 4 x 3 + 3 x 8 + 7 = p 12 + 24 + 7 = p 43 = p Bella have 43 plants on her patio. So, option B is correct. Explanation: Bella has potted plants on her patio. She has 4 pots with 3 begonia plants in each pot. Multiply 4 with 3 the product is 12 begonia plants. She has 3 pots with 8 petunias in each pot. Multiply 3 with 8 the product is 24 petunias plants. She plants 7 more zinnia plants in another pot. Add 12 with 24 and 7 the sum", "over. So he tries again. He plants cloves in five rows and has one left over. So he tries again. He plants cloves in six rows and has one left over. We know that he has fewer than $100$ garlic cloves. How many did he have? You could think about how many cloves he might have had if there were more than $100$. You might like to use this interactivity to investigate the problem. You can alter the number of bulbs using the up and down arrows, and you can drag them into different arrangements using the little white square. This text is usually replaced by the Flash movie.", "# I am spinach $__$ There are $\\color{LimeGreen}{\\textbf{17 raw spinach leaves}}$ in a $\\color{Blue}{\\text{Blue bag}}$ and $\\color{Green}{\\textbf{19 ripe spinach leaves}}$ in a $\\color{Red}{\\text{Red bag}}$. I want to put the $\\color{Blue}{\\text{Blue bag}}$ and the $\\color{Red}{\\text{Red bag}}$ in a $\\color{Purple}{\\text{Purple bag}}$ after removing at least one (and maximum all) leaves from each of the 2 bags. Then how many different possibilities are there for the contents of the $\\color{Purple}{\\text{Purple bag}}$ ? Details and assumptions:- $\\bullet$ The contents of the $\\color{Purple}{\\text{Purple bag}}$ are counted in terms of $\\color{Red}{(Red,}\\color{Blue}{Blue)}$. $\\bullet$ If i remove 3 leaves from $\\color{Blue}{\\text{Blue bag}}$ and 4 leaves from the $\\color{Red}{\\text{Red bag}}$, then the $\\color{Purple}{\\text{Purple bag}}$ will have the contents $\\color{Red}{(Red,}\\color{Blue}{Blue)} = (15,14)$. ($\\color{LimeGreen}{3}$ out of $\\color{LimeGreen}{\\text{17 raw leaves}}$ and $\\color{Green}{4}$ out of $\\color{Green}{\\text{19 ripe leaves}}$ were removed) This problem is part of the set Vegetable Combinatorics ×", "is not automatically yes because bulb temperatures do not add that way. However, the plant may be more than happy to receive 2500K light, so you may get away with it... not because of physics but because of biology. A plant expert would be able to better guide you as to whether the particular fig plant species in question can do well with redder light, or if the \"hotter\" light is essential for its metabolism. • Red light simulates the plant being shaded by leaves above it, so it grows very fast to try to \"get out from under\". (I did this experiment in High School.) Green and Yellow light makes them die - no usable energy because the leaf reflects green light. Blue light results in near-normal growth. – user95006 Jul 15 '16 at 18:01", "all the coloring charts on the reference page that I have ... it really helps to simply put two different colors in a room and let them blend together when you compile the static lighting. From Wolf's WIP Game: Acythian As you can see in the above example, you have two separate light colors. One a sort of magenta while the other is a 'hot kelvin' type orange color. While either of these could be used to set the tone by themselves, by shifting the color dramatically he's achieved a very unique effect that clearly separates the room for a person's eyes to see. Also from Acythian In the above picture you can see a more subtle delineation, allowing a cleaner combination of two similar colors. It also helps draw attention to a critical point, where something vital is hiding (near some glowing mushrooms). use previous example to show blue/white colors and combination from lamp and overhead lamps. In this example we've taken the lamp from the previous segment and put in a white overhead light for general room illumination and then underneath the lamp put in a bright blue light to give the simulation of varied lighting. The result, I think, pays dividends. View the full article ## This Week in GameGuru - 11/26/2018 Well, the American", "other sites Since Whatever Theory has softened his original claim to allow that several colors evident on a normal male, or female of a species, taken under controlled lighting and equipment conditions might be required to positively ID a subject, the numbers swing back in Whatever Theories favor. (snip) That is 16 million times 16 million times 16 milliion times 16 million times 16 milliion, if you take 5 agreed upon standard positions on an animal or on a plant or on a fungus, to come up with its 5 unique RGB values. Which is 1000000000000000000000000000000000000 possible combinations which could very well allow for 9,000,000 species to each have their own number. But still we have not dealt with the albino and the grafted, and the different colored (white, red purple..)flowers of the Impatients, nor the carrots, nor the sick, nor the parasite infested nor the seasonal variations, nor the ... Unless of course if the everything theory is valid, one might be able to also color code the illness and the parasite, and the variety and the condition of the subject. for instance plants might have five different colors, one for the leaf, one for the stem one for the flower, one for the fruit, one for the root, one for the wood or core of the", "are a bunch of groups trying to get glowing plants which can be used as lights. Unlike any of the others I've seen, this project is open source and accessible. All their DNA designs are already available for anyone to look at. Everything else supposedly will be once it's completed. You can take their work, do whatever you want to it yourself, propagate it and transfer it to other people as much as you like. You can duplicate what they've done, create something completely new from it, take advantage of the work without ever sending them a cent. You can wait a year for someone to grow the plants out and offer you a handful of seeds for free. Or, if you'd rather, you can buy a handful of them now for $50 + shipping. How is it a bad thing to make these things accessible to people without a lab to do the genetic modification themselves? Don't you get paid for your work? Quote: Originally posted by Tsjerk Edit; I'm pretty sure this isn't even a real ''glowing'' plant either. looking at the stuff they send it is a fluorescent plant, not a luminescent plant. No. It's actually luminescent. Jealous? I sure as hell am. [Edited on 3-30-2014 by Etaoin Shrdlu] The_Davster A pnictogen Posts: 2861 Registered:", "left over. We know that he has fewer than $100$ garlic cloves. How many did he have? You could think about how many cloves he might have had if there were more than $100$. You might like to use this interactivity to investigate the problem. You can alter the number of bulbs using the up and down arrows, and you can drag them into different arrangements using the little white square. This text is usually replaced by the Flash movie.", "Pharmacopoeia, 205 the administration of preparations of garlic to children is dangerous, and fatalities have been recorded. #### Aloe Barbadensis Botanical Name — Aloe barbadensis Miller Synonyms — A. vera., A. vulgaris Lamk., A. indica Royle Family — Liliaceae Common Names — Aloe vera, curacao aloe, aloe vera gel (AVG) Description — Aloe barbadensis is a perennial herb with rosettes of long pointed leaves from a shortly branched creeping rhizome. The leaves are brittle and exude a clear yellowish viscous sap when broken. They have soft marginal prickles. The flowers are yellow and borne in an elongated compact raceme from the center of the rosette. Propagation is by means of the rhizome branches since it does not produce fruit. 32 Habitat and Distribution — It grows in subtropical regions of the continent. The plant is probably a native of South and East Africa and the Mediterranean regions, although it is widespread throughout the continent. It is not very discriminatory to soil types and climatic conditions, but as a member of the Liliaceae family, it will prefer a well-drained clayey loam, with rainfall of at least 80 in. per annum. Ethnomedicinal Uses — The fresh leaf is used for wound dressing and for the treatment of vertigo. In Nigeria, the exudate from freshly cut leaf is used in the", "# Homework Question • April 29th 2008, 11:34 AM Homework Question Here's the question: A package, say A, of 24 crocus bulbs contains 8 yellow bulbs, 8 white, and 8 purple. Another package, say B, of 24 crocus bulbs contains 6 yellow, and 6 white, 12 purple crocus bulbs. One of the two packages is selected at random. (a) If three bulbs from this package were planted and all three yielded purple flowers, compute the conditional probability that package B was selected. (b) If the three bulbs yielded one yellow flower, one white flower, and one purple flower, compute the probability that package A was selected. I have the answer given in the book [(220/276) and (512/944)], but I cannot find how to get it. I've tried several different methods and can't match the answer. Part a: $P\\left( {B|p} \\right) = \\frac{{P\\left( {p|B} \\right)P(B)}}{{P\\left( {p|B} \\right)P(B) + P\\left( {p|A} \\right)P(A)}}", "0.5 teaspoon) of stroked basil plants compared to 1 g of controlled basil plants will account for 0.5% of daily intake. Given that basil plants are not considered to be the main source of Mg in the human diet, the difference is rather significant. It is important to note, though, that almost all elements were increased, as well as ones that are generally considered to have negative influence on health like Na (salt). According to the reported results, the observed increase in Na of 230–300 $$\\upmu$$g/g, compared to the control, will also account for an increase of 0.008% of the maximum daily recommended intake of salt, for the same 1 g of dried basil plants. Metabolite profiling also revealed differences in the metabolism of plants that have undergone mechanical treatment. From a plant physiology perspective, results suggest that treated plants were clearly subjected to stress conditions, as observed by the increased levels of dehydroascorbic acid (DHA) and GABA, particularly in Experiment 2. An increase in the intrinsic levels of DHA suggest either an increase in ascorbate oxidase activity due to oxidative stress, or a decrease in monodehydroascorbate reductase, responsible for regeneration of ascorbic acid, an oxidant scavenger in plants52. Therefore, higher DHA levels relate to an increase in the formation of reactive oxygen species (ROS), which might occur", "of growing in response. They respond when touched by insects, turn toward sources of light, and some even sniff out other plants. What types of plants are Gibberellin used on? The prayer plant got its name because it folds up its leaves at night, as if in prayer. Plants that give off \"instant jungle vibes,\" like the spider plant, are having a serious moment right now, according to Marino. Get your children to touch these plants and describe what they feel like. When the trap is set, the outer walls are concave and the chamber is under negative hydrostatic pressure. It is also found in Indonesia. Biology Stack Exchange is a question and answer site for biology researchers, academics, and students. If it is not getting enough light, you will know because the leaves will stay folded even during the daytime. How Long for Paint to dry Before Applying Polyurethane. There are many ways to bring the refreshing scent of lemon into the garden without actually planting citrus trees. Poison Ivy is the most common poisonous plant youll encounter and causes an itchy rash for most people who touch it. by. One thing that makes them stand out is their luscious stems and leaves. Claim: Plants have senses and can essentially see, hear, smell, feel, react, and even", "tank. The plant used is Tropica tissue culture as shown below: Traditionally, Alternanthera reineckii ‘Mini’ (AR Mini) is grown in high tech, high light tanks where it will grow into a deep red purplish color. In low tech, lower light tanks, I am trying to see if it can still grow, albeit less red. Maybe a lighter red or olive / bronze color like its cousin Alternanthera Roseafolia (below), would be a good enough result. Tank Parameters and Fertilization Tank size: 5 Gallon (I am using this tank: Ripples Premium Aquarium Glass Fish Tank (with Filter and LED… View original post 653 more words Good phrases for Chinese Composition Writing (Secondary school / Top Primary Students) Using a good phrase (correctly) can instantly add marks for your Chinese Compo (Composition), as it shows that the student is well-versed and adept in using Chinese Idioms/Proverbs. The more exotic the phrase, the better the effect (when used correctly). Common phrases like “风和日丽” have been used thousands of times and hence do not have a beneficial impact on impressing the teacher (or grader). Hence, for the hardworking student who wishes to gain marks in Chinese compo, the best strategy is to memorize some good phrases that have a wide range for applications. Then, apply the phrases appropriately in the Chinese composition." ]

[ "# Overpowering tens unit Discrete Mathematics Level 1 How many 2 digit numbers are there, such that the units digit is strictly smaller than the tens digit? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "+0 # counting 0 95 1 How many 3-digit numbers have the property that the units digit is at least twice the tens digit plus 5? Jun 26, 2022 #1 +2448 0 There are 9 choices for the hundreds digit. The only possible cases for the tens and units digits are: (0, 5 - 9), (1, 7 - 9), (2, 9) This makes for 9 cases for the units and tens digits. Thus, there are $$9 \\times 9 = \\color{brown}\\boxed{81}$$ cases. Jun 26, 2022", "of N is 6 more than the tens digit 2) N is 4 less than 4 times the units digit We need to determine whether the positive two-digit integer N is less than 40. Statement One Alone: The units digit of N is 6 more than the tens digit. With the information in statement one, we know the units digit is the larger of the two digits in N. So let’s say the units digit is 9 (the largest digit possible); then the tens digit will be 3 (since 9 is 6 more than 3). Thus this makes N = 39, which is less than 40. Now let’s say the units digit is 8; then the tens digit will be 2 and thus N = 28, which is still less than 40. Let’s say the units digit is 7; then the tens digit will be 1 and thus N = 17, which is again less than 40. At this point, we can’t make the units digits any smaller; if we did, the tens digits would be 0 or negative, but we know N is a positive two-digit integer. Statement one is sufficient to answer the question. Statement Two Alone: N is 4 less than 4 times the units digit. Again, let’s test some possible numerical values for the", "the tens.", "33, 333, 343, 3, etc.. as long as the units digit is 3 it is enough to find out the units digit for x^2 Sufficient. B Re: If x is a positive integer, what is the units digit of x^2? &nbs [#permalink] 05 Oct 2018, 09:43 Display posts from previous: Sort by", "tens-digit is odd, then it must be divisible by an odd number of primes of the form 11, 13, 17, 19 mod 20. Any even number of these would produce an even tens digit. -", "property to multiply 2-digit numbers by 1-digit is a mathematical operation that has types. Of parentheses or brackets to group numbers four, two times four first the associative Calculator... Are going to get 40 strategies to multiply 2-digit numbers by 1-digit to group.!", "# Understand three-digit place value by counting, grouping, and using base ten blocks (2.NBT.A.1) Understand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones: 100 is a bundle of ten tens, and numbers such as 300, 500, and 900 are bundles of hundreds.", "= 2, so that's the ten's digit. Your number is 26. -Dan 6. ## i know that one Originally Posted by mredhi390 My tens digit is 3 more than 5.My tens digit is double the number of my ones digit.What number am I?", "the hundreds, tens, and units digits of n, respectively. if m and v are three digit positive integers such that f(m) = 9 f(v), then m - v = note: 2^x denote x is power of 2 anyone please give me solution of this problem Math Expert Joined: 02 Sep 2009 Posts: 34497 Followers: 6297 Kudos [?]: 79910 [0], given: 10022 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 11 Aug 2013, 10:30 Riyad47 wrote: the function f is defined for positive three digit integer n by f(n) = 2^x3^y5^z, where x,y,z are the hundreds, tens, and units digits of n, respectively. if m and v are three digit positive integers such that f(m) = 9 f(v), then m - v = note: 2^x denote x is power of 2 anyone please give me solution of this problem Discussed here: the-function-f-is-defined-for-each-positive-three-digit-100847.html _________________ Senior Manager Joined: 10 Jul 2013 Posts: 335 Followers: 3 Kudos [?]: 277 [0], given: 102 Re: New Set of Mixed Questions!!! [#permalink] ### Show Tags 11 Aug 2013, 12:14 Bunuel wrote: The next set of medium/hard PS questions. I'll post OA's with detailed explanations on Friday. Please, post your solutions along with the answers. 1. The distance from the Y-axis to point K is 1/3 of the distance from the X-axis to point", "unit^3 in there.", "power. That is, $$3^4 = 3 \\times 3 \\times 3 \\times 3 = 81$$ Hence, the unit digit is $$1$$.", "# The tens digit is quite odd... For how many positive integers $$n$$ where $$n \\leq 1000$$ are there such that the tens digit of $$n^2$$ is odd? ×", "# A number theory problem by Kamala Ramakrishnan What is the units digit of the following pattern: $3^{1}+ 3^{2} + 3^{3} + \\ldots + 3^{500}$ ×", "the tens digit (40, in this case) to the number — it’s 53. - Stick the units on the end (making 539). Great! We’re literally halfway there. [", "digits and so on. (units in the first time step, tens in the second time step etc.)", "0 What is the millions digit of 319672191? 9 What is the hundred millions digit of 130028023? 1 What is the tens digit of 325238308? 0 What is the tens digit of 46773645? 4 What is the ten thousands digit of 52813169? 1 What is the millions digit of 480863464? 0 What is the ten thousands digit of 44576563? 7 What is the hundred millions digit of 2470399601? 4 What is the millions digit of 1168405263? 8 What is the tens digit of 71323290? 9 What is the thousands digit of 45249196? 9 What is the ten thousands digit of 12029225? 2 What is the units digit of 2903539242? 2 What is the ten millions digit of 29207109? 2 What is the units digit of 606731458? 8 What is the hundreds digit of 2565500474? 4 What is the ten millions digit of 70986354? 7 What is the tens digit of 218335679? 7 What is the hundreds digit of 612049677? 6 What is the billions digit of 2222363461? 2 What is the tens digit of 534404060? 6 What is the units digit of 1428032287? 7 What is the millions digit of 1029928434? 9 What is the ten millions digit of 5933789688? 3 What is the units digit of 4234152066? 6 What is the ten thousands digit of 831200705? 0", "777 How many 3-digit numbers are a multiple of 7? Details and assumptions The number $$12=012$$ is a 2-digit number, not a 3-digit number. ×", "More Mods What is the units digit for the number 123^(456) ? N000ughty Thoughts How many noughts are at the end of these giant numbers? Mod 3 Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3. Squaresearch Stage: 4 Challenge Level: Consider numbers of the form $u_n = 1! + 2! + 3! +...+n!$. How many such numbers are perfect squares?", "281 Is the positive two-digit integer N less than 40 ? 1) The units digit of N is 6 more than the tens digit 2) N is 4 less than 4 times the units digit We need to determine whether the positive two-digit integer N is less than 40. Statement One Alone: The units digit of N is 6 more than the tens digit. With the information in statement one, we know the units digit is the larger of the two digits in N. So let’s say the units digit is 9 (the largest digit possible); then the tens digit will be 3 (since 9 is 6 more than 3). Thus this makes N = 39, which is less than 40. Now let’s say the units digit is 8; then the tens digit will be 2 and thus N = 28, which is still less than 40. Let’s say the units digit is 7; then the tens digit will be 1 and thus N = 17, which is again less than 40. At this point, we can’t make the units digits any smaller; if we did, the tens digits would be 0 or negative, but we know N is a positive two-digit integer. Statement one is sufficient to answer the question. Statement Two Alone: N is 4 less" ]

[ "&4 &9 &4 &8 &1 \\\\ &+ & &4 &9 &1 &0 &0 \\\\ \\hline & & & & & 5&8 &1\\\\ \\hline \\end{array} In the thousands column we get 9 + 9 = 18. So we bring down 8 and carry the 1 to the ten thousands place.\\begin{array}{c} & & &\\text{}^1 4 &9 &4 &8 &1 \\\\ &+ & &4 &9 &1 &0 &0 \\\\ \\hline & & & & 8& 5&8 &1\\\\ \\hline \\end{array} For the ten thousands place we get 1+4+4=9.\\begin{array}{c} & & &\\text{}^1 4 &9 &4 &8 &1 \\\\ &+ & &4 &9 &1 &0 &0 \\\\ \\hline & & & 9& 8& 5&8 &1\\\\ \\hline \\end{array} So the answer is:49\\,481 + 49\\,100 = 98\\,581 Example 3 Find the value of 55\\,872-2619. Worked Solution Create a strategy Use the standard algorithm method. Apply the idea Write it in a vertical algorithm.\\begin{array}{c} & & &5 &5 &8 &7 &2 \\\\ &- & & &2 &6 &1 &9 \\\\ \\hline & \\\\ \\hline \\end{array} In the ones column we can see that 2 is less than 9, so we need to trade 1 ten from the tens place. So we get 12-9=3 in the ones column and 7 tens becomes 6 tens in the first row.\\begin{array}{c} & && 5&5 &8 &6 &\\text{}^1 2 \\\\ &- &&", "&8 &2 &7 \\\\ \\hline & && & & & &9\\\\ \\hline \\end{array} In the tens column, we have 8-2=6:\\begin{array}{c} & && 5&6 &1 &8 &\\text{}^1 6 \\\\ &- && 4&4 &8 &2 &7 \\\\ \\hline & && & & & 6&9\\\\ \\hline \\end{array} In the hundreds column we can see that 1 is less than 8, so we need to trade 1 thousand from the thousands place. So we get 11-8=3 in the hundreds column and 6 thousands becomes 5 thousands in the first row.\\begin{array}{c} & && 5&5 &\\text{}^1 1 &8 &\\text{}^1 6 \\\\ &- && 4&4 &8 &2 &7 \\\\ \\hline & && & & 3& 6&9\\\\ \\hline \\end{array} In the thousands column, we have 5-4=1:\\begin{array}{c} & && 5&5 &\\text{}^1 1 &8 &\\text{}^1 6 \\\\ &- && 4&4 &8 &2 &7 \\\\ \\hline & && & 1& 3& 6&9\\\\ \\hline \\end{array} And for the ten thousands place 5-4=1:\\begin{array}{c} & && 5&5 &\\text{}^1 1 &8 &\\text{}^1 6 \\\\ &- && 4&4 &8 &2 &7 \\\\ \\hline & && 1& 1& 3& 6&9\\\\ \\hline \\end{array} So 56\\,196 - 44\\,827 = 11\\,369. Idea summary When subtracting, we must put the first number, our total, at the top of our algorithm. Unlike addition, which we can solve in any order, subtraction can only be solved in the order it is", "& & &3\\\\ \\hline \\end{array} In the tens column we get 9+3=12. So we bring down 2 and carry the 1 to the hundreds place.\\begin{array}{c} & & &1 &9 &\\text{}^ 1 2 &9 &2 \\\\ &+ & &3 &4 &1 &3 &1 \\\\ \\hline & & & & & &2 &3\\\\ \\hline \\end{array} In the hundreds column we get 1+2+1=4.\\begin{array}{c} & & &1 &9 &\\text{}^ 1 2 &9 &2 \\\\ &+ & &3 &4 &1 &3 &1 \\\\ \\hline & & & & &4 &2 &3\\\\ \\hline \\end{array} In the thousands column we get 9 + 4 = 13. So we bring down 3 and carry the 1 to the ten thousands place.\\begin{array}{c} & & &\\text{}^1 1 &9 &\\text{}^ 1 2 &9 &2 \\\\ &+ & &3 &4 &1 &3 &1 \\\\ \\hline & & & & 3&4 &2 &3\\\\ \\hline \\end{array} For the ten thousands place we get 1+1+3=5.\\begin{array}{c} & & &\\text{}^1 1 &9 &\\text{}^ 1 2 &9 &2 \\\\ &+ & &3 &4 &1 &3 &1 \\\\ \\hline & & & 5& 3&4 &2 &3\\\\ \\hline \\end{array} So the answer is:19\\,292 + 34\\,131 = 53\\,423 Idea summary When we add, we always start from the digit furthest to the right, and work left. Any time we have more than 9 in any place, we regroup to", "&2 &7 &3 &9 &8 \\\\ \\hline & & & & &8 &1 \\\\ \\hline \\end{array} For the hundreds place: 4 - 3 = 1. \\begin{array}{c} & &8 &1 &4 &\\text{}^1 7 &9 \\\\ &- &2 &7 &3 &9 &8 \\\\ \\hline & & & &1 &8 &1 \\\\ \\hline \\end{array} In the thousands column we can see that 1 is less than 7, so we need to trade 1 ten thousand from the ten thousands column. This gives us 11 - 7 = 4 in the thousands column and 8 ten thousands becomes 7 ten thousands in the first row. \\begin{array}{c} & &7 &\\text{}^1 1 &4 &\\text{}^1 7 &9 \\\\ &- &2 &7 &3 &9 &8 \\\\ \\hline & & &4 &1 &8 &1 \\\\ \\hline \\end{array} For the ten thousands place: 7 - 2 = 5.\\begin{array}{c} & &7 &\\text{}^1 1 &4 &\\text{}^1 7 &9 \\\\ &- &2 &7 &3 &9 &8 \\\\ \\hline & &5 &4 &1 &8 &1 \\\\ \\hline \\end{array} So 81\\,579 - 27\\, 398 = 54\\, 181. Idea summary When subtracting, especially when using the vertical algorithm, place value is really important. We want to make sure our numbers are lined up correctly before we subtract, and we start from the place to the far right. Outcomes VCMNA209 Select and apply efficient mental", "algorithms to work with larger numbers as well, as we do in this video. ### Examples #### Example 3 Find the value of 19\\,292 + 34\\,131. Worked Solution Create a strategy Use the standard algorithm method. Apply the idea Write it in a vertical algorithm.\\begin{array}{c} & & &1 &9 &2 &9 &2 \\\\ &+ & &3 &4 &1 &3 &1 \\\\ \\hline & \\\\ \\hline \\end{array} Add the units column first.\\begin{array}{c} & & &1 &9 &2 &9 &2 \\\\ &+ & &3 &4 &1 &3 &1 \\\\ \\hline & & & & & & &3\\\\ \\hline \\end{array} In the tens column we get 9+3=12. So we bring down 2 and carry the 1 to the hundreds place.\\begin{array}{c} & & &1 &9 &\\text{}^ 1 2 &9 &2 \\\\ &+ & &3 &4 &1 &3 &1 \\\\ \\hline & & & & & &2 &3\\\\ \\hline \\end{array} In the hundreds column we get 1+2+1=4.\\begin{array}{c} & & &1 &9 &\\text{}^ 1 2 &9 &2 \\\\ &+ & &3 &4 &1 &3 &1 \\\\ \\hline & & & & &4 &2 &3\\\\ \\hline \\end{array} In the thousands column we get 9 + 4 = 13. So we bring down 3 and carry the 1 to the ten thousands place.\\begin{array}{c} & & &\\text{}^1 1 &9 &\\text{}^ 1 2 &9 &2 \\\\ &+ &", "& \\\\ \\hline \\end{array} Begin with the units place value. So 9 + 5 = 14. Place 4 under the units column in the answer section and carry forward 1 to the tens column. \\begin{array}{c} & &\\text{}^1 1 &9 \\\\ &+ &2 &5 \\\\ \\hline & & &4 \\\\ \\hline \\end{array} Add the numbers in the tens column. Since 1 has been carried over from the units place, we need to add 1 + 1 + 2 = 4. \\begin{array}{c} & &\\text{}^1 1 &9 \\\\ &+ &2 &5 \\\\ \\hline & &4 &4 \\\\ \\hline \\end{array} So 19 + 25 = 44. Idea summary The largest value a digit can be is 9. If we add and end up with more than 9, we need to trade, or regroup, to the next digit to the left. ### Outcomes #### MA2-5NA uses mental and written strategies for addition and subtraction involving two-, three-, four and five-digit numbers", "& & & & &2 \\\\ \\hline \\end{array} For the tens column we have 3-1=2: \\begin{array}{c} & & &6 &2 &3 &\\text{}^1 1 \\\\ &- & &2 &3 &1 &9 \\\\ \\hline & & & & &2 &2 \\\\ \\hline \\end{array} In the hundreds column we can see that 2 is less than 3, so we need to trade 1 thousand from the thousands column. This gives us 12-3=9 and and 6 thousands becomes 5 thousands in the first row.\\begin{array}{c} & & &5 &\\text{}^1 2 &3 &\\text{}^1 1 \\\\ &- & &2 &3 &1 &9 \\\\ \\hline & & & &9 &2 &2 \\\\ \\hline \\end{array} For the thousands column we have 5-2=3:\\begin{array}{c} & & &5 &\\text{}^1 2 &3 &\\text{}^1 1 \\\\ &- & &2 &3 &1 &9 \\\\ \\hline & & &3 &9 &2 &2 \\\\ \\hline \\end{array} So 6241-2319=3922. Idea summary If we don't have enough to subtract, we can regroup from the next place to the left. ### Outcomes #### MA2-5NA uses mental and written strategies for addition and subtraction involving two-, three-, four and five-digit numbers", "&2 \\\\ \\hline & & &4 &6 \\\\ \\hline \\end{array} For the hundreds place: 3 -1 = 2. \\begin{array}{c} & &3 &9 &8 \\\\ &- &1 &5 &2 \\\\ \\hline & &2 &4 &6 \\\\ \\hline \\end{array} So 398 - 152 = 246. Idea summary We can subtract two numbers using a vertical algorithm. We start at the units column and move left. Subtraction of large numbers Let's take a look at how a vertical algorithm helps us solve subtraction with large numbers. Examples Example 4 Find the value of 81\\,579 -27\\,398. Worked Solution Create a strategy Use the subtraction algorithm method. Apply the idea Write it in a vertical algorithm.\\begin{array}{c} & &8 &1 &5 &7 &9 \\\\ &- &2 &7 &3 &9 &8 \\\\ \\hline \\\\ \\hline \\end{array} Begin with the units column: 9 -8 = 1.\\begin{array}{c} & &8 &1 &5 &7 &9 \\\\ &- &2 &7 &3 &9 &8 \\\\ \\hline & & & & & &1 \\\\ \\hline \\end{array} In the tens column we can see that 7 is less than 9, so we need to trade 1 hundred from the hundreds column. This gives us 17 - 9 = 8 in the tens column and 5 hundreds becomes 4 hundreds in the first row. \\begin{array}{c} & &8 &1 &4 &\\text{}^1 7 &9 \\\\ &-", "If the largest digit is the second or third digit then, as we have already counted all the cases when it is equal to another digit, we must count all pairs of $b\\in\\{1,\\ldots,8\\}$ and $a\\in\\{b+1,\\ldots, 9\\}$, twice (for the two placements). The total count of is then: \\begin{align} & =\\sum\\limits_{a=1}^9 \\sum\\limits_{b=0}^a 1 + 2\\times \\sum\\limits_{b=1}^8 \\sum\\limits_{a-b=1}^{9-b} 1 \\\\ & = \\sum\\limits_{a=1}^9 (a+1) + 2\\times \\sum\\limits_{b=1}^8 (9-b) \\\\ & = \\tfrac{9(9+1)}{2} + 9 + 2(9\\cdot 8 -\\tfrac{8(8+1)}{2}) \\\\ & = 126 \\end{align} Various cases are: $$\\begin{array}{|c|c|}\\hline (a,b,c)|a+b=c&\\text{total numbers}\\\\\\hline (1,1,2),(2,2,4),...(4,4,8)&3!/2!\\times4\\\\\\hline (1,2,3)\\text{ to }(1,8,9)&3!\\times7\\\\ (2,3,5)\\text{ to }(2,7,9)&3!\\times5\\\\ (3,4,7)\\text{ to }(3,6,9)&3!\\times3\\\\ (4,5,9)&3!\\times1\\\\\\hline (1,0,1)\\text{ to }(9,0,9)&2\\times9\\\\\\hline \\end{array}$$ Total ways: $$3!/2!\\times4+3!\\times(1+3+5+7)+2\\times9=12+6\\times4^2+18=126$$ • missing that $0$ is a digit too ;) – AlexR Aug 7 '14 at 4:03 • And $(0,0,0)$ ;) – AlexR Aug 7 '14 at 4:06 • @AlexR do you think 000 is a three digit number? – RE60K Aug 7 '14 at 4:07 • Fun fact: allowing $000$ and $011$ etc. results in $1+12+123$ combinations. – AlexR Aug 7 '14 at 4:14 • Hey easy dude, control your tongue. if 1 was at the hundreth place and 0 at the tenth. Do you get a 3 digit number when you swap them? – MonK Aug 7 '14 at 5:44 My answer is similar to that of @shooting-squirrel. Edit: Now it gives", "\\hline 6&60=66-6&-6 \\\\ \\hline 7&70=77-7&-7 \\\\ \\hline 8&80=88-8&-8 \\\\ \\hline 9&90=99-9&-9 \\\\ \\hline \\end{array} Second, let's observe the place values of hundreds place, which could be transformed to the difference of a number that is divisible by 11 and a place value of tens place. Substituting the results about tens place gives the remainders that are equal to the digits on the hundreds place. \\begin{array}{|c|c|c|} \\hline 1&100=110-10&1 \\\\ \\hline 2&200=220-20&2 \\\\ \\hline 3&300=330-30&3 \\\\ \\hline 4&400=440-40&4 \\\\ \\hline 5&500=550-50&5 \\\\ \\hline 6&600=660-60&6 \\\\ \\hline 7&700=770-70&7 \\\\ \\hline 8&800=880-80&8 \\\\ \\hline 9&900=990-90&9 \\\\ \\hline \\end{array} Repeating the process on place values of thousands place, or more reviews the same rule. For even places, the remainders are the opposite numbers of the digits on the place and for odd places, the remainders are the same as the digits on the place, if divided by 11. Next, let’s first evaluate a 4-digit number that is denoted as \\overline{abcd}, in which a, b, c, d are four digits on thousands, hundreds, tens and ones places respectively. Since a number can be expressed as the sum of its place values, \\overline{abcd} is rewritten as \\overline{abcd}=1000a+100b+10c+d (1) For place value on tens place, 10c = 11c-c (2) For the place value on hundreds place, \\begin{aligned} 100b &= 110b-10b \\\\ &=110b-(11b-b) \\\\ &=(110b-11b)+b \\end{aligned} (3)", "the units column. We can see that 7 is less than 8, so we need to trade 1 ten from the tens place. So we get 17 - 8 = 9 in the units column and 9 tens becomes 8 tens in the first row.\\begin{array}{c} & & &9 &8 &\\text{ }^1 7 \\\\ &- & &6 &4 &8 \\\\ \\hline & & & & &9 \\\\ \\hline \\end{array} For the tens place: 8 - 4 = 4.\\begin{array}{c} & & &9 &8 &\\text{ }^1 7 \\\\ &- & &6 &4 &8 \\\\ \\hline & & & &4 &9 \\\\ \\hline \\end{array} For the hundreds place: 9 - 6 = 3.\\begin{array}{c} & & &9 &8 &\\text{ }^1 7 \\\\ &- & &6 &4 &8 \\\\ \\hline & & &3 &4 &9 \\\\ \\hline \\end{array} So 997 - 648 = 349. Idea summary We always start from the ones place, when we work down our page. If we don't have enough ones to subtract, we need to trade from the tens place. If we don't have enough tens to subtract, we need to trade from the hundreds place. ## Add and subtract 4 digit numbers What happens with larger numbers? We can use a standard algorithm to solve addition and subtraction, as we see in this video. ### Examples #### Example", "hundreds place: 3 - 2 = 1.\\begin{array}{c} & &4 &2 &3 &6 &\\text{}^13 \\\\ &- &2 &5 &2 &1 &5 \\\\ \\hline & & & &1 &5 &8 \\\\ \\hline \\end{array} In the thousands column, we can see that 2 is less than 5, so we need to trade 1 ten thousand from the ten thousands column. This gives us 12 - 5 = 7 in the thousands column and 4 ten thousands becomes 3 ten thousands in the first row. \\begin{array}{c} & &3 &\\text{}^1 2 &3 &6 &\\text{}^13 \\\\ &- &2 &5 &2 &1 &5 \\\\ \\hline & & &7 &1 &5 &8 \\\\ \\hline \\end{array} For the ten thousands place: 3 - 2 = 1.\\begin{array}{c} & &3 &\\text{}^1 2 &3 &6 &\\text{}^13 \\\\ &- &2 &5 &2 &1 &5 \\\\ \\hline & &1 &7 &1 &5 &8 \\\\ \\hline \\end{array} 42\\,373 - 25\\,215 = 17\\,158 Idea summary We can use a vertical algorithm to subtract large numbers. This will require more steps to subtract each place value. ## Subtraction with story problems What happens if we have a story problem? We need to look for the keywords or clues to see what we need to do, as we see in this video. ### Examples #### Example 4 A farmer has 83 sheep in a paddock. 9", "# 5.08 Budgets Lesson ## Are you ready? In previous lessons we have practiced addition and subtraction with large numbers . Let's practice with the following problem. ### Examples #### Example 1 Find the value of 6396 - 129. Worked Solution Create a strategy Use the subtraction algorithm method. Apply the idea Write it in a vertical algorithm.\\begin{array}{c} & & &6 &3 &9 &6 \\\\ &- & & &1 &2 &9 \\\\ \\hline & \\\\ \\hline \\end{array} Begin with the units column. We can see that 6 is less than 9, so we need to trade 1 ten from the tens place. So we get 16 - 9 = 7 in the units column and 9 tens becomes 2 tens in the first row.\\begin{array}{c} & & &6 &3 &8 &\\text{ }^1 6 \\\\ &- & & &1 &2 &9 \\\\ \\hline & & & & & &7 \\\\ \\hline \\end{array} For the tens place: 8 - 2 = 6.\\begin{array}{c} & & &6 &3 &8 &\\text{ }^1 6 \\\\ &- & & &1 &2 &9 \\\\ \\hline & & & & &6 &7 \\\\ \\hline \\end{array} For the hundreds place: 3 - 1 = 2.\\begin{array}{c} & & &6 &3 &8 &\\text{ }^1 6 \\\\ &- & & &1 &2 &9 \\\\ \\hline & & & &2 &6 &7 \\\\", "&2 &4 &6 \\\\ &+ & &3 &2 &1 &3 \\\\ \\hline & & & &4 &5 &9 \\\\ \\hline \\end{array} Add the thousands column: 4 + 3 = 7 \\begin{array}{c} & &3 &4 &2 &4 &6 \\\\ &+ & &3 &2 &1 &3 \\\\ \\hline & & &7 &4 &5 &9 \\\\ \\hline \\end{array} Add the ten thousands column: 3 + 0 = 3 \\begin{array}{c} & &3 &4 &2 &4 &6 \\\\ &+ & &3 &2 &1 &3 \\\\ \\hline & & 3 &7 &4 &5 &9 \\\\ \\hline \\end{array} 34\\,246 + 3213 = 37\\,459 Example 2 Find the value of 49\\,481 + 49\\,100. Worked Solution Create a strategy Use the standard algorithm method. Apply the idea Write it in a vertical algorithm.\\begin{array}{c} & & &4 &9 &4 &8 &1 \\\\ &+ & &4 &9 &1 &0 &0 \\\\ \\hline & \\\\ \\hline \\end{array} Add the units column first.\\begin{array}{c} & & &4 &9 &4 &8 &1 \\\\ &+ & &4 &9 &1 &0 &0 \\\\ \\hline & & & & & & &1\\\\ \\hline \\end{array} In the tens column, we get 8+0=8.\\begin{array}{c} & & &4 &9 &4 &8 &1 \\\\ &+ & &4 &9 &1 &0 &0 \\\\ \\hline & & & & & &8 &1\\\\ \\hline \\end{array} In the hundreds column, we get 4+1=5.\\begin{array}{c} & &", "So we write the 8 in the ones place and carry the 1 to the tens place: \\begin{array}{c} &&8&{}^16&9 \\\\ &\\times &&&2 \\\\ \\hline &&&&8 \\\\ \\hline \\end{array} Move left and multiply to get 2\\times 6=12. Then we add the 1 to get 13. So we write the 3 in the tens place and carry the 1 to the hundreds place: \\begin{array}{c} &&{}^18&{}^16&9 \\\\ &\\times &&&2 \\\\ \\hline &&&3&8 \\\\ \\hline \\end{array} Move left and multiply to get 2\\times 8 = 16. Then we add the 1 to get 17. So we write the 1 in the thousands place and the 7 in the hundreds place: \\begin{array}{c} &&&{}^18&{}^16&9 \\\\ &\\times & &&&2 \\\\ \\hline &&1&7&3&8 \\\\ \\hline \\end{array} 869\\times2=1738 Idea summary The standard algorithm method is useful for multiplying large numbers. ## Long multiplication The area model is great, but we can also perform long multiplication. Let's see how they compare. ### Examples #### Example 3 Find 739 \\times 63. Worked Solution Create a strategy Use long multiplication. Apply the idea First we will multiply 739 by 3. 3 \\times 9 = 27 so we put the 7 in the ones place and carry the 2 to the tens place. \\begin{array}{c} &&7&{}^23&9 \\\\ &\\times &&6&3 \\\\ \\hline &&&& 7 \\\\ \\hline \\end{array} 3 \\times 3=9 then add the", "&9 \\\\ \\hline & && & & & 5 &3\\\\ \\hline \\end{array} In the hundreds column, we have 8-6=2:\\begin{array}{c} & && 5&5 &8 &6 &\\text{}^1 2 \\\\ &- && &2 &6 &1 &9 \\\\ \\hline & && & & 2& 5 &3\\\\ \\hline \\end{array} In the thousands column, we have 5-2=3:\\begin{array}{c} & && 5& 5 &8 &6 &\\text{}^1 2 \\\\ &- && &2 &6 &1 &9 \\\\ \\hline & && & 3& 2& 5 &3\\\\ \\hline \\end{array} And for the ten thousands place 5-0=5:\\begin{array}{c} & && 5& 5 &8 &6 &\\text{}^1 2 \\\\ &- && &2 &6 &1 &9 \\\\ \\hline & && 5& 3& 2& 5 &3\\\\ \\hline \\end{array} So the answer is:55\\,872 - 2619 = 53\\,253 Idea summary There are different strategies we can use to solve addition or subtraction questions. Even if our numbers get bigger, we can still use the same strategies, just with more steps. ### Outcomes #### VCMNA209 Select and apply efficient mental and written strategies and appropriate digital technologies to solve problems involving all four operations with whole numbers and make estimates for these computations", "Using the compensation strategy to add 16 to 49 we can add 20 and then take away 4. So now we can jump to right by 2 tens from 49 to get 49 + 20=69. And now we subtract 4 by jumping to the left by 4 to get 69 - 4=65. 49 + 16 = 65 #### Example 3 Find the value of 219 + 29. Worked Solution Create a strategy Apply the idea Write it in a vertical algorithm.\\begin{array}{c} & & &2 &1 &9 \\\\ &+ & & &2 &9 \\\\ \\hline & \\\\ \\hline \\end{array} Add the units value first. So 9 + 9 = 18. Bring down the 8 and carry on 1 to the tens column.\\begin{array}{c} & & &2 &\\text{}^1 1 &9 \\\\ &+ & & &2 &9 \\\\ \\hline & & & & &8 \\\\ \\hline \\end{array} So for the tens place we have 1 + 1 + 2 = 4: \\begin{array}{c} & & &2 &\\text{}^1 1 &9 \\\\ &+ & & &2 &9 \\\\ \\hline & & & &4 &8 \\\\ \\hline \\end{array} Hundreds place: 2 + 0 = 2 \\begin{array}{c} & & &2 &\\text{}^1 1 &9 \\\\ &+ & & &2 &9 \\\\ \\hline & & &2 &4 &8 \\\\ \\hline \\end{array} So 219 + 29 = 248. Idea summary", "1 thousand from the thousands place. So we get 12-4=8 in the hundreds column and 7 thousands becomes 6 thousands in the first row.\\begin{array}{c} & & &6 &\\text{}^1 2 &5 &\\text{}^1 3 \\\\ &- & &2 &4 &1 &8 \\\\ \\hline & & & &8 &4 &5\\\\ \\hline \\end{array} And for the thousands place 6 - 2 = 4:\\begin{array}{c} & & &6 &\\text{}^1 2 &5 &\\text{}^1 3 \\\\ &- & &2 &4 &1 &8 \\\\ \\hline & & &4 &8 &4 &5\\\\ \\hline \\end{array} So 7263 - 2418 = 4845. Idea summary Each time we move one place to the left, the value of the digit is ten times bigger. This means we are able to regroup, or trade, to help us with addition and subtraction. ### Outcomes #### MA2-5NA uses mental and written strategies for addition and subtraction involving two-, three-, four and five-digit numbers", "7 +9 = 16, so we bring down the 6 in the thousands column and carry the 1 to the ten thousands column:\\begin{array}{c} & &\\text{}^1 &7 &\\text{}^1 6 &8 &1 \\\\ &+ & &9 &2 &7 &3 \\\\ \\hline & & &6 &9 &5 &4\\\\ \\hline \\end{array} Add the ten thousands column. \\begin{array}{c} & &\\text{}^1 &7 &\\text{}^1 6 &8 &1 \\\\ &+ & &9 &2 &7 &3 \\\\ \\hline & &1 &6 &9 &5 &4\\\\ \\hline \\end{array} So 7681 + 9273 = 16\\, 954. #### Example 3 Find the value of 55\\,872-2619. Worked Solution Create a strategy Use the standard algorithm method. Apply the idea Write it in a vertical algorithm.\\begin{array}{c} & & &5 &5 &8 &7 &2 \\\\ &- & & &2 &6 &1 &9 \\\\ \\hline & \\\\ \\hline \\end{array} In the ones column we can see that 2 is less than 9, so we need to trade 1 ten from the tens place. So we get 12-9=3 in the ones column and 7 tens becomes 6 tens in the first row.\\begin{array}{c} & && 5&5 &8 &6 &\\text{}^1 2 \\\\ &- && &2 &6 &1 &9 \\\\ \\hline & && & & & &3\\\\ \\hline \\end{array} In the tens column, we have 6-1=5:\\begin{array}{c} & && 5&5 &8 &6 &\\text{}^1 2 \\\\ &- && &2 &6 &1", "in the units place. \\begin{array}{c}& &&2&4&3 \\\\ &\\times& &&1&5 \\\\ \\hline& &1&2&1&5 \\\\ &&&&& 0 \\\\ \\hline \\end{array} 1\\times 3=3 so we put a 3 in the tens place: \\begin{array}{c}& &&2&4&3 \\\\ &\\times& &&1&5 \\\\ \\hline& &1&2&1&5 \\\\ &&&&3& 0 \\\\ \\hline \\end{array} 1\\times 4=4 so we put a 4 in the hundreds place: \\begin{array}{c}& &&2&4&3 \\\\ &\\times& &&1&5 \\\\ \\hline& &1&2&1&5 \\\\ &&&4&3& 0 \\\\ \\hline \\end{array} 1\\times 2=2 so we put a 2 in the thousands place: \\begin{array}{c}& &&2&4&3 \\\\ &\\times& &&1&5 \\\\ \\hline& &1&2&1&5 \\\\ &&2&4&3& 0 \\\\ \\hline \\end{array} \\begin{array}{c} && &2 &4 &3 \\\\ &\\times && &1 &5 \\\\ \\hline &&1 &2 &1 &5 \\\\ &+& 2 &4 &3 & 0 \\\\ \\hline &&3&6&4&5 \\end{array} 243\\times 15=3645 Idea summary When multiplying by a two digit number using a vertical algorithm: • Multiply by the units digit first. • Put a 0 in the units column and then multiply by the tens digit. ### Outcomes #### VCMNA183 Solve problems involving multiplication of large numbers by one- or two-digit numbers using efficient mental, written strategies and appropriate digital technologies #### VCMNA185 Use efficient mental and written strategies and apply appropriate digital technologies to solve problems" ]

[ ". . $\\vdots$ $\\text{ten 9's}\\begin{Bmatrix}9&0 \\\\ 9&1 \\\\ 9& 2 \\\\ \\vdots \\\\ 9&9 \\end{Bmatrix}\\;\\;\\text{0 to 9 = 45}$ In the units-column, we have the sum of the digits 0 to 9 (45) ... nine times. Sum of the units digits: . $9 \\times 45 \\:=\\:405$ In the tens-column, we have: . $\\text{ten 1's + ten 2's + ten 3's + . . . + ten 9's}$ . . $\\;=\\;10(1) + 10(2) + 10(3) + \\cdots + 10(9) \\;=\\;10(1 + 2 + 3 +\\cdots + 9) \\;=\\;10(45)$ Sum of the tens digits: . $450$ Therefore, the sum of the digits is: . $405 + 450 \\;=\\;{\\color{blue}855}$ 12. Originally Posted by Chris L T521 $S_n=\\frac{n}{2}(a+z)$", "+0 # counting 0 95 1 How many 3-digit numbers have the property that the units digit is at least twice the tens digit plus 5? Jun 26, 2022 #1 +2448 0 There are 9 choices for the hundreds digit. The only possible cases for the tens and units digits are: (0, 5 - 9), (1, 7 - 9), (2, 9) This makes for 9 cases for the units and tens digits. Thus, there are $$9 \\times 9 = \\color{brown}\\boxed{81}$$ cases. Jun 26, 2022", "What digits are possible in the tens position of an odd square number? (This number eliminates the second number in the set). Complete these sentences with what you have discovered: * In a square number, the last digit (units digit) can only be _____, _______, _______, _______, _______ or _______. * The second last digit (tens place) of an odd square number is always _______. $\\bf{Cube \\hspace {0.1in} Numbers}$ Cube numbers behave rather differently. A bit of experimentation will show that cube numbers can end in ANY digit (units place). This digit depends on the last digit (units place) of the original number being cubed. Complete this table: $\\left| \\begin{array}{cc} \\mbox {if a number ends in} & \\mbox{its cube will end in}\\\\ 0 & \\\\ 1 & \\\\ 2 & \\\\ 3 & \\\\ 4 & \\\\ 5 & \\\\ 6 & \\\\ 7 & \\\\ 8 & \\\\ 9 & \\end{array}\\right|$ $\\bf{Fourth \\hspace{0.1in}Powers}$ Fourth powers are in fact just square numbers that have been squared. For example, $7^{4}=7^{2} \\times 7^{2}= 49^{2}=2401$. Since $4^{2}=16$ and $9^{2}=81$, the last digit of a fourth power can only be 0, 1, 5 or 6. $\\bf{Fifth \\hspace{0.1in}powers}$ Fifth powers have a magic of their own. Do a bit of experimentation to find out what it is. In p, particular, I suggest you", "can express any number in the base ten number system as the sum of powers of ten. Worked example 1.1: Writing a decimal number as the sum of powers of 10 Write the number 3,721,568 as the sum of powers of 10. 1. Step 1: Write the number in a place value table. \\begin{array}{|c|c|c|c|c|c|c|} \\hline ^\\text{millions} & ^\\text{hundred thousands} & ^\\text{ten thousands} & ^\\text{thousands} & ^\\text{hundreds} & ^\\text{ tens } & ^\\text{ units } \\newline 1,000,000 & 100,000 & 10,000 & 1,000 & 100 & 10 & 1 \\newline (10^6) & (10^5) & (10^4) & (10^3) & (10^2) & (10^1) & (10^0) \\newline \\hline 3 & 7 & 2 & 1 & 5 & 6 & 8 \\newline \\hline \\end{array} 2. Step 2: Write the number as the sum of the values of its digits. 3. Step 3: Write the number as the sum of powers of ten. \\begin{align} \\;&3,721,568 \\\\ &=(3\\times10^6)+(7\\times10^5)+(2\\times10^4)+(1\\times10^3)+(5\\times10^2)+(6\\times10^1)+(8\\times10^0) \\end{align} In the base two number system, we work with powers of 2. The first ten powers of 2 are: \\begin{array}{|l|r|c|} \\hline \\textbf{Name} & \\textbf{Number} & \\textbf{Power of 2} \\newline \\hline \\text{One} & 1 & 2^0 \\newline \\hline \\text{Two} & 2 & 2^1 \\newline \\hline \\text{Four} & 4 & 2^2 \\newline \\hline \\text{Eight} & 8 & 2^3 \\newline \\hline \\text{Sixteen} & 16 & 2^4 \\newline \\hline", "< 250 < 7 3, tens. 1 9 2 7 3, so unit digit in the cube root that. Digit in the cube root of two all I do is sqrt ( 2 ) then I get my.! Digit of the number is 7, so tens digit will be 6 cube of...", "### A $3$-digit number is of the form ‘high-low-high’ $-$ that is the tens digit is smaller than both the hundreds digit and the units (or ‘ones’) digit. How many such $3$-digit numbers are there? $285$ Step by Step Explanation: 1. A $3-$digit number is of the form ‘high-low-high’, so, the tens digit of the $3$-digit number cannot be $9,$ as the units and the hundreds, digit needs to be larger than tens digit and $9$ is the largest digit. Therefore, the smallest tens digit of the $3$-digit number of the required form is $0$ and the largest tens digit of the $3$-digit number is $8.$ 2. If the tens digit is $0,$ then the hundreds digit can be any digit from $1$ to $9,$ and the units digit can also be any digit from $1$ to $9.$ So, there are $9 \\times 9$ possible numbers of the required form with $0$ at tens place. If the tens digit is $1,$ then the hundreds digit can be any digit from $2$ to $9,$ and the units digit can also be any digit from $2$ to $9.$ So, there are $8 \\times 8$ possible numbers of the required form with $1$ at tens place. 3. Similarly, possible numbers with $2$ at tens place is $7 \\times 7,$ possible numbers", "the third number in this set). Now check out the pairs of digits that your odd square numbers end with. What digits are possible in the tens position of an odd square number? (This number eliminates the second number in the set). Complete these sentences with what you have discovered: * In a square number, the last digit (units digit) can only be _____, _______, _______, _______, _______ or _______. * The second last digit (tens place) of an odd square number is always _______. $\\bf{Cube \\hspace {0.1in} Numbers}$ Cube numbers behave rather differently. A bit of experimentation will show that cube numbers can end in ANY digit (units place). This digit depends on the last digit (units place) of the original number being cubed. Complete this table: $\\left| \\begin{array}{cc} \\mbox {if a number ends in} & \\mbox{its cube will end in}\\\\ 0 & \\\\ 1 & \\\\ 2 & \\\\ 3 & \\\\ 4 & \\\\ 5 & \\\\ 6 & \\\\ 7 & \\\\ 8 & \\\\ 9 & \\end{array}\\right|$ $\\bf{Fourth \\hspace{0.1in}Powers}$ Fourth powers are in fact just square numbers that have been squared. For example, $7^{4}=7^{2} \\times 7^{2}= 49^{2}=2401$. Since $4^{2}=16$ and $9^{2}=81$, the last digit of a fourth power can only be 0, 1, 5 or 6. $\\bf{Fifth \\hspace{0.1in}powers}$ Fifth powers have a magic", "number in this set). Now check out the pairs of digits that your odd square numbers end with. What digits are possible in the tens position of an odd square number? (This number eliminates the second number in the set). Complete these sentences with what you have discovered: * In a square number, the last digit (units digit) can only be _____, _______, _______, _______, _______ or _______. * The second last digit (tens place) of an odd square number is always _______. $\\bf{Cube \\hspace {0.1in} Numbers}$ Cube numbers behave rather differently. A bit of experimentation will show that cube numbers can end in ANY digit (units place). This digit depends on the last digit (units place) of the original number being cubed. Complete this table: $\\left| \\begin{array}{cc} \\mbox {if a number ends in} & \\mbox{its cube will end in}\\\\ 0 & \\\\ 1 & \\\\ 2 & \\\\ 3 & \\\\ 4 & \\\\ 5 & \\\\ 6 & \\\\ 7 & \\\\ 8 & \\\\ 9 & \\end{array}\\right|$ $\\bf{Fourth \\hspace{0.1in}Powers}$ Fourth powers are in fact just square numbers that have been squared. For example, $7^{4}=7^{2} \\times 7^{2}= 49^{2}=2401$. Since $4^{2}=16$ and $9^{2}=81$, the last digit of a fourth power can only be 0, 1, 5 or 6. $\\bf{Fifth \\hspace{0.1in}powers}$ Fifth powers have a magic of their", "third number in this set). Now check out the pairs of digits that your odd square numbers end with. What digits are possible in the tens position of an odd square number? (This number eliminates the second number in the set). Complete these sentences with what you have discovered: * In a square number, the last digit (units digit) can only be _____, _______, _______, _______, _______ or _______. * The second last digit (tens place) of an odd square number is always _______. $\\bf{Cube \\hspace {0.1in} Numbers}$ Cube numbers behave rather differently. A bit of experimentation will show that cube numbers can end in ANY digit (units place). This digit depends on the last digit (units place) of the original number being cubed. Complete this table: $\\left| \\begin{array}{cc} \\mbox {if a number ends in} & \\mbox{its cube will end in}\\\\ 0 & \\\\ 1 & \\\\ 2 & \\\\ 3 & \\\\ 4 & \\\\ 5 & \\\\ 6 & \\\\ 7 & \\\\ 8 & \\\\ 9 & \\end{array}\\right|$ $\\bf{Fourth \\hspace{0.1in}Powers}$ Fourth powers are in fact just square numbers that have been squared. For example, $7^{4}=7^{2} \\times 7^{2}= 49^{2}=2401$. Since $4^{2}=16$ and $9^{2}=81$, the last digit of a fourth power can only be 0, 1, 5 or 6. $\\bf{Fifth \\hspace{0.1in}powers}$ Fifth powers have a magic of", "five from the eight dots left in the hundreds column. $$\\begin{split} 8\\; &12 \\\\ 9\\; &2 \\; \\; 1 \\\\ -\\; 5\\; &5 \\; \\; 1 \\\\ \\hline 3\\; &7 \\; \\; 0 \\end{split}$$", "3. Hence, the unit digit of the cube of the number 1787 is …..3….. Note: The unit digit of any number raised to the power of ‘n’ depends on the unit digit of the original number. However the unit digit keeps changing only till a number is raised to a maximum power of 4. After that the unit digit starts repeating itself. For example let us consider a number 43. So, ${\\left( {43} \\right)^1}$ has $3 \\times 1 = 3$ as its unit digit. ${\\left( {43} \\right)^2}$ has $3 \\times 3 = 9$ as its unit digit. ${\\left( {43} \\right)^3}$ has $3 \\times 3 \\times 3 = 27$,that is 7 as its unit digit. ${\\left( {43} \\right)^4}$ has $3 \\times 3 \\times 3 \\times 3 = 81$, that is 1 as its unit digit. ${\\left( {43} \\right)^5}$, has $3 \\times 3 \\times 3 \\times 3 \\times 3 = 243$, that is 3 as its unit digit. So, as soon as the power increases by 1 from 4 and becomes 5, the unit digit also starts repeating itself from 3 onwards.", "Posts: 51223 Re: The three-digit positive integer x has the hundreds, [#permalink] ### Show Tags 09 Dec 2014, 08:03 1 1 JoostGrijsen wrote: Bunuel wrote: mohnish104 wrote: The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l, and m, respectively. If (2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m), what is the value of x – y? A) 21 B) 200 C) 210 D) 300 E) 310 x = abc, where a, b, and c, are the hundreds, tens, and units digits, respectively. y= klm, where k, l, and m, are the hundreds, tens, and units digits, respectively. $$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$ --> $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$ --> $$a-k=2$$, $$b-l=1$$, and $$c-m=0$$ --> $$x-y=210$$. I am sorry but i dont understand the slightest thing of what's going on here. How does that suddenly translate to 2^2*3? really confused here. $$(2^a)(3^b)(5^c) = 12(2^k)(3^l)(5^m)$$; $$\\frac{(2^a)(3^b)(5^c)}{(2^k)(3^l)(5^m)} = 12$$; $$2^{a-k}*3^{b-l}*5^{c-m}=2^2*3$$. Similar questions to practice: k-and-l-are-each-four-digit-positive-integers-with-thousands-91004.html for-any-four-digit-number-abcd-abcd-3-a-5-b-7-c-11-d-126522.html k-and-l-are-each-four-digit-positive-integers-with-thousands-126646.html the-function-f-is-defined-for-each-positive-three-digit-100847.html for-a-three-digit-number-xyz-where-x-y-and-z-are-the-59284.html Hope it helps. _________________ Director Status: Professional GMAT Tutor Affiliations: AB, cum laude, Harvard University (Class of '02) Joined: 10 Jul 2015 Posts: 671 Location: United States (CA) Age: 39 GMAT 1: 770 Q47 V48 GMAT 2: 730 Q44 V47 GMAT 3: 750 Q50 V42 GRE 1: Q168 V169", "last digit (units digit) can only be _____, _______, _______, _______, _______ or _______. * The second last digit (tens place) of an odd square number is always _______. $\\bf{Cube \\hspace {0.1in} Numbers}$ Cube numbers behave rather differently. A bit of experimentation will show that cube numbers can end in ANY digit (units place). This digit depends on the last digit (units place) of the original number being cubed. Complete this table: $\\left| \\begin{array}{cc} \\mbox {if a number ends in} & \\mbox{its cube will end in}\\\\ 0 & \\\\ 1 & \\\\ 2 & \\\\ 3 & \\\\ 4 & \\\\ 5 & \\\\ 6 & \\\\ 7 & \\\\ 8 & \\\\ 9 & \\end{array}\\right|$ $\\bf{Fourth \\hspace{0.1in}Powers}$ Fourth powers are in fact just square numbers that have been squared. For example, $7^{4}=7^{2} \\times 7^{2}= 49^{2}=2401$. Since $4^{2}=16$ and $9^{2}=81$, the last digit of a fourth power can only be 0, 1, 5 or 6. $\\bf{Fifth \\hspace{0.1in}powers}$ Fifth powers have a magic of their own. Do a bit of experimentation to find out what it is. In p, particular, I suggest you create tables of second powers, third powers, fourth powers and fifth powers of all numbers from zero to 20. Check, compare…actually, it is fun to “compare rate of growth of powers with increasing integers”…this idea involves", "repeated twice in the product. For example, $11\\times 1 = 11$, as 11 is multiplied by the number 1, 1 is repeated in the answer which is 11. Similarly, $11\\times 6 =66$, here 6 is repeated. The whole pattern is presented below, and the repeated digits are shown in green color. 11 Times Table Table Outcome 11 x 1 11 11 x 2 22 11 x 3 33 11 x 4 44 11 x 5 55 11 x 6 66 11 x 7 77 11 x 8 88 11 x 9 99 Pattern for the 10th and higher multiples of 11: This method presents the pattern followed by the 10th and the higher multiples of the number 11. Suppose 11 is multiplied by 10 (note that the unit digit of 10 is 0 and the tens digit is 1); the product of $11 \\times 10$ is equal to 110 (unit digit 0, tens digit 1, and hundreds digit 1). The unit digit of the product is the same as the unit digit of the number multiplied by 11. The tens digit of the product is the summation of the unit and the tens digit. In our example, 10 is being multiplied by 11, so the tens digit of the product is $0+1 = 1$. Lastly, the hundreds digit", "'units' point to the century number. Now xor 'units' with twice the 'tens', mod 4. This works because 10==±2 mod 4, so the lower bit of the 'tens' can just toggle bit 1 of the 'units'. We use the result as an index into our remainders table, printing y only if the modular result is zero. # Befunge-98, (41 bytes) &:4%#v_:aa*%#v_28*%| \"y\",;<;@,\"n\";>; ;#[ Simplicity is awesome. # sed, 55 s/00$// y/0123456789/yNnNyNnNyN/ /N.$/y/ny/yn/ s/.\\B//g • First line divides exact centuries by 100. • Second line gives 'N' to odd digits, 'y' to 4s, and 'n' to non-4s. • Third line swaps 'y' and 'n' if there's an odd penultimate digit (because 10 is 2 mod 4) • Final line deletes all but the last character Note that non-leap years may be printed as n or N depending on whether they are even or odd. I consider this a creative interpretation of the rule which allows alternatives to 'yes' and 'no' without specifying that they have to be consistent. # Python2 - 37 g=lambda x:(x%4or x%400and x%100<1)<1 Note that if a is a nonnegative integer, then a<1 is a short way of writing not bool(a). The last <1 thus effectively converts the expression in the parentheses to a boolean and negates the result. Applying the function g to an integer", "last digit of the number which is the tens digit of $7k$ plus twice the units digit of $7k$ is a $3$. If $k=1$, then $7k=7$, and the tens digit plus twice the units digit is $0+14=14$ which has last digit 4. If $k=2$, then $7k=14$, and the tens digit plus twice the units digit is $1+8=9$ which has last digit 9. If $k=3$, then $7k=21$, and the tens digit plus twice the units digit is $2+2=4$ which has last digit 4. If $k=4$, then $7k=28$, and the tens digit plus twice the units digit is $2+16=18$ which has last digit 8. If $k=5$, then $7k=35$, and the tens digit plus twice the units digit is $3+10=13$ which has last digit 3. So $k=5$, and the product is $4235$. Using the 11 times table Two-digit numbers with identical digits are precisely the multiples of 11. So 77 has been multiplied by 11$n$ for some value of $n$. 77$\\times$11$n$ = (77$\\times$11)$n$ = 847$n$. So 847$n$ = _ _ 3 _ . The 800 will not contribute to this digit, so we can consider the 47 times table. 47$\\times$2 = 94 47$\\times$3 = 141 47$\\times$4 = 188 47$\\times$5 = 235 So $n$ = 5, and the product is 847$\\times$5 = 4235. You can find more short problems, arranged by curriculum", "The hundreds' digit of 5^n is 6. $$5^3 = 125$$ $$5^4 = 625$$ $$5^5 = 3125$$ $$5^6 = 15625$$ So, for every even power of 5, hundred’s digit = 6 —> Possible values of n = {4, 6, 8, 10, 12, . . . } —> Insufficient Combining (1) & (2), Possible values of n = {4, 14, 24, . . . . } —> Insufficient Option E Posted from my mobile device Director Joined: 07 Mar 2019 Posts: 902 Location: India GMAT 1: 580 Q43 V27 WE: Sales (Energy and Utilities) Re: If n is a positive integer greater than 2, what is the value of n ? [#permalink] ### Show Tags 17 Jan 2020, 10:22 1 If n is a positive integer greater than 2, what is the value of n ? n > 2, n = ? (1) The tens' digit of $$11^n$$ is 4. $$11^4 = 14641$$ for n = 4 $$11^14 = ... 41$$ for n = 14 (Pattern of 1, 2, 3, 4, 5, 6, 7, 8, 9 & 0 at tens' place is repeated for powers 1, 2, 3, 4, 5, 6, 7, 8, 9 & 0 respectively and 11 to 20 and further ..) INSUFFICIENT. (2) The hundreds' digit of $$5^n$$ is 6. $$5^4 = 625$$ for n = 4", "the units column. We can see that 7 is less than 8, so we need to trade 1 ten from the tens place. So we get 17 - 8 = 9 in the units column and 9 tens becomes 8 tens in the first row.\\begin{array}{c} & & &9 &8 &\\text{ }^1 7 \\\\ &- & &6 &4 &8 \\\\ \\hline & & & & &9 \\\\ \\hline \\end{array} For the tens place: 8 - 4 = 4.\\begin{array}{c} & & &9 &8 &\\text{ }^1 7 \\\\ &- & &6 &4 &8 \\\\ \\hline & & & &4 &9 \\\\ \\hline \\end{array} For the hundreds place: 9 - 6 = 3.\\begin{array}{c} & & &9 &8 &\\text{ }^1 7 \\\\ &- & &6 &4 &8 \\\\ \\hline & & &3 &4 &9 \\\\ \\hline \\end{array} So 997 - 648 = 349. Idea summary We always start from the ones place, when we work down our page. If we don't have enough ones to subtract, we need to trade from the tens place. If we don't have enough tens to subtract, we need to trade from the hundreds place. ## Add and subtract 4 digit numbers What happens with larger numbers? We can use a standard algorithm to solve addition and subtraction, as we see in this video. ### Examples #### Example", "of 10. You may use a place value table to help you with this: \\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \\hline \\text{billion} & \\text{hm} & \\text{tm} & \\text{million} &\\text{hth} & \\text{tth} & \\text{thousand} & \\text{ h } & \\text{ t } & \\text{ u } \\newline \\hline & 2 & 0 & 1 & 1 & 9 & 5 & 4 & 3 & 1 \\newline \\hline \\end{array} The 5 represents the thousands. 3. Step 3: Consider the digits to the right of the one you identified in Step 2. If they are less than the value in Step 1, the digit you identified remains the same. If they are equal to or more than the value in Step 1, the digit you identified increases by 1. The digits to the right of the 5 are 431. 431<500 The 5 remains the same. 4. Step 4: Write down the rounded number with the digit you determined in Step 3, and all the digits to the right of it as zeroes. 201,195,000 If we round off 201,195,431 to the nearest thousand, we actually want to know whether this number is closer to 201,195,000 or 201,196,000. The number 201,195,431 is 431 units away from 201,195,000 and 569 units away from 201,196,000. That is why it is rounded off to 201,195,000. ### Exercise 7.7: Round off", "# Units digit of an expression To find the last digit of an expression which is in the format of ab we have to learn the concept of cyclicity. Digits 0, 1, 5, 6: These digits always give the same units digit. ${0^n} = 0; \\, {1^n} = 1; \\, {5^n} = 5; \\, {6^n} = 6$ Digits 4, 9: For 4, If the power is odd you will get $4$ as units digit. If the power is even, you will get $6$ as units digit. $\\begin{array}{*{20}{c}} {{4^{odd}} = 4}\\\\ {{4^{even}} = 6} \\end{array}$ For 9, If the power is odd you will get $9$ as units digit. If the power is even, you will get $1$ as units digit. $\\begin{array}{*{20}{c}} {{9^{odd}} = 9}\\\\ {{9^{even}} = 1} \\end{array}$ Digits 2, 3, 7, 8: Digits 2, 3, 7, 8 follows cyclicity of $4$. For example, 21 = 2, 22 = 4, 23 = 8, 24 = 6, 25 = 2, 26 = 4 and the cycle repeats. Also, $6 \\times \\text{Even} = \\text{Even}$ $\\begin{array}{l} 6 \\times 2 = 2\\\\ 6 \\times 4 = 4\\\\ 6 \\times 6 = 6\\\\ 6 \\times 8 = 8 \\end{array}$ Firstly, we divide the power by 4 and check the remainder. If it is a multiple of 4, we get the following units digits" ]

[ "# Choosing sides Angel, Bob and Chris each randomly choose different faces of a dodecahedron. The probability that none of the three faces are adjacent, is $$\\frac{a}{b}$$ where $$a$$ and $$b$$ are coprime positive integers. What is $$a+b$$ ? Image credit: http://mathcentral.uregina.ca ×", "Core Skills to excel on the GMAT. A number cube has six faces numbered 1 through 6. If the cube is rolle Author Message TAGS: Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 51215 A number cube has six faces numbered 1 through 6. If the cube is rolle [#permalink] Show Tags 31 Jul 2018, 00:08 00:00 Difficulty: 25% (medium) Question Stats: 65% (01:31) correct 35% (01:25) wrong based on 54 sessions HideShow timer Statistics A number cube has six faces numbered 1 through 6. If the cube is rolled twice, what is the probability that at least one of the rolls will result in a number greater than 4? (A) $$\\frac{2}{9}$$ (B) $$\\frac{1}{3}$$ (C) $$\\frac{4}{9}$$ (D) $$\\frac{5}{9}$$ (E) $$\\frac{2}{3}$$ _________________ examPAL Representative Joined: 07 Dec 2017 Posts: 841 Re: A number cube has six faces numbered 1 through 6. If the cube is rolle [#permalink] Show Tags 31 Jul 2018, 00:16 Bunuel wrote: A number cube has six faces numbered 1 through 6. If the cube is rolled twice, what is the probability that at least one of the rolls will result in a number greater than 4? (A) $$\\frac{2}{9}$$ (B) $$\\frac{1}{3}$$ (C) $$\\frac{4}{9}$$ (D) $$\\frac{5}{9}$$ (E) $$\\frac{2}{3}$$ In questions asking us to calculate 'at least one of' something, it is usually easier to calculate the", "μ = 7400 and estimated asked on November 17, 2016 100. ## Probability A number cube has six faces numbered 1 to 6. John tossed two number cubes several times and added the number each time. asked on January 31, 2013", "duplicated {1;1}, {2,2}.... {6,6}; Otherwise we don't have 36, we only have 30 = 6*5 possible outcome. Thank you! These are the 36 cases: {1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6} {2, 1}, {2, 2}, {2, 3}, {2, 4}, {2, 5}, {2, 6} {3, 1}, {3, 2}, {3, 3}, {3, 4}, {3, 5}, {3, 6} {4, 1}, {4, 2}, {4, 3}, {4, 4}, {4, 5}, {4, 6} {5, 1}, {5, 2}, {5, 3}, {5, 4}, {5, 5}, {5, 6} {6, 1}, {6, 2}, {6, 3}, {6, 4}, {6, 5}, {6, 6} _________________ Karishma Veritas Prep GMAT Instructor Manager Joined: 14 Oct 2015 Posts: 249 GPA: 3.57 Re: Two number cubes with faces numbered 1 to 6 are rolled. What is the [#permalink] ### Show Tags 27 Jan 2018, 17:30 1 vingmat001 wrote: Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that that sum of the rolls is 8. A. 1/12 B. 1/11 C. 1/9 D. 5/36 E. 1/6 I am struggling to understand a concept - This seems like an error in MGMAT book - But i want to get it clarified and get a better understanding. The questions is: Two number cubes with faces numbered 1 to 6 are rolled. What is the probability that", "a 6. Geometrically, this last shell is three faces of the cube. Each previous shell can also be viewed as having three faces (with the possible exception of $S_1$.) To be more precise, $$S_i = \\{(a,b,c)|\\max\\{a,b,c\\}=i\\}$$ for $i =1\\ldots6$. The probability of tossing into shell $i$ is $$P(S_i)=\\frac{i^3-(i-1)^3}{6^3}$$ Each shell $S_i$ has subsets $S_{i,j}$ where $j$ is the second highest roll (possibly equal to $i$). Geometrically, $S_{i,j}$ is composed of line segments running along the three faces of $S_i$. $S_{i,i}$ is composed of the three edges converging to the main corner in $S_i$. For $j$ less than $i$, $S_{i,j}$ is composed of three \"arrows\", one along each face of $S_i$, each pointing to the main corner of $S_i$. The image below illustrates the cube of possilbe attacker rolls, $S_6$, $S_{6,6}$, and $S_{6,3}$. We find that for the attacker's roll, with $j$ less than $i$, \\begin{align*} P(S_{i,i}) & = \\frac{3i-2}{i^3-(i-1)^3}\\frac{i^3-(i-1)^3}{6^3}=\\frac{3i-2}{6^3}\\\\ P(S_{i,j}) & = \\frac{6j-3}{i^3-(i-1)^3}\\frac{i^3-(i-1)^3}{6^3}=\\frac{6j-3}{6^3}\\ \\end{align*} Now we begin to consider the defender's tosses, conditional upon what $S_{i,j}$ the attacker has rolled into. Similar to the cube that we envisioned for the attacker, the defender has a $6\\times 6$ square of possible rolls. Let's start counting the number of armies that the attacker loses. Let $X$ be this random variable, which can take values $0$, $1$, or $2$. Suppose the", "by the total number of possible rolls. The total number of possible cube rolls is 6 × 6 = 36. Make a list: Sum of 2 can happen 1 way: 1 + 1. Sum of 3 can happen 2 ways: 1 + 2 or 2 + 1. Sum of 5 can happen 4 ways: 1 + 4, 2 + 3, 3 + 2, 4 + 1. Sum of 7 can happen 6 ways: 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1. Sum of 11 can happen 2 ways: 5 + 6, 6 + 5. That’s a total of 1 + 2 + 4 + 6 + 2 = 15 ways to roll a prime sum. Thus, the probability is $$\\frac{15}{36}=\\frac{5}{12}$$. _________________ Sandy If you found this post useful, please let me know by pressing the Kudos Button Try our free Online GRE Test Re: Two number cubes with six faces numbered with the integers f [#permalink] 21 Aug 2018, 18:25 Display posts from previous: Sort by", "Re: Alejandro has a six-sided die with faces numbered 1 through [#permalink] 21 Dec 2018, 10:20 The probability that the two rolls are even is $$\\frac{3}{6} * \\frac{3}{6} which is \\frac{1}{4}$$ possible even nos are :2, 4 and 6. The probability that neither roll is a multiple of 3 becomes $$\\frac{4}{6} * \\frac{4}{6}$$ which is $$\\frac{4}{9}$$ Re: Alejandro has a six-sided die with faces numbered 1 through [#permalink] 21 Dec 2018, 10:20 Display posts from previous: Sort by", "### Let's Investigate Triangles Vincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make? ### Noah Noah saw 12 legs walk by into the Ark. How many creatures did he see? ### Five Coins Ben has five coins in his pocket. How much money might he have? # I'm Eight ##### Age 5 to 11Challenge Level Tayla from Marion Primary School in Australia and Luna from Marner Primary School in England, sent in what they called “all the possible additions equalling 8”: 5+3 =8 3+5 =8 6+2=8 2+6=8 1+7=8 7+1=8 4+4=8 and then the subtractions: 9-1=8 10-2=8 11-3=8 12-4=8 13-5=8 14-6=8 PomPom from the Australian International School Malaysia extended the idea and used a division making 12:”¨ 6+6=12, -20+32= 12, 24$\\div$2= 12 Micaela from the Westridge School for Girls went further by using more than two numbers to generate the solution:”¨ For the number 8 I did 100 minus 48 minus 32 minus 4 minus 8 which equaled eight. Now for a random age of 10 I did 87 minus 16 minus 11 then subtract 50. That's both of my answers. Adelaide and Janeen also from Westridge used all four operations:”¨ 0+8=8 Eight minus eight equals zero, so eight plus zero", "through 11. The 3d 8 use looks rather like a three-sided football,each side being a 1×2 unit rectangular face incised at opposite ends with an opposing 90° arc. /) (/ sort of like this. 9. Mike Haumesser says: Mathematically that’s D=[D-1]/2 (3d) -[D-3]/2. Sorry my diagram above didn’t translate well. 10. GreenLego says: Hi Mr Haumesser. I don’t think 3(3d)-2 will work to simulate 1 to 7. There are 27 possible combinations (3^3) but there are only 7 possible outcomes. Since 7 doesn’t divide into 27, there will be uneven distribution of 1 to 7. In fact there is only 1/27 chance of getting a 1 (1,1,1) and also only 1/27 chance of getting a 7 (3,3,3). And there is a 7/27 chance of getting a 4, more than any other number.", "Answer D. ---- Kindly press Kudos if the explanation is clear. Thank you For one roll of a certain number cube with six faces, numbered 1 throu &nbs [#permalink] 22 Oct 2017, 20:13 Display posts from previous: Sort by", "sold in sets, matching in color, of five or six different shapes: the five Platonic solids, whose faces are regular polygons, and optionally the pentagonal trapezohedron, whose faces are ten kites, each with two different edge lengths and three different angles; the die's vertices also are of two different kinds. Normally opposing faces of dice will add up to one more than the number of faces, but in the case of the d4, d5, and standard d10 (among others), this is simply not possible. Sides Shape Notes 4tetrahedron Each face has three numbers: they are arranged such that the upright number (which counts) is the same on all three visible faces. Alternatively, all of the sides have the same number in the lowest edge and no number on the top. This die does not roll well and thus it is usually thrown into the air instead. 6cubeA common die. The sum of the numbers on opposite faces is seven. 8octahedronEach face is triangular; looks like two square pyramids attached base-to-base. Usually, the sum of the opposite faces is 9. 10pentagonal trapezohedronEach face is a kite. The die has two sharp corners, where five kites meet, and ten blunter corners, where three kites meet. The ten faces usually bear numbers from zero to nine, rather than one to ten", "### 3.7.62. Conway packing problem Denotes the fact that a constraint can be used for solving the Conway packing problem, which consists of placing 6 orthotopes of size $4×2×1$, 6 orthotopes of size $3×2×2$ and 5 unit cubes within a $5×5×5$ cube.", "is equal to 6: Noah Dana-Picard 2007-12-28", "# A number cube has sides numbered 1though 6 the probability of rolling a 2 is This is 1/2)2=1/4 hope this helps Only authorized users can leave an answer!", "### I'm Eight Find a great variety of ways of asking questions which make 8. ### Let's Investigate Triangles Vincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make? ### Noah Noah saw 12 legs walk by into the Ark. How many creatures did he see? # Break it Up! ## Break it Up! You have a stick of 7 interlocking cubes (or a tower of 7 Lego blocks). You cannot change the order of the cubes. You break off a bit of it leaving it in two pieces. Here are three of the ways in which you can do it: In how many different ways can it be done? Now try with a stick of 8 cubes and a stick of 6 cubes: Now predict how many ways there will be with 5 cubes. Try it! Were you right? How many ways with 20 cubes? 50 cubes? 100 cubes? ANY number of cubes? ### Why do this problem? This problem makes a very good introduction to algebraic thinking, however, it also has a very accessible starting point so everyone can begin. ### Possible approach You could start by giving all the group some interlocking cubes (or Lego blocks) and asking", "## anonymous 5 years ago If you roll a number cube 12 times, about how many times would you expect to roll a 5 or 6? A. 2 B. 5 C. 4 D. 10 1. anonymous $P(5 or 6)=P(5)+P(6)-P(5 and 6)=P(5)+P(6)$$=\\frac{1}{6}+\\frac{1}{6}=\\frac{1}{3}$ 2. anonymous So 1/3 of 12 times you'd expect to get a 5 or a 6; i.e. 4 times.", "(B) $$\\frac{1}{3}$$ (C) $$\\frac{4}{9}$$ (D) $$\\frac{5}{9}$$ (E) $$\\frac{2}{3}$$ _________________ examPAL Representative Joined: 07 Dec 2017 Posts: 1155 Re: A number cube has six faces numbered 1 through 6. If the cube is rolle [#permalink] ### Show Tags 31 Jul 2018, 01:16 2 Bunuel wrote: A number cube has six faces numbered 1 through 6. If the cube is rolled twice, what is the probability that at least one of the rolls will result in a number greater than 4? (A) $$\\frac{2}{9}$$ (B) $$\\frac{1}{3}$$ (C) $$\\frac{4}{9}$$ (D) $$\\frac{5}{9}$$ (E) $$\\frac{2}{3}$$ In questions asking us to calculate 'at least one of' something, it is usually easier to calculate the complement. This is a Logical approach approach. The complement of 'at least one roll greater than four' is 'none of the rolls are greater than four' which is the same as 'all of the rolls are smaller than 5'. The probability to get a number smaller than 5 is 4/6 and the probability that both rolls are smaller than 5 is (4/6)^2 = (2/3)^2 = 4/9. So our answer is 1 - 4/9 = 5/9. _________________ Senior Manager Joined: 22 Feb 2018 Posts: 407 Re: A number cube has six faces numbered 1 through 6. If the cube is rolle [#permalink] ### Show Tags 31 Jul 2018, 03:22 Bunuel wrote: A", "Each face has three numbers: they are arranged such that the upright number (which counts) is the same on all three visible faces. Alternatively, all of the sides have the same number in the lowest edge and no number on the top. This die does not roll well and thus it is usually thrown into the air instead. 6 cube A common die. The sum of the numbers on opposite faces is seven. 8 octahedron Each face is triangular; looks like two square pyramids attached base-to-base. Usually, the sum of the opposite faces is 9. 10 pentagonal trapezohedron Each face is a kite. The die has two sharp corners, where five kites meet, and ten blunter corners, where three kites meet. The ten faces usually bear numbers from zero to nine, rather than one to ten (zero being read as \"ten\" in many applications). Often all odd numbered faces converge at one sharp corner, and the even ones at the other. The sum of the numbers on opposite faces is usually 9 (numbered 0–9) or 11 (number 1–10). 12 dodecahedron Each face is a regular pentagon. The sum of the numbers on opposite faces is usually 13. 20 icosahedron Faces are equilateral triangles. Icosahedrons have already been made in Roman/ Ptolemaic times, but it is not known if", "which is a real bummer for me. Edit: The accepted answer said it best: Obviously three hexagons sharing a common vertex cannot be perfectly regular, otherwise the surface would be flat in a neighbourhood of the common vertex! So it's quite a bummer. We cannot have arbitrarily large number of hexagons if we want them to be regular hexagons. Looks like the closest with can get is the soccer ball shape (truncated icosahedron), which actually only has two hexagons meeting at a vertex, not three, so they can be regular. • I haven't read the article, but I hope you're not saying \"they are clearly not the same size\" based on their appearance on a grainy picture. Also note that if two neighbouring polygons do not have the same side length, there will be some difficulties sewing them together. – Arthur Oct 27 '16 at 5:37 • You can get a soccer ball or a dodecahedron. You're allowing 4 types of vertices: 5-5-5, 5-5-6, 5-6-6, and 6-6-6. The last is flat, and you know 2 of them. Only 5-5-6 is left, and that won't lead to a globe. – Ed Pegg Oct 27 '16 at 5:53 • @Arthur The \"grainy picture\" is actually pretty accurate. I based it on the picture and my investigation/reading the linked articles. if", "face has three numbers: they are arranged such that the upright number (which counts) is the same on all three visible faces. Alternatively, all of the sides have the same number in the lowest edge and no number on the top. This die does not roll well and thus it is usually thrown into the air instead. 6 cube A common die. The sum of the numbers on opposite faces is seven. 8 octahedron Each face is triangular; looks something like two Egyptian pyramids attached at the base. Usually, the sum of the opposite faces is 9. 10 pentagonal trapezohedron Each face is kite-shaped; five of them meet at the same sharp corner (as at the top of the diagram in this row), and five at another equally sharp one; about halfway between them, a different group of three faces converges at each of ten blunter corners. The ten faces usually bear numbers from zero to nine, rather than one to ten (zero being read as \"ten\" in many applications), and often all odd numbered faces converge at the same sharp corner, and the even ones at the other. 12 dodecahedron Each face is a regular pentagon. 20 icosahedron Faces are equilateral triangles. Typically, opposite faces add to twenty-one. A Roman icosahedron die from the 2nd century AD" ]

[ "balloons and 3 purple balloons. Jeff randomly draws one gumball and then one balloon. What is the probability that he will draw a red gumball and a purple balloon? A. $\\frac{3}{10}$ B. $\\frac{2}{15}$ C. $\\frac{1}{5}$ D. $\\frac{1}{10}$ 12. Gary has 1 red apple, 1 green apple, and 4 yellow apples in a sack. He also has 2 packages of peanuts, 2 packages of almonds, and 2 packages of cashews in another sack. If he reaches into each sack without looking, what is the probability that he will grab a red apple and a package of cashews? A. $\\frac{1}{18}$ B. $\\frac{1}{4}$ C. $\\frac{1}{12}$ D. $\\frac{1}{36}$ 1. A 2. A 3. B 4. A 5. B 6. A 7. B 8. B 9. D 10. B 11. C 12. A ## Explanations 1. Since each of the six sides of the die has an equal chance of turning up on each roll, the probability of landing on a particular number during any roll of the die is $\\normalsize \\frac{1}{6}$ . Since each flip of the coin has an equal chance of landing on heads or tails, the probability of landing on heads is equal to the probability of landing on tails, which is $\\normalsize \\frac{1}{2}$ . Therefore, the probability of getting a 2 and heads as the outcome is calculated", "# Thread: omg probability is so hard!! =[[ 1. ## omg probability is so hard!! =[[ can u show me how to do this step by step? Jerry has a CD case that contains 4 country music CDs, 1 rock-and-roll CD, 2 rap CDs, and 3 Tejano CDs. What is the probability of Jerry randomly selecting a Tejano CD and then, without replacing it, randomly selecting a rap CD from his case? A B C D 2. Originally Posted by whatxxever_2009 can u show me how to do this step by step? Jerry has a CD case that contains 4 country music CDs, 1 rock-and-roll CD, 2 rap CDs, and 3 Tejano CDs. What is the probability of Jerry randomly selecting a Tejano CD and then, without replacing it, randomly selecting a rap CD from his case? A B C D The probability of drawing a Tejano CD is as follows: There are 3 Tejano CDs, but we have 4+1+2+3=10 CDs total. So the probability of him picking a Tejano CD first would be $\\frac{3}{10}$. If he then picks a Rap CD without replacement, then the total number of possible CDs drops down to 9. Thus, the probability of him drawing a Rap CD after he draws a Tejano CD is $\\frac{3}{10}\\cdot\\frac{2}{9}=\\boxed{\\frac{1}{15}}$ Does this make a little more", "that the student is under 120cm. There are a total of 12 students under 120cm. The probability is \\frac{6}{12}=\\frac{1}{2} . 6. Rachel has 9 balls in a bag. 4 of the balls are blue and the other 5 are red. Rachel picks one ball, notes its colour and replaces the ball. She then picks a second ball. Find the probability the two balls she picks are the same colour. \\frac{16}{81} \\frac{41}{81} \\frac{25}{81} \\frac{40}{81} Rachel can pick two red balls or two blue balls. P(red \\ and \\ red) =\\frac{4}{9} \\times \\frac{4}{9}=\\frac{16}{81} P(blue \\ and \\ blue) =\\frac{5}{9} \\times \\frac{5}{9}=\\frac{25}{81} P(same \\ colour) =\\frac{16}{81}+\\frac{25}{81}=\\frac{41}{81} The total probability is \\frac{41}{81} . ### How to calculate probability GCSE questions 1. Jason picks one of the following cards at random. (a) Find the probability that Jason picks a H. (b) Find the probability that Jason picks an M or an A. (c) Find the probability that Jason does not pick an M. (3 marks) (a) \\frac{1}{11} (1) (b) \\frac{2}{11}+\\frac{2}{11}=\\frac{4}{11} (1) (c) \\frac{9}{11} (1) 2. (a) Yasmin designs a game where players must roll a dice and pick a card from a set of cards containing the numbers 1-10 once each. Players win if they roll a multiple of 3 and pick a card that is a multiple of 5. Find the probability", "What is the probability that Brian makes money on his first roll Brian plays a game in which two fair, six-sided dice are rolled simultaneously. For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money. What is the probability that Brian makes money on his first roll? To find the probability, do we need to find the even numbers only. There are 36 outcomes from the two dies. So is there an easy way to get the arrangements? Let the values of the dice in a roll be $a$ and $b$. You win money if ($1$) $a$ and $b$ are both even, ($2$) $a$ is even and $b$ is odd with $a>b$, or ($3$) $a$ is odd and $b$ is even with $a<b$. For ($1$) there are $3$ even numbers in $1,2,\\dots,6$, so there are $3^2=9$ ways for $a$ and $b$ to both be positive. For ($2$) consider the three cases cases: $a=2 \\implies b=1$; $a=4\\implies b=1,3$; and $a=6\\implies b=1,3,5$. Therefore, there are $6$ ways to get condition ($2$). Condition ($3$) is the same as ($2$) by symmetry, so the probability to win money is $$\\frac{3^2 + (2)(6)}{6^2} = \\frac{21}{36} = \\frac{7}{12}.$$ We can easily generalize this to any $n$-faced pair of", "$\\frac {1}{12}$ So, the probability of both events happening is $\\frac {1}{12}$ . 8. One bag contains 3 red apples and 1 green apple. The other bag contains 1 ham sandwich and 1 turkey sandwich. The probability of Chris randomly drawing a green apple and a ham sandwich is calculated below. $\\text{p(green apple,ham sandwich)}\\ =\\ \\frac{1}{4}\\ \\times\\ \\frac{1}{2}\\ =\\ \\frac{1}{8}$ 9. To find the probability of these two events happening at the same time, first find the individual probability of each. There are 52 cards in the deck Kayla has and 13 of them are diamonds. So, the probability of her pulling a diamond is $\\frac {13}{52}$ or $\\frac{1}{4}$ . There are six sides to the die that she is rolling, each side with a number from one to six without any number being repeated. So, the probability of her rolling a 2 is $\\frac {1}{6}$ . To find the compound probability of these events happening at the same time, multiply the probability of each event by the other. probability of diamond × probability of a 2 = probability of both $\\frac {1}{4}$ × $\\frac {1}{6}$ = $\\frac {1}{24}$ So, the probability of both events happening is $\\frac {1}{24}$ . 10. There are 6 combinations that sum to 7 and 2 combinations that sum to 11. This gives", "Linda makes $t$ throws and Dave at least $t+2$ throws is $(5/6)^{2t} \\cdot (1/6)$. Then, as the events for different $t$ are disjoint, $p$ is simply the sum of these probabilities over all $t$. Hence: \\begin{align*} p & = \\sum_{t=1}^\\infty \\left(\\frac 56\\right)^{2t} \\cdot \\frac 16 \\\\ & = \\frac 16 \\cdot \\left(\\frac 56\\right)^2 \\cdot \\sum_{x=0}^\\infty \\left(\\frac{25}{36}\\right)^x \\\\ & = \\frac 16 \\cdot \\frac{25}{36} \\cdot \\frac 1{1 - \\dfrac{25}{36}} \\\\ & = \\frac 16 \\cdot \\frac{25}{36} \\cdot \\frac{36}{11} \\\\ & = \\frac {25}{66} \\end{align*} Hence the probability we were supposed to compute is $1 - 2p = 1 - 2\\cdot \\frac{25}{66} = 1 - \\frac{25}{33} = \\frac 8{33}$, and the answer is $8+33 = \\boxed{041}$. ### Solution 2 Let $p$ be the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. (We will call this event \"a win\", and the opposite event will be \"a loss\".) Let both players roll their first die. With probability $\\frac 1{36}$, both throw a six and we win. With probability $\\frac{10}{36}$ exactly one of them throws a six. In this case, we win if the remaining player throws a six in their next throw, which happens with probability $\\frac 16$. Finally, with probability $\\frac{25}{36}$ none of them", "the cases unless he has a $7$ to begin with. We have $$\\frac{1}{6} > \\underline{\\frac{5,4,3}{36}} > \\frac{2.5}{36} > \\frac{2,1,0}{36}.$$ We count the underlined part's frequency for the numerator without upsetting the probability greater than it. Let $a$ be the roll we keep. We know $a\\leq3$ since $a=4$ would cause Jason to pick up all the dice. When $a=1$, there are $3$ choices for whether it is rolled $1$st, $2$nd, or $3$rd, and in this case the other two rolls have to be at least $6$ (or he would have only picked up $1$). This give $3 \\cdot 1^{2} =3$ ways. Similarly, $a=2$ gives $3 \\cdot 2^{2} =12$ because the $2$ can be rolled in $3$ places and the other two rolls are at least $5$. $a=3$ gives $3 \\cdot 3^{2} =27$. Summing together gives the numerator of $42$. The denominator is $6^3=216$, so we have $\\frac{42}{216}=\\boxed{\\textbf{(A) } \\frac{7}{36}}$. ## Solution 4 Note that Jason will roll $2$ dies if and only if the pairwise sum of the dies is greater than $7$. Suppose that Jason doesn't roll a $1$. This means that roll was $(1, 6, 6)$ which can be arranged in $3$ ways. Suppose that Jason doesn't roll a $2$. This means the roll was either $(2, 5/6, 5/6)$ which can be arranged in $12$ ways. Suppose", "is the solution given. The probability that Adam and Bella don’t have the same number in their respective minds will be 4/5.Now let us include Christopher in the scene. The probability that Christopher doesn’t have a digit same as either Adam or Bella will be 3/5.Now include Drew, the probability that she does not have any number which is same as either Adam or Bella or Christopher will be 2/5.So the combined probability that none of them has the same number in mind will be:4/5 * 3/5 * 2/5 = 24/125.Now the probability that they have the same number in mind will be:1 – 24/125 = 101/125 – user172377 Oct 21 '16 at 0:36 • Which is correct and why? Where do I go wrong? – user172377 Oct 21 '16 at 0:36 • That analysis correctly answers a different question. That answers the question \"what is the probability that at least two of them are thinking of the same number\". Very different. – lulu Oct 21 '16 at 0:37 You are not wrong. The answer is not $101/125$. The probability every one of the other three is mindful of the same number as Adam is $1/5^3$", "a different way. The probability of getting tails is $\\frac12$. The probability of rolling 4 or less is $\\frac23$. Both of these are independent of one another so the probability of getting tails and rolling 4 or less is $\\frac12\\times\\frac23=\\frac13$. If you didn't get tails and a number less than 4 at the same time, then you either obtained a head or a number greater than 4. So the odds of obtaining a head or a number greater than 4 is $1-\\frac13=\\frac23$. -", "so the probability of choosing a club is $\\frac{1}{4}$ . The complement is the probability of not choosing a club, and it is $1-\\frac{1}{4}=\\frac{3}{4} \\ or \\ 75 \\%$ 3. There are 4 factors of 10: 1, 2, 5, and 10. Therefore the probability of rolling a factor of 10 on a 20-sided die is $\\frac{4}{20}$ . The complement is the probability of not rolling one of the four factors: $1-\\frac{4}{20}=\\frac{16}{20}$ 4. Let $P(F)$ be the probability of pulling a fruit-flavor, then $P(F^\\prime)$ is the probability of not pulling a fruit flavor (and getting something disgusting instead!). There are 10 fruit-flavored beans in the bag of 100 beans, so: $P(F)&=\\frac{10}{100} \\\\P(F^\\prime)&=1-P(F)=\\frac{90}{100}$ Therefore, you have a 90% probability of being unhappy with your choice. 5. The factors of ten or twelve are: 1, 2, 3, 4, 5, 6, 10, and 12. Since all six numbers from a standard die are in that set, the probability of rolling one of them is 100% or 1.0. Therefore, the probability of not rolling one of them is $1-1=0$ . #### Practice 1. If you randomly pull a single card from a standard deck, what is the probability that the card is anything other than a king? 2. What is the probability of not pulling a face card from a standard deck? 3.", "die ${2,2,4,4,9,9}$: The probability of Charlie rolling a 1 is 2 from 6 and the probability of Alison rolling a greater number is 6 from 6. The probability of Charlie rolling a 6 is 2 from 6 and the probability of Alison rolling a greater number is 2 from 6 The probability of Charlie rolling a 8 is 2 from 6 and the probability of Alison rolling a greater number is 2 from 6 Therefore the total number of ways Alison can win is (6 + 2 + 2) = 10 out of 18. The blue die is ${3,3,5,5,7,7}$. The probability of Charlie rolling a 1 is 2 from 6 and the probability of Alison rolling a greater number is 6 from 6. The probability of Charlie rolling a 6 is 2 from 6 and the probability of Alison rolling a greater number is 2 from 6. The probability of Charlie rolling a 8 is 2 from 6 and the probability of Alison rolling a greater number is 0 from 6 Therefore the total probability of Alison winning is (6 + 2 + 0) = 8 out of 18. There are similar arguments to show that Alison should choose the blue die if Charlie chooses the green die.", "you choose one card from a deck of cards, you are changing the number of cards left and therefore the probabilities of each outcome for the next draw. Picking one card followed by a second card from a deck of cards are not independent events. 2. Events A and B are shown below. Event A – rolling a die and landing on an even number. Event B – picking a diamond from a deck of cards. Felicity rolls a die and picks a card from a deck of cards. Find the probability of both events A and B occurring. \\frac{3}{4} \\frac{1}{8} \\frac{1}{2} \\frac{1}{4} The probability of a die landing on an even number is \\frac{1}{2}. The probability of picking a diamond from a deck of cards is \\frac{1}{4}. The probability of events A and B both occurring is \\frac{1}{2} \\times \\frac{1}{4} = \\frac{1}{8}. 3. The probability that my train is late on any given day is 0.3. Find the probability that my train is late three days in a row. 0.3 0.9 0.027 0.27 The probability that the train is late on any one day is 0.3. The probability that the train is late on day 1 and day 2 and day 3 is 0.3 \\times 0.3 \\times 0.3 = 0.027. 4. Sam has different options for her", "\\approx 0.5 \\%$$ Is this correct? 18. Dec 17, 2003 ### chroot Staff Emeritus Whoooops, thanks master_coda and Stephen, you guys are quick! - Warren 19. Dec 17, 2003 ### Doug The horse may be cold but there's another way to beat it The probability of getting a 6 on the first roll is: $$\\frac{1}{6}$$ The probability of getting a 6 on the second roll given that we didn't get a 6 on the first is: $$\\frac{5}{6}\\cdot\\frac{1}{6}$$ The probability of getting a 6 on the third roll given that we didn't get a 6 on the first two is: $$\\frac{5}{6}\\cdot\\frac{5}{6}\\cdot\\frac{1}{6}$$ Since we don't care which roll we get the 6 we take the sum: $$\\frac{1}{6}+\\frac{5}{6}\\cdot\\frac{1}{6}+\\frac{5}{6}\\cdot\\frac{5}{6}\\cdot\\frac{1}{6} =\\frac{1}{6}+\\frac{5}{36}+\\frac{25}{216} =\\frac{36}{216}+\\frac{30}{216}+\\frac{25}{216} =\\frac{91}{216}\\approx 42 \\%$$ Doug 20. Dec 18, 2003", "that Jason doesn't roll a $3$. This means that the roll was either $(3, 4/5/6, 4/5/6)$ which can be arranged in $27$ ways. Jason can't roll a $4, 5, 6$ because he can't get a pairwise sum of $7$ from those numbers. Thus he rolls two dies exactly $27 + 12 + 3 = 42$ out of the $6^3$ possible rolls. So the answer is $\\frac{42}{216} = \\boxed{\\textbf{(A) } \\frac{7}{36}}$. ~coolmath_2018 ## Video Solutions ### Video Solution 1 https://youtu.be/B8pt8jF04ZM - Happytwin", "User: Probability of events that are not mutually exclusive If two events, A and B, are not mutually exclusive, then the probabilitythat either A or B occurs is the sum of their probabilities decreased by the probability of both occurring. P(A or B)=p(A)+p(B)-p(A and B) Of 30 boys, 12 are on the Honor Roll, 10 play sports, and 15 are on the Honor Roll and play sport. What is the probability that a randomly selected student plays sports or is on the Honor Roll? Since some students play varsity sports and are on the Honor Roll, the events are not mutually exclusive . $p(sports)=\\frac{10}{30}$ $p(Honor\\;Roll)=\\frac{12}{30}$ $p(sports\\;and\\;Honor\\;Roll)=\\frac{15}{30}$ p(sport or Honor Roll) = p(sport)+p(Honor Roll)-p(sport and Honor Roll) = $\\frac{10}{30}+\\frac{12}{30}-\\frac{15}{30}$ = $\\frac{7}{30}$", "lunch. The probabilities of her choosing the different options are shown below. Find the probability that Sam chooses a cheese sandwich and grapes. 0.56 1.5 0.15 0.06 The probability that Sam chooses a cheese sandwich is 0.7 and the probability that Sam chooses grapes is 0.8. The probability that Sam chooses a cheese sandwich and grapes is 0.7 \\times 0.8=0.56. 5. Sheila and Pete are playing darts. The probability that Sheila hits a bullseye is 0.1 and the probability that they both hit a bullseye is 0.02. Find the probability that Pete hits a bullseye. 0.002 0.2 0.12 0.3 If we call the probability that Pete hits a bullseye \\text{P(P)} then \\begin{aligned} 0.1 \\times \\text{P(P)}&=0.02\\\\\\\\ \\text{P(P)}&=0.02 \\div 0.1\\\\\\\\ &=0.2 \\end{aligned} 6. Marwa rolls a biased die twice. The probability that she gets two sixes is \\frac{4}{25}. Find the probability that Marwa rolls a six on any one roll. \\frac{2}{25} \\frac{2}{5} \\frac{8}{25} \\frac{16}{125} If we call the probability of rolling a six P(6) then \\begin{aligned} \\text{P(6)} \\times \\text{P(6)}&=\\frac{4}{25}\\\\\\\\ \\text{(P(6))}^{2}&=\\frac{4}{25}\\\\\\\\ \\text{P(6)}&=\\sqrt{\\frac{4}{25}}\\\\\\\\ \\text{P(6)}&=\\frac{2}{5} \\end{aligned} ### Independent events GCSE questions 1. Eleri rolls a die and tosses two coins. Find the probability that Eleri gets a 4 on the die and heads on both coins. (2 marks) \\frac{1}{6} \\times \\frac{1}{2} \\times \\frac{1}{2} (1) =\\frac{1}{24} (1) 2. (a) A marble is picked from", "Bitesize Probabilities can be written as fractions, decimals or percentages on a scale from 0 ... There is one way of rolling a 4 and there are six possible outcomes, so the ... To find the probability of the event of rolling an odd number on a dice, find ... What is the probability of selecting a vowel at random from the word PROBABILITY? For more information, see Probability - Edexcel - Revision 2 - GCSE Maths - BBC Bitesize Qalaxia Info Bot 0 Ways to arrange colors (video) | Khan Academy Thinking about how many ways you can pick four colors from a group of 6. ... B-A-R-N is a completely different word than N-A-B-R, even though the same letters were ... and bananas is exactly the same salad as one made of oranges, bananas, .... And I can get any color when I mix 3 basic colors, Red, Green and Blue and ... Qalaxia Info Bot 0 I found an answer from www.history.com 9 Things You May Not Know About Gerald Ford - HISTORY Jul 12, 2013 ... 1. His birth name wasn't Gerald R. Ford. Ford was born Leslie Lynch King ... Upon graduation, Ford received offers from two professional football ... On December 18, 1944, the Monterey was one of many Navy ships ...", "divisible by 5. = $$\\frac{2}{12}$$ = $$\\frac{1 \\cdot 2}{6 \\cdot 2}$$ = $$\\frac{1}{6}$$ The probability of picking a number that is divisible by 5 is $$\\frac{1}{6}$$. This event is unlikely to occur. Find each probability. Write your answer in the simplest form. (Example 2) Question 6. Spinning a spinner that has 5 equal sections marked 1 through 5 and landing on an even number. Use a tree diagram to find the sample space. P(land on even number) = $$\\frac{2}{5}$$ Question 7. Picking a diamond from a standard deck of playing cards which has 13 cards in each of four suits: spades, hearts, diamonds and clubs. A standard deck of playing cards has 13 cards in diamonds. Find the probability of picking a diamond. = $$\\frac{13}{52}$$ = $$\\frac{1 \\cdot 13}{4 \\cdot 13}$$ = $$\\frac{1}{4}$$ The probability of picking a diamond from a standard deck of playing cards is $$\\frac{1}{4}$$. Use the complement to find each probability. (Example 3) Question 8. What is the probability of not rolling a 5 on a standard number cube? The sum of the probabilities of an event and ¡ts complement equals 1. First, find the probability of rolling a number 5 on a standard number cube. Sample space for a standard number cube is {1, 2, 3, 4, 5, 6}. = $$\\frac{1}{6}$$ The probability", "minds will be 4/5.Now let us include Christopher in the scene. The probability that Christopher doesn’t have a digit same as either Adam or Bella will be 3/5.Now include Drew, the probability that she does not have any number which is same as either Adam or Bella or Christopher will be 2/5.So the combined probability that none of them has the same number in mind will be: 4/5 * 3/5 * 2/5 = 24/125.Now the probability that they have the same number in mind will be:1 – 24/125 = 101/125 That is the solution to: “What is the probability that some of them are mindful of the same number?” A different problem entirely. “All of them do it” is not the complement of “None dome of them do it”.", "and Dave at least $t+2$ throws is $(5/6)^{2t} \\cdot (1/6)$. Then, as the events for different $t$ are disjoint, $p$ is simply the sum of these probabilities over all $t$. Hence: \\begin{align*} p & = \\sum_{t=1}^\\infty \\left(\\frac 56\\right)^{2t} \\cdot \\frac 16 \\\\ & = \\frac 16 \\cdot \\left(\\frac 56\\right)^2 \\cdot \\sum_{x=0}^\\infty \\left(\\frac{25}{36}\\right)^x \\\\ & = \\frac 16 \\cdot \\frac{25}{36} \\cdot \\frac 1{1 - \\dfrac{25}{36}} \\\\ & = \\frac 16 \\cdot \\frac{25}{36} \\cdot \\frac{36}{11} \\\\ & = \\frac {25}{66} \\end{align*} Hence the probability we were supposed to compute is $1 - 2p = 1 - 2\\cdot \\frac{25}{66} = 1 - \\frac{25}{33} = \\frac 8{33}$, and the answer is $8+33 = \\boxed{041}$. ### Solution 2 Let $p$ be the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. (We will call this event \"a win\", and the opposite event will be \"a loss\".) Let both players roll their first die. With probability $\\frac 1{36}$, both throw a six and we win. With probability $\\frac{10}{36}$ exactly one of them throws a six. In this case, we win if the remaining player throws a six in their next throw, which happens with probability $\\frac 16$. Finally, with probability $\\frac{25}{36}$ none of them throws a six. Now" ]

[ "out with his boys during Family Math. Mike’s videos show his boys thinking through simpler discrete cases of the problem and their eventual case-by-case solution to the nxnxn question. The last video on the page gives a lovely alternative solution completely bypassing the case-by-case analyses. This is also “Solution 1” on Alexander’s page. EXTENDING THE PROBLEM: When I first encountered the problem, I wanted to think about it before reading any solutions. As with Mike’s boys, I was surprised early on when the probability for a 2x2x2 cube was $\\displaystyle \\frac{1}{2}$ and the probability for a 3x3x3 cube was $\\displaystyle \\frac{1}{3}$. That was far too pretty to be a coincidence. My solution exactly mirrored the nxnxn case-by-case analysis in Mike’s videos: the probability of rolling a painted red face up from a randomly selected smaller cube is $\\displaystyle \\frac{1}{n}$. Surely something this simple and clean could be generalized. Since a cube can be considered a “3-dimensional square”, I re-phrased Alexander’s question into two dimensions. The trickier part was thinking what it would mean to “roll” a 2-dimensional shape. The outside of an nxn square is painted red and is chopped into $n^2$ unit squares. The latter are thoroughly mixed up and put into a bag. One small square is withdrawn at random from the bag and spun on", "of Noah’s 50 results is selected at random, what is the probability that the coin was heads? 3. If one of Noah’s 50 results is selected at random, what is the probability that the number cube was 5? ### Solution For access, consult one of our IM Certified Partners. ### Problem 3 A student surveys 30 people as part of a project for a statistics class. Here are the survey questions. • Are you left-handed or right-handed? • Are you left-eye dominant or right-eye dominant? The results of the survey are summarized in the two-way table. right-eye dominant left-eye dominant right-handed 14 11 left-handed 3 2 What is the probability that a person from the survey chosen at random is right-handed under the condition that they are right-eye dominant? A: $$\\frac{25}{30}$$ B: $$\\frac{14}{30}$$ C: $$\\frac{14}{25}$$ D: $$\\frac{14}{17}$$ ### Solution For access, consult one of our IM Certified Partners. ### Problem 4 Priya flips a fair coin and then rolls a standard number cube. What is the probability that she rolled a 3 under the condition that she flipped heads? A: $$\\frac{1}{2}$$ B: $$\\frac{1}{6}$$ C: $$\\frac{1}{12}$$ D: $$\\frac{3}{12}$$ ### Solution For access, consult one of our IM Certified Partners. (From Unit 8, Lesson 8.) ### Problem 5 Andre flips one fair coin and then flips another fair coin. 1.", "Alice (B) Ben (C) Cindy (D) Dave (E) Ellen AMC 8, 2004, Problem 20 Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are $6$ empty chairs, how many people are in the room? (A) $12$ (B) $18$ (C) $24$ (D) $27$ (E) $36$ AMC 8, 2004, Problem 21 Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ (C) $\\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2004, Problem 23 Tess runs counterclockwise around rectangular block $JKLM$. She lives at corner $J$. Which graph could represent her straight-line distance from home? AMC 8, 2003, Problem 12 When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by $6 ?$ (A) $\\frac{1}{3}$ (B) $\\frac{1}{2}$ (C) $\\frac{2}{3}$ (D) $\\frac{5}{6}$ (E) $1$ AMC 8, 2003, Problem 13 Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The", "a “3-dimensional square”, I re-phrased Alexander’s question into two dimensions. The trickier part was thinking what it would mean to “roll” a 2-dimensional shape. The outside of an nxn square is painted red and is chopped into $n^2$ unit squares. The latter are thoroughly mixed up and put into a bag. One small square is withdrawn at random from the bag and spun on a flat surface. What is the probability that the spinner stops with a red side facing you? Shown below is a 4×4 square, but in all sizes of 2-dimensional squares, there are three possible types in the bag: those with 2, 1, or 0 sides painted. I solved my first variation case by case. In any nxn square, • There are 4 corner squares with 2 sides painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4}{n^2} \\cdot \\frac{2}{4} = \\frac{2}{n^2}$. • There are $4(n-2)$ edge squares not in a corner with 1 side painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4(n-2)}{n^2} \\cdot \\frac{1}{4} = \\frac{n-2}{n^2}$. • All other squares have 0 sides painted, so the probability of picking one of those squares and then spinning a red side is 0. • Adding the probabilities for", "Adding the probabilities for the separate cases gives the total probability: $\\displaystyle \\frac{2}{n^2}+\\frac{n-2}{n^2}=\\frac{1}{n}$ After reading Mike’s and Alexander’s posts, I saw a much easier approach. • Paint all 4 edges of an nxn square, and divide each painted edge into n painted unit segments. This creates $4 \\cdot n$ total painted small segments. • Decompose the nxn original square into $n^2$ unit squares. Each unit square has 4 edges giving $4 \\cdot n^2$ total edges. • Because every edge of every unit square is equally likely to be spun, the total probability of randomly selecting a smaller square and spinning a red side is $\\displaystyle \\frac{4n}{4n^2}=\\frac{1}{n}$. The dimensions of the “square” don’t seem to matter! WARNING: Oliver Wendell Holmes noted, “A mind that is stretched by a new experience can never go back to its old dimensions.” The math after this point has the potential to stretch… EXTENDING THE PROBLEM MORE: I now wondered whether this was a general property. In the 2-dimensional square, 1-dimensional edges were painted and the square was spun to find the probability of a red edge facing. With the originally posed cube, 2-dimensional faces were painted and the cube was tossed to find the probability of an upward-facing red face. These cases suggest that when a cube of some dimension with edge length", "Adding the probabilities for the separate cases gives the total probability: $\\displaystyle \\frac{2}{n^2}+\\frac{n-2}{n^2}=\\frac{1}{n}$ After reading Mike’s and Alexander’s posts, I saw a much easier approach. • Paint all 4 edges of an nxn square, and divide each painted edge into n painted unit segments. This creates $4 \\cdot n$ total painted small segments. • Decompose the nxn original square into $n^2$ unit squares. Each unit square has 4 edges giving $4 \\cdot n^2$ total edges. • Because every edge of every unit square is equally likely to be spun, the total probability of randomly selecting a smaller square and spinning a red side is $\\displaystyle \\frac{4n}{4n^2}=\\frac{1}{n}$. The dimensions of the “square” don’t seem to matter! WARNING: Oliver Wendell Holmes noted, “A mind that is stretched by a new experience can never go back to its old dimensions.” The math after this point has the potential to stretch… EXTENDING THE PROBLEM MORE: I now wondered whether this was a general property. In the 2-dimensional square, 1-dimensional edges were painted and the square was spun to find the probability of a red edge facing. With the originally posed cube, 2-dimensional faces were painted and the cube was tossed to find the probability of an upward-facing red face. These cases suggest that when a cube of some dimension with edge length", "Adding the probabilities for the separate cases gives the total probability: $\\displaystyle \\frac{2}{n^2}+\\frac{n-2}{n^2}=\\frac{1}{n}$ After reading Mike’s and Alexander’s posts, I saw a much easier approach. • Paint all 4 edges of an nxn square, and divide each painted edge into n painted unit segments. This creates $4 \\cdot n$ total painted small segments. • Decompose the nxn original square into $n^2$ unit squares. Each unit square has 4 edges giving $4 \\cdot n^2$ total edges. • Because every edge of every unit square is equally likely to be spun, the total probability of randomly selecting a smaller square and spinning a red side is $\\displaystyle \\frac{4n}{4n^2}=\\frac{1}{n}$. The dimensions of the “square” don’t seem to matter! WARNING: Oliver Wendell Holmes noted, “A mind that is stretched by a new experience can never go back to its old dimensions.” The math after this point has the potential to stretch… EXTENDING THE PROBLEM MORE: I now wondered whether this was a general property. In the 2-dimensional square, 1-dimensional edges were painted and the square was spun to find the probability of a red edge facing. With the originally posed cube, 2-dimensional faces were painted and the cube was tossed to find the probability of an upward-facing red face. These cases suggest that when a cube of some dimension with edge length", "# Thread: omg probability is so hard!! =[[ 1. ## omg probability is so hard!! =[[ can u show me how to do this step by step? Jerry has a CD case that contains 4 country music CDs, 1 rock-and-roll CD, 2 rap CDs, and 3 Tejano CDs. What is the probability of Jerry randomly selecting a Tejano CD and then, without replacing it, randomly selecting a rap CD from his case? A B C D 2. Originally Posted by whatxxever_2009 can u show me how to do this step by step? Jerry has a CD case that contains 4 country music CDs, 1 rock-and-roll CD, 2 rap CDs, and 3 Tejano CDs. What is the probability of Jerry randomly selecting a Tejano CD and then, without replacing it, randomly selecting a rap CD from his case? A B C D The probability of drawing a Tejano CD is as follows: There are 3 Tejano CDs, but we have 4+1+2+3=10 CDs total. So the probability of him picking a Tejano CD first would be $\\frac{3}{10}$. If he then picks a Rap CD without replacement, then the total number of possible CDs drops down to 9. Thus, the probability of him drawing a Rap CD after he draws a Tejano CD is $\\frac{3}{10}\\cdot\\frac{2}{9}=\\boxed{\\frac{1}{15}}$ Does this make a little more", "that the student is under 120cm. There are a total of 12 students under 120cm. The probability is \\frac{6}{12}=\\frac{1}{2} . 6. Rachel has 9 balls in a bag. 4 of the balls are blue and the other 5 are red. Rachel picks one ball, notes its colour and replaces the ball. She then picks a second ball. Find the probability the two balls she picks are the same colour. \\frac{16}{81} \\frac{41}{81} \\frac{25}{81} \\frac{40}{81} Rachel can pick two red balls or two blue balls. P(red \\ and \\ red) =\\frac{4}{9} \\times \\frac{4}{9}=\\frac{16}{81} P(blue \\ and \\ blue) =\\frac{5}{9} \\times \\frac{5}{9}=\\frac{25}{81} P(same \\ colour) =\\frac{16}{81}+\\frac{25}{81}=\\frac{41}{81} The total probability is \\frac{41}{81} . ### How to calculate probability GCSE questions 1. Jason picks one of the following cards at random. (a) Find the probability that Jason picks a H. (b) Find the probability that Jason picks an M or an A. (c) Find the probability that Jason does not pick an M. (3 marks) (a) \\frac{1}{11} (1) (b) \\frac{2}{11}+\\frac{2}{11}=\\frac{4}{11} (1) (c) \\frac{9}{11} (1) 2. (a) Yasmin designs a game where players must roll a dice and pick a card from a set of cards containing the numbers 1-10 once each. Players win if they roll a multiple of 3 and pick a card that is a multiple of 5. Find the probability", "# CTK Insights • ## Pages 20 May ### Probabilities in a Painted Cube A wooden cube - after being painted all over - has been cut into $3\\times 3\\times 3$ smaller cubes. These were thoroughly mixed in a bag, from which one was produced and tossed. What is the probability that a painted side turned up? On impulse, one would approach the problem in a more or less standard way. There are $8$ corner cubes with $3$ painted sides, $12$ mid-edge cube with $2$ painted sides, and $6$ face-central cubes with only $1$ side painted. The total probability is then $\\displaystyle\\frac{8}{27}\\cdot\\frac{3}{6}+\\frac{12}{27}\\cdot\\frac{2}{6}+\\frac{6}{27}\\cdot\\frac{1}{6}=\\frac{1}{3}.$ Now generalize: cut the cube into $n\\times n\\times n$, $n\\gt 1$ smaller cubes and ask the same question. The problem is not awfully difficult but needs some figuring out. Following the foregoing pattern, we eventually arrive at $\\displaystyle\\frac{8}{n^3}\\cdot\\frac{3}{6}+\\frac{12\\cdot (n-2)}{n^3}\\cdot\\frac{2}{6}+\\frac{6\\cdot (n-2)^2}{n^3}\\cdot\\frac{1}{6}$ This expression simplifies, as you can verify, to $\\displaystyle\\frac{1}{n},$ which, by the way, confirms the answer for $n=3.$ The above is a useful and not too difficult exercise, but there is a delightful shortcut that avoids most of the counting of cubes and their sides. $n^3$ cubes have a total of $6\\cdot n^3$ sides. Of these, $6\\cdot n^2$ are painted. All sides have the same probability of turning up, therefore, a painted side will turn up", "a 6×6×6 cube. Picture a plane \"bisecting\" the cube through two opposite vertices. The desirable outcomes are in the the \"larger half\". This is a tetradedon with sides of 6. It contains: $\\frac{6\\cdot7\\cdot8}{3!} = 56$ outcomes. Roll four times. .There are: . $6^4$ outcomes. The ouctomes are displayed in a 6×6×6×6 hypercube. The \"larger half\" contains: . $\\frac{6\\cdot7\\cdot8\\cdot9}{4!} \\:=\\:126$ outcomes. Let's skip to six rolls. .There are: . $6^6$ outcomes. The outcomes are in a 6×6×6×6×6×6 hypercube. The \"larger half\" contains: . $\\frac{6\\cdot7\\cdot8\\cdot9\\cdot10\\cdot11}{6!} \\;=\\;462$ outcomes. Therefore: . $\\text{Prob} \\;=\\;\\frac{462}{46,656} \\;=\\;\\frac{77}{7,776}$ 4. Very interesting Soroban! Now we can generalize, if we toss the dice $n$ times, the resulting probability would be $\\frac{\\displaystyle\\dbinom{n+5}{5}}{\\displaystyle 6^n}$", "# Math Help - Cube probability 1. ## Cube probability On choosing three vertex of a cube at random, what's the probability that the three are in a same face? 2. I would think it would be $(\\frac{1}{6})(\\frac{3}{4})=\\frac{1}{8}$ It may be more involved than that. 3. Originally Posted by Krizalid On choosing three vertex of a cube at random, what's the probability that the three are in a same face? The counting principle, of course. The first pick if free, we may pick any vertex. There are 8 vertices on the cube, we've picked one leaving 7. Of these only 6 share a face with the one we've picked. So we have $1 \\cdot \\frac{6}{7}$ so far. Now, we've picked two vertices. 1) If the two vertices are on the same edge, there are 4 possible of 6 vertices left to pick, so the probability will be $1 \\cdot \\frac{6}{7} \\cdot \\frac{4}{6} = \\frac{4}{7}$ 2) If the two vertices are not on the same edge, they are on a diagonal across one face. In this case there is only one possible face that these could be on, so there are only two possible vertices out of 6, so the probability will be $1 \\cdot \\frac{6}{7} \\cdot \\frac{2}{6} = \\frac{2}{7}$ Only one of these choices is possible, so we", "balloons and 3 purple balloons. Jeff randomly draws one gumball and then one balloon. What is the probability that he will draw a red gumball and a purple balloon? A. $\\frac{3}{10}$ B. $\\frac{2}{15}$ C. $\\frac{1}{5}$ D. $\\frac{1}{10}$ 12. Gary has 1 red apple, 1 green apple, and 4 yellow apples in a sack. He also has 2 packages of peanuts, 2 packages of almonds, and 2 packages of cashews in another sack. If he reaches into each sack without looking, what is the probability that he will grab a red apple and a package of cashews? A. $\\frac{1}{18}$ B. $\\frac{1}{4}$ C. $\\frac{1}{12}$ D. $\\frac{1}{36}$ 1. A 2. A 3. B 4. A 5. B 6. A 7. B 8. B 9. D 10. B 11. C 12. A ## Explanations 1. Since each of the six sides of the die has an equal chance of turning up on each roll, the probability of landing on a particular number during any roll of the die is $\\normalsize \\frac{1}{6}$ . Since each flip of the coin has an equal chance of landing on heads or tails, the probability of landing on heads is equal to the probability of landing on tails, which is $\\normalsize \\frac{1}{2}$ . Therefore, the probability of getting a 2 and heads as the outcome is calculated", "plastic manufacturing technique (the game is made by Vic-Toy, “A Division of Invicta Plastics Ltd.”). 3. Alison I’m so pleased that my tweet about Ox Blocks generated some interest! I’ve been playing around with probability simulations to support the activities on http://nrich.maths.org/probability, so perhaps I ought to take the time to write an online version of Ox Blocks… • Peter Rowlett Alison: if you were pleased to pique my interest, try this one: 4. Alan Depends how you roll it. If you roll such that the hole stays horizontal then blank will never come up. If you roll such that the hole goes end over end then blank will come up from time to time. Did you carefully randomise these orientations?!!! • Ralph Hartley You can bias a regular die the same way. To work, any die has to bounce around enough to approach its equilibrium distribution. Making one moment of inertia different than the other two makes the probability of rolling from “axis vertical” to “axis horizontal” different than “horizontal-horizontal”, but the probability of each transition is still the same as its reverse. The equilibrium probabilities are still uniform. Also, note that the moments of inertia are not *much* different. The maximum ratio (achieved by a thin walled tube) is only 1.2. The purpose of the holes", "$\\frac {1}{12}$ So, the probability of both events happening is $\\frac {1}{12}$ . 8. One bag contains 3 red apples and 1 green apple. The other bag contains 1 ham sandwich and 1 turkey sandwich. The probability of Chris randomly drawing a green apple and a ham sandwich is calculated below. $\\text{p(green apple,ham sandwich)}\\ =\\ \\frac{1}{4}\\ \\times\\ \\frac{1}{2}\\ =\\ \\frac{1}{8}$ 9. To find the probability of these two events happening at the same time, first find the individual probability of each. There are 52 cards in the deck Kayla has and 13 of them are diamonds. So, the probability of her pulling a diamond is $\\frac {13}{52}$ or $\\frac{1}{4}$ . There are six sides to the die that she is rolling, each side with a number from one to six without any number being repeated. So, the probability of her rolling a 2 is $\\frac {1}{6}$ . To find the compound probability of these events happening at the same time, multiply the probability of each event by the other. probability of diamond × probability of a 2 = probability of both $\\frac {1}{4}$ × $\\frac {1}{6}$ = $\\frac {1}{24}$ So, the probability of both events happening is $\\frac {1}{24}$ . 10. There are 6 combinations that sum to 7 and 2 combinations that sum to 11. This gives", "### Home > GC > Chapter 2 > Lesson 2.1.1 > Problem2-9 2-9. Mei puts the shapes below into a bucket and asks Brian to pick one out. 1. What is the probability that he pulls out a quadrilateral with parallel sides? Identify the shapes that are quadrilaterals with a $Q$. Identify the shapes that have parallel sides with a $P$. Write a fraction where the numerator is the number of shapes that have both parallel sides and four sides, and where the denominators the total number of shapes. $\\frac{3}{6}=\\frac{1}{2}$ 2. What is the probability that he pulls out a shape with rotation symmetry?", "to count those cases? – Matthew Conroy Feb 8 '18 at 4:46 • Three answers with three different results...lol its obvious probability is not a science not a clear one at least lol – Isham Feb 8 '18 at 5:41 • Yeah, counting problems always kick my ass because the you can get 3 different answers that are all absolutely correct. – Skulloking Feb 8 '18 at 5:58 For a) think about this: how many pairs there are to make a roll of the kind $aaabb$, that is, how many pairs of $ab$? There are $6\\cdot 5=30$ kinds of these pairs. However each throw of the kind $aaabb$ can appear in any order (by example as $ababa$ or $abbaa$), that is, for each pair $ab$ there are $\\binom52=10$ different ways that a throw can be ordered. Thus the total probability is $30\\cdot10\\cdot \\frac1{6^5}$ For the part b) it happen similarly: you want to choose $5$ different numbers of $6$, and these numbers can appear in any order, that is, it can appear as $abcde$ or $acdeb$. How many distinct groups of $5$ numbers, taken from $6$ exists? there are $\\binom65=6$, and each one can appear in $5!$ different ways, then the probability is $6\\cdot 5!\\cdot\\frac1{6^5}$. In part b) you have the same result than me but it is", "all 12 drummers before we draw the partridge). It’s also clear that there are $78!$ (more than $10^{115}$) possible draws of all figurines, so attempting any brute-force solution is a non-starter. As noted in the introduction, across all $78!$ possible sequences there is an equal number of sequences with the partridge in any of the 78 possible positions. This means that drawing 5 figurines, say, is exactly as likely as drawing 75 figurines. Both size draws occur with probability $\\frac{1}{78}$. These numbers are all quite big, so let’s work with fewer days of Christmas to get a feel for the problem. What about $\\mathop{\\mathbb{E}}[\\text{M}_{1}]$? There’s a single partridge and no other figures, so the answer is $0$. Not very interesting, so let’s add the two turtle doves and look at $\\mathop{\\mathbb{E}}[\\text{M}_{2}]$. Now there are three figurines in total and $3!$ possible draws: $P_1, T_1, T_2\\\\P_1, T_2, T_1\\\\ T_1, P_1, T_2\\\\ T_2, P_1, T_1\\\\T_1, T_2, P_1\\\\ T_2, T_1, P_1$ There are two draws with no turtle-doves, two with a one turtle-dove, and two with two turtle-doves. As mentioned above, each size of draw is equally likely (the probability is $\\frac{1}{6}$) so: $\\mathop{\\mathbb{E}}[\\text{M}_{2}] = 0 \\cdot \\frac{2}{6} + 1 \\cdot \\frac{2}{6} + 2 \\cdot \\frac{2}{6} = 0 + \\frac{1}{3} + \\frac{2}{3} = 1$ As well as providing a tiny scrap", "a 6. Geometrically, this last shell is three faces of the cube. Each previous shell can also be viewed as having three faces (with the possible exception of $S_1$.) To be more precise, $$S_i = \\{(a,b,c)|\\max\\{a,b,c\\}=i\\}$$ for $i =1\\ldots6$. The probability of tossing into shell $i$ is $$P(S_i)=\\frac{i^3-(i-1)^3}{6^3}$$ Each shell $S_i$ has subsets $S_{i,j}$ where $j$ is the second highest roll (possibly equal to $i$). Geometrically, $S_{i,j}$ is composed of line segments running along the three faces of $S_i$. $S_{i,i}$ is composed of the three edges converging to the main corner in $S_i$. For $j$ less than $i$, $S_{i,j}$ is composed of three \"arrows\", one along each face of $S_i$, each pointing to the main corner of $S_i$. The image below illustrates the cube of possilbe attacker rolls, $S_6$, $S_{6,6}$, and $S_{6,3}$. We find that for the attacker's roll, with $j$ less than $i$, \\begin{align*} P(S_{i,i}) & = \\frac{3i-2}{i^3-(i-1)^3}\\frac{i^3-(i-1)^3}{6^3}=\\frac{3i-2}{6^3}\\\\ P(S_{i,j}) & = \\frac{6j-3}{i^3-(i-1)^3}\\frac{i^3-(i-1)^3}{6^3}=\\frac{6j-3}{6^3}\\ \\end{align*} Now we begin to consider the defender's tosses, conditional upon what $S_{i,j}$ the attacker has rolled into. Similar to the cube that we envisioned for the attacker, the defender has a $6\\times 6$ square of possible rolls. Let's start counting the number of armies that the attacker loses. Let $X$ be this random variable, which can take values $0$, $1$, or $2$. Suppose the", "number cube has six faces numbered 1 through 6. If the cube is rolled twice, what is the probability that at least one of the rolls will result in a number greater than 4? (A) $$\\frac{2}{9}$$ (B) $$\\frac{1}{3}$$ (C) $$\\frac{4}{9}$$ (D) $$\\frac{5}{9}$$ (E) $$\\frac{2}{3}$$ OA: D The probability that at least one of the rolls will result in a number greater than 4 can be found out by taking these 3 cases Case 1: {5,6} on 1st roll and {1,2,3,4} on 2nd roll $$Probability_{case1} =\\frac{2}{6}*\\frac{4}{6}=\\frac{8}{36}$$ Case 2: {1,2,3,4} on 1st roll and {5,6} on 2nd roll $$Probability_{case2} =\\frac{4}{6}*\\frac{2}{6}=\\frac{8}{36}$$ Case 3: {5,6} on 1st roll and {5,6} on 2nd roll $$Probability_{case3} =\\frac{2}{6}*\\frac{2}{6}=\\frac{4}{36}$$ $$Total Probability = Probability_{case1}+Probability_{case2}+Probability_{case3}$$ $$=\\frac{8}{36}+\\frac{8}{36}+\\frac{4}{36}$$ $$=\\frac{20}{36} =\\frac{5}{9}$$ Retired Moderator Joined: 30 Jan 2015 Posts: 789 Location: India Concentration: Operations, Marketing GPA: 3.5 Re: A number cube has six faces numbered 1 through 6. If the cube is rolle [#permalink] ### Show Tags 31 Jul 2018, 03:43 1 Bunuel wrote: A number cube has six faces numbered 1 through 6. If the cube is rolled twice, what is the probability that at least one of the rolls will result in a number greater than 4? Possible value in one throw of the dice = { 1, 2, 3, 4, 5, 6 } Values less than 5 = 4" ]

[ "+0 c&p halp polz +7 60 1 +258 A fair 6-sided die is rolled once. If I roll $n$, then I win $6-n$ dollars. What is the expected value of my win, in dollars? Feb 20, 2022 #1 +1384 +1 The expected value is the average of all the possibilities: You can win 0, 1, 2 ,3, 4, or 5 dollars, so the expected is $$\\color{brown}\\boxed{15\\over6}$$ Feb 20, 2022", "the odds against rolling a six with a fair die are 5 to 1. The probability of rolling a six with a fair die is the single number 1/6 or approximately 16.7%. The gambling and statistical uses of odds are closely interlinked. If a bet is a fair one, as in a wager between friends, then the odds offered to the gamblers will perfectly reflect relative probabilities. A fair bet that a fair die will roll a six will pay the gambler $5 for a$1 wager (and return the bettor his or her wager) in the case of a six and nothing in any other case. The terms of the bet are fair, because on average, five rolls result in something other than a six, at a cost of $5, for every roll that results in a six and a net payout of$5. The profit and the expense exactly offset one another and so there is no disadvantage to gambling over the long run. If the odds being offered to the gamblers do not correspond to probability in this way then one of the parties to the bet has an advantage over the other. Casinos, for example, offer odds that place themselves at an advantage, which is how they guarantee themselves a profit and survive as businesses. The", "1. Fair Play Value calculation A coin is weighted so it comes up head 3/4 of the time and tails 1/4 of the time. You play a game where this coin is flipped. If it comes up heads, you must pay 1 dollar. If it comes up tails, you win 2 dollars. What is the Expected Value of your winnings in this game? I solved this portion and had the answered verified : 3/4 x -$1 + 1/4 x$2 = -0.25 I need help with calculating the Fair Play value. The second part goes as follows:", "+0 # Need help with this Expected Value prob 0 113 5 +200 A fair 6-sided die is rolled. If I roll , n then I win n^2 dollars. What is the expected value of my win? Express your answer as a dollar value rounded to the nearest cent. ColdplayMX Aug 14, 2018 #1 0 The expected value of rolling one 6-sided die is: (1+2+3+4+5+6) / 6 =3.5 Your dollar winnings should be: 3.5^2 =\\$12.25 Guest Aug 14, 2018 #3 +1198 +1 Solution: $$\\text {Expected Value } = (1/6)*(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = \\15.17$$ GA GingerAle Aug 15, 2018 #2 +200 +1 15.17. ColdplayMX Aug 15, 2018 #4 +1198 +1 Many math text books give answers to selected odd number questions. Are you imitating your text book? GA GingerAle Aug 15, 2018 #5 +93866 +1 Hi Ginger, it is nice to see you back again. :) Melody Aug 15, 2018", "# You play the following game: You bet $1, a fair die is rolled and if it shows 6 you win$4, otherwise you lose your dollar. If you must choose the number of rounds in advance, how should you choose it to maximize your chance of being ahead (having won more than you have lost) when you quit, and what is the probability of this? Question You play the following game: You bet $1, a fair die is rolled and if it shows 6 you win$4, otherwise you lose your dollar. If you must choose the number of rounds in advance, how should you choose it to maximize your chance of being ahead (having won more than you have lost) when you quit, and what is the probability of this?", "the 1st question. I am still working on the 2nd. I came across another question though Suppose you flip one fair coin and roll one fair six-sided die. Let X be the product of the numbers of heads (i.e., 0 or 1) times the number showing on the die. Compute E(X) Here, if $X$ represents the product of the number of heads with rolled die value, we let $x_i=i$, $i=1,\\ldots,6$. In this case though, $\\mathbb{P}(X=x_i)=\\dfrac{1}{2}\\cdot\\dfrac{1}{6} =\\dfrac{1}{12}$ because the chance of getting a heads is $\\dfrac{1}{2}$ and the chance of rolling a 1 through 6 is $\\dfrac{1}{6}$ since we're working with a fair die. Therefore, $E[X]=\\sum_{i=1}^6 x_i\\mathbb{P}(X=x_i) = \\sum_{i=1}^6\\frac{i}{12}=\\ldots$ I hope this makes sense! #### chisigma ##### Well-known member I am so lost on how to do these questions:. 1.Suppose you start with eight pennies and flip one fair coin. If the coin comes up heads, you get to keep all your pennies; if the coin comes up tails, you have to give half of them back. Let X be the total number of pennies you have at the end. Compute E(X)... There are two not well specified points... a) if You remain with one penny the game is over or You can continue and lose half penny, a quarter of penny and so one?... b) is the number", "# Expected Value and Standard Dev. #### HuskerJay ##### New Member Scenario: You pay $10 and roll a die. If you get a 6, you win$50. If not, you get to roll again. If you get a 6 this time, you get your \\$10 back.", "× Discrete Mathematics # Linearity of Expectation If you roll a six-sided die, the expected value for the number rolled is $$3.5.$$ If you roll 5 six-sided dice, what is the expected value for the sum of the numbers rolled? A standard deck of 52 cards is shuffled randomly, and then the cards are flipped over one-by-one. If the ace of spades is card number $$S$$ and the ace of hearts is card number $$H$$, what is the expected value of $$S+H?$$ Six friends are doing a \"secret santa gift exchange\". Their names are placed into a hat, and each person chooses a name randomly (without replacement) from the hat. However, this leads to the possibility that someone chooses their own name and doesn’t get a gift! What is the expected value for the number of people who will choose their own name? Patrick has a box with 4 yellow balls and 4 green balls. He randomly pulls out a ball, and if it is yellow, he paints it green. He then puts it back in the box and continues this process until all of the balls are green. Find the expected value for the number of times Patrick will pull out a ball. Andy the Ant is on a number line at the location 0. Each minute,", "# Possible Six-Flip Results When $6$ indistinguishable fair coins are thrown, how many different outcomes are there? Details and Assumptions: • Two outcomes are the same if they contain the same number of heads. ×", "much more powerful sense. In order to simulate a loaded die, we build a set of biased coins, one for each side of the die, and then roll a fair die to determine which coin to flip. Based on the die roll, if the appropriate coin comes up heads, we pick the given side, and if the coin comes up tails, we roll the die again and repeat. Let's recap the important points so far. First, the dimensions of these rectangles are as follows - for each side, the height of the \"yes\" rectangle is given by $\\frac{p_i}{p_{max}}$, and the height of the \"no\" rectangle is given by $\\frac{p_{max} - p_i}{p_{max}}$. This normalizes the total heights of the rectangles to be 1. Second, each rectangle has width $1$, though honestly this value doesn't matter. Finally, our algorithm is as follows: until we pick some outcome, roll the fair die to determine which column we are in (in other words, which biased coin to flip). Next, flip the appropriate biased coin. If it comes up heads, choose the outcome corresponding to the chosen column. Otherwise, repeat this process. #### Algorithm: Fair Die/Biased Coin Loaded Die • Initialization: 1. Find the maximum value of $p_i$; call it $p_{max}$. 2. Allocate an array $Coins$ of length $n$, corresponding to the heights", "# Yet Another Unfair Game You were invited to play a game. The game is played with a 6-sided fair dice. You choose two different numbers from the dice, and then the dice is rolled. If the result is one of the numbers you chose, you win a dollar, otherwise,​ you lose one dollar. You knew it is an unfair game, but you wanted to try your luck, so you decided to play with two dollars. You ended up losing your two dollars, but the game owner decided to lend you as much money as you want to spend in the game. You decided to play until you recover your two dollars, and without owing anything to the owner. Let $$P$$ be the probability that you achieve your goal, find $$\\left\\lfloor {10^{10} e^P + 0.5} \\right\\rfloor$$. ×", "# Expected value game - cards and dies A) What is the expected value of a card (from a standard deck of cards) assuming that aces are worth 1 point and face cards are worth 10? B) A game consists of drawing a card and rolling a die simultaneously. If the value of the die is greater than the value of the card (the card values are as in part A above) then you win a number of dollars equal to the value on the die. If the die and the card have equal value, nobody wins anything. If the card is of higher value than the die, you lose 1 dollar. For example, suppose you roll a 4 on the die and draw a 6 of clubs. Then you lose 1 dollar. If, on the other hand, you roll a 5 on the die and draw a 2 of spades, then you win 5 dollars. What is the expected value of the game? - Since this is homework, it would be great if you could please say what you have tried, where you are running into difficulty, etc. – Matthew Conroy Nov 28 '12 at 4:58 What is the probability a card has value $1$? Value $2$? And so on. (The answer for $10$ is different from", "a fair 13-sided die by simply rolling the die and stopping as soon as we have a score between 1 and 13. To see how to simulate a card hand, see the Introduction to Finite Sampling Models. A general method of simulating random variables is based on the quantile function.", "be $2$ because $5+2=7$. $3$ will still be on the right. 2. Right-handed die Roll it to the right so the $6$ is on the bottom. That means $1$ is on the top because $6+1=7$. $3$ is on the right and $2$ stays at the front. 3. Right-handed die Roll it backwards $2$ times so the $5$ is at the back and the 6 is on the bottom. The top is $1$ because $6+1=7$, and the front is $2$ because $5+2=7$. $3$ stays on the right. 4. Left-handed die Roll it forwards so the $6$ is on the bottom. The top is $1$ because $6+1=7$. The $5$ moves from the top to the front. Turn it left so the five moves to the left hand side and the $3$ moves to the front. The right hand side is $2$ because $5+2=7$. This is a great step-by-step way of approaching the problem, April. Well done.", "# One die and 1, 2 or 3 coins A fair die is tossed. If the die lands on 1, 2 or 3, then 3 fair coins are tossed. If the die lands on 4 or 5, then 2 fair coins are tossed. If the die lands on 6, then 1 fair coin is tossed. Given that the resulting coin tosses produced no tails, what is the probability that the die landed on 6? The probability can be expressed as $$\\dfrac ab$$ for coprime positive integers $$a$$ and $$b$$, find the value of $$a+b$$. ×", "side of an equilateral triangle inscribed in the circle? ### Franc-carreau or Fair-square Franc-carreau is a simple game of chance, like the roll-a-penny game often seen at fairs and fêtes. A coin is tossed or rolled down a wooden chute onto a large board ruled into square segments. If the player’s coin lands completely within a square, he or she wins a coin of equal value. If the coin crosses a dividing line, it is lost. The playing board for Franc-Carreau is shown above, together with a winning coin (red) contained within a square and a loosing one (blue) crossing a line. As the precise translation of franc-carreau appears uncertain, the name “fair square” would seem appropriate. The question is: What size should the coin be to ensure a 50% chance of winning?", "it's a fair coin, your probability of winning is 50%.", "You play the following game: Roll a fair 6-sided die as many times as you like. • If it's any integer between 1 to 5 inclusive, you win $100. • If it's a 6, you lose all your money and the game ends. You can continue to roll, winning$100 every time it's not a 6. You can also decide to quit at any time, having not yet rolled a 6, and keep all of your winnings. What is the optimal number of rolls at which you should \"call it quits\"? i.e. For what number, $$n$$, should you decide ahead of time, \"I'm gonna continue to roll until I've done $$n$$ rolls (assuming I haven't lost all my money yet) and then I should quit.\" Give your answer as the sum of all the numbers for which you maximize your expected value of the game. So, for example, if there are two such numbers for which you maximize your expected payoff, $$m$$ and $$n$$, please give your answer as the sum $$m+n$$. ×", "+1 vote 46 views Toss a fair coin twice. You win $1$ dollar if at least one of the two tosses comes out heads. (b) Approximately how many times do you need to play so that you win at least $250$ dollar with probability at least $0.99$. | 46 views", "expect to roll a $7$7? Round your answer to the nearest whole number. eight-sided die ##### Question 13 The following twelve-sided die is rolled. 1. What is the chance of rolling five or more? 2. What is the chance of rolling less than five? ### Outcomes #### MGSE7.SP.5 Understand that the probability of a chance event is a number between 0 and 1 that expresses the likelihood of the event occurring. Larger numbers indicate greater likelihood. A probability near 0 indicates an unlikely event, a probability around 1/2 indicates an event that is neither unlikely nor likely, and a probability near 1 indicates a likely event." ]

[ "or 6. Thus, the probability of E occurring is: \\begin{align}P( E ) &= \\dfrac{\\text{number of favorable outcomes}}{\\text{Number of total outcomes}}\\\\ &= \\dfrac{2}{6}\\\\ &= \\dfrac{1}{3}\\end{align} $$\\therefore$$ the probability= $$\\dfrac{1}{3}$$ Example 2 Shawn throws a die 400 times and he records the score of getting 5 as 30 times. What could be the probability of a) getting a score of 5 b) getting a score under 5 Solution a) P(getting a score of 5) \\begin{align}&=\\dfrac{\\text{number of times getting 5}}{\\text{total times}}\\\\\\\\&=\\dfrac{30}{400}\\\\\\\\&= \\dfrac{3}{40}\\end{align} b) P(getting a score under 5) \\begin{align}&=\\dfrac{\\text{number of times getting under 5}}{\\text{total times}}\\\\\\\\&=\\dfrac{370}{400}\\\\\\\\&= \\dfrac{37}{40}\\end{align} $$\\therefore$$ a) P(getting 5) $$= \\dfrac{3}{40}$$ b) P(getting under 5) $$= \\dfrac{37}{40}$$ Example 3 Patrick rolls a 6-sided die. What will be the probability that he will get a prime number? Solution The total possible outcomes in our sample space is 6 The numbers in the sample space is {1,2,3,4,5,6} The prime numbers among these are {2,3, 5} Probability of getting prime numbers \\begin{align}&= \\dfrac{\\text{favorable outcomes}}{\\text{total possible outcomes}}\\\\ &= \\dfrac{3}{ 6}\\\\&=\\dfrac{1}{2}\\end{align} $$\\therefore$$ Probability(getting a prime number) $$=\\dfrac{1}{2}$$ ## Interactive Questions Here are a few activities for you to practice. Select/Type your answer and click the \"Check Answer\" button to see the result. Think Tank 1. You have rolled a die 3 times in a row. You make a 3 digit number. The second digit is", "Algebra 2 Common Core The probability of getting a $2$ when a standard number cube is rolled is $\\dfrac{1}{6}.$ Also, the probability of getting a $3$ when a standard number cube is rolled is $\\dfrac{1}{6}.$ Adding these probabilities, then the probability of getting a $2$ or a $3$ is \\begin{align*}\\require{cancel} & \\dfrac{1}{6}+\\dfrac{1}{6} \\\\&= \\dfrac{2}{6} \\\\&= \\dfrac{\\cancel2^1}{\\cancel6^3} &\\text{(divide by $2$)} \\\\&= \\dfrac{1}{3} .\\end{align*} Hence, Choice B.", "the second is 0.1576414707. These are certainly very close though! The tiny difference is because $$P(X \\geq 5)$$ includes $$P(X = 11)$$ and $$P(X = 12)$$, while $$P(5 \\leq X \\leq 10)$$ does not. Further, $$P(X = 11)$$ represents the probability that he correctly answers 11 of the questions correctly and latex $$P(X = 12)$$ represents the probability that he answers all 12 of the questions correctly. Both events are very unlikely since he is guessing! In fact: \\begin{align}P(X = 11) &= \\text{binompdf(12,0.25,11)} \\\\ &\\approx \\boxed{2.14 \\times 10^{-6}}\\end{align} and \\begin{align} P(X = 12) &= \\text{binompdf(12,0.25,12)} \\\\ &\\approx \\boxed{5.96 \\times 10^{-8}}\\end{align} Since these are so tiny, including them in the first probability only increases the probability a little bit. Rounding to 4 decimal places, we didn’t even catch the difference.", "19 Explanation: The correct answer is (C). Recall the power rule: If $f(x) = x^n, f'(x) = nx^{n-1}$ Here, $f'(x) = 2 * 2x^1 - 3x^2$ $= 4x - 3x^2$ $f'(3) = 4(3) − 3(3)^2$ $= 12 − 27$ $= −15$ Question 20 ### What is the probability of rolling a prime number in each roll in 4 consecutive rolls of a 6 sided die? A $\\dfrac{1}{4}$ B $\\dfrac{1}{8}$ C $\\dfrac{1}{16}$ D $\\dfrac{1}{32}$ Question 20 Explanation: The correct answer is (C). The prime numbers in the range are: 2, 3, 5. On any roll, the probability of rolling a prime number is: $\\dfrac{3 \\text{ primes}}{6 \\text{ total}}= \\dfrac{1}{2}$ The probability of rolling this 4 times in a row is: $\\dfrac{1}{2} * \\dfrac{1}{2} * \\dfrac{1}{2} * \\dfrac{1}{2} = \\dfrac{1}{16}$ Question 21 ### $p = −3x + 5y$ $q = x + 2y$ What is the value of $p − 2q$? A $5x + y$ B $-5x + y$ C $-y + 5x$ D $5 + y$ Question 21 Explanation: The correct answer is (B). Substitute the values of $p$ and $q$ into the expression: $(−3x + 5y) − 2(x + 2y)$ Distribute and combine like terms to simplify: $= −3x + 5y − 2x − 4y$ $= −5x + y$ Question 22 ### A lab requires 0.76 L of", "Outcomes}}{\\text{No}\\text{.of total outcomes}} \\\\ & =\\dfrac{1}{6} \\\\ \\end{align} Now, when the other die is rolled, the probability of getting 2 would again be $\\dfrac{1}{6}$. Therefore, when this pair of dice is rolled, the probability of obtaining 2 on each die is $=\\left( \\text{Probability of getting 2 on one die} \\right)\\times \\left( \\text{Probability of getting 2 on the other die} \\right)$ $=\\dfrac{1}{6}\\times \\dfrac{1}{6}$ $=\\dfrac{1}{36}$ Hence, option (d) is correct. Note: This question can also be solved as follows: When two dice are rolled, the total number of outcomes are $6\\times 6=36$ Now, we know that there would be just 1 out of 36 outcomes when we would get 2 on both dice. Therefore, the probability of getting 2 on each die \\begin{align} & =\\dfrac{\\text{No}\\text{.of Favorable Outcomes}}{\\text{No}\\text{.of total outcomes}} \\\\ & =\\dfrac{1}{36} \\\\ \\end{align} which is the required value.", "##### Exercise 1 1. Let $$X$$ be the number of winning wrappers in one week (7 days). Then $$X \\stackrel{\\mathrm{d}}{=} \\mathrm{Bi}(7,\\frac{1}{6})$$. 2. $$\\Pr(X = 0) = \\bigl(\\dfrac{5}{6}\\bigr)^7 \\approx 0.279$$. 3. The purchases are assumed to be independent, and hence the outcome of the first six days is irrelevant. The probability of a winning wrapper on any given day (including on the seventh day) is $$\\dfrac{1}{6}$$. 4. Let $$Y$$ be the number of winning wrappers in six weeks (42 days). Then $$Y \\stackrel{\\mathrm{d}}{=} \\mathrm{Bi}(42,\\frac{1}{6})$$. Hence, \\begin{align*} \\Pr(Y \\geq 3) &= 1 - \\Pr(Y \\leq 2) \\\\ &= 1 - \\bigl[\\Pr(Y=0) + \\Pr(Y=1) + \\Pr(Y=2)\\bigr] \\\\ &= 1 - \\bigl[\\bigl(\\tfrac{5}{6}\\bigr)^{42} + 42 \\times \\tfrac{1}{6} \\times \\bigl(\\tfrac{5}{6}\\bigr)^{41} + 861 \\times \\bigl(\\tfrac{1}{6}\\bigr)^2 \\times \\bigl(\\tfrac{5}{6}\\bigr)^{40}\\bigr] \\\\ &\\approx 1 - \\bigl[0.0005 + 0.0040 + 0.0163\\bigr] \\approx 0.979. \\end{align*} This could also be calculated with the help of the $$\\sf \\text{BINOM.DIST}$$ function in Excel. 5. Let $$U$$ be the number of winning wrappers in $$n$$ purchases. Then $$U \\stackrel{\\mathrm{d}}{=} \\mathrm{Bi}(n,\\frac{1}{6})$$ and so \\begin{align*} \\Pr(U \\geq 1) \\geq 0.95 \\ &\\iff\\ 1 - \\Pr(U = 0) \\geq 0.95 \\\\ &\\iff\\ \\Pr(U = 0) \\leq 0.05 \\\\ &\\iff\\ \\bigl(\\tfrac{5}{6}\\bigr)^n \\leq 0.05 \\\\ &\\iff\\ n\\,\\log_e\\bigl(\\tfrac{5}{6}\\bigr) \\leq \\log_e(0.05) \\\\ &\\iff\\ n \\geq \\dfrac{\\log_e(0.05)}{\\log_e\\bigl(\\tfrac{5}{6}\\bigr)} \\approx 16.431. \\end{align*} Hence, he needs to purchase a bar every day for 17 days. #####", "can also be shown as in the formula below: $$Var\\left( X \\right) =\\sum { { \\left( x-\\mu \\right) }^{ 2 }p\\left( x \\right) }$$ Example Given the experiment of rolling a single die, calculate $$Var(X)$$. $$E(X) = 1 \\ast ({1}/{6})+2 \\ast ({1}/{6})+ 3 \\ast ({1}/{6})+ 4 \\ast ({1}/{6})+ 5 \\ast ({1}/{6})+ 6 \\ast ({1}/{6}) = 3.5$$ $$E(X^2) = 1 \\ast ({1}/{6})+2^2 \\ast ({1}/{6})+ 3^2 \\ast ({1}/{6})+ 4^2 \\ast ({1}/{6})+ 5^2 \\ast ({1}/{6})+ 6^2 \\ast ({1}/{6}) = {91}/{6}$$ $$Var \\left(X\\right) = \\left(91/6 \\right) – \\left(3.5 \\right)^2 = {35}/{12} = 2.92$$ We could also calculate $$Var(X)$$ as: $$E \\left(X \\right) = \\mu =3.5$$ \\begin{align*} Var \\left(X \\right) = & (1-3.5)^2 \\ast (1/6) + (2-3.5)^2 \\ast (1/6)+(3-3.5)^2 \\ast (1/6)+(4-3.5)^2 \\ast (1/6)+ \\\\ & (5-3.5)^2 \\ast (1/6)+(6-3.5)^2 \\ast (1/6) = 2.92 \\\\ \\end{align*} The variance of a continuous random variable is shown in the formula below: $$Var\\left( X \\right) =\\int _{ -\\infty }^{ \\infty }{ { x }^{ 2 }f\\left( x \\right) dx } -{ \\left[ \\int _{ -\\infty }^{ \\infty }{ { x }f\\left( x \\right) dx } \\right] }^{ 2 }$$ where $$f(x)$$ is the probability density function of $$x$$. As in the discrete case, it can also be written as: $$Var\\left( X \\right) =E\\left( { X }^{ 2 } \\right) -E{ \\left( X \\right) }^{ 2 }$$ And also,", "= \\dfrac 56$, the probability of failure, i.e. NOT rolling a $5$ (or any other specific number) Plugging it all in, we get $\\approx .01302$, or about $1.302 \\%$. It's probably more accurate to find the probability of getting a $5$ at least $50 \\%$ of the time, because you would be equally or even more impressed with any success rate $\\ge 50 \\%$. To do this, you want to find the probability of $(x=5)$ OR $(x=6)$ OR $\\cdots$ OR $(x=10)$, which is equal to $P(5) + P(6) + \\cdots P(10)$. Plugging this in, we get $\\approx .01546$, or about $1.546 \\%$. To find the probability of getting any $2$ numbers, Daugmented gives a nice idea in the comments. But we can still brute force our way through with the Binomial formula: For getting $2$ numbers exactly $50$ of the time, we have $n = 10$, the number of trials $x=5$, the number of successes, i.e. $50 \\%$ of $10$ $p = \\dfrac 26 = \\dfrac 13$, the probability of success, i.e. rolling a $5$ or a $6$ (or any other pair of numbers) $q = \\dfrac 46 = \\dfrac 23$, the probability of failure, i.e. NOT rolling a $5$ or a $6$ (or any other pair of numbers) The probability of getting a pair of numbers exactly", "Courses Courses for Kids Free study material Free LIVE classes More # A child has a die whose six faces show the letters as given below:The die is thrown once. What is the probability of getting (i) $A?$ (ii) $D?$ Last updated date: 22nd Mar 2023 Total views: 307.8k Views today: 6.86k Verified 307.8k+ views Hint: Use basic definition of probability: $P\\left( E \\right)=\\dfrac{\\text{Total number favourable to cases}}{\\text{Total outcomes (sample space)}}$ As we know that $P\\left( E \\right)=\\dfrac{\\text{ Number of outcomes required}}{\\text{sample space (Total no}\\text{. of outcomes)}}$ Here, As we can observe that we have six faces on a die with six letters, i.e., the number of total outcomes will be 6. Let categorise the number of times letters can come by counting the number of faces at which they are present: \\begin{align} & A=2 \\\\ & B=1 \\\\ & C=1 \\\\ & D=1 \\\\ & E=1 \\\\ \\end{align} Hence, for any face’s letter probability $P\\left( E \\right)=\\dfrac{\\text{Outcome of that particular letter}}{\\text{Total number of outcomes }}$ (i) Probability for getting $A$ \\begin{align} & P\\left( E \\right)=\\dfrac{\\text{No}\\text{. of times A can come to a face}}{\\text{Total number of outcomes possible }} \\\\ & P\\left( E \\right)=\\dfrac{2}{6}=\\dfrac{1}{3} \\\\ \\end{align} (ii) Probability for getting $D$ \\begin{align} & P\\left( E \\right)=\\dfrac{\\text{No}\\text{. of times D can come to a face}}{\\text{Total number of outcomes possible", "at 17:36 We have to find the expected number of turns for the game to end. That is given by: $$\\sum_{r=0}^\\infty rP_r\\;\\;\\;\\;\\;\\;\\;\\text{(by definition)}$$ where $P_r$ is the probability for the game to end in the $r^\\text{th}$ turn. Now, let's see the probability for the game to end in the zeroth turn (that is, the players have just begun playing). The game will end in the zeroth turn iff either the first player wins or the second player wins. Thus, the probability is $P_0 = \\dfrac 16 + \\left(\\dfrac 56\\right)\\dfrac 16$ Similarly, $P_1 = \\left(\\dfrac 56\\right)^2\\dfrac 16 + \\left(\\dfrac 56\\right)^3\\dfrac 16$ It's easy to see that $P_r = \\left(\\dfrac 56\\right)^{2r}\\dfrac 16 + \\left(\\dfrac 56\\right)^{2r+1}\\dfrac 16 = \\dfrac{275}{1296}\\left(\\dfrac{25}{36}\\right)^{r-1}$ Thus, the expected number of turns of the game is: \\begin{align} &\\sum_{r=0}^\\infty r\\dfrac{275}{1296}\\left(\\dfrac{25}{36}\\right)^{r-1} \\\\\\\\ &=\\dfrac{275}{1296} \\sum_{r=0}^\\infty r\\left(\\dfrac{25}{36}\\right)^{r-1}\\\\\\\\ &=\\dfrac{275}{1296}\\dfrac{1}{\\left(1-\\dfrac{25}{36}\\right)^2} = \\boxed{\\boxed{\\dfrac{25}{11}}} \\end{align} where in the last step, I've used the fact that: $$\\sum_{r=0}^\\infty rx^{r-1} = \\dfrac{1}{(1-x)^2} \\forall |x| < 1$$ So, the expected number of turns turns out to be (sic) is $\\dfrac{25}{11}$. EDIT: As suggested by Ross Millikan, it's better to start the counting from $1$. In that case, there's a slight change. Our $P_r$ becomes $\\dfrac{11}{36} \\left(\\dfrac{25}{36}\\right)^{r-1}$. Hence, the 'new' expected number of turns is $\\dfrac{36}{11}$. - The game seems to end whenever either player throws $6$, not when both players", "and one spade $$= {}^{13}{C_3} \\times {}^{13}{C_1}$$ Thus, the probability of obtaining $$3$$ diamonds and one spade $$= \\frac{{{}^{13}{C_3} \\times {}^{13}{C_1}}}{{{}^{52}{C_4}}}$$ Chapter 16 Ex.16.ME Question 3 A die has two faces each with number $$'1'$$, three faces each with number $$'2'$$ and one face with number $$'3'$$. If die is rolled once, determine (i) $$P\\left( 2 \\right)$$ (ii) $$P\\left( {1\\;or\\;3} \\right)$$ (iii)$$P\\left( {not\\;3} \\right)$$ Solution Total number of faces $$= 6$$ Number of faces with number $$'2' = 3$$ Therefore, \\begin{align}P\\left( 2 \\right) &= \\frac{3}{6}\\\\&= \\frac{1}{2}\\end{align} Since, \\begin{align}P\\left( {1\\;or\\;3} \\right) &= P\\left( {not\\;2} \\right)\\\\&= 1 - P\\left( 2 \\right)\\\\&= 1 - \\frac{1}{2}\\\\&= \\frac{1}{2}\\end{align} Number of faces with number $$'3' = 1$$ Therefore, $P\\left( 3 \\right) = \\frac{1}{6}$ Thus, \\begin{align}P\\left( {not{\\rm{ }}3} \\right) &= 1 - P\\left( 3 \\right)\\\\&= 1 - \\frac{1}{6}\\\\&= \\frac{5}{6}\\end{align} Chapter 16 Ex.16.ME Question 4 In a certain lottery $${\\rm{1}}0,000$$ tickets are sold, and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket? (b) two tickets? (c) ten tickets? Solution (a) If we buy one ticket, then Total number of tickets sold $$= 10000$$ Number of prizes awarded $$= 10$$ \\begin{align}P\\left( \\text{getting a prize} \\right)& = \\frac{{10}}{{10000}}\\\\&= \\frac{1}{{1000}}\\end{align} Therefore, \\begin{align}P\\left( \\text{not getting a prize} \\right) &= 1 - \\frac{1}{{1000}}\\\\&= \\frac{{999}}{{1000}}\\end{align} (b) If we buy two", "to the Baron, but I suspect that I did not have his undivided attention at the time. Now, if we label Sir R-----'s roll with $$x_r$$ and the Baron's with $$x_b$$ then we can therefore formulate his expected winnings as \\begin{align*} w_k &= \\Pr(x_r \\geqslant k) \\times \\mathrm{E}\\left[\\max(x_r-x_b, 0) | x_r \\geqslant k\\right] + \\Pr(x_r < k) \\times w_k -1 \\tfrac{12}{100}\\\\ w_k \\times \\left(1 - \\Pr(x_r < k)\\right)&= \\Pr(x_r \\geqslant k) \\times \\mathrm{E}\\left[\\max(x_r-x_b, 0) | x_r \\geqslant k\\right] - \\tfrac{112}{100}\\\\ w_k \\times \\Pr(x_r \\geqslant k) &= \\Pr(x_r \\geqslant k) \\times \\mathrm{E}\\left[\\max(x_r-x_b, 0) | x_r \\geqslant k\\right] - \\tfrac{28}{25}\\\\ w_k &= \\mathrm{E}\\left[\\max(x_r-x_b, 0) | x_r \\geqslant k\\right] -\\frac{28}{25 \\times \\Pr(x_r \\geqslant k)} \\end{align*} where $$\\mathrm{E}[X|C]$$ is the expected value of $$X$$ given that the condition $$C$$ holds true. The probabilities that Sir R-----'s score were equal to any particular value of $$k$$ are easily figured to be $\\begin{matrix} k & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\\\ \\Pr(x_r = k) & \\tfrac{1}{36} & \\tfrac{2}{36} & \\tfrac{3}{36} & \\tfrac{4}{36} & \\tfrac{5}{36} & \\tfrac{6}{36} & \\tfrac{5}{36} & \\tfrac{4}{36} & \\tfrac{3}{36} & \\tfrac{2}{36} & \\tfrac{1}{36} \\end{matrix}$ from which we can trivially deduce that the probabilities that it were greater than or equal to any given value of $$k$$", "of outcomes}}$ 1. Sample space will be $\\Rightarrow S=\\left\\{ 10,11,12,13,14,20,21,22,23,24,30,31,32,33,34,40,41,42,43,44 \\right\\}=20$ So, the number of sample points will be $20$. Probability of getting a prime number, $\\left\\{ 11,13,23,31,41,43 \\right\\}=6$, favorable outcome is $6$. $\\Rightarrow p\\left( P \\right)=\\dfrac{6}{20} =\\dfrac{3}{10}$ 2. Probability of getting a multiple of $4$, favorable outcome will be $\\left\\{ 14,20,24,32,40,44 \\right\\}=6$ $\\Rightarrow p\\left( M \\right)=\\dfrac{6}{20} =\\dfrac{3}{10}$ 3. Probability of getting a number multiple of $11$, favorable outcome will be $\\left\\{ 11,22,33,44 \\right\\}=4$ $\\Rightarrow p\\left( M \\right)=\\dfrac{4}{20} =\\dfrac{1}{5}$ 16.The faces of a die bear numbers $0,1,2,3,4,5$. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero. Ans: To find probability use $p=\\dfrac{\\text{favorable outcomes}}{\\text{total number of outcomes}}$ Sample space will be $\\Rightarrow S=\\left\\{ \\left( 0,0 \\right)\\left( 0,1 \\right)\\left( 0,2 \\right)\\left( 0,3 \\right)\\left( 0,4 \\right)\\left( 0,5 \\right) \\left( 1,0 \\right)\\left( 1,1 \\right)\\left( 1,2 \\right)\\left( 1,3 \\right)\\left( 1,4 \\right)\\left( 1,5 \\right) \\left( 2,0 \\right)\\left( 2,2 \\right)\\left( 2,3 \\right)\\left( 2,4 \\right)\\left( 2,5 \\right) \\left( 3,0 \\right)\\left( 3,1 \\right)\\left( 3,2 \\right)\\left( 3,3 \\right)\\left( 3,4 \\right)\\left( 3,5 \\right) \\left( 4,0 \\right)\\left( 4,1 \\right)\\left( 4,2 \\right)\\left( 4,3 \\right)\\left( 4,4 \\right)\\left( 4,5 \\right) \\left( 5,0 \\right)\\left( 5,1 \\right)\\left( 5,2 \\right)\\left( 5,3 \\right)\\left( 5,4 \\right)\\left( 5,5 \\right)\\right\\}=36,$ In situation $1$: when the die is rolled twice the face has $0$ twice times. So, probability of", "\\\\ \\end{align} In the next step, take constants to one side and the variables to the other side. i.e. 21 from left side goes to right side and becomes -21 and 62n from left side goes to right side and becomes -62n. \\begin{align} & 31-21=63n-62n \\\\ & 10=n \\\\ \\end{align} Hence, we can say that the value of $n=10$. Therefore, the correct answer for this question is option (a). Note: Here, $P\\left( {{E}_{1}} \\right)=\\dfrac{1}{2}$, since, it is the probability of a getting a head from fair $(n+1)$coins. It is because the probability of getting a head is $\\dfrac{1}{2}$. Similarly, $P\\left( {{E}_{2}} \\right)=1$, since we are choosing it from biased n coins with total probability 1. It is only biased towards the head so there is no point in getting a tail.", "50} \\\\ & \\Rightarrow \\dfrac{26}{425} \\\\ \\end{align} The probability that the lost card is diamond by $P\\left( \\dfrac{{{E}_{1}}}{A} \\right)$. By using Baye’s theorem, \\begin{align} & P\\left( \\dfrac{{{E}_{1}}}{A} \\right)=\\dfrac{P\\left( {{E}_{1}} \\right).P\\left( \\dfrac{A}{{{E}_{1}}} \\right)}{P\\left( {{E}_{1}} \\right).P\\left( \\dfrac{A}{{{E}_{1}}} \\right)+P\\left( {{E}_{2}} \\right).P\\left( \\dfrac{A}{{{E}_{2}}} \\right)}...............\\left( 1 \\right) \\\\ & Here, \\\\ & P\\left( {{E}_{1}} \\right)=\\dfrac{1}{4} \\\\ & P\\left( \\dfrac{A}{{{E}_{1}}} \\right)=\\dfrac{22}{425} \\\\ & P\\left( {{E}_{2}} \\right)=\\dfrac{3}{4} \\\\ & P\\left( \\dfrac{A}{{{E}_{2}}} \\right)=\\dfrac{26}{425} \\\\ \\end{align} Putting these values in equation (1), we get, \\begin{align} & P\\left( \\dfrac{{{E}_{1}}}{A} \\right)=\\dfrac{\\dfrac{1}{4}\\times \\dfrac{22}{425}}{\\dfrac{1}{4}\\times \\dfrac{22}{425}+\\dfrac{3}{4}\\times \\dfrac{26}{425}} \\\\ & \\Rightarrow \\dfrac{\\dfrac{1}{425}\\left( \\dfrac{22}{4} \\right)}{\\dfrac{1}{425}\\left( \\dfrac{22}{4}+\\dfrac{26\\times 3}{4} \\right)}=\\dfrac{\\dfrac{11}{2}}{25}=\\dfrac{11}{50} \\\\ \\end{align} Now, $P\\left( \\dfrac{{{E}_{1}}}{A} \\right)=required\\ probability=\\dfrac{11}{50}=P$ $\\therefore 50P=\\dfrac{11}{50}\\times 50=11$ Note: Students can make mistakes by not considering two cases, where the first case is to consider the lost card to be spade, then calculate the probability of selection of two spade cards. Now, consider the second case, that the lost card is not spade and then calculate the probability of selection of two spade cards.", "is the probability of the event not occurring. Now, remember that in Chapter 2, you learned about the formula for nCr\\begin{align*}{_n}C_r\\end{align*}. The formula is shown below: nCr=n!r!(nr)!\\begin{align*}{_n}C_r = \\frac{n!}{r!(n-r)!}\\end{align*} Also, recall that the symbol ! means factorial. As a review, the factorial function (!) just means to multiply a series of consecutive descending natural numbers. Examples: 4!6!1!=4×3×2×1=24=6×5×4×3×2×1=720=1\\begin{align*}4! & = 4 \\times 3 \\times 2 \\times 1 = 24\\\\ 6! & = 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1 = 720\\\\ 1! & = 1\\end{align*} Note: it is generally agreed that 0!=1\\begin{align*}0! = 1\\end{align*}. Technology Tip: You can find the factorial function using: \\begin{align*}\\boxed{\\text{MATH}} \\ \\boxed{\\blacktriangleright} \\ \\boxed{\\blacktriangleright} \\ \\boxed{\\blacktriangleright} \\ (\\text{PRB}) \\boxed{\\blacktriangledown} \\ \\boxed{\\blacktriangledown} \\ \\boxed{\\blacktriangledown} \\ (\\boxed{4})\\end{align*} Now let’s try a few problems with the binomial distribution formula. Example 4 A fair die is rolled 10 times. Let \\begin{align*}X\\end{align*} be the number of rolls in which we see a 2. (a) What is the probability of seeing a 2 in any one of the rolls? (b) What is the probability of seeing a 2 exactly once in the 10 rolls? Solution: (a) \\begin{align*}P(X) = \\frac{1}{6} = 0.167\\end{align*} (b) \\begin{align*}p = 0.167\\!\\\\ q =1 - 0.167 = 0.833\\!\\\\ n = 10\\!\\\\ a = 1\\end{align*} \\begin{align*}P(X = a) & = {_n}C_a \\times p^a \\times q^{(n", "added. – MCT Jun 22 '16 at 4:41 Let $N\\sim\\mathcal{Geo}_1(1/6)$ be the count of rolls until you roll a six. This is a geometric random variable with support $\\{1,2,\\ldots\\}$. Then you know $\\mathsf E(N)$. Let $X_i\\sim\\mathcal U\\{1,2,3,4,5,6\\}$ be the value of the $i$-th roll. Noting that roll $N$ is the first time a $6$ appears, you can calculate $\\mathsf E(X_i\\mid i<N)$ and $\\mathsf E(X_i\\mid i=N)$. Then the expectation you want is given by the Law of Iterated Expectation: \\begin{align}\\mathsf E\\left(\\sum_{i=1}^{N} X_i\\right) ~=&~ \\mathsf E\\big((N-1)\\,\\mathsf E(X_i\\mid i<N)+\\mathsf E(X_i\\mid i=N)\\big)\\\\ =&~ 5\\cdot 3+ 6\\\\ =&~ 21\\end{align}", "is $$g^{(6y)} - \\binom{6}{1}g^{(5y)} + \\binom{6}{2}g^{(4y)} - \\binom{6}{3}g^{(3y)} + \\binom{6}{4}g^{(2y)} - \\binom{6}{5}{g(y)} + 1$$ $$\\text{Now in terms of }~ y^{11} \\text{ for the 11 rolls of the die.}$$ $$\\left(\\dfrac{6^{11}}{11!}\\right) + 6 \\left(\\dfrac{5^{11}}{11!}\\right) + 15 \\left(\\dfrac{4^{11}}{11!}\\right) + 20 \\left(\\dfrac{3^{11}}{11!}\\right) + 15 \\left(\\dfrac{2^{11}}{11!}\\right) +6\\left(\\dfrac{1^{11}}{11!}\\right)$$ $$\\text{Factor out the } \\left(\\dfrac{1}{11!}\\right)\\\\ \\left(\\dfrac{1}{11!}\\right) [(6^{11} - 6(5^{11}) + 15(4^{11}) - 20(3^{11}) + 15(2^{11}) - 6(1^{11})]\\\\ = \\left(\\dfrac{1}{11!}\\right) \\left(129230640\\right) \\\\$$ Note the similarity to the modification of the sterling number formula, where (6!) is factored out. $$\\dfrac{129230640}{11^6} = 0.0.3562064186099 \\\\ \\text { (Probability of rolling at least one of each number (1-6) for 11 rolls of a fair die)}$$ ------------ A Markov matrix may also be used to solve this type of problem. Nauseated Mar 2, 2016 #16 +1037 +5 Errata: Corrected to indicate exponent in term of general expansion $$- \\binom{6}{5}g^{(y)} + 1$$ Corrected odd term subtraction $$\\left(\\dfrac{6^{11}}{11!}\\right) - 6 \\left(\\dfrac{5^{11}}{11!}\\right) + 15 \\left(\\dfrac{4^{11}}{11!}\\right) - 20 \\left(\\dfrac{3^{11}}{11!}\\right) + 15 \\left(\\dfrac{2^{11}}{11!}\\right) -6\\left(\\dfrac{1^{11}}{11!}\\right)$$ ------------ Wolfram alpha link for Sum[Binomial[6, i] (-1)^i (6 - i)^11, {i, 0, 10}] https://www.wolframalpha.com/input/?i=sum+%28binom%286,i%29*%28-1%29^i*%286-i%29^11%29+from+i%3D0+to+10 . Nauseated Mar 2, 2016 #17 +91450 0 Thanks very much Alan for the explanation and history of the Monte-Carlo simulation technique. Thanks also Nauseated for your simplification explanation and for your reference to that older great post. :) Melody Mar 6, 2016 #18 +91450 0 BERTIE (or", "must take the probabilities that these arrangements of the Baron's dice should be observed \\begin{align*} \\Pr(\\{a,b,c,d\\}) &= \\left(1 \\times \\tfrac34 \\times \\tfrac12 \\times \\tfrac14\\right) \\times 1 = \\tfrac{3}{32}\\\\ \\Pr(\\{a,a,b,c\\}) &= \\left(1 \\times \\tfrac14 \\times \\tfrac34 \\times \\tfrac12\\right) \\times 6 = \\tfrac{9}{16}\\\\ \\Pr(\\{a,a,b,b\\}) &= \\left(1 \\times \\tfrac14 \\times \\tfrac34 \\times \\tfrac14\\right) \\times 3 = \\tfrac{9}{64}\\\\ \\Pr(\\{a,a,a,b\\}) &= \\left(1 \\times \\tfrac14 \\times \\tfrac14 \\times \\tfrac34\\right) \\times 4 = \\tfrac{3}{16}\\\\ \\Pr(\\{a,a,a,a\\}) &= \\left(1 \\times \\tfrac14 \\times \\tfrac14 \\times \\tfrac14\\right) \\times 1 = \\tfrac{1}{64} \\end{align*} which when multiplied by their respective expectations yield \\begin{align*} \\Pr(\\{a,b,c,d\\}) \\times \\mathrm{E}_4[\\{a,b,c,d\\}] &= \\frac{3}{32} \\times \\frac{175}{64} = \\frac{525}{2,048}\\\\ \\Pr(\\{a,a,b,c\\}) \\times \\mathrm{E}_4[\\{a,a,b,c\\}] &= \\frac{9}{16} \\times \\frac{37}{16} = \\frac{333}{256}\\\\ \\Pr(\\{a,a,b,b\\}) \\times \\mathrm{E}_4[\\{a,a,b,b\\}] &= \\frac{9}{64} \\times \\frac{121}{64} = \\frac{1,089}{4,096}\\\\ \\Pr(\\{a,a,a,b\\}) \\times \\mathrm{E}_4[\\{a,a,a,b\\}] &= \\frac{3}{16} \\times \\frac{215}{128} = \\frac{645}{2,048}\\\\ \\Pr(\\{a,a,a,a\\}) \\times \\mathrm{E}_4[\\{a,a,a,a\\}] &= \\frac{1}{64} \\times 1 = \\frac{1}{64} \\end{align*} the sum of which is the value that we require $\\frac{525}{2,048} + \\frac{333}{256} + \\frac{1,089}{4,096} + \\frac{645}{2,048} + \\frac{1}{64} = \\frac{1,050+5,328+1,089+1,290+64}{4,096} = \\frac{8,821}{4,096}$ Sir R-----'s expected winnings are consequently $\\frac{8,821}{4,096} \\times 13 - 28 = \\frac{8,821 \\times 13 - 28 \\times 4,096}{4,096} = \\frac{114,673-114,688}{4,096} = -\\frac{15}{4,096}$ and I should have recommended that he declined the Baron's invitation! $$\\Box$$", "$5:\\left \\{6\\right \\}$ $6:\\left \\{\\textrm{not possible}\\right\\}$ Total number of desirable outcomes $5+4+3+2+1=15$ Number of possible outcomes $6\\times 6=36$ Probability that the second number is larger than the first = $15/36$ Answer: $15/36$ Alternative solution First, the die is thrown and the first number $\\left \\{1,2,3,4,5,6\\right \\}$ comes up each with a probability equal $1/6$. Then, the die is thrown again and we want the second number larger than the first. $1:\\left \\{2,3,4,5,6\\right \\}$ probability = $5/6$ $2:\\left \\{3,4,5,6\\right \\}$ probability = $4/6$ $3:\\left \\{4,5,6\\right \\}$ probability = $3/6$ $4:\\left \\{5,6\\right \\}$ probability =$2/6$ $5:\\left \\{6\\right \\}$ probability = $1/6$ $6:\\left \\{\\textrm{not possible}\\right \\}$ probability = $0/6$ Probability of rolling the die two times and the second number is larger than the first $\\dfrac{1}{6}\\times \\dfrac{5}{6}+\\dfrac{1}{6}\\times \\dfrac{4}{6}+\\dfrac{1}{6}\\times \\dfrac{3}{6}+\\dfrac{1}{6}\\times \\dfrac{2}{6}+\\dfrac{1}{6}\\times \\dfrac{1}{6}+\\dfrac{1}{6}\\times \\dfrac{0}{6}=\\dfrac{1}{6}\\left (\\dfrac{5+4+3+2+1+0}{6}\\right )$ $=\\dfrac{1}{6}\\times \\dfrac{15}{6}$ $=15/36$ Values of Integers If $m$ and $n$ are integers such that $2m-n=3$, what are the possible values of $m-2n$? (a) $-3$ only (b) $0$ only (c) only multiples of $3$ (d) any integer (e) none of these Source: NCTM Mathematics Teacher, September 2006 Solution Let $x$ be an integer such that $x=m-2n$. We have the following two equations $2m-n=3\\qquad\\qquad (1)$ $m-2n=x\\qquad\\qquad (2)$ Multiply Eq. $(1)$ by $-2$ and add to Eq. $(2)$ $-4m+2n=-6$ $m-2n=x$ ———————— $-3m=x-6$ Divide both sides by $-3$ $m=-x/3+2$ For $m$" ]

[ "+0 c&p halp polz +7 60 1 +258 A fair 6-sided die is rolled once. If I roll $n$, then I win $6-n$ dollars. What is the expected value of my win, in dollars? Feb 20, 2022 #1 +1384 +1 The expected value is the average of all the possibilities: You can win 0, 1, 2 ,3, 4, or 5 dollars, so the expected is $$\\color{brown}\\boxed{15\\over6}$$ Feb 20, 2022", "+0 # PLS HELP ASAP!!!!! Counting problem!!! 0 189 1 A standard six-sided die is rolled $6$ times. You are told that among the rolls, there was one $1,$ two $2$'s, and three $3$'s. How many possible sequences of rolls could there have been? (For example, $3,2,3,1,3,2$ is one possible sequence.) Feb 28, 2021 This is the same as asking to permute the string $$333221$$ in distinguishable ways. That would be equal to $$6!=6\\cdot5\\cdot4\\cdot3\\cdot2$$, but we're overcounting by a factor of $$3! = 3\\cdot2$$ multiplied by $$2!=2$$, so we need to divide that from the original 6!. That would be equal to $$\\frac{6\\cdot5\\cdot4\\cdot3\\cdot2}{3\\cdot2\\cdot2} = \\boxed{60}$$", "# A fair six-sided die is rolled repeatedly I need help with this homework problem. A fair six-sided die is rolled repeatedly. Each time the die is rolled, the number showing is written down. Let $X$ be the number of rolls until some number shows twice. The possible values of $X$ are $2,3,4,5,6,$ or $7$. For each of the following parts, explain your reasoning and express your answer as a fraction in reduced form. Here is the solution I tried to derive. • $\\Pr(3) = \\dfrac{30}{216}$ $$6 \\times (6-1) \\times 1 = 30$$ $$6^3 = 216$$ • $\\Pr(4) = \\dfrac{120}{1296}$ $$6 \\times (6-1) \\times (6-2) \\times 1 = 120$$ $$6^4=1926$$ • $\\Pr(5) = \\dfrac{360}{7776}$ $$6 \\times (6-1) \\times (6-2) \\times (6-3) \\times 1 = 360$$ $$6^5=7776$$ Can someone help me out please? Thank • The question is not clear – Manish Kundu Feb 6 '18 at 16:50 • It looks like you are on the right track. What's the problem? – herb steinberg Feb 6 '18 at 17:27 I'll let $P(n)$ denote the probability of $X = n$ for brevity. If you want the distribution of $X$, then it's clear that $P(2) = 1/6$, and there's a recurrence $$P(n+1) = \\frac{n}{6}\\left( 1 - \\sum_{i=2}^n P(i) \\right)$$ because the probability that the first $n$ rolls have no repeats is", "the 1st question. I am still working on the 2nd. I came across another question though Suppose you flip one fair coin and roll one fair six-sided die. Let X be the product of the numbers of heads (i.e., 0 or 1) times the number showing on the die. Compute E(X) Here, if $X$ represents the product of the number of heads with rolled die value, we let $x_i=i$, $i=1,\\ldots,6$. In this case though, $\\mathbb{P}(X=x_i)=\\dfrac{1}{2}\\cdot\\dfrac{1}{6} =\\dfrac{1}{12}$ because the chance of getting a heads is $\\dfrac{1}{2}$ and the chance of rolling a 1 through 6 is $\\dfrac{1}{6}$ since we're working with a fair die. Therefore, $E[X]=\\sum_{i=1}^6 x_i\\mathbb{P}(X=x_i) = \\sum_{i=1}^6\\frac{i}{12}=\\ldots$ I hope this makes sense! #### chisigma ##### Well-known member I am so lost on how to do these questions:. 1.Suppose you start with eight pennies and flip one fair coin. If the coin comes up heads, you get to keep all your pennies; if the coin comes up tails, you have to give half of them back. Let X be the total number of pennies you have at the end. Compute E(X)... There are two not well specified points... a) if You remain with one penny the game is over or You can continue and lose half penny, a quarter of penny and so one?... b) is the number", "8k views Suppose a fair six-sided die is rolled once. If the value on the die is $1, 2,$ or $3,$ the die is rolled a second time. What is the probability that the sum total of values that turn up is at least $6$ ? 1. $\\dfrac{10}{21}$ 2. $\\dfrac{5}{12}$ 3. $\\dfrac{2}{3}$ 4. $\\dfrac{1}{6}$ edited | 8k views 0 Hello sir, why we have taken sample space 36 although we not throw dice again when we got 4,5,6 . +7 $\\\\1\\rightarrow 5,6\\\\ 2\\rightarrow4,5,6\\\\ 3\\rightarrow3,4,5,6\\\\ 4\\rightarrow\\times\\\\ 5\\rightarrow\\times\\\\ 6\\rightarrow \\checkmark$ $\\dfrac{1}{6}\\times \\dfrac{2}{6}+\\dfrac{1}{6}\\times \\dfrac{3}{6}+\\dfrac{1}{6}\\times \\dfrac{4}{6}+\\dfrac{1}{6}\\times \\dfrac{6}{6}=\\dfrac{5}{12}$ 0 How $6$ is valid? On first dice only $1,2,3$ can come. Here our sample space consists of $3 + 3\\times 6 = 21$ events- $(4), (5), (6), (1,1), (1,2)\\ldots(3,6).$ Favorable cases $=(6), (1,5), (1,6), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3,5), (3,6).$ Required Probability $= \\dfrac {\\text{No. of favorable cases}}{\\text{Total cases }}= \\dfrac{10}{21}$ But this is wrong way of doing. Because due to $2$ tosses for some and $1$ for some, individual probabilities are not the same. i.e., while $(6)$ has $\\dfrac{1}{6}$ probability of occurrence, $(1,5)$ has only $\\dfrac{1}{36}$ probability. So, our required probability $\\Rightarrow \\dfrac{1}{6} + \\left(9 \\times \\dfrac{1}{36}\\right) =\\dfrac{5}{12}.$ Correct Answer: $B$ by Veteran edited 0 in favorable cases why 6 is considered ? +4 great explanation @Arjun. 0", "# Expected number of rolls until a number appears $k$ times Roll a fair die, what is the expected number of rolls until a number appear $$k$$ times? Not necessarily consecutive. Let $$N$$ be the number of rolls until a number appear $$k$$ times. For $$k=2$$, we know that the largest possible value for $$N$$ is $$7$$. Hence we have \\begin{align} &P(N=1)=0\\\\ &P(N=2)=1/6\\\\ &P(N=3)=5/6\\cdot 2/6\\\\ &P(N=4)=5/6\\cdot 4/6\\cdot 3/6\\\\ &P(N=5)=5/6\\cdot 4/6\\cdot 3/6 \\cdot 4/6\\\\ &P(N=6)=5/6\\cdot 4/6\\cdot 3/6 \\cdot 2/6\\cdot 5/6\\\\ &P(N=7)=5/6\\cdot 4/6\\cdot 3/6 \\cdot 2/6\\cdot 1/6 \\end{align} However I do not know how to generalise it for any $$k$$. Could someone please help? • Is this the birthday problem? en.wikipedia.org/wiki/Birthday_problem Jul 6, 2020 at 21:51 • @eric_kernfeld it is the birthday problem when $k=2$ Jul 6, 2020 at 21:56 • @eric_kernfeld Sure. In the simple case $k=2$. I only ask for the expectation of $N$. But following the birthday problem, I will have to calculate/approximate the entire distribution of $N$. I was hoping for a simpler method. This becomes more important for general $k$. Jul 6, 2020 at 22:09 It's sometimes useful to recast a problem in terms that yield better search engine results. Here is an alternative formulation of your problem: We throw balls at random into $$n=6$$ urns, with equal probability. How many balls do we expect", "be $2$ because $5+2=7$. $3$ will still be on the right. 2. Right-handed die Roll it to the right so the $6$ is on the bottom. That means $1$ is on the top because $6+1=7$. $3$ is on the right and $2$ stays at the front. 3. Right-handed die Roll it backwards $2$ times so the $5$ is at the back and the 6 is on the bottom. The top is $1$ because $6+1=7$, and the front is $2$ because $5+2=7$. $3$ stays on the right. 4. Left-handed die Roll it forwards so the $6$ is on the bottom. The top is $1$ because $6+1=7$. The $5$ moves from the top to the front. Turn it left so the five moves to the left hand side and the $3$ moves to the front. The right hand side is $2$ because $5+2=7$. This is a great step-by-step way of approaching the problem, April. Well done.", "and none of these will be $7$. Otherwise they can be any roll. so we end up with $P[\\text{win}] = 2 \\sum \\limits_{n=2}^\\infty \\dfrac 1 9\\left(\\sum \\limits_{k=1}^{n-1}\\dbinom{n-1}{k}\\left(\\dfrac 1 9\\right)^k\\left(\\dfrac{11}{18}\\right)^{n-1-k}\\right)$ Digesting this a bit. The $2$ in front is because roll $n$ can be either $5$ or $9$, and the probability is identical for both. The $\\dfrac 1 9$ is the probability of roll $n$ being the $5$ (or the $9$) What's inside the parens is the sum of probabilities that there are $k$ occurrences of $5$ or $9$ whichever isn't roll $n$ This can be simplified to $P[\\text{win}] = 2 \\sum\\limits_{n=2}^\\infty \\left(\\left(\\dfrac{13}{18}\\right)^{n-1} - \\left(\\dfrac{11}{18}\\right)^{n-1}\\right)$ and then finally to $P[\\text{win}] =\\dfrac{8}{35}$ I leave it to you to digest all of this and reproduce the simplifications in more detail. ps. $p=\\dfrac {8}{35}$ matches my sim of this very well so I'm pretty confident in this answer. Falsetto November 6th, 2017 08:43 PM THANKYOU!! Exactly what i was hoping to find. And also the answer i had was correct as well :) But a formula like yours is so much better. I will study it in detail tomorrow when i wake up. Thanks for your time and help. Really appreciate it! Falsetto November 15th, 2017 05:58 PM Trying to digest what you did I figured out that there is a lot", "is $$g^{(6y)} - \\binom{6}{1}g^{(5y)} + \\binom{6}{2}g^{(4y)} - \\binom{6}{3}g^{(3y)} + \\binom{6}{4}g^{(2y)} - \\binom{6}{5}{g(y)} + 1$$ $$\\text{Now in terms of }~ y^{11} \\text{ for the 11 rolls of the die.}$$ $$\\left(\\dfrac{6^{11}}{11!}\\right) + 6 \\left(\\dfrac{5^{11}}{11!}\\right) + 15 \\left(\\dfrac{4^{11}}{11!}\\right) + 20 \\left(\\dfrac{3^{11}}{11!}\\right) + 15 \\left(\\dfrac{2^{11}}{11!}\\right) +6\\left(\\dfrac{1^{11}}{11!}\\right)$$ $$\\text{Factor out the } \\left(\\dfrac{1}{11!}\\right)\\\\ \\left(\\dfrac{1}{11!}\\right) [(6^{11} - 6(5^{11}) + 15(4^{11}) - 20(3^{11}) + 15(2^{11}) - 6(1^{11})]\\\\ = \\left(\\dfrac{1}{11!}\\right) \\left(129230640\\right) \\\\$$ Note the similarity to the modification of the sterling number formula, where (6!) is factored out. $$\\dfrac{129230640}{11^6} = 0.0.3562064186099 \\\\ \\text { (Probability of rolling at least one of each number (1-6) for 11 rolls of a fair die)}$$ ------------ A Markov matrix may also be used to solve this type of problem. Nauseated Mar 2, 2016 #16 +1037 +5 Errata: Corrected to indicate exponent in term of general expansion $$- \\binom{6}{5}g^{(y)} + 1$$ Corrected odd term subtraction $$\\left(\\dfrac{6^{11}}{11!}\\right) - 6 \\left(\\dfrac{5^{11}}{11!}\\right) + 15 \\left(\\dfrac{4^{11}}{11!}\\right) - 20 \\left(\\dfrac{3^{11}}{11!}\\right) + 15 \\left(\\dfrac{2^{11}}{11!}\\right) -6\\left(\\dfrac{1^{11}}{11!}\\right)$$ ------------ Wolfram alpha link for Sum[Binomial[6, i] (-1)^i (6 - i)^11, {i, 0, 10}] https://www.wolframalpha.com/input/?i=sum+%28binom%286,i%29*%28-1%29^i*%286-i%29^11%29+from+i%3D0+to+10 . Nauseated Mar 2, 2016 #17 +91450 0 Thanks very much Alan for the explanation and history of the Monte-Carlo simulation technique. Thanks also Nauseated for your simplification explanation and for your reference to that older great post. :) Melody Mar 6, 2016 #18 +91450 0 BERTIE (or", "+0 # help 0 136 2 We roll a fair 6-sided die 5 times. What is the probability that exactly 3 of the 5 rolls are either a 1 or a 2? Feb 4, 2019 #1 +721 -1 There is a 1/3 chance it will be a 1 or 2 and 2/3 chance it won't. So... (1/3)^3 x (2/3)^2 = 4/243 You are very welcome! :P (i was lazy for latex :) Feb 4, 2019 #2 +22569 +9 We roll a fair 6-sided die 5 times. What is the probability that exactly 3 of the 5 rolls are either a 1 or a 2? $$\\begin{array}{|rcll|} \\hline && \\dbinom{5}{3}\\left(\\dfrac{2}{6}\\right)^3\\left(\\dfrac{4}{6}\\right)^2 \\\\\\\\ &=& \\dbinom{5}{3}\\left(\\dfrac{1}{3}\\right)^3\\left(\\dfrac{2}{3}\\right)^2 \\\\\\\\ &=& \\dfrac{5}{3}\\cdot\\dfrac{4}{2}\\cdot\\dfrac{3}{1}\\cdot \\dfrac{1}{3^3}\\cdot \\dfrac{4}{3^2} \\\\\\\\ &=& 10\\cdot \\dfrac{4}{3^5} \\\\\\\\ &=& 10\\cdot \\dfrac{4}{243} \\\\\\\\ &=& \\dfrac{40}{243} \\\\\\\\ &=& 0.16460905350\\ (16.5 \\%) \\\\ \\hline \\end{array}$$ Feb 5, 2019", "+0 # Need help with this Expected Value prob 0 113 5 +200 A fair 6-sided die is rolled. If I roll , n then I win n^2 dollars. What is the expected value of my win? Express your answer as a dollar value rounded to the nearest cent. ColdplayMX Aug 14, 2018 #1 0 The expected value of rolling one 6-sided die is: (1+2+3+4+5+6) / 6 =3.5 Your dollar winnings should be: 3.5^2 =\\$12.25 Guest Aug 14, 2018 #3 +1198 +1 Solution: $$\\text {Expected Value } = (1/6)*(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = \\15.17$$ GA GingerAle Aug 15, 2018 #2 +200 +1 15.17. ColdplayMX Aug 15, 2018 #4 +1198 +1 Many math text books give answers to selected odd number questions. Are you imitating your text book? GA GingerAle Aug 15, 2018 #5 +93866 +1 Hi Ginger, it is nice to see you back again. :) Melody Aug 15, 2018", "$2$ and Candy rolls $2$ or more: $\\dfrac{1}{36}\\cdot\\dfrac{5}{6}=\\dfrac{5}{216}$ • The probability that Andy rolls $3$ and Candy rolls $3$ or more: $\\dfrac{2}{36}\\cdot\\dfrac{4}{6}=\\dfrac{8}{216}$ • The probability that Andy rolls $4$ and Candy rolls $4$ or more: $\\dfrac{3}{36}\\cdot\\dfrac{3}{6}=\\dfrac{9}{216}$ • The probability that Andy rolls $5$ and Candy rolls $5$ or more: $\\dfrac{4}{36}\\cdot\\dfrac{2}{6}=\\dfrac{8}{216}$ • The probability that Andy rolls $6$ and Candy rolls $6$: $\\dfrac{5}{36}\\cdot\\dfrac{1}{6}=\\dfrac{5}{216}$ Hence the probability of the complementary event is: $$1-\\dfrac{5+8+9+8+5}{216}=\\dfrac{181}{216}$$ That's not right since Candy might not roll a six. The most straightforward way to answer this is to simply lay out a table of all the possibilities for the two rollers and then add up all the cells where Andy wins. Another way to do it though is rewrite the problem. P(A > C) is the same as P(A - C > 0). Then, you can use the fact that subtracting a die is the same as adding one and then subtracting 7 (try it out). So P(A - C > 0) is the same as P(A + C - 7 > 0) or P(A + C > 7). A + C is just rolling three dice, so the answer to your question is the same as the answer to \"what is the probability of getting more than 7 on three dice\". Think of the three", "## A game of chance involves rolling a 16-sided die once. If a number from 1 to 3 comes up, you win 4 dollars. If the number 4 or 5 comes up, y Question A game of chance involves rolling a 16-sided die once. If a number from 1 to 3 comes up, you win 4 dollars. If the number 4 or 5 comes up, you win 7 dollars. If any other number comes up, you lose. If it costs 5 dollars to play, what is your expected net winnings in progress 0 5 months 2021-08-29T20:59:35+00:00 1 Answers 0 views 0 -$1.8 or loss of$1.8 The probability of winning $4 is 3/16 The probability of winning$7 is 2/16 The probability of losing $5 is 11/16 E(X) = (4 * 3/16) + (7 * 2/16) + (-5 * 11/16) = 12/16 + 14/16 – 55/16 = -29/16 = -1.8 Hence, your expected net “winnings” would be -$1.8. That’s an expected loss of \\$1.8.", "rolling a fair (standard) die 2n times and having the total score being exactly the expected value 7n. Then a(n) satisfies the following third-order linear recurrence equation with polynomial coefficients −4(5+2n)(2n+3)(2n+1)(7n+19)(5n+11)(7n+20)(7n+13)(n+2)(n+1)a(n) +4(7n + 20)(5 + 2n)(2n + 3)(n + 2) June 19, 2007 15:9 WSPC - Proceedings Trim Size: 9in x 6in WWCA06 Five Applications of Wilf-Zeilberger Theory to Enumeration and Probability 15 (25480n5 + 223496n4 + 755066n3 + 1223233n2 + 946889n+ 279936)a(n+1)− (5n + 6)(5 + 2n)(6 + 7n)(499359n6 + 6777015n5 + 38079431n4 +113390385n3 + 188723986n2 + 166469280n + 60800544)a(n + 2)+ 30(5n + 14)(5n + 13)(5n + 12)(7n + 12) ·(5n + 11)(5n + 6)(7n + 13)(6 + 7n)(n + 3)a(n + 3) = 0, and the Birkhoff-Trijinksi method implies that the asymptotics is: (.197833497170804)n−1/2(1 − (111/1400)/n − (12037/5488000)/n2 +(1367631/1097600000)/n3 + . . . ). Readers can produce their own output for the scenario of their choice. Second Application: How many ways to have r people chip in to pay a bill of n cents In George Pólya’s classic ‘picture writing’ paper,4 he considers the problem of figuring out how many ways can one person pay a bill of n cents using any number of coins. If the denominations are {d1 , . . . , dk } ({1, 5, 10, 25, 50, 100} in the", "the coin is biased (and biased with a probability of about $$80\\%$$ for tossing a head). Even with just this vague understanding of matters, without a clear definition of probability in hand, we can nut our way through some challenging probability problems just by imagining repeating an experiment a large number times. A Typical Example: Imagine rolling a die and tossing a coin. What are the chances of seeing a $$6$$ followed by a HEAD? Answer: Do imagine running this double experiment a large number times, say, $$600$$ times. (That is, roll a die and then toss a coin and record the results as a pair six hundred times.) We would expect to see a roll of a $$6$$ about $$100$$ times, and of those, close to half would be followed by a toss of a HEAD and half the toss of a TAIL. That is, we’d expect to see a $$6$$ followed by a HEAD about $$50$$ times. This is $$\\dfrac{50}{600}=\\dfrac{1}{12}$$ of the time. Thus we conclude: $$P\\left(6 \\; and\\; HEAD\\right) = \\dfrac{1}{12}$$. Alternatively, rather than work with specific numbers, draw a rectangle to represent a large number of runs of the experiment. Each possible roll of the die is represented as $$\\frac{1}{6}$$ of the rectangle. The portion corresponding to rolling a $$6$$ is further divided into", "1), 10)$$ 3. Return the result $$r + 1$$, which will be approximately uniformly distributed on the values $$[1:10]$$. (Alternatively, since $$r$$ is a number in $$[0:9]$$, you could just interpret $$0$$ as $$10$$ instead of adding 1 to $$r$$). Note: I have the minus 1's in the algorithm to help explain how it works, but they aren't necessary. For a truly fast algorithm, you can simply do 1. Let $$r \\leftarrow d_1$$ 2. For $$n = 2, \\dots , N$$, let $$r \\leftarrow \\text{mod} (r + 6(d_n), 10)$$ 3. Return the result $$r + 1$$. ## An Example Run Let's choose $$N = 4$$. Then suppose we roll the 6-sided die once and our first roll is $$d_1 = 6$$. We set $$r \\leftarrow d_1 - 1 = 5$$. Now roll a second time and supposed we get $$d_2 = 4$$. We compute $$6(d_2 - 1) = 18$$. Because we are only interested in the sum mod 10, we can \"forget\" the tens place at each step. Interpret all \"=\" as equal mod 10 from this point forward. So the update for $$r$$ is $$r \\leftarrow r + 6(3) = 5 + 18 = 5 + 8 = 13 = 3$$. We repeat this process. If the third roll is $$d_3 = 1$$, then $$6(d_3 - 1)", "$$\\dfrac{3^2\\cdot 2^1\\cdot 1^2 \\cdot \\dfrac{5!}{2!1!2!}}{6^5} = \\dfrac{5}{72}$$ • Can you please explain why two terms are being multiplied in numerator of last value ? – puffles Sep 14 '18 at 13:29 • I will use 1 to mean a value of 1, 2, or 3 on the die and 4 to mean a value of 4 or 5 on the die. So, 11466 would be an order of the dice that are rolled where the first two dice show numbers between 1 and 3, the third shows either 4 or 5, and the last two are both 6. This is a valid outcome. Other valid outcomes are: 11646, 11664, 14166,... The term $$\\dfrac{5!}{2!1!2!}$$ is the number of ways to order the digits 11466. – InterstellarProbe Sep 14 '18 at 13:45 • And why is it being multiplied with 3^2*2^1*1^2? I hope I'm not bothering too much :D I just need to get my concepts cleared – puffles Sep 14 '18 at 13:47 • How many ways can you get 11466? That would be $3^2\\cdot 2^1\\cdot 1^2$. How many ways can you get 14166? That would be $3\\cdot 2\\cdot 3\\cdot 1^2 = 3^2\\cdot 2^1\\cdot 1^2$. Multiplication is commutative. So, for each order, the number of ways to achieve the die rolls remains constant. So, we multiply the number of", "- 0.6562=14.7838$ C. Step 1: $0.35 x 16=5.6$ Step 2: $16 - 5.6=10.4$ Step 3: $0.425 x 10.4=4.42$ Step 4: $16 - 4.42=11.58$ D. Step 1: $0.35 x 16=5.6$ Step 2: $16 - 5.6=10.4$ Step 3: $0.425 x 10.4=4.42$ Step 4: $10.4 - 4.42=5.98$ 50. What is the value of the expression $(-\\dfrac{8}{9}) \\div (-\\dfrac{2}{3}) \\times (-4 \\dfrac{1}{2})$? A. $-6$ B. $-\\dfrac{8}{27}$ C. $\\dfrac{8}{27}$ D. $6$ 51. A board game has a spinner divided into sections of equal size. Each section is labeled with a number between $1$ and $5$. Which number is a reasonable estimate of the number of times the spinner will land on a section labeled $5$ over the course of $150$ spins? A. $15$ B. $25$ C. $40$ D. $60$", "14 15 Suppose a fair six-sided die is rolled once. If the value on the die is $1, 2,$ or $3,$ the die is rolled a second time. What is the probability that the sum total of values that turn up is at least $6$ ? $\\dfrac{10}{21}$ $\\dfrac{5}{12}$ $\\dfrac{2}{3}$ $\\dfrac{1}{6}$ 16 Assume that you are flipping a fair coin, i.e. probability of heads or tails is equal. Then the expected number of coin flips required to obtain two consecutive heads for the first time is. $4$ $3$ $6$ $10$ $5$ 17 The probability of three consecutive heads in four tosses of a fair coin is. $\\left(\\dfrac{1}{4}\\right)$ $\\left(\\dfrac{1}{8}\\right)$ $\\left(\\dfrac{1}{16}\\right)$ $\\left(\\dfrac{3}{16}\\right)$ $\\text{None of the above.}$ 18 For each element in a set of size $2n$, an unbiased coin is tossed. The $2n$ coin tosses are independent. An element is chosen if the corresponding coin toss was a head. The probability that exactly $n$ elements are chosen is $\\frac{^{2n}\\mathrm{C}_n}{4^n}$ $\\frac{^{2n}\\mathrm{C}_n}{2^n}$ $\\frac{1}{^{2n}\\mathrm{C}_n}$ $\\frac{1}{2}$ 19 Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day? $\\dfrac{1}{7^7}\\\\$ $\\dfrac{1}{7^6}\\\\$ $\\dfrac{1}{2^7}\\\\$ $\\dfrac{7}{2^7}\\\\$ 20 Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ________ . 1 vote 21 After your complaint about", "+0 # help with math question! 0 294 1 You roll a standard 6-sided die. Find the expected value for each experiment. (a) You earn 6 points for each dot on the side that lands up. _____ points (b) You earn 6 points if it lands on a composite number and lose 6 points if it does not land on a composite number. _____ points Sep 28, 2018 #1 +5226 0 a) $$E[X]=\\sum \\limits_{k=1}^6~\\dfrac 1 6 (6k) = \\dfrac{6\\cdot 7}{2} = 21$$ b) $$E[X] = \\dfrac 1 6(6 + 6 + 6 - 6 + 6 - 6) = 2$$ (1,2,3,5 are !composite, 4 and 6 are composite) Sep 28, 2018" ]

[ "cents each, and Spain $5$ cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.) The average price of his $'70$s stamps is closest to (A) $3.5$ cents (B) $4$ cents (C) $4.5$ cents (D) $5$ cents (E) $5.4$ cents AMC 8, 2002, Problem 11 A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth? (A) $11$ (B) $12$ (C) $13$ (D) $14$ (E) $15$ AMC 8, 2002, Problem 12 A board game spinner is divided into three regions labeled $A, B$ and $C$. The probability of the arrow stopping on region $A$ is $\\frac{1}{3}$ and on region $B$ is $\\frac{1}{2}$. The probability of the arrow stopping on region $C$ is (A) $\\frac{1}{12}$ (B) $\\frac{1}{6}$ (C) $\\frac{1}{5}$ (D) $\\frac{1}{3}$ (E) $\\frac{2}{5}$ AMC 8, 2002, Problem 15 Which of the following polygons has the largest area? (A) $A$ (B) $B$ (C) $C$ (D) $D$ (E) $E$ AMC 8, 2002, Problem 16 Right isosceles triangles are constructed on the sides of a $3-4-5$ right triangle, as shown. A capital letter represents the area of each triangle. Which one of", "Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd? (A) $\\frac{17}{36}$ (B) $\\frac{35}{72}$ (C) $\\frac{1}{2}$ (D) $\\frac{37}{72}$ (E) $\\frac{19}{36}$ AMC 8, 2006, Problem 4 Initially, a spinner points west. Chenille moves it clockwise $2 \\frac{1}{4}$ revolutions and then counterclockwise $3 \\frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves? (A) north (B) east (C) south (D) west (E) northwest AMC 8, 2006, Problem 10 Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12 .$ What should his graph look like? AMC 8, 2006, Problem 17 Jeff rotates spinners $P, Q$ and $R$ and adds the resulting numbers. What is the probability that his sum is an odd number? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ (C) $\\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2006, Problem 20 A singles tournament had six players. Each player played every other player only once, with no ties. If Helen", "of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3$ ? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ $(C) \\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2007, Problem 25 On the dart board shown in the Figure, the outer circle has radius $6$ and the inner circle has a radius of $3$ . Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd? (A) $\\frac{17}{36}$ (B) $\\frac{35}{72}$ (C) $\\frac{1}{2}$ (D) $\\frac{37}{72}$ (E) $\\frac{19}{36}$ AMC 8, 2006, Problem 4 Initially, a spinner points west. Chenille moves it clockwise $2 \\frac{1}{4}$ revolutions and then counterclockwise $3 \\frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves? (A) north (B) east (C) south (D) west (E) northwest AMC 8, 2006, Problem 10 Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12", "probability of a randomly thrown dart that hits the square board in shaded region is 0.215 Know more about Andhra Pradesh Board and access various study materials including syllabus, exam timetable, etc. at BYJU’s.", "faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "# Math Help - Spinner 1. ## Spinner A spinner contains eight regions, numbered 1 through 8. The arrow has an equally likely chance of landing on any of the eight regions. If the arrow lands on the line, it is spun again. What is the probability that the arrow lands on an odd number? 2. Originally Posted by symmetry A spinner contains eight regions, numbered 1 through 8. The arrow has an equally likely chance of landing on any of the eight regions. If the arrow lands on the line, it is spun again. What is the probability that the arrow lands on an odd number? The odd numbers, 1-8, are 1, 3, 5, 7, thus there are 4 odd numbers out of the 8, and therefore the probability that the arrow lands on an odd number is 4/8 or 1/2. 3. ## ok I thank you for your great help.", "### Home > INT2 > Chapter 3 > Lesson 3.1.5 > Problem3-61 3-61. The spinner at right has three regions: $\\5,\\0,$ and $\\20$. To play the game, you spin one time and then you win the amount in the region that the spinner lands on. 1. What is the expected value for the game? Compare the given angle measurements to $360^\\circ$. Multiply these ratios by the amount possible to win in each section. $\\text{EV}=\\5\\cdot \\frac{1}{4}+\\0\\cdot \\frac{135}{360}+\\20\\cdot \\frac{135}{360}$ 2. If you have to pay $\\10$ to play, is it a fair game?", "# Dart-throwing Monkeys Geometry Level 2 A circular dart board has a radius of 9cm, the board is split into 3 regions. The first region is enclosed in a circle centered in the center of the dart board with a radius of 2cm. The second region is in between the circle that defines the first region and another circle also centered at the center of the dart board and with a radius of $$x$$cm. The third region is the rest of the dart board that is not occupied by the first or second region. Monica the Monkey throws a dart at the dartboard. The Probability that the dart hits the second Region is $$\\frac 59$$. Find the value of $$x$$. ×", "## Section10.1An Introduction to Probability We continue with an informal discussion intended to motivate the more structured development that will follow. Consider the “spinner” shown in Figure 10.1. Suppose we give it a good thwack so that the arrow goes round and round. We then record the number of the region in which the pointer comes to rest. Then observers, none of whom have studied combinatorics, might make the following comments: 1. The odds of landing in region $$1$$ are the same as those for landing in region $$3\\text{.}$$ 2. You are twice as likely to land in region $$2$$ as in region $$4\\text{.}$$ 3. When you land in an odd numbered region, then 60% of the time, it will be in region $$5\\text{.}$$ We will now develop a more formal framework that will enable us to make such discussions far more precise. We will also see whether Alice is being entirely fair to Bob in her proposed game to one hundred. We begin by defining a probability space as a pair $$(S,P)$$ where $$S$$ is a finite set and $$P$$ is a function that whose domain is the family of all subsets of $$S$$ and whose range is the set $$[0,1]$$ of all real numbers which are non-negative and at most one. Furthermore, the following two key", "= $\\frac{1}{2}$ • The probability of landing on heads is $P$(tails) = $\\frac{1}{2}$ • Dice: (a six-sided die, if singular) • When rolling a die, it will land with one of its six flat faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$ #### Lessons • Introduction Introduction to Probability Outcomes for Coins, Dice, and Spinners: a) What is probability? b) What are outcomes in probability? 1. Outcomes for flipping a coin 2. Outcomes for rolling", "### Home > A2C > Chapter 9 > Lesson 9.3.2 > Problem9-161 9-161. Each dartboard below is a target at the county fair dart-throwing game. What is the probability of hitting the darkened region of each target? Assume you always hit the board but the location on the board is random. 1. Notice that half of the circle is shaded. $\\frac{1}{2}$ 1. See part (a). 1. Remember that 60° is one sixth of a full circle. 1. See part (a). 1. Each of the smaller squares are a quarter of the next one larger than it. $\\frac{1}{16}$ 1. Subtract the value of a circle with a radius of one from the 2 by 2 square. 4 − 1(π) $1-\\frac{\\pi}{4}$", "There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "# Expected value (basic) Expected value uses probability to tell us what outcomes to expect in the long run. ## Problem 1: Board game spinner A board game uses the spinner shown below to determine how many spaces a player will move forward on each turn. The probability is $\\dfrac12$ that the player moves forward $1$ space, and moving forward $2$ or $3$ spaces each have $\\dfrac14$ probability. Kayla is a basketball player who makes $50\\%$ of her $2$-point shots and $20\\%$ of her $3$-point shots.", "= random.uniform(0,1) if P_1 < num < 1: #area corresponding to x_1 wins += 1 #wins num = 0 break else: num2 = random.uniform(0,1) if P_2 < num2 < 1: #area corresponding to x_2 break #loses print(wins/10**7) The correct answer is 0.5821 however I am getting 0.655.. Where am I doing wrong ? Last edited: ## Answers and Replies Related Programming and Computer Science News on Phys.org Mark44 Mentor **Problem:** Here’s a little Monte Carlo challenge problem, Consider the following game, which uses the two spinner disks. Suppose a player spins one or the other of the pointers on the disks according to the following rules: (1) if the player spins pointer i and it stops in the region with area pij, he moves from disk i to disk j (i and j are either 1 or 2); (2) if a pointer stops in the region with area xi, the game ends; (3) if the game ends in the region with area x1, the player wins, but if the pointer stops in the region with area x2 the player loses. What is the probability the player, starting with disk 1, wins? Assume the area of each disk is one, so that x1+p11+p12 =1, as well as that x2+p21+p22 =1 Run your code for the case of p11 =0.2,", "# B. Quiz League ### гаралт стандарт гаралт A team quiz game called \"What? Where? When?\" is very popular in Berland. The game is centered on two teams competing. They are the team of six Experts versus the team of the Audience. A person from the audience asks a question and the experts are allowed a minute on brainstorming and finding the right answer to the question. All it takes to answer a typical question is general knowledge and common logic. The question sent be the audience are in envelops lain out in a circle on a round table. Each envelop is marked by the name of the asker's town. Each question is positioned in a separate sector. In the centre of the table is a spinning arrow. Thus, the table rather resembles a roulette table with no ball but with a spinning arrow instead. The host sets off the spinning arrow to choose a question for the experts: when the arrow stops spinning, the question it is pointing at is chosen. If the arrow points at the question that has already been asked, the host chooses the next unanswered question in the clockwise direction. Your task is to determine which will be the number of the next asked question if the arrow points at sector number $k$.", "# A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, ..., 8 (Fig. 9), which are equally likely outcomes. - Mathematics A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, ..., 8 (Fig. 9), which are equally likely outcomes. What is the probability that the arrow will point at (i) an odd number (ii) a number greater than 3 (iii) a number less than 9. #### Solution Arrow can come to rest at any of the numbers 1, 2, 3, 4, 5, 6, 7 and 8. Total number of events = 8 (i) There are four odd numbers 1, 3, 5 and 7. Probability that the arrow will point at an odd number is given by P (Arrow point at odd number)4/8=1/2 (ii) There are five numbers greater than 3, that is, 4, 5, 6, 7 and 8. Probability that the arrow will point at a number greater than 3 is given by P (Arrow point at a number greater than 3) = 5/8 (iii) All the numbers are less than 9. Probability that", "# What is the probability that the total score after throwing darts is divisible by $3$. To play a game of darts Michael throws three darts at the dart board shown. The number of points $$(1,$$ $$5$$ or $$10)$$ for each of the three regions is indicated. His score is the sum of the points for the three darts. If the radii of the three concentric circles are $$1,$$ $$2$$ and $$3$$ units, and each dart Michael throws hits this dart board at random, what is the probability that his score is evenly divisible by $$3?$$ Express your answer as a common fraction. After taking the values modulo $$3$$, we have $$1, 2, 1$$. I am pretty sure that the only way we can get divisible by $$3$$ in this problem is if we have modulos $$1, 1, 1$$ or $$2, 2, 2$$ for the darts. This means that the probability is $${\\left(\\dfrac23\\right)}^3+{\\left(\\dfrac13\\right)}^3=\\dfrac13$$. I feel as if I am missing something, or am I correct? Thanks! EDIT: \"At random\" means that the likelihood of a dart landing in a region is the total area of that region divided by the total area of the dart-board. • \"Each dart hits the board 'at random'\" This is not clear. Are we meant to assume that each of the three regions", "A game consists of spinning an arrow around a stationary disk as shown below. When the arrow comes to rest, there are eight equally likely outcomes. It could come to rest in any one of the sectors numbered $1, 2, 3,4, 5, 6, 7$ or $8$ as shown. Two such disks are used in a game where their arrows are independently spun. What is the probability that the sum of the numbers on the resulting sectors upon spinning the two disks is equal to 8 after the arrows come to rest? 1. $\\dfrac{1}{16}$ 1. $\\dfrac{5}{64}$ 1. $\\dfrac{3}{32}$ 1. $\\dfrac{7}{64}$ Given the, two disks When we select only one number from any of the two disks, then the probability $= \\frac{1}{8}$ When we select one number from the first disk and the one number from the second disk simultaneously, then the probability $= \\frac{1}{8} \\times \\frac{1}{8} = \\frac{1}{64}$ The numbers on the resulting sectors upon spinning the two disks is equal to $8$ after the arrows come to rest $= \\underbrace{(1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1)}_{\\text{Number of favorable outcomes = 7}}$ $\\therefore$ The required probability $= \\dfrac{7}{64}$ Correct Answer $:\\text{D}$ ${\\color{Magenta}{\\textbf{PS:}}}$ In probability, two events are independent if the incidence of one event does not affect the probability of the other event. If the incidence of one event", "# Will Archer A Survive? Probability Level 3 Three archers, A, B, and C, are standing equidistant from each other, forming an equilateral triangle. Archer A, B, and C has $\\frac { 1 }{ 3 }$, $\\frac { 2 }{ 3 }$, and $\\frac { 3 }{ 3 }$ probability of hitting the target they aimed, respectively. The three archers will play a survival game. The objective of the game for all players is to kill the other two archers and be the only survivor. The order of shooting will be in alphabetical order (A, B, then C). Assuming that all archers will die if he is hit by an arrow aimed at him, and that all archers will make the best moves possible to maximize their chances of winning (surviving), what is the probability that archer A will survive and win?", "Problem 21 Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even? $[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5)); label(\"A\",(-1,1)); label(\"B\",(2,1)); label(\"1\",(-.4,.4)); label(\"2\",(.4,.4)); label(\"3\",(.4,-.4)); label(\"4\",(-.4,-.4)); label(\"1\",(2.6,.4)); label(\"2\",(3.4,.4)); label(\"3\",(3,-.5)); [/asy]$ $\\textbf{(A)}\\ \\frac14\\qquad \\textbf{(B)}\\ \\frac13\\qquad \\textbf{(C)}\\ \\frac12\\qquad \\textbf{(D)}\\ \\frac23\\qquad \\textbf{(E)}\\ \\frac34$ ## Problem 22 At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is $\\frac25$. What fraction of the people in the room are married men? $\\textbf{(A)}\\ \\frac13\\qquad \\textbf{(B)}\\ \\frac38\\qquad \\textbf{(C)}\\ \\frac25\\qquad \\textbf{(D)}\\ \\frac{5}{12}\\qquad \\textbf{(E)}\\ \\frac35$ ## Problem 23 Tess runs counterclockwise around rectangular block $JKLM$. She lives at corner $J$. Which graph could represent her straight-line distance from home? $[asy] unitsize(5mm); pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2); draw(J--K--L--M--cycle); label(\"J\",J,NW); label(\"K\",K,SW); label(\"L\",L,SE); label(\"M\",M,NE); [/asy]$ ## Problem 24 In the figure, $ABCD$ is a rectangle and $EFGH$ is a parallelogram. Using the measurements given in the figure, what is the length $d$ of the segment that is perpendicular to $\\overline{HE}$ and $\\overline{FG}$? $[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair D=(0,0), C=(10,0), B=(10,8), A=(0,8); pair E=(4,8), F=(10,3), G=(6,0), H=(0,5); draw(A--B--C--D--cycle); draw(E--F--G--H--cycle); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"E\",E,N); label(\"F\",(10.8,3)); label(\"G\",G,S); label(\"H\",H,W); label(\"4\",A--E,N); label(\"6\",B--E,N);" ]

[ "probability of stopping; hence (2/3)(1/2)=1/3", "through $C$ is the sum from $i=0$ to $3$ of the probabilities that he enters intersection $C_i$ in the adjoining figure and goes east. The number of paths from $A$ to $C_i$ is ${{2+i} \\choose 2}$, because each such path has $2$ eastward block segments and they can occur in any order. The probability of taking any one of these paths to $C_i$ and then going east is $(\\tfrac{1}{2})^{3+i}$ because there are $3+i$ intersections along the way (including $A$ and $C_i$) where an independent choice with probability $\\tfrac{1}{2}$ is made. So the answer is $$\\sum\\limits_{i=0}^3 {{2+i} \\choose 2} \\left( \\frac{1}{2} \\right)^{3+i} = \\frac{1}{8} + \\frac{3}{16} + \\frac{6}{32} + \\frac{10}{64} = \\boxed{\\frac{21}{32}}.$$", "$\\frac16 \\left(1-1/9^{n-1}\\right)$ and of moving to state 1 is $\\frac56 \\left(1-1/9^{n-1}\\right)$. Then solve to find the limit of the probabilities of being in states 3 or 5. For state 1 you have a choice: either you can say the probability of moving to state 2 is $1/9$ and of moving to state 4 is $1/6$ with the probability of staying in state 1 is $1 - 1/9 -1/6 =13/18$, or you can short circuit this by saying for next change of state has probability of moving to state 2 of $\\frac{1/9}{1/9+1/6} = 2/5$ and of next moving to state 4 of $3/5$. As an example to check your result, I think for example if $m=2$ and $n=3$ then the probability of stopping with two consecutive $5$s is $251/340$ and of stopping with three consecutive $7$s is $89/340$. -", "# 3. Find the. (a) Probability of the pointer stopping on D in (Question 1-(a))? (a) Probability of the pointer stopping on D $=\\frac{1}{5}=0.2$", "your RHS. However, if that was so, then $p_2^* = P(A_2|F_1^c) = \\frac{P(F_1^c|A_2) \\cdot P(A_2)}{P(F_1^c)} = \\frac{1 \\cdot P(A_2)}{1} = p_2$, which doesn't make any sense. :( • Sep 3rd 2011, 09:11 PM CaptainBlack Re: Conditional probability question (with confusing wording) Quote: Originally Posted by Combinatus Thanks for your response! I hadn't heard of posterior probabilities before, but reading a bit on Wikipedia makes it apparent that it's what the problem is asking for. Still, I'm confused about something, and would like to ask a dumb question about the part quoted below. That looks a lot like my P(\"Plane is in region 1 after failing to find it in region 1\") = $P(A_1 | F_1^c) = a_1 P(A_1) / P(F_1^c)$ statement. If it is equivalent, then $P(F_1^c)$ should be 1. But why? Perhaps it makes sense to think of it as \"the probability of of not finding the plane in region 1, before looking for it\"? Or, am I missing something completely? --- ( There is a term missing in my earlier post: $p_1^*=\\frac{p_1 a}{p_1a+(1-p_1)}$ That is I forgot to divide by the probability that the plane is not found when we look in region 1, which is the sum of the probabilities of it being in region 1 and not seen plus the probability that it is", "## Precalculus (6th Edition) Blitzer The probability is $\\frac{5}{6}$. We know that there are six outcomes, so $\\text{n}\\left( \\text{S} \\right)=\\text{ 6}$. The event of stopping on yellow can be represented by ${{\\left( \\text{E} \\right)}_{\\text{yellow}}}\\text{ }=\\text{ }\\left\\{ 1 \\right\\}$ There is one outcome in this event, so $\\text{n}{{\\left( \\text{E} \\right)}_{\\text{yellow}}}=\\text{ 1}$. The probability of stopping on yellow is given below, $\\text{P}{{\\left( \\text{E} \\right)}_{\\text{yellow}}}\\text{ = }\\frac{\\text{n}{{\\left( \\text{E} \\right)}_{\\text{yellow}}}}{\\text{n}\\left( \\text{S} \\right)}\\text{ = }\\frac{1}{6}\\text{ }$ And the probability of not stopping on yellow is given below, \\begin{align} & {{\\text{P}}_{\\text{not yellow}}}\\text{ = 1}-\\text{ }{{\\text{P}}_{\\text{yellow}}} \\\\ & \\text{= 1}-\\text{ }\\frac{1}{6}\\text{ } \\\\ & \\text{= }\\frac{5}{6} \\end{align}", "volume of such region will evidently be $$x_m^{m-1}$$, so that the total volume of such a pyramid is $$V = \\sum\\limits_{1\\, \\le \\,k\\, \\le \\,r} {k^{\\,m - 1} }$$ The m-tuples in which $$x_m$$ is strictly higher than the other components are of course different (non-overlapping) with those in which the highest component is $$x_1$$ or $$x_2$$ and so on. Therefore we have $$m$$ of such regions. Now we have the basic elements to solve your question. In your particular case, $$m=3$$ and $$r=5$$, the probability that a decision be taken is $$P_{decision} = {{3\\sum\\limits_{1\\, \\le \\,k\\, \\le \\,5} {k^{\\,2} } } \\over {6^{\\,3} }} = {{165} \\over {216}} = {{55} \\over {72}}$$ • A decision can be made when the two lesser values are equal: $0\\le x_1\\le x_2 < x_3 \\le r$ – Daniel Mathias Feb 20 at 21:15 • @DanielMathias: you are right: thanks for signalling!. I amended the answer – G Cab Feb 20 at 23:39 The probability that a round ends without desicion is $$\\underbrace{3\\left(\\frac16\\right)^2\\left(\\frac56+\\frac46+\\frac36+\\frac26+\\frac16\\right)}_{\\frac{15}{72}}+ \\underbrace{6\\left(\\frac16\\right)^3}_{\\frac{2}{72}}=\\frac{17}{72}.$$ Here the first summand stands for two equal scores with a value larger than the third one. The factor $$3$$ stands for the number of ways to choose one person from three people. The values $$\\frac56,\\frac46,\\frac36,\\frac26,\\frac16$$ in the parentheses are the probabilities of the third score to be", "equal to $3$. probability = number of outcomes possible / total number of outcomes $\\frac{3}{6} = \\frac{1}{2}$", "the probability that all the remaining $$n-1$$ points will be in the clockwise semicircle is $$\\frac{1}{2^{n-1}}$$. This estimate is for a given starting point numbered at 1. To compute the probability for all $$N$$ points, we apply the same logic to each of the points in sequence. This yields $$\\frac{n}{2^{n-1}}$$ which is the probability that all $$n$$ points lie…", "## Elementary Statistics (12th Edition) $1.17 \\times 10^{-7}$ Yes. We know that $probability=\\frac{number \\ of \\ good \\ outcomes}{number\\ of\\ all\\ outcomes}$. Therefore $probability=\\frac{117}{1000000000}=1.17 \\times 10^{-7}$. Yes. Also, ait ravel has a larger fatality rate than travelling by car. Which maks this comparison unfair is that we use cars for shorter distances too, whilst aeroplanes not so much (e.g . there are not many airports).", "P(A ∩ B) option 1 is correct. 2) P(A̅ B̅) = 1 – P(A B) option 2 is incorrect. 3) $$P\\left( {\\bar A \\cup \\bar B} \\right) = P\\left( {\\overline {A \\cup B} } \\right)$$ # The probability that a person stopping at a gas station will ask to have his tyres checked is 0.12, the probability that he will ask to have his oil checked is 0.29 and the probability that he will ask to have them both checked is 0.07. The probability that a person who has his tyres checked will also have oil checked is 1. 0.34 2. 0.58 3. 0.24 4. 0.41 Option 2 : 0.58 ## Detailed Solution Concept: Conditional probability: It gives the probability of happening of any event if the other has already occurred. $${\\rm{P}}\\left( {\\frac{{{{\\rm{E}}_1}}}{{{{\\rm{E}}_2}}}} \\right) = {\\rm{Probability\\;of\\;getting\\;the\\;event\\;}}{{\\rm{E}}_1}{\\rm{\\;when\\;}}{{\\rm{E}}_2}{\\rm{is\\;already\\;occured}}.$$ $$P\\left( {\\frac{{{E_1}}}{{{E_2}}}} \\right) = \\frac{{P\\left( {{E_1} \\cap {E_2}} \\right)}}{{P\\left( {{E_2}} \\right)}}$$ Calculation: Given: P (E1) = Probability of stopping at the gas station and ask for tyre checked = 0.12 P (E2) = Probability of stopping at the gas station and ask for oil checked = 0.29 P (E1∩ E2) = Probability of both checked = 0.07 $${\\rm{P}}\\left( {\\frac{{{{\\rm{E}}_2}}}{{{{\\rm{E}}_1}}}} \\right) = {\\rm{Probability\\;of\\;person\\;who\\;has\\;his\\;tyre\\;checked\\;will\\;also\\;have\\;oil\\;checked}}$$ ∵ $$P\\left( {\\frac{{{E_2}}}{{{E_1}}}} \\right) = \\frac{{P\\left( {{E_1} \\cap {E_2}} \\right)}}{{P\\left( {{E_1}} \\right)}}$$ ∴ $$P\\left( {\\frac{{{E_2}}}{{{E_1}}}} \\right) = \\frac{{0.07}}{{0.12}} = 0.58$$", "10} \\right)}}{{\\left( {1 \\times 2 \\times 3} \\right) \\times 7!}} = 120$ Let us substitute these values in $P\\left( E \\right) = \\dfrac{{{}^5{C_3}}}{{{}^{10}{C_3}}}$. Therefore, we get $P\\left( E \\right) = \\dfrac{{10}}{{120}} = \\dfrac{1}{{12}}$. Therefore, the probability that none of the trucks chosen will meet emission standards is $\\dfrac{1}{{12}}$. Therefore, option C is correct. Note: For any event $A$, we can write $0 \\leqslant P\\left( A \\right) \\leqslant 1$ where $P\\left( A \\right)$ is the probability of event $A$. The sum of probabilities of all possible outcomes is always $1$. These are the properties of basic probability theory.", "For example, considering the fact that 38% of people live in the South but that “pop” is relatively rare to that region, what’s the posterior probability that the interviewee lives in the South? Per Bayes’ Rule (2.4), we can calculate this probability by $$$P(S | A) = \\frac{P(S)L(S|A)}{P(A)} . \\tag{2.6}$$$ We already have two of the three necessary pieces of the puzzle, the prior probability $$P(S)$$ and likelihood $$L(S|A)$$. Consider the third, the marginal probability that a person uses the term “pop” across the entire U.S., $$P(A)$$. By extending the Law of Total Probability (2.5), we can calculate $$P(A)$$ by combining the likelihoods of using “pop” in each region, while accounting for the regional populations. Accordingly, there’s a 28.26% chance that a person in the U.S. uses the word “pop”: $\\begin{split} P(A) & = L(M|A)P(M) + L(N|A)P(N) + L(S|A)P(S) + L(W|A)P(W) \\\\ & = 0.6447 \\cdot 0.21 + 0.2734 \\cdot 0.17 + 0.0792 \\cdot 0.38 + 0.2943 \\cdot 0.24\\\\ & \\approx 0.2826 . \\\\ \\end{split}$ Then plugging into (2.6), there’s a roughly 10.65% posterior chance that the interviewee lives in the South: $P(S | A) = \\frac{0.38 \\cdot 0.0792}{0.2826}\\; \\approx 0.1065 .$ We can similarly update our understanding of the interviewee living in the Midwest, Northeast, or West. Table 2.6 summarizes the resulting posterior model of region alongside", "1 dimension I just need an interactive and non interactive walk meaning I need only 14 programs. I am tempted to make a 3 dimensional random walk that walks to all points in an origin centered cube. In other words the next step can be any combination of $(-1,0,1)$ for any of the three axis. Thus giving 6 choices with walking towards a face, 8 choices of walking to a corner, and 12 choices of walking to an edge. This could be $p=\\frac{1}{26}$ or in my alternative version we would weight the probability based on the vector length of the walk. The probability would be inversely related to the vector length of the step. Thus setting the probability of walking to a face as unity with probability $p_{f}=\\frac{1}{u}$ and then set the others to the inversion of their vector lengths. Thus walking along a diagonal would be walking to an edge and would have a probability of $p_{e} = \\frac{1}{u \\sqrt{2}}$ and walking to a corner would have probability $p_{c} =\\frac{1}{u \\sqrt{3}}$. Then a basic law of probability would be used to create this equation $1= 6 \\frac{1}{u} + 12 \\frac{1}{u \\sqrt{2}} + 8 \\frac{1}{u \\sqrt{3}}$ which would solve for $u=6+12 \\frac{1}{\\sqrt{2}} + 8 \\frac{1}{\\sqrt{3}} = 19.10408353$ . ### The 3 dimensional walks have all worked now I", "obvious to me, the three probabilities seem to be multiples of $\\frac1{351}$ – Henry Apr 7 at 9:26 • I have edited my post, if you wouldn't mind checking – Riley Lenway Apr 7 at 9:56 • As far as I can tell $\\frac {\\frac 16}{\\frac 16 + \\frac {1}{36} + \\frac {1}{216}}=\\frac{36}{43}\\approx 0.8372$ is slightly too low – Henry Apr 7 at 10:03 Let's define $$P_n$$ as the probability of stopping after exactly $$n$$ rolls, and $$Q_n$$ as the probability of stopping by the time we reach $$n$$ rolls. We can express the probability of stopping on the $$n$$th roll after a $$1$$ as $$A_n=\\frac{1}{6}(1-Q_{n-1})$$ (and $$A_1=\\frac{1}{6}$$, after a pair of 2s as $$B_n=\\frac{1}{36}(1-Q_{n-2})$$ (and $$B_1=0, B_2=\\frac{1}{36}$$), and after a trio of 3s as $$C_n=\\frac{1}{216}(1-Q_{n-3})$$ (and $$C_2=C_1=0, C_3=\\frac{1}{216}$$). We can now build the rest our recursive definitions: $$P_n=A_n+B_n+C_n$$ $$Q_n=\\sum_{j=1}^{n}P_j$$ The chances of ending after a 1 are $$\\sum_{j=1}^\\infty A_j$$, after a pair of 2s are $$\\sum_{j=1}^\\infty B_j$$, and after a trio of 3s are $$\\sum_{j=1}^\\infty C_j$$. Using a computer for computation, I got that these are $$\\frac{36}{43},\\frac{6}{43}$$ and $$\\frac{1}{43}$$ respectively. I could conjecture a reason for this, but it would be only a guess • $B_n$ is only $1/36$ of those whose $n-2$th roll is not 2. Otherwise it contributes to $B_{n-1}$ – Empy2 Apr 8 at", "X_{i-1}=w) = P(X_1=v|X_0=w).$$ Is this correct? BTW, the original equation comes from a stopping rule called the \"filling rule\". - The $p_{w,v}$ is meant to be the probability of going from $w$ at time $i-1$ to $v$ at time $i$. Once the system decides not to stop at time $i-1$, where it goes next is independent of everything (by definition of a stopping time), so that $P(X_i=v|X_{i-1}=w) = P(X_i=v|X_{i-1}=w,T\\ge i)=p_{w,v}$. This gives you what you need.", "that this is absurd, but imagine the probability of finding the your spot at any point is $P$. Now try to calculate the probability of finding your point within a distance $d$ of $x=0$. Since your probability is defined as $\\frac{\\text{probability of event happening}}{\\text{sum of all probabilities}}$, we calculate the probability as $$\\frac{\\int_{-d}^dPdx}{\\int_{-\\infty}^{\\infty}P dx} = \\frac{2d}{\\infty} = 0.$$ Therefore, given that you have traveled any distance $d$, there is $0$ probability that you will find your point. -", "I computed all the different possible cases, calculated the number of stops for each case, and calculated the expected value. Lo and behold your formula nailed it. Not that you probably care, but where I think I goofed up was thinking of this more like a random process. Instead of thinking that each passenger randomly picks a stop at the beginning, I was thinking of each passenger going to a stop and deciding whether to get off or not, with probability $1/n$. But that is not a good model of what is happening. For one thing, to work it this way, the probabilities of getting off would have to increase to 1, and not stay constant at $1/n$ (\"last stop, everybody off). I interpreted $Pr(X_i=0) = (1-1/n)^k$ as the probability that of k people on the bus that they don't get off at that stop. And my thought went that there were only k people on the bus near the beginning and certainly not that many toward the end. So it was this formula that I was questioning. I humbly submit that you did this problem correctly. But darn it, you should see the formula I had come up with! It is correct, but it is ugly. And no, it's not hard to swallow: a) I am quite", "not smoke. If the probability of death due to lung cancer in the region is 0.006, what is the probability of death due to lung cancer given that a person is a smoker? 1. a. 1/140 2. b. 1/70 3. c. 3/140 4. d. 1/10 Solution: 1. Answer: (c) E -> death due to lung cancer E1 -> Person is a smoker E2 -> Person is a non-smoker P(E1) = 1/5 P(E2) = 4/5 P(E) = 0.006 P(E/E1) = 10P(E/E2) Now, P(E) = P(E1).P(E/E1)+P(E2).P(E/E2) 0.006 = (1/5)P(E/E1)+(4/5)(1/10)P(E/E1) => P(E/E1) = 3/140 Question 67. A person goes to the office by car or scooter or bus or train, the probability of which are 1/7, 3/7, 2/7 and 1/7 respectively. The probability that he reaches office late, if he takes a car, scooter, bus or train is 2/9, 1/9, 4/9 and 1/9 respectively. Given that he reached office in time, the probability that he travelled by car is 1. a. 1/7 2. b. 2/7 3. c. 3/7 4. d. 4/7 Solution: 1. Answer: (a) Using Baye's Theorem Let Ei -> Probability of travelling by car, scooter, bus, train P(E/E1)-> Probability of reaching on time = (1/7)(7/8) ÷(1/7)(7/9)+(3/7)(8/9)+(2/7)(5/9)+(1/7)(8/9) = 1/7 Question 68. The value of $$\\int \\frac{(x-2)dx}{\\left \\{ (x-2)^{2}(x+3)^{7} \\right \\}^{\\frac{1}{3}}}$$ is 1. a. (3/20)((x-2)/(x+3))4/3+c 2. b. (3/20)((x-2)/(x+3))3/4+c 3. c. (5/12)((x-2)/(x+3))4/3+c", "is 702/1024 = 0.6855, and the mean outcome is +1.953. If we looked the mean value of outcome per trial in the previous calculation, i.e. using $\\frac{+5}{5}$, $\\frac{+3}{5}$, $\\frac{+1}{5}$, $\\frac{-1}{5}$, $\\frac{-1}{3}$ and $\\frac{-1}{1}$ then we would get +0.184. These are the senses in which there is bias by stopping early in the second scheme, and the bias is in the predicted direction. But it is not the full story. Why do whuber and probabilityislogic think stopping early should produce unbiased results? We know the expected outcome of the trials in the second scheme is +1.953. The expected number of trials turns out to be 3.906. So dividing one by the other we get +0.5, exactly as before and what was described as unbiased. - you are taking the perspective of \"pre-data\" world. What you say is true, that the stopping rule matters but only before you consider the data. This is because the stopping rule provides information about the data, but not about the true probabilities. So once the data is in, the stopping rule no longer matters. Note that the true probabilities are unknown in the actual experiment. So you also need to consider situations when the probabilities are, say $P(+)=\\frac{1}{4}$ and $P(-)=\\frac{3}{4}$, as well as any other possible combination. – probabilityislogic Mar 25 '11 at 21:59" ]

[ "cents each, and Spain $5$ cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.) The average price of his $'70$s stamps is closest to (A) $3.5$ cents (B) $4$ cents (C) $4.5$ cents (D) $5$ cents (E) $5.4$ cents AMC 8, 2002, Problem 11 A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth? (A) $11$ (B) $12$ (C) $13$ (D) $14$ (E) $15$ AMC 8, 2002, Problem 12 A board game spinner is divided into three regions labeled $A, B$ and $C$. The probability of the arrow stopping on region $A$ is $\\frac{1}{3}$ and on region $B$ is $\\frac{1}{2}$. The probability of the arrow stopping on region $C$ is (A) $\\frac{1}{12}$ (B) $\\frac{1}{6}$ (C) $\\frac{1}{5}$ (D) $\\frac{1}{3}$ (E) $\\frac{2}{5}$ AMC 8, 2002, Problem 15 Which of the following polygons has the largest area? (A) $A$ (B) $B$ (C) $C$ (D) $D$ (E) $E$ AMC 8, 2002, Problem 16 Right isosceles triangles are constructed on the sides of a $3-4-5$ right triangle, as shown. A capital letter represents the area of each triangle. Which one of", "Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd? (A) $\\frac{17}{36}$ (B) $\\frac{35}{72}$ (C) $\\frac{1}{2}$ (D) $\\frac{37}{72}$ (E) $\\frac{19}{36}$ AMC 8, 2006, Problem 4 Initially, a spinner points west. Chenille moves it clockwise $2 \\frac{1}{4}$ revolutions and then counterclockwise $3 \\frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves? (A) north (B) east (C) south (D) west (E) northwest AMC 8, 2006, Problem 10 Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12 .$ What should his graph look like? AMC 8, 2006, Problem 17 Jeff rotates spinners $P, Q$ and $R$ and adds the resulting numbers. What is the probability that his sum is an odd number? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ (C) $\\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2006, Problem 20 A singles tournament had six players. Each player played every other player only once, with no ties. If Helen", "of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3$ ? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ $(C) \\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2007, Problem 25 On the dart board shown in the Figure, the outer circle has radius $6$ and the inner circle has a radius of $3$ . Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd? (A) $\\frac{17}{36}$ (B) $\\frac{35}{72}$ (C) $\\frac{1}{2}$ (D) $\\frac{37}{72}$ (E) $\\frac{19}{36}$ AMC 8, 2006, Problem 4 Initially, a spinner points west. Chenille moves it clockwise $2 \\frac{1}{4}$ revolutions and then counterclockwise $3 \\frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves? (A) north (B) east (C) south (D) west (E) northwest AMC 8, 2006, Problem 10 Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12", "faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "fraction Answer: $$\\frac{2}{5}$$ AMC-8, 2016 problem 21 Challenges and Thrills in Pre College Mathematics ## Try with Hints There are 5 Chips, 3 red and 2 green Can you now finish the problem .......... We draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. one at a time without replacement Can you finish the problem........ There are 5 Chips, 3 red and 2 green we draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. if we draw all the green chip boxes then the last box be red or if we draw all red boxes then the last box be green but we draw randomly. there are 3 red boxes and 2 green boxes Therefore the probability that the 3 reds are drawn=$$\\frac{2}{5}$$ # Problem from Probability | AMC 8, 2004 | Problem no. 21 Try this beautiful problem from Probability from AMC 8, 2004. ## Problem from Probability | AMC-8, 2004 | Problem 21 Spinners A and B are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even? • $$\\frac{2}{3}$$ • $$\\frac{1}{3}$$ • $$\\frac{1}{4}$$", "= $\\frac{1}{2}$ • The probability of landing on heads is $P$(tails) = $\\frac{1}{2}$ • Dice: (a six-sided die, if singular) • When rolling a die, it will land with one of its six flat faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$ #### Lessons • Introduction Introduction to Probability Outcomes for Coins, Dice, and Spinners: a) What is probability? b) What are outcomes in probability? 1. Outcomes for flipping a coin 2. Outcomes for rolling", "There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "# Math Help - Spinner 1. ## Spinner A spinner contains eight regions, numbered 1 through 8. The arrow has an equally likely chance of landing on any of the eight regions. If the arrow lands on the line, it is spun again. What is the probability that the arrow lands on an odd number? 2. Originally Posted by symmetry A spinner contains eight regions, numbered 1 through 8. The arrow has an equally likely chance of landing on any of the eight regions. If the arrow lands on the line, it is spun again. What is the probability that the arrow lands on an odd number? The odd numbers, 1-8, are 1, 3, 5, 7, thus there are 4 odd numbers out of the 8, and therefore the probability that the arrow lands on an odd number is 4/8 or 1/2. 3. ## ok I thank you for your great help.", "## Section10.1An Introduction to Probability We continue with an informal discussion intended to motivate the more structured development that will follow. Consider the “spinner” shown in Figure 10.1. Suppose we give it a good thwack so that the arrow goes round and round. We then record the number of the region in which the pointer comes to rest. Then observers, none of whom have studied combinatorics, might make the following comments: 1. The odds of landing in region $$1$$ are the same as those for landing in region $$3\\text{.}$$ 2. You are twice as likely to land in region $$2$$ as in region $$4\\text{.}$$ 3. When you land in an odd numbered region, then 60% of the time, it will be in region $$5\\text{.}$$ We will now develop a more formal framework that will enable us to make such discussions far more precise. We will also see whether Alice is being entirely fair to Bob in her proposed game to one hundred. We begin by defining a probability space as a pair $$(S,P)$$ where $$S$$ is a finite set and $$P$$ is a function that whose domain is the family of all subsets of $$S$$ and whose range is the set $$[0,1]$$ of all real numbers which are non-negative and at most one. Furthermore, the following two key", "prime number are: 2, 3 and 5 (b) not a prime number are : 1 , 4 and 6 (ii) (a) a number greater than 5 is : 6 (b) a number not greater than 5 are: 1, 2, 3, 4 and 5 Find the 1. Probability of the pointer stopping on D ? 2. Probability of getting an ace from a well shuffled deck of 52 playing cards? 3. Probability of getting a red apple. (See figure below) Ans- (i) The possible outcomes on spinning a wheel are : A, B, C, and D Therefore, the pointer can stop at one of the regions. Total number of regions is 5. The pointer will stop at region D only once. Hence, probability that the pointer will stop at region D = $\\frac{1}{5}.$ (ii) Total number of cards in a well shuffled deck = 52 Total number of Ace cards = 4. Therefore, the probability of getting an ace card = $\\frac{4}{52}\\text{ }=\\frac{1}{13}.$ (iii)Total number of apples = 7 Total number of red apples = 4 Total number of green apples = 3 Therefore, the probability of getting a red apple = $\\frac{4}{7}$ Q. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is", "What is the probability that $x<y$? • $\\frac{1}{8}$ • $\\frac{1}{6}$ • $\\frac{2}{3}$ ### Key Concepts Number system Cube Answer: $\\frac{1}{8}$ AMC-10A (2003) Problem 12 Pre College Mathematics ## Try with Hints The given vertices are $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$.if we draw a figure using the given points then we will get a rectangle as shown above.Clearly lengtht of $OC$= $4$ and length of $AO$=$1$.Therefore area of the rectangle is $4 \\times 1=4$.now we have to find out the probability that $x<y$.so we draw a line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.can you find out the area with the condition $x<y$? can you finish the problem…….. Now the line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.Therefore it will form a Triangle $\\triangle AOD$ (as shown above) whose $AO=1$ and $AD=1$.Therefore area of $\\triangle AOD=\\frac{1}{2}$ i.e (red region).Now can you find out the probability with the condition $x<y$? can you finish the problem…….. Therefore the required probability ($x<y$) is $\\frac{\\frac{1}{2}}{4}$=$\\frac{1}{8}$ Categories ## Problem from Probability | AMC 8, 2004 | Problem no. 21 Try this beautiful problem from Probability from AMC 8, 2004. ## Problem from Probability | AMC-8, 2004 | Problem 21 Spinners A and B are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability", "spinner to land on either a violet region or a region labeled $a$, or both. • There are $3$ violet regions and $3$ regions labelled $a$. • There is a $1$ region where it’s both violet and labeled $a$. This shows that the incident is mutually inclusive. Hence, we use $P(A \\text{ or } B) = P(A) + P(B) – P(A \\text{ and } B)$ \\begin{aligned}P(V \\text{ or } a) &= P(V) + P(a) – P(V \\text{ and } a)\\\\&=\\dfrac{3}{7} + \\dfrac{3}{7} – \\dfrac{1}{7}\\\\&= \\dfrac{5}{7}\\end{aligned} a. This means that the probability is equal to $\\dfrac{5}{7}$. It’s impossible to land on a red region and a blue one all at the same time. This means that these two events are mutually exclusive. For these types of events, we add their individual probabilities. b. This means that the probability is equal to $\\dfrac{1}{7} + \\dfrac{2}{7} = \\dfrac{3}{7}$. ### Practice Questions 1. A canvas bag contains $12$ pink cubes, $20$ green cubes, and $22$ purple cubes. One cube is removed from the bag and then replaced. Another cube is drawn from the bag, and repeat this one more time. What is the probability that the first cube is green, the second cube is purple, and the third is another green cube? 2. In a book club of $50$ enthusiastic readers, $26$", "### Home > INT2 > Chapter 3 > Lesson 3.1.5 > Problem3-61 3-61. The spinner at right has three regions: $\\5,\\0,$ and $\\20$. To play the game, you spin one time and then you win the amount in the region that the spinner lands on. 1. What is the expected value for the game? Compare the given angle measurements to $360^\\circ$. Multiply these ratios by the amount possible to win in each section. $\\text{EV}=\\5\\cdot \\frac{1}{4}+\\0\\cdot \\frac{135}{360}+\\20\\cdot \\frac{135}{360}$ 2. If you have to pay $\\10$ to play, is it a fair game?", "a “3-dimensional square”, I re-phrased Alexander’s question into two dimensions. The trickier part was thinking what it would mean to “roll” a 2-dimensional shape. The outside of an nxn square is painted red and is chopped into $n^2$ unit squares. The latter are thoroughly mixed up and put into a bag. One small square is withdrawn at random from the bag and spun on a flat surface. What is the probability that the spinner stops with a red side facing you? Shown below is a 4×4 square, but in all sizes of 2-dimensional squares, there are three possible types in the bag: those with 2, 1, or 0 sides painted. I solved my first variation case by case. In any nxn square, • There are 4 corner squares with 2 sides painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4}{n^2} \\cdot \\frac{2}{4} = \\frac{2}{n^2}$. • There are $4(n-2)$ edge squares not in a corner with 1 side painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4(n-2)}{n^2} \\cdot \\frac{1}{4} = \\frac{n-2}{n^2}$. • All other squares have 0 sides painted, so the probability of picking one of those squares and then spinning a red side is 0. • Adding the probabilities for", "# What is the probability that the total score after throwing darts is divisible by $3$. To play a game of darts Michael throws three darts at the dart board shown. The number of points $$(1,$$ $$5$$ or $$10)$$ for each of the three regions is indicated. His score is the sum of the points for the three darts. If the radii of the three concentric circles are $$1,$$ $$2$$ and $$3$$ units, and each dart Michael throws hits this dart board at random, what is the probability that his score is evenly divisible by $$3?$$ Express your answer as a common fraction. After taking the values modulo $$3$$, we have $$1, 2, 1$$. I am pretty sure that the only way we can get divisible by $$3$$ in this problem is if we have modulos $$1, 1, 1$$ or $$2, 2, 2$$ for the darts. This means that the probability is $${\\left(\\dfrac23\\right)}^3+{\\left(\\dfrac13\\right)}^3=\\dfrac13$$. I feel as if I am missing something, or am I correct? Thanks! EDIT: \"At random\" means that the likelihood of a dart landing in a region is the total area of that region divided by the total area of the dart-board. • \"Each dart hits the board 'at random'\" This is not clear. Are we meant to assume that each of the three regions", "A game consists of spinning an arrow around a stationary disk as shown below. When the arrow comes to rest, there are eight equally likely outcomes. It could come to rest in any one of the sectors numbered $1, 2, 3,4, 5, 6, 7$ or $8$ as shown. Two such disks are used in a game where their arrows are independently spun. What is the probability that the sum of the numbers on the resulting sectors upon spinning the two disks is equal to 8 after the arrows come to rest? 1. $\\dfrac{1}{16}$ 1. $\\dfrac{5}{64}$ 1. $\\dfrac{3}{32}$ 1. $\\dfrac{7}{64}$ Given the, two disks When we select only one number from any of the two disks, then the probability $= \\frac{1}{8}$ When we select one number from the first disk and the one number from the second disk simultaneously, then the probability $= \\frac{1}{8} \\times \\frac{1}{8} = \\frac{1}{64}$ The numbers on the resulting sectors upon spinning the two disks is equal to $8$ after the arrows come to rest $= \\underbrace{(1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1)}_{\\text{Number of favorable outcomes = 7}}$ $\\therefore$ The required probability $= \\dfrac{7}{64}$ Correct Answer $:\\text{D}$ ${\\color{Magenta}{\\textbf{PS:}}}$ In probability, two events are independent if the incidence of one event does not affect the probability of the other event. If the incidence of one event", "the shaded region = 42 - 2 π(2)22 = 4 units [Area of square - Area of 2 semi circles.] Area of the board = π × 162 = 768 units Probability of dart hitting the target = 4 / 768 = 1 / 192 3. Which of the following geometric probabilities is impossible? a. 0.45 b. 1.2 c. 0 d. 1 #### Solution: As probability always lies between 0 and 1, geometric probability can never be greater than 1. 4. What is the probability of getting 75 points when the spinner is spun? a. $\\frac{1}{3}$ b. 1 c. $\\frac{5}{18}$ d. $\\frac{2}{9}$ #### Solution: P(getting 75 points) = Angle subtended by arc CD total angle P (getting 75 points) = 100o360o [Substitute.] P(getting 75 points) = 5 / 18 [Simplify.] 5. A dart thrown at a rectangular board ABCD is equally likely to land on any point. Find the probability of the dart hitting the inner rectangle PQRS. [Given $a$ = 21 cm, $b$ = 16 cm, $c$ = 18 cm, $d$ = 10 cm.] a. $\\frac{29}{15}$ b. $\\frac{15}{28}$ c. $\\frac{4}{7}$ d. $\\frac{28}{15}$ #### Solution: Area of the rectangle ABCD = AB × BC = 21 × 16 = 336 cm 2 [Area of rectangle = length × breadth.] Area of the rectangle PQRS = PQ × QR", "= random.uniform(0,1) if P_1 < num < 1: #area corresponding to x_1 wins += 1 #wins num = 0 break else: num2 = random.uniform(0,1) if P_2 < num2 < 1: #area corresponding to x_2 break #loses print(wins/10**7) The correct answer is 0.5821 however I am getting 0.655.. Where am I doing wrong ? Last edited: ## Answers and Replies Related Programming and Computer Science News on Phys.org Mark44 Mentor **Problem:** Here’s a little Monte Carlo challenge problem, Consider the following game, which uses the two spinner disks. Suppose a player spins one or the other of the pointers on the disks according to the following rules: (1) if the player spins pointer i and it stops in the region with area pij, he moves from disk i to disk j (i and j are either 1 or 2); (2) if a pointer stops in the region with area xi, the game ends; (3) if the game ends in the region with area x1, the player wins, but if the pointer stops in the region with area x2 the player loses. What is the probability the player, starting with disk 1, wins? Assume the area of each disk is one, so that x1+p11+p12 =1, as well as that x2+p21+p22 =1 Run your code for the case of p11 =0.2,", "### Home > A2C > Chapter Ch4 > Lesson 4.2.4 > Problem4-127 4-127. 1. A dart hits each of these dart boards at random. What is the probability that the dart will not land in the shaded area? Homework Help ✎ P(not in shaded area) = P(white) $P(\\text{white})=\\frac{\\text{area of success}}{\\text{total area}}=\\frac{\\text{area of white}}{\\text{area of large circle}}$ Area = πr2 $P(\\text{not in shaded area})=\\frac{1}{4}$ See part (a).", "# Will Archer A Survive Extended... Three archers, A, B, and C, are standing equidistant from each other, forming an equilateral triangle. The chances that A, B, and C successfully hit a target they aim at are $$\\frac { 1 }{ 3 }$$, $$\\frac { 2 }{ 3 }$$, and $$\\frac { 3 }{ 3 } ,$$ respectively. The three archers will play a survival game. The objective of the game for all players is to kill the other two archers and be the only survivor. The order of shooting is to be chosen AT RANDOM. Assuming that all archers will die if he is hit by an arrow aimed at him, and that all archers will make the best possible optimal moves to maximize their chances of winning (surviving), the probability that Archer A wins can be expressed as $$\\dfrac{a}{b}$$ where $$a$$ and $$b$$ are co-prime positive integers. Find the value of $$a+b$$. ×" ]

[ "cents each, and Spain $5$ cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.) The average price of his $'70$s stamps is closest to (A) $3.5$ cents (B) $4$ cents (C) $4.5$ cents (D) $5$ cents (E) $5.4$ cents AMC 8, 2002, Problem 11 A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth? (A) $11$ (B) $12$ (C) $13$ (D) $14$ (E) $15$ AMC 8, 2002, Problem 12 A board game spinner is divided into three regions labeled $A, B$ and $C$. The probability of the arrow stopping on region $A$ is $\\frac{1}{3}$ and on region $B$ is $\\frac{1}{2}$. The probability of the arrow stopping on region $C$ is (A) $\\frac{1}{12}$ (B) $\\frac{1}{6}$ (C) $\\frac{1}{5}$ (D) $\\frac{1}{3}$ (E) $\\frac{2}{5}$ AMC 8, 2002, Problem 15 Which of the following polygons has the largest area? (A) $A$ (B) $B$ (C) $C$ (D) $D$ (E) $E$ AMC 8, 2002, Problem 16 Right isosceles triangles are constructed on the sides of a $3-4-5$ right triangle, as shown. A capital letter represents the area of each triangle. Which one of", "Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd? (A) $\\frac{17}{36}$ (B) $\\frac{35}{72}$ (C) $\\frac{1}{2}$ (D) $\\frac{37}{72}$ (E) $\\frac{19}{36}$ AMC 8, 2006, Problem 4 Initially, a spinner points west. Chenille moves it clockwise $2 \\frac{1}{4}$ revolutions and then counterclockwise $3 \\frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves? (A) north (B) east (C) south (D) west (E) northwest AMC 8, 2006, Problem 10 Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12 .$ What should his graph look like? AMC 8, 2006, Problem 17 Jeff rotates spinners $P, Q$ and $R$ and adds the resulting numbers. What is the probability that his sum is an odd number? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ (C) $\\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2006, Problem 20 A singles tournament had six players. Each player played every other player only once, with no ties. If Helen", "of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3$ ? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ $(C) \\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2007, Problem 25 On the dart board shown in the Figure, the outer circle has radius $6$ and the inner circle has a radius of $3$ . Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd? (A) $\\frac{17}{36}$ (B) $\\frac{35}{72}$ (C) $\\frac{1}{2}$ (D) $\\frac{37}{72}$ (E) $\\frac{19}{36}$ AMC 8, 2006, Problem 4 Initially, a spinner points west. Chenille moves it clockwise $2 \\frac{1}{4}$ revolutions and then counterclockwise $3 \\frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves? (A) north (B) east (C) south (D) west (E) northwest AMC 8, 2006, Problem 10 Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12", "# Math Help - Spinner 1. ## Spinner A spinner contains eight regions, numbered 1 through 8. The arrow has an equally likely chance of landing on any of the eight regions. If the arrow lands on the line, it is spun again. What is the probability that the arrow lands on an odd number? 2. Originally Posted by symmetry A spinner contains eight regions, numbered 1 through 8. The arrow has an equally likely chance of landing on any of the eight regions. If the arrow lands on the line, it is spun again. What is the probability that the arrow lands on an odd number? The odd numbers, 1-8, are 1, 3, 5, 7, thus there are 4 odd numbers out of the 8, and therefore the probability that the arrow lands on an odd number is 4/8 or 1/2. 3. ## ok I thank you for your great help.", "faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "## Section10.1An Introduction to Probability We continue with an informal discussion intended to motivate the more structured development that will follow. Consider the “spinner” shown in Figure 10.1. Suppose we give it a good thwack so that the arrow goes round and round. We then record the number of the region in which the pointer comes to rest. Then observers, none of whom have studied combinatorics, might make the following comments: 1. The odds of landing in region $$1$$ are the same as those for landing in region $$3\\text{.}$$ 2. You are twice as likely to land in region $$2$$ as in region $$4\\text{.}$$ 3. When you land in an odd numbered region, then 60% of the time, it will be in region $$5\\text{.}$$ We will now develop a more formal framework that will enable us to make such discussions far more precise. We will also see whether Alice is being entirely fair to Bob in her proposed game to one hundred. We begin by defining a probability space as a pair $$(S,P)$$ where $$S$$ is a finite set and $$P$$ is a function that whose domain is the family of all subsets of $$S$$ and whose range is the set $$[0,1]$$ of all real numbers which are non-negative and at most one. Furthermore, the following two key", "= $\\frac{1}{2}$ • The probability of landing on heads is $P$(tails) = $\\frac{1}{2}$ • Dice: (a six-sided die, if singular) • When rolling a die, it will land with one of its six flat faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$ #### Lessons • Introduction Introduction to Probability Outcomes for Coins, Dice, and Spinners: a) What is probability? b) What are outcomes in probability? 1. Outcomes for flipping a coin 2. Outcomes for rolling", "There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "to the area of the region. When two darts hit this board, the score is the sum of the point values of the regions hit. What is the probability that the score is odd? $$\\dfrac{17}{36}$$ $$\\dfrac{35}{72}$$ $$\\dfrac{1}{2}$$ $$\\dfrac{37}{72}$$ $$\\dfrac{19}{36}$$ ###### Solution(s): The area of the outer circle is $$6^2\\pi = 36\\pi,$$ and the area of the inner circle is $$3^2\\pi = 9\\pi.$$ Therefore, the area of the outer ring is $$36\\pi - 9\\pi = 27\\pi.$$ This means that the probability of hitting an outer segment is $$\\dfrac{9\\pi}{36\\pi} = \\dfrac{1}{4}.$$ Similarly, the probability of hitting an inner segment is $$\\dfrac{3\\pi}{36\\pi} = \\dfrac{1}{12}.$$ Summing over all possibilities, the probability of hitting a $$1$$ is $\\dfrac{1}{12} + 2 \\cdot \\dfrac{1}{4} = \\dfrac{7}{12}.$ Similarly, the probability of hitting a $$2$$ is $2 \\cdot \\dfrac{1}{12} + \\dfrac{1}{4} = \\dfrac{5}{12}.$ The only way to get an odd score is to hit one $$1$$ and one $$2$$. The probability of hitting these numbers in either order is $2 \\cdot \\dfrac{5}{12} \\cdot \\dfrac{7}{12} = \\dfrac{35}{72}.$ Thus, B is the correct answer.", "A game consists of spinning an arrow around a stationary disk as shown below. When the arrow comes to rest, there are eight equally likely outcomes. It could come to rest in any one of the sectors numbered $1, 2, 3,4, 5, 6, 7$ or $8$ as shown. Two such disks are used in a game where their arrows are independently spun. What is the probability that the sum of the numbers on the resulting sectors upon spinning the two disks is equal to 8 after the arrows come to rest? 1. $\\dfrac{1}{16}$ 1. $\\dfrac{5}{64}$ 1. $\\dfrac{3}{32}$ 1. $\\dfrac{7}{64}$ Given the, two disks When we select only one number from any of the two disks, then the probability $= \\frac{1}{8}$ When we select one number from the first disk and the one number from the second disk simultaneously, then the probability $= \\frac{1}{8} \\times \\frac{1}{8} = \\frac{1}{64}$ The numbers on the resulting sectors upon spinning the two disks is equal to $8$ after the arrows come to rest $= \\underbrace{(1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1)}_{\\text{Number of favorable outcomes = 7}}$ $\\therefore$ The required probability $= \\dfrac{7}{64}$ Correct Answer $:\\text{D}$ ${\\color{Magenta}{\\textbf{PS:}}}$ In probability, two events are independent if the incidence of one event does not affect the probability of the other event. If the incidence of one event", "# What is the probability that the total score after throwing darts is divisible by $3$. To play a game of darts Michael throws three darts at the dart board shown. The number of points $$(1,$$ $$5$$ or $$10)$$ for each of the three regions is indicated. His score is the sum of the points for the three darts. If the radii of the three concentric circles are $$1,$$ $$2$$ and $$3$$ units, and each dart Michael throws hits this dart board at random, what is the probability that his score is evenly divisible by $$3?$$ Express your answer as a common fraction. After taking the values modulo $$3$$, we have $$1, 2, 1$$. I am pretty sure that the only way we can get divisible by $$3$$ in this problem is if we have modulos $$1, 1, 1$$ or $$2, 2, 2$$ for the darts. This means that the probability is $${\\left(\\dfrac23\\right)}^3+{\\left(\\dfrac13\\right)}^3=\\dfrac13$$. I feel as if I am missing something, or am I correct? Thanks! EDIT: \"At random\" means that the likelihood of a dart landing in a region is the total area of that region divided by the total area of the dart-board. • \"Each dart hits the board 'at random'\" This is not clear. Are we meant to assume that each of the three regions", "= random.uniform(0,1) if P_1 < num < 1: #area corresponding to x_1 wins += 1 #wins num = 0 break else: num2 = random.uniform(0,1) if P_2 < num2 < 1: #area corresponding to x_2 break #loses print(wins/10**7) The correct answer is 0.5821 however I am getting 0.655.. Where am I doing wrong ? Last edited: ## Answers and Replies Related Programming and Computer Science News on Phys.org Mark44 Mentor **Problem:** Here’s a little Monte Carlo challenge problem, Consider the following game, which uses the two spinner disks. Suppose a player spins one or the other of the pointers on the disks according to the following rules: (1) if the player spins pointer i and it stops in the region with area pij, he moves from disk i to disk j (i and j are either 1 or 2); (2) if a pointer stops in the region with area xi, the game ends; (3) if the game ends in the region with area x1, the player wins, but if the pointer stops in the region with area x2 the player loses. What is the probability the player, starting with disk 1, wins? Assume the area of each disk is one, so that x1+p11+p12 =1, as well as that x2+p21+p22 =1 Run your code for the case of p11 =0.2,", "# A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, ..., 8 (Fig. 9), which are equally likely outcomes. - Mathematics A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, ..., 8 (Fig. 9), which are equally likely outcomes. What is the probability that the arrow will point at (i) an odd number (ii) a number greater than 3 (iii) a number less than 9. #### Solution Arrow can come to rest at any of the numbers 1, 2, 3, 4, 5, 6, 7 and 8. Total number of events = 8 (i) There are four odd numbers 1, 3, 5 and 7. Probability that the arrow will point at an odd number is given by P (Arrow point at odd number)4/8=1/2 (ii) There are five numbers greater than 3, that is, 4, 5, 6, 7 and 8. Probability that the arrow will point at a number greater than 3 is given by P (Arrow point at a number greater than 3) = 5/8 (iii) All the numbers are less than 9. Probability that", "# Expected value (basic) Expected value uses probability to tell us what outcomes to expect in the long run. ## Problem 1: Board game spinner A board game uses the spinner shown below to determine how many spaces a player will move forward on each turn. The probability is $\\dfrac12$ that the player moves forward $1$ space, and moving forward $2$ or $3$ spaces each have $\\dfrac14$ probability. Kayla is a basketball player who makes $50\\%$ of her $2$-point shots and $20\\%$ of her $3$-point shots.", "prime number are: 2, 3 and 5 (b) not a prime number are : 1 , 4 and 6 (ii) (a) a number greater than 5 is : 6 (b) a number not greater than 5 are: 1, 2, 3, 4 and 5 Find the 1. Probability of the pointer stopping on D ? 2. Probability of getting an ace from a well shuffled deck of 52 playing cards? 3. Probability of getting a red apple. (See figure below) Ans- (i) The possible outcomes on spinning a wheel are : A, B, C, and D Therefore, the pointer can stop at one of the regions. Total number of regions is 5. The pointer will stop at region D only once. Hence, probability that the pointer will stop at region D = $\\frac{1}{5}.$ (ii) Total number of cards in a well shuffled deck = 52 Total number of Ace cards = 4. Therefore, the probability of getting an ace card = $\\frac{4}{52}\\text{ }=\\frac{1}{13}.$ (iii)Total number of apples = 7 Total number of red apples = 4 Total number of green apples = 3 Therefore, the probability of getting a red apple = $\\frac{4}{7}$ Q. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is", "definition is that of conditional probability. The probability of event x conditional on knowing that event y has occurred (or more shortly, the probability of x given y) is defined as p(x|y) ≡ p(x, y) p(y) Here’s one way to think about conditional probability : Imagine we have a dart board, split into 20 sections. We may think of a drunk dart thrower, and describe our lack of knowledge about the throwing by saying that the probability that a dart occurs in any one of the 20 regions is p(region i) = 1/20. Imagine then, that our friend, Randy, the random dart thrower is blindfolded, and throws a dart. With pint in hand, a friend of Randy tells him that he hasn’t hit the 20 region. What is the probability that Randy has hit the 5 region? Well, if Randy hasn’t hit the 20 region, then only the regions 1 to 19 remain and, since there is no preference for Randy for any of these regions, then the probability is 1/19. To see how we would calculate this with the rules of probability : p(region 5, not region 20) p( not region 20) 1/20 1 p(region 5) = = = p(not region 20) 19/20 19 p(region 5| not region 20) = giving the intuitive result. Degree of Belief", "of 4 m. Probability and Area A dartboard has three regions, A, B, and C. For an area that multiple darts hit, the probability would be 1 - the odds that those darts would hit the center. Some of the worksheets for this concept are Probability darts, Geometry practice geometric probability name, Gcse mathematics, Chapter 4 generating functions, Edexcel gcse mathematics a, Pre employment welcome to functional skills with city guilds, Grade 8 mathematics practice test, 1 concepts. Play this game to review Probability. A dart is thrown at random onto a board that has the shape of a circle as shown below. It stops on a multiple of 3 twenty times. Statistics probability with cards version 1 directions. So, P (A/E 1 ) = 3/10, similarly P(A/E 2 ) = 8/10, and P(A/E 3 ) = 4/10. The answer lies in the fact that some events have a probability of 1, but still may not happen. Brown’s classroom, 40% of the students are boys. Using more darts produces a more sharply peaked distribution, such that one is more likely to get the correct answer. what is the probability for a family with three children to have a boy and 2 girls (Assuming the probability of having a boy and a girl is equal)??. The answer lies in", "### Home > INT2 > Chapter 3 > Lesson 3.1.5 > Problem3-61 3-61. The spinner at right has three regions: $\\5,\\0,$ and $\\20$. To play the game, you spin one time and then you win the amount in the region that the spinner lands on. 1. What is the expected value for the game? Compare the given angle measurements to $360^\\circ$. Multiply these ratios by the amount possible to win in each section. $\\text{EV}=\\5\\cdot \\frac{1}{4}+\\0\\cdot \\frac{135}{360}+\\20\\cdot \\frac{135}{360}$ 2. If you have to pay $\\10$ to play, is it a fair game?", "spinner to land on either a violet region or a region labeled $a$, or both. • There are $3$ violet regions and $3$ regions labelled $a$. • There is a $1$ region where it’s both violet and labeled $a$. This shows that the incident is mutually inclusive. Hence, we use $P(A \\text{ or } B) = P(A) + P(B) – P(A \\text{ and } B)$ \\begin{aligned}P(V \\text{ or } a) &= P(V) + P(a) – P(V \\text{ and } a)\\\\&=\\dfrac{3}{7} + \\dfrac{3}{7} – \\dfrac{1}{7}\\\\&= \\dfrac{5}{7}\\end{aligned} a. This means that the probability is equal to $\\dfrac{5}{7}$. It’s impossible to land on a red region and a blue one all at the same time. This means that these two events are mutually exclusive. For these types of events, we add their individual probabilities. b. This means that the probability is equal to $\\dfrac{1}{7} + \\dfrac{2}{7} = \\dfrac{3}{7}$. ### Practice Questions 1. A canvas bag contains $12$ pink cubes, $20$ green cubes, and $22$ purple cubes. One cube is removed from the bag and then replaced. Another cube is drawn from the bag, and repeat this one more time. What is the probability that the first cube is green, the second cube is purple, and the third is another green cube? 2. In a book club of $50$ enthusiastic readers, $26$", "is possible when the number on the face of the dice is any one of the outcomes 1, 2, 3, 4, 5. Question 3 Find the. (a) Probability of the pointer stopping on D in (Question 1 − (a))? (b) Probability of getting an ace from a well shuffled deck of 52 playing cards? (c) Probability of getting a red apple. (See figure below) Sol : (i) The pointer can stop at one of the following regions. A, A, B, C, D Out of these 5 cases, it is possible only in 1 case that the pointer will stop at region D. Therefore, probability that the pointer will stop at region D = (ii) There are 52 cards in a deck of cards and there are 4 ace cards in 1 deck of cards. Probability of getting an ace card = (iii) There are a total of 7 apples, out of which, 4 are red and 3 are green. Probability of getting a red apple = Question 4 Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of. (i) getting a number 6? (ii) getting a number less than 6?" ]

[ "of total probability, get all of the conditional probabilities $P(Y = 5 | X = x)$ using the trick above. You already know that $P(X = x) = 1/6$ for $x \\in \\{1, \\ldots, 6\\}$. So you want to compute the sum: $$\\sum_{x = 1}^6 P(Y = 5 | X = x) P(X = x) = \\frac{1}{6} \\sum_{x = 1}^6 P(Y = 5 | X = x)$$ I've already computed the answer: it is $\\frac{29}{120}$. You can show the work.", "x_0) = \\frac{\\Pr(X > x_0 + x)}{\\Pr(X > x_0)}$. Review Bayes Theorem. Get the required probabilities by integrating the pdf.", "of course that the sum of these probabilities is $1$, as it should be. What I needed to use was:", "equal to $3$. probability = number of outcomes possible / total number of outcomes $\\frac{3}{6} = \\frac{1}{2}$", "the separate probabilities of their occurrence and sum to get the probability $$\\frac{1}{4}(1-p)+p = \\frac{1}{4}+\\frac{3}{4}p.$$", "However If we let each $X_1,X_2,X_3,X_4 = 30$ with probabilities $1/30,2/30,3/30,4/30$ and zero otherwise the probability the sum is $\\ge 30$ is $1-(29/30)(28/30)(27/30)(26/30)\\approx 0.3$ – spaceisdarkgreen Jan 30 '17 at 9:29", "to find the probability distribution $$P$$ of $$X$$. From the characterisation given in the beginning of the paragraph we thus have $$P(x)=\\frac{1}{n-x+1}\\prod_{i=1}^{x-1}\\frac{n-i}{n-i+1}.$$ For $$x=1$$ we define the empty product to be $$1$$. With this probability distribution you can apply the usual calculations.", "the probability assigned to any subset of the values taken on by $X$ must be $0$, contradicting the fact that the sum of the probabilities over the range of $X$ must be $1$.", "How do you find the probability of an event given the odds of the event occurring is 2:1? 1 Answer Mar 16, 2018 The desired probability is $\\frac{2}{3}$ Explanation: Probabilities always range between $0$ and $1$. The odds on the other hand are defined as the probability that the event will occur divided by the probability that the event will not occur. As odds of the event occurring is $2 : 1$ the probability that the event will occur is twice the probability that the event will not occur. But if probabilty of happening of event is $p$, probability of event not happening is $1 - p$ Hence $\\frac{p}{1 - p} = 2$ or $p = 2 - 2 p$ or $3 p = 2$ i.e. $p = \\frac{2}{3}$ and desired probability is $\\frac{2}{3}$.", "+ pe^x = e^x$ $p(1 + e^x) = e^x$ $p = \\frac{e^x}{1 + e^x}$ This final line means we can calculate the probability of getting a one using x which in practice is our total, summed value of everything on the right hand side of the equation.", "terms after $X_3$ sum to less than $\\frac{1}{8}$. So the probability $P(Z > \\frac{1}{2} + \\frac{1}{8}) = \\frac{1}{2}\\cdot(1-\\frac{1}{2}\\cdot\\frac{1}{2})$, so the asked for probability is its complement.", "Sum probability The numbers $a$ and $b$ are given such that $0 \\leq a \\leq 4$ and $0 \\leq b \\leq 8$. The probability that $a + b \\leq 4$ can be represented as a fraction of $\\frac {m} {n}$, where $m$ and $n$ are coprime natural numbers. What is the sum of $m + n$?", "probability. Now let $$y_1:=x_1;\\, y_i:=x_i-x_{i-1}-d,\\,\\forall i\\ge2$$. The required condition is then $$\\big\\{y_i\\ge0,\\,\\forall i \\bigwedge \\sum_{i=1}^N y_i\\le 1-(N-1)d\\big\\}$$. The probability of this set is $$\\frac1{N!}(1-(N-1)d)_+^N$$. The desired probability is then $$(1-(N-1)d)_+^N$$.", "$$\\sum_{y=1}^x c(x+y) = cx^2 + c\\frac{x(x+1)}{2}~.$$ Now, sum this function of $x$ from $1$ to $3$ and equate the resulting expression to $1$ to get $c$. Evaluate $c$ when the total probability mass equals 1.$$1{~=~\\sum_{x=1}^3\\sum_{y=1}^x c(x+y)\\\\~=~ c\\sum_{x=1}^3(x^2+\\sum_{y=1}^x y)\\\\~=~c\\sum_{x=1}^3(x^2+x(x+1)/2)\\\\~=~ \\frac c2\\sum_{x=1}^3(3x^2+x)\\\\ ~~\\vdots}$$", "such as drawing a five or a hears from a deck of cards (5 of hearts fits both groups), the probability of getting A or B consists of the sum of their individual probabilities minus the probability of both events happening. The expected value is the sum of all the different outcomes, each weighted by its probability and payoff. Let's say we a dice game where the payoff is $1 if you roll a 1;$2 if you roll a 2, etc. Each possible outcome has a $$\\frac{1}{6}$$ probability, so the expected value is $$\\frac{1}{6}(1) + \\frac{1}{6}(2) + \\frac{1}{6}(3) + \\frac{1}{6}(4) + \\frac{1}{6}(5) + \\frac{1}{6}(6) = 3.50.$$ The law of large numbers tells us that as the number of independent trials increases, the average of the outcomes will get closer and closer to its expected value. A probability density function plots the assorted outcomes along the x-axis and the expected probability of each outcome on the y-axis; the weighted probabilities - each outcome multiplied by its expected frequency - will add up to 1. With more trials, the plot will become skinnier and skinnier. A decision tree maps out each source of uncertainty and the probabilities associated with all possible outcomes. The end of the tree gives us all the possible payoffs and the probability of each. ## Problems", "## Precalculus (10th Edition) $P(1)=P(3)=P(5)=\\frac{2}{9}$, $P(2)=P(4)=P(6)=\\frac{1}{9}.$ Here if the probability of getting an even number is $x$, then $P(2)=P(4)=P(6)=x$ and $P(1)=P(3)=P(5)=2x$. We know that $P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1\\\\2x+x+2x+x+2x+x=1\\\\9x=1\\\\x=\\frac{1}{9}.$ Therefore $P(1)=P(3)=P(5)=\\frac{2}{9}$, $P(2)=P(4)=P(6)=\\frac{1}{9}.$", "we get, $(\\frac{(_3C_3)(_{10-3}C_{5-3})}{_{10}C_5}$) Solving we get: Probability of 3 female dogs in selection = 0.083 Algebra TutorMe Question: Solve the following equation for x, $( |-4x + 2| = 1$) Manny H. To solve this problem we need to understand how to break down absolute value equations into 2 separate equations. For example, given $$|x + a| = n$$, we can break it down as follows: $(x + a = n$) $(and$) $(-(x + a) = n$) then solve for 2 different x values. So for this problem, we will have $$-4x + 2 = 1$$ and $$-(-4x +2) = 1$$. Solving: $( -4x + 2 = 1$) $( -4x = -1$) $( x = 1/4$) and $( -(-4x + 2) = 1$) $( 4x - 2 = 1$) $( 4x = 3$) $( x = 3/4$) So, our solutions are $$x = 1/4, 3/4$$ Send a message explaining your needs and Manny will reply soon. Contact Manny", "says: This is ${\\color{red}Pr(X \\geq 9)}$ . No harm done though, the OP just needs to subtract from 1 the result of the calculation below. = $\\frac{4}{36} + \\frac{3}{36} + \\frac{2}{36} + \\frac{1}{36}$ = .....rest i will leave up to u to compute b)Probability of sum less than 4 or greater than 9 Take Pr(A) = Probability of sum being less than 4 ie(1,1) (1,2) (2,1) Pr(A) = $\\frac{3}{36}$ = $\\frac{1}{12}$ Take Pr(B) = Probability of sum being greater than 9 ie. (4,6) (5,5) (6,4) (5,6) (6,5) (6,6) $\\therefore$ Pr(B) = $\\frac{6}{36} = \\frac{1}{6}$ Now question is prob of sum less than 4 or greater than 9 and here or means $\\cup$ $\\therefore$ Pr(A $\\cup$ B) = Pr(A) + Pr(B) .......since they both are mutually exclusive = $\\frac{1}{12} + \\frac{1}{6}$ = .............and u can do it from here on by urself c)Probability that the sum is atleast 10, given that the sum is atleast 9 here Pr(A) = sum atleast 10...........which we need to find out Pr(B) = given that the sum is atleast 9 Pr(A $\\cap$ B) = reading which are both in Pr(X $\\le$ 9) and when Pr(X $\\le$ 10) ie (4,5) (5,5) (5,4) (6,5) (5,6) (6,6) = ......... $\\therefore$ $Pr(A \\cap B)$ = $\\frac{1}{6}$ there using conditional prob Pr(A|B) = $\\frac{Pr(A \\cap B)}{Pr(B)}$ = ...........substitute", "6), (6, 4), (5, 5) To get the sum as 11, possible outcomes = (5, 6), (6, 5) To get the sum as 12, possible outcomes = (6, 6) (ii)Probability of each of these sums will not be $\\frac{1}{11}$ as these sums are not equally likely.", "until I came up with that answer. That's the problem 4. ## Re: Discrete Random Variables - pmf Originally Posted by Matt1993 See all I did was let the pmfs equal zero and then I just tried values until I came up with that answer. That's the problem That's close. The axioms of probability (see wiki) require in particular 2 things from the probabilities: 1. Each probability is at least 0: $p \\ge 0$. . 2. The sum of all probabilities is 1: $\\sum p = 1$. $( 1 - \\theta )/4 \\ge 0$ $( 1 - 3\\theta) /4 \\ge 0$ $( 1 + 3\\theta)/4 \\ge 0$ $(1+ \\theta)/4 \\ge 0$ $( 1 -\\theta )/4 + ( 1 - 3\\theta) /4 + ( 1 + 3\\theta)/4 + (1+ \\theta)/4 = 1$" ]

[ "What is the probability that $x<y$? • $\\frac{1}{8}$ • $\\frac{1}{6}$ • $\\frac{2}{3}$ ### Key Concepts Number system Cube Answer: $\\frac{1}{8}$ AMC-10A (2003) Problem 12 Pre College Mathematics ## Try with Hints The given vertices are $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$.if we draw a figure using the given points then we will get a rectangle as shown above.Clearly lengtht of $OC$= $4$ and length of $AO$=$1$.Therefore area of the rectangle is $4 \\times 1=4$.now we have to find out the probability that $x<y$.so we draw a line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.can you find out the area with the condition $x<y$? can you finish the problem…….. Now the line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.Therefore it will form a Triangle $\\triangle AOD$ (as shown above) whose $AO=1$ and $AD=1$.Therefore area of $\\triangle AOD=\\frac{1}{2}$ i.e (red region).Now can you find out the probability with the condition $x<y$? can you finish the problem…….. Therefore the required probability ($x<y$) is $\\frac{\\frac{1}{2}}{4}$=$\\frac{1}{8}$ Categories ## Problem from Probability | AMC 8, 2004 | Problem no. 21 Try this beautiful problem from Probability from AMC 8, 2004. ## Problem from Probability | AMC-8, 2004 | Problem 21 Spinners A and B are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability", "cents each, and Spain $5$ cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.) The average price of his $'70$s stamps is closest to (A) $3.5$ cents (B) $4$ cents (C) $4.5$ cents (D) $5$ cents (E) $5.4$ cents AMC 8, 2002, Problem 11 A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth? (A) $11$ (B) $12$ (C) $13$ (D) $14$ (E) $15$ AMC 8, 2002, Problem 12 A board game spinner is divided into three regions labeled $A, B$ and $C$. The probability of the arrow stopping on region $A$ is $\\frac{1}{3}$ and on region $B$ is $\\frac{1}{2}$. The probability of the arrow stopping on region $C$ is (A) $\\frac{1}{12}$ (B) $\\frac{1}{6}$ (C) $\\frac{1}{5}$ (D) $\\frac{1}{3}$ (E) $\\frac{2}{5}$ AMC 8, 2002, Problem 15 Which of the following polygons has the largest area? (A) $A$ (B) $B$ (C) $C$ (D) $D$ (E) $E$ AMC 8, 2002, Problem 16 Right isosceles triangles are constructed on the sides of a $3-4-5$ right triangle, as shown. A capital letter represents the area of each triangle. Which one of", "Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd? (A) $\\frac{17}{36}$ (B) $\\frac{35}{72}$ (C) $\\frac{1}{2}$ (D) $\\frac{37}{72}$ (E) $\\frac{19}{36}$ AMC 8, 2006, Problem 4 Initially, a spinner points west. Chenille moves it clockwise $2 \\frac{1}{4}$ revolutions and then counterclockwise $3 \\frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves? (A) north (B) east (C) south (D) west (E) northwest AMC 8, 2006, Problem 10 Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12 .$ What should his graph look like? AMC 8, 2006, Problem 17 Jeff rotates spinners $P, Q$ and $R$ and adds the resulting numbers. What is the probability that his sum is an odd number? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ (C) $\\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2006, Problem 20 A singles tournament had six players. Each player played every other player only once, with no ties. If Helen", "fraction Answer: $$\\frac{2}{5}$$ AMC-8, 2016 problem 21 Challenges and Thrills in Pre College Mathematics ## Try with Hints There are 5 Chips, 3 red and 2 green Can you now finish the problem .......... We draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. one at a time without replacement Can you finish the problem........ There are 5 Chips, 3 red and 2 green we draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. if we draw all the green chip boxes then the last box be red or if we draw all red boxes then the last box be green but we draw randomly. there are 3 red boxes and 2 green boxes Therefore the probability that the 3 reds are drawn=$$\\frac{2}{5}$$ # Problem from Probability | AMC 8, 2004 | Problem no. 21 Try this beautiful problem from Probability from AMC 8, 2004. ## Problem from Probability | AMC-8, 2004 | Problem 21 Spinners A and B are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even? • $$\\frac{2}{3}$$ • $$\\frac{1}{3}$$ • $$\\frac{1}{4}$$", "of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3$ ? (A) $\\frac{1}{4}$ (B) $\\frac{1}{3}$ $(C) \\frac{1}{2}$ (D) $\\frac{2}{3}$ (E) $\\frac{3}{4}$ AMC 8, 2007, Problem 25 On the dart board shown in the Figure, the outer circle has radius $6$ and the inner circle has a radius of $3$ . Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd? (A) $\\frac{17}{36}$ (B) $\\frac{35}{72}$ (C) $\\frac{1}{2}$ (D) $\\frac{37}{72}$ (E) $\\frac{19}{36}$ AMC 8, 2006, Problem 4 Initially, a spinner points west. Chenille moves it clockwise $2 \\frac{1}{4}$ revolutions and then counterclockwise $3 \\frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves? (A) north (B) east (C) south (D) west (E) northwest AMC 8, 2006, Problem 10 Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12", "faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$", "= $\\frac{1}{2}$ • The probability of landing on heads is $P$(tails) = $\\frac{1}{2}$ • Dice: (a six-sided die, if singular) • When rolling a die, it will land with one of its six flat faces facing up (on top). There are six sides labelled from 1 to 6. • So, there are 6 possible outcomes to a die roll: 1, 2, 3, 4, 5, or 6 • The probability of landing on any one side (1-6) is one out of six chances • $P$ (1) =$P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = $\\frac{1}{6}$ • Spinners: (can have any number of equal parts; labels with numbers, letters, etc.) • When using a spinner, the arrow will land on one of the marked regions. In this case, there are 4 different coloured regions. • So, there are 4 possible outcomes for this spinner: red, yellow, green, or blue • The probability of landing on any one of the colors is one out of four chances • $P$ (Red) = $P$ (Yellow) = $P$ (Green) = $P$ (Blue) = $\\frac{1}{4}$ #### Lessons • Introduction Introduction to Probability Outcomes for Coins, Dice, and Spinners: a) What is probability? b) What are outcomes in probability? 1. Outcomes for flipping a coin 2. Outcomes for rolling", "a “3-dimensional square”, I re-phrased Alexander’s question into two dimensions. The trickier part was thinking what it would mean to “roll” a 2-dimensional shape. The outside of an nxn square is painted red and is chopped into $n^2$ unit squares. The latter are thoroughly mixed up and put into a bag. One small square is withdrawn at random from the bag and spun on a flat surface. What is the probability that the spinner stops with a red side facing you? Shown below is a 4×4 square, but in all sizes of 2-dimensional squares, there are three possible types in the bag: those with 2, 1, or 0 sides painted. I solved my first variation case by case. In any nxn square, • There are 4 corner squares with 2 sides painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4}{n^2} \\cdot \\frac{2}{4} = \\frac{2}{n^2}$. • There are $4(n-2)$ edge squares not in a corner with 1 side painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4(n-2)}{n^2} \\cdot \\frac{1}{4} = \\frac{n-2}{n^2}$. • All other squares have 0 sides painted, so the probability of picking one of those squares and then spinning a red side is 0. • Adding the probabilities for", "A game consists of spinning an arrow around a stationary disk as shown below. When the arrow comes to rest, there are eight equally likely outcomes. It could come to rest in any one of the sectors numbered $1, 2, 3,4, 5, 6, 7$ or $8$ as shown. Two such disks are used in a game where their arrows are independently spun. What is the probability that the sum of the numbers on the resulting sectors upon spinning the two disks is equal to 8 after the arrows come to rest? 1. $\\dfrac{1}{16}$ 1. $\\dfrac{5}{64}$ 1. $\\dfrac{3}{32}$ 1. $\\dfrac{7}{64}$ Given the, two disks When we select only one number from any of the two disks, then the probability $= \\frac{1}{8}$ When we select one number from the first disk and the one number from the second disk simultaneously, then the probability $= \\frac{1}{8} \\times \\frac{1}{8} = \\frac{1}{64}$ The numbers on the resulting sectors upon spinning the two disks is equal to $8$ after the arrows come to rest $= \\underbrace{(1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1)}_{\\text{Number of favorable outcomes = 7}}$ $\\therefore$ The required probability $= \\dfrac{7}{64}$ Correct Answer $:\\text{D}$ ${\\color{Magenta}{\\textbf{PS:}}}$ In probability, two events are independent if the incidence of one event does not affect the probability of the other event. If the incidence of one event", "# Math Help - Spinner 1. ## Spinner A spinner contains eight regions, numbered 1 through 8. The arrow has an equally likely chance of landing on any of the eight regions. If the arrow lands on the line, it is spun again. What is the probability that the arrow lands on an odd number? 2. Originally Posted by symmetry A spinner contains eight regions, numbered 1 through 8. The arrow has an equally likely chance of landing on any of the eight regions. If the arrow lands on the line, it is spun again. What is the probability that the arrow lands on an odd number? The odd numbers, 1-8, are 1, 3, 5, 7, thus there are 4 odd numbers out of the 8, and therefore the probability that the arrow lands on an odd number is 4/8 or 1/2. 3. ## ok I thank you for your great help.", "spinner to land on either a violet region or a region labeled $a$, or both. • There are $3$ violet regions and $3$ regions labelled $a$. • There is a $1$ region where it’s both violet and labeled $a$. This shows that the incident is mutually inclusive. Hence, we use $P(A \\text{ or } B) = P(A) + P(B) – P(A \\text{ and } B)$ \\begin{aligned}P(V \\text{ or } a) &= P(V) + P(a) – P(V \\text{ and } a)\\\\&=\\dfrac{3}{7} + \\dfrac{3}{7} – \\dfrac{1}{7}\\\\&= \\dfrac{5}{7}\\end{aligned} a. This means that the probability is equal to $\\dfrac{5}{7}$. It’s impossible to land on a red region and a blue one all at the same time. This means that these two events are mutually exclusive. For these types of events, we add their individual probabilities. b. This means that the probability is equal to $\\dfrac{1}{7} + \\dfrac{2}{7} = \\dfrac{3}{7}$. ### Practice Questions 1. A canvas bag contains $12$ pink cubes, $20$ green cubes, and $22$ purple cubes. One cube is removed from the bag and then replaced. Another cube is drawn from the bag, and repeat this one more time. What is the probability that the first cube is green, the second cube is purple, and the third is another green cube? 2. In a book club of $50$ enthusiastic readers, $26$", "the shaded region = 42 - 2 π(2)22 = 4 units [Area of square - Area of 2 semi circles.] Area of the board = π × 162 = 768 units Probability of dart hitting the target = 4 / 768 = 1 / 192 3. Which of the following geometric probabilities is impossible? a. 0.45 b. 1.2 c. 0 d. 1 #### Solution: As probability always lies between 0 and 1, geometric probability can never be greater than 1. 4. What is the probability of getting 75 points when the spinner is spun? a. $\\frac{1}{3}$ b. 1 c. $\\frac{5}{18}$ d. $\\frac{2}{9}$ #### Solution: P(getting 75 points) = Angle subtended by arc CD total angle P (getting 75 points) = 100o360o [Substitute.] P(getting 75 points) = 5 / 18 [Simplify.] 5. A dart thrown at a rectangular board ABCD is equally likely to land on any point. Find the probability of the dart hitting the inner rectangle PQRS. [Given $a$ = 21 cm, $b$ = 16 cm, $c$ = 18 cm, $d$ = 10 cm.] a. $\\frac{29}{15}$ b. $\\frac{15}{28}$ c. $\\frac{4}{7}$ d. $\\frac{28}{15}$ #### Solution: Area of the rectangle ABCD = AB × BC = 21 × 16 = 336 cm 2 [Area of rectangle = length × breadth.] Area of the rectangle PQRS = PQ × QR", "prime number are: 2, 3 and 5 (b) not a prime number are : 1 , 4 and 6 (ii) (a) a number greater than 5 is : 6 (b) a number not greater than 5 are: 1, 2, 3, 4 and 5 Find the 1. Probability of the pointer stopping on D ? 2. Probability of getting an ace from a well shuffled deck of 52 playing cards? 3. Probability of getting a red apple. (See figure below) Ans- (i) The possible outcomes on spinning a wheel are : A, B, C, and D Therefore, the pointer can stop at one of the regions. Total number of regions is 5. The pointer will stop at region D only once. Hence, probability that the pointer will stop at region D = $\\frac{1}{5}.$ (ii) Total number of cards in a well shuffled deck = 52 Total number of Ace cards = 4. Therefore, the probability of getting an ace card = $\\frac{4}{52}\\text{ }=\\frac{1}{13}.$ (iii)Total number of apples = 7 Total number of red apples = 4 Total number of green apples = 3 Therefore, the probability of getting a red apple = $\\frac{4}{7}$ Q. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is", "= random.uniform(0,1) if P_1 < num < 1: #area corresponding to x_1 wins += 1 #wins num = 0 break else: num2 = random.uniform(0,1) if P_2 < num2 < 1: #area corresponding to x_2 break #loses print(wins/10**7) The correct answer is 0.5821 however I am getting 0.655.. Where am I doing wrong ? Last edited: ## Answers and Replies Related Programming and Computer Science News on Phys.org Mark44 Mentor **Problem:** Here’s a little Monte Carlo challenge problem, Consider the following game, which uses the two spinner disks. Suppose a player spins one or the other of the pointers on the disks according to the following rules: (1) if the player spins pointer i and it stops in the region with area pij, he moves from disk i to disk j (i and j are either 1 or 2); (2) if a pointer stops in the region with area xi, the game ends; (3) if the game ends in the region with area x1, the player wins, but if the pointer stops in the region with area x2 the player loses. What is the probability the player, starting with disk 1, wins? Assume the area of each disk is one, so that x1+p11+p12 =1, as well as that x2+p21+p22 =1 Run your code for the case of p11 =0.2,", "$$2, 3$$ and $$5.$$ (b) Outcomes of event of not getting a prime number are $$1, 4$$ and $$6.$$ (ii) (a) Outcomes of event of getting a number greater than $$5$$ is $$6.$$ (b) Outcomes of event of not getting a number greater than $$5$$ are $$1, 2, 3, 4$$ and $$5.$$ ## Chapter 5 Ex.5.3 Question 3 Find the: (a) Probability of the pointer stopping on $$D$$ in (Question 1 (a))? (b) Probability of getting an ace from a well shuffled deck of $$52$$ playing cards? (c) Probability of getting a red apple? (See figure alongside) ### Solution What is known? (i) Spinning wheel image (ii) Deck of $$52$$ playing cards (iii) Image depicting Red and Green apples What is unknown? (i) Probability of the pointer stopping on $$D.$$ (ii) Probability of getting an ace from a well shuffled deck of $$52$$ playing cards. (iii) Probability of getting a red apple Reasoning: Probability \\begin{align}= \\! \\frac{{{\\text{Number of favourable outcomes}}}}{{{\\text{Total number of outcomes}}}}\\end{align} Steps: (a) In a spinning wheel, there are five pointers $$A, A, B, C, D.$$ So, there are five outcomes. Pointer stops at $$D$$ which is $$1$$ outcome. So, the probability of the pointer stopping on \\begin{align}D = \\frac{1}{5}\\end{align} (b) There are $$4$$ aces in a deck of $$52$$ playing cards. So, there are $$4$$", "# What is the probability that the total score after throwing darts is divisible by $3$. To play a game of darts Michael throws three darts at the dart board shown. The number of points $$(1,$$ $$5$$ or $$10)$$ for each of the three regions is indicated. His score is the sum of the points for the three darts. If the radii of the three concentric circles are $$1,$$ $$2$$ and $$3$$ units, and each dart Michael throws hits this dart board at random, what is the probability that his score is evenly divisible by $$3?$$ Express your answer as a common fraction. After taking the values modulo $$3$$, we have $$1, 2, 1$$. I am pretty sure that the only way we can get divisible by $$3$$ in this problem is if we have modulos $$1, 1, 1$$ or $$2, 2, 2$$ for the darts. This means that the probability is $${\\left(\\dfrac23\\right)}^3+{\\left(\\dfrac13\\right)}^3=\\dfrac13$$. I feel as if I am missing something, or am I correct? Thanks! EDIT: \"At random\" means that the likelihood of a dart landing in a region is the total area of that region divided by the total area of the dart-board. • \"Each dart hits the board 'at random'\" This is not clear. Are we meant to assume that each of the three regions", "girls or three girls. So, the probability of each is $$\\frac{1}{4}$$. Is this correct? Justify your answer. Solution: False, because the outcomes are not equally, likely. For no girl, outcome is bbb, for one girl, it is bgb, gbb, bbg, for two girls, it is bgg, ggb, gbg and for all girls, it is ggg. Probability Extra Questions Class 10 Question 4. A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig. 15.1). Are the outcomes 1, 2 and 3 equally likely to occur? Give reason. Solution: False, because the outcome 3 is more likely than the other numbers. Extra Questions Of Probability Class 10 Question 5. Two coins are tossed simultaneously. Find the probability of getting exactly one head. Solution: Possible outcomes are {HH, HT, TH, TT}. (exactly one head) = $$\\frac{2}{4}$$ = $$\\frac{1}{2}$$ Extra Questions On Probability Class 10 Question 6. From a well shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen. Solution: Number of black queens in a pack of cards = 2 ∴ P (black queen) = $$\\frac{2}{52}$$ = $$\\frac{1}{26}$$ Probability Extra Questions Question 7. If P (E) = 0.05, what is the probability of ‘not E’ ? Solution: As we know", "## Algebra 1 $\\frac{1}{4}$ To find a probability $P(x)$, we want to divide the number of events in the group $x$, and divide that number by the total number of events. In this case, there is $1$ blue section on the spinner, and $4$ sections in total, so P(blue) is $\\frac{1}{4}$.", "### Home > A2C > Chapter Ch4 > Lesson 4.2.4 > Problem4-127 4-127. 1. A dart hits each of these dart boards at random. What is the probability that the dart will not land in the shaded area? Homework Help ✎ P(not in shaded area) = P(white) $P(\\text{white})=\\frac{\\text{area of success}}{\\text{total area}}=\\frac{\\text{area of white}}{\\text{area of large circle}}$ Area = πr2 $P(\\text{not in shaded area})=\\frac{1}{4}$ See part (a)." ]

[ "+0 0 31 2 Monica tosses a fair 6-sided die. If the roll is a prime number, then she wins that amount of dollars (so that, for example, if she rolls 3, then she wins 3 dollars). If the roll is composite, she wins nothing. Otherwise, she loses 3 dollars. What is the expected value of her winnings on one die toss? Express your answer as a dollar value to the nearest cent. Jul 17, 2021 #1 +86 +3 Answer: $$1.17$$ Solution: Case 1: Monica rolls a 1. If Monica rolls a 1, then she loses 3 dollars. Case 2: Monica rolls a 2. If Monica rolls a 2, then she gains 2 dollars. Case 3: Monica rolls a 3. If Monica rolls a 3, then she gains 3 dollars. Case 4: Monica rolls a 4. If Monica rolls a 4, then she gains nothing. Case 5: Monica rolls a 5. If Monica rolls a 5, then she gains 5 dollars. Case 6: Monica rolls a 6. If Monica rolls a 6, then she gains nothing. Adding all them up and then placing it over six (there are six possibillities) gives $$-3+2+3+0+5+0\\over6$$ Which is equal to $$\\frac76$$, which rounded to the nearest cent, gives $$1.17$$. Jul 18, 2021 #2 0 Correct! Guest Jul 18, 2021", "the odds against rolling a six with a fair die are 5 to 1. The probability of rolling a six with a fair die is the single number 1/6 or approximately 16.7%. The gambling and statistical uses of odds are closely interlinked. If a bet is a fair one, as in a wager between friends, then the odds offered to the gamblers will perfectly reflect relative probabilities. A fair bet that a fair die will roll a six will pay the gambler $5 for a$1 wager (and return the bettor his or her wager) in the case of a six and nothing in any other case. The terms of the bet are fair, because on average, five rolls result in something other than a six, at a cost of $5, for every roll that results in a six and a net payout of$5. The profit and the expense exactly offset one another and so there is no disadvantage to gambling over the long run. If the odds being offered to the gamblers do not correspond to probability in this way then one of the parties to the bet has an advantage over the other. Casinos, for example, offer odds that place themselves at an advantage, which is how they guarantee themselves a profit and survive as businesses. The", "in the books as his winnings for that round. He should keep a running tally, so players know how much money they have. At the beginning of any turn, BEFORE he rolls the dice, if a player has accumulated enough money to buy a prime number that is in the center of the table, then the player may state, \"I'd like to buy a prime,\" and specifies the one he wants to buy. At that point, the accountant subtracts the appropriate amount of money from his total, and the player takes control of that prime, moving it to his place to indicate his ownership. He then rolls the dice as usual on his turn, and records any winnings as usual. From then on, however, when any other player rolls and declares a multiple of his prime, the other player has to SPLIT his winnings with the owner of the prime. Example: Jane has bought the deed for 7. John later rolls a 4 and a 2 on his turn, so he has the choice to declare $42 or$24. If John declares \"$42,\" then John has to split his earnings with Jane because 7 is a factor of 42. Each will get$21 recorded in their columns. On the other hand, if John declares \"$24,\" then he doesn't have to", "a die. Dave rolls first. He wins control over the TV if he rolls a number greater than four. If he doesn't win, Mae gets her chance and wins if she rolls a prime number. If she doesn't roll a prime number, their parents take the remote and nobody wins. 1. What is Dave's chance of winning? 2. What is Mae's probability of winning? 3. Is this a fair way to settle the argument for both Dave and Mae? Yes A No B ##### QUESTION 3 Given the following options: (A) a sure gain of$\\$500$$500 (B) a 5050% chance of gaining \\1000$$1000 and a$50\\$50% chance of gaining nothing 1. Calculate the financial expectation of Option A. 2. Calculate the financial expectation of Option B. 3. Which option has greater variation? Option A A Option B B", "# Expected value game - cards and dies A) What is the expected value of a card (from a standard deck of cards) assuming that aces are worth 1 point and face cards are worth 10? B) A game consists of drawing a card and rolling a die simultaneously. If the value of the die is greater than the value of the card (the card values are as in part A above) then you win a number of dollars equal to the value on the die. If the die and the card have equal value, nobody wins anything. If the card is of higher value than the die, you lose 1 dollar. For example, suppose you roll a 4 on the die and draw a 6 of clubs. Then you lose 1 dollar. If, on the other hand, you roll a 5 on the die and draw a 2 of spades, then you win 5 dollars. What is the expected value of the game? - Since this is homework, it would be great if you could please say what you have tried, where you are running into difficulty, etc. – Matthew Conroy Nov 28 '12 at 4:58 What is the probability a card has value $1$? Value $2$? And so on. (The answer for $10$ is different from", "× Discrete Mathematics # Linearity of Expectation If you roll a six-sided die, the expected value for the number rolled is $$3.5.$$ If you roll 5 six-sided dice, what is the expected value for the sum of the numbers rolled? A standard deck of 52 cards is shuffled randomly, and then the cards are flipped over one-by-one. If the ace of spades is card number $$S$$ and the ace of hearts is card number $$H$$, what is the expected value of $$S+H?$$ Six friends are doing a \"secret santa gift exchange\". Their names are placed into a hat, and each person chooses a name randomly (without replacement) from the hat. However, this leads to the possibility that someone chooses their own name and doesn’t get a gift! What is the expected value for the number of people who will choose their own name? Patrick has a box with 4 yellow balls and 4 green balls. He randomly pulls out a ball, and if it is yellow, he paints it green. He then puts it back in the box and continues this process until all of the balls are green. Find the expected value for the number of times Patrick will pull out a ball. Andy the Ant is on a number line at the location 0. Each minute,", "# Probability Dice Game! Help me! [closed] \"Megan is selected to play a Megamillions game in which she can win a potentially unlimited amount of money. In this game, she repeatedly rolls a fair six-sided die, and the game ends as soon as she rolls a six. If roll 0 is a six, then she wins nothing; if roll 1 is a six, then she wins a total of 1^2 = 1 dollar; if roll 2 is a six, then she wins a total of 2^2=$4; and so on. What is the expected value of her winnings, in dollars?\" If anyone can help me figure this out that would be amazing! Obviously there is a 1/6 chance that she rolls a six, but I'm not sure where to go from there! ## closed as off-topic by Austin Mohr, user1551, user175968, Em., Edward JiangMay 16 '16 at 1:32 This question appears to be off-topic. The users who voted to close gave this specific reason: • \"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.\" – Austin", "26. Arianna will roll a standard die 10 times in which she will record the value of each roll. What is a trial of this experiment? 27. Martin will draw 3 cards from a standard 52-card deck without replacement 5 different times. For each 3-card draw, he will record the number of red cards and the number of black cards. What is a trial of this experiment? 28. True or False? Dependent is all events which are not included in the specified event. 29. Which of the following gives the definition of trial? 30. Jacqueline will spin a fair spinner with the numbers 0123, and 4 a total of 3 times. If Event A = spinner lands on numbers all greater than 2 and Event B = total sum of 9, which of the following best describes events A and B?", "# Martingales and expectations Three players, $A$, $B$ and $C$, repeatedly toss a standard six-sided die. If the die lands 1 or 2 on a particular toss, then $B$ and $C$ each give one dollar to $A$. If the die lands 3 or 4 on a particular toss, then $A$ and $C$ each give one dollar to $B$. If the die lands 5 or 6 on a particular toss, then $A$ and $B$ each give one dollar to $C$. Players $A$, $B$ and $C$ start the game with respective numbers of dollars $a$, $b$ and $c$. Let $X_n, Y_n$ and $Z_n$ denote the respective amounts of money that players $A,B$, and $C$ have after the $n$-th toss, $n \\geq 1$, and let $X_0 = a$, $Y_0 = b$, and $Z_0 = c$. a) I'm trying to prove that $W_n = X_nY_nZ_n + n(a + b + c - 2)$ is a martingale. b) And the expected # of tosses required to finish the game. For part a, we're essentially looking for a fair game where no knowledge of past events can help to predict future winnings. So we can condition on what? - What part(s) of my answer is (are) not clear to you? – Did Apr 20 '12 at 16:05 Hint for (a): Let $S=\\{(2,-1,-1),(-1,2,-1),(-1,-1,2)\\}$. Assume that", "+0 c&p halp polz +7 60 1 +258 A fair 6-sided die is rolled once. If I roll $n$, then I win $6-n$ dollars. What is the expected value of my win, in dollars? Feb 20, 2022 #1 +1384 +1 The expected value is the average of all the possibilities: You can win 0, 1, 2 ,3, 4, or 5 dollars, so the expected is $$\\color{brown}\\boxed{15\\over6}$$ Feb 20, 2022", "means she loses 4 time (the coin is flipped 5 times in total). So this way K will have 10 + 1 - 4 = 7 dollars (less than we need) If K wins 2 time (which means she loses 3 times), she will have 10 + 2 - 3 = 9 dollars (less than we need) If K wins all 5 times she'll end up with$15 (more than we need). Now you see why we check only the combinations when K wins 3 (K will have $11) and 4 (K will have$13) times. Hope this helps. d2touge wrote: Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David$1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than$10 but less than $15. a) 5/16 b) 15/32 c) 1/2 d) 21/32 e) 11/16 So to answer this, we find the total combinations when you flip the coin 5 times. --> 2^5 = 32. The explanation tells us to find the combinations when kate wins 3 times and 4 times. --> 5C3 and 5C4. we get 10 and 5 respectively. Lastly, we simply find the probability (5/32) + (10/32)", "+1 vote 46 views Toss a fair coin twice. You win $1$ dollar if at least one of the two tosses comes out heads. (b) Approximately how many times do you need to play so that you win at least $250$ dollar with probability at least $0.99$. | 46 views", "# One die and 1, 2 or 3 coins A fair die is tossed. If the die lands on 1, 2 or 3, then 3 fair coins are tossed. If the die lands on 4 or 5, then 2 fair coins are tossed. If the die lands on 6, then 1 fair coin is tossed. Given that the resulting coin tosses produced no tails, what is the probability that the die landed on 6? The probability can be expressed as $$\\dfrac ab$$ for coprime positive integers $$a$$ and $$b$$, find the value of $$a+b$$. ×", "# Watch the bookie win When you roll a die, you have a 6:1 chance of any given option. There's a bookmaker taking bets on die rolls. He owns £20 at the start of the game. The bookmaker offers odds of 4:1, but gamblers take it, because there's money on offer. There are 10 gamblers. A, B, C, D, E, F, G, H, I, and J. Each gambler owns 10 pounds sterling. Each round, the 10 gamblers each bet £1 on a die roll, if they have any money left. • If they win their bet, £4 is transferred from the bookmaker to the gambler. • If they lose, £1 is transferred from the gambler to the bookmaker. The game ends in either of these conditions: • The bookmaker runs out of money, or is in debt. • All of the gamblers have run out of money. We should see the result presented as a series of tables, like so: Turn 1 Roll: 2 | | Bookie | A | B | C | D | E | F | G | H | I | J | | Start money | 20 | 10 | 10 | 10 | 10 | 10 | 10 | 10 | 10 | 10 | 10 | | Bet | n/a", "his 1956 paper A New Interpretation of Information Rate1. Kelly asked: if you have a private stream of information related to a bet, not known to others and so not reflected in the odds, you have an edge; if you have an edge, how much should you wager? If you only have one shot at a positive expectation bet, you wager everything you can afford. But what if you have a stream of these bets and can wager previous winnings2? If you wager everything, you’ll eventually lose it all and walk away with nothing. But what amount between nothing and all? Kelly determined that the fraction of your bankroll to wager is equal to $\\frac{edge}{odds}$. The odds are the multiple of your wager your bankroll increases by if you win. When you roll a die, fair odds are 5 to 1: if you roll your point you win $5 (and keep the dollar you’ve bet), if you don’t you lose$1. In this case the odds are 5 to 1, or just 5. If it’s a fair die, the probability of rolling your point is one-sixth, and you have no edge. Your expected value is getting your wager back. But imagine it’s a loaded die and it rolls 2 one-fifth of the time and something else four-fifths of the", "the 1st question. I am still working on the 2nd. I came across another question though Suppose you flip one fair coin and roll one fair six-sided die. Let X be the product of the numbers of heads (i.e., 0 or 1) times the number showing on the die. Compute E(X) Here, if $X$ represents the product of the number of heads with rolled die value, we let $x_i=i$, $i=1,\\ldots,6$. In this case though, $\\mathbb{P}(X=x_i)=\\dfrac{1}{2}\\cdot\\dfrac{1}{6} =\\dfrac{1}{12}$ because the chance of getting a heads is $\\dfrac{1}{2}$ and the chance of rolling a 1 through 6 is $\\dfrac{1}{6}$ since we're working with a fair die. Therefore, $E[X]=\\sum_{i=1}^6 x_i\\mathbb{P}(X=x_i) = \\sum_{i=1}^6\\frac{i}{12}=\\ldots$ I hope this makes sense! #### chisigma ##### Well-known member I am so lost on how to do these questions:. 1.Suppose you start with eight pennies and flip one fair coin. If the coin comes up heads, you get to keep all your pennies; if the coin comes up tails, you have to give half of them back. Let X be the total number of pennies you have at the end. Compute E(X)... There are two not well specified points... a) if You remain with one penny the game is over or You can continue and lose half penny, a quarter of penny and so one?... b) is the number", "44 views Toss a fair coin twice. You win $1$ dollar if at least one of the two tosses comes out heads. (a) Assume that you play this game $300$ times. What is, approximately, the probability that you win at least $250$ dollar ? | 44 views +1 vote", "+0 # Study guide help 2 0 1 343 2 +623 1) Frank wants to earn as many points as possible in one turn in a game. Two number cubes whose sides are numbered 1 through 6 are rolled. He is given two options for the manner in which points are awarded in the turn. OPTION A: If the sum of the rolls is a prime number, Frank receives 12 points. OPTION B: If the sum of the rolls is a multiple of 4, Frank receives 24 points. Which statement best explains the option he should choose? A) Frank should choose Option A. The mathematical expectation of this option is 6 and is the greater mathematical expectation of the two options. B) Frank should choose Option B. The mathematical expectation of this option is 5 and is the greater mathematical expectation of the two options. C) Frank should choose Option B. The mathematical expectation of this option is 6 and is the greater mathematical expectation of the two options. D) Frank should choose Option A. The mathematical expectation of this option is 5 and is the greater mathematical expectation of the two options. 2) Danielle may choose one of two options for the method in which she may be awarded a money prize. OPTION A: Spin a spinner", "of gold, and was redefined in terms of a basket of European currencies in the seventies. Seems like a strange number, 0.888671 grams of gold. It’s 1/35 troy ounce. The US dollar, under the gold standard, was convertible to gold at $35 per troy ounce. So basically the predecessor to the euro was defined to be worth one US dollar. Rolling the dice From the March 11, 2022 “Riddler”: We’re playing a game where you have to pick four whole numbers. Then I will roll four fair dice. If any two of the dice add up to any one of the numbers you picked, then you win! Otherwise, you lose. For example, suppose you picked the numbers 2, 3, 4 and 12, and the four dice came up 1, 2, 4 and 5. Then you’d win, because two of the dice (1 and 2) add up to at least one of the numbers you picked (3). To maximize your chances of winning, which four numbers should you pick? And what are your chances of winning? Some first thoughts: • You want numbers that are common as the sums of two dice – middling numbers, numbers near seven. • The problem has a reflection symmetry. The dice values $x_1, x_2, x_3, x_4$ win with the target sums $y_1, y_2,", "Posts: 47983 ### Show Tags 31 Oct 2014, 02:56 5 1 Bunuel wrote: Tough and Tricky questions: Probability. Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny$1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability that Kate has more than$10 but less than $15? (A) 5/16 (B) 1/2 (C) 12/30 (D) 15/32 (E) 3/8 total number of possible outcomes when coin will be flipped 5 times= 2^5=32 now, for kate to have more than$10, she must have min. 3 favorable and max. 4 favorable draws in the 5 throws. (because if she wins in all the 5 draws, then she will have $15) kate will win, every time tails (T) appears on the coin. case 1) kate wins in 3 draws and losses in two. so we have 3 T's and 2H's(T,T,T,H,H), which can arrange themselves in 5!/(3!)(2!)=10 case 2) kate wins in 4 draws and losses in the 1 draw. so we have 4 T's and 1H (T,T,T,T,H), which can arrange themselves in 5!/4!=5 thus total no. of favorable ways=10+5=15 and thus required probability=15/32 Intern Joined: 07 May 2015 Posts: 17 Location: India Concentration: International Business, Finance GPA: 3.5 WE:" ]

[ "19 Explanation: The correct answer is (C). Recall the power rule: If $f(x) = x^n, f'(x) = nx^{n-1}$ Here, $f'(x) = 2 * 2x^1 - 3x^2$ $= 4x - 3x^2$ $f'(3) = 4(3) − 3(3)^2$ $= 12 − 27$ $= −15$ Question 20 ### What is the probability of rolling a prime number in each roll in 4 consecutive rolls of a 6 sided die? A $\\dfrac{1}{4}$ B $\\dfrac{1}{8}$ C $\\dfrac{1}{16}$ D $\\dfrac{1}{32}$ Question 20 Explanation: The correct answer is (C). The prime numbers in the range are: 2, 3, 5. On any roll, the probability of rolling a prime number is: $\\dfrac{3 \\text{ primes}}{6 \\text{ total}}= \\dfrac{1}{2}$ The probability of rolling this 4 times in a row is: $\\dfrac{1}{2} * \\dfrac{1}{2} * \\dfrac{1}{2} * \\dfrac{1}{2} = \\dfrac{1}{16}$ Question 21 ### $p = −3x + 5y$ $q = x + 2y$ What is the value of $p − 2q$? A $5x + y$ B $-5x + y$ C $-y + 5x$ D $5 + y$ Question 21 Explanation: The correct answer is (B). Substitute the values of $p$ and $q$ into the expression: $(−3x + 5y) − 2(x + 2y)$ Distribute and combine like terms to simplify: $= −3x + 5y − 2x − 4y$ $= −5x + y$ Question 22 ### A lab requires 0.76 L of", "= \\dfrac{1}{6}$ While, the probability of getting a two is $P (2) = \\dfrac{1}{6}$ And the probability of getting a prime number is $P (prime) = \\dfrac{3}{6} = \\dfrac{1}{2}$ These are all theoretical. We assume that when we roll a die, there is an equal chance of all the sides coming out. So, this is just the number of the favorable outcomes over the total outcomes. This is different from experimental probability. Experimental probability is based on experiment and data. If we roll a die theoretically where we have an equal chance of getting all the side, we will have the same frequency. So, if we roll a die sixty times, we will have $1$, $10$ times, we’ll have $2$, $10$ times, then we will have $3$, $10$ times, we’ll have $4$, $10$ times, we will have $5$, $10$ times, and we will have $6$, $10$ times. That’s theoretical. We expect it to be evenly distributed. But that’s not always the case. For whatever reason, when we roll a die sixty times, we got $1$, $15$ times, then we got $2$, $8$ times, we got $3$, $7$ times, and then we got $4$, $20$ times, we got $5$, $2$ times, and we got $6$, $8$ times. Things didn’t go as planned. For whatever reason, $4$ came out a", "$\\dfrac{5}{44}$ 7. $\\dfrac{77}{225}$,$\\dfrac{209}{225}$ 8. $\\dfrac{13}{32}$ 9. $\\dfrac{271}{285}$ 10. $x=3,\\dfrac{3}{26}$,$\\dfrac{46}{455}$ 11. $\\dfrac{3}{8}$,$\\dfrac{13}{20}$ Group (2013) 1. Draw a tree diagram to list all possible two-digit numerals which can be formed by using the digits $2,3,5$ and 6 without repetition. If one of these numerals is chosen at random, find the probability that it is divisible by $13 .$ Find also the probability that it is either a prime number or a perfect square. (5 marks) 2. Box $A$ contains 4 pieces of paper numbered as $1,2,3$ and 4. Box $B$ contains 3 pieces of paper numbered as 5,6 and 7. One piece of paper is chosen at random from box $A$ and then one piece of paper is chosen at random from box $B$. Draw a tree diagram to list all possible outcomes of the experiment. Find the probability that the product of the two numbers chosen is at most 10 . Find also the probability that the sum of the two chosen numbers is greater than their product. (5 marks) 3. Make a table of order pairs (first die, second die) for rolling two dice. Use the table to find the probability that the sum of the scores is odd, the product of the scores is greater than 15 and the product of the score is a multiple of", "$|{A}^{\\prime }\\cap {B}^{\\prime }\\cap {C}^{\\prime }|$: Excluding sums of 3, 4, and 8 eliminates $2+3+5=10$ possible outcomes for each of the five rolls. Hence, $|{A}^{\\prime }\\cap {B}^{\\prime }\\cap {C}^{\\prime }|=\\left(36-10{\\right)}^{5}={26}^{5}$. By the Inclusion-Exclusion Principle, the number of outcomes in which sums of 3, 4, and 8 all appear when five dice are rolled is ${36}^{5}-{34}^{5}-{33}^{5}-{31}^{5}+{31}^{5}+{29}^{5}+{28}^{5}-{26}^{5}$. Do you have a similar question?", "= \\dfrac 56$, the probability of failure, i.e. NOT rolling a $5$ (or any other specific number) Plugging it all in, we get $\\approx .01302$, or about $1.302 \\%$. It's probably more accurate to find the probability of getting a $5$ at least $50 \\%$ of the time, because you would be equally or even more impressed with any success rate $\\ge 50 \\%$. To do this, you want to find the probability of $(x=5)$ OR $(x=6)$ OR $\\cdots$ OR $(x=10)$, which is equal to $P(5) + P(6) + \\cdots P(10)$. Plugging this in, we get $\\approx .01546$, or about $1.546 \\%$. To find the probability of getting any $2$ numbers, Daugmented gives a nice idea in the comments. But we can still brute force our way through with the Binomial formula: For getting $2$ numbers exactly $50$ of the time, we have $n = 10$, the number of trials $x=5$, the number of successes, i.e. $50 \\%$ of $10$ $p = \\dfrac 26 = \\dfrac 13$, the probability of success, i.e. rolling a $5$ or a $6$ (or any other pair of numbers) $q = \\dfrac 46 = \\dfrac 23$, the probability of failure, i.e. NOT rolling a $5$ or a $6$ (or any other pair of numbers) The probability of getting a pair of numbers exactly", "x is not 1 or 6. We keep the 5 and reroll the last die, for a 1/3 chance to win (with either 1 or 6) -> 7/36*1/3 chance to win - If we do not get a 5 there is a 9/36 chance to get a 1 and a 7/36 chance to get a 6 but no 1, combined 16/36. In both cases we reroll one die for a 1/6 chance with the third roll. - If we don't get 1, 5 or 6, which means just 2,3,4 (9/36 chance) then we reroll both and have a 4/36 chance with the second roll. All cases are mutually exclusive. Sum: 4/36 + 7/36*1/3+16/36*1/6+9/36*4/36 = 5/18 = 10/36 Rerolling 3 and 6 is a little bit worse than rerolling 3 alone. #### mfb Mentor Who wins the game assuming an optimal strategy, the first player or his opponent? This is a nice puzzle. You excluded n from being added, therefore the first player must add 1. 3 is a prime again, the second players adds 1 and we get 4. This first player now can add 1 or 2, reaching 5 or 6. If the first player goes to 5 then the second player must go to 6. If 6 is a winning position, the first player will play", "4 \\\\In a deck of 52 cards, there are 13 diamonds and 13 spades.\\\\ Number of ways of drawing 3 diamonds and one spade =^{13}C_3*^{13}C_1 \\\\ Thus, the probability of obtaining 3 diamonds and one spade =\\dfrac{^{13}C_3*^{13}C_1}{^{52}C_4} 45 A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine (i) P(2) (ii) P(1 or 3) (iii) P(not 3) Solution : Total number of faces = 6\\\\ (i) Number of faces with number ‘2’ = 3\\\\ \\therefore P(2)=\\dfrac{3}{6}=\\dfrac{1}{2} \\\\ (ii)P(1 or 3)=P(not 2)=1-P(2)=1-\\dfrac{1}{2}=\\dfrac{1}{2} (iii)Number of faces with number '3'=1 \\\\ \\therefore P(3)=\\dfrac{1}{6} \\\\ Thus, P(not 3)=1-P(3)=\\dfrac{1}{6}=\\dfrac{5}{6} 46 In a certain lottery, 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy\\\\ (a) one ticket\\\\ (b) two tickets?\\\\ (c) 10 tickets?\\\\ Solution : Total number of tickets sold = 10,000\\\\ Number of prizes awarded = 10\\\\ (i)If we buy one ticket, then\\\\ P(getting a prize)=\\dfrac{10}{10000}=\\dfrac{1}{1000} \\\\ \\therefore P (not getting a prize)=1-\\dfrac{1}{1000}=\\dfrac{999}{1000} \\\\ (ii)If we buy two tickets, then Number of tickets not awarded = 10,000 -10 = 9990\\\\ P(not getting a prize)=\\dfrac{^{9990}C_2}{^{10000}C_2} \\\\ (iii) If we buy 10 tickets, then \\\\ P(not getting a prize)=\\dfrac{^{9990}C_{10}}{^{10000}C_{10}} 47 Out of 100 students, two", "event: the die has no way of reacting to what has gone before. So, on each roll of the die, the probability of getting a $6$ is $P(6) = \\dfrac{1}{6}$. The probability of getting three $6$s in a row is $P(6,6,6) = P(6) \\times P(6) \\times P(6) = \\dfrac{1}{6} \\times \\dfrac{1}{6} \\times \\dfrac{1}{6} = \\dfrac{1}{216}$. ### Example 3: Dice Game To win a certain dice game, you need to roll 5 of the same number with a single die in a row. What is the probability of winning by rolling five $6$s in a row, starting from a given roll of the die? Solution: On each roll of the die, there is a $\\dfrac{1}{6}$ chance of getting a $6$. Each roll is independent, so the probability of getting five $6$s in a row is $\\dfrac{1}{6} \\times \\dfrac{1}{6} \\times \\dfrac{1}{6} \\times \\dfrac{1}{6} \\times \\dfrac{1}{6} = \\dfrac{1}{7776} \\approx 0.0001286,$ which is not very big! The probability of getting 5 of any given number is also $\\dfrac{1}{7776}$, but the probability of winning the game on a given set of 5 rolls is $6 \\times \\dfrac{1}{7776} = \\dfrac{1}{2296}$ as there are six different ways that you can get 5 of the same number. Note: It is very unlikely to roll the same number a large number of times on a die. But,", "if a school sold more, it couldn't have been bigger, which corresponds to choice B. 12. A pair of fair $$6$$-sided dice is rolled $$n$$ times. What is the least value of $$n$$ such that the probability that the sum of the numbers face up on a roll equals $$7$$ at least once is greater than $$\\dfrac{1}{2}?$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ ###### Solution(s): To compute this, we can also find the least $$n$$ such that the probability of not rolling a $$7$$ is less than $$\\dfrac 12.$$ Each roll has an independent probability of $$\\dfrac 16$$ of getting $$7,$$ so it has a $$\\dfrac 56$$ probability of not landing on $$7.$$ Thus, the probability of none of the rolls being $$7$$ is $$\\left(\\dfrac 56\\right)^n.$$ We must find the least $$n$$ such that $$\\left(\\dfrac 56\\right)^n < \\dfrac 12.$$ If $$n=3,$$ then the probability is $$\\dfrac{125}{216},$$ which is greater than $$\\dfrac 12.$$ If $$n=4,$$ then the probability is $$\\dfrac{625}{1296},$$ which is less than $$\\dfrac 12.$$ This makes the answer $$4.$$ 13. The positive difference between a pair of primes is equal to $$2,$$ and the positive difference between the cubes of the two primes is $$31106.$$ What is the sum of the digits of the least prime that is greater than those two primes? $$\\ 8$$ $$\\ 10$$ $$\\", "\\frac{16}{20} = \\frac{4}{5}.$$ Then the probability that 2 is among the four chosen prime numbers is $1 - \\frac{4}{5}$, or $\\frac{1}{5}$. Alternatively, we could count the number of four-tuples that have a 2. This is the same as taking 4 times the number of ways to choose the other three numbers: $19\\cdot 18\\cdot 17$. (The factor of 4 appears since we could put the 2 in any of four positions). Now divide this by the total number of four-tuples, $20 \\cdot 19\\cdot 18\\cdot 17$, to arrive at the same answer as above: $$\\frac{4\\cdot 19\\cdot 18\\cdot 17} {20 \\cdot 19\\cdot 18\\cdot 17} = \\frac{4}{20} = \\frac{1}{5}.$$ 🔗 Question 2 Suppose you have an 8 sided die with sides labeled 1 to 8. You roll the die once and observe the number $f$ showing on the first roll. Then you have the option of rolling a second time. If you decline the option to roll again, you win $f$ dollars. If instead you exercise the option to roll a second time, and if $s$ shows on the second roll, then you win $f + s$ dollars if $f+s < 10$ and 0 dollars if $f+s \\geq 10$. What is the best strategy? Solution A strategy is given by specifying what to do in response to each possible first roll, $f$.", "other possibilities. Therefore, the conditional space becomes: First Die 1 2 3 4 5 6 Second Die 1 1, 1 2, 1 3, 1 4, 1 5, 1 6, 1 2 1, 2 2, 2 3, 2 4, 2 5, 2 6, 2 3 1, 3 2, 3 3, 3 4, 3 5, 3 6, 3 4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4 5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5 6 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6 In set notation... S={(1,6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} Out of the six pairs that have a sum of 7, we can see only one pair, (2,5), has a 5 on the second die. Therefore, $$P(C|D)=\\dfrac{1}{6}$$ Approach 2: From the previous example, we know $$P(D)=\\frac{6}{36}$$, $$P(D\\cap C)=\\frac{1}{36}$$. Therefore, using the formula above, $$P(C|D)=\\dfrac{P(C\\cap D)}{P(D)}=\\dfrac{\\frac{1}{36}}{\\frac{6}{36}}=\\dfrac{1}{6}$$. ## Important Note Regarding Conditional Probabilities For any two events, $A$ and $B$, 1. $$P(A|B)=1-P(A^\\prime|B)$$ 2. $$P(A)=P(A\\cap B)+P(A\\cap B^\\prime)=P(A|B)P(B)+P(A|B^\\prime)P(B^\\prime)$$ The Venn diagrams below will help you visualize why this is true. ##### $$P(A\\cap B)$$ The first diagram shows $$P(A\\cap B)$$. $$+$$ ##### $$P(A\\cap B^\\prime)$$ ...when added to $P(A\\cap B^\\prime)$... $$=$$ ##### $$P(A)$$ ...gives the $$P(A)$$. # 2.6 - Independent Events 2.6 - Independent Events", "# An unbiased cubic die marked with 1, 2, 2, 3, 3, 3 is rolled 3 times. The probability of getting a total score of 4 or 6 is Question : An unbiased cubic die marked with 1, 2, 2, 3, 3, 3 is rolled 3 times. The probability of getting a total score of 4 or 6 is (A) 16/216 (B) 50/216 (C) 60/216 (D) none of these Solution : Die marked with 1, 2, 2, 3, 3, 3 is throw 3 times P(1) = 1/6, P(2) = 2/6, P(3) = 3/6 P(S) = P(4 or 6) = P(112 (3 cases) or 123 (6 cases) or 222) =$\\ 3\\times \\dfrac{1}{6}\\times \\dfrac{1}{6}\\times \\dfrac{2}{6}+6\\times \\dfrac{1}{6}\\times \\dfrac{2}{6}\\times \\dfrac{3}{6}+\\dfrac{2}{6}\\times \\dfrac{2}{6}\\times \\dfrac{2}{6}$ =$\\dfrac{6+36+8}{216}=\\dfrac{50}{216}$ Tags: No tags", "## Finite Math and Applied Calculus (6th Edition) Losing = (doubles) OR [( green is odd) AND ( red is even)] The green and red dice are distinct so the set of possible outcomes is $S=\\{(1,1),(1,2),...(1,6),(2,1),(2,2),...(6,6)\\}$ (first: green die, second: red die) $n(S)=36$ Let A = set of outcomes which are doubles A=$\\{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\\}$ n(A)=6 Let B =set of outcomes where green die is odd and red die is even B=$\\{(1,2), (1,4), (1,6), (3,2), (3,4), (3,6),(5,2), (5,4), (5,6)\\}$ $n(B)=9$ $A\\cap B=\\emptyset$ Let L = set of losing combinations $L=A\\cup B$ $n(L)=n(A\\cup B)=n(A)+n(B)-n(A\\cap B)$ $=6+9-0=15$ Since $n(S)=36$, and $L^{\\prime}$ is the set of winning combinations, $n(L^{\\prime})=n(S)-n(L)=36-15=21$ (there are 21 winning combinations)", "# A discrete mathematics problem by Soumava Pal $$A$$, $$B$$ and $$C$$ take turns rolling a fair die (with equal probability of it showing any of $$\\{1,2,3,4,5,6\\}$$, and the as soon as one of them gets a six, the game ends and he wins. $$A$$ gets the first chance to roll the die ,followed by $$B$$, and then $$C$$, followed by $$A$$ again if none of them gets a six. Let the probability of $$A$$ winning the game be $$\\dfrac{a}{b}$$, that of $$B$$ be $$\\dfrac{c}{d}$$ and that of $$C$$ winning the game be $$\\dfrac{e}{f}$$, with $$a,b,c,d,e,f$$ being positive integers, with $$\\gcd(a,b)=\\gcd(c,d)=\\gcd(e,f)=1$$. Find $$a+b+c+d+e+f$$. ×", "has a probability of $$\\dfrac{19}{20}$$. The final condition is that the roll of the first die is strictly less than $$2\\times$$ the roll of the second die, so we create a summation that traverses all of the possible rolls on the first die from $$2-20$$. The odds of each roll on the first die is $$\\dfrac{1}{19}$$ and the odds that it is strictly less than a roll on the second die is $$1-\\dfrac{1}{2} \\times \\dfrac{n}{20}$$. In other words it is $$1 -$$ the number of times the first roll would be greater than or equal to the second roll. This brings the full equation to: $$\\dfrac{10}{20} + \\dfrac{1}{20} + \\dfrac{10}{20}\\times\\dfrac{19}{20}\\sum_{n=2}^{20}\\dfrac{1}{19}\\times(1-\\dfrac{1}{2}\\times\\dfrac{n}{10})$$ With the above equation I obtained the probability 0.76375, which again is the probability that the second roll is strictly greater than the first, but which we can use to obtain the probability that the first roll will be greater than or equal to the second as such: $$1 - 0.76375 = 0.23625$$ But when I map out the correct number of victories for the first roll over the second I get: |---------------------|------------------|---------------------|------------------| | First Roll | # Victories | First Roll | # Victories | |---------------------|------------------|---------------------|------------------| | 1 | 0 | 11 | 5 | |---------------------|------------------|---------------------|------------------| | 2 | 1 | 12 | 6 | |---------------------|------------------|---------------------|------------------| |", "1 \\times \\dfrac{1}{6} \\times \\dfrac{1}{6} \\times \\dfrac{5}{6} = \\dfrac{20}{216}$$ Thus, our final answer is the sum of the probabilities of events $A$ and $B$. We get $\\dfrac{1}{216} + \\dfrac{20}{216} = \\boxed{\\dfrac{21}{216}} \\approx 9.72\\%$. The number of combinations of rolls of the 4 die is, correctly, $6*6*6*6 = 6^4$. The number of ways in which all 4 die roll the same value is 6 (either they all roll 1 or 2 or 3 and so on...) The number of ways in which Exactly 3 die roll the same value is more complicated. Consider first the number of ways in which the first 3 die roll the same value but the fourth die rolls another. Label the die $W_1 , W_2, W_3, W_4$ for convenience. If $W_1 , W_2, W_3$ all roll 1's for example, then the number of combinations in which $W_4$ is not 1 is just 5. (The combinations are just (1,1,1,2), (1,1,1,3), (1,1,1,4) (1,1,1,5) and (1,1,1,6)). Since there are 6 possible values for the first 3 die, there are 6*5= 30 ways in which $W_1 , W_2, W_3$ are all the same and $W_4$ is different. There are also many ways to choose 3 die out of the 4 namely, ${ 4 \\choose 3} = 4$. Putting it all together: P(at least 3 dice are the same)", "a prime roll number or a girl with a composite roll number or an even roll number. Solution:", "so the probability of hitting a $6$ on the first trial is $\\dfrac{1}{6}$ So the probability of getting a $6$ is $\\dfrac{1}{6}$. Probability of not $6$ is $1 – \\dfrac{1}{6} = \\dfrac{5}{6}$. #### Part One For winning $100$, it is compulsory to score a $6$ in the first trial, and the probability of $6$ is $\\dfrac{1}{6}$. For succeeding $50$, it is required not to score $6$ in the first roll and $6$ in the second roll, and the probability of not getting a $6$ is $\\dfrac{5}{6}$ and the probability of $6$ is $\\dfrac{1}{6}$, so the probability, in this scenario, would be $\\dfrac{1}{6} \\times \\dfrac{5}{6}$, which is equals to $\\dfrac{5}{36}$. For $0$, it’s required not to score $6$ in both the rolls, so the probability, in this circumstance, becomes $\\dfrac{5}{6} \\times \\dfrac{5}{6}$, that is equal to $\\dfrac{25}{36}$. Figure 1 #### Part b: Formula for expected value is given as: $E(x) = \\sum Value.P(x)$ $= (100)(\\dfrac{1}{6}) + (50)(\\dfrac{5}{36}) + (0)(\\dfrac{25}{36})$ ## Numerical Result The expected amount is: $E(x) = \\23.61$ ## Example You roll a die. If it comes up a $6$, you win $100$. If not, you get to roll again. If you get $6$ the $2^{nd}$ time, you win $50$. If not, you get to roll again. If you get a $6$ the $3^{rd}$ time, you win $25$.", "# Probability of dice sum simply more than 100 Can a person please overview me to a means through which I can address the adhering to trouble. There is a die and also 2 gamers. Moving quits as quickly as some goes beyond 100 (not consisting of 100 itself). Therefore you have the adhering to selections : 101, 102, 103, 104, 105, 106. Which need to I pick offered front runner. I'm assuming Markov chains, yet exists a less complex means? Many thanks. MODIFY : I created dice as opposed to die. There is simply one die being rolled 0 2019-12-02 02:48:26 Source Share Answers: 4 It is a wonderful trouble. The opportunity of striking $101$ first is remarkably huge. Allow $a(x)$ be the opportunity of striking $101$ first, beginning at $x$. We address recursively by establishing $a(106)=a(105)=a(104)=a(103)=a(102)=0$ and also $a(101)=1$. After that, for $x$ from $100$ to $0$, placed $a(x)={1\\over 6}\\sum_{j=1}^6 a(x+j)$. By the revival theory, $a(x)$ merges to $1/\\mu$, where $\\mu=(1+2+3+4+5+6)/6=7/2$. The value $a(0)$ is really near this restriction. Actually, the entire striking circulation is about $[6/21,5/21,4/21,3/21,2/21,1/21]$. Included : Let $a^\\prime(x)$ be the opportunity of striking $102$ first, beginning at $x$. After that $a^\\prime(x)$ pleases the very same reappearance as $a(x)$ for $x\\leq 100$, yet with various border problems $a^\\prime(106)=a^\\prime(105)=a^\\prime(104)=a^\\prime(103)=a^\\prime(101)=0$ and also $a^\\prime(102)=1$. Consequently $$a^\\prime(x)=a(x-1)-a(100)a(x)$$ and", "# Probability Mania A fair six-faced die is rolled six times. The probability that the numbers on the die appear either in ascending or descending order is of the form $\\dfrac{A}{B}$ where $$A$$ and $$B$$ are co-prime integers. Find the value of $\\dfrac{B}{24}-A$ Assumptions - The order may not be strictly ascending or descending. For example, $$1-1-2-4-4-6$$ is a valid sequence of numbers. ×" ]

[ "+0 0 31 2 Monica tosses a fair 6-sided die. If the roll is a prime number, then she wins that amount of dollars (so that, for example, if she rolls 3, then she wins 3 dollars). If the roll is composite, she wins nothing. Otherwise, she loses 3 dollars. What is the expected value of her winnings on one die toss? Express your answer as a dollar value to the nearest cent. Jul 17, 2021 #1 +86 +3 Answer: $$1.17$$ Solution: Case 1: Monica rolls a 1. If Monica rolls a 1, then she loses 3 dollars. Case 2: Monica rolls a 2. If Monica rolls a 2, then she gains 2 dollars. Case 3: Monica rolls a 3. If Monica rolls a 3, then she gains 3 dollars. Case 4: Monica rolls a 4. If Monica rolls a 4, then she gains nothing. Case 5: Monica rolls a 5. If Monica rolls a 5, then she gains 5 dollars. Case 6: Monica rolls a 6. If Monica rolls a 6, then she gains nothing. Adding all them up and then placing it over six (there are six possibillities) gives $$-3+2+3+0+5+0\\over6$$ Which is equal to $$\\frac76$$, which rounded to the nearest cent, gives $$1.17$$. Jul 18, 2021 #2 0 Correct! Guest Jul 18, 2021", "# what will be the probability? Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6? my approach was,the number of likely combinations are (1,6),(2,6),(3,6),(2,5),(1,5),(3,5),(2,4),(3,4),(3,3) and total combinations will be 3(for 1,2,3 in first roll)*6( in second roll).so probability should be 9/18.but answer was not accepted.please help me understanding the correct approach. Alicia is playing the game. She wins if the sum total of values is at least $6$. If the first toss results in a $4$ or a $5$, she won't get a chance to retoss, and she won't win. Instant win: If she gets a $6$ on the first toss, then she has won quickly. This has probability $\\dfrac{1}{6}$. We now examine what happens if the first toss is a $1$, $2$, or $3$. Examine the cases one at a time. First toss is a 1: If at first she gets a $1$, (probability $\\frac{1}{6}$), then she gets to retoss. To win, she needs a $5$ or a $6$ on the second toss. The probability this sequence of events happens is $\\dfrac{1}{6}\\cdot \\dfrac{2}{6}$. First toss is a 2: In this case, to win", "19 Explanation: The correct answer is (C). Recall the power rule: If $f(x) = x^n, f'(x) = nx^{n-1}$ Here, $f'(x) = 2 * 2x^1 - 3x^2$ $= 4x - 3x^2$ $f'(3) = 4(3) − 3(3)^2$ $= 12 − 27$ $= −15$ Question 20 ### What is the probability of rolling a prime number in each roll in 4 consecutive rolls of a 6 sided die? A $\\dfrac{1}{4}$ B $\\dfrac{1}{8}$ C $\\dfrac{1}{16}$ D $\\dfrac{1}{32}$ Question 20 Explanation: The correct answer is (C). The prime numbers in the range are: 2, 3, 5. On any roll, the probability of rolling a prime number is: $\\dfrac{3 \\text{ primes}}{6 \\text{ total}}= \\dfrac{1}{2}$ The probability of rolling this 4 times in a row is: $\\dfrac{1}{2} * \\dfrac{1}{2} * \\dfrac{1}{2} * \\dfrac{1}{2} = \\dfrac{1}{16}$ Question 21 ### $p = −3x + 5y$ $q = x + 2y$ What is the value of $p − 2q$? A $5x + y$ B $-5x + y$ C $-y + 5x$ D $5 + y$ Question 21 Explanation: The correct answer is (B). Substitute the values of $p$ and $q$ into the expression: $(−3x + 5y) − 2(x + 2y)$ Distribute and combine like terms to simplify: $= −3x + 5y − 2x − 4y$ $= −5x + y$ Question 22 ### A lab requires 0.76 L of", "the chance of the building being opened on time is $$1 - (0.05 \\times 0.1 \\times 0.2 \\times 0.25 \\times 0.5 \\times 0.9) \\approx 0.9999$$. 2. This assessment entails the multiplication of probabilities, which assumes independence. It seems plausible that circumstances that precipitate a loss-of-coolant accident might also predispose towards the failure of safety functions: for example, an earthquake or human sabotage. ##### Exercise 12 1. The player loses at the first roll if 1, 4 or 6 is obtained. The probability of this occurring is 0.5. 2. To win $6, the player needs to be paid$10. This only occurs if the first roll is 5 and the second roll is 5; the probability of this is $$\\dfrac{1}{36}$$. 3. The probability of winning is obtained by adding up the probabilities of outcomes for which the payout is greater than \\$4. There are eight such outcomes: (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3) and (5,5), each with probability equal to $$\\dfrac{1}{36}$$. So the probability of winning equals $$\\dfrac{8}{36} = \\dfrac{2}{9}$$. ##### Exercise 13 1. $$\\Pr(C'|T_{-}) \\approx 0.9984$$. So, if the test is negative, there is a very high probability that the woman does not have breast cancer. 2. The positive predictive value is calculated as follows: 3. \\begin{align*} \\Pr(C|T_{+}) &= \\dfrac{\\Pr(C) \\Pr(T_{+}|C)}{\\Pr(C) \\Pr(T_{+}|C) + \\Pr(C') \\Pr(T_{+}|C')}\\\\ &= \\dfrac{0.01 \\times", "= \\dfrac{1}{6}$ While, the probability of getting a two is $P (2) = \\dfrac{1}{6}$ And the probability of getting a prime number is $P (prime) = \\dfrac{3}{6} = \\dfrac{1}{2}$ These are all theoretical. We assume that when we roll a die, there is an equal chance of all the sides coming out. So, this is just the number of the favorable outcomes over the total outcomes. This is different from experimental probability. Experimental probability is based on experiment and data. If we roll a die theoretically where we have an equal chance of getting all the side, we will have the same frequency. So, if we roll a die sixty times, we will have $1$, $10$ times, we’ll have $2$, $10$ times, then we will have $3$, $10$ times, we’ll have $4$, $10$ times, we will have $5$, $10$ times, and we will have $6$, $10$ times. That’s theoretical. We expect it to be evenly distributed. But that’s not always the case. For whatever reason, when we roll a die sixty times, we got $1$, $15$ times, then we got $2$, $8$ times, we got $3$, $7$ times, and then we got $4$, $20$ times, we got $5$, $2$ times, and we got $6$, $8$ times. Things didn’t go as planned. For whatever reason, $4$ came out a", "she needs a $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is $\\dfrac{1}{6}\\cdot \\dfrac{3}{6}$. First toss is a 3: In this case she needs a $3$, $4$, $5$, or $6$ on the second toss. The probability this sequence of events happens is $\\dfrac{1}{6}\\cdot \\dfrac{4}{6}$. Thus the probabilility Alicia wins is $\\dfrac{1}{6}+ \\dfrac{1}{6}\\cdot \\dfrac{2}{6}+ \\dfrac{1}{6}\\cdot \\dfrac{3}{6}+ \\dfrac{1}{6}\\cdot \\dfrac{4}{6}$. Simplify. • yes sir, i also figured it out in same way, but unfortunately not when needed. – Abhishek Pandey Aug 23 '13 at 8:30 You forgot 4, 5, 6 rolls. And, you need to take into account the fact that the odds of a 4, 5, or 6 is not the same as, say, (1, 5). Getting a 4 happens 1/6th of the time. Getting a (1, 5) happens 1/18th of the time. • how you are saying there will be 21 possibilities. i am not getting please elaborate. – Abhishek Pandey Aug 17 '13 at 12:47 • yes got it, i was not considering 4,5,6 and our likely case (6,0),answer should be 10/21.am i right? – Abhishek Pandey Aug 17 '13 at 13:03 • @AbhishekPandey No, the 21 possibilities are not all equally likely. Half the time you roll a 4, 5, or 6 on the first roll. The other half the", "# Alternate moves, first one to roll a 6 wins: what's the probability of winning? Question: Players A and B play a sequence of independent games. Player A starts and throws a fair die and wins on a \"six\". If she fails then player B throws the same die and wins on a \"six\" or fails otherwise. If she fails then player A gets the die and throws again, and so on. The first player to throw a \"six\" wins. Find the probability that player A wins. Here is what I have done: A wins right away: $\\dfrac{1}{6}$ A fails B wins: $\\dfrac{5}{6}\\times \\dfrac{1}{6}$ A fails B fails A wins: $\\dfrac{5}{6}\\times \\dfrac{5}{6}\\times \\dfrac{1}{6}$ Unlike another similar question which the player B can win on \"five\" and \"six\" after player A fails on the first throw, in this case, the games do not seem to have an endpoint. How should I calculate the probability that player A wins then? Player A can win straightaway with probability $\\frac{1}{6}$ and with probability $\\frac{5}{6}$ we are in exactly the same situation from player B's perspective. Let's name $p$ the probability of winning (in any number of moves) for the player who is about to roll the die. Then: $$p = \\frac{1}{6} \\cdot 1 + \\frac{5}{6}(1-p) = \\frac{6-5p}{6} \\iff 6p = 6-5p \\iff", "so the probability of hitting a $6$ on the first trial is $\\dfrac{1}{6}$ So the probability of getting a $6$ is $\\dfrac{1}{6}$. Probability of not $6$ is $1 – \\dfrac{1}{6} = \\dfrac{5}{6}$. #### Part One For winning $100$, it is compulsory to score a $6$ in the first trial, and the probability of $6$ is $\\dfrac{1}{6}$. For succeeding $50$, it is required not to score $6$ in the first roll and $6$ in the second roll, and the probability of not getting a $6$ is $\\dfrac{5}{6}$ and the probability of $6$ is $\\dfrac{1}{6}$, so the probability, in this scenario, would be $\\dfrac{1}{6} \\times \\dfrac{5}{6}$, which is equals to $\\dfrac{5}{36}$. For $0$, it’s required not to score $6$ in both the rolls, so the probability, in this circumstance, becomes $\\dfrac{5}{6} \\times \\dfrac{5}{6}$, that is equal to $\\dfrac{25}{36}$. Figure 1 #### Part b: Formula for expected value is given as: $E(x) = \\sum Value.P(x)$ $= (100)(\\dfrac{1}{6}) + (50)(\\dfrac{5}{36}) + (0)(\\dfrac{25}{36})$ ## Numerical Result The expected amount is: $E(x) = \\23.61$ ## Example You roll a die. If it comes up a $6$, you win $100$. If not, you get to roll again. If you get $6$ the $2^{nd}$ time, you win $50$. If not, you get to roll again. If you get a $6$ the $3^{rd}$ time, you win $25$.", "\\frac{16}{20} = \\frac{4}{5}.$$ Then the probability that 2 is among the four chosen prime numbers is $1 - \\frac{4}{5}$, or $\\frac{1}{5}$. Alternatively, we could count the number of four-tuples that have a 2. This is the same as taking 4 times the number of ways to choose the other three numbers: $19\\cdot 18\\cdot 17$. (The factor of 4 appears since we could put the 2 in any of four positions). Now divide this by the total number of four-tuples, $20 \\cdot 19\\cdot 18\\cdot 17$, to arrive at the same answer as above: $$\\frac{4\\cdot 19\\cdot 18\\cdot 17} {20 \\cdot 19\\cdot 18\\cdot 17} = \\frac{4}{20} = \\frac{1}{5}.$$ 🔗 Question 2 Suppose you have an 8 sided die with sides labeled 1 to 8. You roll the die once and observe the number $f$ showing on the first roll. Then you have the option of rolling a second time. If you decline the option to roll again, you win $f$ dollars. If instead you exercise the option to roll a second time, and if $s$ shows on the second roll, then you win $f + s$ dollars if $f+s < 10$ and 0 dollars if $f+s \\geq 10$. What is the best strategy? Solution A strategy is given by specifying what to do in response to each possible first roll, $f$.", "$2$ and Candy rolls $2$ or more: $\\dfrac{1}{36}\\cdot\\dfrac{5}{6}=\\dfrac{5}{216}$ • The probability that Andy rolls $3$ and Candy rolls $3$ or more: $\\dfrac{2}{36}\\cdot\\dfrac{4}{6}=\\dfrac{8}{216}$ • The probability that Andy rolls $4$ and Candy rolls $4$ or more: $\\dfrac{3}{36}\\cdot\\dfrac{3}{6}=\\dfrac{9}{216}$ • The probability that Andy rolls $5$ and Candy rolls $5$ or more: $\\dfrac{4}{36}\\cdot\\dfrac{2}{6}=\\dfrac{8}{216}$ • The probability that Andy rolls $6$ and Candy rolls $6$: $\\dfrac{5}{36}\\cdot\\dfrac{1}{6}=\\dfrac{5}{216}$ Hence the probability of the complementary event is: $$1-\\dfrac{5+8+9+8+5}{216}=\\dfrac{181}{216}$$ That's not right since Candy might not roll a six. The most straightforward way to answer this is to simply lay out a table of all the possibilities for the two rollers and then add up all the cells where Andy wins. Another way to do it though is rewrite the problem. P(A > C) is the same as P(A - C > 0). Then, you can use the fact that subtracting a die is the same as adding one and then subtracting 7 (try it out). So P(A - C > 0) is the same as P(A + C - 7 > 0) or P(A + C > 7). A + C is just rolling three dice, so the answer to your question is the same as the answer to \"what is the probability of getting more than 7 on three dice\". Think of the three", "event: the die has no way of reacting to what has gone before. So, on each roll of the die, the probability of getting a $6$ is $P(6) = \\dfrac{1}{6}$. The probability of getting three $6$s in a row is $P(6,6,6) = P(6) \\times P(6) \\times P(6) = \\dfrac{1}{6} \\times \\dfrac{1}{6} \\times \\dfrac{1}{6} = \\dfrac{1}{216}$. ### Example 3: Dice Game To win a certain dice game, you need to roll 5 of the same number with a single die in a row. What is the probability of winning by rolling five $6$s in a row, starting from a given roll of the die? Solution: On each roll of the die, there is a $\\dfrac{1}{6}$ chance of getting a $6$. Each roll is independent, so the probability of getting five $6$s in a row is $\\dfrac{1}{6} \\times \\dfrac{1}{6} \\times \\dfrac{1}{6} \\times \\dfrac{1}{6} \\times \\dfrac{1}{6} = \\dfrac{1}{7776} \\approx 0.0001286,$ which is not very big! The probability of getting 5 of any given number is also $\\dfrac{1}{7776}$, but the probability of winning the game on a given set of 5 rolls is $6 \\times \\dfrac{1}{7776} = \\dfrac{1}{2296}$ as there are six different ways that you can get 5 of the same number. Note: It is very unlikely to roll the same number a large number of times on a die. But,", "8k views Suppose a fair six-sided die is rolled once. If the value on the die is $1, 2,$ or $3,$ the die is rolled a second time. What is the probability that the sum total of values that turn up is at least $6$ ? 1. $\\dfrac{10}{21}$ 2. $\\dfrac{5}{12}$ 3. $\\dfrac{2}{3}$ 4. $\\dfrac{1}{6}$ edited | 8k views 0 Hello sir, why we have taken sample space 36 although we not throw dice again when we got 4,5,6 . +7 $\\\\1\\rightarrow 5,6\\\\ 2\\rightarrow4,5,6\\\\ 3\\rightarrow3,4,5,6\\\\ 4\\rightarrow\\times\\\\ 5\\rightarrow\\times\\\\ 6\\rightarrow \\checkmark$ $\\dfrac{1}{6}\\times \\dfrac{2}{6}+\\dfrac{1}{6}\\times \\dfrac{3}{6}+\\dfrac{1}{6}\\times \\dfrac{4}{6}+\\dfrac{1}{6}\\times \\dfrac{6}{6}=\\dfrac{5}{12}$ 0 How $6$ is valid? On first dice only $1,2,3$ can come. Here our sample space consists of $3 + 3\\times 6 = 21$ events- $(4), (5), (6), (1,1), (1,2)\\ldots(3,6).$ Favorable cases $=(6), (1,5), (1,6), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3,5), (3,6).$ Required Probability $= \\dfrac {\\text{No. of favorable cases}}{\\text{Total cases }}= \\dfrac{10}{21}$ But this is wrong way of doing. Because due to $2$ tosses for some and $1$ for some, individual probabilities are not the same. i.e., while $(6)$ has $\\dfrac{1}{6}$ probability of occurrence, $(1,5)$ has only $\\dfrac{1}{36}$ probability. So, our required probability $\\Rightarrow \\dfrac{1}{6} + \\left(9 \\times \\dfrac{1}{36}\\right) =\\dfrac{5}{12}.$ Correct Answer: $B$ by Veteran edited 0 in favorable cases why 6 is considered ? +4 great explanation @Arjun. 0", "If not, you lose. Find the Expected amount you win. For winning $100$, P(x) is $\\dfrac{1}{6}$ For winning $50$, P(x) is $\\dfrac{1}{6} \\times \\dfrac{5}{6} = \\dfrac{5}{36}$ For winning $25$, P(x) is $\\dfrac{1}{6} \\times \\dfrac{5}{6} \\times \\dfrac{5}{6} = \\dfrac{25}{216}$ For winning $0$, P(x) is $\\dfrac{5}{6} \\times \\dfrac{5}{6} \\times \\dfrac{5}{6} = \\dfrac{125}{216}$ In the end, the expected amount is the sum of the multiplication of the results and their probabilities: $E(x) = \\sum Value.P(x)$ $= (100)(\\dfrac{1}{6}) + (50)(\\dfrac{5}{36}) + (25)(\\dfrac{25}{216}) + (0)(\\dfrac{125}{216})$ This is the expected amount after the given number of trials: $E(x) = \\25.50$ Images/mathematical drawings are created with GeoGebra.", "of one thing happening given that you know something else has already happened. We use P(A|B) to denote “probability of A happening given B has happened”. In the case of a tree diagram, conditional probability works nicely – if we were to ask, “What is the probability that Monica wins the second match given that Penny has already won the first?”, all we need to do is look to the branch that represents the event of Monica winning the 2nd match after a Penny win. So, the probability of Monica winning the second given that Penny wins the first is 0.2. In other words, $\\text{P(Monica wins 2nd | Penny wins 1st)} = 0.2$ ## Probability and Tree Diagrams Questions The key to answering this question is recognising that these are exhaustive probabilities (the bead you choose has to be either red, blue, or green), so they must add up to 1. Therefore, $0.25+5x+4x=1$ Subtract 0.25 from both sides to get $9x=0.75$ So, we get that $x=0.75\\div9=\\dfrac{1}{12}$. Now we have $x$, we know that the probability of getting a blue bead is $5x=\\dfrac{5}{12}$, and the probability of getting a green bead is $4x=\\dfrac{4}{12}$. 1. a) If we calculate the relative frequency of a car being black using Hannah’s data, we get $\\dfrac{32}{32+48}=0.4=40\\%$. So, according to her data, her statement", "answer. (4 marks) (c) Michelle selects the option with higher expected dollars got and she plays this game for three rounds. She thinks that her chance of getting no dollars is less than ${0.07}$. Do you agree? Explain your answer. (2 marks) (a) ${P}{\\left(\\text{same hole}\\right)}$ $={1}\\times\\dfrac{{1}}{{6}}\\times\\dfrac{{1}}{{6}}=\\dfrac{{1}}{{36}}$ 1A ${P}{\\left(\\text{3 in different holes}\\right)}$ $={1}\\times\\dfrac{{5}}{{6}}\\times\\dfrac{{4}}{{6}}=\\dfrac{{5}}{{9}}$ 1A Expected dollars $=\\dfrac{{1}}{{36}}\\times{120}+\\dfrac{{5}}{{9}}\\times{90}$ $=\\dfrac{{160}}{{3}}$ 1A (b) For option 2, ${P}{\\left(\\text{same hole}\\right)}$ $={1}\\times\\dfrac{{1}}{{6}}\\times\\dfrac{{1}}{{6}}\\times\\dfrac{{1}}{{6}}=\\dfrac{{1}}{{216}}$ ${P}{\\left(\\text{3 in the same hole}\\right)}$ $={{C}_{{3}}^{{4}}}{\\left({1}\\times\\dfrac{{5}}{{6}}\\times\\dfrac{{1}}{{6}}\\times\\dfrac{{1}}{{6}}\\right)}$ 1A $=\\dfrac{{5}}{{54}}$ ${P}{\\left(\\text{4 adjacent holes}\\right)}$ $={6}{{P}_{{4}}^{{4}}}{\\left(\\dfrac{{1}}{{6}}\\right)}^{{4}}$ 1A $=\\dfrac{{1}}{{9}}$ Expected dollars $=\\dfrac{{1}}{{216}}\\times{270}+\\dfrac{{5}}{{54}}\\times{120}+\\dfrac{{1}}{{9}}\\times{120}$ 1M $=\\dfrac{{925}}{{36}}$ $<\\dfrac{{160}}{{3}}$ The expected dollars of option 1 is higher. Thus, option 1 should be selected. 1A (c) For option 1, ${P}{\\left(\\text{no dollars}\\right)}$ $={1}-\\dfrac{{1}}{{36}}-\\dfrac{{5}}{{9}}$ $=\\dfrac{{5}}{{12}}$ ${P}{\\left(\\text{no dollars in 3 rounds}\\right)}$ $={\\left(\\dfrac{{5}}{{12}}\\right)}^{{3}}$ 1M $={0.07233796296296294}$ $>{0.07}$ Thus, Michelle is disagreed. 1A # 專業備試計劃 Level 4+ 保證及 5** 獎賞 ePractice 會以電郵、Whatsapp 及電話提醒練習 ePractice 會定期提供溫習建議 Level 5** 獎勵：會員如在 DSE 取得數學 Level 5** ，將獲贈一套飛往英國、美國或者加拿大的來回機票，唯會員須在最少 180 日內每天在平台上答對 3 題 MCQ。 Level 4 以下賠償：會員如在 DSE 未能達到數學 Level 4 ，我們將會全額退回所有會費，唯會員須在最少 180 日內每天在平台上答對 3 題 MCQ。 # FAQ ePractice 是甚麼？ ePractice 是一個專為中四至中六而設的網站應用程式，旨為協助學生高效地預備 DSE 數學（必修部分）考試。由於 ePractice 是網站應用程式，因此無論使用任何裝置、平台，都可以在瀏覽器開啟使用。更多詳情請到簡介頁面。 ePractice 可以取代傳統補習嗎？ 1. 會員服務期少於兩個月；或 2. 交易額少於 HK\\$100。 Initiating... HKDSE 數學試題練習平台", "4 \\\\In a deck of 52 cards, there are 13 diamonds and 13 spades.\\\\ Number of ways of drawing 3 diamonds and one spade =^{13}C_3*^{13}C_1 \\\\ Thus, the probability of obtaining 3 diamonds and one spade =\\dfrac{^{13}C_3*^{13}C_1}{^{52}C_4} 45 A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine (i) P(2) (ii) P(1 or 3) (iii) P(not 3) Solution : Total number of faces = 6\\\\ (i) Number of faces with number ‘2’ = 3\\\\ \\therefore P(2)=\\dfrac{3}{6}=\\dfrac{1}{2} \\\\ (ii)P(1 or 3)=P(not 2)=1-P(2)=1-\\dfrac{1}{2}=\\dfrac{1}{2} (iii)Number of faces with number '3'=1 \\\\ \\therefore P(3)=\\dfrac{1}{6} \\\\ Thus, P(not 3)=1-P(3)=\\dfrac{1}{6}=\\dfrac{5}{6} 46 In a certain lottery, 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy\\\\ (a) one ticket\\\\ (b) two tickets?\\\\ (c) 10 tickets?\\\\ Solution : Total number of tickets sold = 10,000\\\\ Number of prizes awarded = 10\\\\ (i)If we buy one ticket, then\\\\ P(getting a prize)=\\dfrac{10}{10000}=\\dfrac{1}{1000} \\\\ \\therefore P (not getting a prize)=1-\\dfrac{1}{1000}=\\dfrac{999}{1000} \\\\ (ii)If we buy two tickets, then Number of tickets not awarded = 10,000 -10 = 9990\\\\ P(not getting a prize)=\\dfrac{^{9990}C_2}{^{10000}C_2} \\\\ (iii) If we buy 10 tickets, then \\\\ P(not getting a prize)=\\dfrac{^{9990}C_{10}}{^{10000}C_{10}} 47 Out of 100 students, two", "the odds against rolling a six with a fair die are 5 to 1. The probability of rolling a six with a fair die is the single number 1/6 or approximately 16.7%. The gambling and statistical uses of odds are closely interlinked. If a bet is a fair one, as in a wager between friends, then the odds offered to the gamblers will perfectly reflect relative probabilities. A fair bet that a fair die will roll a six will pay the gambler $5 for a$1 wager (and return the bettor his or her wager) in the case of a six and nothing in any other case. The terms of the bet are fair, because on average, five rolls result in something other than a six, at a cost of $5, for every roll that results in a six and a net payout of$5. The profit and the expense exactly offset one another and so there is no disadvantage to gambling over the long run. If the odds being offered to the gamblers do not correspond to probability in this way then one of the parties to the bet has an advantage over the other. Casinos, for example, offer odds that place themselves at an advantage, which is how they guarantee themselves a profit and survive as businesses. The", "We are only given conditional probabilities for the second set of lights. However, we can obtain the probability $$\\Pr(G_2)$$ that she meets green at the second set by considering the two events $$G_1 \\cap G_2$$ and $$R_1 \\cap G_2$$. These are mutually exclusive, and $$G_2 = (G_1 \\cap G_2) \\cup (R_1 \\cap G_2)$$. Hence $$\\Pr(G_2) = 0.18 + 0.28 = 0.46$$. Note that the events $$G_1$$ and $$G_2$$ are not independent, since $\\Pr(G_1 \\cap G_2) = 0.18 \\qquad \\text{and} \\qquad \\Pr(G_1) \\Pr(G_2) = 0.3 \\times 0.46 = 0.138,$ and therefore $$\\Pr(G_1 \\cap G_2) \\neq \\Pr(G_1) \\Pr(G_2)$$. Exercise 12 A player pays $4 to play the following game. A fair die is rolled once. If the outcome is not a prime number, the player loses. If a prime number is rolled (2, 3 or 5), the die is rolled again. If the outcome of the second roll is not a prime number, the player loses. If a prime number is rolled the second time, the player is paid an amount (in dollars) equal to the sum of the outcomes of the two rolls of the die. Draw the applicable tree diagram and answer the following: 1. What is the probability of losing at the first roll? 2. What is the probability of winning$6? 3. What is the probability of", "the roll$)$ = ${(\\dfrac{5}{6})}^{i-1}\\dfrac{1}{6}$ therefore $$P = \\sum_{i=3}^{\\infty}{(\\dfrac{5}{6})}^{i-1}\\dfrac{1}{6} = \\sum_{i=0}^{\\infty}{(\\dfrac{5}{6})}^{i}\\dfrac{1}{6} - \\dfrac{1}{6} -\\dfrac{5}{6}\\dfrac{1}{6} = \\dfrac{1}{6}\\dfrac{1}{1-\\dfrac{5}{6}}-\\dfrac{1}{6}-\\dfrac{5}{36}=1-\\dfrac{11}{36}=\\dfrac{25}{36}$$ • This is correct but the problem can be solved easily without an infinite sum. – David K Feb 19 '15 at 18:33 • How to solve without using infinite sum? – Zeus Feb 19 '15 at 18:37 • See my answer already posted. You have only a small error in your calculation. – David K Feb 19 '15 at 18:41 • Did you see my asumption that each probability is exclusive? that means the probability to get the sum on the third or later is the complement of getting the sum on the first two, therefore $P($getting the roll on the third or more$) = 1 - P($getting the roll on the first two$) = 1 -\\dfrac{1}{6} - \\dfrac{5}{36}=\\dfrac{25}{36}$ – 3d0 Feb 19 '15 at 18:41 • That is correct, $1-\\frac16-\\frac{5}{36}$ is one way to do it without an infinite sum. (3do: the \"error\" I mentioned was in the OP, not in your calculations.) – David K Feb 19 '15 at 19:10 One of the two dices has red eyes, the other one blue eyes. On a single throw of the two dices, the probability that the number of blue eyes is $7$ minus the number of red eyes is ${1\\over6}$. Therefore", "$P(A|{{E}_{1}})=\\dfrac{{}^{12}{{C}_{2}}}{{}^{51}{{C}_{2}}}=\\dfrac{12!}{2!\\times 10!}\\times \\dfrac{2!\\times 49!}{5!}=\\dfrac{22}{425}$ When the card lost is not a diamond card, then there are 13 diamond cards out of 51 cards. From 13 diamond cards 2 cards can be drawn in ${}^{13}{{C}_{2}}$ ways. Therefore the probability of getting two diamond cards, when the card lost is not a diamond, is given by $P(A|{{E}_{2}})$. $P(A|{{E}_{2}})=\\dfrac{{}^{13}{{C}_{2}}}{{}^{51}{{C}_{2}}}=\\dfrac{13!}{2!\\times 11!}\\times \\dfrac{2!\\times 49!}{5!}=\\dfrac{26}{425}$ The probability that the lost card is diamond is given by $P({{E}_{1}}|A)$. $P({{E}_{1}}|A)=\\dfrac{P({{E}_{1}})P(A|{{E}_{1}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})}$ $=\\dfrac{\\dfrac{1}{4}\\times \\dfrac{22}{425}}{\\dfrac{1}{4}\\times \\dfrac{22}{425}+\\dfrac{3}{4}\\times \\dfrac{26}{425}}$ $=\\dfrac{11}{2}\\div 25$ $=\\dfrac{11}{50}$ 13. Probability that A speaks truth is $\\dfrac{4}{5}$ . A coin is tossed. A reports that a head appears. The probability that actually there was head is A) $\\dfrac{4}{5}$ B) $\\dfrac{1}{2}$ C) $\\dfrac{1}{5}$ D) $\\dfrac{2}{5}$ Ans: Let ${{E}_{1}}$ and ${{E}_{2}}$ be the events such that ${{E}_{1}}$ : A speaks truth ${{E}_{2}}$ : A speaks false Let X be the event that a head appears. $P({{E}_{1}})=\\dfrac{4}{5}$ $\\therefore P({{E}_{2}})=1-P({{E}_{1}})=1-\\dfrac{4}{5}=\\dfrac{1}{5}$ If a coin is tossed, then it may result in either head(H) or tail(T). The probability of getting a head is $\\dfrac{1}{2}$ whether A speaks truth or not. $\\therefore P(X|{{E}_{1}})=P(X|{{E}_{2}})=\\dfrac{1}{2}$ The probability that there actually a head is given by $P({{E}_{1}}|X)$. $P({{E}_{1}}|X)=\\dfrac{P({{E}_{1}})P(X|{{E}_{1}})}{P({{E}_{1}})P(X|{{E}_{1}})+P({{E}_{2}})P(X|{{E}_{2}})}$ $=\\dfrac{\\dfrac{4}{5}\\times \\dfrac{1}{2}}{\\dfrac{4}{5}\\times \\dfrac{1}{2}+\\dfrac{1}{5}\\times \\dfrac{1}{2}}$ $=\\dfrac{\\dfrac{4}{5}\\times \\dfrac{1}{2}}{\\dfrac{1}{2}\\left( \\dfrac{4}{5}+\\dfrac{1}{5} \\right)}$ $=\\dfrac{4}{5}\\div 1$ $=\\dfrac{4}{5}$ 14. If A and B are two events such that $A\\subset B$ and $P(B)\\ne 0$, then which of the" ]

[ "attended the wedding, and averaged three drinks per person.", "arrangement matters). And so while some attendees (the “core group”) might end up having fun, the party doesn’t really work for most participating parties. So, the next time you’re hosting a party, do yourself and your guests a favour and ensure that you don’t end up with between 6 and 10 people at the party. Either less or more is fine! You might want to read this other post I’ve written on coordinating guest lists for birthday parties. ## Meeting types There are essentially three kinds of meetings – those that are entirely “in person”, those that are entirely “on call” and hybrids. I argue that the quality of conversation in the third kind of meeting is significantly inferior to that of the first two types. In person meetings are those where all participants are in one room. These are perhaps the best kind of meetings (except when you know it’s likely to turn antagonistic), for you can maintain eye contact with the others and as long as the number of meeters is small, there is social pressure on meeters to not be distracted, and thus the meeting is likely to conclude its agenda productively and quickly. Conference calls allow you to multitask while you are on the meeting – the positive thing is that you can choose", "in the second. How many chairs can be around one table? • Kate shares Kate shares a 64-ounce bottle of apple cider with five friends. Each person's serving will be the same number of ounces. Between what two whole number of ounces will each person's serving to be? Explain using division.", "trip. if 39 people went on that trip, how many are in the club...", "interesting - and very telling! 6 of 29 hold licenses (20%). 80% don't. I think that explains a lot of what I've read on this forum! 2", "69.74*22)/26 = (69.74(26) - 3*48)/26 = 69.74 - 5.53 = 64.21 Approx 64.24 Option C A total of 22 men and 26 women were at a party. The average (arithmeti [#permalink] 02 Jul 2017, 05:48 Display posts from previous: Sort by", "months each with 60 more attendees.", "Probability # Combinations: Level 2 Challenges At a party, everyone shook hands with everybody else exactly once. There were 66 handshakes. How many people were at the party? Details and Assumptions: • You can't shake your own hand. Robbie is sitting for 9 subject papers in the GCSE O level examinations. In order to pass the examination, the number of subjects that he gets a pass grade in must be more than the number of subjects that he gets a fail grade. The number of ways in which Robbie can pass his GCSE O level examinations is: I was on vacation in England, and wanted to visit the Tower of London. The roads were laid out on a grid map, and the castle was 4 blocks north and 5 blocks east. If I were to travel only north and east, how many routes did I have to get to the castle? I was on vacation in England, and wanted to visit the Tower of London. The roads were laid out on a grid map, and and the castle was 4 blocks north and 5 blocks east. Unfortunately, there was some police activity near the hotel, and that intersection was blocked off so I could not go through it. If I were to travel only north and east, how", "clear the room and then have the one plus five. Each coordinator must confirm which are the five. Could we do this expeditiously, please? Event is not active Filed under:", "## Stats At a certain event, 16 people attend, and 5 will be chosen at random to receive door prizes. The prizes are all the same, so the order in which the people are chosen does not matter. How many different groups of 5 people can be chosen?", "as many females as males [#permalink] 18 Oct 2016, 09:59 Similar topics Replies Last post Similar Topics: At a party, there were five times as many females as males. There were 1 10 Jan 2016, 08:45 In a certain town, there are four times as many people who were born 3 11 Dec 2015, 09:28 7 A certain college party is attended by both male and female students. 7 08 May 2015, 06:04 13 At a picnic there were 3 times as many adults as children 6 26 Nov 2012, 22:56 5 How many ways are there for 3 males and 3 females to sit 5 18 Dec 2010, 05:18 Display posts from previous: Sort by", "5 people are to be seated around a 6 13 Jun 2011, 23:04 8 At a dinner party, 5 people are to be seated around a 11 22 Jul 2011, 14:10 10 At a dinner party, 5 people are to be seated around a circul 7 30 Dec 2013, 08:48 78 At a dinner party, 5 people are to be seated around a 12 08 Jul 2013, 07:41 Display posts from previous: Sort by", "sit together. Formula: Treat the persons who sit together as one and arrange. Then, arrange the group internally. Calculation: Men can be treated as one group. So, five boys and 1 group of men can be arranged in 6! ways. Three men can be arranged in 3! ways. Total no. of cases = 6! × 3! = 720 × 6 = 4320 ways ∴ Required no. of ways = 4320 # There are 7 persons in a party. The number of hands shaken all together in the party is: 1. 21 2. 6 3. 10 4. 2 Option 1 : 21 ## Detailed Solution Given: Number of person in a party = 7 $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ Where, n = number of the person r = minimum number of person required for a handshake Calculation: $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ $$\\Rightarrow {7_{{C_2}}} = \\;\\frac{{7!}}{{2! \\times \\left( {7 - 2} \\right)!}}$$ $$\\Rightarrow \\frac{{7 \\times 6 \\times 5!}}{{2! \\times 5!}}$$ ⇒ 42/2 ⇒ 21 ∴ the number of hands shaken is 21. # A boy contains 5 pink , 3 blue & 6 black ball. How many solution of ball can be formed by taking at least one ball form the boy 1. 157 2. 187 3. 167 4. 197 5.", "themselves and a shaves b and b shaves a, with three barbers, a b and c, a shaves b, b shaves c, and c shaves a. This thing keeps going further for four five or even 100 barbers. With zero barbers it is also not a problem, as everyone has to shave themselves. With one barber only there is a problem. There can never be a negative number of barbers.", "# How many meetings would it take for 12 people to meet in 4 groups of 3 until they met everyone? I have a group of 12 people that I would like to meet in four groups of three each month. How many minimum months would it take such that each person has been in at least one group with every other person? Below is the brute force method I used to get it to seven months: • Thanks for the link. That A) explains what you want to do, and B) serves as context showing that you have worked on this yourself. I expect the votes to close to subside :-) – Jyrki Lahtonen Dec 22 '20 at 15:00 • It takes at least six months. Each person has to meet eleven others and meets two people a month. The question is whether you can avoid too much duplication so you need another month. – Ross Millikan Dec 22 '20 at 15:00 • Nope. After following that scheme for four months, the not-yet-met people are split into four groups of four. And you cannot arrange all them to meet in two meetings. Need a different approach. Sorry about thinking aloud. – Jyrki Lahtonen Dec 22 '20 at 15:34 • @Sdavid552- I think it's because it doesn't have", "# Party acquaintances There are $$n$$ people at a party. Suppose they are friendly people, and each person has 8 acquaintances. Now, Bob, a passer-by notices this special property: (i) Each pair of people who are mutually acquainted has exactly 4 common acquaintances. (ii) Each pair of people who are mutually unacquainted has exactly 2 common acquaintances Find the number of possible values of $$n$$.", "# Dance party with 100 men and 20 women At a dance party there are 100 men and 20 women. Each man selects a group of women as potential dance partners, but in such a way that given any group of 20 men it is always possible to pair 20 men with 20 women, with each man paired with someone on their list. What is the smallest number L where L is the sum of the total number of women on each mans list that will guarantee this. • This doesn't sound like a very good dance party at all. – Servaes Oct 10 '16 at 21:04 • Why the downvotes? It's actually a not bad question. L = 20 is definitely a upper bound because having a list of 20 women of 20 women actually means that every man can pair with every woman. L = 1 is definitely too low because if every man chooses just one woman then there have to be at least two men that chose the same woman and every group of 20 men that contains those two men is then in trouble. – Jeyekomon Oct 10 '16 at 21:32 • My book gives the hint: First show that there is a way to choose the dance lists that works with 1620", "+0 # help please how do i find the minimum 0 27 1 +20 23 people attend a party. Each person shakes hands with at least one other person. What is the minimum possible number of handshakes? I got 253 but that's incorrect (the math was right but the formula was wrong). $$x=\\text{number of handshakes}$$ $$\\frac{23(23-1)}{2}=x \\\\ \\frac{23\\times 22}{2}=x \\\\ x=\\colorbox{red}{253}\\color{red}{\\leftarrow\\text{THIS IS WRONG}}$$ lolzforlife Jul 5, 2018 edited by Guest Jul 5, 2018 edited by Guest Jul 5, 2018 #1 +20 +2", "it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Manager Joined: 06 Mar 2014 Posts: 233 Location: India GMAT Date: 04-30-2015 Followers: 0 Kudos [?]: 33 [0], given: 73 Re: At a party, there were five times as many females as males [#permalink] 24 Sep 2014, 06:01 Bunuel wrote: vasurajiv wrote: At a party, there were five times as many females as males and three times as many adults as children. Which of the following could NOT be the number of people at the party? A. 384 B. 258 C. 216 D. 120 E. 72 Five times as many females as males --> F = 5M. Three times as many adults as children --> (F + M) = 3C. The number of people at the party = F + M + C = 3C + C = 4C. The number of people at the party must be a multiple of 4. The only answer choice which is NOT a multiple of 4 is B. Answer: B. Here adults is assumed to be M+F. However, even children can also be categorized into Male/Female? Adults is", "# Near the end of a party, everyone shakes hands... If everyone shakes hands with each other person we essentially have a complete graph. If there are 12 original guests, then there are ${{12} \\choose 2} = 66$ handshakes. So what about the late comer?" ]

[ "• In how many different ways, can the letters of the word 'VENTURE' be arranged? A) 840 B) 5040 C) 1260 D) 2520 D) 2520 The required different ways = $$\\large\\frac{7!}{2!} = \\frac{7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2!}{2!} = 2520$$ ##### Similar Questions 1). How many different signals , can be made by 5 flags from 8 flags of different colors? A). 6270 B). 1680 C). 20160 D). 6720 2). A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets? A). 12 B). 64 C). 256 D). 60 3). In how many different ways, can the letters of the word 'ASSASSINATION' be arranged, so that all S are together? A). 10! B). 14!/(4!) C). 151200 D). 3628800 4). There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible? A). 120 B). 100000 C). 6720 D). 30240 5). In a meeting between two countries, each country has 12 delegates, all the delegates of one country shakes hands with all delegates of the other country. Find the number of handshakes possible? A). 72 B). 144 C). 288 D). 234 6). Find the", "and 4 girls. Solution: There are 6 boys and 4 girls. A team of 3 boys and 2 girls is to be selected. ∴ 3 boys can be selected from 6 boys in 6C3 ways. 2 girls can be selected from 4 girls in 4C2 ways. ∴ Number of ways the team can be selected Question 9. After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting. Solution: Let there be n participants present in the meeting. A handshake occurs between 2 persons. ∴ Number of handshakes = nC2 Given 66 handshakes were exchanged. 66 = nC2 66 = $$\\frac{\\mathrm{n} !}{2 !(\\mathrm{n}-2) !}$$ 66 × 2 = $$\\frac{n(n-1)(n-2) !}{(n-2) !}$$ 132 = n (n – 1) n(n – 1) = 12 × 11 Comparing on both sides, we get n = 12 ∴ 12 participants were present at the meeting. Question 10. If 20 points are marked on a circle, how many chords can be drawn? Solution: To draw a chord we need to join two points on the circle. There are 20 points on a circle. ∴ Total number of chords possible from these points Question 11. Find the number of diagonals of an n-sided polygon. In particular, find the number", "$S$ greater than $k$, so there are $k(n-k)$ ways to choose one of each so as to make a $3$-element set with $k$ as its middle member. Summing over $k$, we see that $$\\sum_{k=1}^{n-1}k(n-k)=\\binom{n+1}3\\;.$$ If $n$ is the number of people at each party, we now know that $\\binom{n+1}3=165$, i.e., that $$\\frac{(n+1)n(n-1)}{3!}=165\\;,$$ or $(n+1)n(n-1)=990$. The lefthand side is roughly $n^3$, and $10^3=1000$ is pretty close to $990$, so we try $n=10$, and it works: $11\\cdot10\\cdot9=990$. The number of handshakes in the first party is actually a little easier to compute. At each stage there is one handshake for each pair of people present, so When $k$ people are present, there are $\\binom{k}2$ handshakes. The number of people present runs from $n$ down to $2$, so the total number of handshakes is $$\\sum_{k=2}^n\\binom{k}2=\\sum_{k=2}^n\\frac{k(k-1)}2=\\frac12\\left(\\sum_{k=2}^nk^2-\\sum_{k=2}^nk\\right)\\;.$$ You can evaluate this using the same summation formulas as before or, in a pinch, by actually doing the arithmetic, since we know now that $n=10$. The slick way is to use a combinatorial identity known as the hockey stick identity, which says (in this particular case) that $$\\sum_{k=2}^n\\binom{k}2=\\binom{n+1}3\\;.\\tag{1}$$ And with this approach we see that the number of handshakes at the two parties will always be the same so long as the parties start with the same number of guests and use these handshaking schemes!", "To prove this particular case of the hockey stick identity, I just notice that the lefthand side of $(1)$ is yet a third way to count the $3$-element subsets of the set $S$ above: it’s what you get when you count them according to their largest element instead of their middle element. The largest element can be anything from $2$ through $n$. If $k$ is the largest element, the other two elements must be chosen from the $k$ members of $S$ less than $k$, and there are $\\binom{k}2$ ways to do that. Summing over $k$ then yields the identity $(1)$. If there are $N$ people at a party they exchange $\\binom{N}{2}$ handshakes among them. You can understand that as follows: everyone exchanges exactly one handshake with everybody else. So you can think of each handshake as a $2$-element subset of the set of $n$ people. Thus, counting the number of handshakes is equivalent to counting the number of $2$-element subsets. So, for your first party, since people leave one-after-the-other, the total number of handshakes is: $$\\sum_{k=2}^{N}\\binom{k}{2}=\\binom{2}{2}+\\binom{3}{2}+...+\\binom{N-1}{2}+\\binom{N}{2}$$ Now, use your result for the second party, solve for $N$ $$\\sum_{i=1}^{N-1}i(N-i)= \\\\ =1\\times 9+2\\times 8+3\\times 7+4\\times 6+5\\times 5 +6\\times 4+7\\times 3+8\\times2+9\\times1=165 \\Rightarrow \\\\ \\Rightarrow N=10$$ and plug it in the above formula to calculate $\\sum_{k=2}^{N}\\binom{k}{2}=\\sum_{k=2}^{10}\\binom{k}{2}$, to get the result for the", "from 6 boys and 4 girls. Solution: There are 6 boys and 4 girls. A team of 3 boys and 2 girls is to be selected. ∴ 3 boys can be selected from 6 boys in 6C3 ways. 2 girls can be selected from 4 girls in 4C2 ways. ∴ Number of ways the team can be selected = 6C3 × 4C2 = $$\\frac{6 !}{3 ! 3 !} \\times \\frac{4 !}{2 ! 2 !}$$ = $$\\frac{6 \\times 5 \\times 4 \\times 3 !}{3 \\times 2 \\times 1 \\times 3 !} \\times \\frac{4 \\times 3 \\times 2 !}{2 \\times 2 !}$$ = 20 × 6 = 120 ∴ The team of 3 boys and 2 girls can be selected in 120 ways. Question 9. After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting. Solution: Let there be n participants present in the meeting. A handshake occurs between 2 persons. ∴ Number of handshakes = nC2 Given 66 handshakes were exchanged. ∴ 66 = nC2 ∴ 66 = $$\\frac{n !}{2 !(n-2) !}$$ ∴ 66 × 2 = $$\\frac{\\mathrm{n}(\\mathrm{n}-1)(\\mathrm{n}-2) !}{(\\mathrm{n}-2) !}$$ ∴ 132 = n(n – 1) ∴ n(n – 1) = 12 × 11 Comparing on both sides, we get n = 12", "/ 2 = 10 With this information, we can conclude that: s = number of students h = handshakes [ s x (s-1) ] / 2 = h And this is the formula we can use to solve all student handshake problems such as the one mentioned at the top of this post. If we plug the value 66 into this formula, we will discover how many students there were in the cafeteria whereby 66 handshakes took place. At the beginning of this post, I said that I would reveal the answer to the main question. To reveal it though, I will have to solve a quadratic equation by completing the square. I will also have to turn the variable ‘h’ into 66. Let’s do this… $\\frac { s\\left( s-1 \\right) }{ 2 } =66\\\\ \\\\ s\\left( s-1 \\right) =132\\\\ \\\\ { s }^{ 2 }-s=132\\\\ \\\\ { \\left( s-\\frac { 1 }{ 2 } \\right) }^{ 2 }-{ \\left( \\frac { 1 }{ 2 } \\right) }^{ 2 }=132\\\\ \\\\ { \\left( s-\\frac { 1 }{ 2 } \\right) }^{ 2 }=132+{ \\left( \\frac { 1 }{ 2 } \\right) }^{ 2 }$ ${ \\left( s-\\frac { 1 }{ 2 } \\right) }^{ 2 }=132+\\frac { 1 }{ 4 } \\\\ \\\\ { \\left( s-\\frac { 1", "(Paul picks wrongly) = 23×23+13×13=59 10. 36 people {a1, a2… a36} meet and shake hands in a circular fashion. In other words, there are totally 36 handshakes involving the pairs, {a1, a2}, {a2, a3}, …, {a35, a36}, {a36, a1}. Then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is a) 12 b) 11 c) 13 d) 18 Ans: {a1, a2}, {a2, a3},{a3, a4}, {a4, a5},{a5, a6}, {a6, a7} …, {a35, a36}, {a36, a1} From the above arrangement, If we separate a3, a6, a9, …..a36. Total 12 persons the reamining persons must have shaked hand with atleast one person. So answer is 12. 11. There are two boxes, one containing 10 red balls and the other containing 10 green balls. You are allowed to move the balls between the boxes so that when you choose a box at random and a ball at random from the chosen box, the probability of getting a red ball is maximized. This maximum probability is If rearrangement is not allowed, then actual probability of picking up a red ball = 12(10)+12(0)=12 As we are allowed to move the ball, we keep only 1 red in the first box, and shirt the remaining 9 to the second. So = 12(1)+919(0)=1419", "× Back to all chapters # Binomial Theorem If a coin comes up heads you win $10, but if it comes up tails you win$0. How likely are you to win exactly \\$30 in five flips? # Definition of Binomial Coefficient Evaluate $$\\binom{8}{3}$$. Evaluate ${44 \\choose 42}.$ What is the value of $$n$$ if ${n \\choose 3}=35?$ What value of $$n$$ satisfies ${n \\choose 2}=528?$ 5 people became friends at a Summer Camp. On the last day, they shook each other's hands to say farewell. How many handshakes were there? ×", "– I I. Choose the correct answer. Answer all the questions: [20 × 1 = 20] Question 1. If n(A) = 2 and n(B∪C) = 3 then n[(A × B) ∪ (A × C)] is ……………….. (a) 23 (b) 32 (c) 6 (d) 5 (c) 6 Question 2. For any two sets A and B, A∩(A∪B) = ……………………. (a) B (b) ∅ (c) A (d) none of these (c) A Question 3. cos 1° + cos 2° + cos 3° + cos 4° + cos 179° = ………………… (a) 0 (b) 1 (c) -1 (d) 89 (a) 0 Question 4. The value of log9 27 is …………………… (a) $$\\frac{2}{3}$$ (b) $$\\frac{3}{2}$$ (c) $$\\frac{3}{4}$$ (d) $$\\frac{4}{3}$$ (b) $$\\frac{3}{2}$$ Question 5. The value of $$\\frac{\\sin 3 \\theta+\\sin 5 \\theta+\\sin 7 \\theta+\\sin 9 \\theta}{\\cos 3 \\theta+\\cos 5 \\theta+\\cos 7 \\theta+\\cos 9 \\theta}$$ = ……………….. (a) tan3θ (b) tan6θ (c) cot3θ (d) cot6θ (b) tan6θ Question 6. In 3 fingers the number of ways 4 rings can be worn in ……………………. ways. (a) 43 – 1 (b) 34 (c) 68 (d) 64 (d) 64 Question 7. Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is ………………. (a) 11 (b) 12 (c) 10 (d) 6 (b) 12", "if this is not obvious). Now use the cross-products to obtain: 4m - 4 = 3m + 12 Solve for m and you'll see that there are 16 men. Substitute back into the second equation and you'll find the number of women. Using my versions of the equations (to avoid fractions), we might have chosen instead to eliminate m first, replacing m in the first equation with $$m = 4(w-1)$$ to obtain $$4(w-1)-1=3w$$. This simplifies to $$4w-5=3w$$, so that $$w=5$$, and $$m = 4(w-1) = 4(5-1) = 16$$. Therefore there are 16 men and 5 women (including the Li’s), for a total of 21 people at the party. Here I only have to show Mr. and Mrs. Li’s handshakes, with men in blue and women in red: Mr. Li shook hands with 15 men and 5 women (a 3:1 ratio), while Mrs. Li shook hands with 16 men and 4 women (a 4:1 ratio). It worked. You might also want to use this information to determine the TOTAL number of handshakes. You can find it by examining Pascal's Triangle. I had fun with this one! Thanks! The mention of Pascal’s Triangle refers to the fact that it contains binomial coefficients, which count combinations; in particular, we need $${{21}\\choose{2}} = \\frac{21\\cdot 20}{2\\cdot 1} = 210$$ handshakes in all. If", "C 12. The number of five digit telephone numbers having at least one of their digits repeated is (a) 90000 (b) 10000 (c) 30240 (d) 69760 13. The product of r consecutive positive integers is divisible by (a) r! (b) r!+1 (c) (r+1) (d) none of these 14. There are 10 lamps in a hall. Each one of them can be switched on independently. The number of ways in which the hall can be illuminated is (a) 102 (b) 1023 (c) 210 (d) 10! 15. The number of ways of disturbing 7 identical balls in 3 distinct boxes, so that no box is empty is (a) 7 (b) 6 (c) 35 (d) 15 16. If $\\frac { 1 }{ 7! } +\\frac { 1 }{ 8! } =\\frac { A }{ 9! } ,$ then the value of A is (a) 72 (b) 82 (c) 9 (d) 92 17. Assertion (A): Every body in a room shakes hands with everybody else. The total number of persons in the room is n. The number of hand shakes is $\\frac{n(n-1)}{2}$ Reason (R): The number of handshakes is nC2 (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explantion of A (c)", "How many handshakes are there? Solution Each unique handshake corresponds to a unique pair of persons. Also, note that the order of the two people in the pair does not matter. For example, if X and Y are two people, then XY and YX won’t be counted separately; only the pair {X, Y} will be counted. Thus, we need to find the number of ways in which 2 people can be selected out of 10. Clearly, the answer is: $^{10}{C_2} = \\frac{{10!}}{{2!\\left( {10 - 2} \\right)!}} = \\frac{{10!}}{{2!8!}} = 45$ $$\\therefore$$ 45 handshakes in total! Example 2 How many diagonals are there in a polygon with n sides? Solution If we select any two vertices of the polygon and join them, we will get either a diagonal or a side of the polygon. The number of ways of selecting two vertices out of n is $$^n{C_2} = \\frac{{n\\left( {n - 1} \\right)}}{2}$$. Out of these selections, n correspond to the sides of the polygon. Thus, the number of diagonals is: $\\frac{{n\\left( {n - 1} \\right)}}{2} - n = \\frac{{n\\left( {n - 3} \\right)}}{2}$ $$\\therefore$$ number of diagonals are $$\\frac{{n\\left( {n - 3} \\right)}}{2}$$ Example 3 A football team has a squad of 15 members. In how many ways can a playing team of 11 players be selected if", "number of shake hands is 66. The number of persons in the room is………. (a) 11 (b) 12 (c) 10 (d) 6 Solution: (b) 12 Question 8. The H.M of two positive number whose AM and G.M. are 16, 8 respectively is (a) 10 (b) 6 (c) 5 (d) 4 Solution: (d) 4 Question 9. The co-efficient of the term independent of x in the expansion of (2x + $$\\frac{1}{3x}$$)6 is………. (a) $$\\frac{160}{27}$$ (b) $$\\frac{160}{37}$$ (c) $$\\frac{80}{3}$$ (d) $$\\frac{80}{9}$$ Solution: (a) $$\\frac{160}{27}$$ Question 10. The value of $$\\left|\\begin{array}{lll} a & 0 & 0 \\\\ 0 & b & 0 \\\\ 0 & 0 & c \\end{array}\\right|^{2}$$ is………. (a) abc (b) -abc (c) 0 (d) a²b²c² Solution: (d) a²b²c² Question 11. (a) Δ (b) kΔ (c) 3kΔ (d) k³Δ Solution: (d) k³Δ Question 12. A vector makes equal angle with the positive direction of the co-ordinate axes then each angle is equal to……….. (a) cos-1 ($$\\frac{1}{3}$$) (b) cos-1 ($$\\frac{2}{3}$$) (c) cos-1 ($$\\frac{1}{√3}$$) (d) cos-1 ($$\\frac{2}{√3}$$) Solution: (c) cos-1 ($$\\frac{1}{√3}$$) Question 13. If the centroids of AABC and A’B’C’ are respectively G and G’ then $$\\bar{AA’}$$ + $$\\bar{BB’}$$ + $$\\bar{CC’}$$………… (a) $$\\bar{GG’}$$ (b) 3$$\\bar{GG’}$$ (c) 2$$\\bar{GG’}$$ (d) 0 Solution: (b) 3$$\\bar{GG’}$$ Question 14. If f(x) = \\left\\{\\begin{aligned} k x^{2} & \\text { for } \\quad x \\leq 2 \\\\ 3", "student handshake problems such as the one mentioned at the top of this post. If we plug the value 66 into this formula, we will discover how many students there were in the cafeteria whereby 66 handshakes took place. At the beginning of this post, I said that I would reveal the answer to the main question. To reveal it though, I will have to solve a quadratic equation by completing the square. I will also have to turn the variable ‘h’ into 66. Let’s do this… $\\frac { s\\left( s-1 \\right) }{ 2 } =66\\\\ \\\\ s\\left( s-1 \\right) =132\\\\ \\\\ { s }^{ 2 }-s=132\\\\ \\\\ { \\left( s-\\frac { 1 }{ 2 } \\right) }^{ 2 }-{ \\left( \\frac { 1 }{ 2 } \\right) }^{ 2 }=132\\\\ \\\\ { \\left( s-\\frac { 1 }{ 2 } \\right) }^{ 2 }=132+{ \\left( \\frac { 1 }{ 2 } \\right) }^{ 2 }$ ${ \\left( s-\\frac { 1 }{ 2 } \\right) }^{ 2 }=132+\\frac { 1 }{ 4 } \\\\ \\\\ { \\left( s-\\frac { 1 }{ 2 } \\right) }^{ 2 }=\\frac { 4\\left( 100+30+2 \\right) }{ 4 } +\\frac { 1 }{ 4 } \\\\ \\\\ { \\left( s-\\frac { 1 }{ 2 } \\right) }^{ 2 }=\\frac { 400+120+8 }{ 4 }", "handshaking into a graph where the people are vertex and handshakes are represented by edges, we will get one or more disjoint cycles. Each cycle must have at least $3$ people so that everyone shakes hand with two other. So we have $4$ cases. 1. We can take all of the people and put them in a cycle of length $9$. 2. We can divide them into $5$-cycle and $4$-cycle. 3. We can divide them into $6$-cycle and $3$-cycle. 4. We can divide them into three $3$-cycles For the first case, this is just cyclic permutation and it can be done in $\\frac{9!}{2\\times9}$ since rotation and reflection results in same handshaking. For the second case, we can divide $9$ people into group of $5$ people and $4$ people in ${9\\choose4}=\\frac{9!}{4!\\cdot5!}$ ways. For each grouping, there are $\\frac{5!}{2\\times5}\\cdot\\frac{4!}{2\\times4}$ ways to put them in the cycles. So, by multiplying this two results, we get total $\\frac{9!}{2\\times4\\times2\\times5}$ ways Similarly, for the third case, there are total $\\frac{9!}{2\\times3\\times2\\times6}$ ways to do that. And for the fourth case, there are just $\\frac{9!}{2\\times3\\times2\\times3\\times2\\times3}$ ways similarly. So, $N=9!\\left(\\frac{1}{2\\times9}+\\frac{1}{2^2\\times4\\times5}+\\frac{1}{2^2\\times3\\times6}+\\frac{1}{2^3\\times3\\times3\\times3}\\right)$ $\\Longrightarrow \\boxed{N=31416}$ In the 4th case, we need to divide by $3!$. And the rest are correct. And don't forget to divide $N$ by 1000 to find the remainder! $N=9!\\left(\\frac{1}{2\\times9}+\\frac{1}{2^2\\times4\\times5}+\\frac{1}{2^2\\times3\\times6}+\\frac{1}{2^3\\times3\\times3\\times3\\times3!}\\right)$ $\\Longrightarrow N=30016$ So, answer should be $16$.", "of 12 people? We have: How many handshakes will take place between six people in a room when they each shakes hands with all the other people in the room one time? Here, Notice that no consideration is given to the order or arrangement of the items but simply the combinations. Another way of viewing combinations is as follows. Consider the number of combinations of 5 letters taken 3 at a time. This produces: Now assume you permute (arrange) the r = 3 letters in each of the 10 combinations in all possible ways. Each group would produce r! permutations. Letting x = 5C3 for the moment, we would therefore have a total of x(r!) different permutations. This total, however, represents all the possible permutations (arrangements) of n things taken r at a time, which is shown under arrangement numbers and defined as nPr. Therefore, Using the committee of 3 out of 12 people example from above, Consider the following: How many different ways can you enter a 4 door car? It is clear that there are 4 different ways of entering the car. Another way of expressing this is: If we ignore the presence of the front seats for the purpose of this example, how many different ways can you exit the car assuming that you do", "sit together. Formula: Treat the persons who sit together as one and arrange. Then, arrange the group internally. Calculation: Men can be treated as one group. So, five boys and 1 group of men can be arranged in 6! ways. Three men can be arranged in 3! ways. Total no. of cases = 6! × 3! = 720 × 6 = 4320 ways ∴ Required no. of ways = 4320 # There are 7 persons in a party. The number of hands shaken all together in the party is: 1. 21 2. 6 3. 10 4. 2 Option 1 : 21 ## Detailed Solution Given: Number of person in a party = 7 $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ Where, n = number of the person r = minimum number of person required for a handshake Calculation: $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ $$\\Rightarrow {7_{{C_2}}} = \\;\\frac{{7!}}{{2! \\times \\left( {7 - 2} \\right)!}}$$ $$\\Rightarrow \\frac{{7 \\times 6 \\times 5!}}{{2! \\times 5!}}$$ ⇒ 42/2 ⇒ 21 ∴ the number of hands shaken is 21. # A boy contains 5 pink , 3 blue & 6 black ball. How many solution of ball can be formed by taking at least one ball form the boy 1. 157 2. 187 3. 167 4. 197 5.", "the variables. (B) \\ \\ 15 (A) If there are 14 people, there will be 14×132=91 \\frac{14 \\times 13 } { 2} = 91 214×13​=91 handshakes. Trial-and-error doesn’t necessarily produce the best result, but it’s often good enough. 3(6) − 14 = 4 Putting p = 1 Just before that, 5 leaves fall, so there are 10+5=1510+5=1510+5=15 leaves on the tree. Set it out like this. Putting p = 2 6 − 14 = 4 Putting m = 2 It is true. Since we've been told to give the answer correct to 1 decimal place, we have to try $$x = 3.15$$ to determine if the answer is closer to 3.1 or 3.2. The question should indicate the degree of accuracy required. In the examples to follow, we test all choices for your benefit. Since we've been told to give the answer correct to 1 decimal place, we have to try. ab−2a−2b−2=0 ab - 2a - 2b - 2 = 0 ab−2a−2b−2=0. (E) If there are 21 people, there will be 21×202=210 \\frac{21 \\times 20 } { 2} = 210 221×20​=210 handshakes. −14 = 4 ©. Another way to prevent getting this page in the future is to use Privacy Pass. Just before that, half of the leaves fall, so there are 2⋅5=102\\cdot 5=102⋅5=10 leaves on the tree.", "constant pace. $1$ $2$ $3$ $4$ 11 Suppose there are $n$ guests at a party (and no hosts). As the night progresses, the guests meet each other and shake hands. The same pair of guests might shake hands multiple times. for some parties stretch late into the night , and it is hard to keep track.Still, they don't shake hands with ... $2 \\mid \\text{Even} \\mid - \\mid \\text{Odd} \\mid$ $2 \\mid \\text{Odd} \\mid - \\mid \\text{Even} \\mid$ 12 Given a positive integer $m$, we define $f(m)$ as the highest power of $2$ that divides $m$. If $n$ is a prime number greater than $3$, then $f(n^3-1) = f(n-1)$ $f(n^3-1) = f(n-1) +1$ $f(n^3-1) = 2f(n-1)$ None of the above is necessarily true 13 One needs to choose six real numbers $x_1, x_2, . . . , x_6$ such that the product of any five of them is equal to other number. The number of such choices is $3$ $33$ $63$ $93$ 14 How many triplets of real numbers $(x,y,z)$ are simultaneous solutions of the equations $x+y=2$ and $xy-z^2=1$? $0$ $1$ $2$ infinitely many 15 Car parking along St. John street is charged at flat X dollar for any amount of time up to these hours, and 1/5 of X dollar each hour or fraction of an hour", "constant pace. $1$ $2$ $3$ $4$ 11 Suppose there are $n$ guests at a party (and no hosts). As the night progresses, the guests meet each other and shake hands. The same pair of guests might shake hands multiple times. for some parties stretch late into the night , and it is hard to keep track.Still, they don't shake hands with ... $2 \\mid \\text{Even} \\mid - \\mid \\text{Odd} \\mid$ $2 \\mid \\text{Odd} \\mid - \\mid \\text{Even} \\mid$ 12 Given a positive integer $m$, we define $f(m)$ as the highest power of $2$ that divides $m$. If $n$ is a prime number greater than $3$, then $f(n^3-1) = f(n-1)$ $f(n^3-1) = f(n-1) +1$ $f(n^3-1) = 2f(n-1)$ None of the above is necessarily true 13 One needs to choose six real numbers $x_1, x_2, . . . , x_6$ such that the product of any five of them is equal to other number. The number of such choices is $3$ $33$ $63$ $93$ 14 How many triplets of real numbers $(x,y,z)$ are simultaneous solutions of the equations $x+y=2$ and $xy-z^2=1$? $0$ $1$ $2$ infinitely many 15 Car parking along St. John street is charged at flat X dollar for any amount of time up to these hours, and 1/5 of X dollar each hour or fraction of an hour" ]

[ "sit together. Formula: Treat the persons who sit together as one and arrange. Then, arrange the group internally. Calculation: Men can be treated as one group. So, five boys and 1 group of men can be arranged in 6! ways. Three men can be arranged in 3! ways. Total no. of cases = 6! × 3! = 720 × 6 = 4320 ways ∴ Required no. of ways = 4320 # There are 7 persons in a party. The number of hands shaken all together in the party is: 1. 21 2. 6 3. 10 4. 2 Option 1 : 21 ## Detailed Solution Given: Number of person in a party = 7 $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ Where, n = number of the person r = minimum number of person required for a handshake Calculation: $${n_{{C_r}}} = \\;\\frac{{n!}}{{r! \\times \\left( {n - r} \\right)!}}$$ $$\\Rightarrow {7_{{C_2}}} = \\;\\frac{{7!}}{{2! \\times \\left( {7 - 2} \\right)!}}$$ $$\\Rightarrow \\frac{{7 \\times 6 \\times 5!}}{{2! \\times 5!}}$$ ⇒ 42/2 ⇒ 21 ∴ the number of hands shaken is 21. # A boy contains 5 pink , 3 blue & 6 black ball. How many solution of ball can be formed by taking at least one ball form the boy 1. 157 2. 187 3. 167 4. 197 5.", "• In how many different ways, can the letters of the word 'VENTURE' be arranged? A) 840 B) 5040 C) 1260 D) 2520 D) 2520 The required different ways = $$\\large\\frac{7!}{2!} = \\frac{7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2!}{2!} = 2520$$ ##### Similar Questions 1). How many different signals , can be made by 5 flags from 8 flags of different colors? A). 6270 B). 1680 C). 20160 D). 6720 2). A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets? A). 12 B). 64 C). 256 D). 60 3). In how many different ways, can the letters of the word 'ASSASSINATION' be arranged, so that all S are together? A). 10! B). 14!/(4!) C). 151200 D). 3628800 4). There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible? A). 120 B). 100000 C). 6720 D). 30240 5). In a meeting between two countries, each country has 12 delegates, all the delegates of one country shakes hands with all delegates of the other country. Find the number of handshakes possible? A). 72 B). 144 C). 288 D). 234 6). Find the", "Probability # Combinations: Level 2 Challenges At a party, everyone shook hands with everybody else exactly once. There were 66 handshakes. How many people were at the party? Details and Assumptions: • You can't shake your own hand. Robbie is sitting for 9 subject papers in the GCSE O level examinations. In order to pass the examination, the number of subjects that he gets a pass grade in must be more than the number of subjects that he gets a fail grade. The number of ways in which Robbie can pass his GCSE O level examinations is: I was on vacation in England, and wanted to visit the Tower of London. The roads were laid out on a grid map, and the castle was 4 blocks north and 5 blocks east. If I were to travel only north and east, how many routes did I have to get to the castle? I was on vacation in England, and wanted to visit the Tower of London. The roads were laid out on a grid map, and and the castle was 4 blocks north and 5 blocks east. Unfortunately, there was some police activity near the hotel, and that intersection was blocked off so I could not go through it. If I were to travel only north and east, how", "of 12 people? We have: How many handshakes will take place between six people in a room when they each shakes hands with all the other people in the room one time? Here, Notice that no consideration is given to the order or arrangement of the items but simply the combinations. Another way of viewing combinations is as follows. Consider the number of combinations of 5 letters taken 3 at a time. This produces: Now assume you permute (arrange) the r = 3 letters in each of the 10 combinations in all possible ways. Each group would produce r! permutations. Letting x = 5C3 for the moment, we would therefore have a total of x(r!) different permutations. This total, however, represents all the possible permutations (arrangements) of n things taken r at a time, which is shown under arrangement numbers and defined as nPr. Therefore, Using the committee of 3 out of 12 people example from above, Consider the following: How many different ways can you enter a 4 door car? It is clear that there are 4 different ways of entering the car. Another way of expressing this is: If we ignore the presence of the front seats for the purpose of this example, how many different ways can you exit the car assuming that you do", "attended the wedding, and averaged three drinks per person.", "(since A repeats) total permutations= 8! x 4!/8=7! * 4! = 120960 A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw ? Total Combinations of 9 balls into 3= {}^9C_3 Total combination where black ball doesn't show up = {}^{(number of other balls)}C_3 = []^6C_3 Number of combinations where at least one black ball shows up = {}^9C_3 - {}^6C_3 = 84-20 = 64 How many groups of 6 persons can be formed from 8 men and 7 women ? {}^{15}C_3 8 men entered a lounge simultaneously. If each person shakes hand with the other, then find the total no. of handshakes? shake hand is group of 2 No order, A shakes B same as B shakes A So {}^8C_2 In how many different ways, can the letters of the word 'INHALE' be arranged ? INHALE has 6 letters, none repeating. Total arrangements = 6! In how many different ways, can the letters of the word 'BANKING' be arranged ? BANKING has 7 letters, N repeating Total Arrangements = \\frac{7!}{2!} In a meeting between two countries, each country has 12 delegates. All the delegates of one country shake hands", "from 6 boys and 4 girls. Solution: There are 6 boys and 4 girls. A team of 3 boys and 2 girls is to be selected. ∴ 3 boys can be selected from 6 boys in 6C3 ways. 2 girls can be selected from 4 girls in 4C2 ways. ∴ Number of ways the team can be selected = 6C3 × 4C2 = $$\\frac{6 !}{3 ! 3 !} \\times \\frac{4 !}{2 ! 2 !}$$ = $$\\frac{6 \\times 5 \\times 4 \\times 3 !}{3 \\times 2 \\times 1 \\times 3 !} \\times \\frac{4 \\times 3 \\times 2 !}{2 \\times 2 !}$$ = 20 × 6 = 120 ∴ The team of 3 boys and 2 girls can be selected in 120 ways. Question 9. After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting. Solution: Let there be n participants present in the meeting. A handshake occurs between 2 persons. ∴ Number of handshakes = nC2 Given 66 handshakes were exchanged. ∴ 66 = nC2 ∴ 66 = $$\\frac{n !}{2 !(n-2) !}$$ ∴ 66 × 2 = $$\\frac{\\mathrm{n}(\\mathrm{n}-1)(\\mathrm{n}-2) !}{(\\mathrm{n}-2) !}$$ ∴ 132 = n(n – 1) ∴ n(n – 1) = 12 × 11 Comparing on both sides, we get n = 12", "take to make 10 complete rotations? #### 11th Standard Maths Combinations and Mathematical Induction English Medium Free Online Test One Mark Questions 2020 - 2021 - by Question Bank Software - View & Read • 1) The sum of the digits at the 10th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is • 2) The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in physics, first in chemistry and first in English is • 3) The number of ways in which a host lady invite 8 people for a party of 8 out of 12 people of whom two do not want to attend the party together is • 4) Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is • 5) The product of r consecutive positive integers is divisible by _________ #### 11th Standard Maths Important Question - by Question Bank Software - View & Read • 1) By taking suitable sets A, B, C, verify the following results: C-(B-A) = (C$\\cap$ A) $\\cup$ (C$\\cap$B') • 2) Discuss the following relations for reflexivity, symmetricity", "# At a party each man dances with 4 women and each woman dances with 3 men. If 9 men attended the party, how many women attended the party? At a party each man dances with 4 women and each woman dances with 3 men. If 9 men attended the party, how many women attended the party? I have no idea how to approach this problem. • How many ways can a women choose 3 men in a set of 9 ? that would help. – user451844 Jul 13 '17 at 13:22 If you assume that a man dances with a woman, how many dances will be performed? 9 men -> dance 9 * 4 times with a woman, so 36 dances Each woman performs 3 dances, so 12 women. Count dance pairs . . . Since there are $9$ men and each man dances with $4$ women (presumably this means each man dances exactly $4$ times), there are exactly $36$ dance pairs. Let $w$ be the number of women. Since each woman dances with $3$ men (presumably this means each woman dances exactly $3$ times), there are exactly $3w$ dance pairs. Thus, $3w=36$, so $w=12$. • Obviously, we agree, but I don't think once each really matters. 9 women can dance 3 times with the same men,", "+0 # help please how do i find the minimum 0 27 1 +20 23 people attend a party. Each person shakes hands with at least one other person. What is the minimum possible number of handshakes? I got 253 but that's incorrect (the math was right but the formula was wrong). $$x=\\text{number of handshakes}$$ $$\\frac{23(23-1)}{2}=x \\\\ \\frac{23\\times 22}{2}=x \\\\ x=\\colorbox{red}{253}\\color{red}{\\leftarrow\\text{THIS IS WRONG}}$$ lolzforlife Jul 5, 2018 edited by Guest Jul 5, 2018 edited by Guest Jul 5, 2018 #1 +20 +2", "the RSS feed at http://feedpress.me/futilitycloset. If you have any questions or comments you can reach us at [email protected]. Thanks for listening! # All Hands on Deck? A reader named Hamp Stevens sent this conundrum to Martin Gardner, who published it in his Mathematical Magic Show (1965). Can these 25 cards be arranged to form five poker hands, each of them a straight or better (that is, straight, flush, full house, four of a kind, straight flush, or royal flush)? If it’s possible, find the five hands; if it’s not, prove that it’s impossible. “This ingenious puzzle is quickly solved if you go about it correctly,” Gardner wrote. “A single card is the key.” # Black and White By George Edward Carpenter. White to mate in two moves. # Hard Target In a special football game, a team scores 7 points for a touchdown and 3 points for a field goal. What’s the largest mathematically unreachable number of points that a team can score (in an infinitely long game)? A problem by Atlantic College mathematician Paul Belcher: Anna and Bert invite n other couples to a dinner party. Before the meal begins, some people shake hands. No one shakes hands with their own partner, no one shakes hands with themselves, and no two people shake hands with each other", "still being together. So those arrangements are 4!/2! (since A repeats) total permutations= 8! x 4!/8=7! * 4! = 120960 A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw ? Total Combinations of 9 balls into 3= {}^9C_3 Total combination where black ball doesn't show up = {}^{(number of other balls)}C_3 = []^6C_3 Number of combinations where at least one black ball shows up = {}^9C_3 - {}^6C_3 = 84-20 = 64 How many groups of 6 persons can be formed from 8 men and 7 women ? {}^{15}C_3 8 men entered a lounge simultaneously. If each person shakes hand with the other, then find the total no. of handshakes? shake hand is group of 2 No order, A shakes B same as B shakes A So {}^8C_2 In how many different ways, can the letters of the word 'INHALE' be arranged ? INHALE has 6 letters, none repeating. Total arrangements = 6! In how many different ways, can the letters of the word 'BANKING' be arranged ? BANKING has 7 letters, N repeating Total Arrangements = \\frac{7!}{2!} In a meeting between two countries, each country has 12 delegates.", "C 12. The number of five digit telephone numbers having at least one of their digits repeated is (a) 90000 (b) 10000 (c) 30240 (d) 69760 13. The product of r consecutive positive integers is divisible by (a) r! (b) r!+1 (c) (r+1) (d) none of these 14. There are 10 lamps in a hall. Each one of them can be switched on independently. The number of ways in which the hall can be illuminated is (a) 102 (b) 1023 (c) 210 (d) 10! 15. The number of ways of disturbing 7 identical balls in 3 distinct boxes, so that no box is empty is (a) 7 (b) 6 (c) 35 (d) 15 16. If $\\frac { 1 }{ 7! } +\\frac { 1 }{ 8! } =\\frac { A }{ 9! } ,$ then the value of A is (a) 72 (b) 82 (c) 9 (d) 92 17. Assertion (A): Every body in a room shakes hands with everybody else. The total number of persons in the room is n. The number of hand shakes is $\\frac{n(n-1)}{2}$ Reason (R): The number of handshakes is nC2 (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explantion of A (c)", "if this is not obvious). Now use the cross-products to obtain: 4m - 4 = 3m + 12 Solve for m and you'll see that there are 16 men. Substitute back into the second equation and you'll find the number of women. Using my versions of the equations (to avoid fractions), we might have chosen instead to eliminate m first, replacing m in the first equation with $$m = 4(w-1)$$ to obtain $$4(w-1)-1=3w$$. This simplifies to $$4w-5=3w$$, so that $$w=5$$, and $$m = 4(w-1) = 4(5-1) = 16$$. Therefore there are 16 men and 5 women (including the Li’s), for a total of 21 people at the party. Here I only have to show Mr. and Mrs. Li’s handshakes, with men in blue and women in red: Mr. Li shook hands with 15 men and 5 women (a 3:1 ratio), while Mrs. Li shook hands with 16 men and 4 women (a 4:1 ratio). It worked. You might also want to use this information to determine the TOTAL number of handshakes. You can find it by examining Pascal's Triangle. I had fun with this one! Thanks! The mention of Pascal’s Triangle refers to the fact that it contains binomial coefficients, which count combinations; in particular, we need $${{21}\\choose{2}} = \\frac{21\\cdot 20}{2\\cdot 1} = 210$$ handshakes in all. If", "could hold. The maximum of about 42 and minimum of about 7 indicate that there is o lot of variability in the number of Tootsie Pops that people can hold, with the numbers covering a range of about 35 Tootsie Pops. The “box” port of the box plot shows that about half of the peopk con hold within about 2 Tootsie Pops of the median, which was 20 Tootsie Pops. Exercise 4. Here is Jayne’s description of what she sees in the box plot. Do you agree or disagree with her description? Explain your reasoning. “One person could hold as many as 42 Tootsie Pops. The number of Tootsie Pops people could hold was really different and spread about equally from 7 to 42. About one-half of the people could hold more than 20 Tootsie Pops.” You cannot tell that they are evenly spread—the “box” part of the box plot contains about half of the values for the number of Tootsie Pops. However, the box is only four units long. That means half of the people were bunched over those four numbers. Exercise 5. Here is a different box plot of the same data on the number of Tootsie Pops 94 people could hold. a. Why do you suppose there are five values that are shown as separate", "# Have 40 people at 4 tables: how to switch tables so that the minimum number of people possible are sitting with someone at their original table Just a few simple premises: • 40 people, 10 people per table • At some point you want every one to switch tables, and have a few people as possible sitting at a table with someone they sat with the first time around. This is not a homework problem--this is an actual issue I'm encountering while trying to organize a networking event. There are two main questions: 1) Is it even possible to devise a reassignment scheme were everyone is sitting with an entirely new crowd? 2) Assuming the answer to (1) is no, what's a reasonable algorithm for reassigning that will make it so that as many people as possible are sitting with new people at their new table. The reason I think the answer to (1) is \"No\", is because I did 100k simulations of random table reassignment for each person (code below), from that I found there were zero times where nobody was sitting at a table with someone they sat with in the first round. On average, the maximally \"redundant\" table (i.e. the one that had the most people who were previously sat together) had about 4.65", "train were late, we can calculate how many took the train and were on time $58-31=27\\text{ by train and on time}$ The completed frequency tree should therefore look like this: a) First of all, if there are 75 people in total and 63 are right-handed, we can perform a simple calculation to work out how many people are left-handed: $75-63=12\\text{ left-handed people}$ Secondly, we know that of the left-handed people, one third are right-footed. So the number of people who are left-handed and right-footed can be calculated as follows: $12\\div 3=4 \\text{ left-handed and right-footed people}$ Furthermore, since there are 12 left-handed people in total, we can now work out how many of them are left-footed: $12-4=8\\text{ left-handed and left-footed people }$ So, we just have two remaining bubbles, the ones branching off from the 63 right-handed people. The final piece of information given to us in the question tells us that 27 people in total are left-footed. Given that we now know that 8 of the left-handed people are left-footed, we can calculate the number of people who are right-handed and left-footed: $27-8=19\\text{ right-handed and left-footed people }$ Finally, if 19 of the 63 right-handed people are left-footed, we can calculate how many are right-handed and right-footed $63-19=44\\text{ right-handed and right-footed people }$ The completed frequency tree", "different family, then find the total number of handshakes that took place in the party. (C) 729 (B) 800 (A) 801 (D) 850 (E) 851 VP Joined: 19 Oct 2018 Posts: 1174 Location: India Re: Chris and his wife invited a total of ten families on their marriage [#permalink] ### Show Tags 24 Nov 2019, 13:36 Each member of host family shakes hands with 40 persons, and each member of invited family shakes hand with 38 members. Total number of handshakes=$$\\frac{2*40+ 40*38}{2}=800$$ CaptainLevi wrote: Chris and his wife invited a total of ten families on their marriage anniversary. While the host family had just the two members, each family invited consisted of four members. If every person in the party shook hands exactly once with every other person belonging to a different family, then find the total number of handshakes that took place in the party. (C) 729 (B) 800 (A) 801 (D) 850 (E) 851 Re: Chris and his wife invited a total of ten families on their marriage [#permalink] 24 Nov 2019, 13:36 Display posts from previous: Sort by", "the index finger • Eleven → middle link of the index finger • Twelve → lower link of the index finger ### Two-handed counting up to 60 After the first dozen has been counted using the thumb as a pointer with the three phalanges of the remaining four fingers of the same hand (4 × 3 = 12), the counting capacity of one hand is initially exhausted. • The other hand is clenched in a fist. In order to remember that a dozen has been counted, one now extends a finger, e.g. B. the thumb. • Now you continue to count by starting again at one with your first hand . At twelve , the second dozen is full. • In order to remember that two dozen have been counted, one now extends the next finger of the other hand, e.g. B. after the thumb out the index finger. • With the five fingers of the first hand you can count five times a dozen, so 5 × 12 = 60. • Now you can count the next dozen again with the first hand, i.e. count up to 72 with two hands (12 on the first plus 60 on the other hand). This finger counting system still exists in parts of Turkey , Iraq , India and Indochina", "# Dance party with 100 men and 20 women At a dance party there are 100 men and 20 women. Each man selects a group of women as potential dance partners, but in such a way that given any group of 20 men it is always possible to pair 20 men with 20 women, with each man paired with someone on their list. What is the smallest number L where L is the sum of the total number of women on each mans list that will guarantee this. • This doesn't sound like a very good dance party at all. – Servaes Oct 10 '16 at 21:04 • Why the downvotes? It's actually a not bad question. L = 20 is definitely a upper bound because having a list of 20 women of 20 women actually means that every man can pair with every woman. L = 1 is definitely too low because if every man chooses just one woman then there have to be at least two men that chose the same woman and every group of 20 men that contains those two men is then in trouble. – Jeyekomon Oct 10 '16 at 21:32 • My book gives the hint: First show that there is a way to choose the dance lists that works with 1620" ]

[ "maybe it will be a good idea to convert every frame to a number and inspect these sequences? Worth a try: python for i in range(6): colors[i] = [x for (x, j, k) in colors[i]] print('flag' + str(i + 1) + '.gif:', colors[i]) After conversion we get these sequences: flag1.gif: [3, 5, 5, 3, 3, 5, 3, 5, 3, 3, 5, 5, 3, 3, 5, 3, 3, 3, 5, 5, 3, 5, 5, 5, 3, 5, 5, 3, 3, 5, 3, 5, 3, 5, 5, 3, 3, 3, 3, 5, 3, 3, 5, 5, 3, 5, 3, 5, 3, 3, 5, 5, 3, 5, 3, 3, 3, 3, 5, 5, 3, 5, 5, 5, 3, 3, 5, 5, 3, 3, 3, 3, 3, 5, 5, 3, 3, 5, 3, 3, 3, 3, 5, 5, 3, 3, 3, 5] flag2.gif: [3, 18, 18, 3, 18, 10, 3, 18, 15, 5, 13, 13, 3, 20, 3, 5, 13, 18, 3, 18, 18, 5, 13, 15, 3, 18, 3, 5, 13, 18, 5, 13, 5] flag3.gif: [10, 20, 18, 10, 18, 15, 18, 8, 18, 15, 18, 8, 10, 18, 10, 13, 18, 15, 18, 15, 10, 13] flag4.gif: [33, 43, 74, 59, 33, 66, 46, 59, 18, 23, 64, 10, 8, 38, 26, 66, 20, 3, 84, 84, 84, 84,", "# Problem #2070 2070 Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $R$ the area of the red square. Which of the following is correct? $[asy] unitsize(3mm); fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue); fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red); path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle; path divider=(-2,2)--(-3,3)--cycle; fill(onewhite,white); fill(rotate(90)*onewhite,white); fill(rotate(180)*onewhite,white); fill(rotate(270)*onewhite,white); [/asy]$ $\\text{(A)}\\ B = W \\qquad \\text{(B)}\\ W = R \\qquad \\text{(C)}\\ B = R \\qquad \\text{(D)}\\ 3B = 2R \\qquad \\text{(E)}\\ 2R = W$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Difference between revisions of \"British Flag Theorem\" The British flag theorem says that if a point P is chosen inside rectangle ABCD then $AP^{2}+PC^{2}=BP^{2}+DP^{2}$. $[asy] size(200); pair A,B,C,D,P; A=(0,0); B=(200,0); C=(200,150); D=(0,150); P=(124,85); draw(A--B--C--D--cycle); label(\"A\",A,(-1,0)); dot(A); label(\"B\",B,(0,-1)); dot(B); label(\"C\",C,(1,0)); dot(C); label(\"D\",D,(0,1)); dot(D); dot(P); label(\"P\",P,(1,1)); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label(\"w\",(124,0),(0,-1)); label(\"x\",(200,85),(1,0)); label(\"y\",(124,150),(0,1)); label(\"z\",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85)); [/asy]$ The theorem also applies to points outside the rectangle, although the proof is harder to visualize in this case. ## Proof In Figure 1, by the Pythagorean theorem, we have: • $AP^{2} = Aw^{2} + Az^{2}$ • $PC^{2} = wB^{2} + zD^{2}$ • $BP^{2} = wB^{2} + Az^{2}$ • $PD^{2} = zD^{2} + Aw^{2}$ Therefore: • $AP^{2} + PC^{2} = Aw^{2} + Az^{2} + wB^{2} + zD^{2} = wB^{2} + Az^{2} + zD^{2} + Aw^{2} =\\nolinebreak BP^{2} +\\nolinebreak PD^{2}$ The above result i.e. theorem holds true even if the point P is selected on the boundary of rectangle or even outside the rectangle.", "(i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5), white); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5), white); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5), white); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5), white); label(\"$b^2$\", (0.5,-2.5), white); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5), white); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5), white); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5), white); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] Sorry but I don't know how to show the grids. 1. 👍 2. 👎 3. 👁 1. each row is a geometric progression. Starting with n=0, row n is the GP b^n (1 + a + a^2 + ...) so the sum Sn of row n is b^n * 1/(1-a) Now you have another GP, which is the sum of all the row sums. That is 1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b) So the grand total of the whole grid is 1/((1-a)(1-b)) 1. 👍 2. 👎 2. wertwetrwetw 1. 👍 2. 👎 ## Similar Questions 1. ### alebra Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation then find the domain. (f*g)(x) a.2x^3-x^2-4x+3; domain:all real numbers b.2x^3+x^2-3x; domain: all real numbers c.2x^3+x^2+4x-3; domain: negative real numbers 2. ### Precalc To which set(s) of numbers does the number sqrt -16 belong? Select all that apply. real numbers complex numbers*** rational", "# 14.6. Chapter Exercises¶ The code below is supposed to change the white in the French flag to aqua (Aqua is made by mixing blue and green with no red). Arrange and indent the blocks to make a correct program. You will not use them all. Write a program below to change all the greenish pixels in this image to the color: red 30, green 100, blue 150 (a slightly grayish blue). Hint: The greenish pixels are not a pure green. They have almost as much red as they do green. A sample color value might be: 111 red, 115 green, 65 blue. However, they are not all exactly the same color. You will need to figure out a recipe to select the right colors from the image. Write a program below that makes the white pixels in the right half of the image below turn black. The left half of the image should stay the same. Although perfect white is (255, 255, 255), the pixels in the image are not all perfectly white. Write your code to turn any pixel that has red, green, and blue values of above 230 to black (0, 0, 0). Hint: Start by trying to make all the white pixels in the image turn black. then add a condition to check the", "(-10,-12) -- (10,-12) -- ( 9,-8) -- ( 3, -4) -- cycle; \\fill [gray!10] (0,-10) -- (-1,-5) -- (-3,-5) -- (-2,-12) -- (2,-12) -- (3,-4) -- (-1,-5) -- cycle; \\fill [brown] (-3,7) -- (-5,6) -- (-6,-1) -- (-5,-3) -- (-4,-4) -- (4,-4) -- (5,-3) -- (6,-1) -- (5,6) -- (3,7) -- cycle; \\fill [pink!75!yellow!75!brown] (-2, 0) -- (-2,-4) -- (-2,-4) -- (-1,-5) -- (0,-10) -- (1,-5) -- ( 2,-4) -- ( 2,-4) -- ( 2, 0) -- cycle; \\fill [white] (-2,-4) -- (-3,-4) -- (-5,-6) -- (-1,-5) -- cycle; \\fill [white] ( 2,-4) -- ( 3,-4) -- ( 5,-6) -- ( 1,-5) -- cycle; \\fill [pink!75!yellow] (0, 5) -- (-4, 5) -- (-4,-1) -- (-1,-3) -- (1,-3) -- ( 4,-1) -- ( 4, 5) -- cycle; \\fill [brown] (-5,6) -- (-5,-2) -- (-4,2) -- (-3,5) -- (-1,4) -- (3,5) -- (4,2) -- (5,2) -- (5,6) -- cycle; \\end{tikzpicture} \\end{document} It's pretty easy to avoid sharp angles if you use .. controls +(<polar coordinate>) and +(<polar coordinate>) .. and make sure that the in and out angles are the same at each point. My attempt: \\documentclass[tikz, border=3pt]{standalone} \\begin{document} \\begin{tikzpicture}[thick] \\draw (0,0) circle (.53); \\draw (-130:.53) .. controls +(200:.5) and +(-45:-.3) .. (-1,-1.6) .. controls +(-45:.3) and +(45:-.3) .. (1,-1.6) .. controls +(45:.3) and +(-20:.5) .. (-50:.53); \\end{tikzpicture} \\end{document} This", "# Difference between revisions of \"British Flag Theorem\" The British flag theorem says that if a point P is chosen inside rectangle ABCD then $AP^{2}+PC^{2}=BP^{2}+DP^{2}$. $[asy] size(200); pair A,B,C,D,P; A=(0,0); B=(200,0); C=(200,150); D=(0,150); P=(124,85); draw(A--B--C--D--cycle); label(\"A\",A,(-1,0)); dot(A); label(\"B\",B,(0,-1)); dot(B); label(\"C\",C,(1,0)); dot(C); label(\"D\",D,(0,1)); dot(D); dot(P); label(\"P\",P,(1,1)); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label(\"w\",(124,0),(0,-1)); label(\"x\",(200,85),(1,0)); label(\"y\",(124,150),(0,1)); label(\"z\",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85)); [/asy]$ The theorem also applies if the point $P$ is selected outside or on the boundary of the rectangle, although the proof is harder to visualize in this case. ## Proof In Figure 1, by the Pythagorean theorem, we have: • $AP^{2} = Aw^{2} + Az^{2}$ • $PC^{2} = wB^{2} + zD^{2}$ • $BP^{2} = wB^{2} + Az^{2}$ • $PD^{2} = zD^{2} + Aw^{2}$ Therefore: • $AP^{2} + PC^{2} = Aw^{2} + Az^{2} + wB^{2} + zD^{2} = wB^{2} + Az^{2} + zD^{2} + Aw^{2} =\\nolinebreak BP^{2} +\\nolinebreak PD^{2}$", "flag's .check value is only 64 when idx is 64. r and c seems to correspond with coordinates on the map. Let's try to calculate the coordinate of the flag when idx is 64. var r = Math.floor(64 / 5) + 1; var c = (64 % 5) + 1; Therefore, r is 13 and c is 5. Let's try to plot this on a map. [ -4, -3, -3, -3, -3, -3, -2], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, flag, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, stair, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -6, -7, -7, -7, -7, -7, -8]", "# Dutch National Flag The Dutch national flag problem is a famous computer science related programming problem proposed by Edsger Dijkstra. The flag of the Netherlands consists of three colours: red, white and blue. Given balls of these three colours arranged randomly in a line (the actual number of balls does not matter), the task is to arrange them such that all balls of the same colour are together and their collective colour groups are in the correct order. Here is a problem that is stimulated by the idea: Q. Write a function that takes an array A and index i into A, and rearranges the elements such that all elements less than A[i] appear first, followed by elements equal to A[i], followed by elements greater than A[i]. Your algorithm should have O(1) space complexity and O(|A|) time complexity. Soln: ```#Function to swap two variables def swapNormal(A,pos1,pos2): t = 0 t=A[pos1] A[pos1] = A[pos2] A[pos2] = t #initializing an array of randomly generated numbers j = 0 while j < 10: A.append(random.randint(1,100)) j+= 1 print \"The array of numbers is:\",A #asking for the position of the pivotal element i = input(\"Please enter the index number\") pivot = A[i] print \"The pivot element is:\",pivot def dutchNationalFlag(A,pivot): equal = 0 small = 0 large = len(A)-1 while (equal <= large):", "# Dutch National Flag The Dutch national flag problem is a famous computer science related programming problem proposed by Edsger Dijkstra. The flag of the Netherlands consists of three colours: red, white and blue. Given balls of these three colours arranged randomly in a line (the actual number of balls does not matter), the task is to arrange them such that all balls of the same colour are together and their collective colour groups are in the correct order. Here is a problem that is stimulated by the idea: Q. Write a function that takes an array A and index i into A, and rearranges the elements such that all elements less than A[i] appear first, followed by elements equal to A[i], followed by elements greater than A[i]. Your algorithm should have O(1) space complexity and O(|A|) time complexity. Soln: ```#Function to swap two variables def swapNormal(A,pos1,pos2): t = 0 t=A[pos1] A[pos1] = A[pos2] A[pos2] = t #initializing an array of randomly generated numbers j = 0 while j < 10: A.append(random.randint(1,100)) j+= 1 print \"The array of numbers is:\",A #asking for the position of the pivotal element i = input(\"Please enter the index number\") pivot = A[i] print \"The pivot element is:\",pivot def dutchNationalFlag(A,pivot): equal = 0 small = 0 large = len(A)-1 while (equal <= large):", "# Problem #2099 2099 The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing? $[asy] unitsize(10mm); defaultpen(fontsize(10pt)); pen finedashed=linetype(\"4 4\"); filldraw((1,1)--(2,1)--(2,2)--(4,2)--(4,3)--(1,3)--cycle,grey,black+linewidth(.8pt)); draw((0,1)--(0,3)--(1,3)--(1,4)--(4,4)--(4,3)-- (5,3)--(5,2)--(4,2)--(4,1)--(2,1)--(2,0)--(1,0)--(1,1)--cycle,finedashed); draw((0,2)--(2,2)--(2,4),finedashed); draw((3,1)--(3,4),finedashed); label(\"1\",(1.5,0.5)); draw(circle((1.5,0.5),.17)); label(\"2\",(2.5,1.5)); draw(circle((2.5,1.5),.17)); label(\"3\",(3.5,1.5)); draw(circle((3.5,1.5),.17)); label(\"4\",(4.5,2.5)); draw(circle((4.5,2.5),.17)); label(\"5\",(3.5,3.5)); draw(circle((3.5,3.5),.17)); label(\"6\",(2.5,3.5)); draw(circle((2.5,3.5),.17)); label(\"7\",(1.5,3.5)); draw(circle((1.5,3.5),.17)); label(\"8\",(0.5,2.5)); draw(circle((0.5,2.5),.17)); label(\"9\",(0.5,1.5)); draw(circle((0.5,1.5),.17));[/asy]$ $\\mathrm{(A) \\ } 2\\qquad \\mathrm{(B) \\ } 3\\qquad \\mathrm{(C) \\ } 4\\qquad \\mathrm{(D) \\ } 5\\qquad \\mathrm{(E) \\ } 6$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Difference between revisions of \"2003 AMC 8 Problems/Problem 22\" ## Problem The following figures are composed of squares and circles. Which figure has a shaded region with largest area? $[asy]/* AMC8 2003 #22 Problem */ size(3inch, 2inch); unitsize(1cm); pen outline = black+linewidth(1); filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline); filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline); filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1)); filldraw(Circle((1,1), 1), white, outline); filldraw(Circle((3.5,.5), .5), white, outline); filldraw(Circle((4.5,.5), .5), white, outline); filldraw(Circle((3.5,1.5), .5), white, outline); filldraw(Circle((4.5,1.5), .5), white, outline); filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline); label(\"A\", (1, 2), N); label(\"B\", (4, 2), N); label(\"C\", (7, 2), N); draw((0,-.5)--(.5,-.5), BeginArrow); draw((1.5, -.5)--(2, -.5), EndArrow); label(\"2 cm\", (1, -.5)); draw((3,-.5)--(3.5,-.5), BeginArrow); draw((4.5, -.5)--(5, -.5), EndArrow); label(\"2 cm\", (4, -.5)); draw((6,-.5)--(6.5,-.5), BeginArrow); draw((7.5, -.5)--(8, -.5), EndArrow); label(\"2 cm\", (7, -.5)); draw((6,1)--(6,-.5), linetype(\"4 4\")); draw((8,1)--(8,-.5), linetype(\"4 4\"));[/asy]$ $\\textbf{(A)}\\ \\text{A only}\\qquad\\textbf{(B)}\\ \\text{B only}\\qquad\\textbf{(C)}\\ \\text{C only}\\qquad\\textbf{(D)}\\ \\text{both A and B}\\qquad\\textbf{(E)}\\ \\text{all are equal}$ ## Solution First we have to find the area of the shaded region in each of the figures. In figure $\\textbf{A}$ the area of the shaded region is the area of the circle subtracted from the area of the square. That is $2^2-1^2 \\pi=4-\\pi$. In figure $\\textbf{B}$ the area of the shaded region is the sum of the areas of the 4 circles subtracted from the area of the square. That is $2^2-4((\\frac{1}{2})^2 \\pi)=4-4(\\frac{\\pi}{4})=4-\\pi$. In figure $\\textbf{C}$ the area of", "back, (R = red, B = blue, O = orange, Y = yellow, G = green, and W = white). They are hinged together as shown, then folded to form a cube. The face opposite the white face is $[asy] draw((0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,2)--(2,2)--(2,3)--(0,3)--cycle); draw((1,3)--(1,2)--(2,2)--(2,1)--(3,1)--(3,2)); label($ $\\text{(A)}\\ \\text{B} \\qquad \\text{(B)}\\ \\text{G} \\qquad \\text{(C)}\\ \\text{O} \\qquad \\text{(D)}\\ \\text{R} \\qquad \\text{(E)}\\ \\text{Y}$ Jan 2: Three dice with faces numbered 1 through 6 are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is $[asy] draw((0,0)--(2,0)--(3,1)--(3,7)--(1,7)--(0,6)--cycle); draw((3,7)--(2,6)--(0,6)); draw((3,5)--(2,4)--(0,4)); draw((3,3)--(2,2)--(0,2)); draw((2,0)--(2,6)); dot((1,1)); dot((.5,.5)); dot((1.5,.5)); dot((1.5,1.5)); dot((.5,1.5)); dot((2.5,1.5)); dot((.5,2.5)); dot((1.5,2.5)); dot((1.5,3.5)); dot((.5,3.5)); dot((2.25,2.75)); dot((2.5,3)); dot((2.75,3.25)); dot((2.25,3.75)); dot((2.5,4)); dot((2.75,4.25)); dot((.5,5.5)); dot((1.5,4.5)); dot((2.25,4.75)); dot((2.5,5.5)); dot((2.75,6.25)); dot((1.5,6.5)); [/asy]$ $\\text{(A)}\\ 21 \\qquad \\text{(B)}\\ 22 \\qquad \\text{(C)}\\ 31 \\qquad \\text{(D)}\\ 41 \\qquad \\text{(E)}\\ 53$ Jan 3: Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom and sides. $[asy]/* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(1,2)--(1.5,2)); draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5)); draw((1.5,3.5)--(2.5,3.5)); draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5)); draw((3,4)--(3,3)--(2.5,2.5)); draw((3,3)--(4,3)--(4,2)--(3.5,1.5)); draw((4,3)--(3.5,2.5)); draw((2.5,.5)--(3,1)--(3,1.5));[/asy]$ $\\text{(A)}\\ 18 \\qquad \\text{(B)}\\ 24 \\qquad \\text{(C)}\\ 26 \\qquad \\text{(D)}\\ 30 \\qquad \\text{(E)}\\ 36$ Jan 4: Fourteen white cubes are put together to form the figure on the right. The complete surface of", "on both tests? $[asy] draw((2,0)--(7,0)--(7,5)--(2,5)--cycle); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((6,0)--(6,5)); draw((2,1)--(7,1)); draw((2,2)--(7,2)); draw((2,3)--(7,3)); draw((2,4)--(7,4)); draw((.2,6.8)--(1.8,5.2)); draw(circle((4.5,1.5),.5),linewidth(.6 mm)); label(\"0\",(2.5,.2),N); label(\"0\",(3.5,.2),N); label(\"2\",(4.5,.2),N); label(\"1\",(5.5,.2),N); label(\"0\",(6.5,.2),N); label(\"0\",(2.5,1.2),N); label(\"0\",(3.5,1.2),N); label(\"1\",(4.5,1.2),N); label(\"1\",(5.5,1.2),N); label(\"1\",(6.5,1.2),N); label(\"1\",(2.5,2.2),N); label(\"3\",(3.5,2.2),N); label(\"5\",(4.5,2.2),N); label(\"2\",(5.5,2.2),N); label(\"0\",(6.5,2.2),N); label(\"1\",(2.5,3.2),N); label(\"4\",(3.5,3.2),N); label(\"3\",(4.5,3.2),N); label(\"0\",(5.5,3.2),N); label(\"0\",(6.5,3.2),N); label(\"2\",(2.5,4.2),N); label(\"2\",(3.5,4.2),N); label(\"1\",(4.5,4.2),N); label(\"0\",(5.5,4.2),N); label(\"0\",(6.5,4.2),N); label(\"F\",(1.5,.2),N); label(\"D\",(1.5,1.2),N); label(\"C\",(1.5,2.2),N); label(\"B\",(1.5,3.2),N); label(\"A\",(1.5,4.2),N); label(\"A\",(2.5,5.2),N); label(\"B\",(3.5,5.2),N); label(\"C\",(4.5,5.2),N); label(\"D\",(5.5,5.2),N); label(\"F\",(6.5,5.2),N); label(\"Test 1\",(-.5,5.2),N); label(\"Test 2\",(2.6,6),N); [/asy]$ $\\text{(A)}\\ 12\\% \\qquad \\text{(B)}\\ 25\\% \\qquad \\text{(C)}\\ 33\\frac{1}{3}\\% \\qquad \\text{(D)}\\ 40\\% \\qquad \\text{(E)}\\ 50\\%$ ## Problem 13 The perimeter of the polygon shown is $[asy] draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label(\"6\",(0,3),W); label(\"8\",(4,6),N); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((0.5,6)--(0.5,5.5)--(0,5.5)); draw((7.5,6)--(7.5,5.5)--(8,5.5)); draw((7.5,3)--(7.5,3.5)--(8,3.5)); draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); [/asy]$ $\\text{(A)}\\ 14 \\qquad \\text{(B)}\\ 20 \\qquad \\text{(C)}\\ 28 \\qquad \\text{(D)}\\ 48$ $\\text{(E)}\\ \\text{cannot be determined from the information given}$ ## Problem 14 If $200\\leq a \\leq 400$ and $600\\leq b\\leq 1200$, then the largest value of the quotient $\\frac{b}{a}$ is $\\text{(A)}\\ \\frac{3}{2} \\qquad \\text{(B)}\\ 3 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 300 \\qquad \\text{(E)}\\ 600$ ## Problem 15 Sale prices at the Ajax Outlet Store are $50\\%$ below original prices. On Saturdays an additional discount of $20\\%$ off the sale price is given. What is the Saturday price of a coat whose original price is <dollar/>$180$? $\\text{(A)}$ <dollar/>$54$ $\\text{(B)}$ <dollar/>$72$ $\\text{(C)}$ <dollar/>$90$ $\\text{(D)}$ <dollar/>$108$ $\\text{(D)}$ <dollar/>$110$ ## Problem 16 A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "(6.5,0.5)); label(\"\\dots\", (1.5,0.5)); label(\"\\dots\", (0.5,1.5)); label(\"\\dots\", (0.5,2.5)); label(\"\\dots\", (0.5,3.5)); label(\"\\dots\", (0.5,4.5)); label(\"\\dots\", (0.5,5.5)); label(\"\\dots\", (0.5,6.5)); label(\"\\dots\", (6.5,0.5)); label(\"\\dots\", (6.5,1.5)); label(\"\\dots\", (6.5,2.5)); label(\"\\dots\", (6.5,3.5)); label(\"\\dots\", (6.5,4.5)); label(\"\\dots\", (6.5,5.5)); label(\"\\dots\", (0.5,6.5)); label(\"\\dots\", (1.5,6.5)); label(\"\\dots\", (2.5,6.5)); label(\"\\dots\", (3.5,6.5)); label(\"\\dots\", (4.5,6.5)); label(\"\\dots\", (5.5,6.5)); label(\"\\dots\", (6.5,6.5)); label(\"17\", (1.5,1.5)); label(\"18\", (1.5,2.5)); label(\"19\", (1.5,3.5)); label(\"20\", (1.5,4.5)); label(\"21\", (1.5,5.5)); label(\"16\", (2.5,1.5)); label(\"5\", (2.5,2.5)); label(\"6\", (2.5,3.5)); label(\"7\", (2.5,4.5)); label(\"22\", (2.5,5.5)); label(\"15\", (3.5,1.5)); label(\"4\", (3.5,2.5)); label(\"1\", (3.5,3.5)); label(\"8\", (3.5,4.5)); label(\"23\", (3.5,5.5)); label(\"14\", (4.5,1.5)); label(\"3\", (4.5,2.5)); label(\"2\", (4.5,3.5)); label(\"9\", (4.5,4.5)); label(\"24\", (4.5,5.5)); label(\"13\", (5.5,1.5)); label(\"12\", (5.5,2.5)); label(\"11\", (5.5,3.5)); label(\"10\", (5.5,4.5)); label(\"25\", (5.5,5.5)); [/asy]$ $\\textbf{(A)} ~367 \\qquad\\textbf{(B)} ~368 \\qquad\\textbf{(C)} ~369 \\qquad\\textbf{(D)} ~379 \\qquad\\textbf{(E)} ~380$ ## Problem 9 The point $P(a,b)$ in the $xy$-plane is first rotated counterclockwise by $90^\\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$. The image of $P$ after these two transformations is at $(-6,3)$. What is $b - a ?$ $\\textbf{(A)} ~1 \\qquad\\textbf{(B)} ~3 \\qquad\\textbf{(C)} ~5 \\qquad\\textbf{(D)} ~7 \\qquad\\textbf{(E)} ~9$ ## Problem 10 An inverted cone with base radius $12 \\text{cm}$ and height $18\\text{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of $24\\text{cm}$. What is the height in centimeters of the water in the cylinder? $\\textbf{(A) }1.5 \\qquad \\textbf{(B) }3 \\qquad \\textbf{(C) }4 \\qquad \\textbf{(D) }4.5 \\qquad \\textbf{(E) }6$ ##", "(6.5,0.5)); label(\"\\dots\", (1.5,0.5)); label(\"\\dots\", (0.5,1.5)); label(\"\\dots\", (0.5,2.5)); label(\"\\dots\", (0.5,3.5)); label(\"\\dots\", (0.5,4.5)); label(\"\\dots\", (0.5,5.5)); label(\"\\dots\", (0.5,6.5)); label(\"\\dots\", (6.5,0.5)); label(\"\\dots\", (6.5,1.5)); label(\"\\dots\", (6.5,2.5)); label(\"\\dots\", (6.5,3.5)); label(\"\\dots\", (6.5,4.5)); label(\"\\dots\", (6.5,5.5)); label(\"\\dots\", (0.5,6.5)); label(\"\\dots\", (1.5,6.5)); label(\"\\dots\", (2.5,6.5)); label(\"\\dots\", (3.5,6.5)); label(\"\\dots\", (4.5,6.5)); label(\"\\dots\", (5.5,6.5)); label(\"\\dots\", (6.5,6.5)); label(\"17\", (1.5,1.5)); label(\"18\", (1.5,2.5)); label(\"19\", (1.5,3.5)); label(\"20\", (1.5,4.5)); label(\"21\", (1.5,5.5)); label(\"16\", (2.5,1.5)); label(\"5\", (2.5,2.5)); label(\"6\", (2.5,3.5)); label(\"7\", (2.5,4.5)); label(\"22\", (2.5,5.5)); label(\"15\", (3.5,1.5)); label(\"4\", (3.5,2.5)); label(\"1\", (3.5,3.5)); label(\"8\", (3.5,4.5)); label(\"23\", (3.5,5.5)); label(\"14\", (4.5,1.5)); label(\"3\", (4.5,2.5)); label(\"2\", (4.5,3.5)); label(\"9\", (4.5,4.5)); label(\"24\", (4.5,5.5)); label(\"13\", (5.5,1.5)); label(\"12\", (5.5,2.5)); label(\"11\", (5.5,3.5)); label(\"10\", (5.5,4.5)); label(\"25\", (5.5,5.5)); [/asy]$ $\\textbf{(A)} ~367 \\qquad\\textbf{(B)} ~368 \\qquad\\textbf{(C)} ~369 \\qquad\\textbf{(D)} ~379 \\qquad\\textbf{(E)} ~380$ ## Problem 9 The point $P(a,b)$ in the $xy$-plane is first rotated counterclockwise by $90^\\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$. The image of $P$ after these two transformations is at $(-6,3)$. What is $b - a ?$ $\\textbf{(A)} ~1 \\qquad\\textbf{(B)} ~3 \\qquad\\textbf{(C)} ~5 \\qquad\\textbf{(D)} ~7 \\qquad\\textbf{(E)} ~9$ ## Problem 10 An inverted cone with base radius $12 \\text{cm}$ and height $18\\text{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of $24\\text{cm}$. What is the height in centimeters of the water in the cylinder? $\\textbf{(A) }1.5 \\qquad \\textbf{(B) }3 \\qquad \\textbf{(C) }4 \\qquad \\textbf{(D) }4.5 \\qquad \\textbf{(E) }6$ ##", "outline); filldraw(Circle((3.5,.5), .5), white, outline); filldraw(Circle((4.5,.5), .5), white, outline); filldraw(Circle((3.5,1.5), .5), white, outline); filldraw(Circle((4.5,1.5), .5), white, outline); filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline); label(\"A\", (1, 2), N); label(\"B\", (4, 2), N); label(\"C\", (7, 2), N); draw((0,-.5)--(.5,-.5), BeginArrow); draw((1.5, -.5)--(2, -.5), EndArrow); label(\"2 cm\", (1, -.5)); draw((3,-.5)--(3.5,-.5), BeginArrow); draw((4.5, -.5)--(5, -.5), EndArrow); label(\"2 cm\", (4, -.5)); draw((6,-.5)--(6.5,-.5), BeginArrow); draw((7.5, -.5)--(8, -.5), EndArrow); label(\"2 cm\", (7, -.5)); draw((6,1)--(6,-.5), linetype(\"4 4\")); draw((8,1)--(8,-.5), linetype(\"4 4\"));[/asy]$ $\\textbf{(A)}\\ \\text{A only}\\qquad\\textbf{(B)}\\ \\text{B only}\\qquad\\textbf{(C)}\\ \\text{C only}\\qquad\\textbf{(D)}\\ \\text{both A and B}\\qquad\\textbf{(E)}\\ \\text{all are equal}$ ## Problem 23 In the pattern below, the cat (denoted as a large circle in the figures below) moves clockwise through the four squares and the mouse (denoted as a dot in the figures below) moves counterclockwise through the eight exterior segments of the four squares. $[asy]defaultpen(linewidth(0.8)); size(350); path p=unitsquare; int i; for(i=0; i<5; i=i+1) { draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); } path cat=Circle((0.5,0.5), 0.3); draw(shift(0,1)*cat^^shift(4,1)*cat^^shift(7,0)*cat^^shift(9,0)*cat^^shift(12,1)*cat); dot((1.5,0)^^(5,0.5)^^(8,1.5)^^(10.5,2)^^(12.5,2)); label(\"1\", (1,2), N); label(\"2\", (4,2), N); label(\"3\", (7,2), N); label(\"4\", (10,2), N); label(\"5\", (13,2), N); [/asy]$ If the pattern is continued, where would the cat and mouse be after the 247th move? $\\textbf{(A)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(1,0)*cat); dot((0,0.5)); [/asy]$ $\\textbf{(B)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(1,1)*cat); dot((0,0.5)); [/asy]$ $\\textbf{(C)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(1,0)*cat); dot((0,1.5));", "contains 200 rows. Each row is a list containing 136 pixels. Each pixel is a list of form [r,g,b], where r, g, and b are red, blue, and green intensity values from 0 to 255. Try the following expressions: len(b) len(b[0]) b[0][0] b[100][100] Knowing that an image is a list of rows, write a function flip_image(image) that creates a new image that is image flipped upside-down and plots the result. Put this function in flip_image.py Then try flipping the nautical alphabet image upside-down. ### 1.2 Student Activities 1. In the file c_flag.py, write a function c_flag to create a little 5-by-6 pixel image encoding the flag for the letter C in the nautical flag alphabet. Refer to the RGB Color Table to get the codes for red and navy blue. Since this image is very small, you will want to magnify it for display purposes by writing draw_image(c, c_flag(), 0, 0, 20). 2. You may choose to do either 2 or 3. Your choice. 3. In the file swap_bg.py, write a function swap_bg(image) that swaps the blue and green values of every pixel in the image, and plots the result. Test it on the nautical flag image. What happens if you call swap_bg repeatedly on the same image? 4. In the file grayscale.py, write a function grayscale(image) that", "(1.8,6); \\filldraw [fill=cyan, thin, draw=black] (1.8,5) rectangle (2.1,6); \\filldraw [fill=cyan, thin, draw=black] (2.1,5) rectangle (2.4,6); \\filldraw [fill=cyan, thin, draw=black] (2.4,5) rectangle (2.7,6); \\filldraw [fill=cyan, thin, draw=black] (2.7,5) rectangle (3,6); \\filldraw [fill=cyan, thin, draw=black] (3,5) rectangle (3.3,6); \\filldraw [fill=magenta, thin, draw=black] (3.3,5) rectangle (3.6,6); \\filldraw [fill=magenta, thin, draw=black] (3.6,5) rectangle (3.9,6); \\filldraw [fill=magenta, thin, draw=black] (3.9,5) rectangle (4.2,6); \\filldraw [fill=magenta, thin, draw=black] (4.2,5) rectangle (4.5,6); \\filldraw [fill=magenta, thin, draw=black] (4.5,5) rectangle (4.8,6); \\filldraw [fill=magenta, thin, draw=black] (4.8,5) rectangle (5.1,6); \\filldraw [fill=magenta, thin, draw=black] (5.1,5) rectangle (5.4,6); \\filldraw [fill=magenta, thin, draw=black] (5.4,5) rectangle (5.7,6); \\filldraw [fill=magenta, thin, draw=black] (5.7,5) rectangle (6,6); \\filldraw [fill=magenta, thin, draw=black] (6,5) rectangle (6.3,6); \\filldraw [fill=magenta, thin, draw=black] (6.3,5) rectangle (6.6,6); \\filldraw [fill=magenta, thin, draw=black] (6.6,5) rectangle (6.9,6); \\filldraw [fill=magenta, thin, draw=black] (6.9,5) rectangle (7.2,6); \\filldraw [fill=magenta, thin, draw=black] (7.2,5) rectangle (7.5,6); \\filldraw [fill=magenta, thin, draw=black] (7.5,5) rectangle (7.8,6); \\draw [very thick] (0.3,5) rectangle (1.8,6); \\draw [very thick] (1.8,5) rectangle (3.3,6); \\draw [very thick] (3.3,5) rectangle (4.8,6); \\draw [very thick] (4.8,5) rectangle (6.3,6); \\draw [very thick] (6.3,5) rectangle (7.8,6); \\coordinate[label=left: Container B] (B) at (0,5.-1); \\filldraw [fill=cyan, thin, draw=black] (0.3,3.6) rectangle (0.6,4.6); \\filldraw [fill=cyan, thin, draw=black] (0.6,3.6) rectangle (0.9,4.6); \\filldraw [fill=cyan, thin, draw=black] (0.9,3.6) rectangle (1.2,4.6); \\filldraw [fill=magenta, thin, draw=black] (1.2,3.6) rectangle (1.5,4.6); \\filldraw [fill=magenta, thin, draw=black] (1.5,3.6) rectangle (1.8,4.6); \\filldraw [fill=magenta, thin, draw=black] (1.8,3.6) rectangle" ]

[ "used, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag. Thus, the number of different signals that can be made here is: $4\\cdot \\left(4-1\\right)=4\\cdot 3=12$ In the case of using 3 flags, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag, leaving 2 flags remaining as the possible third flag. Thus, the number of different signals that can be made here is: $4\\cdot \\left(4-1\\right)\\cdot \\left(4-2\\right)=4\\cdot 3\\cdot 2=24$ In the final case of using all 4 flags, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag, leaving 2 flags remaining as the possible third flag, leaving only 1 flag as the final flag to be chosen. Thus the number of different signals that can be made here is: $4\\cdot \\left(4-1\\right)\\cdot \\left(4-2\\right)\\cdot \\left(4-3\\right)=4\\cdot 3\\cdot 2\\cdot 1=24$ Adding all of these possible signals will give us the answer: $12+24+24=60$ For question 2, we can think of a tree-diagram to represent this. Starting with 3 separate trees: Science, Geography, and Phys. Ed at the top. From here, there are 4 possible classes to branch off of the first class: Art,", "be chosen out of all possible $4$. once you chose the colour of the first one you only have $3$ choices for the second one - cause it can't be the same which gives you $4 \\cdot 3$ possibilities for the first two stripes. continuing this way you end up with $4 \\cdot 3^5$ ways", ". Hence, there are: . $10\\cdot6 \\:=\\:{\\color{blue}60}$ ways to choose 2 boys and a girl. Therefore: . $P(\\text{2 boys, 1 girl}) \\:=\\:\\frac{60}{165} \\:=\\:\\boxed{\\frac{4}{11}}$", "faces as we go along, noting that there are ${6\\choose 2} = 15$ pairs of colors for the top and bottom faces. Using just one color for the four sides: no change, so $6$ ways to color the sides and $6\\cdot 15 = 90$ colorings altogether. (Swapping ends doesn’t affect the sides.) Using two colors: no change, so $60$ ways and $60\\cdot 15 = 900$ colorings; swapping ends affects the sides, but they can be restored by a rotation. (Running total: $990$) Using three colors: As before, there are $6$ ways to choose the color that appears twice and then $10$ ways to choose the other two colors, and the two faces of the same color may be either adjacent or opposite. If they are opposite, it makes no difference how the remaining two colors are applied to the other two sides, so that sub-case yields $60$ ways and $60\\cdot 15 = 900$ colorings. If they are adjacent, however, the order in which the other two colors are applied to the other two sides does matter, and we get $120$ ways. Since these $120$ ways of coloring the sides are not preserved by swapping ends, each apparently extends to $30$ different ways of coloring the prism, for a total of $120\\cdot 30 = 3600$ colorings. However, this actually", "use the same kind of reasoning yet again. There are $12$ possible choices for president. Once the president is chosen, there are $11$ possible choices for vice president, since the president and vice president cannot be the same person. But there is no such restriction on the secretary or the treasurer: each of them can be any of the $12$ people, so each can be chosen in $12$ ways. The grand total is therefore $12\\cdot11\\cdot12\\cdot12=11\\cdot12^3$. 4. There are three cases, depending on whether one, two, or three yellow flags are used. • Suppose that just one yellow flag is used, so that we’re arranging $5$ flags. There are $\\binom52$ ways to choose which $2$ positions in the string of $5$ will contain the red flags. That leaves $3$ open positions, and there are $\\binom32$ ways to pick $2$ of them for the $2$ green flags. The lone yellow flag will then fill the remaining slot $-$ no choice there. The total number of such $5$-flag signals is therefore $\\binom52\\binom32=10\\cdot3=30$. • Suppose that two yellow flags are used. We reason in exactly the same way. There are $\\binom62$ ways to choose the $2$ positions to be filled by red flags, and then there are $\\binom42$ ways to choose which $2$ of the remaining $4$ slots will get the green", "4 cards of that value: . ${4\\choose2} = {\\color{blue}6}$ ways. . . So there are: . $2\\cdot 12\\cdot 6 \\:=\\:{\\color{blue}144}$ ways. Case 2: Draw a Triple. There are 12 choices for the value of the Triple. You must get 3 of the 4 cards of that value: . ${4\\choose3} = {\\color{blue}4}$ ways. . . So there are: . $12\\cdot4 \\:=\\:{\\color{blue}48}$ ways. Hence, there are: . $144 + 48 \\:=\\:{\\color{red}192}$ ways to get a Full house. $P(\\text{Full House}\\,|\\,\\text{Pair}) \\:=\\:{\\color{red}\\frac{192}{19,\\!600}} \\;=\\;\\boxed{\\frac{12}{1225}}$", "flags; the yellow flags get the leftovers. The total number of such $6$-flag signals is therefore $\\binom62\\binom42=15\\cdot6=90$. • I’ll leave it to you to finish off the problem by calculating the number of $7$-flag signals; just follow the same reasoning. When you’re done, add up the numbers from the three cases to get the final answer. 5. There are $13$ denominations altogether, from ace through king, so there are $13$ ways to choose the denomination of the triplet. The pair has to be of a different denomination, so there are $12$ ways to pick its denomination. Thus, there are $13\\cdot12$ possible types of full house. However, each type (e.g., three $10$’s and two jacks) can occur in several different ways. There are $4$ cards of each denomination, one for each of the four suits, so once its denomination is known, there are $\\binom43$ ways to choose the three cards for the triplet. Similarly, there are $\\binom42$ ways to choose two cards of the pair’s denomination. Thus, each type of full house comprises $\\binom43\\binom42$ different full houses of that type, and the grand total of possible full houses is $$13\\cdot12\\cdot\\binom43\\binom42=13\\cdot12\\cdot4\\cdot6=3744\\;.$$ - 1. In this case we should use formula for number of permutations(http://en.wikipedia.org/wiki/Permutation): $A_n^k = \\frac {n!}{n-k!}$. This formula uses when we have $k$ places and $n$ objects we", "(n - i)}=\\dfrac{n!}{(n-k)!}$$ If we plug $$n=4$$ (our pool of distinct flags) and $$k=2$$ (the number of flags to sample in order to build a signal), according to the problem statement, the formula will yield: $$4P_2 = \\dfrac{4!}{(4-2)!}=\\dfrac{4!}{2!} = \\dfrac{4\\cdot 3\\cdot 2\\cdot 1}{2\\cdot 1}= 4\\cdot 3 = 12$$ As you correctly pointed out that $$12$$ is the right answer. Now for the intuition behind the problem and in general when we have to deal with ordered permutations, we can think of the problem as follows: Let's suppose that we have two placeholders for any of the available flags, denoted as $$F_1 \\, F_2$$ . Furthermore, since the order of the flags matters $$F_1 \\, F_2 \\ne F_2 \\, F_1$$, this leaves us with $$n=4$$ possible options for the $$F_1$$ placeholder and $$n-1=3$$ for the $$F_2$$ placeholder. This is because initially the whole set of flags is available for selection e.g. $$F=\\{R,B,G,O\\}$$, picking any flag from this set leaves us only three of the rest flags available for selection: • Case 1: First choice is $$\\color{red}R$$ from the whole set of flags $$\\{R,B,G,O\\}$$ $$R \\, F_2$$ this entails that the second flag must be sampled from the following set $$\\{B,G,O\\}$$. This leads to, the following three permutations: $$\\mathbf{\\color{red}R}\\color{blue}B, \\, \\mathbf{\\color{red}R}\\color{green}G, \\, \\mathbf{\\color{red}R}\\color{orange}O$$ • Case 2: First choice is $$\\color{blue}B$$,", "time. – Thomas Andrews Jul 15 '16 at 12:49 ## 1 Answer Use inclusion/exclusion principle: • Include the number of ways to choose cards from at most $\\color\\red4$ suits: $\\binom{4}{\\color\\red4}\\cdot\\binom{8\\cdot\\color\\red4}{k}$ • Exclude the number of ways to choose cards from at most $\\color\\red3$ suits: $\\binom{4}{\\color\\red3}\\cdot\\binom{8\\cdot\\color\\red3}{k}$ • Include the number of ways to choose cards from at most $\\color\\red2$ suits: $\\binom{4}{\\color\\red2}\\cdot\\binom{8\\cdot\\color\\red2}{k}$ • Exclude the number of ways to choose cards from at most $\\color\\red1$ suits: $\\binom{4}{\\color\\red1}\\cdot\\binom{8\\cdot\\color\\red1}{k}$ Hence the total number of ways is: $$\\sum\\limits_{n=0}^{3}(-1)^{n}\\cdot\\binom{4}{4-n}\\cdot\\binom{8\\cdot(4-n)}{k}$$ • While your overall result is correct, the formulas for 2 and 3 suits are incorrect. For example for both, 2 and 3, you triple count the number of ways to draw 1 color only. (( 4 chose 2) yields the cases AB, AC, AD which all contain color A. Now, (16 choose 4) includes the cases where you choose from color A only.) – Nils Oct 16 '19 at 15:53", "we can choose $$O$$ in $$3$$ ways, and we can choose $$E$$ from remaining colors in $$2$$ ways. So, the number of coloring in this case is $$3 \\cdot 2 \\cdot 1^5 \\cdot 1^5 = 6$$. We get $$192-6=186$$ as our answer.", ". ${\\color{blue}4\\cdot{48\\choose9}\\cdot{39\\choose13,1 3,13}}$ ways for one player to get the four Aces. Three Aces in one player's hand, one Ace in another's There are 4 choices of players to get the three Aces. There are: ${\\color{red}{4\\choose3}}$ ways for him to get three Aces. For his other 10 cards, there are: ${\\color{red}{48\\choose10}}$ ways. There are 3 choices of players to get the one Ace. There is one way for him to get the one remaining Ace. For his other 12 cards, there are: ${\\color{red}{38\\choose12}}$ ways. The remaining 26 cards are distributed to the other two players. There are: ${\\color{red}{26\\choose13,13}}$ ways. Hence, there are: . ${\\color{blue}4\\cdot{4\\choose3}\\cdot{48\\choose10}\\c dot3\\cdot{38\\choose12}\\cdot{26\\choose13,13}}$ ways Get the idea? Edit: corrected an error . . . sorry for the blunder. . 3. Originally Posted by Soroban Hence, there are: . ${\\color{blue}4\\cdot{4\\choose3}\\cdot{48\\choose10}\\c dot3\\cdot{48\\choose12}\\cdot{26\\choose13,13}}$ ways Hi, thanks! but for the guy with only ace. should it be: . ${\\color{blue}4\\cdot{4\\choose3}\\cdot{48\\choose10}\\c dot3\\cdot\\color{red}{39\\choose12}\\cdot\\color{blue} {26\\choose13,13}}$ 4. Hello again, chopet! Thank you for pointing out my error . . . After the first player gets his 13 cards (including 3 Aces), . . there are 3 choices for the next player. There is one way for him to get the remaining Ace. His other 12 cards come from the remaining 38 cards. So there are: ${\\color{red}{38\\choose12}}$ ways. There are: . ${\\color{blue}3\\cdot{38\\choose12}}$ ways for the second", "for red. $4$ choices for the second appearing slot reserved for red... on up to only $1$ choice for the final appearing slot reserved for red, for a total of $5\\cdot 4\\cdot 3\\cdot 2\\cdot 1 = 5!$ ways to place the red toys into their reserved slots. • From left to right, among the spaces reserved for blue toys, pick which blue toy is used. $4$ choices for the first appearing slot reserved for blue, $3$ choices for the second appearing slot reserved for blue, etc... for a total of $4\\cdot 3\\cdot 2\\cdot 1 = 4!$ ways to place the blue toys into their reserved slots. Applying multiplication principle, there are $\\binom{4}{2}\\cdot 5!\\cdot 4!$ ways to arrange the toys respecting the conditions.", "then use deduction to manually find the number of ways you can fill in the rest of the $4\\times 5$ rectangle. Turns out there are only $11$ configurations, and since the $4$ colours can be interchanged in $4!$ ways there are $11\\cdot 4! = 264$ colourings with those restrictions. – N. Shales Aug 19 '17 at 3:37 Clearly, there are $4! = 24$ ways in a $2\\times 2$ rectangle. Now add one column to the right, making it a $2 \\times 3$ rectangle. We already colored the $2 \\times 2$ square in the left of this $2 \\times 3$ rectangle (in $24$ ways), but the remaining two boxes are not colored yet. This two new boxes can be colored in $2 \\cdot 1 =2$ ways. Therefore, a $2 \\times 3$ rectangle can be colored in $24 \\cdot 2 = 48$ ways. Similarly, adding one more column to the right will give two extra boxes, and there will be $48 \\cdot 2=96$ ways to color a $2 \\times 4$ rectangle. Finally, adding one more column to the right, we will have $96\\cdot 2 = 192$ ways to color a $2 \\times 5$ rectangle. Observe that, after adding one row to a $2 \\times 5$ rectangle, although we will have $5$ new uncolored boxes, when the uncolored box on the", "10\\cdot 9 + 10 - 1}{30\\cdot 120}.$$ Dividing by $9$, we get $$\\frac{5\\cdot 2\\cdot 8 - 10 + 1}{10\\cdot 40} = \\frac{71}{400} \\implies \\boxed{\\textbf{(D) }471}$$. ## Solution 3 Use complementary counting. Denote $T_n$ as the total number of ways to put $n$ colors to $n$ boxes by 3 people out of which $f_n$ ways are such that no box has uniform color. Notice $T_n = (n!)^3$. From this setup we see the question is asking for $1-\\frac{f_5}{(5!)^3}$. To find $f_5$ we want to exclude the cases of a) one box of the same color, b) 2 boxes of the same color, c) 3 boxes of same color, d) 4 boxes of the same color, and e) 5 boxes of the same color. Cases d) and e) coincide. From this, we have $$f_5=T_5 -{\\binom{5}{1}\\binom{5}{1}\\cdot f_4 - \\binom{5}{2}\\binom{5}{2}\\cdot 2!\\cdot f_3 - \\binom{5}{3}\\binom{5}{3}\\cdot 3!\\cdot f_2 - 5!}$$ In case b), there are $\\binom{5}{2}$ ways to choose 2 boxes that have the same color, $\\binom{5}{2}$ ways to choose the two colors, 2! ways to permute the 2 chosen colors, and $f_3$ ways so that the remaining 3 boxes don’t have the same color. The same goes for cases a) and c). In case e), the total number of ways to permute 5 colors is $5!$. Now, we just need to calculate $f_2$,", "We start with cases having the maximum of each type of flag and then decrease the maximum of each flag type by 1 till we reach 1 count for each flag type. In each case we get a group of 5 flags which we simply arrange 5 at a time using standard formula. The sum of all ways i.e. arrangements of 5 flags is the sum of all cases, which is 170. This is also the answer given at the back of the book. Case 1: 3W 3W+2P : 5!/(3!.2!) = 10 ways 3W+2B : 5!/(3!.2!) = 10 ways 3W+1P+1B : 5!/(3!.1!.1!) = 20 ways Case 2: 3P 3P+2W : 5!/(3!.2!) = 10 ways 3P+2B : 5!/(3!.2!) = 10 ways 3P+1B+1W: 5!/(3!.1!.1!) = 20 ways Case 3: 2B 2B+3W 2B+3P 2B+1P+2W : 5!/(2!.2!.1!) = 30 ways 2B+2P+1W : 5!/(2!.2!.1!) = 30 ways Case 4: 1W 1W+3P+1B 1W+2P+2B Case 5: 2W 2W+2P+1B: 5!/(2!.2!.1!) = 30 ways 2W+1P+2B Case 6: 1W 1W+3P+1B 1W+2P+2B Case 7: 2P 2P+1W+2B 2P+2W+1B Case 8: 1P 1P+2W+2B 1P+3W+1B Case 9: 1B 1B+2W+2P 1B+3W+1B 1B+1W+3B", "are then 4 choices for the next tie, 3 for the tie after that, and so on. Thus $$3\\cdot 4! = 72$$ choices. #### 4. Give a counting question where the answer is $$8\\cdot 3 \\cdot 3 \\cdot 5\\text{.}$$ Give another question where the answer is $$8 + 3 + 3 + 5\\text{.}$$ Solution. You own 8 purple bow ties, 3 red bow ties, 3 blue bow ties and 5 green bow ties. How many ways can you select one of each color bow tie to take with you on a trip? $$8 \\cdot 3 \\cdot 3 \\cdot 5$$ ways. How many choices do you have for a single bow tie to wear tomorrow? $$8 + 3 + 3 + 5$$ choices. #### 5. Consider five digit numbers $$\\alpha = a_1a_2a_3a_4a_5\\text{,}$$ with each digit from the set $$\\{1,2,3,4\\}\\text{.}$$ 1. How many such numbers are there? 2. How many such numbers are there for which the sum of the digits is even? 3. How many such numbers contain more even digits than odd digits? Solution. 1. $$4^5$$ numbers. 2. $$4^4\\cdot 2$$ numbers (choose any digits for the first four digits - then pick either an even or an odd last digit to make the sum even). 3. We need 3 or more even digits. 3 even digits: $${5 \\choose", "the same algorithm(1-4) for the remaining $5$. We can't choose balls of only $1$ or $2$ colors. For choosing $3$ balls, we first pickup $1$ of each color in ${4\\choose3}=4$ ways and then we have $3$ ways of choosing the remaining $2$. This gives us $4\\cdot3 = 12$ ways. For the case of $4$ colors, we pickup 4 balls of each color in $1$ way and then we have $4$ ways of picking the last one. This gives us a total of $4+12+4 = \\boxed{20}$ ways.", "very left is colored, the color of the remaining boxes are determined. Since the box on the very left can be colored in two colors, there are $192 \\cdot 2 = 384$ ways to color a $3\\times 5$ rectangle. Finally, we add one more row, and there are $384 \\cdot 2 = 768$ ways. EDIT: Apparently, I cannot delete my answer if it is accepted. As the comments below suggested, there are some cases where this method fails to create a $4 \\times 5$ rectangle. • This answer is simply wrong, the reason being that you cannot always complete the third and the fourth row in a legal way. An example: $$\\matrix{0&a&b&c&b \\cr b&c&0&a&0\\cr a&0&&&\\cr}$$ – Christian Blatter Aug 19 '17 at 8:27 • @ChristianBlatter you are right. I am deleting my answer – ThePortakal Aug 19 '17 at 8:29 Let's consider separately cases where there are $4,2$ and $3$ different colours in the first column. Case 1: $4$ different colors. This has only one pattern since the outermost colors must switch places with the center two colors and there is only one way this can happen. Colors can be chosen in $4!=24$ different ways. $$\\begin{bmatrix}a&c&a&c&a\\\\b&d&b&d&b\\\\c&a&c&a&c\\\\d&b&d&b&d\\end{bmatrix}$$ Case 2: $2$ different colors. First column can be constructed in $4\\cdot3=12$ ways. There can only be $2$ colors per column, but", "correct. However, $6^2$ for the third end-swapping case is wrong: it really is $6^3$, as you originally had it. If the rotation is $\\pi/2$ clockwise, the ends swap, the top and right side swap, and the bottom and left side swap, so you get to choose three colors, not two. Here’s an elementary enumeration along the lines suggested by André Nicolas. Set the prism on end. Suppose first that the top and bottom faces get the same color; we’ll count the number of distinguishable ways to color the four sides. Using just one color: $6$ ways. Using two colors, $3$ sides of one color and $1$ of the other: There are $6$ ways to choose the color for the single face and then $5$ ways to choose the other color, for a total of $30$ ways. (Running total: $36$) Using two colors, $2$ sides of each color: There are ${6 \\choose 2} = 15$ ways to choose the two colors. Once the colors are chosen, they can be applied in just $2$ distinguishable ways: either opposite sides are the same color, or they are different colors. The total number of ways is therefore again $30$. (Running total: $66$) Using three colors: In this case one color must appear twice, the other two once each. There are $6$ ways", "a president, a vice-president, a secretary, and 4 senators? Choose the president ($20$ choices), then the vice-president ($19$ choices), then the secretary ($18$ choices), then the $4$ senators (${}_{17} C_4$ choices). So there are $$20 \\cdot 19 \\cdot 18 \\cdot {}_{17} C_4 = 16279200 \\textrm{ ways}$$ 14. A club has 12 freshmen and 8 sophomores. 1. How many ways can we choose a 4-person committee of 2 freshmen and 2 sophomores? 2. How many ways can we choose a 4-person committee with at least 2 freshmen? 15. A jar contains 8 red, 9 blue, and 10 white balls 1. In how many ways can someone pick 3 balls and get one of each color? 2. In how many ways can someone pick 5 balls and get 3 red and 2 blue? 3. In how many ways can someone pick 5 balls and get exactly 3 red balls? 4. In how many ways can someone pick 2 balls, both of the same color?" ]

[ "maybe it will be a good idea to convert every frame to a number and inspect these sequences? Worth a try: python for i in range(6): colors[i] = [x for (x, j, k) in colors[i]] print('flag' + str(i + 1) + '.gif:', colors[i]) After conversion we get these sequences: flag1.gif: [3, 5, 5, 3, 3, 5, 3, 5, 3, 3, 5, 5, 3, 3, 5, 3, 3, 3, 5, 5, 3, 5, 5, 5, 3, 5, 5, 3, 3, 5, 3, 5, 3, 5, 5, 3, 3, 3, 3, 5, 3, 3, 5, 5, 3, 5, 3, 5, 3, 3, 5, 5, 3, 5, 3, 3, 3, 3, 5, 5, 3, 5, 5, 5, 3, 3, 5, 5, 3, 3, 3, 3, 3, 5, 5, 3, 3, 5, 3, 3, 3, 3, 5, 5, 3, 3, 3, 5] flag2.gif: [3, 18, 18, 3, 18, 10, 3, 18, 15, 5, 13, 13, 3, 20, 3, 5, 13, 18, 3, 18, 18, 5, 13, 15, 3, 18, 3, 5, 13, 18, 5, 13, 5] flag3.gif: [10, 20, 18, 10, 18, 15, 18, 8, 18, 15, 18, 8, 10, 18, 10, 13, 18, 15, 18, 15, 10, 13] flag4.gif: [33, 43, 74, 59, 33, 66, 46, 59, 18, 23, 64, 10, 8, 38, 26, 66, 20, 3, 84, 84, 84, 84,", "# Problem #2070 2070 Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $R$ the area of the red square. Which of the following is correct? $[asy] unitsize(3mm); fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue); fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red); path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle; path divider=(-2,2)--(-3,3)--cycle; fill(onewhite,white); fill(rotate(90)*onewhite,white); fill(rotate(180)*onewhite,white); fill(rotate(270)*onewhite,white); [/asy]$ $\\text{(A)}\\ B = W \\qquad \\text{(B)}\\ W = R \\qquad \\text{(C)}\\ B = R \\qquad \\text{(D)}\\ 3B = 2R \\qquad \\text{(E)}\\ 2R = W$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# Difference between revisions of \"British Flag Theorem\" The British flag theorem says that if a point P is chosen inside rectangle ABCD then $AP^{2}+PC^{2}=BP^{2}+DP^{2}$. $[asy] size(200); pair A,B,C,D,P; A=(0,0); B=(200,0); C=(200,150); D=(0,150); P=(124,85); draw(A--B--C--D--cycle); label(\"A\",A,(-1,0)); dot(A); label(\"B\",B,(0,-1)); dot(B); label(\"C\",C,(1,0)); dot(C); label(\"D\",D,(0,1)); dot(D); dot(P); label(\"P\",P,(1,1)); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label(\"w\",(124,0),(0,-1)); label(\"x\",(200,85),(1,0)); label(\"y\",(124,150),(0,1)); label(\"z\",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85)); [/asy]$ The theorem also applies to points outside the rectangle, although the proof is harder to visualize in this case. ## Proof In Figure 1, by the Pythagorean theorem, we have: • $AP^{2} = Aw^{2} + Az^{2}$ • $PC^{2} = wB^{2} + zD^{2}$ • $BP^{2} = wB^{2} + Az^{2}$ • $PD^{2} = zD^{2} + Aw^{2}$ Therefore: • $AP^{2} + PC^{2} = Aw^{2} + Az^{2} + wB^{2} + zD^{2} = wB^{2} + Az^{2} + zD^{2} + Aw^{2} =\\nolinebreak BP^{2} +\\nolinebreak PD^{2}$ The above result i.e. theorem holds true even if the point P is selected on the boundary of rectangle or even outside the rectangle.", "# 14.6. Chapter Exercises¶ The code below is supposed to change the white in the French flag to aqua (Aqua is made by mixing blue and green with no red). Arrange and indent the blocks to make a correct program. You will not use them all. Write a program below to change all the greenish pixels in this image to the color: red 30, green 100, blue 150 (a slightly grayish blue). Hint: The greenish pixels are not a pure green. They have almost as much red as they do green. A sample color value might be: 111 red, 115 green, 65 blue. However, they are not all exactly the same color. You will need to figure out a recipe to select the right colors from the image. Write a program below that makes the white pixels in the right half of the image below turn black. The left half of the image should stay the same. Although perfect white is (255, 255, 255), the pixels in the image are not all perfectly white. Write your code to turn any pixel that has red, green, and blue values of above 230 to black (0, 0, 0). Hint: Start by trying to make all the white pixels in the image turn black. then add a condition to check the", "(i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5), white); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5), white); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5), white); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5), white); label(\"$b^2$\", (0.5,-2.5), white); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5), white); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5), white); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5), white); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] Sorry but I don't know how to show the grids. 1. 👍 2. 👎 3. 👁 1. each row is a geometric progression. Starting with n=0, row n is the GP b^n (1 + a + a^2 + ...) so the sum Sn of row n is b^n * 1/(1-a) Now you have another GP, which is the sum of all the row sums. That is 1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b) So the grand total of the whole grid is 1/((1-a)(1-b)) 1. 👍 2. 👎 2. wertwetrwetw 1. 👍 2. 👎 ## Similar Questions 1. ### alebra Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation then find the domain. (f*g)(x) a.2x^3-x^2-4x+3; domain:all real numbers b.2x^3+x^2-3x; domain: all real numbers c.2x^3+x^2+4x-3; domain: negative real numbers 2. ### Precalc To which set(s) of numbers does the number sqrt -16 belong? Select all that apply. real numbers complex numbers*** rational", "(-10,-12) -- (10,-12) -- ( 9,-8) -- ( 3, -4) -- cycle; \\fill [gray!10] (0,-10) -- (-1,-5) -- (-3,-5) -- (-2,-12) -- (2,-12) -- (3,-4) -- (-1,-5) -- cycle; \\fill [brown] (-3,7) -- (-5,6) -- (-6,-1) -- (-5,-3) -- (-4,-4) -- (4,-4) -- (5,-3) -- (6,-1) -- (5,6) -- (3,7) -- cycle; \\fill [pink!75!yellow!75!brown] (-2, 0) -- (-2,-4) -- (-2,-4) -- (-1,-5) -- (0,-10) -- (1,-5) -- ( 2,-4) -- ( 2,-4) -- ( 2, 0) -- cycle; \\fill [white] (-2,-4) -- (-3,-4) -- (-5,-6) -- (-1,-5) -- cycle; \\fill [white] ( 2,-4) -- ( 3,-4) -- ( 5,-6) -- ( 1,-5) -- cycle; \\fill [pink!75!yellow] (0, 5) -- (-4, 5) -- (-4,-1) -- (-1,-3) -- (1,-3) -- ( 4,-1) -- ( 4, 5) -- cycle; \\fill [brown] (-5,6) -- (-5,-2) -- (-4,2) -- (-3,5) -- (-1,4) -- (3,5) -- (4,2) -- (5,2) -- (5,6) -- cycle; \\end{tikzpicture} \\end{document} It's pretty easy to avoid sharp angles if you use .. controls +(<polar coordinate>) and +(<polar coordinate>) .. and make sure that the in and out angles are the same at each point. My attempt: \\documentclass[tikz, border=3pt]{standalone} \\begin{document} \\begin{tikzpicture}[thick] \\draw (0,0) circle (.53); \\draw (-130:.53) .. controls +(200:.5) and +(-45:-.3) .. (-1,-1.6) .. controls +(-45:.3) and +(45:-.3) .. (1,-1.6) .. controls +(45:.3) and +(-20:.5) .. (-50:.53); \\end{tikzpicture} \\end{document} This", "# Difference between revisions of \"British Flag Theorem\" The British flag theorem says that if a point P is chosen inside rectangle ABCD then $AP^{2}+PC^{2}=BP^{2}+DP^{2}$. $[asy] size(200); pair A,B,C,D,P; A=(0,0); B=(200,0); C=(200,150); D=(0,150); P=(124,85); draw(A--B--C--D--cycle); label(\"A\",A,(-1,0)); dot(A); label(\"B\",B,(0,-1)); dot(B); label(\"C\",C,(1,0)); dot(C); label(\"D\",D,(0,1)); dot(D); dot(P); label(\"P\",P,(1,1)); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label(\"w\",(124,0),(0,-1)); label(\"x\",(200,85),(1,0)); label(\"y\",(124,150),(0,1)); label(\"z\",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85)); [/asy]$ The theorem also applies if the point $P$ is selected outside or on the boundary of the rectangle, although the proof is harder to visualize in this case. ## Proof In Figure 1, by the Pythagorean theorem, we have: • $AP^{2} = Aw^{2} + Az^{2}$ • $PC^{2} = wB^{2} + zD^{2}$ • $BP^{2} = wB^{2} + Az^{2}$ • $PD^{2} = zD^{2} + Aw^{2}$ Therefore: • $AP^{2} + PC^{2} = Aw^{2} + Az^{2} + wB^{2} + zD^{2} = wB^{2} + Az^{2} + zD^{2} + Aw^{2} =\\nolinebreak BP^{2} +\\nolinebreak PD^{2}$", "back, (R = red, B = blue, O = orange, Y = yellow, G = green, and W = white). They are hinged together as shown, then folded to form a cube. The face opposite the white face is $[asy] draw((0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,2)--(2,2)--(2,3)--(0,3)--cycle); draw((1,3)--(1,2)--(2,2)--(2,1)--(3,1)--(3,2)); label($ $\\text{(A)}\\ \\text{B} \\qquad \\text{(B)}\\ \\text{G} \\qquad \\text{(C)}\\ \\text{O} \\qquad \\text{(D)}\\ \\text{R} \\qquad \\text{(E)}\\ \\text{Y}$ Jan 2: Three dice with faces numbered 1 through 6 are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is $[asy] draw((0,0)--(2,0)--(3,1)--(3,7)--(1,7)--(0,6)--cycle); draw((3,7)--(2,6)--(0,6)); draw((3,5)--(2,4)--(0,4)); draw((3,3)--(2,2)--(0,2)); draw((2,0)--(2,6)); dot((1,1)); dot((.5,.5)); dot((1.5,.5)); dot((1.5,1.5)); dot((.5,1.5)); dot((2.5,1.5)); dot((.5,2.5)); dot((1.5,2.5)); dot((1.5,3.5)); dot((.5,3.5)); dot((2.25,2.75)); dot((2.5,3)); dot((2.75,3.25)); dot((2.25,3.75)); dot((2.5,4)); dot((2.75,4.25)); dot((.5,5.5)); dot((1.5,4.5)); dot((2.25,4.75)); dot((2.5,5.5)); dot((2.75,6.25)); dot((1.5,6.5)); [/asy]$ $\\text{(A)}\\ 21 \\qquad \\text{(B)}\\ 22 \\qquad \\text{(C)}\\ 31 \\qquad \\text{(D)}\\ 41 \\qquad \\text{(E)}\\ 53$ Jan 3: Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom and sides. $[asy]/* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(1,2)--(1.5,2)); draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5)); draw((1.5,3.5)--(2.5,3.5)); draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5)); draw((3,4)--(3,3)--(2.5,2.5)); draw((3,3)--(4,3)--(4,2)--(3.5,1.5)); draw((4,3)--(3.5,2.5)); draw((2.5,.5)--(3,1)--(3,1.5));[/asy]$ $\\text{(A)}\\ 18 \\qquad \\text{(B)}\\ 24 \\qquad \\text{(C)}\\ 26 \\qquad \\text{(D)}\\ 30 \\qquad \\text{(E)}\\ 36$ Jan 4: Fourteen white cubes are put together to form the figure on the right. The complete surface of", "# Difference between revisions of \"2003 AMC 8 Problems/Problem 22\" ## Problem The following figures are composed of squares and circles. Which figure has a shaded region with largest area? $[asy]/* AMC8 2003 #22 Problem */ size(3inch, 2inch); unitsize(1cm); pen outline = black+linewidth(1); filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline); filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline); filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1)); filldraw(Circle((1,1), 1), white, outline); filldraw(Circle((3.5,.5), .5), white, outline); filldraw(Circle((4.5,.5), .5), white, outline); filldraw(Circle((3.5,1.5), .5), white, outline); filldraw(Circle((4.5,1.5), .5), white, outline); filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline); label(\"A\", (1, 2), N); label(\"B\", (4, 2), N); label(\"C\", (7, 2), N); draw((0,-.5)--(.5,-.5), BeginArrow); draw((1.5, -.5)--(2, -.5), EndArrow); label(\"2 cm\", (1, -.5)); draw((3,-.5)--(3.5,-.5), BeginArrow); draw((4.5, -.5)--(5, -.5), EndArrow); label(\"2 cm\", (4, -.5)); draw((6,-.5)--(6.5,-.5), BeginArrow); draw((7.5, -.5)--(8, -.5), EndArrow); label(\"2 cm\", (7, -.5)); draw((6,1)--(6,-.5), linetype(\"4 4\")); draw((8,1)--(8,-.5), linetype(\"4 4\"));[/asy]$ $\\textbf{(A)}\\ \\text{A only}\\qquad\\textbf{(B)}\\ \\text{B only}\\qquad\\textbf{(C)}\\ \\text{C only}\\qquad\\textbf{(D)}\\ \\text{both A and B}\\qquad\\textbf{(E)}\\ \\text{all are equal}$ ## Solution First we have to find the area of the shaded region in each of the figures. In figure $\\textbf{A}$ the area of the shaded region is the area of the circle subtracted from the area of the square. That is $2^2-1^2 \\pi=4-\\pi$. In figure $\\textbf{B}$ the area of the shaded region is the sum of the areas of the 4 circles subtracted from the area of the square. That is $2^2-4((\\frac{1}{2})^2 \\pi)=4-4(\\frac{\\pi}{4})=4-\\pi$. In figure $\\textbf{C}$ the area of", "on both tests? $[asy] draw((2,0)--(7,0)--(7,5)--(2,5)--cycle); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((6,0)--(6,5)); draw((2,1)--(7,1)); draw((2,2)--(7,2)); draw((2,3)--(7,3)); draw((2,4)--(7,4)); draw((.2,6.8)--(1.8,5.2)); draw(circle((4.5,1.5),.5),linewidth(.6 mm)); label(\"0\",(2.5,.2),N); label(\"0\",(3.5,.2),N); label(\"2\",(4.5,.2),N); label(\"1\",(5.5,.2),N); label(\"0\",(6.5,.2),N); label(\"0\",(2.5,1.2),N); label(\"0\",(3.5,1.2),N); label(\"1\",(4.5,1.2),N); label(\"1\",(5.5,1.2),N); label(\"1\",(6.5,1.2),N); label(\"1\",(2.5,2.2),N); label(\"3\",(3.5,2.2),N); label(\"5\",(4.5,2.2),N); label(\"2\",(5.5,2.2),N); label(\"0\",(6.5,2.2),N); label(\"1\",(2.5,3.2),N); label(\"4\",(3.5,3.2),N); label(\"3\",(4.5,3.2),N); label(\"0\",(5.5,3.2),N); label(\"0\",(6.5,3.2),N); label(\"2\",(2.5,4.2),N); label(\"2\",(3.5,4.2),N); label(\"1\",(4.5,4.2),N); label(\"0\",(5.5,4.2),N); label(\"0\",(6.5,4.2),N); label(\"F\",(1.5,.2),N); label(\"D\",(1.5,1.2),N); label(\"C\",(1.5,2.2),N); label(\"B\",(1.5,3.2),N); label(\"A\",(1.5,4.2),N); label(\"A\",(2.5,5.2),N); label(\"B\",(3.5,5.2),N); label(\"C\",(4.5,5.2),N); label(\"D\",(5.5,5.2),N); label(\"F\",(6.5,5.2),N); label(\"Test 1\",(-.5,5.2),N); label(\"Test 2\",(2.6,6),N); [/asy]$ $\\text{(A)}\\ 12\\% \\qquad \\text{(B)}\\ 25\\% \\qquad \\text{(C)}\\ 33\\frac{1}{3}\\% \\qquad \\text{(D)}\\ 40\\% \\qquad \\text{(E)}\\ 50\\%$ ## Problem 13 The perimeter of the polygon shown is $[asy] draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label(\"6\",(0,3),W); label(\"8\",(4,6),N); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((0.5,6)--(0.5,5.5)--(0,5.5)); draw((7.5,6)--(7.5,5.5)--(8,5.5)); draw((7.5,3)--(7.5,3.5)--(8,3.5)); draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); [/asy]$ $\\text{(A)}\\ 14 \\qquad \\text{(B)}\\ 20 \\qquad \\text{(C)}\\ 28 \\qquad \\text{(D)}\\ 48$ $\\text{(E)}\\ \\text{cannot be determined from the information given}$ ## Problem 14 If $200\\leq a \\leq 400$ and $600\\leq b\\leq 1200$, then the largest value of the quotient $\\frac{b}{a}$ is $\\text{(A)}\\ \\frac{3}{2} \\qquad \\text{(B)}\\ 3 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 300 \\qquad \\text{(E)}\\ 600$ ## Problem 15 Sale prices at the Ajax Outlet Store are $50\\%$ below original prices. On Saturdays an additional discount of $20\\%$ off the sale price is given. What is the Saturday price of a coat whose original price is <dollar/>$180$? $\\text{(A)}$ <dollar/>$54$ $\\text{(B)}$ <dollar/>$72$ $\\text{(C)}$ <dollar/>$90$ $\\text{(D)}$ <dollar/>$108$ $\\text{(D)}$ <dollar/>$110$ ## Problem 16 A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar", "# Problem #2099 2099 The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing? $[asy] unitsize(10mm); defaultpen(fontsize(10pt)); pen finedashed=linetype(\"4 4\"); filldraw((1,1)--(2,1)--(2,2)--(4,2)--(4,3)--(1,3)--cycle,grey,black+linewidth(.8pt)); draw((0,1)--(0,3)--(1,3)--(1,4)--(4,4)--(4,3)-- (5,3)--(5,2)--(4,2)--(4,1)--(2,1)--(2,0)--(1,0)--(1,1)--cycle,finedashed); draw((0,2)--(2,2)--(2,4),finedashed); draw((3,1)--(3,4),finedashed); label(\"1\",(1.5,0.5)); draw(circle((1.5,0.5),.17)); label(\"2\",(2.5,1.5)); draw(circle((2.5,1.5),.17)); label(\"3\",(3.5,1.5)); draw(circle((3.5,1.5),.17)); label(\"4\",(4.5,2.5)); draw(circle((4.5,2.5),.17)); label(\"5\",(3.5,3.5)); draw(circle((3.5,3.5),.17)); label(\"6\",(2.5,3.5)); draw(circle((2.5,3.5),.17)); label(\"7\",(1.5,3.5)); draw(circle((1.5,3.5),.17)); label(\"8\",(0.5,2.5)); draw(circle((0.5,2.5),.17)); label(\"9\",(0.5,1.5)); draw(circle((0.5,1.5),.17));[/asy]$ $\\mathrm{(A) \\ } 2\\qquad \\mathrm{(B) \\ } 3\\qquad \\mathrm{(C) \\ } 4\\qquad \\mathrm{(D) \\ } 5\\qquad \\mathrm{(E) \\ } 6$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "flag's .check value is only 64 when idx is 64. r and c seems to correspond with coordinates on the map. Let's try to calculate the coordinate of the flag when idx is 64. var r = Math.floor(64 / 5) + 1; var c = (64 % 5) + 1; Therefore, r is 13 and c is 5. Let's try to plot this on a map. [ -4, -3, -3, -3, -3, -3, -2], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, flag, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -5, 1, 1, stair, 1, 1, -1], [ -5, 1, 1, 1, 1, 1, -1], [ -6, -7, -7, -7, -7, -7, -8]", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "(6.5,0.5)); label(\"\\dots\", (1.5,0.5)); label(\"\\dots\", (0.5,1.5)); label(\"\\dots\", (0.5,2.5)); label(\"\\dots\", (0.5,3.5)); label(\"\\dots\", (0.5,4.5)); label(\"\\dots\", (0.5,5.5)); label(\"\\dots\", (0.5,6.5)); label(\"\\dots\", (6.5,0.5)); label(\"\\dots\", (6.5,1.5)); label(\"\\dots\", (6.5,2.5)); label(\"\\dots\", (6.5,3.5)); label(\"\\dots\", (6.5,4.5)); label(\"\\dots\", (6.5,5.5)); label(\"\\dots\", (0.5,6.5)); label(\"\\dots\", (1.5,6.5)); label(\"\\dots\", (2.5,6.5)); label(\"\\dots\", (3.5,6.5)); label(\"\\dots\", (4.5,6.5)); label(\"\\dots\", (5.5,6.5)); label(\"\\dots\", (6.5,6.5)); label(\"17\", (1.5,1.5)); label(\"18\", (1.5,2.5)); label(\"19\", (1.5,3.5)); label(\"20\", (1.5,4.5)); label(\"21\", (1.5,5.5)); label(\"16\", (2.5,1.5)); label(\"5\", (2.5,2.5)); label(\"6\", (2.5,3.5)); label(\"7\", (2.5,4.5)); label(\"22\", (2.5,5.5)); label(\"15\", (3.5,1.5)); label(\"4\", (3.5,2.5)); label(\"1\", (3.5,3.5)); label(\"8\", (3.5,4.5)); label(\"23\", (3.5,5.5)); label(\"14\", (4.5,1.5)); label(\"3\", (4.5,2.5)); label(\"2\", (4.5,3.5)); label(\"9\", (4.5,4.5)); label(\"24\", (4.5,5.5)); label(\"13\", (5.5,1.5)); label(\"12\", (5.5,2.5)); label(\"11\", (5.5,3.5)); label(\"10\", (5.5,4.5)); label(\"25\", (5.5,5.5)); [/asy]$ $\\textbf{(A)} ~367 \\qquad\\textbf{(B)} ~368 \\qquad\\textbf{(C)} ~369 \\qquad\\textbf{(D)} ~379 \\qquad\\textbf{(E)} ~380$ ## Problem 9 The point $P(a,b)$ in the $xy$-plane is first rotated counterclockwise by $90^\\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$. The image of $P$ after these two transformations is at $(-6,3)$. What is $b - a ?$ $\\textbf{(A)} ~1 \\qquad\\textbf{(B)} ~3 \\qquad\\textbf{(C)} ~5 \\qquad\\textbf{(D)} ~7 \\qquad\\textbf{(E)} ~9$ ## Problem 10 An inverted cone with base radius $12 \\text{cm}$ and height $18\\text{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of $24\\text{cm}$. What is the height in centimeters of the water in the cylinder? $\\textbf{(A) }1.5 \\qquad \\textbf{(B) }3 \\qquad \\textbf{(C) }4 \\qquad \\textbf{(D) }4.5 \\qquad \\textbf{(E) }6$ ##", "(6.5,0.5)); label(\"\\dots\", (1.5,0.5)); label(\"\\dots\", (0.5,1.5)); label(\"\\dots\", (0.5,2.5)); label(\"\\dots\", (0.5,3.5)); label(\"\\dots\", (0.5,4.5)); label(\"\\dots\", (0.5,5.5)); label(\"\\dots\", (0.5,6.5)); label(\"\\dots\", (6.5,0.5)); label(\"\\dots\", (6.5,1.5)); label(\"\\dots\", (6.5,2.5)); label(\"\\dots\", (6.5,3.5)); label(\"\\dots\", (6.5,4.5)); label(\"\\dots\", (6.5,5.5)); label(\"\\dots\", (0.5,6.5)); label(\"\\dots\", (1.5,6.5)); label(\"\\dots\", (2.5,6.5)); label(\"\\dots\", (3.5,6.5)); label(\"\\dots\", (4.5,6.5)); label(\"\\dots\", (5.5,6.5)); label(\"\\dots\", (6.5,6.5)); label(\"17\", (1.5,1.5)); label(\"18\", (1.5,2.5)); label(\"19\", (1.5,3.5)); label(\"20\", (1.5,4.5)); label(\"21\", (1.5,5.5)); label(\"16\", (2.5,1.5)); label(\"5\", (2.5,2.5)); label(\"6\", (2.5,3.5)); label(\"7\", (2.5,4.5)); label(\"22\", (2.5,5.5)); label(\"15\", (3.5,1.5)); label(\"4\", (3.5,2.5)); label(\"1\", (3.5,3.5)); label(\"8\", (3.5,4.5)); label(\"23\", (3.5,5.5)); label(\"14\", (4.5,1.5)); label(\"3\", (4.5,2.5)); label(\"2\", (4.5,3.5)); label(\"9\", (4.5,4.5)); label(\"24\", (4.5,5.5)); label(\"13\", (5.5,1.5)); label(\"12\", (5.5,2.5)); label(\"11\", (5.5,3.5)); label(\"10\", (5.5,4.5)); label(\"25\", (5.5,5.5)); [/asy]$ $\\textbf{(A)} ~367 \\qquad\\textbf{(B)} ~368 \\qquad\\textbf{(C)} ~369 \\qquad\\textbf{(D)} ~379 \\qquad\\textbf{(E)} ~380$ ## Problem 9 The point $P(a,b)$ in the $xy$-plane is first rotated counterclockwise by $90^\\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$. The image of $P$ after these two transformations is at $(-6,3)$. What is $b - a ?$ $\\textbf{(A)} ~1 \\qquad\\textbf{(B)} ~3 \\qquad\\textbf{(C)} ~5 \\qquad\\textbf{(D)} ~7 \\qquad\\textbf{(E)} ~9$ ## Problem 10 An inverted cone with base radius $12 \\text{cm}$ and height $18\\text{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of $24\\text{cm}$. What is the height in centimeters of the water in the cylinder? $\\textbf{(A) }1.5 \\qquad \\textbf{(B) }3 \\qquad \\textbf{(C) }4 \\qquad \\textbf{(D) }4.5 \\qquad \\textbf{(E) }6$ ##", "# Dutch National Flag The Dutch national flag problem is a famous computer science related programming problem proposed by Edsger Dijkstra. The flag of the Netherlands consists of three colours: red, white and blue. Given balls of these three colours arranged randomly in a line (the actual number of balls does not matter), the task is to arrange them such that all balls of the same colour are together and their collective colour groups are in the correct order. Here is a problem that is stimulated by the idea: Q. Write a function that takes an array A and index i into A, and rearranges the elements such that all elements less than A[i] appear first, followed by elements equal to A[i], followed by elements greater than A[i]. Your algorithm should have O(1) space complexity and O(|A|) time complexity. Soln: ```#Function to swap two variables def swapNormal(A,pos1,pos2): t = 0 t=A[pos1] A[pos1] = A[pos2] A[pos2] = t #initializing an array of randomly generated numbers j = 0 while j < 10: A.append(random.randint(1,100)) j+= 1 print \"The array of numbers is:\",A #asking for the position of the pivotal element i = input(\"Please enter the index number\") pivot = A[i] print \"The pivot element is:\",pivot def dutchNationalFlag(A,pivot): equal = 0 small = 0 large = len(A)-1 while (equal <= large):", "# Dutch National Flag The Dutch national flag problem is a famous computer science related programming problem proposed by Edsger Dijkstra. The flag of the Netherlands consists of three colours: red, white and blue. Given balls of these three colours arranged randomly in a line (the actual number of balls does not matter), the task is to arrange them such that all balls of the same colour are together and their collective colour groups are in the correct order. Here is a problem that is stimulated by the idea: Q. Write a function that takes an array A and index i into A, and rearranges the elements such that all elements less than A[i] appear first, followed by elements equal to A[i], followed by elements greater than A[i]. Your algorithm should have O(1) space complexity and O(|A|) time complexity. Soln: ```#Function to swap two variables def swapNormal(A,pos1,pos2): t = 0 t=A[pos1] A[pos1] = A[pos2] A[pos2] = t #initializing an array of randomly generated numbers j = 0 while j < 10: A.append(random.randint(1,100)) j+= 1 print \"The array of numbers is:\",A #asking for the position of the pivotal element i = input(\"Please enter the index number\") pivot = A[i] print \"The pivot element is:\",pivot def dutchNationalFlag(A,pivot): equal = 0 small = 0 large = len(A)-1 while (equal <= large):", "outline); filldraw(Circle((3.5,.5), .5), white, outline); filldraw(Circle((4.5,.5), .5), white, outline); filldraw(Circle((3.5,1.5), .5), white, outline); filldraw(Circle((4.5,1.5), .5), white, outline); filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline); label(\"A\", (1, 2), N); label(\"B\", (4, 2), N); label(\"C\", (7, 2), N); draw((0,-.5)--(.5,-.5), BeginArrow); draw((1.5, -.5)--(2, -.5), EndArrow); label(\"2 cm\", (1, -.5)); draw((3,-.5)--(3.5,-.5), BeginArrow); draw((4.5, -.5)--(5, -.5), EndArrow); label(\"2 cm\", (4, -.5)); draw((6,-.5)--(6.5,-.5), BeginArrow); draw((7.5, -.5)--(8, -.5), EndArrow); label(\"2 cm\", (7, -.5)); draw((6,1)--(6,-.5), linetype(\"4 4\")); draw((8,1)--(8,-.5), linetype(\"4 4\"));[/asy]$ $\\textbf{(A)}\\ \\text{A only}\\qquad\\textbf{(B)}\\ \\text{B only}\\qquad\\textbf{(C)}\\ \\text{C only}\\qquad\\textbf{(D)}\\ \\text{both A and B}\\qquad\\textbf{(E)}\\ \\text{all are equal}$ ## Problem 23 In the pattern below, the cat (denoted as a large circle in the figures below) moves clockwise through the four squares and the mouse (denoted as a dot in the figures below) moves counterclockwise through the eight exterior segments of the four squares. $[asy]defaultpen(linewidth(0.8)); size(350); path p=unitsquare; int i; for(i=0; i<5; i=i+1) { draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); } path cat=Circle((0.5,0.5), 0.3); draw(shift(0,1)*cat^^shift(4,1)*cat^^shift(7,0)*cat^^shift(9,0)*cat^^shift(12,1)*cat); dot((1.5,0)^^(5,0.5)^^(8,1.5)^^(10.5,2)^^(12.5,2)); label(\"1\", (1,2), N); label(\"2\", (4,2), N); label(\"3\", (7,2), N); label(\"4\", (10,2), N); label(\"5\", (13,2), N); [/asy]$ If the pattern is continued, where would the cat and mouse be after the 247th move? $\\textbf{(A)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(1,0)*cat); dot((0,0.5)); [/asy]$ $\\textbf{(B)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(1,1)*cat); dot((0,0.5)); [/asy]$ $\\textbf{(C)}$ $[asy]defaultpen(linewidth(0.8)); size(60); path p=unitsquare; int i=0; draw(shift(3i,0)*(p^^shift(1,0)*p^^shift(0,1)*p^^shift(1,1)*p)); path cat=Circle((0.5,0.5), 0.3); draw(shift(1,0)*cat); dot((0,1.5));", "contains 200 rows. Each row is a list containing 136 pixels. Each pixel is a list of form [r,g,b], where r, g, and b are red, blue, and green intensity values from 0 to 255. Try the following expressions: len(b) len(b[0]) b[0][0] b[100][100] Knowing that an image is a list of rows, write a function flip_image(image) that creates a new image that is image flipped upside-down and plots the result. Put this function in flip_image.py Then try flipping the nautical alphabet image upside-down. ### 1.2 Student Activities 1. In the file c_flag.py, write a function c_flag to create a little 5-by-6 pixel image encoding the flag for the letter C in the nautical flag alphabet. Refer to the RGB Color Table to get the codes for red and navy blue. Since this image is very small, you will want to magnify it for display purposes by writing draw_image(c, c_flag(), 0, 0, 20). 2. You may choose to do either 2 or 3. Your choice. 3. In the file swap_bg.py, write a function swap_bg(image) that swaps the blue and green values of every pixel in the image, and plots the result. Test it on the nautical flag image. What happens if you call swap_bg repeatedly on the same image? 4. In the file grayscale.py, write a function grayscale(image) that", "label(\"N\",(5.5,2.5),blue); label(\"N\",(6.5,2.5),blue); label(\"N\",(4.5,1.5),blue); label(\"N\",(5.5,1.5),blue); label(\"N\",(6.5,1.5),blue); label(\"N\",(7.5,1.5),blue); label(\"N\",(5.5,0.5),blue); label(\"N\",(6.5,0.5),blue); label(\"N\",(7.5,0.5),blue); label(\"Y\",(8.5,0.5),blue); label(\"Y\",(10.5,2.5),blue); label(\"N\",(11.5,2.5),blue); label(\"N\",(12.5,2.5),blue); label(\"N\",(13.5,2.5),blue); label(\"N\",(11.5,1.5),blue); label(\"N\",(12.5,1.5),blue); label(\"N\",(13.5,1.5),blue); label(\"Y\",(14.5,1.5),blue); label(\"Y\",(17.5,2.5),blue); label(\"Y\",(18.5,2.5),blue); label(\"Y\",(19.5,2.5),blue); label(\"N\",(20.5,2.5),blue); label(\"N\",(20.5,1.5),blue); label(\"Y\",(21.5,1.5),blue); label(\"Y\",(22.5,1.5),blue); label(\"Y\",(23.5,1.5),blue); label(\"Y\",(26.5,2.5),blue); label(\"N\",(27.5,2.5),blue); label(\"N\",(28.5,2.5),blue); label(\"N\",(29.5,2.5),blue); label(\"N\",(27.5,1.5),blue); label(\"N\",(28.5,1.5),blue); label(\"N\",(29.5,1.5),blue); label(\"Y\",(30.5,1.5),blue); label(\"Y\",(32.5,2.5),blue); label(\"N\",(33.5,2.5),blue); label(\"N\",(34.5,2.5),blue); label(\"N\",(35.5,2.5),blue); label(\"N\",(33.5,1.5),blue); label(\"N\",(34.5,1.5),blue); label(\"N\",(35.5,1.5),blue); label(\"N\",(36.5,1.5),blue); label(\"N\",(34.5,0.5),blue); label(\"N\",(35.5,0.5),blue); label(\"N\",(36.5,0.5),blue); label(\"Y\",(37.5,0.5),blue); for (real i=0; i<=42; ++i) { draw((i,4)--(i,3)); } draw((0,4)--(42,4)); draw((0,3)--(42,3)); [/asy]$ 2. As a confirmation, we can verify that each student will be able to deduce what $S$ is upon hearing the four replies of \"no\" in unison. For example, if $S=\\{47,48,49,50\\},$ then all students will know that no one gets $46$ or $51,$ otherwise that student will reply yes (as discussed). Therefore, all students will conclude that $S$ has only one possibility. ~MRENTHUSIASM" ]

[ "= 50^{\\circ}$?", ".25^2$ $P(x=17)=\\dbinom{20}{3}.75^{17}\\times .25^3$ add these up 15. anonymous Ah, alright! 16. anonymous this is what i get, but check the calculations http://www.wolframalpha.com/input/?i=.75^20%2B20*.75^19*.25%2B190*.75^18*.25^2%2B1140*.75^17*.25^3 17. anonymous Yeah. Same! Thank you very much! Find more explanations on OpenStudy", "\\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} & 360 \\cdot x^{44} & 360 \\cdot x^{45} & 120 \\cdot x^{46}\\\\ & & & &", "(\\mathbf{A}\\mathbf{B})^{-1} \\dbinom12 = \\dbinom13 } \\\\ \\hline \\end{array}$$ Apr 17, 2020", "can run the same method to compute $$a\\pmod{11!}$$.", "+0 # Help. I have been stuck 0 114 4 Given that $$\\binom{15}{8}=6435$$$$\\binom{16}{9}=11440$$, and $$\\binom{16}{10}=8008$$, find $$\\binom{15}{10}$$. I know that I should be able to do this but I can't figure it out. Aug 10, 2019 #1 +23295 +3 Given that $$\\dbinom{15}{8}=6435$$, $$\\dbinom{16}{9}=11440$$ , and $$\\dbinom{16}{10}=8008$$, find $$\\dbinom{15}{10}$$. Formula: $$\\dbinom{n}{k} + \\dbinom{n}{k+1} = \\dbinom{n+1}{k+1}$$ $$\\begin{array}{|lrcll|} \\hline & \\mathbf{\\dbinom{n}{k} + \\dbinom{n}{k+1}} &=& \\mathbf{\\dbinom{n+1}{k+1}} \\\\\\\\ (1) & \\dbinom{15}{8} + \\dbinom{15}{9} &=& \\dbinom{16}{9} \\\\ (2) & \\dbinom{15}{9} + \\dbinom{15}{10} &=& \\dbinom{16}{10} \\\\ \\hline (1)-(2): & \\dbinom{15}{8} + \\dbinom{15}{9} -\\left( \\dbinom{15}{9} + \\dbinom{15}{10}\\right) &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & \\dbinom{15}{8} - \\dbinom{15}{10} &=& \\dbinom{16}{9}-\\dbinom{16}{10} \\\\ & 6435 - \\dbinom{15}{10} &=& 11440-11440 \\\\ & 6435 - \\dbinom{15}{10} &=& 3432 \\\\ & \\dbinom{15}{10} &=& 6435 - 3432 \\\\ & \\mathbf{\\dbinom{15}{10}} &=& \\mathbf{3003} \\\\ \\hline \\end{array}$$ Aug 11, 2019 #2 0 Thank you so much!! I can't keep all the formulas in my head sometimes. Aug 11, 2019 #3 +1655 +3 There are really too many formulas... tommarvoloriddle Aug 12, 2019 #4 +105540 +1 You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from. There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level. Most times this does", "the 50 coming from? 50 is not part of the problem, and there is no division involved. $d = \\sqrt{300^2 + 400^2} $ calculate it.", ")^{10} \\\\\\\\ && ( a + b + c )^{n} \\\\ &=& \\sum \\limits_{k=0}^{n} \\sum \\limits_{i=0}^{k} \\dbinom{n}{k} \\dbinom{k}{i} a^{n-k} b^{k-i} c^{i} \\qquad | \\qquad (n-k)+(k-i)+(i) = n \\\\\\\\ && ( x^4 + x^5 + x^6 )^{10} \\\\ &=& \\sum \\limits_{k=0}^{10} \\sum \\limits_{i=0}^{k} \\dbinom{10}{k} \\dbinom{k}{i} (x^4)^{10-k} (x^5)^{k-i} (x^6)^{i} \\qquad | \\qquad (10-k)+(k-i)+(i) = 10 \\\\\\\\ \\end{array}$$ $$\\begin{array}{lcll} (k=0,i=0): & \\dbinom{10}{0} \\dbinom{0}{0} (x^4)^{10} (x^5)^{0} (x^6)^{0} &=& x^{40} \\\\ \\hline (k=1,i=0): & \\dbinom{10}{1} \\dbinom{1}{0} (x^4)^{9} (x^5)^{1} (x^6)^{0} &=& 10 \\cdot x^{41} \\\\ (k=1,i=1): & \\dbinom{10}{1} \\dbinom{1}{1} (x^4)^{9} (x^5)^{0} (x^6)^{1} &=& 10 \\cdot x^{42} \\\\ \\hline (k=2,i=0): & \\dbinom{10}{2} \\dbinom{2}{0} (x^4)^{8} (x^5)^{2} (x^6)^{0} &=& 45 \\cdot x^{42} \\\\ (k=2,i=1): & \\dbinom{10}{2} \\dbinom{2}{1} (x^4)^{8} (x^5)^{1} (x^6)^{1} &=& 90 \\cdot x^{43} \\\\ (k=2,i=2): & \\dbinom{10}{2} \\dbinom{2}{2} (x^4)^{8} (x^5)^{0} (x^6)^{2} &=& 45 \\cdot x^{44} \\\\ \\hline (k=3,i=0): & \\dbinom{10}{3} \\dbinom{3}{0} (x^4)^{7} (x^5)^{3} (x^6)^{0} &=& 120 \\cdot x^{43} \\\\ (k=3,i=1): & \\dbinom{10}{3} \\dbinom{3}{1} (x^4)^{7} (x^5)^{2} (x^6)^{1} &=& 360 \\cdot x^{44} \\\\ (k=3,i=2): & \\dbinom{10}{3} \\dbinom{3}{2} (x^4)^{7} (x^5)^{1} (x^6)^{2} &=& 360 \\cdot x^{45} \\\\ (k=3,i=3): & \\dbinom{10}{3} \\dbinom{3}{3} (x^4)^{7} (x^5)^{0} (x^6)^{3} &=& 120 \\cdot x^{46} \\\\ \\cdots \\end{array}$$ $$\\small{ \\begin{array}{l||r|r|r|r|r|r|r|} &x^{40} & \\\\ & & 10 \\cdot x^{41} & 10 \\cdot x^{42} &\\\\ & & & 45 \\cdot x^{42} & 90 \\cdot x^{43} & 45 \\cdot x^{44} & \\\\ & & & & 120 \\cdot x^{43} &", "+ 30^2}$$ Fr = 50 N", "result $\\boxed{492}$.", "+ 50 \\,=\\,75$$ quadrilaterals.", "50 - 25 = 25{\\text{ }}{m^2}$.", "## Intermediate Algebra for College Students (7th Edition) $x^{10}-5x^8+10x^6-10x^4+5x^2-1$ Apply Binomial Theorem. $(a+b)^n=\\sum_{r=0}^n\\dbinom{n}{r}(a)^{n-r}b^r$ $(x^2-1)^5=\\dbinom{5}{0}(x^2)^5(-1)^0+\\dbinom{5}{1}(x^2)^4(-1)^1+\\dbinom{5}{2}(x^2)^3(-1)^2+\\dbinom{5}{3}(x^2)^2(-1)^3+\\dbinom{5}{4}(x^2)^1(-1)^4+\\dbinom{5}{5}(x^2)^0(-1)^5$ or, $=x^{10}-5x^8+10x^6+10x^4(-1)+5x^2+(1)(1)(-1)$ or, $=x^{10}-5x^8+10x^6-10x^4+5x^2-1$", "of $$\\mathbf{3}$$ is $$\\mathbf{53}$$ heureka Sep 18, 2018", "50\\sqrt{2} \\;=\\;300\\sqrt{2}\\text{ m}^2$", "= \\{48\\}$$", "is $$\\mathbf{0}$$.", "1)}{2 - 1}$$ = 2(31) = 62", "# Thread: dbl int and desmos graoh 1. ## dbl int and desmos graoh $\\tiny{15.3.53}\\\\ \\displaystyle \\int_{0}^{\\frac{\\pi}{6}} \\int_{0}^{\\sec{\\theta}} 6r^{3} drd\\theta$ just want to see if this desmos graph is correct wanted to shade the wedge area but ??? also got $0.96225$ for calulated answer but they wanted exact", "## Numbers Greater Than 5,000,000 Using all the digits $3,4,5,5,5,6$ and $6$, how many distinct integers greater than $5,\\!000,\\!000$ can be formed? Source: NCTM Mathematics Teacher, January 2006 Solution The first digit must be a $5$ or a $6$. Case 1: $5$ _ _ _ _ _ _ Since copies of the same digits cannot be distinguished from one another, once we have positions to put them in we can copy them into the positions in any order. We construct an ordering for the digits $3,4,5,5,6,6$ in a four-step process 1) choose a subset of $2$ positions for the two $6$’s 2) choose a subset of $2$ positions for the two $5$’s 3) choose a subset of $1$ position for the $4$ 4) choose a subset of $1$ position for the $3$ $\\dbinom{6}{2}\\dbinom{4}{2}\\dbinom{2}{1}\\dbinom{1}{1}=180$ Case 2: $6$ _ _ _ _ _ _ Similarly, we construct an ordering for the digits $3,4,5,5,5,6$ 1) choose a subset of $1$ position for the $6$ 2) choose a subset of $3$ positions for the $5$’s 3) choose a subset of $1$ position for the $4$ 4) choose a subset of $1$ position for the $3$ $\\dbinom{6}{1}\\dbinom{5}{3}\\dbinom{2}{1}\\dbinom{1}{1}=120$ $180+120=300$ $300$ distinct integers greater than $5,\\!000,\\!000$ can be formed using all the digits $3,4,5,5,5,6$, and $6$. Answer: $300$" ]

[ "2\\dbinom 21}{\\dbinom 52}=\\dfrac{3}{5}$$", "$\\dfrac{x+y}{2}$, the desired answer is $\\boxed{450}$.", "is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$.", ". . . . . $=\\; \\dfrac{60}{9.8} \\;=\\;\\dfrac{300}{49} \\;=\\;6\\frac{6}{49}\\text{ seconds.}$", "= \\dfrac{{13 \\times 12! \\times 4 \\times 3! \\times 50! \\times 2 \\times 1}}{{52 \\times 51 \\times 50! \\times 12! \\times 3!}} \\\\ P = \\dfrac{{13 \\times 4 \\times 2}}{{52 \\times 51}} = \\dfrac{2}{{51}} \\\\$", "31+x = 42\\\\ x = \\boxed{{\\textbf{(D)}\\ 11}}$", "gives us $\\dbinom{n+1}{2}$ ways, for a total of $3\\dbinom{n+1}{2}$ ways. The last case gives us $\\dbinom{n+1}{1}=n+1$ ways. Therefore, the total number of possible ways where there are no isolated men is $\\[\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1).\\]$ The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or $\\[2\\dbinom{n+1}{2}+(n+1).\\]$ Thus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that $\\[\\dfrac{2\\dbinom{n+1}{2}+(n+1)}{\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1)}\\le\\dfrac{1}{100}.\\]$ After simplification, we arrive at $\\[\\dfrac{6(n+1)}{n^2+8n+6}\\le\\dfrac{1}{100}.\\]$ Simplifying again, we see that we seek the smallest positive integer value of $n$ such that $n(n-592)\\ge594$. Clearly $n>592$, or the left side will not even be positive; we quickly see that $n=593$ is too small but $n=\\boxed{594}$ satisfies the inequality.", "i.e. numbering $52$ Thus $Pr =$ $\\dbinom{52}{2} \\over \\dbinom{157}{2}$", "\\dfrac{570}{3^8} = \\dfrac{190}{3^{7}}$. Our answer is the numerator, $\\boxed{190}$.", "- 36 = \\boxed{\\textbf{(E)}\\; 64}$", "\\dfrac{40}{42}$$ giving us the desired answer of $20+21=\\boxed{41}$.", "\\dfrac{629988.3164 \\times 9 \\times 10^{-4}}{0.5}$ N = 1133.98 = 1134 T.", "Die kombinatorische Analysis, als Vorbereitungslehre zum Studium der theoretischen höheren Mathematik. It appears to have become the de facto standard in recent years. As a result, $\\dbinom n k$ is frequently voiced the binomial coefficient $n$ over $k$. Other notations include: $C \\left({n, k}\\right)$ ${}^n C_k$ ${}_n C_k$ $C^n_k$ ${C_n}^k$ all of which can cause a certain degree of confusion. ## Examples ### $2$ from $5$ The number of ways of choosing $2$ objects from a set of $5$ is: $\\dbinom 5 2 = \\dfrac {5 \\times 4} {2 \\times 1} = \\dfrac {5!} {2! \\, 3!} = 10$ ### $3$ from $7$ The number of ways of choosing $3$ objects from a set of $7$ is: $\\dbinom 7 3 = \\dfrac {7 \\times 6 \\times 5} {3 \\times 2 \\times 1} = \\dfrac {7!} {3! \\, 4!} = 35$ ### $3$ from $8$ The number of ways of choosing $3$ objects from a set of $8$ is: $\\dbinom 8 3 = \\dfrac {8 \\times 7 \\times 6} {3 \\times 2 \\times 1} = \\dfrac {8!} {3! \\, 5!} = 56$ ### $4$ from $52$ The number of ways of choosing $4$ objects from a set of $52$ (for example, cards from a deck) is: $\\dbinom {52} 4 = \\dfrac {52 \\times 51 \\times 50 \\times 49} {4", "pair, $\\dfrac{255}{495}=0.515151515151$. We all agree. Look at this totally different approach.", "\\dfrac{\\dbinom{26}{2}\\dbinom{26}{1}}{\\dbinom{52}{3}} = \\dfrac{13}{17}$$ Rom Dec 31, 2018", "\\dfrac{(x+1)(x+4)}{(x+3)(x+3)} \\cdot \\dfrac{(x+3)(x-3)}{x+1} = \\boxed{\\dfrac{x^2+x-12}{x+3}} $ Ah.....oops! Thank you! How did I forget that! Thank you again for your help earboth! drakmord", "= \\boxed{\\textbf{(B)}\\ 24}.$", "@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19 @AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22 Nope, that seems likely to be the most concise way to do it. Count ways to pick $n\\in\\{1,2,3,4\\}$ unique numbers, and to arrange them with $n-1$ \"$>$\" signs, to meet the criteria. $$\\begin{array}{|l:l|} \\hline \\rm a{=}a{=}a{=}a & \\dbinom 3 0 \\dbinom 6 1 \\\\\\hdashline\\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \\dbinom 3 1\\dbinom 6 2 \\\\\\hdashline\\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \\dbinom 3 2 \\dbinom 6 3 \\\\\\hdashline\\rm a{>}b{>}c{>}d & \\dbinom 3 3 \\dbinom 6 4 \\\\ \\hline\\end{array}$$ $$\\dfrac{\\sum_{n=1}^4 \\dbinom{3}{n-1}\\dbinom{6}{n}}{6^4} = \\dfrac{\\dbinom 6 1 + 3\\dbinom 6 2 + 3 \\dbinom 6 3 + \\dbinom 6 4}{6^4}$$ - The problem reduces to finding the cardinality of the following set $$A=\\{(a_1,a_2,a_3,a_4)\\in\\{1,\\ldots,6\\}^4: \\ a_1\\leq a_2\\leq a_3 \\leq a_4\\}.$$ Define $$B=\\{(a_1,a_2,a_3,a_4)\\}\\in\\{1,\\ldots 9\\}^4: \\ a_1 < a_2 < a_3 < a_4\\}$$ and $$f:A\\to B, \\quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \\choose 4}.$$ - @Shailesh This method actually does give the correct answer ($126/6^4$); If we", "for $1 \\le n \\le 31$), and so our answer is $\\boxed{\\textbf{(B) } 31}$", "Oct 28 '14 at 21:20 Here is an alternative solution to the (very good) answer already given here: The number of ways that you can choose $2$ out of $6$ values is $\\dbinom{6}{2}=15$. Given $2$ values A and B, the number of combinations that you can get in $6$ rolls is: • Value A appearing $1$ time and value B appearing $5$ times: $\\dbinom{6}{1}=6$ • Value A appearing $2$ times and value B appearing $4$ times: $\\dbinom{6}{2}=15$ • Value A appearing $3$ times and value B appearing $3$ times: $\\dbinom{6}{3}=20$ • Value A appearing $4$ times and value B appearing $2$ times: $\\dbinom{6}{4}=15$ • Value A appearing $5$ times and value B appearing $1$ time : $\\dbinom{6}{5}=6$ So you can get $6+15+20+15+6=62$ combinations containing A and B. And you can get $15\\cdot62=930$ combinations containing any $2$ out of $6$ values. The total number of combinations that you can get in $6$ rolls is simply $6^6=46656$. Therefore, the probability of having exactly $2$ values in $6$ rolls is $\\dfrac{930}{46656}\\approx0.0199$. • Just a note: the $62$ calculated here through sumation is precisely the $2^6 - 2$ calculated in my answer. This is due to the fact that $\\sum_{i=1}^6{6\\choose i} = 2^6$. – 5xum Oct 28 '14 at 12:12 • @5xum: I agree (except for the fact that you need to" ]

[ "2\\dbinom 21}{\\dbinom 52}=\\dfrac{3}{5}$$", "i.e. numbering $52$ Thus $Pr =$ $\\dbinom{52}{2} \\over \\dbinom{157}{2}$", "@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19 @AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22 Nope, that seems likely to be the most concise way to do it. Count ways to pick $n\\in\\{1,2,3,4\\}$ unique numbers, and to arrange them with $n-1$ \"$>$\" signs, to meet the criteria. $$\\begin{array}{|l:l|} \\hline \\rm a{=}a{=}a{=}a & \\dbinom 3 0 \\dbinom 6 1 \\\\\\hdashline\\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \\dbinom 3 1\\dbinom 6 2 \\\\\\hdashline\\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \\dbinom 3 2 \\dbinom 6 3 \\\\\\hdashline\\rm a{>}b{>}c{>}d & \\dbinom 3 3 \\dbinom 6 4 \\\\ \\hline\\end{array}$$ $$\\dfrac{\\sum_{n=1}^4 \\dbinom{3}{n-1}\\dbinom{6}{n}}{6^4} = \\dfrac{\\dbinom 6 1 + 3\\dbinom 6 2 + 3 \\dbinom 6 3 + \\dbinom 6 4}{6^4}$$ - The problem reduces to finding the cardinality of the following set $$A=\\{(a_1,a_2,a_3,a_4)\\in\\{1,\\ldots,6\\}^4: \\ a_1\\leq a_2\\leq a_3 \\leq a_4\\}.$$ Define $$B=\\{(a_1,a_2,a_3,a_4)\\}\\in\\{1,\\ldots 9\\}^4: \\ a_1 < a_2 < a_3 < a_4\\}$$ and $$f:A\\to B, \\quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \\choose 4}.$$ - @Shailesh This method actually does give the correct answer ($126/6^4$); If we", "he received from B Jul 4, 2021 at 3:53 • I made an edit with the clarifications Jul 4, 2021 at 4:06 Here's my solution based on expectations. First, we calculate the expected value of the smallest, second-smallest, third-smallest... number on the 5 cards drawn. The calculation steps are shown below: $$E(\\text{smallest})=\\dfrac{1\\times \\dbinom{9}{4}+2\\times \\dbinom{8}{4}+\\cdots+6\\times \\dbinom{4}{4}}{\\dbinom{10}{5}}=\\dfrac{11}{6}.$$ Explanation: the numerator basically sums the smallest terms in all 5-card draws, and the denominator is the total number of cases. Similarly, we have $$E(\\text{second-smallest})=\\dfrac{2\\times \\dbinom{8}{3}\\times \\dbinom{1}{1}+3\\times \\dbinom{7}{3}\\times \\dbinom{2}{1}+\\cdots+7\\times \\dbinom{3}{3}\\times \\dbinom{6}{1}}{\\dbinom{10}{5}}=\\dfrac{11}{3};$$ $$E(\\text{third-smallest})=\\dfrac{3\\times \\dbinom{7}{2}\\times \\dbinom{2}{2}+4\\times \\dbinom{6}{2}\\times \\dbinom{3}{2}+\\cdots+8\\times \\dbinom{2}{2}\\times \\dbinom{7}{2}}{\\dbinom{10}{5}}=\\dfrac{11}{2};$$ $$E(\\text{fourth-smallest})=\\dfrac{4\\times \\dbinom{6}{1}\\times \\dbinom{3}{3}+5\\times \\dbinom{5}{1}\\times \\dbinom{4}{3}+\\cdots+9\\times \\dbinom{1}{1}\\times \\dbinom{8}{3}}{\\dbinom{10}{5}}=\\dfrac{22}{3};$$ $$E(\\text{fifth-smallest, or largest})=\\dfrac{5\\times \\dbinom{4}{4}+6\\times \\dbinom{5}{4}+\\cdots+10\\times \\dbinom{9}{4}}{\\dbinom{10}{5}}=\\dfrac{55}{6}.$$ As a quick double check, adding them all together yields $$27.5$$, which is the expected sum of the 5 cards. Next, let's think about A's strategy. In essence, he is discarding his $$N$$ smallest cards to get random cards. Thus, if the expected value of cards that he is discarding is smaller than the expected value of cards he is getting, then A is benefitting. The more A benefits, the higher his chances of winning. Then, clearly $$E(\\text{smallest})$$, $$E(\\text{second-smallest})$$ are smaller than $$5.5$$, the expected value on a random card. $$E(\\text{third-smallest})$$ is equal to $$5.5$$, and $$E(\\text{fourth-smallest})$$, $$E(\\text{fifth-smallest, or largest})$$ are greater than $$5.5$$. This means that discarding the smallest card and", "us $\\dbinom{n+1}{2}$ ways, for a total of $3\\dbinom{n+1}{2}$. The last case gives us $\\dbinom{n+1}{1}=n+1$ ways. Therefore, the total number of possible ways where there are no isolated men is $\\[\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1).\\]$ The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or $\\[2\\dbinom{n+1}{2}+(n+1).\\]$ Thus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that $\\[\\dfrac{\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1)}{2\\dbinom{n+1}{2}+(n+1)}\\le\\dfrac{1}{100}.\\]$ the numerator simplifies to $(n+1)^2$. The denominator simplifies to $\\dfrac{(n+6)(n+2)(n+1)}{6}$ so the whole faction simplifies to $\\dfrac{6(n+1)}{(n+6)(n+2)}$ Since $\\dfrac{n+1}{n+2}$ is slightly less than 1 when $n$ is large, $\\dfrac{6}{n+6}$ will be close to $\\dfrac{1}{100}$. They equal each other when $n = 594$. If we let $n= 595$ or $593$, we will notice that the answer is $\\boxed{594}$", "\\dfrac{\\dbinom{26}{2}\\dbinom{26}{1}}{\\dbinom{52}{3}} = \\dfrac{13}{17}$$ Rom Dec 31, 2018", "# One green In the container are 45 white and 15 balls. We randomly select 5 balls. What is the probability that it will be a maximum one green? Correct result: p = 0.633 #### Solution: C_{{ 5}}(60) = \\dbinom{ 60}{ 5} = \\dfrac{ 60! }{ 5!(60-5)!} = \\dfrac{ 60 \\cdot 59 \\cdot 58 \\cdot 57 \\cdot 56 } { 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 5461512 \\ \\\\ C_{{ 0}}(15) = \\dbinom{ 15}{ 0} = \\dfrac{ 15! }{ 0!(15-0)!} = \\dfrac{ 1 } { 1 } = 1 \\ \\\\ C_{{ 1}}(15) = \\dbinom{ 15}{ 1} = \\dfrac{ 15! }{ 1!(15-1)!} = \\dfrac{ 15 } { 1 } = 15 \\ \\\\ C_{{ 5}}(45) = \\dbinom{ 45}{ 5} = \\dfrac{ 45! }{ 5!(45-5)!} = \\dfrac{ 45 \\cdot 44 \\cdot 43 \\cdot 42 \\cdot 41 } { 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 1221759 \\ \\\\ C_{{ 4}}(45) = \\dbinom{ 45}{ 4} = \\dfrac{ 45! }{ 4!(45-4)!} = \\dfrac{ 45 \\cdot 44 \\cdot 43 \\cdot 42 } { 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 148995 \\ \\\\ \\ \\\\ p = \\dfrac\\dbinom{ {15}{ 0}\\dbinom{ 45}{ 5} + \\dbinom{ 15}{ 1}\\dbinom{ 45}{ 4} }\\dbinom{ { 45+15}{ 5} } = 0.633 Our examples were largely sent or", "+0 # help 0 58 2 +418 compute$$\\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4}$$ Logic Oct 21, 2018 #1 +103 +1 Well, 4 choose 0 = 1, 4 choose 1 = 4, 4 choose 2 = 6, 4 choose 3 = 4 choose 1 = 4, 4 choose 4 = 4 choose 0 = 1, 1+4+6+4+1 = 16 16, and good luck! MATHEXPERTISE Oct 21, 2018 #2 +20108 +2 compute: $$\\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4}$$ $$\\begin{array}{|rcll|} \\hline (1+1)^4 &=& \\dbinom{4}{0}\\cdot 1^4\\cdot 1^0 + \\dbinom{4}{1}\\cdot 1^3\\cdot 1^1 + \\dbinom{4}{2}\\cdot 1^2\\cdot 1^2 + \\dbinom{4}{3}\\cdot 1^1\\cdot 1^3 + \\dbinom{4}{4}\\cdot 1^0\\cdot 1^4 \\\\\\\\ (1+1)^4 &=& \\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4} \\\\\\\\ 2^4 &=& \\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4} \\\\\\\\ 16 &=& \\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4} \\\\ \\hline \\end{array}$$ heureka Oct 22, 2018", "is: $\\set {\\dbinom {a \\ b} {1 \\ 2}, \\dbinom {a \\ b} {1 \\ 3}, \\dbinom {a \\ b} {2 \\ 1}, \\dbinom {a \\ b} {2 \\ 3}, \\dbinom {a \\ b} {3 \\ 1}, \\dbinom {a \\ b} {3 \\ 2} }$ thus demonstrating that there are $\\dfrac {3!} {\\paren {3 - 2}!} = \\dfrac {3!} {1!} = \\dfrac 6 1 = 6$ injections from $A$ to $B$.", "= \\dfrac{{13 \\times 12! \\times 4 \\times 3! \\times 50! \\times 2 \\times 1}}{{52 \\times 51 \\times 50! \\times 12! \\times 3!}} \\\\ P = \\dfrac{{13 \\times 4 \\times 2}}{{52 \\times 51}} = \\dfrac{2}{{51}} \\\\$", "(base change formula yields $\\log_{10}2=\\ln 2/\\ln 10$): $$\\begin{array}{ll} (2\\times 10^6)^{-455} & =10^{-455(6+\\ln2/\\ln10)} \\\\ & =10^{-2866.9686\\cdots} \\\\ & =10^{0.0313\\cdots}\\times10^{-2867} \\\\ & =1.07486\\cdots\\times10^{-2867}.\\end{array}$$ – anon Aug 15 '13 at 17:40 You don't need a computer to put $\\left( \\dfrac {1}{10^7} \\right)^{455}$. This is just $\\dfrac {1}{10^{7 \\cdot 455}} = \\boxed {\\dfrac {1}{10^{3185}} = 10^{-3185}}$. • Probably because you put too much faith in OP's original understanding of the problem (and didn't check it yourself), and subsequently failed to correct the answer in light of new information. Specifically, the $10^{-7}$ figure is incorrect. The other two answers using that figure were deleting by their owners. – anon Aug 17 '13 at 10:26 Actually you need to calculate $$(1/2 * 10^{-6} )^{455}$$ Which is by no means easy : $$1.07486017721073420028655449423203634073126755189808458... × 10^{-2867}$$ (Using Wolfram alpha)", "# One green In the container are 45 white and 15 balls. We randomly select 5 balls. What is the probability that it will be a maximum one green? Result p = 0.633 #### Solution: C_{{ 5}}(60) = \\dbinom{ 60}{ 5} = \\dfrac{ 60! }{ 5!(60-5)!} = \\dfrac{ 60 \\cdot 59 \\cdot 58 \\cdot 57 \\cdot 56 } { 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 5461512 \\ \\\\ C_{{ 0}}(15) = \\dbinom{ 15}{ 0} = \\dfrac{ 15! }{ 0!(15-0)!} = \\dfrac{ 1 } { 1 } = 1 \\ \\\\ C_{{ 1}}(15) = \\dbinom{ 15}{ 1} = \\dfrac{ 15! }{ 1!(15-1)!} = \\dfrac{ 15 } { 1 } = 15 \\ \\\\ C_{{ 5}}(45) = \\dbinom{ 45}{ 5} = \\dfrac{ 45! }{ 5!(45-5)!} = \\dfrac{ 45 \\cdot 44 \\cdot 43 \\cdot 42 \\cdot 41 } { 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 1221759 \\ \\\\ C_{{ 4}}(45) = \\dbinom{ 45}{ 4} = \\dfrac{ 45! }{ 4!(45-4)!} = \\dfrac{ 45 \\cdot 44 \\cdot 43 \\cdot 42 } { 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 148995 \\ \\\\ \\ \\\\ p = \\dfrac\\dbinom{ {15}{ 0}\\dbinom{ 45}{ 5} + \\dbinom{ 15}{ 1}\\dbinom{ 45}{ 4} }\\dbinom{ { 45+15}{ 5} } = 0.633 Our examples were largely sent or created", "# One green In the container are 45 white and 15 balls. We randomly select 5 balls. What is the probability that it will be a maximum one green? Correct result: p = 0.6329 #### Solution: C_{{ 5}}(60) = \\dbinom{ 60}{ 5} = \\dfrac{ 60! }{ 5!(60-5)!} = \\dfrac{ 60 \\cdot 59 \\cdot 58 \\cdot 57 \\cdot 56 } { 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 5461512 \\ \\\\ C_{{ 0}}(15) = \\dbinom{ 15}{ 0} = \\dfrac{ 15! }{ 0!(15-0)!} = \\dfrac{ 1 } { 1 } = 1 \\ \\\\ C_{{ 1}}(15) = \\dbinom{ 15}{ 1} = \\dfrac{ 15! }{ 1!(15-1)!} = \\dfrac{ 15 } { 1 } = 15 \\ \\\\ C_{{ 5}}(45) = \\dbinom{ 45}{ 5} = \\dfrac{ 45! }{ 5!(45-5)!} = \\dfrac{ 45 \\cdot 44 \\cdot 43 \\cdot 42 \\cdot 41 } { 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 1221759 \\ \\\\ C_{{ 4}}(45) = \\dbinom{ 45}{ 4} = \\dfrac{ 45! }{ 4!(45-4)!} = \\dfrac{ 45 \\cdot 44 \\cdot 43 \\cdot 42 } { 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 148995 \\ \\\\ \\ \\\\ p = \\dfrac\\dbinom{ {15}{ 0}\\dbinom{ 45}{ 5} + \\dbinom{ 15}{ 1}\\dbinom{ 45}{ 4} }\\dbinom{ { 45+15}{ 5} } = 0.6329 We would be pleased if you", "# One green In the container are 45 white and 15 balls. We randomly select 5 balls. What is the probability that it will be a maximum one green? Correct result: p = 0.6329 #### Solution: C_{{ 5}}(60) = \\dbinom{ 60}{ 5} = \\dfrac{ 60! }{ 5!(60-5)!} = \\dfrac{ 60 \\cdot 59 \\cdot 58 \\cdot 57 \\cdot 56 } { 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 5461512 \\ \\\\ C_{{ 0}}(15) = \\dbinom{ 15}{ 0} = \\dfrac{ 15! }{ 0!(15-0)!} = \\dfrac{ 1 } { 1 } = 1 \\ \\\\ C_{{ 1}}(15) = \\dbinom{ 15}{ 1} = \\dfrac{ 15! }{ 1!(15-1)!} = \\dfrac{ 15 } { 1 } = 15 \\ \\\\ C_{{ 5}}(45) = \\dbinom{ 45}{ 5} = \\dfrac{ 45! }{ 5!(45-5)!} = \\dfrac{ 45 \\cdot 44 \\cdot 43 \\cdot 42 \\cdot 41 } { 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 1221759 \\ \\\\ C_{{ 4}}(45) = \\dbinom{ 45}{ 4} = \\dfrac{ 45! }{ 4!(45-4)!} = \\dfrac{ 45 \\cdot 44 \\cdot 43 \\cdot 42 } { 4 \\cdot 3 \\cdot 2 \\cdot 1 } = 148995 \\ \\\\ \\ \\\\ p = \\dfrac\\dbinom{ {15}{ 0}\\dbinom{ 45}{ 5} + \\dbinom{ 15}{ 1}\\dbinom{ 45}{ 4} }\\dbinom{ { 45+15}{ 5} } = 0.6329 We would be pleased if you", "+0 # help 0 236 2 +1040 compute$$\\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4}$$ Oct 21, 2018 #1 +129 +1 Well, 4 choose 0 = 1, 4 choose 1 = 4, 4 choose 2 = 6, 4 choose 3 = 4 choose 1 = 4, 4 choose 4 = 4 choose 0 = 1, 1+4+6+4+1 = 16 16, and good luck! Oct 21, 2018 #2 +23043 +9 compute: $$\\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4}$$ $$\\begin{array}{|rcll|} \\hline (1+1)^4 &=& \\dbinom{4}{0}\\cdot 1^4\\cdot 1^0 + \\dbinom{4}{1}\\cdot 1^3\\cdot 1^1 + \\dbinom{4}{2}\\cdot 1^2\\cdot 1^2 + \\dbinom{4}{3}\\cdot 1^1\\cdot 1^3 + \\dbinom{4}{4}\\cdot 1^0\\cdot 1^4 \\\\\\\\ (1+1)^4 &=& \\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4} \\\\\\\\ 2^4 &=& \\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4} \\\\\\\\ 16 &=& \\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4} \\\\ \\hline \\end{array}$$ Oct 22, 2018", "+0 # help 0 139 2 +586 compute$$\\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4}$$ Oct 21, 2018 #1 +128 +1 Well, 4 choose 0 = 1, 4 choose 1 = 4, 4 choose 2 = 6, 4 choose 3 = 4 choose 1 = 4, 4 choose 4 = 4 choose 0 = 1, 1+4+6+4+1 = 16 16, and good luck! Oct 21, 2018 #2 +21848 +9 compute: $$\\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4}$$ $$\\begin{array}{|rcll|} \\hline (1+1)^4 &=& \\dbinom{4}{0}\\cdot 1^4\\cdot 1^0 + \\dbinom{4}{1}\\cdot 1^3\\cdot 1^1 + \\dbinom{4}{2}\\cdot 1^2\\cdot 1^2 + \\dbinom{4}{3}\\cdot 1^1\\cdot 1^3 + \\dbinom{4}{4}\\cdot 1^0\\cdot 1^4 \\\\\\\\ (1+1)^4 &=& \\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4} \\\\\\\\ 2^4 &=& \\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4} \\\\\\\\ 16 &=& \\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4} \\\\ \\hline \\end{array}$$ Oct 22, 2018", "roll ten dice. I like to reduce the case approach to two multinomial coefficients for [choose] $\\times$ [place] to make it more mechanical and thus less error prone. $5-1-1-1-1-1: \\dbinom6{1,5}\\dbinom{10}{5,1,1,1,1,1}$ $4-2-1-1-1-1: \\dbinom6{1,1,4}\\dbinom{10}{4,2,1,1,1,1}$ $3-3-1-1-1-1: \\dbinom6{2,4}\\dbinom{10}{3,3,1,1,1,1}$ $3-2-2-1-1-1: \\dbinom6{1,2,3}\\dbinom{10}{3,2,2,1,1,1}$ $2-2-2-2-1-1: \\dbinom6{4,2}\\dbinom{10}{2,2,2,2,1,1}$ Add up and divide by $6^{10}$ PS Or you might like to write it as [choose] $\\times$ [permute], viz. $5-1-1-1-1-1: \\dbinom6{1,5}\\times\\dfrac{10!}{5!}$ $4-2-1-1-1-1: \\dbinom6{1,1,4}\\times\\dfrac{10!}{4!2!}$ $3-3-1-1-1-1: \\dbinom6{2,4}\\times\\dfrac{10!}{3!3!}$ $3-2-2-1-1-1: \\dbinom6{1,2,3}\\times\\dfrac{10}{3!2!2!}$ $2-2-2-2-1-1: \\dbinom6{4,2}\\times\\dfrac{10!}{2!2!2!2!}$ Based on the comments below, I try a different approach. I consider $1-$the number of cases where all numbers are not there. Choose 5 out of 6 numbers. Each dice can be any number = $5^{10}$ So, the total number of ways is $6^{10} - 6*5^{10}$ • Double Counting! Suppose you pick the first die and its a one, there's another one in the six you didn't pick. Suppose you picked that place, and its a one, and the one among the six you didn't pick is in first place. You've counted these at least twice. – Graham Kemp Feb 16 '16 at 1:37 • This method double counts just as I warned above in the comments. – Gregory Grant Feb 16 '16 at 1:37 • @RobertMingHao Be careful this is a wrong solution, it double counts many possibilities. You should convince yourself of why this is wrong there's an important", "females must also be picked, for a total of $\\dbinom{12}{12}$. Therefore, these cases can be combined to $$\\dbinom{11}{0} \\cdot \\left(\\dbinom{12}{1} + \\dbinom{12}{12}\\right)$$ Since $\\dbinom{12}{12} = \\dbinom{12}{0}$, and $\\dbinom{12}{0} + \\dbinom{12}{1} = \\dbinom{13}{1}$, we can further simplify this to $$\\dbinom{11}{0} \\cdot \\dbinom{13}{1}$$ All other cases proceed similarly. For example, the case with one men or ten men is equal to $\\dbinom{11}{1} \\cdot \\dbinom{13}{2}$. Now, if we factor out a $13$, then all cases except the first two have a factor of $121$, so we can factor this out too to make our computation slightly easier. The first two cases (with $13$ factored out) give $1+66=67$, and the rest gives $121(10+75+270+504) = 103,939$. Adding the $67$ gives $104,006$. Now, we can test for prime factors. We know there is a factor of $2$, and the rest is $52,003$. We can also factor out a $7$, for $7,429$, and the rest is $17 \\cdot 19 \\cdot 23$. Adding up all the prime factors gives $2+7+13+17+19+23 = \\boxed{081}$. ### Video Solution: (Solves using both methods - Casework and Vandermonde's Identity) ## Solution 3 (Vandermonde's identity) Applying Vandermonde's identity by setting $m=12$, $n=11$, and $r=11$, we obtain $\\binom{23}{11}\\implies$ $\\boxed{081}$. ~Lcz ### Short Proof Consider the following setup: $[asy] size(1000, 100); for(int i=0; i<23; ++i){ dot((i, 0)); } draw((10.5, -1.5)--(10.5, 1.5), dashed); [/asy]$ The", "\\dbinom{99}{5} + \\dbinom{99}{7}+ \\dots +\\dbinom{99}{49}\\\\\\\\ &=& \\dbinom{99}{0} + \\dbinom{99}{1} + \\dbinom{99}{2}+ \\dbinom{99}{3} + \\dbinom{99}{4}+ \\dots \\\\ && + \\dbinom{99}{46} + \\dbinom{99}{47} + \\dbinom{99}{48}+\\dbinom{99}{49}\\\\\\\\ &=& \\dfrac{2^{99}}{2} \\\\\\\\ &=& \\mathbf{ 2^{98} } \\\\ \\hline \\end{array}$$ May 19, 2021 edited by heureka May 19, 2021 #4 +121056 +1 Thanks, heureka !!!! CPhill May 19, 2021 #5 +114485 +1 Thanks Ilovepizza and Heureka, Where does 2^99 come from? I understand everything else. May 19, 2021 #6 +26213 +2 Where does $$2^{99}$$ come from? Pascal's triangle: See the sum of the each line: $$\\small{ \\begin{array}{|l|rclcl|} \\hline n \\\\ \\hline 0 &\\dbinom{0}{0} &=& 1 &=& 2^0 \\\\ 1 &\\dbinom{1}{0}+\\dbinom{1}{1} &=& 1+1 &=& 2^1 \\\\ 2 &\\dbinom{2}{0}+\\dbinom{2}{1}+\\dbinom{2}{2} &=& 1+2+1 &=& 2^2 \\\\ 3 &\\dbinom{3}{0}+\\dbinom{3}{1}+\\dbinom{3}{2}+\\dbinom{3}{3} &=& 1+3+3+1 &=& 2^3 \\\\ 4 &\\dbinom{4}{0}+\\dbinom{4}{1}+\\dbinom{4}{2}+\\dbinom{4}{3}+\\dbinom{4}{4} &=& 1+4+6+4+1 &=& 2^4 \\\\ 5 &\\dbinom{5}{0}+\\dbinom{5}{1}+\\dbinom{5}{2}+\\dbinom{5}{3}+\\dbinom{5}{4}+\\dbinom{5}{5} &=& 1+5+10+10+5+1 &=& 2^5 \\\\ \\ldots \\\\ \\color{red}99 & \\dbinom{99}{0} + \\dbinom{99}{1} + \\dots + \\dbinom{99}{98}+\\dbinom{99}{99} &=&1+99+\\dots + 99 + 1 &=& \\color{red}2^{99} \\\\ \\hline \\end{array} }$$ $$\\begin{array}{|rcll|} \\hline && \\color{red}(1+1)^n \\\\\\\\ &=&\\sum \\limits_{k=0}^{n}\\dbinom{n}{k} \\\\ \\\\ &=& \\dbinom{n}{0}+\\dbinom{n}{1} + \\dots + \\dbinom{n}{n-1}+\\dbinom{n}{n} \\\\\\\\ &=& \\color{red}2^n \\\\ \\hline \\end{array}$$ May 20, 2021 #7 +114485 0 Thanks very much Heureka. Hopefully I will remember next time. :// Melody May 20, 2021", "we have a total of $$\\displaystyle 2G+3$$ pens in the box. Lets look at the probability: $$\\displaystyle \\dfrac{\\dbinom{G+3}{2}+\\dbinom{G}{2}}{\\dbinom{2G+3}{2}}=\\dfrac{27}{55}$$ $$\\displaystyle \\dfrac{\\dbinom{G+3}{2}+\\dbinom{G}{2}}{\\dbinom{2G+3}{2}}=\\dfrac{\\dfrac{(G+3)(G+2)}{2}+\\dfrac{G(G-1)}{2}}{\\dfrac{(2G+3)(2G+2)}{2}}=\\dfrac{27}{55}$$ Simple algebra should let one solve for $$\\displaystyle G$$ But remember that $$\\displaystyle G+B>12$$." ]

[ "7\")", "1. polygon How many diagonals does a regular seven-sided polygon contain? 2. Hello, sri340! How many diagonals does a regular seven-sided polygon contain? Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ 3. Originally Posted by Soroban Hello, sri340!Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ ... But if you take a \"diagonal\" to mean a line connecting two non-adjacent vertices then the answer will be less (14, in fact).", "= 7", "# polygon • February 1st 2010, 01:53 PM sri340 polygon How many diagonals does a regular seven-sided polygon contain? • February 1st 2010, 08:28 PM Soroban Hello, sri340! Quote: How many diagonals does a regular seven-sided polygon contain? Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ • February 2nd 2010, 08:19 AM Opalg Quote: Originally Posted by Soroban Hello, sri340!Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ ... But if you take a \"diagonal\" to mean a line connecting two non-adjacent vertices then the answer will be less (14, in fact).", "many diagonals does a 63-sided convex polygon have? [#permalink] 08 Aug 2017, 13:58 Display posts from previous: Sort by", "7 28 Jun 2012, 17:58 How many diagonals does a polygon with 21 sides have, if one 1 19 Dec 2007, 17:48 How many diagonals does a polygon with 21 sides have, if one 3 06 Aug 2007, 17:32 How many diagonals does a polygon with 21 sides have, if one 5 09 Feb 2006, 17:00 a parallelogram has 2 sides: 14, 18 one of the diagonals is 13 24 Sep 2005, 06:03 Display posts from previous: Sort by", "# What is this polyhedron? Geometry Level pending Suppose a convex polyhedron has $$12$$ square faces, $$8$$ regular hexagon faces, and $$6$$ regular octagon faces. At each vertex of the polyhedron, one square, one hexagon, and one octagon meet. How many space diagonals does this polyhedron have? ×", "Comment Share Q) # A polygon has 54 diagonals. Then the number of its sides is : $\\begin {array} {1 1} (1)\\;7 & \\quad (2)\\;9 \\\\ (3)\\;10 & \\quad (4)\\;12 \\end {array}$", "Forum Only ) Manager Joined: 02 Jul 2016 Posts: 99 Re: How many diagonals does a regular 11 -sided polygon contain? [#permalink] Show Tags 28 Nov 2016, 06:48 1 This post received KUDOS 1 This post was BOOKMARKED Straight 15 second answer!!!!!!!!!!!!!!!!! Its D i.e. 44 The formula for the number of diagonals for an n sided polygon is represented as nC2-n where n represents the number of sides of the polygon Here, n=11 so by placing the value of n we get the number of diagonals as 44 Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2116 Re: How many diagonals does a regular 11 -sided polygon contain? [#permalink] Show Tags 02 Dec 2016, 06:56 1 This post received KUDOS Expert's post Bunuel wrote: How many diagonals does a regular 11-sided polygon contain? A. 35 B. 38 C. 40 D. 44 E. 55 Given a vertex, in order to form a diagonal, we need to choose a vertex other than the vertex itself and the two adjacent vertices; thus, we have 8 possibilities for a diagonal from any given vertex. This idea is true for all 11 vertices; however, we must divide out the overlap since each diagonal is counted twice. Thus, the number of diagonals created from", "diagonals and rhombus sides.", "an 11-sided polygon is (11 x 8) / 2 = 44 diagonals. Answer: D _________________ Jeffery Miller Head of GMAT Instruction GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Intern Joined: 16 Jul 2016 Posts: 2 Re: How many diagonals does a regular 11 -sided polygon contain? [#permalink] Show Tags 02 Dec 2016, 07:24 No. of diagonals of a regular n sided polygon is nc2-n Sent from my Lenovo A7010a48 using GMAT Club Forum mobile app Intern Joined: 15 Oct 2016 Posts: 34 Re: How many diagonals does a regular 11 -sided polygon contain? [#permalink] Show Tags 03 Mar 2018, 23:12 1 This post received KUDOS Bunuel wrote: How many diagonals does a regular 11-sided polygon contain? A. 35 B. 38 C. 40 D. 44 E. 55 Approach 1: Excluding the side lengths from the total line segments = nc2 - n Approach 2: Every vertex can yield n-3 diagonals (excluding itself and two adjacent points). So, n vertices would yield n(n-3) diagonals. However, it has double counting since the diagonal from A to B is essentially the same as that from B to A. Hence, no. of unique diagonals n(n-3)/2 Re: How many diagonals does a regular 11 -sided polygon contain? [#permalink] 03 Mar 2018, 23:12 Display posts from previous: Sort by How", "# Geometry Geometry Level 1 If a polygon has $27$ diagonals, then the number of sides of the polygon is ×", "Figure 7.10.", "7.4.", "6 sides of the irregular polygon is a six-sided polygon where side.", "5 9 14 20 27 35 44 54 sides____ 3 4 5 6 7_ 8_ 9_ 10_ 11_ 12 _ Re: A polygon has 12 edges. How many different diagonals does it [#permalink] 01 Jan 2019, 23:28 Display posts from previous: Sort by", "sides yields the below. (7)7", "2) = 4.39334 cm Question: Answers:Think of a box or a cube. Those are examples of rectangular prisms. There are six faces (the sides of the solid). There are eight vertices (the \"corners\" of the solid). Answer: 6 faces 8 vertices Question:i asked to draw a polygon.and how many sides it is. Answers:A polygon will have as many sides as you want, at least three, for a triangle, for example. An octagon has 8 sides. Poly = many, gone = angle, octo = 8, in Greek ;-)", "diagonals does a regular 11 -sided polygon contain? [#permalink] ### Show Tags 23 Nov 2016, 07:14 2 Top Contributor 3 Bunuel wrote: How many diagonals does a regular 11-sided polygon contain? A. 35 B. 38 C. 40 D. 44 E. 55 A nice/fast approach that doesn't involve any counting techniques is to recognize that, for each of the 11 points (vertices), there are 8 possible points that we can connect to to create a diagonal. ASIDE: there are 8 possible points because we cannot create a diagonal by connecting 2 adjacent points. So, there are 8 possible diagonals for each of the 11 points These means that there are 88 possible diagonals in total (8x11=88). However, we need to recognize that this method counts each diagonal twice. For example, it counts the diagonal AB and the diagonal BA as 2 separate diagonals. So, to account for this duplication, we'll divide 88 by 2 to get 44 Answer: Cheers, Brent _________________ Brent Hanneson – GMATPrepNow.com Sign up for our free Question of the Day emails ##### Most Helpful Community Reply Ask GMAT Experts Forum Moderator Status: Preparing for GMAT Joined: 25 Nov 2015 Posts: 963 Location: India GPA: 3.64 Re: How many diagonals does a regular 11 -sided polygon contain? [#permalink] ### Show Tags 23 Nov 2016, 10:09 6", "How many more diagonals are in an octagon than in a hexagon? $20-9=11$ Answer: $11$" ]

[ "2) = 4.39334 cm Question: Answers:Think of a box or a cube. Those are examples of rectangular prisms. There are six faces (the sides of the solid). There are eight vertices (the \"corners\" of the solid). Answer: 6 faces 8 vertices Question:i asked to draw a polygon.and how many sides it is. Answers:A polygon will have as many sides as you want, at least three, for a triangle, for example. An octagon has 8 sides. Poly = many, gone = angle, octo = 8, in Greek ;-)", "choose 4 different from them. Let they be P<Q<R<S. You can construct two pairs of line segments: (PQ, RS) and (PR, QS). If the endpoint of the segments are a1, a2 and b1, b2, (so as a1<a2 and b1<b2) b1 and b2 has to be at opposite sides of the segment (a1,a2) so as the segment (b1,b2) intersects (a1,a2) inside the polygon. That means b1<a2. So you have only a single pair of intersecting diagonals for every 4 points, (PR) and (QS), that is one intersection. It was said that the intersections do not coincide. So you have $(^n_4)$ intersections. See picture. You choose the vertices 1,3,5,7. The diagonals (1,5) and (3,7) intersect each other inside the polygon, the other pair of diagonals do not. ehild File size: 16.8 KB Views: 217 8. Jan 6, 2013 ### sankalpmittal Thanks ILS and ehild , for such insights ! And apologies for late reply. I come online but rarely...", "1. polygon How many diagonals does a regular seven-sided polygon contain? 2. Hello, sri340! How many diagonals does a regular seven-sided polygon contain? Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ 3. Originally Posted by Soroban Hello, sri340!Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ ... But if you take a \"diagonal\" to mean a line connecting two non-adjacent vertices then the answer will be less (14, in fact).", "# What is this polyhedron? Geometry Level pending Suppose a convex polyhedron has $$12$$ square faces, $$8$$ regular hexagon faces, and $$6$$ regular octagon faces. At each vertex of the polyhedron, one square, one hexagon, and one octagon meet. How many space diagonals does this polyhedron have? ×", "# Geometry Geometry Level 1 If a polygon has $27$ diagonals, then the number of sides of the polygon is ×", "# Geometry Geometry Level 1 If a polygon has $$27$$ diagonals, then the number of sides of the polygon is × Problem Loading... Note Loading... Set Loading...", "Choose $4$ vertices from the $2n+1$ available. Exactly one pair of diagonals determined by dividing these vertices into two pairs meet in the interior of the polygon. Thus the required number of pairs of diagonals is $\\binom{2n+1}{4}$.", "A convex polyhedron $$P$$ has 26 vertices, 60 edges, and 36 faces, 24 of which are triangular, and 12 of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $$P$$ have? (第二十二届AIME1 2004 第3题)", "the other two nodes such that the \"big\" polygon has 18 edges. I will try these approaches today. – Jimeree Sep 6 '12 at 11:04", "# Number of triangulationss of a polygon that each triangle has at least one edge with polygon What is the number of triangulations of a polygon that each triangle has at least one edge with polygon? I want Triangulation of a convex n-gon so that all triangles share a side with the polygon. For example for $n=6$ we have $12$. • Oct 30 '17 at 15:03 • @samjoe for n=6 we have 12 Oct 30 '17 at 15:04 • @samjoe I want Triangulation of a convex n-gon so that all triangles share a side with the polygon Oct 30 '17 at 15:05 • Sorry for the confusion! I didn't understand correctly what was being asked. I recommend you to edit your question and add the links there too! Sorry again. Oct 30 '17 at 15:24 ## 1 Answer There are $n-2$ triangles in a triangulation of an $n$-sided polygon. Each of them must share at least one side with the polygon, and there are $n$ sides to go around, which means that $n-4$ of them share one side (call these triangles type 1), and $2$ of them share two sides (call these two triangles type 2). Pick one of the type-2 triangles to be the \"starting\" triangle. The starting triangle has one exposed diagonal, which is shared by", "# polygon • February 1st 2010, 01:53 PM sri340 polygon How many diagonals does a regular seven-sided polygon contain? • February 1st 2010, 08:28 PM Soroban Hello, sri340! Quote: How many diagonals does a regular seven-sided polygon contain? Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ • February 2nd 2010, 08:19 AM Opalg Quote: Originally Posted by Soroban Hello, sri340!Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ ... But if you take a \"diagonal\" to mean a line connecting two non-adjacent vertices then the answer will be less (14, in fact).", "# A Specific Dissection Three vertices are chosen randomly from the vertices of a regular 7-sided polygon. If the probability that they from the vertices of an isosceles triangle can be expressed as $$\\dfrac{a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers, find $$a+b$$. ×", "vertices that are not adjacent to each other. How many different diagonals does it have? View Answer Find the interior and exterior angles of a 9-sided polygon. A diagonal connects any two non-consecutive vertices of a polygon. (Remember a rectangle is a type of parallelogram so rectangles get all of the parallelogram properties) If MO = 26 and the diagonals bisect each other, then MZ = ½(26) = 13 How many diagonals does a triangle have? It doesn't matter how long the length of each side is or how it's shaped - it will always be three.... How Many Perpendicular Lines Does A Square Have? Therefore, number of diagonals = \\\\frac{63 * 60}{2}\\\\) = 1890. By definition, it has n sides. Show Answer. Here are some related questions which you might be interested in reading. Since there would be no diagonal drawn back to itself, and the diagonals to each adjacent vertex would lie on top of the adjacent sides, the number of diagonals from a single vertex is three less the the number of sides, or n-3. 1) Regular pentagon P has all five diagonals drawn. Relevance. A rhombus is a type of parallelogram, and what distinguishes its shape is that all four of its sides are congruent.. Question 2: How many sides does a triangle have?", "# Number of Diagonals A diagonal of a polygon is a line segment that connects two vertices that are not next to each other. Use the canvas below to find the total number of line segments, and the number of diagonals for each polygon. How many segments will there be in a 20-sided polygon (icosagon)? How many of them will be the diagonals of the icosagon? Try to write a general term for the total number of line segments and the number of diagonals of any polygon. • What is your strategy to count the total number of segments for each polygon? What did you realize? • Is there a number pattern for the total number of segments? How many of them are the sides and how many are the diagonals? • What is the number of segments you draw from each vertex? How is it related to the total number? • What is the number of diagonals you can draw per vertex? • How can we avoid double counting when finding the total number of diagonals? ## Solution There are several ways of writing a general rule for an n-sided polygon. You may recognize the pattern for the total number of segments: 3-6-10-15-21-28-36-45 .. It is the sequence of triangular numbers starting from 3. One way to", "Convex polyhedron with five, six, or seven vertices at distinct corners of a cube What are the names of the convex polyhedron with five, six, or seven vertices, where all vertices lie at distinct corners of a cube? I'm particularly interested in the five vertex case. - With five vertices, depending on which ones you choose, you can get an oblique square pyramid, an oblique rectangular pyramid, or an irregular triangular bipyramid with $S_3$ symmetry. With six vertices, you can get a right isosceles triangular prism, an equilateral triangular right antiprism, or an irregular octahedron with $S_2$ symmetry. I don't think the seven-vertex figure has any particular name. Neither of these names are precise enough to express the fact that the vertices are chosen among the vertices of a cube. -", "that are parallel. All rectangles are four-sided polygons. And, as we have learned, they are also called quadrilaterals. Quad means “four.” Count the number of angles or sides in the last figure. It has seven angles and sides. This makes it a heptagon. Hept-means “seven.” Sometimes, we can use polygons when problem solving. Let’s look at this example. Example Morgan stitched a geometric figure into part of a quilt she is sewing. She measured the angles to help her know where to stitch. The sum of the angle measures was $1,080^\\circ$. What kind of polygon did she stitch? In this example, we have no easy way of knowing how many angles or sides the figure has. What do we know? We know that the sum of the interior angles, however many there are, is $1,080^\\circ$. Only one polygon has angles that always add up to this amount. Check the table above. Morgan must have stitched an octagon. Now we know that the distinguishing properties of an octagon are not only that it has eight sides and angles, but that its eight angles must have a sum of $1,080^\\circ$. 8I. Lesson Exercises Use what you have learned to identify each polygon described: 1. The sum of the interior angles is $180^\\circ$. 2. It has seven sides. 3. It has", "more than one vertex, we call them vertices. The vertices in this polygon are labeled $$A$$, $$B$$, $$C$$, $$D$$, and $$E$$. • volume Volume is the number of cubic units that fill a three-dimensional region, without any gaps or overlaps. For example, the volume of this rectangular prism is 60 units3, because it is composed of 3 layers that are each 20 units3.", "and Crosses. 18.) Opposite faces of a die always add to 7.", "use similar reasoning to count the cube's vertices. Every square has four vertices, but now every vertex is shared by three faces. So, the cube has $\\frac{6 \\times 4}{3} = 8$ vertices. We could also have reasoned that each of the cube's twelve edges has two vertices at its ends, and when the cube is glued together, each vertex is shared by three edges. This way of counting vertices also gives us $\\frac{12 \\times 2}{3} = 8$ vertices. By reasoning about how the parts of other $3\\text{D}$ shapes are connected and using some careful counting techniques, we can determine how many of each part the shape has. Try it on the challenge below. # Today's Challenge On the left is a cube and on the right is a tetrahedron. Initially, the cube has more faces. Which has more faces after being transformed? $\\\\[1em]$", "strictly opposite each other. Here we can select a pair of vertices in 8*7/2! = 28 ways. If we draw a line between our pair of vertices, it will be a diagonal unless it's one of the 8 edges of the octagon. So we can draw 28 - 8 = 20 diagonals, and the answer is 20/28 = 5/7. _________________ GMAT Tutor in Toronto If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com Director Joined: 19 Oct 2018 Posts: 791 Location: India Re: ABCDEFGH is an octagon. Two vertices of the octagon are selected [#permalink] Show Tags 19 Jun 2019, 09:15 My bad!! Gotta edit my solution. IanStewart wrote: nick1816 wrote: Number of ways you can select 2 verices= 8C2=28 Number of ways these vertices form a diagonal= (1/2)*(8C1 * 1)=4 the probability that these vertices form a diagonal of the octagon. = 4/28=1/7 A diagonal of a polygon is any line connecting two vertices that is not an edge, so an octagon has quite a few diagonals -- I think you're only counting lines connecting vertices that are strictly opposite each other. Here we can select a pair of vertices in 8*7/2! = 28 ways. If" ]

[ "1. polygon How many diagonals does a regular seven-sided polygon contain? 2. Hello, sri340! How many diagonals does a regular seven-sided polygon contain? Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ 3. Originally Posted by Soroban Hello, sri340!Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ ... But if you take a \"diagonal\" to mean a line connecting two non-adjacent vertices then the answer will be less (14, in fact).", "# polygon • February 1st 2010, 01:53 PM sri340 polygon How many diagonals does a regular seven-sided polygon contain? • February 1st 2010, 08:28 PM Soroban Hello, sri340! Quote: How many diagonals does a regular seven-sided polygon contain? Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ • February 2nd 2010, 08:19 AM Opalg Quote: Originally Posted by Soroban Hello, sri340!Take the 7 vertices, 2 at a time: . $_7C_2 \\:=\\:\\frac{7!}{2!\\,5!}$ ... But if you take a \"diagonal\" to mean a line connecting two non-adjacent vertices then the answer will be less (14, in fact).", "2) = 4.39334 cm Question: Answers:Think of a box or a cube. Those are examples of rectangular prisms. There are six faces (the sides of the solid). There are eight vertices (the \"corners\" of the solid). Answer: 6 faces 8 vertices Question:i asked to draw a polygon.and how many sides it is. Answers:A polygon will have as many sides as you want, at least three, for a triangle, for example. An octagon has 8 sides. Poly = many, gone = angle, octo = 8, in Greek ;-)", "# What is this polyhedron? Geometry Level pending Suppose a convex polyhedron has $$12$$ square faces, $$8$$ regular hexagon faces, and $$6$$ regular octagon faces. At each vertex of the polyhedron, one square, one hexagon, and one octagon meet. How many space diagonals does this polyhedron have? ×", "# Geometry Geometry Level 1 If a polygon has $27$ diagonals, then the number of sides of the polygon is ×", "vertices that are not adjacent to each other. How many different diagonals does it have? View Answer Find the interior and exterior angles of a 9-sided polygon. A diagonal connects any two non-consecutive vertices of a polygon. (Remember a rectangle is a type of parallelogram so rectangles get all of the parallelogram properties) If MO = 26 and the diagonals bisect each other, then MZ = ½(26) = 13 How many diagonals does a triangle have? It doesn't matter how long the length of each side is or how it's shaped - it will always be three.... How Many Perpendicular Lines Does A Square Have? Therefore, number of diagonals = \\\\frac{63 * 60}{2}\\\\) = 1890. By definition, it has n sides. Show Answer. Here are some related questions which you might be interested in reading. Since there would be no diagonal drawn back to itself, and the diagonals to each adjacent vertex would lie on top of the adjacent sides, the number of diagonals from a single vertex is three less the the number of sides, or n-3. 1) Regular pentagon P has all five diagonals drawn. Relevance. A rhombus is a type of parallelogram, and what distinguishes its shape is that all four of its sides are congruent.. Question 2: How many sides does a triangle have?", "7 28 Jun 2012, 17:58 How many diagonals does a polygon with 21 sides have, if one 1 19 Dec 2007, 17:48 How many diagonals does a polygon with 21 sides have, if one 3 06 Aug 2007, 17:32 How many diagonals does a polygon with 21 sides have, if one 5 09 Feb 2006, 17:00 a parallelogram has 2 sides: 14, 18 one of the diagonals is 13 24 Sep 2005, 06:03 Display posts from previous: Sort by", "There are several formulas for the rhombus that have to do with its: Sides (click for more detail). From each of the N vertices,... How Many Congruent Sides Does A Scalene Triangle Have? The number of diagonals is n(n-1)/2 - n. Hope this helps, Stephen La Rocque. A square has two diagonals. Number of diagonals can be drawn from 1 vertex in n sided polygon is (n - 3) Total number of diagonals that can be drawn is n (n-3)/2 as per ur question n is 20. as one of its vertices doesn't connect. What is the sum of the measures of the angles of a convex quadrilateral? Answer. And. It can have no diagonals. 0 0. Show that all four triangles created by these diagonals – ΔAOD, ΔCOD, ΔAOB and ΔCOB – have equal areas. The three... How Many Diagonals Does A 9 Sided Shape Have ? (a) Since sum of linear pair angles is. How many diagonals has a 15-sided polygon? In any convex polygon, you can count the number of diagonals as follows. Why? This problem is adapted from the South East Asian Mathematics Competition (b) Find. a triangle has three diagonals as well as three sides Can a scalene triangle be a triangle? 1 C. 2 D. depends on the trapezoid. An angle only", "Comment Share Q) # A polygon has 54 diagonals. Then the number of its sides is : $\\begin {array} {1 1} (1)\\;7 & \\quad (2)\\;9 \\\\ (3)\\;10 & \\quad (4)\\;12 \\end {array}$", "# Geometry Geometry Level 1 If a polygon has $$27$$ diagonals, then the number of sides of the polygon is × Problem Loading... Note Loading... Set Loading...", "5 9 14 20 27 35 44 54 sides____ 3 4 5 6 7_ 8_ 9_ 10_ 11_ 12 _ Re: A polygon has 12 edges. How many different diagonals does it [#permalink] 01 Jan 2019, 23:28 Display posts from previous: Sort by", "Math Secondary School How many diagonals are in triangle?? Since the diagonals of a rectangle are congruent MO = 26. d How many diagonals … To see how many diagonals intersections exist, we just need to know that we need 2 diagonals for one intersection,so we need 4 vertex in total there are $$\\binom{7}{4}=35$$ diagonals intersections. Figure – 12: Triangle counting in Fig – 12 = 45. Therefore, a quadrilateral has two diagonals, joining opposite pairs of vertices. A rectangle has two diagonals, and each is the same length. The formula is: 0.5*(n2-3n) = number of diagonals when n is the number of sides of the polygon So i though there were $$7\\cdot35\\cdot34$$ triangles sharing one vertex with the heptagon and having the other two on diagonals intersections. How many diagonals does a 27-gon have? If you know side lengths of the rectangle, you can easily find the length of the diagonal using the Pythagorean Theorem, since a diagonal divides a rectangle into two right triangles. 2. Diagonals bisect vertex angles. 7. How much percent is anil's income more than aman's income ?​, Please aap relevant answers to delete mat karo....konsi dusmani nikal rahe ho​, ask me the answer it's urgent please guys ​, manjoor h..........xDhehehe.....mujhe to ye pd kr hnsi hi aa gyi​, solve the following quadratic", "many diagonals does a 63-sided convex polygon have? [#permalink] 08 Aug 2017, 13:58 Display posts from previous: Sort by", "from one It also has three sides and three vertices. The angle sum of a convex polygon with number of sides 7 is From each vertex, N-4 diagonals can be drawn, where N is the number of vertices. How many sides does the polygon have? The number of triangles is n-2 (above). How many diagonals does each of the following have? Answer: (a) A convex quadrilateral has two diagonals. Diagonals bisect vertex angles. The sum of the interior angles of a 7-gon is. The three perpendicular bisectors of a triangle's three sides intersect at the circumcenter (the center of the circle through the three vertices). The diagonals have the following properties: The two diagonals are congruent (same length). You know how the angles of a triangle always add up to 180 0?Why is that? Examine the table. Can Someone Describe The Term \"Learning\" In Organizational Behavior? So from every vertex we can draw 6 diagonals, and we can do this from all 9 vertices for a total of 54 diagonals. You have learned a lot about particularly important parts of polygons, their diagonals. Well, the diagonals I think will be, 1/2 * 100 * (100 - 3) = 4850. the triangles I have no idea, quite a few I'd guess. What is the measure of an interior angle?", "choose 4 different from them. Let they be P<Q<R<S. You can construct two pairs of line segments: (PQ, RS) and (PR, QS). If the endpoint of the segments are a1, a2 and b1, b2, (so as a1<a2 and b1<b2) b1 and b2 has to be at opposite sides of the segment (a1,a2) so as the segment (b1,b2) intersects (a1,a2) inside the polygon. That means b1<a2. So you have only a single pair of intersecting diagonals for every 4 points, (PR) and (QS), that is one intersection. It was said that the intersections do not coincide. So you have $(^n_4)$ intersections. See picture. You choose the vertices 1,3,5,7. The diagonals (1,5) and (3,7) intersect each other inside the polygon, the other pair of diagonals do not. ehild File size: 16.8 KB Views: 217 8. Jan 6, 2013 ### sankalpmittal Thanks ILS and ehild , for such insights ! And apologies for late reply. I come online but rarely...", "How many more diagonals are in an octagon than in a hexagon? $20-9=11$ Answer: $11$", "each point will have access to 36-1 points as we need to exclude the vertices of square, hexagon, octagon and -1 for the point itself. So now we get 35 for one point for the next we will have 34, 33 and so on. So total 35+34+33+32.. +1 1 Like #13 Answer is 840 @Mayank_2019_1 bro 1 Like #14 For which one ? #15 Okay got it ... Didn't do that properly. What I missed was that when each point is connected to 35 different points then we don't get the points that are there in the plane of square, hexagon, octagon. and then we can't subtract 1 from that and say that the next point will have 34 connections. so the solution: As explained above the polyhedron will have 48 vertices. By the same logic we will have 483/2 = 72 edges. We will first count all the connections so we will have 48+47+46+45+...+1 = 1128 connections Now we need to exclude the ones which are in the plane of a point. So first we exclude all the edges we have 72 edges. For 1 square we have 2 diagonals so 12 2=24 For 1 hexagon we will have 3+3+2+1 = 9 diagonals so 89=72 For 1 octagon we will have 5+5+4+3+2+1=20 diagonals so 6 20=120 So", "A convex polyhedron $$P$$ has 26 vertices, 60 edges, and 36 faces, 24 of which are triangular, and 12 of which are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagonals does $$P$$ have? (第二十二届AIME1 2004 第3题)", "# How many triangles exist in the diagonals intersections of an heptagon? In a heptagon (not necessarily regular) with all its diagonals marked, how many triangles exist with two vertex at two diagonals intersections and sharing one vertex with the heptagon? To see how many diagonals intersections exist, we just need to know that we need 2 diagonals for one intersection,so we need 4 vertex in total there are $$\\binom{7}{4}=35$$ diagonals intersections. So i though there were $$7\\cdot35\\cdot34$$ triangles sharing one vertex with the heptagon and having the other two on diagonals intersections. Actually i don't find this correct at all because there are non-collinear diagonals intersections that you can select, and you're counting them as triangles. So i was trying to count the diagonals intersections in each diagonal and there are some with 4 and 6 but i don't think this to be efficiently at all. So i was wondering for any hint,or answer. thanks. ## 1 Answer Assuming that you're fine with triangles whose sides are not, themselves, parts of the diagonals, that will be fine for a heptagon whose vertices are in general convex position: we just need to subtract off triangles with two points on a diagonal. To do this as generally as possible: if we fix a vertex $P$, the number of degenerate", "a type of parallelogram so rectangles get all of the parallelogram properties) If MO = 26 and the diagonals bisect each other, then MZ = ½(26) = 13 b How many diagonals does a quadrilateral have? Type – 4 : Counting triangles with in embedded Triangle. A triangle has zero diagonals. In the case of a pentagon, which \"n\" will be 5, the formula as expected is equal to 5. A triangle has no diagonals How many diagonals does a triangle have? But because a polygon can’t have a negative number of sides, n must be 15. There are a number of theorems that we need to look at before we doing the proof. All polygons, except for the triangle, have a number of diagonals. Is equal to 5 the rhombus that have to do with its: sides ( click for more detail.. Meet at a vertex of the quadrilateral is produced from what animal 's coat diagonals well... €” Here 's How to how many diagonals in triangle them but the vertices must not be to. And AB respectively and AB respectively: sides ( click for more )., hexagon = 4 triangles, pentagon = 3 triangles, pentagon = 3 triangles, BCD DAB! Are a number of diagonals of a rectangle divides it into two right triangles, and." ]

[ "# Problem #1818 1818 The regular octagon $ABCDEFGH$ has its center at $J$. Each of the vertices and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal. In how many ways can this be done? $\\textbf{(A)}\\ 384 \\qquad\\textbf{(B)}\\ 576 \\qquad\\textbf{(C)}\\ 1152 \\qquad\\textbf{(D)}\\ 1680 \\qquad\\textbf{(E)}\\ 3456$ $[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label(\"A\",A,NNW); label(\"B\",B,NNE); label(\"C\",C,ENE); label(\"D\",D,ESE); label(\"E\",E,SSE); label(\"F\",F,SSW); label(\"G\",G,WSW); label(\"H\",H,WNW); label(\"J\",J,SE); [/asy]$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# 2013 AMC 10B Problems/Problem 22 ## Problem The regular octagon $ABCDEFGH$ has its center at $J$. Each of the vertices and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal. In how many ways can this be done? $\\textbf{(A)}\\ 384 \\qquad\\textbf{(B)}\\ 576 \\qquad\\textbf{(C)}\\ 1152 \\qquad\\textbf{(D)}\\ 1680 \\qquad\\textbf{(E)}\\ 3456$ $[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label(\"A\",A,NNW); label(\"B\",B,NNE); label(\"C\",C,ENE); label(\"D\",D,ESE); label(\"E\",E,SSE); label(\"F\",F,SSW); label(\"G\",G,WSW); label(\"H\",H,WNW); label(\"J\",J,SE); [/asy]$ ## Solution 1 First of all, note that $J$ must be $1$, $5$, or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have: $[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label(\"A\",A,NNW); label(\"B\",B,NNE); label(\"C\",C,ENE); label(\"D\",D,ESE); label(\"E\",E,SSE); label(\"F\",F,SSW); label(\"G\",G,WSW); label(\"H\",H,WNW); label(\"J (1, 5, 9)\",J,SE); [/asy]$ We also notice that $A+E = B+F = C+G =", "and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal. In how many ways can this be done? $\\textbf{(A)}\\ 384 \\qquad\\textbf{(B)}\\ 576 \\qquad\\textbf{(C)}\\ 1152 \\qquad\\textbf{(D)}\\ 1680 \\qquad\\textbf{(E)}\\ 3456$ $[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label(\"A\",A,NNW); label(\"B\",B,NNE); label(\"C\",C,ENE); label(\"D\",D,ESE); label(\"E\",E,SSE); label(\"F\",F,SSW); label(\"G\",G,WSW); label(\"H\",H,WNW); label(\"J\",J,SE); [/asy]$ Solution ## Problem 23 In triangle $ABC$, $AB = 13$, $BC = 14$, and $CA = 15$. Distinct points $D$, $E$, and $F$ lie on segments $\\overline{BC}$, $\\overline{CA}$, and $\\overline{DE}$, respectively, such that $\\overline{AD} \\perp \\overline{BC}$, $\\overline{DE} \\perp \\overline{AC}$, and $\\overline{AF} \\perp \\overline{BF}$. The length of segment $\\overline{DF}$ can be written as $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$? $\\textbf{(A)}\\ 18\\qquad\\textbf{(B)}\\ 21\\qquad\\textbf{(C)}\\ 24\\qquad\\textbf{(D)}\\ 27\\qquad\\textbf{(E)}\\ 30$ ## Problem 24 A positive integer $n$ is nice if there is a positive integer $m$ with exactly four positive divisors (including $1$ and $m$) such that the sum of the four divisors is equal to $n$. How many numbers in the set $\\{ 2010,2011,2012,\\dotsc,2019 \\}$ are", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "(0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)", "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "label(\"d\",(15*k-1,8),N); label(\"i\",(15*k-1,7),N); label(\"s\",(15*k-1,6),N); label(\"t\",(15*k-1,5),N); label(\"a\",(15*k-1,4),N); label(\"n\",(15*k-1,3),N); label(\"c\",(15*k-1,2),N); label(\"e\",(15*k-1,1),N); label(\"time\",(15*k+8,0),S); } for (int k=0; k<2; ++k) { label(\"d\",(15*k-1,8-15),N); label(\"i\",(15*k-1,7-15),N); label(\"s\",(15*k-1,6-15),N); label(\"t\",(15*k-1,5-15),N); label(\"a\",(15*k-1,4-15),N); label(\"n\",(15*k-1,3-15),N); label(\"c\",(15*k-1,2-15),N); label(\"e\",(15*k-1,1-15),N); label(\"time\",(15*k+8,0-15),S); } label(\"(A)\",(5,9),N); label(\"(B)\",(20,9),N); label(\"(C)\",(35,9),N); label(\"(D)\",(5,-6),N); label(\"(E)\",(20,-6),N); [/asy]$ ## Problem 19 Around the outside of a $4$ by $4$ square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters. Another square, $ABCD$, has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles. The area of the square $ABCD$ is $[asy] pair A,B,C,D; A = origin; B = (4,0); C = (4,4); D = (0,4); draw(A--B--C--D--cycle); draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 32 \\qquad \\text{(C)}\\ 36 \\qquad \\text{(D)}\\ 48 \\qquad \\text{(E)}\\ 64$ ## Problem 20 Let $W,X,Y$ and $Z$ be four different digits selected from the set $\\{ 1,2,3,4,5,6,7,8,9\\}.$ If the sum $\\dfrac{W}{X} + \\dfrac{Y}{Z}$ is to be as small as possible, then $\\dfrac{W}{X} + \\dfrac{Y}{Z}$ must equal $\\text{(A)}\\ \\dfrac{2}{17} \\qquad \\text{(B)}\\ \\dfrac{3}{17} \\qquad \\text{(C)}\\ \\dfrac{17}{72} \\qquad \\text{(D)}\\ \\dfrac{25}{72} \\qquad \\text{(E)}\\ \\dfrac{13}{36}$ ## Problem 21 A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure", "(A) at (-1,0); \\coordinate[label=right:B] (B) at (1,0); \\coordinate[label=C] (C) at (0,{sqrt(3)}); \\coordinate[label=below: D] (D) at (0,0); \\coordinate[label=left:E] (E) at (0,1); \\coordinate[label=left:F] (F) at (0,{2-sqrt(3)}); \\coordinate[label=below:$A_1$] (A1) at ({2/sqrt(3)-1},0); \\draw (A) -- (B) -- (C) -- cycle; \\draw (D) -- (C); \\draw[fill] (E) circle (.01); \\draw (F) -- (A1); \\end{tikzpicture} Or did I misunderstand? How can we draw a line parallel to $AC$ from $F$ ? Isn't $F$ on $AC$ ? Or do you mean from $E$ instead of $F$ ? Because: \\begin{tikzpicture}[scale=3] \\coordinate[label=left:A] (A) at (-1,0); \\coordinate[label=right:B] (B) at (1,0); \\coordinate[label=C] (C) at (0,{sqrt(3)}); \\coordinate[label=below: D] (D) at (0,0); \\coordinate[label=left:E] (E) at (0,1); \\coordinate[label=left:F] (F) at (0,{2-sqrt(3)}); \\coordinate[label=below:$A_1$] (A1) at ({1-2/sqrt(3)},0); \\coordinate[label=below:$B_1$] (B1) at ({2/sqrt(3)-1},0); \\draw (A) -- (B) -- (C) -- cycle; \\draw (D) -- (C); \\draw[fill] (E) circle (.01); \\draw (F) -- (A1); \\draw (F) -- (B1); \\end{tikzpicture} Why will we not reach zero? We have a new equilateral triangle that has non-zero size. After 2 more iterations, we will have yet again a new equilateral triangle that has non-zero size, won't we? #### mathmari ##### Well-known member MHB Site Helper I thought that we get the following construction: Isn't this correct? #### Klaas van Aarsen ##### MHB Seeker Staff member I thought that we get the following construction: Isn't this correct? Ah okay. My mistake.", "= (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6 (Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find", "dot(\"S\", S, E); dot(\"K\",K,E); dot(\"L\",L,W); // dot(\"R\",R,NE); dot(\"Q\",Q,NW); dot(\"G\", G, SE); dot(\"Q\",Q,NW); dot(\"P\",P,NE); dot(\"M\",M,S); draw(arc1,dashed); draw(arc2, dashed); draw(circle2); draw(A--D--B--cycle); // draw(A--C, dashed); draw(A--H); draw(C--S--H--cycle); draw(C--T--H); draw(D--C--B); draw(D--L--K--B, dashed); draw(T--G, dashed); draw(K--H, dashed); draw(D--Q--P--B,dashed); draw(Q--H--P,dashed); draw(C--L--H, dashed); draw(C--K,dashed); draw(T--S,dashed); // draw(anglemark(H,S,K)); // draw(anglemark(K,H,S)); label(\"x\",C+(0,0.1),N); label(\"y\",C+(-0.1,0.3),N); label(\"w\",H+(-0.03,0.1), N); label(\"z\", H+(0.06,0.12), N ); label(\"u\",T+(-0.1,-0.2), S); label(\"v\", S+(0,-0.2), S); label(\"t\",T+(0.1,-0.3), S); label(\"s\", S+(-0.08,-0.2), S); [/asy]$ Denote $\\angle{HSB}=v$, $\\angle{HTD}=u$, $\\angle{HSC}=s$, $\\angle{HTC}=t$, $\\angle{HCS}=x$, $\\angle{HCT}=y$, $\\angle{AHS}=z$, $\\angle{AHT}=w$. Since $\\angle{CHS}-\\angle{CSB} =90$ and $\\angle{CHT}-\\angle{CTD}=90$, we have $\\angle{CSA}=x+90$, $\\angle{CTA}=y+90$. Since $\\angle{CHS} = \\angle{CSB}+90$, the tangent of the circumcircle of $\\triangle{CSH}$ at point $S$ is perpendicular to $SB$; therefore, the circumcenter of $\\triangle{CSH}$ (point $K$) is on $AB$. Similarly, the circumcenter of $\\triangle{CTH}$ (point $L$) is on $AD$. In addition, $KL$ is the perpendicular bisector of $CH$. Extend $CB$ to meet circumcircle of $\\triangle{CSH}$ at $P$, and extend $CD$ to meet circumcircle of $\\triangle{CTH}$ at $Q$. Then, since $\\angle{ADC}=\\angle{ABC}=90$, $AD$ and $AB$ are the perpendicular bisector of $CQ$ and $CP$, respectively; hence $A$ is the circumcenter of $\\triangle{PCQ}$. Since $B$ and $D$ are midpoints on $CP$ and $CQ$, $PQ \\parallel BD$; also, $AH \\perp BD$, so $AH \\perp PQ$. Since $A$ is the circumcenter, $AH$ is also the perpendicular bisector of $PQ$. Hence, $$HP=HQ$$ We have $$u=\\angle{CHT}-90=(90-w+\\angle{DHC}) - 90$$ $$v=\\angle{CHS}-90 = (90-z+\\angle{BHC}) - 90$$ Hence, $u+v = -w-z+\\angle{DHC}+\\angle{BHC} =", "my armpit Dec 29 '13 at 20:11 • See, we have a constraint of symmetricity? Do you have others that's what I asked. – percusse Dec 29 '13 at 20:21 Here is the Asymptote version, which does not use the general Asymptote commands to split a path (guide), it just performs a simple subdivision of the cubic Bezier segment: // split.asy: size(5cm); pair A,B,C,D; A=(3,3); B=(1,1); C=(-1,1); D=(-3,3); pair P,B1,C1,B2,C2; P=(A+D+3(B+C))/8; B1 = (A+B)/2; C1 = ((A+C)/2+B)/2; C2 = (D+C)/2; B2 = ((D+B)/2+C)/2; guide g=A..controls B and C..D; guide gl=A..controls B1 and C1..P; guide gr=P..controls B2 and C2..D; draw(g); draw(gl,deepgreen+1.6bp+opacity(0.5)); draw(gr,red+1.6bp+opacity(0.5)); draw((0,3)--(0,-1),red); dot(A--B--C--D--P,UnFill); dot(B1--C1,deepgreen,UnFill); dot(B2--C2,red,UnFill); label(\"$A$\",A,SE); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$D$\",D,SW); label(\"$P$\",P,NE); label(\"$B_1$\",B1,E,deepgreen); label(\"$C_1$\",C1,E,deepgreen); label(\"$B_2$\",B2,W,red); label(\"$C_2$\",C2,W,red); To get a standalone split.pdf, run asy -f pdf split.asy. • I really need the converting formulae. – kiss my armpit Dec 29 '13 at 19:29 • @Stiff Jokes: Btw, the Bezier curve is a very interesting object, you might enjoy studying it in a spare time. – g.kov Dec 29 '13 at 19:43 A translated version for @g.kov's Asymptote answer in PSTricks. \\documentclass[pstricks,border=24pt]{standalone} \\usepackage{pst-eucl} \\begin{document} \\begin{pspicture}[showgrid=true](-3,-1)(3,3) \\pstGeonode (3,3){A} (1,1){B} (-1,1){C} (-3,3){D} \\psbezier(A)(B)(C)(D) \\psline[linecolor=red](0,3)(0,-1) \\nodexn{.125(A)+.125(D)+.375(B)+.375(C)}{R'} \\nodexn{.5(A)+.5(B)}{P'} \\nodexn{.25(A)+.5(B)+.25(C)}{Q'} \\pstGeonode (P'){P} (Q'){Q} (R'){R} \\psbezier[linecolor=red](A)(P)(Q)(R) \\end{pspicture} \\end{document}", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "# Difference between revisions of \"2020 AIME I Problems/Problem 9\" ## Problem Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$ ## Solution $[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; draw(b1 -- a1 -- c1); draw(b2 -- a2 -- c2); draw(b3 -- a3 -- c3); dot(a1); dot(a2); dot(a3); label(\"a_1\", a1, NE); label(\"a_2\", a2, NE); label(\"a_3\", a3, NE); label(\"b_1\", b1, S); label(\"b_2\", b2, S); label(\"b_3\", b3, S); label(\"c_1\", c1, W); label(\"c_2\", c2, W); label(\"c_3\", c3,", "label(\"D\",D,NNW); label(\"C\",C,SW); label(\"H\",H,SE); draw(e--H,dashed); label(\"O\",(0,0),NE); label(\"1\",(C--O),N); label(\"1\",(H--O),N); label(\"2\",(A--C),N); label(\"2\",(H--B),N); label(\"3\",(O--D),NE); label(\"3\",(O--e),NE); label(\"2\\sqrt{2}\",(D--C),W); label(\"2\\sqrt{2}\",(H--e),E); draw(rightanglemark(e,(e.x,0),A,2)); draw(rightanglemark(D,C,B,2));[/asy]$ Let the center of the circle be $O$. Without loss of generality, let the radius of the circle be equal to $3$. Thus, $AO=3$ and $OB=3$. As a consequence of $2(AC)=BC$, $AC=2$ and $CO=1$. Also, we know that $DO$ and $OE$ are both equal to $3$ due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to $\\sqrt{3^2-1^2}$ or $2\\sqrt{2}$. Now we know that the area of $[ABD]$ is equal to $\\frac{(3+2+1)(2\\sqrt{2})}{2}$ or $6\\sqrt{2}$. Know we need to find the area of $[DCE]$. By simple inspection $[COD]$ $\\cong$ $[HOE]$ due to angles being equal and CPCTC. Thus $HE=2\\sqrt{2}$ and $OH=1$. Know we know the area of $[CHE]=\\frac{(1+1)(2\\sqrt{2})}{2}$ or $2\\sqrt{2}$. We also know that the area of $[OHE]=\\frac{(1)(2\\sqrt{2})}{2}$ or $\\sqrt{2}$. Thus the area of $[COE]=2\\sqrt{2}-\\sqrt{2}$ or $\\sqrt{2}$. We also can calculate the area of $[DOC]$ to be $\\frac{(1)(2\\sqrt{2})}{2}$ or $\\sqrt{2}$. Thus $[DCE]$ is equal to $[COE]$ + $[DOC]$ or $\\sqrt{2}+\\sqrt{2}$ or $2\\sqrt{2}$. The ratio between $[DCE]$ and $[ABD]$ is equal to $\\frac{2\\sqrt{2}}{6\\sqrt{2}}$ or $\\frac{1}{3}$ $\\Longrightarrow \\mathrm{(C)}$. Solution 5 We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for $D (x,y)$, and notice how $E$ is a", "= \\left\\lfloor \\frac{\\text{deg}(V)}{2} \\right\\rfloor$ where deg$(V)$ is the number of edges connected to $V$. This is because to visit any of these vertices, we would have to visit and exit on two different edges (This is a pretty standard principle in graph theory). We will label each vertex with this number. $[asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label(\"A\", (0,2), W); label(\"L\", (3,0), E); label(\"1\", (1,2), N); label(\"1\", (2,2), N); label(\"1\", (3,2), N); label(\"1\", (0,1), W); label(\"2\", (1,1), NW); label(\"2\", (2,1), NW); label(\"1\", (3,1), E); label(\"1\", (0,0),S); label(\"1\", (1,0),S); label(\"1\", (2,0),S); [/asy]$ Additionally, notice that if we visit $n$ vertices (not necessarily distinct; i.e. if we visit a vertex twice we count it as two vertices) along our path, we must traverse $n-1$ edges (this is also pretty easy to prove). Thus, if we want to traverse 13 edges total, we have to visit 14 vertices. We will visit $A$ and $L$ for sure, leaving us with 12", "- 40\\sqrt{10}$. Through power of a point, we can find out that $(CE)(DE)=20\\sqrt{10} - 20$, so $CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\\sqrt{10}) + 2(20\\sqrt{10} - 20) = \\boxed{\\textbf{(E) } 100}$. ~Math_Wiz_3.14 ## Solution 4 (Reflections) $[asy] draw(circle((0,0),7.07)); dot((-7.07,0)); label(\"A\",(-7.07,0),W); dot((7.07,0)); label(\"B\",(7.07,0),E); dot((0,0)); label(\"O\",(0,0),N); dot((-2.24,-6.71)); label(\"C\",(-2.24,-6.71),SSW); dot((6.71,2.24)); label(\"D\",(6.71,2.24),NE); draw((-2.24,-6.71)--(6.71,2.24)); dot((6.71,-2.24)); label(\"D'\",(6.71,-2.24),SE); draw((4.51,0)--(6.71,-2.24)); dot((4.51,0)); label(\"E\",(4.51,0),NNW); draw((-7.07,0)--(7.07,0)); draw((0,0)--(-2.24,-6.71)); draw((0,0)--(6.71,-2.24)); draw((-2.24,-6.71)--(6.71,-2.24)); [/asy]$ Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$, we have that $\\angle BED'=45$. Since $\\angle AEC=45$, $\\angle CED'=90$. Since $DE=ED'$, by the Pythagorean Theorem, our desired solution is just $CD'^2$. Looking next to circle arcs, we know that $\\angle AEC=\\frac{\\overarc{AC}+\\overarc{BD}}{2}=45$, so $\\overarc{AC}+\\overarc{BD}=90$. Since $\\overarc{BD'}=\\overarc{BD}$, and $\\overarc{AC}+\\overarc{BD'}+\\overarc{CD'}=180$, $\\overarc{CD'}=90$. Thus, $\\angle COD'=90$. Since $OC=OD'=5\\sqrt{2}$, by the Pythagorean Theorem, the desired $CD'^2= \\boxed{\\textbf{(E) } 100}$. ~sofas103 ## See Also 2020 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's", "(i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5), white); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5), white); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5), white); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5), white); label(\"$b^2$\", (0.5,-2.5), white); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5), white); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5), white); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5), white); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] Sorry but I don't know how to show the grids. 1. 👍 2. 👎 3. 👁 1. each row is a geometric progression. Starting with n=0, row n is the GP b^n (1 + a + a^2 + ...) so the sum Sn of row n is b^n * 1/(1-a) Now you have another GP, which is the sum of all the row sums. That is 1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b) So the grand total of the whole grid is 1/((1-a)(1-b)) 1. 👍 2. 👎 2. wertwetrwetw 1. 👍 2. 👎 ## Similar Questions 1. ### alebra Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation then find the domain. (f*g)(x) a.2x^3-x^2-4x+3; domain:all real numbers b.2x^3+x^2-3x; domain: all real numbers c.2x^3+x^2+4x-3; domain: negative real numbers 2. ### Precalc To which set(s) of numbers does the number sqrt -16 belong? Select all that apply. real numbers complex numbers*** rational", "# 2020 AIME I Problems/Problem 9 ## Problem Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$ ## Solution 1 $[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; draw(b1 -- a1 -- c1); draw(b2 -- a2 -- c2); draw(b3 -- a3 -- c3); dot(a1); dot(a2); dot(a3); label(\"a_1\", a1, NE); label(\"a_2\", a2, NE); label(\"a_3\", a3, NE); label(\"b_1\", b1, S); label(\"b_2\", b2, S); label(\"b_3\", b3, S); label(\"c_1\", c1, W); label(\"c_2\", c2, W); label(\"c_3\", c3, W); [/asy]$ First,", "A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"D\",D,SW); label(\"E\",DD,SE); label(\"F\",CC,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Since $\\sqrt{96+529}<\\sqrt{96+961}$, $25$ is the shorter length, so the answer is $\\boxed{\\textbf{(B) }25}$. Solution 2 size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"$A$\",A,NW); label(\"$B$\",B,NE); label(\"$C$\",C,SE); label(\"$D$\",D,SW); label(\"$D$\",D,SW); label(\"$21$\",(C+D)/2,S); (Error compiling LaTeX. E = (4,7); ^ 1b1ea1b866e85d255ff55009031082b8f710a74e.asy: 9.1: modifying non-public field outside of structure)" ]

[ "$6 \\cdot 5 \\cdot 4 \\cdot 4 \\cdot 4 = 1920$ combinations. In the case that $C$ has a different color from $A$ and $D$ has the same color as $A$, $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices, $D$ has $1$ choice, and $E$ has $5$ choices, resulting in a possible of $6 \\cdot 5 \\cdot 4 \\cdot 1 \\cdot 5 = 600$ combinations. Adding all those combinations up, we get $600+1920+600=\\boxed{3120 \\ \\mathbf{(C)}}$. ## Solution 2 First, notice that there can be $3$ cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors. Case $1$: There are $6!$ ways of assigning each vertex a different color. $6! = 720$ Case $2$: There are $\\frac {6!}{2!} * 5$ ways. After picking four colors, we can rotate our pentagon in $5$ ways to get different outcomes. $\\frac {6!}{2} * 5 = 1800$ Case $3$: There are $\\frac {\\frac {6!}{3!} * 10}{2!}$ ways of arranging the final case. We can pick $3$ colors for our pentagon. There are $5$ spots for the first pair of colors. Then, there are $2$ possible ways we can put the final pair in the last $3$", "D+H$. WLOG, assume that $J = 1$. Thus the pairs of vertices must be $9$ and $2$, $8$ and $3$, $7$ and $4$, and $6$ and $5$. There are $4! = 24$ ways to assign these to the vertices. Furthermore, there are $2^{4} = 16$ ways to switch them (i.e. do $2$ $9$ instead of $9$ $2$). Thus, there are $16(24) = 384$ ways for each possible J value. There are $3$ possible J values that still preserve symmetry: $384(3) = \\boxed{\\textbf{(C) }1152}$ ## Solution 2 As in solution 1, $J$ must be $1$, $5$, or $9$ giving us 3 choices. Additionally $A+E = B+F = C+G = D+H$. This means once we choose $J$ there are $8$ remaining choices. Going clockwise from $A$ we count, $8$ possibilities for $A$. Choosing $A$ also determines $E$ which leaves $6$ choices for $B$, once $B$ is chosen it also determines $F$ leaving $4$ choices for $C$. Once $C$ is chosen it determines $G$ leaving $2$ choices for $D$. Choosing $D$ determines $H$, exhausting the numbers. Additionally, there are three possible values for $J$. To get the answer we multiply $2\\cdot4\\cdot6\\cdot8\\cdot3=\\boxed{\\textbf{(C) }1152}$. ## Side Note to Solution 1 & 2 Solution 1 and 2 states that $J=1, 5, 9$ without rigor analysis. Here it is: Let $S=A+J+E=B+J+F=C+J+G=D+J+H$ $4S=A+B+C+D+E+F+G+H+4J$ $4S=45+3J$ $4S=3(15+J)$ Because", "solutions to $i+j <10, i >1, j > 1$. Putting $x = i-1, y = j-1$, we need the solutions $x+y < 8$ where $x >0, y > 0$. By stars and bars method this is $\\binom{1}{1}+\\binom{2}{1}+\\binom{3}{1}+\\binom{4}{1}+\\binom{5}{1} +\\binom{6}{1}= 21$. Thus the answer for part 1 is $21\\times 12=252$. The polygon cannot be arbitrary. Odd things can happen, for example with a $12$-sided cross. Let our polygon be regular. By quadrilateral we mean convex quadrilateral. (1) As in your solution, there are $12$ ways to choose the side in common with the $12$-gon. The \"opposite\" side's vertices are chosen from the $8$ remaining candidate vertices. There are $\\binom{8}{2}$ ways to choose $2$ vertices. But $7$ of these pairs are adjacent, leaving $21$ choices, for a total of $(12)(21)$. (2) We can choose $3$ consecutive vertices in $12$ ways, and for each way choose a non-consecutive in $7$ ways. Thus far we have a total of $84$. Now we count the cases where the edges shared with the $12$-gon are opposite. Choose an edge of the $12$-gon, and colour it blue. There are $12$ ways to do this. Now choose a non-adjacent edge and colour it red. There are $7$ ways to do this. The product $84$ double-counts our quadrilaterals. So there are $42$ of this type, for a total", "and there are $12$ edges, so $b = 12$. To find $c$, each three-cycle is distinguished by the vertex, and there are $8$ edges. However, since the two three-cycles are indistinguishable, $c = 8/2 = 4$. Clearly $d = 1$. Finally, $$8(1)(1)(1) + 12(1)(2) + 4(2)(2) + (32) = 80$$ Each bug has $4$ possibilities to choose from, so the probability is $\\frac{80}{4^6} = \\frac{5}{256}$. ## Solution 3 We use the idea of cycles. Case 1: 2 3-cycles We have 4 ways to divide the faces into cycles (do casework on which two ants share a cycle with ant A), then $2^2$ ways to choose the direction of the cycle. There are $4 \\cdot 4 = 16.$ Case 2: 3 2-cycles There are $4 \\cdot 2 = 8$ ways to do so. Case 3: 1 4-cycle, 1 2-cycle There are $2 \\cdot 4 = 8$ ways here because we can only choose the 2-cycle without ants who are on the top of the octahedron. Case 4: 1 6-cycle There are $4 \\cdot 2 = 8$ ways here because we do casework on which place the ant A goes and which ant goes to vertex A (which is at the top). Overall, there are $40$ cases. Answer is $40/2^{12} = 5/256$ ~Williamgolly", "a total of $12 \\cdot 4 = 8 \\cdot 6 = 6 \\cdot 8 = 48$ vertices. This means that for each segment we have $48$ choices of vertices for the first endpoint of the segment. Since each vertex is the meeting point of a square, octagon, and hexagon, then there are $3$ other vertices of the square that are not the first one, and connecting the first point to any of these would result in a segment that lies on a face or edge. Similarly, there are $5$ points on the adjacent hexagon and $7$ points on adjoining octagon that, when connected to the first point, would result in a diagonal or edge. However, the square and hexagon share a vertex, as do the square and octagon, and the hexagon and octagon. Subtracting these from the $47$ vertices we have left to choose from, and adding the $3$ that we counted twice, we get $$48 \\cdot (47 - 3 - 5 - 7 + 3) = 48 \\cdot 35 = 1680$$ We over-counted, however, as choosing vertex $A$ then $B$ is the same thing as choosing $B$ then $A$, so we must divide $1680 / 2 = \\boxed{840}$.", "-2 \\\\ \\mbox{11} & j & j & \\cdots & 0 \\\\ \\mbox{12} & i,j & j & x_i\\cdots & 0 \\\\ \\mbox{13} & \\_ & i,j & -x_i-x_j\\cdots & 2 \\\\ \\mbox{14} & i & i,j & -x_j\\cdots & 0 \\\\ \\mbox{15} & j & i,j & -x_i\\cdots & 0 \\\\ \\mbox{16} & i,j & i,j & \\cdots & 0 \\end{array}$$ the total coefficient of $$x_ix_j$$ over all the cases is 0. Therefore $$\\sum_{S,T\\subseteq I_d}(x_S-x_T)^2$$ only contains square terms and by symmetry must then be a multiple of $$\\sum_{i=1}^d x_i^2 \\quad\\blacksquare$$ Regarding the fact that $$v_d$$ above is not minimal $$\\ell_\\infty$$-wise, counterexamples are provided via the Atkinson-Negro-Santoro sequence: $$d=4: [3,5,6,7]$$ $$d=5: [6,9,11,12,13]$$ $$d=6: [11,17,20,22,23,24]$$ etc. • I should mention that the Atkinson-Negro-Santoro derived sequences mentioned above are by no means the best in the $\\ell_\\infty$ metric. For example $[39,59,70,77,78,79,81,84]$ is a better one for $d=8$. Feb 14 at 13:28 Given a unit vector $u \\in \\mathbb R^d$, the \"heights\" of vertices of the $n$-cube where $u$ is regarded as the vertical direction are the sums of subsets of the entries of $u$. Thus the minimum separation is the minimum difference between the sums of two distinct subsets of these entries. If you take $$u = [1,2,\\ldots,2^{d-1}]/\\sqrt{1^2 + 2^2 + \\ldots (2^{d-1})^2} = \\sqrt{\\dfrac{3}{4^d-1}}[1,2,\\ldots,2^{d-1}]$$ you get uniform separation", "$1-8, 2-7, 3-6, 4-5$. thinking about $4$ end points number need to have sum of $18$. It is easy to notice only $1-7-6-4$ vs $8-2-3-5$ would work. So if we fix one direction $1-8 ($or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$ Now, the problem is same as the problem to arrange $4$ points in a two-dimensional square. which is $4!/4$=$\\boxed{\\textbf{(C) }6.}$ ## Solution 2 Again, all faces sum to $18.$ If $x,y,z$ are the vertices next to one, then the remaining vertices are $17-x-y, 17-y-z, 17-x-z, x+y+z-16.$ Now it remains to test possibilities. Note that we must have $x+y+z>17.$ Without loss of generality, let $x $3,7,8:$ Does not work. $4,6,8:$ Works. $5,6,7:$ Does not work. $5,6,8:$ Works. $5,7,8:$ Does not work. $6,7,8:$ Works. So our answer is $3\\cdot 2=\\boxed{\\textbf{(C) }6.}$ ## Solution 3 We know the sum of each face is $18.$ If we look at an edge of the cube whose numbers sum to $x$, it must be possible to achieve the sum $18-x$ in two distinct ways, looking at the two faces which contain the edge. If $8$ and $6$ were on the same face, it is possible to achieve the desired sum only with the numbers $1$ and $3$ since the values must be distinct. Similarly,", "a multiple of 10. As in the previous cases, this implies that $S_{j}-(a_{1}-a_{10})=10$. That is, $a_{2}+a_{3}+ \\ldots + a_{j}+a_{10}=10$. Hence, $\\{a_{2},a_{3}, ldots, a_{j},a_{10} \\}$ and $\\{ a_{1}, a_{j+1}, \\ldots, a_{9}\\}$ balance each other. Thus, in all cases, the given 10 objects can be split into two groups that balance each other. Problem: At each of the eight corners of a cube write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so written down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero? Solution: Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}$ be the numbers written at the corners. Then, the final sum is given by $\\sum_{i=1}^{8}x_{i}+x_{1}x_{2}x_{3}x_{4}+x_{5}x_{6}x_{7}x_{8}+x_{1}x_{4}x_{5}x_{8}+x_{2}x_{3}x_{6}x_{7}+x_{1}x_{2}x_{5}x_{6}+x_{3}x_{4}x_{7}+x_{8}$. Because there are fourteen terms in the above sum, and each of the terms is +1 or -1, the sum will be zero only if some seven terms are +1 each and the remaining 7 terms are -1 each. But, the product of the fourteen terms is $(x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{8})^{4}=(\\pm 1)^{4}=+1$. Therefore, it is impossible to have an odd number of -1’s in the above sum. We conclude that the desired arrangement is not possible. Problem:", "that vertex. Then $$V= V_3 + V_4 + \\ldots,\\quad (3)$$ and also as every edge has two vertices, we have $$2E = 3V_3+4V_4+5V_5+\\ldots . \\quad (4)$$ This gives the inequality $2E \\geq 3V$ because \\begin{equation*}2E = 3V_3+4V_4+5V_5+\\ldots \\geq 3V_3+3V_4+3V_5+\\ldots = 3V.\\end{equation*} We are going to show that $$F_3+F_4+F_5 \\geq 4, \\quad F_3+V_3 \\geq 8. \\quad (5)$$ This tells us, for example, that not all of $F_3$, $F_4$ and $F_5$ can be zero, so there must be at least one face with exactly three, four of five sides. Thus it is indeed impossible to make a polyhedron with each face having at least six sides. We know that \\begin{equation*}F-E+V = 2, \\quad 2E \\geq 3F, \\quad 2E \\geq 3V,\\end{equation*} so that \\begin{equation*}2+E = F+V \\leq F + 2E/3.\\end{equation*} This shows that $2+E/3 \\leq F$ and hence that $12+2E \\leq 6F$. If we substitute for $F$ and for $E$ their expressions in terms of the $F_k$ given in (1) and (2), we get \\begin{equation*}12 + 3F_3 + 4F_4 + 5F_5 + \\ldots \\leq 6F_3 + 6F_4 + 6F_5 + \\ldots.\\end{equation*} This gives \\begin{equation*}12+F_7+2F_8+3F_9+\\ldots \\leq 3F_3+2F_4+F_5\\end{equation*} so that, finally, $12 \\leq 3F_3 + 2F_4 + F_5$. This is the first inequality in (5). To prove the second inequality in (5), we start with Euler's formula $F-E+V=2$ and write this as $(F-E/2)+(V-E/2)", "is antiparallel to edge-$5$, while edge-$3$ is parallel to edge-$4$. And that in turn determines the sides they are on: if the joined edges are antiparallel, they are on the same side, folded back, while if they are parallel, they are on opposite sides, connected across the divide. Another way to think of this rule is by labelling each edge uniquely: start at $0$, work our way anticlockwise (say) increasing the labels through all the numbers mod $N$. Orient the edges so they point in the direction of increasing numbers. Then edge-$u$ is identified with edge-$v$ precisely when $v=\\frac{-1}{u}$, and they are always antiparallel (which is now a rather ordinary consequence of the fact that the numbering now gives the faces a consistent orientation). 1 10 2 9 3 8 4 7 5 6 $1\\cdot 10=10=-1$ mod $11$. $2\\cdot 5=10=-1$ mod $11$. $3\\cdot 7=21=-1$ mod $11$. $4\\cdot 8=32=-1$ mod $11$. $6\\cdot 9=54=-1$ mod $11$. Let’s illustrate this for a few more cases. Here’s $N=17$: 1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1 This, again, has genus $1$. The cycle of edges $3\\rightarrow 5\\rightarrow 6\\rightarrow 7$ on the left is joined to the same cycle on the right, but again with a $180^\\circ$ twist. Since $4\\cdot 4=-1$ mod $17$, this edge", "by $4$: $$0+4+8+12+\\cdots+4(n-1)=2n(n-1)$$ Add $1$ to each term ($n$ in total): $$1+5+9+13+\\cdots+(4n-3)=\\color{red}{2n^2-n}\\quad\\blacksquare$$ By counting dots (of Hexagonal number) in two ways. On the left: there are four \"outer\" sides of each hexagon $j = 2, \\dots, n$, where each side has $j$ dots but 3 of them are shared, in addition the trivial hexagon $j=1$ is one dot. Therefore, in total there are $\\sum_{j=1}^n4j-3$ dots. On the right: there are $2n-1$ rows (since a hexagon has both top and bottom sides, except the trivial case), and each row adds $n$ dots. Hence $n(2n-1) = 2n^2-n$ dots. *Despite being geometric, induction may well be needed to justify the counting arguments, as Bill Dubuque noted. There are some very clever answers here, but what if you don't spot the clever trick? It would be nice to be able to mechanically grind out the answer, and in order to do that, you can use the discrete calculus. So here the problem is to evaluate $\\sum_{1\\le i\\le n}{4i-3}$. First, expand all powers of the summation variable (here, $i$) into sums of falling powers (by inspection in simple cases, or using Stirling cycle numbers). This is trivial here, since $i = i^{\\underline 1}$, so $$\\sum_{1\\le i\\le n}{4i-3} = \\sum_{1\\le i\\le n}{4i^{\\underline 1}-3}.$$ Now, summation of falling powers is just like integration of ordinary", "\\boxed{123} $$\\boxed{123}$$ Sort by: Use coordinates. To avoid fractions, I'll scale everything by a factor of $$12$$. WLOG, let $$A(0,0), B(12,0), C(12,12), D(0,12)$$. Then, $$X(6,0), Y(12,6), X(6,12), W(0,6)$$. Let $$V_1, \\ldots , V_8$$ be the vertices of the octagon going counterclockwise, with $$V_1$$ being closest to $$W$$. Note that there is $$8$$-fold symmetry about the center $$M(6,6)$$. Line $$AZ$$ is $$y = 2x$$. Line $$DX$$ is $$y = 12-2x$$. Their intersection is $$V_1(3,6)$$. By symmetry about $$M(6,6)$$, the other odd numbered vertices are $$V_3(6,3), V_5(9,6), V_7(6,9)$$. Line $$BW$$ is $$y = -\\tfrac{1}{2}x+6$$. Line $$DX$$ is $$y = 12-2x$$. Their intersection is $$V_2(4,4)$$. By symmetry about $$M(6,6)$$, the other even numbered vertices are $$V_4(8,4), V_6(8,8), V_8(4,8)$$. From here, it is easy to calculate the area of one \"slice\" of the scaled octagon: $$[\\Delta MV_1V_2] = \\tfrac{1}{2} \\cdot 3 \\cdot 2 = 3$$. Therefore, the scaled octagon has area $$8 \\cdot 3 = 24$$. Hence, the original octagon has area $$\\dfrac{24}{12^2} = \\boxed{\\dfrac{1}{6}}$$. - 4 years, 10 months ago Coincidentally, this was a problem from the 1989 Dutch finals, it was put in a collection of the 100 best problems from Dutch olympiads over the years. I tried to avoid coordinates, as it's clear that that's going to work, but I tried to find a more elegant solution, e.g. by", "and putting in our assumption that $x_2>x_1$, we get $29\\sqrt2\\times (x_2-x_1) = 58\\sqrt2\\times z$. Dividing both sides by $29\\sqrt2$, we get $x_2-x_1 = 2z$ As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well. • There are 48 ordered pairs $(x_2,x_1)$ such that their positive difference is 2. • There are 46 ordered pairs $(x_2,x_1)$ such that their positive difference is 4. ... • Finally, there are 2 ordered pairs $(x_2,x_1)$ such that their positive difference is 48. Summing them up, we get that there are $2+4+\\dots + 48 = \\boxed{600}$ triangles.", "# Fill in numbers on the cube! You are given a cube. You are told to fill in each vertex with the numbers $$4,5,6,...,11$$, with no repetition. What is the probability that for each two vertices that are connected by a common edge, he two numbers written on them are co-prime? Source: HK Prelim 2019 Q20 • Note that if it is an ongoing competition you are not supposed to post the problem here. Not before it has ended. – Florian F Jul 3 at 14:33 • @FlorianF It is from 2019. – Culver Kwan Jul 3 at 23:35 That means that up to rotation and reflection, there are only 2 ways to arrange the numbers. The group of symmetries of the cube has size 48, so after rotations and reflections there are $$2\\cdot48=96$$ valid arrangements. There are $$8!=40320$$ ways to arrange the numbers, of which $$96$$ are valid. The probability is therefore $$\\frac{96}{40320}=\\frac{1}{420}$$.", "2(5+19) = 48.$ As such, the largest triangle we can make under these constraints is a triangle with sides $$5,19,23$$, and therefore a perimeter of $$47,$$ as required. Therefore, the smallest possible number greater than any perimeter is $$48.$$ Thus, the correct answer is D. 9. On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $$20$$ days? $$39$$ $$40$$ $$210$$ $$400$$ $$401$$ ###### Solution(s): We want to find $1+3 + \\cdots + 39.$ This sum can be rewritten as \\begin{align*}&(1+39) + (3+ 37) \\\\&+\\cdots +(19+21),\\end{align*} which we can see has $$10$$ terms. Further notice that each term is equal to $$40.$$ Therefore, the sum is $10\\cdot40=400.$ Thus, the correct answer is D . 10. How many integers between $$1000$$ and $$9999$$ have four distinct digits? $$3024$$ $$4536$$ $$5040$$ $$6480$$ $$6561$$ ###### Solution(s): First, there are $$9$$ digits to choose for the thousands digit since $$0$$ can't be chosen. Then, after that, there are $$9$$ ways to choose the hundreds digit, $$8$$ ways to choose the tens digit, and $$7$$ ways to choose the ones digit.", "4$$ and $$y = 9$$, there are three walls (one of which uses two balls) and four buckets. Hence, in that case, $$n = x$$ and $$m = y – x – (x – 1) + 1 = y – 2x + 2$$. Secondly, since we are pulling four ($$x$$) balls, we can have up to two ($$\\lfloor{\\frac{x}{2}}\\rfloor$$) pairs of consecutive numbers. This means that our final formula will need to make use of sums, adding up the various cases for $$n(C_2)$$. In this specific case, let’s use $$n(C_{21})$$ for the case where we have one consecutive pair and $$n(C_{22})$$ for the case where we have two. The third consideration is how the pairs are placed. If there is one pair among a three walls, it could be the first wall, the second wall, or the third wall. Here we will be using the combination (binomial) function yet again. If we have $$w$$ walls and $$v$$ pairs, then there are $$\\frac{w!}{v!(w-v)!}$$ ways of distributing those pairs. In the specific case for $$C_{21}$$, $$w = 3$$ and $$v = 1$$, so $$\\frac{w!}{v!(w-v)!}=\\frac{3!}{1!(3-1)!}=3$$. Thus $n(C_{21}) = 3 \\cdot \\frac{(y-x+1)!}{(y-2x+2)!(x-1)!} = \\frac{3 \\cdot (9-4+1)!}{(9-2\\cdot 4+2)!(4-1)!} = \\frac{3 \\cdot 6!}{3!\\cdot 3!} = 60$ Meanwhile, for $$C_{22}$$, $$w = 2$$ and $$v = 2$$, giving us a multiplier of 1. $$n =x-1$$ and", "have intersection of size $$1$$. There are $$8$$ ways to have a (maximal) clique of three pairwise intersecting lines. Of these cliques, $$4$$ correspond to the triples and $$4$$ correspond to the elements. Depending on how the labeling was done, one kind is the points and the other is the planes. Either way, we have the three spreads (rulings) $$ab|cd$$, $$ac|bd$$ and $$ad|bc.$$ For the Petersen graph, use lines and points of PG(1,3) to label $$10$$ vertices. Make an edge for incident point-line pairs and also for label disjoint (i.e common ruling) line-line pairs. That is the graph. PG(3,2) has $$2^3+2^2+2+1=15$$ points and $$35$$ lines with each point on $$7$$ lines. There are also $$15$$ Fano planes. It turns out that the lines of PG(3,2) can be labeled (in various ways) with the $$35$$ triples from the set $$S=\\{a,b,c,d,e,f\\}$$ in such a way that two lines are coincident precisely when the labels have one element in common. Let me avoid, or at least delay, the issues of describing how to do this and finding the points and planes of PG(3,2) from the structure of the line graph. For the Hoffman-Singleton Graph, use lines and points of PG(2,3) to label $$50$$ vertices. Make an edge for incident point-line pairs and also for label disjoint line-line pairs That is the", "of $9$ points. Select any $3$ from them. Total number of ways are: $\\binom{9}{3} = 84$. Number of collinear point selection cases: $\\binom{2}{3} +\\binom{3}{3} +\\binom{4}{3} = 5$ ways. Totally $84-5 = 79$ ways. Hope it helps. With all vertices on different sides, you form $2\\cdot3\\cdot4$ triangles. With two vertices on the same side, you form $1,3$ and $6$ pairs respectively; then you can match every pair with a vertex on another side, giving $$1\\cdot(3+4)+3\\cdot(2+4)+6\\cdot(2+3).$$ Total $79$. (I assume that we can rule out the $0+1+4$ flat triangles obtained with three vertices on the same side.) The approach by @Rohan is better: from all $9\\cdot8\\cdot7/6$ possible triangles, discard the $0+1+4$ degenerate ones. • Why do you have $2$ times $(3+4)$, when you've noted that there's only one way to choose two points on the side with only two points? (I was about to write almost the same). – Henrik Dec 2 '16 at 9:24 • @Henrik: thanks, you spotted my mistake. – Yves Daoust Dec 2 '16 at 9:25", "ways. Fixing $2$ pairs $(x,x)$ (the diagonal minus $3$) and map them to the other fitting pairs $(x,f(x))$. You can do this in $\\frac{(5\\cdot4\\cdot6\\cdot4\\cdot2)}{2!3!} + \\frac{(5\\cdot4\\cdot6\\cdot4\\cdot2)}{2!2!} + \\frac{(5\\cdot4\\cdot6\\cdot2\\cdot1)}{2!2!} = 380$ ways. Lastly, fixing $1$ pair $(x,x)$ (the diagonal minus $4$) and map them to the other fitting pairs $(x,f(x))$. You can do this in $\\tfrac{5!}{4!} + 4\\cdot\\tfrac{5!}{3!} + 5! = 205$ ways. So $1 + 20 + 150 + 380 + 205 = \\framebox{756}$ Solution 2 We perform casework on the number of fixed points (the number of points where $f(x) = x$). To better visualize this, use the grid from Solution 1. Case 1: 5 fixed points - Obviously, there must be $1$ way to do so. Case 2: 4 fixed points - $\\binom 54$ ways to choose the $4$ fixed points. For the sake of conversation, let them be $(1, 1), (2, 2), (3, 3), (4, 4)$. - The last point must be $(5, 1), (5, 2), (5, 3),$ or $(5, 4)$. - There are $\\binom 54 \\cdot 4 = 20$ solutions for this case. Case 3: 3 fixed points - $\\binom 53$ ways to choose the $3$ fixed points. For the sake of conversation, let them be $(1, 1), (2, 2), (3, 3)$. - Subcase 3.1: None of $4$ or $5$ map to each other", "71 face are $6$ and $4$. That yields 2 labelings, since the $64$ edge can be oriented in two different ways. I think that's it, up to rotations and reflections. Now get reflections out of it. Sep 24, 2021 at 5:06 Note that the sum of the numbers on each face must be $$18$$ because $$\\frac{1+2+\\cdots+8}{2}=18$$. So now consider the opposite edges (two edges that are parallel but not on the same face of the cube); they must have the same sum value too. Now think about the points $$1$$ and $$8$$. If they are not on the same edge, they must be endpoints of opposite edges, and we should have $$1+X=8+Y$$. However, this scenario would yield no solution for $$[2,7]$$, which is a contradiction. The points $$1$$ and $$8$$ are therefore on the same side and all edges parallel must also sum to $$9 .$$ Now we have $$4$$ parallel sides $$1-8,2-7,3-6,4-5$$. Thinking about $$4$$ endpoints, we realize they need to sum to 18. It is easy to notice only $$1-7-6-4$$ and $$8-2-3-5$$ would work. So if we fix one direction $$1-8($$ or $$8-1)$$ all other 3 parallel sides must lay in one particular direction. $$(1-8,7-2,6-3,4-5)$$ or $$(8-1,2-7,3-6,5-4)$$ Now, the problem is the same as arranging 4 points in a two-dimensional square, which is $$\\frac{4 !}{4}=6$$. It" ]

[ "# Problem #1818 1818 The regular octagon $ABCDEFGH$ has its center at $J$. Each of the vertices and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal. In how many ways can this be done? $\\textbf{(A)}\\ 384 \\qquad\\textbf{(B)}\\ 576 \\qquad\\textbf{(C)}\\ 1152 \\qquad\\textbf{(D)}\\ 1680 \\qquad\\textbf{(E)}\\ 3456$ $[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label(\"A\",A,NNW); label(\"B\",B,NNE); label(\"C\",C,ENE); label(\"D\",D,ESE); label(\"E\",E,SSE); label(\"F\",F,SSW); label(\"G\",G,WSW); label(\"H\",H,WNW); label(\"J\",J,SE); [/asy]$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# 2013 AMC 10B Problems/Problem 22 ## Problem The regular octagon $ABCDEFGH$ has its center at $J$. Each of the vertices and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal. In how many ways can this be done? $\\textbf{(A)}\\ 384 \\qquad\\textbf{(B)}\\ 576 \\qquad\\textbf{(C)}\\ 1152 \\qquad\\textbf{(D)}\\ 1680 \\qquad\\textbf{(E)}\\ 3456$ $[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label(\"A\",A,NNW); label(\"B\",B,NNE); label(\"C\",C,ENE); label(\"D\",D,ESE); label(\"E\",E,SSE); label(\"F\",F,SSW); label(\"G\",G,WSW); label(\"H\",H,WNW); label(\"J\",J,SE); [/asy]$ ## Solution 1 First of all, note that $J$ must be $1$, $5$, or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have: $[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label(\"A\",A,NNW); label(\"B\",B,NNE); label(\"C\",C,ENE); label(\"D\",D,ESE); label(\"E\",E,SSE); label(\"F\",F,SSW); label(\"G\",G,WSW); label(\"H\",H,WNW); label(\"J (1, 5, 9)\",J,SE); [/asy]$ We also notice that $A+E = B+F = C+G =", "and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal. In how many ways can this be done? $\\textbf{(A)}\\ 384 \\qquad\\textbf{(B)}\\ 576 \\qquad\\textbf{(C)}\\ 1152 \\qquad\\textbf{(D)}\\ 1680 \\qquad\\textbf{(E)}\\ 3456$ $[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label(\"A\",A,NNW); label(\"B\",B,NNE); label(\"C\",C,ENE); label(\"D\",D,ESE); label(\"E\",E,SSE); label(\"F\",F,SSW); label(\"G\",G,WSW); label(\"H\",H,WNW); label(\"J\",J,SE); [/asy]$ Solution ## Problem 23 In triangle $ABC$, $AB = 13$, $BC = 14$, and $CA = 15$. Distinct points $D$, $E$, and $F$ lie on segments $\\overline{BC}$, $\\overline{CA}$, and $\\overline{DE}$, respectively, such that $\\overline{AD} \\perp \\overline{BC}$, $\\overline{DE} \\perp \\overline{AC}$, and $\\overline{AF} \\perp \\overline{BF}$. The length of segment $\\overline{DF}$ can be written as $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$? $\\textbf{(A)}\\ 18\\qquad\\textbf{(B)}\\ 21\\qquad\\textbf{(C)}\\ 24\\qquad\\textbf{(D)}\\ 27\\qquad\\textbf{(E)}\\ 30$ ## Problem 24 A positive integer $n$ is nice if there is a positive integer $m$ with exactly four positive divisors (including $1$ and $m$) such that the sum of the four divisors is equal to $n$. How many numbers in the set $\\{ 2010,2011,2012,\\dotsc,2019 \\}$ are", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "(0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)", "3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6(Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find $EF=BF=2\\sqrt{15}$ $[\\bigtriangleup EBF]=\\frac{(2\\sqrt{15})^2}{2}=\\boxed{(B) 30}\\blacksquare$ -Trex4days ##", "label(\"d\",(15*k-1,8),N); label(\"i\",(15*k-1,7),N); label(\"s\",(15*k-1,6),N); label(\"t\",(15*k-1,5),N); label(\"a\",(15*k-1,4),N); label(\"n\",(15*k-1,3),N); label(\"c\",(15*k-1,2),N); label(\"e\",(15*k-1,1),N); label(\"time\",(15*k+8,0),S); } for (int k=0; k<2; ++k) { label(\"d\",(15*k-1,8-15),N); label(\"i\",(15*k-1,7-15),N); label(\"s\",(15*k-1,6-15),N); label(\"t\",(15*k-1,5-15),N); label(\"a\",(15*k-1,4-15),N); label(\"n\",(15*k-1,3-15),N); label(\"c\",(15*k-1,2-15),N); label(\"e\",(15*k-1,1-15),N); label(\"time\",(15*k+8,0-15),S); } label(\"(A)\",(5,9),N); label(\"(B)\",(20,9),N); label(\"(C)\",(35,9),N); label(\"(D)\",(5,-6),N); label(\"(E)\",(20,-6),N); [/asy]$ ## Problem 19 Around the outside of a $4$ by $4$ square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters. Another square, $ABCD$, has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles. The area of the square $ABCD$ is $[asy] pair A,B,C,D; A = origin; B = (4,0); C = (4,4); D = (0,4); draw(A--B--C--D--cycle); draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,NW); [/asy]$ $\\text{(A)}\\ 16 \\qquad \\text{(B)}\\ 32 \\qquad \\text{(C)}\\ 36 \\qquad \\text{(D)}\\ 48 \\qquad \\text{(E)}\\ 64$ ## Problem 20 Let $W,X,Y$ and $Z$ be four different digits selected from the set $\\{ 1,2,3,4,5,6,7,8,9\\}.$ If the sum $\\dfrac{W}{X} + \\dfrac{Y}{Z}$ is to be as small as possible, then $\\dfrac{W}{X} + \\dfrac{Y}{Z}$ must equal $\\text{(A)}\\ \\dfrac{2}{17} \\qquad \\text{(B)}\\ \\dfrac{3}{17} \\qquad \\text{(C)}\\ \\dfrac{17}{72} \\qquad \\text{(D)}\\ \\dfrac{25}{72} \\qquad \\text{(E)}\\ \\dfrac{13}{36}$ ## Problem 21 A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure", "(A) at (-1,0); \\coordinate[label=right:B] (B) at (1,0); \\coordinate[label=C] (C) at (0,{sqrt(3)}); \\coordinate[label=below: D] (D) at (0,0); \\coordinate[label=left:E] (E) at (0,1); \\coordinate[label=left:F] (F) at (0,{2-sqrt(3)}); \\coordinate[label=below:$A_1$] (A1) at ({2/sqrt(3)-1},0); \\draw (A) -- (B) -- (C) -- cycle; \\draw (D) -- (C); \\draw[fill] (E) circle (.01); \\draw (F) -- (A1); \\end{tikzpicture} Or did I misunderstand? How can we draw a line parallel to $AC$ from $F$ ? Isn't $F$ on $AC$ ? Or do you mean from $E$ instead of $F$ ? Because: \\begin{tikzpicture}[scale=3] \\coordinate[label=left:A] (A) at (-1,0); \\coordinate[label=right:B] (B) at (1,0); \\coordinate[label=C] (C) at (0,{sqrt(3)}); \\coordinate[label=below: D] (D) at (0,0); \\coordinate[label=left:E] (E) at (0,1); \\coordinate[label=left:F] (F) at (0,{2-sqrt(3)}); \\coordinate[label=below:$A_1$] (A1) at ({1-2/sqrt(3)},0); \\coordinate[label=below:$B_1$] (B1) at ({2/sqrt(3)-1},0); \\draw (A) -- (B) -- (C) -- cycle; \\draw (D) -- (C); \\draw[fill] (E) circle (.01); \\draw (F) -- (A1); \\draw (F) -- (B1); \\end{tikzpicture} Why will we not reach zero? We have a new equilateral triangle that has non-zero size. After 2 more iterations, we will have yet again a new equilateral triangle that has non-zero size, won't we? #### mathmari ##### Well-known member MHB Site Helper I thought that we get the following construction: Isn't this correct? #### Klaas van Aarsen ##### MHB Seeker Staff member I thought that we get the following construction: Isn't this correct? Ah okay. My mistake.", "= (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"60\", (A+G+D)/3); label(\"30\", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea) ## Solution 5 (Area Ratios) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\\dfrac{EF}{AE} = \\dfrac{x}{60} = \\dfrac{120-x}{180}.$$Solving gives $x = \\boxed{\\textbf{(B) }30}$. (Credit to scrabbler94 for the idea) ## Solution 6 (Coordinate Bashing) Let $ADB$ be a right triangle, and $BD=CD$ Let $A=(-2\\sqrt{30}, 0)$ $B=(0, 4\\sqrt{30})$ $C=(4\\sqrt{30}, 0)$ $D=(0, 0)$ $E=(0, 2\\sqrt{30})$ $F=(\\sqrt{30}, 3\\sqrt{30})$ The line $\\overleftrightarrow{AE}$ can be described with the equation $y=x-2\\sqrt{30}$ The line $\\overleftrightarrow{BC}$ can be described with $x+y=4\\sqrt{30}$ Solving, we get $x=3\\sqrt{30}$ and $y=\\sqrt{30}$ Now we can find", "dot(\"S\", S, E); dot(\"K\",K,E); dot(\"L\",L,W); // dot(\"R\",R,NE); dot(\"Q\",Q,NW); dot(\"G\", G, SE); dot(\"Q\",Q,NW); dot(\"P\",P,NE); dot(\"M\",M,S); draw(arc1,dashed); draw(arc2, dashed); draw(circle2); draw(A--D--B--cycle); // draw(A--C, dashed); draw(A--H); draw(C--S--H--cycle); draw(C--T--H); draw(D--C--B); draw(D--L--K--B, dashed); draw(T--G, dashed); draw(K--H, dashed); draw(D--Q--P--B,dashed); draw(Q--H--P,dashed); draw(C--L--H, dashed); draw(C--K,dashed); draw(T--S,dashed); // draw(anglemark(H,S,K)); // draw(anglemark(K,H,S)); label(\"x\",C+(0,0.1),N); label(\"y\",C+(-0.1,0.3),N); label(\"w\",H+(-0.03,0.1), N); label(\"z\", H+(0.06,0.12), N ); label(\"u\",T+(-0.1,-0.2), S); label(\"v\", S+(0,-0.2), S); label(\"t\",T+(0.1,-0.3), S); label(\"s\", S+(-0.08,-0.2), S); [/asy]$ Denote $\\angle{HSB}=v$, $\\angle{HTD}=u$, $\\angle{HSC}=s$, $\\angle{HTC}=t$, $\\angle{HCS}=x$, $\\angle{HCT}=y$, $\\angle{AHS}=z$, $\\angle{AHT}=w$. Since $\\angle{CHS}-\\angle{CSB} =90$ and $\\angle{CHT}-\\angle{CTD}=90$, we have $\\angle{CSA}=x+90$, $\\angle{CTA}=y+90$. Since $\\angle{CHS} = \\angle{CSB}+90$, the tangent of the circumcircle of $\\triangle{CSH}$ at point $S$ is perpendicular to $SB$; therefore, the circumcenter of $\\triangle{CSH}$ (point $K$) is on $AB$. Similarly, the circumcenter of $\\triangle{CTH}$ (point $L$) is on $AD$. In addition, $KL$ is the perpendicular bisector of $CH$. Extend $CB$ to meet circumcircle of $\\triangle{CSH}$ at $P$, and extend $CD$ to meet circumcircle of $\\triangle{CTH}$ at $Q$. Then, since $\\angle{ADC}=\\angle{ABC}=90$, $AD$ and $AB$ are the perpendicular bisector of $CQ$ and $CP$, respectively; hence $A$ is the circumcenter of $\\triangle{PCQ}$. Since $B$ and $D$ are midpoints on $CP$ and $CQ$, $PQ \\parallel BD$; also, $AH \\perp BD$, so $AH \\perp PQ$. Since $A$ is the circumcenter, $AH$ is also the perpendicular bisector of $PQ$. Hence, $$HP=HQ$$ We have $$u=\\angle{CHT}-90=(90-w+\\angle{DHC}) - 90$$ $$v=\\angle{CHS}-90 = (90-z+\\angle{BHC}) - 90$$ Hence, $u+v = -w-z+\\angle{DHC}+\\angle{BHC} =", "my armpit Dec 29 '13 at 20:11 • See, we have a constraint of symmetricity? Do you have others that's what I asked. – percusse Dec 29 '13 at 20:21 Here is the Asymptote version, which does not use the general Asymptote commands to split a path (guide), it just performs a simple subdivision of the cubic Bezier segment: // split.asy: size(5cm); pair A,B,C,D; A=(3,3); B=(1,1); C=(-1,1); D=(-3,3); pair P,B1,C1,B2,C2; P=(A+D+3(B+C))/8; B1 = (A+B)/2; C1 = ((A+C)/2+B)/2; C2 = (D+C)/2; B2 = ((D+B)/2+C)/2; guide g=A..controls B and C..D; guide gl=A..controls B1 and C1..P; guide gr=P..controls B2 and C2..D; draw(g); draw(gl,deepgreen+1.6bp+opacity(0.5)); draw(gr,red+1.6bp+opacity(0.5)); draw((0,3)--(0,-1),red); dot(A--B--C--D--P,UnFill); dot(B1--C1,deepgreen,UnFill); dot(B2--C2,red,UnFill); label(\"$A$\",A,SE); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$D$\",D,SW); label(\"$P$\",P,NE); label(\"$B_1$\",B1,E,deepgreen); label(\"$C_1$\",C1,E,deepgreen); label(\"$B_2$\",B2,W,red); label(\"$C_2$\",C2,W,red); To get a standalone split.pdf, run asy -f pdf split.asy. • I really need the converting formulae. – kiss my armpit Dec 29 '13 at 19:29 • @Stiff Jokes: Btw, the Bezier curve is a very interesting object, you might enjoy studying it in a spare time. – g.kov Dec 29 '13 at 19:43 A translated version for @g.kov's Asymptote answer in PSTricks. \\documentclass[pstricks,border=24pt]{standalone} \\usepackage{pst-eucl} \\begin{document} \\begin{pspicture}[showgrid=true](-3,-1)(3,3) \\pstGeonode (3,3){A} (1,1){B} (-1,1){C} (-3,3){D} \\psbezier(A)(B)(C)(D) \\psline[linecolor=red](0,3)(0,-1) \\nodexn{.125(A)+.125(D)+.375(B)+.375(C)}{R'} \\nodexn{.5(A)+.5(B)}{P'} \\nodexn{.25(A)+.5(B)+.25(C)}{Q'} \\pstGeonode (P'){P} (Q'){Q} (R'){R} \\psbezier[linecolor=red](A)(P)(Q)(R) \\end{pspicture} \\end{document}", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). $[asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label(\"1\", (5.5, .5)); label(\"1\", (4.5, 1.5)); label(\"1\", (6.5, 1.5)); label(\"1\", (3.5, 2.5)); label(\"1\", (7.5, 2.5)); label(\"2\", (5.5, 2.5)); label(\"1\", (2.5, 3.5)); label(\"3\", (6.5, 3.5)); label(\"3\", (4.5, 3.5)); label(\"4\", (3.5, 4.5)); label(\"3\", (7.5, 4.5)); label(\"6\", (5.5, 4.5)); label(\"10\", (4.5, 5.5)); label(\"9\", (6.5, 5.5)); label(\"19\", (5.5, 6.5)); label(\"9\", (7.5, 6.5)); label(\"28\", (6.5, 7.5)); [/asy]$ The answer is therefore $\\boxed{\\textbf{(A) }28}$. ## Solution 2 Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. $[asy] int N = 7; for (int i", "# Difference between revisions of \"2020 AIME I Problems/Problem 9\" ## Problem Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$ ## Solution $[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; draw(b1 -- a1 -- c1); draw(b2 -- a2 -- c2); draw(b3 -- a3 -- c3); dot(a1); dot(a2); dot(a3); label(\"a_1\", a1, NE); label(\"a_2\", a2, NE); label(\"a_3\", a3, NE); label(\"b_1\", b1, S); label(\"b_2\", b2, S); label(\"b_3\", b3, S); label(\"c_1\", c1, W); label(\"c_2\", c2, W); label(\"c_3\", c3,", "label(\"D\",D,NNW); label(\"C\",C,SW); label(\"H\",H,SE); draw(e--H,dashed); label(\"O\",(0,0),NE); label(\"1\",(C--O),N); label(\"1\",(H--O),N); label(\"2\",(A--C),N); label(\"2\",(H--B),N); label(\"3\",(O--D),NE); label(\"3\",(O--e),NE); label(\"2\\sqrt{2}\",(D--C),W); label(\"2\\sqrt{2}\",(H--e),E); draw(rightanglemark(e,(e.x,0),A,2)); draw(rightanglemark(D,C,B,2));[/asy]$ Let the center of the circle be $O$. Without loss of generality, let the radius of the circle be equal to $3$. Thus, $AO=3$ and $OB=3$. As a consequence of $2(AC)=BC$, $AC=2$ and $CO=1$. Also, we know that $DO$ and $OE$ are both equal to $3$ due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to $\\sqrt{3^2-1^2}$ or $2\\sqrt{2}$. Now we know that the area of $[ABD]$ is equal to $\\frac{(3+2+1)(2\\sqrt{2})}{2}$ or $6\\sqrt{2}$. Know we need to find the area of $[DCE]$. By simple inspection $[COD]$ $\\cong$ $[HOE]$ due to angles being equal and CPCTC. Thus $HE=2\\sqrt{2}$ and $OH=1$. Know we know the area of $[CHE]=\\frac{(1+1)(2\\sqrt{2})}{2}$ or $2\\sqrt{2}$. We also know that the area of $[OHE]=\\frac{(1)(2\\sqrt{2})}{2}$ or $\\sqrt{2}$. Thus the area of $[COE]=2\\sqrt{2}-\\sqrt{2}$ or $\\sqrt{2}$. We also can calculate the area of $[DOC]$ to be $\\frac{(1)(2\\sqrt{2})}{2}$ or $\\sqrt{2}$. Thus $[DCE]$ is equal to $[COE]$ + $[DOC]$ or $\\sqrt{2}+\\sqrt{2}$ or $2\\sqrt{2}$. The ratio between $[DCE]$ and $[ABD]$ is equal to $\\frac{2\\sqrt{2}}{6\\sqrt{2}}$ or $\\frac{1}{3}$ $\\Longrightarrow \\mathrm{(C)}$. Solution 5 We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for $D (x,y)$, and notice how $E$ is a", "= \\left\\lfloor \\frac{\\text{deg}(V)}{2} \\right\\rfloor$ where deg$(V)$ is the number of edges connected to $V$. This is because to visit any of these vertices, we would have to visit and exit on two different edges (This is a pretty standard principle in graph theory). We will label each vertex with this number. $[asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A); } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label(\"A\", (0,2), W); label(\"L\", (3,0), E); label(\"1\", (1,2), N); label(\"1\", (2,2), N); label(\"1\", (3,2), N); label(\"1\", (0,1), W); label(\"2\", (1,1), NW); label(\"2\", (2,1), NW); label(\"1\", (3,1), E); label(\"1\", (0,0),S); label(\"1\", (1,0),S); label(\"1\", (2,0),S); [/asy]$ Additionally, notice that if we visit $n$ vertices (not necessarily distinct; i.e. if we visit a vertex twice we count it as two vertices) along our path, we must traverse $n-1$ edges (this is also pretty easy to prove). Thus, if we want to traverse 13 edges total, we have to visit 14 vertices. We will visit $A$ and $L$ for sure, leaving us with 12", "- 40\\sqrt{10}$. Through power of a point, we can find out that $(CE)(DE)=20\\sqrt{10} - 20$, so $CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\\sqrt{10}) + 2(20\\sqrt{10} - 20) = \\boxed{\\textbf{(E) } 100}$. ~Math_Wiz_3.14 ## Solution 4 (Reflections) $[asy] draw(circle((0,0),7.07)); dot((-7.07,0)); label(\"A\",(-7.07,0),W); dot((7.07,0)); label(\"B\",(7.07,0),E); dot((0,0)); label(\"O\",(0,0),N); dot((-2.24,-6.71)); label(\"C\",(-2.24,-6.71),SSW); dot((6.71,2.24)); label(\"D\",(6.71,2.24),NE); draw((-2.24,-6.71)--(6.71,2.24)); dot((6.71,-2.24)); label(\"D'\",(6.71,-2.24),SE); draw((4.51,0)--(6.71,-2.24)); dot((4.51,0)); label(\"E\",(4.51,0),NNW); draw((-7.07,0)--(7.07,0)); draw((0,0)--(-2.24,-6.71)); draw((0,0)--(6.71,-2.24)); draw((-2.24,-6.71)--(6.71,-2.24)); [/asy]$ Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$, we have that $\\angle BED'=45$. Since $\\angle AEC=45$, $\\angle CED'=90$. Since $DE=ED'$, by the Pythagorean Theorem, our desired solution is just $CD'^2$. Looking next to circle arcs, we know that $\\angle AEC=\\frac{\\overarc{AC}+\\overarc{BD}}{2}=45$, so $\\overarc{AC}+\\overarc{BD}=90$. Since $\\overarc{BD'}=\\overarc{BD}$, and $\\overarc{AC}+\\overarc{BD'}+\\overarc{CD'}=180$, $\\overarc{CD'}=90$. Thus, $\\angle COD'=90$. Since $OC=OD'=5\\sqrt{2}$, by the Pythagorean Theorem, the desired $CD'^2= \\boxed{\\textbf{(E) } 100}$. ~sofas103 ## See Also 2020 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's", "(i = 0; i <= 4; ++i) { draw((i,0)--(i,-4.5)); draw((0,-i)--(4.5,-i)); } label(\"$1$\", (0.5,-0.5), white); label(\"$a$\", (1.5,-0.5)); label(\"$a^2$\", (2.5,-0.5), white); label(\"$a^3$\", (3.5,-0.5)); label(\"$b$\", (0.5,-1.5)); label(\"$ab$\", (1.5,-1.5), white); label(\"$a^2 b$\", (2.5,-1.5)); label(\"$a^3 b$\", (3.5,-1.5), white); label(\"$b^2$\", (0.5,-2.5), white); label(\"$ab^2$\", (1.5,-2.5)); label(\"$a^2 b^2$\", (2.5,-2.5), white); label(\"$a^3 b^2$\", (3.5,-2.5)); label(\"$b^3$\", (0.5,-3.5)); label(\"$ab^3$\", (1.5,-3.5), white); label(\"$a^2 b^3$\", (2.5,-3.5)); label(\"$a^3 b^3$\", (3.5,-3.5), white); label(\"$\\dots$\", (5,-2)); label(\"$\\vdots$\", (2,-5)); [/asy] Sorry but I don't know how to show the grids. 1. 👍 2. 👎 3. 👁 1. each row is a geometric progression. Starting with n=0, row n is the GP b^n (1 + a + a^2 + ...) so the sum Sn of row n is b^n * 1/(1-a) Now you have another GP, which is the sum of all the row sums. That is 1/(1-a) (1 + b + b^2 + ...) whose sum is 1/(1-a) * 1/(1-b) So the grand total of the whole grid is 1/((1-a)(1-b)) 1. 👍 2. 👎 2. wertwetrwetw 1. 👍 2. 👎 ## Similar Questions 1. ### alebra Let f(x)=2x^2+x-3 and g(x)=x-1. Perform the indicated operation then find the domain. (f*g)(x) a.2x^3-x^2-4x+3; domain:all real numbers b.2x^3+x^2-3x; domain: all real numbers c.2x^3+x^2+4x-3; domain: negative real numbers 2. ### Precalc To which set(s) of numbers does the number sqrt -16 belong? Select all that apply. real numbers complex numbers*** rational", "# 2020 AIME I Problems/Problem 9 ## Problem Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$ ## Solution 1 $[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; draw(b1 -- a1 -- c1); draw(b2 -- a2 -- c2); draw(b3 -- a3 -- c3); dot(a1); dot(a2); dot(a3); label(\"a_1\", a1, NE); label(\"a_2\", a2, NE); label(\"a_3\", a3, NE); label(\"b_1\", b1, S); label(\"b_2\", b2, S); label(\"b_3\", b3, S); label(\"c_1\", c1, W); label(\"c_2\", c2, W); label(\"c_3\", c3, W); [/asy]$ First,", "A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label(\"33\",(A+B)/2,N); label(\"21\",(C+D)/2,S); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); label(\"D\",D,SW); label(\"E\",DD,SE); label(\"F\",CC,SW); draw(C--CC); draw(D--DD); label(\"21\",(CC+DD)/2,N); label(\"2\",(A+DD)/2,N); label(\"10\",(CC+B)/2,N); label(\"\\sqrt{96}\",(C+CC)/2,W); label(\"\\sqrt{96}\",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$ The two diagonals are $\\overline{AC}$ and $\\overline{BD}$. Using the Pythagorean theorem again on $\\bigtriangleup AFC$ and $\\bigtriangleup BED$, we can find these lengths to be $\\sqrt{96+529} = 25$ and $\\sqrt{96+961} = \\sqrt{1057}$. Since $\\sqrt{96+529}<\\sqrt{96+961}$, $25$ is the shorter length, so the answer is $\\boxed{\\textbf{(B) }25}$. Solution 2 size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label(\"10\",(A+D)/2,W); label(\"14\",(B+C)/2,E); label(\"$A$\",A,NW); label(\"$B$\",B,NE); label(\"$C$\",C,SE); label(\"$D$\",D,SW); label(\"$D$\",D,SW); label(\"$21$\",(C+D)/2,S); (Error compiling LaTeX. E = (4,7); ^ 1b1ea1b866e85d255ff55009031082b8f710a74e.asy: 9.1: modifying non-public field outside of structure)" ]

[ "+0 # Binomial Probability +2 168 2 +290 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability $$\\frac{3}{5}$$ and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. TheMathCoder Apr 26, 2018 #1 +1201 +1 A solution to this question is here: https://web2.0calc.com/questions/is-anyone-going-to-help#r9 GA GingerAle Apr 26, 2018 #1 +1201 +1 A solution to this question is here: https://web2.0calc.com/questions/is-anyone-going-to-help#r9 GA GingerAle Apr 26, 2018 #2 +290 +3 Thanks GingerAle TheMathCoder Apr 26, 2018", "+0 # Is anyone going to help? +1 306 12 +737 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/5 and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. MIRB16 Apr 13, 2018 #1 +549 +1 how many matches does each team play in total? lynx7 Apr 13, 2018 edited by lynx7 Apr 13, 2018 #2 +737 +1 it only says that they need to win 4 games before the other team MIRB16 Apr 13, 2018 #3 +549 +1 3/5 + 3/5 + 3/5 + 3/5 =12/5 (probability that Cubs will win all four matches) 12/5 * 100 = 240% i am not sure if i am right. lynx7 Apr 13, 2018 #4 +1 If the probability of winning any one game is 3/5, then the probability of winning the World Series should be: (3/5)^4 =0.1296 =~13%. Guest Apr 13, 2018 #5 +1 I think the probability should be: 7C4 x (3/5)^4 x (2/5)^3 =~29% Guest Apr 13, 2018 #6 +1 Let everybody chime in!. I think you have to sum", "+0 # The Cubs are playing the Red Sox in the World Series. 0 116 10 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. Nov 16, 2020 #3 +2140 +2 Solution: $$\\text {The series ends after a team has a fourth win.}\\\\ \\text{Here are the four scenarios with derived probabilities }\\\\ \\text{where the Cubs win the series. }\\\\ \\text{ }\\\\ \\text{1) The Cubs win the first four games.}\\ \\text{ }\\\\ \\hspace{35 mm} \\rho(\\text {Cbs win on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} \\left(\\dfrac{3}{4}\\right)^4 = \\dfrac{81}{256} \\approx 31.64%\\\\$$ $$\\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \\dfrac{81}{256} \\approx 31.64 \\% \\\\ \\text{ }\\\\$$ $$\\text{3) Cubs win series in", "+0 # The Cubs are playing the Red Sox in the World Series. 0 536 10 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. Nov 16, 2020 #3 +2372 +3 Solution: $$\\text {The series ends after a team has a fourth win.}\\\\ \\text{Here are the four scenarios with derived probabilities }\\\\ \\text{where the Cubs win the series. }\\\\ \\text{ }\\\\ \\text{1) The Cubs win the first four games.}\\ \\text{ }\\\\ \\hspace{35 mm} \\rho(\\text {Cbs win on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} \\left(\\dfrac{3}{4}\\right)^4 = \\dfrac{81}{256} \\approx 31.64%\\\\$$ $$\\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \\dfrac{81}{256} \\approx 31.64 \\% \\\\ \\text{ }\\\\$$ $$\\text{3) Cubs win series in", "game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \\dfrac{405}{2048} \\approx 19.78\\% \\\\ \\text{ }\\\\$$ $$\\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\\dfrac{405}{4096} \\approx 9.88\\% \\\\ \\text{ }\\\\$$ $$\\text{ The sum of the individual probabilities gives }\\\\ \\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA Nov 16, 2020 #1 0 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series?", "game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \\dfrac{405}{2048} \\approx 19.78\\% \\\\ \\text{ }\\\\$$ $$\\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\\dfrac{405}{4096} \\approx 9.88\\% \\\\ \\text{ }\\\\$$ $$\\text{ The sum of the individual probabilities gives }\\\\ \\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA Nov 16, 2020 #1 0 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series?", "teams did not win the WS from 1919 to 2003, as was the case with the Cubs and the Red Sox, is a little over 1%. Not something anyone would have predicted in 1918, but had the two met in the 2003 WS (both teams advanced to their respective league's Championship Series that year), the world likely would not have ended either. We should remember though that while the Cubs and the Red Sox received all the attention, there was a third team that did not win a championship between 1919 and 2003: the Chicago White Sox, who, before winning the WS in 2005, last won in 1917. Together this trio of futility does push the limits of incredulity, as the probability of at least three teams concurrently trapped in 85-year droughts appears to be less than 0.04%. By the way, as a die-hard Cleveland Indians fan, what is the probability that, conditioning on the Cubs's drought, some other team hasn't won the WS since 1948, as is the case with the Indians? It is actually about 53%! So sure, the Indians have been very unlucky, but it's almost as if a flip of a coin decided that some team has to be that unlucky, and that team just happens to be the Indians (well, that, and decades", "Cubs' probability of winning the World Series stays constant. Another point is just because the system is underdetermined doesn't mean that we can pick anything for $$w_{cb}$$. For example, $$w_{cb}$$ can't be lower than around $$0.4$$ or there are no solutions. (since the Cubs have a $$0.4$$ chance of winning the World Series regardless of who is playing!) ## Conclusion It's a little disappointing we can't solve for the various $$w$$'s (especially because if we know whether the Cubs or Dodgers win but the Indians and Blue Jays are still playing, then we can solve for all the $$w$$'s!), but now we know for sure! Fancy math equations by MathJax.", "it may be less clear why we subtract the joint probability. The answer is simple - because we counted where they overlap twice (those instances in both A and in B) so we have to subtract out one instance. If, though, the events are mutually exclusive, we do not need to subtract the overlap. Mutually exclusive events are events that cannot occur at the same time, so there is no overlap. Suppose you are from Chicago and will be happy if either the Cubs or the White Sox win the World Series. Those events are mutually exclusive since only one team can win the World Series so to find the union of those probabilities we simply have to add the probability of the Cubs winning to the probability of the White Sox winning.", "in question. Simulation puts the probability of at least one 107-year drought at about 1.5% in this model. In conclusion, if 2015 is the year that the Cubs's drought finally ends, it would be the end of a streak that, while surely painful for generations of Cubs fans, is not really all that mathematically remarkable in and of itself. However, do notice the droughts of the Red Sox (no WS from 1919 to 2003, winning in 2004) and the White Sox (no WS from 1918 to 2004, winning in 2005), and how neatly the Cubs winning the 2015 WS would completely the symmetry. As we speak, the Cubs are down 3 games to none to the Mets. If they still manage to make it to the World Series and win, well, that may indeed be of some numerological interest. *One might ask why, if we know $p$, one team's probability of not winning the WS since 1907, we cannot simply compute the probability of some team not winning the WS since 1907 with the formula $1-(1-p)^{\\text{numTeams}}$. The reason is that teams's records are not independent. To be fair, the dependence is very weak and the aforementioned formula does yield about 6% as well.", "### Why the world will not end if Cubs win the 2015 World Series The Chicago Cubs are currently (as of October 20 2015) playing the New York Mets in the 2015 National League Championship Series, potentially eight victories away from winning the World Series, something that has not happened since 1907. As baseball lore goes, the team is cursed, and a World Series victory would be an event of cosmic proportions, possibly signaling end of times itself. This begs the question: just how unlikely is an 107-year drought? For simplicity, let us first assume that in any given year, each team has equal chance of winning the WS. From the Cubs' point of view, the probability of it not winning the WS since 1907 is about 0.38%, i.e. pretty unlikely. However, what is the probability that some (at least one) team hasn't won the WS since 1907? By my simulation*, it is about 6%, i.e. not probable, but not impossible. This is similar to the Birthday Paradox: if you walk into a room of 30 people, the probability that someone has the same birthday as you is low (~7.6%), but the probability that some pair of people share the same birthday is quite high (~70%). While we are on the topic, the probability that at least 2", "1945 William Sianis and his goat are turned away from a World Series game at Wrigley Field. The curse begins. 1960 The exploding scoreboard debuts at Comiskey 1968 Artificial turf is installed in Comiskey's infield (later replaced with natural grass) 1973 The designated hitter debuts 1984 The Sox win a game that lasts 2 day and goes 25 innings. Cubs clinch their Division title to advance to the playoffs 1988 Wrigley Field's 1st official night game is played Aug 9 1991 The new Comiskey park opens across the street from the previous park 1997 The Cubs and Sox meet for the first time in regular-season play 2000 The Cubs play their season opener in Tokyo. The Sox win their Division title 2003 The Cubs lost game 7, the very game they needed to get to the World Series. Fire authorities in California found a corpse in a burned out section of forest while assessing the damage done by forest fire. The deceased male was dressed in a full west suit, complete with scuba tanks on his back, flippers, and face mask in the middle of the forest. A post-mortem test revealed that the man died not from burns, but from massive internal injuries. Dental records provided identification. Investigators then set about to determine how a fully clad diver", "(ii) Cubs win in $4$ if they win Games 2, 3, 4. This has probability $p^3$. For the probability that $X=4$, add the numbers obtained in (i) and (ii). Next we find the probability that $X=5$. We could do a cases analysis. But we know the answer! It is $1-\\Pr(X=3)-\\Pr(X=4)$. Remark: In the post, there is, among other things, the assertion that the probability Cubs win in $4$ games is $p^3(1-p)$. This is true for unconditional probabilities. But the Cards won the first game, that's a fact. So the probability the Cubs win in $4$ games, given that fact, is simply $p^3$. - Ah! I see, basically I no longer include the probability of the first game because it becomes a certainty-It has already occurred. That is why I was off by a power. I was able to do this question with your help from last time. Thanks @AndréNicolas!!!!!!!! – user1527227 Jun 7 '13 at 19:39 Yes, the Cardinals have won the first game. Now the analysis is no longer symmetrical, since Cards have to win $2$ and Cubs have to win $3$. But general analysis could be done, using ideas like those in my previous answer. Note that calculation is very quick, since as I point out, the (conditional) probability that $X=5$ is known once we know", "\\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA GingerAle Nov 16, 2020 #10 +123309 0 .....The other thing is a concept that was discussed on the radio show Car Talk, the question whether two people together can know less than either of them individually. ..... LOL !!!!! This is sometimes demonstrated on the Forum...... Dec 1, 2020", "3, or 4 games. Similarly, team $\\ell$ would win if they win 3 or 4 games. Yet another way to view this is that team $w$ will not win the series if they win 0 or 1 more game. But again, each of these approaches only yields the correct probability if each likelihood is computed assuming that all four games are played. For example, using the approach of computing one minus the likelihood that team $w$ wins 1 or 0 additional game: As I was writing this post, the Red Sox won Game 3 and took a 2-1 series lead!", "Cubs win MORE THAN 110 games. Cheers.", "9th inning lead in Game 4. The Red Sox walked off that game, and the next. Then in Game 6, a game-tying run by the Yankees was nullified due to interference, which allowed the Red Sox to hold onto the lead and win, and the Red Sox won Game 7 handily. This was the first time any team had come back from a 3-0 series lead in the postseason, and Boston subsequently swept the Cardinals in the World Series. Crazy-assed luck, if you tell me. So you can guess my reaction to the 2007 18-0 Patriots season after the Red Sox swept the World Series earlier that year. PLUS the Patriots defeated my Chargers 21-12 in the Championship game. I was all out for a big F-U to the Patriots in the Super Bowl, and my wish came true as the Giants pulled a major upset victory with a miracle catch and a last minute touchdown. Incredible. Well, since then, the Patriots have not won a single postseason game and the Red Sox have been trippin. In 2007 I vowed never to root for a Boston team again. I might be getting to that turning point where I will put in exceptions… such as when the Patriots are up against loser teams like the Jets. But yeah, bipolar", "9th inning lead in Game 4. The Red Sox walked off that game, and the next. Then in Game 6, a game-tying run by the Yankees was nullified due to interference, which allowed the Red Sox to hold onto the lead and win, and the Red Sox won Game 7 handily. This was the first time any team had come back from a 3-0 series lead in the postseason, and Boston subsequently swept the Cardinals in the World Series. Crazy-assed luck, if you tell me. So you can guess my reaction to the 2007 18-0 Patriots season after the Red Sox swept the World Series earlier that year. PLUS the Patriots defeated my Chargers 21-12 in the Championship game. I was all out for a big F-U to the Patriots in the Super Bowl, and my wish came true as the Giants pulled a major upset victory with a miracle catch and a last minute touchdown. Incredible. Well, since then, the Patriots have not won a single postseason game and the Red Sox have been trippin. In 2007 I vowed never to root for a Boston team again. I might be getting to that turning point where I will put in exceptions… such as when the Patriots are up against loser teams like the Jets. But yeah, bipolar", "# Tournament Probabilities October 2016 ## Problem statement FiveThirtyEight does baseball predictions where, for each team, they give the probabilities of advancing to the playoffs, advancing to the next round, etc. Here were the probabilities as of October 15: Chance of making TeamDivisionWorld SeriesWin World Series CubsNL Central64%40% DodgersNL West36%19% Blue JaysAL East55%24% IndiansAL Central45%17% (note that the Cubs and Dodgers were playing each other in the NLCS, and the Blue Jays and Indians were playing each other in the ALCS) Looking at this, it's clear that the chance to win the World Series is conditional on who your opponent is going to be. Since the model thinks that the Blue Jays are better than the Indians, for example, the Cubs would have a better chance of winning the World Series if the Indians advance than if the Blue Jays do. So clearly, there are some underlying probabilities here: let $$w_{cb}$$ be the probability that the Cubs would beat the Blue Jays in the World Series, assuming they both advance. Similarly, let $$w_{ci}$$ be the probability that the Cubs would beat the Indians, and so on with $$w_{db}$$ and $$w_{di}$$. The question I have is: can we derive the various $$w$$'s from the probabilities given above? ## Setup Let's define the given probabilities from the table with some", "win the World Series. If all of that is news to you, you should also be told that the Cubs have, frankly, sucked for a long, long time. They haven’t won a pennant since 1945 and a World Series since 1908. There is even a short story (The Last Pennant Before Armageddon) tying the Cubs winning a pennant to the end of the world. (To answer the obvious question, the World Series is scheduled to start on October 25th. Barring rain delays, the last possible game would be on November 2nd. The US presidential election is on November 8th.) So if the Cubs make it to the World Series, there will be a lot of excitement around here. If they actually win the Series, Cook County will likely shut down for a month. And that all raises a question: How many Cubs t-shirts can be sold? Read Full Post » ## Zulily, delivery times, and operations strategy About a year ago, we had a post on Zulily and how they managed their order fulfillment. It featured a nifty graphic from the Wall Street Journal showing just how much longer their delivery times were relative to other interet retailers. Now, the Journal has another story — with a spiffy updated graphic — discussing how their delivery times have gotten" ]

[ "game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \\dfrac{405}{2048} \\approx 19.78\\% \\\\ \\text{ }\\\\$$ $$\\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\\dfrac{405}{4096} \\approx 9.88\\% \\\\ \\text{ }\\\\$$ $$\\text{ The sum of the individual probabilities gives }\\\\ \\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA Nov 16, 2020 #1 0 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series?", "game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \\dfrac{405}{2048} \\approx 19.78\\% \\\\ \\text{ }\\\\$$ $$\\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\\dfrac{405}{4096} \\approx 9.88\\% \\\\ \\text{ }\\\\$$ $$\\text{ The sum of the individual probabilities gives }\\\\ \\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA Nov 16, 2020 #1 0 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series?", "+0 # The Cubs are playing the Red Sox in the World Series. 0 116 10 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. Nov 16, 2020 #3 +2140 +2 Solution: $$\\text {The series ends after a team has a fourth win.}\\\\ \\text{Here are the four scenarios with derived probabilities }\\\\ \\text{where the Cubs win the series. }\\\\ \\text{ }\\\\ \\text{1) The Cubs win the first four games.}\\ \\text{ }\\\\ \\hspace{35 mm} \\rho(\\text {Cbs win on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} \\left(\\dfrac{3}{4}\\right)^4 = \\dfrac{81}{256} \\approx 31.64%\\\\$$ $$\\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \\dfrac{81}{256} \\approx 31.64 \\% \\\\ \\text{ }\\\\$$ $$\\text{3) Cubs win series in", "+0 # The Cubs are playing the Red Sox in the World Series. 0 536 10 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. Nov 16, 2020 #3 +2372 +3 Solution: $$\\text {The series ends after a team has a fourth win.}\\\\ \\text{Here are the four scenarios with derived probabilities }\\\\ \\text{where the Cubs win the series. }\\\\ \\text{ }\\\\ \\text{1) The Cubs win the first four games.}\\ \\text{ }\\\\ \\hspace{35 mm} \\rho(\\text {Cbs win on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} \\left(\\dfrac{3}{4}\\right)^4 = \\dfrac{81}{256} \\approx 31.64%\\\\$$ $$\\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \\dfrac{81}{256} \\approx 31.64 \\% \\\\ \\text{ }\\\\$$ $$\\text{3) Cubs win series in", "on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} \\left(\\dfrac{3}{4}\\right)^4 = \\dfrac{81}{256} \\approx 31.64%\\\\$$ $$\\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \\dfrac{81}{256} \\approx 31.64 \\% \\\\ \\text{ }\\\\$$ $$\\text{3) Cubs win series in game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \\dfrac{405}{2048} \\approx 19.78\\% \\\\ \\text{ }\\\\$$ $$\\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\\dfrac{405}{4096} \\approx 9.88\\% \\\\ \\text{ }\\\\$$ $$\\text{ The sum of the individual probabilities gives }\\\\", "+0 # Binomial Probability +2 168 2 +290 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability $$\\frac{3}{5}$$ and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. TheMathCoder Apr 26, 2018 #1 +1201 +1 A solution to this question is here: https://web2.0calc.com/questions/is-anyone-going-to-help#r9 GA GingerAle Apr 26, 2018 #1 +1201 +1 A solution to this question is here: https://web2.0calc.com/questions/is-anyone-going-to-help#r9 GA GingerAle Apr 26, 2018 #2 +290 +3 Thanks GingerAle TheMathCoder Apr 26, 2018", "fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \\dfrac{81}{256} \\approx 31.64 \\% \\\\ \\text{ }\\\\$$ $$\\text{3) Cubs win series in game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \\dfrac{405}{2048} \\approx 19.78\\% \\\\ \\text{ }\\\\$$ $$\\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\\dfrac{405}{4096} \\approx 9.88\\% \\\\ \\text{ }\\\\$$ $$\\text{ The sum of the individual probabilities gives }\\\\ \\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA GingerAle Nov 16, 2020 #10 +114094 0 .....The other thing is a concept that was discussed on the radio show Car", "\\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA GingerAle Nov 16, 2020 #10 +123309 0 .....The other thing is a concept that was discussed on the radio show Car Talk, the question whether two people together can know less than either of them individually. ..... LOL !!!!! This is sometimes demonstrated on the Forum...... Dec 1, 2020", "+0 # Is anyone going to help? +1 306 12 +737 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/5 and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. MIRB16 Apr 13, 2018 #1 +549 +1 how many matches does each team play in total? lynx7 Apr 13, 2018 edited by lynx7 Apr 13, 2018 #2 +737 +1 it only says that they need to win 4 games before the other team MIRB16 Apr 13, 2018 #3 +549 +1 3/5 + 3/5 + 3/5 + 3/5 =12/5 (probability that Cubs will win all four matches) 12/5 * 100 = 240% i am not sure if i am right. lynx7 Apr 13, 2018 #4 +1 If the probability of winning any one game is 3/5, then the probability of winning the World Series should be: (3/5)^4 =0.1296 =~13%. Guest Apr 13, 2018 #5 +1 I think the probability should be: 7C4 x (3/5)^4 x (2/5)^3 =~29% Guest Apr 13, 2018 #6 +1 Let everybody chime in!. I think you have to sum", "they win the first game but lose the second • $lw$ denote the outcome where they lose the first game but win the second • $ll$ denote the outcome where the Cubs lose both games The sample space for the first two games is given by: $$\\Omega = \\{ww, wl, lw, ll\\}$$ ### Before any games are played: One possible $\\sigma$-field is given by: $$S_0 = \\left\\{ \\left\\{ \\emptyset \\right\\}, \\left\\{ww, wl, lw, ll \\right\\} \\right\\}$$ This captures what is knowable before any game is played. You can't distinguish between any of the outcomes. And any random variable $X_0$ observable at time $t=0$ should map outcomes $ww, wl, lw, ll$ to the same value. This is captured formally with the notion of a measurable function. ### After the first game: After the first game is played, there is additional information, hence: $$S_1 = \\left\\{ \\left\\{ \\emptyset \\right\\}, \\left\\{ww, wl \\right\\}, \\left\\{lw, ll \\right\\}, \\left\\{ww, wl, lw, ll \\right\\} \\right\\}$$ That is, you can tell whether the Cubs won or lost the first game, but you can't tell apart $ww$ from $wl$! You can't tell who won the second game. Let $C_1$ be a random variable denoting whether Cubs won game 1. $C_1$ is not measurable with respect to $S_0$ but it is measurable with respect to $S_1$. Let", "give a source for your code! You may hope it helps, but it does not help very much. While this is the correct result, without an explanation of its origination, it’s of little help to a student trying to learn how to solve these types of probability questions. GA GingerAle Apr 14, 2018 #9 +1221 +3 Solution: $$\\text {The series ends after a team has a fourth win.}\\\\ \\text{Here are the four scenarios with derived probabilities }\\\\ \\text{where the Cubs win the series. }\\\\ \\text{1) The Cubs win the first four games.}\\ \\text{ }\\\\ \\hspace{35 mm} \\rho(\\text {Cbs win on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} (3/5)^4 = (81/625) = 12.96\\%\\\\ \\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\binom{4}{3}(3/5)^3*(2/5) *(3/5) = (648/3125) \\approx 20.74\\% \\\\ \\text{ }\\\\ \\text{3) Cubs win series in game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ }", "this means. The other thing is a concept that was discussed on the radio show Car Talk, the question whether two people together can know less than either of them individually. The two hosts on the show were known for self-deprecating humor, and the question was submitted by a fan of the show. The discussion ran for weeks, but if it was ever resolved, I missed that episode. I almost adopted the username MisterBB for this site, in your honor, but I checked it out first on the internet and discovered to my disappointment that it had already been claimed. Sigh... gang aft agley. Goodnight, Gracey. . Guest Nov 23, 2020 #9 +2140 0 Also Pending ................ GingerAle Dec 1, 2020 #3 +2140 +2 Solution: $$\\text {The series ends after a team has a fourth win.}\\\\ \\text{Here are the four scenarios with derived probabilities }\\\\ \\text{where the Cubs win the series. }\\\\ \\text{ }\\\\ \\text{1) The Cubs win the first four games.}\\ \\text{ }\\\\ \\hspace{35 mm} \\rho(\\text {Cbs win on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} \\left(\\dfrac{3}{4}\\right)^4 = \\dfrac{81}{256} \\approx 31.64%\\\\$$ $$\\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the", "(ii) Cubs win in $4$ if they win Games 2, 3, 4. This has probability $p^3$. For the probability that $X=4$, add the numbers obtained in (i) and (ii). Next we find the probability that $X=5$. We could do a cases analysis. But we know the answer! It is $1-\\Pr(X=3)-\\Pr(X=4)$. Remark: In the post, there is, among other things, the assertion that the probability Cubs win in $4$ games is $p^3(1-p)$. This is true for unconditional probabilities. But the Cards won the first game, that's a fact. So the probability the Cubs win in $4$ games, given that fact, is simply $p^3$. - Ah! I see, basically I no longer include the probability of the first game because it becomes a certainty-It has already occurred. That is why I was off by a power. I was able to do this question with your help from last time. Thanks @AndréNicolas!!!!!!!! – user1527227 Jun 7 '13 at 19:39 Yes, the Cardinals have won the first game. Now the analysis is no longer symmetrical, since Cards have to win $2$ and Cubs have to win $3$. But general analysis could be done, using ideas like those in my previous answer. Note that calculation is very quick, since as I point out, the (conditional) probability that $X=5$ is known once we know", "is the following: In order to have the series of length k, the winning team must have a win in its last game. This is because if there are $w$ wins in the $(k-1)$ interval, then the series would be over before $k$ games are played. So for cardinals winning in $k=4$, we have: W _ _ W. So there are $2$ remaining spots left which is represented as $(k-2)$ and we have to place $1$ win. So I would write this as: $\\Pr(\\text{Cardinals Win})=\\binom{k-2}{w-2} \\cdot (1-p)^w \\cdot p^{k-w} \\text{ for } k=3,4,5\\tag{1}$ We could test this with k=5 and get the same derivation: W _ _ _ W. Here we have to place 1 win in 3 spots. Cubs Win: k=5 The cubs lost the first game, must win the last game, and place 2 wins in the remaining spots: L _ _ _ W. I wrote this as: $\\Pr(\\text{Cubs Win})=\\binom{k-2}{w-1} \\cdot (p)^w \\cdot (1-p)^{k-w} \\text{ for } k=4,5\\tag{2}$ Note that for the case k=3, the Cubs lost the first game and they cannot win for that case. Since the event that the teams win are disjoint, the probability for the number of games played is simply the sum of equations (1) and (2). I then solved the equations for each value of $k$: $Pr(\\text{Cubs win in", "the Cubs win the World Series, first they have to advance to the World Series (by beating the Dodgers), then beat the Indians if the Indians advanced, or beat the Blue Jays if the Blue Jays advanced. So, this works out to $$c_2 = c_1\\cdot(b_1w_{cb}+i_1w_{ci})$$Remembering that the $$w$$'s are the variables here and everything else are constants, we can simplify this to $$\\frac{c_2}{c_1}=b_1w_{cb}+i_1w_{ci}$$Similarly for the other teams, we have $$\\frac{d_2}{d_1}=b_1w_{db}+i_1w_{di}$$ $$\\frac{b_2}{b_1}=c_1(1-w_{cb})+d_1(1-w_{db})$$ $$\\frac{i_2}{i_1}=c_1(1-w_{ci})+d_1(1-w_{di})$$ We can write this in matrix form after a little more simplification to get $$\\left( \\begin{array}{cccc} b_1 & i_1 & 0 & 0 \\\\ 0 & 0 & b_1 & i_1 \\\\ -c_1 & 0 & -d_1 & 0 \\\\ 0 & -c_1 & 0 & -d_1 \\end{array} \\right) \\left( \\begin{array}{cccc} w_{cb} \\\\ w_{ci} \\\\ w_{db} \\\\ w_{di} \\end{array} \\right) = \\left( \\begin{array}{cccc} \\frac{c_2}{c_1} \\\\ \\frac{d_2}{d_1} \\\\ \\frac{b_2}{b_1} - c_1 - d_1 \\\\ \\frac{i_2}{i_1} - c_1 - d_1 \\end{array} \\right)$$ Now we can use Gaussian elimination to reduce this to row echelon form and then solve for the $$w$$'s. At this point you can probably plug it in to Mathematica or Maple, but I was working on paper so I did it by hand. So here goes: Add $$\\frac{c_1}{b_1}$$ times row 1 to row 3: $$\\left( \\begin{array}{cccc} b_1 & i_1 & 0 & 0 \\\\ 0", "Cubs' probability of winning the World Series stays constant. Another point is just because the system is underdetermined doesn't mean that we can pick anything for $$w_{cb}$$. For example, $$w_{cb}$$ can't be lower than around $$0.4$$ or there are no solutions. (since the Cubs have a $$0.4$$ chance of winning the World Series regardless of who is playing!) ## Conclusion It's a little disappointing we can't solve for the various $$w$$'s (especially because if we know whether the Cubs or Dodgers win but the Indians and Blue Jays are still playing, then we can solve for all the $$w$$'s!), but now we know for sure! Fancy math equations by MathJax.", "Red Sox ahead. Then you must bet an amount such that if the Red Sox lose, you lose an amount to set you at $0 (because the score is then 3-3), and such that if the Red Sox win, you win an amount to set you at$1. So if the score is 3-2, your total winnings must be +$0.50 and you must bet$0.50. Similarly if the score is 2-3, you must be at -$0.50 and bet$0.50. Proceed backwards in this manner to reach the answer for when the score is 0-0. 7. Aug 31, 2009 ### wofsy Right. You can think of your methodology as pricing a contingent claim on the world series with pay off +- 1 depending on the outcome of the series. This backwards induction is standard in pricing options and other non-path dependent contingent claims. 8. Aug 31, 2009 ### wofsy but there is a probability proof as well.", "their playoff spot and will play their first playoff game a week from today. The Cubs’ road to the World Series title consists of a best-of-five series followed by two best-of-seven series. How many unique strings of wins and losses could the Cubs assemble if they make their way through the playoffs and win their first championship title since 1908? (For example, one possible string would be WWWWWWWWWWW — three straight sweeps. Another would be WWWWWWWLLLWWWW — two sweeps plus a dramatic World Series comeback.)2 If you need a hint, you can try asking me nicely. Want to submit a new Riddler Express? Email me. And now, for Riddler Classic, let’s make it interesting with a puzzle adapted from Keith Wynroe: You are a gambler and a Cubs fan. The Cubs are competing in a seven-game series against the Red Sox — first to four games wins. Your bookie agrees to take any even-odds bets on any of the individual games. Can you construct a series of bets such that the guaranteed outcomes are: You win $100 if the Cubs wins the series and lose$100 if the Red Sox win it? If you need a hint, you can try asking me nicely. Want to submit a new Riddler? Email me. Here’s the solution to last week’s Riddler Express,", "\\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\binom{5}{3}(3/5)^3*(2/5)^2 *(3/5) = (648/3125) \\approx 20.74\\% \\\\ \\text{ }\\\\ \\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\binom{6}{3}(3/5)^3*(2/5)^3 *(3/5) \\approx 16.59\\% \\\\ \\text{ }\\\\ \\text{ The sum of the individual probabilities gives }\\\\ \\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ (20.74+20.74+16.59+12.96) \\bf \\approx 71.0\\%$$ GA GingerAle Apr 14, 2018 #12 +737 +1 Thank you All! MIRB16 Apr 14, 2018", "imgur. The bars in red are the win total values covered by the 95% interval. The blue line represents my estimate of the team's \"True Win Total\" based on its performance - so if the blue line is to the left of the peak, the team is predicted to finish \"lucky\" - more wins than would be expected based on their talent level - and if the blue line is to the right of the peak, the team is predicted to finish \"unlucky\" - fewer wins that would be expected based on their talent level. It's still difficult to predict final win totals even at the beginning of September - intervals have a width of approximately 11-12 games. The Texas Ranges have been lucky this season, with a projected win total over 10 games larger than their estimated true talent level! Conversely, the Tampa Bay Rays have been unlucky, with a projected win total 10 games lower than their true talent level. The Chicago Cubs have a good chance at winning 105+ games. My system believes they are a \"true\" 101 win team. Conversely, the system believes that the worst team is the Atlanta Braves, which are a \"true\" 68 win team (though the Minnesota Twins are projected to have the worst record at 62 wins). ## Terminology" ]

[ "+0 # The Cubs are playing the Red Sox in the World Series. 0 116 10 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. Nov 16, 2020 #3 +2140 +2 Solution: $$\\text {The series ends after a team has a fourth win.}\\\\ \\text{Here are the four scenarios with derived probabilities }\\\\ \\text{where the Cubs win the series. }\\\\ \\text{ }\\\\ \\text{1) The Cubs win the first four games.}\\ \\text{ }\\\\ \\hspace{35 mm} \\rho(\\text {Cbs win on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} \\left(\\dfrac{3}{4}\\right)^4 = \\dfrac{81}{256} \\approx 31.64%\\\\$$ $$\\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \\dfrac{81}{256} \\approx 31.64 \\% \\\\ \\text{ }\\\\$$ $$\\text{3) Cubs win series in", "+0 # The Cubs are playing the Red Sox in the World Series. 0 536 10 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. Nov 16, 2020 #3 +2372 +3 Solution: $$\\text {The series ends after a team has a fourth win.}\\\\ \\text{Here are the four scenarios with derived probabilities }\\\\ \\text{where the Cubs win the series. }\\\\ \\text{ }\\\\ \\text{1) The Cubs win the first four games.}\\ \\text{ }\\\\ \\hspace{35 mm} \\rho(\\text {Cbs win on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} \\left(\\dfrac{3}{4}\\right)^4 = \\dfrac{81}{256} \\approx 31.64%\\\\$$ $$\\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \\dfrac{81}{256} \\approx 31.64 \\% \\\\ \\text{ }\\\\$$ $$\\text{3) Cubs win series in", "+0 # Binomial Probability +2 168 2 +290 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability $$\\frac{3}{5}$$ and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. TheMathCoder Apr 26, 2018 #1 +1201 +1 A solution to this question is here: https://web2.0calc.com/questions/is-anyone-going-to-help#r9 GA GingerAle Apr 26, 2018 #1 +1201 +1 A solution to this question is here: https://web2.0calc.com/questions/is-anyone-going-to-help#r9 GA GingerAle Apr 26, 2018 #2 +290 +3 Thanks GingerAle TheMathCoder Apr 26, 2018", "game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \\dfrac{405}{2048} \\approx 19.78\\% \\\\ \\text{ }\\\\$$ $$\\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\\dfrac{405}{4096} \\approx 9.88\\% \\\\ \\text{ }\\\\$$ $$\\text{ The sum of the individual probabilities gives }\\\\ \\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA Nov 16, 2020 #1 0 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series?", "game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \\dfrac{405}{2048} \\approx 19.78\\% \\\\ \\text{ }\\\\$$ $$\\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\\dfrac{405}{4096} \\approx 9.88\\% \\\\ \\text{ }\\\\$$ $$\\text{ The sum of the individual probabilities gives }\\\\ \\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA Nov 16, 2020 #1 0 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/4 and there are no ties, what is the probability that the Cubs will win the World Series?", "+0 # Is anyone going to help? +1 306 12 +737 The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/5 and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent. MIRB16 Apr 13, 2018 #1 +549 +1 how many matches does each team play in total? lynx7 Apr 13, 2018 edited by lynx7 Apr 13, 2018 #2 +737 +1 it only says that they need to win 4 games before the other team MIRB16 Apr 13, 2018 #3 +549 +1 3/5 + 3/5 + 3/5 + 3/5 =12/5 (probability that Cubs will win all four matches) 12/5 * 100 = 240% i am not sure if i am right. lynx7 Apr 13, 2018 #4 +1 If the probability of winning any one game is 3/5, then the probability of winning the World Series should be: (3/5)^4 =0.1296 =~13%. Guest Apr 13, 2018 #5 +1 I think the probability should be: 7C4 x (3/5)^4 x (2/5)^3 =~29% Guest Apr 13, 2018 #6 +1 Let everybody chime in!. I think you have to sum", "on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} \\left(\\dfrac{3}{4}\\right)^4 = \\dfrac{81}{256} \\approx 31.64%\\\\$$ $$\\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \\dfrac{81}{256} \\approx 31.64 \\% \\\\ \\text{ }\\\\$$ $$\\text{3) Cubs win series in game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \\dfrac{405}{2048} \\approx 19.78\\% \\\\ \\text{ }\\\\$$ $$\\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\\dfrac{405}{4096} \\approx 9.88\\% \\\\ \\text{ }\\\\$$ $$\\text{ The sum of the individual probabilities gives }\\\\", "\\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA GingerAle Nov 16, 2020 #10 +123309 0 .....The other thing is a concept that was discussed on the radio show Car Talk, the question whether two people together can know less than either of them individually. ..... LOL !!!!! This is sometimes demonstrated on the Forum...... Dec 1, 2020", "fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \\dfrac{81}{256} \\approx 31.64 \\% \\\\ \\text{ }\\\\$$ $$\\text{3) Cubs win series in game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 5})* \\rho\\text{(Cbs win 6th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \\dfrac{405}{2048} \\approx 19.78\\% \\\\ \\text{ }\\\\$$ $$\\text{4) Cubs win series in game seven. }\\\\ \\text{For the Cubs to win the series in game seven, }\\\\ \\text{they need to win three games in six trials and then win the seventh game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 6})* \\rho \\text{(Cbs win 7th game}) =\\\\ \\text{ } \\hspace{34 mm} \\dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\\dfrac{405}{4096} \\approx 9.88\\% \\\\ \\text{ }\\\\$$ $$\\text{ The sum of the individual probabilities gives }\\\\ \\text{ the overall probability of the Cubs winning the series.}\\\\ \\text {Sum of individual probabilities: } \\\\ \\left(\\dfrac{81}{256}\\right)+ \\left(\\dfrac{81}{256}\\right) + \\left(\\dfrac{405}{2048}\\right)+ \\left(\\dfrac{405}{4096}\\right) = \\dfrac{3807}{4096} \\bf \\approx 92.94\\%$$ GA GingerAle Nov 16, 2020 #10 +114094 0 .....The other thing is a concept that was discussed on the radio show Car", "Red Sox ahead. Then you must bet an amount such that if the Red Sox lose, you lose an amount to set you at $0 (because the score is then 3-3), and such that if the Red Sox win, you win an amount to set you at$1. So if the score is 3-2, your total winnings must be +$0.50 and you must bet$0.50. Similarly if the score is 2-3, you must be at -$0.50 and bet$0.50. Proceed backwards in this manner to reach the answer for when the score is 0-0. 7. Aug 31, 2009 ### wofsy Right. You can think of your methodology as pricing a contingent claim on the world series with pay off +- 1 depending on the outcome of the series. This backwards induction is standard in pricing options and other non-path dependent contingent claims. 8. Aug 31, 2009 ### wofsy but there is a probability proof as well.", "is the following: In order to have the series of length k, the winning team must have a win in its last game. This is because if there are $w$ wins in the $(k-1)$ interval, then the series would be over before $k$ games are played. So for cardinals winning in $k=4$, we have: W _ _ W. So there are $2$ remaining spots left which is represented as $(k-2)$ and we have to place $1$ win. So I would write this as: $\\Pr(\\text{Cardinals Win})=\\binom{k-2}{w-2} \\cdot (1-p)^w \\cdot p^{k-w} \\text{ for } k=3,4,5\\tag{1}$ We could test this with k=5 and get the same derivation: W _ _ _ W. Here we have to place 1 win in 3 spots. Cubs Win: k=5 The cubs lost the first game, must win the last game, and place 2 wins in the remaining spots: L _ _ _ W. I wrote this as: $\\Pr(\\text{Cubs Win})=\\binom{k-2}{w-1} \\cdot (p)^w \\cdot (1-p)^{k-w} \\text{ for } k=4,5\\tag{2}$ Note that for the case k=3, the Cubs lost the first game and they cannot win for that case. Since the event that the teams win are disjoint, the probability for the number of games played is simply the sum of equations (1) and (2). I then solved the equations for each value of $k$: $Pr(\\text{Cubs win in", "they win the first game but lose the second • $lw$ denote the outcome where they lose the first game but win the second • $ll$ denote the outcome where the Cubs lose both games The sample space for the first two games is given by: $$\\Omega = \\{ww, wl, lw, ll\\}$$ ### Before any games are played: One possible $\\sigma$-field is given by: $$S_0 = \\left\\{ \\left\\{ \\emptyset \\right\\}, \\left\\{ww, wl, lw, ll \\right\\} \\right\\}$$ This captures what is knowable before any game is played. You can't distinguish between any of the outcomes. And any random variable $X_0$ observable at time $t=0$ should map outcomes $ww, wl, lw, ll$ to the same value. This is captured formally with the notion of a measurable function. ### After the first game: After the first game is played, there is additional information, hence: $$S_1 = \\left\\{ \\left\\{ \\emptyset \\right\\}, \\left\\{ww, wl \\right\\}, \\left\\{lw, ll \\right\\}, \\left\\{ww, wl, lw, ll \\right\\} \\right\\}$$ That is, you can tell whether the Cubs won or lost the first game, but you can't tell apart $ww$ from $wl$! You can't tell who won the second game. Let $C_1$ be a random variable denoting whether Cubs won game 1. $C_1$ is not measurable with respect to $S_0$ but it is measurable with respect to $S_1$. Let", "Cubs' probability of winning the World Series stays constant. Another point is just because the system is underdetermined doesn't mean that we can pick anything for $$w_{cb}$$. For example, $$w_{cb}$$ can't be lower than around $$0.4$$ or there are no solutions. (since the Cubs have a $$0.4$$ chance of winning the World Series regardless of who is playing!) ## Conclusion It's a little disappointing we can't solve for the various $$w$$'s (especially because if we know whether the Cubs or Dodgers win but the Indians and Blue Jays are still playing, then we can solve for all the $$w$$'s!), but now we know for sure! Fancy math equations by MathJax.", "their playoff spot and will play their first playoff game a week from today. The Cubs’ road to the World Series title consists of a best-of-five series followed by two best-of-seven series. How many unique strings of wins and losses could the Cubs assemble if they make their way through the playoffs and win their first championship title since 1908? (For example, one possible string would be WWWWWWWWWWW — three straight sweeps. Another would be WWWWWWWLLLWWWW — two sweeps plus a dramatic World Series comeback.)2 If you need a hint, you can try asking me nicely. Want to submit a new Riddler Express? Email me. And now, for Riddler Classic, let’s make it interesting with a puzzle adapted from Keith Wynroe: You are a gambler and a Cubs fan. The Cubs are competing in a seven-game series against the Red Sox — first to four games wins. Your bookie agrees to take any even-odds bets on any of the individual games. Can you construct a series of bets such that the guaranteed outcomes are: You win $100 if the Cubs wins the series and lose$100 if the Red Sox win it? If you need a hint, you can try asking me nicely. Want to submit a new Riddler? Email me. Here’s the solution to last week’s Riddler Express,", "in question. Simulation puts the probability of at least one 107-year drought at about 1.5% in this model. In conclusion, if 2015 is the year that the Cubs's drought finally ends, it would be the end of a streak that, while surely painful for generations of Cubs fans, is not really all that mathematically remarkable in and of itself. However, do notice the droughts of the Red Sox (no WS from 1919 to 2003, winning in 2004) and the White Sox (no WS from 1918 to 2004, winning in 2005), and how neatly the Cubs winning the 2015 WS would completely the symmetry. As we speak, the Cubs are down 3 games to none to the Mets. If they still manage to make it to the World Series and win, well, that may indeed be of some numerological interest. *One might ask why, if we know $p$, one team's probability of not winning the WS since 1907, we cannot simply compute the probability of some team not winning the WS since 1907 with the formula $1-(1-p)^{\\text{numTeams}}$. The reason is that teams's records are not independent. To be fair, the dependence is very weak and the aforementioned formula does yield about 6% as well.", "give a source for your code! You may hope it helps, but it does not help very much. While this is the correct result, without an explanation of its origination, it’s of little help to a student trying to learn how to solve these types of probability questions. GA GingerAle Apr 14, 2018 #9 +1221 +3 Solution: $$\\text {The series ends after a team has a fourth win.}\\\\ \\text{Here are the four scenarios with derived probabilities }\\\\ \\text{where the Cubs win the series. }\\\\ \\text{1) The Cubs win the first four games.}\\ \\text{ }\\\\ \\hspace{35 mm} \\rho(\\text {Cbs win on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} (3/5)^4 = (81/625) = 12.96\\%\\\\ \\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the fifth game. }\\\\ \\text{ } \\hspace{34 mm} \\rho \\text {(Cbs win 3 in 4})*\\rho \\text{(Cbs win 5th game}) =\\\\ \\text{ } \\hspace{35 mm} \\ \\binom{4}{3}(3/5)^3*(2/5) *(3/5) = (648/3125) \\approx 20.74\\% \\\\ \\text{ }\\\\ \\text{3) Cubs win series in game six. }\\\\ \\text{For the Cubs to win the series in game six, }\\\\ \\text{they need to win three games in five trials and then win the sixth game. }\\\\ \\text{ }", "(ii) Cubs win in $4$ if they win Games 2, 3, 4. This has probability $p^3$. For the probability that $X=4$, add the numbers obtained in (i) and (ii). Next we find the probability that $X=5$. We could do a cases analysis. But we know the answer! It is $1-\\Pr(X=3)-\\Pr(X=4)$. Remark: In the post, there is, among other things, the assertion that the probability Cubs win in $4$ games is $p^3(1-p)$. This is true for unconditional probabilities. But the Cards won the first game, that's a fact. So the probability the Cubs win in $4$ games, given that fact, is simply $p^3$. - Ah! I see, basically I no longer include the probability of the first game because it becomes a certainty-It has already occurred. That is why I was off by a power. I was able to do this question with your help from last time. Thanks @AndréNicolas!!!!!!!! – user1527227 Jun 7 '13 at 19:39 Yes, the Cardinals have won the first game. Now the analysis is no longer symmetrical, since Cards have to win $2$ and Cubs have to win $3$. But general analysis could be done, using ideas like those in my previous answer. Note that calculation is very quick, since as I point out, the (conditional) probability that $X=5$ is known once we know", "# Betting on the world series This Riddler classic puzzle is about placing bets on baseball: You are a gambler and a Cubs fan. The Cubs are competing in a seven-game series against the Red Sox — first to four games wins. Your bookie agrees to take any even-odds bets on any of the individual games. Can you construct a series of bets such that the guaranteed outcomes are: You win \\$100 if the Cubs wins the series and lose \\$100 if the Red Sox win it? The challenge here is that we don’t know ahead of time when the series will end. It could end in a four-game blowout, or it could last the full seven games. How should we construct our bets so that the result is the same regardless of series length? Here is my solution: [Show Solution] ## 4 thoughts on “Betting on the world series” 1. Is it more elegant to know that you have to be at a sum of zero if the series is tied at 3 each, at which time you’d need to bet \\$100? Then, if you have listed the sample space as a tree, you can just work backwards. 1. Jon Z says: If you want to eliminate some parameters you can just express it as (n C", "the Cubs win the World Series, first they have to advance to the World Series (by beating the Dodgers), then beat the Indians if the Indians advanced, or beat the Blue Jays if the Blue Jays advanced. So, this works out to $$c_2 = c_1\\cdot(b_1w_{cb}+i_1w_{ci})$$Remembering that the $$w$$'s are the variables here and everything else are constants, we can simplify this to $$\\frac{c_2}{c_1}=b_1w_{cb}+i_1w_{ci}$$Similarly for the other teams, we have $$\\frac{d_2}{d_1}=b_1w_{db}+i_1w_{di}$$ $$\\frac{b_2}{b_1}=c_1(1-w_{cb})+d_1(1-w_{db})$$ $$\\frac{i_2}{i_1}=c_1(1-w_{ci})+d_1(1-w_{di})$$ We can write this in matrix form after a little more simplification to get $$\\left( \\begin{array}{cccc} b_1 & i_1 & 0 & 0 \\\\ 0 & 0 & b_1 & i_1 \\\\ -c_1 & 0 & -d_1 & 0 \\\\ 0 & -c_1 & 0 & -d_1 \\end{array} \\right) \\left( \\begin{array}{cccc} w_{cb} \\\\ w_{ci} \\\\ w_{db} \\\\ w_{di} \\end{array} \\right) = \\left( \\begin{array}{cccc} \\frac{c_2}{c_1} \\\\ \\frac{d_2}{d_1} \\\\ \\frac{b_2}{b_1} - c_1 - d_1 \\\\ \\frac{i_2}{i_1} - c_1 - d_1 \\end{array} \\right)$$ Now we can use Gaussian elimination to reduce this to row echelon form and then solve for the $$w$$'s. At this point you can probably plug it in to Mathematica or Maple, but I was working on paper so I did it by hand. So here goes: Add $$\\frac{c_1}{b_1}$$ times row 1 to row 3: $$\\left( \\begin{array}{cccc} b_1 & i_1 & 0 & 0 \\\\ 0", "this means. The other thing is a concept that was discussed on the radio show Car Talk, the question whether two people together can know less than either of them individually. The two hosts on the show were known for self-deprecating humor, and the question was submitted by a fan of the show. The discussion ran for weeks, but if it was ever resolved, I missed that episode. I almost adopted the username MisterBB for this site, in your honor, but I checked it out first on the internet and discovered to my disappointment that it had already been claimed. Sigh... gang aft agley. Goodnight, Gracey. . Guest Nov 23, 2020 #9 +2140 0 Also Pending ................ GingerAle Dec 1, 2020 #3 +2140 +2 Solution: $$\\text {The series ends after a team has a fourth win.}\\\\ \\text{Here are the four scenarios with derived probabilities }\\\\ \\text{where the Cubs win the series. }\\\\ \\text{ }\\\\ \\text{1) The Cubs win the first four games.}\\ \\text{ }\\\\ \\hspace{35 mm} \\rho(\\text {Cbs win on } 4^{th} \\text{game )= }\\ \\text{ }\\\\ \\hspace{35 mm} \\left(\\dfrac{3}{4}\\right)^4 = \\dfrac{81}{256} \\approx 31.64%\\\\$$ $$\\text{ }\\\\ \\text{2) Cubs win series in game five. }\\\\ \\text{For the Cubs to win the series in game five,}\\\\ \\text{ they need to win three games in four trials and then win the" ]

[ "# How many ways can I place four distinct pawns on the board such that each column and row of the board contains no more than one pawn? If I have a $4\\times 4$ chess board, in how many ways can I place four distinct pawns on the board such that each column and row of the board contains no more than one pawn? I have a feeling this is a basic combination problem with cases. I don't see it yet. We have 4 choices for the column of the pawn in the first row, 3 choices for the column of the pawn in the second row, 2 choices for the column of the pawn in row 3, and 1 choice for the column of the pawn in row 4, for a total of $4!$ places for the positions of the pawns. Since the pawns are distinct, there are $4!$ ways to place them in these chosen positions; so there are $4!\\cdot4!=576$ possibilities. Alternatively, there are 16 places for the first pawn, then 9 places for the second pawn, only 4 choices left for the third pawn, and just 1 choice for the fourth pawn, giving $\\;16\\cdot9\\cdot4\\cdot1=576$ possibilities. • It's an interesting interpretation that four distinct pawns means four distinguishable pawns. I took it to mean merely that", "row, each column, and each long diagonal contains at most 4 pawns? ×", "they stand, but that doesn't mean that there isn't another configuration out there with 18 pawns or more. – Michael Seifert Feb 24 '16 at 16:33 UPDATE: Based on the suggestion by f'' in the comments, my original statements can be greatly improved: To each pawn, we can associate its own square and the four squares surrounding it, making a plus-shaped tile with an area of 5 square units. Any two such tiles that abut will automatically have a distance of at least $\\sqrt{5}$ between their centers (i.e., the pawn locations). Moreover, the total area not on the original chessboard covered by these tiles will be at most 12 square tiles; we can fit at most three such tiles along each edge, meaning that at most three \"squares\" from our plus-shaped tiles can stick out over the edge. The total area covered by tiles is thus at most $9 \\times 9 + 3\\times 4 = 93$ square units, and so $N < 93/5 = 18.6$. Thus, a significantly better version of Professor Half-Brain's Second Theorem is It is not possible to place 19 pawns according to the above rules. Note that the bound is substantially better both because we have increased the \"size\" of the tiles and reduced the maximum amount of area they can cover. Below this", "therefore there must be at least two more distinct ones in the last four squares. Mar 2 at 15:26", "the number of pawns cannot be more than $64 - 16 = 48$. An arrangement of 48 pawns that satisfies the condition of the problem will be obtained if you put a pawn on each cell of the board, except for the cells of the two main diagonals. A similar logic can be applied to the $2n \\times 2n$ chessboard, obtaining $4 \\cdot n^2 - 4 \\cdot n$ as an answer. Four dice are thrown. What is the probability that the product of the four numbers equals $36$?", "here was my original 39 move solution: Two of the four knights are promoted black pawns The idea was to create a position with as many black pawns ready to promote as possible, as each pawn could make up to 4 different moves. However, Paul Panzer's solution managed to squeeze in an extra black move with better positioning and a g-file rook instead of an a-file rook", "# Puzzle Solutions, Issue 03 Warning: Contains SPOILERS! These solutions relate to puzzles found in Issue 03 of the magazine. 792 ### Chessboard Squares There are 64 1×1 squares, 49 2×2 squares, 36 3×3 squares, 25 4×4 squares, 16 5×5 squares, 9 6×6 squares, 4 7×7 squares and 1 8×8 square on a chessboard. This can be shown by counting how many positions the top left corner of the square can sit on. For example, the top left corner of a 5×5 square can be in the first four rows and columns of the board (otherwise the square will go off the board) and 4×4=16. 64+49+36+25+16+9+4+1=204. • ### Book of the Year 2019 We announce the winner of this coveted prize", "# Total no of squares on a Chess Board Is there any formula than calculates the total number of squares on chessboard? For example in a $8\\times8$ chessboard, there are squares of sizes $1\\times1$, $2\\times2$, $\\ldots$, $8\\times8$. So I want to know their total number. Is there a general formula for a $n\\times n$ chessboard? • Yes, it's classroom assignment question but not sure both have same assignment.. – Shyam Shingadiya Jun 11 '17 at 17:29 • Yes the logic is same, but only the representation is different. – Shyam Shingadiya Jun 11 '17 at 17:31 There are $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 = 204$ squares on an $8\\times8$ board. In the general case, the number of squares is $1^2 + \\dots + n^2$ for an $n\\times n$ board, or equivalently, $\\rlap{\\raise{2.5ex}{~\\scriptsize ~n}}{\\rlap{\\lower{2ex}{\\scriptsize{k\\text{=1}}}}{\\large{\\sum}~k^2}}$. As @M Oehm noted, a Faulhaber formula lets us simplify this to $\\large ~~\\frac{n(n+1)(2n+1)}6$. This is because ... For $1\\times1$ squares, there are obviously $n^2$ of them. You can fit $(n-1)$ rows of $2\\times2$ squares in an $n^2$ area, and each row has $(n-1)$ in it. You can fit $(n-2)$ rows of $3\\times3$ squares in an $n^2$ area, and each row has $(n-2)$ in it. ... You can fit $2$ rows of $(n-1)\\times(n-1)$ squares in", "has 8 pawns and black has 6 pawns, the last term of the score would be pawnWt * (wP-bP) = 1 * (8-6) = 2. (assuming the value of a pawn to be 1) That leads to a relative advantage for white only based on pawns, but according to how many of the other pieces are on the board, black might still have an advantage.", "you just show that it works for three randomly chosen sequences of numbers? Doesn’t that prove that it always works? No. Let’s go back to overcoming one’s self. You have figured out what the answer to Problem 1 is, so the answer to Problem 2 seems obvious. It is a gut feeling, an ancient instinct that you are always going to get the right answer if you double the last number and subtract one, right? So, here is a question for you: Uhmmm, let me see… The cube root of 3,141,592… Oh great, now I am melting. Problem 3: I will give you \\$333,333 for all the M&Ms you put on a 4-by-4 wooden chessboard of your choice, but you have to put 3 times as many delicious M&Ms in each consecutive square of the board starting with one M&M – on average, you can sell 1Kg of M&Ms for \\$5 and each M&M weights about 1g (the things you learn here). Oh, and I get to keep the useless chessboard. Deal or no deal?", "# Crowded Pawns Logic Level 4 We can place a maximum of $$M$$ pawns on a $$4\\times4$$ chessboard such that there's no pawn being attacked. In how many ways can we do that? In other words, how many configurations let us place $$M$$ pawns? ×", "A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with $15$ copies of itself, be used to cover an eight by eight chessboard?", "# Chessboard $8\\times8$ and domino $1\\times2$ There is a chessboard of size $8\\times 8$. I am given dominoes of size $1\\times 2$ and of a single color (assume, it has a color). It is possible to place a domino on the board so that it covers exactly two squares. I can't place two dominoes on the same square and I can't place a domino so that it is partially off the board. a) How many ways are there to place a single domino on the board( I place it only horizontally or vertically) My Answer is : 112 b)How many ways are there to place two different dominoes on the board(red domino and blue)? My Answer is : $1024$ (But, I'm not sure at all) c)How many ways are there to place two blue dominoes on the board? I don't know how to approach to this question. I have started to learn Combinatorics this year, and kind of stuck. Thank you, in advance. • Why 32? Why 1024? What reasoning led you to those numbers? – Arthur Dec 7 '17 at 17:11 (a) Treat the centers of $64$ chess positions as vertices and join neighboring vertices by edges. There are $112 = 2\\times 7 \\times 8$ of them. There is a one-one correspondence between possible placement of domino", "# Combinatorics How many ways can you mark 8 squares of an $$8\\times8$$ chessboard so that no two marked squares are in the same row or column, and none of the four corner squares is marked? (Rotations and reflections are considered different. ×", "Number of equal triangles in a chessboard $$1\\times 1$$ is cut and taken out from every corner of a $$8\\times 8$$ chess board. At least, how many equal triangles (equal triangles means congruent triangles, and color is not important) can be drawn on the remaining figure? What is the answer when we cut a $$1\\times 1$$ from a single corner of chessboard? For the first part of the question I cover chessboard by 20 equal triangles after removing corners like this: Is less than this possible? • I think the question is not quite complete. For any integer $n > 0$, a single (grey or white) square can be divided into an $n \\times n$ grid, and each subsquare divided by a diagonal into a pair of triangles, yielding $2n^2$ triangles. Hence the whole board can be divided into $63 \\cdot 2n^2$ equal triangles, for any $n$...but surely that's not what's intended. Can you clarify? – John Hughes Apr 9 '19 at 15:37 • Thanks for the point you brought up. Look at my edit please. – Ghartal Apr 9 '19 at 15:39 • The triangles I described were all congruent (45-45-90), so I don't believe your edit addresses my question. I'll be honest -- I have no idea what you're actually asking for here! – John Hughes", "Fill It Up With Pawns Logic Level 1 What is the maximum number of pawns that can be placed on an $$8\\times 8$$ chessboard such that each row, each column, and each long diagonal contains at most 4 pawns? ×", "# How Many Queens? Logic Level 1 How many queens can be placed on an $8\\times 8$ board such that no two queens are in the same row, column, or diagonal path? ×", "# Multicolored 4x4 With four colors, how many different ways can you color in the 16 squares in a $4\\times4$ grid such that each of the nine $2\\times2$ grids inside the $4\\times4$ grid contains each of the four colors? ×", "+0 # Challenge question 0 96 1 How many ways can a domino be placed on a chessboard? Each \"half\" of the domino must cover exactly one square of the chessboard. Guest Apr 30, 2018 #1 +867 +1 Sorry, my previous answer was wrong. Corner dominoes have two options, but regular squares have 3 or even 4. Here is the correct one. If the domino is oriented horizontally, its position is determined by the leftmost square it covers, for which there are $$7\\cdot8=56$$ possibilities (the rightmost file is excluded). Similar for vertical orientation. Thus, there are $$2\\cdot56=112$$ possibilities in total. I hope this helped, Gavin. GYanggg Apr 30, 2018 edited by GYanggg Apr 30, 2018", "9$$ board and pawns, consider a $$10 \\times 10$$ board and $$2 \\times 2$$ square pieces. We can associate to each pawn configuration on the $$9 \\times 9$$ board a unique configuration of squares in the $$10 \\times 10$$ board; consider the image below. The blue squares are the squares we added to our board to make it $$10 \\times 10$$. The $$2 \\times 2$$ square on the side is a piece for our new board, and the red square marks where the pawn lies for our previous board. It's easy to see this relation is a bijection. For example in the image below, to the left we have the $$9 \\times 9$$ board in which red squares mark pawn positions. To the right is the corresponding $$10 \\times 10$$ board with $$2 \\times 2$$ square pieces; notice the position of red squares is preserved and no pieces overlap. ### Solution of the puzzle Solving Professor Halfbrain's problem now becomes the same as solving the translated problem: How many $$2 \\times 2$$ square pieces can we place on a $$10 \\times 10$$ board without overlappings? Now, our new board has a total of $$100$$ squares, and each piece uses up $$4$$ squares. It's thus clear that no configuration can go beyond $$25$$ pieces. On the other hand, it" ]

[ "# How many ways can I place four distinct pawns on the board such that each column and row of the board contains no more than one pawn? If I have a $4\\times 4$ chess board, in how many ways can I place four distinct pawns on the board such that each column and row of the board contains no more than one pawn? I have a feeling this is a basic combination problem with cases. I don't see it yet. We have 4 choices for the column of the pawn in the first row, 3 choices for the column of the pawn in the second row, 2 choices for the column of the pawn in row 3, and 1 choice for the column of the pawn in row 4, for a total of $4!$ places for the positions of the pawns. Since the pawns are distinct, there are $4!$ ways to place them in these chosen positions; so there are $4!\\cdot4!=576$ possibilities. Alternatively, there are 16 places for the first pawn, then 9 places for the second pawn, only 4 choices left for the third pawn, and just 1 choice for the fourth pawn, giving $\\;16\\cdot9\\cdot4\\cdot1=576$ possibilities. • It's an interesting interpretation that four distinct pawns means four distinguishable pawns. I took it to mean merely that", "not.\" The end result is no. If we were to describe the set of outcomes, we could do so by describing it as a colored sequence of eight spaces from the board. Pick the location and the color for the first placed, then the location and the color for the second placed, and so on to get $64\\cdot 8\\cdot 63\\cdot 7\\cdot 62\\cdot 6\\cdots 57\\cdot 1=\\frac{64!}{56!}8!$ As in the previous case, we do the same with the exception of worryign about color now too, giving $64\\cdot 8\\cdot 49\\cdot 7\\cdot 36\\cdot 6\\cdots 4\\cdot 2 \\cdot 1\\cdot 1=(8!)^3$. This would yield a final probability of$$\\frac{(8!)^3}{\\frac{64!}{56!}8!}$$ which again equals the same as before. Scenario 4: Not only are rooks distinguishable by color, order of selection matters, but the person placing pieces will continue to place $56$ additional multicolored pawns after having placed all eight rooks. In this scenario, we would have a whopping $(64!)^2$ different outcomes in our sample space. The locations and colors of the pawns has literally nothing to do with whether or not the rooks share rows or columns (alternatively this is assuming that we say that pawns don't block movement to keep the spirit of the problem the same). If we were to run with this scenario, not only will we need to account for rooks not sharing", "# Kevin has 5 cubes. Each cube is a different color. Kevin will arrange the cubes side by side in a row. What is the total number of different arrangements of the 5 cubes that Kevin can make? ##### 1 Answer Oct 21, 2016 There are $120$ different arrangements of the five colored cubes. #### Explanation: The first position is one of five possibilities; the second position is therefore one of the four remaining possibilities; the third position is one of the three remaining possibilities; the fourth position will be one of the remaining two possibilities; and the fifth position will be filled by the remaining cube. Therefore, the total number of different arrangements is given by: $5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120$ There are $120$ different arrangements of the five colored cubes.", "3 \\cdot 2 \\cdot 1=362880$. You could also use Permutation notation as follows: $_9P_9 = \\frac{9!}{(9-9!)}=\\frac{9!}{0!}=9!= 362880$. How many different possible ways are there to play in first and second row of numbers? There are still 362880 possibilities for the first row but in the second row you can’t repeat any of the same numbers in the same columns as the numbers in the first row. Therefore there are following possibilities for the second row: $8!= 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 \\cdot 1=40320$. Then, to find the different possibilities for the first and second row you multiply the possibilities for each row, 362880 by 40320, which equals 14,631,321,600. How many different possible ways are there to play the entire game of Sudoku (all nine rows)? To get the total number of possible ways to play in the entire game of Sudoku (all nine rows) multiplies the possibilities in each row. $1^{st}$ row: 362880, $2^{nd}$ row: 40320, $3^{rd}$ row: 5040, $4^{th}$ row: 720, $5^{th}$ row: 120, $6^{th}$ row: 24, $7^{th}$ row: 6, $8^{th}$ row: 2, and $9^{th}$ row: 1. There are a grand total of 2,548,518,711,000,000,000 ways to play the entire game of Sudoku. There are a lot of ways! Playing Cards. The game Big Two, also called Deuces,", "The points outside that square have total weight$15/16$, which is less than$1$, so ... 5 Each of the 16 pawns can move at most 6 times and there are 30 captures possible. Therefore$(16\\cdot6+30)\\cdot 50=6300$is a rough upper bound (for example, not all pawns can make it to the opposite line without sometimes capturing - which would mean that sometimes pawn move and capture occur together). 5 Going to higher dimension will not help; it makes the problem worse. The Hales-Jewett theorem implies that for any fixed$p$and$n$, the first player can win the$p$-in-a-row game on the$n^d$board for all sufficiently large$d$. So adding dimensions makes these games less fair, not more fair. For example: The first player obviously wins 2-in-a-row ... 4 This is a cute little problem. The insight here is to look for an invariant that is preserved by the allowed moves. In this case, the right way to approach it is to think of the original coin at the origin as having a fixed mass of 1. The rule for replacing the coin with two others can be thought of as breaking the coin in half, to be replaced by two coins ... 4 You're correct that this is a nim variant. More generally, all impartial games (those where both players have the same moves available to", "second, given that the first one is placed? After the two coins are placed, how many options are left for the third coin? - Imagine placing the coins one at a time. The first coin is free to go anywhere. Once we place the first coin, there are $39$ available positions for the second coin. But we must avoid the positions that are in the same row or column as the first coin. So imagine rubbing out the row and column of the first coin. That leaves a $3$ by $9$ grid of \"good\" positions for the second coin. So the probability that after two placements, we are OK, is $\\dfrac{27}{39}$. Given that we are OK so far, place the third coin. There are $38$ places it could be put. To count the places that avoid the rows/columns of the first two coins, imagine rubbing out these rows and columns. That leaves a $2$ by $8$ grid, so $16% places. So given that the second placement was OK, the probability that the third is OK is$\\dfrac{16}{38}$. So our probability is$\\dfrac{27}{39}\\cdot \\dfrac{16}{38}$. Now we are finished, but are forced to do some mechanical work so we can be mechanically graded. Another way: We can choose the locations of the coins in$\\dbinom{40}{3}$equally likely ways. Now we want to count the", "Sum of numbers on chessboard. Consider the squares of an $8 \\times 8$ chessboard filled with the numbers $1,2,3,4 \\ldots ,64$ in sequential order. If we choose $8$ squares with the property that there is exactly $1$ from each row and exactly $1$ from each column, and add up the numbers in the chosen squares, show that the sum obtained is always $260$. Please provide full solution and explanation. • Without specifying how the squares were numbered, this is false: put $1$ in square $a1$, $2$ in square $b2$, etc. Then picking the diagonals the sum is only $1 + \\dots + 8 = 36$. – grantfgates May 9 '14 at 13:21 • The numbers are entered in sequential order. – Calvin Lin May 9 '14 at 13:23 If we put the numbers in sequential order, the numbers of the row $i$ are of the form $$8i + j$$ $$i \\in \\lbrace 0,1,2, \\ldots 7\\rbrace$$ $$j \\in \\lbrace 1,2,, \\ldots 8\\rbrace$$ If we choose $8$ squares with the property that there is exactly one from each row and exactly one from each column, with this notation the total is $$8 \\cdot \\biggl( \\sum_{i=0}^{7}i \\biggr) + \\sum_{j=1}^{8} j = 260$$ Hint: work in $\\pmod 8$ Hint: $260=28\\cdot 8 + 36$", "# Determine optimal strategy in board game We have a $25 \\times 25$ board and two players. The fields of the board are numbered like this: From $0$ to $24$ from west to the east, and from $24$ to $0$ from north to the south. The first player places the pawn in one of the fields in the top line, for example, in the column number $1$. Then the second player moves the pawn in one of three direction by any number of fields: To the south, to the west, or to the south-west, and so does the first player, and so on. One of the players wins when he places the pawn in the most south-west corner (that is, column $0$, line $0$). Determine the optimal first move for the second player, that is, the move such that after making it the second player will win for sure. It's a hard problem for me, as I'm not used to dealing with combinatorial game problems. Any hints appreciated. • I believe the same techniques used to show that player 2 has a winning strategy in the game nim will work here (counting path lengths in binary in this case). – Justin Benfield Mar 12 '16 at 10:47 • What do you mean by \"couting path length in binary\"?", "### Description Solve the problem below for $$T$$ test cases. You have a chess board of dimension $$1 \\times M$$ with positions numbered from $$1$$ to $$M$$ from left to right. You have $$N$$ identical pawns, initially lined up from position $$1$$ to $$N$$ on the chess board. You can do any of the following operations as many number of times in any order: • Select the $$i$$-th pawn from the left ($$1 \\le i \\le N - 2$$) such that the right of the $$(i + 2)$$-th pawn is inside the chessboard, and move the $$i$$-th pawn one position to the right of the $$(i+2)$$-th pawn. While there are two pawns on the same position, move one of the pawns one position to the left. • Rotate the board $$180$$ degrees, i.e., a pawn at position $$x$$ will end up at position $$M - x + 1$$. Find the number of distinct configurations of pawns that you can end up with after doing the above operations modulo $$998244353$$. Two configurations are considered distinct if there is at least one position where one configuration has a pawn but not the other. ### Constraints • $$1 \\le T \\le 10000$$ • $$3 \\le N \\le M \\le 200\\;000$$ ### Input The first line contains the number of test cases", "& \\cdot & \\times & \\cdot \\\\ \\cdot & \\cdot & * & \\cdot & \\cdot & \\cdot \\\\ \\end{matrix}$$ The only ... 6 The minimum number of switches required is$100$. This puzzle can be turned into a linear algebra problem over${\\Bbb Z}/2{\\Bbb Z}$by letting the vector$v\\in({\\Bbb Z}/2{\\Bbb Z})^{100}$record which switches are pressed ($1$if pressed,$0$if not), the vector$w\\in({\\Bbb Z}/2{\\Bbb Z})^{100}$record which lights are initially on ($1$if on,$0$if off), ... 6 This is Welter's Game with special starting positions. In chapter 13 of On Numbers and Games, John Horton Conway presents an amazing analysis based on \"animating functions.\" It appears to be his own take on an earlier solution presented by Welter (but without attribution). Although there is not enough space to give a full account of this game, I will ... 6 One example is Sylver's Coinage played like so: Player's alternate selecting positive integers ($1, 2, 3, 4\\dots$). The rule is that no number is allowed to be expressed as sum (with possible duplicates) of the previous. For example, say$\\{4, 8, 5, 7\\}$($8$would have had to been said before$4$) was previously said. Then$6$could be said, but$14$could ... 6 Each of the 16 pawns can move at most 6 times and there are 30 captures possible. Therefore$(16\\cdot6+30)\\cdot 50=6300$is a rough upper bound (for example, not all pawns can", "the number of pawns cannot be more than $64 - 16 = 48$. An arrangement of 48 pawns that satisfies the condition of the problem will be obtained if you put a pawn on each cell of the board, except for the cells of the two main diagonals. A similar logic can be applied to the $2n \\times 2n$ chessboard, obtaining $4 \\cdot n^2 - 4 \\cdot n$ as an answer. Four dice are thrown. What is the probability that the product of the four numbers equals $36$?", "four positions, which means the second position cannot be a $$3$$ and the first position can be any of the possible outcomes for a die roll. Hence, the number of ways of rolling a die six times so that there are exactly four consecutive threes is $$|R_1| + |R_2| + |R_3| = 5 \\cdot 6 + 5^2 + 5 \\cdot 6 = 30 + 25 + 30 = 85$$ How many ways are there of rolling a die six times in a row such that there are at least four consecutive threes? Let $$R_i$$ be the event that there are four consecutive threes beginning in the $$i$$th position, where $$1 \\leq i \\leq 3$$. Notice that the word exactly was not used in the description of $$R_i$$. Observe that to have five consecutive threes, there are two possibilities: 1. There must be four consecutive threes beginning in the first position and four consecutive threes beginning in the second position. 2. There must be four consecutive threes beginning in the first position and four consecutive threes beginning in the third position. To have six consecutive threes, there must be four consecutive threes beginning in the first position and four consecutive threes beginning in the second position and four consecutive threes beginning in the third position. Thus, the set of", "same but order in which the rooks are placed matters. In this case, as the order in which the locations in which rooks are placed are selected matter, we can describe any outcome of how they are placed as sequences of length-$8$ with no repeats with entries all taken from our set of $64$ available spaces. As such, there will be $64\\cdot 63\\cdot 62\\cdots 57=\\frac{64!}{56!}$ equally likely sequences. Each arrangement which satisfies the description \"no rook can attack another\" can be described by a sequence of spaces selected such that no space selected can be attacked any any previous space. Each time we make such a selection, our available choices decreases. Originally there are $64$ choices. Having removed that and those spaces targettable from that space, there are $49$ remaining spaces, and then $36$ and so on, giving a total of $64\\cdot 49\\cdot 36\\cdots 4\\cdot 1=8^2\\cdot 7^2\\cdot 6^2\\cdots 1^2=(8!)^2$ which yields a final probability of $$\\frac{(8!)^2}{\\frac{64!}{56!}}$$ which one should be able to quickly confirm equals the same answer as before. Scenario 3: Rooks look different (e.g. by color), the order in which rooks are taken to be placed is randomized, and the order in which the rooks are placed matters as well. This is to answer your question of \"does it matter if rooks look the same or", "Three coins are randomly placed into different positions on a $4×10$ grid. The probability that no two coins are in the same row or column can be expressed as $a/b$ where $a$ and $b$ are coprime positive integers. What is the sum of $a+b$ ? - Somebody decided to turn to MSE to get their homework done and is rapidly posting it on the site by parts. – Did Apr 2 '13 at 9:39 What is your probability space (i.e. how many different combinations of the three coins are there)? How many options do you have for the first coin? How many for the second, given that the first one is placed? After the two coins are placed, how many options are left for the third coin? - Once we place the first coin, there are $39$ available positions for the second coin. But we must avoid the positions that are in the same row or column as the first coin. So imagine rubbing out the row and column of the first coin. That leaves a $3$ by $9$ grid of \"good\" positions for the second coin. So the probability that after two placements, we are OK, is $\\dfrac{27}{39}$. Given that we are OK so far, place the third coin. There are $38$ places it could be put.", "the box is square (as it is for the final two figures), the subgraph can also be rotated within the box, but if the box is rectangular, then a quarter rotation requires the box to be rotated (or \"transposed\") as well. In any case we can count all possible ways to place the box in the chess board, and respectively the ways to place the subgraph in the box. The leftmost figure, a \"chair\" or \"h\", just fits into an upright $2\\times 1$ box as shown or in quarter rotation a $1\\times 2$ transposed box. The upright box can be placed on the chess board in $(m-2)\\cdot (n-1)$ locations, while the transposed box goes onto $(m-1)\\cdot (n-2)$ locations. Each box allows four distinct reflected/half-rotated versions of the subgraph, so the total number of subgraphs under this specific figure is: $$4(m-2)\\cdot (n-1) + 4(m-1)\\cdot (n-2)$$ The next figure, very like a capital \"F\", has the same size boxes and symmetries within those boxes. Thus it gives the same total number of subgraphs as the first figure. The third figure, perhaps viewed as \"y\" with a little imagination, fits into a $3\\times 1$ box. Thus the upright box has $(m-3)\\cdot (n-1)$ board locations, and the transposed box has $(m-1)\\cdot (n-3)$ locations. After rotations and reflections, there will be a total", "has 4 horizontal locations and 8 vertical locations. . . Hence, it has 32 possible locations. The vertical 5x1 bar also has 32 possible locations. Hence, it has $32 + 32 \\:=\\:{\\color{blue}64}$ ways. Therefore, there are: . $1764 + 840 + 64 \\:=\\:\\boxed{2668}$ ways . . to place the twelve pentominos on a chessboard. I'm grateful they didn't ask for \"six neighbour squares\". There would be $35$ hexominos to consider. Thanks for the praise, ARDASH. I didn't pay much attention, but it probably took an hour (most work on diagrams, using COPY and PASTE over and over). It's certainly obvious by now, isn't it? . . . I truly enjoy doing this. .", "the solutions can be found here. In our case, we can rescale our circles so that they have radius 1; the square in which they must all fit is therefore of side $(8 + \\sqrt{5})/(\\sqrt{5}/2) \\approx 9.15542.$ Unfortunately, it is not known whether 21 circles can fit into a square of this size; the best known packing is into a square of side 9.358... So it seems likely that you cannot place 21 pawns according to the above rules, but this is not a proof. Note that none of these are strict upper bounds; the fact that we must place the pawns on square grid points will mean that we cannot optimally pack our circles/hexagons, and so some of the square's area will be \"wasted\". Basically, for the first two theorems, we have proven an upper bound among all possible pawn placements on the chessboard; any placement for which the pawns are on grid points will necessarily be no greater than this bound. Most likely, the smallest number of pawns that cannot be placed is less than 25 (or 21, if you believe the \"non-proof\" in the previous paragraph.) • When you say \"circles of radius $\\sqrt{5}$\", do you mean \"circles of radius $\\sqrt{5}/2$\"? I may be misunderstanding something. Also, doesn't this require that the pawns be placed", "get ${4\\choose2} = 6$. We get $180 \\cdot 6 = 1080$. • If column 1 and column 2 share 2 filled squares on the same row (${6\\choose2} = 15$ places), they must also share 2 empty squares on the same row (${4\\choose2} = 6$). The last two squares can be arranged in ${2\\choose1} = 2$ positions; this totals to $15 \\cdot 6 \\cdot 2 = 180$. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have ${2\\choose1} = 2$ places, giving $180 \\cdot 2 = 360$. • If column 1 and column 2 share 3 filled squares on the same row (${6\\choose3} = 20$ places), then the squares on columns 3 and 4 are fixed. Thus, there are $400 + 1080 + 360 + 20 = 1860$ number of shadings, and the solution is $860$.", "bids will be some number of passes. Before the first bid there could be zero, one, two or three passes (four possibilities), and between all subsequent bids there can be zero, one or two passes (three possibilities chosen at $$N-1$$ points in the auction). We can account for passes by multiplying by those numbers of possibilities. That leads us to our final, big sum: \\begin{equation*}1+\\sum_{N=1}^{35} {35 \\choose N}\\cdot 4\\cdot 3^{N-1}\\end{equation*} An online calculator can turn this into our numerical final answer. That lone “1+” at the front of the equation is the possibility that the auction ends immediately with all four players passing. For extra credit, I asked how many different auctions there would be if we incorporated the bridge tactics known as doubling and redoubling. The calculation process is very similar, and the new formula — taking into account that there are now 21 intra-bid combinations of passes, doubles and redoubles, plus seven ways the auction ends with passes, doubles and redoubles — looks like this: \\begin{equation*}1+\\sum_{N=1}^{35} {35 \\choose N}\\cdot 4\\cdot 21^{N-1}\\cdot 7\\end{equation*} And the final answer is, of course, bigger. Much, much bigger: 128,745,650,347,030,683,120,231,926,111,609,371,363,122,697,557. But in the grand scale of gnarly game math, the number of possible bridge auctions isn’t all that enormous: It is only about 0.25 percent of the number of possible chess positions.", "in other words, for each of the ${49 \\choose 2}-24$ pairs, there are three other pairs that yield an equivalent board. Thus, the number of inequivalent boards is $\\frac{{49 \\choose 2} - 24}{4} + \\frac{24}{2} = \\boxed{300}$. For a $(2n+1) \\times (2n+1)$ board, this argument generalizes to $n(n+1)(2n^2+2n+1)$ inequivalent configurations. ## Solution 2 (Casework) There are 4 cases: 1. The center square is occupied, in which there are $12$ cases. 2. The center square isn't occupied and the two squares that are opposite to each other with respect to the center square, in which there are $12$ cases. 3. The center square isn't occupied and the two squares can rotate to each other with a $90^{\\circ}$ rotation with each other and with respect to the center square, in which case there are $12$ cases. 4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are $\\dbinom{12}{2} \\cdot \\frac{16}{4} = 264$ cases. Add up all the values for each case to get $\\boxed{300}$ as your answer. ~First" ]

[ "# How many ways can I place four distinct pawns on the board such that each column and row of the board contains no more than one pawn? If I have a $4\\times 4$ chess board, in how many ways can I place four distinct pawns on the board such that each column and row of the board contains no more than one pawn? I have a feeling this is a basic combination problem with cases. I don't see it yet. We have 4 choices for the column of the pawn in the first row, 3 choices for the column of the pawn in the second row, 2 choices for the column of the pawn in row 3, and 1 choice for the column of the pawn in row 4, for a total of $4!$ places for the positions of the pawns. Since the pawns are distinct, there are $4!$ ways to place them in these chosen positions; so there are $4!\\cdot4!=576$ possibilities. Alternatively, there are 16 places for the first pawn, then 9 places for the second pawn, only 4 choices left for the third pawn, and just 1 choice for the fourth pawn, giving $\\;16\\cdot9\\cdot4\\cdot1=576$ possibilities. • It's an interesting interpretation that four distinct pawns means four distinguishable pawns. I took it to mean merely that", "not.\" The end result is no. If we were to describe the set of outcomes, we could do so by describing it as a colored sequence of eight spaces from the board. Pick the location and the color for the first placed, then the location and the color for the second placed, and so on to get $64\\cdot 8\\cdot 63\\cdot 7\\cdot 62\\cdot 6\\cdots 57\\cdot 1=\\frac{64!}{56!}8!$ As in the previous case, we do the same with the exception of worryign about color now too, giving $64\\cdot 8\\cdot 49\\cdot 7\\cdot 36\\cdot 6\\cdots 4\\cdot 2 \\cdot 1\\cdot 1=(8!)^3$. This would yield a final probability of$$\\frac{(8!)^3}{\\frac{64!}{56!}8!}$$ which again equals the same as before. Scenario 4: Not only are rooks distinguishable by color, order of selection matters, but the person placing pieces will continue to place $56$ additional multicolored pawns after having placed all eight rooks. In this scenario, we would have a whopping $(64!)^2$ different outcomes in our sample space. The locations and colors of the pawns has literally nothing to do with whether or not the rooks share rows or columns (alternatively this is assuming that we say that pawns don't block movement to keep the spirit of the problem the same). If we were to run with this scenario, not only will we need to account for rooks not sharing", "# Kevin has 5 cubes. Each cube is a different color. Kevin will arrange the cubes side by side in a row. What is the total number of different arrangements of the 5 cubes that Kevin can make? ##### 1 Answer Oct 21, 2016 There are $120$ different arrangements of the five colored cubes. #### Explanation: The first position is one of five possibilities; the second position is therefore one of the four remaining possibilities; the third position is one of the three remaining possibilities; the fourth position will be one of the remaining two possibilities; and the fifth position will be filled by the remaining cube. Therefore, the total number of different arrangements is given by: $5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120$ There are $120$ different arrangements of the five colored cubes.", "the number of pawns cannot be more than $64 - 16 = 48$. An arrangement of 48 pawns that satisfies the condition of the problem will be obtained if you put a pawn on each cell of the board, except for the cells of the two main diagonals. A similar logic can be applied to the $2n \\times 2n$ chessboard, obtaining $4 \\cdot n^2 - 4 \\cdot n$ as an answer. Four dice are thrown. What is the probability that the product of the four numbers equals $36$?", "# Problem of the week #91- December 22nd, 2013 Status Not open for further replies. #### Jameson Staff member A standard tic-tac-toe board (or noughts and crosses) is a 3x3 grid, with 9 total spaces. There are many different ways that the X's and O's can be placed on the board. If we ignore for now the rule that 3 pieces of one type in a row wins, how many ways can we fill the board completely? Assume that the same piece, X or O, plays first each time so you will always have 1 more of that piece than the other. Note: The problem isn't to count moves that lead to a certain position. 1 filled board can be reached many ways. Just count the number of total filled boards. Also, don't worry about rotations or reflections for you guys who are more advanced. -------------------- #### Jameson Staff member Congratulations to the following members for their correct solutions: 1) MarkFL Solution (from MarkFL): This problem boils down to looking at how many ways to choose 9 things 5 at a time, or equivalently, how to choose 9 things 4 at a time. Thus, the number $N$ of distinct filled boards is given by: $$\\displaystyle N={9 \\choose 5}={9 \\choose 4}=\\frac{9!}{4!\\cdot5!}=\\frac{9\\cdot8\\cdot7\\cdot6}{4\\cdot3\\cdot2}=9\\cdot7\\cdot2=14(10-1)=140-14=126$$ Thus, we find there are 126 distinct filled", "same but order in which the rooks are placed matters. In this case, as the order in which the locations in which rooks are placed are selected matter, we can describe any outcome of how they are placed as sequences of length-$8$ with no repeats with entries all taken from our set of $64$ available spaces. As such, there will be $64\\cdot 63\\cdot 62\\cdots 57=\\frac{64!}{56!}$ equally likely sequences. Each arrangement which satisfies the description \"no rook can attack another\" can be described by a sequence of spaces selected such that no space selected can be attacked any any previous space. Each time we make such a selection, our available choices decreases. Originally there are $64$ choices. Having removed that and those spaces targettable from that space, there are $49$ remaining spaces, and then $36$ and so on, giving a total of $64\\cdot 49\\cdot 36\\cdots 4\\cdot 1=8^2\\cdot 7^2\\cdot 6^2\\cdots 1^2=(8!)^2$ which yields a final probability of $$\\frac{(8!)^2}{\\frac{64!}{56!}}$$ which one should be able to quickly confirm equals the same answer as before. Scenario 3: Rooks look different (e.g. by color), the order in which rooks are taken to be placed is randomized, and the order in which the rooks are placed matters as well. This is to answer your question of \"does it matter if rooks look the same or", "Sum of numbers on chessboard. Consider the squares of an $8 \\times 8$ chessboard filled with the numbers $1,2,3,4 \\ldots ,64$ in sequential order. If we choose $8$ squares with the property that there is exactly $1$ from each row and exactly $1$ from each column, and add up the numbers in the chosen squares, show that the sum obtained is always $260$. Please provide full solution and explanation. • Without specifying how the squares were numbered, this is false: put $1$ in square $a1$, $2$ in square $b2$, etc. Then picking the diagonals the sum is only $1 + \\dots + 8 = 36$. – grantfgates May 9 '14 at 13:21 • The numbers are entered in sequential order. – Calvin Lin May 9 '14 at 13:23 If we put the numbers in sequential order, the numbers of the row $i$ are of the form $$8i + j$$ $$i \\in \\lbrace 0,1,2, \\ldots 7\\rbrace$$ $$j \\in \\lbrace 1,2,, \\ldots 8\\rbrace$$ If we choose $8$ squares with the property that there is exactly one from each row and exactly one from each column, with this notation the total is $$8 \\cdot \\biggl( \\sum_{i=0}^{7}i \\biggr) + \\sum_{j=1}^{8} j = 260$$ Hint: work in $\\pmod 8$ Hint: $260=28\\cdot 8 + 36$", "the box is square (as it is for the final two figures), the subgraph can also be rotated within the box, but if the box is rectangular, then a quarter rotation requires the box to be rotated (or \"transposed\") as well. In any case we can count all possible ways to place the box in the chess board, and respectively the ways to place the subgraph in the box. The leftmost figure, a \"chair\" or \"h\", just fits into an upright $2\\times 1$ box as shown or in quarter rotation a $1\\times 2$ transposed box. The upright box can be placed on the chess board in $(m-2)\\cdot (n-1)$ locations, while the transposed box goes onto $(m-1)\\cdot (n-2)$ locations. Each box allows four distinct reflected/half-rotated versions of the subgraph, so the total number of subgraphs under this specific figure is: $$4(m-2)\\cdot (n-1) + 4(m-1)\\cdot (n-2)$$ The next figure, very like a capital \"F\", has the same size boxes and symmetries within those boxes. Thus it gives the same total number of subgraphs as the first figure. The third figure, perhaps viewed as \"y\" with a little imagination, fits into a $3\\times 1$ box. Thus the upright box has $(m-3)\\cdot (n-1)$ board locations, and the transposed box has $(m-1)\\cdot (n-3)$ locations. After rotations and reflections, there will be a total", "four positions, which means the second position cannot be a $$3$$ and the first position can be any of the possible outcomes for a die roll. Hence, the number of ways of rolling a die six times so that there are exactly four consecutive threes is $$|R_1| + |R_2| + |R_3| = 5 \\cdot 6 + 5^2 + 5 \\cdot 6 = 30 + 25 + 30 = 85$$ How many ways are there of rolling a die six times in a row such that there are at least four consecutive threes? Let $$R_i$$ be the event that there are four consecutive threes beginning in the $$i$$th position, where $$1 \\leq i \\leq 3$$. Notice that the word exactly was not used in the description of $$R_i$$. Observe that to have five consecutive threes, there are two possibilities: 1. There must be four consecutive threes beginning in the first position and four consecutive threes beginning in the second position. 2. There must be four consecutive threes beginning in the first position and four consecutive threes beginning in the third position. To have six consecutive threes, there must be four consecutive threes beginning in the first position and four consecutive threes beginning in the second position and four consecutive threes beginning in the third position. Thus, the set of", "x x x x x - x x x x x x x x - - - - - - - - We can also determine that the nullity of $A$ is $8$, so there are $2^8=256$ solutions in total. How can we describe the entire solution set, you may ask? Well, just put any number of pawns wherever you want on the top row. These act as our $8$ free variables. Then place whatever pawns you need in the second row so that the oddness property is satisfied on the first row. Then place whatever pawns you need in the third row so that the oddness property is satisfied in the second row. Repeat this until you place pawns on the last row and the whole board is a solution. This works for $8\\times 8$ specifically because there are $8$ free variables. In general, if you only want the number of solutions, there is a very nice formula for any $m\\times n$ grid, as shown by K. Sutner in The $\\sigma$-Game and Cellular Automata. See theorem on pg. 28. $2^{\\gcd(m + 1, n + 1) - 1}$ if a solution exists, $0$ otherwise. Start by focusing on the black squares. • • • A • • C B • • E D • G F • •", "& \\cdot & \\times & \\cdot \\\\ \\cdot & \\cdot & * & \\cdot & \\cdot & \\cdot \\\\ \\end{matrix}$$ The only ... 6 The minimum number of switches required is$100$. This puzzle can be turned into a linear algebra problem over${\\Bbb Z}/2{\\Bbb Z}$by letting the vector$v\\in({\\Bbb Z}/2{\\Bbb Z})^{100}$record which switches are pressed ($1$if pressed,$0$if not), the vector$w\\in({\\Bbb Z}/2{\\Bbb Z})^{100}$record which lights are initially on ($1$if on,$0$if off), ... 6 This is Welter's Game with special starting positions. In chapter 13 of On Numbers and Games, John Horton Conway presents an amazing analysis based on \"animating functions.\" It appears to be his own take on an earlier solution presented by Welter (but without attribution). Although there is not enough space to give a full account of this game, I will ... 6 One example is Sylver's Coinage played like so: Player's alternate selecting positive integers ($1, 2, 3, 4\\dots$). The rule is that no number is allowed to be expressed as sum (with possible duplicates) of the previous. For example, say$\\{4, 8, 5, 7\\}$($8$would have had to been said before$4$) was previously said. Then$6$could be said, but$14$could ... 6 Each of the 16 pawns can move at most 6 times and there are 30 captures possible. Therefore$(16\\cdot6+30)\\cdot 50=6300$is a rough upper bound (for example, not all pawns can", "ways to do this. So in all there are 36 ways. So there are $20(3+54+36) = 1860$ different shadings, and the solution is $860$. ### Solution 2 There are ${6\\choose3}$ to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns: • If column 1 and column 2 do not share any two filled squares on the same row, then there are ${6\\choose3}$ combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives ${6\\choose3}^2 = 400$ arrangements. • If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row ($6 - 1 = 5$ places), share another empty square on a row, and have 2 squares each on different rows. This gives $6 \\cdot 5 \\cdot {4\\choose2} = 180$. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we", "+ 2 = 234.$ The reader may wish to carry out the multiplication $$6\\cdot 15,$$ an example requiring extensive carries in the abbreviation process. In preparation for extracting square roots on the chessboard calculator, it is worth noting here that, after obtaining $$18\\cdot13=(16+2)(8+4+1)$$ in Step 1 above, Step 2 amounts to an application of the distributive law to obtain: $(16+2)(8+4+1) = 16\\cdot8 + 2\\cdot8 + 16\\cdot4 + 2\\cdot4 + 16\\cdot1 + 2\\cdot1,$ with the six products on the right-hand side of this equation necessarily arranged in a 3 row by 2 column rectangle on the chessboard. More specifically, removal of one entire row and two entire columns from the second board shown above would result in a 3 row by 2 column rectangle consisting of six contiguous squares with counters in them. Next: division on the chessboard calculator (followed by extraction of square roots)", "the pawns must occupy four distinct spaces. – hardmath Sep 11 '17 at 5:58 • @hardmath Four distinct pawns does mean the pawns are distinguishable. – N. F. Taussig Sep 11 '17 at 8:25 Note: I assumed the pawns, characteristically the weakest and most numerous of chess pieces, are interchangeable (not individually distinguishable), but as user84413 interprets \"four distinct pawns\", they are distinguishable. That leads to an extra factor of permutations (of the individual pieces), and hence to a different answer, $(4!)^2 = 24^2 = 576$. Yes, it is a basic combinatorial problem, with $4!= 24$ solutions. The key to seeing this is to consider the pawns must be one in each column and one in each row. Therefore if we enumerate the pawns by row, their column indexes are a permutation of the row indexes. Hence (for $4$ pawns on a $4\\times 4$ board) we get $4!$ placements. • How do we know it's a 4x4 board? – Derek Sep 10 '17 at 19:50 • @Derek: From the body of the Question: \"If I have a $4\\times 4$ chess board...\" We could treat the case of extra rows and columns by counting subsets, but I think this is what the OP was asking. – hardmath Sep 10 '17 at 19:52 • Wow. I totally skipped over that", "amount of combinations for 2 objects as 10 objects: No Positions: 12, Objects: 2: \"12 choose 2\" - 12! / (10! * 2!) No Positions: 12, Objects: 10: \"12 choose 10\" - 12! / (2! * 10!) Both options give me 66 combinations. That seems off... That may be, because the question seems a bit off. Lets make it simple. Suppose we have a board with twelve peg holes in a row. We have three blue pegs. There are $\\binom{12}{3}=220$ ways to put those pegs into the peg holes. If we have two blue pegs. There are $\\binom{12}{2}=66$ ways to put those pegs into the peg holes. On the other hand. If we have three pegs, one red, one white, and one blue. There are $12\\cdot 11\\cdot 10=1320$ ways to put those pegs into the peg holes. If we have two pegs, one red and one blue. There are $12\\cdot 11=132$ ways to put those pegs into the peg holes. That models your question in two ways. In the first order does not matter. In the second order matters. 9. ## Re: Set of points, 3 objects, get all possible combinations Hi Plato, Thanks for that reply. I understand now. That's alot of combinations! I need this for a input to a program that will take 30 hours", "### Description Solve the problem below for $$T$$ test cases. You have a chess board of dimension $$1 \\times M$$ with positions numbered from $$1$$ to $$M$$ from left to right. You have $$N$$ identical pawns, initially lined up from position $$1$$ to $$N$$ on the chess board. You can do any of the following operations as many number of times in any order: • Select the $$i$$-th pawn from the left ($$1 \\le i \\le N - 2$$) such that the right of the $$(i + 2)$$-th pawn is inside the chessboard, and move the $$i$$-th pawn one position to the right of the $$(i+2)$$-th pawn. While there are two pawns on the same position, move one of the pawns one position to the left. • Rotate the board $$180$$ degrees, i.e., a pawn at position $$x$$ will end up at position $$M - x + 1$$. Find the number of distinct configurations of pawns that you can end up with after doing the above operations modulo $$998244353$$. Two configurations are considered distinct if there is at least one position where one configuration has a pawn but not the other. ### Constraints • $$1 \\le T \\le 10000$$ • $$3 \\le N \\le M \\le 200\\;000$$ ### Input The first line contains the number of test cases", "has 4 horizontal locations and 8 vertical locations. . . Hence, it has 32 possible locations. The vertical 5x1 bar also has 32 possible locations. Hence, it has $32 + 32 \\:=\\:{\\color{blue}64}$ ways. Therefore, there are: . $1764 + 840 + 64 \\:=\\:\\boxed{2668}$ ways . . to place the twelve pentominos on a chessboard. I'm grateful they didn't ask for \"six neighbour squares\". There would be $35$ hexominos to consider. Thanks for the praise, ARDASH. I didn't pay much attention, but it probably took an hour (most work on diagrams, using COPY and PASTE over and over). It's certainly obvious by now, isn't it? . . . I truly enjoy doing this. .", "here was my original 39 move solution: Two of the four knights are promoted black pawns The idea was to create a position with as many black pawns ready to promote as possible, as each pawn could make up to 4 different moves. However, Paul Panzer's solution managed to squeeze in an extra black move with better positioning and a g-file rook instead of an a-file rook", "\\times 2$ squares: Choosing three such squares leaves only one square left, with four places to place it. This is $2 \\cdot 4 = 8$ ways. • For at least four $2 \\times 2$ squares, we clearly only have one way. By the Principle of Inclusion-Exclusion, there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \\times 2$ square. There are $2^9=512$ ways to paint the $3 \\times 3$ square with no restrictions, so there are $512-95=417$ ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a $2 \\times 2$ red square is $\\frac{417}{512}$, and $417+512=\\boxed{929}$.", "# How many possible setups on the chessboard? Jason has $5$ indistinguishable white pawns and $4$ indistinguishable black pawns that he wishes to place on a $3 \\times 3$ chessboard (where each square of the chessboard is distinct) such that there is exactly one pawn per square. He describes a position as drawish if no row or column has pawns of all the same color. How many possible setups are drawish? Attempt: We have that since $A \\cup B = A + B - (A \\cap B)$, that the number of ways to have at least one row or column black is $6 \\binom{6}{2}-9$. Similarly the number of ways to have at least one row white or one column white is $6 \\binom{6}{1}$ (choose a row or column to be white then we pick one other white tile from the other $6$). Therefore, using $A \\cup B = A + B - (A \\cap B)$ we can calculate the number having at least one row or column black or at least one row or column white. This is equal to $A^* \\cup B^* = 6 \\binom{6}{2} - 9 +6 \\binom{6}{1}-6*2*3$ (the $6*2*3$ comes from the fact of choosing a black row or column then a white then choosing $2$ of the remaining $3$ tiles to be black. Thus," ]

[ "• anonymous Bicycle license plates in Flatville each contain three letters. The first is chosen from the set $${C,H,L,P,R}$$ the second from $$A,I,O$$, and the third from $${D,M,N,T\\}$$. When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates than can be made by adding two letters? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# Difference between revisions of \"1999 AMC 8 Problems/Problem 15\" ## Problem Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}. When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters? $\\text{(A)}\\ 24 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 36 \\qquad \\text{(D)}\\ 40 \\qquad \\text{(E)}\\ 60$ ## Solution ### Solution 1 There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \\cdot 3 \\cdot 4 = 60$ license plates. Adding $2$ letters to the start gives $7\\cdot 3 \\cdot 4 = 84$ plates. Adding $2$ letters to the middle gives $5 \\cdot 5 \\cdot 4 = 100$ plates. Adding $2$ letters to the end gives $5 \\cdot 3 \\cdot 6 = 90$ plates. Adding a letter to the start and middle gives $6 \\cdot 4 \\cdot 4 = 96$ plates. Adding a letter to the start and end gives $6 \\cdot 3", "the number of bicycles, how many ...", "## #1577 Tile Code 31 1 s 128 MB ## Description The city of Songpa is now carrying out a project to build a bicycle transportation system called ‘green Songpa.’ By the end of this year, citizens and visitors alike will be able to pick up and drop off bicycles throughout the city. Recently, it was decided to attach a number tag to each bicycle for management use. The bicycles will be under control of the city’s traffic system. The number tag contains a tile code of length n , which consists of 1×2 , 2×1 , and 2×2 tiles placed on a 2×n rectangular plate in a way that every cell of the plate is covered by exactly one tile. The plate is divided into 2n cells of size 1×1 . Of course, no two tiles are allowed to overlap each other. The 2×5 plate and a tile code of length 5 are shown in Figures 1 and 2, respectively. The code will always be read from left to right. However, there is no distinction between the top side and the bottom side of the code. The code may be turned upside down. The code shown in Figure 3 is essentially the same code as in Figure 2. Given a positive integer n , the project director", "# A bicycle license plate requires 3 digits, each ranging from 0-9. How many different combinations are possible? Mar 20, 2018 $10 \\cdot 10 \\cdot 10$ to find the number of combinations. $10 \\cdot 10 \\cdot 10 = 1000$ possible combinations", "# [✓] Solve a Bike Lock problem with 4 dials and 10 letters on each dial? Posted 5 months ago 581 Views | 2 Replies | 2 Total Likes | I have a WordLock bike lock with 4 dials and 10 letters on each dial. I saw a post (https://community.wolfram.com/groups/-/m/t/926136) that someone used this program to find the correct lock combination. I don't know how to use this program, so if someone can tell me the possible combinations that would be great!Dial 1 Letters: S P H M T W D L F B Dial 2 Letters: L E Y H N R U O A I Dial 3 Letters: E N M L R T A O S K Dial 4 Letters: D S N M P Y L K T E 2 Replies Sort By: Posted 1 month ago Hi Marco,I have the exact same lock (same letters on the same dials, in the same order) as OP, and am spinning my wheels trying to figure out how to use the handy-dandy code you helpfully shared from the other post.It would be super helpful if you shared the complete list of 859 words -- I can confirm the word I'm looking for is not one the the 173 you already shared, and I'm quite certain I'll", "post) cheers 18. Sep 30, 2011 ### uart Agreed. 19. Sep 30, 2011 ### awkward Indeed, I misinterpreted your initial post as saying there were 10 _types_ of bicycles rather than 10 bicycles. 20. Oct 1, 2011 ### uart Ok I see. Yes, your answer looks correct for that variant of the problem.", "c), (blue, s).\" Thus, we get 6 distinct objects (Fig 1). From the illustration given above we note that \\ \\ \\ \\ \\ \\ \\ \\ \\ color{blue } (A × B = {\"(red,b), (red,c), (red,s), (blue,b), (blue,c), (blue,s)\"}.) \"Again, consider the two sets :\" color{purple}( ✍️ A = {DL, MP, KA},) where DL, MP, KA represent Delhi, Madhya Pradesh and Karnataka, respectively and color{purple} (B = {01,02, 03}) representing codes for the licence plates of vehicles issued by DL, MP and KA . ● If the three states, Delhi, Madhya Pradesh and Karnataka were making codes for the licence plates of vehicles, with the restriction that the code begins with an element from set A, ● which are the pairs available from these sets and how many such pairs will there be (Fig 2)? ● The available pairs are: (DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03) and the product of set A and set B is given by color{purple}(A × B = {(DL,01), (DL,02), (DL,03), (MP,01), (MP,02), (MP,03), (KA,01), (KA,02), (KA,03)}.) ● It can easily be seen that there will be 9 such pairs in the Cartesian product, since there are 3 elements in each of the sets A and B. This gives us 9 possible codes. \\color{red} \\ox \\color{red} \\mathbf(COMMON \\ CONFUSION) Also", "yellow, blue, green, white so that each flag consisted of three different colors? • Cars plates How many different licence plates can country have, given that they use 3 letters followed by 3 digits?", "bikes shop 2, 10 combinations out of 5 available bikes shop 3, no choices, simply remaining bikes...1 combination total combinations = 2520 5 3 2 shop 1, 252 combinations out of 10 available bikes shop 2, 10 combinations out of 5 available bikes shop 3, no choices, simply remaining bikes...1 combination total combinations = 2520 5 4 1 shop 1, 252 combinations out of 10 available bikes shop 2, 5 combinations out of 5 available bikes shop 3, no choices, simply remaining bikes...1 combination total combinations = 1260 5 5 0 shop 1, 252 combinations out of 10 available bikes shop 2, 1 combinations out of 5 available bikes shop 3, no choices, simply remaining bikes...1 combination total combinations = 252 Grand total n = 45507 8. Sep 28, 2011 ### holyGrill looks awesome gsal, i will find out the answer soon and let you know!! many many thanks, this defo makes sense to me ... 9. Sep 28, 2011 ### uart At the moment I cant see that I've missed any combinations so I suspect it is the correct answer. Might post some more details later. 10. Sep 28, 2011 ### cone Also make sure you test small numbers first and work your way up. If you run out of memory with small numbers, there may be", "Peter Norvig, 15 June 2015 # Let's Code About Bike Locks¶ The June 15, 2015 post on Bike Snob NYC leads with \"Let's talk about bike locks.\" Here's what I want to talk about: in a local bike shop, I saw a combination lock called WordLock®, which replaces digits with letters. I classified this as a Fred lock, \"Fred\" being the term (championed by Bike Snob NYC) for an amateurish bicyclist with the wrong equipment. I tried the combination \"FRED,\" and was amused with the result: FRED BUNS! Naturally I bought the lock (and set the other ones on the rack to FRED BUNS as well). Unfortunately, it turns out the combination on each lock is pre-set; it can't be changed to FRED BUNS (some other models of WordLock® are user-settable). But we can still have fun writing code to answer some questions about the WordLock®: # Questions¶ 1. How many words can the WordLock® make? 2. Can a lock with different letters on the tumblers make more words? 3. How many words can be made simultaneously? For example, with the tumbler set to \"FRED\", the lock above also makes \"BUNS\" in the next line, but with \"SOMN\", fails to make a word in the third line. Could different letters make words in every horizontal line? 4. Is", "It means no one can have two bikes of different brands. Example: A has two bikes, one brand X, one brand Y. This is not allowed by statement (i). That's the way I would analyse this. And, at this point I believe I truly understand statement (i). 20. Jan 9, 2017 ### Marclan A good help but i'm still on my past solution: ∀x ∀y (member(x) ∧ bicycle(y) ∧ owns(x,y) ∃ z brand(y,z)¬∃ z2 (brand(y,z2) ∧ brand(y,z))) And i have just wrote that it's probably wrong because it's only saying that for each bicycle not exist more than one brand associated with it. If only i could know how to write this, i will solve all the exercize.", "so, however, at some time in the past, and may have to buy a new plate when buying a new car. 22. ^ Additionally, the white sticker reading «25» signals low speed to following drivers. 23. ^ Regular size 130mm × 101mm (5⅛″×4″); also available as sticker for e-scooters in 67mm × 55mm (2⅝″×2⅛″) ### Examples 1. ^ Drivers from Offenbach are slandered to be ohne Verstand (without sense) and even misspell it OF 2. ^ In Nürnberger Land district e.g., LAU and HEB can be used in any combination, ESB with N only, PEG with A only, N with one-letter identifiers only, yet excluding the “new” letters B/F/G/I/O/Q and S 3. ^ The Police of North Rhine-Westphalia uses NRW 4, NRW 5 and, for motorbikes, NRW 6 instead of a local code. This is followed by a four-digit number (e.g. NRW 4-1960). 4. ^ starting with 10 = Vatican; 17 = USA; 49 = UK; etc. 5. ^ The rural districts of Heidelberg (HD), Mannheim (MA) and part of Sinsheim (SNH) merged into Rhein-Neckar-Kreis (HD); the districts of Alsfeld (ALS) and Lauterbach (LAT) merged into Vogelsbergkreis (VB). 6. ^ Öhringen (ÖHR) and Künzelsau (KÜN) merged into Hohenlohekreis (KÜN) 7. ^ The rural districts of Straubing and Bogen merged to Straubing-Bogen, keeping SR (and sharing it with Straubing", "650 unique 2 letter combinations which can be used as the first two letters of the license plate. Two combinations that differ only in ordering of their characters are the same combination. 50400 is the number of ways to arrange 10 letters (alphabets) word \"STATISTICS\" by using Permutations (nPr) formula. There are 95^8 possible combinations of 8 characters. Open a browser in PNG free application web site and go to the Merger tool. Believe it or not, the letter Z has not always been the last letter of the alphabet; in the Greek alphabet, it had a respectable place at number seven. Correct answer: for the letters there are 26 possibilities for each of the 3 tracks and for the numbers there are 10 possibilities for each of the 3 tracks. Unless indicated, rates listed are. b) Adding another digit increases the number of plates by. (Compare Problem 5, Topic 26. How many words of length 10 can you make from these, such that • all the A’s are used. One can also use the combination formula for this problem: n C r = n! / (n-r)! r! Therefore: 5 C 3 = 5! / 3! 2! = 10 (Note: an example of a counting problem in which order would matter is a lock or passcode situation. For", "Search a-shop-sells-bicycles-and-tricycles-in-total-there-are-text-cycles-cycles-include-both-bicycles-and-tricycles-and-text # A shop sells bicycles and tricycles in total there are text cycles cycles include both bicycles and tricycles and text ## Top Questions ycles and tricycles) and $$\\text{19}$$ wheels. Determine how many of each there are, if a bicycle has two wheels and a tricycle has three wheels. View More 1.AU MAT 120 Systems of Linear Equations and Inequalities Discussion mathematicsalgebra Physics", "2B + 3T = 2(3T) + 3T = 9T = 81,$$ so the number of trikes is $T=9$. Then, the number of bikes is $$B = 3T = 3(9) = 27.$$ And we can have some assurance we're right by calculating the number of wheels: $$W_B + W_T = 2B + 3T = 2(27) + 3(9) = 54 + 27 = 81.$$ B = 3T. (B for Bycicle, T for Trycicle). 2B + 3T = 81 (wheels) Just replace B for 3T: 2(3T) + 3T = 81. That gives you: 9T = 81. So there are 9 tricycles. Therefore B = 3*9 = 27 (bicycles) 1: 3t = b (three times as many bikes as trikes) 2: 2b + 3t = 81 (each bike has two wheels, trike three wheels, totaling 81) Solving: 3: b - 3t = 0 (from line 1, moving 3t to the other side) 4: 2b - 6t = 0 (multiplying both sides by 2) 5: 9t = 81 (subtracting line 4 from line 2) 6: t = 9 (dividing both sides by 9) 7: b = 27 (substituting 9 into line 1) \\begin{align} &\\text{Number of bicycles} =3 \\times 2x\\text{ wheels}\\\\ &\\text{Number of tricycles} =3x\\text{ wheels}\\\\ &3 \\times 2x\\text{ wheels}\\ +\\ 3x\\text{ wheels}=81\\text{ wheels}\\\\ &9x=81\\\\ &x=9\\\\ &\\text{Number of bicycles} =3 \\times 9\\\\ &\\text{Number of", "code could be in turn followed by a number from 1 to 9 for the same purpose as in the 1901 system. The first car registered in Paris obtained 1 RB, and when 999 RB had been reached the following car obtained 1 RB1.[15] Temporary plates for vehicles in transit on the French territory were created in 1933. They used TT instead of the geographical code. Diplomatic plates were first issued in 1936. They were in dark yellow with white lettering.[15] Initially planned to last 75 years, the system was withdrawn in 1950.[15] The following format kept the geographical structure but identified départements with numbers rather than letters, allowing a greater range of combinations. Schematic representations: 7857 - BX; Plate from Aveyron 4955 - HK9; Plate from Isère 3178 - ZA1; Plate from Vaucluse Geographical codes in use between 1928 and 1950[16] Code Region Code Region Code Region Code Region Code Region AB - AE Ain AF - AM Aisne AN - AQ Allier AR Basses-Alpes AS Hautes-Alpes AT - AU Ardèche AV - AY Ardennes AZ Ariège BA - BM Alpes-Maritimes BN - BS Aube BT - BV Aude BX - BY Aveyron BZ Territoire-de-Belfort CA - CR Bouches-du-Rhône CT - CY Calvados CZ Cantal DB - DF Charente DG - DM Charente-Inférieure DN - DQ Cher", "have made this list there are 27combinations in total hope this helps someone this is how I would pick my bike combinations when I forgot the combo. com FREE DELIVERY possible on eligible purchases. If other characters, such as spaces are allowed, the number of possibilities for each digit grows. For large lists and lists where the comparison information is expensive to calculate, and Python versions before 2. Contain 8 to 72 characters; Cannot be a repeat of your four previous passwords; Must include 2 of the following 3 items: 1 or more numbers; 1 or more uppercase letters and 1 or more lowercase letters; 1 or more non-alphanumeric characters; Check the Remember My Username box to have the system populate the Company ID and Username. Separate Alphanumeric Characters; In one cell there is a combination of number and text(i. Each Alpha lock box features three- or four-digit number, alpha or alphanumeric dial combination access. For first character number of options = 26. Decide on your new 4 letter combination - choose from suggestions or create your own. Let’s first figure out how many possibilities for a license plate with only one character there are. Interleaving the two characters gives the original 12 bits. ly/ Any mix of capitalized letters, lowercase letters, and numbers; If you limit the", "at most one bike bell • at most one luggage rack She writes the rules down concisely: f <=> f1 | f2 | f3 f1 + f2 + f3 = 1 h <=> h1 | h2 | h3 | h4 | h5 h1 + h2 + h3 + h4 + h5= 1 s <=> s1 | s2 | s3 | s4 s1 + s2 + s3 + s4 = 1 wf <=> wf1 | wf2 | wf3 | wf4 | wf5 wf1 + wf2 + wf3 + wf4 + wf5 = 1 wb <=> wb1 | wb2 | wb3 | wb4 | wb5 wb1 + wb2 + wb3 + wb4 + wb5 = 1 b <=> b1 | b2 | b3 b1 + b2 + b3 <= 1 r <=> r1 | r2 | r3 | r4 r1 + r2 + r3 + r4 <= 1 c <=> c1 | c2 | c3 | c4 c1 + c2 + c3 + c4 = 1 The first type of formulas, e.g. f <=> f1 | f2 | f3 is called an equivalence and can be read as an if and only if statement: A frame f is contained in the bike if and only if one of the frames f1, f2 or f3 are contained. The second type of", "use the abbreviations of their names instead of a city code. • The Technisches Hilfswerk (German Federal Agency for Technical Relief) uses its abbreviation THW, so the plates read THW-80000, for example. All numbers on THW plates start either with the digit 8 or 9. • The Wasserstraßen- und Schifffahrtsverwaltung des Bundes (Federal Administration of Waterways and Navigation) uses BW followed by a digit identifying the region of the office (from 1=north to 7=south). • Before the Deutsche Bundesbahn (German Federal Railways) and the Deutsche Bundespost (German Federal Mail) were privatised, they used the abbreviations DB and BP (e.g. DB-12345, BP-12345). • The federal police uses the code BP for Bundespolizei instead of a local code. Before 2006 the code BG, for their former name Bundesgrenzschutz, was used in the scheme BG-12345. This old code still remains valid, but any new vehicles will get the new code BP. Official registered vehicle (here: fire brigade) Official registered vehicle for disaster relief • Until the legal reforms of 2006, official cars such as police, fire fighting and municipality vehicles did not carry a letter after the sticker, such as M-1234. These included: • vehicles of the district government: 1-199, 1000-1999, 10000-19999 • vehicles of the local government (for example: fire brigade): 200-299, 2000-2999, 20000-29999, 300-399 • police: 3000-3999, 7000-7999, 30000-39999," ]

[ "four for the third. This means that $5\\cdot 3\\cdot 4=60$ plates could have been made. If two letters are added to the second set, then $5\\cdot 5\\cdot 4=100$ plates can be made. If one letter is added to each of the second and third sets, then $5\\cdot 4\\cdot 5=100$ plates can be made. None of the other four ways to place the two letters will create as many plates. So, $100-60=\\boxed{40}$ ADDITIONAL plates can be made.So the correct choice is $\\boxed{D}$", "\\cdot 5 = 90$ plates. Adding a letter to the middle and end gives $5 \\cdot 4 \\cdot 5 = 100$ plates. You can get at most $100$ license plates total, giving an additional $100 - 60 = 40$ plates, making the answer $\\boxed {D}$ ### Solution 2 Using the same logic as above, the number of combinations of plates is simply the product of the size of each set of letters. In general, when three numbers have the same fixed sum, their product will be maximal when they are as close together as possible. This is a 3D analogue of the fact that a rectangle with fixed perimeter maximizes its area when the sides are equal (ie when it becomes a square). In this case, no matter where you add the letters, there will be $5 + 3 + 4 + 2 = 14$ letters in total. If you divide them as evenly as possible among the three groups, you get $5, 5, 4$, which is a possible situation. As before, the answer is $5 \\cdot 5 \\cdot 4 - 5 \\cdot 3 \\cdot 4 = 40$, and the correct choice is $\\boxed{D}$ ### Solution 3 Before new letters were added, five different letters could have been chosen for the first position, three for the second, and", "the reasoning. Each of those $10^2$ ways to post the first two letters can be combined with any of $10$ different ways to post the third letter, so there are $10^2\\cdot10=10^3$ ways to post the first three letters. Two more arguments of the same kind lead to the conclusion that there are $10^5$ different ways to post the $5$ letters. Note that I’m assuming that you’re allowed to post more than one letter in the same letter box. If that’s not the case, the reasoning is similar, but the answer is quite different. If you can post at most one letter in a box, there are still $10$ different ways to post the first letter. After you’ve done that, however, there are only $9$ ways to post the second letter, since you can’t use the first letter box again. Thus, you get only $10\\cdot9$ different combinations of letter boxes for the first two letters. Similarly, after they’ve been posted you must pick one of the $8$ remaining letter boxes for the third letter, so each of the $10\\cdot9$ ways of posting the first two letters gives you only $8$ ways of posting the first three. As a result, there are $10\\cdot9\\cdot8$ ways of posting the first three letters. Two more arguments of the same kind lead this time", "for its position. . . There are $3\\cdot1 \\,=\\3$ ways. In the fifth row, place a letter. There are 2 choices for the letter, 3 choices for its position. . . There are $2\\cdot3 \\,=\\,6$ ways. So, there are: . $18\\cdot5\\cdot12\\cdot3\\cdot6 \\:=\\:19,\\!400$ ways . . to place one letter in each row. For the last letter, there 6 choices of boxes. Hence, there are: . $19,\\!400\\cdot6 \\:=\\:116,\\!640$ ways . . to place the six letters in the boxes. But there is duplication caused by . . the two identical L's and two identical E's. Therefore, there are: . $\\frac{116,\\!640}{2!\\,2!} \\:=\\:29,\\!160$ ways . . to place the letters of the word LETTER in the boxes.", "chosen, the second letter may be selected in $25$ ways since it cannot be the letter in the first position. The third letter can be chosen in $25$ ways since it cannot be the letter in the second position. The fourth letter can be chosen in $25$ ways since it cannot be the letter in the third position, which yields $26 \\cdot 25 \\cdot 25 \\cdot 25$ possible four letter words in which no two consecutive letters are identical.", "second letters with first letter B, and so on. We add 26 to itself 26 times. Or quicker: there are $$26 \\cdot 26$$ choices for the first two letters. Now for each choice of the first two letters, we have 26 choices for the third letter. That is, 26 third letters for the first two letters AA, 26 choices for the third letter after starting AB, and so on. There are $$26 \\cdot 26$$ of these $$26$$ third letter choices, for a total of $$(26\\cdot26)\\cdot 26$$ choices for the first three letters. And for each of these $$26\\cdot26\\cdot26$$ choices of letters, we have a bunch of choices for the remaining digits. In fact, there are going to be exactly 1000 choices for the numbers. We can see this because there are 1000 three-digit numbers (000 through 999). This is 10 choices for the first digit, 10 for the second, and 10 for the third. The multiplicative principle says we multiply: $$10\\cdot 10 \\cdot 10 = 1000\\text{.}$$ All together, there were $$26^3$$ choices for the three letters, and $$10^3$$ choices for the numbers, so we have a total of $$26^3 \\cdot 10^3$$ choices of license plates. Careful: “and” doesn't mean “times.” For example, how many playing cards are both red and a face card? Not $$26 \\cdot 12\\text{.}$$ The", "select the duplicated letter, ${4 \\choose 2}=6$ ways to choose the positions for that letter, and $2$ ways to place the other two. So we have $24+3\\cdot 6 \\cdot 2=60$ successes.", "plate does not have the desired property is $\\frac{10\\cdot 10\\cdot 9\\cdot 26\\cdot 26\\cdot 25}{10^3\\cdot 26^3}=\\frac{45}{52}$. We subtract this from 1 to get $1-\\frac{45}{52}=\\frac{7}{52}$ as our probability. Therefore, our answer is $7+52=\\boxed{059}$.", "# Difference between revisions of \"1999 AMC 8 Problems/Problem 15\" ## Problem Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}. When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters? $\\text{(A)}\\ 24 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 36 \\qquad \\text{(D)}\\ 40 \\qquad \\text{(E)}\\ 60$ ## Solution ### Solution 1 There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \\cdot 3 \\cdot 4 = 60$ license plates. Adding $2$ letters to the start gives $7\\cdot 3 \\cdot 4 = 84$ plates. Adding $2$ letters to the middle gives $5 \\cdot 5 \\cdot 4 = 100$ plates. Adding $2$ letters to the end gives $5 \\cdot 3 \\cdot 6 = 90$ plates. Adding a letter to the start and middle gives $6 \\cdot 4 \\cdot 4 = 96$ plates. Adding a letter to the start and end gives $6 \\cdot 3", "# How many four letter words are possible using the first 5 letters of the alphabet if the first letter can not be a and adjacent letters can not be alike? Jan 26, 2016 The first five letters are $A , B , C , D , E$ Consider this box. Each $1 , 2 , 3 , 4$ places represent place of a letter. First place $1$ can be filled in $4$ ways. (Excluding A) First place $2$ can be filled in $4$ ways. First place $1$ can be filled in $3$ ways. First place $1$ can be filled in $2$ ways. First place $1$ can be filled in $1$ ways. Total number of ways$= 4 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 96 w a y s$ Hence $96$ letters can be made.", "$$2{6\\choose 2}+6{5\\choose2}=90$$ admissible placements for these three letters. For each such placement the remaining $5$ letters $RRGED$ can then be placed arbitrarily in the $5$ remaining cells in ${5!\\over2!}=60$ ways. It follows that there are $90\\cdot60=5400$ admissible arrangements in all.", "used, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag. Thus, the number of different signals that can be made here is: $4\\cdot \\left(4-1\\right)=4\\cdot 3=12$ In the case of using 3 flags, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag, leaving 2 flags remaining as the possible third flag. Thus, the number of different signals that can be made here is: $4\\cdot \\left(4-1\\right)\\cdot \\left(4-2\\right)=4\\cdot 3\\cdot 2=24$ In the final case of using all 4 flags, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag, leaving 2 flags remaining as the possible third flag, leaving only 1 flag as the final flag to be chosen. Thus the number of different signals that can be made here is: $4\\cdot \\left(4-1\\right)\\cdot \\left(4-2\\right)\\cdot \\left(4-3\\right)=4\\cdot 3\\cdot 2\\cdot 1=24$ Adding all of these possible signals will give us the answer: $12+24+24=60$ For question 2, we can think of a tree-diagram to represent this. Starting with 3 separate trees: Science, Geography, and Phys. Ed at the top. From here, there are 4 possible classes to branch off of the first class: Art,", "to be numbers and three have to be letters? Because your calculation \"10*9*8*26*25*24\" suggests that repetition is not allowed. However, if repetition is allowed and there must be three letters and three numbers, we can find all unique combinations as follows: $10 \\cdot 10 \\cdot 10 = 1000$ number permutations $26 \\cdot 26 \\cdot 26 = 17576$ letter permutations But we're not done yet! Now we need to think about the positioning of the letters/numbers. So we need to see how many different ways the letters/numbers can be ordered. What we want is to choose three places for, let's say, the letters (then the numbers will automatically take the other three positions). So we can order them in 6 choose 3 different ways. ${6 \\choose 3} = \\frac{6!}{3!(6-3)!}=20$ Then, finally, there are: $1000 \\cdot 17576 \\cdot 20 = 351520000$ unique license plates.", "How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? # How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? J Asked by 2 years ago 240 points a) How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? b) Repeat part (a) under the assumption that no letter or number can be repeated in a single license plate. How many jeffp ### 1 Answer Answered by 2 years ago 8.7k points part (a) There are 26 letters in the alphabet and 10 possible numbers to use. It is a counting problem and hence, $$26^2 10^5 = 67600000$$ part (b) Repetitions are not allowed so for the first choice we have 26 letters to choose from, and then 25 letters.. Likewise for the numbers we can choose 10 numbers for the 3rd position, then 9 numbers, and so on... $$26 \\cdot 25 \\cdot 10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6= 19656000$$ ### Your Answer Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\\beta^2-1=0$$ Use LaTeX to type formulas and markdown", "# Counting different kinds of permutations • What's the number of strings of digits and letters of length $7$ without repetition? • Solution 1. Pick $2$ digits, $4$ consonants, and $1$ vowel and then line them up: $\\binom{10}{2}\\binom{21}{4} \\cdot 5 \\cdot 7!$ • Solution 2. Pick $2$ positions in the string for the digits, $4$ places for the consonants, leaving $1$ for the vowel. Then fill the spots: $\\binom72\\binom54 \\cdot 10 \\cdot 9 \\cdot 21 \\cdot 20 \\cdot 19 \\cdot 18 \\cdot 5.$ I think the methods above also count the permutations of MISSISSIPPI. But what about the problem below: • License plates in some state consist of $3$ different digits followed by $3$ different letters. How many of these plates are there? We choose places first and for each of those we choose symbols: $\\binom63\\binom33P(10, 3)P(26, 3).$ But apparently that's wrong. The actual answer is $P(10, 3)P(26, 3).$ What makes these two problems different? Why does choosing places first in the second problem results in wrong answer? For the pick of digits you have exactly 10 choices for first pick and 9 for the second and then 8 for the thrid giving a total of $$10\\times9\\times8=P(10,3)$$ similarly for the letters you have 26 choices for the first and 25 for the second and 24 for the third", "to the conclusion that there are $10\\cdot9\\cdot8\\cdot7\\cdot6=30,240$ different ways to post the letters if you can use each letter box at most once.", "$n \\geq 4$, at least two 2-cent stamps are used. Remove two of these and add a single 5-cent stamp. We now have $n - 2 \\cdot 2 + 5 = n+1$; a realization of $n+1$ cents. -", "pairs of letters you could choose from the remaining six letters C, D, E, F, G and H. You now have $15$ possible sets of four letters. Each set of four letters can be arranged in $4! = 24$ different ways, giving you $15 \\cdot 24 = 360$ possible strings.", "How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? # How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? J 240 points a) How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? b) Repeat part (a) under the assumption that no letter or number can be repeated in a single license plate. How many jeffp 8.7k points part (a) There are 26 letters in the alphabet and 10 possible numbers to use. It is a counting problem and hence, $$26^2 10^5 = 67600000$$ part (b) Repetitions are not allowed so for the first choice we have 26 letters to choose from, and then 25 letters.. Likewise for the numbers we can choose 10 numbers for the 3rd position, then 9 numbers, and so on... $$26 \\cdot 25 \\cdot 10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6= 19656000$$ Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\\beta^2-1=0$$ Use LaTeX to type formulas and markdown to format text. See example.", "of the other factors; or he could choose $$1$$ from the third factor and any of $$1$$, $$x^5$$, $$x^{10}$$, $$x^{15}$$, or $$x^{20}$$ from the second factor, in each case, choosing whichever term from the first factor brings the exponent up to $$95$$.” This is exactly equivalent to saying, “Well, if he buys a package of $$75$$ plates, then there are only six ways to buy $$95$$ plates in total: he could buy a package of $$20$$ plates and be done; or he could buy $$0$$, $$1$$, $$2$$, $$3$$, or $$4$$ packages of $$5$$ plates, in each case, buying as many single plates as are needed to bring the total up to $$95$$.” So what’s the advantage of the generating function approach? It comes in a couple of ways. First, it solves multiple problems at once: if we actually multiply out the generating function above, we will be able to read off not only how many ways there are of buying $$95$$ plates, but also how many ways there are of buying every number of plates up to $$95$$. (If we hadn’t cut the factors off as we did, we could also work out the answers for any number of plates higher than $$95$$.) So by doing a bunch of multiplication once (and it’s easy to feed into a" ]

[ "# Difference between revisions of \"1999 AMC 8 Problems/Problem 15\" ## Problem Bicycle license plates in Flatville each contain three letters. The first is chosen from the set {C,H,L,P,R}, the second from {A,I,O}, and the third from {D,M,N,T}. When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates that can be made by adding two letters? $\\text{(A)}\\ 24 \\qquad \\text{(B)}\\ 30 \\qquad \\text{(C)}\\ 36 \\qquad \\text{(D)}\\ 40 \\qquad \\text{(E)}\\ 60$ ## Solution ### Solution 1 There are currently $5$ choices for the first letter, $3$ choices for the second letter, and $4$ choices for the third letter, for a total of $5 \\cdot 3 \\cdot 4 = 60$ license plates. Adding $2$ letters to the start gives $7\\cdot 3 \\cdot 4 = 84$ plates. Adding $2$ letters to the middle gives $5 \\cdot 5 \\cdot 4 = 100$ plates. Adding $2$ letters to the end gives $5 \\cdot 3 \\cdot 6 = 90$ plates. Adding a letter to the start and middle gives $6 \\cdot 4 \\cdot 4 = 96$ plates. Adding a letter to the start and end gives $6 \\cdot 3", "• anonymous Bicycle license plates in Flatville each contain three letters. The first is chosen from the set $${C,H,L,P,R}$$ the second from $$A,I,O$$, and the third from $${D,M,N,T\\}$$. When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates than can be made by adding two letters? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "four for the third. This means that $5\\cdot 3\\cdot 4=60$ plates could have been made. If two letters are added to the second set, then $5\\cdot 5\\cdot 4=100$ plates can be made. If one letter is added to each of the second and third sets, then $5\\cdot 4\\cdot 5=100$ plates can be made. None of the other four ways to place the two letters will create as many plates. So, $100-60=\\boxed{40}$ ADDITIONAL plates can be made.So the correct choice is $\\boxed{D}$", "\\cdot 5 = 90$ plates. Adding a letter to the middle and end gives $5 \\cdot 4 \\cdot 5 = 100$ plates. You can get at most $100$ license plates total, giving an additional $100 - 60 = 40$ plates, making the answer $\\boxed {D}$ ### Solution 2 Using the same logic as above, the number of combinations of plates is simply the product of the size of each set of letters. In general, when three numbers have the same fixed sum, their product will be maximal when they are as close together as possible. This is a 3D analogue of the fact that a rectangle with fixed perimeter maximizes its area when the sides are equal (ie when it becomes a square). In this case, no matter where you add the letters, there will be $5 + 3 + 4 + 2 = 14$ letters in total. If you divide them as evenly as possible among the three groups, you get $5, 5, 4$, which is a possible situation. As before, the answer is $5 \\cdot 5 \\cdot 4 - 5 \\cdot 3 \\cdot 4 = 40$, and the correct choice is $\\boxed{D}$ ### Solution 3 Before new letters were added, five different letters could have been chosen for the first position, three for the second, and", "# Counting different kinds of permutations • What's the number of strings of digits and letters of length $7$ without repetition? • Solution 1. Pick $2$ digits, $4$ consonants, and $1$ vowel and then line them up: $\\binom{10}{2}\\binom{21}{4} \\cdot 5 \\cdot 7!$ • Solution 2. Pick $2$ positions in the string for the digits, $4$ places for the consonants, leaving $1$ for the vowel. Then fill the spots: $\\binom72\\binom54 \\cdot 10 \\cdot 9 \\cdot 21 \\cdot 20 \\cdot 19 \\cdot 18 \\cdot 5.$ I think the methods above also count the permutations of MISSISSIPPI. But what about the problem below: • License plates in some state consist of $3$ different digits followed by $3$ different letters. How many of these plates are there? We choose places first and for each of those we choose symbols: $\\binom63\\binom33P(10, 3)P(26, 3).$ But apparently that's wrong. The actual answer is $P(10, 3)P(26, 3).$ What makes these two problems different? Why does choosing places first in the second problem results in wrong answer? For the pick of digits you have exactly 10 choices for first pick and 9 for the second and then 8 for the thrid giving a total of $$10\\times9\\times8=P(10,3)$$ similarly for the letters you have 26 choices for the first and 25 for the second and 24 for the third", "the S's and 4! again for the number of ways of rearranging the four I's. $$n = \\frac{11!}{2! \\cdot 4! \\cdot 4!} = 34,650$$ So there are 34,650 unique combinations of the letters in the name MISSISSIPPI. 5. In how many unique ways can the letters of the word \"JUGGERNAUT\" be arranged? Solution This problem is similar to #4 above. There are two G's and two U's in this word, so in any configuration of the 10-letter string, these could be swapped and we wouldn't know it. The number of unique rearrangements is $$n = \\frac{10!}{2! \\cdot 2!} = 907,200$$ So there are 907,200 unique arrangements of those ten letters. That's amazing. 6. A skateboard shop stocks 12 different decks, 3 types of trucks and 8 types of wheels. How many different skateboards (Deck + trucks + wheels) can be constructed from these parts? Solution The choices of decks, trucks and wheels are all independent, so the number of boards is just \\begin{align} n &= n_d \\cdot n_t \\cdot n_w \\\\[5pt] &= 12 \\cdot 3 \\cdot 8 = 288 \\end{align} 7. This is a classic problem in probability. How many people must gather together in a room to make the probability that two of them will share the same birthday 50% ? Solution One reason to study a", "$\\text{Total number of licences} - 26\\cdot 25\\cdot 24\\cdot 23\\cdot 10\\cdot 9\\cdot 8\\cdot 7$. Where the total number of licences is $26^4\\cdot 10^4$ which is what you calculated. Assuming the letters and numbers can appear in any order: You can obtain this answer by multiplying the answer obtained above by $\\dfrac{8!}{4!4!}$ which is the number of permutations of $8$ things in which $4$ objects are of one kind and $4$ objects are of the other kind. This is the same as the answer given by Omnitic. Please note that as pointed out by Omnitic, the following does not restrict the number of letters and digits to $4$: Assuming the numbers or letters can be filled in any order (and are not restricted to be $4$): The licences have $8$ places which can be filled in with a number or a letter. Thus each place can have one of $26+10=36$ values. Now, number of licences with no letters or numbers repeated: $36\\cdot 35\\cdot 34\\cdot\\cdot\\cdot 30\\cdot29$. So, number of licences with at least one letter or number repeated: $\\text{Total number of licences} - 36\\cdot 35\\cdot 34\\cdot\\cdot\\cdot 30\\cdot29$ The total number is given by $36^8$. - If the licence plate is of the form $LLLLNNNN$ where L stands for letter and N for integer between 0 and 9 inclusive then no. Number", "# A bicycle license plate requires 3 digits, each ranging from 0-9. How many different combinations are possible? Mar 20, 2018 $10 \\cdot 10 \\cdot 10$ to find the number of combinations. $10 \\cdot 10 \\cdot 10 = 1000$ possible combinations", "### Home > A2C > Chapter 12 > Lesson 12.4.2 > Problem12-190 12-190. In the past, many states had license plates composed of three letters followed by three digits ($0\\text{ to }9$). Recently, many states have responded to the increased number of cars by adding one digit ($1\\text{ to }9$) ahead of the three letters. How many more license plates of the second type are possible? What is the probability of being randomly assigned a license plate containing $\\text{ALG }2$? Since repetition is allowed, the total number of license plates of the first type is $26\\cdot26\\cdot26\\cdot10\\cdot10\\cdot10 = 17,576,000$. The total number of license plates of the second type is $9\\cdot26\\cdot26\\cdot26\\cdot10\\cdot10\\cdot10 = 158,184,000$. Find the difference. Find the number of license plates containing $\\text{ALG }2$. $9\\cdot1\\cdot1\\cdot1\\cdot1\\cdot10\\cdot10 =$ $P(\\text{ALG }2)=\\frac{900}{158,184,000}=?$", "second letters with first letter B, and so on. We add 26 to itself 26 times. Or quicker: there are $$26 \\cdot 26$$ choices for the first two letters. Now for each choice of the first two letters, we have 26 choices for the third letter. That is, 26 third letters for the first two letters AA, 26 choices for the third letter after starting AB, and so on. There are $$26 \\cdot 26$$ of these $$26$$ third letter choices, for a total of $$(26\\cdot26)\\cdot 26$$ choices for the first three letters. And for each of these $$26\\cdot26\\cdot26$$ choices of letters, we have a bunch of choices for the remaining digits. In fact, there are going to be exactly 1000 choices for the numbers. We can see this because there are 1000 three-digit numbers (000 through 999). This is 10 choices for the first digit, 10 for the second, and 10 for the third. The multiplicative principle says we multiply: $$10\\cdot 10 \\cdot 10 = 1000\\text{.}$$ All together, there were $$26^3$$ choices for the three letters, and $$10^3$$ choices for the numbers, so we have a total of $$26^3 \\cdot 10^3$$ choices of license plates. Careful: “and” doesn't mean “times.” For example, how many playing cards are both red and a face card? Not $$26 \\cdot 12\\text{.}$$ The", "letters and two digits. Motorcycle plates have two letters followed by three digits. Setting aside vanity plates and ignoring the fact that some three-letter words are avoided, how many license plates are available to individuals in Alberta for their cars or motorcycles? Solution To answer this question, there is a natural division into four cases: regular car plates with three digits; regular car plates with four digits; veteran plates; and motorcycle plates. For a regular car plate with three digits, there are 24 choices for the first letter, followed by 24 choices for the second letter, and 24 choices for the third letter. There are 10 choices for the first digit, 10 choices for the second digit, and 10 choices for the third digit. Using the product rule, the total number of license plates in this category is $$243 \\cdot 103 = 13,824,000$$. For a regular car plate with four digits, there are 20 choices for the first letter, followed by 20 choices for the second letter, and 20 choices for the third letter. There are 10 choices for the first digit, 10 choices for the second digit, 10 choices for the third digit, and 10 choices for the fourth digit. Using the product rule, the total number of license plates in this category is $$203 \\cdot 104", "# There are 10 bicyclists entered in a race. In how many different orders could these 10 bicyclist finish? Apr 13, 2017 #### Answer: 10! is the answer. #### Explanation: this is just like you are given 10 lines on a paper and you have to arrange 10 names on those 10 lines in all different ways . so , starting with the lowermost line, you can put one of 10 names on that line and then on the line above that you can put 1 of 9 names and so on . so the total ways to arrange all the names on those lines in all ways will be : $10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1$ = 10!", "= 80,000,000$$. For a veteran plate, there are 24 choices for the first letter, followed by 24 choices for the second letter. There are 10 choices for the first digit, and 10 choices for the second digit. Using the product rule, the total number of license plates in this category is $$242 \\cdot 102 = 57,600$$. Finally, for a motorcycle plate, there are 24 choices for the first letter, followed by 24 choices for the second letter. There are 10 choices for the first digit, 10 choices for the second digit, and 10 choices for the third digit. Using the product rule, the total number of license plates in this category is $$242 \\cdot 103 = 576,000$$. Using the sum rule, we see that the total number of license plates is $$13,824,000 + 80,000,000 + 57,600 + 576,000$$ which is $$94,457,600$$. It doesn’t always happen that the sum rule is applied first to break the problem down into cases, followed by the product rule within each case. In some problems, these might occur in the other order. Sometimes there may seem to be one “obvious” way to look at the problem, but often there is more than one equally effective analysis, and different analyses might begin with different rules. In Example 2.3.1, we could have begun by noticing", "CD-R disc was introduced manufacturers said the media had a data life in excess of 40 years. Your vehicle is stolen and not recovered . The DVLA V750 Certificate of Entitlement/DVLA V778 Retention Document allows you to keep the number for a specified period e.g. Shortly and have applied to put the number plate retain the number plate for a specified period e.g 'd... Entitlement ' pink A4 size piece of paper you 'll then be asked to that. Vehicle to light goods vehicle 90s when the first policy disc was introduced manufacturers the... Plates cost around $50 and the personalization fee is$ 48 number plate be! But personalized plates take longer, usually about 12 weeks, then the symbol can transfer. About Onedrive version history menu you can only use your retained VRN to a vehicle you not! 180 days our website is fully PCI compliant and is so secure we even have the maximum of. Plate builder for instant results that come with an unbeatable price tag length be! If they die ), read here with V750 ’ s for new plates, read here then the can! As with V750 ’ s for new plates, read here not own can..., simply use our suffix plate builder for instant results that come with an unbeatable price tag length", "$$\\PageIndex{3}$$ A certain shirt comes in four sizes and six colors. One also has the choice of a dragon, an alligator, or no emblem on the pocket. How many different shirts could you order? $$72=4\\cdot 6\\cdot 3$$ Exercise $$\\PageIndex{4}$$ A builder of modular homes would like to impress his potential customers with the variety of styles of his houses. For each house there are blueprints for three different living rooms, four different bedroom configurations, and two different garage styles. In addition, the outside can be finished in cedar shingles or brick. How many different houses can be designed from these plans? Exercise $$\\PageIndex{5}$$ The Pi Mu Epsilon mathematics honorary society of Outstanding University wishes to have a picture taken of its six officers. There will be two rows of three people. How many different way can the six officers be arranged? $$720=6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1$$ Exercise $$\\PageIndex{6}$$ An automobile dealer has several options available for each of three different packages of a particular model car: a choice of two styles of seats in three different colors, a choice of four different radios, and five different exteriors. How many choices of automobile does a customer have? Exercise $$\\PageIndex{7}$$ A clothing manufacturer has put out a mix-and-match collection consisting of two blouses, two pairs of pants, a skirt,", "How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? # How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? J Asked by 2 years ago 240 points a) How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? b) Repeat part (a) under the assumption that no letter or number can be repeated in a single license plate. How many jeffp ### 1 Answer Answered by 2 years ago 8.7k points part (a) There are 26 letters in the alphabet and 10 possible numbers to use. It is a counting problem and hence, $$26^2 10^5 = 67600000$$ part (b) Repetitions are not allowed so for the first choice we have 26 letters to choose from, and then 25 letters.. Likewise for the numbers we can choose 10 numbers for the 3rd position, then 9 numbers, and so on... $$26 \\cdot 25 \\cdot 10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6= 19656000$$ ### Your Answer Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\\beta^2-1=0$$ Use LaTeX to type formulas and markdown", "= 20$. Question 14 The license plate for a special vehicle will consist of three letters, picked from A to Z. Letters can be repeated. How many license plates can be given out for this special vehicle? A $15,600$ B $17,576$ C $18,200$ D $18,576$ Question 14 Explanation: The correct answer is (B). Three letters are picked from A to Z. So, that is $26$ letters to pick from. Problem says that the letters can be repeated – so you can have AAA, AAB, or ABA, etc . The first letter can be picked in $26$ ways, the second letter in $26$ ways, and the third letter in $26$ ways. So, totally $26 \\cdot 26 \\cdot 26 = 17,576$ $3$-letter plates can be given out for this vehicle. Question 15 When $3x^2 + 4x + 1$ is divided by $3x + 1$, the remainder is A $x + 1$ B $0$ C $2x + 1$ D $2x – 1$ Question 15 Explanation: The correct answer is (A). Doing the long division: Question 16 A circus clown walks down a straight rope from a point $9$ feet above the ground as shown. At the same time, a trapeze artist swings down a parabolic path shown. After crossing the trapeze artist for the first time at $9$ feet, approximately how", "plate does not have the desired property is $\\frac{10\\cdot 10\\cdot 9\\cdot 26\\cdot 26\\cdot 25}{10^3\\cdot 26^3}=\\frac{45}{52}$. We subtract this from 1 to get $1-\\frac{45}{52}=\\frac{7}{52}$ as our probability. Therefore, our answer is $7+52=\\boxed{059}$.", "How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? # How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? J 240 points a) How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? b) Repeat part (a) under the assumption that no letter or number can be repeated in a single license plate. How many jeffp 8.7k points part (a) There are 26 letters in the alphabet and 10 possible numbers to use. It is a counting problem and hence, $$26^2 10^5 = 67600000$$ part (b) Repetitions are not allowed so for the first choice we have 26 letters to choose from, and then 25 letters.. Likewise for the numbers we can choose 10 numbers for the 3rd position, then 9 numbers, and so on... $$26 \\cdot 25 \\cdot 10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6= 19656000$$ Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\\beta^2-1=0$$ Use LaTeX to type formulas and markdown to format text. See example.", "permutations repetitions/no repetitions License plates consist of sequence of 3 letters followed by 3 digits. How can they be arranged if (i) no repetition of letters is permitted, how many possible license plates are there? should it be $26P3 \\cdot (10^3)$ or $3! \\cdot (10^3)$? Your first answer is correct. Since repetition of letters is not permitted, there are $26$ ways of choosing the first letter, $25$ ways of choosing the second letter, and $24$ ways of choosing the third letter. Therefore, there are $P(26, 3) = 26 \\cdot 25 \\cdot 24$ ways of selecting the three letters. Since repetition of digits is permitted, there are ten choices for each of the three digits. Thus, the number of license plates that can be formed is $$P(26, 3) \\cdot 10^3 = 26 \\cdot 25 \\cdot 24 \\cdot 10^3 = 15,600,000$$" ]

[ "# What does Bob have? Level pending Bob has 10 000 000 000 lollipops. He eats 100 of them every hour until he runs out. What does he have now? Instead of writing a numerical answer, write as letters. Then, convert these letters to their corresponding numbers, e.g. a is 1, b is 2, c is 3, etc. Then, add the numbers up. That is your answer. Note: • Bob did not die from eating all these • Lollipops contain no lipids • Spaces are ignored in the answer $$\\text{This is an entry in the Troll King Contest}$$ ×", "A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $$\\frac mn,$$ where $$m$$ and $$n$$ are relatively prime integers, find $$m+n.$$ (第二十三届AIME2 2005 第2题)", "## March Break Contest '22 Problem 1 - Bob's Dice Game View as PDF Points: 7 (partial) Time limit: 2.0s Memory limit: 64M Author: Problem type Bob is playing a new game that he has discovered over March Break. To play the game, he has bought a -sided die with faces numbered to which he must roll times in a row. Bob succeeds on the th roll if the number he rolls is at least . In order to increase the number of rolls that he succeeds, Bob will use his portent ability inside the game which will allow him to replace up to two of the rolls with two values that he had rolled beforehand, and . Note that Bob may only choose to replace a roll after rolling the die for that roll. Bob would like to know the number of possible dice roll configurations, out of a total of , such that he can succeed on all rolls with optimal use of his ability. Since this number may be very large, Bob would like to know its value modulo . #### Input Specifications The first line will contain two space-separated integers, and . The next line will contain two space-separated integers, and . The next line will contain space-separated integers, . No further constraints. ####", "# Golden Pancakes Alice and Bob play a game with two stacks of fluffy, golden pancakes, of sizes $p$ and $q$. They alternate turns, with Alice starting. On their turn, a player must eat some number of pancakes from the larger stack. The number they eat must be a multiple of the size of the smaller stack. Since the bottom pancake of each stack is soggy, the first player to finish a stack loses. For example, if the stacks were sized $5$ and $17$, then the current player can eat either $5,10$ or $15$ pancakes from the large stack. For which values of $p$ and $q$ does Alice have a winning strategy? Source: Mathematical Puzzles: A Connoisseur's Collection, Peter Winkler. • None, because that is a lot of pancakes, and she is going to be sick by the end no matter what. – Aggie Kidd Jul 8 '15 at 1:51 • Do I understand correctly: Does it mean that when Alice eats 15 from the second (larger stack), then there are stacks 5 and 2, so Bob has to eat 2 or 4 from the first stack? – Voitcus Jul 8 '15 at 3:52 • I hope the pancakes aren't made from real gold ;). – Bojidar Marinov Jul 8 '15 at 10:05 • In next weeks episode,", "# Roll me a prime Optimus Prime loves prime numbers. So, he decides to keep rolling a fair six-sided die, until he rolls a prime number less than 100. e.g. If his last two rolls were a 6 and then a 1, he would be done since 61 is a prime number. Or, for example, any time he rolls a 3 he would also be done since 3 is a prime number. If the expected number of rolls he must make is $$\\dfrac{a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers, what is $$a+b$$? ×", "a die. Dave rolls first. He wins control over the TV if he rolls a number greater than four. If he doesn't win, Mae gets her chance and wins if she rolls a prime number. If she doesn't roll a prime number, their parents take the remote and nobody wins. 1. What is Dave's chance of winning? 2. What is Mae's probability of winning? 3. Is this a fair way to settle the argument for both Dave and Mae? Yes A No B ##### QUESTION 3 Given the following options: (A) a sure gain of$\\$500$$500 (B) a 5050% chance of gaining \\1000$$1000 and a$50\\$50% chance of gaining nothing 1. Calculate the financial expectation of Option A. 2. Calculate the financial expectation of Option B. 3. Which option has greater variation? Option A A Option B B", "5k views ### The coupon collectors problem [closed] Each box of a certain breakfast cereal contains one of ten different coupons, each with the same probability. We win a prize if we manage to obtain a complete collection of all the different coupons. ... 622 views ### How long does it take to complete a sticker album? We are collecing stickers in chocolate bars and whenever we open a bar we get a random new sticker. There are many different stickers and we try to collect them all. We open the first bar and get a ... 115 views ### Number of attempts to see all faces $1, 2, 3, …, 6$ when rolling a dice? [closed] Recently I have been thinking about the following random experiment: we repeatedly roll a dice until we see all the faces $1, 2, 3, 4, 5, 6$ of the dice at least once. Let $X$ = number of attempts ... 489 views ### Expected number of rolls on a die until each face has appeared at least twice Note: The die is fair, normal 1 - 6 die. So I understand that the expected number of rolls until each face occurs is 14.7 by the following post: Expected time to roll all 1 through 6 on a die but ... 607", "12 and wins. theres a 5/6 chance of this happening. If the dice rolls 1, bob wins. • That was fast. Spot on. – Marius Apr 6 '16 at 10:39 • Isomorphic to nim – Martin Capodici Apr 6 '16 at 22:40", "# Probability that 20 sided die beats 12 sided die with reroll Alice rolls a 12 sided die (the faces labeled 1 through 12) and Bob rolls a 20 sided die (the faces labeled 1 through 20). After seeing their roll (but not the other person's roll), each person can choose to roll again. Bob wins if his final roll is strictly greater than Alice's final roll. What is the probability that Bob wins? I'm assuming that Alice's optimal strategy is to reroll if she gets a roll that is less than 7, and Bob's optimal strategy is to reroll if he gets less than 11. Is this correct? • I don't know. Have you proved it? It seems highly unlikely. If B rolls 10, then he has won unless A gets a 10,11 or 12. If he rolls again he could easily worsen his position. But I have not done the calculations. – almagest Sep 13 '14 at 20:52 • You have not stated that A's die is 1-12 (on the faces) and B's 1-20 and both are fair dice. Obviously there is no hope of approaching this question if we know nothing about the dice. Or to be more accurate, the reasoning will be totally different. – almagest Sep 13 '14 at 21:05 • @almagest Fair", "you get more flips!” “Fine!” answers Alice. “How about this? You flip the coin 2009 times and I’ll flip it 2010 times and if I get more Heads, I win. If you get more Heads, you win. And, if there’s a tie, we’ll say that you win too.” Bob shrugs his shoulders and agrees to play. February 18, 2010 at 23:27 Posted in Round 1 ## Round 1: Problem 2 The figure to the right is made of 16 congruent arcs. Each arc is a quarter of a circle of radius 1. What is the area of this figure? February 18, 2010 at 23:25 Posted in Round 1 ## Round 1: Problem 1 The numbers 1,2,3,…,2010 are listed in this order. What is the least non-negative result that you can form by placing a “+” or “-” sign in front of each number and summing the values? Prove your answer. February 18, 2010 at 23:24 Posted in Round 1 ## Round 5: Problem 6 Let’s play a game. A fair die is rolled repeatedly until two consecutive rolls are equal, at which point the game ends. If we let S be the sum of the results of all of the rolls, is S more likely to be even or odd, or is it equally likely to be both?", "I just read this \"news\" report about Stanford professors who discovered a pattern in prime numbers: https://www.nature.com/news/peculiar-pattern-found-in-random-prime-numbers-1.19550 According to this report (which is admittedly not the original research paper), it sounds like the claim is merely that consecutive primes share a units digit less often than chance. My first thought was \"why would one expect differently?\". If one merely makes a list of \"potential primes\" (namely any number with at least two digits that ends in 1,3,7,or 9) and then assigns each number an independent probability of being prime, then one would expect the units digits in consecutive primes to be equal slightly less often than chance. To see what I mean, let's re-cast this as the following problem: Suppose I roll an ordinary 6-sided die many, many times, take 1000 as an example. I number the rolls 1-1000 in a list, and I put a check mark next to every time the number \"1\" comes up on the die. The probability that the numbers next to a pair of consecutive check marks have the same units digit will be less than 1/10. Why you ask? Because, the number of \"failures\" (rolls that are NOT a \"1\") follows a geometric distribution, which is monotone decreasing. Therefore, the probabilities that, after rolling a 1, it will take 1,2,3,...9 additional", "that in a typical day about 45 to 50 sheets are used, and they have 2 partial rolls and 4 full rolls of the same TP on hand but don't know how many sheets these rolls contain (e.g., without the packaging). From that usage rate, a TP sheet's mean weight, and the rolls' weight, we could estimate time until TP depletion at least crudely without, say, careful modeling of W's distribution or uncertainty due to sampling or measurement error. – Adam Hafdahl Mar 30 '20 at 14:04", "box of cereal. What is the expected number of boxes of cereal you must buy? It turns out to be nln(n), and is very tight around that. 9) I learned more from the talks, more from the meals, and more from my own alone time, but what is above is a good sampling. 10) More chalk-talks then I would have thought. Either chalk or slides can be done well or poorly. 11) Looking forward to the next Dagstuhl! #### 1 comment: 1. Thanks for the review -- I would have come to this one but it was too close to the beginning of our semester...", "and Bob have different total amounts of candy. • It is guaranteed there exists an answer.", "at different rates for each machine. The important part is that they each update the world proportionately to the amount of time that has passed in fixed time steps. In summary: Fast computer iterates through loop many times, slow computer iterates through loop few times. Both of them update the world the same amount, by being based on time in discrete time steps. For a very silly analogy, Alice and Bob are asked to get six eggs from the market, and return in one hour with exactly six eggs. Alice can make it to the market and back in 20 minutes, whereas Bob takes a full hour to go to the market and return. Alice, wanting to show off and be flashy, runs back and forth from point to point, taking two eggs in her basket at a time, and bringing them back. Bob takes his time, and grabs all six eggs, putting handfuls of two in his basket at a time, in one trip in order to make the deadline. After one hour, both return, and each are holding their six eggs. Second, i'm not sure if this is based on my UPS calculations, or the game logic itself, but my UPS runs between 59.99xxx and 60.00xxx, where the xxx are any numbers, usually the same each", "CROW: Something to do with prime numbers. MIKE: Prime … numbers. Likelihood of success? TOM: We have no idea what that was all about. MIKE: … Was … all … about. Okay, Pearl, good luck with your world conquest through prime numbers. [ CASTLE FORRESTER. PROFESSOR BOBO and OBSERVER are working as above; PEARL is in front, by the camera. After a pause OBSERVER does that brain-wave thing, and MIKE’s report pops into her hand. ] PEARL: Excellent! And when we take over this … cabal … we’ll be able to … uh … factorize numbers like … twenty-eight thousand, eight hundred fifty eight in — BOBO: [ Without looking up ] Two times forty-seven times three hundred seven. PEARL: [ Slighty thrown ] Or … one million, five hundred thirty-one thousand, one hundred twelve … BOBO: Two to the third power times eleven times one hundred twenty-seven times one hundred thirty-seven. PEARL: [ Challenging ] 89 thousand, one hundred seventy-five. BOBO: [ Finally looking at her ] Three times five squared times twenty-nine times forty-one. PEARL: [ Testy ] Nine million … three hundred eighty six thousand … seven hundred thirty-one. [ PEARL stares at BOBO. OBSERVER slips his pile of papers onto BOBO’s table while she fumes, and then slips away. After letting her temper build,", "for you. Let's suppose a regular roll has 100 sheets, and I use on average 5 sheets per day. This means you would expect me to replace the roll after 20 days (20 days = 100 sheets / 5 sheets per day). If a Long Roll has 50% more, then it has 100 sheets + 50% of 100 sheets = 150 sheets total. Cool! But I'm still using 5 sheets per day. Therefore I will change the roll after 150 sheets / 5 sheets per day = 30 days. In 60 days time, I will have used three regular rolls of 100 sheets, or two Long Rolls of 150 sheets. That means I'm replacing the Long Rolls two-thirds as often as the regular ones. To replace the rolls half as often, each roll would need to be twice as long, i.e. have 100% more sheets, not 50% more sheets. Note that you can swap out the size of a regular roll and the rate of use, and the maths still works out (as long as you're consistent)---a Long Roll, if it has 50% more sheets than a regular roll, will be replaced two-thirds as often. Other than to have even bigger rolls, the only way I can see to get long rolls replaced half as often as regular", "he knows the 3 numbers without knowing who had what. this changes when Charlie proclaims he knows the 3 numbers as well, Bob and Charlie now know the 3 numbers and who had what. • If Alice is holding 1, 2, or 3, how does she know for certain that the other two numbers are unique? If instead you are implying that she is holding 97 or 98, then if Bob is holding 1, 2, or 3 how does he know what all three numbers are knowing only what number he is holding (1,2, or 3) and that Alice has told them all three numbers are unique? Aug 21 '17 at 12:09", "odds that Bob will win on his second roll\" - 125/1296, for \"Sue misses, Bob misses, Sue misses, Bob hits\" in that order. The problem is that that answer is the odds that Bob will win on the second turn, out of the set of ALL GAMES. What we want is the odds that Bob will win on his second turn, out of the set of GAMES THAT BOB WON. How do we know how many games Bob will win? Well, pretend that Sue and Bob roll simultaneously, but that Sue's die is counted first if it is a 6. This gives us 36 possible results. Of those, 6 are wins for Sue, 5 are wins for Bob (he loses the 6-6 tie), and 25 are \"roll again\". This means that there are 11 possible exit conditions of the loop, all equally likely, and Bob wins 5 of those. Thus, in the set of ALL GAMES, Bob will win 5/11 of them. Therefore: In all games, Bob wins 125/1296 of them on the second turn. Bob wins 5/11 of all games. Therefore, Bob wins on the second turn in (125/1286) / (5/11) = 275/1296 of all games he wins. Which is the same answer everyone else is getting. Using a faster method. -John 99.224.156.92 22:27, 11 February 2009", "# 1981 AHSME Problems/Problem 26 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is $\\frac{1}{6}$, independent of the outcome of any other toss.)" ]

[ "survive 45 days = x o The amount required by Bob to survive 1 day = $$\\frac{x}{45}$$ • The amount required by Amy to survive 36 days = x o The amount required by Amy to survive 1 day = $$\\frac{x}{36}$$ • Amount required by both of them to survive 1 day = $$\\frac{x}{45} + \\frac{x}{ 36} = \\frac{(5x + 4x)}{180} = \\frac{x}{20}$$ o With $$\\frac{x}{20}$$ they both can survive 1 day o With $$x$$, they can survive for 20days If Bob can survive 45 days with $180, it means he uses up$4 per day. Similarly, Amy uses up $5 per day. These values represent the individual rates of Bob and Amy. If they have to survive together, they will consume a total of$9 per day. If a total of $360 is available, they can survive for a total of 40 days ($$\\frac{360}{9}$$). The correct answer option is D. Hope that helps! _________________ Re:$x is enough for Bob to survive 45 days, while for Amy $x is enough to [#permalink] 31 Mar 2020, 23:19 #$x is enough for Bob to survive 45 days, while for Amy \\$x is enough to new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne", "as follows: $$P(X=0)=\\frac{e^{-7}\\times 7^{0}}{0!}\\\\=e^{-7}\\\\=0.00091188\\\\\\approx0.0009$$ Thus, the probability that Bob receives no tweets during his lunch hour 0.0009. (b) Compute the probability that Bob receives at least 4 tweets during his lunch hour: $$P(X\\geq 4)=1-P(X<4)\\\\=1-[P(X=3)+P(X=2)+P(X=1)+P(X=0)]\\\\=1-[\\frac{e^{-7}7^{3}}{3!}+\\frac{e^{-7}7^{2}}{2!}+\\frac{e^{-7}7^{1}}{1!}+\\frac{e^{-7}7^{0}}{0!}]\\\\=1 – 0.08176\\\\=0.91824\\\\\\approx0.9182$$ Thus, the probability that Bob receives at least 4 tweets during his lunch hour is 0.9182. (c) Compute the expected number of tweets Bob receives during the first 30 minutes of his lunch hour as follows: Number of tweets in 60 minutes = 7 Number of tweets in 1 minutes = $$\\frac{7}{60}$$ Number of tweets in 30 minutes = $$\\frac{7}{60}\\times 30=3.5$$ Thus, the expected number of tweets in the first 30 minutes is 3.5. (d) Compute the probability that Bob receives no tweets during the first 30 minutes of his lunch hour as follows: Now the average is, $$\\lambda=3.5$$ $$P(X=0)=\\frac{e^{-3.5}\\times 3.5^{0}}{0!}\\\\=e^{-3.5}\\\\=0.0301974\\\\\\approx0.0302$$ Thus, the probability that Bob receives no tweets during the first 30 minutes of his lunch hour is 0.0302.", "(5/6)(1 - p) => p = 5/11. The absolute probability (i.e. at the outset of a game) of Bob winning on his second roll = P(s)P(b)P(s)P(B) = 125/1296, so the probability of Bob winning on his second roll, given that he won is: (125/1296)/(5/11) = 275/1296 = 0.212... Note, that is the fraction of all the possible games that Bob won on his second roll out of all the games that Bob wins (regardless of which roll was the winning one). --- Here's another way to understand that 275/1296 is the correct answer. I'm simply trying to make the situation more concrete. The probability that Bob wins a game on his second roll is 125/1296. Only a few posters don't even get that bit - this note isn't for them. The probability that Bob wins a game is 5/11. I believe that (almost) everyone accepts that too. If Sue and Bob played sets of 1000000 (one million) games, then on average Bob will win 1000000 * 5/11 = 454545 of them. Bob will win 1000000 * 125/1296 = 96451 games on his second roll. (I've rounded the numbers to integers). So the fraction of games that Bob wins on his second roll out of all the games he wins is (approximately) 96451/454545 ≈ 0.212 Now taking out the", "# Fix $0\\leq\\delta\\leq1.$ Bob rolls a die repeatedly in the hopes of rolling a six. Fix a parameter $$0\\leq\\delta\\leq1.$$ Bob rolls a die repeatedly in the hopes of rolling a six. However, after each failure to roll a six he gives up with probability $$1-\\delta$$ and decides to try again with probability $$\\delta$$. What is the probability that Bob will never roll a six? Let $$A$$ denote the event that Bob does not roll a six, and let $$B$$ be the event that he gives up after a failure. Then $$P(A\\cap B)=1-\\delta$$ and $$P(A\\cap B^{C})=\\delta.$$ Now after the first roll, the probability that Bob did not roll a six is $$5/6$$. I am having difficulty with understanding how the parameter $$\\delta$$ comes into the calculation of the probability that Bob does not get a six given that he failed and tried again. Thank you for time, I appreciate any feedback. • $\\delta$ comes up because Bob may quit before he gets the $6$. For instance, if $\\delta =0$ then the probability he never throws a $6$ is $\\frac 56$ since he is sure to give up after one failure. If $\\delta =1$ then he never gives up so he'll eventually get his $6$ with probability $1$. – lulu Feb 15 at 13:51 HINT : Let $$K$$ denote the", "Probability Level 5 Bob wants to keep a good-streak on Brilliant, so he logs in each day to Brilliant in the month of June. But he doesn't have much time, so he selects the first problem he sees, answers it randomly and logs out, despite whether it is correct or incorrect. Assume that Bob answers all problems with $\\frac{7}{13}$ probability of being correct. He gets only 10 problems correct, surprisingly in a row, out of the 30 he solves. If the probability that happens is $\\frac{p}{q}$, where $p$ and $q$ are coprime positive integers, find the last $3$ digits of $p+q$. ×", "+0 0 92 3 Question: Bob repeatedly rolls a fair 6-sided die. What is the probability that he rolls his first 5 before he rolls his second (not necessarily distinct) even number? What I've Tried: Let 'n' be the number of rolls in the game. This means that n has to be at least 2. I then created cases: n = 2: Both rolls would have to be even so $$\\frac{1}{2}\\cdot\\frac{1}{2}=\\frac{1}{4}$$ n = 3: We have to have an odd number (not a 5) and two even numbers (but one of the even numbers has to be last), so $$\\frac{1}{2}\\cdot\\frac{1}{3}\\cdot2\\cdot\\frac{1}{2} = \\frac{1}{6}$$ After this, I'm not sure what to do. Help would be appreciated! Jun 1, 2021 #1 +795 0 first 5 meaning he rolls the number 5, or rolls 5 times? edited by MathProblemSolver101 Jun 1, 2021 #2 +2250 0 o = rolling a 1 or 3 (1/3 probability) e = rolling a 2, 4, or 6 (1/2 probability) f = rolling a 5 (1/6 probability) So we have 3 cases. Rolling a bunch of \"o\"s before rolling an f. Rolling a bunch of \"o\"s and 1 \"e\" before rolling an f. Rolling an f. Rolling a bunch of \"o\"s before rolling an f. ((1/3) * 1/6) + ((1/3)^2 * 1/6) + ((1/3)^3 * 1/6).... + ((1/3)^infinity", "x 7 = 28 equal to 28. 7 + 7 + 7 + 7 = 4 × 7. Question 4. Bob bought 3 packs of model cars. He gave 4 cars to Ann. Bob has 11 cars left. How many model cars were in each pack? Options: a. 18 b. 11 c. 7 d. 5 d. 5 Explanation: If Bob has 11 cars left after he gave 4 to Ann, that means he had 15 cars in total. If Bob had 15 cars in total, and these 15 cars were divided into 3 packs, that means each pack contained 5 cars. The answer is 5 cars in each pack. Question 5. Randy looked at his watch when he started and finished reading. How long did Randy read? Options: a. 55 minutes b. 45 minutes c. 35 minutes d. 15 minutes b. 45 minutes Explanation: Subtract to find the Randy read time. 4:55 – 4:10 = 45 minutes Question 6. Which statement does the model represent? Options: a. $$\\frac{4}{4}$$ = 1 b. $$\\frac{3}{4}$$ = 1 c. $$\\frac{2}{4}$$ = 1 d. $$\\frac{1}{4}$$ = 1 a. $$\\frac{4}{4}$$ = 1 Explanation: The first image represents the whole circle divided into 4 equal parts. So, each part is $$\\frac{1}{4}$$. Together, all 4 parts represent $$\\frac{4}{4}$$ or one whole circle. The second image represents", "at the highest die, what will the result be on average?--Sonjaaa 11:45, 9 March 2007 (UTC) You might look at order statistic; although I don't know if that information will help you compute it. It might be easier if you just went through all the outcomes on a computer and computer the average, or if you did a lot of random trials and estimated it probabilistically. --Spoon! 11:59, 9 March 2007 (UTC) I computed it in QBasic and got 4.5 for 1d8, 5.8 for 2d8 and 6.5 for 3d8.--Sonjaaa 12:41, 9 March 2007 (UTC) For n 8-sided dice the average of the highest die is ${\\displaystyle 8-{\\frac {(1^{n}+2^{n}+...+7^{n})}{8^{n}}}}$ For two dice the average of the highest is 8-(140/64), which is 5 13/16. For three dice the average of the highest is 8-(784/512), which is 6 15/32. Gandalf61 14:18, 9 March 2007 (UTC) Gandalf's answer looks incorrect. For instance, looking at the case when n=1, his formula says the average result of one eight sided die is 7 7/8. But in fact the average of a single 8-sided die is 4.5 . Dugwiki 16:21, 9 March 2007 (UTC) How did you get 7 7/8? Looking for the case when n=1 Gandalf's formula says the average is ${\\displaystyle 8-{\\frac {(1^{1}+2^{1}+...+7^{1})}{8^{1}}}}$${\\displaystyle =8-{\\frac {(1+2+...+7)}{8}}}$${\\displaystyle =8-{\\frac {28}{8}}}$${\\displaystyle =8-3.5=4.5}$ --CiaPan 16:34, 9 March 2007", "odds that Bob will win on his second roll\" - 125/1296, for \"Sue misses, Bob misses, Sue misses, Bob hits\" in that order. The problem is that that answer is the odds that Bob will win on the second turn, out of the set of ALL GAMES. What we want is the odds that Bob will win on his second turn, out of the set of GAMES THAT BOB WON. How do we know how many games Bob will win? Well, pretend that Sue and Bob roll simultaneously, but that Sue's die is counted first if it is a 6. This gives us 36 possible results. Of those, 6 are wins for Sue, 5 are wins for Bob (he loses the 6-6 tie), and 25 are \"roll again\". This means that there are 11 possible exit conditions of the loop, all equally likely, and Bob wins 5 of those. Thus, in the set of ALL GAMES, Bob will win 5/11 of them. Therefore: In all games, Bob wins 125/1296 of them on the second turn. Bob wins 5/11 of all games. Therefore, Bob wins on the second turn in (125/1286) / (5/11) = 275/1296 of all games he wins. Which is the same answer everyone else is getting. Using a faster method. -John 99.224.156.92 22:27, 11 February 2009", "as noted by @Penguino, you have a $\\frac23$ chance of survival playing second. In variant two, the odds are still in your favor, but at the slightly lower odds of $\\frac6{10}$. Penguino explains the odds for variant one here. The odds for variant two are trickier, and were determined by simulation. • For the exact calculation of probabilities in the second variant of the game see the answer by @masonk and my comment there. Mar 6, 2017 at 14:27 Assume the bullets are in chambers 5 and 6 (you can do this because the number of the cylinders is arbitrary). Then the first player dies on their first turn if the cylinder is at 5 or 6 after the initial random spin, or after their second turn if the cylinder starts on 3, or after their third turn if the cylinder starts on 1. So best to start second with 2/3 survival rate. Rubio's second variant (\"spin chamber after every trigger pull\") is the more difficult one to answer, but it can be solved exactly with a Markov matrix. This matrix, call it M, has four states: Alice's Turn, Alice Died, Bob's Turn, Bob Died 0 0 4/6 0 2/6 1 0 0 4/6 0 0 0 0 0 2/6 1 Solve for the eigenvalues of this matrix", "Death Rolling Analyzing World of Warcraft players funnest past time. Posted by Krystian Wojcicki on Friday, January 7, 2022 Tags: Programming 13 minute read Intro Death rolling is a player made mini-game within World of Warcraft where two players alternate rolling a die between $$[1-\\text{previous players roll}]$$ (the bounds are inclusive) until one player rolls a $$1$$. The following is an example of a game: Bob and Alice agree to death roll for 100 gold Bob agrees to go first Bob rolls 42 [1-100] Alice rolls 12 [1-42] Bob rolls 3 [1-12] Alice rolls 1 [1-3] Alice transfers 100 gold to Bob. Now, as a player trying to maximize your returns should you go first or second? Bet large or small? Naturally, I wanted to find out. Game Death roll against a computer, alternating who goes first. Skip ahead to find out how to size your rolls. Keep in mind you can only roll whole integer amounts. Simulation My initial assumption was for very small bets the second roller would have a large advantage and the first roller would have a minimal $$1/2\\%$$ advantage for larger rolls. The intuition behind this assumption is the following: for small starting rolls, for example starting at $$2$$, there is a large probability that the initial roll will be a $$1$$ resulting", "b bobcats. If he gets 4 more than three times the number of bobcats he has now, he will have 10 bob cats. How many bobcats does Bob have? 4. Originally Posted by Saoirse Mac Write an equation to help you solve this promblem: Two-fifthsof my money is €24. How much have I? Twenty-five per cent of the eggs in a box are broken. If there are three dozen eggs in the box, how many unbroken eggs are there? The length of a rectangular garden is three times its width. Find the perimeter of the garden if the length is 48m. Write a story for this equation: 3b + 4 = 10 Thank you #1 $\\frac{2}{5}M=24 \\implies M=\\frac{24 \\cdot 5}{2}=60$ euros #2 $.25x$ are broken, where x is the number of eggs. Thus, if there are 3 dozen = 36 eggs, then $.25(36)=9$ eggs are broken. #3 Since L=3W, and L=48m, then $W=\\frac{48}{3}=16m$. Thus, $P=2L+2W\\implies P=2(48m)+2(16m)=96m+32m=128m$. #4 What kind of story??? Hope this helped you out a little bit! 5. ## Help with equations Left side = right side term + term = term + term ----------------- (2/5) of n = 24 2n/5 = 24 Multiply both sides by 5, then divide both sides by 2. n = 5(24)/2 n = 60 Is 24 two fifths of 60?", "+0 Can someone confirm pls? 0 319 2 +61 Could someone please confirm? I think the average is 4.2 and the mode is 1, if not could someone correct me? May 15, 2021 2+0 Answers #1 +102 +2 1. If you add all the values up together, you would get $$21$$. Divide that by the number of days and you would get your mean (average): $$\\frac{21}{5} =$$ 4.2 2. Just count which number of cookies Grogg ate most in a day; the mode would be 1. You are correct on all of them! Hope that helped, AnxiousLlama May 15, 2021 #2 +873 +1 (a) $\\frac{1+11+3+5+1}{5}=\\frac{21}{5}=\\boxed{4.2}.$ (b) All other values except for $1$ appear exactly once. Thus the mode is $\\boxed{1}.$ You are correct on both of them. - Jimmy May 15, 2021", "(i. e. $n \\ge 22$) and $n+1$ otherwise (if $n \\le 21$). His expected winnings will be $n(n-1)/60$ if $n \\ge 22$ and $(31+n)(30-n)/60$ if $n \\le 21$. Now, what will Alice choose? Alice chooses in such a way as to minimize Bob's expected winnings, because minizing Bob's expected winnings is the same as maximizing her own. This is because the sum of Alice's expected winnings and Bob's expected winnings is the expected number that comes up when you roll a 30-sided die. (This is 15.5, but that's irrelevant right now.) The function $n(n-1)/60$ is increasing with $n$, so Bob does better as $n$ increases above 22. And $(31+n)(30-n)/60$ decreases with $n$, so Bob does better as $n$ decreases below 21. Bob's worst number is therefore either 21 or 22. Explicitly computing, Bob's expected winnigs are $22(22-1)/60 = 7.7$ if Alice picks $n = 22$, and $(31+21)(30-21)/60 = 7.8$ if Alice picks $n = 21$. Therefore Alice will pick $n = 22$, shd expected to win $15.5 - 7.7 = 7.8$, and Bob expects to win $7.7$. Finally, a player given the choice of going first or second will choose to go first (that is, to be Alice). This depends on the fact that the number of sides of the die is $k = 30$, and in particular", "$\\frac{1}{5} *1 + \\frac{4}{5}*5=4.2$ rounds Time until Eating – In the first choice, there is a $\\frac{1}{5}$ chance of ending in 3 minutes. If that doesn’t happen, there is a subsequent $\\frac{1}{5}$ chance of ending with the second choice with no additional time. If neither of those events happen, there will be 1.6 minutes on the first choice plus an average 5 more rounds, each taking an average 1.5 minutes, for a total average $1.6+5*1.5=9.1$ minutes. So the total average time until both siblings finish simultaneously will be $\\frac{1}{5}*3+\\frac{4}{5}*9.1 = 7.88$ minutes CONCLUSION: My 7.88 minute mean is reasonably to the right of Nick’s 5 minute mode shown above. We’ll see tomorrow if I match the FiveThirtyEight solution. Anyone else want to give it a go? I’d love to hear other approaches.", "by example. Let’s look at two different cases that cover two days. First consider Alice and Bob sharing birthdays, and Carol and Dan sharing birthdays: • Alice’s birthday can fall on any day of the year (probability $$365/365$$), • Bob’s birthday has to fall on Alice’s birthday (probability $$1/365$$), and • Carol’s birthday has to fall on any day other than Alice’s and Bob’s joint birthday (probability $$364/365$$), • Dan’s birthday has to fall on Carol’s birthday (probability $$1/365$$), so the probability of this happening is: $\\frac{365}{365}\\cdot\\frac{1}{365}\\cdot\\frac{364}{365}\\cdot\\frac{1}{365} = \\frac{365\\cdot 364}{365^4}.$ Now consider Alice, Bob and Carol sharing birthdays, with Dan having his birthday on a different day: • Alice’s birthday can fall on any day of the year (probability $$365/365$$), • Bob’s birthday has to fall on Alice’s birthday (probability $$1/365$$), • Carol’s birthday has to fall on Alice’s and Bob’s shared birthday (probability $$1/365$$), and • Dan’s birthday has to fall on any day apart from Alice’s, Bob’s and Carol’s shared birthday (probability $$364/365$$), so the probability of this happening is: $\\frac{365}{365}\\cdot\\frac{1}{365}\\cdot\\frac{1}{365}\\cdot\\frac{364}{365} = \\frac{365\\cdot 364}{365^4}.$ Note in particular that the left hand side of both expressions have the same factors, just in a different order. Whenever we add someone sharing their birthday with the previous person, we add a factor $$1/365$$, and whenever we add someone who", "\\frac1+2+3+4+5+66 = 3.5. See more: What Are The Declining-Balance Method Of Depreciation Produces Average Dice worth Against variety of Rolls: an illustration of the convergence of sequence averages of rolfes of a die to the expected value of 3.5 together the variety of rolls (trials) grows.", "$\\displaystyle \\frac{\\textrm{probability Bob wins on his second turn}}{\\textrm{probability Bob wins (regardless of what turn he wins on)}} =$ $\\displaystyle \\frac{(5^{3}/6^{4})}{(5/11)} = \\Big(\\frac{5^{3}}{6^{4}}\\Big) \\Big(\\frac{11}{5}\\Big) = \\Big(\\frac{5^{2}}{6^{4}}\\Big) \\Big(11\\Big) = \\frac{275}{1296} \\approx .2122$ So the final answer is around 21%. Some commenters on the xkcd blog post got it right, providing varying degrees of justification for their answers, but you’ll mostly find a small set of repeatedly cited wrong answers there. Ok, that’s that. As I said, I’ll soon follow this up with a post covering two problems that expand on the tools explored here: How many rolls of a fair, 6-sided die are expected in order to see a 6? How many to see all six sides? I’ll link it once posted. Post Script: Here’s an alternative solution for finding the denominator—that is, the probability that Bob wins, period. Let “$\\mathfrak{B}$” represent that probability. The key observation here is that if neither Sue nor Bob wins on their first turn, the game resets. So we can get everything we need by looking at those two turns. Here are the relevant scenarios: Either… …Sue wins game on her first turn, in which case Bob then goes on to win the game with probability 0. We can disregard this scenario, as it adds 0 our equation. …or Sue’s first turn misses, then", "and you will find closed form solution for the equilibrium probability between Alice Died and Bob Died. The answer is exactly .6, .4. • Thanks to the symmetry this specific case can be easily solved without general Markov description. If Alice starts, her chance to die in the first move is $1/3$. Let $p$ denote her overall chance to die in the game. If she survives, then Bob, the second player, appears in the same situation as Alice was when starting. That means Bob has chance $p$ to die in the following game. So the overal chance of Bob's death is $2/3\\cdot p$, and that equals $(1-p)$, hence $p=3/5$ – the first player dies with probability of $60\\%$. Mar 6, 2017 at 13:58 • @CiaPan: And that is exactly the same as the second part of my answer. Mar 6, 2017 at 20:27 • @PeregrineRook You're right. I just tried to show the same with less math formalism, but exposing more 'nature of the problem' instead. Mar 14, 2017 at 9:35 As Rubio points out, the question is slightly ambiguous. If we assume that a player spins the cylinder on each turn, then In round 1, Player 1 dies with probability $\\frac13$. So there is a $\\frac23$ probability of there being a round 2. Then, Player 2 dies", "to cases of distributions of people among more birthdays As an example, with $m=2$ and $n=4$ this gives $\\frac{28}{11}$. If you consider the sixteen equally probable distributions of birthdays for four people among two days Day1 Day2 ABCD - ABC D ABD C AB CD ACD B AC BD A BCD BCD A there are $2$ cases of four people sharing a birthday, $8$ cases of three people sharing a birthday, $12$ cases of two people sharing (as well as $8$ of one and $2$ of zero) making the average number of people per shared birthday $\\frac{2\\times 4+8 \\times3 +12\\times 2}{2+8+12}=\\frac{28}{11}$ as predicted. A different approach could say the average should be calculated as the average of $2$ cases with the average being $4$, $8$ cases with the average being $3$ and $6$ cases with the average being $2$, giving a result of $\\frac{2\\times 4+8 \\times3 +6\\times 2}{2+8+6}=\\frac{11}{4}$" ]

[ "Probability Level 5 Bob wants to keep a good-streak on Brilliant, so he logs in each day to Brilliant in the month of June. But he doesn't have much time, so he selects the first problem he sees, answers it randomly and logs out, despite whether it is correct or incorrect. Assume that Bob answers all problems with $\\frac{7}{13}$ probability of being correct. He gets only 10 problems correct, surprisingly in a row, out of the 30 he solves. If the probability that happens is $\\frac{p}{q}$, where $p$ and $q$ are coprime positive integers, find the last $3$ digits of $p+q$. ×", "+0 0 92 3 Question: Bob repeatedly rolls a fair 6-sided die. What is the probability that he rolls his first 5 before he rolls his second (not necessarily distinct) even number? What I've Tried: Let 'n' be the number of rolls in the game. This means that n has to be at least 2. I then created cases: n = 2: Both rolls would have to be even so $$\\frac{1}{2}\\cdot\\frac{1}{2}=\\frac{1}{4}$$ n = 3: We have to have an odd number (not a 5) and two even numbers (but one of the even numbers has to be last), so $$\\frac{1}{2}\\cdot\\frac{1}{3}\\cdot2\\cdot\\frac{1}{2} = \\frac{1}{6}$$ After this, I'm not sure what to do. Help would be appreciated! Jun 1, 2021 #1 +795 0 first 5 meaning he rolls the number 5, or rolls 5 times? edited by MathProblemSolver101 Jun 1, 2021 #2 +2250 0 o = rolling a 1 or 3 (1/3 probability) e = rolling a 2, 4, or 6 (1/2 probability) f = rolling a 5 (1/6 probability) So we have 3 cases. Rolling a bunch of \"o\"s before rolling an f. Rolling a bunch of \"o\"s and 1 \"e\" before rolling an f. Rolling an f. Rolling a bunch of \"o\"s before rolling an f. ((1/3) * 1/6) + ((1/3)^2 * 1/6) + ((1/3)^3 * 1/6).... + ((1/3)^infinity", "# Fix $0\\leq\\delta\\leq1.$ Bob rolls a die repeatedly in the hopes of rolling a six. Fix a parameter $$0\\leq\\delta\\leq1.$$ Bob rolls a die repeatedly in the hopes of rolling a six. However, after each failure to roll a six he gives up with probability $$1-\\delta$$ and decides to try again with probability $$\\delta$$. What is the probability that Bob will never roll a six? Let $$A$$ denote the event that Bob does not roll a six, and let $$B$$ be the event that he gives up after a failure. Then $$P(A\\cap B)=1-\\delta$$ and $$P(A\\cap B^{C})=\\delta.$$ Now after the first roll, the probability that Bob did not roll a six is $$5/6$$. I am having difficulty with understanding how the parameter $$\\delta$$ comes into the calculation of the probability that Bob does not get a six given that he failed and tried again. Thank you for time, I appreciate any feedback. • $\\delta$ comes up because Bob may quit before he gets the $6$. For instance, if $\\delta =0$ then the probability he never throws a $6$ is $\\frac 56$ since he is sure to give up after one failure. If $\\delta =1$ then he never gives up so he'll eventually get his $6$ with probability $1$. – lulu Feb 15 at 13:51 HINT : Let $$K$$ denote the", "boxes loaded by two crews did the day crew load? Assume the number of boxes loaded in dayshift is equal to 4, then the number of boxed loaded in night shift = 3 Assume the worked on dayshift = 5, then workers on night shift = 4 So boxes loaded in day shift = 4 x 5 = 20, and boxes loaded in night shift = 3 x 4 = 12 so fraction of boxes loaded in day shift = $\\displaystyle\\frac{{20}}{{20 + 12}} = \\displaystyle\\frac{5}{8}$ 22) A bakery opened yesterday with its daily supply of 40 dozen rolls. Half of the rolls were sold by noon and 80 % of the remaining rolls were sold between noon and closing time. How many dozen rolls had not been sold when the bakery closed yesterday? Ans: If half of the rolls were sold by noon, the remaining are 50 % (40) = 20. Given 80% of the remaining were sold after the noon to closing time $\\Rightarrow$ 80% (20) = 16 Unsold = 20 - 16 = 4 23) If N=4P, where P is a prime number greater than 2, how many different positive even divisors does n have including n? Ans: N = ${2^2} \\times {{\\rm{P}}^1}$ We know that total factors of a number which is in the format", "# Trust me, its fair Two friends, Alice and Bob, were bored so Bob came up with a game for them to play. The game is fairly simple. They roll and six sided die twice and concatenate the results to get a single two digit number. If this number is divisible by 3 Alice wins and if it is not, Bob wins. (For example, if the first roll is a 4 and the second role is a 2 then the number is $$42=3\\times14$$ and Alice wins). Alice is skeptical about this game because she feels she is at a disadvantage, but Bob assures her that he is using a special weighted die that makes the game fair. Assume that Bob's die is weighted such that $$P(2)=P(3)=P(4)=P(5)=\\frac{1}{6}$$ and $$P(1)=p$$. Find the value of $$p$$ that makes this game fair (i.e. Alice wins with probabilty $$\\frac{1}{2}$$). Enter -1 as your answer if you think that the game cannot be made fair. ×", "balloons and 3 purple balloons. Jeff randomly draws one gumball and then one balloon. What is the probability that he will draw a red gumball and a purple balloon? A. $\\frac{3}{10}$ B. $\\frac{2}{15}$ C. $\\frac{1}{5}$ D. $\\frac{1}{10}$ 12. Gary has 1 red apple, 1 green apple, and 4 yellow apples in a sack. He also has 2 packages of peanuts, 2 packages of almonds, and 2 packages of cashews in another sack. If he reaches into each sack without looking, what is the probability that he will grab a red apple and a package of cashews? A. $\\frac{1}{18}$ B. $\\frac{1}{4}$ C. $\\frac{1}{12}$ D. $\\frac{1}{36}$ 1. A 2. A 3. B 4. A 5. B 6. A 7. B 8. B 9. D 10. B 11. C 12. A ## Explanations 1. Since each of the six sides of the die has an equal chance of turning up on each roll, the probability of landing on a particular number during any roll of the die is $\\normalsize \\frac{1}{6}$ . Since each flip of the coin has an equal chance of landing on heads or tails, the probability of landing on heads is equal to the probability of landing on tails, which is $\\normalsize \\frac{1}{2}$ . Therefore, the probability of getting a 2 and heads as the outcome is calculated", "5k views ### The coupon collectors problem [closed] Each box of a certain breakfast cereal contains one of ten different coupons, each with the same probability. We win a prize if we manage to obtain a complete collection of all the different coupons. ... 622 views ### How long does it take to complete a sticker album? We are collecing stickers in chocolate bars and whenever we open a bar we get a random new sticker. There are many different stickers and we try to collect them all. We open the first bar and get a ... 115 views ### Number of attempts to see all faces $1, 2, 3, …, 6$ when rolling a dice? [closed] Recently I have been thinking about the following random experiment: we repeatedly roll a dice until we see all the faces $1, 2, 3, 4, 5, 6$ of the dice at least once. Let $X$ = number of attempts ... 489 views ### Expected number of rolls on a die until each face has appeared at least twice Note: The die is fair, normal 1 - 6 die. So I understand that the expected number of rolls until each face occurs is 14.7 by the following post: Expected time to roll all 1 through 6 on a die but ... 607", "## March Break Contest '22 Problem 1 - Bob's Dice Game View as PDF Points: 7 (partial) Time limit: 2.0s Memory limit: 64M Author: Problem type Bob is playing a new game that he has discovered over March Break. To play the game, he has bought a -sided die with faces numbered to which he must roll times in a row. Bob succeeds on the th roll if the number he rolls is at least . In order to increase the number of rolls that he succeeds, Bob will use his portent ability inside the game which will allow him to replace up to two of the rolls with two values that he had rolled beforehand, and . Note that Bob may only choose to replace a roll after rolling the die for that roll. Bob would like to know the number of possible dice roll configurations, out of a total of , such that he can succeed on all rolls with optimal use of his ability. Since this number may be very large, Bob would like to know its value modulo . #### Input Specifications The first line will contain two space-separated integers, and . The next line will contain two space-separated integers, and . The next line will contain space-separated integers, . No further constraints. ####", "observed in steps 1 -3. Expected number of rolls: $1/\\frac{3}{6}=2$ 5. We have $\\frac{2}{6}$ probability of seeing a different side than what was previously observed in steps 1-4. Expected number of rolls: $1/\\frac{2}{6}=3$ 6. We have $\\frac{1}{6}$ probability of seeing a different side than what was previously observed in steps 1-5. Expected number of rolls: $1/\\frac{1}{6}=6$ Adding all the expected number of rolls for each definition of success we get 14.7. So we expect to roll a die about 15 times on average before we observe all sides at least once. We can use this same line of thinking to answer such “real-life” questions as how many boxes of cereal would you have to buy to collect the latest series of cheap plastic toys, assuming the toys are randomly added to the cereal in a uniform manner. Of course if you’re desperate to collect the precious toys and have a modicum of patience, the answer is probably 0. You’d just wait to buy the whole set on eBay.", "odds that Bob will win on his second roll\" - 125/1296, for \"Sue misses, Bob misses, Sue misses, Bob hits\" in that order. The problem is that that answer is the odds that Bob will win on the second turn, out of the set of ALL GAMES. What we want is the odds that Bob will win on his second turn, out of the set of GAMES THAT BOB WON. How do we know how many games Bob will win? Well, pretend that Sue and Bob roll simultaneously, but that Sue's die is counted first if it is a 6. This gives us 36 possible results. Of those, 6 are wins for Sue, 5 are wins for Bob (he loses the 6-6 tie), and 25 are \"roll again\". This means that there are 11 possible exit conditions of the loop, all equally likely, and Bob wins 5 of those. Thus, in the set of ALL GAMES, Bob will win 5/11 of them. Therefore: In all games, Bob wins 125/1296 of them on the second turn. Bob wins 5/11 of all games. Therefore, Bob wins on the second turn in (125/1286) / (5/11) = 275/1296 of all games he wins. Which is the same answer everyone else is getting. Using a faster method. -John 99.224.156.92 22:27, 11 February 2009", "many whole bags of mulch does Austin need to buy to completely cover his flower bed? (c) If each bag of mulch costs $$\\ 2.58$$ , how much will he need to spend on mulch? Write your answer to the nearest cent. Q: Tracy plans to serve punch along with the chicken wings. According to her calculatlons, two times the amount of punch needed is more than $$4$$ liters. Write an inequality for $$y$$ , the amount of punch needed to serve at Tracy's party. Q: Brad, Preston, Zack, and Steven each roll a six-sided die a different number of times, record their results, and then calculate the experimental probability of rolling a $$6$$ . If Brad rolled his die $$5$$ times, Preston rolled his die $$25$$ times, Zack rolled his die $$50$$ times, and Steven rolled his die $$15$$ times, whose experimental probability will most likely approach the theoretical probability? A. Preston B. Zack D. Steven Q: Betsey is $$11$$ years older than Waldo. Betsey's age is $$15$$ years less than three times Waldo's age. The system below models the relationship between Betsey's age (b) and waldo's age (w): $$b = w + 11$$ $$b = 3 w - 15$$ Which of the following methods is correct to find Betsey's and Waldo's age? Solve $$w + 11", "whose links you can find by date, here. Choose life. Intern Joined: 18 Aug 2017 Posts: 30 GMAT 1: 670 Q49 V33 On a certain day, a bakery produced a batch of rolls at a total produc [#permalink] ### Show Tags 23 Aug 2017, 21:16 6 1 My approach: Let x is the number of rolls made. Average cost is 300/x. 4/5 of the rolls in the batch were sold, each at a price that was 50 percent greater than the average (arithmetic mean) production cost per roll, so first day revenue = $$\\frac{4x}{5}$$ x $$\\frac{300}{x}$$ x 150% The remaining rolls in the batch were sold the next day, each at a price that was 20 percent less than the price of the day before, so second day revenue = $$\\frac{x}{5}$$ x $$\\frac{300}{x}$$ x 150% x 80% Total Revenue = $$\\frac{4x}{5}$$ x $$\\frac{300}{x}$$ x 150% + $$\\frac{x}{5}$$ x $$\\frac{300}{x}$$ x 150% x 80% = 432 (x will be canceled out) Cost = 300, so Profit = 432-300 = 132 ##### General Discussion Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2817 Re: On a certain day, a bakery produced a batch of rolls at a total produc [#permalink] ### Show Tags 09 Aug 2017, 13:32 2 3 carcass wrote: On a certain day,", "so the probability of choosing a club is $\\frac{1}{4}$ . The complement is the probability of not choosing a club, and it is $1-\\frac{1}{4}=\\frac{3}{4} \\ or \\ 75 \\%$ 3. There are 4 factors of 10: 1, 2, 5, and 10. Therefore the probability of rolling a factor of 10 on a 20-sided die is $\\frac{4}{20}$ . The complement is the probability of not rolling one of the four factors: $1-\\frac{4}{20}=\\frac{16}{20}$ 4. Let $P(F)$ be the probability of pulling a fruit-flavor, then $P(F^\\prime)$ is the probability of not pulling a fruit flavor (and getting something disgusting instead!). There are 10 fruit-flavored beans in the bag of 100 beans, so: $P(F)&=\\frac{10}{100} \\\\P(F^\\prime)&=1-P(F)=\\frac{90}{100}$ Therefore, you have a 90% probability of being unhappy with your choice. 5. The factors of ten or twelve are: 1, 2, 3, 4, 5, 6, 10, and 12. Since all six numbers from a standard die are in that set, the probability of rolling one of them is 100% or 1.0. Therefore, the probability of not rolling one of them is $1-1=0$ . #### Practice 1. If you randomly pull a single card from a standard deck, what is the probability that the card is anything other than a king? 2. What is the probability of not pulling a face card from a standard deck? 3.", "at the highest die, what will the result be on average?--Sonjaaa 11:45, 9 March 2007 (UTC) You might look at order statistic; although I don't know if that information will help you compute it. It might be easier if you just went through all the outcomes on a computer and computer the average, or if you did a lot of random trials and estimated it probabilistically. --Spoon! 11:59, 9 March 2007 (UTC) I computed it in QBasic and got 4.5 for 1d8, 5.8 for 2d8 and 6.5 for 3d8.--Sonjaaa 12:41, 9 March 2007 (UTC) For n 8-sided dice the average of the highest die is ${\\displaystyle 8-{\\frac {(1^{n}+2^{n}+...+7^{n})}{8^{n}}}}$ For two dice the average of the highest is 8-(140/64), which is 5 13/16. For three dice the average of the highest is 8-(784/512), which is 6 15/32. Gandalf61 14:18, 9 March 2007 (UTC) Gandalf's answer looks incorrect. For instance, looking at the case when n=1, his formula says the average result of one eight sided die is 7 7/8. But in fact the average of a single 8-sided die is 4.5 . Dugwiki 16:21, 9 March 2007 (UTC) How did you get 7 7/8? Looking for the case when n=1 Gandalf's formula says the average is ${\\displaystyle 8-{\\frac {(1^{1}+2^{1}+...+7^{1})}{8^{1}}}}$${\\displaystyle =8-{\\frac {(1+2+...+7)}{8}}}$${\\displaystyle =8-{\\frac {28}{8}}}$${\\displaystyle =8-3.5=4.5}$ --CiaPan 16:34, 9 March 2007", "# Golden Pancakes Alice and Bob play a game with two stacks of fluffy, golden pancakes, of sizes $p$ and $q$. They alternate turns, with Alice starting. On their turn, a player must eat some number of pancakes from the larger stack. The number they eat must be a multiple of the size of the smaller stack. Since the bottom pancake of each stack is soggy, the first player to finish a stack loses. For example, if the stacks were sized $5$ and $17$, then the current player can eat either $5,10$ or $15$ pancakes from the large stack. For which values of $p$ and $q$ does Alice have a winning strategy? Source: Mathematical Puzzles: A Connoisseur's Collection, Peter Winkler. • None, because that is a lot of pancakes, and she is going to be sick by the end no matter what. – Aggie Kidd Jul 8 '15 at 1:51 • Do I understand correctly: Does it mean that when Alice eats 15 from the second (larger stack), then there are stacks 5 and 2, so Bob has to eat 2 or 4 from the first stack? – Voitcus Jul 8 '15 at 3:52 • I hope the pancakes aren't made from real gold ;). – Bojidar Marinov Jul 8 '15 at 10:05 • In next weeks episode,", "# Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession. Suppose we roll a fair six sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession From the link below, I learned that $258$ rolls are expected to see 3 sixes appear in succession. So I'm thinking that for a same (any) number, the rolls expected would be $258/6 = 43$. But I'm unsure how to show this and whether it really is correct. How many times to roll a die before getting two consecutive sixes? For $n\\in \\{0,1,2\\}$ Let $E[n]$ denote the answer given that you are starting from a streak of $n$ consecutive rolls. The answer you want is $E=E[0]$, though you are never in state $0$ except at the start. We note $$E[2]=\\frac 16\\times 1+\\frac 56\\times \\left(E[1]+1\\right)$$ $$E[1]=\\frac 16\\times \\left(E[2]+1\\right)+\\frac 56\\times \\left(E[1]+1\\right)$$ $$E=E[0]=E[1]+1$$ this system is easily solved and, barring error (always possible), yields $$\\boxed {E=43}$$ • This is very elegant (+1) - great answer!! – Satish Ramanathan Aug 31 '17 at 16:38 • One issue that isn't really an issue: everything needs to be finite for this to work (cf optional stopping theorem) – cats Aug 31 '17 at", "# Homework Help: Probablity problem using series 1. Apr 13, 2014 ### toothpaste666 1. The problem statement, all variables and given/known data Professors Bob and Fred cannot decide who should buy a new kite. In order to decide, they play a game. They take it is turns to throw a (standard, six-sided) die, with Professor Bob going fi rst. The winner of the game is the fi rst one to throw a 4. For example, Professor Bob wins if he throws a 4 immediately or the results are non-4 for Bob, non-4 for Fred, 4 for Bob or non-4 for Bob, non-4 for Fred, non-4 for Bob, non-4 for Fred, 4 for Bob and so on. Find the probability that Professor Bob wins. (Hint: the calculation requires a certain type of series) 2. Relevant equations $\\frac{a}{1-r}$ 3. The attempt at a solution Bobs first turn he has 1/6 chance of rolling a 4. If it goes to his second turn that means that he did not throw a 4 the first turn and also that Fred did not throw a 4 on his turn so his chances of rolling a 4 on the second turn are 5/6 * 5/6 * 1/6. To analyze the chances of Bob rolling a 4 on his first 3 turns: $\\frac{1}{6} + (\\frac{5}{6}", "# Maximizing score in number-guessing game. This is inspired by a puzzle (related to the two-envelopes problem) that I've seen in several places, including unbounded generalizations. The basic premise is that Alice chooses two real numbers from $[0,1]$ uniformly at random, and writes them down on slips of paper. Bob chooses one slip of paper at random, reads the number, and may either keep that slip or switch for the other. Bob's score is the number he ends up with, and our task is to maximize his expected value. The standard solution is for Bob to choose his own real number uniformly at random, and switch if the number he picks is greater than what's on the slip he holds. However, this is not the only solution. Bob has one piece of data, $x$, and the use of a random number generator. He can keep his initial slip with some probability $f(x)$ and swap with probability $1-f(x)$. Call this function $f(x)$ his strategy; the standard solution corresponds to strategy $f(x)=x$. We calculate his expected winnings as $$E=\\iint_{[0,1]\\times[0,1]} f(x)x+(1-f(x))y\\, dx dy=\\int_0^1f(x)(x-0.5)+0.5\\,dx$$ Trying $f(x)=x$ gives $E=0.58\\overline{3}$, while $f(x)=x^{1.4}$ gives $E\\approx 0.585784$. UPDATE: In the comments Calvin Lin suggests $f(x)=\\begin{cases}0&x<0.5\\\\1&x>0.5\\end{cases}$, with $E=0.6125$. What's the highest $E$ Bob can achieve, and what is the corresponding $f$? The maximum, if Bob knew what", "affect when Bob's wins will come, even if we take only the subset of games where Bob won. -John 74.14.228.170 17:27, 12 February 2009 (UTC) To the 5/36 people: You are understanding the problem and it's wording correctly, the problem with your argument is the false (but reasonable sounding assumption) that you can ignore Sue's rolls just because she lost. If this doesn't make intuitive since, consider the following modified problem: Let's say instead of rolling one dice, sue rolls 100 dice and picks the best roll. Let's also assume that we don't know who won. Since Sue goes first, the odds of Bob winning on the first round are (5/6)^100*1/6=2.01244558 × 10-9 while the odds of bob winning on the second round are (5/6)^200*5/6*1/6=2.0249686 × 10-17. This means bob is roughly 100,000,000 times more likely to win on his first roll than his second roll (intuitively, he might have dodged Sue's 100 dice once but it's pretty inconceivable that he dodged them again). Now, let's assume we know Bob won. This doesn't change the fact that he's still 100,000,000 times more likely to win on his first roll than his second roll. This means that assuming Bob won, the probability that he won on his second roll<.0000001%. Now, since the only difference between my alternative problem at", "# Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession. Suppose we roll a fair six sided die repeatedly. Find the expected number of rolls required to see $$3$$ of the same number in succession From the link below, I learned that $$258$$ rolls are expected to see 3 sixes appear in succession. So I'm thinking that for a same (any) number, the rolls expected would be $$258/6 = 43$$. But I'm unsure how to show this and whether it really is correct. How many times to roll a die before getting two consecutive sixes? For $n\\in \\{0,1,2\\}$ Let $E[n]$ denote the answer given that you are starting from a streak of $n$ consecutive rolls. The answer you want is $E=E[0]$, though you are never in state $0$ except at the start. We note $$E[2]=\\frac 16\\times 1+\\frac 56\\times \\left(E[1]+1\\right)$$ $$E[1]=\\frac 16\\times \\left(E[2]+1\\right)+\\frac 56\\times \\left(E[1]+1\\right)$$ $$E=E[0]=E[1]+1$$ this system is easily solved and, barring error (always possible), yields $$\\boxed {E=43}$$ • This is very elegant (+1) - great answer!! Aug 31, 2017 at 16:38 • One issue that isn't really an issue: everything needs to be finite for this to work (cf optional stopping theorem) – cats Aug 31, 2017 at 18:32 • This" ]

[ "the referral of any bill or resolution to any Standing Committee, the Rules Committee shall hear the same, resolve the issue and report to the Senate within one (1) legislative day its decision which decision may be overruled by a vote of twenty-nine (29) Senators. The several committees shall have such powers and duties as provided for in these rules. It shall not be in order for any committee to consider any proposed committee amendment (other than a technical, clerical, or conforming amendment) which contains any significant matter not within the jurisdiction of the committee proposing such amendment. B. Committee Composition The membership of the above listed committees shall be as follows: The Committees on Judiciary and Finance shall each have twenty-three (23) members. All other standing Committees except the Committee on Ethics and the Committee on Invitations shall have seventeen (17) members. The Committee on Ethics shall be composed of ten (10) members. Of the ten (10) members selecting a seat, five (5) shall be members of the majority party and five (5) shall be members of the minority party. The Committee on Invitations shall be limited to not more than eleven (11) members. The total membership of each Standing Committee shall be composed of members of the two major political parties in proportion to the number", "1. Politics probability question A senate committee consists of five Democrats and five Republicans. If four people from the committee are selected at random, what is the probability that all four are Democrats, given that at least one of them is a Democrat? The probability that all four are Democrats is? 2. Re: Politics probability question you want to show some work for this one?", "seats on committees in proportion to its overall strength. Most committee work is performed by 16 standing committees, each of which has jurisdiction over a field such as finance or foreign relations. Each standing committee may consider, amend, and report bills that fall under its jurisdiction. Furthermore, each standing committee considers presidential nominations to offices related to its jurisdiction. (For instance, the Judiciary Committee considers nominees for judgeships, and the Foreign Relations Committee considers nominees for positions in the Department of State.) Committees may block nominees and impede bills from reaching the floor of the Senate. Standing committees also oversee the departments and agencies of the executive branch. In discharging their duties, standing committees have the power to hold hearings and to subpoena witnesses and evidence. The Senate also has several committees that are not considered standing committees. Such bodies are generally known as select or special committees; examples include the Select Committee on Ethics and the Special Committee on Aging. Legislation is referred to some of these committees, although the bulk of legislative work is performed by the standing committees. Committees may be established on an ad hoc basis for specific purposes; for instance, the Senate Watergate Committee was a special committee created to investigate the Watergate scandal. Such temporary committees cease to exist after fulfilling their", "persons as floor leaders for the parties during the sessions of the Senate Bills The Committee System Congress conducts most of its work through two major types of committees. The most common are standing committees, which are usually permanent. Each standing committee takes responsibility for a particular subject area. Congress creates temporary committees, usually called select or special committees, to write bills on a particular topic or to conduct investigations. For example, both chambers have select committees to authorize and oversee the nation’s intelligence-gathering operations, including the Central Intelligence Agency (CIA) and the National Security Agency. Most standing committees have the power to authorize government action but cannot commit funding to implement the policy. Before most laws can be carried out, they must receive an appropriation of funding, processed and brought to the floor by the House and Senate appropriations committees. This occurs during Congress’s annual budget process. The amount of money in the budget depends on the level of taxes and other revenues brought in by law processed by the Senate Finance Committee and the House Ways and Means Committee. Much of a committee’s work is handled by a sizable number of professional staff assistants. In 1999 there were nearly 1,300 House committee aides and more than 900 Senate committee aides. Some of the staff are experts", "himself is hated by a completely different set of 3 other senators. Thus, given one senator, there may be a maximum of 6 other senators whom he cannot work with. If we have a minimum of 7 committees, there should be at least one committee suitable for the senator $S$ after the assignment of the 6 conflicting senators. ACS WASC ACCREDITED SCHOOL Our Team Our History Jobs", "and you get 2 different combinations, and yet possible combinations are: ax ay bx by it's an unfortunate trick of the problem 4. Originally Posted by Rimas There are 100 members in the U.s. Senate(2 from each state)In how many ways can a committee of 5 senators be formed if no state may be represened more than once. There are combin(50,5) to chose the states. There are 2^5 ways the chose the members. There are combin(50,5)*2^5 ways to do this: 39200 ways.", "+0 # Probability 0 13 1 A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. A subcommittee of 3 members is randomly chosen. What is the probability that the subcommittee consists of 1 Republican and 2 Democrats? Aug 5, 2022 #1 +2275 +1 There are $${13 \\choose 3} = 286$$ ways to choose the committee Of these, there are $${5 \\choose 1} = 5$$ ways to choose a Republican and $${6 \\choose 2} = 15$$ ways to choose a Democrat, making for $$5 \\times 15 = 75$$ choices. So, the probability is $$\\color{brown}\\boxed{75 \\over 286}$$ Aug 5, 2022", "three of whom must be members of the House of Representatives appointed by the Speaker at least one of whom must be a member of the minority party; three of whom must be members of the Senate appointed by the President pro Tempore at least one of whom must be a member of the minority party; and three of whom shall be appointed by the governor from the general public at large, of which one must represent businesses with fewer than fifty employees and one of whom must represent businesses with fewer than five hundred employees. A member of the general public appointed by the governor may not be a member of the General Assembly. (B) The committee must meet as soon as practicable after appointment and organize itself by electing one of its members as chairman and other officers as the committee considers necessary. Afterward, the committee at least annually shall meet and at the call of the chairman or a majority of the members. A quorum consists of five members. (C) Unless the committee finds a person qualified to serve as the executive director of the Department of Workforce, the person may not be appointed. (D) A member of the committee that misses three consecutive scheduled meetings at which a quorum is present must be removed", "different committees each composed of 2 Republicans 14 11 Nov 2009, 07:33 10 In a survey of students, each student selected from a list 15 16 Jun 2008, 20:23 Display posts from previous: Sort by", "# Question A political committee consists of seven Democrats and five Republicans. A subcommittee of six people needs to be formed from this group. Determine the probability that this subcommittee will consist of the following: a. Two Democrats and four Republicans if they were randomly selected b. Four Democrats and two Republicans if they were randomly selected c. Three Democrats and three Republicans if they were randomly selected d. Calculate the mean and standard deviation for the distribution which defines selecting a Republican as a success. e. Verify these results with Excel. Sales1 Views36", "of the Appropriations Committee consists of 19 members. Disregarding party affiliation or any special seats on the Subcommittee, how many different 19-member subcommittees may be chosen from among the 29 Senators on the Appropriations Committee? In the preceding Try It problem we assumed that the 19 members of the Defense Subcommittee were chosen without regard to party affiliation. In reality this would never happen: if Republicans are in the majority they would never let a majority of Democrats sit on (and thus control) any subcommittee. (The same of course would be true if the Democrats were in control.) So let’s consider the problem again, in a slightly more complicated form: ### Example The United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee? This example is worked through below.", "+0 # A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. 0 33 1 +247 A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. A subcommittee of 3 members is randomly chosen. What is the probability that the subcommittee consists of three Republicans? Darkside Sep 7, 2018 #1 +2012 0 first off we regroup things so that we have 5 Republicans and 8 non-Republicans this gets us $$P[R]=\\dfrac{5}{13},~P[!R] = \\dfrac{8}{13}$$ now we have a simple Binomial distribution with parameters $$n=3,~p=\\dfrac{5}{13}$$ $$P[3] = \\left(\\dfrac{5}{13}\\right)^3 = \\dfrac{125}{2197}$$ Rom Sep 7, 2018", "+0 # Help probability 0 318 3 A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. A subcommittee of 3 members is randomly chosen. What is the probability that the subcommittee consists of 1 Republican and 2 Democrats? Jun 5, 2021 #1 +9 +1 A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. A subcommittee of 3 members is randomly chosen. What is the probability that the subcommittee consists of 1 Republican and 2 Democrats? Out of 5+6+2=13 total members we must choose a 3 member subcommittee. There is $$\\binom{13}{3}=286$$ ways to do this. There are 5 ways to choose the Republican and $$6\\cdot5=30$$ ways to choose the Democrats. So, the probability is $$\\frac{35}{286}$$. Please correct me if I'm wrong, I'm not the best at committee problems. Jun 5, 2021 #2 +118203 +2 Nice try Supernova :) But you messed up a bit on the numerator. It is multiply not plus. Becasue.. for EVERY pair of democrats, there are 5 choices of republican - hence multiply. also Choosing R1 then R2 is the same as choosing R2 then R1 so you would have to halve that part of your answer. Alternatively, you could say there are 6C2 ways to chose the democrats and 5 ways to chose the republican so the numerator", "shall happen) shall act as chair. Rank shall be determined by the order members are named in resolutions electing them to the committee. In the case of a vacancy in the elected chair of a committee, the House shall elect another chair. (2) Except in the case of the Committee on Rules, a member of a standing committee may not serve as chair of the same standing committee, or of the same subcommittee of a standing committee, during more than three consecutive Congresses (disregarding for this purpose any service for less than a full session in a Congress). (d)(1) Except as permitted by subparagraph (2), a committee may have not more than five subcommittees. (2) A committee that maintains a subcommittee on oversight may have not more than six subcommittees. The Committee on Appropriations may have not more than 13 subcommittees. The Committee on Oversight and Government Reform may have not more than seven subcommittees. (e) The House shall fill a vacancy on a standing committee by election on the nomination of the respective party caucus or conference. Expense resolutions 6. (a) Whenever a committee, commission, or other entity (other than the Committee on Appropriations) is granted authorization for the payment of its expenses (including staff salaries) for a Congress, such authorization initially shall be procured by one", "the seven numbers?​... ### Which group could be elected to the senate for life? Which group could be elected to the senate for life?... ### 1/2 + 3/8 What the answer super easy points 1/2 + 3/8 What the answer super easy points... -- 0.015511--", "The US Senate has 100 members. After a certain election, there were 4 more Democrats than Republicans with no other parties represented. How many members of each party were there in the Senate? May 1, 2018 48 Republicans and 52 Democrats. Explanation: Let the number of Republicans be $x$. There are $x + 4$ Democrats. Thus, $x + \\left(x + 4\\right) = 100$ $\\implies 2 x = 100 - 4$ $\\implies x = \\frac{96}{2} = 48$ There are 48 + 4 = 52 Democrats.", "+0 # help 0 295 1 A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. A subcommittee of 3 members is randomly chosen. What is the probability that the subcommittee consists of 1 Republican, 1 Democrat, and 1 Independent? Oct 11, 2018 #1 +6115 +2 \\(P[\\{R,D,I\\}] = \\dfrac{\\dbinom{5}{1}\\dbinom{6}{1}\\dbinom{2}{1}}{\\dbinom{13}{3}}=\\dfrac{30}{143}\\) . Oct 11, 2018 edited by Rom Oct 11, 2018 edited by Rom Oct 11, 2018 #1 +6115 +2", "the public should be made to focus upon. This is in opposition to an idea in which the number of Senators should grow roughly in proportion to the size of the population itself, in order to ensure each faction of some fixed size at least one representative in the Senate. Now, the simple fact is that, no matter how much you would like the public to devote sufficient time to be able to understand all of the subtleties of politics, that people will refuse (and rightly so) to spend too much time on this, and so the media will simplify the world into a small number of factions. In the United States today, by looking at how the media talks about political factions, I guess they divide domestic politics into about 6 factions: (religious Republicans, other Republicans, anti-corporate Democrats, other Democrats, greens, \"other\"). So let's say that the magic number is between 5 and 7. Since we will allow each of these factions (except \"other\") to field 2 reps, we have a body of at least 9 to 13. Now, if even the small factions such as the greens have 2 reps, surely some of the larger ones will have more. So we want a Senate of about 13 to 24 members. Nth-root formulas are one alterantive, but", "Combinatorial Analysis HW Help blackhaze5 New Member There are 100 U.S Senators, two from each of the 50 States. If 12 Senators are randomly chosen to form a committee, what is the probabilty that they are alll from different states? If 12 senators are chosen at random from the 100 senators, what is the probabilty that neither of the two senators from Florida is in the committee? I have absolutely no idea how to begin this problem.", "control.) So let's consider the problem again, in a slightly more complicated form: ##### Example 33 The United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee? Solution In this case we need to choose 10 of the 15 Republicans and 9 of the 14 Democrats. There are $$_{15} C_{10}=3003$$ ways to choose the 10 Republicans and $$_{14} C_{9}=2002$$ ways to choose the 9 Democrats. But now what? How do we finish the problem? Suppose we listed all of the possible 10-member Republican groups on 3003 slips of red paper and all of the possible 9-member Democratic groups on 2002 slips of blue paper. How many ways can we choose one red slip and one blue slip? This is a job for the Basic Counting Rule! We are simply making one choice from the first category and one choice from the second category, just like in the restaurant menu problems from earlier. There must be $$3003 \\cdot 2002=6,012,006$$ possible ways of selecting the members of the Defense Subcommittee. ## Probability using Permutations and Combinations We can use permutations and combinations" ]

[ "ways. Then the 6 people can be lined up in $6! = 720$ ways. Answer: . $70 \\times 720 \\:=\\:50,\\!400$", "final arrangement can be done, $$= 5P_3 \\times 4!$$ $$= \\frac{5!}{(5 - 3)!} \\times 4!$$ $$= \\frac{5!}{2!} \\times 4!$$ $$= \\frac{5 \\times 4 \\times 3 \\times 2!}{2!} \\times 24$$ $$= 1440 \\ Ways$$ 1. How many ways 8 persons can be sit around a dinning table in 8 chairs? 1. 5040 2. 4152 3. 4020 4. 3822 Solution: For circular permutation total number of ways of arrangement = $$(n - 1)!$$ So, here 8 persons can be sit around a dinning table in 8 chairs by $$(8 - 1)!$$ ways.$$= 7!$$ $$= 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1$$ $$= 5040 \\ Ways$$ 1. How many three digit numbers can be formed by using 0, 1, 2, 3, 4, if repetetion is not allowed? 1. 32 2. 36 3. 42 4. 48 Solution: Zero can not come in the first place, only 4 numbers (1, 2, 3, 4) can be come in first place. So rest of the numbers can be come in next places. The three digit numbers can be formed, $$= 4 \\times 4 \\times 3$$ $$= 48$$ 1. Find the number of ways the arrangement of 9 people in 4 chairs? 1. 3024 2. 2856 3. 2405 4. 2425 Solution: $$nP_r = \\frac{n!}{(n - r)!}$$ $$9P_4 = \\frac{9!}{(9 -", "1. ## Line 'em up! There are 34 people in line to get into a theater. Admission cost is 5 dollars. Of the 34 people, 28 have only a 5 dollar bill and 6 have only a 10 dollar bill. To complicate matters, the ticket office has an empty cash register. In how many ways can the people line up so that change is available when needed? Example with 4 people, 2 (a and b) with 5's and 2 (c and d) with 10's: a b c d a b d c b a c d b a d c a c b d a d b c b c a d b d a c So 8 ways... 2. ## Re: Line 'em up! You only need to look at the first 11 positions. There are 34! arrangements without restriction. There are $6\\times 33!$ arrangements that fail because the first person has only a 10. There are $28\\times 6\\times 5 \\times 31!$ ways that the first person has a 5 and the second and third both have 10s. Next we want 5,10,5,10,10,anything (because any other arrangement that fails in the first 5 people was already counted above) $28\\times 6\\times 27 \\times 5 \\times 4 \\times 29!$ We see the pattern: $$\\displaystyle 34!- \\sum_{n=1}^6\\dfrac{28!}{(29-n)!}\\cdot \\dfrac{6!}{(6-n)!}\\cdot (35-2n)!$$ The chances", "\\right)!}$, we get $^{5}{{C}_{2}}=\\dfrac{5!}{3!2!}=\\dfrac{120}{12}=10$. Hence the total number of ways of seating of four boys and two girls so that the girls are not together = 480. Hence P(E’) $=\\dfrac{480}{720}=\\dfrac{2}{3}$. Hence we have P(E) $=\\dfrac{1}{3}$.", "one can be solved in more than 1 way: Sisters sit separately: 1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24 Or 2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case=8. Total=24+8=32. Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32 2. What is the probability that a 3-digit positive integer picked at random will have one or more \"7\" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10 Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25 3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere? (A) $$10(\\sqrt{3}- 1)$$ (B) $$5$$ (C) $$10(\\sqrt{2} - 1)$$ (D) $$5(\\sqrt{3} - 1)$$ (E) $$5(\\sqrt{2} - 1)$$ Shortest distance=(diagonal of cube-diameter of sphere)/2= $$\\frac{10*\\sqrt{3}-10}{2}=5(\\sqrt{3}-1)$$ 4. A contractor estimated that his 10-man crew could complete the construction in 110", "879 , 897 , 978 , 987$ Combined with the 6 ways to arrange the groups, we now have ${6}^{4}$ valid permutations so far. And since we're at a round table, we allow for the 3 arrangements where first group could be \"half\" on one end and \"half\" on the other: $\\text{A A A G G G I I I}$ $\\text{A A G G G I I I A}$ $\\text{A G G G I I I A A}$ The number of total ways to get all 3 groups to be seated together is ${6}^{4} \\times 3.$ The number of random ways to arrange all 9 people is 9! The probability of randomly choosing one of the \"successful\" ways is then $\\frac{6 \\times 6 \\times 6 \\times 6 \\times 3}{9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}$ $= \\frac{3}{2 \\times 7 \\times 5 \\times 4}$ $= \\frac{3}{280}$", "number of permutations of D in chair 1 or 5 minus the permutations of both. So, there are $2\\cdot 1\\cdot 3\\cdot 2\\cdot 1 = 12$ permutations where both occur. $120-(48+24-12) = 60$. You have removed twice the cases where A sits in 3 and D sits in 1 or 5. You need to add them back in once. This is the inclusion-exclusion principle. Alternately you can remove all the cases where A sits in 3, then remove the ones where D sits in 1 or 5 but A doesn't sit in 3.", "on a circle, then how many chords can be drawn by joning these points? 1. 10 2. 12 3. 15 4. 16 Solution: Number of chords can be drawn by $$5C_2$$ ways.$$5C_2 = \\frac{5!}{(5 - 2)! \\times 2!}$$ $$= \\frac{5!}{3! \\times 2!}$$ $$= 10 \\ Chords$$ 1. Six people A, B, C, D, E, and F occupy seats in a row such that C and D sit next to each other. In how many ways can these six people can sit? 1. 180 2. 212 3. 226 4. 240 Solution: If we assume C and D as one, then these people can sit by $$5!$$ ways, but C and D can also sit 2 ways within their group. Hence, $$= 5! \\times 2$$ $$= 240 \\ Ways$$ 1. Find the value of $$8C_5 + 7C_4 + 6C_3$$ 1. 111 2. 112 3. 113 4. 110 Solution: $$8C_5 = \\frac{8!}{(8 - 5)! \\times 5!}$$ $$= \\frac{8!}{3! \\times 5!}$$ $$= 56$$ $$7C_4 = \\frac{7!}{(7 - 4)! \\times 4!}$$ $$= \\frac{7!}{3! \\times 4!}$$ $$= 35$$ $$6C_3 = \\frac{6!}{(6 - 3)! \\times 3!}$$ $$= \\frac{6!}{3! \\times 3!}$$ $$= 20$$ Hence,$$= 56 + 35 + 20$$ $$= 111$$ 1. In how many different ways can the letters of the word AUSTRALIA be arranged such that the vowels always come together? 1. 2200 2.", "$D$ together? Duct-tape $C$ and $A$ together. There are 2 ways: . $\\boxed{CA}$ or $\\boxed{AC}$ Duct-tape $N$ and $D$ together. There are 2 ways: . $\\boxed{ND}$ or $\\boxed{DN}$ We have six \"letters\" to arrange: . $\\{A,E,L,R,\\boxed{CA},\\,\\boxed{ND}\\, \\}$ . . They can be arranged in: ${\\color{blue}6!}$ ways. Hence, there are: . $2\\times 2 \\times 6! \\:=\\:2880$ ways to have $C,A$ and $N,D$ together. Therefore, there are: . $10,080 - 2,880 \\;=\\;\\boxed{7,200}$ ways . . in which $C$ and $A$ are together, but $N$ and $D$ are not.", "## Precalculus (10th Edition) $24$ The number of ways arranging $n$ distinct objects is $n!=n\\cdot(n-1)\\cdot(n-2)\\cdot...\\cdot3\\cdot2\\cdot1.$ Hence, the number of ways of arranging $4$ people in a line is: $4!=4\\cdot3\\cdot2\\cdot1=24.$", "and D are to be seated together, we have to consider them as one unit. So the arrangement is to be done in $4$ ways: $(4 – 1)!=3!=6$ Moreover, persons A and D can shuffle their positions in $2$ ways. So the arrangements can be made in $6 \\times 2 = 12$ ways. In the second part, the number of ways two persons can sit is $12$ and the total number of ways was $24$. So the total number of ways in which persons C and E must not sit together is: $=24 – 12=12\\space Ways$", "different arrangements are possible if; a) The boys and girls alternate. There are 2 possible arrangements: . $BGBGBGBG\\,\\text{ and }\\,GBGBGBGB.$ The four boys can be placed in 4! ways. The four girls can be placed in 4! ways. Therefore, there are: . $2 \\times 4! \\times 4! \\;=\\;\\boxed{1152}$ ways. b) 2 particular girls wish to stand together. Suppose the two girls are $X$ and $Y.$ Duct-tape them together. Note that there are 2 possible orders: . $XY\\,\\text{ or }\\,YX.$ Now we have seven \"people\" to arrange: . $\\boxed{XY}\\,,G,G,B,B,B,B$ . . and they can be arranged in ${\\color{blue}7!}$ ways. Therefore, there are: . $2 \\times 71\\;=\\;\\boxed{10,080}$ ways. c) All the boys stand together. Duct-tape the four boys together. . . Note that there are 4! possible orders. Now we have five \"people\" to arrange. . . There are $5!$ ways. Therefore, there are: . $4! \\times 5! \\:=\\:\\boxed{2880}$ ways. d) Find the probability that 3 particular people will be together if the queue forms randomly. First of all, there are 8! possible arrangements. Suppose the three people are $X, Y\\text{ and }Z.$ Duct-tape them together: . $\\boxed{XYZ}$ . . Note that there are 3! orderings. Now we have six \"people\" to arrange, . . and there are 6! ways. Hence, there are: . ${\\color{red}3!\\times6!}$ ways for $X,Y,Z$ to be", "can be two possible cases. Case 1:- Person A is given 2 balls and person B is given 8 balls So, number of ways for this distribution will be $\\dfrac{{10!}}{{\\left( {2!} \\right) \\times \\left( {8!} \\right)}} = \\dfrac{{10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}}{{\\left( {2 \\times 1} \\right) \\times \\left( {8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1} \\right)}} = 45$ Case 2:- Person A is given 8 balls and person B is given two balls. So, number of ways for this distribution will be $\\dfrac{{10!}}{{\\left( {8!} \\right) \\times \\left( {2!} \\right)}} = \\dfrac{{10 \\times 9 \\times 8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1}}{{\\left( {8 \\times 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1} \\right) \\times \\left( {2 \\times 1} \\right)}} = 45$ So, the total number of ways to distribute 10 balls such that one person gets 2 and the other gets 8 balls will be equal to the sum of ways for both cases. Hence, the total number of ways will be 45 + 45 = 90. Note: Whenever we come up with this type of problem then if all objects are identical and", "+0 # How to solve a) and b)? 0 97 3 how to solve these I am perplexed. Thanks in advance. Jun 19, 2022 #1 0 is it 10!*8!*2 = 292,626,432,000 Jun 19, 2022 #2 +2448 0 a) $$10!$$ ways to arrange the boys and $$8!$$ ways to order the girls. There are also 2 ways to order the groups (10 boys, 8 girls; 8 girls, 10 boys). This makes for $$8! \\times 10! \\times 2 = \\color{brown}\\boxed{292,626,432,000}$$ ways. b) $$8!$$ ways to order the girls. Now, treat the 8 girls as 1 person. Now, there are $$11!$$ ways to order everything else. This makes for $$8! \\times 11! = \\color{brown}\\boxed{1,609,445,376,000}$$ ways. Jun 19, 2022 #3 +1 thanks Guest Jun 19, 2022", "in ways. ∴ Total number of ways = (iii) There is only one set of two alike letters, which can be selected in 2C1 ways. Now, from the remaining 5 letters, 2 letters can be chosen in 5C2 ways. Thus, 2 alike letters and 2 different letters can be selected in 2C1$×$5C2 = 20 ways. Now, the letters of each group can be arranged in $\\frac{4!}{2!}$ ways. ∴ Total number of ways = $20×\\frac{4!}{2!}=240$ (iv) There are 6 different letters A, D, M,B, O and R. So, the number of ways of selecting 4 letters is 6C4, i.e. 15, and these letters can be arranged in 4! ways. ∴ Total number of ways = 15 $×$ 4! = 360 ∴ Total number of ways = 20 + 6 + 240 + 360 = 626 #### Question 9: A business man hosts a dinner to 21 guests. He is having 2 round tables which can accommodate 15 and 6 persons each. In how many ways can he arrange the guests? A businessman hosts a dinner for 21 guests. 15 people can be accommodated at one table in 21C15 ways. They can arrange themselves in $\\left(15-1\\right)!=14!$ ways. The remaining 6 people can be accommodated at another table in 6C6 ways. They can arrange themselves in $\\left(6-1\\right)!=5!$ ways. ∴ Total number", "program) and it is wrong the answer they were looking for is $11!$. The problem is i have no idea why The Factorial Rule states that for n different items, there are n! arrangements. 5. The answer is $11!$. Remember that this is a bracelet, not a string. You can place n things around a circle in $(n-1)!$ ways. 6. this is super late because i had to go somewhere but... oh!!! I get it i swear you just add in a formula that relates and things always begin to click with math. Thanks for the help. 7. Originally Posted by wingless The answer is $11!$. Remember that this is a bracelet, not a string. You can place n things around a circle in $(n-1)!$ ways. Wingless, that is a good point. In that case, the 12 rotations around the bracelet of $11!$ arrangements is removed. IMO, the question was lacking details, but this is surely what the creators of the question were going for. 8. For those acquainted with Permutations and Combinations, . . but unfamiliar with the \"Circular Table Problem\", . . this is a classic \"trick question.\" There are six people to be seated in six chairs set around a circular table. In how many ways can they be seated? We must understand that a", "the other 10 people. . . There are: $_{10}C_5 = 252$ ways. The remaing 5 people are assigned to the 2 Doors and 4 Floors. . . There are: $\\begin{pmatrix}6\\\\2,4\\end{pmatrix} = 15$ ways. Hence, there are: $252 \\times 15 = \\boxed{3780}$ ways that John can be a Floater. Therefore, there are: . $2100 + 3360 + 3780 \\:=$ 9240 ways to assign the positions. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ There are: $\\begin{pmatrix}12\\\\2,4,6\\end{pmatrix} = 13,860$ assignments with no restrictions. Let's count the ways that John and Tom are together. Duct-tape them together and we have 11 \"people\" to assign. There are three cases . . . $JT$ is at the Door. Then there are $\\begin{pmatrix}10\\\\4,6\\end{pmatrix} = 210$ ways to assign the other ten people. $JT$ is on the Floor. Then there are $\\begin{pmatrix}10\\\\2,2,6\\end{pmatrix} = 1260$ ways to assign the other ten people. $JT$ is a Floater. Then there are $\\begin{pmatrix}10\\\\2,4,4\\end{pmatrix} = 3150$ ways to assign the other ten people. Hence, there are $210 + 1260 + 3150 = 4620$ ways for John and Tom to serve together. Therefore, there are $13,860 - 4620 \\,=$ 9240 ways that John and Tom do not serve together.", "ways that only leave $1\\times1$ gaps (all larger gaps have to be filled). Now, count the number of empty spots in each arrangement and add them up for all arrangements. You will get 2019. - 1 year, 9 months ago 2019 alsp happens to be the sum of the first 22 perfect powers. - 1 year, 9 months ago", "seated at a circular table so that no 2 women sits together? Options 1. 414 2. 120 3. 240 4. 144 Solution: 4 men may be seated in 3! ways, leaving one seat empty. Then at remaining 4 seats, 4 women can sit in 4! ways. = 3! x 4! = 3 x 2 x 1 x 4 x 3 x 2 x 1 = 6 x 24 = 144 Correct option: 4 ### Type 2: When clockwise and anticlockwise Question 1. How many different garlands can be made using 10 flowers of different colors? Options 1. 181440 2. 362880 3. 145690 4. 5040 Solution: Number of arrangements possible = $\\frac{1}{2}$ × (n-1) ! = $\\frac{1}{2}$ × (10-1) ! = $\\frac{1}{2}$ × 9 ! = $\\frac{1}{2}$ × 362880 = 181440 Correct option: 1 Question 2. How many necklace of 10 beads each can be made from 20 beads of different colors? Options 1. $\\frac{10!}{19^2}$ 2. $\\frac{{19!}^2}{10!}$ 3. $\\frac{19!}{19^2}$ 4. $\\frac{10!}{10^2}$ Solution: In case of necklace the clockwise or anticlockwise arrangements are not different. Therefore, the required ways = $\\frac{^{20}P_{10}}{10 ×2}$ = $\\frac{20! }{10! × 10 ×2}$ = $\\frac{19!}{10!}$ Now the beads can be arranged in the Circular Fashion in (20-1) = 19 ! ways Required number of ways = $\\frac{19!}{10!} \\times 19!$ = $\\frac{{19!}^2}{10!}$ Correct Option : 2", "Then we have 10 \"people\" to arrange: . $\\boxed{ABC}\\;D\\;E\\;F\\;G\\;H\\;I\\;J\\;K\\;L$ . . and there are $10!$ ways. Therefore, there are: . $10! \\times 6$ possible line-ups. • January 19th 2010, 08:33 AM ANDS! Oh man I was sitting here for like 10 minutes trying to figure out just how the hell they managed to get that. I had the exact same thought, \"Why the heck would you divide by the permutation of the boys and not multiply!\"" ]

[ "$$n$$ kindergarteners and $$k$$ 5th graders, how many ways are their of pairing up students? If you want to appeal to an older audience, maybe we are pairing up tutors with students for individual instruction? This is totally not what you want, but still an interesting problem: how many injective functions exist from a set with $$n$$ elements to a set with $$k$$ elements? How about surjective functions? How does this change with $$n$$ and $$k$$? A few possibilities off the top of my head: Students and chairs. How many ways are there for $$n$$ students to sit in $$k$$ chairs. The game of musical chairs might be fun to play around with. One can also consider natural restrictions, such as myopic students who need to sit near the front. Replace \"men\" and \"women\" with faculty from different departments. For example, you might need a curriculum committee made up of mathematicians and physicists. In the senate, there are two senators from each state. How many ways are there of making a committee of 6 senators, no two of which are from the same state? This really only works in the US, but I am sure that there are similar phrasings that could work elsewhere. One could also consider party affiliation here, which could lead to more interesting problems.", "the referral of any bill or resolution to any Standing Committee, the Rules Committee shall hear the same, resolve the issue and report to the Senate within one (1) legislative day its decision which decision may be overruled by a vote of twenty-nine (29) Senators. The several committees shall have such powers and duties as provided for in these rules. It shall not be in order for any committee to consider any proposed committee amendment (other than a technical, clerical, or conforming amendment) which contains any significant matter not within the jurisdiction of the committee proposing such amendment. B. Committee Composition The membership of the above listed committees shall be as follows: The Committees on Judiciary and Finance shall each have twenty-three (23) members. All other standing Committees except the Committee on Ethics and the Committee on Invitations shall have seventeen (17) members. The Committee on Ethics shall be composed of ten (10) members. Of the ten (10) members selecting a seat, five (5) shall be members of the majority party and five (5) shall be members of the minority party. The Committee on Invitations shall be limited to not more than eleven (11) members. The total membership of each Standing Committee shall be composed of members of the two major political parties in proportion to the number", "+0 # Help probability 0 318 3 A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. A subcommittee of 3 members is randomly chosen. What is the probability that the subcommittee consists of 1 Republican and 2 Democrats? Jun 5, 2021 #1 +9 +1 A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. A subcommittee of 3 members is randomly chosen. What is the probability that the subcommittee consists of 1 Republican and 2 Democrats? Out of 5+6+2=13 total members we must choose a 3 member subcommittee. There is $$\\binom{13}{3}=286$$ ways to do this. There are 5 ways to choose the Republican and $$6\\cdot5=30$$ ways to choose the Democrats. So, the probability is $$\\frac{35}{286}$$. Please correct me if I'm wrong, I'm not the best at committee problems. Jun 5, 2021 #2 +118203 +2 Nice try Supernova :) But you messed up a bit on the numerator. It is multiply not plus. Becasue.. for EVERY pair of democrats, there are 5 choices of republican - hence multiply. also Choosing R1 then R2 is the same as choosing R2 then R1 so you would have to halve that part of your answer. Alternatively, you could say there are 6C2 ways to chose the democrats and 5 ways to chose the republican so the numerator", "+0 # Probability 0 13 1 A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. A subcommittee of 3 members is randomly chosen. What is the probability that the subcommittee consists of 1 Republican and 2 Democrats? Aug 5, 2022 #1 +2275 +1 There are $${13 \\choose 3} = 286$$ ways to choose the committee Of these, there are $${5 \\choose 1} = 5$$ ways to choose a Republican and $${6 \\choose 2} = 15$$ ways to choose a Democrat, making for $$5 \\times 15 = 75$$ choices. So, the probability is $$\\color{brown}\\boxed{75 \\over 286}$$ Aug 5, 2022", "the number of seats available to members of each party. The Clerk of the Senate shall then call the roll twice in order of continuous service. Each member, upon his or her name being called during the first call of the roll, shall select four (4) unfilled Standing Committees on which he or she wishes to serve (and shall also select at this same time a seat on each any or all of the Ethics, Invitations and Interstate Cooperation Committees so long as a vacancy exists). Each member must select either the Finance or Judiciary Committee during the first call of the roll. When the prescribed number of seats provided for a particular party within a Standing Committee has been filled, the President shall announce that the seats available for either the Majority or Minority party are filled. When the roll is called for the second time, it shall be called in reverse order of continuous service and each member upon his or her name being called, may select one additional unfilled Standing Committee on which he or she wishes to serve. In the event any member is unable to be present for selection of Standing Committees, that member may authorize in writing any member of the Senate to make selections in his or her behalf. This procedure", "fun to play around with. One can also consider natural restrictions, such as myopic students who need to sit near the front. • Replace \"men\" and \"women\" with faculty from different departments. For example, you might need a curriculum committee made up of mathematicians and physicists. • In the senate, there are two senators from each state. How many ways are there of making a committee of 6 senators, no two of which are from the same state? This really only works in the US, but I am sure that there are similar phrasings that could work elsewhere. One could also consider party affiliation here, which could lead to more interesting problems. • True story: My daughter, who is in kindergarten, has a 5th grade \"reading buddy.\" If there are $n$ kindergarteners and $k$ 5th graders, how many ways are their of pairing up students? If you want to appeal to an older audience, maybe we are pairing up tutors with students for individual instruction? • This is totally not what you want, but still an interesting problem: how many injective functions exist from a set with $n$ elements to a set with $k$ elements? How about surjective functions? How does this change with $n$ and $k$? • This question on MSE uses 9th and 10th graders. \"Juniors\"", "the public should be made to focus upon. This is in opposition to an idea in which the number of Senators should grow roughly in proportion to the size of the population itself, in order to ensure each faction of some fixed size at least one representative in the Senate. Now, the simple fact is that, no matter how much you would like the public to devote sufficient time to be able to understand all of the subtleties of politics, that people will refuse (and rightly so) to spend too much time on this, and so the media will simplify the world into a small number of factions. In the United States today, by looking at how the media talks about political factions, I guess they divide domestic politics into about 6 factions: (religious Republicans, other Republicans, anti-corporate Democrats, other Democrats, greens, \"other\"). So let's say that the magic number is between 5 and 7. Since we will allow each of these factions (except \"other\") to field 2 reps, we have a body of at least 9 to 13. Now, if even the small factions such as the greens have 2 reps, surely some of the larger ones will have more. So we want a Senate of about 13 to 24 members. Nth-root formulas are one alterantive, but", "prize. How many different ways are there to select your spots?\" You have 9 choices for the first spot. After you have made that choice there are 8 spots left so you have 8 choices for the second spot. Then there are 7 spots left for the third choice.. You can do this in 9*8*7= 504 ways. That is the same as $\\displaystyle \\frac{9*8*7*6*5*4*3*2*1}{6*5*4*3*2*1}= \\frac{9!}{6!}= \\frac{9!}{(9- 3)!}= P(9, 3)$ but it is better to think about a problem than to memorize formulas. I think it's a permutation, because order matters. In that case, can I still solve it using 9C3 (9 choose 3)? Or do I need to do something else to account for the position of the answers? Also, this one: \"If a Congressional committee is to be made of six people selected from a group of eight Democrats and five Republicans, how many ways are there to select exactly one Republican?\" The part that throws me off here is the \"exactly one\" republican. It wouldn't be as simple as a 13C6 calculation, but would that be part of it? Thank you! -matt Start by choosing the republican first- there are 5 ways to do that. Now choose the first democrat. There are 8 ways to do that. That leaves 7 democrats so there are 7 ways", "## anonymous 4 years ago Assume that the committee consists of 8 Republicans and 5 Democrats. A subcommittee of 4 is randomly selected from all subcommittees of 4 which contain at least 1 Democrat. What is the probability that the new subcommittee will contain at least 2 Democrats? 1. anonymous good lord 2. anonymous lets see if we can figure out how many subcommittees contain at least one democrat, and i guess the best way to do it is figure out how many contain no democrats and subtract it from the total possible subcommittees 3. anonymous these things ar ekilling me and i have to have these done by 730 am 4. anonymous number of subcommittees with not restriction is $\\dbinom{13}{4}=\\frac{13\\times 12\\times 11\\times 10}{4\\times 3\\times 2}=715$ 5. anonymous number of subcommittees that are all republicans are $\\frac{8\\times 7\\times 6\\times 5}{4!}=70$ so i guess we have $715-70=645$ total subcommittees that contain at least one democrat 6. anonymous that will be our denominator. now we have to figure out, of these 645 subcommittees with at least one democrat, how many have at least two 7. anonymous ok exactly one democrat number of ways would be $\\dbinom{5}{1}\\dbinom{8}{3}$ aka $5\\times 56=280$ 8. anonymous so subtract this off from 645 to get the number that contain more than 1 democrat. you get $645-280=365$", "+0 # A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. 0 33 1 +247 A Senate committee consists of 5 Republicans, 6 Democrats, and 2 Independents. A subcommittee of 3 members is randomly chosen. What is the probability that the subcommittee consists of three Republicans? Darkside Sep 7, 2018 #1 +2012 0 first off we regroup things so that we have 5 Republicans and 8 non-Republicans this gets us $$P[R]=\\dfrac{5}{13},~P[!R] = \\dfrac{8}{13}$$ now we have a simple Binomial distribution with parameters $$n=3,~p=\\dfrac{5}{13}$$ $$P[3] = \\left(\\dfrac{5}{13}\\right)^3 = \\dfrac{125}{2197}$$ Rom Sep 7, 2018", "control.) So let's consider the problem again, in a slightly more complicated form: ##### Example 33 The United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee? Solution In this case we need to choose 10 of the 15 Republicans and 9 of the 14 Democrats. There are $$_{15} C_{10}=3003$$ ways to choose the 10 Republicans and $$_{14} C_{9}=2002$$ ways to choose the 9 Democrats. But now what? How do we finish the problem? Suppose we listed all of the possible 10-member Republican groups on 3003 slips of red paper and all of the possible 9-member Democratic groups on 2002 slips of blue paper. How many ways can we choose one red slip and one blue slip? This is a job for the Basic Counting Rule! We are simply making one choice from the first category and one choice from the second category, just like in the restaurant menu problems from earlier. There must be $$3003 \\cdot 2002=6,012,006$$ possible ways of selecting the members of the Defense Subcommittee. ## Probability using Permutations and Combinations We can use permutations and combinations", "+0 # math problem 0 90 1 We know there are 100 members and currently there are 48 Democrats and 52 Republicans. Use this information to answer the following. Step 1 of 5: What is the probability that a random committee of 6 will contain all Democrats? Feb 24, 2019 $$\\text{The number of Democrats selected out of 6 is a random variable having a binomial distribution}\\\\ \\text{with parameters }n=6,~p = \\dfrac{48}{100}=\\dfrac{12}{25}\\\\ p[6] = \\dbinom{6}{6} p^6 = \\left(\\dfrac{12}{25}\\right)^6 = \\dfrac{2985984}{244140625} \\approx 0.01223$$", "seats on committees in proportion to its overall strength. Most committee work is performed by 16 standing committees, each of which has jurisdiction over a field such as finance or foreign relations. Each standing committee may consider, amend, and report bills that fall under its jurisdiction. Furthermore, each standing committee considers presidential nominations to offices related to its jurisdiction. (For instance, the Judiciary Committee considers nominees for judgeships, and the Foreign Relations Committee considers nominees for positions in the Department of State.) Committees may block nominees and impede bills from reaching the floor of the Senate. Standing committees also oversee the departments and agencies of the executive branch. In discharging their duties, standing committees have the power to hold hearings and to subpoena witnesses and evidence. The Senate also has several committees that are not considered standing committees. Such bodies are generally known as select or special committees; examples include the Select Committee on Ethics and the Special Committee on Aging. Legislation is referred to some of these committees, although the bulk of legislative work is performed by the standing committees. Committees may be established on an ad hoc basis for specific purposes; for instance, the Senate Watergate Committee was a special committee created to investigate the Watergate scandal. Such temporary committees cease to exist after fulfilling their", "practices of the Twins' Table Tennis Club. The coaches did not want any brothers to play at the same table. $\\displaystyle a)$ In how many different ways may they divide the players to play round-the-table games at two different tables? $\\displaystyle b)$ In how many different ways is it possible to divide the players into sets of four to play doubles at three different tables? (The position of the players at the tables does not matter.) (Based on an English problem) (5 pont) solution (in Hungarian), statistics ## Problems with sign 'B' Deadline expired on April 10, 2019. B. 5014. After the elections in Nowhereland, there are $\\displaystyle 50 < n < 100$ representatives in the parliament, all from a single party called the Blue Party. (The Blue Party has a single president.) According to the law, a party in the parliament may be divided into two parties as long as the following conditions are met: • The president of the old party is not allowed to become a member of the newly formed parties. His or her parliament mandate will terminate, thereby reducing the total number of representatives. • Every other member may decide which new party to join. • Each of the new parties must have at least one member among the representatives. • Each of", "each The senate in a certain state is comprised of 51 republicans, 48 democrats, and 1 independent. How many committees can be formed if each committee must have 3 republicans and 2 democrats 11/27/16 The production function for a firm is f(l,k)=20l+25k-l^2-3k^2 The cost to the firm of land kis 2 and 4 per unit, respectively. If the firm wants the total cost The production function for a firm is f(l,k)=20l+25k-l^2-3k^2The cost to the firm of land kis 2 and 4 per unit, respectively. If the firm wants the total cost of input to be 50, find the greatest... more Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.", "— that’s why there’s a Gang of Six and not a Gang of Sixteen or Twenty (yet). Whether it’s pride, pork, or pressure, as Ted DiBiase would say, “Everybody has a price.” This price, for an individual legislator, would be equivalent to the utility of standing his ground. Let’s suppose these utilities look like this for nine representative legislators on a committee. (Notice that I represent utility below the line as a negative utility from the status quo — it is a positive utility for the opposite outcome.) Suppose the other party has a budget of 20 (say, in billions of dollars) to spend on bribing members of this party to join their coalition. The status quo party (call it X) has the first move and can pre-emptively give the committee members some additional utility for staying on this side, and afterward, the other party (Y) has the opportunity to make counteroffers. Now, say that Y needs to ultimately win over 5 of these 9 committee members. As it stands, who would Y bribe? Certainly, Mr. I is already on Y’s side, so Y needs to bribe four more people — probably the least expensive of them, E, F, G, and H. Perhaps party X should devote its energy toward bolstering these weak members, possibly even try to", "# Math Help - Permutations in a circle 1. ## Permutations in a circle Q:Three engineers and nine coucillors have a meeting around a circular table. If three coucillors are between each engineer, find the number of possible seating arrangments. the circle permutations formula is $(n-1)!$ the answer is Arrangements: $2! \\times 9!$ whats wrong with $3!\\times8!$, could someone explain why this is wrong, why is it not adhering to the conditions of the question? Thanks 2. ## Re: Permutations in a circle Hello, aonin! Three engineers and nine coucillors have a meeting around a circular table. If three coucillors are between each engineer, . . find the number of possible seating arrangments. The answer is: . $2! \\times 9!$ The basic arrangement looks like this (not including rotations): . . $\\begin{array}{c} E \\\\ C \\;\\;C \\\\ C\\;\\;\\;\\;\\;\\;C \\\\ C \\;\\;\\;\\;\\;\\;\\;\\;\\;\\;C \\\\ E \\;\\;C\\;\\;C\\;\\;C\\;\\;E \\end{array}$ The first engineer can sit in any chair (it doesn't matter). The second engineer has a choice of either of the other two E-chairs. The third engineer has only one choice: the remaining E-chair. . . Hence, the engineers can be seated in $2!$ ways. Then the nine \"coucillors\" can be seated in $9!$ ways. .Answer: . $2! \\times 9!$ 3. ## Re: Permutations in a circle Thanks soroban, but why are the", "The US Senate has 100 members. After a certain election, there were 4 more Democrats than Republicans with no other parties represented. How many members of each party were there in the Senate? May 1, 2018 48 Republicans and 52 Democrats. Explanation: Let the number of Republicans be $x$. There are $x + 4$ Democrats. Thus, $x + \\left(x + 4\\right) = 100$ $\\implies 2 x = 100 - 4$ $\\implies x = \\frac{96}{2} = 48$ There are 48 + 4 = 52 Democrats.", "1. Politics probability question A senate committee consists of five Democrats and five Republicans. If four people from the committee are selected at random, what is the probability that all four are Democrats, given that at least one of them is a Democrat? The probability that all four are Democrats is? 2. Re: Politics probability question you want to show some work for this one?", "# Probability each state of USA is represented in a committee Each state of USA has $2$ senators. The committee has $50$ sits. What's the probability that all states has represented in committee? I know why $\\Omega=\\binom{100}{50}$, but I can't understand why I have $2^{50}$ probabilities to have all states represented. I can't figure out how many possibilities there are if $1$ senator is in committee. I don't know where to start. • The numerator gives the number of ways to pick a committee with exactly one senator from each state. The denominator gives the number of ways to choose 50 senators from a group of 100 senators, i.e., the total number of ways to form the committee. – David Jul 25 '17 at 13:38 • It's a simple application of $\\text{probability} = \\frac{\\text{favourable}}{\\text{possible}}$ – Arthur Jul 25 '17 at 13:38 By yourself and with the other answers' help you seem to have figured out most of it, but the sticking point is the $2^{50}$. You've got 50 states and 50 seats, so in order to have all represented you can look at a seat and say that it's the Texas seat (Texas needs to have a seat somewhere, but if it had 2 then probably Nevada wouldn't get a seat). Then there's 2 options for who sits" ]

[ "simultaneously. What is the probability of getting two numbers whose product is even? A) [1/2] B) [3/4] C) [3/8] D) [5/16] E) [5/6] Probability of both dice to be odd is = 3/6 * 3/6 = 1/4 So probability of even = 1- probability of odd = 1-1/4 = 3/4 Sent from my BND-AL10 using GMAT Club Forum mobile app CrackVerbal Quant Expert Joined: 23 Apr 2019 Posts: 36 Re: Two dice are thrown simultaneously. What is the probability of getting [#permalink] ### Show Tags 09 Oct 2019, 23:57 Hi, While dealing with complex events in probability, the best way is to use the below equation to set up the question. Probability (Complex event) = Probability (Individual Events) * arrangement The complex event should always be represented by a set of alphabets, for example, if we need to find the probability of getting two heads and a tail when we toss a coin 3 times, then the probability can be represented as P(HHT). P(HHT) = 1/2 * 1/2 * 1/2 * 3!/2! (3!/2! -----> arrangement of the word HHT) P(HHT) = 3/8 Similarly here we need the probability of the product being even. There will be two cases here: 1. One number is even and one number odd, let us represent this as P(EO) OR 2. Both numbers", "# Math Help - Determine the probability of the product of two numbers being even. 1. ## Determine the probability of the product of two numbers being even. Well, the title says it all. Assume that the probability of choosing an odd number and an even number is the same. I thought I would split it into cases: one number is even, both numbers are even, and neither number is even. I get this: (1/2) + (1/4) - (1/4), respectively. So is 1/2 correct? Thanks! Any help is appreciated! 2. ## Re: Determine the probability of the product of two numbers being even. The product of two numbers are even if at least one number is even. Hence I get an answer of $\\dfrac{3}{4}$ 3. ## Re: Determine the probability of the product of two numbers being even. The four possibilities are Odd * Odd Odd * Even Even * Odd Even * Even 3 favorable cases out of 4. Hence the probability is 3/4 4. ## Re: Determine the probability of the product of two numbers being even. Originally Posted by gpksam07 The four possibilities are Odd * Odd Odd * Even Even * Odd Even * Even 3 favorable cases out of 4. Hence the probability is 3/4 I did $1 - P(OO)$ where P(OO) is", "a uniform distribution on the whole set of natural or whole numbers, this is not possible to have. Hence the concept of \"probability of a number being even\" is not uniquely understood. Possible probability measures usually are higher for small numbers and lower for high numbers. If, for example, $0$ is the number with higher probability, and the probability decreases with magnitude, than the even number would have slightly higher probability than odd numbers.", "of randomly selected digits is uniform, and the second is!", "+ 5/4 260) If a number is randomly selected from the set {2,3,4,5}, what is the probability that it is both even and prime? - 0 + 1/4 - 1/2 - 3/4 - 1 - 5/4 261) If a number is randomly selected from the set {2,3,4,5}, what is the probability that it is either even or prime? - 0 - 1/4 - 1/2 - 3/4 + 1 - 5/4", "What is the probability of the sum of two odd numbers to be even We know : Probability P ( E ) = But we know sum of any two odd number is always an even number , So Total desired events n ( E ) = total number of events n ( S ) Therefore, Probability of the sum of two odd numbers to be even = 1 ( Ans )", "set B =$$\\frac{2}{5}$$ $$\\frac{3}{5}*\\frac{2}{5}=\\frac{6}{25}$$ The products are either even or odd. Probability of an even or odd product = 1, so P(even) = 1 - P(odd) $$\\frac{25}{25}-\\frac{6}{25}=\\frac{19}{25}$$ _________________ Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date. Our lives begin to end the day we become silent about things that matter. -- Dr. Martin Luther King, Jr. Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2799 Re: Set A contains three different positive odd integers and two [#permalink] ### Show Tags 05 Apr 2018, 10:04 oss198 wrote: Set A contains three different positive odd integers and two different positive even integers; set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even? (A) 6/25 (B) 2/5 (C) 1/2 (D) 3/5 (E) 19/25 We can use the formula: 1 - P(odd product) = P(even product) We may recall that odd x odd = odd. P(odd from set A) = 3/5 P(odd from set B) = 2/5 P(odd product) = 3/5 x 2/5 = 6/25 P(even product) = 1 - 6/25 = 19/25 _________________ #", "# Probability MCQ Question with Answer ## Probability MCQ with detailed explanation for interview, entrance and competitive exams. Explanation are given for understanding. ### Download Probability MCQ Question Answer PDF Question No : 15 Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ? $\\frac{3}{4}$ $\\frac{1}{4}$ $\\frac{7}{4}$ $\\frac{1}{2}$", "even number From sample space, it is clear that the probability of obtaining a head and an even number is$\\frac{3}{8}$ i.e, {H2,H4,H6} with total no of elements in sample space as 8.", "the probability that an even number will come up exactly 3 times is $$\\frac{5}{{16}}$$ $$\\frac{1}{2}$$ $$\\frac{3}{{16}}$$ $$\\frac{3}{2}$$ ## Questions 49 of 50 Question:The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then $$P(X = 1)$$ is 1/32 1/16 1/8 1/4 ## Questions 50 of 50 Question:A coin is tossed n times. The probability of getting head at least once is greater than 0.8, then the least value of n is", "any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6) Why does this result in the wrong probability? Hi... What you are doing here is \" finding probability of picking an even number from atleast one of the set\" Say the set B were 5,6,8 so two even Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1? But we are looking for xy to be even.. This will include following cases.. 1) even from A and odd number from B... 2/4 * 2/3=1/3 2) even from A and even from B 2/4 *1/3=1/6 3) odd from A and even from B 2/4*1/3=1/6 Total =1/3+1/6+1/6=2/3 Other easier way would be - Subtract prob of odd from 1 Prob of odd .. odd from both = 2/4*2/3=1/3 So Ans=1 -1/3=2/3 Thanks for your response! I guess I'm still a bit confused. For example, if a question was asking: \"A fair six-sided dice is rolled twice. What is the probability of getting at least once?\" Then I understand I could calculate the probability of not getting 3 twice and subtract: $$1 - (\\frac{5}{6}*\\frac{5}{6}) = \\frac{11}{36}$$ Nevertheless, I could still say: 1. If I", "## Algebra 1 Each cube has an equal number of even and odd numbers. Thus, the change of rolling an odd number on each individual cube is 1/2. In order to find the odds of rolling it on both cubes, we multiply this probability: $1/2 \\times 1/2 =1/4$", "can't figure out why my original logic resulted in an incorrect answer. I reasoned: If any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6) Why does this result in the wrong probability? Hi... What you are doing here is \" finding probability of picking an even number from atleast one of the set\" Say the set B were 5,6,8 so two even Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1? But we are looking for xy to be even.. This will include following cases.. 1) even from A and odd number from B... 2/4 * 2/3=1/3 2) even from A and even from B 2/4 *1/3=1/6 3) odd from A and even from B 2/4*1/3=1/6 Total =1/3+1/6+1/6=2/3 Other easier way would be - Subtract prob of odd from 1 Prob of odd .. odd from both = 2/4*2/3=1/3 So Ans=1 -1/3=2/3 I guess I'm still a bit confused. For example, if a question was asking: \"A fair six-sided dice is rolled twice. What is the probability of getting at least once?\" Then I understand I could calculate the probability of not getting 3 twice and subtract: $$1 -", "if P(A) = 1/2 and P(B) = 1? a. 1, 1/2 b. 1/2, 1 c. 1/2, 1/4 d. 1/4, 1/2 18. If we roll a fair dice twice, then what would be the probability that we get a result where an even number follows an odd number? a. 1/2 b. 1/3 c. 1/4 d. 1/6", "1 and 5 being even is 40%. So, the odds of doing both are: P(x is even and x <= 5) = P(x is even given x <= 5) P(x <= 5) = (2/5) * (1/2) = 1/5 Observe that since P(A and B) = P(B given A) P(A) = P(A given B) P(B), you can express the probability of A given B in terms of the reverse: P(A given B) = P(B given A) P(A) / P(B). ### Probability of A or B Image we were picking a number between 1 and 10 (inclusive). What’s the probability of picking an even number or a number between 1 and 5? P(x is even or x <= 5) = P(x is even) + P(x <= 5) - P(x is even and x <= 5) = 1/2 + 1/2 - 1/5 = 4/5 Formula: P(A or B) = P(A) + P(B) - P(A and B) ### Pick all odd numbers There are unique numbers 1 through 7, what’s the probability to pick up the first all odd numbers without replacement? Solution: • We don’t care the order • All odd numbers are: 1, 3, 5, 7, and only one event for this • All events to pick up 4 numbers are: (7654)/(4321) = 35 • So the answer is 1/35", "an incorrect answer. I reasoned: If any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6) Why does this result in the wrong probability? Manager Joined: 21 Oct 2017 Posts: 82 Location: France Concentration: Entrepreneurship, Technology GMAT 1: 750 Q48 V44 GPA: 4 WE: Project Management (Internet and New Media) Re: If x is to be chosen at random from the set {1, 2, 3, 4} and y is to [#permalink] ### Show Tags 02 Dec 2017, 17:14 Same question as above! A little confused why that doesn't work. chetan2u can you please shed some light on this? Thanks! pablogutierrez24 wrote: After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer. I reasoned: If any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6) Why does this result in the wrong probability? _________________ Please Press +1 Kudos if it helps! October 9th, 2017: Diagnostic Exam", "4 # Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ? $\\frac{3}{4}$ $\\frac{1}{4}$ $\\frac{7}{4}$ $\\frac{1}{2}$ A. $\\frac{3}{4}$ So Probability =$\\frac{27}{36}=\\frac{3}{4}$", "please clarify? I reasoned: If any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6) Why does this result in the wrong probability? Manager Joined: 21 Oct 2017 Posts: 68 Concentration: Entrepreneurship, Technology GMAT 1: 750 Q48 V44 WE: Project Management (Internet and New Media) Re: If x is to be chosen at random from the set {1, 2, 3, 4} and [#permalink] ### Show Tags 02 Dec 2017, 18:14 Same question as above! A little confused why that doesn't work. chetan2u can you please shed some light on this? Thanks! pablogutierrez24 wrote: After seeing these replies I understand why 2/3 is the correct answer. But I still can't figure out why my original logic resulted in an incorrect answer. Can someone please clarify? I reasoned: If any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6) Why does this result in the wrong probability? _________________ Please Press +1 Kudos if it helps! October 9th, 2017: Diagnostic Exam - Admit Master (GoGMAT)", "# Even When It Is Even Level pending Let x, y, z be three numbers chosen randomly from the set {3, 4, 5, 6, 7, 8}. The probability that xy+z is even is express in the form a/b. What is the value of a+2b? ×", "probability of being mapped to (here, equal simply means that no natural is given preference as to being mapped to over any other natural). The problem is that, because non-measurable sets are used to do this, no meaningful cumulative distribution function can be established (as might be expected). To see an example of such a selection process, see here: Is this fraction undefined? Infinite Probability Question. Hi :) I've always had a nagging suspicion about drawing (uniform) random numbers from an infinite set. I'm not convinced it's possible: here's an intuition about why I'm sceptical: it would be nice if someone could explain where I go wrong. My counterargument is structured as follows: if it's possible to draw random natural numbers (i.e. from the whole infinite set) then there must be a probability of selecting an EVEN number: I argue that $$Pr(even) = \\frac{1}{2}$$ AND $$Pr(even) = 1$$, causing a contradiction. First we need to define: URD: A \"uniform random draw\" for a set $$S$$, ($$S = \\mathbb{N} = \\{1, 2, 3, ... \\}$$ in this case) is a random selection $$x$$ from $$S$$ such that every $$y$$ in $$S$$ has an equal chance of being picked. This is what we're doing in the video, and it's what we usually mean when we just say \"random\". Consider the" ]

[ "set B =$$\\frac{2}{5}$$ $$\\frac{3}{5}*\\frac{2}{5}=\\frac{6}{25}$$ The products are either even or odd. Probability of an even or odd product = 1, so P(even) = 1 - P(odd) $$\\frac{25}{25}-\\frac{6}{25}=\\frac{19}{25}$$ _________________ Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date. Our lives begin to end the day we become silent about things that matter. -- Dr. Martin Luther King, Jr. Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2799 Re: Set A contains three different positive odd integers and two [#permalink] ### Show Tags 05 Apr 2018, 10:04 oss198 wrote: Set A contains three different positive odd integers and two different positive even integers; set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even? (A) 6/25 (B) 2/5 (C) 1/2 (D) 3/5 (E) 19/25 We can use the formula: 1 - P(odd product) = P(even product) We may recall that odd x odd = odd. P(odd from set A) = 3/5 P(odd from set B) = 2/5 P(odd product) = 3/5 x 2/5 = 6/25 P(even product) = 1 - 6/25 = 19/25 _________________ #", "# Math Help - Determine the probability of the product of two numbers being even. 1. ## Determine the probability of the product of two numbers being even. Well, the title says it all. Assume that the probability of choosing an odd number and an even number is the same. I thought I would split it into cases: one number is even, both numbers are even, and neither number is even. I get this: (1/2) + (1/4) - (1/4), respectively. So is 1/2 correct? Thanks! Any help is appreciated! 2. ## Re: Determine the probability of the product of two numbers being even. The product of two numbers are even if at least one number is even. Hence I get an answer of $\\dfrac{3}{4}$ 3. ## Re: Determine the probability of the product of two numbers being even. The four possibilities are Odd * Odd Odd * Even Even * Odd Even * Even 3 favorable cases out of 4. Hence the probability is 3/4 4. ## Re: Determine the probability of the product of two numbers being even. Originally Posted by gpksam07 The four possibilities are Odd * Odd Odd * Even Even * Odd Even * Even 3 favorable cases out of 4. Hence the probability is 3/4 I did $1 - P(OO)$ where P(OO) is", "simultaneously. What is the probability of getting two numbers whose product is even? A) [1/2] B) [3/4] C) [3/8] D) [5/16] E) [5/6] Probability of both dice to be odd is = 3/6 * 3/6 = 1/4 So probability of even = 1- probability of odd = 1-1/4 = 3/4 Sent from my BND-AL10 using GMAT Club Forum mobile app CrackVerbal Quant Expert Joined: 23 Apr 2019 Posts: 36 Re: Two dice are thrown simultaneously. What is the probability of getting [#permalink] ### Show Tags 09 Oct 2019, 23:57 Hi, While dealing with complex events in probability, the best way is to use the below equation to set up the question. Probability (Complex event) = Probability (Individual Events) * arrangement The complex event should always be represented by a set of alphabets, for example, if we need to find the probability of getting two heads and a tail when we toss a coin 3 times, then the probability can be represented as P(HHT). P(HHT) = 1/2 * 1/2 * 1/2 * 3!/2! (3!/2! -----> arrangement of the word HHT) P(HHT) = 3/8 Similarly here we need the probability of the product being even. There will be two cases here: 1. One number is even and one number odd, let us represent this as P(EO) OR 2. Both numbers", "any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6) Why does this result in the wrong probability? Hi... What you are doing here is \" finding probability of picking an even number from atleast one of the set\" Say the set B were 5,6,8 so two even Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1? But we are looking for xy to be even.. This will include following cases.. 1) even from A and odd number from B... 2/4 * 2/3=1/3 2) even from A and even from B 2/4 *1/3=1/6 3) odd from A and even from B 2/4*1/3=1/6 Total =1/3+1/6+1/6=2/3 Other easier way would be - Subtract prob of odd from 1 Prob of odd .. odd from both = 2/4*2/3=1/3 So Ans=1 -1/3=2/3 Thanks for your response! I guess I'm still a bit confused. For example, if a question was asking: \"A fair six-sided dice is rolled twice. What is the probability of getting at least once?\" Then I understand I could calculate the probability of not getting 3 twice and subtract: $$1 - (\\frac{5}{6}*\\frac{5}{6}) = \\frac{11}{36}$$ Nevertheless, I could still say: 1. If I", "is odd. 3 even numbers. 3 odd numbers. First one odd and second even; first even and second odd = 3 × 3 × 2 = 18 outcomes The outcomes are (1, 2)(1, 4)(1, 6)(3, 2)(3, 4)(3, 6)(5, 2)(5, 4)(5, 6) (2, 1)(2, 3)(2, 5)(4, 1)(4, 3)(4, 5)(6, 1)(6, 3)(6, 5) (b) Both are even. 3 possibilities for the first and 3 for the second = 3 × 3 = 9 outcomes. The outcomes are (2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6) = 9 outcomes Total outcomes when product is even = 18 + 9 = 27 Total outcomes when two different dice are thrown = 6 × 6 = 36 Probability that product is even = $\\frac{\\text{number of outcomes when product is even}}{\\text{Total number of outcomes}}$ = $\\frac{27}{36}$ = $\\frac{3}{4}$ .", "a product of any two of the factors. So: $P = \\frac{2*1}{4*3} = \\frac{1}{6}$. So we have a 1 in 6 chance of getting the required product. -Dan", "it to say, first pick n natural numbers randomly, then check to see if we have property $P$. It makes sense now, thanks. – Joe Tait Aug 3 '12 at 14:22 Perhaps a justification is this. In the first question it is (correctly) claimed that it is impossible to have a uniform distribution on the natural numbers. Thus we can't develop a sensible way of choosing particular numbers at random, when each supposedly has the same probability. The last question though is dealing with the probability of picking a certain class of numbers. In that case the approach is to pick $N$ large, impose a uniform distribution on $[1,N]$ (which we can always do), work out the probability as a function of $N$, and then take the limit $N \\rightarrow \\infty.$ Example: what's the probability of randomly picking an even number? Fix $N$. If $N$ is even then the probability of picking an even number in the uniform distribution on $[1,N]$ is exactly $1/2$. If $N$ is odd then the probability is $$\\frac{\\text{amount of even numbers}}{\\text{total amount}} = \\frac{ (N-1)/2 }{N} = \\frac{1}{2} - \\frac{1}{2N}$$ And so, as we take the limit $N \\rightarrow \\infty$, we say that the probability of choosing an even number is 1/2. This approach is really a \"natural density\" approach; see http://en.wikipedia.org/wiki/Natural_density. The", "craft Nov 5 '15 at 7:50 Calculate $1$ minus the probability of the complementary event: • The probability of getting an odd number in a roll is $\\frac36$ • The probability of getting an odd product in $n$ rolls is $\\left(\\frac36\\right)^n$ • The probability of getting an even product in $n$ rolls is $1-\\left(\\frac36\\right)^n$", "7. Thanks 8. Hello, saberteeth! 1) Two dice are thrown simultaneously. Find the probability that the product of the two numbers on the two dice is an even number. There are $6^2=36$ possible outcomes. The product is odd if both numbers are odd. There are 3 choices of an odd number for the first die: $\\{1,3,5\\}$ . . and 3 choices of odd number for the second die. Hence, there are: . $3\\cdot3\\:=\\:9$ odd-product outcomes. So there are: . $36 - 9 \\:=\\:27$ even-product outcomes. Therefore: . $P(\\text{even product}) \\:=\\:\\frac{27}{36} \\;=\\;\\frac{3}{4}$ 3) What is the probability that the product of two consecutive non-negative integers will be a number ending with 0? Two consecutive integers can have the following pairs of units-digits: . . $(0,1), (1,2), (2,3),(3,4), (4,5), (5,6), (6,7), (7,8), (8,9), (9,0)$ The unit-digits of their product are: . $0,\\;2,\\;6,\\;2,\\;0,\\;0,\\;2,\\;6,\\;2,\\;0$ Got it? 9. Originally Posted by Soroban Hello, saberteeth! There are $6^2=36$ possible outcomes. The product is odd if both numbers are odd. There are 3 choices of an odd number for the first die: $\\{1,3,5\\}$ . . and 3 choices of odd number for the second die. Hence, there are: . $3\\cdot3\\:=\\:9$ odd-product outcomes. So there are: . $36 - 9 \\:=\\:27$ even-product outcomes. Therefore: . $P(\\text{even product}) \\:=\\:\\frac{27}{36} \\;=\\;\\frac{3}{4}$ Two consecutive integers can have the following", "3 1 4 3 2 5 4 1 5 4 2 6 Probability of sum being odd = $$\\frac { 4 }{ 8 }$$ = $$\\frac { 1 }{ 2 }$$ Probability of sum being even = $$\\frac { 4 }{ 8 }$$ = $$\\frac { 1 }{ 2 }$$ Kerala Syllabus 10th Standard Maths Chapter 3 Question 3. A box contains four slips numbers 1, 2, 3, 4 and another contains three slips numbered 1, 2, 3. If one slip is taken from each, what is the probability of the product being odd? The probability of the product being even? First box Second box Product 1 1 1 1 2 2 1 3 3 2 1 2 2 2 4 2 3 6 3 1 3 3 . 2 6 3 3 9 4 1 4 4 2 8 4 3 12 Probability of sum being odd = $$\\frac { 4 }{ 12 }$$ = $$\\frac { 1 }{ 3 }$$ Probability of sum being even = $$\\frac { 8 }{ 12 }$$ = $$\\frac { 2 }{ 3 }$$ 10 Th Maths Text Book Questions And Answers Question 4. From all two-digit numbers with either digit 1, 2 or 3 one number is chosen. i. What is the probability of both digits being the same? ii.", "Mar 2018, 16:16 pushpitkc wrote: If three integers from among the first 10 positive integers are chosen, what is the probability that the product of these integers is even? A. $$\\frac{1}{12}$$ B. $$\\frac{1}{3}$$ C. $$\\frac{1}{6}$$ D. $$\\frac{3}{4}$$ E. $$\\frac{11}{12}$$ Source: Experts Global Odd integer from 1 to 10: 1,3,5,7,9 It would be a lot easier to calculate the probability that the product of 3 integer will be odd: $$P(Odd) = \\frac{5*4*3}{10*9*8}$$ = $$\\frac{1}{12}$$ The probability that product of 3 integer will be even = $$1 - P(Odd) = 1 - \\frac{1}{12} =\\frac{11}{12}$$ Manager Joined: 11 Feb 2017 Posts: 190 Re: If three integers from among the first 10 positive integers are chosen [#permalink] ### Show Tags 22 Mar 2018, 14:39 JeffTargetTestPrep wrote: pushpitkc wrote: If three integers from among the first 10 positive integers are chosen, what is the probability that the product of these integers is even? A. $$\\frac{1}{12}$$ B. $$\\frac{1}{3}$$ C. $$\\frac{1}{6}$$ D. $$\\frac{3}{4}$$ E. $$\\frac{11}{12}$$ We can use the formula: P(even product) = 1 - (probability of an odd product) In order to get an odd product we need 3 odd integers. The probability of selecting 3 odd integers is: 5/10 x 4/9 x 3/8 = 1/2 x 1/3 x 1/2 = 1/12 So the probability of an even product is 1 - 1/12 =", "Jane Street Test The questions below appeared on an online test administered by Jane Street Capital Management to assess whether a person is worthy of a phone interview. 🔗 Question 1 Suppose we choose four numbers at random without replacement from the first 20 prime numbers. What is the probability that the sum of these four numbers is odd? Solution By definition, an even number is an integer multiple of 2; that is, even numbers have the form $2n$, for some integer $n$. An odd number is one more than an even number; that is, odd numbers have the form $2n+1$ for some integer $n$. If all four numbers are odd, then the sum is even. This follows from the fact that the sum of two odd numbers is even: $(2n+1) + (2k+1) = 2(n+k) + 2 = 2(n+k+1)$. And the sum of two even numbers is again even: $2n+2k = 2(n+k)$. Therefore, in order for the sum of the four primes chosen to be odd, the only even prime number, 2, must be among the four numbers we selected. Here are two ways to compute the probability that 2 is one of the four numbers selected: First, consider the probability of not selecting 2: $$P(\\text{2 is not among the chosen primes}) = \\frac{19}{20} \\frac{18}{19} \\frac{17}{18} \\frac{16}{17} =", "If n is the number of colours available, and r is the number used for a given code, the number of possible combinations is $_nC_r = \\frac{n!}{r!(n-r)!}$ No. of colours Number represented by one colour Number represented by two colours Total represented Decision 4 4 $_4C_2=\\frac{4!}{2!(4-2)!} = 6$ 4 + 6 = 10 Not sufficient 5 5 $_5C_2 = \\frac{5!}{2!(5-2)!} = 10$ 5 + 10 = 15 Sufficient 1. A box contains 100 balls, numbered 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers will be odd? 1. ¼ 2. 3/8 3. ½ 4. 5/8 Since there are 50 odd and 50 even balls, the probability of selecting an odd or an even ball is ½. For the sum of the three numbers to be odd, the numbers must all be odd or one number must be odd and the other two even. Probability of selecting three odd balls = $\\frac{1}{2}\\frac{1}{2}\\frac{1}{2} = \\frac{1}{8}$ Probability of selecting one odd and two even balls = P{odd, even, even} + P{even, odd, even} + P{even, even, odd} P{odd,even,even} = P{even, odd, even} = P{even, even, odd} = $\\frac{1}{2}\\frac{1}{2}\\frac{1}{2} = \\frac{1}{8}$ Thus, the probability of one odd and two even balls = $\\frac{1}{8} +", "others are even or odd. However, this misses the probabilities of the 1st being odd, the 1st 2 being odd, the 1st 3 being odd, the 1st and 3rd being odd......etc. 7. Originally Posted by Archie Meade Hi u2_wa, In forming a number from the 4 digits, the even digit may be placed at the end to make it even, thus forming an even number from the digits available. You don't need to stick to the order the digits came in. The book answer gives $\\frac{6}{13}$ which is the probability that the first digit is even, considering that it doesn't matter whether the others are even or odd. However, this misses the probabilities of the 1st being odd, the 1st 2 being odd, the 1st 3 being odd, the 1st and 3rd being odd......etc. What I did to solve is this: There are in total $13P4$ ways. There should $(2,4,6,8,0,2)$ be among $six$ digits in the end to make an even number. Ways to get an even number: $12P3*6$ probability $=\\frac{12P3*6}{13P4}$ $=\\frac{6}{13}$ (The answer given in the book) 8. Yes, that's good work u2_wa, shows you can work it out in alternative ways. You are calculating the probability if we must take the digits in the order they come in. The way the question is worded suggests that", "not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number. If the product is odd, it can always be written as product of two odd numbers. Let’s go back to our question: We have 5 numbers: 1, 2, 3, 4, 5 Our favorable cases constitute those in which either both numbers are odd or the product has 4 as a factor. 3 Odd numbers: 1, 3, 5 2 Even numbers: 2, 4 Number of cases when both numbers are odd = 3C2 = 3 (select 2 of the 3 odd numbers) Number of cases when 4 is a factor of the product = Number of cases such that we select 4 and any other number = 1*4C1 = 4 Total number of favorable cases = 3 + 4 = 7 Note that this includes the case where we take both even numbers. Had there been more even numbers such as 6, we would have included more cases where we pick both even numbers such as 2 and 6 since their product would have 4 as a factor. Required Probability = 7/10 Takeaway: When can we write a number as difference of squares? -", "the probability of obtaining a product that is an even number or a product of greater than 30, adds to a number larger than 1. Given we know the probability of an event must lie between 0 and 1 explain in your own words if this is possible I think there maybe a translation problem here. Taking what is posted verbatim, twelve sided dice numbered $$\\displaystyle 1\\text{ to 12}$$ and product (not sum) of the faces. There are $$\\displaystyle 144$$ possible pairs of which we want to count any pair which has a product which is even or has value greater than thirty. Any pair with an even number yields an even product. Thus remove any pair that contains two odd numbers whose product is less than thirty. That count is $$\\displaystyle \\|(\\{1,3,5\\}\\times\\{1,3,5\\}\\cup\\{(1,7)\\,(7,1)\\,(1,9)\\,(9,1)\\,(3,7)\\,(7,3)\\,(1,11)\\,(11,1)\\}\\|=17$$ OR $$\\displaystyle 144-17=127$$ pairs the product of which is even or less than thirty. Thus what the is probability of obtaining a product that is an even number or a product of greater than 30 ? #### Dr.Peterson ##### Elite Member on any roll of a set of two fair 12 sided dice, the probability of obtaining a product that is an even number or a product of greater than 30, adds to a number larger than 1. Given we know the probability of an", "the number on the first die is even, the product will be even regardless of the second die's value. You get 18 for that (3 even values for die number 1)(all 6 values for die number 2) An odd value on the first die coupled with an even value on the second will also produce an even product. How many of those may we have ? The final probability can then be calculated I guess what you tried to calculate initially was the number of even 2-digit values 5. Originally Posted by Archie Meade 18 is too small a count, saberteeth... If the number on the first die is even, the product will be even regardless of the second die's value. You get 18 for that (3 even values for die number 1)(all 6 values for die number 2) An odd value on the first die coupled with an even value on the second will also produce an even product. How many of those may we have ? The final probability can then be calculated I guess what you tried to calculate initially was the number of even 2-digit values Then 3*3*3 = 27. So, 27/36 (3/4) should be the answer. Am i correct? 6. Yes, E(E)=3(3) products Even E(O)=3(3) products Even O(E)=3(3) products Even total is $3(3)3=3^3$", "# Expectation with $-1$ I am trying to study probability theory by myself. I asked myself a question about expectation, and I have no idea how to solve it. Please help me with solution and my understanding. The question I asked is following: Let $a, b, c, d$ be non-negative natural numbers, say from the set $\\{0, 1, ..., n\\}.$ Calculate the expectation $$E\\left( (-1)^{a+b+c+d}\\right).$$ You should say something about the probability distribution of your four random variables $a,b,c,d$. – Michael Hardy Feb 29 '12 at 3:32 $(-1)^{a+b+c+d} = (-1)^a (-1)^b (-1)^c (-1)^d$. Assuming $a,b,c,d$ are independent (which you didn't say but is probably what is meant), the expected value of the product is the product of the expected values. $(-1)^a$ is $-1$ if $a$ is odd and $1$ if $a$ is even. So if the values $0,1,\\ldots,n$ are all equally likely (which again you didn't say but you probably meant), it all comes down to counting how many odd and even numbers there are in $0,1,\\ldots,n$. ...and that depends on whether $n$ is odd or even. If $n$ is odd, then there are equally many odd and even numbers, so $E((-1)^a)$ would be $0$. If $n$ is even, it's a bit more complicated. – Michael Hardy Feb 29 '12 at 3:34", "4 # Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ? $\\frac{3}{4}$ $\\frac{1}{4}$ $\\frac{7}{4}$ $\\frac{1}{2}$ A. $\\frac{3}{4}$ So Probability =$\\frac{27}{36}=\\frac{3}{4}$", "the product of the pair of numbers that we choose to be divisible by 4, the pair has to have either of the following sets of characteristics. - It includes 2 even numbers. (The prime factors of a product of 2 even numbers will always include (2, 2). Thus, the product of any two even numbers is a multiple of 4.) - It includes an odd number and 4. First calculate the total number of possible pairs that include 2 even numbers. Number of even numbers: 3 per set Total number of ways to a pair of even numbers: 3 x 3 = 9 Now calculate the total number of pairs that include 4 and an odd number. The number 4 from the first set can go with 3 odd numbers from the second set: 3 The number 4 from the second set can go with 3 odd numbers from the first set: 3 Total number of pairs that include 4 and an odd number: 3 + 3 = 6 Total number of pairs the product of which is a multiple of 4: 9 + 6 = 15 Favorable Outcomes/Total Outcomes = 15/36 _________________ # Marty Murray Chief Curriculum and Content Architect [email protected] 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep" ]

[ "set B =$$\\frac{2}{5}$$ $$\\frac{3}{5}*\\frac{2}{5}=\\frac{6}{25}$$ The products are either even or odd. Probability of an even or odd product = 1, so P(even) = 1 - P(odd) $$\\frac{25}{25}-\\frac{6}{25}=\\frac{19}{25}$$ _________________ Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date. Our lives begin to end the day we become silent about things that matter. -- Dr. Martin Luther King, Jr. Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2799 Re: Set A contains three different positive odd integers and two [#permalink] ### Show Tags 05 Apr 2018, 10:04 oss198 wrote: Set A contains three different positive odd integers and two different positive even integers; set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even? (A) 6/25 (B) 2/5 (C) 1/2 (D) 3/5 (E) 19/25 We can use the formula: 1 - P(odd product) = P(even product) We may recall that odd x odd = odd. P(odd from set A) = 3/5 P(odd from set B) = 2/5 P(odd product) = 3/5 x 2/5 = 6/25 P(even product) = 1 - 6/25 = 19/25 _________________ #", "simultaneously. What is the probability of getting two numbers whose product is even? A) [1/2] B) [3/4] C) [3/8] D) [5/16] E) [5/6] Probability of both dice to be odd is = 3/6 * 3/6 = 1/4 So probability of even = 1- probability of odd = 1-1/4 = 3/4 Sent from my BND-AL10 using GMAT Club Forum mobile app CrackVerbal Quant Expert Joined: 23 Apr 2019 Posts: 36 Re: Two dice are thrown simultaneously. What is the probability of getting [#permalink] ### Show Tags 09 Oct 2019, 23:57 Hi, While dealing with complex events in probability, the best way is to use the below equation to set up the question. Probability (Complex event) = Probability (Individual Events) * arrangement The complex event should always be represented by a set of alphabets, for example, if we need to find the probability of getting two heads and a tail when we toss a coin 3 times, then the probability can be represented as P(HHT). P(HHT) = 1/2 * 1/2 * 1/2 * 3!/2! (3!/2! -----> arrangement of the word HHT) P(HHT) = 3/8 Similarly here we need the probability of the product being even. There will be two cases here: 1. One number is even and one number odd, let us represent this as P(EO) OR 2. Both numbers", "# Math Help - Determine the probability of the product of two numbers being even. 1. ## Determine the probability of the product of two numbers being even. Well, the title says it all. Assume that the probability of choosing an odd number and an even number is the same. I thought I would split it into cases: one number is even, both numbers are even, and neither number is even. I get this: (1/2) + (1/4) - (1/4), respectively. So is 1/2 correct? Thanks! Any help is appreciated! 2. ## Re: Determine the probability of the product of two numbers being even. The product of two numbers are even if at least one number is even. Hence I get an answer of $\\dfrac{3}{4}$ 3. ## Re: Determine the probability of the product of two numbers being even. The four possibilities are Odd * Odd Odd * Even Even * Odd Even * Even 3 favorable cases out of 4. Hence the probability is 3/4 4. ## Re: Determine the probability of the product of two numbers being even. Originally Posted by gpksam07 The four possibilities are Odd * Odd Odd * Even Even * Odd Even * Even 3 favorable cases out of 4. Hence the probability is 3/4 I did $1 - P(OO)$ where P(OO) is", "any number in either set is even, then xy will be even. So, probability that even number is selected from first set (2/4) + probability that even number is selected from second set (1/3) = 5/6. (2/4 +1/3=5/6) Why does this result in the wrong probability? Hi... What you are doing here is \" finding probability of picking an even number from atleast one of the set\" Say the set B were 5,6,8 so two even Answer as per the method above 2/4+2/3=7/6 >1!!!! Probability>1? But we are looking for xy to be even.. This will include following cases.. 1) even from A and odd number from B... 2/4 * 2/3=1/3 2) even from A and even from B 2/4 *1/3=1/6 3) odd from A and even from B 2/4*1/3=1/6 Total =1/3+1/6+1/6=2/3 Other easier way would be - Subtract prob of odd from 1 Prob of odd .. odd from both = 2/4*2/3=1/3 So Ans=1 -1/3=2/3 Thanks for your response! I guess I'm still a bit confused. For example, if a question was asking: \"A fair six-sided dice is rolled twice. What is the probability of getting at least once?\" Then I understand I could calculate the probability of not getting 3 twice and subtract: $$1 - (\\frac{5}{6}*\\frac{5}{6}) = \\frac{11}{36}$$ Nevertheless, I could still say: 1. If I", "If n is the number of colours available, and r is the number used for a given code, the number of possible combinations is $_nC_r = \\frac{n!}{r!(n-r)!}$ No. of colours Number represented by one colour Number represented by two colours Total represented Decision 4 4 $_4C_2=\\frac{4!}{2!(4-2)!} = 6$ 4 + 6 = 10 Not sufficient 5 5 $_5C_2 = \\frac{5!}{2!(5-2)!} = 10$ 5 + 10 = 15 Sufficient 1. A box contains 100 balls, numbered 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers will be odd? 1. ¼ 2. 3/8 3. ½ 4. 5/8 Since there are 50 odd and 50 even balls, the probability of selecting an odd or an even ball is ½. For the sum of the three numbers to be odd, the numbers must all be odd or one number must be odd and the other two even. Probability of selecting three odd balls = $\\frac{1}{2}\\frac{1}{2}\\frac{1}{2} = \\frac{1}{8}$ Probability of selecting one odd and two even balls = P{odd, even, even} + P{even, odd, even} + P{even, even, odd} P{odd,even,even} = P{even, odd, even} = P{even, even, odd} = $\\frac{1}{2}\\frac{1}{2}\\frac{1}{2} = \\frac{1}{8}$ Thus, the probability of one odd and two even balls = $\\frac{1}{8} +", "Jane Street Test The questions below appeared on an online test administered by Jane Street Capital Management to assess whether a person is worthy of a phone interview. 🔗 Question 1 Suppose we choose four numbers at random without replacement from the first 20 prime numbers. What is the probability that the sum of these four numbers is odd? Solution By definition, an even number is an integer multiple of 2; that is, even numbers have the form $2n$, for some integer $n$. An odd number is one more than an even number; that is, odd numbers have the form $2n+1$ for some integer $n$. If all four numbers are odd, then the sum is even. This follows from the fact that the sum of two odd numbers is even: $(2n+1) + (2k+1) = 2(n+k) + 2 = 2(n+k+1)$. And the sum of two even numbers is again even: $2n+2k = 2(n+k)$. Therefore, in order for the sum of the four primes chosen to be odd, the only even prime number, 2, must be among the four numbers we selected. Here are two ways to compute the probability that 2 is one of the four numbers selected: First, consider the probability of not selecting 2: $$P(\\text{2 is not among the chosen primes}) = \\frac{19}{20} \\frac{18}{19} \\frac{17}{18} \\frac{16}{17} =", "it to say, first pick n natural numbers randomly, then check to see if we have property $P$. It makes sense now, thanks. – Joe Tait Aug 3 '12 at 14:22 Perhaps a justification is this. In the first question it is (correctly) claimed that it is impossible to have a uniform distribution on the natural numbers. Thus we can't develop a sensible way of choosing particular numbers at random, when each supposedly has the same probability. The last question though is dealing with the probability of picking a certain class of numbers. In that case the approach is to pick $N$ large, impose a uniform distribution on $[1,N]$ (which we can always do), work out the probability as a function of $N$, and then take the limit $N \\rightarrow \\infty.$ Example: what's the probability of randomly picking an even number? Fix $N$. If $N$ is even then the probability of picking an even number in the uniform distribution on $[1,N]$ is exactly $1/2$. If $N$ is odd then the probability is $$\\frac{\\text{amount of even numbers}}{\\text{total amount}} = \\frac{ (N-1)/2 }{N} = \\frac{1}{2} - \\frac{1}{2N}$$ And so, as we take the limit $N \\rightarrow \\infty$, we say that the probability of choosing an even number is 1/2. This approach is really a \"natural density\" approach; see http://en.wikipedia.org/wiki/Natural_density. The", "7. Thanks 8. Hello, saberteeth! 1) Two dice are thrown simultaneously. Find the probability that the product of the two numbers on the two dice is an even number. There are $6^2=36$ possible outcomes. The product is odd if both numbers are odd. There are 3 choices of an odd number for the first die: $\\{1,3,5\\}$ . . and 3 choices of odd number for the second die. Hence, there are: . $3\\cdot3\\:=\\:9$ odd-product outcomes. So there are: . $36 - 9 \\:=\\:27$ even-product outcomes. Therefore: . $P(\\text{even product}) \\:=\\:\\frac{27}{36} \\;=\\;\\frac{3}{4}$ 3) What is the probability that the product of two consecutive non-negative integers will be a number ending with 0? Two consecutive integers can have the following pairs of units-digits: . . $(0,1), (1,2), (2,3),(3,4), (4,5), (5,6), (6,7), (7,8), (8,9), (9,0)$ The unit-digits of their product are: . $0,\\;2,\\;6,\\;2,\\;0,\\;0,\\;2,\\;6,\\;2,\\;0$ Got it? 9. Originally Posted by Soroban Hello, saberteeth! There are $6^2=36$ possible outcomes. The product is odd if both numbers are odd. There are 3 choices of an odd number for the first die: $\\{1,3,5\\}$ . . and 3 choices of odd number for the second die. Hence, there are: . $3\\cdot3\\:=\\:9$ odd-product outcomes. So there are: . $36 - 9 \\:=\\:27$ even-product outcomes. Therefore: . $P(\\text{even product}) \\:=\\:\\frac{27}{36} \\;=\\;\\frac{3}{4}$ Two consecutive integers can have the following", "We know that PROBABILITY = Hence probability of getting an even digit is Hence the correct option is #### Question 3: In Q. No. 1, the probability that the digit is a multiple of 3 is (a) $\\frac{1}{3}$ (b) $\\frac{2}{3}$ (c) $\\frac{1}{9}$ (d) $\\frac{2}{9}$ GIVEN: digits are chosen from 1, 2, 3, 4, 5, 6, 7, 8, 9 are placed in a box and mixed thoroughly. One digit is picked at random. TO FIND: Probability of getting a multiple of 3 Total number of digits is 9 Digits that are multiple of 3 are 3, 6 and 9 Total digits that are multiple of 3 are 3 We know that PROBABILITY = Hence probability of getting a multiple of 3 is Hence the correct option is option #### Question 4: If three coins are tossed simultaneously, then the probability of getting at least two heads, is (a) $\\frac{1}{4}$ (b) $\\frac{3}{8}$ (c) $\\frac{1}{2}$ (d) $\\frac{1}{4}$ GIVEN: Three coins are tossed simultaneously. TO FIND: Probability of getting at least two head. When three coins are tossed then the outcome will be TTT, THT, TTH, THH. HTT, HHT, HTH, HHH Hence total number of outcome is 8. At least two heads means that, THH, HHT, HTH and HHH are favorable events Hence total number of favorable outcome is 4 We know", "numbers are selected randomly from 20 consecutive natural numbers, find the probability that the sum of the two numbers is (i) an even number (ii) an odd number. Solution: (i) Let A be the event that the sum of the numbers is even when two numbers are selected out of 20 consecutive natural numbers. In 20 consecutive natural numbers, we have 10 odd and 10 even natural numbers. ∵ The sum of two odd natural numbers is an even number and the sum of two even natural numbers is also an even number (ii) Probability that the sum of two numbers is an odd number P($$\\overline{\\mathrm{A}}$$) = 1 – P(A) = 1 – $$\\frac{9}{19}$$ = $$\\frac{10}{19}$$ Question 11. A game consists of tossing a coin 3 times and noting its outcome. A boy wins if all tosses give the same outcome and lose otherwise. Find the probability that the boy loses the game. Solution: Let A be the event that the boy wins the game getting the same outcome when a coin is tossed 3 times and S be the sample space. ∴ n(S) = 23 = 8 A = {HHH, TTT} n(A) = 2 P(A) = $$\\frac{n(A)}{n(S)}=\\frac{2}{8}=\\frac{1}{4}$$ ∴ The probability that the boy loses the game = P($$\\overline{\\mathrm{A}}$$) = 1 – P(A) = 1 – $$\\frac{1}{4}$$ =", "if both are odd, $$P(odd, \\ odd)=\\frac{1}{2}*\\frac{1}{2}=\\frac{1}{4}$$, or one is odd and another is even but not multiple of 4 (note we'll have two cases odd from 1st set and even but not 4 from 2nd and vise-versa), $$P(odd, \\ even \\ but\\ not \\ 4)=\\frac{1}{2}*\\frac{2}{6}+\\frac{2}{6}*\\frac{1}{2}=\\frac{1}{3}$$. So, $$P=1-(\\frac{1}{4}+\\frac{1}{3})=\\frac{5}{12}=\\frac{15}{36}$$. _________________ Kudos [?]: 133093 [0], given: 12408 Senior Manager Joined: 23 Oct 2010 Posts: 381 Kudos [?]: 403 [0], given: 73 Location: Azerbaijan Concentration: Finance Schools: HEC '15 (A) GMAT 1: 690 Q47 V38 Re: There are two set each [#permalink] ### Show Tags 11 Aug 2012, 22:34 could you please explain me the following - I found this solution in the internet- Picking 2 numbers from each set : 6c1*6c1=36 Favorable outcomes = 15 (1,4)(2,2),(2,4),(2,6)(3,4)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(5,4)(6,2),(6,4),(6,6) Therefore Required probability = 15/36 my question is why do we take into account ,say, both (2;4) and (4;2)? I understand that they are different,since they are taken from dif.sets, but their products are the same. (always have a problem with such kind of things). _________________ Happy are those who dream dreams and are ready to pay the price to make them come true I am still on all gmat forums. msg me if you want to ask me smth Kudos [?]: 403 [0], given: 73 Director Joined: 22 Mar 2011 Posts: 610 Kudos", "simultaneously. What is the probability of getting [#permalink] ### Show Tags 27 Jan 2018, 07:55 sxsd wrote: Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even? A) [1/2] B) [3/4] C) [3/8] D) [5/16] E) [5/6] Easier way is to find ways of ODD product.. each dice can show 1,3 or 5.. so ways of getting ODD product= $$3*3=9$$ Total ways to choose two numbers= $$6*6=36$$ ways of picking numbers such that product is even = 36-9=27 probability of picking even product = $$\\frac{27}{36}=\\frac{3}{4}$$ B OR second way.. ways of getting even product.. 1)first dice even number - 2,4 or 6- 3 ways for these three even numbers, second dice can be any of 6 numbers - 6 ways total ways = 3*6=18 2) first dice odd number = 1,3, or 5 -- 3 ways for these three odd numbers, second dice has to throw even numbers so 2,4,6 - 3 ways total 3*3 = 9 ways total ways = 18+9=27 total ways of picking any two numbers = 6*6=36 prob = 27/36=3/4 B _________________ Manager Joined: 26 Sep 2017 Posts: 95 Re: Two dice are thrown simultaneously. What is the probability of getting [#permalink] ### Show Tags 28 Jan 2018, 12:19 1 1 sxsd wrote: Two dice are thrown", "is odd. 3 even numbers. 3 odd numbers. First one odd and second even; first even and second odd = 3 × 3 × 2 = 18 outcomes The outcomes are (1, 2)(1, 4)(1, 6)(3, 2)(3, 4)(3, 6)(5, 2)(5, 4)(5, 6) (2, 1)(2, 3)(2, 5)(4, 1)(4, 3)(4, 5)(6, 1)(6, 3)(6, 5) (b) Both are even. 3 possibilities for the first and 3 for the second = 3 × 3 = 9 outcomes. The outcomes are (2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6) = 9 outcomes Total outcomes when product is even = 18 + 9 = 27 Total outcomes when two different dice are thrown = 6 × 6 = 36 Probability that product is even = $\\frac{\\text{number of outcomes when product is even}}{\\text{Total number of outcomes}}$ = $\\frac{27}{36}$ = $\\frac{3}{4}$ .", "Odd) = (8/15)*(7/15) + (7/15)*(8/15) = 112/225 Why cant i solve the above using the normal combinations method? Total outcomes = $$\\frac{15*14}{2}$$ What does this mean? - \"You can select the two numbers from different groups or from the same group. Hence the total number of cases is 15*15. Also, you can select the same number again.\" Math Expert Joined: 02 Aug 2009 Posts: 8283 ### Show Tags 23 Jan 2019, 05:21 1 blitzkriegxX wrote: Quote: Question 1: First 15 positive integers are written on a board. If two numbers are selected one by one from the board at random (the numbers are not necessarily different), what is the probability that the sum of these numbers is odd? Solution 1: Numbers: 1, 2, 3, 4, …, 13, 14, 15 When will the sum of two of these numbers be odd? When one number is odd and the other is even. P(Sum is Odd) = P(First number is Odd)*P(Second number is even) + P(First number is Even)*P(Second number is Odd) P(Sum is Odd) = (8/15)*(7/15) + (7/15)*(8/15) = 112/225 Why cant i solve the above using the normal combinations method? Total outcomes = $$\\frac{15*14}{2}$$ What does this mean? - \"You can select the two numbers from different groups or from the same group. Hence the total number of cases", "# A Seemingly Impossible Problem Given that $w, x, y, z$ take on values $0$ and $1$ with equal probability, what is the probability that $w+x+y+z$ is odd? Which of the following arguments is correct? Furthermore, can you generalize this result? Argument 1: If all 4 numbers are even, the sum is even. If 3 numbers are even, the sum is odd. If 2 numbers are even, the sum is even. If 1 number is even, the sum is odd. If 0 numbers are even, the sum is even. In 2 of the 5 cases, the sum is odd. Hence, $w + x +y + z$ is odd with probability $\\frac{ 2}{5}$. Argument 2: $w + x + y + z$ is either odd or even. In 1 of the 2 cases, the sum is odd. Hence, $w + x +y + z$ is odd with probability $\\frac{ 1}{2}$. Argument 3: By listing out all the $2^4$ cases, we find that there are ${ 4 \\choose 0} + { 4 \\choose 2} + { 4 \\choose 4 } = 8$ cases with an even sum, and ${ 4 \\choose 1} + {4 \\choose 3} = 8$ cases with an odd sum. Hence, $w + x +y + z$ is odd with probability $\\frac{ 1}{2}$. Context: In a recent", "question that I have is if I can permute remaining papers in (n-1)! ways for each, even and odd. Why isn't it n^2*((n-1)!)^2 ? And second, can you please give me a little bit more detailed explanation about middle part of your answer(number of even and odd sums) since I still don't understand it. Thank you. – user423843 Jun 10 '17 at 9:12 1. All even papers must be assigned to even boxes, and odd papers must be assigned to odd boxes. Given that their are $(2n)!$ possible arrangements, this probability equals: $$\\frac{(n!)^2}{(2n)!}$$ 1. In order for two box/paper combinations to be even, one odd paper must be assigned to one odd box and one even paper must be assigned to one even box. For all other combinations, an even paper must be assigned to an odd box and an odd paper must be assigned to an even box. As such, the probability of this happening equals: $$\\frac{{n \\choose 1}^2{n \\choose 1}^2(n-1)!(n-1)!}{(2n)^2} = \\frac{(n!)^2n^2}{(2n)!}$$ 1. The probability of having at least two even boxes equals 1 minus the probability of no even box/paper combinations or one even box/paper combination. However, it is impossible to have only one even box/paper combination: if an odd label is assigned to an odd box, we are left with an unassigned even label", "the product of the pair of numbers that we choose to be divisible by 4, the pair has to have either of the following sets of characteristics. - It includes 2 even numbers. (The prime factors of a product of 2 even numbers will always include (2, 2). Thus, the product of any two even numbers is a multiple of 4.) - It includes an odd number and 4. First calculate the total number of possible pairs that include 2 even numbers. Number of even numbers: 3 per set Total number of ways to a pair of even numbers: 3 x 3 = 9 Now calculate the total number of pairs that include 4 and an odd number. The number 4 from the first set can go with 3 odd numbers from the second set: 3 The number 4 from the second set can go with 3 odd numbers from the first set: 3 Total number of pairs that include 4 and an odd number: 3 + 3 = 6 Total number of pairs the product of which is a multiple of 4: 9 + 6 = 15 Favorable Outcomes/Total Outcomes = 15/36 _________________ # Marty Murray Chief Curriculum and Content Architect [email protected] 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep", "3 1 4 3 2 5 4 1 5 4 2 6 Probability of sum being odd = $$\\frac { 4 }{ 8 }$$ = $$\\frac { 1 }{ 2 }$$ Probability of sum being even = $$\\frac { 4 }{ 8 }$$ = $$\\frac { 1 }{ 2 }$$ Kerala Syllabus 10th Standard Maths Chapter 3 Question 3. A box contains four slips numbers 1, 2, 3, 4 and another contains three slips numbered 1, 2, 3. If one slip is taken from each, what is the probability of the product being odd? The probability of the product being even? First box Second box Product 1 1 1 1 2 2 1 3 3 2 1 2 2 2 4 2 3 6 3 1 3 3 . 2 6 3 3 9 4 1 4 4 2 8 4 3 12 Probability of sum being odd = $$\\frac { 4 }{ 12 }$$ = $$\\frac { 1 }{ 3 }$$ Probability of sum being even = $$\\frac { 8 }{ 12 }$$ = $$\\frac { 2 }{ 3 }$$ 10 Th Maths Text Book Questions And Answers Question 4. From all two-digit numbers with either digit 1, 2 or 3 one number is chosen. i. What is the probability of both digits being the same? ii.", "4 # Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ? $\\frac{3}{4}$ $\\frac{1}{4}$ $\\frac{7}{4}$ $\\frac{1}{2}$ A. $\\frac{3}{4}$ So Probability =$\\frac{27}{36}=\\frac{3}{4}$", "not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number. If the product is odd, it can always be written as product of two odd numbers. Let’s go back to our question: We have 5 numbers: 1, 2, 3, 4, 5 Our favorable cases constitute those in which either both numbers are odd or the product has 4 as a factor. 3 Odd numbers: 1, 3, 5 2 Even numbers: 2, 4 Number of cases when both numbers are odd = 3C2 = 3 (select 2 of the 3 odd numbers) Number of cases when 4 is a factor of the product = Number of cases such that we select 4 and any other number = 1*4C1 = 4 Total number of favorable cases = 3 + 4 = 7 Note that this includes the case where we take both even numbers. Had there been more even numbers such as 6, we would have included more cases where we pick both even numbers such as 2 and 6 since their product would have 4 as a factor. Required Probability = 7/10 Takeaway: When can we write a number as difference of squares? -" ]

[ "; one card is drawn at random. Find the probability that the number on the card drawn is a multiple of : (i) 3 (ii) 5 (iii) 3 and 5 (iv) 3 or 5 Solution: Number of identical cards = 25 Numbers marked on their are 1 to 25 i.e. 1, 2, 3, 4, 5, …. 21, 22, 23, 24, 25 ∴ Number of possible outcome = 25 (i) Multiple of 3 are 3, 6, 9, 12, 15, 18, 21, 24 Which are 8 in numbers. (ii) Multiple of 5 are 5, 10, 15, 20, 25 which are 5 in numbers (iii) Multiple of 3 and 5 are = 15 which is 1 in number (iv) Multiple of 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25 which are 12 in numbers Question 4. A die is thrown once. Find the probability of getting a number : (i) less than 3. (ii) greater than or equal to 4. (iii) less than 8 (iv) greater than 6. Solution: A die has 6 numbers i.e., 1, 2, 3, 4, 5, 6 ∴ Number of possible outcome = 6 (i) Less than 3 are 1, 2 which are 2 in numbers (ii) Greater than or equal to 4 are 4, 5, 6 which are", "is a multiple of $$5$$, she then flips all cards whose card number is a multiple of $$6$$, then $$7$$, then $$8$$, and so on until she flips all cards whose number is a multiple of $$100$$. Once Sarai has finished, how many cards will have their red side facing up?", "# Math Help - Two Digits 1. ## Two Digits 10 cards with one integer between 1 and 5 written on them are inserted into a bag. Each integer will be written on two cards each. Two cards are taken from the bag, the first one takes the tens place, and the second one takes the ones place to create a two-digit-number. If the probability for any card to be taken out of the bag is equal, answer the following; (i) What is the probability for the two digits to be different integers? (ii) What is the probability for the number to be less than 20? (iii) What is the probability for the number to be a multiple of three? I'm going to be completely honest here and say I have no clue how to even get started on this one >_< Any input greatly appreciated. 2. the probability that the first digit is 1 is $\\frac{2}{10}$ the probability that the second one also is 1 is $\\frac{2-1}{10-1}$ since you can consider that one card is chosen a second or so after the first. Since the probability is the same for the other 4 digits, the probability of the two cards being the same is $5\\ \\frac{2}{10}\\ \\frac{1}{9}=\\frac{1}{9}$ 3. For the 2-digit number to be <20, the first", "# probability tricky question Nils has a telephone number with eight different digits. He has made 28 cards with statements of the type “The digit a occurs earlier than the digit b in my telephone number” – one for each pair of digits appearing in his number. How many cards can Nils show you without revealing his number Source: The Niels Henrik Abel mathematics competition • There is no probability involved in the question, in spite of what the title suggests, but a tricky probability question would be to find the probability of revealing the number after showing n randomly chosen cards - with replacement or without replacement. – Pere May 21 '17 at 18:17 • Has a correct answer been given? If so, please don't forget to $\\color{green}{\\checkmark \\small\\text{Accept}}$ it :) – Rubio Jun 19 '17 at 7:30 The question doesn't specify if the cards to show are chosen arbitrarily or chosen at random. Therefore, the question be understood two ways: 1. How many cards can Nils show you without revealing his number, provided that he can chose which cards he shows first? The answer to this question is the one given by JonMark Perry: 27 cards (that is, Nils can show all cards except one card relating two consecutive digits). 1. How many randomly chosen cards", "digit can only be 1. The numbers are 11, 12, 13, 14, 15 However, there are 2 of each number. 11 can occur in 2 ways 12 can occur in 4 ways if you imagine giving the digits different colours. There are 18 possibilities from 10(9) in total. Or, 5 possible indistinguishable 2-digit numbers from 5(5)=25 different possibilities. 4. The numbers range from 11 to 55. The multiples of 3 in this range are 12, 15, 18, 21.......up to 54. Any number ending in a digit >5 cannot appear, hence 12, 15, 21, 24, 33, 42, 45, 51 and 54 are the possible numbers divisible by 3. That's 9 out of 25 5. Hello, davidman! 10 cards with an integer from 1 and 5 written on them are inserted into a bag. Each integer will be written on two cards each. Two cards are taken from the bag (without replacement). The first takes the ten's place, and the second takes the one's place to create a two-digit-number. If the probability for any card to be taken out of the bag is equal, answer the following; (a) What is the probability for the two digits to be different integers? (b) What is the probability for the number to be less than 20? (c) What is the probability for the", "……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: 5 From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: 3 and 5 From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: 3 or 5 A die is thrown once. Find the probability of getting a number: less than 3 A die is thrown once. Find the probability of getting a number: greater than or equal to 4 A die is thrown once. Find the probability of getting a number: less than 8 A die is thrown once. Find the probability of getting a number: greater than 6 A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8? A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die. If two coins are tossed once, what is the", "a multiple of 5) = #### Question 16: There are 25 tickets numbered 1, 2, 3, 4,..., 25 respectively. One ticket is draw at random. What is the probability that the number on the ticket is a multiple of 3 or 5? (a) $\\frac{2}{5}$ (b) $\\frac{11}{25}$ (c) $\\frac{12}{25}$ (d) $\\frac{13}{25}$ (c) Explanation: ​Total number of tickets = 25 Let E be the event of getting a multiple of 3 or 5. Then, Multiples of 3 are 3, 6, 9, 12, 15, 18, 21 and 24. Multiples of 5 are 5, 10, 15, 20 and 25. Number of favourable outcomes = ( 8 + 5 − 1) = 12 ∴ (getting a multiple of 3 or 5 ) = P (E) = #### Question 17: Cards, each marked with one of the numbers 6, 7, 8,..., 15, are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with a number less than 10? (a) $\\frac{3}{5}$ (b) $\\frac{1}{3}$ (c) $\\frac{1}{2}$ (d) $\\frac{2}{5}$ (d) Explanation: All possible outcomes are 6, 7, 8................15. Number of all possible outcomes = 10 Out of these, the numbers that are less than 10 are 6, 7, 8 and 9. Number of favourable outcomes = 4 P(getting a number that is", "2.5 × 8 - 1.05 × 19 = 0.05 or 2.5 × 8 = 1.05 × 19 + 0.05 For each 0.05 we have to add 19 × 1.05 to end up at an integer multiple of 2.5. But we never have to add more than 50 × 0.05 = 2.5. Thus we divide the remaining value on your card by 0.05 (or multiply by 20) and multiply that by 19 which is 380 altogether. From this we take the remainder after division by 50. Cheers Thomas « Next Oldest | Next Newest » User(s) browsing this thread: 1 Guest(s)", "1, 3, 5, …., 49, i.e., 25 ∴ P(an odd number) = $\\frac{25}{49}$ (ii) ‘A multiple of 5’ numbers are 5, 10, 15, ……., 45, i.e., 9 ∴ P(a multiple of 5) = $\\frac{9}{49}$ (iii) “A perfect square” numbers are 1, 4, 9, …….., 49, i.e., 7 ∴ P(a perfect square number) = $\\frac{7}{49}=\\frac{1}{7}$ (iv) “An even prime number” is 2, i.e., only one number ∴ P(an even prime number) = $\\frac{1}{49}$ Question 46. A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (2015D) (i) divisible by 2 or 3, (ii) a prime number. Solution: (i) Numbers divisible by 2 or 3 from 1 to 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 9, 15, 20 = 13 Total outcomes = 20 Possible outcomes = 13 ∴ P(divisible by 2 or 3) = $\\frac{13}{20}$ (ii) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 = 8 Total Outcomes = 20 Possible outcomes = 8 ∴ P(a prime number) = $\\frac{8}{20}=\\frac{2}{5}$ Question 47. A bag contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number", "27 Tue 7 14 21 28 Wed 1 8 15 22 29 Thurs 2 9 16 23 30 Fri 3 10 17 24 31 sat 4 11 18 25 Sun 5 12 19 26 Find the probability that the choose: (1) a Wednesday (2) a friday (3) a tuesday or a saturday. Exercise 25 (B) [Page 393] ### Selina solutions for Concise Maths Class 10 ICSE Chapter 25 Probability Exercise 25 (B) [Page 393] Exercise 25 (B) | Q 1.1 | Page 393 Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be: an even number Exercise 25 (B) | Q 1.2 | Page 393 Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be: a multiple of 3 Exercise 25 (B) | Q 1.3 | Page 393 Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be: an even number and a multiple of 3 Exercise 25 (B) | Q 1.4 | Page 393 Nine cards (identical in all respects) are numbered 2", "8 = n(E)/ n(S) = 6/6 = 1 (iv) On a dice, numbers greater than 6 = 0 So, n(E) =... ### A die is thrown once. Find the probability of getting a number: (i) less than 3 (ii) greater than or equal to 4 Solution: We know that, In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6} So, n(S) = 6 (i) On a dice, numbers less than 3 = {1, 2} So, n(E) = 2 Hence, probability of getting a number... ### From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: (iii) 3 and 5 (iv) 3 or 5 (iii) From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e. {15} So, favorable number of events = n(E) = 1 Hence, probability of selecting a card with a multiple of 3 and... ### From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: (i) 3 (ii) 5 Solution: We know that, there are 25 cards from", "+ 2 + 0 + 4 + 5 = 12, which is divisible by 3 No other choice of 5 digits can have a sum divisible by 3, because there is no other way to make the sum 12 or 15, and we certainly can't have a sum of 9 or 18 So the number of 5-digit numbers that can be formed from the digits {1,2,3,4,5} is Number of ways = 1 x 2 x 3 x 4 x 5 = 120 And the number of 5-digit numbers that can be formed from the digits {1, 2, 0, 4, 5} is figured this way Number of ways = 1 x 2 x 3 x 4 x 4 = 96 Total Number of ways = 120 + 96 = 216 QUESTION: 33 A person draws a card from a pack of playing cards, replaces it and shuffles the pack. He continues doing this until he draws a spade. The chance that he will fail in first two times is Solution: QUESTION: 34 A five digit number is formed by writing the digits 1,2,3,4,5 in a random order without repetitions. Then the probability that the number is divisible by 4, is Solution: QUESTION: 35 You are given a box with 20 cards in it. 10 of these cards have the", "The four digits 5, 6, 7 and 8 are put at random in the spaces of the number below: 3 _ 1 _ 4 _ 0 _ 9 2 • Estimate the probability that the answer will be a multiple of 396.", "are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is: less than 48 Exercise 25 (B) | Q 3.1 | Page 393 From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: 3 Exercise 25 (B) | Q 3.2 | Page 393 From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: 5 Exercise 25 (B) | Q 3.3 | Page 393 From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: 3 and 5 Exercise 25 (B) | Q 3.4 | Page 393 From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of: 3 or 5 Exercise 25 (B) | Q 4.1 | Page 393 A", "multiple of 5 = 8 (5, 10, 15, 20, 25, 30, 35, 40) Probability = $\\\\ \\frac { 8 }{ 40 }$ = $\\\\ \\frac { 1 }{ 5 }$ (a) Question 22 If a number is randomly chosen from the numbers 1,2,3,4, …, 25, then the probability of the number to be prime is (a) $\\\\ \\frac { 7 }{ 25 }$ (b) $\\\\ \\frac { 9 }{ 25 }$ (c) $\\\\ \\frac { 11 }{ 25 }$ (d) $\\\\ \\frac { 13 }{ 25 }$ There are 25 number bearing numbers 1, 2, 3,…,25 Prime numbers are 2, 3, 5, 7, 11, 13, 17 19, 23 = 9 Probability being a prime number = $\\\\ \\frac { 9 }{ 25 }$ (b) Question 23 A box contains 90 cards numbered 1 to 90. If one card is drawn from the box at random, then the probability that the number on the card is a perfect square is (a) $\\\\ \\frac { 1 }{ 10 }$ (b) $\\\\ \\frac { 9 }{ 100 }$ (c) $\\\\ \\frac { 1 }{ 9 }$ (d) $\\\\ \\frac { 1 }{ 100 }$ In a box, there are 90 cards bearing numbers 1 to 90 Perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9 Probability", "with odd number are picked, (ii) a card labelled P or the card with a number which is multiple of 3 are picked. [5 marks] (a) Diagram 8.2 Solution: (a) (b)(i) n (s) = 9 P (a card labelled Q and a card with odd number) = {(Q 5), (Q 9)} $=\\frac{2}{9}$ (b)(ii) P (a card labelled P or a card with a number which is multiple of 3) = {(P 5), (P 6), (P 9), (Q 6), (Q 9), (R 6), (R 9)} $=\\frac{7}{9}$ ### 1 thought on “9.5.4 Probability of Combined Events, SPM Paper (Long Questions)” 1. VERY USEFUL AND GOOD QUESTIONS FOR PRACTICE. THANK YOU.", "## gabbyalicorn one year ago Cards numbered 1 through 30 are placed in a bag and shuffled. One card is chosen at random. What is the probability of randomly selecting a number less than or equal to 5, or a multiple of 7? https://www.dropbox.com/s/n43kflbepqbgbdm/Screenshot%202015-06-16%2018.54.18.png?dl=0 1. gabbyalicorn @jim_thompson5910 @mathstudent55 could you help me with math please? :) 2. misty1212 HI!! 3. misty1212 count ! 4. misty1212 how many numbers are less than or equal to five? (don't think too hard for this one) 5. gabbyalicorn Hello! I think 1 2 3 4 5 6. misty1212 those are the numbers, the question is \"how many\" of them are there? 7. gabbyalicorn oh, 5 8. misty1212 right 9. misty1212 now how many multiples of 7 are less than 30? 10. gabbyalicorn 4 11. misty1212 ok good, so how many total? 12. gabbyalicorn 9 13. misty1212 right 14. misty1212 9 out of 30 total is your answer, or as a fraction $\\frac{9}{30}$ which you probably want to reduce 15. gabbyalicorn so then 3/10 ? 16. misty1212 yes 17. gabbyalicorn :D Yay! Which would be A Thank you so much! 18. misty1212 $\\huge \\color\\magenta\\heartsuit$ 19. gabbyalicorn :D <3 20. anonymous :D 21. anonymous :D 22. anonymous I agree! 23. anonymous xD 24. misty1212 hmm got a big audience for this one must be", "+0 # Counting 0 48 3 Sam writes down the numbers 1, 2, 3, ..., 99 (a) How many digits did Sam write, in total? (b) Sam chooses one of the digits written down, at random. What is the probability that Sam chooses a 0? (c) What is the sum of all the digits that Sam wrote down? Feb 8, 2023 #1 +2602 +1 a) Sam wrote 9 x 1 = 9 digits from 1 - 9 Sam wrote 90 x 2 = 180 digits from 10 - 99 So, he wrote a total of $$180 + 9 = \\color{brown}\\boxed{189}$$ digits Feb 8, 2023 #2 +2602 +1 b) The only digits that are 0 occur in 10, 20, 30, ... , 90 There are 9 of these, so the probability is $${9 \\over 189} = \\color{brown}\\boxed{1 \\over 21}$$ Feb 8, 2023 #3 +2602 +1 c) Note that each digit, except for 0, occurs 20 times - 10 times as a tens place and 10 times as a one place So, the sum is just $$20(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = \\color{brown}\\boxed{900}$$ Feb 8, 2023", "Joined: 12 Sep 2015 Posts: 3726 Re: In a set of 24 cards, each card is numbered with a different positive [#permalink] Show Tags 25 Jun 2018, 07:22 1 1 Top Contributor Bunuel wrote: In a set of 24 cards, each card is numbered with a different positive integer from 1 to 24. One card will be drawn at random from the set. What is the probability that the card drawn will have either a number that is divisible by both 2 and 3 or a number that is divisible by 7 ? A. 3/24 B. 4/24 C. 7/24 D. 8/24 E. 17/24 P(event) = (number of outcomes that meet the given condition)/(total number of possible outcomes) So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = number of integers that are divisible by both 2 and 3 OR are divisible by 7/24 number of integers that are divisible by both 2 and 3 OR are divisible by 7 The integers that meet this condition are: 6, 7, 12, 14, 18, 21, 24 There are 7 such numbers So, P(selected number is divisible by both 2 and 3 OR is divisible by 7) = 7/24 Cheers, Brent _________________ Test confidently with gmatprepnow.com Math Expert Joined: 02 Aug 2009 Posts: 7686 Re: In", "is the probability that a boy from Form Five will be chosen? Solution: P (a boy from Form Five) $=\\frac{24}{90}=\\frac{4}{15}$ Question 8: Diagram below shows some number cards. A card is picked at random. State the probability that a prime number is picked. Solution: prime number = {11, 19, 37} P (a prime number) $=\\frac{3}{6}=\\frac{1}{2}$" ]

[ "not. So, we have to pick numbers by adding $8$ to every multiple of $8$ . The first multiple is $8$ . The next will be $8+8=16$ . Similarly, the other numbers divisible by $8$ are $24,32,40,48,56,64,72,80,88$ and $96$ . So, the total number of numbers between $1$ and $100$ that are divisible by $8$ is $12$ . Now, to find the value of $m$ , we first need to find the probability of choosing a number divisible by $8$ from a set of numbers between $1$ and $100$ . The favorable event is choosing a number between $1$ and $100$ which is divisible by $8$. So , the number of favorable outcomes is $12$. The total possible outcome is choosing any number between $1$ and $100$. So , the number of possible outcomes is $100$ . So , probability $P=\\dfrac{12}{100}=\\dfrac{3}{25}$. Now , in the question , the value of the probability is given as $\\dfrac{m}{25}$ . So, $\\dfrac{m}{25}=\\dfrac{3}{25}$ . Or, $m=3$ . Hence, the value of $m$ is equal to $3$. Note: To find the number of numbers between $1$ and $100$ that are divisible by $8$ , we can divide $100$ by $8$ , and the quotient will be our required value , i.e. $\\dfrac{100}{8}=12(rem=4)$. So, the number of numbers between $1$ and $100$ that are", "Number of card in a box = 25 numbered 1 to 25 (i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 i.e. number of favourable outcomes = 12 Probability of an even number will be P(E) = $\\\\ \\frac { 12 }{ 25 }$ (ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 i.e. number of primes = 9 Probability of primes will be P(E) = $\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$ $\\\\ \\frac { 9 }{ 25 }$ (iii) Multiples of 6 are 6, 12, 18, 24 Number of multiples = 4 Probability of multiples of 6 will be P(E) = $\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$ $\\\\ \\frac { 4 }{ 25 }$ Question 26 A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is : (i) Odd (ii) prime (iii) divisible by 3 (iv) divisible by 3 and 2 both (v) divisible by 3 or 2 (vi) a perfect square number. Number of cards in a box =15 numbered 1 to 15 (i) Odd numbers are 1, 3, 5, 7,", "first place. So, the number divisible by 4 in the above case can be formed in 2 × 2 × 3 = 12 ways. Hence total number of ways by which 5-digit number divisible by 4 = 12 + 18 = 30 Calculation: Favourable outcomes = 30 Total number of outcomes = 4 × 4! = 96 Probability = (Favourable outcomes) / (Total number of outcomes) $$= \\frac{{30}}{{96}} = \\frac{5}{{16}}$$", "45, 50, 55, 60, 65, 70, 75, 80, 85, 90 which are 18 in numbers. Probability of number divisible by 5 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 18 }{ 90 }$$ = $$\\\\ \\frac { 1 }{ 5 }$$ Question 31. Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is (i) an even number (ii) a number less than 14 (iii) a number which is a perfect square (iv) a prime number less than 30. Solution: Number of cards with numbered from 2 to 101 are placed in a box Number of possible outcomes = 100 one card is drawn (i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16,….., 96, 98, 100 which are 50 in numbers. Probability of even number will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 50 }{ 100 }$$ = $$\\\\ \\frac { 1 }{ 2 }$$ (ii) Numbers less than 14 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 which are 12 in numbers", "## Algebra and Trigonometry 10th Edition $\\frac{10}{13}$ We know that $probability=\\frac{\\text{number of favourable outcomes}}{\\text{number of all outcomes}}$ The total number of cards (the number of all outcomes) is $52$ There are $12$ face cards ($3$ in each of the $4$ suits), thus there are $52-12=40$ non-face cards, which are the \"good outcomes\". Hence the probability here is: $\\frac{40}{52}=\\frac{10}{13}$", "placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is (i) divisible by 5 (ii) a number which is a perfect square. Solution: Number of cards which are marked with numbers 13, 14, 15, 16, 17,….to 59, 60 are = 48 (i) Numbers which are divisible by 5 will be 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10 Probability of number divisible by 5 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 10 }{ 48 }$$ = $$\\\\ \\frac { 5 }{ 24 }$$ (ii) Numbers which is a perfect square are 16, 25, 36, 49 which are 4 in numbers. Probability of number which is a perfect square will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 4 }{ 48 }$$ = $$\\\\ \\frac { 1 }{ 12 }$$ Question 11. The box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from the box has (i) an odd number (ii) a perfect square number. Solution:", "of these numbers, number divisible by 3 and 5 means number divisible by 15 is 15. Number of favourable outcomes = 1. $\\therefore \\mathrm{P}$ (getting a number divisible by 3 and 5$)=\\mathrm{P}\\left(\\mathrm{E}_{2}\\right)=\\frac{\\text { Number of outcomes favourable to } \\mathrm{E}_{2}}{\\text { Number of all possible outcomes }}$ $=\\frac{1}{18}$ Thus, the probability of getting a card bearing a number divisible by 3 and 5 is $\\frac{1}{18}$.", "So the number of those cards which are not divisible by 3 is 20 P(getting a number not divisible by 3) = $\\frac { 20 }{ 30 } =\\frac { 2 }{ 3 }$ Question 18: Total number of all possible outcomes = 25 (i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 P(getting an even number) = $\\frac { 12 }{ 25 }$ (ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 P(getting a prime number) = $\\frac { 9 }{ 25}$ (iii) Numbers that are multiples of 6 are 6, 12, 18, 24 P(getting a multiple of 6) = $\\frac { 4 }{ 25 }$ Question 19: Number of all possible outcomes = 19 (i) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 P(getting a prime number) = $\\frac { 8 }{ 19 }$ (ii) Numbers divisible by 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18 P(getting a number divisible by 3 or 5) = $\\frac { 8 }{ 19 }$ (iii) Numbers divisible by 5 and 10 are 5, 10, 15 P(getting a number neither divisible by 5 nor 10) = $1-\\frac { 3 }{ 19 } =\\frac { 16 }{ 19 }$ (iv) Even numbers are", "}{ 35 }$$ Question 10. Cards marked with numbers 13, 14, 15,…..60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is (i) divisible by 5 (ii) a number which is a perfect square. Solution: Number of cards which are marked with numbers 13, 14, 15, 16, 17,….to 59, 60 are = 48 (i) Numbers which are divisible by 5 will be 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10 Probability of number divisible by 5 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 10 }{ 48 }$$ = $$\\\\ \\frac { 5 }{ 24 }$$ (ii) Numbers which is a perfect square are 16, 25, 36, 49 which are 4 in numbers. Probability of number which is a perfect square will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 4 }{ 48 }$$ = $$\\\\ \\frac { 1 }{ 12 }$$ Question 11. The box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from", "Question # A box contains $$25$$ cards numbered from $$1$$ to $$25$$. A card is drawn at random from the bad. Find the probability that the number on the drawn card is divisible by $$2$$ or $$3$$. Solution ## Total numbers of cards $$=25$$The favourable outcome = numbers divisible by $$2=\\left\\{ 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\\right\\}$$The favourable outcome = number divisible by $$3=\\left\\{ 3, 6, 9, 12, 15, 18, 21, 24 \\right\\}$$Total number of favorable outcomes $$=16$$( Note: count $$6, 12, 18$$ and $$24$$ are common, so consider only one set )Probability is given by,$$\\displaystyle P(A)$$$$=\\dfrac{\\text{no. of possible outcomes}}{\\text{No. of total outcomes}}$$Now, $$P($$ drawn card is divisible by $$2$$ or $$3)=16/25$$MathematicsRS AgarwalStandard X Suggest Corrections 0 Similar questions View More Same exercise questions View More People also searched for View More", "probability of the event that a number chosen from $1-100$is a prime number? 1. $\\dfrac{1}{5}$ 2. $\\dfrac{6}{25}$ 3. $\\dfrac{1}{4}$ 4. $\\dfrac{13}{50}$ Ans: Firstly, write the favorable outcomes so, $\\{ 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97\\}=24$ And total number of outcomes is $100$ Formula to find the probability is $p=\\dfrac{\\text{favorable outcomes}}{\\text{total number of outcomes}}$ $p=\\dfrac{24}{100}$ $p =\\dfrac{6}{25}$ Hence, the total probability of event that a number chosen from $1-100$ is a prime number is $\\dfrac{6}{25}$ and the correct option is (B). 4. There are $40$ cards in a bag. Each bears a number from $1-40$. One card is drawn at random. What is the probability that the card bears a number which is a multiple of $5$? 1. $\\dfrac{1}{5}$ 2. $\\dfrac{3}{5}$ 3. $\\dfrac{4}{5}$ 4. $\\dfrac{1}{3}$ Ans: first find out the numbers which are multiple of $5$ so, $\\left\\{ 5,10,15,20,25,30,35,40 \\right\\}=8$, thus the favorable outcomes are $8$. to find probability is $p=\\dfrac{\\text{favorable outcomes}}{\\text{total number of outcomes}}$ Total outcomes are $40$ so probability that the card bears a number which is a multiple of $5$ will be $\\Rightarrow \\dfrac{8}{40}=\\dfrac{1}{5}$. hence, option (A) is correct. 5. If $n\\left( A \\right)=2,p\\left( A \\right)=\\dfrac{1}{5}$, then $n\\left( S \\right)=?$. 1. $10$ 2. $\\dfrac{5}{2}$ 3. $\\dfrac{2}{5}$ 4. $\\dfrac{1}{3}$ Ans: Applying the probability formula $p\\left( A \\right)=\\dfrac{n\\left( A \\right)}{n\\left( S \\right)}$ $\\Rightarrow \\dfrac{1}{5}=\\dfrac{2}{n\\left( S \\right)}$ $\\Rightarrow n\\left( S \\right)=10$ Hence, the correct", "(iii) multiple of 6 The possible outcomes are 1,2, 3,4, 5................25. Number of all possible outcomes = 25 (i) Out of the given numbers, the even numbers are 2, 4, 6, 8,10,12,14,16, 18, 20,22 and 24. Let E1 be the event of getting an even number. Then, number of favourable outcomes = 12​ ∴ P (getting an even number) = $\\frac{12}{25}$ (ii) Out of the given numbers, the prime numbers are 2, 3, 5, 7,11,13, 17, 19 and 23. Let E2 be the event of getting a prime number. Then,​ number of favourable outcomes =​ 9 ∴ P (getting a prime number) = $\\frac{9}{25}$ (iii) Out of the given numbers, the numbers that are multiples of 6 are 6,12,18 and 24. Let E3 be the event of getting a multiple of 6. Then,​ number of favourable outcomes = 4 P (getting a multiple of 6) = $\\frac{4}{25}$ #### Question 19: A box contains 19 balls bearing numbers 1, 2, 3, ..., 19 respectively. A ball is drawn at random from the box. Find the probability that the number on the ball is (i) a prime number (ii) divisible by 3 or 5 (iii) neither divisible by 5 nor by 10 (iv) an even number The possible outcomes are 1,2, 3,4, 5................19. Number of all possible outcomes = 19", "13 Q5Find the probability of drawing a 6 from a stack of cards numbered 0 through 11. A. 112 B. 14 C. 13 D. 16 Step: 1 Number of possible outcomes is 12. [Count numbers 0 to 11.] Step: 2 Number of favorable outcomes is 1. Step: 3 Probability(event) = Number of favorable outcomesNumber of possible outcomes [Formula.] Step: 4 P(6) = 112 [Substitute the values.] Step: 5 Probability of drawing the card numbered 6 is 112. Correct Answer is : 112 Q6What is the probability of choosing an even number from the set of numbers from 1 to 10? A. 1 B. 110 C. 12 D. 15 Step: 1 There are 10 numbers from 1 to 10. So, the number of possible outcomes is 10. Step: 2 Number of favorable outcomes is 5. [The even numbers are 2, 4, 6, 8 and 10.] Step: 3 Probability(event) = Number of favorable outcomesNumber of possible outcomes [Formula.] Step: 4 P(even number) = 510 [Substitute the values.] Step: 5 = 12 [Simplify.] Step: 6 Probability of selecting an even number from the set of numbers from 1 to 10 is 12. Correct Answer is : 12 Q7What is the probability of selecting a digit, which is multiple of 2, from the numbers 2, 7, 4 and 8? A. 34 B.", "equal to $3$. probability = number of outcomes possible / total number of outcomes $\\frac{3}{6} = \\frac{1}{2}$", "8, 10, 12, 14, 16, 18, 20, 22, 24 i.e. number of favourable outcomes = 12 Probability of an even number will be P(E) = $$\\\\ \\frac { 12 }{ 25 }$$ (ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 i.e. number of primes = 9 Probability of primes will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 9 }{ 25 }$$ (iii) Multiples of 6 are 6, 12, 18, 24 Number of multiples = 4 Probability of multiples of 6 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 4 }{ 25 }$$ Question 26. A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is : (i) Odd (ii) prime (iii) divisible by 3 (iv) divisible by 3 and 2 both (v) divisible by 3 or 2 (vi) a perfect square number. Solution: Number of cards in a box =15 numbered 1 to 15 (i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15 Number of odd numbers = 8 Probability of odd numbers will be", "favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$ = $\\\\ \\frac { 9 }{ 18 }$ = $\\\\ \\frac { 1 }{ 2 }$ Question 8 A box contains 20 balls bearing numbers 1, 2, 3, 4,……, 20. A ball is drawn at random from the box. What is the probability that the number on the ball is (i) an odd number (ii) divisible by 2 or 3 (iii) prime number (iv) not divisible by 10? In a box, there are 20 balls containing 1 to 20 number Number of possible outcomes = 20 (i) Numbers which are odd will be, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 = 10 balls. Probability of odd ball will be P(E) = $\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$ = $\\\\ \\frac { 10 }{ 20 }$ = $\\\\ \\frac { 1 }{ 2 }$ (ii) Numbers which are divisible by 2 or 3 will be 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 = 13 balls Probability of ball which is divisible by 2 or 3 will be P(E) = $\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$ = $\\\\ \\frac { 13 }{ 20 }$ (iii) Prime numbers will be 2, 3,", "card is drawn at random from the bag. The probability that the number on this card is divisible by both 2 and 3 is Solution: Total number of outcomes = 25 The number which is divisible by both 2 and 3 are 6, 12, 18, 24. Number of favourable outcomes = 4 Probability of number which is divisible by both 2 and 3 = $$\\frac { 4 }{ 25 }$$ (c) Hope given RD Sharma Class 10 Solutions Chapter 16 Probability MCQS are helpful to complete your math homework. If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.", "× # Proof note - Combinatorics Prove that the operators $$+$$ and $$-$$ cannot be chosen such that : $$\\quad$$ $$\\pm1 \\pm 2 \\pm 3 \\dots \\pm 100 = 101$$ Note by Arian Tashakkor 2 years ago Sort by: Firstly,suppose all the operators are chosen as $$+$$ sign.That gives a total sum of 5050 which is an even number.Now changing each $$+$$ to a $$-$$ an even amount is reduced from the total sum and hence , the sum always remains an even integer!(for example consider changing $$+3$$ to $$-3$$ , the sum is reduced by $$2\\times3=6$$) · 2 years ago Separate the sum into evens and odds. Any number of evens, regardless of the parities, will and to an even number. Any two odds, regardless of parity, will add to an even number. Now the sum has $$50$$ odd numbers, so we can form $$25$$ pairs of odds in any of$$\\frac{50!}{2^{25}}$$ ways. Whichever way we do this, we will have $$25$$ even numbers as a result, which of course add to an even number, again regardless of parity. Thus the given sum, with whatever operators we choose, must be even, and hence can never result in a sum of $$101.$$ The next question to ask is: how many distinct values can be achieved from the given (variable)", "$$n!$$ # of ways and out of these $$n!$$ ways only one, namely $$\\{x_1, \\ x_2, \\ ..., \\ x_n\\}$$ will be in ascending order. So 1 out of n!. $$P=\\frac{1}{n!}$$. Hence, according to the above, the only thing we need to know to answer the question is the size of the subset ($$n$$) we are selecting from set $$A$$. Hi, Got the first point but a doubt on the second one how can you say the subset of n numbers will be selected in n! ways. Suppose a situation where there are 12 distinct numbers and you select 5. Then the possible combinations will be 12C5 ways i.e 12!/(5!*7!) ways Say n = 5 and k = 12 (1, 2, 3, 4, ..., 12). Now, the number of 5-number groups out of 12 is indeed 12C5. Second point there says that the number of different sequences of 5 numbers is 5!. For example, if 5 numbers we are selecting are a, b, c, d, e, then the order of selections can be a, b, c, d, e, or e, d, c, b, a, or, e, a, b, c, d, ... Total = 5!. ok got it SO we can say Probability = possible outcomes/ total outcomes = 12C5/12P5 12C5: shows selection of 5 numbers and out of", "the event the sum of the two numbers is 8′ denoted by E, are : (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (see figure) i.e., the number of outcomes favourable to E = 5. Hence, P(E) = $$\\frac{5}{36}$$ (ii) As you can see from figure, there is no outcome favourable to the event F, ‘the sum of two numbers is 13’. So, P(F) = $$\\frac{0}{36}$$ = 0 (iii) As you can see from figure, all the outcomes are favourable to the event G, ‘sum of two numbers ≤ 12. So, P(G) = $$\\frac{36}{36}$$ = 1. Question 5. A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards throughly. Find the probability that the number on the drawn card is: (i) an odd number. (ii) a multiple of 5. (iii) a perfect square. (iv) an even prime number. Solution: Total number of cards = 49 Total number of outcomes = 49 (i) Odd number Favourable outcomes : 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49 Number of favourable outcomes = 25 Question 6. All the black face cards are removed from a pack of 52 playing cards." ]

[ "Number of card in a box = 25 numbered 1 to 25 (i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 i.e. number of favourable outcomes = 12 Probability of an even number will be P(E) = $\\\\ \\frac { 12 }{ 25 }$ (ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 i.e. number of primes = 9 Probability of primes will be P(E) = $\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$ $\\\\ \\frac { 9 }{ 25 }$ (iii) Multiples of 6 are 6, 12, 18, 24 Number of multiples = 4 Probability of multiples of 6 will be P(E) = $\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$ $\\\\ \\frac { 4 }{ 25 }$ Question 26 A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is : (i) Odd (ii) prime (iii) divisible by 3 (iv) divisible by 3 and 2 both (v) divisible by 3 or 2 (vi) a perfect square number. Number of cards in a box =15 numbered 1 to 15 (i) Odd numbers are 1, 3, 5, 7,", "1, 3, 5, …., 49, i.e., 25 ∴ P(an odd number) = $\\frac{25}{49}$ (ii) ‘A multiple of 5’ numbers are 5, 10, 15, ……., 45, i.e., 9 ∴ P(a multiple of 5) = $\\frac{9}{49}$ (iii) “A perfect square” numbers are 1, 4, 9, …….., 49, i.e., 7 ∴ P(a perfect square number) = $\\frac{7}{49}=\\frac{1}{7}$ (iv) “An even prime number” is 2, i.e., only one number ∴ P(an even prime number) = $\\frac{1}{49}$ Question 46. A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (2015D) (i) divisible by 2 or 3, (ii) a prime number. Solution: (i) Numbers divisible by 2 or 3 from 1 to 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 9, 15, 20 = 13 Total outcomes = 20 Possible outcomes = 13 ∴ P(divisible by 2 or 3) = $\\frac{13}{20}$ (ii) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 = 8 Total Outcomes = 20 Possible outcomes = 8 ∴ P(a prime number) = $\\frac{8}{20}=\\frac{2}{5}$ Question 47. A bag contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number", "divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24. Out of the given numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21 and 24. Out of the given numbers, numbers divisible by both 2 and 3 are 6, 12, 18 and 24. Number of favourable outcomes = 16 ∴ P(getting a card divisible by 2 or 3) = P(E1) = = $\\frac{16}{25}$ Thus, the probability that the number on the drawn card is divisible by 2 or 3 is $\\frac{16}{25}$. (ii) Let E2 be the event of getting a prime number. Out of the given numbers, prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23. Number of favourable outcomes = 9 ∴ P(getting a prime number) = P(E2) = = $\\frac{9}{25}$ Thus, the probability that the number on the drawn card is a prime number is $\\frac{9}{25}$. #### Question 2: A box contains cards numbered 3, 5, 7, 9, .... , 35, 37. A card is drawn at random from the box. Find the probability that the number on the card is a prime number. Given numbers 3, 5, 7, 9, .... , 35, 37 form an AP with a = 3 and d = 2. Let Tn = 37. Then, 3", "8, 10, 12, 14, 16, 18, 20, 22, 24 i.e. number of favourable outcomes = 12 Probability of an even number will be P(E) = $$\\\\ \\frac { 12 }{ 25 }$$ (ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 i.e. number of primes = 9 Probability of primes will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 9 }{ 25 }$$ (iii) Multiples of 6 are 6, 12, 18, 24 Number of multiples = 4 Probability of multiples of 6 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 4 }{ 25 }$$ Question 26. A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is : (i) Odd (ii) prime (iii) divisible by 3 (iv) divisible by 3 and 2 both (v) divisible by 3 or 2 (vi) a perfect square number. Solution: Number of cards in a box =15 numbered 1 to 15 (i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15 Number of odd numbers = 8 Probability of odd numbers will be", "the box has (i) an odd number (ii) a perfect square number. Solution: Cards in a box are from 14 to 99 = 86 No. of total cards = 86 One card is drawn at random Cards bearing odd numbers are 15, 17, 19, 21, …, 97, 99 Which are 43 (i) P(E) = $$\\frac { Number\\quad of\\quad actual\\quad events }{ Number\\quad of\\quad total\\quad events }$$ = $$\\\\ \\frac { 43 }{ 86 }$$ = $$\\\\ \\frac { 1 }{ 2 }$$ (ii) Cards bearing number which are a perfect square = 16, 25, 36, 49, 64, 81 Which are 6 P(E) = $$\\frac { Number\\quad of\\quad actual\\quad events }{ Number\\quad of\\quad total\\quad events }$$ = $$\\\\ \\frac { 6 }{ 86 }$$ = $$\\\\ \\frac { 3 }{ 43 }$$ Question 12. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is four times that of a red ball, find the number of balls in the bags. Solution: Number of red balls = 5 and let number of blue balls = x Total balls in the bag = 5 + x and that of red balls = $$\\\\ \\frac { 5 }{ 5+x }$$ According to the condition, $$\\frac { x }{ 5+x } =4\\times \\frac { 5", "be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 16 }{ 19 }$$ (iv) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18 which are 9 in numbers. Probability of there number will be Number of favourable outcome P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 9 }{ 19 }$$ Question 28. Cards marked with numbers 13, 14, 15, …, 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card drawn is (i) divisible by 5 (ii) a perfect square number. Solution: Number of card bearing numbers 13,14,15, … 60 = 48 One card is drawn at random. (i) Card divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10 Probability = $$\\\\ \\frac { 10 }{ 48 }$$ = $$\\\\ \\frac { 5 }{ 24 }$$ (ii) A perfect square = 16, 25, 36, 49 = 4 Probability = $$\\\\ \\frac { 4 }{ 48 }$$ = $$\\\\ \\frac { 1 }{ 12 }$$ Question 29. Tickets numbered 3, 5, 7, 9,…., 29 are placed in a box and mixed thoroughly.", "90 }$$ = $$\\\\ \\frac { 1 }{ 5 }$$ Question 31. Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is (i) an even number (ii) a number less than 14 (iii) a number which is a perfect square (iv) a prime number less than 30. Solution: Number of cards with numbered from 2 to 101 are placed in a box Number of possible outcomes = 100 one card is drawn (i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16,….., 96, 98, 100 which are 50 in numbers. Probability of even number will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 50 }{ 100 }$$ = $$\\\\ \\frac { 1 }{ 2 }$$ (ii) Numbers less than 14 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 which are 12 in numbers Probability of number less than 14 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 12 }{ 100 }$$ = $$\\\\ \\frac { 3 }{ 25 }$$ (iii) Perfect square are 4,", "So the number of those cards which are not divisible by 3 is 20 P(getting a number not divisible by 3) = $\\frac { 20 }{ 30 } =\\frac { 2 }{ 3 }$ Question 18: Total number of all possible outcomes = 25 (i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 P(getting an even number) = $\\frac { 12 }{ 25 }$ (ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 P(getting a prime number) = $\\frac { 9 }{ 25}$ (iii) Numbers that are multiples of 6 are 6, 12, 18, 24 P(getting a multiple of 6) = $\\frac { 4 }{ 25 }$ Question 19: Number of all possible outcomes = 19 (i) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 P(getting a prime number) = $\\frac { 8 }{ 19 }$ (ii) Numbers divisible by 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18 P(getting a number divisible by 3 or 5) = $\\frac { 8 }{ 19 }$ (iii) Numbers divisible by 5 and 10 are 5, 10, 15 P(getting a number neither divisible by 5 nor 10) = $1-\\frac { 3 }{ 19 } =\\frac { 16 }{ 19 }$ (iv) Even numbers are", "is (i) divisible by 2 or 3, (ii) a prime number. Solution: Total number of outcomes = 25 (i) Let E1 be the event of getting a card divisible by 2 or 3. Out of given numbers, numbers divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24. Out of the given numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21 and 24. Out of the given numbers, numbers divisible by both 2 and 3 are 6, 12, 18 and 24. Number of favorable outcomes = 16 Therefore, P(getting a card divisible by 2 or 3) = P(E1) = $\\frac{Number\\, of\\, outcomes\\, favorable\\, to\\, E_{1}}{Number\\, of\\, all\\, possible\\, outcomes}$ = $\\frac{16}{25}$ Thus, the probability that the number on the drawn card is divisible by 2 or 3 is $\\frac{16}{25}$. (ii) Let E2 be the event of getting a prime number. Out of the given numbers, prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23. Number of favorable outcomes = 9 Therefore, P(getting a prime number) = P(E2) = $\\frac{Number\\, of\\, outcomes\\, favorable\\, to\\, E_{2}}{Number\\, of\\, all\\, possible\\, outcomes}$ = $\\frac{9}{25}$ Thus, the probability that the number on the drawn card is a prime number is $\\frac{9}{25}$. QUESTION 3:A box contains cards numbered 3,", "are 18 in numbers. Probability of number divisible by 5 will be P(E) = $\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$ $\\\\ \\frac { 18 }{ 90 }$ $\\\\ \\frac { 1 }{ 5 }$ Question 31 Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is (i) an even number (ii) a number less than 14 (iii) a number which is a perfect square (iv) a prime number less than 30. Number of cards with numbered from 2 to 101 are placed in a box Number of possible outcomes = 100 one card is drawn (i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16,….., 96, 98, 100 which are 50 in numbers. Probability of even number will be P(E) = $\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$ $\\\\ \\frac { 50 }{ 100 }$ $\\\\ \\frac { 1 }{ 2 }$ (ii) Numbers less than 14 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 which are 12 in numbers Probability of number less than 14 will be P(E) = $\\frac { Number\\quad of\\quad favourable\\quad outcome", "}{ 35 }$$ Question 10. Cards marked with numbers 13, 14, 15,…..60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is (i) divisible by 5 (ii) a number which is a perfect square. Solution: Number of cards which are marked with numbers 13, 14, 15, 16, 17,….to 59, 60 are = 48 (i) Numbers which are divisible by 5 will be 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10 Probability of number divisible by 5 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 10 }{ 48 }$$ = $$\\\\ \\frac { 5 }{ 24 }$$ (ii) Numbers which is a perfect square are 16, 25, 36, 49 which are 4 in numbers. Probability of number which is a perfect square will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 4 }{ 48 }$$ = $$\\\\ \\frac { 1 }{ 12 }$$ Question 11. The box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from", "100 }$ (iv) Prime numbers less than 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 Total number of prime numbers =12 P(getting a prime number less 40) = $\\frac { 12 }{ 100 } =\\frac { 3 }{ 25 }$ Question 23: Cards are marked with numbers 13, 14, 15, �, 60 the number of cards = 48 (i) The numbers on cards which are divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 the number of favorable cases = 10 probability of getting a card with a number divisible by 5 = $\\frac { 10 }{ 48 } =\\frac { 5 }{ 24 }$ (ii) The numbers on cards which are perfect squares are 16, 25, 36, 49 the number of favorable outcomes = 4 probability of getting a card with number which is a perfect square = $\\frac { 4 }{ 48 } =\\frac { 1 }{ 12 }$ Question 24: The cards are marked 5 to 50 the total number of cards = 46 (i) Prime numbers less than 10 are 5 and 7 There are two prime numbers Probability of getting a prime number less than 10 = $\\frac { 2 }{ 46 } =\\frac { 1 }{ 23 }$ (ii) Perfect square", "45, 50, 55, 60, 65, 70, 75, 80, 85, 90 which are 18 in numbers. Probability of number divisible by 5 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 18 }{ 90 }$$ = $$\\\\ \\frac { 1 }{ 5 }$$ Question 31. Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is (i) an even number (ii) a number less than 14 (iii) a number which is a perfect square (iv) a prime number less than 30. Solution: Number of cards with numbered from 2 to 101 are placed in a box Number of possible outcomes = 100 one card is drawn (i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16,….., 96, 98, 100 which are 50 in numbers. Probability of even number will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 50 }{ 100 }$$ = $$\\\\ \\frac { 1 }{ 2 }$$ (ii) Numbers less than 14 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 which are 12 in numbers", "a multiple of 5) = #### Question 16: There are 25 tickets numbered 1, 2, 3, 4,..., 25 respectively. One ticket is draw at random. What is the probability that the number on the ticket is a multiple of 3 or 5? (a) $\\frac{2}{5}$ (b) $\\frac{11}{25}$ (c) $\\frac{12}{25}$ (d) $\\frac{13}{25}$ (c) Explanation: ​Total number of tickets = 25 Let E be the event of getting a multiple of 3 or 5. Then, Multiples of 3 are 3, 6, 9, 12, 15, 18, 21 and 24. Multiples of 5 are 5, 10, 15, 20 and 25. Number of favourable outcomes = ( 8 + 5 − 1) = 12 ∴ (getting a multiple of 3 or 5 ) = P (E) = #### Question 17: Cards, each marked with one of the numbers 6, 7, 8,..., 15, are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with a number less than 10? (a) $\\frac{3}{5}$ (b) $\\frac{1}{3}$ (c) $\\frac{1}{2}$ (d) $\\frac{2}{5}$ (d) Explanation: All possible outcomes are 6, 7, 8................15. Number of all possible outcomes = 10 Out of these, the numbers that are less than 10 are 6, 7, 8 and 9. Number of favourable outcomes = 4 P(getting a number that is", "# A Bag Contains Cards Numbered from 1 to 25. a Card is Drawn at Random from the Bag. the Probability that the Number on this Card is Divisible by Both 2 and 3 is - Mathematics MCQ A bag contains cards numbered from 1 to 25. A card is drawn at random from the bag. The probability that the number on this card is divisible by both 2 and 3 is #### Options • $\\frac{1}{5}$ • $\\frac{3}{25}$ • $\\frac{4}{25}$ • $\\frac{2}{25}$ #### Solution The total number of cards in the bag is 25. ∴ Total number of outcomes = 25 The numbers from 1 to 25 which are divisible by both 2 and 3 are 6, 12, 18 and 24. So, the favourable number of outcomes are 4. ∴ P(number on the drawn card is divisible by both 2 and 3) = $\\frac{\\text{ Favourable number of outcomes }}{\\text{ Total number of outcomes }} = \\frac{4}{25}$ Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 10 Maths Chapter 16 Probability Q 40 | Page 39", "placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is (i) divisible by 5 (ii) a number which is a perfect square. Solution: Number of cards which are marked with numbers 13, 14, 15, 16, 17,….to 59, 60 are = 48 (i) Numbers which are divisible by 5 will be 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10 Probability of number divisible by 5 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 10 }{ 48 }$$ = $$\\\\ \\frac { 5 }{ 24 }$$ (ii) Numbers which is a perfect square are 16, 25, 36, 49 which are 4 in numbers. Probability of number which is a perfect square will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 4 }{ 48 }$$ = $$\\\\ \\frac { 1 }{ 12 }$$ Question 11. The box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from the box has (i) an odd number (ii) a perfect square number. Solution:", "multiple of 5 = 8 (5, 10, 15, 20, 25, 30, 35, 40) Probability = $\\\\ \\frac { 8 }{ 40 }$ = $\\\\ \\frac { 1 }{ 5 }$ (a) Question 22 If a number is randomly chosen from the numbers 1,2,3,4, …, 25, then the probability of the number to be prime is (a) $\\\\ \\frac { 7 }{ 25 }$ (b) $\\\\ \\frac { 9 }{ 25 }$ (c) $\\\\ \\frac { 11 }{ 25 }$ (d) $\\\\ \\frac { 13 }{ 25 }$ There are 25 number bearing numbers 1, 2, 3,…,25 Prime numbers are 2, 3, 5, 7, 11, 13, 17 19, 23 = 9 Probability being a prime number = $\\\\ \\frac { 9 }{ 25 }$ (b) Question 23 A box contains 90 cards numbered 1 to 90. If one card is drawn from the box at random, then the probability that the number on the card is a perfect square is (a) $\\\\ \\frac { 1 }{ 10 }$ (b) $\\\\ \\frac { 9 }{ 100 }$ (c) $\\\\ \\frac { 1 }{ 9 }$ (d) $\\\\ \\frac { 1 }{ 100 }$ In a box, there are 90 cards bearing numbers 1 to 90 Perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9 Probability", "# Aptitude:: Probability @ : Home > Aptitude > Probability > General Questions ### Exercise \" #### A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white? A. $\\frac{3}{4}$ B. $\\frac{4}{7}$ C. $\\frac{1}{8}$ D. $\\frac{3}{7}$ #### One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)? A. $\\frac{1}{13}$ B. $\\frac{3}{13}$ C. $\\frac{1}{4}$ D. $\\frac{9}{52}$ #### Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is: A. $\\frac{3}{20}$ B. $\\frac{29}{34}$ C. $\\frac{47}{100}$ D. $\\frac{13}{102}$ #### A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is: A. $\\frac{1}{22}$ B. $\\frac{3}{22}$ C. $\\frac{2}{91}$ D. $\\frac{2}{77}$ #### A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is: A. $\\frac{1}{13}$ B. $\\frac{2}{13}$ C. $\\frac{1}{26}$ D. $\\frac{1}{52}$ Page 1 of 3 Submit*", "19, 23 i.e. number of primes = 9 Probability of primes will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 9 }{ 25 }$$ (iii) Multiples of 6 are 6, 12, 18, 24 Number of multiples = 4 Probability of multiples of 6 will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 4 }{ 25 }$$ Question 26. A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is : (i) Odd (ii) prime (iii) divisible by 3 (iv) divisible by 3 and 2 both (v) divisible by 3 or 2 (vi) a perfect square number. Solution: Number of cards in a box =15 numbered 1 to 15 (i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15 Number of odd numbers = 8 Probability of odd numbers will be P(E) = $$\\frac { Number\\quad of\\quad favourable\\quad outcome }{ Number\\quad of\\quad possible\\quad outcome }$$ = $$\\\\ \\frac { 8 }{ 15 }$$ (ii) Prime number are 2, 3, 5, 7, 11, 13 Number of primes is 6 Probability of prime number will be", "+0 0 111 2 Cards numbered 1 through 30 are placed in a bag and shuffled. One card is chosen at random. What is the probability of randomly selecting a number less than or equal to 5, or a multiple of 7? Aug 27, 2018 #1 +98141 +2 Less than or equal to 5....there are 5 cards ...1,2,3,4,5 Multiple of 7....there are 4 cards .... 7, 14, 21 , 28 So.... [5 + 4 ] / 30 = 9 / 30 = 3 /10 Aug 27, 2018 #2 +377 +2 In order to find the probability of a certain event, we need to set up the fraction; $$\\frac{p(successful)}{p(possible)}$$. Let's see how many numbers are successful: Less than or equal to 5: $$1, 2, 3, 4, 5$$ Multiple of 7: $$7, 14, 21, 28$$. These nine numbers are successful, so your probability would be $$\\frac{9}{30} = \\frac{3}{10}$$. - Daisy Aug 27, 2018" ]

[ "# Combinatorics and possibly stars and bars I can across this question: Dillian has 5 pieces of paper, each with a different math problem. In how many ways can he give these problems to his 10 friends where each friend can receive more than one problem? I am a little confused because I do not think the answer is simply $$10^5$$. What am I doing wrong? • It looks like $10^5$ to me: for each paper he has $10$ choice of friend to whom to give it, and there are no restrictions. – Brian M. Scott May 21 at 16:07 How many ways can he give the 1st piece of paper to one of his friends? 10. How many ways can he give the 2nd piece of paper to one of his friends? 10. ... How many ways can he give the 5th piece of paper to one of his friends? 10. Each of these decisions is independent of each of the other. So there are $$10 \\times 10 \\times \\dotsb \\times 10$$ ways of distributing his problems. He can give each problem to his friends in 10 different ways and since there are 5 problems the answer is $$10^5$$. • Yes that is right.Apologies – Monocerotis May 21 at 16:35", "division problems. “How many groups?” “How many in each group?” Han has 12 colored pencils. He wants to put 2 colored pencils in each box until he’s out of colored pencils. How many boxes does Han need? Elena has 12 colored pencils. She has 2 boxes and wants to put the same number of colored pencils in each box. How many pencils will be in each box? $$12 \\div 2$$ $$12 \\div 2$$", "p. Write an equation, and solve for p. The total number of pens = 36. Explanation: In the above-given question, given that, Each box contains 6 pens. 6 x p = 6. p = 6 x 6. p = 36. Question 3. Mr. Lucas divides 28 students into 7 equal groups for a project. Draw a tape diagram, and label the number of students in each group as n. Write an equation, and solve for n.", "• Use algebra tiles to model the equation. Model 3x in the left rectangle, and model 21 in the right rectangle. • There are three x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile. • To do this, divide each side of your model into three equal groups. Solve the equation by drawing a model. Question 4. 6 = 3x Answer: 2 Question 5. 4x = 12 x = _______ Answer: 3 Problem Solving Question 6. A chef used 20 eggs to make 5 omelets. Model and solve the equation 5x = 20 to find the number of eggs x in each omelet. _______ eggs Answer: 4 Explanation: A chef used 20 eggs to make 5 omelets. The equation is 5x = 20 x = 50/5 = 4 Thus there are 4 eggs in each omelet. Question 7. Last month, Julio played 3 times as many video games as Scott did. Julio played 18 video games. Write and solve an equation to find the number of video games Scott played. _______ video games Answer: 6 Explanation: Last month, Julio played 3 times as many video games as Scott did. Julio played 18 video games. The equation will be 3x =", "## Problem of the Day #143: Math Problems for BillyAugust 9, 2011 Posted by Alex in : potd , trackback Billy and his five non-identical twins sit in a line. How many ways are there to distribute $18$ identical math problems to the $6$ Billys, given that at most $6$ math problems can be left over?", "Part of what makes this problem difficult is that it is about money, but there are no numbers given. That means the numbers must not be important. So just make them up! You can work forwards: Assume Alex had some specific amount of money when he showed up at school, say $100. Then figure out how much he gives to each person. Or you can work backwards: suppose he has some specific amount left at the end, like$10. Since he gave Chris half of what he had left, that means he had $20 before running into Chris. Now, work backwards and figure out how much each person got. Watch the solution only after you tried this strategy for yourself. If you use the “Make Up Numbers” strategy, it is really important to remember what the original problem was asking! You do not want to answer something like “Everyone got$10.” That is not true in the original problem; that is an artifact of the numbers you made up. So after you work everything out, be sure to re-read the problem and answer what was asked! (Squares on a Chess Board) How many squares, of any possible size, are on a 8 × 8 chess board? (The answer is not 64... It’s a lot bigger!) Remember Pólya’s first step is", "a GREAT tool in helping me learn. Alex", "# Whose camp are you in? 40 students attended the brilliant summer camp at Stanford University. $$k$$ of the students at the camp are working on math problems and the others are working on non-math problems. One of the camp mentors noticed that whenever all the students were partitioned into 2 or more groups of equal size, they could never have the same number of math students in every group. How many different values could $$k$$ have? ×", "### Primary 6 Problem Sums/Word Problems - Try FREE Score : (Single Attempt) #### Question Alex then gave 2/5 of his beads to Wendy. Subsequently, Wendy gave 110 beads to Pamela. (a) How many beads did Alex have at first? (b) How many beads did Pamela have at first? Note to students: If the question above has parts, (e.g. (a) and (b)), given that the answer for part (a) is 10 and the answer for part (b) is 12, give your answer as:10,12 The correct answer is : 6520,1194 (a)________, (b)______", "the same Polo total? What other totals are possible? Is there more than one way of making each one? Many thanks to Alan Parr for this investigation. Alan has written several problem-solving maths adventures for 10 and 11 year-olds. The games are easy to use and very popular with children. Details can be obtained from Alan at [email protected]", "# Elementary School Math Problems-2 Algebra Level 1 A teacher has 30 pens and 45 pencils that he wants to give to his students. If each student were to receive the same number of pens and the same numbers of pencils with none left over, then what is the maximum number of students that this teacher could have? ×", "strategies for economic development change over time?... ### Alex had 4 cakes. He gives his friend 3, how many does he have?​ Alex had 4 cakes. He gives his friend 3, how many does he have?​... ### Find the slope of the line passing through the two points. (–7, 8), (–4, 3) Find the slope of the line passing through the two points. (–7, 8), (–4, 3)...", "people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids. Exploring Number Patterns You Make Stage: 2 Challenge Level: Explore Alex's number plumber. What questions would you like to ask? What do you think is happening to the numbers?", "help me ASAP $Please help me ASAP$... There are 6 students in a small class. to make a team, the names of 3 of them will be drawn from a hat. There are 6 students in a small class. to make a team, the names of 3 of them will be drawn from a hat. how many different teams of 3 students are possible?... Helppp with my geometry PLSS HELP Helppp with my geometry PLSS HELP $Helppp with my geometry PLSS HELP$... How r u supposed to get over your ex bf How r u supposed to get over your ex bf...", "76 tiles. Explanation: Allie had 8 tiles in each of 13 rows, which means 13×8= 104 tiles. And the center section was built by 4 tiles in each of 7 rows, which means 4×7= 28 tiles. So 104-28= 76 tiles more needed to complete the patio. ### Common Core – Problem Solving Multistep Multiplication Problems – Page No. 117 Solve each problem. Question 1. A community park has 6 tables with a chessboard painted on top. Each board has 8 rows of 8 squares. When a game is set up, 4 rows of 8 squares on each board are covered with chess pieces. If a game is set up on each table, how many total squares are NOT covered by chess pieces? 4 × 8 = 32 32 × 6 = 192 squares Question 2. Jonah and his friends go apple picking. Jonah fills 5 baskets. Each basket holds 15 apples. If 4 of Jonah’s friends pick the same amount as Jonah, how many apples do Jonah and his friends pick in all? Draw a diagram to solve the problem. Answer: 375 apples. Explanation: As Jonah fills 5 baskets which holds 15 apples, So Jonah picked 15×5= 75 apples. And 4 of his friends pick same amount of apples, which means 75×4=300. So total apples Jonah and his", "… ,32, 33. The first number I counted was 11, and the last number I counted was 33. How many numbers did I count in all? 4.) A line and a circle are drawn on a piece of paper so that the line passes through the center of the circle. How many times do the line cross the circle? 5.) There is a group of children standing in a circle. To the left of Alex, between Alex and Nick, there are 4 children. To the right of Alex, between Alex and Nick there are 7 children. What is the total number of children in the circle?", "# A mini game with 15 pieces of paper You are given $$15$$ pieces of paper, you are asked to write down any distinct integer numbers on them as you wish. Then these pieces of paper will be turned back, mixed and put on the table in a straight order. After that, you are allowed to ask the sum of numbers in the face down papers for as many papers next to each other as you want, but you cannot ask separate paper sums, they have to be consecutive papers. Though you may even ask a specific number in a piece of paper if you wish. What is the minimum number of questions you need to ask to know every single number on the papers? • I think you should ask for 15 distinct numbers, otherwise I could write 42 on each paper and ask no questions. – Daniel Mathias Mar 25 at 8:29 • @DanielMathias you are right, I fixed that part. – Oray Mar 25 at 8:30 • ^ Upvote for @DanielMathias purely for the choice of 42 as the arbitrary number... – Stiv Mar 25 at 12:17 • Is there some particular reason the puzzle doesn't have 16 paper pieces? Rand's solution could solve that, too: with \"n, o\" as the final sum, 15 pieces", "the same as n. Mr. O.: Nice. So, how is your story problem going to start? Ahlam: There are 24 pencils. Mr. O.: That's a good start. Since this is a division problem, what kind of action are you going to use in your story? Ahlam: Well, they could be given to people. Mr. O.: Tell me more about giving 24 pencils to people. Pointing to the equation. Ahlam: There are 24 pencils that 6 people are going to share them. Mr. O.: OK. What are you trying to find out? Ahlam: The answer. Mr. O.: In this problem with people and pencils, what does the answer mean? What does it tell us? Ahlam: It means how many pencils. Mr. O.: I thought we already knew there were 24 pencils. Ahlam.: No.....the answer tells how many pencils each person gets. The 24 tells how many pencils there are before sharing. Mr. O.: (as he rewrites the equation) What I hear you saying is there are twenty four pencils that will be shared by sixchildren. You are going to find out how many pencils each child will get. 24 pencils ÷ 6 children = number of pencils each person gets. Ahlam: Yes. Mr. O.: Ahrem, are you finding the number of groups or the number in each group? Ahlam:", "He then give a the same number of tickets to each of another 8 friends. How many stickers did he give each of her 5 friends ? Each of his 8 friends ?... ### One positive number is 8 more than twice another. If their product is 234 find the numbers One positive number is 8 more than twice another. If their product is 234 find the numbers...", "36 ÷ 4 = 9. Question 3. Ms. Allyn was getting ready for a math investigation. Each student needed 8 paperclips. She had 32 paperclips. How many students were able to do the investigation? a. Equation Equation: Number of students were able to do the investigation = 32 ÷ 8 = 4. Explanation: Number of paperclips each student needed = 8. Total number of paperclips = 32. Number of students were able to do the investigation = Total number of paperclips ÷ Number of paperclips each student needed = 32 ÷ 8 = 4. b. The answer to this problem tells How many in each group How many groups The answer to this problem tells: How many groups. Explanation: Number of paperclips each student needed = 8. Total number of paperclips = 32. Number of students were able to do the investigation = Total number of paperclips ÷ Number of paperclips each student needed = 32 ÷ 8 = 4. Question 4. The math club was going on a field trip. They were driving 8 school vans. If there were 32 students in the math club, and each van took the same number of students, how many students went in each van? a. Equation Equation: Number of students went in each van = 32 ÷ 8 = 4." ]

[ "# Combinatorics and possibly stars and bars I can across this question: Dillian has 5 pieces of paper, each with a different math problem. In how many ways can he give these problems to his 10 friends where each friend can receive more than one problem? I am a little confused because I do not think the answer is simply $$10^5$$. What am I doing wrong? • It looks like $10^5$ to me: for each paper he has $10$ choice of friend to whom to give it, and there are no restrictions. – Brian M. Scott May 21 at 16:07 How many ways can he give the 1st piece of paper to one of his friends? 10. How many ways can he give the 2nd piece of paper to one of his friends? 10. ... How many ways can he give the 5th piece of paper to one of his friends? 10. Each of these decisions is independent of each of the other. So there are $$10 \\times 10 \\times \\dotsb \\times 10$$ ways of distributing his problems. He can give each problem to his friends in 10 different ways and since there are 5 problems the answer is $$10^5$$. • Yes that is right.Apologies – Monocerotis May 21 at 16:35", "Five of Seven Daniel picks 5 distinct numbers from the set $$\\{1,2,3,4,5,6,7\\}$$ and tells Alex their product. If Alex cannot determine the original numbers, how many distinct sets of 5 numbers could Daniel have chosen? ×", "## Problem of the Day #143: Math Problems for BillyAugust 9, 2011 Posted by Alex in : potd , trackback Billy and his five non-identical twins sit in a line. How many ways are there to distribute $18$ identical math problems to the $6$ Billys, given that at most $6$ math problems can be left over?", "first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.", "After too much computer programming, Alex becomes obsessed with binary. He will count from $0$ to $65,535$, inclusive, converting each number to binary. How many binary numbers will he encounter with $n$ $1$s, where $3 | n$?", "Limited access Alex likes to play Trivia Triumph, an online trivia game. In this game, trivia questions are presented and players have up to $30$ seconds to select an answer from a list of $5$ choices. Each question starts at a point value of $300$ points, but $10$ points are deducted for each second it takes the player to answer the question. Therefore, the number of points, $P$, that a player earns on a single question can be described by the following equation $$P=300-10S$$ ... where $S$ represents the number of seconds it takes the player to select an answer. Whether he knows the answer to a question or not, Alex tends to answer in about the same amount of time. His online statistics shows that the mean amount of time it takes him to answer a question, ${\\mu}_{S}$ is $13.4$ seconds with a variance of $5.29$ seconds. What is the standard deviation of the number of points Alex earns when he gets a question correct? A $-52.9$ B $23$ C $52.9$ D $323$ E $529$ Select an assignment template", "### Primary 6 Problem Sums/Word Problems - Try FREE Score : (Single Attempt) #### Question Alex then gave 2/5 of his beads to Wendy. Subsequently, Wendy gave 110 beads to Pamela. (a) How many beads did Alex have at first? (b) How many beads did Pamela have at first? Note to students: If the question above has parts, (e.g. (a) and (b)), given that the answer for part (a) is 10 and the answer for part (b) is 12, give your answer as:10,12 The correct answer is : 6520,1194 (a)________, (b)______", "### Sidebar pow:problem5s22 Problem 5 (due on Monday, April 11) A group of 100 friends plans a trip to a remote island to play a new exciting game. Upon arrival each of them will be given one of 199 possible nick-names to use during the game. Each player will know the nick-name of all the other players, but not her own. No communication will be possible between the players while they are on the island. However, at the end of the game each player will be asked about her nick-name. Only if at least one player answers correctly, they all will be allowed to return home. Can they prepare a strategy which will guarantee they can return? We received only one solution: from Leo Kargin (a 9-th grader at the Proof school in San Francisco). Leo's solution is based on the same idea as our solution, but his execution of the key step is much better than the one in our original solution. In fact, after additional simplifications, the strategy is rather easy to formulate: Number the nick-names from 1 to 199. Each player knows the 99 numbers representing nick-names of her friends. The player computes the sum $s$ of these numbers modulo 100 (so $0\\leq s<100$) and sets $k=100-s$. Then her guess is the $k$-th smallest of", "### Primary 5 Problem Sums/Word Problems - Try FREE Score : (Single Attempt) #### Question Alex had half as much money as David. David spent $47.30 and still had$26.70 more than Alex. (a) How much money does David have now? (b) Find the average amount of money Alex and David have now? Notes to students: 1. If the question above has parts, (e.g. (a) and (b)), given that the answer for part (a) is 10 and the answer for part (b) is 12, give your answer as:10,12 (a)$____, (b)$____", "### Primary 4 Problem Sums/Word Problems - Try FREE Score : (Single Attempt) #### Question John has 3 boxes of balls. There were 120 balls in each box. He gave 1/4 of the balls to Gina. After that, he gave 1/6 of the remaining balls to Kelly. How many balls did John have left? The correct answer is : 225", "Part of what makes this problem difficult is that it is about money, but there are no numbers given. That means the numbers must not be important. So just make them up! You can work forwards: Assume Alex had some specific amount of money when he showed up at school, say $100. Then figure out how much he gives to each person. Or you can work backwards: suppose he has some specific amount left at the end, like$10. Since he gave Chris half of what he had left, that means he had $20 before running into Chris. Now, work backwards and figure out how much each person got. Watch the solution only after you tried this strategy for yourself. If you use the “Make Up Numbers” strategy, it is really important to remember what the original problem was asking! You do not want to answer something like “Everyone got$10.” That is not true in the original problem; that is an artifact of the numbers you made up. So after you work everything out, be sure to re-read the problem and answer what was asked! (Squares on a Chess Board) How many squares, of any possible size, are on a 8 × 8 chess board? (The answer is not 64... It’s a lot bigger!) Remember Pólya’s first step is", "Primary 4 Problem Sums/Word Problems - Try FREE Score : (Single Attempt) Question There are 3 boxes of balls in a shop. 1/3 of all the balls in the shop are in Box X and the rest are in Box Y and Box Z. Box Y and Box Z had 200 balls and 100 balls respectively. 1/3 of the balls in Box X are sold at $3 each. 1/2 of the balls in Box Z are sold at$5 each. 1/4 of the balls in Box Y are sold at $2 each. How much was collected from the sale of the balls? The correct answer is : 500$", "# Problem of the Week Problem D and Solution Sharing Sweets ## Problem Diana has $$10$$ candies, and she wishes to give all $$10$$ candies to her three friends, Victoria, Manuela, and Alejandra. She does not necessarily want to distribute the candy equally, but she does want each friend to receive at least one candy. In how many ways can she distribute the candies to Victoria, Manuela, and Alejandra? ## Solution We know that there are $$10$$ candies and that each friend must receive at least one. We will consider the following cases: 1. Victoria receives one candy. Then Manuela and Alejandra receive a total of $$10-1=9$$ candies between them. This can be done in $$8$$ possible ways: $(1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1)$ 2. Victoria receives two candies. Then Manuela and Alejandra receive a total of $$10-2=8$$ candies between them. This can be done in $$7$$ possible ways: $(1,7),(2,6),(3,5),(4,4),(5,3),(6,2),(7,1)$ 3. Victoria receives three candies. Then Manuela and Alejandra receive a total of $$10-3=7$$ candies between them. This can be done in $$6$$ possible ways: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ 4. Victoria receives four candies. Then Manuela and Alejandra receive a total of $$10-4=6$$ candies between them. This can be done in $$5$$ possible ways: $(1,5),(2,4),(3,3),(4,2),(5,1)$ 5. Victoria receives five candies. Then Manuela and Alejandra receive a total of $$10-5=5$$ candies between them. This can be done", "Featured resource Check the Clues E: Secondary Cooperative Group Problem Solving Cards for Mathematics Members: $22.40 inc.GST Others:$ 28.00 inc.GST Home > resolve > FAQ > How many Champions will there be? How many Champions will there be? There will be a minimum of 240 Champions around the country, with at least 10 in the smaller jurisdictions (ACT and NT). Beyond that, it is expected that numbers of Champions will be roughly proportional to population.", "## Problem of the Day #395: DessertApril 17, 2012 Posted by Saketh in : potd , trackback Alex just had a sumptuous dinner at his favorite restaurant. He now has $n$ different choices for dessert. With so many good options, he decides to pick randomly using a coin. However, he wants to ensure that each dessert has an equal probability of being picked. Devise a strategy for choosing dessert that minimizes the expected number of flips needed, and express the expected number in terms of $n$. Bonus: Now solve the same problem in the case that Alex’s coin is weighted to land heads more often than tails.", "out to nine people if everyone can receive just one? Plato is absolutely correct! Quote: The next problem asks the same question but in this case, everyone can have any number of candies. For each of the 6 candies, there are 9 choices of people to give it to. Answer: . $9^6 \\:=\\:531,\\!441$ ways.", "of boxes AI can fill = 45 hats. Explanation: In the above-given question, given that, Al needs to put 9 hats in each box. he has 45 hats. 45 / 9 = 5. 9 x 5 = 45. 9 + 9 + 9 + 9 + 9 = 45. so the number of boxes AI can fill = 45 hats. Question 10. There are 48 students in a class. The teacher puts them into 8 equal groups. How many students are in each group? students The number of students in each group = 6 students. Explanation: In the above-given question, given that, There are 48 students in a class. the teacher puts them into 8 equal groups. 48 / 8 = 6. 6 x 8 = 48. so the number of students in each group = 6 students. Question 11. Saima makes 14 muffins to give to her friends. She wants to give 2 muffins to each friend at her party. How many friends can Saima invite to her party? Explain how Saima can figure out how many friends to invite. The number of friends can Saima invite to her party = 7 friends. Explanation: In the above-given question, given that, Saima makes 14 muffins to give to her friends. She wants to give 2 muffins to each", "### Primary 4 Problem Sums/Word Problems - Try FREE Score : (Single Attempt) #### Question John had 300 rubber bands. He gave 2/5 of the rubber bands to Sally. After that, he gave 1/3 of the remaining rubber bands to Mary. How many rubber bands did John have left? The correct answer is : 120", "big, \"Oh yeah?\" In all, there were 20 perfect scores and many more with at least 50 correct digits. Trefethen apparently spent $100 of his own funds, plus another$200 from Oxford Scientific Consulting, to give three of the winners the promised prize. The SIAM 100-Digit Challenge consists of descriptions of how the problems were solved, written by some of the winners. David Bailey provides an interesting foreword.", "### Primary 4 Problem Sums/Word Problems - Try FREE Score : (Single Attempt) #### Question There are 3 boxes of balls in a shop. 13/38 of all the balls in the shop are in Box A and the rest are in Box B and Box C. Box B and Box C had 100 balls and 50 balls respectively. 1/3 of the balls in Box A are sold at $1 each and 1/2 of the balls in Box C are sold at$2 each. How much was collected from the sale of the balls? The correct answer is : 76 \\$" ]

[ "# Combinatorics and possibly stars and bars I can across this question: Dillian has 5 pieces of paper, each with a different math problem. In how many ways can he give these problems to his 10 friends where each friend can receive more than one problem? I am a little confused because I do not think the answer is simply $$10^5$$. What am I doing wrong? • It looks like $10^5$ to me: for each paper he has $10$ choice of friend to whom to give it, and there are no restrictions. – Brian M. Scott May 21 at 16:07 How many ways can he give the 1st piece of paper to one of his friends? 10. How many ways can he give the 2nd piece of paper to one of his friends? 10. ... How many ways can he give the 5th piece of paper to one of his friends? 10. Each of these decisions is independent of each of the other. So there are $$10 \\times 10 \\times \\dotsb \\times 10$$ ways of distributing his problems. He can give each problem to his friends in 10 different ways and since there are 5 problems the answer is $$10^5$$. • Yes that is right.Apologies – Monocerotis May 21 at 16:35", "Five of Seven Daniel picks 5 distinct numbers from the set $$\\{1,2,3,4,5,6,7\\}$$ and tells Alex their product. If Alex cannot determine the original numbers, how many distinct sets of 5 numbers could Daniel have chosen? ×", "division problems. “How many groups?” “How many in each group?” Han has 12 colored pencils. He wants to put 2 colored pencils in each box until he’s out of colored pencils. How many boxes does Han need? Elena has 12 colored pencils. She has 2 boxes and wants to put the same number of colored pencils in each box. How many pencils will be in each box? $$12 \\div 2$$ $$12 \\div 2$$", "After too much computer programming, Alex becomes obsessed with binary. He will count from $0$ to $65,535$, inclusive, converting each number to binary. How many binary numbers will he encounter with $n$ $1$s, where $3 | n$?", "## Problem of the Day #143: Math Problems for BillyAugust 9, 2011 Posted by Alex in : potd , trackback Billy and his five non-identical twins sit in a line. How many ways are there to distribute $18$ identical math problems to the $6$ Billys, given that at most $6$ math problems can be left over?", "Part of what makes this problem difficult is that it is about money, but there are no numbers given. That means the numbers must not be important. So just make them up! You can work forwards: Assume Alex had some specific amount of money when he showed up at school, say $100. Then figure out how much he gives to each person. Or you can work backwards: suppose he has some specific amount left at the end, like$10. Since he gave Chris half of what he had left, that means he had $20 before running into Chris. Now, work backwards and figure out how much each person got. Watch the solution only after you tried this strategy for yourself. If you use the “Make Up Numbers” strategy, it is really important to remember what the original problem was asking! You do not want to answer something like “Everyone got$10.” That is not true in the original problem; that is an artifact of the numbers you made up. So after you work everything out, be sure to re-read the problem and answer what was asked! (Squares on a Chess Board) How many squares, of any possible size, are on a 8 × 8 chess board? (The answer is not 64... It’s a lot bigger!) Remember Pólya’s first step is", "of boxes AI can fill = 45 hats. Explanation: In the above-given question, given that, Al needs to put 9 hats in each box. he has 45 hats. 45 / 9 = 5. 9 x 5 = 45. 9 + 9 + 9 + 9 + 9 = 45. so the number of boxes AI can fill = 45 hats. Question 10. There are 48 students in a class. The teacher puts them into 8 equal groups. How many students are in each group? students The number of students in each group = 6 students. Explanation: In the above-given question, given that, There are 48 students in a class. the teacher puts them into 8 equal groups. 48 / 8 = 6. 6 x 8 = 48. so the number of students in each group = 6 students. Question 11. Saima makes 14 muffins to give to her friends. She wants to give 2 muffins to each friend at her party. How many friends can Saima invite to her party? Explain how Saima can figure out how many friends to invite. The number of friends can Saima invite to her party = 7 friends. Explanation: In the above-given question, given that, Saima makes 14 muffins to give to her friends. She wants to give 2 muffins to each", "p. Write an equation, and solve for p. The total number of pens = 36. Explanation: In the above-given question, given that, Each box contains 6 pens. 6 x p = 6. p = 6 x 6. p = 36. Question 3. Mr. Lucas divides 28 students into 7 equal groups for a project. Draw a tape diagram, and label the number of students in each group as n. Write an equation, and solve for n.", "3 Tyler has some beads on a bracelet. He took 6 beads off to give to Clare. Now there are 7 beads on Tyler's bracelet. How many beads were on Tyler's bracelet before he took the beads off? Show your thinking using drawings, numbers, words, or equations. ### Problem 4 There are 13 players on the soccer field. Some of their friends come to play. Now there are 19 players on the soccer field. Circle the equation that represents the situation. A: $$13 + 19 = \\boxed{\\phantom{32}}$$ B: $$13 + \\boxed{\\phantom{32}} = 19$$ C: $$\\boxed{\\phantom{32}} - 13 = 19$$ ### Problem 5 Kiran and Clare made 13 origami swans together. How many swans did Clare make? Show your thinking using drawings, numbers, words, or equations. ### Problem 6 #### Exploration Andre used these connecting cubes to represent a story problem. What could Andre's story be? ### Problem 7 #### Exploration Write a story problem.", "# how many ways can 5 prizes be given to a group of 20 people with restrictions Sorry if this is super simple, I've just been stuck on this concept for way too long. Each person can only win one prize. How many ways can 5 prizes be handed out to a group of 20 people if: (a) all 5 prizes are identical, also Alex and Sally can't both get prizes. (b) the 5 prizes are ranked 1st-5th, also Alex and Sally can't both get prizes. So I think I've tried to solve this but I think I've made a mistake. New Attempt a) Now I'm trying to find the total number of ways that the 5 prizes could be handed out to 20 people. I think this should be: $${20 \\choose 5}$$ now we need to find the situations in which both Sally and Alex win prizes and subtract that from the total. I think this is wrong but it might just be: $${18 \\choose 3}$$ So the final value would be: $${20 \\choose 5} - {18 \\choose 3}$$ b) I think the trick here is still to multiply the answer from part (a) by '5!'. $$5! ({20 \\choose 5} - {18 \\choose 3})$$ • Can a person receive multiple prizes, e.g. could Alex hypothetically get all", "the area of the square has been shaded? 11. A particular prism has ten faces. How many edges does it have? A: B: C: D: E: 12. The pupils in my class work very quickly. Jasleen answers four questions every $30$ seconds and Ella answers five questions every $40$ seconds. Last week, Jasleen took exactly $1$ hour to answer a large set of questions. How many minutes more than Jasleen did Ella take to answer the same set of questions? A: D: 13. Five line segments coincide at a point as shown. What is the sum of the marked angles? 14. I begin with a three-digit positive integer. I divide it by $9$ and then subtract $9$ from the answer. My final answer is also a three-digit integer. How many different positive integers could I have begun with? A: B: C: D: E: 15. Alex has a pile of $2\\text{p}$. She swapped exactly half of them for the same number of $10\\text{p}$ coins. Now she had $\\text{£}4.20$. How much money did Alex have initially? Back to the top", "# Help, I'm Trapped in a Math Factory! Algebra Level 5 A certain factory produces numbers to be used in math problems. A worker there named Alex is given a batch of $$N$$ numbers. First, he removes one number for inspection, then packages half of the remaining numbers. From the remaining numbers, he removes one and packages half the numbers after that. He repeats this process $$2014$$ more times (remove one, package half the remaining), and at the end, he is left with one number that he keeps for himself. Find the units digit of $$N$$. Notes: After the first sorting, Alex is left with $$\\frac{1}{2}(N-1)$$ numbers. After the second sorting, he is left with $$\\frac { 1 }{ 2 } \\left(\\frac { 1 }2(N-1)-1\\right)$$ numbers, and so on. ×", "70 + 7 = 63. Multiply and match. Question 12. Solve. Question 13. Mrs. Thompson buys 2 books. Each book costs $7. How much does Mrs. Thompson pay in all? 2 ×$7 = $___ Mrs. Thompson pays$__________ in all. Mrs. Thompson pays $14 in all. Explanation: Given that Mrs. Thompson bought 2 books and each book costs$7. So the total cost of Mrs. Thompson pay in all is 2 × $7 which is$14. Question 14. A box contains 7 crayons. Alex packs 10 such boxes into his bog. How many crayons does Alex have in all? 10 × 7 = ______ Alex has ___ crayons in all. Alex has 70 crayons in all. Explanation: Given that a box contains 7 crayons and Alex packs 10 such boxes into his box. So the number of crayons does Alex have in all is 7 × 10 is 70 crayons. Question 15. Mr. Dean gives each student 7 okras in art class. How many okras does he give 4 students? 4 × 7 = ___ He gives 4 students ___ okras.", "### Primary 6 Problem Sums/Word Problems - Try FREE Score : (Single Attempt) #### Question Alex then gave 2/5 of his beads to Wendy. Subsequently, Wendy gave 110 beads to Pamela. (a) How many beads did Alex have at first? (b) How many beads did Pamela have at first? Note to students: If the question above has parts, (e.g. (a) and (b)), given that the answer for part (a) is 10 and the answer for part (b) is 12, give your answer as:10,12 The correct answer is : 6520,1194 (a)________, (b)______", "## For a math project, Elyssa and Zoe want to combine their money and distribute it equally among 12 fifth graders. If Elyssa has Question For a math project, Elyssa and Zoe want to combine their money and distribute it equally among 12 fifth graders. If Elyssa has $8.40 and Zoe has$7.20, how much money does each fifth grader get?", "- a good way to learn the solving techniques by examples. Quick Start. Select a digit on the side of the grid first. Then click on the cells where you want to place the selected digit. After entering your puzzle, press either ... Answer Mathematics, 21.06.2019 23:30 Tatiana wants to give friendship bracelets to her 32 classmates. she already has 5 bracelets, and she can buy more bracelets in packages of 4. write an inequality to determine the number of packages, p, tatiana could buy to have enough bracelets. bigbutt ebony pornmunicode miami beach L1a", "classroom = 2 + 1 + 3 = 6 5. Jack spent 5 dollars for a pen, 3 dollars for a color box, 2 dollars for a pencil box. How much did he spend altogether? Amount of money Jack spent for a pen = $5 Amount of money he spent for a color box =$3 Amount of money he spent for a pencil box = $2 Therefore, total amount of money he spend altogether =$5 + $3 +$2 = 10 6. There were 6 yellow hats, 2 red hats and 6 blue hats. How many hats were there? Number of yellow hats = 6 Number of red hats = 2 Number of blue hats = 6 Therefore, total number of hats were there = 6 + 2 + 6 = 14 7. Alex had 4 books on cars, 5 on airplanes and 7 on boats. How many books did he have in all? Number of books on cars Alex had = 4 Number of books on airplanes = 5 Number of books on boats = 7 Therefore, total number of books he have in all = 4 + 5 + 7 = 16 More examples on statement problem solving on addition: 8. In a game Mary had the best score. She made 8 in her first turn, 6", "to get 3+3+3+3+0=12 4. Students 3 and 2 become friends: We then sum the number of friends that each student has to get 3+3+3+3+0=12 . When we add the sums from each step, we get total = 2+6+12+12=32. We then print 32 on a new line. Solution Working on the solution.", "Explain how you found your answers. You can use appropriate tools to make an array to show the multiplication. Answer: The number of pencils in 5 boxes = 25. the number of pencils in 9 boxes = 45. the number of pencils in 10 boxes = 50. Explanation: In the above-given question, given that, each box contains 5 pencils. 1 box = 1 x 5. 5 boxes = 5 x 5 = 25. 9 boxes = 9 x 5 = 45. 10 boxes = 10 x 5 = 50. Look Back! A different company sells boxes that have 9 colored pencils in each box. How many pencils are in 5 boxes? 9 boxes? 10 boxes? Answer: The number of pencils in 5 boxes = 45. the number of pencils in 9 boxes = 81. the number of pencils in 10 boxes = 90. Explanation: In the above-given question, given that, each box contains 9 pencils. 1 box = 1 x 9. 5 boxes = 5 x 9 = 45. 9 boxes = 9 x 9 = 81. 10 boxes = 10 x 9 = 90. Visual Learning Bridge Essential Question How Do You Use Multiplication Facts to Solve Problems? A. Brendan has archery practice. The target shows the points he gets for hitting a section. How many points", "# A mini game with 15 pieces of paper You are given $$15$$ pieces of paper, you are asked to write down any distinct integer numbers on them as you wish. Then these pieces of paper will be turned back, mixed and put on the table in a straight order. After that, you are allowed to ask the sum of numbers in the face down papers for as many papers next to each other as you want, but you cannot ask separate paper sums, they have to be consecutive papers. Though you may even ask a specific number in a piece of paper if you wish. What is the minimum number of questions you need to ask to know every single number on the papers? • I think you should ask for 15 distinct numbers, otherwise I could write 42 on each paper and ask no questions. – Daniel Mathias Mar 25 at 8:29 • @DanielMathias you are right, I fixed that part. – Oray Mar 25 at 8:30 • ^ Upvote for @DanielMathias purely for the choice of 42 as the arbitrary number... – Stiv Mar 25 at 12:17 • Is there some particular reason the puzzle doesn't have 16 paper pieces? Rand's solution could solve that, too: with \"n, o\" as the final sum, 15 pieces" ]

[ "+0 # Help Probability! +1 48 2 +82 Two integers are relatively prime if they have no common factors other than 1 or -1. What is the probability that a positive integer less than or equal to 30 is relatively prime to 30? Express your answer as a common fraction. MathCuber Jul 8, 2018 #1 -1 $${7 \\over 29}$$ Guest Jul 8, 2018 #2 +1 Since 30 = 2 x 3 x 5, therefore it has common factors with the following numbers: 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30[22 numbers] It is relatively prime with following numbers: 1, 7, 11, 13, 17, 19, 23, 29[8 numbers] Therefore, the probability that a number =< 30 is relatively prime to 30 is: 8 / 30 = 4 / 15 Guest Jul 8, 2018", "+0 # Help Probability! +1 157 2 +314 Two integers are relatively prime if they have no common factors other than 1 or -1. What is the probability that a positive integer less than or equal to 30 is relatively prime to 30? Express your answer as a common fraction. Jul 8, 2018 #1 -1 $${7 \\over 29}$$ . Jul 8, 2018 #2 +1 Since 30 = 2 x 3 x 5, therefore it has common factors with the following numbers: 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30[22 numbers] It is relatively prime with following numbers: 1, 7, 11, 13, 17, 19, 23, 29[8 numbers] Therefore, the probability that a number =< 30 is relatively prime to 30 is: 8 / 30 = 4 / 15 Jul 8, 2018", "You are not signed in. To save solutions, earn points and appear on the leaderboard. register or sign in. # Relatively prime integers #### Description Two integers $$a$$ and $$b$$ are said to be relatively prime if the only positive integer that divides both of them is 1. The positive integers less than 10 which are relatively prime to 10 are: 1, 3, 7, 9 therefore the sum of positive integers less than 10 which are relatively prime is 20. #### Question What is the sum of all positive integers less than 1,000,000 which are relatively prime to 1,000,000? What is the sum of all positive integers less than 1,000,000 which are relatively prime to 1,000,000?", "# Probability and Number Theory-all in one question. The probability that two random integers chosen have no common prime factors is of the form a/b where a and b are positive coprime integers.Find a+b.Take pi=22/7 ×", "pairwise relatively prime. Every pairwise relatively prime finite set is relatively prime; however, the converse is not true: {6, 10, 15} is relatively prime, but not pairwise relative prime (in fact, each pair of integers in the set has a non-trivial common factor). ## Probabilities Given two randomly chosen integers A and B, it is reasonable to ask how likely it is that A and B are coprime. In this determination, it is convenient to use the characterization that A and B are coprime if and only if no prime number divides both of them (see Fundamental theorem of arithmetic). Intuitively, the probability that any number is divisible by a prime (or any integer), p is 1 / p (for example, every 7th integer is divisible by 7.) Hence the probability that two numbers are both divisible by this prime is 1 / p2, and the probability that at least one of them is not is 1 − 1 / p2. Now, for distinct primes, these divisibility events are mutually independent. (This would not, in general, be true if they were not prime.) (For the case of two events, a number is divisible by p and q if and only if it is divisible by pq; the latter has probability 1/pq.) Thus the probability that two numbers are", "be relatively prime? It's analogous to the corresponding definition for integers. For any numbers a and b, there is always some number d such that both a and b are multiples of d. (d = 1 is always a solution.) The greatest such number is called the greatest common divisor or GCD of a and b. The GCD of two numbers might be 1, or it might be some larger number. If it's 1, we say that the two numbers are relatively prime (to each other). For example, the GCD of 100 and 28 is 4, so 100 and 28 are not relatively prime. But the GCD of 100 and 27 is 1, so 100 and 27 are relatively prime. One can prove theorems like these: If p is prime, then either a is a multiple of p, or a is relatively prime to p, but not both. And the equation ap + bq = 1 has a solution (in integers) if and only if p and q are relatively prime. The definition for polynomials is just the same. Take two polynomials over some variable x, say p and q. There is some polynomial d such that both p and q are multiples of d; d(x) = 1 is one such. When the only solutions are trivial polynomials", "is an important statement about coprime ideals. The concept of being relatively prime can also be extended to any finite set of integers S = {a1, a2, .... an} to mean that the greatest common divisor Greatest common divisor In mathematics, the greatest common divisor , also known as the greatest common factor , or highest common factor , of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.For example, the GCD of 8 and 12 is 4.This notion can be extended to... of the elements of the set is 1. If every pair in a (finite or infinite) set of integers is relatively prime, then the set is called pairwise relatively prime. Every pairwise relatively prime finite set is relatively prime; however, the converse is not true: {6, 10, 15} is relatively prime, but not pairwise relative prime (in fact, each pair of integers in the set has a non-trivial common factor). Probabilities Given two randomly chosen integers and , it is reasonable to ask how likely it is that and are coprime. In this determination, it is convenient to use the characterization that and are coprime if and only if no prime number divides both of them (see Fundamental theorem of arithmetic Fundamental theorem of arithmetic In number", "## Relatively Prime Two integers are relatively prime if they share no common factors (divisors) except 1. Using the notation to denote the Greatest Common Divisor, two integers and are relatively prime if . Relatively prime integers are sometimes also called Strangers or Coprime and are denoted . The probability that two Integers picked at random are relatively prime is , where is the Riemann Zeta Function. This result is related to the fact that the Greatest Common Divisor of and , , can be interpreted as the number of Lattice Points in the Plane which lie on the straight Line connecting the Vectors and (excluding itself). In fact, is the fractional number of Lattice Points Visible from the Origin (Castellanos 1988, pp. 155-156). Given three Integers chosen at random, the probability that no common factor will divide them all is (1) where is Apéry's Constant. This generalizes to random integers (Schoenfeld 1976). References Castellanos, D. The Ubiquitous Pi.'' Math. Mag. 61, 67-98, 1988. Guy, R. K. Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 3-4, 1994. Schoenfeld, L. Sharper Bounds for the Chebyshev Functions and , II.'' Math. Comput. 30, 337-360, 1976.", "and only if the numbers 2a − 1 and 2b − 1 are coprime. As a generalization of this, following easily from Euclidean algorithm in base n > 1: $\\gcd (n^a-1,n^b-1)=n^{\\gcd(a,b)}-1.$ ## Cross notation, group If n≥1 and is an integer, the numbers coprime to n, taken modulo n, form a group with multiplication as operation; it is written as (Z/nZ)× or Zn*. ## Generalizations The concept of being relatively prime can also be extended to any finite set of integers S = {a1, a2, .... an} to mean that the greatest common divisor of the elements of the set is 1. If every pair in a (finite or infinite) set of integers is relatively prime, then the set is called pairwise relatively prime. Every pairwise relatively prime finite set is relatively prime; however, the converse is not true: {6, 10, 15} is relatively prime (the only positive integer that divides all of 6, 10 and 15 is 1), but not pairwise relative prime (each pair of integers in the set has a non-trivial common factor). Two ideals A and B in the commutative ring R are called coprime (or comaximal) if A + B = R. This generalizes Bézout's identity: with this definition, two principal ideals (a) and (b) in the ring of integers Z are coprime", "+0 # Probability 0 110 1 A positive integer divisor of 11 is chosen at random. What is the probability that this divisor is prime? Express your answer as a common fraction. Jul 12, 2022 It is $$\\frac{1}{2},$$ since 11 has only two divisors, 1 and 11, and 11 is prime while 1 is not.", "commondivisors,we f i n dt h a t ( 1 5 , 8 1 ): 3 , ( 1 0 0 , 5 ) : 5 , ( I 7 , 2 5 ) : l , ( 0 , 4 4 ): 4 4 , ( - 6 , - 1 5 ) : 3 , and (-17, 289) : 17. We are particularly interested in pairs of integers sharing no common divisorsgreaterthan l. Such pairs of integersare called relatively prime. Definition. The integers a and b are called relatively prime if a and b have greatestcommondivisor (a, b) : l. Example. Since Q5,42) : 1,25 and 42 are relativelyprime. 53 54 GreatestCommonDivisorsand prime Factorization Note that since the divisors of -c are the same as the divisors of a, it follows that (a, b) : (lal, la ll (where lc I denotesthe absolute value of a which equalsa if a )0 and equals -a if a <0). Hence, we can restrict our attentionto greatestcommondivisorsof pairs of positiveintegers. We now provesomepropertiesof greatestcommondivisors. Proposition 2.1. Let a, b, and c be integerswith G, b) : d. Then (;) (ii) b /d , b l d ) : I (atcb, b) : (a, b). Proof. (D Let a and b be integers with (a,b) : d. we will show", "1. Other 2. Other 3. let and let and suppose that and are relatively... # Question: let and let and suppose that and are relatively... ###### Question details Let and . Let and suppose that and are relatively prime. Show that there exist integers such that and . Also show by example that the conclusion may fail if and are not relatively prime.", "[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ] So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4). Clearly, there are 2 such pairs.", "# Are zero and one relatively prime? The definition of relative primality that I was taught was that: Two numbers are relatively prime if the only common positive factor of the two numbers is one. Every integer (except zero) divides zero and the only positive factor of one is one. Thus, the only common positive factor of zero and one is one. Thus, it would seem that zero and one are relatively prime by the definition above. By convention is this not the case? i.e. zero is defined as not being relatively prime with any integer? • Yes, $0$ and $1$ are relatively prime. Your argument shows that they are. – André Nicolas Feb 19 '15 at 23:06 • $1$ is relatively prime to every integer. Even to itself. – ajotatxe Feb 19 '15 at 23:07 • I'm curious as to what brought this up. – Akiva Weinberger Feb 20 '15 at 1:13 • I wanted to know if a pairwise relatively prime set could contain zeroes. – terminex9 Feb 20 '15 at 3:58 $\\gcd(0,n)=n$ for all $n\\in\\mathbb{N}$. For $n=1$ it turns out to be $1$, so if you insist, $0$ and $1$ are relatively prime. Zero is not defined to be not relatively prime with any integer. It just so happens that it is divisible by any", "Common factors and common multiples To add two fractions we need to find a common multiple, or the LCM, of the two given denominators. To cancel fractions, or to simplify ratios, we need to be able to spot common factors and to find HCFs. Two positive integers a, b which have no (positive) common factors other than 1 (that is, with HCF (a,b) = 1) are said to be relatively prime, or coprime. Problem 5 [This problem requires a mixture of serious thought and written proof.] (a) I choose six integers between 10 and 19 (inclusive). (i) Prove that some pair of integers among my chosen six must be relatively prime. (ii) Is it also true that some pair must have a common factor? (b) I choose six integers in the nineties (from 90–99 inclusive). (i) Prove that some pair among my chosen integers must be relatively prime. (ii) Is it also true that some pair must have a common factor? (c) I choose n + 1 integers from a run of 2n consecutive integers. (i) Prove that some pair among the chosen integers must be relatively prime. (ii) Is it also true that some pair must have a common factor? ### 1.1.5 The Euclidean algorithm School mathematics gives the impression that to find the HCF of two", "0.6079271018540266286632767792\\ldots\\$ The question is easily seen to be equivalent to \"What is the probability that two integers are relatively prime?\" This old chestnut of elementary number theory has been addressed before on this site: see here. The idea is incredibly simple and appealing: we ... Only top voted, non community-wiki answers of a minimum length are eligible", "# Relatively Prime Numbers and Polynomials Two numbers are said to be relatively prime if their greatest common factor ( GCF ) is . Example 1: The factors of are . The factors of are . The only common factor is . So, the GCF is . Therefore, are relatively prime. Example 2: The factors of are . The factors of are . The greatest common factor here is . Therefore, are not relatively prime. The definition can be extended to polynomials . In this case, there should be no common variable or polynomial factors, and the scalar coefficients should have a GCF of . Example 3: The polynomial can be factored as . The polynomial can be factored as . are relatively prime, and none of the binomial factors are shared. So, the two polynomials are relatively prime. Example 4: The polynomial can be factored as . The polynomial can be factored as . The two polynomials share a binomial factor: . So are not relatively prime.", "contains C. The Chinese remainder theorem is an important statement about coprime ideals. The concept of being relatively prime can also be extended any finite set of integers S = {a1, a2, .... an} to mean that the greatest common divisor of the elements of the set is 1. If every pair of integers in the set is relatively prime, then the set is called pairwise relatively prime. Every pairwise relatively prime set is relatively prime; however, the converse is not true: {6, 10, 15} is relatively prime, but not pairwise relative prime. (In fact, each pair of integers in the set has a non-trivial common factor.) Probabilities Given two randomly chosen integers $A$ and $B$, it is reasonable to ask how likely it is that $A$ and $B$ are coprime. In this determination, it is convenient to use the characterization that $A$ and $B$ are coprime if and only if no prime number divides both of them (see Fundamental theorem of arithmetic). Intuitively, the probability that any number is divisible by a prime (or any integer), $p$ is $1/p$. Hence the probability that two numbers are both divisible by this prime is $1/p^2$, and the probability that at least one of them is not is $1-1/p^2$. Thus the probability that two numbers are coprime is given by", "what the error refers to.For a given confidence interval , standard deviation , and sample size , the margin of error (for a normal distribution) iswhere is the inverse erf function. ### Relatively prime Two integers are relatively prime if they share no common positive factors (divisors) except 1. Using the notation to denote the greatest common divisor, two integers and are relatively prime if . Relatively prime integers are sometimes also called strangers or coprime and are denoted . The plot above plots and along the two axes and colors a square black if and white otherwise (left figure) and simply colored according to (right figure).Two numbers can be tested to see if they are relatively prime in the Wolfram Language using CoprimeQ[m, n].Two distinct primes and are always relatively prime, , as are any positive integer powers of distinct primes and , .Relative primality is not transitive. For example, and , but .The probability that two integers and picked at random are relatively prime is(1)(OEIS A059956; Cesàro and Sylvester 1883; Lehmer 1900; Sylvester 1909; Nymann 1972; Wells 1986, p. 28; Borwein and Bailey 2003, p. 139;.. ### Line graph A line graph (also called an adjoint, conjugate, covering, derivative, derived, edge, edge-to-vertex dual, interchange, representative, or -obrazom graph) of a simple graph is obtained by associating", "because only 1 is a common factor. Example 2: 14 and 16 The factors of 14 are 1, 2, 7, and 14. The factors of 16 are 1, 2, 4, 8 and 16 Here, 1 and 2 are the common factors of 14 and 16. HCF $(14, 16) = 1 \\times 2 = 2$ Thus, (14, 16) is not relatively prime. ## Properties of Relatively Prime Numbers Here are a few properties of relatively prime numbers. You can easily form a “list of relatively prime numbers” using these properties. • Any two consecutive integers are always relatively prime. Example: (2, 3) ; (11, 12) ; (21, 22) ; (100, 101) and so on. • Any two prime numbers are relatively prime to each other. As every prime number has only two factors 1 and the number itself, the only common factor of two prime numbers will be 1. Example: 11 and 13 are two prime numbers. Factors of 11 are 1, 11, and factors of 13 are 1, 13. The only common factor is 1 and hence they are relatively prime. • If two numbers have 0 and 5 at ones place, then they are not relatively prime to each other. For example: 10 and 15 are not relatively prime since their HCF is 5 (or divisible by" ]

[ "# Math Help - Positive Integers 1. ## Positive Integers How many positive integers less than 2007 are relatively prime to 1001? 2. Factor: $1001 = 7 \\cdot 11 \\cdot 13$. How many positive integers less than 2007 are multiples of at least one of 7, 11, or 13? None of those integers can be relatively prime (also known as coprime) with 1001. 3. Hello, Dragon! How many positive integers less than 2007 are relatively prime to 1001? There are 2006 positive integers less than 2007. How many of these have a common factor with 1001? We find that: . $1001 \\:=\\:7\\!\\cdot\\!11\\!\\cdot\\!13$ There are: . $\\left[\\frac{2006}{7}\\right] = 286$ with a factor of 7. There are: . $\\left[\\frac{2006}{11}\\right] = 182$ with a factor of 11. There are: . $\\left[\\frac{2006}{13}\\right] = 154$ with a factor of 13. There are: . $\\left[\\frac{2006}{77}\\right] = 26$ with a factor of 7 and 11. There are: . $\\left[\\frac{2006}{91}\\right] = 22$ with a factor of 7 and 13. There are: . $\\left[\\frac{2006}{143}\\right] = 14$ with a factor 11 and 13. There are: . $\\left[\\frac{2006}{1001}\\right] = 2$ with a factor of 7, 11 and 13. Formula: $n(A\\,\\cup B\\,\\cup\\ C) \\;=\\;n(A) + n(B) + n(C) - n(A\\,\\cap\\,B) - n(A\\,\\cap\\ C)$ $- n(B\\,\\cap\\,C) + n(A\\,\\cap\\,B\\,\\cap\\,C)$ Hence: . $n(7 \\cup 11 \\cup 13) \\;=\\;286 + 182 + 154 -", "$21$ relatively prime? Solution Since $35 = 7 \\cdot 5$ and $21 = 7 \\cdot 3,$ we see that $7$ is a common divisor of $35$ and $21.$ It can be checked that no positive integer larger than $7$ that divides both $35$ and $21.$ Thus, we see that $\\gcd(35,21) = 7.$Since $\\gcd(35,21) \\neq 1,$ we see that $35$ and $21$ are not relatively prime. Additionally, it is not hard to see that $\\gcd(5,3) = 1,$ and thus $\\frac{35}{7}$ and $\\frac{21}{7}$ are relatively prime. In general, we have the following theorem. Theorem Let $a$ and $b$ be positive integers. Suppose that $d$ is a positive integer such that $d \\mid a$ and $d \\mid b.$ Then, $d = \\gcd(a,b)$ if and only if $\\frac{a}{d}$ and $\\frac{b}{d}$ are relatively prime. Let $a$ and $b$ be positive integers. The smallest positive integer $m$ such that $a$ divides $m$ and $b$ divides $m$ is referred to as the least common multiple of $a$ and $b.$ The least common multiple of $a$ and $b$ is denoted by $\\operatorname{lcm}(a,b).$ Example Find the least common multiple of $35$ and $21.$ Solution Since $105 = 35 \\cdot 3$ and $105 = 21 \\cdot 5,$ we see that $105$ is a common multiple of $35$ and $21.$ It can be checked that no positive integer smaller", "This yields $10{,}000 \\cdot (1-\\frac{1}{2}) \\cdot (1-\\frac{1}{5}) = 4,000$ integers in $\\{1, \\cdots , 10{,}000\\}$ that are relatively prime to $10{,}000$. Edit: We do not want to include $1$ for a possibility of $e$ and so we remove this one possibility. This yields $4{,}000 - 1 = 3{,}999$ such integers.", "Counting integers less than $n$ that are relatively prime to $x\\#$ Let $$x,n$$ be integers such that $$x < n$$. Let $$x\\#$$ be the primorial of $$x$$ so that $$6\\# = 5\\# = 30$$ and $$7\\# = 210$$ Let gcd$$(a,b)$$ be the greatest common divisor of $$a$$ and $$b$$. Let $$S(x,n)$$ be the set of integers such that $$s \\in S(x,n)$$ if and only if gcd$$(s,x\\#)=1$$ and $$s \\le n$$ Let $$|S(x,n)|$$ be the number of elements in the set $$S(x,n)$$ Let $$p$$ be the least prime greater than $$x$$. Does it follow that there are at least $$\\left\\lfloor\\left(\\frac{p-1}{p}\\right)|S(x,n)|\\right\\rfloor$$ elements relatively prime to $$p$$ in $$S(x,n)$$? Here are some examples: • $$S(3,35) = \\{1,5,7,11,13,17,19, 23, 25, 29, 31, 35\\}$$ There are at least $$\\left\\lfloor\\left(\\frac{4}{5}\\right)12\\right\\rfloor = 9$$ • $$S(5,49) = \\{1,7,11,13,17,19,23,29,31,37,41,43,47,49\\}$$ There are at least $$\\left\\lfloor\\left(\\frac{6}{7}\\right)14\\right\\rfloor = 12$$ I suspect the answer is no. Could someone provide an example of $$x,n$$ where there are less than $$\\left\\lfloor\\left(\\frac{p-1}{p}\\right)|S(x,n)|\\right\\rfloor$$ elements relatively prime to $$p$$ in $$S(x,n)$$ Edit: I reworded the question to fix a mistake in my wording. I had said \"greater than $$p$$\", I had meant to say \"relatively prime to $$p$$.\". My examples had always shown \"relatively prime to $$p$$\". I had excluded elements that had $$p$$ as a divisor. • Closely related to $y$-rough numbers below $x$. –", "11$ $1575 = 3^2 \\cdot 5^2 \\cdot 7$ From the above: what primes do these two numbers have in common? As we can see, the common primes are 3 and 5. Looking at the exponents of these common primes in each of the numbers, we look the minimum between the two. In this case, the minimum exponent for 3 is 1, and the minimum exponent for 5 is also 1. Therefore $GCD = 3^1 \\cdot 5^1 = 3 \\cdot 5 = 15$ Now, all we have to do is to simplify the original fraction by 15: $\\frac{165}{1575} = \\frac{165/15}{1575/15} = \\frac{11}{105}$ which corresponds to the fraction in its lowest terms, because it cannot be further simplified. In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us.", "and $\\,9\\,$ are unrelated $8\\,$ and $\\,9\\,$ are strangers When two numbers are relatively prime (unrelated, strangers), then their least common multiple is their product: • the least common multiple of $\\,8\\,$ and $\\,9\\,$ is $\\,8\\cdot 9 = 72\\,$ • $\\text{lcm}(3,5) = 3\\cdot 5 = 15\\,$ • $\\displaystyle\\frac 17 + \\frac1{10} = \\frac{10}{70} + \\frac{7}{70} = \\frac{17}{70}$ Note that the Least Common Denominator (LCD), one important use of the Least Common Multiple (LCM), is $\\,7\\cdot 10 = 70\\,$. Relatively prime is an important mathematical concept, and hence deserves a formal definition: DEFINITION relatively prime Two counting numbers are relatively prime if and only if they have no common factor other than $\\,1\\,$. Equivalently: Two counting numbers are relatively prime if and only if their greatest common factor is $\\,1\\,$. ## LIVES-IN The ‘lives-in’ relationship says that one of the structures lives entirely inside the other. That is, one of the numbers contains all the prime factors of the other. It's easy to recognize the lives-in case, because the smaller number goes into the bigger one evenly. Examples: $6\\,$ lives in $\\,18\\,$ (see image at right) The number $\\,18\\,$ contains all the prime factors of $\\,6\\,$. $3\\,$ lives in $\\,12\\,$ The number $\\,12\\,$ is divisible by $\\,3\\,$. $15\\,$ lives in $\\,30\\,$ The number $\\,30\\,$ is a multiple of $\\,15\\,$.", "# Difference between revisions of \"2005 AMC 10B Problems/Problem 22\" ## Problem For how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1 + 2 + \\cdots + n?$ $\\text{(A) 8 } \\text{(B) 12 } \\text{(C) 16 } \\text{(D) 17 } \\text{(E) 21 }$ ## Solution Since $1 + 2 + \\cdots + n = \\frac{n(n+1)}{2}$, the condition is equivalent to having an integer value for $\\frac{n!} {\\frac{n(n+1)}{2}}$. This reduces, when $n\\ge 1$, to having an integer value for $\\frac{2(n-1)!}{n+1}$. This fraction is an integer unless $n+1$ is an odd prime. There are 8 odd primes less than or equal to 24, so there are $24 - 8 = \\boxed{\\text{(C)}16}$ numbers less than or equal to 24 that satisfy the condition. ~savannahsolver", "# Relatively prime numbers are prime The problem is to find all numbers $n$ such that all numbers $k>1$ smaller than $n$ and coprime with $n$ are prime. • Do you have examples of such $n$ which are not too small? – AdLibitum Jul 3 '17 at 9:58 • The two formulations differ in meaning though: The only number $1<k<n$ coprime to $n=6$ is $k=5$, but $\\phi(6)-1=4\\ne 3=\\pi(6)$ – Hagen von Eitzen Jul 3 '17 at 9:59 • @HagenvonEitzen Yep sorry, it is the non mathematical way... – Maths4F Jul 3 '17 at 10:01 • @HagenvonEitzen But I have to find all $n$ such that it is the case! – Maths4F Jul 3 '17 at 10:01 Let $n$ have the property and let $p$ be the smallest prime not dividing $n$. Then $n<p^2$ as otherwise $\\gcd(p^2,n)=1$ destroys the property. On the other hand, this implies that $n$ is a multiple of the product of all primes $<p$. For $p\\ge13$, by Bertrand's postulate, the prime preceding $p$ is $>\\frac p2$ and the one preceding that is $>\\frac p4$ (and $>5$). Hence $p^2>2\\cdot 3\\cdot5\\cdot \\frac p4\\cdot\\frac p2=\\frac{15}{4}p^2$, contradiction. But also for $p=11$, we find $11^2=121>2\\cdot3\\cdot 5\\cdot 7=210$, contradiction. For $p=7$, we only get $n<49$ and $2\\cdot3\\cdot 5=30\\mid n$,. Instead of a contradiction, this implies $n=30$. We verify that this does", "# 2008 Mock ARML 2 Problems/Problem 2 ## Problem Given that the sum of all positive integers with exactly two proper divisors, each of which is less than $30$, is $2397$, find the sum of all positive integers with exactly three proper divisors, each of which is less than $30$ (a proper divisor of $n$ is a positive integer that divides but is not equal to $n$). ## Solution Let $p,q$ be primes and $r > 1$. Then we note that $1 | p;\\ \\boxed{1,p | p^2};\\ \\boxed{1,p,q | pq};\\ \\boxed{1,p,p^2 | p^3};\\ 1,p,pq,p^2 | p^2q;\\ 1,p,q,r,pq,qr,pr | pqr$. In other words, only perfect squares of primes have exactly two proper divisors (hence $2397 = 2^2 + 3^2 + 5^2 + 7^2 + \\cdots + 29^2$), and the desired sum consists of all products of two distinct primes and perfect cubes of primes. The former can be found by using the well-known expansion $(a_1 + a_2 + \\cdots + a_n)^2 = a_1^2 + a_2^2 + \\cdots + a_n^2 + 2(a_1a_2 + a_1a_3 + \\cdots + a_{n-1}a_n)$, or $$(2+3+5 + \\cdots + 29)^2 = (2^2 + 3^2 + 5^2 + \\cdots + 29^2) + 2(2 \\cdot 3 + \\cdots + 23 \\cdot 29).$$ Thus, this sum is $\\frac{129^2 - 2397}{2} = 7122$. The perfect cubes whose divisors are all", "24 60 12 30 Continue the steps until we have all prime numbers on the left side and at the bottom. 2 24 60 2 12 30 3 6 15 2 5 The LCM is 2 × 2 × 3 × 2 × 5 = 120. Formula If we know the greatest common divisor (GCD) of integers a and b, we can calculate the LCM using the following formula. $\\mathrm{LCM}\\left(a,b\\right)=\\frac{a×b}{\\mathrm{GCD}\\left(a,b\\right)}$ Using back the same example above, we can find the LCM of 24 and 60 as follows. $\\mathrm{LCM}\\left(24,60\\right)=\\frac{24×60}{12}=120$ Of course, you can use this formula to find GCD of two integers if you already know the LCM. Confused and have questions? We’ve got answers. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. If you rather get a 1:1 study help, try 30 minutes of free online tutoring with Chegg Tutors.", "$\\frac{6}{2} = 3$, $\\frac{6}{3} = 2$ and $\\frac{6}{6} = 1$ The term composite means \"something made by combining things\". So, a composite number is an integer made by multiplying prime numbers: • 2 is a prime • 3 is a prime • 4 is a composite: $2 \\cdot 2 = 4$ • 5 is a prime • 6 is a composite: $2 \\cdot 3 = 6$ • 7 is a prime • 8 is a composite: $2 \\cdot 2 \\cdot 2 = 8$ • 9 is a composite: $3 \\cdot 3 = 9$ • etc. Therefore, an integer is either a prime number or a product of prime numbers (a composite number). The famous Greek mathematician Euclid proved that there are infinite many prime numbers. Michael Rabin and Gary Miller have developed an algorithm that decides whether an integer is a prime or composite number by testing the integer with multiple bases denoted by $a$. The algorithm is called the Rabin-Miller primality test. #### Greatest common divisor Before we describe what the greatest common divisor of two integers are, we first define what we mean by a divisor. In this context is a divisor of an integer $x$ some integer that divides $x$ evenly, i.e. the result must not be a float. E.g. if you try to divide", "+0 # How many positive integers less than 500 are the product of exactly 0 1166 1 How many positive integers less than 500 are the product of exactly two distinct (meaning different) primes, each of which is greater than 10? Guest Mar 5, 2015 #1 +18835 +5 How many positive integers less than 500 are the product of exactly two distinct (meaning different) primes, each of which is greater than 10 ? $$\\small{\\text{ \\begin{array}{r|rcl|r} \\hline n & p_1 & & p_2 & p_1\\cdot p_2 \\\\ \\hline 1 & 11 & \\cdot & 13 & = 143 \\\\ 2 & 11 & \\cdot & 17 & = 187 \\\\ 3 & 11 & \\cdot & 19 & = 209 \\\\ 4 & 11 & \\cdot & 23 & = 253 \\\\ 5 & 11 & \\cdot & 29 & = 319 \\\\ 6 & 11 & \\cdot & 31 & = 341 \\\\ 7 & 11 & \\cdot & 37 & = 407 \\\\ 8 & 11 & \\cdot & 41& = 451 \\\\ 9 & 11 & \\cdot & 43 & = 473 \\\\ 10 & 13 & \\cdot & 17 & = 221 \\\\ 11 & 13 & \\cdot & 19 & = 247 \\\\ 12 & 13 & \\cdot & 23 & = 299 \\\\", "and prime to 30 are $1,7,11,13, 17,19,23,29.$ There are 8 of them, hence they are precisely the exponents of ${E}_{8}.$ 3. $W={E}_{7}:$ so $h=18.$ The positive integers less than 18 and prime to 18 are $1,5,7,11,13,17.$ So these are 6 of the exponents, and the other one must be $\\frac{1}{2}h=9,$ by (1) above. 4. $W={E}_{6}:$ $r=6,$ $\\mathrm{Card}\\left(R\\right)=72,$ so $h=12.$ Hence 1, 5, 7, 11 are four of the exponents. In fact, the other two are 4, 8, but one needs some extra information to deduce this. For example if one knew that $|W|={2}^{7}\\cdot {3}^{4}\\cdot 5=\\prod {d}_{i}=\\prod \\left({e}_{i}+1\\right)=2\\cdot 6\\cdot 8\\cdot 12\\left(e+1\\right)\\left(e\\prime +1\\right)$ where $e,e\\prime$ are the two unknown exponents, then since $2\\cdot 6\\cdot 8\\cdot 12={2}^{7}\\cdot {3}^{2}$ we have $(e+1) (e′+1) = 32⋅5 = 45$ which together with $e+e\\prime =12$ gives $ee\\prime =45-1-12=32$ and hence $e,e\\prime =4,8.$ 5. $W={F}_{4}:$$r=4,$ $\\mathrm{Card}\\left(R\\right)=48,$ so $h=12$ again. Hence the exponents are 1, 5, 7, 11. In fact the eigenvalues of the Cartan matrix $A={\\left(⟨{\\alpha }_{i},{\\alpha }_{j}^{\\vee }⟩\\right)}_{1\\le i,j\\le r}$ are by an ingenious argument due to Coxeter [Bourbaki, p.140 Ex. 3,4]. $G=$ simple (or almost simple) compact Lie group, $R$ the root system of $G$ relative to a maximal torus $T.$ Then (over the reals) $G$ has the same cohomology as a product of spheres of dimensions $2{e}_{i}+1,$ i.e., ${H}^{*}\\left(G;ℝ\\right)$ is an exterior algebra", "Olympiad. Problem: (1995 APMO - #2) Let$a_1, a_2, \\ldots, a_n$ be a sequence of integers with values between $2$ and $1995$ such that: (i) Any two of the $a_i$'s are realtively prime, (ii) Each $a_i$ is either a prime or a product of distinct primes. Determine the smallest possible values of $n$ to make sure that the sequence will contain a prime number. Solution: We claim that $n = 14$. A sequence with $13$ values is $41(43), 37(47), 31(53), 29(59), 23(61), 19(67), 17(71), 13(73), 11(79), 7(83), 5(89), 3(97), 2(101)$. Suppose that there exist $14$ composite numbers satisfying the given condition. Consider the smallest prime divisors of each of the $14$ numbers. They must be distinct to fulfill the relatively prime condition. But since $41$ is the $13$th prime, one of the numbers must have smallest prime divisor $\\ge 43$. Then that number is at least $43 \\cdot 47 > 1995$. Contradiction. Hence $n = 14$, as claimed. QED. ## Friday, March 10, 2006 ### Sum Big Stuff. Topic: Algebra/Sequences and Series. Level: Olympiad. Problem: (1997 APMO - #1) Given $S = 1 + \\frac{1}{1+\\frac{1}{3}} + \\frac{1}{1+\\frac{1}{3}+\\frac{1}{6}} + \\cdots + \\frac{1}{1+\\frac{1}{3}+\\frac{1}{6}+\\cdots+\\frac{1}{1993006}}$ where the denominators contain partial sums of the sequence of reciprocals of triangular numbers (i.e. $k = \\frac{n(n+1)}{2}$ for $n = 1,2,\\cdots,1996$). Prove that $S > 1001$. Solution:", "Show Tags Updated on: 06 Feb 2019, 07:11 Let the four numbers be a,b,c and d in increasing order. $$\\frac{a+b+c+d}{4}$$=60 a+b+c+d = 240 1. Median of 3 largest integers is 51. therefore c=51 c+d=190 ---(i) we know that a+b+c+d = 240 ---(ii) (ii) - (i) a+b = 50 There are 2 numbers less than 50 which are a & b. SUFFICIENT 2. Median of 4 integers is 50. $$\\frac{b+c}{2}$$=50 b+c=100 Values can be b =49 and c=51 or b=48 and C= 52 etc Hence there are 2 numbers less than 50 which are a & b. SUFFICIENT _________________ Originally posted by avinkhurana on 06 Feb 2019, 05:46. Last edited by avinkhurana on 06 Feb 2019, 07:11, edited 1 time in total. SVP Joined: 18 Aug 2017 Posts: 1887 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) If the average of four distinct positive integers is 60, how many inte [#permalink] ### Show Tags Updated on: 06 Feb 2019, 07:53 Bunuel wrote: If the average of four distinct positive integers is 60, how many integers of these four are less than 50? (1) The median of the three largest integers is 51 and the sum of two largest integers is 190. (2) The median of the four integers is 50. M08-28 given a+b+c+d= 240 #1", "Problem 41 Consider the following set: $$\\left\\{-5,-4,-\\frac{2}{3}, 0,1, \\sqrt{2}, 2,2.75,6,7\\right\\}$$ Which numbers are integers? Check back soon! ### Problem 42 Consider the following set: $$\\left\\{-5,-4,-\\frac{2}{3}, 0,1, \\sqrt{2}, 2,2.75,6,7\\right\\}$$ Which numbers are rational numbers? Check back soon! ### Problem 43 Consider the following set: $$\\left\\{-5,-4,-\\frac{2}{3}, 0,1, \\sqrt{2}, 2,2.75,6,7\\right\\}$$ Which numbers are irrational numbers? Check back soon! ### Problem 44 Consider the following set: $$\\left\\{-5,-4,-\\frac{2}{3}, 0,1, \\sqrt{2}, 2,2.75,6,7\\right\\}$$ Which numbers are prime numbers? Check back soon! ### Problem 45 Consider the following set: $$\\left\\{-5,-4,-\\frac{2}{3}, 0,1, \\sqrt{2}, 2,2.75,6,7\\right\\}$$ Which numbers are composite numbers? Check back soon! ### Problem 46 Consider the following set: $$\\left\\{-5,-4,-\\frac{2}{3}, 0,1, \\sqrt{2}, 2,2.75,6,7\\right\\}$$ Which numbers are even integers? Check back soon! ### Problem 47 Consider the following set: $$\\left\\{-5,-4,-\\frac{2}{3}, 0,1, \\sqrt{2}, 2,2.75,6,7\\right\\}$$ Which numbers are odd integers? Check back soon! ### Problem 48 Consider the following set: $$\\left\\{-5,-4,-\\frac{2}{3}, 0,1, \\sqrt{2}, 2,2.75,6,7\\right\\}$$ Which numbers are negative numbers? Check back soon! ### Problem 49 Graph each subset of the real numbers on a number line. The natural numbers between 1 and 5 Check back soon! ### Problem 50 Graph each subset of the real numbers on a number line. The composite numbers less than 10 Check back soon! ### Problem 51 Graph each subset of the real numbers on a number line. The prime numbers between 10 and 20", "we also have $$p|r^n$$ (exercise 5), and $$p$$ is prime so $$p|r$$ (by lemma 14). But then $$p$$ is a common divisor of $$r$$ and $$s$$ so $$\\gcd(r,s)\\geq p$$, which is a CONTRADICTION (we chose $$r$$ and $$s$$ to be coprime). Hence no prime number divides $$s$$, so the prime factorization of $$s$$ is the product of no prime numbers, and therefore $$s=1$$, and $$x=\\frac{r} {s}=r$$ which is an integer. ∎ Example Take any positive integer $$n$$. Then $$\\sqrt{n}$$ is either an integer or it is irrational. This is the particular case $$f(x)=x^2-n$$ in proposition 21. For the second way of deriving a new number system from $$\\mathbb{Z}$$, imagine taking the number line and rolling it up just tightly enough for 5 to appear directly over 0 (one layer higher in the roll), 6 above 1, etc. We shall regard two integers as equivalent if one appears directly over the other in this roll, i.e. if their difference is divisible by 5. For example, 14 appears above -1 so we say 14 is congruent to -1 (modulo 5) and write $$14\\equiv -1\\pmod{5}$$. We can also collect together integers that are equivalent with each other, and in this way we obtain the following 5 sets of integers: $\\begin{array}{c} \\{\\ldots,-5,0,5,10,15,\\ldots\\} \\\\ \\{\\ldots,-4,1,6,11,16,\\ldots\\} \\\\ \\{\\ldots,-3,2,7,12,17,\\ldots\\} \\\\ \\{\\ldots,-2,3,8,13,18,\\ldots\\} \\\\ \\{\\ldots,-1,4,9,14,19,\\ldots\\} \\end{array}$ Each", "# Math Help - If n>1 , show that n^4 + 4^n is never a prime? 1. ## If n>1 , show that n^4 + 4^n is never a prime? If n>1, show that n^4 + 4^n is never a prime? 2. $n$ clearly has to be odd, so $\\tfrac{n+1}{2}$ is an integer. Then $n^4+4^n=\\left(n^2+2^n\\right)^2-2^{n+1}n^2=\\left[n^2+2^n+2^{\\frac{n+1}{2}}n\\right]\\left[n^2+2^n-2^{\\frac{n+1}{2}}n\\right]$ $n^2+2^n+2^{\\frac{n+1}{2}}n$ is clearly greater than 1; if you can show that $n^2+2^n-2^{\\frac{n+1}{2}}n>1$ then you’re done ( $n^4+4^n$ would be the product of two integers greater than one and so can’t be prime). 3. Okay, here we go. Prove that $n^2+2^n-2^{\\frac{n+1}{2}}\\cdot n>1$ for all odd integers $n>1$. First, check the cases $n=3$ and $n=5$ separately. $3^2+2^3-2^{\\frac{3+1}{2}}\\cdot3=9+8-12=5>1$ $5^2+2^5-2^{\\frac{5+1}{2}}\\cdot5=25+32-24=33>1$ For odd integers $n\\ge7$, I claim that $2^n>2^{\\frac{n+1}{2}}\\cdot n$. This can be proved by a slight variation of the method of induction. When $n=7$, $2^7=128>112=2^{\\frac{7+1}{2}}\\cdot7$. Suppose $2^n>2^{\\frac{n+1}{2}}\\cdot n$ for some odd integer $n\\ge7$. Now consider $2^{\\frac{(n+2)+1}{2}}\\cdot(n+2)$. (NB: $n+2$ is the next odd integer.) $2^{\\frac{(n+2)+1}{2}}\\cdot(n+2)=2\\cdot2^{\\frac{n+1 }{2}}\\cdot(n+2)$ $\\color{white}.\\hspace{27mm}.$ $=2\\cdot2^{\\frac{n+1}{2}}\\cdot n+2\\cdot2^{\\frac{n+1}{2}}\\cdot2$ $\\color{white}.\\hspace{27mm}.$ $<2\\cdot2^n+2\\cdot2^n\\color{white}\\ .$* $\\color{white}.\\hspace{27mm}.$ $=4\\cdot2^n=2^{n+2}$ * $\\ 3 Hence, $2^n>2^{\\frac{n+1}{2}}\\cdot n$ for all odd integers $n>5$, and it follows that $n^2+2^n-2^{\\frac{n+1}{2}}\\cdot n>n^2>1$.", "be relatively prime integers with a > b > 0 and a 3 −b 3 ( [#permalink] ### Show Tags 17 Apr 2019, 10:52 2 Let a and b be relatively prime integers with a > b > 0 and $$\\frac{a^3 −b^3}{(a−b)^3}$$ =$$\\frac{73}{3}$$ What is a − b? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 a and b are relatively prime and a>b>0. $$\\frac{a^3 −b^3}{(a−b)^3}$$ =$$\\frac{73}{3}$$ $$\\frac{(a-b)(a^2 +ab + b^2)}{(a−b)^3}$$ =$$\\frac{73}{3}$$ Since a>b>0 So, a-b=/=0 $$\\frac{(a^2 + ab + b^2)}{(a−b)^2}$$ =$$\\frac{73}{3}$$ $$\\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}$$ =$$\\frac{73}{3}$$ $$\\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}-1$$ =$$\\frac{73}{3}-1$$ $$\\frac{(3ab)}{(a^2 - 2ab + b^2)}$$ =$$\\frac{70}{3}$$ $$\\frac{(ab)}{(a-b)^2}$$ =$$\\frac{70}{9}$$ $$\\frac{(ab)}{(a-b)^2}$$ =$$\\frac{70x^2}{(3x)^2}$$ so, a-b = 3x and ab = 70x^2 At x =1, a-b = 3, ab = 70 -> (3+b)b= 70 ->b^2 +3b -70 = 0 -> (b+10)(b-7) = 0 -> b= 7, a = 10 , a-b = 3 At x= 2, a-b = 6 , ab = 70*2^2, a= 20, b=14 and so on... But a and b are co-prime and hence a-b =3 , ab = 70, a = 10 & b=7. _________________ CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler UPSC Aspirants : Get my app UPSC Important News Reader from Play store. MBA Social Network : WebMaggu Appreciate by", "(Official MAA) Suppose that $N$ has the required property. Then there are positive integers $k$ and $m$ such that $N = 10^4m + 2020 = k\\cdot m$. Thus $(k - 10^4)m = 2020$, which holds exactly when $m$ is a positive divisor of $2020.$ The number $2020 = 2^2\\cdot 5\\cdot 101$ has $12$ divisors: $1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010$, and $2020.$ The requested sum is therefore the sum of the digits in these divisors plus $12$ times the sum of the digits in $2020,$ which is $$(1+2+4+5+1+2+2+4+8+10+2+4)+12\\cdot4 = 93.$$ ## Solution 3 Note that for all $N \\in S$, $N$ can be written as $N=10000x+2020=20(500x+101)$ for some positive integer $x$. Because $N$ must be divisible by $x$, $\\frac{20(500x+101)}{x}$ is an integer. We now let $x=ab$, where $a$ is a divisor of $20$. Then $\\frac{20(500x+101)}{x}=(\\frac{20}{a})( \\frac{500x}{b}+\\frac{101}{b})$. We know $\\frac{20}{a}$ and $\\frac{500x}{b}$ are integers, so for $N$ to be an integer, $\\frac{101}{b}$ must be an integer. For this to happen, $b$ must be a divisor of $101$. $101$ is prime, so $b\\in \\left \\{ 1, 101 \\right \\}$. Because $a$ is a divisor of $20$, $a \\in \\left \\{ 1,2,4,5,10,20\\right\\}$. So $x \\in \\left\\{1,2,4,5,10,20,101,202,404,505,1010,2020\\right\\}$. Be know that all $N$ end in $2020$, so the sum of the digits of each $N$ is the sum" ]

[ "+0 # Help Probability! +1 48 2 +82 Two integers are relatively prime if they have no common factors other than 1 or -1. What is the probability that a positive integer less than or equal to 30 is relatively prime to 30? Express your answer as a common fraction. MathCuber Jul 8, 2018 #1 -1 $${7 \\over 29}$$ Guest Jul 8, 2018 #2 +1 Since 30 = 2 x 3 x 5, therefore it has common factors with the following numbers: 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30[22 numbers] It is relatively prime with following numbers: 1, 7, 11, 13, 17, 19, 23, 29[8 numbers] Therefore, the probability that a number =< 30 is relatively prime to 30 is: 8 / 30 = 4 / 15 Guest Jul 8, 2018", "# Math Help - Positive Integers 1. ## Positive Integers How many positive integers less than 2007 are relatively prime to 1001? 2. Factor: $1001 = 7 \\cdot 11 \\cdot 13$. How many positive integers less than 2007 are multiples of at least one of 7, 11, or 13? None of those integers can be relatively prime (also known as coprime) with 1001. 3. Hello, Dragon! How many positive integers less than 2007 are relatively prime to 1001? There are 2006 positive integers less than 2007. How many of these have a common factor with 1001? We find that: . $1001 \\:=\\:7\\!\\cdot\\!11\\!\\cdot\\!13$ There are: . $\\left[\\frac{2006}{7}\\right] = 286$ with a factor of 7. There are: . $\\left[\\frac{2006}{11}\\right] = 182$ with a factor of 11. There are: . $\\left[\\frac{2006}{13}\\right] = 154$ with a factor of 13. There are: . $\\left[\\frac{2006}{77}\\right] = 26$ with a factor of 7 and 11. There are: . $\\left[\\frac{2006}{91}\\right] = 22$ with a factor of 7 and 13. There are: . $\\left[\\frac{2006}{143}\\right] = 14$ with a factor 11 and 13. There are: . $\\left[\\frac{2006}{1001}\\right] = 2$ with a factor of 7, 11 and 13. Formula: $n(A\\,\\cup B\\,\\cup\\ C) \\;=\\;n(A) + n(B) + n(C) - n(A\\,\\cap\\,B) - n(A\\,\\cap\\ C)$ $- n(B\\,\\cap\\,C) + n(A\\,\\cap\\,B\\,\\cap\\,C)$ Hence: . $n(7 \\cup 11 \\cup 13) \\;=\\;286 + 182 + 154 -", "+0 # Help Probability! +1 157 2 +314 Two integers are relatively prime if they have no common factors other than 1 or -1. What is the probability that a positive integer less than or equal to 30 is relatively prime to 30? Express your answer as a common fraction. Jul 8, 2018 #1 -1 $${7 \\over 29}$$ . Jul 8, 2018 #2 +1 Since 30 = 2 x 3 x 5, therefore it has common factors with the following numbers: 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30[22 numbers] It is relatively prime with following numbers: 1, 7, 11, 13, 17, 19, 23, 29[8 numbers] Therefore, the probability that a number =< 30 is relatively prime to 30 is: 8 / 30 = 4 / 15 Jul 8, 2018", "# Number of pairs of co-prime positive integers $(a, b)$ such that $\\frac{a}{b} + \\frac{14b}{a}$ is an integer? How many pairs of positive integers $(a, b)$ are there such that $a$ and $b$ have no common factor greater than $1$ and $\\frac{a}{b} + \\frac{14b}{a}$ is an integer? The problem I'm facing in this question is that is there any algebraic or short-cut method to find the pairs. Else I've done it by Hit and Trial but that consumes a lot of time. • Another way to say this is: $$\\frac{a^2+14b^2}{ab} \\text{ is an integer}$$ I'm not sure if this helps, though. – Noble Mushtak Jun 26 '16 at 19:42 ## 2 Answers You are looking for coprime integers $a$ and $b$ for which $$\\frac{a}{b}+14\\frac{b}{a}=\\frac{a^2+14b^2}{ab},$$ is an integer. Then $b$ divides $a^2$, and so $b=1$ because $\\gcd(a,b)=1$. Now the above reduces to $$\\frac{a^2+14}{a},$$ which is an integer if and only if $a$ divides $14$, i.e. $a\\in\\{1,2,7,14\\}$. • Why $b$ divides $a^2$? – Qwerty Jun 26 '16 at 19:49 • Because $b$ divides $a^2+14b^2$, because the shown fraction is an integer. – Inactive - Objecting Extremism Jun 26 '16 at 19:49 • Why not $a$ divides $b^2$? – Qwerty Jun 26 '16 at 19:51 • @Qwerty: you also have $a$ divides $14b^2$, but factors can come from $14$ as well.", "# Can fractions be relatively prime? Two numbers are relatively prime if they do not share any factors, other than 1. Is it possible for fractions to be relatively prime? To reword this, do fractions even have factors? - One can extend gcd and lcm to rationals. – Bill Dubuque May 18 '14 at 0:44 Every rational number is of the form $\\prod_{n=1}^\\infty p_n^{e_n}$ where $p_n$ is the $n$th prime number and each $e_n$ is some integer. The difference between that statement and a similar one about integers is that with integers one has $e_n\\ge0$ for every value of $n$. Two integers are relatively prime if the set of primes with non-zero exponents for one of them is disjoint from the set of primes with nonzero exponents for the other. One could use that same definition with rational numbers and say, for example, that $8/35$ and $11/9$ are relatively prime because $$\\frac{8}{35} = 2^3 5^{-1} 7^{-1} \\text{ and }\\frac{11}{9} = 3^{-2} 11^1$$ and $$\\{2,5,7\\} \\cap \\{3,11\\}=\\varnothing.$$ $\\frac{1}{3}, \\frac{2}{5}$ become $\\frac{5}{15}, \\frac{6}{15}$ which have relatively prime numerators. In this sense fractions are just integers split up into a common number of parts.", "# Give an example of four integers which are relatively prime but not relatively prime in pairs Give an example of four integers which are relatively prime but not relatively prime in pairs I can think of 3 integers that are relatively prime but not relatively prime in pairs: 10,12,15. (10,12,15)=1 (10,12)=2 (12,15)=3 (10,15)=5 But I am not able to think of an example for 4 integers that are relatively prime but no relatively prime in pairs...Can somebody help me with this. • $(10,6,15, 2\\cdot3\\cdot5\\cdot7)$? The trick is to write up the prime factorization of the numbers and make sure that any two has at least one common factor, while all the four don't have a common factor. – Andrew Sep 28 '16 at 12:01 ## 1 Answer For the three number case, many different examples can be found by the following construction: let $p,q,r$ be distinct primes. Then $pq, qr, pr$ will have the property you want. Here's how you might come up with an answer. Let $p,q,r,s$ be distinct primes. What can you say about the numbers $pqr, pqs, prs, qrs$? Well, none of $p,q,r,s$ divide all four of them, however its not too hard to see that any two share a common factor. Explicitly, choosing the first four primes you can see that $30, 42,", "average of the first 30 positive integers is $\\displaystyle \\frac 1{30} \\cdot \\frac {30(30 + 1)}2 = 15.5$ for the second problem your conjecture is correct. to find the actual number, you would use the formula i gave you above , , , , , , , , , , , , , , # averages integers pre a Click on a term to search for related topics.", "You are not signed in. To save solutions, earn points and appear on the leaderboard. register or sign in. # Relatively prime integers #### Description Two integers $$a$$ and $$b$$ are said to be relatively prime if the only positive integer that divides both of them is 1. The positive integers less than 10 which are relatively prime to 10 are: 1, 3, 7, 9 therefore the sum of positive integers less than 10 which are relatively prime is 20. #### Question What is the sum of all positive integers less than 1,000,000 which are relatively prime to 1,000,000? What is the sum of all positive integers less than 1,000,000 which are relatively prime to 1,000,000?", "are looking number of integers less than $$x$$). For example consider $$x=7=prime$$: how many numbers are less than 7 and have no common factors with 7: 1, 2, 3, 4, 5, and 6, so total of $$7-1=6$$ numbers. _________________ Manager Joined: 06 Jun 2013 Posts: 106 Concentration: Technology, General Management GMAT 1: 680 Q48 V35 GPA: 3.47 ### Show Tags 15 Jun 2015, 05:33 Hi Bunuel, Aren't we supposed to return the set of all the positive numbers less than x that does not have a common factor with x OTHER THAN 1. In that case, aren't we supposed to ignore the positive factor 1 and find out how many other numbers less than x are there who do not have common factor with X. With this as my understanding, I selected option (A) -> x-2. eg: say prime number is 7. Numbers less than 7 other than 1 -> 2, 3,4,5,6. Total - 5 and hence X-2. Bunuel wrote: Official Solution: If $$x$$ is a positive integer, $$f(x)$$ is defined as the number of positive integers which are less than $$x$$ and do not have a common factor with $$x$$ other than 1. If $$x$$ is prime, then $$f(x)=$$? A. $$x - 2$$ B. $$x - 1$$ C. $$\\frac{(x + 1)}{2}$$ D. $$\\frac{(x - 1)}{2}$$ E. 2", "5 \\cdot 54673257461630679457$. A first approximation to the abundance of $n$ is $\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{5} = 1.03\\bar{3}$; a second approximation includes the small divisors that are product of the three small primes (i.e. the exact abundance of $2 \\cdot 3 \\cdot 5 = 30$): $\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{10} + \\frac{1}{15} + \\frac{1}{30} = \\frac{7}{5} = 1.4$. The last terms to include in the full abundance are $\\frac{1}{54673257461630679457}$, $\\frac{1.4}{54673257461630679457}$, and $\\frac{1}{30 \\cdot 54673257461630679457}$, which are 1.83e-20, 2.56e-20, and 6.10e-22 respectively, meaning the actual abundance is roughly 1.4000000000000000000445, which is actually beyond the precision that standard double-sized floating point numbers can represent. Knowing now that the smallest factors of a number largely determine its abundance, it's easy to see why most even numbers are abundant: all of them have an abundance of at least $\\frac{1}{2}$, and any further powers of two or a couple more small primes will push the abundance over 1 in a pinch. Conversely, odd numbers have no guaranteed small factor: if an odd number's smallest prime factor is, say, 101, then unless there are hundreds or thousands of other divisors of this number, the abundance will certainly be less than 1. For example, these are the five smallest odd abundant numbers: • $945 = 3^3 \\cdot 5 \\cdot 7$", "that one of the two integers randomly selected from range 20-29, inclusive, is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal) A. 0.06 B. 0.12 C. 0.15 D. 0.18 E. 0.20 Prime integers: 23 and 29. Multiples of 3: 21, 24, and 27. The probability that the first number is prime while the second is a multiple of $$3 = \\frac{2}{10} \\frac{3}{10} = 0.06$$. The probability that the first number is a multiple of 3 while the second is prime $$= \\frac{3}{10} \\frac{2}{10} = 0.06$$. The probability that one of the two integers is prime and the other is a multiple of $$3 = 0.06 + 0.06 = 0.12$$. _________________ Intern Joined: 27 Sep 2015 Posts: 15 ### Show Tags 03 Apr 2016, 06:15 Hi Bunuel I was wondering why the order of the selection matters here ? Thanks Bunuel wrote: Official Solution: What is the probability that one of the two integers randomly selected from range 20-29 is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal) A. 0.06 B. 0.12 C. 0.15 D. 0.18 E. 0.20 Prime integers: 23 and 29. Multiples of 3: 21, 24, and 27. The", "prime $8\\,$ and $\\,9\\,$ are unrelated $8\\,$ and $\\,9\\,$ are strangers When two numbers are relatively prime (unrelated, strangers), then their least common multiple is their product: • The least common multiple of $\\,8\\,$ and $\\,9\\,$ is $\\,8\\cdot 9 = 72\\,.$ • $\\text{lcm}(3,5) = 3\\cdot 5 = 15\\,$ • $\\displaystyle\\frac 17 + \\frac1{10} = \\frac{10}{70} + \\frac{7}{70} = \\frac{17}{70}$ Note that the Least Common Denominator (LCD), one important use of the Least Common Multiple (LCM), is $\\,7\\cdot 10 = 70\\,$. Relatively prime is an important mathematical concept, and hence deserves a formal definition: DEFINITION relatively prime Two counting numbers are relatively prime if and only if they have no common factor other than $\\,1\\,$. Equivalently: Two counting numbers are relatively prime if and only if their greatest common factor is $\\,1\\,$. ## Lives-In The ‘lives-in’ relationship says that one of the structures lives entirely inside the other. That is, one of the numbers contains all the prime factors of the other . It's easy to recognize the lives-in case, because the smaller number goes into the bigger one evenly. ## Examples $6\\,$ lives in $\\,18\\,$ (see image at right). The number $\\,18\\,$ contains all the prime factors of $\\,6\\,$. $3\\,$ lives in $\\,12\\,.$ The number $\\,12\\,$ is divisible by $\\,3\\,$. $15\\,$ lives in $\\,30\\,.$ The number $\\,30\\,$ is", "= \\frac{195}{15}$ $b = 13$ ## Practice Problems 1 ### Which among the following is relatively prime with 16? 2 3 4 8 CorrectIncorrect Two even numbers can never be relatively prime. 2, 4, 8, and 16 all are even numbers. So, the number relatively prime with 16 is 3. 2 ### Which of the following pairs of numbers are relatively prime? 11, 39 30, 51 3, 18 4, 16 CorrectIncorrect Numbers 11 and 39 have no common factors other than 1. So, 11 and 39 are relatively prime numbers. 3 ### If two numbers, a and b, are relatively prime, what is the Highest Common Factor (HCF) of a and b? 0 1 2 3 CorrectIncorrect Two numbers are relatively prime, when their Highest Common Factor (HCF) is 1. So, the Highest Common Factor (HCF) of a and b is 1. 4 ### Which of the following numbers cannot be identified as relatively prime? Consecutive Integers Two even numbers Two prime numbers m, n where GCD(m,n) $= 1$ CorrectIncorrect Correct answer is: Two even numbers Two even numbers can never be relatively prime since they have a common factor 2. The numbers 1 and $-1$ are the only integers coprime to 0.", "+0 # Help PLS, ASAP +2 700 1 +140 The least common multiple of two positive integers is $7!$, and their greatest common divisor is $9$. If one of the integers is $315$, then what is the other? (Note that $7!$ means $7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot 1$.) Mar 19, 2019 edited by CorbellaB.15 Mar 19, 2019 #1 0 The identity lcm[a,b]*gcd(a,b)=ab holds for all positive integers a and b. Let a=315 in this identity, and let b be the number we are looking for. Thus $$7! \\cdot 9=315 \\cdot b$$ so $$b = \\frac{7!\\cdot 9}{315} = \\frac{7!\\cdot 9}{35\\cdot 9} = \\frac{7!}{35} = \\frac{{7}\\cdot 6\\cdot {5}\\cdot 4\\cdot 3\\cdot 2\\cdot 1}{{35}} = 6\\cdot4\\cdot3\\cdot2 = \\boxed{144}.$$ Hope this helps :) Sep 7, 2019", "the actual amount = $$\\frac{16}{2}, \\frac{17}{2}, \\frac{21}{2}$$ = Rs.8; Rs. 8.50; Rs. 10.50 Now the average = $$\\frac{8+8.50+10.50}{3}=\\frac{27.00}{3}$$ = Rs .9 = $$\\frac{\\text { Original average }}{2}$$ iv) The change In the observations is also carried out in the mean. Question 7. Find the mean of the first ten natural numbers. Solution: First ten natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 ∴ Average = $$\\frac{\\text { Sum of the numbers }}{\\text { Number }}$$ = $$\\frac{1+2+3+4+5+6+7+8+9+10}{10}=\\frac{55}{10}$$ = 5.5 ∴ The mean of the first ten natural numbers = 5.5. Question 8. Find the mean of the first five prime numbers. Solution: First five prime numbers are 2, 3, 5, 7 and 11 Average of first five prime numbers = $$\\frac{\\text { Sum of the numbers }}{\\text { Number of primes }}$$ = $$\\frac{2+3+5+7+11}{5}=\\frac{28}{5}$$ = 5.6 Question 9. In a set of four integers, the average of the two smallest integers is 102, the average of the three smallest integers is 103, the average of all four is 104. Which is the greatest of these integers? Solution: Given that The average of four lntegers = 104 = $$\\frac{\\text { Sum of the four integers }}{\\text { Number of integers }}$$ ∴ The sum of the four integers = average × number =", "60, how many integers of these four are less than 50? It's almost always better to express the average in terms of the sum: the average of four distinct positive integers is 60, means that the sum of four distinct positive> integers is $$4*60=240$$. Say four integers are $$a$$, $$b$$, $$c$$ and $$d$$ so that $$0<a<b<c<d$$. So, we have that $$a+b+c+d=240$$. (1) The median of the three largest integers is 51 and the sum of two largest integers is 190. The mdian of $$\\{b,c,d\\}$$ is 51 means that $$c=51$$. Now, if $$b=50$$, then only $$a$$, will be less than 50, but if $$b<50$$, then both $$a$$ and $$b$$, will be less than 50. But we are also given that $$c+d=190$$. Substitute this value in the above equation: $$a+b+190=240$$, which boils down to $$a+b=50$$. Now, since given that all integers are positive then both $$a$$ and $$b$$ must be less than 50. Sufficient. (2) The median of the four integers is 50. The median of a set with even number of terms is the average of two middle terms, so $$median=\\frac{b+c}{2}=50$$. Since given that $$b<c$$ then $$b<50<c$$, so both $$a$$ and $$b$$ are less than 50. Sufficient. Answer: D. Thanks Bunuel. That makes a lot more sense Director Joined: 22 Mar 2011 Posts: 612 WE: Science (Education) Followers: 100", "and only if the numbers 2a − 1 and 2b − 1 are coprime. As a generalization of this, following easily from Euclidean algorithm in base n > 1: $\\gcd (n^a-1,n^b-1)=n^{\\gcd(a,b)}-1.$ ## Cross notation, group If n≥1 and is an integer, the numbers coprime to n, taken modulo n, form a group with multiplication as operation; it is written as (Z/nZ)× or Zn*. ## Generalizations The concept of being relatively prime can also be extended to any finite set of integers S = {a1, a2, .... an} to mean that the greatest common divisor of the elements of the set is 1. If every pair in a (finite or infinite) set of integers is relatively prime, then the set is called pairwise relatively prime. Every pairwise relatively prime finite set is relatively prime; however, the converse is not true: {6, 10, 15} is relatively prime (the only positive integer that divides all of 6, 10 and 15 is 1), but not pairwise relative prime (each pair of integers in the set has a non-trivial common factor). Two ideals A and B in the commutative ring R are called coprime (or comaximal) if A + B = R. This generalizes Bézout's identity: with this definition, two principal ideals (a) and (b) in the ring of integers Z are coprime", "Olympiad. Problem: (1995 APMO - #2) Let$a_1, a_2, \\ldots, a_n$ be a sequence of integers with values between $2$ and $1995$ such that: (i) Any two of the $a_i$'s are realtively prime, (ii) Each $a_i$ is either a prime or a product of distinct primes. Determine the smallest possible values of $n$ to make sure that the sequence will contain a prime number. Solution: We claim that $n = 14$. A sequence with $13$ values is $41(43), 37(47), 31(53), 29(59), 23(61), 19(67), 17(71), 13(73), 11(79), 7(83), 5(89), 3(97), 2(101)$. Suppose that there exist $14$ composite numbers satisfying the given condition. Consider the smallest prime divisors of each of the $14$ numbers. They must be distinct to fulfill the relatively prime condition. But since $41$ is the $13$th prime, one of the numbers must have smallest prime divisor $\\ge 43$. Then that number is at least $43 \\cdot 47 > 1995$. Contradiction. Hence $n = 14$, as claimed. QED. ## Friday, March 10, 2006 ### Sum Big Stuff. Topic: Algebra/Sequences and Series. Level: Olympiad. Problem: (1997 APMO - #1) Given $S = 1 + \\frac{1}{1+\\frac{1}{3}} + \\frac{1}{1+\\frac{1}{3}+\\frac{1}{6}} + \\cdots + \\frac{1}{1+\\frac{1}{3}+\\frac{1}{6}+\\cdots+\\frac{1}{1993006}}$ where the denominators contain partial sums of the sequence of reciprocals of triangular numbers (i.e. $k = \\frac{n(n+1)}{2}$ for $n = 1,2,\\cdots,1996$). Prove that $S > 1001$. Solution:", "$a_1, \\cdots, a_r$ is relatively prime if $(a_1,\\cdots, a_r) = 1$. A collection $a_1,\\cdots,a_r$ is said to be pairwise relatively prime if $(a_i,a_j)=1$ whenever $i \\neq j$. In the examples above, we see that 21 and 10 are relatively prime, and that the collection $\\{10, 12, 15\\}$ is relatively prime. Notice in this last example that the collection is relatively prime even though each pair of integers from the collection is not relatively prime. (As a general rule of thumb, you'll care more about whether a collection is pairwise relatively prime than whether it's relatively prime). Properties of the GCD Having met and played around with greatest common divisors a bit, we'll now introduce a few properties that they enjoy. Removing the GCD First, we'll see what we get when we remove the gcd of two integers. Lemma: For any pair of integers a and b, we have $\\displaystyle \\left(\\frac{a}{(a,b)},\\frac{b}{(a,b)}\\right) = 1$. Proof: Since $(a,b) \\mid a$ and $(a,b) \\mid b$, there exist integers $q_a,q_b$ so that (3) \\begin{align} a = (a,b)q_a \\quad \\mbox{ and } \\quad b = (a,b)q_b. \\end{align} Our goal, then, is to show that $(q_a,q_b) = 1$. For this, suppose that there were some common divisor $d>1$ of both $q_a$ and $q_b$. This would imply that there exist integers $c_a,c_b$ satisfying (4) \\begin{align} q_a", "5 . Definition. We say that the integers a1.e2,...,e1 are mutually relatively prime if (a1, e2,...,an) : l. These integers 4re called pairwise relatively prime if for each pair of integers4; and a; from the set, (ai, a1): l, that is, if each pair of integersfrom the set is relatively prime. It is easy to see that if integersare pairwise relatively prime, they must be mutually relatively prime. However, the converseis false as the following exampleshows. Example. Considerthe integers15, 21, and 35. Since ( 1 5 , 2 r , 3 5 ) (: t s ,( 2 t , 3 5 ) ) :( r 5 , 7 ) : r , we see that the three integersare mutually relatively prime. However, they are not pairwise relatively prime, b e c a u s(et S . z l ) : 3 , ( 1 5 , 3 5 ): 5 , a n d (21,35):7. 2.1 Problems l. Find the greatestcommon divisor of each of the following pairs of integers il 15,35 b) 0,lll c) -12.t8 d) 99, 100 e ) 1l , l 2 l f) 100,102 Show that if a and b are integerswith (a, b) : l, then (a*b, a-b) : I or 2. Show that if a and b are integers, that are" ]

[ "the same time, player two gets a point. If one red and one green marble are pulled at the same time, both players get a point. Is this a fair game?", "zebra fish. In how many different ways can George choose 2 zebra fish to buy? 12. Calculation of CN Calculate: ? 13. Trinity How many different triads can be selected from the group 43 students?", "If you think you have a good answer to this, do let me know, but note that the really hard problem, still I think unsolved, is to prove the best optimum result. # Sunday Times Teaser 3007 – Paving Stones ### by Howard Williams #### Published Sunday May 10 2020 (link) James has decided to lay square block paving stones on his rectangular patio. He has calculated that starting from the outside and working towards the middle that he can lay a recurring concentric pattern of four bands of red stones, then three bands of grey stones, followed by a single yellow stone band. By repeating this pattern and working towards the centre he is able to finish in the middle with a single line of yellow stones to complete the patio. He requires 402 stones to complete the first outermost band. He also calculates that he will require exactly 5 times the number of red stones as he does yellow stones. How many red stones does he require?` # Sunday Times Teaser 3006 – Raffle Tickets ### by Andrew Skidmore #### Published Sunday May 03 2020 (link) At our local bridge club dinner we were each given a raffle ticket. The tickets were numbered from 1 to 80. There were six people on our table and all", "John has 340 marbles. He put them into 4 bags. He put the same number of marbles in each bag. How many marbles are in each bag? 63 views by (1.6k points) 340/4=85 this is the answer", "# Math Help - Colored Stones 1. ## Colored Stones You have 5 differently colored stones-red, orange, blue, green, purple. If the green stone cannot be placed at the front or the back of the sequence, how many possible arrangements can you make? I know that without the above restriction, the amount would be 5!=120. But I don't get how to use the restriction. The back of the book says that the answer is 72. Thanks for any help 2. That would be 5! - 2*4! 3. Originally Posted by paultwang That would be 5! - 2*4! Can you please explain I don't understand the logic behind that. Thanks 4. When green is placed at the back: ROBPG, ROPBG, ..... (4! combinations) When green is placed at the front: GROBP, GROPB, ..... (4! combinations) 5! - 2*4! 5. There are 5 stones. Since you CANNOT place the green stone at the front, there are 4 stones you CAN place there. Assuming you have placed that stone, there are 4 stones left. Since you CANNOT place the green stone at the end (and one of the 4 left is green), there are 3 stones left that you CAN place there. Now you have 3 stones left to put in the remaining 3 places- of course there are 3! ways", "Time Limit : sec, Memory Limit : KB English Problem E: Colored Octahedra A young boy John is playing with eight triangular panels. These panels are all regular triangles of the same size, each painted in a single color; John is forming various octahedra with them. While he enjoys his playing, his father is wondering how many octahedra can be made of these panels since he is a pseudo-mathematician. Your task is to help his father: write a program that reports the number of possible octahedra for given panels. Here, a pair of octahedra should be considered identical when they have the same combination of the colors allowing rotation. Input The input consists of multiple datasets. Each dataset has the following format: Color1 Color2 ... Color8 Each Colori (1 ≤ i ≤ 8) is a string of up to 20 lowercase alphabets and represents the color of the i-th triangular panel. The input ends with EOF. Output For each dataset, output the number of different octahedra that can be made of given panels. Sample Input blue blue blue blue blue blue blue blue red blue blue blue blue blue blue blue red red blue blue blue blue blue blue Output for the Sample Input 1 1 3", "# John's Red and Blue Balls John has a box that contains one red ball and one blue ball. Every minute he reaches into the box and randomly takes out a ball from it (without looking). No matter which ball he took out, he puts it back ${together\\ with}$ another ball of the same color. Thus, after $n$ minutes, the box contains a total of $n+2$ balls. The probability that after $2$ hours, the box contains exactly $77$ blue balls, can be written as $\\frac{a}{b}$, where $a$ and $b$ are coprime positive integers. Find $a+b$. ×", "apples - 7 red and 3 green. 4 30 May 2006, 09:58 Display posts from previous: Sort by", "1 : 6. There is an equal amount of lemon syrup in both cups. Cup B contains 189 mℓ more water than Cup A. What is the amount of lemon syrup in Cup A? Leave the answer in mℓ. 3 m Level 3 Mrs Lu had a total of 520 purple and yellow lilies in the ratio of 8 : 5. After she had given away 6 times as many purple lilies as yellow lilies, the number of purple and yellow lilies left was in the ratio of 10 : 9. How many more purple lilies than yellow lilies did she give away? 5 m Level 3 David, Ethan and Tim have some marbles. The ratios of their marbles are as follows: David : (David + Ethan + Tim) = 1 : 5 Ethan : (David + Tim) = 1 : 3 Tim has 42 marbles more than Ethan. In the end, Tim gives away all he has to Ethan and David so that both Ethan and David end up with the same number of marbles, how many marbles has Ethan received from Tim? 4 m Level 3 Oscar bought some mangosteens and coconuts in the ratio of 2 : 5 respectively. The cost of a mangosteen was $1.50 more than the cost of a coconut. He spent", "of doing that. The total number of ways you can place the 5 stones WITHOUT the green stone being first or last, is (4)(3)(3!)= 4*3*6= 72. That is, of course, exactly the same as 5!- 2(4!).", "values don't exceed N bits, but may be incorrect if they do exceed that\"; \"Your answer must be theoretically correct if your language's integer types had infinite width, but you may use any integer type with at least 8 bits\". – Peter Taylor Apr 30 '18 at 15:41 • @ Peter Taylor Thank you for your help! You are completely right abiut the overflow - I have used J's extended precision to make the test cases. I'm going to update the description. – Galen Ivanov May 1 '18 at 8:35 # Relational Division Given a database of colored marbles in bins, find which bin has the most marbles of a certain color. In database administration, this task is known as relational division. ## Example Input marbles (flexible format, this is just one possibility): red,A blue,B green,C red,B red,A blue,A Input color: red Program output: A ## Format Each 'marble' is represented as a (color, bin) pair, so the entire database of marbles is represented by a list of these pairs. The list can be fed into your program in any reasonable way so long as the association of the input data remains between colors and bins -- for example via CSV on stdin in color,bin pairs split on newlines in string-manipulation programs, or perhaps via a list of", "Between two red cubes must be a different color. 2. Between two blue must be two different colors. 3. Between two green must be three different colors. 4. Between two yellow blocks must be four • Salami How many ways can we choose 5 pcs of salami if we have 6 types of salami for 10 pieces and one type for 4 pieces? • Twos Vojta started writing the number of this year 2019202020192020 into the workbook. .. And so he kept going. When he wrote 2020 digits, he no longer enjoyed it. How many twos did he write? • Octahedron - sum On each wall of a regular octahedron is written one of the numbers 1, 2, 3, 4, 5, 6, 7 and 8, wherein on different sides are different numbers. For each wall John make the sum of the numbers written of three adjacent walls. Thus got eight sums, which also • Mouse Hryzka Mouse Hryzka found 27 identical cubes of cheese. She first put in a large cube out of them and then waited for a while before the cheese cubes stuck together. Then from every wall of the big cube she will eats the middle cube. Then she also eats the cube • MO SK/CZ Z9–I–3 John had the ball that rolled into the", "cubes. How many red and blue cubes would you need?", "## HotPinkGirl one year ago There are four marbles in a bag with the colors red, white, blue, and green. john pulls out one marble and tosses a coin. which shows the sample space for pulling a marble and tossing a coin? 1. HotPinkGirl 2. misty1212 there will be $$5\\times 2=10$$ elements in the sample space 3. misty1212 oops 4. misty1212 four colors, so $$4\\times 2=8$$ 5. HotPinkGirl is it the first one ? 6. misty1212 yes 7. HotPinkGirl thanks : ) 8. misty1212 $\\color\\magenta\\heartsuit$", "Level 1 There are some children. There are 500 boys. The girls are 300 more than the boys. How many children are there altogether? 2 m Level 1 There are some children. There are 500 boys. The number of girls are 300 less than the number of boys. How many children are there altogether? 2 m Level 1 There are some children. There are 500 boys. The number of boys are 300 more than the number of girls. How many children are there altogether? 2 m Level 1 There are some children. There are 500 boys. The number of boys are 300 less than the number of girls. How many children are there altogether? 2 m Level 1 There are 2300 marbles in total. 800 are red marbles. The rest are blue marbles. How many more blue marbles than red marbles are there? 2 m Level 1 There are 2300 marbles in total. 800 are red marbles. The rest are blue marbles. How many less red marbles than blue marbles are there? 2 m Level 1 Adam has $500. Adam has$300 more than Bryan. How much do they have altogether? 2 m Level 1 Adam pays $10 in total for a sticker and a marble. The sticker costs$6 more than the marble. What is the cost of the", "can buy. Cogs will fall for 'em and you will too! Tell the Gag Shop Maggie sent you! ## Trivia • The Marbles do not change their shadow no matter what area it is in. • The Marbles are the last Trap gag to cause a cog to trip and the only one along with the Banana Peel. • The Marbles' image when choosing it in battle shows the Marbles in a sack. However, the sack is not seen in the animation when the gag is being used. • There are six marbles used in this gag: red, yellow, violet, dark blue, light blue, and green. ## In other languages Language Name French Billes Spanish ??? German ??? Brazilian Portuguese ??? Japanese ???", "2 white, and 2 black, that's identical to 2 white, 2 black, and 2 red. Remember, this part of the counting is about the colors chosen --- we would handle the distribution of the tiles on the floor in separate step. The three possible colors are equally represented, so there's only 1 way to do that. In the 3-2-1 case, yes, we will choose three colors again, but now, unlike the 2-2-2 case, order matters. Having 3 reds, 2 blacks, and 1 white is different from having, say 3 reds, 2 whites, and 1 black. Again, I would argue that use of nCr is unnecessary, and a product of nCr's is ridiculous. Use the FCP. We have tiles of three colors, and we are going to choose for the 3-2-1 patterns. In how many ways can we choose the color that will have three tiles? Three. Given that choice, in how many ways can be pick a second color for the two tiles? Two. Once those two are determined, we have nothing else to count --- the last color must have 1 tile. So, the product of the color choices is 3*2 = 6. Does all that make sense? Also, I will say about this problem --- all this stuff about counting the combinations for each individual case", "and gave him 1/4 of this remaining marbles plus 5. Just before arriving at the asked by dyl on December 14, 2010 9. ### maths ANDY HAS TWICE AS MANY MARBLES AS PANDY AND SANDY HAS HALF AS MANY HAS ANDY AND PANDY PUT TOGETHER .ANDY HAS 110 MARBLES WHICH IS 115 MARBLES LESS THAN SANDY.HOW MANY MARBLES EACH OF THEM HAVE asked by Anamika on December 31, 2012 10. ### MATH HENRY, ANDY AND JOE SHARED SOME SWEETS IN THE RATIO 15:12:9 RESPECTIVELY. HENRY HAD 6 MORE SWEETS THAN ANDY. IF ANDY GAVE 9 SWEETS TO JOE, HOW MANY SWEETS WOULD JOE HAVE? asked by FLOR on August 25, 2015 More Similar Questions", "number of notebooks Keecia bought? 19. Sheri can make 2 bracelets in 16 minutes. She needs to make 7 bracelets. a. Set up a proportion that could be used to find $m$, the number of minutes it would take Sheri to make 7 bracelets. b. How many minutes will it take Sheri to make 7 bracelets? 20. A bag contains only two colors of marbles––silver and gold. The ratio of silver marbles to gold marbles in the bag is 5 : 3. There are a total of 30 silver marbles in the bag. a. Set up a proportion that could be used to find $t$, the total number of marbles in the bag Feb 22, 2012 Jan 06, 2016", "uses 4 red beads and 3 blue beads to make a belt. How many beads will Isaac use to make 4 belts? (A) 48 (B) 12 (C) 28 (D) 16 Explanation: Isaac uses 4 red beads and 3 blue beads to make a belt. 4 x 3 = 12 12 x 4 = 48 48 beads will Isaac use to make 4 belts. Question 13. Multi-Step Corey uses 5 yellow tiles and 4 green tiles to make a design. How many tiles will he need to repeat the design 5 times? (A) 45 (B) 9 (C) 25 (D) 20" ]

[ "optimal solution is the one by Marvis. But a variant of your idea is pretty quick. There are $\\binom{k}{2}$ ways to choose the two places that John and Jack will occupy. For every one of these choices, there is only one way to place John and Jack. And now there are $(k-2)!$ ways to place the others, for a total of $$\\binom{k}{2}(k-2)!.$$", "36 E.48 12- Of the following, which number is the greatest? A. 0.092 B. 0.8913 C. 0.8923 D. 0.8896 E. 0.88 13- Solve: $$\\frac{7}{8}−\\frac{3}{4}=$$? A. 0.125 B. 0.375 C. 0.5 D. 0.625 E. 0.775 14- Which of the following is the closest to $$4.02$$? A. 4 B. 4.2 C. 4.3 D. 4.5 E. 3.5 15- Which of the following statements is False? A. $$(7×2+14)×2=56$$ B. $$(2×5+4)÷2=7$$ C. $$3+(3×6)=21$$ D. $$4×(3+9)=48$$ E.$$14÷(2+5)=5$$ 16- When 78 is divided by 5, the remainder is the same as when 45 is divided by A. 2 B. 4 C. 5 D. 7 D. 9 17- John has 2,400 cards and Max has 606 cards. How many more cards does John have than Max? A. 1,794 B. 1,798 C. 1,812 D. 1,828 E. 1,994 18- In the following right triangle, what is the value of $$x$$? A. 15 B. 30 C. 45 D. 60 E. It cannot be determined from the information given 19- What is 20 percent of 120? A. 20 B. 24 C. 30 D. 40 E.44 20- In a basket, the ratio of red marbles to blue marbles is 3 to 2. Which of the following could NOT be the total number of red and blue marbles in the basket? A. 15 B. 32 C. 55 D. 60 E. 70 ##", "want all of the green marbles.. then you would choose $i$ marbles from the batch of $n$ marbles. So if there were a total of 5 marbles ($n$) and two were green, and you wanted all the green's $~(i)$,then you could use the notation $n\\choose i$=$5\\choose 2$. Meaning, out of the $5$ marbles, I only choose $2$. Hope that this cleared up the meaning some for the binomial coefficient. – night owl Jul 3 '11 at 3:07 The way it works if you expand $(a+b)^n$ is that there are $n$ brackets and you have to choose an $a$ or a $b$ from each one. There are 2 choices for each bracket, hence 2^n choices overall. The sum you have given gathers all the terms with $i$ copies of $a$ and $n-i$ copies of $b$. The bracket $\\binom{n}{i}$ is the number of ways of doing this, and is called n-Choose-i (which is the reason for the $C$ in some alternative notations) or a Binomial Coefficient (binomial because the original brackets contain two terms $a$ and $b$). It might be interesting for you to see whether you can get any intuition or insight for why the value of the bracket is what it is. - It is read \"$n$ choose $i$\" and is equal to the number of ways to", "zebra fish. In how many different ways can George choose 2 zebra fish to buy? 12. Calculation of CN Calculate: ? 13. Trinity How many different triads can be selected from the group 43 students?", "Thanks in advance. • A more intuitive way (at least for way) of understanding the solution to the first problem is $\\binom{8}{2}\\binom{6}{2},$ i.e., how many ways can we choose two out of first eight, then six. – Bobson Dugnutt Oct 27 '16 at 7:21 1) If it is so, we subtract from 420 the colorings which have the two blue tiles adjacent. With 8 tiles there are 7 ways to choose 2 adjacent tiles. Then, among the remaining 6 tiles we choose the two tiles to be colored green in $\\frac{6!}{4!2!}=15$ ways. 2) If the color green is not used in part 2, we simply subtract from $\\frac{8!}{6!2!}=28$ the 7 ways to choose 2 adjacent tiles.", "are 18 birds on a tree. 9 of them flew away. Choose the number sentence that helps you to calculate the number of birds left on the tree. a. 9 + $n$ = 19 b. $n$ + 18 = 9 c. 18 - 9 = $n$ d. $n$ + 20 = 18 6. Choose the relation that best represents the graph. a. Cost of a bouquet is $5. b. Each bouquet requires 5 flowers. c. Each bouquet requires 2 flowers. d. Cost of 5 bouquets is$1. 7. Which situation can be described by the expression $n$ - 3? a. John ate $n$ chocolates and Gary ate 3 more than John. b. John ate $n$ chocolates and Gary ate 3 less than John. c. John ate $n$ chocolates and Gary ate 3 times more than John. d. John ate 3 chocolates and Gary ate $n$ more than John. 8. Choose a situation that represents the equation. $n$ + 7 = 37 a. If cost of each marble is $7, then the cost of $n$ marbles are$37. b. John had 37 marbles. He bought $n$ more. Then he had 7 marbles. c. Bob had $n$ marbles. He gave 7 to his sister. Then he had 37 marbles. d. Jim had $n$ marbles. He bought 7 more. Then he had 37", "what we have $5$ of, and for each there are $\\binom{25}{2}$ ways to count what we have $1$ each of (kind of tricky). I assume you know total number of choices with no restrictions. Work on it, I will be away for a while. – André Nicolas Apr 27 '12 at 18:59 Although you are undoubtedly familiar with some of what follows, we review things for completeness. In particular we will see that there is a simple way to count the number of choices if no restrictions are made. Then dealing with your restriction is relatively painless. First we count the number of ways to choose $7$ letters, repetitions allowed. For definiteness, we assume that we are dealing with a $26$-letter alphabet. Modification to deal with an alphabet with other than letters, or multisets of size other than $7$, is straightforward. Think of the $26$ letters of the alphabet as bins. We want to place $7$ identical marbles in these bins. The translation to choices of $7$ letters is immediate. For example, ABALNKM corresponds to putting $2$ marbles into bin A, and $1$ into each of B, L, N, K, M. So we are distributing $7$ marbles between $26$ \"people.\" It is easier to count the number of ways to distribute $26+7$ marbles between $26$ people, with everybody", "value of $$4∎12$$? A. 4 B. 18 C. 22 D. 36 E.48 12- Of the following, which number is the greatest? A. 0.092 B. 0.8913 C. 0.8923 D. 0.8896 E. 0.88 13- Solve: $$\\frac{7}{8}−\\frac{3}{4}=$$? A. 0.125 B. 0.375 C. 0.5 D. 0.625 E. 0.775 14- Which of the following is the closest to $$4.02$$? A. 4 B. 4.2 C. 4.3 D. 4.5 E. 3.5 15- Which of the following statements is False? A. $$(7×2+14)×2=56$$ B. $$(2×5+4)÷2=7$$ C. $$3+(3×6)=21$$ D. $$4×(3+9)=48$$ E.$$14÷(2+5)=5$$ 16- When 78 is divided by 5, the remainder is the same as when 45 is divided by A. 2 B. 4 C. 5 D. 7 D. 9 17- John has 2,400 cards and Max has 606 cards. How many more cards does John have than Max? A. 1,794 B. 1,798 C. 1,812 D. 1,828 E. 1,994 18- In the following right triangle, what is the value of $$x$$? A. 15 B. 30 C. 45 D. 60 E. It cannot be determined from the information given 19- What is 20 percent of 120? A. 20 B. 24 C. 30 D. 40 E.44 20- In a basket, the ratio of red marbles to blue marbles is 3 to 2. Which of the following could NOT be the total number of red and blue marbles in the basket? A.", ". Hence, there are: . $10\\cdot6 \\:=\\:{\\color{blue}60}$ ways to choose 2 boys and a girl. Therefore: . $P(\\text{2 boys, 1 girl}) \\:=\\:\\frac{60}{165} \\:=\\:\\boxed{\\frac{4}{11}}$", "+0 # HELP! 0 334 1 +1206 A cook has 10 red peppers and 5 green peppers. If the cook selects 6 peppers at random, what is the probability that he selects at least 4 green peppers? Express your answer as a common fraction. Apr 28, 2018 #1 +4296 +2 Solution: We can count the number of ways to choose a group of 4 green and 2 red peppers and the number of ways to choose 5 green and 1 red peppers. These are $$\\binom{5}{4}\\binom{10}{2}=5\\cdot45=225$$ and $$\\binom{5}{5}\\binom{10}{1}=10$$. The total number of ways the cook can choose peppers is $$\\binom{15}{6}=5005$$. Therefore, the probability that out of six randomly chosen peppers at least four will be green is $$\\frac{235}{5005}=\\boxed{\\frac{47}{1001}}$$. Apr 28, 2018", "of doing that. The total number of ways you can place the 5 stones WITHOUT the green stone being first or last, is (4)(3)(3!)= 4*3*6= 72. That is, of course, exactly the same as 5!- 2(4!).", "+0 # help 0 34 1 +529 John has 6 green marbles and 4 purple marbles. He chooses a marble at random, writes down its color, and then puts the marble back. He performs this process 5 times. What is the probability that he chooses exactly two green marbles? Logic Dec 7, 2018 #1 +3213 +2 $$\\text{Since you are replacing the marbles after each draw the total number of green marbles}\\\\ \\text{you select has a binomial distribution with parameters }n=5,~p=\\dfrac 3 5\\\\$$ $$P[2] = \\dbinom{5}{2} \\left(\\dfrac 3 5\\right)^2 \\left(\\dfrac 2 5\\right)^3 = \\dfrac{144}{625}$$ Rom Dec 7, 2018 #1 +3213 +2 $$\\text{Since you are replacing the marbles after each draw the total number of green marbles}\\\\ \\text{you select has a binomial distribution with parameters }n=5,~p=\\dfrac 3 5\\\\$$ $$P[2] = \\dbinom{5}{2} \\left(\\dfrac 3 5\\right)^2 \\left(\\dfrac 2 5\\right)^3 = \\dfrac{144}{625}$$ Rom Dec 7, 2018", "to order 7 objects. However, since there’s 3!= 6 ways to switch the yellow tiles around without changing anything (since they’re indistinguishable) and 2! = 2 ways for green tiles. Final Step I am sure that you are almost there for the final calculation but let me help those who are still not there $\\frac {7!}{6 \\cdot 2} = 420$ . So the correct answer is B) 420", "focus on $Y, Y, Y, G, G$ first, which gives us $\\binom{5}{3}=10.$ Now we can insert our brown tile into this which only has $6$ choices(like $Y,B, Y, Y, G, G$ and $Y, Y, Y, G, G, B$ etc.), then insert purple tile which only has $7$ choices(like $B,Y, P, Y, Y, G, G$ and $B, Y, Y, Y, G, G, P$ etc.). Multiply them together we get $10\\cdot 6\\cdot 7=\\boxed{420}.$ ~@azure123456 BZ ~ pi_is_3.14 ~IceMatrix ~savannahsolver", "# Figuring out how many ways 5 marbles can be drawn (combination/permutation problem) A bag contains 24 marbles, 4 red, 12 green, and 8 brown. How many ways can 5 marbles be drawn with all 5 marbles green. I know you can consider that there are just 12 green marbles and 12 not green marbles but I am not sure of the steps to follow You can see it like this way: You know you have $$12$$ green marbles, $$4$$ red marbles and $$8$$ brown marbles, since you need to choose $$5$$ green marbles, you selected them from the $$12$$ possibles, so you need: $$\\binom{12}{5} = \\frac{12!}{5!(12-5)!} = 792$$ Now that you have this done, you need to choose $$0$$ brown marbles from the $$8$$ possibles i.e. $$\\binom{8}{0} = 1$$ and $$0$$ red marbles from $$4$$ possibles, that is: $$\\binom{4}{0} = 1$$ OR you can see it as you need to choose $$0$$ marbles from the $$12$$ not green marbles possibilities ($$8$$ browns plus $$4$$ reds), i.e. $$\\binom{12}{0} = 1$$ Finally, all the possibilities to choose 5 green marbles are: $$\\binom{12}{5} \\binom{8}{0} \\binom{4}{0} = 792*1*1 = 792 = 792 * 1 = \\binom{12}{5} \\binom{12}{0}$$ So the answer is $$792$$ possible ways. $${12\\choose 5}{8\\choose 0}{4\\choose 0}=\\frac{12!}{7!\\cdot5!}$$.", "# Grid of 3 columns. There are 6 different marbles.How many ways can the marbles can be placed in the grid such that no column is empty? There is a grid of 3 columns: the first column has 4 squares, the second has 2 squares, and the third has 2 squares. □□□ □□□ There are 6 marbles of different colors. How many ways can those 6 marbles be placed inside the grid such that each column has at least one marble? My solution is as follows: There will be four cases as: $(4,1,1), (3,2,1), (3,1,2), (2,2,2)$ for number of marbles in respective columns. $$\\bigg(\\binom64 \\cdot 4! \\cdot \\binom21 \\cdot 2 \\cdot 2\\bigg) + 2 \\cdot \\bigg(\\binom63 \\cdot \\binom43 \\cdot 3! \\cdot \\binom32 \\cdot 2 \\cdot 2\\bigg) \\\\ + \\bigg(\\binom62 \\cdot \\binom42 \\cdot 2! \\cdot \\binom42 \\cdot 2 \\cdot 2\\bigg) = 18720$$ (Logic: $\\big(\\binom62 \\cdot \\binom42 \\cdot 2! \\cdot \\binom42 \\cdot 2 \\cdot 2\\big)$ means first choose 2 balls out of 6, then choose 2 squares out of 4 in the first column and permute; second, choose 2 balls out of the 4 remaining balls and permute in the second column; in the third column, permute the 2 remaining balls). Is this right or not? Kindly help me. There are only two configurations where placing $6$ marbles in such", "4 cards of that value: . ${4\\choose2} = {\\color{blue}6}$ ways. . . So there are: . $2\\cdot 12\\cdot 6 \\:=\\:{\\color{blue}144}$ ways. Case 2: Draw a Triple. There are 12 choices for the value of the Triple. You must get 3 of the 4 cards of that value: . ${4\\choose3} = {\\color{blue}4}$ ways. . . So there are: . $12\\cdot4 \\:=\\:{\\color{blue}48}$ ways. Hence, there are: . $144 + 48 \\:=\\:{\\color{red}192}$ ways to get a Full house. $P(\\text{Full House}\\,|\\,\\text{Pair}) \\:=\\:{\\color{red}\\frac{192}{19,\\!600}} \\;=\\;\\boxed{\\frac{12}{1225}}$", "to color $$5$$ tiles with $$3$$ colors. We must subtract the ways that only use $$2$$ colors, so we have $$3^5-3\\cdot2^5$$. Now what about a coloring with one color? It's been counted once originally, and subtracted twice, since one of the two-colorings doesn't include it, so we need to add it back in. This gives a final answer of $$3^5-2\\cdot2^5+3=150$$ Note that this is the same as $$\\binom51\\binom41\\cdot3+\\binom51\\binom42\\cdot3=150$$ • Yes, those colorings are different, but you've taken care of that in the first part of the calculation. Suppose we have $1$ red, $1$ blue, $3$ green. When you calculate $\\binom51\\binom41\\binom33$, you're saying, \"Choose a tile to color red, then a tile to color blue, then color the others green.\" We don't want to also say, \"Choose a tile to color blue, then a tile to color red, etc.\" Personally, I find inclusion-exclusion much less error-prone for problems like this. – saulspatz Mar 18 at 14:46 • Yes, that's right. I can't be very specific about when I like PIE, but I guess it's in problems like this where we have multiple stages of counting, and it's easy to make mistakes in how they interrelate. When I find myself getting confused, I try to fall back on PIE. In this case, I immediately thought you were overcounting by a", "## ksf-2015-ec-20 Peter has ten balls, numbered from 0 to 9. He distributed these balls among three friends: John got three balls, George four and Ann three. Then he asked each of his friends to multiply the numbers on the balls they got, and the results were: 0 for John, 72 for George and 90 for Ann. Question: What is the sum of the numbers on the balls that John received?", "green card. If that's the case, I believe this is the solution: If you have 21 choose 5 cards (because order doesn't matter), there are 20,349 possible hands. $$\\binom{21}{5}=20,349$$ So, look at the odds of getting the most basic winning hand: 1 red and 1 green. There are $\\binom{2}{1} = 2$ ways to get 1 green and $\\binom{4}{1} = 4$ ways to get 1 red (green 1 + red 1, green 1 + red 2, etc.). Then, there are $\\binom{15}{3} = 455$ ways to select from the remaining non-green non-red cards. Thus, there are $\\binom{2}{1} \\cdot \\binom{4}{1} \\cdot \\binom{15}{3} = 3640$ ways to get a winning hand with 1 green, 1 red, and 3 non-green non-reds. If we add that to the other possible hands, we'll know the total number of winning combinations. The possible winning hands and their counts are: • 3640 hands with 1 green, 1 red, and 3 other • 420 hands with 2 green, 1 red, and 2 other • 1260 hands with 1 green, 2 red, and 2 other • 90 hands with 2 green, 2 red, and 1 other • 120 hands with 1 green, 3 red, and 1 other • 4 hands with 2 green, 3 red, and 0 other • 2 hands with 1 green, 4 red, and 0 other" ]

[ "36 E.48 12- Of the following, which number is the greatest? A. 0.092 B. 0.8913 C. 0.8923 D. 0.8896 E. 0.88 13- Solve: $$\\frac{7}{8}−\\frac{3}{4}=$$? A. 0.125 B. 0.375 C. 0.5 D. 0.625 E. 0.775 14- Which of the following is the closest to $$4.02$$? A. 4 B. 4.2 C. 4.3 D. 4.5 E. 3.5 15- Which of the following statements is False? A. $$(7×2+14)×2=56$$ B. $$(2×5+4)÷2=7$$ C. $$3+(3×6)=21$$ D. $$4×(3+9)=48$$ E.$$14÷(2+5)=5$$ 16- When 78 is divided by 5, the remainder is the same as when 45 is divided by A. 2 B. 4 C. 5 D. 7 D. 9 17- John has 2,400 cards and Max has 606 cards. How many more cards does John have than Max? A. 1,794 B. 1,798 C. 1,812 D. 1,828 E. 1,994 18- In the following right triangle, what is the value of $$x$$? A. 15 B. 30 C. 45 D. 60 E. It cannot be determined from the information given 19- What is 20 percent of 120? A. 20 B. 24 C. 30 D. 40 E.44 20- In a basket, the ratio of red marbles to blue marbles is 3 to 2. Which of the following could NOT be the total number of red and blue marbles in the basket? A. 15 B. 32 C. 55 D. 60 E. 70 ##", "in this collection? How many silver coins are in this collection? We are told that there is one coin with each combination of characteristics. Since we are told that the coin is silver, there is only one way to choose its color. There are three choices for its shape and three choices for the imprinted letter. Hence, there are $$1 \\cdot 3 \\cdot 3 = 9$$ silver coins. How many coins have the letter A on them? Since we are told the coin has the letter A on it, there is only one way to choose its letter. There are five ways to choose its color and three ways to choose its shape. Hence, there are $$5 \\cdot 3 \\cdot 1 = 15$$ coins with the letter $$A$$ imprinted on them. How many ways can Jeff choose four balls? What is the probability that Caitlin correctly guesses the four numbers? Method 1: Caitlin must guess all four numbers correctly, which can be done in $$\\binom{4}{4} = 1$$ way, while selecting four of the eighteen numbers. Hence, the probability that Caitlin guesses all four numbers correctly is $$\\frac{\\dbinom{4}{4}}{\\dbinom{18}{4}}$$ Method 2: Let's correct your attempt. Caitlin chooses four different numbers. Therefore, she has $$18$$ ways to choose the first number, $$17$$ ways to choose the second number, $$16$$ ways to", "# Grid of 3 columns. There are 6 different marbles.How many ways can the marbles can be placed in the grid such that no column is empty? There is a grid of 3 columns: the first column has 4 squares, the second has 2 squares, and the third has 2 squares. □□□ □□□ There are 6 marbles of different colors. How many ways can those 6 marbles be placed inside the grid such that each column has at least one marble? My solution is as follows: There will be four cases as: $(4,1,1), (3,2,1), (3,1,2), (2,2,2)$ for number of marbles in respective columns. $$\\bigg(\\binom64 \\cdot 4! \\cdot \\binom21 \\cdot 2 \\cdot 2\\bigg) + 2 \\cdot \\bigg(\\binom63 \\cdot \\binom43 \\cdot 3! \\cdot \\binom32 \\cdot 2 \\cdot 2\\bigg) \\\\ + \\bigg(\\binom62 \\cdot \\binom42 \\cdot 2! \\cdot \\binom42 \\cdot 2 \\cdot 2\\bigg) = 18720$$ (Logic: $\\big(\\binom62 \\cdot \\binom42 \\cdot 2! \\cdot \\binom42 \\cdot 2 \\cdot 2\\big)$ means first choose 2 balls out of 6, then choose 2 squares out of 4 in the first column and permute; second, choose 2 balls out of the 4 remaining balls and permute in the second column; in the third column, permute the 2 remaining balls). Is this right or not? Kindly help me. There are only two configurations where placing $6$ marbles in such", "zebra fish. In how many different ways can George choose 2 zebra fish to buy? 12. Calculation of CN Calculate: ? 13. Trinity How many different triads can be selected from the group 43 students?", "If you think you have a good answer to this, do let me know, but note that the really hard problem, still I think unsolved, is to prove the best optimum result. # Sunday Times Teaser 3007 – Paving Stones ### by Howard Williams #### Published Sunday May 10 2020 (link) James has decided to lay square block paving stones on his rectangular patio. He has calculated that starting from the outside and working towards the middle that he can lay a recurring concentric pattern of four bands of red stones, then three bands of grey stones, followed by a single yellow stone band. By repeating this pattern and working towards the centre he is able to finish in the middle with a single line of yellow stones to complete the patio. He requires 402 stones to complete the first outermost band. He also calculates that he will require exactly 5 times the number of red stones as he does yellow stones. How many red stones does he require?` # Sunday Times Teaser 3006 – Raffle Tickets ### by Andrew Skidmore #### Published Sunday May 03 2020 (link) At our local bridge club dinner we were each given a raffle ticket. The tickets were numbered from 1 to 80. There were six people on our table and all", "Level 1 There are some children. There are 500 boys. The girls are 300 more than the boys. How many children are there altogether? 2 m Level 1 There are some children. There are 500 boys. The number of girls are 300 less than the number of boys. How many children are there altogether? 2 m Level 1 There are some children. There are 500 boys. The number of boys are 300 more than the number of girls. How many children are there altogether? 2 m Level 1 There are some children. There are 500 boys. The number of boys are 300 less than the number of girls. How many children are there altogether? 2 m Level 1 There are 2300 marbles in total. 800 are red marbles. The rest are blue marbles. How many more blue marbles than red marbles are there? 2 m Level 1 There are 2300 marbles in total. 800 are red marbles. The rest are blue marbles. How many less red marbles than blue marbles are there? 2 m Level 1 Adam has $500. Adam has$300 more than Bryan. How much do they have altogether? 2 m Level 1 Adam pays $10 in total for a sticker and a marble. The sticker costs$6 more than the marble. What is the cost of the", "Step 3: Multiply the value of one part by the number of parts Aaron has: $\\textcolor{orange}{£500} \\times \\textcolor{red}{3} = £1500$ So Aaron gets $£1500$. KS3 Level 4-5 ## Skill 6: Difference between Parts of a ratio You may sometimes be given the difference between two parts of the ratio, instead of the total amount. Example: Josh, James and John share sweets in the ratio $\\textcolor{orange}{1}:\\textcolor{blue}{2}:\\textcolor{red}{4}$. Josh has $\\textcolor{limegreen}{9}$ sweets less than John. How many sweets does James have? Step 1: Firstly, work out how many parts of the ratio $\\textcolor{limegreen}{9}$ sweets makes up: $\\textcolor{limegreen}{9} \\, \\text{sweets} = \\text{John's sweets} - \\text{Josh's sweets} = \\textcolor{red}{4} \\, \\text{parts} - \\textcolor{orange}{1} \\, \\text{part} = 3 \\, \\text{parts}$. Step 2: Then divide to find $1$ part: $\\textcolor{limegreen}{9} \\, \\text{sweets} \\div 3 = \\textcolor{purple}{3} \\, \\text{sweets}$ $\\textcolor{purple}{3} \\text{ sweets } = 1 \\text{ part }$ Step 3: Multiply by the number of parts James has to find how many sweets he has: $\\text{James' sweets} = \\textcolor{blue}{2} \\, \\text{parts} = \\textcolor{blue}{2} \\times \\textcolor{purple}{3} \\, \\text{sweets} = \\bf{6 \\, \\text{sweets}}$ KS3 Level 4-5 Level 6-7 ## Skill 7: Changing Ratios You need to be prepared for questions where the ratio changes Example: Billy and Claire share marbles in the ratio $\\textcolor{blue}{5}:\\textcolor{limegreen}{3}$. Billy gives $\\textcolor{orange}{4}$ marbles to Claire and the ratio is now $1:1$. How many", "1 : 6. There is an equal amount of lemon syrup in both cups. Cup B contains 189 mℓ more water than Cup A. What is the amount of lemon syrup in Cup A? Leave the answer in mℓ. 3 m Level 3 Mrs Lu had a total of 520 purple and yellow lilies in the ratio of 8 : 5. After she had given away 6 times as many purple lilies as yellow lilies, the number of purple and yellow lilies left was in the ratio of 10 : 9. How many more purple lilies than yellow lilies did she give away? 5 m Level 3 David, Ethan and Tim have some marbles. The ratios of their marbles are as follows: David : (David + Ethan + Tim) = 1 : 5 Ethan : (David + Tim) = 1 : 3 Tim has 42 marbles more than Ethan. In the end, Tim gives away all he has to Ethan and David so that both Ethan and David end up with the same number of marbles, how many marbles has Ethan received from Tim? 4 m Level 3 Oscar bought some mangosteens and coconuts in the ratio of 2 : 5 respectively. The cost of a mangosteen was $1.50 more than the cost of a coconut. He spent", "+0 # help 0 34 1 +529 John has 6 green marbles and 4 purple marbles. He chooses a marble at random, writes down its color, and then puts the marble back. He performs this process 5 times. What is the probability that he chooses exactly two green marbles? Logic Dec 7, 2018 #1 +3213 +2 $$\\text{Since you are replacing the marbles after each draw the total number of green marbles}\\\\ \\text{you select has a binomial distribution with parameters }n=5,~p=\\dfrac 3 5\\\\$$ $$P[2] = \\dbinom{5}{2} \\left(\\dfrac 3 5\\right)^2 \\left(\\dfrac 2 5\\right)^3 = \\dfrac{144}{625}$$ Rom Dec 7, 2018 #1 +3213 +2 $$\\text{Since you are replacing the marbles after each draw the total number of green marbles}\\\\ \\text{you select has a binomial distribution with parameters }n=5,~p=\\dfrac 3 5\\\\$$ $$P[2] = \\dbinom{5}{2} \\left(\\dfrac 3 5\\right)^2 \\left(\\dfrac 2 5\\right)^3 = \\dfrac{144}{625}$$ Rom Dec 7, 2018", "value of $$4∎12$$? A. 4 B. 18 C. 22 D. 36 E.48 12- Of the following, which number is the greatest? A. 0.092 B. 0.8913 C. 0.8923 D. 0.8896 E. 0.88 13- Solve: $$\\frac{7}{8}−\\frac{3}{4}=$$? A. 0.125 B. 0.375 C. 0.5 D. 0.625 E. 0.775 14- Which of the following is the closest to $$4.02$$? A. 4 B. 4.2 C. 4.3 D. 4.5 E. 3.5 15- Which of the following statements is False? A. $$(7×2+14)×2=56$$ B. $$(2×5+4)÷2=7$$ C. $$3+(3×6)=21$$ D. $$4×(3+9)=48$$ E.$$14÷(2+5)=5$$ 16- When 78 is divided by 5, the remainder is the same as when 45 is divided by A. 2 B. 4 C. 5 D. 7 D. 9 17- John has 2,400 cards and Max has 606 cards. How many more cards does John have than Max? A. 1,794 B. 1,798 C. 1,812 D. 1,828 E. 1,994 18- In the following right triangle, what is the value of $$x$$? A. 15 B. 30 C. 45 D. 60 E. It cannot be determined from the information given 19- What is 20 percent of 120? A. 20 B. 24 C. 30 D. 40 E.44 20- In a basket, the ratio of red marbles to blue marbles is 3 to 2. Which of the following could NOT be the total number of red and blue marbles in the basket? A.", ". ${\\color{blue}4\\cdot{48\\choose9}\\cdot{39\\choose13,1 3,13}}$ ways for one player to get the four Aces. Three Aces in one player's hand, one Ace in another's There are 4 choices of players to get the three Aces. There are: ${\\color{red}{4\\choose3}}$ ways for him to get three Aces. For his other 10 cards, there are: ${\\color{red}{48\\choose10}}$ ways. There are 3 choices of players to get the one Ace. There is one way for him to get the one remaining Ace. For his other 12 cards, there are: ${\\color{red}{38\\choose12}}$ ways. The remaining 26 cards are distributed to the other two players. There are: ${\\color{red}{26\\choose13,13}}$ ways. Hence, there are: . ${\\color{blue}4\\cdot{4\\choose3}\\cdot{48\\choose10}\\c dot3\\cdot{38\\choose12}\\cdot{26\\choose13,13}}$ ways Get the idea? Edit: corrected an error . . . sorry for the blunder. . 3. Originally Posted by Soroban Hence, there are: . ${\\color{blue}4\\cdot{4\\choose3}\\cdot{48\\choose10}\\c dot3\\cdot{48\\choose12}\\cdot{26\\choose13,13}}$ ways Hi, thanks! but for the guy with only ace. should it be: . ${\\color{blue}4\\cdot{4\\choose3}\\cdot{48\\choose10}\\c dot3\\cdot\\color{red}{39\\choose12}\\cdot\\color{blue} {26\\choose13,13}}$ 4. Hello again, chopet! Thank you for pointing out my error . . . After the first player gets his 13 cards (including 3 Aces), . . there are 3 choices for the next player. There is one way for him to get the remaining Ace. His other 12 cards come from the remaining 38 cards. So there are: ${\\color{red}{38\\choose12}}$ ways. There are: . ${\\color{blue}3\\cdot{38\\choose12}}$ ways for the second", ". Hence, there are: . $10\\cdot6 \\:=\\:{\\color{blue}60}$ ways to choose 2 boys and a girl. Therefore: . $P(\\text{2 boys, 1 girl}) \\:=\\:\\frac{60}{165} \\:=\\:\\boxed{\\frac{4}{11}}$", "to color $$5$$ tiles with $$3$$ colors. We must subtract the ways that only use $$2$$ colors, so we have $$3^5-3\\cdot2^5$$. Now what about a coloring with one color? It's been counted once originally, and subtracted twice, since one of the two-colorings doesn't include it, so we need to add it back in. This gives a final answer of $$3^5-2\\cdot2^5+3=150$$ Note that this is the same as $$\\binom51\\binom41\\cdot3+\\binom51\\binom42\\cdot3=150$$ • Yes, those colorings are different, but you've taken care of that in the first part of the calculation. Suppose we have $1$ red, $1$ blue, $3$ green. When you calculate $\\binom51\\binom41\\binom33$, you're saying, \"Choose a tile to color red, then a tile to color blue, then color the others green.\" We don't want to also say, \"Choose a tile to color blue, then a tile to color red, etc.\" Personally, I find inclusion-exclusion much less error-prone for problems like this. – saulspatz Mar 18 at 14:46 • Yes, that's right. I can't be very specific about when I like PIE, but I guess it's in problems like this where we have multiple stages of counting, and it's easy to make mistakes in how they interrelate. When I find myself getting confused, I try to fall back on PIE. In this case, I immediately thought you were overcounting by a", "Thanks in advance. • A more intuitive way (at least for way) of understanding the solution to the first problem is $\\binom{8}{2}\\binom{6}{2},$ i.e., how many ways can we choose two out of first eight, then six. – Bobson Dugnutt Oct 27 '16 at 7:21 1) If it is so, we subtract from 420 the colorings which have the two blue tiles adjacent. With 8 tiles there are 7 ways to choose 2 adjacent tiles. Then, among the remaining 6 tiles we choose the two tiles to be colored green in $\\frac{6!}{4!2!}=15$ ways. 2) If the color green is not used in part 2, we simply subtract from $\\frac{8!}{6!2!}=28$ the 7 ways to choose 2 adjacent tiles.", "are 18 birds on a tree. 9 of them flew away. Choose the number sentence that helps you to calculate the number of birds left on the tree. a. 9 + $n$ = 19 b. $n$ + 18 = 9 c. 18 - 9 = $n$ d. $n$ + 20 = 18 6. Choose the relation that best represents the graph. a. Cost of a bouquet is $5. b. Each bouquet requires 5 flowers. c. Each bouquet requires 2 flowers. d. Cost of 5 bouquets is$1. 7. Which situation can be described by the expression $n$ - 3? a. John ate $n$ chocolates and Gary ate 3 more than John. b. John ate $n$ chocolates and Gary ate 3 less than John. c. John ate $n$ chocolates and Gary ate 3 times more than John. d. John ate 3 chocolates and Gary ate $n$ more than John. 8. Choose a situation that represents the equation. $n$ + 7 = 37 a. If cost of each marble is $7, then the cost of $n$ marbles are$37. b. John had 37 marbles. He bought $n$ more. Then he had 7 marbles. c. Bob had $n$ marbles. He gave 7 to his sister. Then he had 37 marbles. d. Jim had $n$ marbles. He bought 7 more. Then he had 37", "the same time, player two gets a point. If one red and one green marble are pulled at the same time, both players get a point. Is this a fair game?", "## ksf-2015-ec-20 Peter has ten balls, numbered from 0 to 9. He distributed these balls among three friends: John got three balls, George four and Ann three. Then he asked each of his friends to multiply the numbers on the balls they got, and the results were: 0 for John, 72 for George and 90 for Ann. Question: What is the sum of the numbers on the balls that John received?", "optimal solution is the one by Marvis. But a variant of your idea is pretty quick. There are $\\binom{k}{2}$ ways to choose the two places that John and Jack will occupy. For every one of these choices, there is only one way to place John and Jack. And now there are $(k-2)!$ ways to place the others, for a total of $$\\binom{k}{2}(k-2)!.$$", "up of 2 colours, we have 200 colours in total. Using either Statement alone, we haven''t accounted for very many of the colours, so we could have quite a few half-blue marbles or we could have none, and our probability of picking half-blue marbles ...” May 13, 2019 Ian Stewart posted a reply to Cotto Toy Store sells Product X and Product Y at two in the Data Sufficiency forum “Neither Statement is sufficient alone, because if the initial price of X is enormously larger than the initial price of Y, using either Statement alone, the discounted price of X will remain larger, and similarly if the initial price of Y is vastly larger than that of X, so will be Y''s discounted ...” May 13, 2019 Ian Stewart posted a reply to Maria can either buy a basket that contains P pounds of in the Data Sufficiency forum “The basket costs$16.50, and the p pounds of apples cost $0.95p. We want to know which of those figures is larger. Statement 1 is irrelevant, since we don''t care how many apples are in the basket -- we already know the basket costs$16.50. What we need is information about the p pounds of ...” May 13, 2019 Brent@GMATPrepNow posted a reply to A line that passes through (–1, –4)", "+0 # HELP! 0 334 1 +1206 A cook has 10 red peppers and 5 green peppers. If the cook selects 6 peppers at random, what is the probability that he selects at least 4 green peppers? Express your answer as a common fraction. Apr 28, 2018 #1 +4296 +2 Solution: We can count the number of ways to choose a group of 4 green and 2 red peppers and the number of ways to choose 5 green and 1 red peppers. These are $$\\binom{5}{4}\\binom{10}{2}=5\\cdot45=225$$ and $$\\binom{5}{5}\\binom{10}{1}=10$$. The total number of ways the cook can choose peppers is $$\\binom{15}{6}=5005$$. Therefore, the probability that out of six randomly chosen peppers at least four will be green is $$\\frac{235}{5005}=\\boxed{\\frac{47}{1001}}$$. Apr 28, 2018" ]

[ "be the set of sides and diagonals of a regular pentagon. A pair [#permalink] ### Show Tags 20 Mar 2019, 07:23 Bunuel wrote: Let S be the set of sides and diagonals of a regular pentagon. A pair of elements of S are selected at random without replacement. What is the probability that the two chosen segments have the same length? (A) 2/5 (B) 4/9 (C) 1/2 (D) 5/9 (E) 4/5 Pentagon has 5 equal sides and 5 diagonals ; total equal sides 10 so case 1: picking 2 sides of equal length; 5/10 * 4/9 = 2/9 case 2 : picking 2 sides of equal diagonal ;5/10 * 4/9 = 2/9 IMO B Re: Let S be the set of sides and diagonals of a regular pentagon. A pair [#permalink] 20 Mar 2019, 07:23 Display posts from previous: Sort by", "# Don't Cross A Corner! On an $n \\times n$ square grid, two vertices are chosen at random and a segment is drawn connecting them. As $n\\rightarrow \\infty,$ what is the probability that the segment does not cross any vertices of this graph besides the two chosen ones? ×", "Are the lengths from this recursive construction a geometric sequence? In his 1999 review of Edward Tufte's Visual Explanations in the Notices of the AMS (third page), Bill Casselman gives a very pretty proof of the irrationality of the golden mean. More precisely, Casselman shows that the side and diagonal of a regular pentagon cannot be commensurable. The proof assumes the contrary, and thereby constructs a vanishing sequence of commensurable terms, which is a contradiction. Upon closer inspection, however, I realized that there's a step in the proof that I cannot justify to my satisfaction. Casselman does show that the commensurability of the side and the diagonal of a regular pentagon implies the existence of a strictly decreasing sequence of commensurable lengths (e.g. the sequence consisting of the lengths of the sides of successively constructed nested regular pentagons). It is not clear to me, however, that he showed that this sequence converges to 0 (and not to some positive number). Of course, convergence to 0 would follow immediately if we could establish that the sequence of side lengths is a geometric one, i.e. that successive terms in it are in a constant ratio to each other (which has to be $< 1$, since the sequence is strictly decreasing). That this ratio is constant seems to me intuitively obvious", "visual2 Visual Explanations in Mathematics ## 2. Casselman's visual proof The incommensurability of side and diagonal in a regular pentagon In this regular pentagon, call the side length s and the length of the diagonal d. Our job is to show that s and d are incommensurable. First notice that in a regular pentagon, a diagonal is parallel to the opposite side. This follows from the symmetry of the pentagon. Using this fact for two adjacent sides shows that this green quadrilateral is a parallelogram, with opposite sides equal: all four sides of this parallelogram have length s. Now suppose that the side and the diagonal are commensurable: there exists a length h which fits exactly a whole number of times into the diagonal and into the side. The length h is symbolized by the distance between two adjacent dots in this picture. The length of the highlighted segment must also be a whole number of h's, since it is the difference of two such numbers, d and s. Call this length d*. d* = d - s. The length of this highlighted segment is d - 2d*. So this length must also be equal to a whole number of h's. Call this length s*. s* = d - 2d* = 2s - d. Now notice that the", "pentagon $$BFDEG$$ truely is regular, having side length $$b$$. You then even could deduce that $$D$$, $$F$$, and $$(H=I)$$ are collinear and that $$F(H=I)=a$$ as well.", "# Pentagon in a Square???Sounds Strange!!! Aaron takes a square sheet of paper with one corner labelled as $$A$$. Point $$P$$ is chosen at random inside the square and Aaron folds the paper so that $$A$$ and $$P$$ coincide. He cuts the sheet along the crease and discards the piece containing $$A$$. Let $$p$$ be the probability that the remaining piece is a pentagon. Find the integer nearest to $$100p$$. ×", "Drinfel'd's pentagon equation implies the generalized double shuffle relation. As a corollary, an embedding from the Grothendieck-Teichmuller group $GRT_1$ into Racinet's double", "Question 10.4.2 A regular pentagon $ABCDE$ has diagonals $AC$ and $BD$, which intersect at $F$. Show that $\\Delta ABF$ is isosceles. Show hint", "five diagonals determine a smaller regular pentagon inside our original one. The side length of this pentagon is the s* we just considered. In the smaller pentagon, the diagonals are again parallel to the opposite sides. It follows that the highlighted figure is a parallelogram, and that the diagonal in the smaller pentagon has length d*. So in the smaller pentagon, side and diagonal are both multiples of our original h. We can repeat the construction with the smaller pentagon. Since s* is strictly shorter than s, and they are both whole multiples of h, they must differ by at least h. We can repeat the process again and again, and obtain an infinite decreasing sequence of side lengths s > s* > s** ... each of which is at least h smaller than the one before. But this can't work, because there was only a finite number of h's in s to start! This contradiction shows that the hypothesis, that s and d are commensurable, must be false. Welcome to the Feature Column! These web essays are designed for those who have already discovered the joys of mathematics as well as for those who may be uncomfortable with mathematics.", "all possible triangles formed by joining vertices of an n-sided regular polygon. If $$T_{n+1}$$ – $$T_n$$ = 10, then the value of n is From a group of 10 persons consisting of 5 lawyers, 3 doctors and 2 engineers, four persons are selected at random. The probability that selection contains one of each category is", "The length of an edge is a. b) Find the shortest collection of roads in the shape indicated for towns The T-connector worked perfectly! Students have to say a sentence (he is a firefighter) in order to place their teams chips in the box they want. The row can be horizontal, vertical or diagonal. Why red and blue boxes in close proximity seems to shift position vertically under a dark background. A simple way to set this up is to make a \"7\" with your checkers, such that finishing the horizontal or diagonal connect-four on the right will be two spaces on top of each other. 4.Take corner squares over edges. pieces of Perspex with four spacers attached at the corners of a square. The dimensions of the square are found by calculating the distance between various corner points.Recall that we can find the distance between any two points if we know their coordinates. Regular Pentagon sides in terms of interior pentagon and segments connecting vertices. The volume is V=a³, the surface ist O = 6a². Plugging this into the original equation, I have $$L = \\frac{3}{\\sqrt{3}} + a$$ Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Find", "set of squares. Implying that $Y>X$ with probability $1$. See McKay's original paper:", "at each iteration, the first candidate $$(p_j, p_k)$$ satisfying the search conditions will have minimal distance $$|p_j - p_k|$$ for the current value of $$p_k$$. ## Solution Implementation¶ from lib.sequence import Pentagonals def solve(): \"\"\" Compute the answer to Project Euler's problem #44 \"\"\" pentagonals = Pentagonals() visited_pentagonals = [] # will reverse-iterate over this list {p_j} pentagonals_set = set() # for testing whether sums or differences of pentagonal numbers are pentagonal numbers for p_k in pentagonals: for p_j in reversed(visited_pentagonals): if (p_k - p_j) in pentagonals_set and (p_k + p_j) in pentagonals: return p_k - p_j # p_k > p_j => |p_k - p_j| = p_k = p_j visited_pentagonals.append(p_k) solutions.problem44.solve()", "Greeks would have presented the proof, but it is very similar in spirit, and we make no apologies for the anachronism. Let’s get started. Suppose, for the sake of an eventual contradiction, that we have a regular pentagon whose side length is commensurable with the side length of the inscribed pentagram. Let $s$ denote the side length of the pentagon and $t$ denote the side length of the pentagram (i.e., the diagonal of the pentagon). In modern language, our assumption is that $\\frac{t}{s}$ is rational, i.e., that there are positive integers $p$ and $q$ such that $\\frac{t}{s} = \\frac{p}{q}$. By scaling the pentagon, we can in fact assume that $t = p$ and $s = q$. So let this be our starting assumption: there is a regular pentagon whose side length $s$ and diagonal length $t$ are both positive integers. Let us label the vertices of the pentagon by the letters A,B,C,D,E. The center of the pentagram forms another regular pentagon, whose vertices we shall call a,b,c,d,e. This is shown in the diagram below. Note that $s$ is the length of the line segment connecting A and B (we will denote this length by |AB|), and $t$ is |AC|. Let $s'$ denote the side length of the inner pentagon, i.e., |ab|, and let $t'$ denote the length of", "A square has sides of length 2. Set $$S$$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $$S$$ enclose a region whose area to the nearest hundredth is $$k.$$ Find $$100k.$$ (第二十二届AIME1 2004 第4题)", "# The Enveloping Lines Part (a) 2 random points are chosen in a unit square. These points are then linked to form a line such as the one displayed in the picture above. What is the expected value for the length of the line segment?", "Institutions: Global | TUT | UCT | UJ | UZ | Wits 208 views Let $AB$ and $BC$ be two consecutive sides of a regular pentagon inscribed in a unit circle (that is, a circle of radius 1 unit). Find the value of $(AB · AC)^2$ | 208 views", "there are 11 vertices. S is the set of all lines that contain two distinct vertices of P. so, total lines in S, or you can say total elements in S are : $$\\binom{11}2 \\qquad\\text{no of ways of choosing 2 vertices out of 11} = 55$$ total pairs of lines = $\\binom{55}2 = 1485$ Now, number of pairs of parallel lines for every edge possible = $$\\binom{\\lfloor n/2\\rfloor - 1}2 = \\binom42= 6.$$ so for $11$ edges, = $66$ note that for even number of sides in polygon, the number of pairs of parallel lines will be half of those calculated by above formula, because every pair will be repeated for parallel edges. probability that the chosen pair will be parallel, is $66/1485 = 0.0444444$ Use Poisson's distribution with $\\lambda$ as $.044444$, probability = $0.04251$. • \"no edge has more than 1 parallel line associated with it\" just isn't so. Consider the edge between verts 5 and 6; then lines 4-7, 3-8, etc. are all parallel to that edge. – Steven Stadnicki Aug 31 '13 at 17:11 • @steven, thanks for that... – Sumedh Aug 31 '13 at 19:05 But you are choosing three lines. The probability that you have calculated is for choosing a pair of parallel lines. The question asked is you randomly choose 3 lines", "The Mean Width of Circumscribed Random Polytopes For a given convex body $K$ in ${\\mathbb R}^d$, a random polytope $K^{(n)}$ is defined (essentially) as the intersection of $n$ independent closed halfspaces containing $K$ and having an isotropic and (in a specified sense) uniform distribution. We prove upper and lower bounds of optimal orders for the difference of the mean widths of $K^{(n)}$ and $K$ as $n$ tends to infinity. For a simplicial polytope $P$, a precise asymptotic formula for the difference of the mean widths of $P^{(n)}$ and $P$ is obtained. Keywords:random polytope, mean width, approximationCategories:52A22, 60D05, 52A27", "of the pentagon is the same length. is the shape a regular pentagon? Each side of the pentagon is the same length. is the shape a regular pentagon?..." ]

[ "# Difference between revisions of \"2013 AMC 10B Problems/Problem 12\" ## Problem Let $S$ be the set of sides and diagonals of a regular pentagon. A pair of elements of $S$ are selected at random without replacement. What is the probability that the two chosen segments have the same length? $\\textbf{(A) }\\frac{2}5\\qquad\\textbf{(B) }\\frac{4}9\\qquad\\textbf{(C) }\\frac{1}2\\qquad\\textbf{(D) }\\frac{5}9\\qquad\\textbf{(E) }\\frac{4}5$ ## Solution ### Solution 1 In a regular pentagon, there are 5 sides with the same length, and 5 diagonals with the same length. Picking an element at random will leave 4 elements with the same length as the element picked, with 9 total elements remaining. Therefore, the probability is $\\boxed{\\textbf{(B) }\\frac{4}{9}}$. ### Solution 2 Alternatively, we can divide this problem into two cases. Case 1: Side; In this case, there is a $\\frac{5}{10}$ chance of picking a side, and a $\\frac{4}{9}$ chance of picking another side. Case 2: Diagonal; This case is similar to the first, for again, there is a $\\frac{5}{10}$ chance of picking a diagonal, and a $\\frac{4}{9}$ chance of picking another diagonal. Summing these cases up gives us a probability of $\\boxed{\\textbf{(B) }\\frac{4}{9}}$.", "be the set of sides and diagonals of a regular pentagon. A pair [#permalink] ### Show Tags 20 Mar 2019, 07:23 Bunuel wrote: Let S be the set of sides and diagonals of a regular pentagon. A pair of elements of S are selected at random without replacement. What is the probability that the two chosen segments have the same length? (A) 2/5 (B) 4/9 (C) 1/2 (D) 5/9 (E) 4/5 Pentagon has 5 equal sides and 5 diagonals ; total equal sides 10 so case 1: picking 2 sides of equal length; 5/10 * 4/9 = 2/9 case 2 : picking 2 sides of equal diagonal ;5/10 * 4/9 = 2/9 IMO B Re: Let S be the set of sides and diagonals of a regular pentagon. A pair [#permalink] 20 Mar 2019, 07:23 Display posts from previous: Sort by", "Thread: probability --just can't figure it out 1. probability --just can't figure it out I haven't done this type of math in over 13 years and need some clarification since Im taking an online statistics course; hard to reach teacher for help! A) From the information provided, create the sample space of possible outcomes: A coin and a pentagonal die are tossed. B)Suppose you are playing a game of chance. If you bet $10 on a certain event, you will collect$500 (including your \\$10 bet) if you win. Find the odds used for determining the payoff. Now for part A)---the pentagonal die is supposed to have 5 sides, correct? And B) I have no idea. lol. 2. PART A) Sample Space is pretty simple. Pentagonal die has 5 sides. Coin has 2 sides. # of sides = # of possibilities for each item The sample space is just a box showing the possibilities. eg. Sample space for 2 coins. (H, H) (H, T) (T, H) (T, T) Notice how each coin has 2 sides, therefore 2 possibilities. So the box is going to be a 2x2 box. With 2x2 = 4 entries. So, the sample space for your question should have 2 x 5 = 10 entries. Just some busy work.", "arrangements of side lengths are the same: 2, 3, 4, 5, 6 6, 2, 3, 4, 5 In a typical permutations question, they would be counted as di¤erent. Here, the side lengths are the same relative to each other. In fact, for every possible pentagon, there are 5 \"typical permutation\" ways of counting it; in addition to the above, there are: 5, 6, 2, 3, 4 4, 5, 6, 2, 3 3, 4, 5, 6, 2 Thus, 5! overstates the number of possible pentagons by a factor of 5, so the total number of pentagons is: 5! 5 = 4! = 4 \u0002 3 \u0002 2 \u0002 1 = 24, choice (C). Intern Joined: 08 Aug 2016 Posts: 36 Re: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. [#permalink] ### Show Tags 13 Nov 2016, 04:52 vrgmat wrote: Bunuel, Can you please verify if the OA given is correct. In Jeff Sackman Extreme challenge PDF the answer given is 24. The OE is as follows: Start by thinking of this as a typical permutations problem. How many ways could the numbers 2, 3, 4, 5, and 6 be arranged in di¤erent orders? The answer to that question would be 5!: the number permutations of 5 di¤erent objects. However, since \"two", "is that picking the different types of cube has different probabilities. It is easier to consider the 64 cubes distinguishable (number them) and then figure out how many of the choices meet your requirement. In this case, it also deals with the fact that picking a 2 face cube the first time changes the probability that the second pick will be a 2 face cube. In fact, my space would be $64 \\cdot 63$, not $64 \\choose 2$. How many of those are both 2 face cubes?", "$C_5$ (which are in fact the diagonals of the pentagon). That is your complement. And what you have there is another pentagon, drawn as a star pentagon. – N. S. Dec 3 '13 at 20:51 @Wanderer If $C_5$ is the graph $1-2-3-4-5-1$ then its complement is $1-3-5-2-4-1$. So $f(1)=1, f(2)=3, f(3)=5, f(4)=2, f(5)=5$ is your isomorphism. [Completelly irrelevant, but this function is exactly $f(x)=2x-1 \\pmod{5}$.] – N. S. Dec 3 '13 at 20:52 helpful and perfect answer. thank you! – Wanderer Dec 3 '13 at 20:59 oh, one more thing. where do we get the initial formula from? (the one with the degrees) – Wanderer Dec 3 '13 at 21:02", "Let us first find the relationship between Pattern Number and the Number of pentagons. Number of pentagons in Pattern $$1 = 1$$ Number of pentagons in Pattern $$2 = 2$$ Number of pentagons in Pattern $$3 = 3$$ Therefore, as a general rule, we find that Pattern Number $$=$$ Number of pentagons Hence, a pattern having $$43$$ pentagons is pattern $$43$$ Now, in (a) we determined that, Number of beads $$= 5 + \\text{( Pattern Number} - 1\\,) \\times 4$$ Number of beads in Pattern Number $$43$$ \\begin{align*}​ &= 5 + (43 - 1) \\times 4 \\\\ &= 5 + 42 \\times 4 \\\\ &= 5 + 168 \\\\ &= 173 \\end{align*} Number of beads Andy had at first $$= 151$$ $$173 - 151 = 22$$ Andy needs $$22$$ more beads to make the pattern of $$43$$ pentagons. $$22$$ ## Conclusion • Grouping In number patterns involving grouping, we group the patterns into sets. To find the letter or object in a specific position, we will need to do division. The quotient will tell us the number of times the complete set appears. The remainder will tell us the position of the letter or object in the set. • Common Difference In number patterns involving common differences, we need to first identify what is the difference between each", "think I am not interpreting the below mentioned line correctly- \"two pentagons are considered different only when the positions of the side lengths are different relative to each other,\" that eliminates some of those 5! permutations. For instance, the following two arrangements of side lengths are the same: 2, 3, 4, 5, 6 6, 2, 3, 4, 5 Senior CR Moderator Status: Long way to go! Joined: 10 Oct 2016 Posts: 1354 Location: Viet Nam The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. [#permalink] ### Show Tags 29 Nov 2016, 23:45 2 honchos wrote: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lengths are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lengths ? (A) 5 (B) 12 (C) 24 (D) 32 (E) 120 Source: Jeff Sackman Tricky one. 5 sides of that pentagon are $$a,b,c,d,e$$ The totale numbers of possibles pentagons are $$5 \\times 4 \\times 3 \\times 2 \\times 1 =5!$$ Note that each possible selection of $$(a,b,c,d,e)$$, there are 4 other possible selections that are formed by arranged around each element. For example", "# Easy Permutation-Combination Solved QuestionAptitude Discussion Q. How many diagonals can be drawn in a pentagon? ✔ A. 5 ✖ B. 7 ✖ C. 8 ✖ D. 10 Solution: Option(A) is correct A pentagon has $5$ sides. We obtain the diagonals by joining the vertices in pairs. Total number of sides and diagonals: $= {^5C_2}$ $= \\dfrac{5×4}{2×1}$ $= 5×2$ $= 10$ This includes its $5$ sides also. ⇒ Diagonals $= 10 - 5 = 5$ Hence the number of diagonals, $= 10 - 5$ $= 5$", "these 5 selections are considered the same pentagons $$\\begin{split} 2,3,4,5,6\\\\ 6,2,3,4,5\\\\ 5,6,2,3,4\\\\ 4,5,6,2,3\\\\ 3,4,5,6,2\\\\ \\end{split}$$ Hence, in order to count the possible different selections, we need to divide the result by 5: $$5!/5=4!$$ Note that each pentagon could be fliped over, so both are considered the same. For example, these 2 selections are considered the same pentagons $$\\begin{split} 2,3,4,5,6\\\\ 6,5,4,3,2\\\\ \\end{split}$$ So, we also need to divide the result by 2 to avoid flipped over ones: $$4!/2=12$$ The final result are 12. The answer is B. If the result are 24, this means that it covers the flipped over pentagons. _________________ Senior Manager Joined: 03 Apr 2013 Posts: 274 Location: India Concentration: Marketing, Finance GMAT 1: 740 Q50 V41 GPA: 3 Re: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. [#permalink] ### Show Tags 18 Jun 2017, 23:22 honchos wrote: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lengths are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lengths ? (A) 5 (B) 12 (C) 24 (D) 32 (E) 120 Source: Jeff Sackman I am confident", "$11$: just add a $2k$-element periodic cycle hinted at in the comments. Unfortunately, this does not work with $9$ balls, but I believe the above idea can be extended to work for $9$ balls as well. I think it should be possible to find a periodic configuration with $5$ balls + extra ball, and then use it to construct a periodic configuration with a tringle connected to a pentagon (much like the above was a triangle connected to another triangle). I think the above shows that, if there is no periodic configuration with $5$ balls, then it a rather special case and a peculiarity of number $5$ - not a general property of odd numbers. - Nice construction, but you didn't notice the condition that No three of the lines should intersect in one point. Without this condition, making an example for $5$ is also easy, just consider $5$ points on sides and one diagonal of a square. – Goodarz Mehr Sep 16 '12 at 11:19 You're right, I completely overlooked it. – Jakub Konieczny Sep 16 '12 at 11:35 It think you should note at the top of the post that it doesn't answer the question; people might not read the comments before reading through the entire answer. – joriki Sep 18 '12 at 11:27 @joriki :", "without overlapping. For example, the value of $360/\\theta_5$ indicates that there is a slight overlap when we construct the fifth side of the alleged pentagon. What this column shows is that if we perform the construction accurately, when we have constructed the $11$th side of the alleged $11$-gon there is a gap almost large enough to fit half of another isoceles triangle. The alleged $15$-gon has very nearly enough space to fit a $16$th side, and the alleged $16$-gon has space for a $17$th side with room to spare. The discrepancies just get worse as we try more sides. The alleged $23$-gon winds up with at least $25$ sides (two more than it is supposed to have). The construction that is supposed to give $30$ sides gives $33.$ You cannot fix these errors by drawing more carefully. The errors are inherent in the trigonometry of the construction. For come purposes, your engineer may not care about the two-degree overlap of the pentagon construction, but surely it is not acceptable to get $17$ sides when you wanted $16$ or $33$ when you wanted $30.$ If we're going to go to so much trouble to get such bad results, we might as well just use a calculator to find the inradius of the polygon, lay off that distance along the", "### Home > GC > Chapter 12 > Lesson 12.2.4 > Problem12-94 12-94. Polly has a pentagon with angle measures $3x-26º$, $2x + 70º$, $5x-10º$, $3x$, and $2x + 56º$. Find the probability that if one vertex is selected at random, then the measure of its angle is more than or equal to $90º$. $(3x-26º)+(2x+70º)+(5x-10º)+(3x)+(2x+56º)=540º$ Solve for $x$, which will then allow you to find the value of each angle. Add together the angles that are $≥ 90º$, then divide the sum by $5$ (the total number of vertices). $\\text{Probability}=\\frac{4}{5}$", "# Easy Permutation-Combination Solved QuestionAptitude Discussion Q. How many diagonals can be drawn in a pentagon? ✔ A. 5 ✖ B. 7 ✖ C. 8 ✖ D. 10 Solution: Option(A) is correct A pentagon has $5$ sides. We obtain the diagonals by joining the vertices in pairs. Total number of sides and diagonals: $= {^5C_2}$ $= \\dfrac{5×4}{2×1}$ $= 5×2$ $= 10$ This includes its $5$ sides also. ⇒ Diagonals $= 10 - 5 = 5$ Hence the number of diagonals, $= 10 - 5$ $= 5$ ## (1) Comment(s) Raj Singh () From one vertex, you can draw 3 diagonals. From second vertex again you can draw 3 more diagonals. From third vertex again you can draw 3 more diagonals. Overall, you can draw 5x3=15 diagonals in a pentagon.", "the edge, each with a $$\\frac{1}{32}$$ chance of happening. After filling the required diagonals, we have the configuration as shown. This has the long triangle with a base at the bottom having all of one stroke, so there are no possible configurations. Case 3: The configuration $$3-2$$ This can occur with probability of $$\\frac{5}{16}$$ as there are two stroke types that can be the $$1$$ and $$5$$ positions for the edge, each with a $$\\frac{1}{32}$$ chance of happening. After filling the required diagonals, we have the configuration as shown. The other two diagonals have to be solid to have the longer triangles with the upper right and left edge have at least one distinct side. This makes the long triangle with the bottom side all solid, so there are no possible configurations. Case 3: The configuration $$2-1-1-1$$ This can occur with probability of $$\\frac{5}{16}$$ as there are two strokes that can be the $$2$$ and $$5$$ positions for the adjacent pair, each with a $$\\frac{1}{32}$$ chance of happening. After filling the required diagonals, we have the configuration as shown. The other two diagonals have to be solid to have the longer triangles with the upper right and left edge have at least one distinct side. This makes the long triangle with the bottom side all solid, so there", "white add up to -4). It is pretty easy to work out the exact values of the corner subgames To find the value of the total game we have to sum up these values which can either be done by hand (use this and this to get started and use the inductive rule $G+H = \\\\{ G^L+H,G+H^L \\\\vert G^R+H,G+H^R \\\\}$) or using combinatorial game suite to verify that this sum is equal to $\\\\{ \\\\{ 3 \\\\vert 2 \\\\} \\\\vert -1 \\\\}$ which is a fuzzy game (that is, confused with zero or a first player's win). To find the actual winning moves just try out the Left (bLack) and Right (white) options in the corner games to find out that there is a unique winning move for white and there are just 2 winning moves for bLack, all indicated in the pictures below. A _quintomino_ is a regular pentagon having all its sides colored by five different colours. Two quintominoes are the same if they can be transformed into each other by a symmetry of the pentagon (that is, a cyclic rotation or a flip of the two faces). It is easy to see that there are exactly 12 different quintominoes. On the other hand, there are also exactly 12 pentagonal faces of a dodecahedron whence the", "combinations correct? So since I had a set of 100, and I want a subset of 10, there are $$\\frac{100!}{10!*90!}$$ different ways of ordering that set... So is the correct answer... $$(\\frac{1}{50})^{10} * \\frac{100!}{10!*90!} = 0.00018$$ ? Thanks for your help! Last edited: Dec 19, 2005 2. Dec 20, 2005 ### NateTG Hmm. Lets say we have 1000 objects, and 20 are in a particular subset - call it set A, and we want to know what the odds are of picking a subset B of 20 elements at random so that exactly 2 of these elements is in A. Clearly, there are 1000 choose 20 ways to choose B. Now, if B is to have two elements from A, then there are 20 choose 2 ways to pick those two elements, and, since B will also have 18 elements from the remaining 1980 objects, there are 1980 chose 18 ways to pick those. Thus, the probablity of getting exactly 2 elements would be: $$\\frac{20 \\rm{C}2 \\times 1980 \\rm{C}18}{2000 \\rm{C} 20}$$ Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook", "What is the probability of having a pentagon in 6 points If the probability that $5$ random points in the plane whose horizontal coordinate and vertical coordinate are uniformly distributed on the interval $\\left(0,1\\right)$ occur to be the vertices of a convex pentagon is $\\frac{49}{144}$, what is the probability that a subset of $6$ random points in the plane whose horizontal coordinate and vertical coordinate are uniformly distributed on the interval $\\left(0,1\\right)$ occurs to be the vertices of a convex pentagon? Thanks a lot. - By the way, I mean the exact value of the probability in the question. Thanks. – kejma Jan 18 '13 at 5:27 Where does the $49/144$ come from? Maybe the place where that's proved would be a good place to start on the 6-point problem. – Gerry Myerson Jan 18 '13 at 6:21 It would have made sense to provide a link to this related question of yours, both initially and in particular in response to @Gerry's question. – joriki Jan 18 '13 at 7:26 I think this is considerably harder than the corresponding question with $5$ and $4$ points that you posed in the comments under my answer to the other question, because if the convex hull is a quadrilateral the remaining two points may or may not be part of a", "and 20 diagonals that are the third shortest(3 vertices apart). Thus, the probability is $${{3 \\times 20} \\over 170 } = {\\color{brown}\\boxed{6 \\over 17}}$$ Jul 8, 2022", "# What is the probability of having a pentagon in 6 points If the probability that $5$ random points in the plane whose horizontal coordinate and vertical coordinate are uniformly distributed on the interval $\\left(0,1\\right)$ occur to be the vertices of a convex pentagon is $\\frac{49}{144}$, what is the probability that a subset of $6$ random points in the plane whose horizontal coordinate and vertical coordinate are uniformly distributed on the interval $\\left(0,1\\right)$ occurs to be the vertices of a convex pentagon? Thanks a lot. • By the way, I mean the exact value of the probability in the question. Thanks. – kejma Jan 18 '13 at 5:27 • Where does the $49/144$ come from? Maybe the place where that's proved would be a good place to start on the 6-point problem. – Gerry Myerson Jan 18 '13 at 6:21 • It would have made sense to provide a link to this related question of yours, both initially and in particular in response to @Gerry's question. – joriki Jan 18 '13 at 7:26 • I think this is considerably harder than the corresponding question with $5$ and $4$ points that you posed in the comments under my answer to the other question, because if the convex hull is a quadrilateral the remaining two points may or may not" ]

[ "be the set of sides and diagonals of a regular pentagon. A pair [#permalink] ### Show Tags 20 Mar 2019, 07:23 Bunuel wrote: Let S be the set of sides and diagonals of a regular pentagon. A pair of elements of S are selected at random without replacement. What is the probability that the two chosen segments have the same length? (A) 2/5 (B) 4/9 (C) 1/2 (D) 5/9 (E) 4/5 Pentagon has 5 equal sides and 5 diagonals ; total equal sides 10 so case 1: picking 2 sides of equal length; 5/10 * 4/9 = 2/9 case 2 : picking 2 sides of equal diagonal ;5/10 * 4/9 = 2/9 IMO B Re: Let S be the set of sides and diagonals of a regular pentagon. A pair [#permalink] 20 Mar 2019, 07:23 Display posts from previous: Sort by", "# Difference between revisions of \"2013 AMC 10B Problems/Problem 12\" ## Problem Let $S$ be the set of sides and diagonals of a regular pentagon. A pair of elements of $S$ are selected at random without replacement. What is the probability that the two chosen segments have the same length? $\\textbf{(A) }\\frac{2}5\\qquad\\textbf{(B) }\\frac{4}9\\qquad\\textbf{(C) }\\frac{1}2\\qquad\\textbf{(D) }\\frac{5}9\\qquad\\textbf{(E) }\\frac{4}5$ ## Solution ### Solution 1 In a regular pentagon, there are 5 sides with the same length, and 5 diagonals with the same length. Picking an element at random will leave 4 elements with the same length as the element picked, with 9 total elements remaining. Therefore, the probability is $\\boxed{\\textbf{(B) }\\frac{4}{9}}$. ### Solution 2 Alternatively, we can divide this problem into two cases. Case 1: Side; In this case, there is a $\\frac{5}{10}$ chance of picking a side, and a $\\frac{4}{9}$ chance of picking another side. Case 2: Diagonal; This case is similar to the first, for again, there is a $\\frac{5}{10}$ chance of picking a diagonal, and a $\\frac{4}{9}$ chance of picking another diagonal. Summing these cases up gives us a probability of $\\boxed{\\textbf{(B) }\\frac{4}{9}}$.", "• The adversary must find a short $s$ or a short $a$? Additionally, $a$ is chosen uniformly at random from the set of non-invertible elements? – Hilder Vitor Lima Pereira May 8 '19 at 7:41 • Answer to Vitor: To find a short $s$. $a$ is not a random element and public. – user67451 Jun 2 '19 at 7:08", "Are the lengths from this recursive construction a geometric sequence? In his 1999 review of Edward Tufte's Visual Explanations in the Notices of the AMS (third page), Bill Casselman gives a very pretty proof of the irrationality of the golden mean. More precisely, Casselman shows that the side and diagonal of a regular pentagon cannot be commensurable. The proof assumes the contrary, and thereby constructs a vanishing sequence of commensurable terms, which is a contradiction. Upon closer inspection, however, I realized that there's a step in the proof that I cannot justify to my satisfaction. Casselman does show that the commensurability of the side and the diagonal of a regular pentagon implies the existence of a strictly decreasing sequence of commensurable lengths (e.g. the sequence consisting of the lengths of the sides of successively constructed nested regular pentagons). It is not clear to me, however, that he showed that this sequence converges to 0 (and not to some positive number). Of course, convergence to 0 would follow immediately if we could establish that the sequence of side lengths is a geometric one, i.e. that successive terms in it are in a constant ratio to each other (which has to be $< 1$, since the sequence is strictly decreasing). That this ratio is constant seems to me intuitively obvious", "$C_5$ (which are in fact the diagonals of the pentagon). That is your complement. And what you have there is another pentagon, drawn as a star pentagon. – N. S. Dec 3 '13 at 20:51 @Wanderer If $C_5$ is the graph $1-2-3-4-5-1$ then its complement is $1-3-5-2-4-1$. So $f(1)=1, f(2)=3, f(3)=5, f(4)=2, f(5)=5$ is your isomorphism. [Completelly irrelevant, but this function is exactly $f(x)=2x-1 \\pmod{5}$.] – N. S. Dec 3 '13 at 20:52 helpful and perfect answer. thank you! – Wanderer Dec 3 '13 at 20:59 oh, one more thing. where do we get the initial formula from? (the one with the degrees) – Wanderer Dec 3 '13 at 21:02", "probability exactly $2/(d+1)$. In quickrelevant, this same probability holds for pairs that are separated by one of the decision boundaries. But pairs of elements that are within the same homogeneous block are less likely to be compared, because of the possibility that the recursion will stop before it separates or compares them. If a pair of elements lies within a single block, has distance $d$ separating them, and is $e$ units from both ends of the block, then it will only be compared if one of the two elements is first to be chosen in at least one of the two intervals of length $d+e$ extending from the two elements towards an end of their block. This happens with probability at most $4/(d+e)$, because there are two ways of choosing which element to pick first as the pivot and two ways of choosing which extended interval it is first in. If we sum up these probabilities, for pairs involving a single element that is $e$ units from its nearest block boundary among a set of $n$ elements, we get $O\\bigl(\\log(n/e)\\bigr)$ as the expected contribution to the total time for that one element. If we sum the contributions from the elements within a block, for a block of length $\\ell_i$, we get a total expected contribution from that block", "the edge, each with a $$\\frac{1}{32}$$ chance of happening. After filling the required diagonals, we have the configuration as shown. This has the long triangle with a base at the bottom having all of one stroke, so there are no possible configurations. Case 3: The configuration $$3-2$$ This can occur with probability of $$\\frac{5}{16}$$ as there are two stroke types that can be the $$1$$ and $$5$$ positions for the edge, each with a $$\\frac{1}{32}$$ chance of happening. After filling the required diagonals, we have the configuration as shown. The other two diagonals have to be solid to have the longer triangles with the upper right and left edge have at least one distinct side. This makes the long triangle with the bottom side all solid, so there are no possible configurations. Case 3: The configuration $$2-1-1-1$$ This can occur with probability of $$\\frac{5}{16}$$ as there are two strokes that can be the $$2$$ and $$5$$ positions for the adjacent pair, each with a $$\\frac{1}{32}$$ chance of happening. After filling the required diagonals, we have the configuration as shown. The other two diagonals have to be solid to have the longer triangles with the upper right and left edge have at least one distinct side. This makes the long triangle with the bottom side all solid, so there", "Thread: probability --just can't figure it out 1. probability --just can't figure it out I haven't done this type of math in over 13 years and need some clarification since Im taking an online statistics course; hard to reach teacher for help! A) From the information provided, create the sample space of possible outcomes: A coin and a pentagonal die are tossed. B)Suppose you are playing a game of chance. If you bet $10 on a certain event, you will collect$500 (including your \\$10 bet) if you win. Find the odds used for determining the payoff. Now for part A)---the pentagonal die is supposed to have 5 sides, correct? And B) I have no idea. lol. 2. PART A) Sample Space is pretty simple. Pentagonal die has 5 sides. Coin has 2 sides. # of sides = # of possibilities for each item The sample space is just a box showing the possibilities. eg. Sample space for 2 coins. (H, H) (H, T) (T, H) (T, T) Notice how each coin has 2 sides, therefore 2 possibilities. So the box is going to be a 2x2 box. With 2x2 = 4 entries. So, the sample space for your question should have 2 x 5 = 10 entries. Just some busy work.", "there are 11 vertices. S is the set of all lines that contain two distinct vertices of P. so, total lines in S, or you can say total elements in S are : $$\\binom{11}2 \\qquad\\text{no of ways of choosing 2 vertices out of 11} = 55$$ total pairs of lines = $\\binom{55}2 = 1485$ Now, number of pairs of parallel lines for every edge possible = $$\\binom{\\lfloor n/2\\rfloor - 1}2 = \\binom42= 6.$$ so for $11$ edges, = $66$ note that for even number of sides in polygon, the number of pairs of parallel lines will be half of those calculated by above formula, because every pair will be repeated for parallel edges. probability that the chosen pair will be parallel, is $66/1485 = 0.0444444$ Use Poisson's distribution with $\\lambda$ as $.044444$, probability = $0.04251$. • \"no edge has more than 1 parallel line associated with it\" just isn't so. Consider the edge between verts 5 and 6; then lines 4-7, 3-8, etc. are all parallel to that edge. – Steven Stadnicki Aug 31 '13 at 17:11 • @steven, thanks for that... – Sumedh Aug 31 '13 at 19:05 But you are choosing three lines. The probability that you have calculated is for choosing a pair of parallel lines. The question asked is you randomly choose 3 lines", "# Easy Permutation-Combination Solved QuestionAptitude Discussion Q. How many diagonals can be drawn in a pentagon? ✔ A. 5 ✖ B. 7 ✖ C. 8 ✖ D. 10 Solution: Option(A) is correct A pentagon has $5$ sides. We obtain the diagonals by joining the vertices in pairs. Total number of sides and diagonals: $= {^5C_2}$ $= \\dfrac{5×4}{2×1}$ $= 5×2$ $= 10$ This includes its $5$ sides also. ⇒ Diagonals $= 10 - 5 = 5$ Hence the number of diagonals, $= 10 - 5$ $= 5$", "Let us first find the relationship between Pattern Number and the Number of pentagons. Number of pentagons in Pattern $$1 = 1$$ Number of pentagons in Pattern $$2 = 2$$ Number of pentagons in Pattern $$3 = 3$$ Therefore, as a general rule, we find that Pattern Number $$=$$ Number of pentagons Hence, a pattern having $$43$$ pentagons is pattern $$43$$ Now, in (a) we determined that, Number of beads $$= 5 + \\text{( Pattern Number} - 1\\,) \\times 4$$ Number of beads in Pattern Number $$43$$ \\begin{align*}​ &= 5 + (43 - 1) \\times 4 \\\\ &= 5 + 42 \\times 4 \\\\ &= 5 + 168 \\\\ &= 173 \\end{align*} Number of beads Andy had at first $$= 151$$ $$173 - 151 = 22$$ Andy needs $$22$$ more beads to make the pattern of $$43$$ pentagons. $$22$$ ## Conclusion • Grouping In number patterns involving grouping, we group the patterns into sets. To find the letter or object in a specific position, we will need to do division. The quotient will tell us the number of times the complete set appears. The remainder will tell us the position of the letter or object in the set. • Common Difference In number patterns involving common differences, we need to first identify what is the difference between each", "five diagonals determine a smaller regular pentagon inside our original one. The side length of this pentagon is the s* we just considered. In the smaller pentagon, the diagonals are again parallel to the opposite sides. It follows that the highlighted figure is a parallelogram, and that the diagonal in the smaller pentagon has length d*. So in the smaller pentagon, side and diagonal are both multiples of our original h. We can repeat the construction with the smaller pentagon. Since s* is strictly shorter than s, and they are both whole multiples of h, they must differ by at least h. We can repeat the process again and again, and obtain an infinite decreasing sequence of side lengths s > s* > s** ... each of which is at least h smaller than the one before. But this can't work, because there was only a finite number of h's in s to start! This contradiction shows that the hypothesis, that s and d are commensurable, must be false. Welcome to the Feature Column! These web essays are designed for those who have already discovered the joys of mathematics as well as for those who may be uncomfortable with mathematics.", "## moneybird Group Title A pattern of Figures is shown below. Figure 1 is a regular pentagon with side length 1. Figure 2 is a regular pentagon of side length 2 drawn around Figure 1 so that the two shapes share the top vertex, T, and the sides on either side of T overlap. The pattern continues so that each n>1, Figure n is a regular pentagon of side length n drawn around the previous Figure so that the two shapes share the top vertex, T, and the sides on either side of T overlap. The ink length of each Figure is the sum of the lengths of all of the line segments in the Figure. 2 years ago 2 years ago 1. moneybird |dw:1322004729461:dw| 2. moneybird Determine the general equation of ink length for Figure n. 3. asnaseer thinking... looks like at each step you add 5n and remove 2(n-1) 4. moneybird This question is from today's Canadian Intermediate Mathematics Contest 5. Tomas.A all sides are equal of that pentagon or not? 6. moneybird yes it's a regular pentagon 7. Tomas.A sorry I am not familiar with terminology in english 8. asnaseer ok, I think it is:$\\frac{5n(n+1)}{2}-n(1-n)$ 9. asnaseer sorry - I think it should be \"+\" after the fraction 10. asnaseer $\\frac{5n(n+1)}{2}+n(1-n)$ 11. moneybird My answer", ")$ , three numbers are chosen at random . Find the probability that the sum of the chosen numbers is divisible by 3. 13) Two players $P_{1}$ and $P_{2}$ are playing the fina of a chess championship , which consisits of a series of matches . Probability of $P_{1}$ winning a match is $\\dfrac{2}{3}$ and for $P_{2}$ is $\\dfrac{1}{3}$ . The winner will be the one who is ahead by 2 games as compared to the other player and wins atleast $6$ games . Now , if the player $P_{2}$ wins first four matches , find the probability of $P_{1}$ winning the championship. 14) Let $X$ be a set containing n elements . Find the number of all ordered triplets $(A, B , C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C$. 15) A point is selected at random from the interior of the pentagon with vertices $A = (0,2) , B = (4 , 0) , C = (2 \\pi + 1 , 0) , D = ( 2 \\pi + 1 , 4) , E = (0 , 4)$ . What is the probability that angle $APB$ is obtuse . 16) There are two bags , each containing $5$ red and $3$ black balls", "think I have a an argument that formalizes my intuition. (It is, admittedly, pretty tedious, but I guess this is inevitable, since it is formalizing a reasoning that is intuitively obvious to most people.) Let $p_0$ be the construction's initial regular pentagon, let $s_0$ be the \"base\" of $p_0$ (namely, the side of $p_0$ that runs horizontally along its bottom), and let $t_0$ be the triangle formed by $s_0$ and the two diagonals of $p_0$ incident on it. Now, for every $i \\in \\{1, 2, 3, \\dots\\}$, let $p_i$ be the first regular pentagon produced by the construction inside of $p_{i - 1}$. Let $s_i$ be the (unique) side of $p_i$ that is parallel to $s_{i - 1}$, and let $t_i$ be the triangle formed by $s_i$ and the two diagonals of $p_i$ that are incident on it. Now, define $F$ as the (composite) figure consisting of all the triangles $t_0, t_1, t_2, \\dots$. Likewise, define $G$ as the (composite) figure consisting of all the triangles $t_1, t_2, t_3, \\dots$. It is easy to see that the $i$-th triangle of $F$ (namely $t_{i-1}$) is similar to the $i$-th triangle of $G$ (namely $t_i$). From this correspondence, and the details of the construction, it follows that $F$ is similar to $G$.1 Therefore, there is a \"scaling factor\" $\\rho$ such", "Greeks would have presented the proof, but it is very similar in spirit, and we make no apologies for the anachronism. Let’s get started. Suppose, for the sake of an eventual contradiction, that we have a regular pentagon whose side length is commensurable with the side length of the inscribed pentagram. Let $s$ denote the side length of the pentagon and $t$ denote the side length of the pentagram (i.e., the diagonal of the pentagon). In modern language, our assumption is that $\\frac{t}{s}$ is rational, i.e., that there are positive integers $p$ and $q$ such that $\\frac{t}{s} = \\frac{p}{q}$. By scaling the pentagon, we can in fact assume that $t = p$ and $s = q$. So let this be our starting assumption: there is a regular pentagon whose side length $s$ and diagonal length $t$ are both positive integers. Let us label the vertices of the pentagon by the letters A,B,C,D,E. The center of the pentagram forms another regular pentagon, whose vertices we shall call a,b,c,d,e. This is shown in the diagram below. Note that $s$ is the length of the line segment connecting A and B (we will denote this length by |AB|), and $t$ is |AC|. Let $s'$ denote the side length of the inner pentagon, i.e., |ab|, and let $t'$ denote the length of", "equivalently such that the average value of $a_i$ is $(n+1)/2$, which is of course the average value of a random element of $[n]$. A random $n$-sided die in this model is simply an $n$-sided die chosen uniformly at random from the set of all such dice. Given $n$-sided dice $A=(a_1,\\dots,a_n)$ and $B=(b_1,\\dots,b_n)$, we say that $A$ beats $B$ if $|\\{(i,j):a_i>b_j\\}|>|\\{(i,j):a_i If the two sets above have equal size, then we say that $A$ ties with $B$. ### A first reduction When looking at this problem, it is natural to think about the following directed graph: the vertex set is the set of all $n$-sided dice and we put an arrow from $A$ to $B$ if $A$ beats $B$. We believe (and even believe we can prove) that ties are rare. Assuming that to be the case, then the conjecture above is equivalent to the statement that if $A,B,C$ are three vertices chosen independently at random in this graph, then the probability that $ABC$ is a directed cycle is what you expect for a random tournament, namely 1/8. One can also make a more general conjecture, namely that the entire (almost) tournament is quasirandom in a sense defined by Chung and Graham, which turns out to be equivalent to the statement that for almost all pairs $(A,B)$ of dice,", "$c$ are all between $1$ and $6$, inclusive. We check the divisibility by taking this modulo $8$. If the result is equivalent to $0$, then it is divisible. Setting up the equivalence condition: 100a+10b+c \\equiv 0 ... 2 In a decagon, there are $10\\choose2$ ways to pick pairs of vertices that are 1 side apart. Connecting these pairs gives pairs of parallel diagonals. But for each pair, only 5 of the other 9 pairs don't share a share a vertex or result in a diagonal that is just a side of the decagon, so there are $\\frac59{10\\choose2}$ ways here. Now we check pairs of ... 2 For the first case, you have 37 elements in all, but are excluding $z$, leaving 36. If both $x$ and $y$ are included, you get to choose only the other 34 elements, giving $2^{34}$ possible sets. If you decide to include $x$, by the same reasoning above you get $2^{35}$ subsets, if you include $y$ again $2^{35}$; but that counts the sets containing $x$ and ... 2 This problem has nothing to do with probability, but your idea of picking a king frog is useful. Starting with the king frog, enumerate both frogs and squares in a spiral (like the starting point for Ulam's spiral). Then simply have frogs 1 and 2", "pentagon $$BFDEG$$ truely is regular, having side length $$b$$. You then even could deduce that $$D$$, $$F$$, and $$(H=I)$$ are collinear and that $$F(H=I)=a$$ as well.", "by the base $BC$ is $s.h.d^{n}/2$. Now, if $d=1$, the total area of all the constructed triangles after each iteration remains constant at $s.h/2$, although the total length of all the sides opposing the base $BC$ increases monotonically. Moreover, if $1>d>0$, it would appear that the base $BC$ of the original equilateral triangle will always be the limiting’ configuration of the sides opposing the base $BC$. Fig 1 This is indeed so if $0, since (see fig. 1 above) the total length of all the sides opposing the base $BC$ at the $n$‘th iteration—say $l_{n}$—yields a Cauchy sequence whose limiting value is, indeed, the length $s$ of the base $BC$. Fig 2 However, if $d=1/2$, the total length (see fig. 2 above) of all the sides opposing their base on $BC$ is always $2s$! Fig. 3 Finally, if $1>d>1/2$, the total length (see fig. 3 above) of all the sides opposing their base on $BC$ is a monotonically increasing value. Consider now: Case 1 Treating the above iteration as the fragmentation of a land-holding, how would one interpret the limit’ of such an interpretation (which is postulated as existing in a putative completion’ of Euclidean space)? Case 2 Let $s$ be one light-year and consider how long it would take a light signal to travel from $B$ to" ]

[ "[0], given: 182 Re: Jaime has n comic books. He sells one half of them and then gives one [#permalink] 14 Oct 2017, 09:43 Display posts from previous: Sort by", "titles, you'd think it would all be about different \"Spider-Men\" characters but, NOPE, same guy, just different series. Marvel sort-of solved this issue with the Ultimate series, but I hear that started suffering from the same issues. And, that's including all the mishaps where they keep using the same characters (I don't mean \"same characters\" but different universe, I mean 100% the same characters in and from the same universe) having all of this comics resetting to zero every other year, which creates even more confusion. (And, I don't buy the \"You don't have to read everything that came prior because they designed the comic around being an 'entrance' for newcomers\" excuse because the series immediately starts using characters that haven't been seen in decades or referencing past events that predates the current happenings in the comics that are suppose to be for newcomers). >>311278 >If they were smart, they'd move entirely to a graphic novel model, since practically all stories are \"written for the trade (paperback)\" anyway, which means they're all like six issues long. Instead of putting out over 52 issues per month twelve months a year, they'd be better off putting out like two graphic novels per week. It's not just the magazine model. it's also the money involved. If they sell 52 comics for", "The bin wasn’t too hard to assemble, but the on-line video was little help; it was pretty vague and didn’t address the things that weren’t intuitively obvious. I would have preferred printed directions. I noticed there was a certain amount of static electricity present and that attracted dust. The next one gets assembled in the cat-free comic room. The removable dividers are ideal for keeping your books upright even when the box is not-quite-full. Once assembled, the bin feels sturdy and looks nice. A set of them would give you good storage that uses space efficiently. It’s not airtight, so I don’t think condensation is likely to be an issue. A set would likely be cost-prohibitive for a large collection, but I think a few bins for high-end books would be a good investment. # The Credit Where it’s Due Department: Why no Byrne Variants? A happy mail call today. I got my Alex Ross variants of the new Fantastic Four #1. These look great. Both honor one of the most iconic covers of all time; the first FF #1 from 1961. I’m not usually one to be enthused about variant covers, but I’m really happy to get these. It’s a fitting tribute to Lee and Kirby who created the team and set the standard for everything that", "# Permutation Question #### Bushy ##### Member There are three separate bundles of reading material comprising of 4 comics, 2 novels and 3 magazines. They are placed together to form one pile. In how many ways can this be done if there are no restrictions on where the individual items are to be placed? I say 9! = 362880 Determine the number of permutations if the order of the comic books in each bundle does not change I'm not sure what they mean by 'each bundle' here #### Jameson If you're arranging these into one pile and the order of the comic books doesn't change then I would say it's $$\\displaystyle \\frac{9!}{4!}$$. Dividing by $4!$ accounts for all the different ways you can order the comic books but keep everything else the same. Since we want just one order I would say this is the answer but again this question is worded in a way where I'm not sure.", "$5 a piece, that's$260 gained for all 52 comics selling one copy. Compare that to 9 graphic novels (Got that number from supposing that they compile all 52 comics into 6 different graphic novel lines, leaving us with \"9 series) selling for a maximum of $20, for a total of$180 earned for selling 1 copy of all 9 series. On paper, the former \"magazine\" model method brings in more money, and that's all that comic publishers really care about. And, if they were to start following the \"graphic novel\" method, they'd probably fuck it up and demand that the novels sell for $30 a pop so that they can maintain their near$300 figure. And, keep in mind, manga volumes are selling for $10. >>311379 >It's based around digital monsters, that come from another dimension, and have a master that they befriend and serve in order to gain power so that they can fight bigger and bigger threats. And Spider-Man is about a young guy who gets bitten by a special spider and then gets spider-powers which he uses to fight crime. That doesn't mean Miles Morales is the same as Peter Parker. But at least at the start they followed the pretense of Miles trying to live up to (Ultimate) Peter's legacy. >And, what about the fact that", "book is the fourth book below purple book  Purple book is the third book above red book  Yellow book is between red book and white book  Purple book is at the either end of the stack Amy chooses the second book from the top. Which color is that book? A) Blue B) Green C) Yellow D) Red E) White 15. Jason builds a wall with two types of lego pieces as follows. After that, he removes some lego pieces from the wall, and this is what he sees when he looks directly from the left side and from the right side of it. What is the total number of lego pieces that Jason removes from the wall? A) 4 B) 6 C) 7 D) 8 E) 12", "2017, 08:43 Jaime has n comic books. He sells one half of them -- Jaime now has n/2 comic books left Then gives one third of the ones he has left to a friend -- Jaime now has 2/3 of the n/2 comic books left = n/3 comic books left If he has 10 comic books left => n/3 = 10 => n=30 Re: Jaime has n comic books. He sells one half of them and then gives one &nbs [#permalink] 14 Oct 2017, 08:43 Display posts from previous: Sort by", "but one cover, but that also is no longer the case. This is because of another trend in comics. There is now a proliferation of variants. Variants are copies of the same issue of a series that are absolutely identical except for the cover. To me, this is getting excessive. There are variants to encourage dealers to order more; there are variants with some special gimmick on the cover; there are variants to celebrate a special event and variants to popularize a particular character. Many, like the “Special Inauguration Day Edition” of Amazing Spider-Man #583 are rare enough to be worth considerably more than the standard version but most probably have no additional value at all. When the latest volume of Fantastic Four hit the stands last August, there were at least 42 different variants of the first issue, some of which are shown above (you can see three more here). This title has averaged roughly nine variants per issue since its inception less than a year ago. So, if you can’t just go by number and you can’t just go by cover, what do you do? One answer is to make sure you know the publication date. If you’re looking at back issues in a comic store, you can always check the indicia. As long as your", "computer and build—at the age of 58—a database of my teenaged purchases. As you might expect, my work consisted mostly of writing little scripts that collected, organized, and assembled the data. I’ll describe those scripts in a later post, but here I want to talk in more general terms about the overall process, the obstacles I encountered, and the dead-end paths I traveled down. First, let’s talk about the size and nature of my collection. Back when one of the Guardians of the Galaxy movies came out, I had pulled several issues of Defenders and Marvel Presents1 out of the box, and they were still stacked on a bookshelf. Measuring the height of that stack led me to guess that the density of the box was about 10–12 issues per inch of height. So maybe 250–300 comic books in my collection. This turned out to be a significant underestimate, probably because the books in the box are squeezed more tightly together by the weight of the overburden.2 As for the nature of my collection, it’s basically all Marvel. I knew I had a handful (four, as it turned out) of non-Marvel titles, including a Jack Kirby Sandman, but my database wouldn’t need too much adjustment if I initially treated it as 100% Marvel. So my initial plan was", "# Moderate Permutation-Combination Solved QuestionAptitude Discussion Q. There are five comics numbered from 1 to 5. In how many ways can they be arranged, so that part-1 and part-3 are never together? ✖ A. 48 ✔ B. 72 ✖ C. 120 ✖ D. 210 Solution: Option(B) is correct The total number of ways in which 5 part can be arranged $= 5! =120$. The total number of ways in which part-1 and part-3 are always together: $= 4!×2!$ $= 48$ Therefore, the total number of arrangements, in which they are not together is: $=120-48$ $= \\textbf{72}$", ". . . . . . . . . . 15.07 361540 Origami Games. Fullman . . . . . . . . . . . . . . . . . . . . . . 15.07 361550 Origami Magic. Fullman. . . . . . . . . . . . . . . . . . . . . . . 15.07 065854 Origami Monsters. Ard . . . . . . . . . . . . . . . . . . . . . . . . 15.07 065853 Origami Planes. Ard. . . . . . . . . . . . . . . . . . . . . . . . . . . 15.07 Set S22254 of 5 books @ \\$75.35 New BTSB Titles In Blue 98 Showcase Magazine Bound to Stay Bound Books Series & Sets Draw 50 Series. Ames. Doubleday. Ages 8-15 057010 Draw 50 Airplanes, Aircraft, & Spacecraft. . . . . . . 15.02 057014 Draw 50 Aliens, UFOs, Galaxy Ghouls, Milky Way Marauders, And Other Extraterrestrial Creatures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.02 057017", "books have over a thousand pages. They have no cover art. Compendium One was also released as a limited edition hardcover in 2012 at the Emerald City Comicon and San Diego Comicon with 1000 pieces available. Title ISBN Release Date Collected Material Template-specific style sheet: 1-60706-076-0 May 6, 2009 Collects #1–48 ### Novels Title ISBN Release Date Authors The Walking Dead: Rise of the Governor Template-specific style sheet: 9780312547738 October 11, 2011 Robert Kirkman and Jay Bonansinga Template-specific style sheet: 9780312547745 October 16, 2012 Robert Kirkman and Jay Bonansinga ## References 1. The Walking Dead Volume 5: The Best Defense 2. Sullivan, Michael Patrick. WEEK OF THE DEAD I: Robert Kirkman, Comic Book Resources, May 19, 2008 3. Sullivan, Michael Patrick. WEEK OF THE DEAD II: Charlie Adlard, Comic Book Resources, May 20, 2008 4. Tony Moore's The Walking Dead Cover Art Gallery 5. The Walking Dead #1 at the Comic Book DB. Retrieved Dec 23, 2011. 6. The Walking Dead #2 at the Comic Book DB. Retrieved Dec 23, 2011. 7. The Walking Dead #6 at the Comic Book DB. Retrieved Dec 23, 2011. 8. The Walking Dead #9 at the Comic Book DB Retrieved Dec 23, 2011. 9. The Walking Dead #11 at the Comic Book DB Retrieved Dec 23, 2011. 10. The Walking Dead #16", "have missed. Another neat fact I learned in doing this is that I have had 137 separate orders with Amazon since 2004... most of which had multiple books :/ I need to seek group help or something... Steve", "Oct 2015, 02:12 Similar topics Replies Last post Similar Topics: 5 Jim and Julia bought a bag of jellybeans, and ate some of them. If the 2 22 May 2015, 02:11 6 Danny bought some packets of pens, each packet at a cost of$4, and so 7 03 Feb 2015, 09:01 3 A bookshelf holds both paperback and hardcover books. The 3 11 Oct 2013, 16:05", "4Packets are not in order. 4 > 4Packets are not in order. 5 > 5Packets are not in order. 7 > 7Packets are not in order. 8 > 8Packets are not in order. 9 > 9Packets are not in order. 10 > 10Packets are not in order. 10 > 10Packets are not in order. 14 > 14Packets are not in order. 15 > 15Packets are not in order. 16 > 16Packets are not in order. 17 > 17Packets are not in order. 19 > 19Packets are not in order. 20 > 20Packets are not in order. 21 > 21Packets are not in order. 22 > 22Packets are not in order. 23 > 23Packets are not in order. 24 > 24Packets are not in order. 25 > 25Packets are not in order. 26 > 26Packets are not in order. 28 > 28Packets are not in order. 29 > 29Packets are not in order. 31 > 31Packets are not in order. 32 > 32Packets are not in order. 33 > 33Packets are not in order. 34 > 34Packets are not in order. 35 > 35Packets are not in order. 36 > 36Packets are not in order. 37 > 37Packets are not in order. 38 > 38Packets are not in order. 39 > 39Packets are not in order. 40 > 40Packets", "the order of events until a year later when all the story arcs are done and you can look at them in retrospect. It's all fucked and they need to cut back massively and also just ditch the magazine model entirely. Also, part of the issue is that you really don't need to read \"everything\" that came before, but you do need a certain level of knowledge of past stuff. And good luck finding where precisely that knowledge should come from. Once you reach a certain point you can read most of whatever and understand it, but good luck reaching that point. > If they sell 52 comics for$5 a piece, that's $260 gained for all 52 comics selling one copy. Compare that to 9 graphic novels (Got that number from supposing that they compile all 52 comics into 6 different graphic novel lines, leaving us with \"9 series I looked at it with the assumption that a graphic novel is about the length of six issues, which is basically the standard right now. Instead of each series doing 12 issues per year (though really they tend to do 13 with annuals, and the annuals are a little longer), they could say each of those 52 series does two graphic novels per year, or 104 graphic novels per", "one before, long enough ago I kind of accept it. ## Reading the Comics, February 9, 2019: Garfield Outwits Me Edition Comic Strip Master Command decreed that this should be a slow week. The greatest bit of mathematical meat came at the start, with a Garfield that included a throwaway mathematical puzzle. It didn’t turn out the way I figured when I read the strip but didn’t actually try the puzzle. Jim Davis’s Garfield for the 3rd is a mathematics cameo. Working out a problem is one more petty obstacle in Jon’s day. Working out a square root by hand is a pretty good tedious little problem to do. You can make an estimate of this that would be not too bad. 324 is between 100 and 400. This is worth observing because the square root of 100 is 10, and the square root of 400 is 20. The square of 16 is 256, which is easy for me to remember because this turns up in computer stuff a lot. But anyway, numbers from 300 to 400 have square roots that are pretty close to but a little less than 20. So expect a number between 17 and 20. But after that? … Well, it depends whether 324 is a perfect square. If it is a perfect square,", "# So I'm a Spider, So What?, Vol. 3 (manga) (Kobo eBook) \\$6.99 Available Now ### Description You wanna piece of me?! I thought I'd seen the last of those dang monkeys, but now their evolved friends are here for payback! Getting past this howling mob is gonna take all the skills I've learned so far (and a few new ones). If I can just make it past these guys, I'll finally be able to get back up to the middle level of the labyrinth--which has gotta be easier than down here, right? Wait...What do you mean it's on FIRE?! Product Details ISBN-13: 9781975353728 Publisher: Yen Press Publication Date: July 23rd, 2018 Language:", "133 DC Comics. Superman Adventures. . . . . . . . . . . . . . . . 124 DC Super Friends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 DC Super Heroes. Dark Knight . . . . . . . . . . . . . . . . . . . . 116 DC Super Heroes. Man Of Steel . . . . . . . . . . . . . . . . . . . 116 DC Super Heroes Origins . . . . . . . . . . . . . . . . . . . . . . . . . 116 DC Super Heroes. Tales Of The Batcave. . . . . . . . . . . . 116 D.W. Books. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 Dear America . . . . . . . . . . . . . . . . .", "There are 14 girls and 11 boys in the class. How many ways can a four-member team be chosen so that there are exactly two boys in it? • Five girls Five girls traveling by bus. Each holds in each hand two baskets. In each basket are four cats. Each cat has three kittens. Two girls exits bus. How many legs are in the bus? • Classroom In a class are 32 pupils. Boys are 8 less than girls. How many boys and girls are in the classroom? • A library A library has 12,500 fiction books and 19,000 non fiction books. Currently 2/5 of the fiction books are checked out. Currently 2/5 of the non fiction books are checked out. Of the books checked out, only 1/10 are due back this week. How many books are due • Marbles Dave had 40 marbles. Junjun has 2 1/5 more than Dave’s marbles. How many marbles do they have altogether? • Equivalent fractions 2 Write the equivalent multiplication expression. 2 1/6÷3/4" ]

[ "equal in value? e) State the domain and range for each comic. 2. Paulette bought a Bobby Orr rookie card for$300. The value of the card appreciates (increases) by 30% each year. a) Complete a table of values to show the first five years of the investment. b) Determine the common ratio for the successive terms. c) Determine the equation of the exponential function that models this investment. 3. Due to the closure of the pulp and paper mill, the population of the small town is decreasing at a rate of 12% annually. If there are now 2400 people living in the town, what will the town's projected population be in eight years? 1. a) The purchase price of each comic book is the $y$ -intercept. The $y$ -intercept is the initial value of the books. The Spiderman comic book cost $30.00 and the Superman comic book cost$5.00. b) The Superman comic book shows exponential growth. c) The Spiderman comic book shows exponential decay. d) In 2005 both comic books were equal in value. The graphs intersect at approximately (5, $12.50), where 5 represents five years after the books were purchased. e) The domain and range for each comic is $D=\\{x| x \\in R\\}$ and $R=\\{y|y>0, \\ y \\in R\\}$ 2. $&300(.30)=90 && 390(.30)=117 && 507(.30)=152.10\\\\&300+90=390 && 390+117=507", "$5 a piece, that's$260 gained for all 52 comics selling one copy. Compare that to 9 graphic novels (Got that number from supposing that they compile all 52 comics into 6 different graphic novel lines, leaving us with \"9 series) selling for a maximum of $20, for a total of$180 earned for selling 1 copy of all 9 series. On paper, the former \"magazine\" model method brings in more money, and that's all that comic publishers really care about. And, if they were to start following the \"graphic novel\" method, they'd probably fuck it up and demand that the novels sell for $30 a pop so that they can maintain their near$300 figure. And, keep in mind, manga volumes are selling for $10. >>311379 >It's based around digital monsters, that come from another dimension, and have a master that they befriend and serve in order to gain power so that they can fight bigger and bigger threats. And Spider-Man is about a young guy who gets bitten by a special spider and then gets spider-powers which he uses to fight crime. That doesn't mean Miles Morales is the same as Peter Parker. But at least at the start they followed the pretense of Miles trying to live up to (Ultimate) Peter's legacy. >And, what about the fact that", "what year were both comic books equal in value? 5. State the domain and range for each comic. 1. The purchase price of each comic book is the \\begin{align*}y\\end{align*}-intercept. The \\begin{align*}y\\end{align*}-intercept is the initial value of the books. The Spiderman comic book cost30.00 and the Superman comic book cost $5.00. 2. The Superman comic book shows exponential growth. 3. The Spiderman comic book shows exponential decay. 4. In 2005 both comic books were equal in value. The graphs intersect at approximately (5,$12.50), where 5 represents five years after the books were purchased. 5. The domain and range for each comic is \\begin{align*}D=\\{x| x \\in R\\}\\end{align*} and \\begin{align*}R=\\{y|y>0, \\ y \\in R\\}\\end{align*} Paulette bought a Bobby Orr rookie card for 300. The value of the card appreciates (increases) by 30% each year. 1. Complete a table of values to show the first five years of the investment. 2. Determine the common ratio for the successive terms. 3. Determine the equation of the exponential function that models this investment. 1. First, calculate the values and create the table: \\begin{align*}&300(.30)=90 && 390(.30)=117 && 507(.30)=152.10\\\\ &300+90=390 && 390+117=507 && 507+152.10=659.10\\\\ \\\\ &659.10(.30)=197.73 && 856.83(.30)=257.05\\\\ &659.10+197.73=856.83 && 856.83+257.05=1113.88\\end{align*} \\begin{align*}&\\text{Time} \\ (years) \\ \\ 0 \\qquad 1 \\qquad 2 \\qquad \\quad 3 \\qquad \\quad 4 \\qquad \\quad \\ 5\\\\ &\\text{Value (\\)} \\qquad 300 \\quad", "# In how many different ways can a 9-panel comic grid be used? While writing my answer to Why does “Watchmen” use the 9-panel grid? I used this picture to indicate the many ways it can be divided into groups (which may be used for the panels of a comic, as was the case in “Watchmen”): Afterwards, it has been pointed out to me in a comment that there are some combinations that are not present in this picture: The $$8181$$ variations on the $$99$$ panel grid in that diagram don’t exhaust the possibilities — there are certainly many others. For example, this one is one of the many that aren’t shown there. Another comment says that the 9-panel grid can be used in $$40964096$$ different ways: The are $$1212$$ interior borders, each of which can be included or excluded in a particular layout. That’s $$212=40962^{12} = 4096$$ possibilities. A two-page spread treats two $$3×33 \\times 3$$ grids as a single $$6×36 \\times 3$$ grid with $$2121$$ internal borders, for $$2,097,1522,097,152$$ possibilities. How can I calculate that? I tried the following: • There a $$33$$ ways to group the third row. • That leaves us with $$44$$ panels in 2nd and 3rd rows. I count $$44$$ ways to group those. • The $$1212$$ combinations so far must be", "comic. 2. Paulette bought a Bobby Orr rookie card for$300. The value of the card appreciates (increases) by 30% each year. a) Complete a table of values to show the first five years of the investment. b) Determine the common ratio for the successive terms. c) Use technology to determine the equation of the exponential function that models this investment. 3. Due to the closure of the pulp and paper mill, the population of the small town is decreasing at a rate of 12% annually. If there are now 2400 people living in the town, what will the town's projected population be in eight years. 1. a) The purchase price of each comic book is the $y$ -intercept. The $y$ -intercept is the initial value of the books. The Spiderman comic book cost $30.00 and the Superman comic book cost$5.00. b) The Superman comic book shows exponential growth. c) The Spiderman comic book shows exponential decay. d) In 2005 both comic books were equal in value. The graphs intersect at approximately (5, $12.50), where 5 represents five years after the books were purchased. e) The domain and range for each comic is $D=\\{x| x \\varepsilon R\\}$ and $R=\\{y|y>0, \\ y \\varepsilon R\\}$ 2. $&300(.30)=90 && 390(.30)=117 && 507(.30)=152.10\\\\&300+90=390 && 390+117=507 && 507+152.10=659.10\\\\\\\\&659.10(.30)=197.73 && 856.83(.30)=257.05\\\\&659.10+197.73=856.83 && 856.83+257.05=1113.88$ a) $&\\text{Time}", "one bag of chips, and one comic book. In year one, the basket costs $8.00. In year two, the price of the same basket is$7.00. From year one to year two, there is ____ at an annual rate of _________(how do i calculate the rate?) In year one, $40.00 will buy_______ baskets, and in year two,$40.00 will buy ______. Suppose the price level reflects the number of dollars needed to buy a basket of goods containing one can of soda, one bag of chips, and one comic book. In year one, the basket costs $8.00. In year two, the price of the same basket is$7.00. From year one to year two, there is ____ at an annual rat...", "Question 7. Model with Mathematics A comic store sells two rare comic books for$1,824. The price of one comic book is 11 times as much as the price of the other. • Write an equation to model the price of the less expensive comic book. Then find the price. • Write an equation to model the price of the more expensive comic book. Then find the price. L + 11 L = $1,824, The price of the less expensive comic book is$152, E = 11 L, The price of the more expensive comic book is $1,672, Explanation: Given a comic store sells two rare comic books for$1,824. L + E = $1,824, Equation to model the price of the less expensive comic book is L + 11 L =$1,824, 12L = $1,824, L =$1,824 ÷ 12 = $152, The equation to model the price of the more expensive comic book. E = 11 L, as L =$152 , E = 11 X 152 = \\$1,672.", "State the domain and range for each comic. 2. Paulette bought a Bobby Orr rookie card for300. The value of the card appreciates (increases) by 30% each year. a) Complete a table of values to show the first five years of the investment. b) Determine the common ratio for the successive terms. c) Use technology to determine the equation of the exponential function that models this investment. 3. Due to the closure of the pulp and paper mill, the population of the small town is decreasing at a rate of 12% annually. If there are now 2400 people living in the town, what will the town's projected population be in eight years. 1. a) The purchase price of each comic book is the \\begin{align*}y\\end{align*}-intercept. The \\begin{align*}y\\end{align*}-intercept is the initial value of the books. The Spiderman comic book cost $30.00 and the Superman comic book cost$5.00. b) The Superman comic book shows exponential growth. c) The Spiderman comic book shows exponential decay. d) In 2005 both comic books were equal in value. The graphs intersect at approximately (5, 12.50), where 5 represents five years after the books were purchased. e) The domain and range for each comic is \\begin{align*}D=\\{x| x \\varepsilon R\\}\\end{align*} and \\begin{align*}R=\\{y|y>0, \\ y \\varepsilon R\\}\\end{align*} 2. \\begin{align*}&300(.30)=90 && 390(.30)=117 && 507(.30)=152.10\\\\ &300+90=390 && 390+117=507 && 507+152.10=659.10\\\\", ". $820 - 277 \\:=\\:\\boxed{543}$ ways to get different languages. By the way, these are not permuations. .", "to the conclusion that there are $10\\cdot9\\cdot8\\cdot7\\cdot6=30,240$ different ways to post the letters if you can use each letter box at most once.", "spend on both comic books and magazines. b) How much does Peter need to spend if he wants 9 comic books and 3 magazines? c) Write a new expression for Peter's expenditure on books if the bookstore raises the price of a comic book to $15 and a magazine to$10.", "Question At a comic book convention vendor 1 was selling a set of 5 comics for $53.10. Vendor 2 was selling a set of 3 comics for$31.71. Which vendor has the higher unit price?", "1CD = 4PC CD cost in PNs? From above, PC = $$\\frac{2}{3}$$ PN So 1CD = (4 *$$\\frac{2}{3}$$PN) 1 CD = $$\\frac{8}{3}$$ PN (3) One book cost the same as 6 pens 1B = 6PN (4) Book and CD cost in terms of PENS 1B = 6PN 1CD = $$\\frac{8}{3}$$PN (5) Ratio of book cost to CD cost? $$\\frac{B}{CD}=\\frac{6}{(\\frac{8}{3})}=(6*\\frac{3}{8})=\\frac{18}{8}=\\frac{9}{4}$$ _________________ In the depths of winter, I finally learned that within me there lay an invincible summer. Current Student Joined: 09 Mar 2017 Posts: 89 Location: Netherlands Concentration: General Management, Finance GMAT 1: 670 Q47 V36 Re: The cost of 4 pens is equal to the cost of 6 pencils. The cost of a CD [#permalink] ### Show Tags 09 Jul 2018, 17:57 We can solve it by making the unit of comparison the same. 4 pens=6 pencils so 1 pen=1.5 pencils. Cost of CD=4 pencils Cost of Book=6 pens or 9 pencils. Cost of Book to Cost of CD ratio=9:4. Ans is A. Sent from my SM-G935F using GMAT Club Forum mobile app Senior Manager Joined: 04 Aug 2010 Posts: 277 Schools: Dartmouth College Re: The cost of 4 pens is equal to the cost of 6 pencils. The cost of a CD [#permalink] ### Show Tags 09 Jul 2018, 17:59 selim wrote: The cost of 4 pens", "= 95$$ The expression to represent shipping cost to each area comes from multiplying the cost per book by the number of books. $$US \\,shipping = 4.90p$$ $$Int’l\\, shipping = 16.50(200 – p)$$ Add the three elements of the cost together to write an expression representing the total cost to distribute the books. $$95 + 4.90p + 16.50(200 – p)$$ Use the distributive property. $$95 + 4.90p +3300 – 16.50p$$ Combine like terms. $$3395 – 11.60p$$ Answer: $$3395 – 11.60p$$ Sarina can use this polynomial to find out the cost of shipping $$200$$ books when $$p$$ of them are shipped within the U.S. (domestically). Example Problem A carpet designer creates a carpet that uses four colors according to the pattern and dimensions below. Express the area of the carpet as a polynomial. Find the area of each colored piece of carpet by multiplying the length by the width. Red: $$x \\cdot x = x^{2}$$ Blue: $$2y(x + 3y) = 2xy + 6y^{2}$$ White: $$2y \\cdot x = 2xy$$ Green: $$x(x + 3y) = x^{2} + 3x$$ Find the area of the whole carpet by combining the areas of the four pieces. $$x^{2} + 2xy + 6y^{2} + 2xy + x^{2} + 3xy$$ Combine like terms. $$2x^{2} + 7xy + 6y^{2}$$ Answer: $$2x^{2} + 7xy + 6y^{2}$$ Summary One", "and the four of a kind is four '5's). There are $7\\choose 4$ ways to select ''slots'' in which to place your 4 of a kind. The remaining three slots will then contain the three of a kind. So, there are $6\\cdot5\\cdot{7\\choose 4}$ different ways to obtain a three of a kind and 4 of a kind. Since outcomes are equally likely here, the chance of obtaining a three of a kind and 4 of a kind is as you stated.", "# Permutation Question #### Bushy ##### Member There are three separate bundles of reading material comprising of 4 comics, 2 novels and 3 magazines. They are placed together to form one pile. In how many ways can this be done if there are no restrictions on where the individual items are to be placed? I say 9! = 362880 Determine the number of permutations if the order of the comic books in each bundle does not change I'm not sure what they mean by 'each bundle' here #### Jameson If you're arranging these into one pile and the order of the comic books doesn't change then I would say it's $$\\displaystyle \\frac{9!}{4!}$$. Dividing by $4!$ accounts for all the different ways you can order the comic books but keep everything else the same. Since we want just one order I would say this is the answer but again this question is worded in a way where I'm not sure.", "comic books purchases in the year 2000. Both comics were expected to be good investments, but one of them did not perform as expected. Use the graphs to answer the questions. License: CC BY-NC 3.0 1. What was the purchase price of each comic book? 2. Which comic book shows exponential growth? 3. Which comic book shows exponential decay? 4. In what year were both comic books equal in value? 5. State the domain and range for each comic. 1. The purchase price of each comic book is the -intercept. The -intercept is the initial value of the books. The Spiderman comic book cost$30.00 and the Superman comic book cost $5.00. 2. The Superman comic book shows exponential growth. 3. The Spiderman comic book shows exponential decay. 4. In 2005 both comic books were equal in value. The graphs intersect at approximately (5,$12.50), where 5 represents five years after the books were purchased. 5. The domain and range for each comic is and Paulette bought a Bobby Orr rookie card for $300. The value of the card appreciates (increases) by 30% each year. 1. Complete a table of values to show the first five years of the investment. 2. Determine the common ratio for the successive terms. 3. Determine the equation of the exponential function that models this", "of $7$-groups is fully determined. Or if there are still more groups ... – Hagen von Eitzen Apr 25 '14 at 17:35 • Well, they may only be fully determined if there is only one if we only know the order of the group, if I understand you correctly. – Andrew Thompson Apr 25 '14 at 17:42 • For a group of order $105$, the number of $5$-Sylows is exactly $1$, same for $7$-Sylows. – Mikko Korhonen Apr 25 '14 at 17:47 • @MikkoKorhonen, why isn't $21$ an alternative for $5$-Sylow? $7 \\cdot 3 \\equiv 1 \\pmod 5$. – Andrew Thompson Apr 25 '14 at 19:45", "2017, 04:06 1 DavidTutorexamPAL wrote: Bunuel wrote: If John brought n comic books at$2 each, then brought m comic books at $7 each, and m- n comic books at$13 each, then the average (arithmetic mean) cost, in dollars per comic book, is equal to A. 22(m + n)/m B. (22m - 11n)/(2m) C. 11m D. (6m - 15n)/(2m - 2n) E. 22mn/(m + n) As all our values are small, we'll go for a straightforward calculation. This is a Precise approach. Total money spent is n*2 + m*7 + (m-n)*13 = m*20 - n*11 Total comic books purchased = n + m + (m-n) = 2m Average price per book is (m*20 - n*11)/2m. Bunuel are you sure the question is correct? If I'm not mistaken the coefficient of m should be 20 and not 22 in all relevant answers. (Or the question stem should be different) My mistake, you are right. Edited. Thank you. _________________ Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 3504 Location: India GPA: 3.5 Re: If John brought n comic books at $2 each, then brought m comic books [#permalink] ### Show Tags 28 Dec 2017, 09:30 Bunuel wrote: If John brought n comic books at$2 each, then brought m comic books at $7 each, and m-", "$$\\frac{2}{5}$$ ounces of beads. How many necklaces can she make if each uses 2 $$\\frac{7}{10}$$ ounces of beads? ______ necklaces 12 necklaces Explanation: Alexis has 32 $$\\frac{2}{5}$$ ounces of beads. If each uses 2 $$\\frac{7}{10}$$ ounces of beads, 32 $$\\frac{2}{5}$$ × 2 $$\\frac{7}{10}$$ 32 $$\\frac{2}{5}$$ = 162/5 2 $$\\frac{7}{10}$$ = 27/10 162/5 × 27/10 = 12 necklaces Question 4. Joseph has $32.40. He wants to buy several comic books that each cost$2.70. How many comic books can he buy? ______ comic books 12 comic books Explanation: Joseph has $32.40. He wants to buy several comic books that each cost$2.70. $32.40/$2.70 = 12 comic books Question 5. A rectangle is 2 $$\\frac{4}{5}$$ meters wide and 3 $$\\frac{1}{2}$$ meters long. What is its area? ______ $$\\frac{□}{□}$$ m2 9$$\\frac{4}{5}$$ m2 Explanation: 2 $$\\frac{4}{5}$$ = 14/5 3 $$\\frac{1}{2}$$ = 7/2 14/5 × 7/2 = 9 4/5 Question 6. A rectangle is 2.8 meters wide and 3.5 meters long. What is its area? ______ m2 9.8 m2 Explanation: A rectangle is 2.8 meters wide and 3.5 meters long. 2.8 × 3.5 = 9.8 ### Chapter 2 Review/Test – Page No. 131 Question 1. Write the values in order from least to greatest. Type below: __________ 0.45, 0.5, 5/8, 3/4 Explanation: 3/4 = 0.75 5/8 = 0.625 0.45, 0.5 0.45 < 0.5 < 0.625" ]

[ "equal in value? e) State the domain and range for each comic. 2. Paulette bought a Bobby Orr rookie card for$300. The value of the card appreciates (increases) by 30% each year. a) Complete a table of values to show the first five years of the investment. b) Determine the common ratio for the successive terms. c) Determine the equation of the exponential function that models this investment. 3. Due to the closure of the pulp and paper mill, the population of the small town is decreasing at a rate of 12% annually. If there are now 2400 people living in the town, what will the town's projected population be in eight years? 1. a) The purchase price of each comic book is the $y$ -intercept. The $y$ -intercept is the initial value of the books. The Spiderman comic book cost $30.00 and the Superman comic book cost$5.00. b) The Superman comic book shows exponential growth. c) The Spiderman comic book shows exponential decay. d) In 2005 both comic books were equal in value. The graphs intersect at approximately (5, $12.50), where 5 represents five years after the books were purchased. e) The domain and range for each comic is $D=\\{x| x \\in R\\}$ and $R=\\{y|y>0, \\ y \\in R\\}$ 2. $&300(.30)=90 && 390(.30)=117 && 507(.30)=152.10\\\\&300+90=390 && 390+117=507", "what year were both comic books equal in value? 5. State the domain and range for each comic. 1. The purchase price of each comic book is the \\begin{align*}y\\end{align*}-intercept. The \\begin{align*}y\\end{align*}-intercept is the initial value of the books. The Spiderman comic book cost30.00 and the Superman comic book cost $5.00. 2. The Superman comic book shows exponential growth. 3. The Spiderman comic book shows exponential decay. 4. In 2005 both comic books were equal in value. The graphs intersect at approximately (5,$12.50), where 5 represents five years after the books were purchased. 5. The domain and range for each comic is \\begin{align*}D=\\{x| x \\in R\\}\\end{align*} and \\begin{align*}R=\\{y|y>0, \\ y \\in R\\}\\end{align*} Paulette bought a Bobby Orr rookie card for 300. The value of the card appreciates (increases) by 30% each year. 1. Complete a table of values to show the first five years of the investment. 2. Determine the common ratio for the successive terms. 3. Determine the equation of the exponential function that models this investment. 1. First, calculate the values and create the table: \\begin{align*}&300(.30)=90 && 390(.30)=117 && 507(.30)=152.10\\\\ &300+90=390 && 390+117=507 && 507+152.10=659.10\\\\ \\\\ &659.10(.30)=197.73 && 856.83(.30)=257.05\\\\ &659.10+197.73=856.83 && 856.83+257.05=1113.88\\end{align*} \\begin{align*}&\\text{Time} \\ (years) \\ \\ 0 \\qquad 1 \\qquad 2 \\qquad \\quad 3 \\qquad \\quad 4 \\qquad \\quad \\ 5\\\\ &\\text{Value (\\)} \\qquad 300 \\quad", "$5 a piece, that's$260 gained for all 52 comics selling one copy. Compare that to 9 graphic novels (Got that number from supposing that they compile all 52 comics into 6 different graphic novel lines, leaving us with \"9 series) selling for a maximum of $20, for a total of$180 earned for selling 1 copy of all 9 series. On paper, the former \"magazine\" model method brings in more money, and that's all that comic publishers really care about. And, if they were to start following the \"graphic novel\" method, they'd probably fuck it up and demand that the novels sell for $30 a pop so that they can maintain their near$300 figure. And, keep in mind, manga volumes are selling for $10. >>311379 >It's based around digital monsters, that come from another dimension, and have a master that they befriend and serve in order to gain power so that they can fight bigger and bigger threats. And Spider-Man is about a young guy who gets bitten by a special spider and then gets spider-powers which he uses to fight crime. That doesn't mean Miles Morales is the same as Peter Parker. But at least at the start they followed the pretense of Miles trying to live up to (Ultimate) Peter's legacy. >And, what about the fact that", "comic. 2. Paulette bought a Bobby Orr rookie card for$300. The value of the card appreciates (increases) by 30% each year. a) Complete a table of values to show the first five years of the investment. b) Determine the common ratio for the successive terms. c) Use technology to determine the equation of the exponential function that models this investment. 3. Due to the closure of the pulp and paper mill, the population of the small town is decreasing at a rate of 12% annually. If there are now 2400 people living in the town, what will the town's projected population be in eight years. 1. a) The purchase price of each comic book is the $y$ -intercept. The $y$ -intercept is the initial value of the books. The Spiderman comic book cost $30.00 and the Superman comic book cost$5.00. b) The Superman comic book shows exponential growth. c) The Spiderman comic book shows exponential decay. d) In 2005 both comic books were equal in value. The graphs intersect at approximately (5, $12.50), where 5 represents five years after the books were purchased. e) The domain and range for each comic is $D=\\{x| x \\varepsilon R\\}$ and $R=\\{y|y>0, \\ y \\varepsilon R\\}$ 2. $&300(.30)=90 && 390(.30)=117 && 507(.30)=152.10\\\\&300+90=390 && 390+117=507 && 507+152.10=659.10\\\\\\\\&659.10(.30)=197.73 && 856.83(.30)=257.05\\\\&659.10+197.73=856.83 && 856.83+257.05=1113.88$ a) $&\\text{Time}", "# Moderate Permutation-Combination Solved QuestionAptitude Discussion Q. There are five comics numbered from 1 to 5. In how many ways can they be arranged, so that part-1 and part-3 are never together? ✖ A. 48 ✔ B. 72 ✖ C. 120 ✖ D. 210 Solution: Option(B) is correct The total number of ways in which 5 part can be arranged $= 5! =120$. The total number of ways in which part-1 and part-3 are always together: $= 4!×2!$ $= 48$ Therefore, the total number of arrangements, in which they are not together is: $=120-48$ $= \\textbf{72}$", "all? Sam has 6 toy spiders. Each spider has 8 legs. 6 × 8 = 48 Thus 48 legs are there in all. Question 4. Ms. Adams spends $42. She buys books that cost$6 each. How many books does she buy? Given, Ms. Adams spends $42. She buys books that cost$6 each. 42/6 = 7 Find the unknown number for each Fast Array drawing. Question 5.", "# Thread: How many possible collection of toys? 1. ## How many possible collection of toys? John will be given 4 chances to randomly pick toys from a bag containing 2 Ironmans, 3 Supermans, and 2 Batmans. How many possible collection of toys can he pull out? HELP please ... I tried to solve this many times but can't get the correct answer. BTW, 35 is the answer. 2. ## Re: How many possible collection of toys? Hello, Kaloda! John will be given 4 chances to randomly pick toys from a bag containing: 2 Ironmans, 3 Supermans, and 2 Batmans. How many possible collections of toys can he pull out? There is no formula for this problem. We must construct an exhaustive list. $\\text{Let }(x,y,z)\\text{ represent: }x\\text{ Ironmans, }y\\text{ Supermans, }z\\text{ Batmans.}$ $\\begin{array}{ccccccc}(2,1,1): & {2\\choose2}{3\\choose1}{2\\choose1} &=& 1\\cdot3\\cdot2 &=& 6 \\\\ \\\\[-4mm] (1,2,1): & {2\\choose1}{3\\choose2}{2\\choose1} &=& 2\\cdot3\\cdot2 &=& 12 \\\\ \\\\[-4mm] (1,1,2): & {2\\choose1}{3\\choose1}{2\\choose2} &=& 2\\cdot3\\cdot1 &=& 6 \\\\ \\\\ (2,2,0): & {2\\choose2}{3\\choose2}{2\\choose0} &=& 1\\cdot3\\cdot1 &=& 3 \\\\ \\\\[-4mm] (2,0,2): & {2\\choose2}{3\\choose0}{2\\choose2} &=& 1\\cdot1\\cdot1 &=& 1 \\\\ \\\\[-4mm] (0,2,2): & {2\\choose0}{3\\choose2}{2\\choose2} &=& 1\\cdot3\\cdot1 &=& 3 \\\\ \\\\ (1,3,0): & {2\\choose1}{3\\choose3}{2\\choose0} &=& 2\\cdot1\\cdot1 &=& 2 \\\\ \\\\[-4mm] (0,3,1): & {2\\choose0}{3\\choose3}{2\\choose1} &=& 1\\cdot1\\cdot2 &=& 2 \\\\ \\hline &&& \\text{Total:} && 35 \\end{array}$ 3. ## Re: How many possible collection of toys?", "= 95$$ The expression to represent shipping cost to each area comes from multiplying the cost per book by the number of books. $$US \\,shipping = 4.90p$$ $$Int’l\\, shipping = 16.50(200 – p)$$ Add the three elements of the cost together to write an expression representing the total cost to distribute the books. $$95 + 4.90p + 16.50(200 – p)$$ Use the distributive property. $$95 + 4.90p +3300 – 16.50p$$ Combine like terms. $$3395 – 11.60p$$ Answer: $$3395 – 11.60p$$ Sarina can use this polynomial to find out the cost of shipping $$200$$ books when $$p$$ of them are shipped within the U.S. (domestically). Example Problem A carpet designer creates a carpet that uses four colors according to the pattern and dimensions below. Express the area of the carpet as a polynomial. Find the area of each colored piece of carpet by multiplying the length by the width. Red: $$x \\cdot x = x^{2}$$ Blue: $$2y(x + 3y) = 2xy + 6y^{2}$$ White: $$2y \\cdot x = 2xy$$ Green: $$x(x + 3y) = x^{2} + 3x$$ Find the area of the whole carpet by combining the areas of the four pieces. $$x^{2} + 2xy + 6y^{2} + 2xy + x^{2} + 3xy$$ Combine like terms. $$2x^{2} + 7xy + 6y^{2}$$ Answer: $$2x^{2} + 7xy + 6y^{2}$$ Summary One", "# Permutation Question #### Bushy ##### Member There are three separate bundles of reading material comprising of 4 comics, 2 novels and 3 magazines. They are placed together to form one pile. In how many ways can this be done if there are no restrictions on where the individual items are to be placed? I say 9! = 362880 Determine the number of permutations if the order of the comic books in each bundle does not change I'm not sure what they mean by 'each bundle' here #### Jameson If you're arranging these into one pile and the order of the comic books doesn't change then I would say it's $$\\displaystyle \\frac{9!}{4!}$$. Dividing by $4!$ accounts for all the different ways you can order the comic books but keep everything else the same. Since we want just one order I would say this is the answer but again this question is worded in a way where I'm not sure.", "Share your story! « Reply #3865 on: April 10, 2013, 12:44:19 PM » My husband and I collect comic books. The closest shop is about 40 miles away, but a local book store stocks News stand covers. These are usually worth a bit more than the other first prints for the book, so we'll go in there and pick up any that are still in good condition- the book store is rough on their comics, so more often than not, we leave without getting anything. One day we went in and found a really nice copy of one comic and decided to pick it up. We took it up to the cashier and she rang us up, put the book in the bag, and grabbed our receipt. She then proceeded to put the receipt on top of the bag to show up the link to a survey at the bottom. It was one of those \"rate the service\" kind of things... And she took her pen... And said \"make sure you give me a 10!\" And drew the number 10 and circled it on top of the comic book. Which effectively left a giant 10 indent on the front of the book. Comic book grading is incredibly strict...and her drawing on the book had severely reduced the value", "State the domain and range for each comic. 2. Paulette bought a Bobby Orr rookie card for300. The value of the card appreciates (increases) by 30% each year. a) Complete a table of values to show the first five years of the investment. b) Determine the common ratio for the successive terms. c) Use technology to determine the equation of the exponential function that models this investment. 3. Due to the closure of the pulp and paper mill, the population of the small town is decreasing at a rate of 12% annually. If there are now 2400 people living in the town, what will the town's projected population be in eight years. 1. a) The purchase price of each comic book is the \\begin{align*}y\\end{align*}-intercept. The \\begin{align*}y\\end{align*}-intercept is the initial value of the books. The Spiderman comic book cost $30.00 and the Superman comic book cost$5.00. b) The Superman comic book shows exponential growth. c) The Spiderman comic book shows exponential decay. d) In 2005 both comic books were equal in value. The graphs intersect at approximately (5, 12.50), where 5 represents five years after the books were purchased. e) The domain and range for each comic is \\begin{align*}D=\\{x| x \\varepsilon R\\}\\end{align*} and \\begin{align*}R=\\{y|y>0, \\ y \\varepsilon R\\}\\end{align*} 2. \\begin{align*}&300(.30)=90 && 390(.30)=117 && 507(.30)=152.10\\\\ &300+90=390 && 390+117=507 && 507+152.10=659.10\\\\", "spend on both comic books and magazines. b) How much does Peter need to spend if he wants 9 comic books and 3 magazines? c) Write a new expression for Peter's expenditure on books if the bookstore raises the price of a comic book to $15 and a magazine to$10.", "$$\\frac{2}{5}$$ ounces of beads. How many necklaces can she make if each uses 2 $$\\frac{7}{10}$$ ounces of beads? ______ necklaces 12 necklaces Explanation: Alexis has 32 $$\\frac{2}{5}$$ ounces of beads. If each uses 2 $$\\frac{7}{10}$$ ounces of beads, 32 $$\\frac{2}{5}$$ × 2 $$\\frac{7}{10}$$ 32 $$\\frac{2}{5}$$ = 162/5 2 $$\\frac{7}{10}$$ = 27/10 162/5 × 27/10 = 12 necklaces Question 4. Joseph has $32.40. He wants to buy several comic books that each cost$2.70. How many comic books can he buy? ______ comic books 12 comic books Explanation: Joseph has $32.40. He wants to buy several comic books that each cost$2.70. $32.40/$2.70 = 12 comic books Question 5. A rectangle is 2 $$\\frac{4}{5}$$ meters wide and 3 $$\\frac{1}{2}$$ meters long. What is its area? ______ $$\\frac{□}{□}$$ m2 9$$\\frac{4}{5}$$ m2 Explanation: 2 $$\\frac{4}{5}$$ = 14/5 3 $$\\frac{1}{2}$$ = 7/2 14/5 × 7/2 = 9 4/5 Question 6. A rectangle is 2.8 meters wide and 3.5 meters long. What is its area? ______ m2 9.8 m2 Explanation: A rectangle is 2.8 meters wide and 3.5 meters long. 2.8 × 3.5 = 9.8 ### Chapter 2 Review/Test – Page No. 131 Question 1. Write the values in order from least to greatest. Type below: __________ 0.45, 0.5, 5/8, 3/4 Explanation: 3/4 = 0.75 5/8 = 0.625 0.45, 0.5 0.45 < 0.5 < 0.625", "see now. The same reasoning is used to solve b) (permutations again), which comes to $1 \\times 1 \\times 8 \\times 7 \\times 6 \\times 5 = 1,680$ combinations times $6 * 5$ ways to permute each combination, so the answer is $30 * 1680 = 50,400$. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. a row from a group of 10 people where the bride and groom are part of the group of 10 such that (a) at least one of them (the bride and the groom) must be in the pic? Course Hero is not sponsored or endorsed by any college or university. To learn more, see our tips on writing great answers. How easy is it to recognize that a creature is under the Dominate Monster spell? Bride and groom take two places each of 1 way. Two people to the … Answer = 2c1 * 8c5 * 6! Making statements based on opinion; back them up with references or personal experience. Thus, there are $80640=(2\\cdot6)\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4$ permutations in this case. How many ways can a photographer at a wedding arrange 6 people in. As you noted, \"arrange ... in a row\" indicates that order does matter, i.e., we are counting permutations. Mathematics Stack Exchange is a question and", "Question 7. Model with Mathematics A comic store sells two rare comic books for$1,824. The price of one comic book is 11 times as much as the price of the other. • Write an equation to model the price of the less expensive comic book. Then find the price. • Write an equation to model the price of the more expensive comic book. Then find the price. L + 11 L = $1,824, The price of the less expensive comic book is$152, E = 11 L, The price of the more expensive comic book is $1,672, Explanation: Given a comic store sells two rare comic books for$1,824. L + E = $1,824, Equation to model the price of the less expensive comic book is L + 11 L =$1,824, 12L = $1,824, L =$1,824 ÷ 12 = $152, The equation to model the price of the more expensive comic book. E = 11 L, as L =$152 , E = 11 X 152 = \\$1,672.", "value to place the first digit. The first digit of the quotient will be in the hundreds place. STEP 2 Divide the hundreds. STEP 3 Divide the tens. STEP 4 Divide the ones. Question 8. The flower shop received a shipment of 248 pink roses and 256 red roses. The shop owner uses 6 roses to make one arrangement. How many arrangements can the shop owner make if he uses all the roses? _____ arrangement Answer: 84 arrangements Explanation: Number of pink roses = 248 Number of red roses = 256 Total number of roses = 504 Number of roses in each arrangement = 6 Number of arrangements = 504 ÷ 6 = 84 ### Page No. 262 Use the table for 9–11. Question 9. Four teachers bought 10 origami books and 100 packs of origami paper for their classrooms. They will share the cost of the items equally. How much should each teacher pay? _____$ Answer: $210 Explanation: Number of origami books = 10 Cost of each origami book =$24 Total cost of origami books = $24 x 10 =$240 Number of origami papers = 100 Cost of each origami book = $6 Total cost of origami books =$6 x 100 = $600 Total cost of items =$240 + $600 =$840 Number of teachers = 4 Cost", "Level 1 PSLE The total cost of a pencil and a book is $35. The cost of the pencil is 25 the cost of the book. What is the cost of the book? 2 m Level 2 PSLE The average mass of 3 boys, Abel, Benny, Charlie, is 44 kg. Abel is heavier than Benny by 6 kg. The total mass of Abel and Benny is the same as the mass of Charlie. What is the mass of Abel? 2 m Level 2 PSLE May had some money. She used 14 of it on a dress and 16 of it on a gift. The dress and the gift cost$133.50 altogether. How much money had she left? 2 m Level 2 PSLE Derrick paid $153 for 3 shirts and 1 sweater. The price of a shirt was 38 of the price of the sweater. How much did Derrick pay for the sweater? 2 m Level 2 PSLE The pie chart shows the different genres of CDs at a stall. 15 of the CDs were Hip-Hop and 14 of the CDs were either Rock or Jazz. There were 3 times as many Rock CDs as Jazz CDs. What fraction of the CDs were Rock CDs? 2 m Level 2 A handbag cost$6 more than a dress but $10 less than", "## Boring Book After finishing a very boring book, Fred notices that the page numbers contained exactly $29$ zeros and exactly $137$ ones. The book started with page $1$. How many pages were in the book? Source: NCTM Mathematics Teacher, December 2005 SOLUTION We arrange the numbers in rows of $10$ numbers as follows $1,2,3,\\cdots,9$ $10,11,12,13,\\cdots,19$ $20,21,22,23,\\cdots,29$ $30,31,32,33,\\cdots,39$ $\\cdots$ $90,91,92,93,\\cdots,99$ $100,101,102,103,\\cdots,109$ $110,111,112,113,\\cdots,119$ $\\cdots$ $180,181,182,183,\\cdots,189$ $190,191,192,193,\\cdots,199$ Each row starting from row $10$ contains $1$ zero except row $100$ which contains $11$ zeros. The number of rows containing $1$ zero is $18$. The total number of zeros from page $1$ to page $199$ is $18\\times 1+1\\times 11=29$ Thus, the book has at most $199$ pages. On the other hand, $9$ rows have $1$ one, $10$ rows have $11$ ones, while row $110$ has $21$ ones. The total number of ones from page $1$ to page $199$ is $9\\times 1+10\\times 11+1\\times 21=140$ Since the book has $137$ ones, the last page must be $196$. The book has $196$ pages. Answer: $196$", "$$\\PageIndex{3}$$ A certain shirt comes in four sizes and six colors. One also has the choice of a dragon, an alligator, or no emblem on the pocket. How many different shirts could you order? $$72=4\\cdot 6\\cdot 3$$ Exercise $$\\PageIndex{4}$$ A builder of modular homes would like to impress his potential customers with the variety of styles of his houses. For each house there are blueprints for three different living rooms, four different bedroom configurations, and two different garage styles. In addition, the outside can be finished in cedar shingles or brick. How many different houses can be designed from these plans? Exercise $$\\PageIndex{5}$$ The Pi Mu Epsilon mathematics honorary society of Outstanding University wishes to have a picture taken of its six officers. There will be two rows of three people. How many different way can the six officers be arranged? $$720=6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1$$ Exercise $$\\PageIndex{6}$$ An automobile dealer has several options available for each of three different packages of a particular model car: a choice of two styles of seats in three different colors, a choice of four different radios, and five different exteriors. How many choices of automobile does a customer have? Exercise $$\\PageIndex{7}$$ A clothing manufacturer has put out a mix-and-match collection consisting of two blouses, two pairs of pants, a skirt,", "the order of events until a year later when all the story arcs are done and you can look at them in retrospect. It's all fucked and they need to cut back massively and also just ditch the magazine model entirely. Also, part of the issue is that you really don't need to read \"everything\" that came before, but you do need a certain level of knowledge of past stuff. And good luck finding where precisely that knowledge should come from. Once you reach a certain point you can read most of whatever and understand it, but good luck reaching that point. > If they sell 52 comics for$5 a piece, that's $260 gained for all 52 comics selling one copy. Compare that to 9 graphic novels (Got that number from supposing that they compile all 52 comics into 6 different graphic novel lines, leaving us with \"9 series I looked at it with the assumption that a graphic novel is about the length of six issues, which is basically the standard right now. Instead of each series doing 12 issues per year (though really they tend to do 13 with annuals, and the annuals are a little longer), they could say each of those 52 series does two graphic novels per year, or 104 graphic novels per" ]

[ "+0 # Complementary Counting Problem! 0 242 2 +128 The probability of snow for each of the next three days is 2/3. What is the probability that it will snow at least once during those three days? Express your answer as a common fraction. Thanks for the help guys! Mar 20, 2019 edited by EvanWei123 Mar 20, 2019 #1 +391 +2 By using Complementary Counting, you can find the probability that no days snow, and subtract that from $$1$$. The probability that no day snows is $$\\frac{1}{3}*\\frac{1}{3}*\\frac{1}{3}=\\frac{1}{27}$$, so the probability of it snowing at least one day is $$1 - \\frac{1}{27}= \\frac{26}{27}$$. - Daisy Mar 20, 2019 #2 +128 0 Thank You Daisy! Mar 23, 2019", "+0 # The probability of snow for each of the next three days is $\\frac{2}{3}$. What is the probability that it will snow at least once during tho 0 134 3 The probability of snow for each of the next three days is . What is the probability that it will snow at least once during those three days? Express your answer as a common fraction. Aug 9, 2021 #1 +1 Case 1: once: 2/3 * 1/3 * 1/3 * 3 Case 2: twice: 2/3 * 2/3 * 1/3 * 3 C 2 case 3: 3 times: (2/3)^3 2/27*3 + 4/27*3 + 8/27 = 6/27 + 12/27 + 8/27 = 26/27. Aug 9, 2021 #2 +23 +2 We could use Complementary Counting for this. The probability that it will not slow at all is $$\\frac{1}{3} \\times \\frac{1}{3} \\times \\frac{1}{3} = \\frac{1}{27}$$ Then we minus the total by $$\\frac{1}{27}$$ $$1 - \\frac{1}{27} = \\frac{26}{27}$$ Aug 9, 2021 #3 +1 we got the same answer, in different ways :D Guest Aug 9, 2021", "# Math Help - Impossible Question 1. ## Impossible Question Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out. if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day? It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas? 2. Originally Posted by megno Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out. if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day? It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas? Suppose that the probability that it snows on any day is p. Now consider today. There is a probability", "2581 Re: If the probability of snow is the same every day this week, what is th [#permalink] ### Show Tags 03 Nov 2017, 08:31 Top Contributor Bunuel wrote: If the probability of snow is the same every day this week, what is the probability of snow on Saturday? (1) The probability of snow on Monday is 1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. Target question: What is the probability of snow on Saturday? Given: The probability of snow is the same every day this week Statement 1: The probability of snow on Monday is 1/7. If the probability of snow is THE SAME every day this week, then the probability of snow on Saturday is also 1/7 Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: The probability having of no snow on both Thursday and Friday is 36/49. Let p = probability of snow on Saturday. Since the probability of snow is the same every day this week, we know that p = probability of snow on Thursday and p = probability of snow on Friday. Also, (1-p) = probability of NO SNOW on Thursday and (1-p) = probability of NO SNOW on Friday. Statement 2 says: P(no snow on Thursday AND", "marble is $P((G,G))$ and we know it’s probability. Consequently, $\\{$we didn’t pick a single blue marble$\\}=1-P((G,G))=\\displaystyle{1-\\frac{6}{55}=\\frac{49}{55}}$ #### Example If it snows on a given day, the probability that it snows the next day is $\\frac{1}{3}$. If it doesn’t, the probability that it snows the next day is $\\frac{1}{6}$. The probability it will snow tomorrow is $\\frac{2}{5}$. What is the probability that it will snow the day after tomorrow? To illustrate all the possible outcomes, we’ll use tree diagram. First of all, we need to determine what happens tomorrow. The probability it will snow tomorrow is $\\frac{2}{5}$, so the probability that it won’t snow tomorrow is $\\frac{3}{5}$. Consequently, our starting point will be today, and first two events will be about tomorrow. If it snows on a given day, the probability it will snow the next day is $\\frac{1}{3}$. Alternatively, the probability it won’t snow the next day is $\\frac{2}{3}$. Furthermore, if it doesn’t snow on a given day, the probability it will snow the next day is $\\frac{1}{6}$. Consequently, the probability it won’t snow the next day is $\\frac{5}{6}$. When we put all the events and their probabilities in the tree diagram, we get: All we need to do is calculate the probability that it will snow the day after tomorrow. $P($snow on the day after tomorrow$)=P($snow, snow$)$", "# The probability of it snowing in neverland is 1/3. what is the probability that it will snow monday and not snow tuesday As we all know, snowstorms frequently last more than one day, sometimes 2, sometimes 3. This means that probability of snowing Monday is not independent from the probability of snowing the next day (a Tuesday). Assuming in Neverland, the snowstorms only last one day, and the probability of snowing is completely INDEPENDENT from one day to another, then we can use the multiplication rule Probability of snowing Monday AND Tuesday =Probability of snowing Monday * Probability of snowing Tuesday =P(Monday)*P(Tuesday) =(1/3)*(1/3) =1/9 (Only in Neverland!) Only authorized users can leave an answer!", "+0 0 276 6 +128 On each of the first three days of January, there is a 1/3 chance that it will snow where Bob lives. On each of the next four days there is a 1/4 chance that it will snow. What is the probabilty that it snows at least once during the first week of January? Mar 18, 2019 #1 +128 0 oops im sorry. I double posted. Mar 18, 2019 #2 +6045 +1 . Mar 18, 2019 edited by Rom Mar 18, 2019 #3 +954 +3 Let's find the chance that it *doesn't* snow. There's a 2/3 chance for the first three days and 3/4 chance for the last four. So... $$(\\frac{2}{3})^3\\times (\\frac{3}{4})^4=\\frac{3}{32}$$ Now, $$1-\\frac{3}{32}=\\frac{29}{32}$$ so there is a 29/32 chance it will snow at least once. You are very welcome! :P Mar 18, 2019 #4 +2 Hi Rom Can you explain to me why the following reasoning is wrong ? The probability of snow on the first day is 1/3, so the probability that it doesn't snow is 2/3. Same for the next two days, so the probability of no snow during the first three days is (2/3)^3. Similarly the probability of no snow during the next four days is (3/4)^4. So the probablity of no snow during the week is (2/3)^3.(3/4)^4 = 3/32.", "We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy. +0 # Complementary Counting Problem! 0 191 2 +93 The probability of snow for each of the next three days is 2/3. What is the probability that it will snow at least once during those three days? Express your answer as a common fraction. Thanks for the help guys! Mar 20, 2019 edited by EvanWei123 Mar 20, 2019 ### 2+0 Answers #1 +391 +2 By using Complementary Counting, you can find the probability that no days snow, and subtract that from $$1$$. The probability that no day snows is $$\\frac{1}{3}*\\frac{1}{3}*\\frac{1}{3}=\\frac{1}{27}$$, so the probability of it snowing at least one day is $$1 - \\frac{1}{27}= \\frac{26}{27}$$. - Daisy Mar 20, 2019 #2 +93 0 Thank You Daisy! Mar 23, 2019", "+0 # probability +1 100 1 On each of the first three days of January, there is a 1/2 chance that it will snow where Bob lives. On each of the next four days, there is a 1/3 chance that it will snow. What is the probability that it snows at least once during the first week of January? Jan 13, 2021 #1 +283 +1 We can use complementary counting for this problem by calculating the probability that it will not snow at all first. That is equal to $$(\\frac{1}{2})^3\\cdot(\\frac{2}{3})^4=\\frac{1}{8}\\cdot\\frac{16}{81}=\\frac{2}{81}$$ But we're trying to find the probability that it will snow at least once, so the answer is $$1-\\frac{2}{81}=\\frac{79}{81}$$ Here's a very similar problem: https://web2.0calc.com/questions/please-help-complementry-probabilty-problem Jan 13, 2021", "## A community for students. Sign up today Here's the question you clicked on: ## anonymous 5 years ago the probability that it would snow on sunday is 3/5 what is the probabilty that it would snow on monday, if it snowed on sunday ? • This Question is Closed 1. anonymous there must be some more information 2. anonymous ooh sorry the probability that it would snow on sunday is 3/5 . the probability that it will snow on both sunday and monday is 3/10 . what is the probabilty that it would snow on monday, if it snowed on sunday 3. anonymous $P(A|B)=\\frac{P(A\\cap B)}{P(B)}$ 4. anonymous $P(A|B)=\\frac{\\frac{3}{10}}{\\frac{3}{5}}=\\frac{3}{10}\\times \\frac{5}{3}=\\frac{1}{2}$ 5. anonymous thank you 6. anonymous welcome #### Ask your own question Sign Up Find more explanations on OpenStudy Privacy Policy", "## Algebra and Trigonometry 10th Edition $0.1$ The probability that it will not snow is the complement of the probability that it will snow. Thus $P(no \\ snow)=1-P(snow)=1-0.9=0.1$.", "# Math Help - Probability test question 1. ## Probability test question I had this question on a test: The probability of snowing in one city is 0.4, and the probability of snowing in another city is 0.7. Assume independence. What is the probability that it snows in exactly one of the two cities? The way I approached it was to say that the probability that it snows in exactly one of the two cities is: Let A denote probability of snow in the first city. Let B denote probability of snow in the second city. P(A and 'not B') or P('not A' and B) = (0.4)(0.3) + (0.6)(0.7) - (0.4)(0.3)(0.6)(0.7) = 0.4896 Is this correct? 2. There are four possibilities: It snows in neither city P1 = (0.6)(0.3) It snows in both cities P2 = (0.4)(0.7) It snows in the first city only P3 = (0.4)(0.3) It snows in the second city only P4 = (0.6)(0.7) The probability that it snows in exactly one city is simply P3 + P4. 3. Originally Posted by BrownianMan The probability of snowing in one city is 0.4, and the probability of snowing in another city is 0.7. Assume independence. What is the probability that it snows in exactly one of the two cities? Let A denote probability of snow in", "+0 Probability 0 146 3 On each of the first three days of January, there is a 1/2 chance that it will snow where Bob lives. On each of the next four days, there is a 1/3 chance that it will snow. What is the probability that it snows at least once during the first week of January? Apr 14, 2022 #1 +2540 +1 It is easier to count for what we don't what and subtract that from 1. The probability that there will 0 days of snow is: $${1 \\over 2}^3 \\times {2 \\over 3}^4 = {2 \\over 81}$$ Thus, the probability that it will snow for at least 1 day of the week is: $$1 - {2 \\over 81} = \\color{brown}\\boxed{79 \\over 81}$$ Apr 14, 2022 #2 +579 +1 Why are there so many counting and prob questions today???? Vinculum Apr 14, 2022", "or $P($no snow, snow$)$. To calculate the probability of these two events we simply multiply along the branches. $P($snow, snow$)= \\displaystyle{\\frac{2}{15}}$, $P($no snow, snow$)=\\displaystyle{\\frac{1}{10}}$ Similarly as before, ‘or’ means $+$ since these are disjoint events. To get the probability of snow on the day after tomorrow, we add up probabilities of the events above. Therefore, $P($snow on the day after tomorrow$)= \\displaystyle{\\frac{2}{15} +\\frac{1}{10}=\\frac{7}{30}}$", "+0 0 92 1 On each of the first three days of January, there is a $\\frac{1}{3}$ chance that it will snow where Bob lives. On each of the next four days, there is a $\\frac{1}{4}$ chance that it will snow. What is the probability that it snows at least once during the first week of January? Jul 12, 2021", "### Show Tags 05 Aug 2016, 05:24 Bunuel wrote: If the probability of snow is the same every day this week, what is the probability of snow on Saturday? (1) The probability of snow on Monday is 1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. From stimulus P (Snow )= $$\\frac{1}{7}$$ (same on Sun, Mon, Tue, Wed, Thur, Fri, Sat) P (No Snow)= $$\\frac{6}{7} ($$same on Sun, Mon, Tue, Wed, Thur, Fri, Sat) AIM :- FIND P(Snow) on Saturday (1) The probability of snow on Monday is$$\\frac{1}{7}$$ Since P(Snow) on Monday = P (Snow) on Saturday So P(Snow) on Saturday = $$\\frac{1}{7}$$ SUFFICIENT (2) The probability having of no snow on both Thursday and Friday is $$\\frac{36}{49}$$ We know from Stimulus that P (No snow) on Thursday= P (No Snow) on Friday therefore P (No Snow) Thursday and P(No Snow)Friday $$= \\frac{36}{49} = \\frac{6}{7}*\\frac{6}{7}$$ Therefore P (Snow) on thursday or friday or saturday (since it is same)$$= 1-\\frac{6}{7} = \\frac{1}{7}$$ SUFFICIENT _________________ Posting an answer without an explanation is \"GOD COMPLEX\". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired. CEO Joined: 12 Sep 2015 Posts:", "the first city. Let B denote probability of snow in the second city. P(A and 'not B') or P('not A' and B) = (0.4)(0.3) + (0.6)(0.7) - (0.4)(0.3)(0.6)(0.7) = 0.4896 Is this correct? No the part in red is not correct. Think about it. Can A and notA both happen?", "is having only two events, that is, red and white. The event of stepping on to a red square cannot be disintegrated any further. b) These are mutually exclusive events as they cannot happen together. Also, they are mutually exhaustive events as they together equals to the whole sample space. Getting a head or getting tail, separately, are simple events. c) The event is $\\{red\\ top,\\ blue\\ jeans\\}$. It can be further disintegrated to simpler events $\\{red\\ top\\}$ and $\\{blue jeans\\}$. Hence, it is a compound event. Problem 3: The probability of having snowfall in London today is $0.3$. What will be the probability that there is no snowfall in London today? Solution: To have a snowfall and to not have a snowfall are complementary events. Hence, the probability of not having snowfall is the complement of the probability of having snowfall. Probability of not having snowfall = $1\\ -\\ 0.3$ = $0.7$ Problem 4: There are $2$ red and $6$ green balls in a bag. If two balls are taken out one by one with replacement what is the probability that we get $1$ red and $1$ green ball. Solution: These two events are independent of each other. Probability of getting a red ball, $P(A)$ = $\\frac{2}{8}$ Probability of getting a red ball, $P(B)$ = $\\frac{6}{8}$ Probability", "{.188} & {.406} \\\\ {.391} & {.203} & {.406} \\\\ \\end{array} } \\right]$ So if it rains today, then in three days there is a probability of .203 it will be nice. If it snows today it will snow in three days is .406. But you see it all becomes the same (steady state) on the sixth day and thereafter. Does that help? 5. Oh... so that means once the steady state vector is reached then prediction is not possible... is that correct? Also, can you help me understand how the matrix is being constructed in the example given in link Markov Chains Ex. 2: Another Weather Forecast thanks once again... 6. Well yes it is possible. It is just the same as on the sixth day hence. 7. Thanks!! I think I have got the example now... Thanks a lot again", "1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. _________________ Math Expert Joined: 02 Aug 2009 Posts: 5938 Re: If the probability of snow is the same every day this week, what is th [#permalink] ### Show Tags 28 Feb 2016, 19:06 Bunuel wrote: If the probability of snow is the same every day this week, what is the probability of snow on Saturday? (1) The probability of snow on Monday is 1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. Hi, info :- prob is the same each day lets see the statements:- (1) The probability of snow on Monday is 1/7. straight forward. prob is the same every day, so on saturday too prob=1/7 Suff (2) The probability having of no snow on both Thursday and Friday is 36/49. let the Prob of snowing be x, so prob of not snowing= 1-x.. so on two days= (1-x)(1-x)=36/49.. (1-x)^2=(6/7)^2.. so 1-x==6/7 or -6/7.. but xcannot be>1,so 1-x=6/7 or x=1/7 Suff D _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html GMAT online Tutor Director Joined: 04 Jun 2016 Posts: 609 GMAT 1: 750 Q49 V43 Re: If the probability of snow is the same every day this week, what is th [#permalink]" ]

[ "+0 # Complementary Counting Problem! 0 242 2 +128 The probability of snow for each of the next three days is 2/3. What is the probability that it will snow at least once during those three days? Express your answer as a common fraction. Thanks for the help guys! Mar 20, 2019 edited by EvanWei123 Mar 20, 2019 #1 +391 +2 By using Complementary Counting, you can find the probability that no days snow, and subtract that from $$1$$. The probability that no day snows is $$\\frac{1}{3}*\\frac{1}{3}*\\frac{1}{3}=\\frac{1}{27}$$, so the probability of it snowing at least one day is $$1 - \\frac{1}{27}= \\frac{26}{27}$$. - Daisy Mar 20, 2019 #2 +128 0 Thank You Daisy! Mar 23, 2019", "We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy. +0 # Complementary Counting Problem! 0 191 2 +93 The probability of snow for each of the next three days is 2/3. What is the probability that it will snow at least once during those three days? Express your answer as a common fraction. Thanks for the help guys! Mar 20, 2019 edited by EvanWei123 Mar 20, 2019 ### 2+0 Answers #1 +391 +2 By using Complementary Counting, you can find the probability that no days snow, and subtract that from $$1$$. The probability that no day snows is $$\\frac{1}{3}*\\frac{1}{3}*\\frac{1}{3}=\\frac{1}{27}$$, so the probability of it snowing at least one day is $$1 - \\frac{1}{27}= \\frac{26}{27}$$. - Daisy Mar 20, 2019 #2 +93 0 Thank You Daisy! Mar 23, 2019", "# Math Help - Impossible Question 1. ## Impossible Question Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out. if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day? It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas? 2. Originally Posted by megno Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out. if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day? It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas? Suppose that the probability that it snows on any day is p. Now consider today. There is a probability", "+0 # probability +1 100 1 On each of the first three days of January, there is a 1/2 chance that it will snow where Bob lives. On each of the next four days, there is a 1/3 chance that it will snow. What is the probability that it snows at least once during the first week of January? Jan 13, 2021 #1 +283 +1 We can use complementary counting for this problem by calculating the probability that it will not snow at all first. That is equal to $$(\\frac{1}{2})^3\\cdot(\\frac{2}{3})^4=\\frac{1}{8}\\cdot\\frac{16}{81}=\\frac{2}{81}$$ But we're trying to find the probability that it will snow at least once, so the answer is $$1-\\frac{2}{81}=\\frac{79}{81}$$ Here's a very similar problem: https://web2.0calc.com/questions/please-help-complementry-probabilty-problem Jan 13, 2021", "+0 # The probability of snow for each of the next three days is $\\frac{2}{3}$. What is the probability that it will snow at least once during tho 0 134 3 The probability of snow for each of the next three days is . What is the probability that it will snow at least once during those three days? Express your answer as a common fraction. Aug 9, 2021 #1 +1 Case 1: once: 2/3 * 1/3 * 1/3 * 3 Case 2: twice: 2/3 * 2/3 * 1/3 * 3 C 2 case 3: 3 times: (2/3)^3 2/27*3 + 4/27*3 + 8/27 = 6/27 + 12/27 + 8/27 = 26/27. Aug 9, 2021 #2 +23 +2 We could use Complementary Counting for this. The probability that it will not slow at all is $$\\frac{1}{3} \\times \\frac{1}{3} \\times \\frac{1}{3} = \\frac{1}{27}$$ Then we minus the total by $$\\frac{1}{27}$$ $$1 - \\frac{1}{27} = \\frac{26}{27}$$ Aug 9, 2021 #3 +1 we got the same answer, in different ways :D Guest Aug 9, 2021", "marble is $P((G,G))$ and we know it’s probability. Consequently, $\\{$we didn’t pick a single blue marble$\\}=1-P((G,G))=\\displaystyle{1-\\frac{6}{55}=\\frac{49}{55}}$ #### Example If it snows on a given day, the probability that it snows the next day is $\\frac{1}{3}$. If it doesn’t, the probability that it snows the next day is $\\frac{1}{6}$. The probability it will snow tomorrow is $\\frac{2}{5}$. What is the probability that it will snow the day after tomorrow? To illustrate all the possible outcomes, we’ll use tree diagram. First of all, we need to determine what happens tomorrow. The probability it will snow tomorrow is $\\frac{2}{5}$, so the probability that it won’t snow tomorrow is $\\frac{3}{5}$. Consequently, our starting point will be today, and first two events will be about tomorrow. If it snows on a given day, the probability it will snow the next day is $\\frac{1}{3}$. Alternatively, the probability it won’t snow the next day is $\\frac{2}{3}$. Furthermore, if it doesn’t snow on a given day, the probability it will snow the next day is $\\frac{1}{6}$. Consequently, the probability it won’t snow the next day is $\\frac{5}{6}$. When we put all the events and their probabilities in the tree diagram, we get: All we need to do is calculate the probability that it will snow the day after tomorrow. $P($snow on the day after tomorrow$)=P($snow, snow$)$", "# The probability of it snowing in neverland is 1/3. what is the probability that it will snow monday and not snow tuesday As we all know, snowstorms frequently last more than one day, sometimes 2, sometimes 3. This means that probability of snowing Monday is not independent from the probability of snowing the next day (a Tuesday). Assuming in Neverland, the snowstorms only last one day, and the probability of snowing is completely INDEPENDENT from one day to another, then we can use the multiplication rule Probability of snowing Monday AND Tuesday =Probability of snowing Monday * Probability of snowing Tuesday =P(Monday)*P(Tuesday) =(1/3)*(1/3) =1/9 (Only in Neverland!) Only authorized users can leave an answer!", "+0 0 276 6 +128 On each of the first three days of January, there is a 1/3 chance that it will snow where Bob lives. On each of the next four days there is a 1/4 chance that it will snow. What is the probabilty that it snows at least once during the first week of January? Mar 18, 2019 #1 +128 0 oops im sorry. I double posted. Mar 18, 2019 #2 +6045 +1 . Mar 18, 2019 edited by Rom Mar 18, 2019 #3 +954 +3 Let's find the chance that it *doesn't* snow. There's a 2/3 chance for the first three days and 3/4 chance for the last four. So... $$(\\frac{2}{3})^3\\times (\\frac{3}{4})^4=\\frac{3}{32}$$ Now, $$1-\\frac{3}{32}=\\frac{29}{32}$$ so there is a 29/32 chance it will snow at least once. You are very welcome! :P Mar 18, 2019 #4 +2 Hi Rom Can you explain to me why the following reasoning is wrong ? The probability of snow on the first day is 1/3, so the probability that it doesn't snow is 2/3. Same for the next two days, so the probability of no snow during the first three days is (2/3)^3. Similarly the probability of no snow during the next four days is (3/4)^4. So the probablity of no snow during the week is (2/3)^3.(3/4)^4 = 3/32.", "2581 Re: If the probability of snow is the same every day this week, what is th [#permalink] ### Show Tags 03 Nov 2017, 08:31 Top Contributor Bunuel wrote: If the probability of snow is the same every day this week, what is the probability of snow on Saturday? (1) The probability of snow on Monday is 1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. Target question: What is the probability of snow on Saturday? Given: The probability of snow is the same every day this week Statement 1: The probability of snow on Monday is 1/7. If the probability of snow is THE SAME every day this week, then the probability of snow on Saturday is also 1/7 Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: The probability having of no snow on both Thursday and Friday is 36/49. Let p = probability of snow on Saturday. Since the probability of snow is the same every day this week, we know that p = probability of snow on Thursday and p = probability of snow on Friday. Also, (1-p) = probability of NO SNOW on Thursday and (1-p) = probability of NO SNOW on Friday. Statement 2 says: P(no snow on Thursday AND", "### Show Tags 05 Aug 2016, 05:24 Bunuel wrote: If the probability of snow is the same every day this week, what is the probability of snow on Saturday? (1) The probability of snow on Monday is 1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. From stimulus P (Snow )= $$\\frac{1}{7}$$ (same on Sun, Mon, Tue, Wed, Thur, Fri, Sat) P (No Snow)= $$\\frac{6}{7} ($$same on Sun, Mon, Tue, Wed, Thur, Fri, Sat) AIM :- FIND P(Snow) on Saturday (1) The probability of snow on Monday is$$\\frac{1}{7}$$ Since P(Snow) on Monday = P (Snow) on Saturday So P(Snow) on Saturday = $$\\frac{1}{7}$$ SUFFICIENT (2) The probability having of no snow on both Thursday and Friday is $$\\frac{36}{49}$$ We know from Stimulus that P (No snow) on Thursday= P (No Snow) on Friday therefore P (No Snow) Thursday and P(No Snow)Friday $$= \\frac{36}{49} = \\frac{6}{7}*\\frac{6}{7}$$ Therefore P (Snow) on thursday or friday or saturday (since it is same)$$= 1-\\frac{6}{7} = \\frac{1}{7}$$ SUFFICIENT _________________ Posting an answer without an explanation is \"GOD COMPLEX\". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired. CEO Joined: 12 Sep 2015 Posts:", "# Math Help - Probability test question 1. ## Probability test question I had this question on a test: The probability of snowing in one city is 0.4, and the probability of snowing in another city is 0.7. Assume independence. What is the probability that it snows in exactly one of the two cities? The way I approached it was to say that the probability that it snows in exactly one of the two cities is: Let A denote probability of snow in the first city. Let B denote probability of snow in the second city. P(A and 'not B') or P('not A' and B) = (0.4)(0.3) + (0.6)(0.7) - (0.4)(0.3)(0.6)(0.7) = 0.4896 Is this correct? 2. There are four possibilities: It snows in neither city P1 = (0.6)(0.3) It snows in both cities P2 = (0.4)(0.7) It snows in the first city only P3 = (0.4)(0.3) It snows in the second city only P4 = (0.6)(0.7) The probability that it snows in exactly one city is simply P3 + P4. 3. Originally Posted by BrownianMan The probability of snowing in one city is 0.4, and the probability of snowing in another city is 0.7. Assume independence. What is the probability that it snows in exactly one of the two cities? Let A denote probability of snow in", "or $P($no snow, snow$)$. To calculate the probability of these two events we simply multiply along the branches. $P($snow, snow$)= \\displaystyle{\\frac{2}{15}}$, $P($no snow, snow$)=\\displaystyle{\\frac{1}{10}}$ Similarly as before, ‘or’ means $+$ since these are disjoint events. To get the probability of snow on the day after tomorrow, we add up probabilities of the events above. Therefore, $P($snow on the day after tomorrow$)= \\displaystyle{\\frac{2}{15} +\\frac{1}{10}=\\frac{7}{30}}$", "## A community for students. Sign up today Here's the question you clicked on: ## anonymous 5 years ago the probability that it would snow on sunday is 3/5 what is the probabilty that it would snow on monday, if it snowed on sunday ? • This Question is Closed 1. anonymous there must be some more information 2. anonymous ooh sorry the probability that it would snow on sunday is 3/5 . the probability that it will snow on both sunday and monday is 3/10 . what is the probabilty that it would snow on monday, if it snowed on sunday 3. anonymous $P(A|B)=\\frac{P(A\\cap B)}{P(B)}$ 4. anonymous $P(A|B)=\\frac{\\frac{3}{10}}{\\frac{3}{5}}=\\frac{3}{10}\\times \\frac{5}{3}=\\frac{1}{2}$ 5. anonymous thank you 6. anonymous welcome #### Ask your own question Sign Up Find more explanations on OpenStudy Privacy Policy", "## Algebra and Trigonometry 10th Edition $0.1$ The probability that it will not snow is the complement of the probability that it will snow. Thus $P(no \\ snow)=1-P(snow)=1-0.9=0.1$.", "1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. _________________ Math Expert Joined: 02 Aug 2009 Posts: 5938 Re: If the probability of snow is the same every day this week, what is th [#permalink] ### Show Tags 28 Feb 2016, 19:06 Bunuel wrote: If the probability of snow is the same every day this week, what is the probability of snow on Saturday? (1) The probability of snow on Monday is 1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. Hi, info :- prob is the same each day lets see the statements:- (1) The probability of snow on Monday is 1/7. straight forward. prob is the same every day, so on saturday too prob=1/7 Suff (2) The probability having of no snow on both Thursday and Friday is 36/49. let the Prob of snowing be x, so prob of not snowing= 1-x.. so on two days= (1-x)(1-x)=36/49.. (1-x)^2=(6/7)^2.. so 1-x==6/7 or -6/7.. but xcannot be>1,so 1-x=6/7 or x=1/7 Suff D _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html GMAT online Tutor Director Joined: 04 Jun 2016 Posts: 609 GMAT 1: 750 Q49 V43 Re: If the probability of snow is the same every day this week, what is th [#permalink]", "p it snowed yesterday, and (1-p) that it did not. Therefore the probability that it snows today is: p*0.3+(1-p)*0.2=p Solve this for p to find p=2/9~=0.222. RonL 3. Originally Posted by megno Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out. if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day? It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas? This is an example of a two states Markov chain. Let X(n) be the state of the process, i.e the weather, at day n. Then we have a state space S={0,1} 0: Not snowing 1: Snowing Transition probabilities between the states from day to day is $P(X(n+1)=0|X(n)=0)=p_{00}=0.8$ $P(X(n+1)=1|X(n)=0)=p_{01}=0.2$ $P(X(n+1)=1|X(n)=1)=p_{11}=0.3$ $P(X(n+1)=0|X(n)=1)=p_{01}=0.7$ That is we have a transition probability matrix P $P=\\left( \\begin{array}{cc} 0.8 & 0.2 \\\\ 0.7 & 0.3 \\end{array} \\right)$ What we're interested in is to find the long run behaviour of the process. (There are certain conditions to be satisfied in order for", "example, the first row shows that 2,147 observations were counted that had the combination where it was observed and predicted to have not snowed that day. These 2,147 observations would be instances of correct classification. The second row shows that 245 observations occurred where it was observed to not have snowed, but the model predicted it would snow that day. All of the 245 observations were misclassified based on the classification model. From this table, the overall model accuracy can be calculated by summing up the cases that matched and dividing by the total number of observations. This computation would look like: $accuracy = \\frac{\\textrm{matching predictions}}{\\textrm{total observations}} = \\frac{(2147 + 479)}{(2147 + 245 + 529 + 479)} = .772 = 77.2%$ This means that the overall classification accuracy for this example was just over 77%, meaning that about 77% of days the model was able to correctly classify whether it snowed or not. This computation can also be done programmatically. To do this, a new attribute named, same_class, can be added to the data that is given a value of 1 if the observed data matches the predicted class and a value of 0 otherwise. Descriptive statistics, such as the mean and sum, can be computed on this new vector to represent the accuracy as a proportion and", "+0 0 92 1 On each of the first three days of January, there is a $\\frac{1}{3}$ chance that it will snow where Bob lives. On each of the next four days, there is a $\\frac{1}{4}$ chance that it will snow. What is the probability that it snows at least once during the first week of January? Jul 12, 2021", "\\right) \\approx \\frac{1-a}{1+b-a}$$ and so becomes increasingly independent of the probability of it raining on the first day. Substituting in the values given in the question $a=0.7$ and $b=0.2$ $$\\Rightarrow \\mathrm{P} \\left( D_n \\right) = \\frac{1-0.7}{1+0.2-0.7} = \\frac{0.3}{0.5} = 0.6$$", "+0 Probability 0 146 3 On each of the first three days of January, there is a 1/2 chance that it will snow where Bob lives. On each of the next four days, there is a 1/3 chance that it will snow. What is the probability that it snows at least once during the first week of January? Apr 14, 2022 #1 +2540 +1 It is easier to count for what we don't what and subtract that from 1. The probability that there will 0 days of snow is: $${1 \\over 2}^3 \\times {2 \\over 3}^4 = {2 \\over 81}$$ Thus, the probability that it will snow for at least 1 day of the week is: $$1 - {2 \\over 81} = \\color{brown}\\boxed{79 \\over 81}$$ Apr 14, 2022 #2 +579 +1 Why are there so many counting and prob questions today???? Vinculum Apr 14, 2022" ]

[ "+0 # Complementary Counting Problem! 0 242 2 +128 The probability of snow for each of the next three days is 2/3. What is the probability that it will snow at least once during those three days? Express your answer as a common fraction. Thanks for the help guys! Mar 20, 2019 edited by EvanWei123 Mar 20, 2019 #1 +391 +2 By using Complementary Counting, you can find the probability that no days snow, and subtract that from $$1$$. The probability that no day snows is $$\\frac{1}{3}*\\frac{1}{3}*\\frac{1}{3}=\\frac{1}{27}$$, so the probability of it snowing at least one day is $$1 - \\frac{1}{27}= \\frac{26}{27}$$. - Daisy Mar 20, 2019 #2 +128 0 Thank You Daisy! Mar 23, 2019", "+0 # The probability of snow for each of the next three days is $\\frac{2}{3}$. What is the probability that it will snow at least once during tho 0 134 3 The probability of snow for each of the next three days is . What is the probability that it will snow at least once during those three days? Express your answer as a common fraction. Aug 9, 2021 #1 +1 Case 1: once: 2/3 * 1/3 * 1/3 * 3 Case 2: twice: 2/3 * 2/3 * 1/3 * 3 C 2 case 3: 3 times: (2/3)^3 2/27*3 + 4/27*3 + 8/27 = 6/27 + 12/27 + 8/27 = 26/27. Aug 9, 2021 #2 +23 +2 We could use Complementary Counting for this. The probability that it will not slow at all is $$\\frac{1}{3} \\times \\frac{1}{3} \\times \\frac{1}{3} = \\frac{1}{27}$$ Then we minus the total by $$\\frac{1}{27}$$ $$1 - \\frac{1}{27} = \\frac{26}{27}$$ Aug 9, 2021 #3 +1 we got the same answer, in different ways :D Guest Aug 9, 2021", "# Math Help - Impossible Question 1. ## Impossible Question Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out. if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day? It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas? 2. Originally Posted by megno Okay so maybe it's not impossible, but neither I nor any of my other classmates were able to figure it out. if on any given day it snows, there is a .3 probability that it will snow the following day. if on any given day it doesn't snow, there is a .8 probability that it will not snow the following day. what is the probability that it will snow on any given day? It was a question on our final, and I simply guessed (it was multiple choice). Anyone have any ideas? Suppose that the probability that it snows on any day is p. Now consider today. There is a probability", "marble is $P((G,G))$ and we know it’s probability. Consequently, $\\{$we didn’t pick a single blue marble$\\}=1-P((G,G))=\\displaystyle{1-\\frac{6}{55}=\\frac{49}{55}}$ #### Example If it snows on a given day, the probability that it snows the next day is $\\frac{1}{3}$. If it doesn’t, the probability that it snows the next day is $\\frac{1}{6}$. The probability it will snow tomorrow is $\\frac{2}{5}$. What is the probability that it will snow the day after tomorrow? To illustrate all the possible outcomes, we’ll use tree diagram. First of all, we need to determine what happens tomorrow. The probability it will snow tomorrow is $\\frac{2}{5}$, so the probability that it won’t snow tomorrow is $\\frac{3}{5}$. Consequently, our starting point will be today, and first two events will be about tomorrow. If it snows on a given day, the probability it will snow the next day is $\\frac{1}{3}$. Alternatively, the probability it won’t snow the next day is $\\frac{2}{3}$. Furthermore, if it doesn’t snow on a given day, the probability it will snow the next day is $\\frac{1}{6}$. Consequently, the probability it won’t snow the next day is $\\frac{5}{6}$. When we put all the events and their probabilities in the tree diagram, we get: All we need to do is calculate the probability that it will snow the day after tomorrow. $P($snow on the day after tomorrow$)=P($snow, snow$)$", "We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy. +0 # Complementary Counting Problem! 0 191 2 +93 The probability of snow for each of the next three days is 2/3. What is the probability that it will snow at least once during those three days? Express your answer as a common fraction. Thanks for the help guys! Mar 20, 2019 edited by EvanWei123 Mar 20, 2019 ### 2+0 Answers #1 +391 +2 By using Complementary Counting, you can find the probability that no days snow, and subtract that from $$1$$. The probability that no day snows is $$\\frac{1}{3}*\\frac{1}{3}*\\frac{1}{3}=\\frac{1}{27}$$, so the probability of it snowing at least one day is $$1 - \\frac{1}{27}= \\frac{26}{27}$$. - Daisy Mar 20, 2019 #2 +93 0 Thank You Daisy! Mar 23, 2019", "+0 # probability +1 100 1 On each of the first three days of January, there is a 1/2 chance that it will snow where Bob lives. On each of the next four days, there is a 1/3 chance that it will snow. What is the probability that it snows at least once during the first week of January? Jan 13, 2021 #1 +283 +1 We can use complementary counting for this problem by calculating the probability that it will not snow at all first. That is equal to $$(\\frac{1}{2})^3\\cdot(\\frac{2}{3})^4=\\frac{1}{8}\\cdot\\frac{16}{81}=\\frac{2}{81}$$ But we're trying to find the probability that it will snow at least once, so the answer is $$1-\\frac{2}{81}=\\frac{79}{81}$$ Here's a very similar problem: https://web2.0calc.com/questions/please-help-complementry-probabilty-problem Jan 13, 2021", "+0 0 276 6 +128 On each of the first three days of January, there is a 1/3 chance that it will snow where Bob lives. On each of the next four days there is a 1/4 chance that it will snow. What is the probabilty that it snows at least once during the first week of January? Mar 18, 2019 #1 +128 0 oops im sorry. I double posted. Mar 18, 2019 #2 +6045 +1 . Mar 18, 2019 edited by Rom Mar 18, 2019 #3 +954 +3 Let's find the chance that it *doesn't* snow. There's a 2/3 chance for the first three days and 3/4 chance for the last four. So... $$(\\frac{2}{3})^3\\times (\\frac{3}{4})^4=\\frac{3}{32}$$ Now, $$1-\\frac{3}{32}=\\frac{29}{32}$$ so there is a 29/32 chance it will snow at least once. You are very welcome! :P Mar 18, 2019 #4 +2 Hi Rom Can you explain to me why the following reasoning is wrong ? The probability of snow on the first day is 1/3, so the probability that it doesn't snow is 2/3. Same for the next two days, so the probability of no snow during the first three days is (2/3)^3. Similarly the probability of no snow during the next four days is (3/4)^4. So the probablity of no snow during the week is (2/3)^3.(3/4)^4 = 3/32.", "2581 Re: If the probability of snow is the same every day this week, what is th [#permalink] ### Show Tags 03 Nov 2017, 08:31 Top Contributor Bunuel wrote: If the probability of snow is the same every day this week, what is the probability of snow on Saturday? (1) The probability of snow on Monday is 1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. Target question: What is the probability of snow on Saturday? Given: The probability of snow is the same every day this week Statement 1: The probability of snow on Monday is 1/7. If the probability of snow is THE SAME every day this week, then the probability of snow on Saturday is also 1/7 Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: The probability having of no snow on both Thursday and Friday is 36/49. Let p = probability of snow on Saturday. Since the probability of snow is the same every day this week, we know that p = probability of snow on Thursday and p = probability of snow on Friday. Also, (1-p) = probability of NO SNOW on Thursday and (1-p) = probability of NO SNOW on Friday. Statement 2 says: P(no snow on Thursday AND", "# The probability of it snowing in neverland is 1/3. what is the probability that it will snow monday and not snow tuesday As we all know, snowstorms frequently last more than one day, sometimes 2, sometimes 3. This means that probability of snowing Monday is not independent from the probability of snowing the next day (a Tuesday). Assuming in Neverland, the snowstorms only last one day, and the probability of snowing is completely INDEPENDENT from one day to another, then we can use the multiplication rule Probability of snowing Monday AND Tuesday =Probability of snowing Monday * Probability of snowing Tuesday =P(Monday)*P(Tuesday) =(1/3)*(1/3) =1/9 (Only in Neverland!) Only authorized users can leave an answer!", "## A community for students. Sign up today Here's the question you clicked on: ## anonymous 5 years ago the probability that it would snow on sunday is 3/5 what is the probabilty that it would snow on monday, if it snowed on sunday ? • This Question is Closed 1. anonymous there must be some more information 2. anonymous ooh sorry the probability that it would snow on sunday is 3/5 . the probability that it will snow on both sunday and monday is 3/10 . what is the probabilty that it would snow on monday, if it snowed on sunday 3. anonymous $P(A|B)=\\frac{P(A\\cap B)}{P(B)}$ 4. anonymous $P(A|B)=\\frac{\\frac{3}{10}}{\\frac{3}{5}}=\\frac{3}{10}\\times \\frac{5}{3}=\\frac{1}{2}$ 5. anonymous thank you 6. anonymous welcome #### Ask your own question Sign Up Find more explanations on OpenStudy Privacy Policy", "# Math Help - Probability test question 1. ## Probability test question I had this question on a test: The probability of snowing in one city is 0.4, and the probability of snowing in another city is 0.7. Assume independence. What is the probability that it snows in exactly one of the two cities? The way I approached it was to say that the probability that it snows in exactly one of the two cities is: Let A denote probability of snow in the first city. Let B denote probability of snow in the second city. P(A and 'not B') or P('not A' and B) = (0.4)(0.3) + (0.6)(0.7) - (0.4)(0.3)(0.6)(0.7) = 0.4896 Is this correct? 2. There are four possibilities: It snows in neither city P1 = (0.6)(0.3) It snows in both cities P2 = (0.4)(0.7) It snows in the first city only P3 = (0.4)(0.3) It snows in the second city only P4 = (0.6)(0.7) The probability that it snows in exactly one city is simply P3 + P4. 3. Originally Posted by BrownianMan The probability of snowing in one city is 0.4, and the probability of snowing in another city is 0.7. Assume independence. What is the probability that it snows in exactly one of the two cities? Let A denote probability of snow in", "or $P($no snow, snow$)$. To calculate the probability of these two events we simply multiply along the branches. $P($snow, snow$)= \\displaystyle{\\frac{2}{15}}$, $P($no snow, snow$)=\\displaystyle{\\frac{1}{10}}$ Similarly as before, ‘or’ means $+$ since these are disjoint events. To get the probability of snow on the day after tomorrow, we add up probabilities of the events above. Therefore, $P($snow on the day after tomorrow$)= \\displaystyle{\\frac{2}{15} +\\frac{1}{10}=\\frac{7}{30}}$", "### Show Tags 05 Aug 2016, 05:24 Bunuel wrote: If the probability of snow is the same every day this week, what is the probability of snow on Saturday? (1) The probability of snow on Monday is 1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. From stimulus P (Snow )= $$\\frac{1}{7}$$ (same on Sun, Mon, Tue, Wed, Thur, Fri, Sat) P (No Snow)= $$\\frac{6}{7} ($$same on Sun, Mon, Tue, Wed, Thur, Fri, Sat) AIM :- FIND P(Snow) on Saturday (1) The probability of snow on Monday is$$\\frac{1}{7}$$ Since P(Snow) on Monday = P (Snow) on Saturday So P(Snow) on Saturday = $$\\frac{1}{7}$$ SUFFICIENT (2) The probability having of no snow on both Thursday and Friday is $$\\frac{36}{49}$$ We know from Stimulus that P (No snow) on Thursday= P (No Snow) on Friday therefore P (No Snow) Thursday and P(No Snow)Friday $$= \\frac{36}{49} = \\frac{6}{7}*\\frac{6}{7}$$ Therefore P (Snow) on thursday or friday or saturday (since it is same)$$= 1-\\frac{6}{7} = \\frac{1}{7}$$ SUFFICIENT _________________ Posting an answer without an explanation is \"GOD COMPLEX\". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired. CEO Joined: 12 Sep 2015 Posts:", "## Algebra and Trigonometry 10th Edition $0.1$ The probability that it will not snow is the complement of the probability that it will snow. Thus $P(no \\ snow)=1-P(snow)=1-0.9=0.1$.", "+0 Probability 0 146 3 On each of the first three days of January, there is a 1/2 chance that it will snow where Bob lives. On each of the next four days, there is a 1/3 chance that it will snow. What is the probability that it snows at least once during the first week of January? Apr 14, 2022 #1 +2540 +1 It is easier to count for what we don't what and subtract that from 1. The probability that there will 0 days of snow is: $${1 \\over 2}^3 \\times {2 \\over 3}^4 = {2 \\over 81}$$ Thus, the probability that it will snow for at least 1 day of the week is: $$1 - {2 \\over 81} = \\color{brown}\\boxed{79 \\over 81}$$ Apr 14, 2022 #2 +579 +1 Why are there so many counting and prob questions today???? Vinculum Apr 14, 2022", "1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. _________________ Math Expert Joined: 02 Aug 2009 Posts: 5938 Re: If the probability of snow is the same every day this week, what is th [#permalink] ### Show Tags 28 Feb 2016, 19:06 Bunuel wrote: If the probability of snow is the same every day this week, what is the probability of snow on Saturday? (1) The probability of snow on Monday is 1/7. (2) The probability having of no snow on both Thursday and Friday is 36/49. Hi, info :- prob is the same each day lets see the statements:- (1) The probability of snow on Monday is 1/7. straight forward. prob is the same every day, so on saturday too prob=1/7 Suff (2) The probability having of no snow on both Thursday and Friday is 36/49. let the Prob of snowing be x, so prob of not snowing= 1-x.. so on two days= (1-x)(1-x)=36/49.. (1-x)^2=(6/7)^2.. so 1-x==6/7 or -6/7.. but xcannot be>1,so 1-x=6/7 or x=1/7 Suff D _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html GMAT online Tutor Director Joined: 04 Jun 2016 Posts: 609 GMAT 1: 750 Q49 V43 Re: If the probability of snow is the same every day this week, what is th [#permalink]", "+0 0 92 1 On each of the first three days of January, there is a $\\frac{1}{3}$ chance that it will snow where Bob lives. On each of the next four days, there is a $\\frac{1}{4}$ chance that it will snow. What is the probability that it snows at least once during the first week of January? Jul 12, 2021", "is having only two events, that is, red and white. The event of stepping on to a red square cannot be disintegrated any further. b) These are mutually exclusive events as they cannot happen together. Also, they are mutually exhaustive events as they together equals to the whole sample space. Getting a head or getting tail, separately, are simple events. c) The event is $\\{red\\ top,\\ blue\\ jeans\\}$. It can be further disintegrated to simpler events $\\{red\\ top\\}$ and $\\{blue jeans\\}$. Hence, it is a compound event. Problem 3: The probability of having snowfall in London today is $0.3$. What will be the probability that there is no snowfall in London today? Solution: To have a snowfall and to not have a snowfall are complementary events. Hence, the probability of not having snowfall is the complement of the probability of having snowfall. Probability of not having snowfall = $1\\ -\\ 0.3$ = $0.7$ Problem 4: There are $2$ red and $6$ green balls in a bag. If two balls are taken out one by one with replacement what is the probability that we get $1$ red and $1$ green ball. Solution: These two events are independent of each other. Probability of getting a red ball, $P(A)$ = $\\frac{2}{8}$ Probability of getting a red ball, $P(B)$ = $\\frac{6}{8}$ Probability", "of snow? – Łukasz Grad Apr 16 '17 at 15:13 • The event may be binary, but since the forecasts are probabilistic, typically \"accuracy\" would be more complicated than \"# correct/# forecasts\" (see here). Also, the suggested answer was geometric mean of probabilities or odds-ratios? (i.e. arithmetic mean of logits) – GeoMatt22 Apr 16 '17 at 15:19 • This question makes no (Bayesian) sense to me unless 'accuracy' is better defined. – Memming Apr 16 '17 at 20:19 With probabilistic forecasts, accuracy cannot be simply defined as (number correct forecasts)/(total number of forecasts), or better, you must define what you mean by a \"correct forecast\". For example you can set a threshold: if your forecast gives P(\"snow\")>0.5, you predict \"snow\", otherwise \"no snow\". With this threshold, then if it snows tomorrow, source 1's forecast is correct, and the other two are wrong. At this point, you can use the same procedure as in question https://stats.stackexchange.com/a/34141/58675 to compute a probability of snow tomorrow, conditional to the fact that source 1 predicts \"snow\", and source 2 and 3 don't ($P(y=1|\\hat{y}_1=1,\\hat{y}_2=0, \\hat{y}_3=0)$. However, with respect to the original question, we have already introduced an arbitrary threshold. Not only that: as you can see in the linked answer, to get $P(y=1|\\hat{y}_1=1,\\hat{y}_2=0, \\hat{y}_3=0)$ we need also the prior probabilities $P(y=1), P(y=0)$. Assignign these", "example, the first row shows that 2,147 observations were counted that had the combination where it was observed and predicted to have not snowed that day. These 2,147 observations would be instances of correct classification. The second row shows that 245 observations occurred where it was observed to not have snowed, but the model predicted it would snow that day. All of the 245 observations were misclassified based on the classification model. From this table, the overall model accuracy can be calculated by summing up the cases that matched and dividing by the total number of observations. This computation would look like: $accuracy = \\frac{\\textrm{matching predictions}}{\\textrm{total observations}} = \\frac{(2147 + 479)}{(2147 + 245 + 529 + 479)} = .772 = 77.2%$ This means that the overall classification accuracy for this example was just over 77%, meaning that about 77% of days the model was able to correctly classify whether it snowed or not. This computation can also be done programmatically. To do this, a new attribute named, same_class, can be added to the data that is given a value of 1 if the observed data matches the predicted class and a value of 0 otherwise. Descriptive statistics, such as the mean and sum, can be computed on this new vector to represent the accuracy as a proportion and" ]

[ "# Difference between revisions of \"2007 AMC 10A Problems/Problem 12\" ## Problem Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible? $\\text{(A)}\\ 56 \\qquad \\text{(B)}\\ 58 \\qquad \\text{(C)}\\ 60 \\qquad \\text{(D)}\\ 62 \\qquad \\text{(E)}\\ 64$ ## Solution Each tourist has to pick in between the $2$ guides, so for $6$ tourists there are $2^6$ possible groupings. However, since each guide must take at least one tourist, we subtract the $2$ cases where a guide has no tourist. Thus the answer is $2^6 - 2 = \\boxed{62}\\ \\mathrm{(D)}$. ## Solution 2 Without loss of generality, let's call one of the tour guides tour guide A, and the other tour guide B. To count the number of total groupings of guides and tourists possible, we can count the number of ways some number of tourists go to tour guide A. Thus, we can see that the total groupings is: $$\\binom{6}{1} + \\binom{6}{2} + \\binom{6}{3} + \\binom{6}{4} + \\binom{6}{5} = 62$$ ## See also 2007 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4", "when the students go Randis school requires that there are 2 adults chaperones for every 18 students when the students go on a field trip to the museum. if there are 99 students going to the museum, how many adult chaperones are needed?... ### Definition of ecology​ Definition of ecology​...", "already taken, only services remain. Many tourists will pay sums for half a days guiding that would otherwise require several weeks of agricultural toil to earn. Others stay away. In the absence of strong governance, groups emerge to capture the opportunity and without official authority, ultimately resort to stand over tactics to get their way. This is the environment that has bred the guiding cartel whose members can expect to do well as long as the monopoly can be maintained. One guide told me they have 63 members and work is divided amongst the members with each guide going to the bottom of the list after each job. My informant said they averaged 20 climbers a day. While pricing is variable, they have strong pricing discipline, in my case, only dropping the guiding offer to Rp280,000 after things had become so unpleasant that hiring a guide at any price was unlikely. This can only be possible when they are effective in suppressing competition. Batur's climbers are a mere quarter of 1 percent of Bali's visitors of 2.8 million in 2011. Guides from elsewhere will usually bring tourists only as far as Kintamani for the view from the caldera rim and will not offer the climbing opportunity. The road into the caldera is tough on their vehicles, so much", "of 7 people. (B) Add 1 to the quotient. If Ben gives tours to all 80 people, how many tours will he give? A tour can have no more than 7 people. To show all 80 people around, Ben will have to give 1 more tour. So, ben will give _________ tours in all for 80 people. So, ben will give 12 tours in all for 80 people Explanation: If Ben gives tours to all 80 people, A tour can have no more than 7 people. To show all 80 people around, Ben will have to give 1 more tour. 80 ÷ 7 = 11 3 people left. Estimate of 12 tours can be done. (C) Use only the remainder. Ben gives tours to all 80 people. After he completes the tours for groups of 7 people, how many people are in his last tour? The remainder is 3. So, Ben’s last tour will have ___________ people. 3 people only Explanation: Ben gives tours to all 80 people. After he completes the tours for groups of 7 people, how many people are in his last tour The remainder is 3. 80 ÷ 7 = 11 3 people left. Try This! Soccer players and their coaches are driven to soccer games in vans. Each van holds 9 people. How", "will impact many but from which only a few elite are likely to benefit. The best assets are privatised by the politically influential and the poor barely get by labouring at agriculture. The only ways to escape hardship are to leave or exploit and with the best assets already taken, only services remain. Many tourists will pay sums for half a days guiding that would otherwise require several weeks of agricultural toil to earn. Others stay away. In the absence of strong governance, groups emerge to capture the opportunity and without official authority, ultimately resort to stand over tactics to get their way. This is the environment that has bred the guiding cartel whose members can expect to do well as long as the monopoly can be maintained. One guide told me they have 63 members and work is divided amongst the members with each guide going to the bottom of the list after each job. My informant said they averaged 20 climbers a day. While pricing is variable, they have strong pricing discipline, in my case, only dropping the guiding offer to Rp280,000 after things had become so unpleasant that hiring a guide at any price was unlikely. This can only be possible when they are effective in suppressing competition. Batur's climbers are a mere quarter of", "(It turns out there there is no solution otherwise, but that is a digression.) Each musician must attend at least 2 concerts, or else they would see only 3 other musicians onstage. But 6 musicians attending 2 concerts each takes up all 12 audience spots, so every musician is at exactly 2 concerts. Each musician thus sees exactly six musicians onstage, and since five of them must be different, one is a repeat, and the viewing event is wasted. We knew there would be some waste, since there are 36 viewing avents available and only 30 can be useful, but now we know that each spectator wastes exactly one event. A happy side effect of splitting the musicians evenly between the stage and the audience in every concert is that we can exploit the symmetry: if we have a solution to the problem, then we can obtain a dual solution by exchanging the performers and the audience in each concert. The conclusion of the previous paragraph is that in any solution, each spectator wastes exactly one event; the duality tells us that each performer is the subject of exactly one wasted event. Now suppose the same two musicians, say A and B, perform together twice. We know that some spectator must see A twice; this spectator sees B", "bus and listening off and on to the tour guide. Hey, there’s nothing wrong with that, but we do encourage you to get off the bus and make contact with the natives as often as you can.", "visitor who spends at least twenty-four hours in the place due to recreation, vacation, health and business, among other reasons; and an excursionist is a temporary visitor who spends less than twenty-four hours at the visited place. Although some tourists may be found fishing, it is not their main objective there, which distinguishes them from the sport fishers. As the sport fishers were sequentially positioned along the river, we sampled them in a systematic manner. We interviewed 20% of them in the sampling area at each day. They were numbered following the ordering along the river or the boats' positions. Among the first six fishers, the first interviewee was randomly chosen by means of a die. After him, we proceeded by choosing every fifth fishers in the sequence until completing the sample size. Only 18 years old or older fishers were selected for the interview. The tourists and excursionists available to the interview were found in a more unpredictable way. The tourists were contacted mainly in the hotels were they stayed. The hotels were surveyed during the weekends, but in several occasions no tourist was available. The excursionists were found walking or staying nearby restaurants or the waterfall. Although their selection was arbitrarily defined without concerning any kind of characteristic, the dependence on their acceptance to the interview", "2015 04-16 # Tourism Planning Several friends are planning to take tourism during the next holiday. They have selected some places to visit. They have decided which place to start their tourism and in which order to visit these places. However, anyone can leave halfway during the tourism and will never back to the tourism again if he or she is not interested in the following places. And anyone can choose not to attend the tourism if he or she is not interested in any of the places. Each place they visited will cost every person certain amount of money. And each person has a positive value for each place, representing his or her interest in this place. To make things more complicated, if two friends visited a place together, they will get a non negative bonus because they enjoyed each other’s companion. If more than two friends visited a place together, the total bonus will be the sum of each pair of friends’ bonuses. Your task is to decide which people should take the tourism and when each of them should leave so that the sum of the interest plus the sum of the bonuses minus the total costs is the largest. If you can’t find a plan that have a result larger than 0, just tell", "22. Modeling Real Life Tours of a factory can have no more than 9 guests. There are 76 guests in line to tour the factory. • How many tours are full? • How many tours are needed? • How many guests are on the last tour? Answer: a) The number of tours that are full = 8 tours b) The number of tours needed = 9 tours c) The number of guests on the last tour = 4 guests Explanation: Given that there are 76 guests in line to tour a factory It is also given that there are no more than 9 guests on the tours of a factory. We have to observe that to make all the guests full on all the trips without leftovers, we will need 9 trips. i.e.., 81 guests. So, now we have to find the number of guests in each tour of a crayon factory by finding the quotient and remainder of 76 ÷ 9. Now, 76 ÷ 9 From this, we can see The number of guests in each tour that are full = 8 The number of guests leftover on the last trip = 4 Hence, from the above, We can conclude that a) 8 Tours are full. b) 9 Tours are needed. c) 4 guests are on the", "<meta http-equiv=\"refresh\" content=\"1; url=/nojavascript/\"> # 1.4: Order of Operations Difficulty Level: At Grade Created by: CK-12 ## Introduction Dividing up the Groups The first few days of Teen Adventure were spent at the base of the White Mountains in the Lafayette Place Campground. There the teens worked on survival skills, map skills, and basic first aid. The students were kept together and Kelly made some terrific new friends. Her favorite part was all of the team building and trust building exercises that they learned. She even surprised herself by being able to complete some tasks that she wasn’t sure that she could do. Kelly and the other students learned to rely on themselves and each other over those first few days. The night before their first hike, the group leaders told them that they would be divided up into hiking groups. Each week they would meet altogether once again to talk about their adventures, but during the actual week each group would be on a different trail. There would be three groups out of the 30 students. The 6 leaders would be divided into the 3 groups. For the first few days, there would be a manager along and two first-aid persons. After the first few days, the manager and the two first aid persons would head back", "less than four times: If the gulf is, say, 15 units long, that’s our entire bound right there, we cannot do a tour that does not use this edge for less than 60 units. And so we’re done, we have a lower bound on this branch that’s greater than the upper bound on our other branch, so any optimal tour must include that edge. If you think this stuff is cool, I recommend the book In Pursuit of the Traveling Salesman by Bill Cook, who taught the class I learned about this in. 1. assuming P!=NP2 2. and come on, P!=NP, we all know it, it's just that nobody is brave enough to say it3", "skills, map skills, and basic first aid. The students were kept all together and Kelly made some terrific new friends. Her favorite part was all of the team building and trust building exercises that they learned. She even surprised herself by being able to complete some tasks that she wasn’t sure that she could do. Kelly and the other students learned to rely on themselves and each other over those first few days. The night before their first hike, the group leaders told them that they would be divided up into hiking groups. Each week they would meet altogether once again to talk about their adventures, but during the actual week each group would be on a different trail. There would be three groups out of the 30 students. The 6 leaders would be divided into the 3 groups. For the first few days, there would be a manager along and two first-aid persons. After the first few days, the manager and the two first aid-persons would head back to the base office to be checked in with weekly or in case of emergency. So they would start with one number of people and after the first few days the number would change. Kelly is curious about how many people will start and how many will be together", "# 3 travelers and 9 diamonds I think this is the most appropriate place to share this. I haven't figured out the answer yet, but I think I'm close. I couldn't find anything like this on the Internet, so I'm posting it here. There are three travelers. They are inside a cave. Deep down in this cave, there are three doors, each door containing at least one diamond, and there are a total of 9 diamonds. Each traveler picks a door and gets the diamonds behind it. The travelers loot their diamonds, and, before exiting the cave, all three must say simultaneously if they can deduce the number of diamonds that each of the other two have. They never lie, and all of them said they couldn't deduce it. After these statements were said, one of the travelers realized that now he knows the answer. So the question is: How were the diamonds split between the three travelers? (the order doesn't matter) I was going to write what I've managed to do so far, but I think it isn't a good idea to give away too much information even if its concealed by spoilers. This riddle was suggested by my discrete mathematics professor. It isn't homework and she didn't mention where it came from. • Welcome to Puzzling.SE!", "has 18 members: 10 girls and 6 boys, 2 leaders. How many different patrols can be created, if one patrol is 2 boys, 3 girls and 1 leader?", "that it will be a maximum one green? • Division Division has 18 members: 10 girls and 6 boys, 2 leaders. How many different patrols can be created, if one patrol is 2 boys, 3 girls and 1 leader?", "sub-hypercubes. Your mission - should you choose to accept - is to: Find a Knight's Tour Four ...", "# travel agency Small travel agency offers 5 different tours at honeymoon. What is the probability that the bride and groom choose the same tour (they choose independently)? Correct result: p = 20 % #### Solution: $p=5/(5 \\cdot \\ 5) \\cdot \\ 100=20 \\%$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! Tips to related online calculators Would you like to compute count of combinations? ## Next similar math problems: • The test The test contains four questions, and there are five different answers to each of them, of which only one is correct, the others are incorrect. What is the probability that a student who does not know the answer to any question will guess the right answer • Division Division has 18 members: 10 girls and 6 boys, 2 leaders. How many different patrols can be created, if one patrol is 2 boys, 3 girls and 1 leader? On the menu are 12 kinds of meal. How many ways can we choose four different meals into the", "number of monks divisible by six\". By standing at the exit, the monks communicate, to everyone nearby, something about their knowledge that is not just the fact that they know that they have a spot. – Anon Jul 30 '16 at 9:54", "on the trip? 2. How many children can go if a second 40-person bus is added? 3. Four of the adult chaperones decide to arrive separately by car. Now how many children can go in the two buses? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes Show Hide Details Description Tags: Subjects:" ]

[ "# Difference between revisions of \"2007 AMC 10A Problems/Problem 12\" ## Problem Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible? $\\text{(A)}\\ 56 \\qquad \\text{(B)}\\ 58 \\qquad \\text{(C)}\\ 60 \\qquad \\text{(D)}\\ 62 \\qquad \\text{(E)}\\ 64$ ## Solution Each tourist has to pick in between the $2$ guides, so for $6$ tourists there are $2^6$ possible groupings. However, since each guide must take at least one tourist, we subtract the $2$ cases where a guide has no tourist. Thus the answer is $2^6 - 2 = \\boxed{62}\\ \\mathrm{(D)}$. ## Solution 2 Without loss of generality, let's call one of the tour guides tour guide A, and the other tour guide B. To count the number of total groupings of guides and tourists possible, we can count the number of ways some number of tourists go to tour guide A. Thus, we can see that the total groupings is: $$\\binom{6}{1} + \\binom{6}{2} + \\binom{6}{3} + \\binom{6}{4} + \\binom{6}{5} = 62$$ ## See also 2007 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4", "# Probabilities and Stirling numbers of the second kind I would like to solve the following problem, from the book Understanding Probability by Henk Tijms. Fifteen tourists are stranded in a city with four hotels, all of which are located near each other in the city center. Each hotel has enough rooms available to accomodate all 15 tourists. Each tourist randomly chooses a hotel, independently of the choices made by the others. What is the probability that not all four hotels will be chosen? My idea is as follows. Denote by $A, B, C, D$ the four hotels, and by $P(A\\cap B\\cap C\\cap D)$ the probability that all four hotels are chosen by at least one tourist. I would start by computing $P(A\\cap B\\cap C\\cap D)$, and then take the complementary probability. Here is where I am stuck. A possible approach would be that of writing down a table such as: $$\\begin{array}{ccccc} A & B & C & D & \\text{probability} & \\text{count} \\\\ \\hline 1 & 1 & 1 & 12 & \\binom{15}{1}\\binom{14}{1}\\binom{13}{1}/4^{15} & 4 \\\\ 1 & 1 & 2 & 11 & \\binom{15}{1}\\binom{14}{1}\\binom{13}{2}/4^{15} & 4 \\\\ \\dots \\end{array}$$ However, this approach is clearly too time-demanding (and error-prone) in this case, so I guess that the solution should be related to the number of ways to", "# a question on probability A hotel can accommodate 50 customers, experiences show that $0.1$ of those who make a reservation will not show up. Suppose that the hotel accepts 55 reservations. Calculate the probability that the hotel will be able to accommodate all of the customers that show up. I only tried to use the Poisson distribution, but I am sure it can not apply here, how this question can be solved easier? thanks - – Sasha Dec 22 '11 at 6:57 You need to make use of the binomial distribution here. It is not hard to evaluate the answer using binomial distribution itself. The probability that the hotel will be able to accommodate all of the customers that show up is the probability that at-most $50$ customers show up. Hence, $$\\mathbb{P}(\\text{at-most }50 \\text{ customers show up}) = 1 - \\mathbb{P}( \\text{number of customers } \\geq 51)$$ Hence, \\begin{align} \\mathbb{P}( \\text{number of customers}) & = \\binom{55}{55}(0.9)^{55} + \\binom{55}{54}(0.9)^{54} (0.1)^1 + \\binom{55}{53}(0.9)^{53} (0.1)^2 \\\\ & + \\binom{55}{52}(0.9)^{52} (0.1)^3 + \\binom{55}{51}(0.9)^{51} (0.1)^4\\\\ &\\approx 0.345 \\end{align} Hence, the required probability is $$\\approx 1 - 0.345 = 0.645$$ - For a very similar problem (with $105$ passengers holding reservations on a $100$-seat flight and $0.1$ probability of not showing up), see here and the complete solution is here. It even uses", "# Combinations problem - Finding the number I would like to ask for your confirmation to my thought in regards with the following exercise: One man has 30 different statues, 27 genuine and 3 fake. He sold 10 of these statues in museum A, 10 in museum B and 10 in museum C. What is the probability that each of these museums bought fake statue from the man? I found the probability for each museum, through combinations, that is : $$\\frac{{3 \\choose 1} {27 \\choose 9}+{3 \\choose 2 } {27 \\choose 8}+{3 \\choose 3} {27 \\choose 7}}{{30 \\choose 10}}$$ Is my way of thinking correct? Thank you very much in advance. • I don’t think it is. You are adding up also the cases where one museum recieves more than one fake statue, leaving not enough for the other two. – Tavish Apr 14 '20 at 7:56 • Can you explain your thinking? What are the cases you're counting with the terms in the numerator, and what is the $\\binom{30}{10}$ in the denominator counting? – Karl Apr 14 '20 at 8:13 • I think @Tavish is correct. Each museum gets a fake staue means that they all get exactly one fake statue because there are $3$ statues. This means that the probability should be $$\\frac{\\binom{3}{1}.\\binom{27}{9} + \\binom{2}{1}.\\binom{18}{9} +", "Binomial coefficient real life example. \"At a university, $15$ juniors and $20$ seniors volunteer to serve as a special committee that requires $8$ members. A lottery is used to select the committee from among the volunteers. Suppose the chosen students consists of six juniors and two seniors. (a) For a test of homogeneity, what are the expected counts? This question I understand. (b) If the selection had been random, what is the probability of the committee having exactly two seniors? My answer was that the probability is binomial, with Binom$(k=2, n=8, p=0.57)$, but this is apparently wrong. Instead the correct answer is: $$\\frac{\\binom{20}{2}\\binom{15}{6}}{\\binom{35}{8}}$$. Can anyone explain the difference between this and standard binomial distribution? • I suppose that you made a type and that there are $15$ junior volunteers, not $13$. Am I right? – José Carlos Santos Jun 17 '18 at 7:12 • It's not a straight binomial since the trials are not independent. Knowing that the first choice was a senior changes the probability that the second was a senior. – lulu Jun 17 '18 at 7:12 • Ahh. Thanks! I understand now :-) – Mathe Jun 17 '18 at 7:20 • This is the hypergeometric distribution – Lord Shark the Unknown Jun 17 '18 at 7:21 • To be sure of useful answers, you should", "third country. We have four objects to arrange, the two blocks of three consecutive people from the same country, the pair of adjacent people from the third country, and the other person. There are $$\\binom{3}{2}$$ ways to select the two countries from which all three people sit in consecutive seats, $$\\binom{3}{2}$$ ways to select which two of the three delegates from the third country sit in adjacent seats, $$3!$$ distinguishable ways to arrange the four objects (up to rotation), $$3!$$ ways to arrange the three people within each block of three people, and $$2!$$ ways to arrange the pair of adjacent people within the block of two people. Hence, there are $$\\binom{3}{2}\\binom{3}{2}3!3!3!2!$$ such seating arrangements. Six pairs in which a pair of delegates from the same country are adjacent: This can only occur if the delegates from each country sit in a block of three consecutive seats. Thus, we have three objects to arrange, the three blocks of three consecutive people from each country. The three objects can be arranged in $$2!$$ distinguishable ways around the table (up to rotation). The three people within each block can be arranged in $$3!$$ ways. Hence, there are $$2!3!3!3!$$ such seating arrangements. By the Inclusion-Exclusion Principle, the number of admissible seating arrangements is $$8! - \\binom{3}{1}\\binom{3}{2}7!2! + \\binom{3}{1}6!3! + \\binom{3}{2}\\binom{3}{2}\\binom{3}{2}6!2!2! -", "\\binom{1}{1}.\\binom{9}{9}}{\\binom{30}{10}}$$ – Aditya Apr 14 '20 at 8:16 • @Karl, the $\\binom{30}{10}$ is the total number of outcomes.(The probability is asked in the question). – Aditya Apr 14 '20 at 8:18 • Probability that a particular museum purchased fake statue(s) is that, correct. For probability that all museums purchased a fake statue, it is $\\frac{3!\\times\\binom{27}{9\\ 9\\ 9}}{\\binom{30}{10\\ 10\\ 10}}$ – Rezha Adrian Tanuharja Apr 14 '20 at 8:18 As discussed in comments, your approach is not correct. A correct method would be to assume without loss of generality that the statues are identical except for fakeness and lined up in random order, whereupon the first, middle and last groups of $$10$$ statues are given to the three museums. The number of distinct ways to select where the fakes lie in the line is $$\\binom{30}3$$. The number of ways to select the fakes so that each museum gets one fake is $$10^3$$. The probability is $$\\frac{10^3}{\\binom{30}3}=\\frac{50}{203}$$. • I really hate this argument with identification of different things. But fortunetly we can avoid it here! – Aqua Apr 14 '20 at 8:54 • @Aqua I prefer doing this because it tends to give small numbers, which I can more easily run through my head. – Parcly Taxel Apr 14 '20 at 8:56 • @Aqua In other words, here I am", "# Thread: Combination problem involving book selling. 1. ## Combination problem involving book selling. 16. A student has to sell 2 books from a collection of 6 math, 7 science, and 4 economics books. How many choices are possible if (b) the books are to be on different subjects? So for the first pass one can choose one book out of one of the three piles so, $\\binom{6}{1}+\\binom{7}{1}+\\binom{4}{1}=17$. For the second pass one can only choose a book from a pool that wasn't selected from the first pool. Leaving 3 options; 1. If you selected from $\\binom{6}{1}$ then $\\binom{7}{1}+\\binom{4}{1}=11$ 2. If you selected from $\\binom{7}{1}$ then $\\binom{6}{1}+\\binom{4}{1}=10$ 3. If you selected from $\\binom{4}{1}$ then $\\binom{6}{1}+\\binom{7}{1}=13$ I'm not entirely sure what I've done wrong here. I think I should just be able to add everything from pass one and pass two to get the correct answer but the listed answer is 94 and $51\\neq94$ Could someone clue me into what I'm doing wrong here? 2. ## Re: Combination problem involving book selling. Originally Posted by bkbowser 16. A student has to sell 2 books from a collection of 6 math, 7 science, and 4 economics books. How many choices are possible if (b) the books are to be on different subjects? I'm not entirely sure what I've done wrong", "when selecting three courses is \\begin{align*} \\sum_{k = 1}^{3} \\binom{5}{k}\\binom{4}{3 - k} & = \\binom{5}{1}\\binom{4}{2} + \\binom{5}{2}\\binom{4}{1} + \\binom{5}{3}\\binom{4}{0}\\\\ & = 5 \\cdot 6 + 10 \\cdot 4 + 10 \\cdot 1\\\\ & = 30 + 40 + 10\\\\ & = 80 \\end{align*} Method 2: Complementary counting. The number of ways of selecting at least one vegetarian course can be found by subtracting the number of ways of selecting only meat courses from the total number of ways of selecting three courses. There are $$\\binom{5 + 4}{3} = \\binom{9}{3}$$ ways to select three of the nine courses. There are $$\\binom{4}{3}$$ ways to select three of the four meat courses. Hence, the number of ways of selecting at least one vegetarian course is $$\\binom{9}{3} - \\binom{4}{3} = 84 - 4 = 80$$ You are counting each meal with two vegetarian courses twice, once for each way you could designate one of the two courses as the vegetarian course. You are counting each meal with three vegetarian courses three times, once for each way you could designate one of the three courses as the vegetarian course. Observe that $$\\binom{5}{1}\\binom{4}{2} + \\color{red}{\\binom{2}{1}}\\binom{5}{2}\\binom{4}{1} + \\color{red}{\\binom{3}{1}}\\binom{5}{3}\\binom{4}{0} = 140$$ Then why did you obtain $$280$$? Suppose that the meal Jane selects consists of vegetarian chili, avocado and tomato salad, and green salad. If you", "with \"BBW\" There are $\\binom 41=4$ equally probable arrangements like this ( you just have to decide where to put the second \"W\" in the 4 remaining places) To win on her thirrd selection the arrangement must start with \"BBBBW\" There are $\\binom 21=2$ equally probable arrangements like this ( you just have to decide where to put the second \"W\" in the 2 remaining places) so $$P(WIN) = \\frac{ \\binom 61+\\binom 41+\\binom 21 }{ \\binom 72}=\\frac{12}{21}$$ • This was great. Thank you very much – Pedro Alonso Jul 3 '18 at 1:59", "7: 7 possibilities; pick one of the remaining 6: 6 possibilities. Which gives you $8\\times 7\\times 6$ possibilities over all ;) Sorry for spoiling :( You still need to get rid of the ordering here though – pinkwerther Oct 27 '14 at 20:26 So the answer is $\\binom{8}{3} = 56$ not 84.", "# Counting (Permutation/ Combination) (CIE/0606/2021/m/22Q8) A photographer takes 12 different photographs. There are 3 of sunsets, 4 of oceans, and 5 of mountains. (a) The photographs are arranged in a line on a wall. (i) How many possible arrangements are there if there are no restrictions?$[1]$ (ii) How many possible arrangements are there if the first photograph is of a sunset and the last photograph is of an ocean? $\\quad[2]$ (iii) How many possible arrangements are there if all the photographs of mountains are next to each other? (b) Three of the photographs are to be selected for a competition. (i) Find the number of different possible selections if no photograph of a sunset is chosen. $\\quad[2]$ (ii) Find the number of different possible selections if one photograph of each type (sunset, ocean, mountain) is chosen. *****\\\\ *********math solution************* $\\begin{array}{lll}\\text{(a)}&\\text{(i)}&\\text{There are total of 12 photographs.}\\\\&& \\text{Number of arrangements }=12 !=479001600 \\\\&\\text {(ii)}&\\text{First place (3 sunset photographs)}= 3 \\text { ways } \\\\&&\\text {Last place (4 ocean photographs) }=4 \\text { ways } \\\\&&\\text {Total number of arrangements }= 3 \\times 10 ! \\times 4=43545600 \\\\&\\text{(iii) }&\\text {Sunset and ocean photographs }=7 \\\\&&\\text {Number of arrangements }=7 ! \\\\&&\\text {Number of arrangements for mountains }=5 ! \\\\&&\\text {There are } 6 \\text { places of between } 7 \\text", "there are $\\binom{10}2=45$ to pick the other two members of the team. That gives us another $15\\cdot45=675$ teams. • $3$ singles can be picked in $\\binom63=20$ ways, and we can pick any three of the other ten people, so we get another $20\\binom{10}3=20\\cdot120=2400$ teams. Now it gets a little trickier, since we start running into restrictions on whom we can choose from the couples. • $2$ singles can be picked in $\\binom62=15$ ways. There are $\\binom{10}4=210$ ways to pick $4$ people from the couples, but some of these are forbidden: specifically, we may not pick two couples. Since there are $5$ couples altogether, there are $\\binom52=10$ pairs of couples. That’s $10$ sets of four people that we aren’t allowed to choose, leaving $210-10=200$ sets that we are allowed to choose. Thus, we can form $15\\cdot200=3000$ teams with $2$ singles. • One single can be picked in $6$ ways. There are $\\binom{10}5=252$ ways to choose $5$ people from the couples, but here again some are not allowed. Specifically, we have to throw out those groups that consist of two-and-a-half couples. As before, there are $\\binom52=10$ ways to pick two couples, and there are then $6$ ways to pick one more person from the remaining three couples. That makes $10\\cdot6=60$ ways to fill out the team and gives us $6\\cdot60=360$", "you are viewing a single comment's thread. [–] 3 points4 points (6 children) sorry, this has been archived and can no longer be voted on However, the original problem states the chef makes 7 dishes of \"anything he wants,\" so he is not really hoping to maximize his chances of matching the table. In fact, it is more likely that he chooses 7 of the same dish since it will be easier for him to cook, which of course is the least likely of the 120 combos since it has only one permutation (assuming equality of preference in the 4 dishes). edit: Supposing the chef chooses 7 dishes to cook at random how likely is he to be correct? This now becomes a difficult problem I think. edit: answer is .01844597. I don't know if there's a simple formula for it or not. I did it by hand. [–] 1 point2 points (0 children) sorry, this has been archived and can no longer be voted on The closest I can get to a closed form for this version is [; \\frac{1}{4^{14}} \\sum_{k=0}^7 \\binom{7}{k}^2 \\binom{2k}{k} \\binom{2(7-k)}{7-k}. ;] It comes out to 4951552/268435456 which is the same value you got. Maybe someone can simplify the expression further? [–] 0 points1 point (3 children) sorry, this has been archived and can", "which one of two selections can be carried out. For instance, if you have 3 pants and 4 shirts, then you have $3\\times 4$ ways of selecting what pair of pants and shirt to wear; if you have 3 routes you can take by car to get to school and 4 routes you can take by bike, then you have $3+4$ routes you can take to get to school by either car or bike. The binomial coefficient on the right hand side, $\\binom{n}{n-r}$, counts the number of ways in which you can select $n-r$ items from $n$ possibilities; note that this is the same as $\\binom{n}{r}$, the number of ways of selecting $r$ items from $n$ possibilities: because selecting the $n-r$ that you want is equivalent to selecting the $r$ that you don't want. So the right hand side counts the number of ways of selecting $r$ items from $n$ possibilities, $2^{n-r}$ times. Let's look at the sum on the left hand side: the $k$th summand is $$\\binom{n}{k}\\binom{k}{r}.$$ The first factor counts the number of ways of selecting $k$ items from $n$ possibilities; the second factor counts the number of ways of selecting $r$ items from $k$ items. One possible interpretation is: you have $n$ things; first you select $k$ of them, and then you select $r$ items", "we can make from 8 people, since the remaining 4 people will go in the second car. Therefore, there are: $\\binom{8}{4} = \\frac{8!}{4!(8-4)!} = 70$ possible combinations. If you’re not familiar with this notation, you can read $$\\binom{8}{4}$$ as choose 4 people out of 8 people, of which there are 70 possibilities. That means that if we wanted to optimize the distribution of people into the two cars – for example, if we wanted to group up the best friends together, or minimize the total weight of people in car1, or some other objective –, we would need to look at 70 solutions. This problem is purely combinatorial. Now let’s add some constraints. Our 8 friends are coming back from the bar. Out of the 8 friends, 3 of them didn’t drink and are therefore allowed to drive. Thus, the number of possible arrangements of friends in the car has been reduced, as each car needs a driver. For one car, we need to select 1 driver out of 3, and 3 remaining passengers out of 7. However, the other car will need a driver, so really there are 6 passengers to choose from. Finally, for every arrangement there is a duplicate arrangement with the cars swapped. The number of possibilities is therefore: $\\binom{3}{1} \\binom{6}{3} \\times 2 =", "2 2 1 1 1 2 1 2 1 1 3 2 1 0 1 3 1 2 0 1 2 2 2 0 As you can see, there are 10 combinations of (a,b,c,d,e) possible. For each of these possibilities, the number of hands is $\\binom{1}{a} \\binom{21}{b} \\binom{4}{c} \\binom{3}{d} \\binom{31}{e}$ Sum up these products for the 10 (a,b,c,d,e) possibilities. (I used a spreadsheet.) If you do it correctly, I think you will find the sum is 1,980,720. So the probability of drawing an acceptable hand is $\\frac{1,980,720}{382,606,920}$, which is approximately 0.00512. If you are not familiar with the notation above, $\\binom{n}{m} = \\frac{n!}{m! (n-m)!}$ is the number of combinations of n objects taken m at a time.", "144 views Compute the probability that if 10 married couples are seated at random at a round table, then no wife sits next to her husband 1 wife sits next to her husband. pick one of the 10 couples=$\\binom{10}{1}$. These couples can interchange their position such that they sit next to each other in $2^1$ ways now let the couple form a unit. so there are 18+1=19 people who have to be arranged in circular permutation. this can be done in (19-1)! ways so for one couple → $\\binom{10}{1}*18!*2^1$ for 2 couples → $\\binom{10}{2}*17!*2^2$ [here the two couples can interchange their position so 2*2 permutations can be done] here the two couples forms 2 units. so total 2+16=18 people have to be arranged in circular permutation. this can be done in 17! ways 3 couples → $\\binom{10}{3}*16!*2^3$ 4 couples → $\\binom{10}{4}*15!*2^4$ 5 couples → $\\binom{10}{5}*14!*2^5$ 6 couples → $\\binom{10}{6}*13!*2^6$ 7 couples → $\\binom{10}{7}*12!*2^7$ 8 couples → $\\binom{10}{8}*11!*2^8$ 9 couples → $\\binom{10}{9}*10!*2^9$ 10 couples → $\\binom{10}{10}*9!*2^{10}$ Atleast one couple sits together =$\\binom{10}{1}*18!*2^1$ – $\\binom{10}{2}*17!*2^2$ + $\\binom{10}{3}*16!*2^3$ – $\\binom{10}{4}*15!*2^4$ + $\\binom{10}{5}*14!*2^5$ – $\\binom{10}{6}*13!*2^6$ + $\\binom{10}{7}*12!*2^7$ – $\\binom{10}{8}*11!*2^8$ + $\\binom{10}{9}*10!*2^9$ – $\\binom{10}{10}*9!*2^{10}$ =N probability that atleast one couple sits together=$\\frac{N}{19!}$ so probability that no couple sits together=$1-\\frac{N}{19!}$ is this correct? edited | 144 views 0 See my answer , For calculating", "group: $\\binom42\\binom21 = 12$ • $1$ from the last group and $2$ from the first group:$\\binom41\\binom22 = 4$ • Total ways of forming midfield if C is included = $4+12+4 = 20$ If C is not selcted for midfield, four others have to be chosen in similar fashion, total ways of forming midfield = $\\binom44 + \\binom43\\binom21 + \\binom42\\binom22 = 15$ Can you now proceed to a conclusion ? Btw, you have used \"odds\" at one place, \"chance\" at another. Pl. don't use \"odds\" as a synonym for \"probability\", they are not the same. • Ah, in my haze of confusion I must've gotten \"odds\" and \"chance\" mixed up, thank you for the correction. As for the task: I tried to do the total chance for each of the restrictions: $One random player$: 9C4 - 8C4 = 56. $Mary and Sue$: (9C4 - 8C4) - (8C3 - 7C3) = 56 - 35 = 21. Etc. And then add together those answers to get the total number of combinations. I now see where I went wrong: I should've taken the combinations for each group in question, rather the entire selection. And by adding combinations + C (favorable outcomes) and dividing them by combinations - C, you get the $chance$ for C. – Doe J Feb 3 '18 at 15:48", "cards from remaining 21 non-spade cards. Similarly, when he has one spade, he can rearrange rest of the cards in $\\binom{21}{12}$ ways. Of course, East has five choices to pick that one particular spade. Therefore, East has $\\binom{5}{1}\\binom{21}{12}$ ways to have one spade! So, this works: $\\frac{\\binom{5}{3}\\binom{21}{10}}{\\binom{5}{0}\\binom{21}{13}+\\binom{5}{1}\\binom{21}{12}+\\binom{5}{2}\\binom{21}{11}+\\binom{5}{3}\\binom{21}{10}+\\binom{5}{4}\\binom{21}{9}+\\binom{5}{5}\\binom{21}{8}} = 0.339$. Two morals from this story: 1) In most introductory probability questions, we make the mistake of counting incorrectly. 2) There are often multiple ways of solving the same problem. Try the problem yourself in your own way and double check with the answer to ensure your approach is correct. Do not read probability book like a novel." ]

[ "# Difference between revisions of \"2007 AMC 10A Problems/Problem 12\" ## Problem Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible? $\\text{(A)}\\ 56 \\qquad \\text{(B)}\\ 58 \\qquad \\text{(C)}\\ 60 \\qquad \\text{(D)}\\ 62 \\qquad \\text{(E)}\\ 64$ ## Solution Each tourist has to pick in between the $2$ guides, so for $6$ tourists there are $2^6$ possible groupings. However, since each guide must take at least one tourist, we subtract the $2$ cases where a guide has no tourist. Thus the answer is $2^6 - 2 = \\boxed{62}\\ \\mathrm{(D)}$. ## Solution 2 Without loss of generality, let's call one of the tour guides tour guide A, and the other tour guide B. To count the number of total groupings of guides and tourists possible, we can count the number of ways some number of tourists go to tour guide A. Thus, we can see that the total groupings is: $$\\binom{6}{1} + \\binom{6}{2} + \\binom{6}{3} + \\binom{6}{4} + \\binom{6}{5} = 62$$ ## See also 2007 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4", "÷ 9 = 4. Solve. Question 14. Jack has $18 to spend on fish for an aquarium. He buys 6 fish. Each fish costs the same amount. How much does each fish cost? Answer: The cost of each fish is$3. Explanation: Given that Jack has $18 to spend on fish for an aquarium and he bought 6 fish. As each fish costs the same amount. So the cost of each fish is 18÷6 which is$3. Question 15. Nykeba is assigning tour guides for people going on museum tours. Each guide leads 4 people around the museum. There are 20 people going on a tour. How many tour guides does Nykeba assign? The number of tour guides does Nykeba assigned is 5 people. Explanation: Given that Nykeba is assigning tour guides for people going on museum tours and each guide leads 4 people around the museum as there are 20 people going on a tour. So the number of tour guides does Nykeba assign is 20÷4 = 5. Question 16. Open Ended Write a word problem that could be solved using these related multiplication and division equations. 12 ÷ = 3 and 3 × = 12", "of the same destinations together]. [PS. If \"important\" were interpreted to mean important to the tourist (which I don't think is unreasonable), then your proposed answer to (a) would be correct: there are $\\binom{10}{6}$ unordered sets of destinations, and the tourist visits them according to her preferences.] (a) is $10\\choose6$, combinations (b) is ${10\\choose 6} \\times 6!$ because there are $6!$ orders to visit the cities", "If there are 72 tourists, what is the cost of hiring guides? Let $$M=$$ the age of sister Molly now. Worksheets > Math > Grade 4 > Word problems. We have to divide by. Word problems for systems of linear equations are troublesome for most of the students in understanding the situations and bringing the word problem into equations. Also, remember that if the problem calls for a pure solution or concentrate, use 100%. If the product of a number and –7 is reduced by 3, the resulting number is 33 less than twice the opposite of that number. One ounce of solution Y contains ingredients a and b in a ratio of 1:2. ingredient a}\\\\3\\times 54=162\\,\\,\\,\\,\\text{oz}\\text{. Do you see how if we divide the number of tourists by 10, and go up to the next integer, we’ll get the number of tour guides we need? For$42.50 total, they can buy p boxes of pizza. Most tricky and tough algebra word problems are covered here. Solve the inequality and graph the results. Each box of pizza costs $8.50. Since she sold 20 less than she bought, she sold 50 – 20 = 30 programs. She sold all but 20 of them for$3 each and made a profit of $15 total. $$\\displaystyle \\frac{{86+92+78+83+99+99}}{6}=\\frac{{540}}{6}\\,=90\\,\\,\\,\\,\\,\\surd$$. Example: number of girls to total", "of 7 people. (B) Add 1 to the quotient. If Ben gives tours to all 80 people, how many tours will he give? A tour can have no more than 7 people. To show all 80 people around, Ben will have to give 1 more tour. So, ben will give _________ tours in all for 80 people. So, ben will give 12 tours in all for 80 people Explanation: If Ben gives tours to all 80 people, A tour can have no more than 7 people. To show all 80 people around, Ben will have to give 1 more tour. 80 ÷ 7 = 11 3 people left. Estimate of 12 tours can be done. (C) Use only the remainder. Ben gives tours to all 80 people. After he completes the tours for groups of 7 people, how many people are in his last tour? The remainder is 3. So, Ben’s last tour will have ___________ people. 3 people only Explanation: Ben gives tours to all 80 people. After he completes the tours for groups of 7 people, how many people are in his last tour The remainder is 3. 80 ÷ 7 = 11 3 people left. Try This! Soccer players and their coaches are driven to soccer games in vans. Each van holds 9 people. How", "largest number of tours that can be run at the same time? Solution Answer: $2n$ First we will show an arrangement of $2n$ tours that satisfy the given constraints, and then we shall show that it is the largest attainable number. For each island, have a tour that starts at that island and circle it closely to the shore so that we don't greatly alter the shape of the island. Altogether, there are $n$ of such tours. Then, label the island $I_1, \\dots, I_n$. For each $i = 2, \\dots, n$, have a tour that starts at $I_1$, circles $I_1, \\dots, I_i$ and comes back to $I_1$. There are $n-1$ such tour. Finally we can have a tour that doesn't circle any island. In total, there are $2n$ tours. The proof for maximality is similar to our construction. There can be at most one tour that doesn't circle any island, and there can be at most $n$ tours that circle exactly one island. Now we show that there are at most $n-1$ tours that circle multiple islands. Let's call a tour that circles multiple islands a \"complex\" tour. Suppose $T_1$ and $T_2$ are both complex tours and both circle an island in common $I_1$. Furthermore, WLOG, we can assume that $T_1$ circle an island $I_2$ that $T_2$ doesn't.", "center. • How many tours are full? • How many tours are needed? • How many guests are on the last tour? Answer: a) 4 Tours are full. b) 5 Tours are needed. c) 3 guests are on the last tour. Explanation: Given that there are 31 guests in line to tour the space center. It is also given that there are no more than 7 guests in the tours of a space center. We have to observe that to make all the guests full on all the trips without leftovers, we will need 5 trips. i.e.., 35 guests. So, now we have to find the number of guests in each tour of a crayon factory by finding the quotient and remainder of 31 ÷ 7. Now, 31 ÷ 7 From this, we can see The number of guests in each tour that are full = 4 The number of guests leftover on the last trip = 3 Hence, from the above, We can conclude that a) 4 Tours are full. b) 5 Tours are needed. c) 3 guests are on the last tour. Question 9. Modeling Real Life You need 3 googly eyes to make one monster puppet. You have 28 googly eyes. How many monster puppets can you make? Answer: You can make 9 monster puppets.", "only, also impossible. Thus, we have just proved the following statement: Given two complex tours, either they are disjoint or one is inside the other. From here, it's matter of induction to show that there can be at most $n-1$ complex tours given $n$ islands. Take the smallest complex tour $T$ that does not contain another complex tour. $T$ circles at least two islands, and any other complex tour either contains $T$ or is disjoint from $T$. So as far as other complex tours are concerned, we can regard $T$ as an island on its own, thereby reducing the number islands by at least 1. Since there are now at most $n-1$ \"islands\", then there are at most $n-2$ complex tours. Altogether with $T$, we have at most $n-1$ tours.", "## Setting Mr. G. works as a tourist guide in Bangladesh. His current assignment is to show a group of tourists a distant city. As in all countries, certain pairs of cities are connected by two-way roads. Each pair of neighboring cities has a bus service that runs only between those two cities and uses the road the directly connects them. Each bus service has a particular limit on the maximum number of passengers it can carry. Mr. G. has a map showing the cities and the roads connecting them, as well as the service limit for each bus service. It is not always possible for him to take all the tourists to the destination city in a single trip. For example, consider seven cities with the following connections between them: City City Passenger limit ------------------------------- 1 2 30 1 3 15 1 4 10 2 4 25 2 5 60 3 4 40 3 6 20 4 7 35 5 7 20 6 7 30 It will take at least five trips for Mr. G. to take 99 tourists from city 1 to city 7, since he has to ride the bus with each group (the first four trips will consist of two ways each, the last trip of one way). The best route to take is", "Adam will be 23 years old. When Adam was as old as Ben, Ben was two years old. How old is today Ben and Adam? • Tourists Tourists hire a guide. Everyone will pay the same amount. If there are two less of them, they will pay 14 crowns more, and if there are three more, they will pay 14 crowns less. How many tourists are there and how much have the guides paid. • Two grandmothers Two grandmothers went to sell eggs at the market, and they had a total of 100. When they sold all the eggs, they made the same money. The first grandmother says to the second: \"If I sold my eggs for your price, I would earn 15 crowns. \" The other grandmot • Crowns 1180 crowns are divided into three people: A got 20% less than B and C by 15% more than B. How many got A, B, and C? • Four friends Four friends shared the money. Vasek got 1/4 of the total amount. Tonda received 1/3 of the rest of the money, Joe got a half from the second residue and Jirka left 80. How much money get together? • Logistics Head of logistics department informed the meeting of management of the same item is currently not available in", "There are two leaders in the tourist group, with an average age of 30 years and several children with an average age of 10 years. The total age of the group is 12 years. How many children are in the group? 15. Sum of the seventeen numbers The sum of the 17 different natural numbers is 154. Determine the sum of the two largest ones. 16. The tourist The tourist wanted to walk the route 16 km at a specific time. He, therefore, came out at the necessary constant speed. After a 4 km walk, however, he fell unplanned into the lake, where he almost drowned. It took him 20 minutes to get to the shore and re 17. Large family I have as many brothers as sisters and each my brother has twice as many sisters as brothers. How many children do parents have?", "another group of $6$ people. This is nothing but choosing $6$ people from a group of $18$ people. This is nothing but $\\binom{18}{6} = 18564 \\Rightarrow 564$. ~coolmath_2018", "number of tours needed so all the students were able to take a tour? Answer: The least number of tours was needed so all the students were able to take a tour is 3 with 1 leftover. Explanation: Given that five fourth grade classes from an elementary school took a trip to the United States Capitol and there were 25 students in each class and at the Capitol, so the number of students will be 25 × 5 which is 25 students. And a maximum of 40 students was allowed on a tour at one time, so the least number of tours was needed so all the students were able to take a tour is 125 ÷ 40 which is 3 with 1 leftover. Assessment Practice Question 14. Which division equation is represented by the drawing below? A. 72 ÷ 6 = 12 B. 62 ÷ 3 = 24 C. 64 ÷ 3 = 24 D. 72 ÷ 3 = 24 Answer: 72 ÷ 3 = 24. Explanation: In the above image, we can see that three circles with 24 in each circle. So the equation will be 72 ÷ 3 = 24. Question 15. What is the missing divisor? 2,244 ÷ n = 374 A. 3 B. 4 C. 6 D. 7 Answer: The missing divisor is", "22. Modeling Real Life Tours of a factory can have no more than 9 guests. There are 76 guests in line to tour the factory. • How many tours are full? • How many tours are needed? • How many guests are on the last tour? Answer: a) The number of tours that are full = 8 tours b) The number of tours needed = 9 tours c) The number of guests on the last tour = 4 guests Explanation: Given that there are 76 guests in line to tour a factory It is also given that there are no more than 9 guests on the tours of a factory. We have to observe that to make all the guests full on all the trips without leftovers, we will need 9 trips. i.e.., 81 guests. So, now we have to find the number of guests in each tour of a crayon factory by finding the quotient and remainder of 76 ÷ 9. Now, 76 ÷ 9 From this, we can see The number of guests in each tour that are full = 8 The number of guests leftover on the last trip = 4 Hence, from the above, We can conclude that a) 8 Tours are full. b) 9 Tours are needed. c) 4 guests are on the", "• # question_answer A father with 8 children takes them 3 at a time to the Zoological gardens, as often as he can without taking the same 3 children together more than once. The number of times each child will go to the garden is A) 56 B) 21 C) 112 D) None of these Solution : Each child will go as often as he (or she) can be accompanied by two others. Therefore the required number is${{=}^{4}}{{C}_{1}}{{\\times }^{6}}{{C}_{4}}=60$. You need to login to perform this action. You will be redirected in 3 sec", "situations in real life that can be modelled as a maths problem. Now let’s do some problems that use some of the translations above. As you can see, this problem is massive! The problems here only involve one variable; later we’ll work on some that involve more than one. We see that there are 330 ounces of ingredient a in solution Y. Take note that since x + 10 is the quantity of adult tickets, you must put One guide can only take 10 tourists and additional tour guides may be hired if needed. ingredient a}\\\\3\\times 54=162\\,\\,\\,\\,\\text{oz}\\text{. She sold 10 more adult tickets than children equation to match the problem. But what if you had 14? by the price of$5 you have to distribute the 5. This is important stuff - when it comes time to spend YOUR money - Let $$T=$$ the number of liters we need from the 20% concentrate, and then $$80-T$$ will be the number of liters from the 60% concentrate. You’d have 10 boys and 4 girls, since 10 is 5 times 2, and 4 is 2 times 2. How many minutes were charged on this bill? What is the number? In other words, we need to see how many boys out of 28 will keep a ratio of 5 boys to 7", "# Students are chosen in groups of 6 to tour a local business. How many ways can 6 students be selected from 3 classes totaling 53 students? $22.16 \\times {10}^{9}$ The way to find how many possibilities there are is by taking the number of items - $53$ - and putting it to the power of how many are chosen - $6$ -. For example a $3$ digit code which could have the numbers $0 \\to 9$ would have ${10}^{3}$ possibilities. ${53}^{6} = 22.16 \\ldots \\times {10}^{9}$", "3 in8C3 ways. Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways. Therefore, the total number of ways in which 8 students can travel is: $8C3+8C4$=56 + 70= 126", "inside. The magical powers of the temple doubles the flowers a devotee carries every time he/she passes under a doorway. Each devotee has to pass on straight through the doorway and cannot retrace his steps till he comes to the innermost idol. 27. Ram carries X flowers at each idol he places an identical number of flowers Y. He returns from the temple without a single flower. X was most probably (a)2 (b)5 (c)6 (d)7 28. In the situation above Y was most probably (a) 8(b)5(c)6(d)7 29. If Sita took 8 flowers to the temple and offered 4 flowers each to the first two idols then by the time she faces the third idol she has (a)40 flowers (b)36 flowers (c)52 flowers (d) 56 flowers aditiSECTION 4 - COMPUTATIONAL 30. 2 passengers have together 560 kgs of luggage and are charged for the excess above the weight allowed at 10$and 26$. If all the luggage had belonged to one of them he would have to pay 46$. The amount of luggage each passenger is allowed without any charge is (a)100 kg (b)150 kg (c)160 kg (d)Insufficient data 31. 6 pigs cost the same as 9 sheep. 27 sheep cost the same as 30 goats. 50 goats cost the same as 3 elephants. If two elephants cost$4800, then the", "pay their $5 to bowl. Therefore, each student will need to pay $$\\displaystyle \\frac{{500}}{x}+5$$, and since we don’t want any student to pay more than$15, the inequality that represents this situation is $$\\displaystyle \\frac{{500}}{x}+5\\le 15$$. To see how many students would have to attend to keep the cost at $15 per person, solve for $$x$$. In this example, we can multiply both sides by $$x$$ without worrying about reversing the inequality sign, since $$x$$ can only be positive. $$\\displaystyle \\frac{{500}}{x}+5\\le 15;\\,\\,\\,\\,\\frac{{500}}{x}\\le 10;\\,\\,\\,\\,500\\le 10x;\\,\\,\\,\\,x\\ge 50$$. At least 50 students would have to attend. Integer Function Problem (a little bit more advanced…) The fee for hiring a tour guide to explore Italy is$1000. One guide can only take 10 tourists and additional tour guides may be hired if needed. What is the cost of hiring tour guides, as a function of the number of tourists who go on the tour? If there are 72 tourists, what is the cost of hiring guides? Solution: Let’s think about this by using some real numbers. From 1–10 tourists the fee is $$1\\times 1000=\\1000$$, for 11–20 tourists, the fee is $$2\\times 2000=\\2000$$, and so on. Do you see how if we divide the number of tourists by 10, and go up to the next integer, we’ll get the number of tour guides we need?" ]

[ "$[asy] import olympiad; size(250); defaultpen(linewidth(0.7)+fontsize(11pt)); real r = 31, t = -10; pair A = origin, B = dir(r-60), C = dir(r); pair X = -0.8 * dir(t), Y = 2 * dir(t); pair D = foot(B,X,Y), E = foot(C,X,Y); draw(A--B--C--A^^X--Y^^B--D^^C--E); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,N); label(\"D\",D,dir(B--D)); label(\"E\",E,dir(C--E)); [/asy]$ ### Problem 28 The largest prime factor of $999999999999$ is greater than $2006$. Determine the remainder obtained when this prime factor is divided by $2006$. ### Problem 29 The altitudes in triangle $ABC$ have lengths 10, 12, and 15. The area of $ABC$ can be expressed as $\\tfrac{m\\sqrt n}p$, where $m$ and $p$ are relatively prime positive integers and $n$ is a positive integer not divisible by the square of any prime. Find $m + n + p$. $[asy] import olympiad; size(200); defaultpen(linewidth(0.7)+fontsize(11pt)); pair A = (9,8.5), B = origin, C = (15,0); draw(A--B--C--cycle); pair D = foot(A,B,C), E = foot(B,C,A), F = foot(C,A,B); draw(A--D^^B--E^^C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); [/asy]$ ### Problem 30 Triangle $ABC$ is equilateral. Points $D$ and $E$ are the midpoints of segments $BC$ and $AC$ respectively. $F$ is the point on segment $AB$ such that $2BF=AF$. Let $P$ denote the intersection of $AD$ and $EF$, The value of $EP/PF$ can be expressed as $m/n$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$. $[asy] import olympiad;", "CoCalc Public FilesFinal_Projects_F20 / Burnside_Russo_2.ipynb Views : 76 Compute Environment: Ubuntu 20.04 (Default) # Big Wheel Numbers Determine the chances of getting between a 0.85 and a 1.00 in two spins on The Price is Right Big Wheel game. The Big Wheel contains 20 values in 5 cent increments ranging from 5 to 100. The contestant can spin the wheel twice, and the value of each spin is added to their total price money amount. To determine to chances of getting a sum between 85 cents and 1.00 dollar with two spins, we can implement a function to generate all possible combinations of values and from there find the number of ways to obtain a total value of 0.85, 0.9, 0.95, or 1.00. In [ ]: import numpy as np values=[5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100] #20 possible spin values on the big wheel spinCombinations=[[spin1,spin2] for spin1 in values for spin2 in values] # all possible combinations of spins SpinSum=[[spin1+spin2] for spin1 in values for spin2 in values] #all possible sums of two spins SpinStreak=0 for i in range(len(SpinSum)): #evaluates combined sum of all combinations of two spins for Spins in SpinSum[i]: if (Spins>=85 and Spins<=100): print(spinCombinations[i], '=', SpinSum[i], 'cents') SpinStreak+=1 #if sum is between 85 and 100 then the spinStreak is appended print('Total number of combinations >=85 and <=100: ', SpinStreak) print('Total", "- 0.6562=14.7838$ C. Step 1: $0.35 x 16=5.6$ Step 2: $16 - 5.6=10.4$ Step 3: $0.425 x 10.4=4.42$ Step 4: $16 - 4.42=11.58$ D. Step 1: $0.35 x 16=5.6$ Step 2: $16 - 5.6=10.4$ Step 3: $0.425 x 10.4=4.42$ Step 4: $10.4 - 4.42=5.98$ 50. What is the value of the expression $(-\\dfrac{8}{9}) \\div (-\\dfrac{2}{3}) \\times (-4 \\dfrac{1}{2})$? A. $-6$ B. $-\\dfrac{8}{27}$ C. $\\dfrac{8}{27}$ D. $6$ 51. A board game has a spinner divided into sections of equal size. Each section is labeled with a number between $1$ and $5$. Which number is a reasonable estimate of the number of times the spinner will land on a section labeled $5$ over the course of $150$ spins? A. $15$ B. $25$ C. $40$ D. $60$", "# 1991 AJHSME Problems/Problem 22 ## Problem Each spinner is divided into $3$ equal parts. The results obtained from spinning the two spinners are multiplied. What is the probability that this product is an even number? $[asy] draw(circle((0,0),2)); draw(circle((5,0),2)); draw((0,0)--(sqrt(3),1)); draw((0,0)--(-sqrt(3),1)); draw((0,0)--(0,-2)); draw((5,0)--(5+sqrt(3),1)); draw((5,0)--(5-sqrt(3),1)); draw((5,0)--(5,-2)); fill((0,5/3)--(2/3,7/3)--(1/3,7/3)--(1/3,3)--(-1/3,3)--(-1/3,7/3)--(-2/3,7/3)--cycle,black); fill((5,5/3)--(17/3,7/3)--(16/3,7/3)--(16/3,3)--(14/3,3)--(14/3,7/3)--(13/3,7/3)--cycle,black); label(\"1\",(0,1/2),N); label(\"2\",(sqrt(3)/4,-1/4),ESE); label(\"3\",(-sqrt(3)/4,-1/4),WSW); label(\"4\",(5,1/2),N); label(\"5\",(5+sqrt(3)/4,-1/4),ESE); label(\"6\",(5-sqrt(3)/4,-1/4),WSW); [/asy]$ $\\text{(A)}\\ \\frac{1}{3} \\qquad \\text{(B)}\\ \\frac{1}{2} \\qquad \\text{(C)}\\ \\frac{2}{3} \\qquad \\text{(D)}\\ \\frac{7}{9} \\qquad \\text{(E)}\\ 1$ ## Solution Instead of computing this probability directly, we can find the probability that the product is odd, and subtract that from $1$. The product of two integers is odd if and only if each of the two integers is odd. The probability the first spinner yields an odd number is $2/3$ and the probability the second spinner yields an odd number is $1/3$, so the probability both yield an odd number is $(2/3)(1/3)=2/9$. The desired probability is thus $1-2/9=7/9\\rightarrow \\boxed{\\text{D}}$.", "randomness. Sunday, March 29, 2009. Suppose there is a spin 1/2 particle in a state$\\chi = {1 \\over \\sqrt{5}} \\begin{bmatrix} 1\\\\ 2\\\\ \\end{bmatrix} $. Therefore, the charity can be very confident about gaining a net contribution of at least$500 from 1,000 plays of the game. The randomness comes from atmospheric noise, which for many purposes is better than the pseudo-random number algorithms typically used in computer programs. Each branch in a tree diagram represents a possible outcome. You count the number of sections with the result you want and then you place it over the. Each of the four outcomes on this wheel is equally likely and outcomes are independent from one spin to the next. I doubt that you could fabricate any wheel which could have a tangential speed 0. If you win $100, cash out$50 and play with the rest, for example. A stainless steel razor blade scrapes away the top layer to expose a 0. P(x) = €5 + €1. Let q be the probability of losing e. The player with the higher (or highest) total on this extra spin wins. You spin a wheel and it randomly lands on $1,$2, or END. The student then writes down the equations or answers that are formed by the word wheel. Solution The wheel has 38 equally", "FALSE to your expand.grid call; otherwise, expand.grid will save the combinations as factors, an unfortunate choice that will disrupt the score function. Solution. To create a data frame of each combination of three symbols, you need to run expand.grid and give it three copies of wheel. The result will be a data frame with 343 rows, one for each unique combination of three slot symbols: combos <- expand.grid(wheel, wheel, wheel, stringsAsFactors = FALSE) combos ## Var1 Var2 Var3 ## 1 DD DD DD ## 2 7 DD DD ## 3 BBB DD DD ## 4 BB DD DD ## 5 B DD DD ## 6 C DD DD ## ... ## 341 B 0 0 ## 342 C 0 0 ## 343 0 0 0 Now, let’s calculate the probability of getting each combination. You can use the probabilities contained in the prob argument of get_symbols to do this. These probabilities determine how frequently each symbol is chosen when your slot machine generates symbols. They were calculated after observing 345 plays of the Manitoba video lottery terminals. Zeroes have the largest chance of being selected (0.52) and cherries the least (0.01): get_symbols <- function() { wheel <- c(\"DD\", \"7\", \"BBB\", \"BB\", \"B\", \"C\", \"0\") sample(wheel, size = 3, replace = TRUE, prob = c(0.03, 0.03, 0.06, 0.1, 0.25,", "# SOLUTION: Wheel of Fortune: In Exercises 35-38 use the small replica of the Wheel of Fortune. If the wheel is spun at random, determine the probability of the sector indicated stopping unde Algebra -> Algebra -> Probability-and-statistics -> SOLUTION: Wheel of Fortune: In Exercises 35-38 use the small replica of the Wheel of Fortune. If the wheel is spun at random, determine the probability of the sector indicated stopping unde Log On Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Algebra: Probability and statistics Solvers Lessons Answers archive Quiz In Depth Click here to see ALL problems on Probability-and-statistics Question 639136: Wheel of Fortune: In Exercises 35-38 use the small replica of the Wheel of Fortune. If the wheel is spun at random, determine the probability of the sector indicated stopping under the pointer. A number greater than $700.Answer by solver91311(16872) (Show Source): You can put this solution on YOUR website! Count the number of sectors that have a value greater than$700. Do not count sectors that ARE $700 because$700 is not greater than \\$700. This is the numerator of your fraction. Count the number of sectors altogether. This is your denominator. The fraction (once you have reduced it to lowest terms) is", "### Home > PC > Chapter 4 > Lesson 4.2.2 > Problem4-89 4-89. On the midway of MathAmerica Carnival land is a wheel of fortune. A transparent wheel with a radius of one meter is in front of a shaded winning region that is stationary behind the wheel. Spin the wheel and if point P lands in the region, you win. 1. If the point moves $5$ radians around the circle, how many degrees has it moved? 2. Show the new position of the point after the rotation in part (a). 3. The point has traveled a total of $1200$ degrees. How far has it traveled? 4. If the point rests in the shaded region when the wheel stops, the player wins. Will the player win? $\\frac{\\text{degrees}}{180\\degree}=\\frac{\\text{radians}}{\\pi}$ $\\frac{1200}{360}$ number of times around $+$ the remainder. $\\text{Circumference }=(2\\pi)\\text{radius}$ That is short of the shaded region.", "n) = ((100-n)/99)^{N-1} ((n-1)/99)$ Which is effectively saying we want the first $N-1$ rolls to be greater than the first roll, and the last one to be the complement, or:$1-(100-n)/99 = (n-1)/99$. So now that we have this, assuming it's correct, I'm stuck on the following: Find E[N], assuming an infinite number of contestants. I know that the expected value formula is the following: $sum_{x=2}^{\\infty} xp_x(x)$ where x is the index of the contestant. However, the contestant's probability also relies on the value of the initial value of the first spin. Is the next logical step to decompose into a double sum and evaluate that? Or is it something completely different? • First, there’s a piece of information missing from the problem: what is the value of $N$ if the first contestant’s spin is the smallest? ($1$ seems reasonable.) In addition, you’re confusing $n$ in $P(N\\gt n)$ with the result of the first contestant’s spin. You can see that there’s something wrong with your solution so far by considering the case of only two contestants: by symmetry, $P(n>1)$ must be $1/2$, that is, the second contestant has a smaller spin half of the time, but your solution produces $(1-1)/99=0$. – amd Jan 3 '17 at 3:31 There are a few problems with the first part, so I'll show", "will enter and visualize the probability distribution for 20-line Cleopatra. (Note that while there is a bonus/free spin feature, the probability distribution is constructed such that the payout structure is the same, but free spins are not included in simulations. This makes life easier.) In [1]: import matplotlib.pyplot as plt import numpy as np import scipy.stats as st %matplotlib inline In [2]: # Possible intervals you could fall into # [0, 0.2] is losing, the rest are winning intervals # If you fall in an interval, multiply your total bet by a random number in the interval to get your winnings # For example, for a 30 cent bet: you land in [2., 5.]. Select a random number \"x\" from a uniform dist. in [2,5]. # Your winnings are 0.3 * x in dollars intervals20 = np.array([ [0., 0.2], [0.2, 0.5], [0.5, 1.], [1., 2.], [2., 5.], [5., 10.], [10., 20.], [20., 50.], [50., 100.], [100., 200.], [200., 500.], [500., 1000.]]) # This is a scaled version of the winning probability distribution from Casino Guru. # This corresponds to all entries in \"intervals20\" except the first entry, which is the losing interval. # The multiplication factor 0.798... ensures that expected return is 95.025% cleopatra20_prob = 0.7987446566693283 * np.array( [929482, 740452, 563289, 1031867, 149001, 92050, 58006, 17450, 3538, 505,", "A miniature roulette wheel is divided into 10 equal sectors [#permalink] ### Show Tags 11 Sep 2018, 09:29 ok , i know the previous answers are straight to the point ; my take : 1-P(having an odd product)=1-P(the product of 3 odd integers) P(the product of 3 odd integers) = number of choices of having 3 odd integers / number of all choices number of choices of having 3 odd integers = OOO = 5^3 (Repetition and order matter) number of all choices = 10^3 1-P (odd product) =(10^3-5^3)/10^3 =7/8=0.875 GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 935 Re: A miniature roulette wheel is divided into 10 equal sectors [#permalink] ### Show Tags 24 Jan 2019, 07:17 pacifist85 wrote: A miniature roulette wheel is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly determines the winning sector by settling in that sector. If the wheel is spun three times, approximately what is the probability that the product of the three winning sectors’ integers will be even? A 88% B 75% C 67% D 63% E 50% $$?\\,\\,\\, \\cong \\,\\,\\,1 - P\\left( {\\underbrace {{\\text{all}}\\,\\,{\\text{odd}}\\,\\,{\\text{sectors}}}_{{\\text{unfavorable}}}} \\right)$$ $$P\\left( {{\\rm{all}}\\,\\,{\\rm{odd}}\\,\\,{\\rm{sectors}}} \\right)\\,\\,\\,\\mathop = \\limits^{{\\rm{independency}}} \\,\\,\\,{1 \\over 2} \\cdot {1 \\over 2} \\cdot {1 \\over 2}", "helps to regain vehicle stability. The charity would like to know the probability of obtaining a net contribution of at least S500 in 1,000 plays of the game. Dice Roller. 57% of spins, or 892 of them, to win. The contestants win cans of turtle wax based on which number they land on. The columns of values you enter in the worksheet go from inner to outer. Download or preview 1 pages of PDF version of Basic periodic table - black and white (DOC: 158. Universal Wheel of Chance Brief Description: Thisexhibitisalarge wheel%with%the%numbers%from0%to%99%thatvisitors%spin. Respectively, the table layout features an additional 'double-0' area along with the regular '0' and the other 36 black and red numbers. On the double zero wheel, it has a house edge of 7. The Money Wheel was launched at the Jackpot Casino in Red Deer, Alberta, in June 2011. This is a wheel divided into 12 segments, of which five are £0 and seven have a value, most £1 but there is one £512. Double-click the set number to use the names in the Slot Machine. Spin the wheel by clicking or tapping the spin icon. Roulette is a casino game named after the French word meaning little wheel. - the smallest wheel has 24 spots, all marked with numbers 1-24. Using the prize", "Go Back to the Home Page :-) Classroom Timers - Fun Timers for classrooms and meetings :-) Holiday Timers - More Fun Timers - But these are Holiday. * Let x represent the total score for a single player. Assume that the cost to spin the wheel once is $5. Wheel to Spin Plus Instructions. One, it guarantees that the highest-scoring elements will be most likely to reproduce. Probability Spinner Spin the wheel and calculate both theoretical and experimental probabilitiesif you dare. The chance of winning is 4 out of 52, while the chance against winning is 48 out of 52 (52-4=48). Imagine a roulette wheel with the stops labelled as 0. See the image below for an example of how easy it is to modify your prizes. But you can see that the odds of that are happening even with just even numbers that have 1:1 payouts are pretty much the same. You can also design your own wheel and a sign that describes the probabilities for your wheel. We enter the values 4 times in the column. These can be used as random number generators in lessons. The ball will be spun in the opposite direction. Each index should have equal probability of returning pick(1) should return 0. Now I have spun the wheel 12 times", "of the groups were fairly strong, there had to be something about the way the question was posed that was throwing them off. I got it, but something was off for them. • The questions the students were asking were all about winning or losing. For example, if they chose purple, but the spinner landed on blue, what would happen? The assumption they had in their heads was that they would either get \\$200 or nothing. Of course they would choose to wait until there was a better than 50:50 chance before switching to purple. The part about maximizing the winnings wasn't what they understood from the task. • When I modified the language in the sketch to say when do you 'choose' purple instead of 'bet' on the \\$200 between the Algebra 2 group and the Geometry group, there wasn't a significant change in the results. They still tended to choose percentages that were close to the 50:50 range. I made an updated sketch that allowed students to do just that, available here in my Geogebra repository. It lets the user choose the moment for switching, simulates 500 spins, and shows the amount earned if the person stuck to either color. I tried it out on an unsuspecting student that stayed after school for some help, one", "### Home > PC > Chapter 4 > Lesson 4.2.2 > Problem4-89 4-89. 1. On the midway of MathAmerica Carnival land is a wheel of fortune. A transparent wheel with a radius of one meter is in front of a shaded winning region that is stationary behind the wheel. Spin the wheel and if point P lands in the region, you win. Homework Help ✎ 1. If the point moves 5 radians around the circle, how many degrees has it moved? 2. Show the new position of the point after the rotation in part (a). 3. The point has traveled a total of 1200 degrees. How far has it traveled? 4. If the point rests in the shaded region when the wheel stops, the player wins. Will the player win? $\\frac{\\text{degrees}}{180\\degree}=\\frac{\\text{radians}}{\\pi}$ $\\frac{1200}{360}=\\text{ number of times around + the remainder.}$ $\\text{Circumference }=(2\\pi)\\text{radius}$ That is short of the shaded region.", "+0 # A spinner used in a board game is divided into 8 equally sized sectors. 0 293 1 A spinner used in a board game is divided into 8 equally sized sectors. Three of these sectors indicate that the player should move his token forward on the board, two of these sectors indicate that the player should move his token backward, and the remaining sectors award the player bonus points but do not move his token on the board. The total area of the sectors that do not allow the player to move his token is 35.9 inches squared. What is the radius of the spinner? Enter your answer, rounded to the nearest tenth of an inch, in the box. So far I have, $$3/8( 360)= 135 degress$$ $$Area=1/2r^2sin135$$ $$35.9=1/2r^2(\\sqrt 2/2)$$ Apr 3, 2020 #1 +21015 0 3 of the 8 sectors do not allow the player to move the token. This area is 35.9 in2. (3/8)·pi·r2 = 35.9 in2 r = sqrt( 35.9 / pi · 8/3 ) Apr 3, 2020", "$IJ$, and still again with $GK$. If he now plays $CD$, I have nothing better than $DH$ (scoring one), but, as I have to play again, I am compelled, whatever I do, to give him all the rest. So he will win by $8$ to $1$ -- a bad defeat for me. Now, what should I have played instead of that disastrous $FG$? There is room for a lot of skilful play in the game, and it can never end in a draw. #### $209$ - A Wheel Fallacy The wheel shown in the diagram makes one complete revolution in passing from $A$ to $B$. It is therefore obvious that the line $AB$ is exactly equal in length to the circumference of the wheel. Now the inner circle (the large hub in the diagram) also makes one complete revolution along the dotted line $CD$ and, since the line CD is equal to the line $AB$, the circumference of the larger and smaller circles are the same. This is clearly not true. Wherein lies the fallacy? #### $210$ - A Famous Paradox When a bicycle is in motion, does the upper part of each wheel move faster than the bottom part near the ground? #### $211$ - Another Wheel Paradox When a railway train is in motion, it is", "200 5 5 ## 8 V6 200 6 5 ## 9 V3 150 3 4 ## 10 V4 100 4 4 ## # ... with 24 more rows As you can see even when two events pay the same, there is order: Wild, Wild, Wild, Wild, Sym2 will pay for “4 Wilds in a row” and not “5 Sym2 in a row” even though they both pay x500, because the former appears first. # On Our Way to Simulation Land Let’s simulate a spin. This means one of 32 locations for each of the 5 reels: spin <- sample(1:32, 5, replace = TRUE) spin ## [1] 1 15 6 15 26 Wow, that was easy. If we agree the location sampled represents what will appear in the middle row, we can see how the grid would look like: reelsMat <- t(as.matrix(reels[, 2:6])) firstRowSpin <- ifelse(spin - 1 < 1, 32, spin - 1) thirdRowSpin <- ifelse(spin + 1 > 32, 1, spin + 1) resGrid <- rbind(reelsMat[cbind(seq_along(firstRowSpin), firstRowSpin)], reelsMat[cbind(seq_along(spin), spin)], reelsMat[cbind(seq_along(thirdRowSpin), thirdRowSpin)]) resGrid ## [,1] [,2] [,3] [,4] [,5] ## [1,] \"Sym6\" \"Wild\" \"Sym7\" \"Sym7\" \"Sym8\" ## [2,] \"Scatter\" \"Sym6\" \"Sym9\" \"Sym6\" \"Sym6\" ## [3,] \"Sym9\" \"Sym5\" \"Sym6\" \"Sym2\" \"Sym1\" Note first that the same location can be sampled for more than one reel, that is why we", "Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ Manager Joined: 19 Sep 2017 Posts: 126 Location: United Kingdom GPA: 3.9 WE: Account Management (Other) Re: A miniature roulette wheel is divided into 10 equal sectors [#permalink] ### Show Tags 14 Mar 2019, 07:31 pacifist85 wrote: A miniature roulette wheel is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly determines the winning sector by settling in that sector. If the wheel is spun three times, approximately what is the probability that the product of the three winning sectors’ integers will be even? A 88% B 75% C 67% D 63% E 50% Hi Folks, In order to find the probability that the product of the three winning sectors’ integers will be even, we need to find what are the chances that at least in one out of three times, the ball would stop at an even number. Why? because Even * Odd = Even. So, At least one = 1-None None means all three are odd. i.e. P(odd) = #Chances of odd/#total selections = 1/2 for all three times = (1/2)^3 = 1/8 Atleast one = 1-1/8 = 7/8 Since the question statement has asked percentage, Atleast one % = 7/8 * 100 =", "# Prize Wheels N/A Pokémon Essentials Version This script lets you create Prize Wheels that players can spin to win one of ten prizes! Your prize pool can be Pokemon, items, money, or a mix of the three! (Yes I got this idea from Webkinz/Neopets what are you a cop) Code Paste in a new section above Main! Ruby: Prizes = [ [:POKEBALL,:REPEATBALL,:FASTBALL,:LEVELBALL,:NESTBALL,:DUSKBALL,:DIVEBALL,:HEAVYBALL,:GREATBALL,:MASTERBALL], [:CHARMANDER,:CYNDAQUIL,:EEVEE,:PIKACHU,:CHIKORITA,:BULBASAUR,:SQUIRTLE,:TOTODILE,:MARILL,:RATTATA], [\"100\",\"200\",\"300\",\"400\",\"500\",\"600\",\"700\",\"800\",\"900\",\"1000\"], [:POKEBALL,\"100\",:CHARMANDER,:POKEBALL,\"100\",:CHARMANDER,:POKEBALL,\"100\",:CHARMANDER,:POKEBALL,\"100\",:CHARMANDER,], ] WheelStyles = [ #bg graphic, wheel graphic, radius, minimum spins, #spin SFX, volume, pitch #winning sfx, volume, pitch [\"hatchbg\",\"prizewheel\",110,3,\"battle ball shake\",80,150,\"mining reveal full\",100,100] ] def Wheel(prizelist,cost,style=0) PrizeWheel.new(prizelist,cost,style) end class PrizeWheel def pbUpdate end def initialize(prizelist,cost,style=0) @viewport = Viewport.new(0, 0, Graphics.width, Graphics.height) @prizes=Prizes[prizelist] @sprites = {} @cost=cost @style=WheelStyles[0] @bg=@style[0] @wheel=@style[1] @minspins=@style[3] @prizespin=0 @angle=[72,36,0,313,287,251,216,180,144,108] @xyangle=[198,234,270,306,340,16,52,90,126,162] @sprites[\"bg\"] = Sprite.new(@viewport) @sprites[\"bg\"].bitmap = Bitmap.new(\"Graphics/Pictures/#{@bg}\") @sprites[\"downarrow\"] = AnimatedSprite.new(\"Graphics/Pictures/downarrow\",8,28,40,2,@viewport) @sprites[\"downarrow\"].x = (Graphics.width/2)-15 @sprites[\"downarrow\"].y = 10 @sprites[\"downarrow\"].z = 5 @sprites[\"downarrow\"].play @sprites[\"wheel\"] = Sprite.new(@viewport) @sprites[\"wheel\"].bitmap = Bitmap.new(\"Graphics/Pictures/#{@wheel}\") @sprites[\"wheel\"].center_origins @sprites[\"wheel\"].x=Graphics.width/2 @sprites[\"wheel\"].y=Graphics.height/2 for i in 0...10 if getID(PBItems,@prizes[i])>0 @sprites[\"prize#{i}\"]=ItemIconSprite.new(0,0,0,@viewport) @sprites[\"prize#{i}\"].item=getID(PBItems,@prizes[i]) @sprites[\"prize#{i}\"].center_origins elsif getID(PBSpecies,@prizes[i])>0 @sprites[\"prize#{i}\"]=PokemonSpeciesIconSprite.new(getID(PBSpecies,@prizes[i]),@viewport) @sprites[\"prize#{i}\"].ox=32 @sprites[\"prize#{i}\"].oy=32 else @sprites[\"prize#{i}\"] = Sprite.new(@viewport) @sprites[\"prize#{i}\"].bitmap = Bitmap.new(\"Graphics/Pictures/Money\") @sprites[\"prize#{i}\"].center_origins end @sprites[\"prize#{i}\"].angle = @angle[i] @sprites[\"prize#{i}\"].x=(Graphics.width/2) + Math.cos(@xyangle[i].degrees)*@style[2] @sprites[\"prize#{i}\"].y=(Graphics.height/2) + Math.sin(@xyangle[i].degrees)*@style[2] end main end def main loop do Graphics.update Input.update pbUpdate if Input.trigger?(Input::C) if @cost>0 confirmtext=\"Spin the wheel for $#{@cost}?\" else confirmtext=\"Spin the wheel?\" end if pbConfirmMessage(\"#{confirmtext}\") if$Trainer.money>=@cost $Trainer.money-=@cost else pbMessage(_INTL(\"You don't have enough money...\")) break end spins=rand(360) spins+=360*(@minspins)" ]

[ "probability that the $\\heartsuit$ Ace lands in one of these slots is $\\frac{39}{51}$. Similarly, when it comes to deal the $\\diamondsuit$ Ace, there are $50$ open slots left, of which $26$ are \"favourable.\" And given that the first three Aces have landed nicely, the probability the fourth lands nicely is $\\frac{13}{49}$.", "### Home > INT2 > Chapter 3 > Lesson 3.1.5 > Problem3-61 3-61. The spinner at right has three regions: $\\5,\\0,$ and $\\20$. To play the game, you spin one time and then you win the amount in the region that the spinner lands on. 1. What is the expected value for the game? Compare the given angle measurements to $360^\\circ$. Multiply these ratios by the amount possible to win in each section. $\\text{EV}=\\5\\cdot \\frac{1}{4}+\\0\\cdot \\frac{135}{360}+\\20\\cdot \\frac{135}{360}$ 2. If you have to pay $\\10$ to play, is it a fair game?", "using several binary variables. • When the number of spins is 50, the number of possible configurations is $2^{50}$. That means $2^{50}$ terms in that equation. This seems to be impractical for 50 spins. Jul 25, 2021 at 6:29", "wins. C) So does that mean there is an average 125 spins to get all apples? The closest to 1 in 100 chance? I am confused. Well, I believe you have the same probability for each slot. $\\left(\\frac{a^3}{10}\\right)^3=\\frac1{100}$ so $a=\\sqrt[3]{10}$ so it must occupy $\\sqrt[3]{10}$ slots, whatever that means. If you put $1$ apple in one slot, you would get $\\frac1{1000},$ but if you put $2$ apples, then there are $2\\cdot2\\cdot2=8$ combinations for a total of $\\frac1{125}.$ Note: If you allow different (assuming integral) numbers of slots, it can be shown it must be a permutation if $(1,5,2)$ or $(1,1,10)$ slots. EDIT: There are a few more questions asked. A) Yes. There are $10$ combinations for each slot for a total of $10\\cdot10\\cdot10=1000$ combinations. B) Yes. The probability for one slot is $\\frac2{10}=\\frac15.$ The probability of three is $\\left(\\frac15\\right)^3=\\frac1{125}=\\frac8{1000}.$ C) Yes, the expected value of spins required is $125.$ It is the closest unless you allow one slot to have more apple slots than another. • Do you mean $\\frac1{1000}$ if there was just 1 apple? – Joe Sep 7 '18 at 4:40 • Yes... edited :) – Jason Kim Sep 7 '18 at 4:50", "+0 -1 96 3 +397 deleted. Nov 15, 2019 edited by sinclairdragon428 Nov 20, 2019 #2 +118 +3 Actually out of $$5\\cdot5\\cdot5=125$$possible outcomes, there are 6 outcomes that result in winnings of $1700. They are $$(300, 400, 1000), (300, 1000, 400), (400, 300, 1000), (400, 1000, 300), (1000, 300, 400)$$and $$(1000, 400, 300)$$ where the ordered triplets indicate the results of the first, second, and third spin, respectively. So the probability we are looking for is $$\\frac{6}{125}$$. Nov 15, 2019 ### 3+0 Answers #1 0 The only way to get 1700 is to land on 300, 400, then 1000, so the probability is 1/(5*4*3) = 1/60. Nov 15, 2019 #2 +118 +3 Best Answer Actually out of $$5\\cdot5\\cdot5=125$$possible outcomes, there are 6 outcomes that result in winnings of$1700. They are $$(300, 400, 1000), (300, 1000, 400), (400, 300, 1000), (400, 1000, 300), (1000, 300, 400)$$and $$(1000, 400, 300)$$ where the ordered triplets indicate the results of the first, second, and third spin, respectively. So the probability we are looking for is $$\\frac{6}{125}$$. Gadfly Nov 15, 2019 #3 +397 -1 thanks! sinclairdragon428 Nov 15, 2019", "the board once, you get a payment of $200$ dollars (first case) for each time the player lands on a railway space. So if she is really unlucky, you get $800$ dollars and begin to dream about a vacation. We also make the probably unreasonable assumption that landing on any one of the spaces is equally likely, and that the events are independent. We also make the assumption that it is not possible to land twice on the same space in one tour of the board, something that is in fact not true, since sadly, in Monopoly as in life, one sometimes makes negative progress. Four railways, one payer: Let $X_1,X_2,X_3,X_4$ be the incomes we get from each of the four railways. Then our total income is $X_1+\\cdots+X_4$, which has expectation $E(X_1)+\\cdots+E(X_4)$. This is $4\\left(\\frac{1}{40}\\cdot 200\\right)$. Three railways, two payers: Let $X$ be the income we get from the first payer, and $Y$ the income we get from the second. We want $E(X+Y)$, which is $E(X)+E(Y)$. By an analysis like the one above, we have $E(X)=3\\left(\\frac{1}{40}\\cdot 100\\right)$. But $E(Y)$ is the same, so your expected income is $(2)(3)\\left(\\frac{1}{40}\\cdot 100\\right)$. Remark: So deal $1$ is better, if we neglect the cost of acquiring that fourth railway. It is more interesting to solve the problem under the assumption we can", "thus making this all more impressive? Turns out that if we know we’ll have about 60 spins, we can make the smallest slice 1/25th of the spinner and still be confident we’ll hit it at least once. Cool. If we want to keep the Twister board, and have the smallest slice be 1/16th of the circle, we actually have a 90% chance of hitting it at least twice. So that makes things even more predictable (for the teacher), while still making it less predictable (to the kids) than the previous spinner. More messing around led to this suggested spinner: 1/2 chance of$100, 5/16 chance of $200, 1/8 chance of$400, and 1/16 chance of $2500. The chance that “$2500″ is the right bet after 60 spins of this spinner is about 88%, so again you can make your bet confidently — but this time, the “right” answer doesn’t look as obvious. In short, I’d recommend using this spinner for around 60 spins, rather than Dan’s spinner for 20 spins. It’s not guaranteed to be “optimal” but it’s far more reliable than the original suggestion. If anyone tries it, I’d be curious to hear how it went! Spinner Doctor The setup Dan Meyer, a (former?) math teacher with some extraordinary ideas, has a nifty concept for teaching expected values: “So", "Spin a spinner numbered 1-8 and you land on 8 twice in a row what is the probability? Apr 29, 2018 The probability of landing on 8 twice is $\\frac{1}{64}$. Explanation: If you spin the spinner once, there are 8 possibilities each time (1-8). The possibility of landing on the number 8 is 1 of these 8 possibilities. Its probability is therefore $\\frac{1}{8}$. Each spin is independent. In other words, the spins do not affect each other. The number you land on during the first spin does not impact the second spin's number. The probability of landing on 8 the second time is the same as the first time: $\\frac{1}{8}$. Because the two events are independent, we find the probability of them both occurring by multiplying them together. $\\frac{1}{8} \\cdot \\frac{1}{8} = \\frac{1}{64}$ The probability of landing on 8 twice is $\\frac{1}{64}$. Another way to think about this is to consider all of the total possibilities. If there are 8 possibilities for the first spin, there are 8 possibilities for the second spin. $8 \\cdot 8$ gives 64 total combinations for the two spins. Below are just some of the possibilities. • 1,1 • 1,2 • 1,3 • 1,4 • 1,5 • 1,6 • 1,7 • 1,8 • 2,1 • 2,2 There are 64 total possibilities and 8,8", "smallest slice, while keeping predictability high, and thus making this all more impressive? Turns out that if we know we’ll have about 60 spins, we can make the smallest slice 1/25th of the spinner and still be confident we’ll hit it at least once. Cool. If we want to keep the Twister board, and have the smallest slice be 1/16th of the circle, we actually have a 90% chance of hitting it at least twice. So that makes things even more predictable (for the teacher), while still making it less predictable (to the kids) than the previous spinner. More messing around led to this suggested spinner: 1/2 chance of$100, 5/16 chance of $200, 1/8 chance of$400, and 1/16 chance of $2500. The chance that “$2500” is the right bet after 60 spins of this spinner is about 88%, so again you can make your bet confidently — but this time, the “right” answer doesn’t look as obvious. In short, I’d recommend using this spinner for around 60 spins, rather than Dan’s spinner for 20 spins. It’s not guaranteed to be “optimal” but it’s far more reliable than the original suggestion. If anyone tries it, I’d be curious to hear how it went!", "## NEW – Spin for Cash: Touch the Spinner wheel and Win It Opinions: SCAM or LEGIT? | Proxies123.com Hello everyone, have you tried this Spin for Cash application: Touch the Spinner & Win It wheel? This application allows the user to play by turning the wheel and earn coins. They have 6 ways to earn coins. ✓ lucky luck ✓ Unlimited centrifugation ✓ invite and win ✓ Install and win The minimum payment is 40,000 coins with a value of \\$ 1 and the payment method is through the account PAYPAL & JAZZCASH Link here if you want to try I won cash using the Spin For Cash application. You can also win by downloading the application from the link below and enter the reference code at login-P05IJ3RX The reference code will help you get a bonus of 100 coins. ## c # – How do I add random objects to a random terrain using lightning bolts, and spin them with the terrain? I have a game that requires, random generation of trees and rocks for a world that is generated as you move. I have a rough installation system where I install the rigid bodies in a random location with a height of approximately 50, and they fall and create instances of trees and rocks, where", "but my teacher knew which one would win!” But the lesson just isn’t true with this spinner and time-frame: here, the highest-expected-value choice is actually NOT the one most likely to have earned the most money after only 20 or 30 spins. And the wow-factor is not guaranteed: none of the choices is much more likely to win than the others in only 20-30 spins, so the teacher can’t know the winning bet in advance. It’s like you’re a magician doing a card trick that only works a third of the time. You can still have a good discussion about the math, but it’s just not as cool. I’d like to re-design the spinner so that the lesson is true, and the wow-factor still happens, after only a month of spins. WAit, is there really a problem? First, what’s wrong with the spinner? By my eyeball, the expected values per spin are$100/2 = $50;$300/3 = $100;$600/9 = $67ish;$5000/27 = $185ish; and$12000/54 = $222ish. So in the LONG run, if you spin this spinner a million times, the “$12000″ has the highest expected value and is almost surely the best bet. No question. But in Dan’s suspense-building setup, you only spin once a day for a month, for a total of 20ish spins (since weekends are out). With only", "# Thread: red followed by a yellow 1. ## red followed by a yellow there are 3 red spaces, 1 blue space, 3 yellow spaces, and one green space for a total of 8 spaces on a spinner. What is the probability of getting a red spin followed by a yellow spin in two spins? I am new to probability, so I wanted to check out a couple of my homework questions. Here is what I think: There are n(s) = 8*8 = 64 possible spins in two total spins n(r,y) = 3C1 * 3C1 = 9 possible reds followed by a yellow so the probability of spinning a red then yellow in two spins is 9/64??? 2. Hello, ihavvaquestion! The probability that the first spin is Red is: $\\frac{3}{8}$ The probability that the second spin is Yellow is: $\\frac{3}{8}$ Therefore: . $P(\\text{Red, then Yellow}) \\;=\\;\\frac{3}{8}\\cdot\\frac{3}{8} \\;=\\;\\frac{9}{64}$", "1) ^ {ind (D_ rho)}$$, where $$D_ rho$$ is the Dirac operator associated with the spin structure and $$ind (D_ rho)$$ It is the index of the operator. My question could also be formulated as if the index of a Dirac operator is an integral part of some local quantity that is \"canonically\" associated with the spin structure. NEW – Spin for Cash: Touch the Spinner wheel and Win It Opinions: SCAM or LEGIT? | Proxies123.com Hello everyone, have you tried this Spin for Cash application: Touch the Spinner & Win It wheel? This application allows the user to play by turning the wheel and earn coins. They have 6 ways to earn coins. ✓ lucky luck ✓ Unlimited centrifugation ✓ invite and win ✓ Install and win The minimum payment is 40,000 coins with a value of \\$ 1 and the payment method is through the account PAYPAL & JAZZCASH Link here if you want to try I won cash using the Spin For Cash application. You can also win by downloading the application from the link below and enter the reference code at login-P05IJ3RX The reference code will help you get a bonus of 100 coins. c # – How do I add random objects to a random terrain using lightning bolts, and spin them with", "group. Now reword the conclusion as \"Out of 253 pairs of people, there is a 50% chance that one pair will share a birthday.\" ## Explanation 2 by visualizing a spinning prize wheel Picture a giant spinner wheel, like at a carnival. There are 367 pegs making 366 slots for the pointer to land on. We'll pretend the 366 slot is 1/4 the size of the others to signify Feb. 29. Once someone lands on a slot, it's colored in before the next person spins. For the first 10 or so spins you have 1/366, 2/366, 3/366, etc... chance of landing on a colored slot, quite low odds. However, at say the 60th person around $$~1/6$$ of the wheel will be colored. Using these crude numbers, wouldn't you expect to hit a $$~1/6$$ chance sometime in the next 10 spins? Landing on the non-colored slots would equate to roughly $$(5/6)^{10}$$ which is about a $$~16\\%$$ chance just in those 10 spins. This would mean in those last hypothetical 10 spins you would have $$~84\\%$$ chance of landing on at least 1 previously landed-on slot. In this example the 'randomness' of the spin indicates the perceived 'randomness' of each person's birthday in the selected group. ## Explanation 3 by visualizing the Pigeonhole Principle Once you understand what the \"problem\"", "ace, we first need to have the first $29$ slots ace-free. The first ace has $52$ slots to go into, of which $23$ are not in the first $29$, so it will be in the correct zone with probability $\\frac{23}{52}$. Assuming that this first ace misses the forbidden zone, the second ace has $51$ slots, of which $22$ are not in the first $29$, and so on. After placing all four aces, the probability of an ace-free first $29$ is $P_1$:$$P_1=\\frac{23\\times22\\times21\\times20}{52\\times51\\times50\\times49}$$ Assume that the first $29$ slots are indeed ace-free. What is the probability that all four aces have missed Slot #$30$? The first ace has, we know, landed in one of $23$ slots; $22$ of them are not Slot #$30$, so this first ace misses Slot #$30$ with probability $\\frac{22}{23}$. The second ace has only $22$ slots, of which $21$ are not Slot #$30$ and so on. The probability that all four aces miss Slot #$30$ is $P_2$:$$P_2=\\frac{22\\times21\\times20\\times19}{23\\times22\\times21\\times20}=\\frac{19}{23}$$So the probability that one of the four aces will hit Slot #$30$ is $1-P_2$, or $\\frac{4}{23}$ So the overall probability of finding the first ace in Slot #$30$ is $$P=P_1 \\times \\frac4{23}=\\frac{23\\times22\\times21\\times20\\times4}{52\\times51\\times50\\times49\\times23}$$", "of slot for each ball, and these choices are made independently for each ball, so they can be made in $$15^4=50,625$$ ways. Equivalently, you’re just counting the possible $$4$$-tuples of slot numbers, where $$\\langle 1,1,1,1\\rangle$$ represents your first combination, $$\\langle 1,1,1,15\\rangle$$ represents your second combination, $$\\langle 1,6,6,1\\rangle$$ represents your fourth combination, and so on.", "spinner a million times, the “$12000″ has the highest expected value and is almost surely the best bet. No question. But in Dan’s suspense-building setup, you only spin once a day for a month, for a total of 20ish spins (since weekends are out). With only 20 spins, the results are too unpredictable with the given spinner — none of the five choices is especially likely to be the winner. How do we know? Instead of thinking “the action is spinning the spinner once, and we’re going to do this action twenty times,” let’s look at it another way: “the action is spinning the spinner twenty times in a row, and we’re going to do this action once.” That’s what really matters to the classroom teacher running this exercise: you get one shot to confidently place my bet at the start of the month; after a single month of daily spins, will the kids be wowed by seeing that you placed the right bet? I ran a simulation in R (though sometime I’d like to tackle this analytically too): Take 20 random draws from a multinomial distribution with the same probabilities as Dan’s spinner. Multiply the results by the values of each bet. > nr.spins <- 20 > spins=rmultinom(1,size=nr.spins,prob=c(1/2,1/3,1/9,1/27,1/54)) > spins [,1] [1,] 11 [2,] 7 [3,] 2", "20 spins, and then it becomes unpredictable. It turns out the smallest sector has to be at least about 1/9th of the spinner if you want to be 90% confident of hitting that sector at least once in those 20 spins. (Let $y \\sim \\mathrm{Binomial}(p=1/9, n=20)$ . Then $latex p(y<0) = 1-p(y=0) = 0.905$.) If we round that up to 1/8th instead, we can easily use a Twister spinner (which has 16 equal sectors). After playing with some different options, using the same simulation approach as the previous post, I found that the following spinner seems to work decently: 1/2 chance of $100, 3/8 chance of$150, and 1/8 chance of $1500. After 20 spins of this spinner, there's about a 87% chance that the \"$1500\" will have been the winning bet, so you can be pretty confident about making the right bet a month in advance. Unfortunately, predicting that spinner correctly is kind of unimpressive. The “$1500” is a fairly big slice, so it doesn’t look too risky. ### spin just a few more times and use a safer spinner! What if we spin it just a few more than 20 times — say 60 times, so two or three times each day? That’s not too much data to keep track of. Will that let us shrink the", "payout total payout of 89.6%. My sloppy calculation of the retriggers yields .078749 free spins per regular spin for a payout of 90.7497%. I don't think that is correct. Thanks for any solutions on how to calculate this. Consider a single regular spin. With probability $0.25$ it pays $3$, and with probability $0.005$ it pays out the pay from $10$ free spins, with probability $0.001$ it pays out the pay from $15$ free spins and with probability $0.0004$ it pays out the pay from $20$ free spins. So the expected pay is: $ p=0.25 \\times 3 + 0.005 \\times 10 \\times x + 0.001 \\times 15 \\times x + 0.0004 \\times 20 \\times x $ where $x$ is the expected pay as a result of single free spin (including all the extra free spins that may be won). Now $x=0.25 \\times 8 + 0.005 \\times 10 \\times x + 0.001 \\times 15 \\times x + 0.0004 \\times 20 \\times x$, or: $x=025 \\times 8 + 0.073x$, so $x=2.1575$ CB • November 26th 2008, 05:17 PM danno sweet! That is exactly what I needed, it did end up matching to my calculations but yours is precise rather than an approximation. Thanks Again!", "CoCalc Public FilesFinal_Projects_F20 / Burnside_Russo_2.ipynb Views : 76 Compute Environment: Ubuntu 20.04 (Default) # Big Wheel Numbers Determine the chances of getting between a 0.85 and a 1.00 in two spins on The Price is Right Big Wheel game. The Big Wheel contains 20 values in 5 cent increments ranging from 5 to 100. The contestant can spin the wheel twice, and the value of each spin is added to their total price money amount. To determine to chances of getting a sum between 85 cents and 1.00 dollar with two spins, we can implement a function to generate all possible combinations of values and from there find the number of ways to obtain a total value of 0.85, 0.9, 0.95, or 1.00. In [ ]: import numpy as np values=[5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100] #20 possible spin values on the big wheel spinCombinations=[[spin1,spin2] for spin1 in values for spin2 in values] # all possible combinations of spins SpinSum=[[spin1+spin2] for spin1 in values for spin2 in values] #all possible sums of two spins SpinStreak=0 for i in range(len(SpinSum)): #evaluates combined sum of all combinations of two spins for Spins in SpinSum[i]: if (Spins>=85 and Spins<=100): print(spinCombinations[i], '=', SpinSum[i], 'cents') SpinStreak+=1 #if sum is between 85 and 100 then the spinStreak is appended print('Total number of combinations >=85 and <=100: ', SpinStreak) print('Total" ]

[ "1) ^ {ind (D_ rho)}$$, where $$D_ rho$$ is the Dirac operator associated with the spin structure and $$ind (D_ rho)$$ It is the index of the operator. My question could also be formulated as if the index of a Dirac operator is an integral part of some local quantity that is \"canonically\" associated with the spin structure. NEW – Spin for Cash: Touch the Spinner wheel and Win It Opinions: SCAM or LEGIT? | Proxies123.com Hello everyone, have you tried this Spin for Cash application: Touch the Spinner & Win It wheel? This application allows the user to play by turning the wheel and earn coins. They have 6 ways to earn coins. ✓ lucky luck ✓ Unlimited centrifugation ✓ invite and win ✓ Install and win The minimum payment is 40,000 coins with a value of \\$ 1 and the payment method is through the account PAYPAL & JAZZCASH Link here if you want to try I won cash using the Spin For Cash application. You can also win by downloading the application from the link below and enter the reference code at login-P05IJ3RX The reference code will help you get a bonus of 100 coins. c # – How do I add random objects to a random terrain using lightning bolts, and spin them with", "## NEW – Spin for Cash: Touch the Spinner wheel and Win It Opinions: SCAM or LEGIT? | Proxies123.com Hello everyone, have you tried this Spin for Cash application: Touch the Spinner & Win It wheel? This application allows the user to play by turning the wheel and earn coins. They have 6 ways to earn coins. ✓ lucky luck ✓ Unlimited centrifugation ✓ invite and win ✓ Install and win The minimum payment is 40,000 coins with a value of \\$ 1 and the payment method is through the account PAYPAL & JAZZCASH Link here if you want to try I won cash using the Spin For Cash application. You can also win by downloading the application from the link below and enter the reference code at login-P05IJ3RX The reference code will help you get a bonus of 100 coins. ## c # – How do I add random objects to a random terrain using lightning bolts, and spin them with the terrain? I have a game that requires, random generation of trees and rocks for a world that is generated as you move. I have a rough installation system where I install the rigid bodies in a random location with a height of approximately 50, and they fall and create instances of trees and rocks, where", "- 0.6562=14.7838$ C. Step 1: $0.35 x 16=5.6$ Step 2: $16 - 5.6=10.4$ Step 3: $0.425 x 10.4=4.42$ Step 4: $16 - 4.42=11.58$ D. Step 1: $0.35 x 16=5.6$ Step 2: $16 - 5.6=10.4$ Step 3: $0.425 x 10.4=4.42$ Step 4: $10.4 - 4.42=5.98$ 50. What is the value of the expression $(-\\dfrac{8}{9}) \\div (-\\dfrac{2}{3}) \\times (-4 \\dfrac{1}{2})$? A. $-6$ B. $-\\dfrac{8}{27}$ C. $\\dfrac{8}{27}$ D. $6$ 51. A board game has a spinner divided into sections of equal size. Each section is labeled with a number between $1$ and $5$. Which number is a reasonable estimate of the number of times the spinner will land on a section labeled $5$ over the course of $150$ spins? A. $15$ B. $25$ C. $40$ D. $60$", "### Home > INT2 > Chapter 3 > Lesson 3.1.5 > Problem3-61 3-61. The spinner at right has three regions: $\\5,\\0,$ and $\\20$. To play the game, you spin one time and then you win the amount in the region that the spinner lands on. 1. What is the expected value for the game? Compare the given angle measurements to $360^\\circ$. Multiply these ratios by the amount possible to win in each section. $\\text{EV}=\\5\\cdot \\frac{1}{4}+\\0\\cdot \\frac{135}{360}+\\20\\cdot \\frac{135}{360}$ 2. If you have to pay $\\10$ to play, is it a fair game?", "CoCalc Public FilesFinal_Projects_F20 / Burnside_Russo_2.ipynb Views : 76 Compute Environment: Ubuntu 20.04 (Default) # Big Wheel Numbers Determine the chances of getting between a 0.85 and a 1.00 in two spins on The Price is Right Big Wheel game. The Big Wheel contains 20 values in 5 cent increments ranging from 5 to 100. The contestant can spin the wheel twice, and the value of each spin is added to their total price money amount. To determine to chances of getting a sum between 85 cents and 1.00 dollar with two spins, we can implement a function to generate all possible combinations of values and from there find the number of ways to obtain a total value of 0.85, 0.9, 0.95, or 1.00. In [ ]: import numpy as np values=[5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100] #20 possible spin values on the big wheel spinCombinations=[[spin1,spin2] for spin1 in values for spin2 in values] # all possible combinations of spins SpinSum=[[spin1+spin2] for spin1 in values for spin2 in values] #all possible sums of two spins SpinStreak=0 for i in range(len(SpinSum)): #evaluates combined sum of all combinations of two spins for Spins in SpinSum[i]: if (Spins>=85 and Spins<=100): print(spinCombinations[i], '=', SpinSum[i], 'cents') SpinStreak+=1 #if sum is between 85 and 100 then the spinStreak is appended print('Total number of combinations >=85 and <=100: ', SpinStreak) print('Total", "bestpick/nr.sims [1] 0.0574 0.6977 0.2371 > tiedpick/nr.sims [1] 0.00200 0.00783 0.00596 That’s okay, but there’s still only about a 70% chance of the highest-expected-value (“\\$300″ here) being the winner after 20 spins… and anyway it’s much easier to guess “correctly” here, no math required, so it’s not as impressive if the teacher does guess right. Hmmm. Gotta think a bit harder about whether it’s possible to construct a spinner that’s both (1) predictable and (2) non-obvious, given only 20 or so spins. Let me know if you have any thoughts. Edit: I propose a better solution in the next post. R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more...", "6 7 Since all $16$ products $1\\cdot 1, 1\\cdot 5, 1\\cdot 6, 1\\cdot 7, 2\\cdot 1, 2 \\cdot 5, \\dots, 4 \\cdot 7$ are distinct, and all numbers are between $1$ and $7$, these sets are valid. The common size of each set is $4$, so this sample output gets $4$ points. ### Tasks 1. $N = 2, M = 400$ 2. $N = 2, M = 4,000$ 3. $N = 2, M = 40,000$ 4. $N = 3, M = 4,000$ 5. $N = 4, M = 1,000$ 6. $N = 8, M = 200$ # Circle Covers (2021 Optimization 2) Aliens, aka CMIMC Programming Organizers, are trying to abduct the contestants of CMIMC Programming! As you are all socially distancing, from the point of view of the aliens, you are just points in a plane. To abduct the contestants, the aliens will drop crop circles of varying radii on groups of you. Given $N$ points on a 2D plane, and $M$ lengths denoting the radii of circles, your goal is to place these $M$ circles on the plane in a way that maximizes the number of points covered by some circle. ### Input Format The first line contains a single integer $N$, denoting the number of points. $N$ lines of input follow, each containing a space-separated", "group. Now reword the conclusion as \"Out of 253 pairs of people, there is a 50% chance that one pair will share a birthday.\" ## Explanation 2 by visualizing a spinning prize wheel Picture a giant spinner wheel, like at a carnival. There are 367 pegs making 366 slots for the pointer to land on. We'll pretend the 366 slot is 1/4 the size of the others to signify Feb. 29. Once someone lands on a slot, it's colored in before the next person spins. For the first 10 or so spins you have 1/366, 2/366, 3/366, etc... chance of landing on a colored slot, quite low odds. However, at say the 60th person around $$~1/6$$ of the wheel will be colored. Using these crude numbers, wouldn't you expect to hit a $$~1/6$$ chance sometime in the next 10 spins? Landing on the non-colored slots would equate to roughly $$(5/6)^{10}$$ which is about a $$~16\\%$$ chance just in those 10 spins. This would mean in those last hypothetical 10 spins you would have $$~84\\%$$ chance of landing on at least 1 previously landed-on slot. In this example the 'randomness' of the spin indicates the perceived 'randomness' of each person's birthday in the selected group. ## Explanation 3 by visualizing the Pigeonhole Principle Once you understand what the \"problem\"", "the board once, you get a payment of $200$ dollars (first case) for each time the player lands on a railway space. So if she is really unlucky, you get $800$ dollars and begin to dream about a vacation. We also make the probably unreasonable assumption that landing on any one of the spaces is equally likely, and that the events are independent. We also make the assumption that it is not possible to land twice on the same space in one tour of the board, something that is in fact not true, since sadly, in Monopoly as in life, one sometimes makes negative progress. Four railways, one payer: Let $X_1,X_2,X_3,X_4$ be the incomes we get from each of the four railways. Then our total income is $X_1+\\cdots+X_4$, which has expectation $E(X_1)+\\cdots+E(X_4)$. This is $4\\left(\\frac{1}{40}\\cdot 200\\right)$. Three railways, two payers: Let $X$ be the income we get from the first payer, and $Y$ the income we get from the second. We want $E(X+Y)$, which is $E(X)+E(Y)$. By an analysis like the one above, we have $E(X)=3\\left(\\frac{1}{40}\\cdot 100\\right)$. But $E(Y)$ is the same, so your expected income is $(2)(3)\\left(\\frac{1}{40}\\cdot 100\\right)$. Remark: So deal $1$ is better, if we neglect the cost of acquiring that fourth railway. It is more interesting to solve the problem under the assumption we can", "randomness. Sunday, March 29, 2009. Suppose there is a spin 1/2 particle in a state$\\chi = {1 \\over \\sqrt{5}} \\begin{bmatrix} 1\\\\ 2\\\\ \\end{bmatrix} $. Therefore, the charity can be very confident about gaining a net contribution of at least$500 from 1,000 plays of the game. The randomness comes from atmospheric noise, which for many purposes is better than the pseudo-random number algorithms typically used in computer programs. Each branch in a tree diagram represents a possible outcome. You count the number of sections with the result you want and then you place it over the. Each of the four outcomes on this wheel is equally likely and outcomes are independent from one spin to the next. I doubt that you could fabricate any wheel which could have a tangential speed 0. If you win $100, cash out$50 and play with the rest, for example. A stainless steel razor blade scrapes away the top layer to expose a 0. P(x) = €5 + €1. Let q be the probability of losing e. The player with the higher (or highest) total on this extra spin wins. You spin a wheel and it randomly lands on $1,$2, or END. The student then writes down the equations or answers that are formed by the word wheel. Solution The wheel has 38 equally", "of captain. Example 2: At a school fair, you are given a number of tokens. In one stall at the fair, there is a spinner with $8$ sectors. If the spinner lands on a red sector, you win $3$ tokens. If you land on a green sector, you win $5$ tokens. If you land on any other sector, you lose $2$ tokens. Is this game fair? The spinner has $8$ sectors and each is equally likely possibility. Sample space is {red sector, green sector, $6$ other sectors} Write the probability distribution for a single spin of spinner and the amount of tokens you win. $\\begin{array}{|cccc|}\\hline x=\\text{color}& \\text{Red}& \\text{Green}& \\begin{array}{l}\\text{Others}\\\\ \\left(\\text{Blue}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{and}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{Yellow}\\right)\\end{array}\\\\ P\\left(x\\right)& \\frac{1}{8}& \\frac{1}{8}& \\frac{6}{8}\\\\ \\text{Tokens}& 3& 5& -2\\\\ \\hline\\end{array}$ Use the weighted average formula. $\\begin{array}{l}E\\left(x\\right)=3\\cdot \\frac{1}{8}+5\\cdot \\frac{1}{8}+\\left(-2\\right)\\cdot \\frac{6}{8}\\\\ =\\frac{3}{8}+\\frac{5}{8}-\\frac{12}{8}\\\\ =-\\frac{4}{8}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{or}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}-0.5\\end{array}$ The expected value is not zero, and the game is not fair. So you will lose about $0.5$ tokens for a single spin.", "will enter and visualize the probability distribution for 20-line Cleopatra. (Note that while there is a bonus/free spin feature, the probability distribution is constructed such that the payout structure is the same, but free spins are not included in simulations. This makes life easier.) In [1]: import matplotlib.pyplot as plt import numpy as np import scipy.stats as st %matplotlib inline In [2]: # Possible intervals you could fall into # [0, 0.2] is losing, the rest are winning intervals # If you fall in an interval, multiply your total bet by a random number in the interval to get your winnings # For example, for a 30 cent bet: you land in [2., 5.]. Select a random number \"x\" from a uniform dist. in [2,5]. # Your winnings are 0.3 * x in dollars intervals20 = np.array([ [0., 0.2], [0.2, 0.5], [0.5, 1.], [1., 2.], [2., 5.], [5., 10.], [10., 20.], [20., 50.], [50., 100.], [100., 200.], [200., 500.], [500., 1000.]]) # This is a scaled version of the winning probability distribution from Casino Guru. # This corresponds to all entries in \"intervals20\" except the first entry, which is the losing interval. # The multiplication factor 0.798... ensures that expected return is 95.025% cleopatra20_prob = 0.7987446566693283 * np.array( [929482, 740452, 563289, 1031867, 149001, 92050, 58006, 17450, 3538, 505,", "= 36. 58 + 36 = 94. so the money she earns on both days =$94. Eureka Math Grade 2 Module 4 Lesson 10 Exit Ticket Answer Key Question 1. Solve using the algorithm. Draw chips and bundle when you can. 27 + 137 27 + 137 = 164. Explanation: In the above-given question, given that, Draw and bundle place value disks on the place value chart. 1 x 100 = 100. 3 x 10 = 30. 7 x 1 = 7. 2 x 10 = 20. 7 x 1 = 7. 100 + 30 + 7 = 137. 20 + 7 = 27. 137 + 27 = 164. Question 2. Using the previous problem, fill in the blanks. Use place value language to explain how you used bundling to rename the solution. Before bundling a ten __1___ hundreds __3___ tens __7___ ones After bundling a ten ___1__ hundreds __3___ tens __4___ ones 27 + 137 = 164. Explanation: In the above-given question, given that, Draw and bundle place value disks on the place value chart. 1 x 100 = 100. 3 x 10 = 30. 7 x 1 = 7. 2 x 10 = 20. 7 x 1 = 7. 100 + 30 + 7 = 137. 20 + 7 = 27. 137 + 27 = 164.", "+0 -1 96 3 +397 deleted. Nov 15, 2019 edited by sinclairdragon428 Nov 20, 2019 #2 +118 +3 Actually out of $$5\\cdot5\\cdot5=125$$possible outcomes, there are 6 outcomes that result in winnings of $1700. They are $$(300, 400, 1000), (300, 1000, 400), (400, 300, 1000), (400, 1000, 300), (1000, 300, 400)$$and $$(1000, 400, 300)$$ where the ordered triplets indicate the results of the first, second, and third spin, respectively. So the probability we are looking for is $$\\frac{6}{125}$$. Nov 15, 2019 ### 3+0 Answers #1 0 The only way to get 1700 is to land on 300, 400, then 1000, so the probability is 1/(5*4*3) = 1/60. Nov 15, 2019 #2 +118 +3 Best Answer Actually out of $$5\\cdot5\\cdot5=125$$possible outcomes, there are 6 outcomes that result in winnings of$1700. They are $$(300, 400, 1000), (300, 1000, 400), (400, 300, 1000), (400, 1000, 300), (1000, 300, 400)$$and $$(1000, 400, 300)$$ where the ordered triplets indicate the results of the first, second, and third spin, respectively. So the probability we are looking for is $$\\frac{6}{125}$$. Gadfly Nov 15, 2019 #3 +397 -1 thanks! sinclairdragon428 Nov 15, 2019", "### Home > CCG > Chapter Ch10 > Lesson 10.1.4 > Problem10-45 10-45. The spinner at right is designed so that if you randomly spin the spinner and land in the shaded sector, you win \\$$1,000,000$. Unfortunately, if you land in the unshaded sector, you win nothing. Assume point $C$ is the center of the spinner. 1. If $m∠ACB = 90º$, how many times would you have to spin to reasonably expect to land in the shaded sector? How did you get your answer? $4$ times 2. What if $m∠ACB = 1º$? How many times would you have to spin to reasonably expect to land in the shaded sector? 3. Suppose $P(\\text{winning}\\space 1,000,000) = \\frac { 1 } { 5 }$ for each spin. What must $m∠ACB$ equal? Show how you got your answer. $\\frac{360º}{5}=72º$", "9 Programs In this chapter, you will build a real, working slot machine that you can play by running an R function. When you’re finished, you’ll be able to play it like this: play() ## 0 0 DD ## $0 play() ## 7 7 7 ##$80 The play function will need to do two things. First, it will need to randomly generate three symbols; and, second, it will need to calculate a prize based on those symbols. The first step is easy to simulate. You can randomly generate three symbols with the sample function—just like you randomly “rolled” two dice in Project 1: Weighted Dice. The following function generates three symbols from a group of common slot machine symbols: diamonds (DD), sevens (7), triple bars (BBB), double bars (BB), single bars (B), cherries (C), and zeroes (0). The symbols are selected randomly, and each symbol appears with a different probability: get_symbols <- function() { wheel <- c(\"DD\", \"7\", \"BBB\", \"BB\", \"B\", \"C\", \"0\") sample(wheel, size = 3, replace = TRUE, prob = c(0.03, 0.03, 0.06, 0.1, 0.25, 0.01, 0.52)) } You can use get_symbols to generate the symbols used in your slot machine: get_symbols() ## \"BBB\" \"0\" \"C\" get_symbols() ## \"0\" \"0\" \"0\" get_symbols() ## \"7\" \"0\" \"B\" get_symbols uses the probabilities observed in a group of video", "Go Back to the Home Page :-) Classroom Timers - Fun Timers for classrooms and meetings :-) Holiday Timers - More Fun Timers - But these are Holiday. * Let x represent the total score for a single player. Assume that the cost to spin the wheel once is $5. Wheel to Spin Plus Instructions. One, it guarantees that the highest-scoring elements will be most likely to reproduce. Probability Spinner Spin the wheel and calculate both theoretical and experimental probabilitiesif you dare. The chance of winning is 4 out of 52, while the chance against winning is 48 out of 52 (52-4=48). Imagine a roulette wheel with the stops labelled as 0. See the image below for an example of how easy it is to modify your prizes. But you can see that the odds of that are happening even with just even numbers that have 1:1 payouts are pretty much the same. You can also design your own wheel and a sign that describes the probabilities for your wheel. We enter the values 4 times in the column. These can be used as random number generators in lessons. The ball will be spun in the opposite direction. Each index should have equal probability of returning pick(1) should return 0. Now I have spun the wheel 12 times", "### Home > CCG > Chapter 3 > Lesson 3.2.3 > Problem3-79 3-79. Congratulations! You are going to be a contestant on a new game show with a chance to win some money. You will spin the two spinners shown below to see how much money you will win. Test your ideas creating the situation below using the 3-79 Double Spinner tool (CPM). 1. Make a tree diagram of all the possible outcomes of spinning the two spinners. At the ends of the branches, on the far right, write the amount you would win for each combination of spins. 2. Are each of the outcomes in the sample space equally likely? Do any outcomes appear more than once? In each spinner, is any result more likely than the other(s)? 3. What is the probability that you will take home $\\200$? What is the probability that you will take home more than $\\500$? $\\frac{1}{6},\\frac{3}{6}$ 4. What is the probability that you will double your winnings? Does the probability that you will double your winnings depend on the result of the first spinner? Look at the second spinner by itself. 5. What if the amounts on the first spinner were $\\100$, $\\200$, and $\\1500$? What is the probability that you would take home $\\200$? Justify your conclusion. Draw a second", "geometric series to find $\\sum _{k=1}^{7}-0.2\\cdot {\\left(-5\\right)}^{k-1}.$ ${S}_{7}=-2604.2$ Find the sum of the infinite geometric series $\\sum _{k=1}^{\\infty }\\frac{1}{3}\\cdot {\\left(-\\frac{1}{5}\\right)}^{k-1}.$ Rachael deposits \\$3,600 into a retirement fund each year. The fund earns 7.5% annual interest, compounded monthly. If she opened her account when she was 20 years old, how much will she have by the time she’s 55? How much of that amount was interest earned? Total in account: $\\text{}140,355.75;$ Interest earned: $\\text{}14,355.75$ In a competition of 50 professional ballroom dancers, 22 compete in the fox-trot competition, 18 compete in the tango competition, and 6 compete in both the fox-trot and tango competitions. How many dancers compete in the fox-trot or tango competitions? A buyer of a new sedan can custom order the car by choosing from 5 different exterior colors, 3 different interior colors, 2 sound systems, 3 motor designs, and either manual or automatic transmission. How many choices does the buyer have? $5×3×2×3×2=180$ To allocate annual bonuses, a manager must choose his top four employees and rank them first to fourth. In how many ways can he create the “Top-Four” list out of the 32 employees? A rock group needs to choose 3 songs to play at the annual Battle of the Bands. How many ways can they choose their set if have 15 songs to", "geometric series to find $\\sum _{k=1}^{7}-0.2\\cdot {\\left(-5\\right)}^{k-1}.$ ${S}_{7}=-2604.2$ Find the sum of the infinite geometric series $\\sum _{k=1}^{\\infty }\\frac{1}{3}\\cdot {\\left(-\\frac{1}{5}\\right)}^{k-1}.$ Rachael deposits \\$3,600 into a retirement fund each year. The fund earns 7.5% annual interest, compounded monthly. If she opened her account when she was 20 years old, how much will she have by the time she’s 55? How much of that amount was interest earned? Total in account: $\\text{}140,355.75;$ Interest earned: $\\text{}14,355.75$ In a competition of 50 professional ballroom dancers, 22 compete in the fox-trot competition, 18 compete in the tango competition, and 6 compete in both the fox-trot and tango competitions. How many dancers compete in the fox-trot or tango competitions? A buyer of a new sedan can custom order the car by choosing from 5 different exterior colors, 3 different interior colors, 2 sound systems, 3 motor designs, and either manual or automatic transmission. How many choices does the buyer have? $5×3×2×3×2=180$ To allocate annual bonuses, a manager must choose his top four employees and rank them first to fourth. In how many ways can he create the “Top-Four” list out of the 32 employees? A rock group needs to choose 3 songs to play at the annual Battle of the Bands. How many ways can they choose their set if have 15 songs to" ]

[ "unit cube: 200 uniformly random points, which are joined pairwise if they are less than 0.2 apart", "exactly 100 square units. Is there more than one? Can you find them all? Drilling Many Cubes Stage: 2 and 3 Challenge Level: A useful visualising exercise which offers opportunities for discussion and generalising, and which could be used for thinking about the formulae needed for generating the results on a spreadsheet. Cubes Within Cubes Revisited Stage: 3 Challenge Level: Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need? Square Coordinates Stage: 3 Challenge Level: A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides? Frogs Stage: 3 Challenge Level: How many moves does it take to swap over some red and blue frogs? Do you have a method? Painted Cube Stage: 3 Challenge Level: Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces? Convex Polygons Stage: 3 Challenge Level: Show that among the interior angles of a convex polygon there cannot be more than three acute", "# Painted Cube Faces Probability Level 3 Consider an $n \\times n \\times n$ cube $(n \\geq 3)$ whose outer surface is painted blue. The unit cubes at the corners each have 3 blue faces, while the unit cubes in the center each have 0 blue faces. For what value of $n$ will the number of unit cubes with 0 blue faces be twice the number of unit cubes with 1 blue face? ×", "number of unit cubes 3 painted faces, 2 painted faces, 1 painted face and 0 painted face given a side length n units. To determine how many unit cubes are painted given a particular size, it will help us if we know the properties of a cube. Let us recall that a cube has 8 vertices, 6 faces and 12 edges. Figure 3 – Parts of a cube. For the sake of discussion, we will name and color the unit cubes (see Figure 4) and group them depending on their positions – whether they are at the edges, vertices or faces of the cube. • Vertex Cubes (Red) – are the unit cubes located at the vertices of the cube. It is evident only 3 of their faces are painted. • Edge Cubes (Green) – are the unit cubes located at the 12 edges of the cube. Note that only two of their faces are painted. • Wall Cubes (Blue) – are the unit cubes at the faces of the cube. We will call the wall cubes because we will use the word “face” in another context. • Core Cubes (No color) – are cubes that are at the core of the cube that was not painted. Figure 4 – The cube showing number of painted faces depending", "of 64 white snap cubes. Someone spray paints all 6 faces of the large cube blue. After the paint dries, they disassemble the large cube into a pile of 64 snap cubes. 1. How many of those 64 snap cubes have exactly 2 faces that are blue? 2. What are the other possible numbers of blue faces the cubes can have? How many of each are there? 3. Try this problem again with some larger-sized cubes that use more than 64 snap cubes to build. What patterns do you notice? ### 14.4: What’s the Prism’s Height? There are 4 different prisms that all have the same volume. Here is what the base of each prism looks like. 1. Order the prisms from shortest to tallest. Explain your reasoning. 2. If the volume of each prism is 60 units3, what would be the height of each prism? 3. For a volume other than 60 units3, what could be the height of each prism? 4. Discuss your thinking with your partner. If you disagree, work to reach an agreement. ### Summary Any cross section of a prism that is parallel to the base will be identical to the base. This means we can slice prisms up to help find their volume. For example, if we have a rectangular prism that", "reasons, half the triangles are colored blue and half are colored black. But it’s not just pretty: there’s also math here. If we take any black triangle and reflect it across any mirror, we get a blue triangle… and vice versa. Instead of taking a sphere and slicing it with mirrors, we can start with the cube itself. Here’s what we get: It’s not quite as pretty (especially because I drew it), but it makes certain games easier to play. These games involve picking one triangle and calling it our favorite. It doesn’t matter which. But we have to pick one… so how about this: Each different symmetry of the cube sends this triangle to a different triangle. This instantly lets us count the symmetries: there are 48, since each of the cube’s 6 faces has been chopped into 8 triangles. But even better, we get a vivid picture of the symmetries of the cube! Let’s see how this works. Any triangle in the Coxeter complex has three corners: • one corner is a vertex of the cube, • one corner is the center of an edge of the cube, • one corner is the center of a face of the cube. Here’s how it works for our favorite triangle: Now the real fun starts. We can move", "edge cubes away from the other two layers. Consequently it is a simple process to \"solve\" a Cube by taking it apart and reassembling it in a solved state. There are twelve edge pieces which show two coloured sides each, and eight corner pieces which show three colours. Each piece shows a unique colour combination, but not all combinations are present (for example, if red and orange are on opposite sides of the solved Cube, there is no edge piece with both red and orange sides). The location of these cubes relative to one another can be altered by twisting an outer third of the Cube 90°, 180° or 270°, but the location of the coloured sides relative to one another in the completed state of the puzzle cannot be altered: it is fixed by the relative positions of the centre squares and the distribution of colour. However, Cubes with alternative colour arrangements also exist; for example, they might have the yellow face opposite the green, and the blue face opposite the white (with red and orange opposite faces remaining unchanged). Douglas R. Hofstadter, in the July 1982 issue of Scientific American, pointed out that Cubes could be coloured in such a way as to emphasise the corners or edges, rather than the faces as the standard colouring", "call the bigger or uncut cubes “cube”and the smaller cubes “unit cubes”. How many unit cubes can be formed from a painted cube with length 4 units? 5 units? n units? Figure 1 – Cubes with side lengths 2 and 3 cut into unit cubes. Before scrolling down, investigate the number of painted faces of each yellow unit cube. In Figure 2, the unit cubes have been drawn from different perspectives to make visualization easier. How many cubes are there with 3 painted faces? 2 painted faces? 1 painted face? 0 painted face? Figure 2 – Cube cut into 27 unit cubes shown in different perspectives. Without drawing, can you determine the number of painted faces of a cube with side length 4 units? How about 5 units? 6 units? Challenge: Find a formula for the number of unit cubes 3 painted faces, 2 painted faces, 1 painted face and 0 painted face given a side length n units. To determine how many unit cubes are painted given a particular size, it will help us if we know the properties of a cube. Let us recall that a cube has 8 vertices, 6 faces and 12 edges. Figure 3 – Parts of a cube. For the sake of discussion, we will name and color the unit cubes (see", "# Probability problem. There are 16 disks in a box. Question: There are 16 disks in a box. Five of them are painted red, five of them are painted blue, and six are painted red on one side, and blue on the other side. We are given a disk at random, and see that one of its sides is red. Is the other side of this disk more likely to be red or blue? - To fully specify the problem, you need to specify how we see that one of the sides is red. The answer would be different if a) we randomly choose a side to look at or b) we can somehow tell (e.g. by looking from a distance) whether one of the sides of a disk is red or not or c) the person who drew the disk gives it to us with a red side showing, possibly according to some strategy such as always showing a red side if possible. – joriki Apr 9 '12 at 20:22 @joriki : \"We are given a disk at random, and see that one of its sides is red\" is enough for me to believe that I am given a random disk and I see a random side of the disk? I don't think we should over exaggerate", "will be used. Explanation: Count the number of unit cubes in the first figure. There are 10 unit cubes in figure 1 so match the figure 1 to 10 unit cubes. Count the number of unit cubes in the second figure. There are 12 unit cubes in figure 2 so match figure 2 to 12 unit cubes. Count the number of unit cubes in the third figure. There are 9 unit cubes in figure 3 so match figure 3 to 9 unit cubes. Question 6. Chuck is making a poster about polyhedrons for his math class. He will draw figures and organize them in different sections of the poster. Part A Chuck wants to draw three-dimensional figures whose lateral faces are rectangles. He says he can draw prisms and pyramids. Do you agree? i. yes ii. no Explanation: The lateral faces of a pyramid are triangles. The lateral faces of a prism are rectangles. Question 6. Part B Chuck says that he can draw a cylinder on his polyhedron poster because it has a pair of bases that are congruent. Is Chuck correct? i. yes ii. no Explanation: A cylinder does have 2 congruent bases, but a cylinder is not a polyhedron. A cylinder has 1 curved surface, while a polyhedron has faces that are polygons Chapter Review/Test", "four blue shapes labelled A, B, C and D are one unit each, what is the combined area of all the blue shapes? Explain any reasoning you have used.", "vertical. There are 3 scenarios in which four vertical faces are red and two blue: pick any pair of opposing faces of the cube to be blue, and the rest red. Then we can choose to place the cube blue-face-down, so the red faces come out vertical. There are 3 such pairs (since there are 6 faces). There are 6 scenarios in which five faces are red and one blue: pick any particular face to be blue, then we can choose to place the cube blue-face-down (or blue-face-up), and the red faces come out vertical. There is 1 scenario in which six faces are red. The red faces will come out vertical no matter how we place the cube. So there should be a total of 10 scenarios in which we can place the cube with four vertical red faces. As you alluded to, the same argument goes through for blue faces, giving 20 scenarios in total. So the probability should be $\\frac{20}{64}$ • why are the scenarios in which the the vertical faces do not have the same color counted in the desirable outcomes – jamarcus tyrone Dec 27 '17 at 1:05 • @jamarcus can you explain what you mean? – greenturtle3141 Dec 27 '17 at 1:26 • @greenturtle3141 The scenarios in which the vertical faces are", "# Red-Red Blocks A toy factory creates cube blocks in which each side is randomly colored red or blue. You have all 10 distinct (up to rotation) cubes placed in a bag. You reach into the bag and pick up one of these blocks, noticing that the one (and only one) side that you see is red. What is the probability that the opposite side is also red? ×", "# Geometry posted by on . A 12 by 12 by 12 cube has all its faces painted blue. It is then broken down to 1728 individual 1 by 1 by 1 cubes. The probability that a randomly chosen 1 by 1 by 1 cube has exactly 2 faces painted blue is \\frac {a} {b} , where a, b are coprime. What is the value of a+ b? • Geometry - , There will be 12*(12 - 2) = 120 cubes with exactly 2 faces painted blue. ie 120/1728 = 5/72 ie a + b = 5 + 72 =77", "colour and a blue between different colours then remove the original beads'. Keep repeating this. What happens? ### Counting Binary Ops ##### Stage: 4 Challenge Level: How many ways can the terms in an ordered list be combined by repeating a single binary operation. Show that for 4 terms there are 5 cases and find the number of cases for 5 terms and 6 terms. ### Transitivity ##### Stage: 5 Suppose A always beats B and B always beats C, then would you expect A to beat C? Not always! What seems obvious is not always true. Results always need to be proved in mathematics. ##### Stage: 4 Challenge Level: A walk is made up of diagonal steps from left to right, starting at the origin and ending on the x-axis. How many paths are there for 4 steps, for 6 steps, for 8 steps? ### Russian Cubes ##### Stage: 4 Challenge Level: How many different cubes can be painted with three blue faces and three red faces? A boy (using blue) and a girl (using red) paint the faces of a cube in turn so that the six faces are painted. . . . ### One Basket or Group Photo ##### Stage: 2, 3, 4 and 5 Challenge Level: Libby Jared helped to set up NRICH and", "should occur halfway between the x– and y‑intercepts of the linear graph. Yet, he had an intuitive insight that just happens to be true. As a result, what took me about five minutes of deriving and manipulating took him about five seconds. The second problem arose at the grocery store. Among our purchases was a box of sugar cubes, which contained, surprisingly, 126 cubes.This number is surprising in the sense that it’s not a number you’ll see very often, except for an occasional appearance in the ninth row of Pascal’s Triangle, or maybe if you’re a chemist searching for stable atoms. A question that could have arisen from this situation involves surface area and volume: A rectangular prism with integer dimensions has a volume of 126 cubic units. What is the least possible surface area? That’s not the question that was shared with Alex and Eli, though. (The answer, if you care, is 162 square units, which results from a 3 × 6 × 7 arrangement — which, in fact, is the exact arrangement of cubes in the box above. I suspect this is not a coincidence.) The problem that I shared with my sons involved probability: Imagine that the arrangement of cubes is removed from the box intact, and all six faces of the prism are painted", "avoid confusion, we will simply call the bigger or uncut cubes “cube”and the smaller cubes “unit cubes”. How many unit cubes can be formed from a painted cube with length 4 units? 5 units? n units? Figure 1 – Cubes with side lengths 2 and 3 cut into unit cubes. Before scrolling down, investigate the number of painted faces of each yellow unit cube. In Figure 2, the unit cubes have been drawn from different perspectives to make visualization easier. How many cubes are there with 3 painted faces? 2 painted faces? 1 painted face? 0 painted face? Figure 2 – Cube cut into 27 unit cubes shown in different perspectives. Without drawing, can you determine the number of painted faces of a cube with side length 4 units? How about 5 units? 6 units? Challenge: Find a formula for the number of unit cubes 3 painted faces, 2 painted faces, 1 painted face and 0 painted face given a side length n units. To determine how many unit cubes are painted given a particular size, it will help us if we know the properties of a cube. Let us recall that a cube has 8 vertices, 6 faces and 12 edges. Figure 3 – Parts of a cube. For the sake of discussion, we will name and", "# Need help with a probability and a cube question. • December 13th 2007, 03:25 PM help1 Need help with a probability and a cube question. A 4\"x4\"x4\" cube is painted and then cut into sixty-four 1\"x1\"x1\" cubes. A unit cube is then randomly selected and rolled. What is the probability that the top face of the rolled cube is painted? Express your answer as a common fraction. Now, originally, I tohught the answer was 96/384, but I realized it was wrong since some sides (the corner cubes) have more than one painted side. • December 13th 2007, 04:02 PM Plato Quote: Originally Posted by help1 A 4\"x4\"x4\" cube is painted and then cut into sixty-four 1\"x1\"x1\" cubes. A unit cube is then randomly selected and rolled. What is the probability that the top face of the rolled cube is painted? Express your answer as a common fraction. Now, originally, I tohught the answer was 96/384, but I realized it was wrong since some sides (the corner cubes) have more than one painted side. There are cubes with no painted sides. There are cubes with one painted side. There are cubes with two painted sides. There are cubes with three painted sides. How many of each? • December 13th 2007, 04:05 PM galactus This is actually a cool", "the Concept of Functions Problem: A cube is painted on all faces and cut into smaller cubes of the same size. Investigate the number of painted faces of the smaller cubes. Discussion In Figure 1, shown are the cubes with side lengths two units and three units. The light blue cube has been painted and cut into eight smaller unit cubes, while the yellow cube has been painted yellow and cut into 27 smaller unit cubes. To avoid confusion, we will simply call the bigger or uncut cubes “cube”and the smaller cubes “unit cubes”. How many unit cubes can be formed from a painted cube with length 4 units? 5 units? n units? Figure 1 – Cubes with side lengths 2 and 3 cut into unit cubes. Before scrolling down, investigate the number of painted faces of each yellow unit cube. In Figure 2, the unit cubes have been drawn from different perspectives to make visualization easier. How many cubes are there with 3 painted faces? 2 painted faces? 1 painted face? 0 painted face? Figure 2 – Cube cut into 27 unit cubes shown in different perspectives. Without drawing, can you determine the number of painted faces of a cube with side length 4 units? How about 5 units? 6 units? Challenge: Find a formula for the", "a very simple process to solve a Cube by taking it apart and reassembling it in a solved state. There are twelve edge pieces which show two coloured sides each, and eight corner pieces which show three colours. Each piece shows a unique colour combination, but not all combinations are present (for example, if red and orange are on opposite sides of the solved Cube, there is no edge piece with both red and orange sides). The location of these cubes relative to one another can be altered by twisting an outer third of the Cube 90°, 180° or 270°, but the location of the coloured sides relative to one another in the completed state of the puzzle cannot be altered: it is fixed by the relative positions of the centre squares and the distribution of colour combinations on edge and corner pieces. For most recent Cubes, the colours of the stickers are red opposite orange, yellow opposite white, and green opposite blue. However, Cubes with alternative colour arrangements also exist; for example, they might have the yellow face opposite the green, and the blue face opposite the white (with red and orange opposite faces remaining unchanged). Douglas R. Hofstader, in the July 1982 Scientific American, pointed out that Cubes could be coloured in such a way as" ]

[ "on the faces in any order. That gives $$\\binom{8}{6}\\cdot6!=20160$$ ways. The probability then becomes $$\\frac{144}{20160}=\\frac{1}{140}$$.", "# Math Help - Cube probability 1. ## Cube probability On choosing three vertex of a cube at random, what's the probability that the three are in a same face? 2. I would think it would be $(\\frac{1}{6})(\\frac{3}{4})=\\frac{1}{8}$ It may be more involved than that. 3. Originally Posted by Krizalid On choosing three vertex of a cube at random, what's the probability that the three are in a same face? The counting principle, of course. The first pick if free, we may pick any vertex. There are 8 vertices on the cube, we've picked one leaving 7. Of these only 6 share a face with the one we've picked. So we have $1 \\cdot \\frac{6}{7}$ so far. Now, we've picked two vertices. 1) If the two vertices are on the same edge, there are 4 possible of 6 vertices left to pick, so the probability will be $1 \\cdot \\frac{6}{7} \\cdot \\frac{4}{6} = \\frac{4}{7}$ 2) If the two vertices are not on the same edge, they are on a diagonal across one face. In this case there is only one possible face that these could be on, so there are only two possible vertices out of 6, so the probability will be $1 \\cdot \\frac{6}{7} \\cdot \\frac{2}{6} = \\frac{2}{7}$ Only one of these choices is possible, so we", "\\times 2$ squares: Choosing three such squares leaves only one square left, with four places to place it. This is $2 \\cdot 4 = 8$ ways. • For at least four $2 \\times 2$ squares, we clearly only have one way. By the Principle of Inclusion-Exclusion, there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \\times 2$ square. There are $2^9=512$ ways to paint the $3 \\times 3$ square with no restrictions, so there are $512-95=417$ ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a $2 \\times 2$ red square is $\\frac{417}{512}$, and $417+512=\\boxed{929}$.", "# CTK Insights • ## Pages 20 May ### Probabilities in a Painted Cube A wooden cube - after being painted all over - has been cut into $3\\times 3\\times 3$ smaller cubes. These were thoroughly mixed in a bag, from which one was produced and tossed. What is the probability that a painted side turned up? On impulse, one would approach the problem in a more or less standard way. There are $8$ corner cubes with $3$ painted sides, $12$ mid-edge cube with $2$ painted sides, and $6$ face-central cubes with only $1$ side painted. The total probability is then $\\displaystyle\\frac{8}{27}\\cdot\\frac{3}{6}+\\frac{12}{27}\\cdot\\frac{2}{6}+\\frac{6}{27}\\cdot\\frac{1}{6}=\\frac{1}{3}.$ Now generalize: cut the cube into $n\\times n\\times n$, $n\\gt 1$ smaller cubes and ask the same question. The problem is not awfully difficult but needs some figuring out. Following the foregoing pattern, we eventually arrive at $\\displaystyle\\frac{8}{n^3}\\cdot\\frac{3}{6}+\\frac{12\\cdot (n-2)}{n^3}\\cdot\\frac{2}{6}+\\frac{6\\cdot (n-2)^2}{n^3}\\cdot\\frac{1}{6}$ This expression simplifies, as you can verify, to $\\displaystyle\\frac{1}{n},$ which, by the way, confirms the answer for $n=3.$ The above is a useful and not too difficult exercise, but there is a delightful shortcut that avoids most of the counting of cubes and their sides. $n^3$ cubes have a total of $6\\cdot n^3$ sides. Of these, $6\\cdot n^2$ are painted. All sides have the same probability of turning up, therefore, a painted side will turn up", "in the box. – Arpit Bajpai May 5 '12 at 6:01 Would that lead you to claim that the chance of picking any given color is $\\frac 14$? This is not correct. Put 10 blue marbles and one red marble in a hat, draw 20 times with replacement, and see if you get around 10 reds. – Ross Millikan May 5 '12 at 14:33 We can also work directly with the probabilities. Pick the cubes a little cube at random, then pick another. The probability that the first cube picked has exactly $2$ red faces is $\\frac{24}{64}$. Given that the first cube picked had exactly $2$ red faces, the probability that the second cube picked has exactly $2$ red faces is $\\frac{23}{63}$. Thus the probability we get two cubes with exactly $2$ red faces is $\\frac{24}{64}\\cdot\\frac{23}{63}$. Imagine that we have a total of $64$ cubes, with $24$ identical yellow ones, and the rest identical and blue. Put the $64$ cubes in a bag, shake well, and blindfolded pick a cube at random. Is the probability of picking a yellow equal to $1$? – André Nicolas May 5 '12 at 5:11 In the previous comment, shouldn't you ask if the probability to pick a yellow one is $\\frac 12$? – Ross Millikan May 5 '12 at 5:16 The problem", "point $C$ is one of the two points on the side that contains $AB$. Hence the probability of $ABC$ intersecting the inside of the cube is $2/3$. In the third case, already the diagonal $AB$ contains points inside the cube, hence this case will be good regardless of the choice of $C$. Summing up all cases, the resulting probability is: $$\\frac 37\\cdot\\frac 13 + \\frac 37\\cdot \\frac 23 + \\frac 17\\cdot 1 = \\boxed{\\frac 47}$$ ## Solution 2 There are $\\binom{8}{3}=56$ ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability. There are four ways to choose three points from the vertices of a single face. Since there are six faces, $4 \\times 6 = 24$. Thus, the probability of what we don't want is $\\frac{24}{56} = \\frac{3}{7}$. Using complementary probability, $$1- \\frac 37 = \\boxed{\\frac 47}$$", "So the probability of rolling a painted side after selecting a small cube at random is: $\\displaystyle \\frac{\\text{surface area}}{6\\cdot\\text{volume}}$. For example, say we have a 3x4x5 rectangle made up of 60 1x1x1 cubes, and we paint the outside. If we break it apart and roll a cube at random, the probability that a painted side will show up is: $\\displaystyle \\frac{2(3(4)+4(5)+3(5))}{6\\cdot3(4)(5)}=\\frac{47}{180}$. You would get the same answer if you went through the analysis of interior, face, edge, and corner pieces: there are $(3-2)\\times(4-2)\\times(5-2)=6$ interior pieces (subtract two from each dimension, because you don’t want the two outside pieces), 24 edge pieces (for an edge of side $n$, there are $n-2$ edge pieces; there are 4 edges of each length, so we have $4((3-2)+(4-2)+(5-2))=24$), 8 corners (like always), and 22 face pieces (what’s remaining of the 60 cubes must be the face pieces!) , for a probability of $\\displaystyle \\underbrace{\\frac{6}{60}\\left(\\frac{0}{6}\\right)}_\\text{interior piece}+\\underbrace{\\frac{22}{60}\\left(\\frac{1}{6}\\right)}_\\text{face piece}+\\underbrace{\\frac{24}{60}\\left(\\frac{2}{6}\\right)}_\\text{edge piece}+\\underbrace{\\frac{8}{60}\\left(\\frac{3}{6}\\right)}_\\text{corner}=\\frac{22+48+24}{360}=\\frac{47}{180}$. This is quite complicated, even though we made life a little easier by counting the face pieces last; they are hardest to count directly, so we counted the other pieces first and took the remainder to be face pieces. For a cube with side length $n$, the surface area is $6n^2$ and the volume is $n^3$. So the probability of rolling a painted side would be", "four coordinate positions instead of the required two. Let !!E!! be an even vertex in an !!n!!-cube. We will change two of its coordinates, to obtain another even vertex, and call other vertex !!E'!!. The point !!E!! has !!n!! coordinates, and there are !!\\binom n2!! ways to choose the two coordinate positions to switch to obtain !!E'!!. We have !!\\frac122^n!! choices for !!E!!, then !!\\binom n2!! choices of which pair of coordinates to switch to get !!E'!!, and then !!E!! and !!E'!! together determine a single face of the !!n!!-cube. But that double-counts the faces, since pair !!\\{ E, E'\\}!! determines the same face as !!\\{ E', E\\}!!. So to count the faces we need to divide the final answer by !!2!!. Counting by this method tells us that the number of faces in an !!n!!-cube is is: $$\\frac122^n\\cdot \\binom n2 \\cdot \\frac12$$ where the !!\\frac122^n!! counts the number of ways to choose an even vertex !!E!!, the !!\\binom n2!! counts the number of ways to choose an even vertex !!E'!! that shares a face with !!E!!, and the extra !!\\frac12!! adjusts for the double-counting in the orderof !!E!! and !!E'!!. We can simplify this to $$2^{n-2}\\binom n2.$$ When !!n=3!! we get $$2^{3-2}\\binom 32 = 2\\cdot 3 = 6$$ and it is correct for all !!n!!, since !!\\binom", "up to rotation and reflection, there are only 2 ways to arrange the numbers. The group of symmetries of the cube has size 48, so after rotations and reflections there are $$2\\cdot48=96$$ valid arrangements. Now for the probability: There are $$8!=40320$$ ways to arrange the numbers, of which $$96$$ are valid. The probability is therefore $$\\frac{96}{40320}=\\frac{1}{420}$$. • Correct! Checkmark incoming! Jul 4, 2020 at 12:18", "$6 \\cdot 5 \\cdot 4 \\cdot 4 \\cdot 4 = 1920$ combinations. In the case that $C$ has a different color from $A$ and $D$ has the same color as $A$, $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices, $D$ has $1$ choice, and $E$ has $5$ choices, resulting in a possible of $6 \\cdot 5 \\cdot 4 \\cdot 1 \\cdot 5 = 600$ combinations. Adding all those combinations up, we get $600+1920+600=\\boxed{3120 \\ \\mathbf{(C)}}$. ## Solution 2 First, notice that there can be $3$ cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors. Case $1$: There are $6!$ ways of assigning each vertex a different color. $6! = 720$ Case $2$: There are $\\frac {6!}{2!} * 5$ ways. After picking four colors, we can rotate our pentagon in $5$ ways to get different outcomes. $\\frac {6!}{2} * 5 = 1800$ Case $3$: There are $\\frac {\\frac {6!}{3!} * 10}{2!}$ ways of arranging the final case. We can pick $3$ colors for our pentagon. There are $5$ spots for the first pair of colors. Then, there are $2$ possible ways we can put the final pair in the last $3$", "way that is different from both pairs. Therefore, there are $$\\binom{4}{2} \\cdot 16 \\cdot 15 \\cdot 14$$ such cases. 3. Two different pairs of identical quadrants: There are three ways to choose which of the other quadrants will be painted in the same way as the upper left quadrant, $16$ ways to choose how to paint those two quadrants, and $15$ ways to choose the pattern for the other two quadrants. Hence, there are $$3 \\cdot 16 \\cdot 15$$ such cases. 4. Three identical quadrants: There are $\\binom{4}{3}$ ways to choose which three of the quadrants, $16$ ways to choose how to paint those three quadrants, and $15$ ways to choose how to paint the remaining quadrant. Hence, there are $$\\binom{4}{3} \\cdot 16 \\cdot 15$$ such cases. 5. Four identical quadrants: As you observed, there are $16$ options for painting all four quadrants in the same way. Observe that $$16 \\cdot 15 \\cdot 14 \\cdot 13 + \\binom{4}{2} \\cdot 16 \\cdot 15 \\cdot 14 + 3 \\cdot 16 \\cdot 15 + \\binom{4}{3} \\cdot 16 \\cdot 15 + 16 = 2^{16}$$ Though I did not added and calculated it all, here are dew flaws that I saw:- First of all 240 = 16*15 Second, you did not include case where 3 of them are same", "below is a 4×4 square, but in all sizes of 2-dimensional squares, there are three possible types in the bag: those with 2, 1, or 0 sides painted. I solved my first variation case by case. In any nxn square, • There are 4 corner squares with 2 sides painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4}{n^2} \\cdot \\frac{2}{4} = \\frac{2}{n^2}$. • There are $4(n-2)$ edge squares not in a corner with 1 side painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4(n-2)}{n^2} \\cdot \\frac{1}{4} = \\frac{n-2}{n^2}$. • All other squares have 0 sides painted, so the probability of picking one of those squares and then spinning a red side is 0. • Adding the probabilities for the separate cases gives the total probability: $\\displaystyle \\frac{2}{n^2}+\\frac{n-2}{n^2}=\\frac{1}{n}$ After reading Mike’s and Alexander’s posts, I saw a much easier approach. • Paint all 4 edges of an nxn square, and divide each painted edge into n painted unit segments. This creates $4 \\cdot n$ total painted small segments. • Decompose the nxn original square into $n^2$ unit squares. Each unit square has 4 edges giving $4 \\cdot n^2$ total edges. • Because every edge of every unit square is equally likely", "is excluded. This gives us the problem of arranging one red cube, three blue cubes, and four green cubes. The number opossible arrangements is $\\frac{8!}{4!\\cdot3!}=280$. Note that we do not need to multiply by the number of red cubes because there is no way to distinguish between the first red cube and the second. Case 2: The blue cube is excluded. This gives us the problem of arranging two red cubes, two blue cubes, and four green cubes. The number of possible arrangements is $\\frac{8!}{2!\\cdot2!\\cdot4!}=420$. Case 3: The green cube is excluded. This gives us the problem of arranging two red cubes, three blue cubes, and three green cubes. The number of possible arrangements is $\\frac{8!}{2!\\cdot3!\\cdot3!}=560$. Adding up the individual cases from above gives the answer as $280+420+560=\\boxed{\\textbf{(D) } 1,260}$.", "how to evaluate combinations of different configurarations of a rubik's cube 1. how to calculate the total combination number of an rubik's cube ? 2. and what is the number of combination if we fix one face of it with all matched colours , specifically saying blue one ,?? 3. no. of combination of two adjacent faces , say blue and yellow are with their matched colours ( ,if it is possible ) 4. no. of combination with 3 adjacent faces are matched with their colours (say the faces are red,blue and yellow ,, again if it is possible) 5. same thing with 4 adjacent faces ,say blue yellow ,, red and white ,are matched (again if it is possible ) 1. There are 8 corner cubies, each with 3 orientations; 12 edge cubies, each with 2 orientations; you cannot rotate a single corner cubie; you cannot flip a single edge cubie; you cannot have odd number of permutations. This leads to: $$\\frac{8!3^812!2^12}{3\\cdot2\\cdot2}\\approx4.3\\times 10^{19}.$$ 2. 4 corner cubies and 8 edge cubies are flexible, so it is: $$\\frac{4!3^48!2^8}{3\\cdot2\\cdot2}\\approx1.7\\times 10^9.$$ 3. 2 corner cubies and 5 edge cubies are flexible, so it is: $$\\frac{2!3^25!2^5}{3\\cdot2\\cdot2}=5760.$$ 4. 1 corner cubies and 3 edge cubies are flexible, so it is: $$\\frac{1!3^13!2^3}{3\\cdot2\\cdot2}=12.$$ 5. Only one edge cubie is not matched, so it cannot", "faces as we go along, noting that there are ${6\\choose 2} = 15$ pairs of colors for the top and bottom faces. Using just one color for the four sides: no change, so $6$ ways to color the sides and $6\\cdot 15 = 90$ colorings altogether. (Swapping ends doesn’t affect the sides.) Using two colors: no change, so $60$ ways and $60\\cdot 15 = 900$ colorings; swapping ends affects the sides, but they can be restored by a rotation. (Running total: $990$) Using three colors: As before, there are $6$ ways to choose the color that appears twice and then $10$ ways to choose the other two colors, and the two faces of the same color may be either adjacent or opposite. If they are opposite, it makes no difference how the remaining two colors are applied to the other two sides, so that sub-case yields $60$ ways and $60\\cdot 15 = 900$ colorings. If they are adjacent, however, the order in which the other two colors are applied to the other two sides does matter, and we get $120$ ways. Since these $120$ ways of coloring the sides are not preserved by swapping ends, each apparently extends to $30$ different ways of coloring the prism, for a total of $120\\cdot 30 = 3600$ colorings. However, this actually", "successful outcomes}}{\\text{total number of outcomes}} = \\frac{24}{54} = \\frac{4}{9}.$$ No, because you have picked a random face of the cube to look at and seen it is black. That makes it more likely that you have a corner cube than one of the other ones with some painted faces. Maybe a more careful statement of the problem is we choose a random cube and a random face of that cube. Given that the face we choose is black, what is the chance the cube is a corner cube? The simple solution is that we have selected a random black face. There are $$54$$ black faces, $$24$$ of which are on corner cubes, so the chance we have a corner cube is $$\\frac {24}{54}$$ • Thank you very much sirs! I gave acceptance to the one I see first. Hope I was not unfair! – Sal.Cognato Feb 16 at 17:18 You can also use Bayes' theorem. The prior probability of choosing a corner cube is $$8 \\over 27$$. The probability of a corner cube showing a black face is $$1 \\over 2$$. The overall probability of a black face is $${54 \\over 162}=1/3$$, so $$P({\\rm corner }|{\\rm black})={1/2 \\over1/3} \\times 8/27 =4/9$$", "= 56 56 x 1 = 56 Question 5. $$\\frac{5}{12}$$ of 144 is _________. Explanation: $$\\frac{5}{12}$$ of 144 = $$\\frac{5×144}{12}$$ = $$\\frac{720}{12}$$ = 60 Question 6. 7 × 8 – 4 × 3 = _____________ Explanation: According to BODMAS rule Brackets, Order, Division, Multiplication, Addition, Subtraction. 7 × 8 – 4 × 3 = 56 – 12 = 44 Question 7. .9 = % Explanation: .9 = % Let the unknown number be x 0.9 = x% x = 0.9 x 100 x = 90 Question 8. Find z. $$\\frac{z}{21}$$ = 7 z = _________ z = 147 Explanation: $$\\frac{z}{21}$$ = 7 z = 7 x 21 z = 147 Question 9. Write 6 : 58 P.M. in a 24-hour time format. 18:58 P.M Explanation: 6 : 58 P.M. in a 24-hour time format is 18:58 where as, Write 6 : 58 P.M. is mentioned in 12 hours format. Question 10. How many faces are on the solid? 5 faces Explanation: Given solid shape has four side faces and, one bottom face. So, total 5 faces. Question 11. Two red cubes, two green cylinders, and three yellow rectangular prisms were placed into a bag. If you reach into the bag and pull out an object, what is the probability that it will not be a red block? $$\\frac{2}{7}$$", "divide the final answer by !!2!!. Counting by this method tells us that the number of faces in an !!n!!-cube is is: $$\\frac122^n\\cdot \\binom n2 \\cdot \\frac12$$ where the !!\\frac122^n!! counts the number of ways to choose an even vertex !!E!!, the !!\\binom n2!! counts the number of ways to choose an even vertex !!E'!! that shares a face with !!E!!, and the extra !!\\frac12!! adjusts for the double-counting in the orderof !!E!! and !!E'!!. We can simplify this to $$2^{n-2}\\binom n2.$$ When !!n=3!! we get $$2^{3-2}\\binom 32 = 2\\cdot 3 = 6$$ and it is correct for all !!n!!, since !!\\binom n2 = 0!! when !!n<2!!: $$\\begin{array}{r|rrrrr} n& 0&1&2&3&4&5 \\\\ \\text{faces} = 2^{n-2}\\binom n2 & 0 & 0 & 1 & 6 & 24 & 80 \\\\ \\end{array}$$ That's not quite the !!\\binom{2^{n-1}}2!! or the !!\\left({{(n-1)2^{n-2}}\\atop 2}\\right)!! we were looking for, but it at least does produce a bijective map between faces and pairs of something.", "vertical. There are 3 scenarios in which four vertical faces are red and two blue: pick any pair of opposing faces of the cube to be blue, and the rest red. Then we can choose to place the cube blue-face-down, so the red faces come out vertical. There are 3 such pairs (since there are 6 faces). There are 6 scenarios in which five faces are red and one blue: pick any particular face to be blue, then we can choose to place the cube blue-face-down (or blue-face-up), and the red faces come out vertical. There is 1 scenario in which six faces are red. The red faces will come out vertical no matter how we place the cube. So there should be a total of 10 scenarios in which we can place the cube with four vertical red faces. As you alluded to, the same argument goes through for blue faces, giving 20 scenarios in total. So the probability should be $\\frac{20}{64}$ • why are the scenarios in which the the vertical faces do not have the same color counted in the desirable outcomes – jamarcus tyrone Dec 27 '17 at 1:05 • @jamarcus can you explain what you mean? – greenturtle3141 Dec 27 '17 at 1:26 • @greenturtle3141 The scenarios in which the vertical faces are", "draw 10 of these and note inside each hexagon where the portal takes you. I realize I could just assign this randomly, but then it might not be a correct 5-cube. And gosh darn it, it matters that it is! - The 0D corners of a 5-cube can be represented by 5 bit binary words. There are 32 of them. The 1D edges between the corners are then pairs of corners that differ in exactly 1 bit. There are $\\frac {32\\cdot 5}2=80$ of them, as you can choose any corner, choose one direction to move in (the bit to change) but each edge is counted twice, once from each end. The 2D squares can be found by taking a corner, selecting two bits to change, and changing them to all four possibilities. There are $\\frac {32 \\cdot 10}4=80$ of them and so on. This gives $10$ 4D tesseract faces and $40$ 3D cells. As each tesseract has $8$ 3D cubical faces, when you enter one you can go straight through or make any of six right-angle turns. There is only one tesseract you can't get to from a given one-the opposite one. Each tesseract face has one coordinate fixed and 16 corners that come from all possibilities of the other four bits. The cells of a tesseract come" ]

[ "# Math Help - Cube probability 1. ## Cube probability On choosing three vertex of a cube at random, what's the probability that the three are in a same face? 2. I would think it would be $(\\frac{1}{6})(\\frac{3}{4})=\\frac{1}{8}$ It may be more involved than that. 3. Originally Posted by Krizalid On choosing three vertex of a cube at random, what's the probability that the three are in a same face? The counting principle, of course. The first pick if free, we may pick any vertex. There are 8 vertices on the cube, we've picked one leaving 7. Of these only 6 share a face with the one we've picked. So we have $1 \\cdot \\frac{6}{7}$ so far. Now, we've picked two vertices. 1) If the two vertices are on the same edge, there are 4 possible of 6 vertices left to pick, so the probability will be $1 \\cdot \\frac{6}{7} \\cdot \\frac{4}{6} = \\frac{4}{7}$ 2) If the two vertices are not on the same edge, they are on a diagonal across one face. In this case there is only one possible face that these could be on, so there are only two possible vertices out of 6, so the probability will be $1 \\cdot \\frac{6}{7} \\cdot \\frac{2}{6} = \\frac{2}{7}$ Only one of these choices is possible, so we", "below is a 4×4 square, but in all sizes of 2-dimensional squares, there are three possible types in the bag: those with 2, 1, or 0 sides painted. I solved my first variation case by case. In any nxn square, • There are 4 corner squares with 2 sides painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4}{n^2} \\cdot \\frac{2}{4} = \\frac{2}{n^2}$. • There are $4(n-2)$ edge squares not in a corner with 1 side painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4(n-2)}{n^2} \\cdot \\frac{1}{4} = \\frac{n-2}{n^2}$. • All other squares have 0 sides painted, so the probability of picking one of those squares and then spinning a red side is 0. • Adding the probabilities for the separate cases gives the total probability: $\\displaystyle \\frac{2}{n^2}+\\frac{n-2}{n^2}=\\frac{1}{n}$ After reading Mike’s and Alexander’s posts, I saw a much easier approach. • Paint all 4 edges of an nxn square, and divide each painted edge into n painted unit segments. This creates $4 \\cdot n$ total painted small segments. • Decompose the nxn original square into $n^2$ unit squares. Each unit square has 4 edges giving $4 \\cdot n^2$ total edges. • Because every edge of every unit square is equally likely", "# CTK Insights • ## Pages 20 May ### Probabilities in a Painted Cube A wooden cube - after being painted all over - has been cut into $3\\times 3\\times 3$ smaller cubes. These were thoroughly mixed in a bag, from which one was produced and tossed. What is the probability that a painted side turned up? On impulse, one would approach the problem in a more or less standard way. There are $8$ corner cubes with $3$ painted sides, $12$ mid-edge cube with $2$ painted sides, and $6$ face-central cubes with only $1$ side painted. The total probability is then $\\displaystyle\\frac{8}{27}\\cdot\\frac{3}{6}+\\frac{12}{27}\\cdot\\frac{2}{6}+\\frac{6}{27}\\cdot\\frac{1}{6}=\\frac{1}{3}.$ Now generalize: cut the cube into $n\\times n\\times n$, $n\\gt 1$ smaller cubes and ask the same question. The problem is not awfully difficult but needs some figuring out. Following the foregoing pattern, we eventually arrive at $\\displaystyle\\frac{8}{n^3}\\cdot\\frac{3}{6}+\\frac{12\\cdot (n-2)}{n^3}\\cdot\\frac{2}{6}+\\frac{6\\cdot (n-2)^2}{n^3}\\cdot\\frac{1}{6}$ This expression simplifies, as you can verify, to $\\displaystyle\\frac{1}{n},$ which, by the way, confirms the answer for $n=3.$ The above is a useful and not too difficult exercise, but there is a delightful shortcut that avoids most of the counting of cubes and their sides. $n^3$ cubes have a total of $6\\cdot n^3$ sides. Of these, $6\\cdot n^2$ are painted. All sides have the same probability of turning up, therefore, a painted side will turn up", "up to rotation and reflection, there are only 2 ways to arrange the numbers. The group of symmetries of the cube has size 48, so after rotations and reflections there are $$2\\cdot48=96$$ valid arrangements. Now for the probability: There are $$8!=40320$$ ways to arrange the numbers, of which $$96$$ are valid. The probability is therefore $$\\frac{96}{40320}=\\frac{1}{420}$$. • Correct! Checkmark incoming! Jul 4, 2020 at 12:18", "is excluded. This gives us the problem of arranging one red cube, three blue cubes, and four green cubes. The number opossible arrangements is $\\frac{8!}{4!\\cdot3!}=280$. Note that we do not need to multiply by the number of red cubes because there is no way to distinguish between the first red cube and the second. Case 2: The blue cube is excluded. This gives us the problem of arranging two red cubes, two blue cubes, and four green cubes. The number of possible arrangements is $\\frac{8!}{2!\\cdot2!\\cdot4!}=420$. Case 3: The green cube is excluded. This gives us the problem of arranging two red cubes, three blue cubes, and three green cubes. The number of possible arrangements is $\\frac{8!}{2!\\cdot3!\\cdot3!}=560$. Adding up the individual cases from above gives the answer as $280+420+560=\\boxed{\\textbf{(D) } 1,260}$.", "So the probability of rolling a painted side after selecting a small cube at random is: $\\displaystyle \\frac{\\text{surface area}}{6\\cdot\\text{volume}}$. For example, say we have a 3x4x5 rectangle made up of 60 1x1x1 cubes, and we paint the outside. If we break it apart and roll a cube at random, the probability that a painted side will show up is: $\\displaystyle \\frac{2(3(4)+4(5)+3(5))}{6\\cdot3(4)(5)}=\\frac{47}{180}$. You would get the same answer if you went through the analysis of interior, face, edge, and corner pieces: there are $(3-2)\\times(4-2)\\times(5-2)=6$ interior pieces (subtract two from each dimension, because you don’t want the two outside pieces), 24 edge pieces (for an edge of side $n$, there are $n-2$ edge pieces; there are 4 edges of each length, so we have $4((3-2)+(4-2)+(5-2))=24$), 8 corners (like always), and 22 face pieces (what’s remaining of the 60 cubes must be the face pieces!) , for a probability of $\\displaystyle \\underbrace{\\frac{6}{60}\\left(\\frac{0}{6}\\right)}_\\text{interior piece}+\\underbrace{\\frac{22}{60}\\left(\\frac{1}{6}\\right)}_\\text{face piece}+\\underbrace{\\frac{24}{60}\\left(\\frac{2}{6}\\right)}_\\text{edge piece}+\\underbrace{\\frac{8}{60}\\left(\\frac{3}{6}\\right)}_\\text{corner}=\\frac{22+48+24}{360}=\\frac{47}{180}$. This is quite complicated, even though we made life a little easier by counting the face pieces last; they are hardest to count directly, so we counted the other pieces first and took the remainder to be face pieces. For a cube with side length $n$, the surface area is $6n^2$ and the volume is $n^3$. So the probability of rolling a painted side would be", "successful outcomes}}{\\text{total number of outcomes}} = \\frac{24}{54} = \\frac{4}{9}.$$ No, because you have picked a random face of the cube to look at and seen it is black. That makes it more likely that you have a corner cube than one of the other ones with some painted faces. Maybe a more careful statement of the problem is we choose a random cube and a random face of that cube. Given that the face we choose is black, what is the chance the cube is a corner cube? The simple solution is that we have selected a random black face. There are $$54$$ black faces, $$24$$ of which are on corner cubes, so the chance we have a corner cube is $$\\frac {24}{54}$$ • Thank you very much sirs! I gave acceptance to the one I see first. Hope I was not unfair! – Sal.Cognato Feb 16 at 17:18 You can also use Bayes' theorem. The prior probability of choosing a corner cube is $$8 \\over 27$$. The probability of a corner cube showing a black face is $$1 \\over 2$$. The overall probability of a black face is $${54 \\over 162}=1/3$$, so $$P({\\rm corner }|{\\rm black})={1/2 \\over1/3} \\times 8/27 =4/9$$", "in the box. – Arpit Bajpai May 5 '12 at 6:01 Would that lead you to claim that the chance of picking any given color is $\\frac 14$? This is not correct. Put 10 blue marbles and one red marble in a hat, draw 20 times with replacement, and see if you get around 10 reds. – Ross Millikan May 5 '12 at 14:33 We can also work directly with the probabilities. Pick the cubes a little cube at random, then pick another. The probability that the first cube picked has exactly $2$ red faces is $\\frac{24}{64}$. Given that the first cube picked had exactly $2$ red faces, the probability that the second cube picked has exactly $2$ red faces is $\\frac{23}{63}$. Thus the probability we get two cubes with exactly $2$ red faces is $\\frac{24}{64}\\cdot\\frac{23}{63}$. Imagine that we have a total of $64$ cubes, with $24$ identical yellow ones, and the rest identical and blue. Put the $64$ cubes in a bag, shake well, and blindfolded pick a cube at random. Is the probability of picking a yellow equal to $1$? – André Nicolas May 5 '12 at 5:11 In the previous comment, shouldn't you ask if the probability to pick a yellow one is $\\frac 12$? – Ross Millikan May 5 '12 at 5:16 The problem", "point $C$ is one of the two points on the side that contains $AB$. Hence the probability of $ABC$ intersecting the inside of the cube is $2/3$. In the third case, already the diagonal $AB$ contains points inside the cube, hence this case will be good regardless of the choice of $C$. Summing up all cases, the resulting probability is: $$\\frac 37\\cdot\\frac 13 + \\frac 37\\cdot \\frac 23 + \\frac 17\\cdot 1 = \\boxed{\\frac 47}$$ ## Solution 2 There are $\\binom{8}{3}=56$ ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability. There are four ways to choose three points from the vertices of a single face. Since there are six faces, $4 \\times 6 = 24$. Thus, the probability of what we don't want is $\\frac{24}{56} = \\frac{3}{7}$. Using complementary probability, $$1- \\frac 37 = \\boxed{\\frac 47}$$", "a “3-dimensional square”, I re-phrased Alexander’s question into two dimensions. The trickier part was thinking what it would mean to “roll” a 2-dimensional shape. The outside of an nxn square is painted red and is chopped into $n^2$ unit squares. The latter are thoroughly mixed up and put into a bag. One small square is withdrawn at random from the bag and spun on a flat surface. What is the probability that the spinner stops with a red side facing you? Shown below is a 4×4 square, but in all sizes of 2-dimensional squares, there are three possible types in the bag: those with 2, 1, or 0 sides painted. I solved my first variation case by case. In any nxn square, • There are 4 corner squares with 2 sides painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4}{n^2} \\cdot \\frac{2}{4} = \\frac{2}{n^2}$. • There are $4(n-2)$ edge squares not in a corner with 1 side painted. The probability of picking one of those squares and then spinning a red side is $\\displaystyle \\frac{4(n-2)}{n^2} \\cdot \\frac{1}{4} = \\frac{n-2}{n^2}$. • All other squares have 0 sides painted, so the probability of picking one of those squares and then spinning a red side is 0. • Adding the probabilities for", "1. ## Probability In a box there are 22 cubes, 12 red and 10 blue. 7 cubes are chosen at random. Find the probability for the following events: a. Among the seven cubes there are 4 blue cubes and 3 red cubes There are others, but I've tried to do them, and for a I get something like.. 0.00780 or something like that. Which I think might be wrong.. So if anyone could just show me what to do, I think I can do the rest myself, thanks 2. Hi Fnus, The probability of getting one blue cube is $\\displaystyle \\frac{10}{22}$ . Due to the fact that the cubes aren't being replaced the probability of getting another blue cube is $\\displaystyle \\frac{9}{21}$ and so on and so fourth. This gives an answer of $\\displaystyle \\frac{10}{1539} \\approx 0.0065$ 3. Hmm, I think that was what I did.. And I'd multiply them, right? Maybe I just pressed something wrong. I'll try again, thanks. 4. Here is the correct answer: $\\displaystyle \\frac {{10 \\choose 4}{12 \\choose 3}}{{22 \\choose 7}} =0.270897832817337$ 5. Originally Posted by Moo I thought it was that... Oohh...I'm gonna have to deal with that guy and all these other weird distributions this fall in my hardcore stats class...once I survive probability --Chris 6. Originally Posted by Plato $\\displaystyle", "how to evaluate combinations of different configurarations of a rubik's cube 1. how to calculate the total combination number of an rubik's cube ? 2. and what is the number of combination if we fix one face of it with all matched colours , specifically saying blue one ,?? 3. no. of combination of two adjacent faces , say blue and yellow are with their matched colours ( ,if it is possible ) 4. no. of combination with 3 adjacent faces are matched with their colours (say the faces are red,blue and yellow ,, again if it is possible) 5. same thing with 4 adjacent faces ,say blue yellow ,, red and white ,are matched (again if it is possible ) 1. There are 8 corner cubies, each with 3 orientations; 12 edge cubies, each with 2 orientations; you cannot rotate a single corner cubie; you cannot flip a single edge cubie; you cannot have odd number of permutations. This leads to: $$\\frac{8!3^812!2^12}{3\\cdot2\\cdot2}\\approx4.3\\times 10^{19}.$$ 2. 4 corner cubies and 8 edge cubies are flexible, so it is: $$\\frac{4!3^48!2^8}{3\\cdot2\\cdot2}\\approx1.7\\times 10^9.$$ 3. 2 corner cubies and 5 edge cubies are flexible, so it is: $$\\frac{2!3^25!2^5}{3\\cdot2\\cdot2}=5760.$$ 4. 1 corner cubies and 3 edge cubies are flexible, so it is: $$\\frac{1!3^13!2^3}{3\\cdot2\\cdot2}=12.$$ 5. Only one edge cubie is not matched, so it cannot", "$6 \\cdot 5 \\cdot 4 \\cdot 4 \\cdot 4 = 1920$ combinations. In the case that $C$ has a different color from $A$ and $D$ has the same color as $A$, $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices, $D$ has $1$ choice, and $E$ has $5$ choices, resulting in a possible of $6 \\cdot 5 \\cdot 4 \\cdot 1 \\cdot 5 = 600$ combinations. Adding all those combinations up, we get $600+1920+600=\\boxed{3120 \\ \\mathbf{(C)}}$. ## Solution 2 First, notice that there can be $3$ cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors. Case $1$: There are $6!$ ways of assigning each vertex a different color. $6! = 720$ Case $2$: There are $\\frac {6!}{2!} * 5$ ways. After picking four colors, we can rotate our pentagon in $5$ ways to get different outcomes. $\\frac {6!}{2} * 5 = 1800$ Case $3$: There are $\\frac {\\frac {6!}{3!} * 10}{2!}$ ways of arranging the final case. We can pick $3$ colors for our pentagon. There are $5$ spots for the first pair of colors. Then, there are $2$ possible ways we can put the final pair in the last $3$", "faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is $7$? • $\\frac{3}{8}$ • $\\frac{2}{11}$ • $\\frac{2}{3}$ • $\\frac{1}{3}$ • $\\frac{2}{9}$ ### Key Concepts Probability Combinatorics Cube Answer: $\\frac{2}{11}$ AMC-10A (2009) Problem 22 Pre College Mathematics ## Try with Hints We assume that the colours of the numbers are different.there are two dices and each of them 1 to 6.after throw,the probability of getting some pair of colors is the same for any two colors. Therefore there are $\\ 12 \\choose 2$=$66$ ways to pick to of the colours… can you finish the problem…….. Now given condition is that the sum will be $7$.So $7$ can be obtained by $1 +6$,$2+5$,$3+4$ and Each number in the bag has two different colors, Therefore each of these three options corresponds to four pairs of colors.SO $7$ comes from $3.4$=$12$ pairs….. can you finish the problem…….. So our required probability will be $\\frac{12}{66}=\\frac{2}{11}$. Categories ## Probability in Coordinates | AMC-10A, 2003 | Problem 12 Try this beautiful problem from Probability based on Coordinates. ## Probability in Coordinates – AMC-10A, 2003- Problem 12 A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$.", "edges giving $4 \\cdot n^2$ total edges. • Because every edge of every unit square is equally likely to be spun, the total probability of randomly selecting a smaller square and spinning a red side is $\\displaystyle \\frac{4n}{4n^2}=\\frac{1}{n}$. The dimensions of the “square” don’t seem to matter! WARNING: Oliver Wendell Holmes noted, “A mind that is stretched by a new experience can never go back to its old dimensions.” The math after this point has the potential to stretch… EXTENDING THE PROBLEM MORE: I now wondered whether this was a general property. In the 2-dimensional square, 1-dimensional edges were painted and the square was spun to find the probability of a red edge facing. With the originally posed cube, 2-dimensional faces were painted and the cube was tossed to find the probability of an upward-facing red face. These cases suggest that when a cube of some dimension with edge length n is painted, is decomposed into unit cubes of the original dimension, and is spun/tossed to show a cube of one smaller dimension, then the probability of getting a painted smaller-dimensional cube of is always $\\displaystyle \\frac{1}{n}$, independent of the dimensions occupied by the cube. Going beyond the experiences of typical middle or high school students, I calculated this probability for a 4-dimensional hypercube (a tesseract). • The", "vertical. There are 3 scenarios in which four vertical faces are red and two blue: pick any pair of opposing faces of the cube to be blue, and the rest red. Then we can choose to place the cube blue-face-down, so the red faces come out vertical. There are 3 such pairs (since there are 6 faces). There are 6 scenarios in which five faces are red and one blue: pick any particular face to be blue, then we can choose to place the cube blue-face-down (or blue-face-up), and the red faces come out vertical. There is 1 scenario in which six faces are red. The red faces will come out vertical no matter how we place the cube. So there should be a total of 10 scenarios in which we can place the cube with four vertical red faces. As you alluded to, the same argument goes through for blue faces, giving 20 scenarios in total. So the probability should be $\\frac{20}{64}$ • why are the scenarios in which the the vertical faces do not have the same color counted in the desirable outcomes – jamarcus tyrone Dec 27 '17 at 1:05 • @jamarcus can you explain what you mean? – greenturtle3141 Dec 27 '17 at 1:26 • @greenturtle3141 The scenarios in which the vertical faces are", "then $$3$$ colors opposite the blue. There are two colors left, which can be placed in $$2$$ ways onto the remaining two opposite fields. It follows that there are $$4\\cdot3\\cdot 2=24$$ different cubes that can be formed, given the two fields red and blue neighboring each other. 4: Since the cube can be rotated, assume B is fixed to the front face of the cube. Hence, there are 4 possible ways R can be oriented around B. Since there are rotations, it is redundant to multiply by the 6 faces on the cube. Hence 4 ways. 1) That says nothing about the other colors; just that red is either above, below, to the left, or to the right, of the blue. 2) Because those are rotations THEY are redundant. You can count this as $$1$$ way. But no you have to figure out the different ways to put the remaining four colors. 10: If I map out the cube by drawing 4 horizontal boxes and 1 box below and 1 box above, and assign B to each face, I count the number of ways R can be next to B for each box, and arrive at 10. Huh???????? 24: Similar concept to 4, just multiplied by 6. Coincidentally that is the correct answer but for entirely the wrong", "disks [#permalink] ### Show Tags 08 Jun 2016, 01:49 TheLostBear wrote: Bunuel wrote: sharmasneha wrote: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3? (A) 1/16 (B) 41/128 (C) 87/128 (D) 225/256 (E) 255/256 $$P(0 < blue < 4) = 1 - P(blue \\ = \\ 0 \\ or \\ 4) =$$ $$=1 - (\\frac{75}{100}*\\frac{75}{100}*\\frac{75}{100}*\\frac{75}{100} + \\frac{25}{100}*\\frac{25}{100}*\\frac{25}{100}*\\frac{25}{100}) = \\frac{87}{128}$$. Hi Bunuel, Since the problem does not explicitly state that we will replace the disks that are taken out of the bags, shouldn't we have accounted for the non-replacement? I was thinking along the line of this example; we pick 2 blocks from a box holding 10 blocks, 3 of which are red. The probability of getting 2 red blocks then is 3/10*2/9= 1/15. Using this approach on the question of interest, I arrived at a very ugly figure which was definitely not the OA. Thank you. If there is a replacement it'll be explicitly mentioned. Also, we are choosing only 1 disk from each bag, so replacement does not play part in", "on the faces in any order. That gives $$\\binom{8}{6}\\cdot6!=20160$$ ways. The probability then becomes $$\\frac{144}{20160}=\\frac{1}{140}$$.", "Adding the probabilities for the separate cases gives the total probability: $\\displaystyle \\frac{2}{n^2}+\\frac{n-2}{n^2}=\\frac{1}{n}$ After reading Mike’s and Alexander’s posts, I saw a much easier approach. • Paint all 4 edges of an nxn square, and divide each painted edge into n painted unit segments. This creates $4 \\cdot n$ total painted small segments. • Decompose the nxn original square into $n^2$ unit squares. Each unit square has 4 edges giving $4 \\cdot n^2$ total edges. • Because every edge of every unit square is equally likely to be spun, the total probability of randomly selecting a smaller square and spinning a red side is $\\displaystyle \\frac{4n}{4n^2}=\\frac{1}{n}$. The dimensions of the “square” don’t seem to matter! WARNING: Oliver Wendell Holmes noted, “A mind that is stretched by a new experience can never go back to its old dimensions.” The math after this point has the potential to stretch… EXTENDING THE PROBLEM MORE: I now wondered whether this was a general property. In the 2-dimensional square, 1-dimensional edges were painted and the square was spun to find the probability of a red edge facing. With the originally posed cube, 2-dimensional faces were painted and the cube was tossed to find the probability of an upward-facing red face. These cases suggest that when a cube of some dimension with edge length" ]

[ "red balls and blue balls and some containers, how would you distribute those balls among the containers such that the probability of picking a red ball is maximized, assuming that the user randomly chooses a container and then randomly picks a ball from that?", "are simultaneously picked at random from a jar containing 2 red balls, 2 green balls, and 6 yellow balls. What is the probability that exactly two of the selected balls will be red?", "# arrow_back There is a container full of coloured bottles, red, blue, green and orange. Some of the bottles are picked out and displaced. Sumit did this 1000 times and got the following results: 75 views There is a container full of coloured bottles, red, blue, green and orange. Some of the bottles are picked out and displaced. Sumit did this 1000 times and got the following results: - No. of blue bottles picked out: 300 - No. of red bottles: 200 - No. of green bottles: 450 - No. of orange bottles: 50 a) What is the probability that Sumit will pick a green bottle? b) If there are 100 bottles in the container, how many of them are likely to be green? a) What is the probability that Sumit will pick a green bottle? Ans: For every 1000 bottles picked out, 450 are green. Therefore, $P($ green $)=450 / 1000=0.45$ b) If there are 100 bottles in the container, how many of them are likely to be green? Ans: The experiment implies that 450 out of 1000 bottles are green. Therefore, out of 100 bottles, 45 are green. by Gold Status (30,543 points) ## Related questions Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Arrange the following group of colored lights in the order of", "# An Amazon interview question (not completely solved) Posted on January 24, 2011 by phimuemue Taken from this site: Given red balls and blue balls and some containers, how would you distribute those balls among the containers such that the probability of picking a red ball is maximized, assuming that the user randomly chooses a container and then randomly picks a ball from that? ## Solution (probably) The solution to this problem works as follows: Take all your red balls and put them one by one in one container after another until you have no balls left. If there are containers left (i.e. without red balls), take your blue balls and distribute them as you like into the remaining containers. If there are no containers left, take all your blue balls and put them in one arbitrary container. To prove this (at least one case), we first do some definitions: W.l.o.g I assume , i.e. all containers can be filled with at least one ball. Let's call the number of containers . denotes the event that container is choosen. denotes the event that a red ball is chosen from the -th container. The probability of choosing the -th container is (since we have containers in total) . Let denote the number of red balls in container and denote", "# A red container contains 7 items of which 2 are new and 5 are old items. A purple container contains 4 items of which 1 is new and 3 are old items. An item is drawn at random from each box. 1. What is the probability that both items are old items? ## What is the probability that one item is new and one item old? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 2 Mar 13, 2018 Two old items: $\\frac{15}{28} \\approx 0.5357$ One old item: $\\frac{11}{28} \\approx 0.3929$ #### Explanation: The draw from the red container for an old item has a probability of $\\frac{5}{7}$ of being successful. The draw from the purple container for an old item has a probability of $\\frac{3}{4}$ of being successful. This means that there is a $\\frac{5}{7} \\times \\frac{3}{4} = \\frac{15}{28} \\approx 0.5357$ probability that both items will be old. For finding where one item is old, we can look at the probability of the old item being drawn from the red container and a new item from the purple container $\\left(\\frac{5}{7} \\times \\frac{1}{4} = \\frac{5}{28}\\right)$ and adding that to the probability of the old item being drawn from the", "# how to distribute n red and m blue balls in some containers to maximize probability of random picking a red one from them? This is an interview question. Given n red balls and m blue balls and some containers, how would you distribute those balls among the containers such that the probability of picking a red ball is maximized, assuming that the user randomly chooses a container and then randomly picks a ball from that. My solution: suppose we have c containers, distribute n/c read balls to each c. If c == 1, put all of them together, it is n/(m+n) If c == 2, put 1 red in c1 and all left red and all blue ones in c2 in this way , we have: 1/2 + 1/2 *(n-1) /(m+n-1) > 1/2 If c == 3, put a red in c1, put a red in c2, put left red and all blue in c3, we have: 1/3 + 1/3 + 1/3 * (n-2)/(m+n-2) > 2/3 If c == n, put a red in each of p -1, and all left red and blue in pth container, we have: ( Sum of (1/n) from 1 to n-1 ) + ( 1/n * 1/(m+1) ) (n-1)/n + 1/n * 1/(m+1) == 1 (almost) As n is large, the", "# how to distribute n red and m blue balls in some containers to maximize probability of random picking a red one from them? This is an interview question. Given n red balls and m blue balls and some containers, how would you distribute those balls among the containers such that the probability of picking a red ball is maximized, assuming that the user randomly chooses a container and then randomly picks a ball from that. My solution: suppose we have c containers, distribute n/c read balls to each c. If c == 1, put all of them together, it is n/(m+n) If c == 2, put 1 red in c1 and all left red and all blue ones in c2 in this way , we have: 1/2 + 1/2 *(n-1) /(m+n-1) > 1/2 If c == 3, put a red in c1, put a red in c2, put left red and all blue in c3, we have: 1/3 + 1/3 + 1/3 * (n-2)/(m+n-2) > 2/3 If c == n, put a red in each of p -1, and all left red and blue in pth container, we have: ( Sum of (1/n) from 1 to n-1 ) + ( 1/n * 1/(m+1) ) (n-1)/n + 1/n * 1/(m+1) == 1 (almost) As n is large, the", "have no way of determining whether or not a green jelly bean exists in this jar without some more information. But now here’s the kicker: with this jar comes a note that simply says “if we draw a jelly bean at random from this jar, the probability of selecting a green jelly bean is nonzero.” With this simple note we gain an important piece of information: there must be a green jelly bean in this jar if the probability of drawing one is nonzero. We cannot look too far into the note. We don’t know if there are a plethora of green jelly beans or just one. We don’t know if the green jelly beans are pristine, kidney shaped pieces of apple flavored heaven or misshapen, moldy lumps of questionable origin. But we know that a green jelly bean exists. And that is the crux of the probabilistic method. The Probabilistic Method: Suppose we have a group of combinatorial objects $C$, and we want to determine whether or not one of these objects has the property $P$. If we select an object randomly (or following any arbitrary probability distribution) from $C$ and this object has $P$ with nonzero probability, then an object in $C$ with property $P$ exists. With this simple introduction to the method in hand, let’s", "the number of blue balls in container . It is obvious that . So the probability of choosing a red ball from container if we chose container already, is . No we can use a case differentiation: ### Case 1: In this case, we have containers containing exactly 1 red ball and containers containing either one or more blue balls. So the probability of picking a red ball according to the proposed experiment, is . Proving that this probability is maximal is not very hard: We know, that at most summands can be positive (since containers with no red balls yield a probability of 0 and we have exactly red balls). We know further, that . Therefore we know, that the maximal probability in this case is as stated above. ### Case 2: (unsolved) Here we have at least one red ball in each container. More precisely, we have containers with exactly one red ball, and one container holding red and blue balls. The probability of picking a red ball can then be computed as . The above can be simplified to . To prove that this is the optimal probability is quite difficult (as far as I can judge it). We have to fill all containers. For all holds that . If we distribute the balls according", "# A fruit bowl contains 4 green apples and 7 red apples. What is the probability that a randomly selected apple will be green? ##### 1 Answer Apr 2, 2018 $\\frac{4}{11}$ #### Explanation: Probability is the green apples over the total number of apples", "# A fruit bowl contains 4 green apples and 7 red apples. What is the probability that a randomly selected apple will be green? Apr 2, 2018 $\\frac{4}{11}$ #### Explanation: Probability is the green apples over the total number of apples", "selected } ) & = 4 \\times \\frac { 4 like this: lesson... Probabilities of Conditional events 3, 4 and 5 ) = 1/169, approximately! To fit in the containers and have to fit in the containers have... Provide a free, world-class education to anyone, anywhere, so asnwer! Of each color ), such as marbles or poker chips results with more than once permutations...", "the area of $A$. That's what the definition is saying. You have seen many examples where you have quite naturally written down conditional probabilities without using the division rule. In these examples, an experiment had several stages (for example, cards were being dealt one by one) and you were working with conditional probabilities of events at later stages given the results of earlier stages. The division rule helps when the conditioning makes you go backwards in time, for example by giving you the result of a later stage and asking your opinion about what might have happened earlier. ### A Random Container I have two containers: a jar and a box. Each container has five balls. • The jar contains three red balls and two green balls. • The box contains one red ball and four green balls. I pick a container at random and then a ball at random from that container. Given that the ball is red, what is the chance I picked the box? At a glance, you should guess that the answer is less than half. Given that the ball is red, you should be leaning towards guessing that I picked the jar because it has a higher proportion of red balls than the box does. To make this intuition precise, it helps to", "# One container contains 6 red and 4 white balls, while a second container contains 7 red and 3 white balls One container contains 6 red and 4 white balls, while a second container contains 7 red and 3 white balls. A ball is chosen at random from the first container and placed in the second. Then a ball is chosen at random from the second container and placed in first. a. What is the probability to choose a red ball of the second container? b. What is the probability the ball was red chosen from the first container if know that from the second selected white? - Solution for question 1: Probability that a red ball will be chosen on first step: $\\frac{6}{10}$ In this case, we will have 8 red and 3 white balls in the second container, and the probability to chose a red ball on second step is $\\frac{8}{11}$. Overall probability in this case gives $\\frac{6}{10}*\\frac{8}{11}$ = $\\frac{48}{110}$. Let's mark it as p$_1$. Probability that a white ball will be chosen on first step: $\\frac{4}{10}$ Similarly to the previous case, we will get probability to choose a red ball from the second container is p$_2$=$\\frac{4}{10}*\\frac{7}{11}=\\frac{28}{110}$ Because our cases do not intersect, we have total probability p=p$_1$+p$_2$=$\\frac{76}{110}$=$\\frac{38}{55}$. Solutin for question 2: Here we shall use", "• anonymous A bag contains 4 red marbles, 6 blue marbles, and 8 green marbles. One marble is to be chosen at random. What is the probability that the chosen marble will be green? Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "# Probability problem • September 20th 2010, 01:47 PM ashamrock415 Probability problem One box contains 6 red balls and 4 green balls, and a second box contains 7 red balls and 3green balls. A ball is randomly selected from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning? • September 20th 2010, 02:01 PM Plato Quote: Originally Posted by ashamrock415 One box contains 6 red balls and 4 green balls, and a second box contains 7 red balls and 3green balls. A ball is randomly selected from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning? What must sequence must happen to reach that conclusion? Here is a hint: $R_1R_2\\text{ or }G_1G_2$.", "a large container is filled with red, yel [#permalink] ### Show Tags 22 May 2014, 11:18 In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed? A. 5 B. 4 C. 3 D. 2 E. 0 I am not able to comprehend the question correctly. There could be a lot of combinations for removing the beads. How could we find any particular answer? Math Expert Joined: 02 Sep 2009 Posts: 44398 Re: In a certain game, a large container is filled with red, yel [#permalink] ### Show Tags 22 May 2014, 11:29 Expert's post 2 This post was BOOKMARKED amarmeena1992 wrote: In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed? A. 5 B. 4 C. 3 D. 2 E. 0 I am not able to comprehend the question correctly.", "be created, if one patrol is 2 boys, 3 girls and 1 leader? 16. One green In the container are 45 white and 15 balls. We randomly select 5 balls. What is the probability that it will be a maximum one green? 17. PIN - codes How many five-digit PIN - code can we create using the even numbers?", "# Question Continuation of Exercise 2-65. Three containers are selected, at random, without replacement, from the batch. (a) What is the probability that the third one selected is defective given that the first and second one selected were defective? (b) What is the probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay? (c) What is the probability that all three are defective? Sales0 Views219", "# Math Help - Probability problem 1. ## Probability problem One box contains 6 red balls and 4 green balls, and a second box contains 7 red balls and 3green balls. A ball is randomly selected from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning? 2. Originally Posted by ashamrock415 One box contains 6 red balls and 4 green balls, and a second box contains 7 red balls and 3green balls. A ball is randomly selected from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning? What must sequence must happen to reach that conclusion? Here is a hint: $R_1R_2\\text{ or }G_1G_2$." ]

[ "# One container contains 6 red and 4 white balls, while a second container contains 7 red and 3 white balls One container contains 6 red and 4 white balls, while a second container contains 7 red and 3 white balls. A ball is chosen at random from the first container and placed in the second. Then a ball is chosen at random from the second container and placed in first. a. What is the probability to choose a red ball of the second container? b. What is the probability the ball was red chosen from the first container if know that from the second selected white? - Solution for question 1: Probability that a red ball will be chosen on first step: $\\frac{6}{10}$ In this case, we will have 8 red and 3 white balls in the second container, and the probability to chose a red ball on second step is $\\frac{8}{11}$. Overall probability in this case gives $\\frac{6}{10}*\\frac{8}{11}$ = $\\frac{48}{110}$. Let's mark it as p$_1$. Probability that a white ball will be chosen on first step: $\\frac{4}{10}$ Similarly to the previous case, we will get probability to choose a red ball from the second container is p$_2$=$\\frac{4}{10}*\\frac{7}{11}=\\frac{28}{110}$ Because our cases do not intersect, we have total probability p=p$_1$+p$_2$=$\\frac{76}{110}$=$\\frac{38}{55}$. Solutin for question 2: Here we shall use", "group” interpretation shows why. Here are diagrams using that interpretation showing $3 \\div \\frac12 = 2 \\cdot 3$ and $1 \\frac34 \\div \\frac12 = 2 \\cdot 1 \\frac34$.In fact, the structure of this context is so powerful, we can see why dividing any number by $\\frac12$ would double that number: $$x \\div \\frac12 = 2 \\cdot x = x \\cdot \\frac21$$ This is true for dividing by any unit fraction, for example $\\frac15$:In the diagram above, we can see that $1\\frac34$ is $\\frac15$ of a container, so a full container is $1\\frac34 \\div \\frac15$. Looking at the diagram, we can see why it must be that the full container is $5 \\cdot 1 \\frac34 = 1 \\frac34 \\cdot \\frac51$. With a little more work to make sense of it, we can use this interpretation to see why we multiply by the reciprocal when we divide by any fraction, for example $\\frac25$:In the diagram above, we can see that $1\\frac34$ is $\\frac25$ of a container, so a full container is $1\\frac34 \\div \\frac25$. We can see in the diagram that $\\frac12$ of $1\\frac34$ is $\\frac15$ of the container, so our first step is to multiply by $\\frac12$: $$1\\frac34 \\cdot \\frac12$$ Now, just as before, to find the full container, we multiply by 5: $\\left (1\\frac34 \\cdot \\frac12 \\right) \\cdot", "container is being filled with benzene at a constant rate. It took 2 hours to fill 3/8 of its total volume. How much more time will it take to finish filling the container? We can recognize this as a proportion problem since we have two different variables: hours and fraction of container filled, each with two different values. Since we can logically deduce that the more time that passes, the more of our container will be filled, this must be a direct proportion (when one variable increases, so too does the other, and at a constant ratio). A direct proportion can be solved by setting up the following equation: $\\frac{x_{1}}{y_{1}}=\\frac{x_{2}}{y_{2}}$ Here, our first time is 2 hours and our second time is unknown. Our first fraction of the container filled is 3/8 and our second is 5/8: $x_{1}=2$ $y_{1}=\\frac{3}{8}$ $x_{2}=unknown$ $y_{2}=1-\\frac{3}{8}=\\frac{5}{8}$ For clarification, we get that fraction 5/8 since the question asks how much time it would take to finish filling the container. Since we have filled 3/8 already, 5/8 remains. Then we set up the following: $\\frac{2}{\\frac{3}{8}}=\\frac{x_{2}}{\\frac{5}{8}}$ To divide by a fraction, we multiply each of our numerators by the reciprocal of the denominator: $\\frac{16}{3}=\\frac{8y_{1}}{5}$ By cross multiplying, we get: $24y_{1}=80$ $y_{1}=3\\frac{1}{3}$ 1/3 of an hour is 20 minutes, giving us a final answer of 3 hours", "## 2) If there are 4 green jelly beans, 2 white jelly beans, and 5 purple jelly beans in a jar, what is the probability of Question 2) If there are 4 green jelly beans, 2 white jelly beans, and 5 purple jelly beans in a jar, what is the probability of choosing a green jelly bean, eating it, and then choosing a purple jelly bean? in progress 0 9 mins 2023-01-12T23:25:55+00:00 2 Answers 0 views 0 1. Step-by-step explanation: The probability of the first event is $$\\frac{4}{11}$$ The probability of the second one is $$\\frac{5}{10}$$ So $$\\frac{4}{11} \\times \\frac{1}{2} = \\frac{2}{11}$$", "# A red container contains 7 items of which 2 are new and 5 are old items. A purple container contains 4 items of which 1 is new and 3 are old items. An item is drawn at random from each box. 1. What is the probability that both items are old items? ## What is the probability that one item is new and one item old? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 2 Mar 13, 2018 Two old items: $\\frac{15}{28} \\approx 0.5357$ One old item: $\\frac{11}{28} \\approx 0.3929$ #### Explanation: The draw from the red container for an old item has a probability of $\\frac{5}{7}$ of being successful. The draw from the purple container for an old item has a probability of $\\frac{3}{4}$ of being successful. This means that there is a $\\frac{5}{7} \\times \\frac{3}{4} = \\frac{15}{28} \\approx 0.5357$ probability that both items will be old. For finding where one item is old, we can look at the probability of the old item being drawn from the red container and a new item from the purple container $\\left(\\frac{5}{7} \\times \\frac{1}{4} = \\frac{5}{28}\\right)$ and adding that to the probability of the old item being drawn from the", "# A bag contains 12 sweets, of which 5 are red, 4 are green and 3 are yellow. 3 sweets are chosen without replacement. What is the probability that he chooses 1 sweet of each color? Nov 27, 2017 $\\frac{3}{11}$ #### Explanation: There are 2 ways to look at this problem: As a \"standard\" probability question, or as a combinations question. \"Standard\" Probability Consider the possibility that the draw takes place and we draw a red, followed by a green, followed by a yellow. Since there are originally 12 sweets, of which 5 are red. we know the probability of drawing a red is $\\frac{5}{12}$. After we draw the red, there are 4 green out of 11 sweets (since the red is out), and so the probability of drawing the green is $\\frac{4}{11}$. After we draw the green, there are 3 yellow out of 10 sweets, leaving the probability of drawing the yellow as $\\frac{3}{10}$. Using the Multiplication Principle, we can find the probability of the sequence RGY to be: $\\frac{5}{12} \\cdot \\frac{4}{11} \\cdot \\frac{3}{10} = \\frac{60}{1320} = \\frac{1}{22}$. However, that's not the only path towards our desired end. We could also draw the sweets in the order RYG and still be acceptable. In that case, following the same logic, the probability of going RYG is $\\frac{5}{12} \\cdot", "So, for all green: $$P =\\frac{10}{20}*\\frac{9}{19}*\\frac{8}{18}*\\frac{7}{17}*\\frac{6}{16}*\\frac{5}{15} = \\frac{7}{1292}$$ Since you have the same number of red balls, it will be the same probability. Therefore, the total probability is: $$\\frac{7}{1292}*2 = \\frac{14}{1292}$$ Now we simply subtract this from the total probability, $1$: $$1 - \\frac{14}{1292} = \\frac{639}{646}$$ Comment if you have any questions. • How would the problem change if we instead had 3 colors and you wanted at least 1 of each, lets say 10 red, 10 green, and 10 white? – Rovert Renchirk Aug 15 '14 at 4:02 • You would do something similarly where you subtract the probabilities of getting all white all green and all red. – Varun Iyer Aug 15 '14 at 11:00 There are $\\binom{20}{6}$ ways to select the balls, and the number of ways to select them so that you do not get at least one red and at least one green is given by $2\\binom{10}{6}$, since there are $\\binom{10}{6}$ ways to select 6 red balls (or 6 green balls). Therefore the probability is $\\displaystyle1-\\frac{2\\binom{10}{6}}{\\binom{20}{6}}=1-\\frac{420}{38760}=1-\\frac{7}{646}=\\frac{639}{646}$.", "the third one there are three broken light bulbs out of eight. What is the probability that if we choose a box at random and we take a light bulb, this is broken? We will write $$C1$$, $$C2$$, $$C3$$ if we choose the boxes 1, 2, and 3, respectively. Since the election is random, we have probability $$\\dfrac{1}{3}$$ of choosing each box. The event that we are interested in is $$F =$$ \"broken light bulb\", or if we write $$\\overline{F} =$$ \"working bulb\". We represent it in a tree: In dark orange we observe the branches of the tree that lead to the event that we are interested in, i.e. \"broken light bulbs\". $$(C1, C2, C3)$$ is a partition of the sample space, since we will always choose one of three boxes (in other words, its union is the whole), and we cannot choose more than one box (i.e., they are two by two incompatible). We apply, then, the law of total probability: $$\\begin{array}{rl} P(F) =& P(C1)\\cdot P(F/C1) + P(C2)\\cdot P(F/C2) + P(C3)\\cdot P(F/C3) \\\\ =& \\dfrac{1}{3}\\cdot\\dfrac{4}{10}+\\dfrac{1}{3}\\cdot\\dfrac{1}{6}+ \\dfrac{1}{3}\\cdot\\dfrac{3}{8} = \\dfrac{4}{30}+\\dfrac{1}{18}+\\dfrac{3}{24}= \\dfrac{113}{360} \\end{array}$$$ As we can see, to apply the theorem of the total probability is the same as to compute the probability using the tree. The only thing we have to care about is to compute the probabilities", "higher numbered box. Start from the first box. There are $4$ possible results in the box: Red, Green, Red and Green, or none, with an equal probability of $\\frac{1}{4}$ for each. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution, so if $p$ is the probability that Red wins, we can write $p = \\frac{1}{4} + \\frac{1}{4}p$: there is a $\\frac{1}{4}$ probability that \"Red\" wins immediately, a $0$ probability in the cases \"Green\" or \"Red and Green\", and in the \"None\" case (which occurs with $\\frac{1}{4}$ probability), we then start again, giving the same probability $p$. Hence, solving the equation, we get $p = \\boxed{\\textbf{(C) } \\frac{1}{3}}$. Solution 5 Write out the infinite geometric series as $\\frac{1}{2}$, $\\frac{1}{4}, \\frac{1}{8}, \\frac{1}{16}, \\cdots$. To find the probablilty that red goes in a higher-numbered bin than green, we can simply remove all odd-index terms (i.e term $1$, term $3$, etc.), and then sum the remaining terms - this is in fact precisely equivalent to the method of Solution 2. Writing this out as another infinite geometric sequence, we are left with $\\frac{1}{4}, \\frac{1}{16}, \\frac{1}{64}, \\cdots$. Summing, we get $$\\sum_{i=1}^{\\infty} \\frac{1}{4^i} = \\boxed{\\textbf{(C) } \\frac{1}{3}}$$ Solution 6 Probability that the green ball is in a bin below", "the probability halves. Thus, our answer is $\\frac{1}{6} + \\frac{1}{12} + \\frac{1}{24} + \\cdots$, which, by the geometric series sum formula, is $\\boxed{\\textbf{(C) } \\frac{1}{3}}$. -fidgetboss_4000 ## Solution 5 (quick, conceptual) Define a win as a ball appearing in higher numbered box. Start from the first box. There are $4$ possible results in the box: Red, Green, Red and Green, or none, with an equal probability of $\\frac{1}{4}$ for each. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution, so if $p$ is the probability that Red wins, we can write $p = \\frac{1}{4} + \\frac{1}{4}p$: there is a $\\frac{1}{4}$ probability that \"Red\" wins immediately, a $0$ probability in the cases \"Green\" or \"Red and Green\", and in the \"None\" case (which occurs with $\\frac{1}{4}$ probability), we then start again, giving the same probability $p$. Hence, solving the equation, we get $p = \\boxed{\\textbf{(C) } \\frac{1}{3}}$. ## Solution 6 Write out the infinite geometric series as $\\frac{1}{2}$, $\\frac{1}{4}, \\frac{1}{8}, \\frac{1}{16}, \\cdots$. To find the probablilty that red goes in a higher-numbered bin than green, we can simply remove all odd-index terms (i.e term $1$, term $3$, etc.), and then sum the remaining terms - this is in fact precisely equivalent to the method", "is the container it is from; thus $$p\\left(\\left. X_{n+1} = k \\right| X_n = m \\right) = \\left\\lbrace \\begin{array}{11} 1 & m = N \\text{ and } k = N - 1 \\\\ 1 & m = 0 \\text{ and } k = 1 \\\\ \\frac{N - m}{N} & 0 < m < N \\text{ and } k = m + 1 \\\\ \\frac{m}{N} & 0 < m < N \\text{ and } k = m - 1 \\\\ 0 & \\text{otherwise} \\end{array} \\right.$$ This becomes more apparent by noting that $\\frac{m}{N}$ is the probability of choosing a molecule from container A, and $\\frac{N-m}{N}$ is the converse probability of choosing a molecule from container B", "in the 3 flavor box = $$\\frac{5}{6} * \\frac{4}{5} * \\frac{3}{4} = \\frac{1}{2}$$. Therefore the probability of having 2 flavors and no grape = $$\\frac{1}{3} * \\frac{1}{2}= \\frac{1}{6}$$. the probability of box containing 4 flavors = $$\\frac{1}{3}$$(Since there are 3 box containing 2 flavors, 3 flavors or 4 flavors) Probability of no grape in the 4 flavor box = $$\\frac{5}{6} * \\frac{4}{5} * \\frac{3}{4} * \\frac{2}{3} =\\frac{1}{3}$$. Therefore the probability of having 4 flavors and no grape = $$\\frac{1}{3} * \\frac{1}{3} = \\frac{1}{9}$$. Therefore probability of no grape in 2, 3 or 4 flavor box = $$\\frac{2}{9} + \\frac{1}{6} + \\frac{1}{9} = \\frac{1}{2}$$. Therefore the probability of having grape in any given box = $$1 - \\frac{1}{2} = \\frac{1}{2}$$. _________________ If you found this post useful, please let me know by pressing the Kudos Button Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html Director Joined: 09 Nov 2018 Posts: 509 Followers: 0 Kudos [?]: 21 [0], given: 1 Re: A certain candy store sells jellybeans in the following six [#permalink] 18 Nov 2018, 18:03 Supreme Moderator Joined: 01 Nov 2017 Posts: 370 Followers: 5 Kudos [?]: 107 [0], given: 4 Re: A certain candy store sells jellybeans in the following six [#permalink] 18 Nov 2018, 20:02 Expert's post AE wrote: Carcass wrote: A certain candy store sells jellybeans in the", "green, Cube = yellow $|$ Label = $B$ or Cube = blue) = 3/20. Or even more briefly, P(B=g, C=y $|$ L=$B$ or C=b) = 3/20. Similarly, given a particular decision rule, instead of writing The probability of incorrectly predicting red when using that decision rule is 48%. we can write P(B=g, X=r) = 48%. And so on. I use $X$ instead of $P$ to stand for the prediction, since we already use $P$ to stand for probability. Properties of decision rules In Section 2.5, we looked at a property of decision rules, the error probability. Remember: given a crate, a decision rule's error probability is the probability of predicting incorrectly when using it. Let's define some more properties —eight more of them in fact. To keep track of them all, we'll put them in a table: X = r X = g condition on B B=r true red probability, a false green probability, b false green rate, $\\frac{b}{a+b}$ B=g false red probability, c true green probability, d false red rate, $\\frac{c}{c+d}$ condition on X red prediction error, $\\frac{c}{a+c}$ green prediction error, $\\frac{b}{b+d}$ error probability, $b + c$ Things are simpler than they might appear. Fix a crate. A decision rule's true red probability is the probability of correctly predicting red, P(B=r, X=r). Its true green probability is", "broken, and in the third one there are three broken light bulbs out of eight. What is the probability that if we choose a box at random and we take a light bulb, this is broken? We will write $C1$, $C2$, $C3$ if we choose the boxes 1, 2, and 3, respectively. Since the election is random, we have probability $\\dfrac{1}{3}$ of choosing each box. The event that we are interested in is $F =$ \"broken light bulb\", or if we write $\\overline{F} =$ \"working bulb\". We represent it in a tree: In dark orange we observe the branches of the tree that lead to the event that we are interested in, i.e. \"broken light bulbs\". $(C1, C2, C3)$ is a partition of the sample space, since we will always choose one of three boxes (in other words, its union is the whole), and we cannot choose more than one box (i.e., they are two by two incompatible). We apply, then, the law of total probability: $$\\begin{array}{rl} P(F) =& P(C1)\\cdot P(F/C1) + P(C2)\\cdot P(F/C2) + P(C3)\\cdot P(F/C3) \\\\ =& \\dfrac{1}{3}\\cdot\\dfrac{4}{10}+\\dfrac{1}{3}\\cdot\\dfrac{1}{6}+ \\dfrac{1}{3}\\cdot\\dfrac{3}{8} = \\dfrac{4}{30}+\\dfrac{1}{18}+\\dfrac{3}{24}= \\dfrac{113}{360} \\end{array}$$ As we can see, to apply the theorem of the total probability is the same as to compute the probability using the tree. The only thing we have to care about is to", "the probability the first two picks are green, then red, in that order, is $\\frac{5}{15}\\cdot\\frac{7}{14}$. Given that the first two choices were green and red, in that order, the probability the next is blue is $\\frac{3}{13}$. We conclude that the probability of GRB is $\\frac{5}{15}\\cdot\\frac{7}{14}\\cdot \\frac{3}{13}$. There is a total of $3$ colour sequences through which we end up with $1$ of each colour. It turns out that each of them has probability $\\frac{5}{15}\\cdot\\frac{7}{14}\\cdot \\frac{3}{13}$. So our required probability is $$6\\cdot\\frac{5}{15}\\cdot\\frac{7}{14}\\cdot \\frac{3}{13}.$$ Why were these six probabilities all equal? One way to think about it is as follows. Suppose that, blindfolded, we use a scoop to pull out $3$ balls, and these happen to represent all the colours. Now we reach into the scoop and pull out the balls one at a time. Of course all $3!$ colour orders will be equally likely. Another more arithmetical way of seeing it is that the denominators will be, in order, $15$, $15$, and $13$, since we are picking from a diminishing number. Each time we consider a partiular colour, the number of choices will be the number of balls of that colour. So the numbers $7$, $3$, and $5$ must each appear exactly once in the numerators. • Been wanting to know what do you mean by the (7,1) is", "# Project Euler #100: Arranged probability If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, $P(BB) = \\frac {15} {21} \\times \\frac {14} {20} = \\frac 1 2$. The next such arrangement, for which there is exactly 50% chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs. By finding the first arrangement to contain over $10^{12} = 1,000,000,000,000$ discs in total, determine the number of blue discs that the box would contain. This is a fun problem that seems out of reach initially but can be reduced to something more simple. ## Compute the probability🔗 Assuming there are $b$ blue discs and $r$ red discs, the probability of choosing one blue disc at random is $P_1 = \\frac {b} {b+r}$. The probability of choosing another blue disc from the remaining set is $P_2 = \\frac {b-1} {b+r-1}$. So the probability of choosing 2 blue discs is $P(BB) = P_1 \\cdot P_2 = \\frac {b} {b+r} \\cdot \\frac {b-1} {b+r-1}$ We need to find the smallest solution of: $$\\begin{cases} \\frac {b} {b+r} \\cdot \\frac {b-1} {b+r-1} = \\frac 1 2 \\newline b +", "following table represents the number of paint cans in stock for each colour. $\\begin{array}{|c|c|} \\text{Colour} & \\text{\\# in stock} \\\\ \\hline \\text{Red} & 10 \\\\ \\text{Yellow} & 5 \\\\ \\text{Blue} & 20 \\\\ \\text{Orange} & 1 \\\\ \\text{Green} & 3 \\\\ \\text{Purple} & 1\\\\ \\hline \\end{array}$ It is important to note the total number of paint cans in stock is the sum of the second column $10+5+20+1+3+1 = 40$ Now let's find some probabilities 1. Find the probability of picking a can of Yellow paint 2. Find the probability of picking a can of Red paint 3. Find the probability of picking a can of paint which is not Blue Spoiler: Using some letter abbreviations to make life easier 1. $P(\\text{Yellow}) = P(Y)= \\frac{5}{40} = \\frac{1}{8}$ 2. $P(R) = \\frac{10}{40} = \\frac{1}{4}$ 3. In this question we need to add together all cans that are not blue $P(\\text{Not B} ) = \\frac{10+5+1+3+1}{40} = \\frac{20}{40}= \\frac{1}{2}$ Well I hope that was easy enough because it's time to introduce Complementary Probability. Complementary Probability The complement of an event is defined to be everything that is not the event in question. In the last set of questions we were asked Find the probability of picking a can of paint which is not Blue. This is the complement of everything that is", "the green disk was selected first. What if it wasn't? You calculated the probability of getting to that result by a specific route, but this excludes other ways to get to that same result. One way to approach this would be a kind of tree approach--ultimately, the choices, in order, could be GOO or OGO or OOG, and all three of those would have to be calculated and added. P(GOO) = 1/20 (what you calculated) P(OGO) = (4/10)*(3/9)*(3/8) = 1/20 P(OOG) = (4/10)*(3/9)*(3/8) = 1/20 P(total) = 3/20 Another approach would be to apply counting techniques. See: GMAT Probability and Counting Techniques Think of it this way. Suppose the discs are give letters, A - J. Discs A, B, & C are green. Discs D-G are orange, and rest are blue. Question #1: how many different sets of three altogether could we pick? 10C3 = $$\\frac{10!}{(7!)(3!)}$$ = $$\\frac{10*9*8}{3*2*1}$$ = 10*3*4 = 120. That's the denominator. Question #2: how many different sets of three involve just one green and two oranges? Three ways to choose the single green member, and 4C2 = 6 ways to choose the two orange members. By the Fundamental Counting Principle, number of ways = 3*6 = 18. That's the numerator. Probability = 18/120 = 3/20 Here are a couple more blogs you may find", "and Boxes Example Consider example where someone picks a box, picks a ball from that box, then picks another ball from the same box without replacing the first ball. What is the probability of pulling out a green ball followed by a blue ball? $P(2=blue|1=green)$ The Baye’s net for this problem is. Ball 1 Probabilities Box Green Yellow Blue 1 34 14 0 2 25 0 35 Ball 2 Probabilities, assuming ball 1 was green Box Green Yellow Blue 1 23 13 0 2 14 0 34 The formula below follows from the marginalization rule and the chain rule: $$P(2=blue|1=green)=P(2=blue|1=green,box=1)P(box=1|1=green)$$ $$+P(2=blue|1=green,box=2)P(box=2|1=green)$$ The formulae below follows from Baye’s rule: $$P(box=1|1=green)=P(1=green|box=1)\\frac{Pr(box=1)}{Pr(1=green)}$$ $$P(box=2|1=green)=P(1=green|box=2)\\frac{Pr(box=2)}{Pr(1=green)}$$ The following comes from averaging the likelihood of the first draw being a green, given that the likelihood of choosing either box is $\\frac{1}{2}$: $$P(1=green)=\\frac{3}{4}*\\frac{1}{2}+\\frac{2}{5}*\\frac{1}{2}=\\frac{23}{40}$$ $$P(box=1|1=green)=\\frac{3}{4} \\frac{\\frac{1}{2}}{\\frac{23}{40}}=\\frac{15}{23}$$ $$P(box=2|1=green)=\\frac{2}{5} \\frac{\\frac{1}{2}}{\\frac{23}{40}}=\\frac{8}{23}$$ Plugging back in to the first equation: $$P(2=blue|1=green)=(0)(\\frac{15}{23})+(\\frac{3}{4})(\\frac{8}{23})=\\frac{6}{23}$$ ### Naive Bayes - Algorithmic Approach Naive Bayes is “naive” because it assumes that attributes are independent of one another, conditional on the label. If a message is a spam email, it is likely to contain certain words. Which True False $P(spam)$ .4 .6 $P(“Viagra”|spam)$ .3 .001 $P(“Price”|spam)$ .2 .1 $P(“Udacity”|spam)$ .0001 .1 $$P(spam|“Viagra”,\\ not\\ “Prince”,\\ not\\ “Udacity”)=$$ $$=\\frac{P(“Viagra”,\\ not\\ “Prince”,\\ not\\ “Udacity”)|spam}{…}$$ $$=\\frac{P(“Viagra”|spam)P(not\\ “Prince”)|spam)P(not\\ “Udacity”|spam)P(spam)}{…}$$ $$=\\frac{(.3)(.8)(.9999)(.4)}{…}$$ ### Naive Bayes -", "# Total probability, Bayes There are 2 identical containers. The first one has 5 white and 8 black balls, the second one has 7 white and 9 black balls. 3 balls are randomly taken out of the first container and 4 balls are randomly taken from the second container. All of the balls (7) are put in a third container. If a ball is taken out of the third container, what is the probability that it will be white? ## 1 Answer The event \"ball taken out at the end is white\" can happen in two disjoint ways: (i) The ball came originally from the first container, and is white or (ii) The ball came originally from the second container, and is white. The probability that the chosen ball came from the first container is $\\frac{3}{7}$. Given that it came from the first container, the probability that it is white is $\\frac{5}{13}$. So the probability it came from the first container and is white is $\\frac{3}{7}\\cdot \\frac{5}{13}$. Continue." ]

[ "# One container contains 6 red and 4 white balls, while a second container contains 7 red and 3 white balls One container contains 6 red and 4 white balls, while a second container contains 7 red and 3 white balls. A ball is chosen at random from the first container and placed in the second. Then a ball is chosen at random from the second container and placed in first. a. What is the probability to choose a red ball of the second container? b. What is the probability the ball was red chosen from the first container if know that from the second selected white? - Solution for question 1: Probability that a red ball will be chosen on first step: $\\frac{6}{10}$ In this case, we will have 8 red and 3 white balls in the second container, and the probability to chose a red ball on second step is $\\frac{8}{11}$. Overall probability in this case gives $\\frac{6}{10}*\\frac{8}{11}$ = $\\frac{48}{110}$. Let's mark it as p$_1$. Probability that a white ball will be chosen on first step: $\\frac{4}{10}$ Similarly to the previous case, we will get probability to choose a red ball from the second container is p$_2$=$\\frac{4}{10}*\\frac{7}{11}=\\frac{28}{110}$ Because our cases do not intersect, we have total probability p=p$_1$+p$_2$=$\\frac{76}{110}$=$\\frac{38}{55}$. Solutin for question 2: Here we shall use", "# A red container contains 7 items of which 2 are new and 5 are old items. A purple container contains 4 items of which 1 is new and 3 are old items. An item is drawn at random from each box. 1. What is the probability that both items are old items? ## What is the probability that one item is new and one item old? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 2 Mar 13, 2018 Two old items: $\\frac{15}{28} \\approx 0.5357$ One old item: $\\frac{11}{28} \\approx 0.3929$ #### Explanation: The draw from the red container for an old item has a probability of $\\frac{5}{7}$ of being successful. The draw from the purple container for an old item has a probability of $\\frac{3}{4}$ of being successful. This means that there is a $\\frac{5}{7} \\times \\frac{3}{4} = \\frac{15}{28} \\approx 0.5357$ probability that both items will be old. For finding where one item is old, we can look at the probability of the old item being drawn from the red container and a new item from the purple container $\\left(\\frac{5}{7} \\times \\frac{1}{4} = \\frac{5}{28}\\right)$ and adding that to the probability of the old item being drawn from the", "Charlotte choosing red \\text{AND} Joe choosing red =\\frac{2}{3} \\times \\frac{1}{3}=\\frac{2}{9}. Probability of Charlotte choosing green \\text{AND} Joe choosing green = \\frac{1}{3} \\times \\frac{1}{3}=\\frac{1}{9}. Probability of both choosing red OR both choosing green =\\frac{2}{9}+\\frac{1}{9}=\\frac{3}{9}=\\frac{1}{3}. Option 1: using the table. There are 9 combined outcomes for these events. There are six combinations where Charlotte and Joe choose a different colour (RG, RG, RY, RY, GR, GY). Hence the answer is six out of nine. \\frac{6}{9}=\\frac{2}{3} Option 2: using probabilities summing to one. From part (d), the probability that they both choose the same colour is \\frac{1}{3}. Using the property that the probabilities of an exhaustive set of mutually exclusive events sum to one, the answer can be found by subtracting the probability that they choose the same colour from 1. 1-\\frac{1}{3}=\\frac{2}{3}. Note: As shown, all answers to example 3 can be calculated using the \\text{AND} probability rule. However, if you have drawn a sample space diagram it is often quicker to use the information from the table. ### Example 4: combined events and permutations Abdul tosses 3 fair coins. What is the probability that he will get exactly one head? There are three events – the toss of each fair coin. The outcome of each event can be heads or tails. Probability of tails = 0.5. The probability of getting", "2 of the disks selected are green? A. 21/55 B. 28/55 C. 34/55 D. 5/8 E. 139/220 _________________ Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score ! Math Expert Joined: 02 Sep 2009 Posts: 50016 The only contents of a container are 4 blue disks and 8 gree [#permalink] ### Show Tags 07 Jan 2014, 04:51 1 3 sanjoo wrote: The only contents of a container are 4 blue disks and 8 green disks. If 3 disks are selected one after the other, and at random and without replacement from the container, what is the probability that 1 of the disks selected is blue, and 2 of the disks selected are green? A. 21/55 B. 28/55 C. 34/55 D. 5/8 E. 139/220 APPROACH #1: $$P(BGG) = \\frac{C^1_4*C^2_8}{C^3_{12}}=\\frac{4*28}{220}=\\frac{28}{55}$$. APPROACH #2: $$P(BGG) =\\frac{3!}{2!}*\\frac{4}{12}*\\frac{8}{11}*\\frac{7}{10}=\\frac{28}{55}$$. We need to multiply by 3!/2! because BGG scenario can occur in several ways: BGG, GBG, GGB (permutations of 3 letters BGG). Hope it's clear. _________________ Intern Joined: 17 Oct 2013 Posts: 41 Schools: HEC Dec\"18 GMAT Date: 02-04-2014 Re: The only contents of a container are 4 blue disks and 8 gree [#permalink] ### Show Tags 07 Jan 2014, 10:30 1 sanjoo wrote: The only contents of a container are 4 blue disks and 8 green disks. If 3 disks", "# An Amazon interview question (not completely solved) Posted on January 24, 2011 by phimuemue Taken from this site: Given red balls and blue balls and some containers, how would you distribute those balls among the containers such that the probability of picking a red ball is maximized, assuming that the user randomly chooses a container and then randomly picks a ball from that? ## Solution (probably) The solution to this problem works as follows: Take all your red balls and put them one by one in one container after another until you have no balls left. If there are containers left (i.e. without red balls), take your blue balls and distribute them as you like into the remaining containers. If there are no containers left, take all your blue balls and put them in one arbitrary container. To prove this (at least one case), we first do some definitions: W.l.o.g I assume , i.e. all containers can be filled with at least one ball. Let's call the number of containers . denotes the event that container is choosen. denotes the event that a red ball is chosen from the -th container. The probability of choosing the -th container is (since we have containers in total) . Let denote the number of red balls in container and denote", "# A bag contains 12 sweets, of which 5 are red, 4 are green and 3 are yellow. 3 sweets are chosen without replacement. What is the probability that he chooses 1 sweet of each color? Nov 27, 2017 $\\frac{3}{11}$ #### Explanation: There are 2 ways to look at this problem: As a \"standard\" probability question, or as a combinations question. \"Standard\" Probability Consider the possibility that the draw takes place and we draw a red, followed by a green, followed by a yellow. Since there are originally 12 sweets, of which 5 are red. we know the probability of drawing a red is $\\frac{5}{12}$. After we draw the red, there are 4 green out of 11 sweets (since the red is out), and so the probability of drawing the green is $\\frac{4}{11}$. After we draw the green, there are 3 yellow out of 10 sweets, leaving the probability of drawing the yellow as $\\frac{3}{10}$. Using the Multiplication Principle, we can find the probability of the sequence RGY to be: $\\frac{5}{12} \\cdot \\frac{4}{11} \\cdot \\frac{3}{10} = \\frac{60}{1320} = \\frac{1}{22}$. However, that's not the only path towards our desired end. We could also draw the sweets in the order RYG and still be acceptable. In that case, following the same logic, the probability of going RYG is $\\frac{5}{12} \\cdot", "It's a bit more annoying with 2 and 3 balls. Now, I have to calculate the odds of every combination of two reds and three greens. However, I think the math ends up being the same. The denominator is always $11\\cdot10\\cdot9\\cdot8\\cdot7$ or $\\frac{11!}{6!}$ $=55440$. The numerator for the first red is 7, the second is 6, the third is 5. The numerator for green is 4, 3, 2. So $P(RRGGG)$ $=P($any specific permutation of 2 red, 3 green$)$ $=\\frac{7\\cdot6\\cdot4\\cdot3\\cdot2}{11!\\div6!}$. Then, there are many permutations of 2R3G. 4 ways the first red is the first draw, 3 ways it's the second draw, 2 ways it's the third draw, and 1 way it's the fourth draw. $4+3+2+1$ $=10$ total permutations. So $P(\\text{2 red})$ should be $10\\cdot\\frac{7\\cdot6\\cdot4\\cdot3\\cdot2}{11!\\div6!}$ $=\\frac{2}{11}$. There should be $10$ permutations of 3R2G (same math, except imagine ways to get 2G instead of 2R), which means $P(\\text{3 red})$ $=10\\cdot\\frac{7\\cdot6\\cdot5\\cdot4\\cdot3}{11!\\div6!}$ $=\\frac{5}{11}$. Recap: $P(0\\text{ red})$ $=0$, $P(1 \\text{ red})$ $=\\frac{1}{66}$, $P(2 \\text{ red})$ $=\\frac{2}{11}$, $P(3 \\text{ red})$ $=\\frac{5}{11}$, $P(4 \\text{ red})$ $=\\frac{10}{33}$, $P(5 \\text{ red})$ $=\\frac{1}{22}$. $P(\\text{total})$ $=0$ $+\\frac{1}{66}$ $+\\frac{2}{11}$ $+\\frac{5}{11}$ $+\\frac{10}{33}$ $+\\frac{1}{22}$ $=1$, which is good. Converting from Specific to General From this point, I can hopefully find a pattern of how to solve generic questions of this nature. Switching away from the nomenclature given in the title, I have", "of each of these, and they are mutually exclusive, so you can add them. @Supa: that would make the probability greater than 1, which cannot be true. The chance to get $3$ green is $\\frac 58 \\cdot \\frac 47 \\cdot \\frac 36=\\frac {60}{336}$. Each fraction is the chance to get a green marble with the bag holding what is left at that point. – Ross Millikan Jan 7 '13 at 21:03", "colored balls in a box, including 1 Green [#permalink] ### Show Tags 02 Jan 2014, 04:59 1 KUDOS Expert's post theGame001 wrote: Bunuel wrote: plock3vr wrote: Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different? The probability of each case will be the same: {GXX} - $$P=\\frac{1}{10}*\\frac{8}{9}*\\frac{7}{8}=\\frac{7}{90}$$; {XGX} - $$P=\\frac{8}{10}*\\frac{1}{9}*\\frac{7}{8}=\\frac{7}{90}$$; {XXG} - $$P=\\frac{8}{10}*\\frac{7}{9}*\\frac{1}{8}=\\frac{7}{90}$$; The sum: $$\\frac{7}{90}+\\frac{7}{90}+\\frac{7}{90}=\\frac{7}{30}$$. Hope it helps. In the second case {XGX} - $$P=\\frac{8}{10}*\\frac{1}{9}*\\frac{7}{8}=\\frac{7}{90}$$, how did we get 8/10? If we are considering any ball other than green, why is it not 9/10? Because we don't need the yellow ball, X is any but green and yellow. _________________ Manager Joined: 14 Jan 2013 Posts: 154 Concentration: Strategy, Technology GMAT Date: 08-01-2013 GPA: 3.7 WE: Consulting (Consulting) Followers: 3 Kudos [?]: 252 [1] , given: 30 Re: There are 10 solid colored balls in a box,including 1 Green [#permalink] ### Show Tags 14 Jan 2014, 15:54 1 KUDOS Bunuel wrote: imhimanshu wrote: There are 10 solid colored balls in a box,including 1 Green and 1 Yellow. If 3 of the balls in the box", "# Total probability, Bayes There are 2 identical containers. The first one has 5 white and 8 black balls, the second one has 7 white and 9 black balls. 3 balls are randomly taken out of the first container and 4 balls are randomly taken from the second container. All of the balls (7) are put in a third container. If a ball is taken out of the third container, what is the probability that it will be white? ## 1 Answer The event \"ball taken out at the end is white\" can happen in two disjoint ways: (i) The ball came originally from the first container, and is white or (ii) The ball came originally from the second container, and is white. The probability that the chosen ball came from the first container is $\\frac{3}{7}$. Given that it came from the first container, the probability that it is white is $\\frac{5}{13}$. So the probability it came from the first container and is white is $\\frac{3}{7}\\cdot \\frac{5}{13}$. Continue.", "the rule above, you can see that the probability is $Probability\\;=\\;\\frac{Number\\;of\\;days\\;beginning\\;with\\;the\\;letter\\;S}{Total\\;number\\;of\\;days\\;of\\;a\\;week}=\\frac{2}{7}$ The probability that a day chosen at random will begin with the letter S is $\\frac{2}{7}$ A spinner has 4 equal sectors colored yellow, blue, green and red. After spinning the spinner, what is the probability of landing on each color? The possible outcomes of this experiment are yellow, blue, green, and red. $P(yellow)\\;=\\;\\frac{Number\\;of\\;ways\\;to\\;land\\;on\\;yellow}{Total\\;number\\;of\\;colors}=\\frac{1}{4}$ $P(blue)\\;=\\;\\frac{Number\\;of\\;ways\\;to\\;land\\;on\\;blue}{Total\\;number\\;of\\;colors}=\\frac{1}{4}$ $P(green)\\;=\\;\\frac{Number\\;of\\;ways\\;to\\;land\\;on\\;green}{Total\\;number\\;of\\;colors}=\\frac{1}{4}$ $P(red)\\;=\\;\\frac{Number\\;of\\;ways\\;to\\;land\\;on\\;red}{Total\\;number\\;of\\;colors}=\\frac{1}{4}$", "# how to distribute n red and m blue balls in some containers to maximize probability of random picking a red one from them? This is an interview question. Given n red balls and m blue balls and some containers, how would you distribute those balls among the containers such that the probability of picking a red ball is maximized, assuming that the user randomly chooses a container and then randomly picks a ball from that. My solution: suppose we have c containers, distribute n/c read balls to each c. If c == 1, put all of them together, it is n/(m+n) If c == 2, put 1 red in c1 and all left red and all blue ones in c2 in this way , we have: 1/2 + 1/2 *(n-1) /(m+n-1) > 1/2 If c == 3, put a red in c1, put a red in c2, put left red and all blue in c3, we have: 1/3 + 1/3 + 1/3 * (n-2)/(m+n-2) > 2/3 If c == n, put a red in each of p -1, and all left red and blue in pth container, we have: ( Sum of (1/n) from 1 to n-1 ) + ( 1/n * 1/(m+1) ) (n-1)/n + 1/n * 1/(m+1) == 1 (almost) As n is large, the", "# how to distribute n red and m blue balls in some containers to maximize probability of random picking a red one from them? This is an interview question. Given n red balls and m blue balls and some containers, how would you distribute those balls among the containers such that the probability of picking a red ball is maximized, assuming that the user randomly chooses a container and then randomly picks a ball from that. My solution: suppose we have c containers, distribute n/c read balls to each c. If c == 1, put all of them together, it is n/(m+n) If c == 2, put 1 red in c1 and all left red and all blue ones in c2 in this way , we have: 1/2 + 1/2 *(n-1) /(m+n-1) > 1/2 If c == 3, put a red in c1, put a red in c2, put left red and all blue in c3, we have: 1/3 + 1/3 + 1/3 * (n-2)/(m+n-2) > 2/3 If c == n, put a red in each of p -1, and all left red and blue in pth container, we have: ( Sum of (1/n) from 1 to n-1 ) + ( 1/n * 1/(m+1) ) (n-1)/n + 1/n * 1/(m+1) == 1 (almost) As n is large, the", "given that the outcome is odd is: \\begin{align*} \\mathbb{P}(A\\vert{B})&= \\frac{\\mathbb{P}(A,B)}{\\mathbb{P}(B)}\\\\ &=\\frac{\\mathbb{P}(A)}{\\mathbb{P}(B)}\\\\ &=\\frac{1/6}{3/6}\\\\ &=\\frac{1}{3} \\end{align*} Here, $\\mathbb{P}(A,B)=\\mathbb{P}(A)$ in this case because the probability that the outcome is a $3$ and an odd number is equal to the probability that the outcome is $3$. Example. Defective products and complaints The probability that a product is defective is $0.5$. The probability that a product is defective and a customer complains is $0.2$. What is the probability that a customer complains given that the product is defective? Solution. We first define the following events: • event $A$ - customer complains. • event $B$ - product is defective. The probability that a customer complains given that the product is defective is: $$\\mathbb{P}(A|B) =\\frac{\\mathbb{P}(A,B)}{\\mathbb{P}(B)} =\\frac{0.2}{0.5} =0.4$$ Definition. Conditional probability of 3 events If $A$, $B$ and $C$ are events, then the conditional probability of $C$ given $A$ and $B$ is: $$P(C|A,B)= \\frac{\\mathbb{P}(A,B,C)} {\\mathbb{P}(A)\\cdot\\mathbb{P}(B\\vert{A})}$$ Intuition. Suppose we have a bag containing $7$ balls, of which $3$ are green and $4$ are red. We draw $3$ balls from this bag successively at random without replacement. Let's define the following events: • $\\color{green}G_1$, $\\color{green}G_2$ and $\\color{green}G_3$ are the events of selecting a green ball on the 1st, 2nd and 3rd draw respectively. • $\\color{red}R_1$, $\\color{red}R_2$ and $\\color{red}R_3$ are the events of selecting a red ball on the", "be the same? I always write them out to see whether they are the same or not. Thanks. _________________ If my post was helpful, press Kudos. If not, then just press Kudos !!! Math Expert Joined: 02 Sep 2009 Posts: 49303 Re: There are 10 solid colored balls in a box, including 1 Green and 1 [#permalink] ### Show Tags 12 May 2015, 03:44 Ergenekon wrote: Bunuel wrote: plock3vr wrote: Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different? The probability of each case will be the same: {GXX} - $$P=\\frac{1}{10}*\\frac{8}{9}*\\frac{7}{8}=\\frac{7}{90}$$; {XGX} - $$P=\\frac{8}{10}*\\frac{1}{9}*\\frac{7}{8}=\\frac{7}{90}$$; {XXG} - $$P=\\frac{8}{10}*\\frac{7}{9}*\\frac{1}{8}=\\frac{7}{90}$$; The sum: $$\\frac{7}{90}+\\frac{7}{90}+\\frac{7}{90}=\\frac{7}{30}$$. Hope it helps. Bunuel, is there any way to know beforehand that all scenarios will be the same? I always write them out to see whether they are the same or not. Thanks. 1 G and 2 X's has the same probability no matter the order. _________________ Director Joined: 04 Dec 2015 Posts: 701 Location: India Concentration: Technology, Strategy Schools: ISB '19, IIMA , IIMB, XLRI WE: Information Technology (Consulting) Re: There are 10 solid", "# Balls and Boxes – Combinatorics This is studying for my final, not homework. Six balls are dropped at random into ten boxes. What is the probability that no box will contain two or more balls? So I know that to select 6 balls from ten we do: ${10\\choose 6}$, but I'm not sure how to check whether or not each box contains less than two balls. Is it just: ${6\\choose 1}\\over{10\\choose 6}$ - There are $10^6$ different ways in which the six balls could land in the ten boxes. Now we'll count the number of ways that have each ball in a different box. If each ball is in a different box, they must occupy $6$ of the $10$ boxes. There are $\\binom{10}6$ different sets of six boxes that they could occupy, and they can be permuted in $6!$ ways amongst those boxes, so there are altogether $$\\binom{10}6\\cdot 6!=\\frac{10!}{4!}$$ ways in which they can all fall into different boxes. The desired probability is therefore $$\\frac{\\frac{10!}{4!}}{10^6}=\\frac{10!}{4!\\cdot10^6}=\\frac{9!}{4!\\cdot10^5}=0.1512\\;.$$ Drop a ball into a box. Now drop the next ball. There are $9$ empty boxes, so the probability of no collision is $\\frac{9}{10}$. Then drop the next one. The probability of no collision is $\\frac{8}{10}$. And so on. So the overall probability of no collision is $$\\frac{9}{10}\\cdot\\frac{8}{10}\\cdot\\frac{7}{10}\\cdot\\frac{6}{10}\\cdot\\frac{5}{10}.$$", "it helps. _________________ Manager Joined: 13 Jul 2013 Posts: 68 GMAT 1: 570 Q46 V24 Re: There are 10 solid colored balls in a box, including 1 Green and 1 [#permalink] ### Show Tags 02 Jan 2014, 01:59 Bunuel wrote: plock3vr wrote: Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different? The probability of each case will be the same: {GXX} - $$P=\\frac{1}{10}*\\frac{8}{9}*\\frac{7}{8}=\\frac{7}{90}$$; {XGX} - $$P=\\frac{8}{10}*\\frac{1}{9}*\\frac{7}{8}=\\frac{7}{90}$$; {XXG} - $$P=\\frac{8}{10}*\\frac{7}{9}*\\frac{1}{8}=\\frac{7}{90}$$; The sum: $$\\frac{7}{90}+\\frac{7}{90}+\\frac{7}{90}=\\frac{7}{30}$$. Hope it helps. In the second case {XGX} - $$P=\\frac{8}{10}*\\frac{1}{9}*\\frac{7}{8}=\\frac{7}{90}$$, how did we get 8/10? If we are considering any ball other than green, why is it not 9/10? Math Expert Joined: 02 Sep 2009 Posts: 49303 Re: There are 10 solid colored balls in a box, including 1 Green and 1 [#permalink] ### Show Tags 02 Jan 2014, 05:59 1 theGame001 wrote: Bunuel wrote: plock3vr wrote: Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do", "(1st green * 2nd green * 3rd red) but it is not necessary 3rd is red. May be 1st or 2nd ball is red. So we have 3 cases. We can write as $\\binom{3}{2}$ or C(3,2) for number of possible cases i.e $\\frac{3!}{2!*1*!} = 3$ – Kanwaljit Singh Dec 29 '16 at 5:47 • If still any doubt you can ask. – Kanwaljit Singh Dec 29 '16 at 5:48 The probability that the first ball selected is green is $P_1 = \\frac {3}{9}$. After selecting a green balls, there are now $2$ green and a total of $8$ balls. Now the probability of selecting a green balls is $P_2 = \\frac {2}{8}$. Now there are a total of $7$ balls and a green ball. Now the probability is $P_3=\\frac {1}{7}$. Thus, $$P_{\\text {req}} = P_1 \\times P_2 \\times P_3 = \\frac {3}{9} \\times \\frac {2}{8} \\times \\frac {1}{7} = \\frac {1}{84}$$ which is the same as that got using the hypergeometric distribution formula. Hope it helps. • but if the question is: If three balls are randomly chosen without replacement, what is the probability that two balls are green? Will it be: $(3/9)\\times\\{(2/8)+(3/8)\\}$? – user81411 Dec 29 '16 at 5:11 $$P = \\frac{3}{9} \\cdot \\frac{2}{8}\\cdot \\frac{1}{7} = \\frac{1}{3} \\cdot \\frac{1}{4} \\cdot \\frac{1}{7} = \\frac{1}{84}$$", "you just need to count the number of outcomes that belong to the event you're considering, and divide by the total number of outcomes. • Thank you for the explanation, as I started writing it down I came to the same conclusion but it's very helpful to see the problem from a different perspective! – Vasya Apr 18 at 12:25 You can comprehend the calculation in a simpler way with smaller numbers. Daniel randomly chooses balls from the group of $$3$$ red and $$2$$ green. What is the probability that he picks $$2$$ red and $$2$$ green if balls are drawn without replacement. Indeed we have to regard the order. There are $$\\frac{4!}{2!\\cdot 2!}=6$$ ways to draw 2 red and 2 green balls: $$\\color{green}g\\color{green}g\\color{red}r\\color{red}r, \\color{green}g\\color{red}r\\color{green}g\\color{red}r, \\color{green}g\\color{red}r\\color{red}r\\color{green}g, \\color{red}r\\color{green}g\\color{green}g\\color{red}r, \\color{red}r\\color{green}g\\color{red}r\\color{green}g, \\color{red}r\\color{red}r\\color{green}g\\color{green}g$$ Each way has the same probability: $$\\frac{3}{5}\\cdot \\frac{2}{4}\\cdot \\frac{2}{3}\\cdot \\frac{1}{2} \\quad (ggrr)$$ Multiplying with 6 (ways) we get $$6\\cdot \\frac{3}{5}\\cdot \\frac{2}{4}\\cdot \\frac{2}{3}\\cdot \\frac{1}{2}=\\frac{3}5=0.6$$ Using binomial coefficients we get $$\\frac{\\binom{3}{2}\\cdot \\binom{2}{2}}{\\binom{5}{4}}=\\frac{3\\cdot 1}{5}=\\frac35=0.6$$ And we get the same result. There is a principle called \"conservation of expected evidence\" that says that if you have events A and B, then when you calculate the probability of A without knowing whether B happens, the result should be the same as the expected value of the probability over the possible results of B. In", "+1 vote 70 views A box contains six red balls and four green balls. Four balls are selected at random from the box. What is the probability that two of the selected balls are red and two are green ? 1. $\\large\\frac{3}{7}$ 2. $\\large\\frac{4}{7}$ 3. $\\large\\frac{5}{7}$ 4. $\\large\\frac{6}{7}$ recategorized | 70 views +1 vote $4$ balls can be chosen from $10$ balls in $^{10}C_4$ ways. (Total number of ways) The desirable(favourable) case is choosing $2$ red balls from $6$ red balls and choosing $2$ green balls from $4$ green balls. So the required probability would be: $\\frac{^{6}C_2\\times ^{4}C_2}{^{10}C_4}$ $=$ $\\frac{3}{7}$ Option A is correct. by Active" ]