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QpyXMtfVkQTUnuxoZ3qaz | maths | permutations-and-combinations | conditional-permutations | The number of four-digit numbers strictly greater
than 4321 that can be formed using the digits
0,1,2,3,4,5 (repetition of digits is allowed) is : | [{"identifier": "A", "content": "306"}, {"identifier": "B", "content": "288"}, {"identifier": "C", "content": "310"}, {"identifier": "D", "content": "360"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263995/exam_images/sztfmutoowxk1xy02qdm.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265675/exam_images/gagx61scdkiyqg7gijyq.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263477/exam_images/gvf2kbeefhosl9fgwiul.webp"><source media="(max-width: 860px)" srcset="https://imagex.cdn.examgoal.net/7kejek12cwvv5/e1820e51-959b-40f7-8768-b52f93a03f4e/1d0d6c10-e146-11e9-a7bc-b7b08ce8d541/file-7kejek12cwvv6-860w.jpg"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265433/exam_images/cfevw9axkaw11ccg7uvl.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Evening Slot Mathematics - Permutations and Combinations Question 137 English Explanation 1"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264654/exam_images/orf01lahtof8wxakxvjo.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263467/exam_images/l0dsxmxugqsm3anof7fa.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267400/exam_images/glgmv1vg27yh38ibnblh.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266543/exam_images/pw4odyrxb3tvv94tn2xj.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266444/exam_images/lzgxhaamyintp77qz02x.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Evening Slot Mathematics - Permutations and Combinations Question 137 English Explanation 2"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267593/exam_images/fnflxpimvmqdlxogp24z.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263741/exam_images/eqgwouh5gfvmon3daivs.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264382/exam_images/zrhxrybzplvc1xdx6omn.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263630/exam_images/cght9rk6mog200xbpc4u.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265599/exam_images/a06mbwqkrmeayldp4uew.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Evening Slot Mathematics - Permutations and Combinations Question 137 English Explanation 3"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267639/exam_images/cnmvue2e458erlfgpqr3.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264214/exam_images/jhk6vl727jcczej84wzd.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263652/exam_images/brrw9edcixvrhnbkxbam.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263691/exam_images/nzysrlihszm4mrggd4db.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263667/exam_images/au1aj9g5bmhlq9ar6avi.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Evening Slot Mathematics - Permutations and Combinations Question 137 English Explanation 4"></picture>
<br>So total 4 digit numbers possibe = 216 + 72 + 18 + 4 = 310 | mcq | jee-main-2019-online-8th-april-evening-slot |
T7pubLMDX4L70jGUBQ3rsa0w2w9jwy0oxdn | maths | permutations-and-combinations | conditional-permutations | The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by
11 and no digit is repeated is : | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "48"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267428/exam_images/uaonbq0dqm3qxf5khxpq.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265874/exam_images/qiafu8cvzerixiwxgsev.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265081/exam_images/qmstaeian1un25etchun.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265213/exam_images/dktnxgjwsqoz38e5qkf4.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267165/exam_images/zraheupjhbidvxeinfg5.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267579/exam_images/zbetk9ejsp72v7nlkgcq.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264276/exam_images/alcfixgbowl2nahuzsan.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263587/exam_images/gylahnddae97tru1yytr.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Morning Slot Mathematics - Permutations and Combinations Question 135 English Explanation"></picture><br>
digit 0, 1, 2, 5, 7, 9<br><br>
$$\left( {{a_1} + {a_3} + {a_5}} \right) - \left( {{a_2} + {a_4} + {a_6}} \right) = 11\,K$$<br><br>
so (1, 2, 9) (0, 5, 7)<br><br>
Now number of ways to arranging them<br><br>
= 3! × 3! + 3! × 2 × 2<br><br>
= 6 × 6 + 6 × 4<br><br>
= 6 × 10<br><br>
= 60 | mcq | jee-main-2019-online-10th-april-morning-slot |
E0UshBABGmQdzdNVQPjgy2xukg4n2d68 | maths | permutations-and-combinations | conditional-permutations | The number of words (with or without meaning)
that can be formed from all the letters of the
word “LETTER” in which vowels never come
together is ________ . | [] | null | 120 | Consonants $$ \to $$ LTTR
<br>Vowels $$ \to $$ EE
<br><br>Total No of words = $${{6!} \over {2!2!}}$$ = 180
<br><br>Total no of words if vowels are together
<br>= $${{5!} \over {2!}}$$ = 60
<br><br>$$ \therefore $$ Total no of words where<br> vowels never come together = 180 – 60 = 120. | integer | jee-main-2020-online-6th-september-evening-slot |
peQrKRXgQeisz7wOkTjgy2xukfuv9z75 | maths | permutations-and-combinations | conditional-permutations | Two families with three members each and one family with four members are to be seated in a row.
In how many ways can they be seated so that the same family members are not separated? | [{"identifier": "A", "content": "2! 3! 4!"}, {"identifier": "B", "content": "(3!)<sup>3</sup>.(4!) "}, {"identifier": "C", "content": "3! (4!)<sup>3</sup>"}, {"identifier": "D", "content": "(3!)<sup>2</sup>.(4!)"}] | ["B"] | null | F<sub>1</sub> $$ \to $$ 3 members
<br>F<sub>2</sub> $$ \to $$ 3 members
<br>F<sub>3</sub> $$ \to $$ 4 members
<br><br>Total arrangements of three families = 3!
<br><br>Arrangement between members of F<sub>1</sub> family = 3!
<br><br>Arrangement between members of F<sub>2</sub> family = 3!
<br><br>Arrangement between members of F<sub>3</sub> family = 4!
<br><br>$$ \therefore $$ Total numbers of ways can they be seated so that the same family members are not separated
<br><br>= 3! $$ \times $$ 3! $$ \times $$ 3! $$ \times $$ 4!
<br><br>= (3!)<sup>3</sup>.(4!) | mcq | jee-main-2020-online-6th-september-morning-slot |
C8HGleU1NgkvV0IfgIjgy2xukfjjwd85 | maths | permutations-and-combinations | conditional-permutations | The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the letters of the word ’SYLLABUS’ such that two letters are distinct and two letters are alike, is :
| [] | null | 240 | In 'SYLLABUS' word
<br><br>1. Two S letters
<br><br>2. Two L letters
<br><br>3. One Y letter
<br><br>4. One A letter
<br><br>5. One B letter
<br><br>6. One U letter
<br><br>Number of ways we can select two alike
letters = <sup>2</sup>C<sub>1</sub>
<br><br>Then number of ways we can select two distinct
letters = <sup>5</sup>C<sub>2</sub>
<br><br>Then total arrangement of
selected letters = $${{4!} \over {2!}}$$
<br><br>So total number of words, with or without meaning, that can be formed
<br><br>= <sup>2</sup>C<sub>1</sub> $$ \times $$ <sup>5</sup>C<sub>2</sub> $$ \times $$ $${{4!} \over {2!}}$$ = 240 | integer | jee-main-2020-online-5th-september-morning-slot |
FnMRlmZon3hFFMIfYv7k9k2k5ior0st | maths | permutations-and-combinations | conditional-permutations | If the number of five digit numbers with distinct
digits and 2 at the 10<sup>th</sup> place is 336 k, then k
is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "7"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265718/exam_images/p2ps2ybooqoc7gyxtqdc.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 9th January Morning Slot Mathematics - Permutations and Combinations Question 126 English Explanation">
<br><br>Except 1 and 2 there are eight options on first place and only one option at fourth place possible which is digit '2'.
<br><br>Remaining three places can be filled by any three digits out of 1, 3, 4, 5, 6, 7, 8 and 9.
<br><br>$$ \therefore $$ No. of five digits numbers = 8 $$ \times $$ 8 $$ \times $$ 7 $$ \times $$ 1 $$ \times $$ 6 = 336 $$ \times $$ 8
<br><br>Also 336k = 336 $$ \times $$ 8
<br><br>$$ \Rightarrow $$ k = 8 | mcq | jee-main-2020-online-9th-january-morning-slot |
IGIb0SRhLxdVea31gf1kls5ho0j | maths | permutations-and-combinations | conditional-permutations | The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is _____________. | [] | null | 32 | The numbers are lying between 100 and 1000
then each number is of three digits.
<br><br>The possible combination of 3 digits numbers
are
<br><br>1, 2, 3; 1, 2, 4; 1, 2, 5; 1, 3, 4; 1, 3, 5; 1, 4, 5;
2, 3, 4; 2, 3, 5; 2, 4, 5; and 3, 4, 5.
<br><br>The possible combination of numbers which are divisible by 3 are 1, 2,
3; 3, 4, 5; 1, 3, 5 and 2, 3, 4.
<br>(If sum of digits of a number is divisible by 3 then the number is divisible by 3)
<br><br>$$ \therefore $$ Total number of numbers = 4 × 3! = 24
<br><br>The possible combination of numbers divisible by 5 are 1, 2, 5; 2, 3, 5; 3, 4, 5; 1, 3, 5;
1, 4, 5 and 2, 4, 5.
<br>(If the last digit of a number is 0 or 5 then the number is divisible by 5)
<br><br>$$ \therefore $$ Total number of numbers = 6 × 2! = 12
<br><br>The possible combination of number divisible by both 3 and 5 are 1, 3, 5 and 3, 4, 5.
<br><br>$$ \therefore $$ Total number of numbers = 2 $$ \times $$ 2! = 4
<br><br>$$ \therefore $$ Total required number = 24 + 12 - 4 = 32 | integer | jee-main-2021-online-25th-february-morning-slot |
v0HhheYanRw9k3ecDT1kmlj6yhm | maths | permutations-and-combinations | conditional-permutations | The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is : | [{"identifier": "A", "content": "26664"}, {"identifier": "B", "content": "122664"}, {"identifier": "C", "content": "122234"}, {"identifier": "D", "content": "22264"}] | ["A"] | null | Total possible numbers using 1, 2, 2 and 3 is
<br><br>= $${{4!} \over {2!}}$$ = 12
<br><br>When unit place is 1, the total possible numbers using remaining 2, 2 and 3 are
<br><br>= $${{3!} \over {2!}}$$ = 3
<br><br>When unit place is 2, the total possible numbers using remaining 1, 2 and 3 are
<br><br>= 3! = 6
<br><br>When unit place is 3, the total possible numbers using remaining 1, 2 and 2 are
<br><br>= $${{3!} \over {2!}}$$ = 3
<br><br>$$ \therefore $$ Sum of unit places of all (3 + 6 + 3) 12 numbers is
<br><br>= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3)
<br><br>Similarly,
<br><br>When 10<sup>th</sup> place is 1, the total possible numbers using remaining 2, 2 and 3 are
<br><br>= $${{3!} \over {2!}}$$ = 3
<br><br>When 10<sup>th</sup> place is 2, the total possible numbers using remaining 1, 2 and 3 are
<br><br>= 3! = 6
<br><br>When 10<sup>th</sup> place is 3, the total possible numbers using remaining 1, 2 and 2 are
<br><br>= $${{3!} \over {2!}}$$ = 3
<br><br>$$ \therefore $$ Sum of 10<sup>th</sup> places of all (3 + 6 + 3) 12 numbers is
<br><br>= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 10
<br><br>Similarly,
<br><br>Sum of 100<sup>th</sup> places of all (3 + 6 + 3) 12 numbers is
<br><br>= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 100
<br><br>and Sum of 1000<sup>th</sup> places of all (3 + 6 + 3) 12 numbers is
<br><br>= ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 1000
<br><br>$$ \therefore $$ Total sum = ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) + ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 10
<br><br>+ ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 100 + ( 1$$ \times $$3 + 2$$ \times $$6 + 3$$ \times $$3) $$ \times $$ 1000
<br><br>= (3 + 12 + 9) (1 + 10 + 100 + 1000) = 1111 $$ \times $$ 24 = 26664 | mcq | jee-main-2021-online-18th-march-morning-shift |
vTTgxkoqnfAoseQRCc1kmlj9jej | maths | permutations-and-combinations | conditional-permutations | The number of times the digit 3 will be written when listing the integers from 1 to 1000 is : | [] | null | 300 | In single digit numbers = 1
<br><br>In double digit numbers = 10 + 9 = 19
<br><br>In triple digit numbers = 100 + 90 + 90 = 280
<br><br>Total = 300 times | integer | jee-main-2021-online-18th-march-morning-shift |
1kruajm92 | maths | permutations-and-combinations | conditional-permutations | If the digits are not allowed to repeat in any number formed by using the digits 0, 2, 4, 6, 8, then the number of all numbers greater than 10,000 is equal to _____________. | [] | null | 96 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264389/exam_images/aiakh8f4ehgt1n8uagk2.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 22th July Evening Shift Mathematics - Permutations and Combinations Question 100 English Explanation"><br>= 4 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 = 96 | integer | jee-main-2021-online-22th-july-evening-shift |
1ktbiwpdb | maths | permutations-and-combinations | conditional-permutations | The number of three-digit even numbers, formed by the digits 0, 1, 3, 4, 6, 7 if the repetition of digits is not allowed, is ______________. | [] | null | 52 | (i) When '0' is at unit place<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264026/exam_images/psga0gi4ukcamvr6fhaw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Morning Shift Mathematics - Permutations and Combinations Question 96 English Explanation 1"><br><br>Number of numbers = 20<br><br>(ii) When 4 or 6 are at unit place <br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267736/exam_images/adpwow8qovowjur8hlhv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Morning Shift Mathematics - Permutations and Combinations Question 96 English Explanation 2"><br><br>Number of numbers = 32<br><br>Total three digit even number = 20 + 32 = 52 | integer | jee-main-2021-online-26th-august-morning-shift |
1kteplg92 | maths | permutations-and-combinations | conditional-permutations | A number is called a palindrome if it reads the same backward as well as forward. For example 285582 is a six digit palindrome. The number of six digit palindromes, which are divisible by 55, is ____________. | [] | null | 100 | <table class="tg">
<thead>
<tr>
<td class="tg-baqh">5</td>
<td class="tg-baqh">a</td>
<td class="tg-baqh">b</td>
<td class="tg-baqh">b</td>
<td class="tg-baqh">a</td>
<td class="tg-baqh">5</td>
</tr>
</thead>
</table>
<br><br>For divisible by 55 it shall be divisible by 11 and 5
both, for divisibility by 5 unit digit shall be 0 or 5 but
as the number is six digit palindrome unit digit is 5.
<br><br>A number is divisible by 11 if the difference between sum of the digits in the odd places and the sum of the digits in the even places is a multiple of 11 or zero.
<br><br>Sum of the digits in the even place = a + b + 5
<br><br>Sum of the digits in the odd places = a + b + 5
<br><br>Difference between the two sums = (a + b + 5 ) - (a + b + 5) = 0
<br><br>0 is divisible by 11.
<br><br>Hence, 5abba5 is divisible by 11.
<br><br>So, required number = 10 $$\times$$ 10 = 100 | integer | jee-main-2021-online-27th-august-morning-shift |
1ktisnbyi | maths | permutations-and-combinations | conditional-permutations | The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is ___________. | [] | null | 576 | Total possible words = 6! = 720
<br><br>When 4 consonants are together (V, W, L, S)
<br>such cases = 3! ⋅ 4! = 144
<br><br>All consonants should not be together<br><br>= Total $$-$$ All consonants together,<br><br>= 6! $$-$$ 3! 4! = 576 | integer | jee-main-2021-online-31st-august-morning-shift |
1ktobhmqv | maths | permutations-and-combinations | conditional-permutations | All the arrangements, with or without meaning, of the word FARMER are written excluding any word that has two R appearing together. The arrangements are listed serially in the alphabetic order as in the English dictionary. Then the serial number of the word FARMER in this list is ___________. | [] | null | 77 | First find all possible words and then subtract words
from each case that have both R together.
<br><br>FARMER (6)<br><br>A, E, F, M, R, R<br><br><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;width:100%}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-0lax{text-align:left;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-0lax">A</th>
<th class="tg-0lax"></th>
<th class="tg-0lax"></th>
<th class="tg-0lax"></th>
<th class="tg-0lax"></th>
<th class="tg-0lax"></th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0lax">E</td>
<td class="tg-0lax"></td>
<td class="tg-0lax"></td>
<td class="tg-0lax"></td>
<td class="tg-0lax"></td>
<td class="tg-0lax"></td>
</tr>
<tr>
<td class="tg-0lax">F</td>
<td class="tg-0lax">A</td>
<td class="tg-0lax">E</td>
<td class="tg-0lax"></td>
<td class="tg-0lax"></td>
<td class="tg-0lax"></td>
</tr>
<tr>
<td class="tg-0lax">F</td>
<td class="tg-0lax">A</td>
<td class="tg-0lax">M</td>
<td class="tg-0lax"></td>
<td class="tg-0lax"></td>
<td class="tg-0lax"></td>
</tr>
<tr>
<td class="tg-0lax">F</td>
<td class="tg-0lax">A</td>
<td class="tg-0lax">R</td>
<td class="tg-0lax">E</td>
<td class="tg-0lax"></td>
<td class="tg-0lax"></td>
</tr>
<tr>
<td class="tg-0lax">F</td>
<td class="tg-0lax">A</td>
<td class="tg-0lax">R</td>
<td class="tg-0lax">M</td>
<td class="tg-0lax">E</td>
<td class="tg-0lax">R</td>
</tr>
</tbody>
</table><br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwoqbke8/bc58d634-a369-4a42-b528-57c54b3ada02/7f0aa000-534e-11ec-9cbb-695a838b20fb/file-1kwoqbke9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kwoqbke8/bc58d634-a369-4a42-b528-57c54b3ada02/7f0aa000-534e-11ec-9cbb-695a838b20fb/file-1kwoqbke9.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 50vh" alt="JEE Main 2021 (Online) 1st September Evening Shift Mathematics - Permutations and Combinations Question 91 English Explanation"> | integer | jee-main-2021-online-1st-september-evening-shift |
1l546nn3x | maths | permutations-and-combinations | conditional-permutations | <p>Let b<sub>1</sub>b<sub>2</sub>b<sub>3</sub>b<sub>4</sub> be a 4-element permutation with b<sub>i</sub> $$\in$$ {1, 2, 3, ........, 100} for 1 $$\le$$ i $$\le$$ 4 and b<sub>i</sub> $$\ne$$ b<sub>j</sub> for i $$\ne$$ j, such that either b<sub>1</sub>, b<sub>2</sub>, b<sub>3</sub> are consecutive integers or b<sub>2</sub>, b<sub>3</sub>, b<sub>4</sub> are consecutive integers. Then the number of such permutations b<sub>1</sub>b<sub>2</sub>b<sub>3</sub>b<sub>4</sub> is equal to ____________.</p> | [] | null | 18915 | <p>There are 98 sets of three consecutive integer and 97 sets of four consecutive integers.</p>
<p>Using the principle of inclusion and exclusion,</p>
<p>Number of permutations of $b_{1} b_{2} b_{3} b_{4}=$ Number of permutations when $b_{1} b_{2} b_{3}$ are consecutive + Number of permutations when $b_{2} b_{3} b_{4}$ are consecutive - Number of permutations when $b_{1} b_{2}$ $b_{3} b_{4}$ are consecutive</p>
<p>$=97 \times 98+97 \times 98-97=97 \times 195=18915$.</p> | integer | jee-main-2022-online-29th-june-morning-shift |
1l54udtan | maths | permutations-and-combinations | conditional-permutations | <p>The total number of four digit numbers such that each of first three digits is divisible by the last digit, is equal to ____________.</p> | [] | null | 1086 | If unit digit is 1 then $\rightarrow 9 \times$ s $10 \times 10=900$ numbers <br/><br/>If unit digit is 2 then $\rightarrow 4 \times 5 \times 5=100$ numbers <br/><br/>If unit digit is 3 then $\rightarrow 3 \times 4 \times 4=48$ numbers<br/><br/> If unit digit is 4 then $\rightarrow 2 \times 3 \times 3=18$ numbers<br/><br/> If unit digit is 5 then $\rightarrow 1 \times 2 \times 2=4$ numbers <br/><br/>If unit digit is 6 then $\rightarrow 1 \times 2 \times 2=4$ numbers
<br/><br/>
For $7,8,9 \rightarrow 4+4+4=12$ Numbers
<br/><br/>
Total $=1086$ Numbers | integer | jee-main-2022-online-29th-june-evening-shift |
1l5668ose | maths | permutations-and-combinations | conditional-permutations | <p>The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is :</p> | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "48"}, {"identifier": "C", "content": "60"}, {"identifier": "D", "content": "72"}] | ["D"] | null | To make a no. divisible by 3 we can use the digits
1,2,5,6,7 or 1,2,3,5,7.<br/><br/>
Using 1,2,5,6,7, number of even numbers is
= 4 × 3 × 2 × 1 × 2 = 48<br/><br/>
Using 1,2,3,5,7, number of even numbers is
= 4 × 3 × 2 × 1 × 1 = 24<br/><br/>
Required answer is 72. | mcq | jee-main-2022-online-28th-june-morning-shift |
1l59le9dg | maths | permutations-and-combinations | conditional-permutations | <p>The total number of three-digit numbers, with one digit repeated exactly two times, is ______________.</p> | [] | null | 243 | <p>$$C - 1:$$ All digits are non-zero</p>
<p>$${}^9{C_2}\,.\,2\,.\,{{3!} \over 2} = 216$$</p>
<p>$$C - 2$$ : One digit is 0</p>
<p>$$0,\,0,\,x \Rightarrow {}^9{C_1}\,.\,1 = 9$$</p>
<p>$$0,x,\,x \Rightarrow {}^9{C_1}\,.\,2 = 18$$</p>
<p>Total $$ = 216 + 27 = 243$$</p> | integer | jee-main-2022-online-25th-june-evening-shift |
1l5ajmbj0 | maths | permutations-and-combinations | conditional-permutations | <p>The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is _____________.</p> | [] | null | 63 | For odd number unit place shall be $1,3,5,7$ or 9 .<br/><br/>
$\therefore$ x y 1, x y 3, x y 5, x y 7, x y 9 are the type of numbers. numbers.<br/><br/>
If $x \,y\, 1$ then $x+y=6,13,20$... Cases are required<br/><br/>
i.e., $6+6+0+\ldots=12$ ways<br/><br/>
If $x \, y \,3$ then<br/><br/>
$x+y=4,11,18, \ldots$ Cases are required<br/><br/>
i.e., $4+8+1+0 \ldots=13$ ways<br/><br/>
Similarly for $x \,y \,5$, we have $x+y=2,9,16, \ldots$<br/><br/>
i.e., $2+9+3=14$ ways<br/><br/>
for $x \,y$ we have<br/><br/>
$x+y=0,7,14, \ldots$<br/><br/>
i.e., $0+7+5=12$ ways<br/><br/>
And for $x$ y we have<br/><br/>
$$
x+y=5,12,19 \ldots
$$<br/><br/>
i.e., $5+7+0 \ldots=12$ ways<br/><br/>
$\therefore$ Total 63 ways | integer | jee-main-2022-online-25th-june-morning-shift |
1l5bb1vut | maths | permutations-and-combinations | conditional-permutations | <p>The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is _____________.</p> | [] | null | 576 | Digits are $1,2,3,4,5,7,9$<br/><br/>
Multiple of $11 \rightarrow$ Difference of sum at even and odd place is divisible by 11 .<br/><br/>
Let number of the form <b>abcdefg</b><br/><br/>
$$
\begin{aligned}
&\therefore(\mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g})-(\mathrm{b}+\mathrm{d}+\mathrm{f})=11 \mathrm{x} \\\\
&\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}=31 \\\\
&\therefore \text { either } \mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g}=21 \text { or } 10 \\\\
&\therefore \mathrm{b}+\mathrm{d}+\mathrm{f}=10 \text { or } 21
\end{aligned}
$$<br/><br/>
<b>Case-1</b><br/><br/>
$$
\begin{aligned}
&a+c+e+g=21 \\\\
&b+d+f=10 \\\\
&(b, d, f) \in\{(1,2,7)(2,3,5)(1,4,5)\} \\\\
&(a, c, e, g) \in\{(1,4,7,9),(3,4,5,9),(2,3,7,9)\}
\end{aligned}
$$<br/><br/>
<b>Case-2</b><br/><br/>
$$
\begin{aligned}
&\mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g}=10 \\\\
&\mathrm{~b}+\mathrm{d}+\mathrm{f}=21 \\\\
&(\mathrm{a}, \mathrm{b}, \mathrm{e}, \mathrm{g}) \in\{1,2,3,4)\} \\\\
&(\mathrm{b}, \mathrm{d}, \mathrm{f}) \&\{(5,7,9)\} \\\\
&\therefore \text { Total number in case } 2=3 ! \times 4 !=144 \\\\
&\therefore \text { Total numbers }=144+432=576
\end{aligned}
$$ | integer | jee-main-2022-online-24th-june-evening-shift |
1l5w0pg9y | maths | permutations-and-combinations | conditional-permutations | <p>The number of 6-digit numbers made by using the digits 1, 2, 3, 4, 5, 6, 7, without repetition and which are multiple of 15 is ____________.</p> | [] | null | 360 | <p>A number is multiple of 15 when the number is divisible by 5 and sum of digits of the number is divisible by 3.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l65uqr2z/3aa8732e-2bc7-4c8a-a45b-f53f925f4291/0f4fb6b0-0ee7-11ed-a7de-eff776fdb55c/file-1l65uqr30.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l65uqr2z/3aa8732e-2bc7-4c8a-a45b-f53f925f4291/0f4fb6b0-0ee7-11ed-a7de-eff776fdb55c/file-1l65uqr30.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Permutations and Combinations Question 76 English Explanation"></p>
<p>Among 1, 2, 3, 4, 5, 6, 7 unit place is filled with 5 so it is multiple of 5.</p>
<p>Now to make it divisible by 3, take remaining 5 digits such a way that sum becomes divisible by 3.</p>
<p>Remaining 5 digits can be</p>
<p>(1) 1, 2, 3, 4, 6</p>
<p>Here sum = 1 + 2 + 3 + 4 + 6 + 5 = 21 (divisible by 3)</p>
<p>This 5 digits can be filled in those 5 placed without repetition in 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 = 51 = 120 ways</p>
<p>(2) 2, 3, 4, 6, 7</p>
<p>Here sum = 2 + 3 + 4 + 6 + 7 + 5 = 27 (divisible by 3)</p>
<p>$$\therefore$$ Number of ways = 51 = 120</p>
<p>(3) 1, 2, 3, 6, 7</p>
<p>Here sum = 1 + 2 + 3 + 6 + 7 + 5 = 24 (divisible by 3)</p>
<p>$$\therefore$$ Number of ways = 51 = 120</p>
<p>$$\therefore$$ Total possible 6 digit numbers divisible by 15</p>
<p>= 120 + 120 + 120 = 360</p> | integer | jee-main-2022-online-30th-june-morning-shift |
1l6gjaknm | maths | permutations-and-combinations | conditional-permutations | <p>The number of 5-digit natural numbers, such that the product of their digits is 36 , is __________.</p> | [] | null | 180 | <p>Factors of 36 = 2<sup>2</sup> . 3<sup>2</sup> . 1</p>
<p>Five-digit combinations can be</p>
<p>(1, 2, 2, 3, 3) (1, 4, 3, 3, 1), (1, 9, 2, 2, 1)</p>
<p>(1, 4, 9, 11) (1, 2, 3, 6, 1) (1, 6, 6, 1, 1)</p>
<p>i.e., total numbers</p>
<p>$${{5!} \over {2!2!}} + {{5!} \over {2!2!}} + {{5!} \over {2!2!}} + {{5!} \over {3!}} + {{5!} \over {2!}} + {{5!} \over {3!2!}}$$</p>
<p>$$ = (30 \times 3) + 20 + 60 + 10 = 180$$.</p> | integer | jee-main-2022-online-26th-july-morning-shift |
1l6hzoa3g | maths | permutations-and-combinations | conditional-permutations | <p>Numbers are to be formed between 1000 and 3000 , which are divisible by 4 , using the digits $$1,2,3,4,5$$ and 6 without repetition of digits. Then the total number of such numbers is ____________.</p> | [] | null | 30 | Here 1<sup>st</sup> digit is 1 or 2 only<br><br>
<b>Case-I</b><br><br>
If first digit is 1<br><br>
Then last two digits can be 24, 32, 36, 52, 56, 64 <br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97twac2/f9689a1f-abde-44d3-bf0b-1a1289bdd7bc/f788ec10-4b61-11ed-80b9-4154b7faa509/file-1l97twac3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97twac2/f9689a1f-abde-44d3-bf0b-1a1289bdd7bc/f788ec10-4b61-11ed-80b9-4154b7faa509/file-1l97twac3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th July Evening Shift Mathematics - Permutations and Combinations Question 73 English Explanation 1"><br>
<b>Case – II </b><br><br>
If first digit is 2 then last two digit can be 16, 36,
56, 64 <br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97tyhd1/d57389e9-902a-407f-843c-e06ca2c6e0b2/3494f950-4b62-11ed-80b9-4154b7faa509/file-1l97tyhd2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97tyhd1/d57389e9-902a-407f-843c-e06ca2c6e0b2/3494f950-4b62-11ed-80b9-4154b7faa509/file-1l97tyhd2.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th July Evening Shift Mathematics - Permutations and Combinations Question 73 English Explanation 2"><br>
Total ways = 12 + 18 = 30 ways | integer | jee-main-2022-online-26th-july-evening-shift |
1l6rflsnq | maths | permutations-and-combinations | conditional-permutations | <p>The number of natural numbers lying between 1012 and 23421 that can be formed using the digits $$2,3,4,5,6$$ (repetition of digits is not allowed) and divisible by 55 is _________.</p> | [] | null | 6 | <b>Case-I</b><br/><br/> When number is 4-digit number $(\overline{a b c d})$ here $d$ is fixed as 5
<br/><br/>
So, $(a, b, c)$ can be $(6,4,3),(3,4,6),(2,3,6)$, $(6,3,2),(3,2,4)$ or $(4,2,3)$
<br/><br/>
$\Rightarrow 6$ numbers
<br/><br/>
<b>Case-II</b><br/><br/> No number possible | integer | jee-main-2022-online-29th-july-evening-shift |
1ldo7hjc5 | maths | permutations-and-combinations | conditional-permutations | <p>The total number of six digit numbers, formed using the digits 4, 5, 9 only and divisible by 6, is ____________.</p> | [] | null | 81 | A number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 and sum of all digits of the number is divisible by 3 .
<br/><br/>Units, place must be occupied by 4 and hence, at
least one 4 must be there.
<br/><br/>Possible combination of 4, 5, 9 are as follows :
<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;width:100%}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-0pky">4</th>
<th class="tg-0pky">5</th>
<th class="tg-0pky">9</th>
<th class="tg-0pky">Total number<br> of Number</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0pky">1</td>
<td class="tg-0pky">1</td>
<td class="tg-0pky">4</td>
<td class="tg-0pky">$${{5!} \over {4!}} = 5$$</td>
</tr>
<tr>
<td class="tg-0pky">1</td>
<td class="tg-0pky">4</td>
<td class="tg-0pky">1</td>
<td class="tg-0pky">$${{5!} \over {4!}} = 5$$</td>
</tr>
<tr>
<td class="tg-0pky">2</td>
<td class="tg-0pky">2</td>
<td class="tg-0pky">2</td>
<td class="tg-0pky">$${{5!} \over {2!2!}} = 30$$</td>
</tr>
<tr>
<td class="tg-0pky">3</td>
<td class="tg-0pky">0</td>
<td class="tg-0pky">3</td>
<td class="tg-0pky">$${{5!} \over {2!3!}} = 10$$</td>
</tr>
<tr>
<td class="tg-0pky">3</td>
<td class="tg-0pky">3</td>
<td class="tg-0pky">0</td>
<td class="tg-0pky">$${{5!} \over {2!3!}} = 10$$</td>
</tr>
<tr>
<td class="tg-0pky">4</td>
<td class="tg-0pky">1</td>
<td class="tg-0pky">1</td>
<td class="tg-0pky">$${{5!} \over {3!}} = 20$$</td>
</tr>
<tr>
<td class="tg-0pky">6</td>
<td class="tg-0pky">0</td>
<td class="tg-0pky">0</td>
<td class="tg-0pky">$${{5!} \over {5!}} = 1$$</td>
</tr>
</tbody>
</table>
<br/><br/>Total = 5 + 5 + 30 + 10 + 10 + 20 + 1 = 81. | integer | jee-main-2023-online-1st-february-evening-shift |
1ldoo9cl4 | maths | permutations-and-combinations | conditional-permutations | <p>The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is ___________.</p> | [] | null | 50400 | Vowels : A,A,A,I,I,O
<br/><br/>Consonants : S,S,S,S,N,N,T
<br/><br/>$$ \therefore $$ Total number of ways in which vowels come together
<br/><br/>$=\frac{8 !}{4 ! 2 !} \times \frac{6 !}{3 ! 2 !}=50400$ | integer | jee-main-2023-online-1st-february-morning-shift |
1ldptkt5n | maths | permutations-and-combinations | conditional-permutations | <p>Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to ____________.</p> | [] | null | 710 | Numbers which are divisible by 3 (4 digit) and less than or equal to 2800
<br/><br/>$=\frac{2799-1002}{3}+1=600$
<br/><br/>Numbers which are divisible by 11 (4 digit) and less than or equal to 2800
<br/><br/>$=\frac{2794-1001}{11}+1=164$
<br/><br/>Numbers which are divisible by 33 (4 digit) and less than or equal to 2800
<br/><br/>$=\frac{2772-1023}{33}+1=54$
<br/><br/>$\therefore$ Total numbers = $600+164-54=710$ | integer | jee-main-2023-online-31st-january-morning-shift |
ldqzur7j | maths | permutations-and-combinations | conditional-permutations | The number of seven digits odd numbers, that can
be formed using all the<br/><br/>seven digits 1, 2, 2, 2, 3, 3,
5 is ____________. | [] | null | 240 | <p>$$.......1 \to {{6!} \over {2!3!}} = 60$$</p>
<p>$$.......3 \to {{6!} \over {3!}} = 120$$</p>
<p>$$.......5 \to {{6!} \over {3!2!}} = 60$$</p>
<p>Total = 240</p> | integer | jee-main-2023-online-30th-january-evening-shift |
1ldr7rs32 | maths | permutations-and-combinations | conditional-permutations | <p>Number of 4-digit numbers (the repetition of digits is allowed) which are made using the digits 1, 2, 3 and 5, and are divisible by 15, is equal to ___________.</p> | [] | null | 21 | <p>We have to make 4 digit numbers using the
digits, 1, 2, 3 and 5.
<br><br>The unit digit of the 4 digit number will be 5.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lf61sqp4/9a9d5f7b-a4f5-43a8-8dd7-0303f78b9d4c/98f24a80-c130-11ed-8b61-5f07d5ca97fb/file-1lf61sqp5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lf61sqp4/9a9d5f7b-a4f5-43a8-8dd7-0303f78b9d4c/98f24a80-c130-11ed-8b61-5f07d5ca97fb/file-1lf61sqp5.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 30th January Morning Shift Mathematics - Permutations and Combinations Question 57 English Explanation"></p>
<p>Now, the sum (x + y + z) should be of the (3p + 1).
<br><br>Therefore, the possible cases are
<br><br>(x, y, z) = (1, 1, 5), (1, 1, 2), (2, 2, 3), (2, 3, 5), (3, 3, 1) and
(5, 5, 3).</p>
<p>So, total arrangements are
<br><br>For $(1,1,5) \rightarrow \frac{3 !}{2 !}=3 ;$
<br><br>For $(1,1,2) \rightarrow \frac{3 !}{2 !}=3 ;$
<br><br>For $(2,2,3) \rightarrow \frac{3 !}{2 !}=3 ;$
<br><br>For $(2,3,5) \rightarrow 3 !=6 ;$
<br><br>For $(3,3,1) \rightarrow \frac{3 !}{2 !}=3 ;$
<br><br>For $(5,5,3) \rightarrow \frac{3 !}{2 !}=3 ;$
<br><br>So, total number of arrangements $=3+3+3+6+3+3=$ 21</p> | integer | jee-main-2023-online-30th-january-morning-shift |
1ldswvvgd | maths | permutations-and-combinations | conditional-permutations | <p>If all the six digit numbers $$x_1\,x_2\,x_3\,x_4\,x_5\,x_6$$ with $$0< x_1 < x_2 < x_3 < x_4 < x_5 < x_6$$ are arranged in the increasing order, then the sum of the digits in the $$\mathrm{72^{th}}$$ number is _____________.</p> | [] | null | 32 | $1 \ldots \ldots \ldots \ldots \ldots \rightarrow{ }^{8} C_{5}=56$
<br/><br/>
23 $\ldots\ldots\ldots\ldots\ldots\rightarrow{ }^{6} C_{4}=\frac{15}{71}$
<br/><br/>
$72^{\text {th }}$ number $=245678$
<br/><br/>
Sum $=32$ | integer | jee-main-2023-online-29th-january-morning-shift |
1ldu5hsr5 | maths | permutations-and-combinations | conditional-permutations | <p>The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1, 3, 5, 7, 9 without repetition, is :</p> | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "12"}] | ["C"] | null | Numbers between 5000 & 10000<br><br>
Using digits 1, 3, 5, 7, 9<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lef5alut/f8dc7152-4573-493d-a2cc-0b09243a1255/1b9fc550-b265-11ed-a126-5dfa1a9d5fb8/file-1lef5aluu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lef5alut/f8dc7152-4573-493d-a2cc-0b09243a1255/1b9fc550-b265-11ed-a126-5dfa1a9d5fb8/file-1lef5aluu.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 25th January Evening Shift Mathematics - Permutations and Combinations Question 48 English Explanation"><br>
Total Numbers = 3 × 4 × 3 × 2 = 72 | mcq | jee-main-2023-online-25th-january-evening-shift |
lgnwkhha | maths | permutations-and-combinations | conditional-permutations | The total number of three-digit numbers, divisible by 3, which can be formed using the digits $1,3,5,8$, if repetition of digits is allowed, is : | [{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "20"}] | ["B"] | null | The number of three-digit numbers divisible by 3 by considering the possible sums of digits that are divisible by 3. Your approach is as follows:
<br/><br/>1. Sum of digits is 3: $(1, 1, 1)$ - 1 possible number
<br/><br/>2. Sum of digits is 9: $(1, 3, 5)$ and $(3, 3, 3)$ -
<br/><br/> Let's consider the cases separately :
<br/><br/><b>a.</b> Sum of digits is 9: $(1, 3, 5)$
For this case, we can arrange the digits in $3!$ ways :
135, 153, 315, 351, 513, and 531.
<br/><br/><b>b.</b> Sum of digits is 9: $(3, 3, 3)$
For this case, since all the digits are the same, there is only 1 possible number :
333.
<br/><br/>Now, the total number of possible numbers when the sum of digits is 9 is :
<br/><br/>$$
3! + 1 = 6 + 1 = 7
$$
<br/><br/>3. Sum of digits is 12: $(1, 3, 8)$ - $3!$ possible numbers
<br/><br/>4. Sum of digits is 15: $(5, 5, 5)$ - 1 possible number
<br/><br/>5. Sum of digits is 18: $(5, 5, 8)$ - $\frac{3!}{2!}$ possible numbers (since 5 is repeated)
<br/><br/>6. Sum of digits is 21: $(5, 8, 8)$ - $\frac{3!}{2!}$ possible numbers (since 8 is repeated)
<br/><br/>7. Sum of digits is 24: $(8, 8, 8)$ - 1 possible number
<br/><br/>Adding up the possible numbers for each case, we get:
<br/><br/>$$
1 + 7 + 3! + 1 + \frac{3!}{2!} + \frac{3!}{2!} + 1 = 1 + 7 + 6 + 1 + 3 + 3 + 1 = 22
$$
<br/><br/>So, there are a total of 22 three-digit numbers divisible by 3 that can be formed using the digits $1, 3, 5, 8$ with repetition allowed. | mcq | jee-main-2023-online-15th-april-morning-shift |
lgnys0w5 | maths | permutations-and-combinations | conditional-permutations | A person forgets his 4-digit ATM pin code. But he remembers that in the code all the digits are different, the greatest digit is 7 and the sum of the first two digits is equal to the sum of the last two digits. Then the maximum number of trials necessary to obtain the correct code is ___________. | [] | null | 72 | Let the 4-digit ATM pin code be represented by the digits $$abxy$$, where all digits are different, the greatest digit is 7, and the sum of the first two digits is equal to the sum of the last two digits: $$a + b = x + y$$.
<br><br>Since the greatest digit is 7, the possible digits for the pin code are $$0, 1, 2, 3, 4, 5, 6,$$ and $$7$$.
<br><br>1. Let's denote the sum of the first two digits and the sum of the last two digits as $$\lambda$$. Since the largest digit is 7, we can analyze different possible values for $$\lambda$$.
<br><br>We will analyze different possible values for $$\lambda$$ and the corresponding possible pairs of digits:
<br><br>- $$\lambda = 7$$: The pairs of digits that have a sum of 7 are $(0,7),(7,0),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. Since 7 must be present, we consider the pairs with 7: $(0,7),(7,0)$.
<br><br>When the pair $(0,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. For each of these pairs, there are 2 possible pin codes: $$07xy$$ and $$07yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.
<br><br>When the pair $(7,0)$ is chosen, the other two digits can also be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. For each of these pairs, there are 2 possible pin codes: $$70xy$$ and $$70yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.
<br><br><br>So, there are 24 possible pin codes for $$\lambda = 7$$.<br>
<br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgv2wsca/cbc5754b-29f9-4539-a370-acd512a8c4f5/c2331800-e2c0-11ed-920b-3fff2b1cb8d0/file-1lgv2wscg.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgv2wsca/cbc5754b-29f9-4539-a370-acd512a8c4f5/c2331800-e2c0-11ed-920b-3fff2b1cb8d0/file-1lgv2wscg.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 41 English Explanation 1">
<br><br>- $$\lambda = 8$$: The pairs of digits that have a sum of 8 are $(1,7),(7,1),(2,6),(6,2),(3,5),(5,3)$. Since 7 must be present, we consider the pairs with 7: $(1,7),(7,1)$.
<br><br>When the pair $(1,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$. For each of these pairs, there are 2 possible pin codes: $$17xy$$ and $$17yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.
<br><br>When the pair $(7,1)$ is chosen, the other two digits can also be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$. For each of these pairs, there are 2 possible pin codes: $$71xy$$ and $$71yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.
<br><br>So, there are 16 possible pin codes for $$\lambda = 8$$.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgv2zu9s/f697b5a7-ed4b-4299-82b8-a7a0d3445b64/171d6000-e2c1-11ed-920b-3fff2b1cb8d0/file-1lgv2zu9t.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgv2zu9s/f697b5a7-ed4b-4299-82b8-a7a0d3445b64/171d6000-e2c1-11ed-920b-3fff2b1cb8d0/file-1lgv2zu9t.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 41 English Explanation 2">
<br><br>- $$\lambda = 9$$: The pairs of digits that have a sum of 9 are $(2,7),(7,2),(3,6),(6,3),(4,5),(5,4)$. Since 7 must be present, we consider the pairs with 7: $(2,7),(7,2)$.
<br><br>When the pair $(2,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$. For each of these pairs, there are 2 possible pin codes: $$27xy$$ and $$27yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.
<br><br>When the pair $(7,2)$ is chosen, the other two digits can also be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$. For each of these pairs, there are 2 possible pin codes: $$72xy$$ and $$72yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.
<br><br>So, there are 16 possible pin codes for $$\lambda = 9$$.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgv31bpl/c0980147-5172-44a2-a58d-c6cb14a3f1ab/4066ae30-e2c1-11ed-99f0-2347181d377e/file-1lgv31bpy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgv31bpl/c0980147-5172-44a2-a58d-c6cb14a3f1ab/4066ae30-e2c1-11ed-99f0-2347181d377e/file-1lgv31bpy.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 41 English Explanation 3">
<br><br>- $$\lambda = 10$$: The pairs of digits that have a sum of 10 are $(3,7),(7,3),(4,6),(6,4)$. Since 7 must be present, we consider the pairs with 7: $(3,7),(7,3)$.
<br><br>When the pair $(3,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(4,6),(6,4)$. For each of these pairs, there are 2 possible pin codes: $$37xy$$ and $$37yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.
<br><br>When the pair $(7,3)$ is chosen, the other two digits can also be any one of the remaining pairs: $(4,6),(6,4)$. For each of these pairs, there are 2 possible pin codes: $$73xy$$ and $$73yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.
<br><br>So, there are 8 possible pin codes for $$\lambda = 10$$.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgv32tqq/b0d7959c-9018-4ef6-a138-3892ec03995b/6a221f20-e2c1-11ed-99f0-2347181d377e/file-1lgv32tqr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgv32tqq/b0d7959c-9018-4ef6-a138-3892ec03995b/6a221f20-e2c1-11ed-99f0-2347181d377e/file-1lgv32tqr.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 41 English Explanation 4">
<br><br>- $$\lambda = 11$$: The pairs of digits that have a sum of 11 are $(4,7),(7,4),(5,6),(6,5)$. Since 7 must be present, we consider the pairs with 7: $(4,7),(7,4)$.
<br><br>When the pair $(4,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(5,6),(6,5)$. For each of these pairs, there are 2 possible pin codes: $$47xy$$ and $$47yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.
<br><br>When the pair $(7,4)$ is chosen, the other two digits can also be any one of the remaining pairs: $(5,6),(6,5)$. For each of these pairs, there are 2 possible pin codes: $$74xy$$ and $$74yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.
<br><br>So, there are 8 possible pin codes for $$\lambda = 11$$.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgv34b44/8ad108c1-684d-453a-8d6e-de913d462fc1/935ca040-e2c1-11ed-99f0-2347181d377e/file-1lgv34b45.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgv34b44/8ad108c1-684d-453a-8d6e-de913d462fc1/935ca040-e2c1-11ed-99f0-2347181d377e/file-1lgv34b45.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Permutations and Combinations Question 41 English Explanation 5">
<br><br>- $$\lambda = 12$$: The only pair of digits that have a sum of 12 is $(5,7)$. Since 7 must be present, this case is not possible.
<br><br>- $$\lambda = 13$$: The only pair of digits that have a sum of 13 is $(6,7)$. Since 7 must be present, this case is not possible.
<br><br>- $$\lambda = 14$$: The only pair of digits that have a sum of 14 is $(7,7)$. Since all the digits must be different, this case is not possible.
<br><br>Adding up all the possible pin codes for each value of $$\lambda$$, we get:
<br><br>$$16 + 24 + 24 + 16 + 8 = 72$$
<br><br>Therefore, the maximum number of trials necessary to obtain the correct code is 72. | integer | jee-main-2023-online-15th-april-morning-shift |
1lgoy7m2y | maths | permutations-and-combinations | conditional-permutations | <p>Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits $$1,2,3,4,5$$ with repetition, is _________.</p> | [] | null | 16 | A number is divisible by 6 if it is divisible by both 2 and 3. A number is divisible by 2 if its last digit is even, which means it must be either 2 or 4 from the given digits. A number is divisible by 3 if the sum of its digits is divisible by 3.
<br/><br/>$$
\begin{aligned}
& (2,1,3),(2,3,4),(2,5,5),(2,2,5),(2,2,2) \\\\
& (4,1,1),(4,4,1),(4,4,4),(4,3,5) \\\\
& 2,1,3 \Rightarrow 312,132 \\\\
& 2,3,4 \Rightarrow 342,432,234,324 \\\\
& 2,5,5 \Rightarrow 552 \\\\
& 2,2,5 \Rightarrow 252,522 \\\\
& 2,2,2 \Rightarrow 222 \\\\
& 4,1,1 \Rightarrow 114 \\\\
& 4,4,1 \Rightarrow 414,144 \\\\
& 4,4,4 \Rightarrow 444 \\\\
& 4,3,5 \Rightarrow 354,534
\end{aligned}
$$
<br/><br/>Total 16 numbers. | integer | jee-main-2023-online-13th-april-evening-shift |
1lgrefqkf | maths | permutations-and-combinations | conditional-permutations | <p>The number of five digit numbers, greater than 40000 and divisible by 5 , which can be formed using the digits $$0,1,3,5,7$$ and 9 without repetition, is equal to :</p> | [{"identifier": "A", "content": "132"}, {"identifier": "B", "content": "72"}, {"identifier": "C", "content": "120"}, {"identifier": "D", "content": "96"}] | ["C"] | null | Since the five-digit number must be greater than 40000, the only options for the first digit are 5, 7, or 9. That leaves 3 remaining choices for the first digit.
<br/><br/>Since the number has to be divisible by 5, the last digit must be 0 or 5. If the first digit is 5, the last digit can only be 0, since digits cannot be repeated. If the first digit is 7 or 9, the last digit can be 0 or 5, so there are 2 choices.
<br/><br/>Five possible configurations for the first and last digits :
<br/><br/>- 5xxx0
<br/><br/>- 7xxx0
<br/><br/>- 7xxx5
<br/><br/>- 9xxx0
<br/><br/>- 9xxx5
<br/><br/>For each of these configurations, there are three places in the middle that can be filled with the remaining 4 unused digits.
<br/><br/>Since repetition isn't allowed, there are 4 options for the second digit, 3 for the third, and 2 for the fourth, which can be represented as <sup>4</sup>P<sub>3</sub>, or permutations of 4 items taken 3 at a time. This is equal to 4 $$ \times $$ 3 $$ \times $$ 2 = 24.
<br/><br/>So, for each of the 5 configurations, there are 24 ways to arrange the middle three digits. Therefore, the total number of five-digit numbers that meet the criteria is 5 $$ \times $$ 24 = 120.
| mcq | jee-main-2023-online-12th-april-morning-shift |
1lgrgicm5 | maths | permutations-and-combinations | conditional-permutations | <p>Let the digits a, b, c be in A. P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?</p> | [] | null | 1260 | <p>The problem involves forming nine-digit numbers from three digits a, b, c which are in Arithmetic Progression (AP), used three times each, such that at least once, three consecutive digits are in AP.</p>
<p>We have the two possible sequences for the AP :</p>
<ol>
<li>a, b, c</li>
<li>c, b, a</li>
</ol>
<p>This shows the flexibility in ordering the three digits that are in AP in our nine-digit number.</p>
<p>The next step is to choose the location of this sequence of three numbers within our nine-digit number.</p>
<p>Since there are nine places in our number and our sequence takes up three places, we have seven different starting points for our sequence : it can start at the first place, the second place, and so on, up to the seventh place.</p>
<p>Therefore, the number of ways to select 3 consecutive places out of the 9 places for the AP sequence is 7.</p>
<p>However, we also have to account for the fact that our sequence can be in one of two orders (a, b, c or c, b, a). So, we multiply the number of starting points by 2 to get $^7C_1 \times 2 = 14$ ways to arrange the sequence within our nine-digit number.</p>
<p>The remaining 6 digits (two 'a', two 'b', two 'c") can be arranged in $\frac{6!}{2!2!2!}$ ways.</p>
<p>Therefore, the total number of such nine-digit numbers is $^7C_1 \times 2 \times \frac{6!}{2!2!2!} = 1260$.</p>
| integer | jee-main-2023-online-12th-april-morning-shift |
1lgxwdlm3 | maths | permutations-and-combinations | conditional-permutations | <p>The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is ___________.</p> | [] | null | 4898 | Given that digits are $1,2,3,4,5,6,7$
<br/><br/>Total permutations $=7$!
<br/><br/>Let $p=$ Number which containing string 153
<br/><br/>$q=$ Number which containing string 2467
<br/><br/>$$
\begin{array}{ll}
& \therefore n(p)=5! \times 1 \\\\
& \Rightarrow n(q)=4! \times 1 \\\\
& \Rightarrow n(p \cap q)=2!
\end{array}
$$
<br/><br/>$$
\begin{aligned}
& \therefore n(p \cup q)=n(p)+n(q)-n(p \cap q) \\\\
& = 5 !+4 !-2 !=120+24-2=142
\end{aligned}
$$
<br/><br/>$\therefore n$ (neither string 143 nor string 2467)
<br/><br/>$$
=7 !-142=5040-142=4898
$$ | integer | jee-main-2023-online-10th-april-morning-shift |
1lgylnv0d | maths | permutations-and-combinations | conditional-permutations | <p>If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which $$\mathrm{C}$$ and $$\mathrm{S}$$ do not come together, is $$(6 !) \mathrm{k}$$, then $$\mathrm{k}$$ is equal to :</p> | [{"identifier": "A", "content": "5670"}, {"identifier": "B", "content": "1890"}, {"identifier": "C", "content": "2835"}, {"identifier": "D", "content": "945"}] | ["A"] | null | $$
\text { Total number of words }=\frac{11 !}{2 ! 2 ! 2 !}
$$
<br/><br/>Number of words in which $\mathrm{C}$ and $\mathrm{S}$ are together
<br/><br/>$$
=\frac{10 !}{2 ! 2 ! 2 !} \times 2 \text { ! }
$$
<br/><br/>So, required number of words
<br/><br/>$$
\begin{aligned}
& =\frac{11 !}{2 ! 2 ! 2 !}-\frac{10 !}{2 ! 2 !} \\\\
& =\frac{11 \times 10 !}{2 ! 2 ! 2 !}-\frac{10 !}{2 ! 2 !} \\\\
& =\frac{10 !}{2 ! 2 !}\left[\frac{11}{2}-1\right]=\frac{10 !}{2 ! 2 !} \times \frac{9}{2} \\\\
& =5670 \times 6 ! \\\\
& \Rightarrow k(6 !)=5670 \times 6 ! \\\\
& \Rightarrow k=5670
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lgzxfup5 | maths | permutations-and-combinations | conditional-permutations | <p>The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is :</p> | [{"identifier": "A", "content": "16800"}, {"identifier": "B", "content": "14800"}, {"identifier": "C", "content": "18000"}, {"identifier": "D", "content": "33600"}] | ["A"] | null | In the given word, <br/><br/>vowels are : I, E, E, E, E <br/><br/>Consonants are : N, D, P, N, D, N, C <br/><br/>So, number of words $=\frac{8 !}{3 ! 2 !} \times \frac{5 !}{4 !}$ <br/><br/>$=\frac{8 \times 7 \times 6 \times 5 \times 4}{2} \times 5=16800$
<br/><br/><b>Concept :</b>
<br/><br/>Out of $n$ objects, if $r$ things are same, so number of ways $=\frac{n !}{r !}$ | mcq | jee-main-2023-online-8th-april-morning-shift |
1lh2yrfzn | maths | permutations-and-combinations | conditional-permutations | <p>The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is __________.</p> | [] | null | 432 | Given, word is UNIVERSE
<br/><br/>Here, vowels are E, I, U and consonants are N, R, S, V
<br/><br/>$\therefore$ Required number of 4-letters words, with or without meaning,
<br/><br/>each consisting of 2 vowels and 2 consonants
<br/><br/>$$
\begin{aligned}
& ={ }^3 C_2 \times{ }^4 C_2 \times 4 ! \\\\
& ={ }^3 C_1 \times{ }^4 C_2 \times 24 \\\\
& =3 \times 6 \times 24=432
\end{aligned}
$$ | integer | jee-main-2023-online-6th-april-evening-shift |
1lguwuo8m | maths | permutations-and-combinations | dearrangement | <p>In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is _________.</p> | [] | null | 44 | This problem can be solved using the concept of derangements, which is a permutation of objects where no object appears in its original position. In this case, we have 5 students who should not sit in their allotted seats.
<br/><br/>The formula for calculating the number of derangements (also known as subfactorials) is :
<br/><br/>D(n) $$
=n !\left[\frac{1}{0!}-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\ldots \ldots . .+(-1)^n \frac{1}{n !}\right]
$$
<br/><br/>Where n is the number of students, in this case, 5.
<br/><br/>Using the formula, let's calculate the derangements for 5 students :
<br/><br/>D(5) = $5! \left(\frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right)$
<br/><br/>D(5) = $120 \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$
<br/><br/>D(5) = $120 \left(0 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$
<br/><br/>D(5) = $120 \left(\frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$
<br/><br/>D(5) = $120 \left(\frac{60}{120} - \frac{20}{120} + \frac{5}{120} - \frac{1}{120}\right)$
<br/><br/>D(5) = $120 \left(\frac{44}{120}\right)$
<br/><br/>D(5) = 44
<br/><br/>So, there are 44 ways in which none of the students sits on the allotted seat. | integer | jee-main-2023-online-11th-april-morning-shift |
38m7NMVKw7MZYgve | maths | permutations-and-combinations | divisibility-of-numbers | The sum of integers from 1 to 100 that are divisible by 2 or 5 is : | [{"identifier": "A", "content": "3000"}, {"identifier": "B", "content": "3050"}, {"identifier": "C", "content": "3600"}, {"identifier": "D", "content": "3250"}] | ["B"] | null | According to this question, any number between 1 to 100 should be divisible by 2 or 5 but not by 2$$ \times $$5 = 10.
<br><br>Possible numbers between 1 to 100 divisible by 2 are 2, 4, 6, .... , 100
<br><br>This is an A.P where first term = 2, last term = 100 and total terms = 50.
<br><br>$$ \therefore $$ Sum of the numbers divisible by 2
<br><br> = $${{50} \over 2}\left[ {2 + 100} \right]$$
<br><br>= 25$$ \times $$102
<br><br>= 2550
<br><br>Possible numbers between 1 to 100 divisible by 5 are 5, 10, 15, .... , 100
<br><br>$$ \therefore $$ Sum of the numbers divisible by 5
<br><br> = $${{20} \over 2}\left[ {5 + 100} \right]$$
<br><br>= 10$$ \times $$105
<br><br>= 1050
<br><br> And possible numbers between 1 to 100 divisible by 10 are 10, 20, 30, .... , 100
<br><br>$$ \therefore $$ Sum of the numbers divisible by 10
<br><br> = $${{10} \over 2}\left[ {10 + 100} \right]$$
<br><br>= 5$$ \times $$110
<br><br>= 550
<br><br>$$ \therefore $$ Required sum = 2550 + 1050 - 550 = 3050 | mcq | aieee-2002 |
vKUBJgogUkTqMHWSIq45x | maths | permutations-and-combinations | divisibility-of-numbers | The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is : | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "30"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "48"}] | ["B"] | null | Here number should be divisible by 3, that means sum of numbers should be divisible by 3.
<br><br>Possible 4 digits among 0, 1, 2, 3, 4 which are divisible by 3 are
<br><br>(1)$$\,\,\,\,$$ (0, 2, 3, 4) Sum of digits = 0 + 2 + 3 +4 = 9 (divisible by 3)
<br><br>(2) $$\,\,\,\,$$ (0, 1, 2, 3) Sum of digits = 0 + 1 + 2 + 3 = 6 (divisible by 3)
<br><br><b><u>Case 1</u> :</b>
<br><br>When 4 digits are (0, 2, 3, 4) then
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265112/exam_images/crszvfmrucm793mhf0u4.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 16th April Morning Slot Mathematics - Permutations and Combinations Question 151 English Explanation 1"><br><br>$$\therefore\,\,\,\,$$ Total possible numbers = $$^3{C_1}$$ $$ \times $$ $$^3{C_1}$$ $$ \times $$ $$^2{C_1}$$ $$ \times $$ $$^1{C_1}$$
<br><br>= 3 $$ \times $$ 3 $$ \times $$ 2 $$ \times $$ 1 = 18
<br><br><b><u>Case 2</u> :</b>
<br><br>When 4 digits are (0, 1, 2, 3) then,
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263381/exam_images/wbv3ybxvwrwbq6hh5suh.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 16th April Morning Slot Mathematics - Permutations and Combinations Question 151 English Explanation 2">
<br><br>$$\therefore\,\,\,\,$$ Total possible number in this case = $$^2{C_1}$$ $$ \times $$ $$^3{C_1}$$ $$ \times $$ $$^2{C_1}$$ $$ \times $$ $$^1{C_1}$$
<br><br>= 2 $$ \times $$ 3 $$ \times $$ 2 $$ \times $$ 1 = 12
<br><br>$$\therefore\,\,\,\,$$ Total possible numbers will be = 18 + 12 = 30 | mcq | jee-main-2018-online-16th-april-morning-slot |
OjrcXgele0CMwAVPCy1kluvyg3b | maths | permutations-and-combinations | divisibility-of-numbers | A natural number has prime factorization given by n = 2<sup>x</sup>3<sup>y</sup>5<sup>z</sup>, where y and z are such <br/>that y + z = 5 and y<sup>$$-$$1</sup> + z<sup>$$-$$1</sup> = $${5 \over 6}$$, y > z. Then the number of odd divisions of n, including 1, is : | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "6x"}] | ["C"] | null | y + z = 5 ....... (1)<br><br>$${1 \over y} + {1 \over z} = {5 \over 6}$$<br><br>$$ \Rightarrow {{y + z} \over {yz}} = {5 \over 6}$$<br><br>$$ \Rightarrow {5 \over {yz}} = {5 \over 6}$$<br><br>$$ \Rightarrow $$ yz = 6<br><br>Also, (y $$-$$ z)<sup>2</sup> = (y + z)<sup>2</sup> $$-$$ 4yz<br><br>$$ \Rightarrow $$ (y $$-$$ z)<sup>2</sup> = (y + z)<sup>2</sup> $$-$$ 4yz<br><br>$$ \Rightarrow $$ (y $$-$$ z)<sup>2</sup> = 25 $$-$$ 4(6) = 1<br><br>$$ \Rightarrow $$ y $$-$$ z = 1 ..... (2)<br><br>from (1) and (2), y = 3 and z = 2<br><br>for calculating odd divisor of p = 2<sup>x</sup> . 3<sup>y</sup> . 5<sup>z</sup><br><br>x must be zero<br><br>P = 2<sup>0</sup> . 3<sup>3</sup> . 5<sup>2</sup> <br><br>$$ \Rightarrow $$ Total possible cases = (3<sup>0</sup>5<sup>0</sup> + 3<sup>1</sup>5<sup>0</sup> + 3<sup>2</sup>5<sup>0</sup> + 3<sup>3</sup>5<sup>0</sup> + .... + 3<sup>3</sup>5<sup>2</sup>)<br><br>$$ \therefore $$ Total odd divisors must be (3 + 1) ( 2 + 1) = 12 | mcq | jee-main-2021-online-26th-february-evening-slot |
1krygm1ew | maths | permutations-and-combinations | divisibility-of-numbers | Let n be a non-negative integer. Then the number of divisors of the form "4n + 1" of the number (10)<sup>10</sup> . (11)<sup>11</sup> . (13)<sup>13</sup> is equal to __________. | [] | null | 924 | N = 2<sup>10</sup> $$\times$$ 5<sup>10</sup> $$\times$$ 11<sup>11</sup> $$\times$$ 13<sup>13</sup><br><br>Now, power of 2 must be zero,<br><br>power of 5 can be anything,<br><br>power of 13 can be anything<br><br>But, power of 11 should be even.<br><br>So, required number of divisors is <br><br>1 $$\times$$ 11 $$\times$$ 14 $$\times$$ 6 = 924 | integer | jee-main-2021-online-27th-july-evening-shift |
1l58h031t | maths | permutations-and-combinations | divisibility-of-numbers | <p>The total number of 3-digit numbers, whose greatest common divisor with 36 is 2, is ___________.</p> | [] | null | 150 | <p>$$\because$$ x $$\in$$ [100, 999], x $$\in$$ N</p>
<p>Then $${x \over 2}$$ $$\in$$ [50, 499], $${x \over 2}$$ $$\in$$ N</p>
<p>Number whose G.C.D. with 18 is 1 in this range have the required condition. There are 6 such number from 18 $$\times$$ 3 to 18 $$\times$$ 4. Similarly from 18 $$\times$$ 4 to 18 $$\times$$ 5 ......., 26 $$\times$$ 18 to 27 $$\times$$ 18</p>
<p>$$\therefore$$ Total numbers = 24 $$\times$$ 6 + 6 = 150</p>
<p>The extra numbers are 53, 487, 491, 493, 497 and 499.</p> | integer | jee-main-2022-online-26th-june-evening-shift |
1ldsg0itd | maths | permutations-and-combinations | divisibility-of-numbers | <p>The total number of 4-digit numbers whose greatest common divisor with 54 is 2, is __________.</p> | [] | null | 3000 | <p>$$\gcd (a,54) = 2$$ when a is a 4 digit no.</p>
<p>And $$54 = 3 \times 3 \times 3 \times 2$$</p>
<p>So, $$a=$$ all even no. of 4 digits $$-$$ Even multiple of 3 (4 digits)</p>
<p>$$ = 4500 - 1500$$</p>
<p>$$ = 3000$$</p> | integer | jee-main-2023-online-29th-january-evening-shift |
uKpYZD2xI25YxVqSvGYd0 | maths | permutations-and-combinations | factorial | The sum $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $$ is equal to : | [{"identifier": "A", "content": "(11)!"}, {"identifier": "B", "content": "10 $$ \\times $$ (11!)"}, {"identifier": "C", "content": "101 $$ \\times $$ (10!)"}, {"identifier": "D", "content": "11 $$ \\times $$ (11!)\n"}] | ["B"] | null | $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right)} \times r!$$
<br><br>$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2} - 2r} \right]\,.\,r!} $$
<br><br>$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2}\,.\,r!\,\, - \,\,2r\,.\,r!} \right]} $$
<br><br>$$ = \sum\limits_{r = 1}^{10} {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,\,2\sum\limits_{r = 1}^{10} {r\,.\,r!} } $$
<br><br>$$ = \sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,r\,\,.\,\,r!} \right]} - \sum\limits_{r = 1}^{10} {r\,\,.\,\,r!} $$
<br><br>$$ = \left[ {\left( {2.2! - 1.1!} \right) + \left( {3.3! - 2.2!} \right) + .... + \left( {11.11! - 10.10!} \right)} \right]$$
<br><br> $$ - \sum\limits_{r = 1}^{10} {r\,.\,r!} $$
<br><br>$$=$$ 11.11! $$-$$ 1.1! $$-$$ $$\sum\limits_{r = 1}^{10} {r\,.\,r!} $$
<br><br>$$=$$ (11.11! $$-$$ 1) $$-$$ $$\sum\limits_{r = 1}^{10} {\left( {r + 1 - 1} \right)} \,.\,r!$$
<br><br>$$=$$ (11.11! $$-$$ 1) $$-$$ $$\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)r! - r!} \right]} $$
<br><br>$$=$$ (11.11! $$-$$ 1) $$-$$ $$\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)! - r!} \right]} $$
<br><br>$$=$$ (11.11! $$-$$ 1) $$-$$ [(2! $$-$$ 1!) + (3! $$-$$ 2!) + . . . .+ (11! $$-$$ 10!)]
<br><br>$$=$$ (11.11! $$-$$ 1) $$-$$ (11! $$-$$ 1)
<br><br>$$=$$ 11.11! $$-$$ 11!
<br><br>$$=$$ 11! (11 $$-$$ 1)
<br><br>$$=$$ $$10\,.\,\left( {11!} \right)$$ | mcq | jee-main-2016-online-10th-april-morning-slot |
jaoe38c1lscnqca5 | maths | permutations-and-combinations | factorial | <p>Let $$\alpha=\frac{(4 !) !}{(4 !)^{3 !}}$$ and $$\beta=\frac{(5 !) !}{(5 !)^{4 !}}$$. Then :</p> | [{"identifier": "A", "content": "$$\\alpha \\in \\mathbf{N}$$ and $$\\beta \\in \\mathbf{N}$$\n"}, {"identifier": "B", "content": "$$\\alpha \\in \\mathbf{N}$$ and $$\\beta \\notin \\mathbf{N}$$\n"}, {"identifier": "C", "content": "$$\\alpha \\notin \\mathbf{N}$$ and $$\\beta \\in \\mathbf{N}$$\n"}, {"identifier": "D", "content": "$$\\alpha \\notin \\mathbf{N}$$ and $$\\beta \\notin \\mathbf{N}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \alpha=\frac{(4 !) !}{(4 !)^{3 !}}, \beta=\frac{(5 !) !}{(5 !)^{4 !}} \\
& \alpha=\frac{(24) !}{(4 !)^6}, \beta=\frac{(120) !}{(5 !)^{24}}
\end{aligned}$$</p>
<p>Let 24 distinct objects are divided into 6 groups of 4 objects in each group.</p>
<p>No. of ways of formation of group $$=\frac{24 !}{(4 !)^6 .6 !} \in \mathrm{N}$$<?p>
<p>Similarly,</p>
<p>Let 120 distinct objects are divided into 24 groups of 5 objects in each group.</p>
<p>No. of ways of formation of groups</p>
<p>$$=\frac{(120) !}{(5 !)^{24} \cdot 24 !} \in \mathrm{N}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd4qjvk | maths | permutations-and-combinations | factorial | <p>If for some $$m, n ;{ }^6 C_m+2\left({ }^6 C_{m+1}\right)+{ }^6 C_{m+2}>{ }^8 C_3$$ and $${ }^{n-1} P_3:{ }^n P_4=1: 8$$, then $${ }^n P_{m+1}+{ }^{\mathrm{n}+1} C_m$$ is equal to</p> | [{"identifier": "A", "content": "380"}, {"identifier": "B", "content": "376"}, {"identifier": "C", "content": "372"}, {"identifier": "D", "content": "384"}] | ["C"] | null | <p>$$\begin{aligned}
& { }^6 \mathrm{C}_{\mathrm{m}}+2\left({ }^6 \mathrm{C}_{\mathrm{m}+1}\right)+{ }^6 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& { }^7 \mathrm{C}_{\mathrm{m}+1}+{ }^7 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& { }^8 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& \therefore \mathrm{m}=2 \\
& \text { And }{ }^{\mathrm{n}-1} \mathrm{P}_3:{ }^n \mathrm{P}_4=1: 8 \\
& \frac{(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}=\frac{1}{8} \\
& \therefore \mathrm{n}=8 \\
& \therefore{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{m}+1}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{m}}={ }^8 \mathrm{P}_3+{ }^9 \mathrm{C}_2 \\
& =8 \times 7 \times 6+\frac{9 \times 8}{2} \\
& =372
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
txW7MxHrrC595ct0 | maths | permutations-and-combinations | number-of-combinations | If $${}^n{C_r}$$ denotes the number of combination of n things taken r at a time, then the expression $$\,{}^n{C_{r + 1}} + {}^n{C_{r - 1}} + 2\, \times \,{}^n{C_r}$$ equals | [{"identifier": "A", "content": "$$\\,{}^{n + 1}{C_{r + 1}}$$ "}, {"identifier": "B", "content": "$${}^{n + 2}{C_r}$$ "}, {"identifier": "C", "content": "$${}^{n + 2}{C_{r + 1}}$$ "}, {"identifier": "D", "content": "$$\\,{}^{n + 1}{C_r}$$ "}] | ["C"] | null | Arrange it this way,
<br><br>$$^n{C_{r + 1}} + 2.{}^n{C_r} + {}^n{C_{r - 1}}$$
<br><br>$$ = {}^n{C_{r + 1}} + {}^n{C_r} + {}^n{C_r} + {}^n{C_{r - 1}}$$
<br><br>$$\left[ \, \right.$$ Now use the rule,
<br><br> $$\left. {{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}} \right]$$
<br><br>$$ = {}^{n + 1}{C_{r + 1}} + {}^{n + 1}Cr$$
<br><br>$$ = {}^{n + 2}{C_{r + 1}}$$ | mcq | aieee-2003 |
cxhMyG1Oo7EJnOxq | maths | permutations-and-combinations | number-of-combinations | The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is | [{"identifier": "A", "content": "$${}^8{C_3}$$ "}, {"identifier": "B", "content": "21"}, {"identifier": "C", "content": "$${3^8}$$ "}, {"identifier": "D", "content": "5 "}] | ["B"] | null | To distribute n objects among p people where everyone should get atleast one object, then number of ways to distribute those n objects
<br><br>= $${}^{n - 1}{C_{p - 1}}$$
<br><br>For this question, n = 8 and p = 3
<br><br>$$ \therefore $$ Number of ways = $${}^{8 - 1}{C_{3 - 1}}$$ = $${}^7{C_2}$$ = 21 | mcq | aieee-2004 |
WAd9iWoQGJdqmxc0 | maths | permutations-and-combinations | number-of-combinations | At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is | [{"identifier": "A", "content": "5040"}, {"identifier": "B", "content": "6210"}, {"identifier": "C", "content": "385"}, {"identifier": "D", "content": "1110"}] | ["C"] | null | A voter can give vote to either 1 candidate or 2 candidates or 3 candidates or 4 candidates.
<br><br>Case 1 : When he give vote to only 1 candidate then no ways = $${}^{10}{C_1}$$
<br><br>Case 2 : When he give vote to 2 candidates then no ways = $${}^{10}{C_2}$$
<br><br>Case 3 : When he give vote to 3 candidates then no ways = $${}^{10}{C_3}$$
<br><br>Case 4 : When he give vote to 4 candidates then no ways = $${}^{10}{C_4}$$
<br>So, total no of ways he can give votes
<br>= $${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}$$
<br>= 385
<br><br><b>Note :</b> Here we use addition rule as he can vote any one of those four rules. Whenever there is "or" choices, we use addition rule. | mcq | aieee-2006 |
wZ6fkRvOqG6GRUaw | maths | permutations-and-combinations | number-of-combinations | The set S = {1, 2, 3, ........., 12} is to be partitioned into three sets A, B, C of equal size. Thus $$A \cup B \cup C = S,\,A \cap B = B \cap C = A \cap C = \phi $$. The number of ways to partition S is | [{"identifier": "A", "content": "$${{12!} \\over {{{(4!)}^3}}}\\,\\,$$ "}, {"identifier": "B", "content": "$${{12!} \\over {{{(4!)}^4}}}\\,\\,$$ "}, {"identifier": "C", "content": "$${{12!} \\over {3!\\,\\,{{(4!)}^3}}}$$ "}, {"identifier": "D", "content": "$${{12!} \\over {3!\\,\\,{{(4!)}^4}}}$$ "}] | ["A"] | null | <p>The total number of ways is</p>
<p>$${}^{12}{C_4} \times {}^{12 - 4}{C_4} \times {}^{12 - 4 - 4}{C_4} = {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = {{12!} \over {{{(4!)}^3}}}$$</p> | mcq | aieee-2007 |
pW3FNSb5ohyzwnDR | maths | permutations-and-combinations | number-of-combinations | In a shop there are five types of ice-cream available. A child buys six ice-cream.
<br/> <b> Statement - 1: </b> The number of different ways the child can buy the six ice-cream is $${}^{10}{C_5}$$.
<br/> <b> Statement - 2: </b> The number of different ways the child can buy the six ice-cream is equal to the number of different ways of arranging 6 A and 4 B's in a row. | [{"identifier": "A", "content": "Statement - 1 is false, Statement - 2 is true "}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1"}, {"identifier": "C", "content": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1"}, {"identifier": "D", "content": "Statement - 1 is true, Statement - 2 is false"}] | ["A"] | null | <b>Note :</b> n items can be distribute among p persons are $${}^{n + p - 1}{C_{p - 1}}$$ ways.
<br><br>Here n = 6 ice-cream
<br><br>p = 5 types of ice-cream
<br><br>Each ice-cream belongs to one of the 5 ice-cream type. So chosen 6 ice-crean can be divide into 5 types of ice-cream.
<br><br>$$ \therefore $$ The number of different ways the child can buy the six ice-cream is = $${}^{6 + 5 - 1}{C_{5 - 1}}$$ = $${}^{10}{C_4}$$
<br><br>$$ \therefore $$ <b>Statement - 1 is false.</b>
<br><br>Number of different ways of arranging 6 A and 4 B's in a row
<br><br>= $${{10!} \over {6!4!}} = {}^{10}{C_4}$$
<br><br>$$ \therefore $$ <b>Statement - 2 is true.</b> | mcq | aieee-2008 |
gu4xEM8HE3wz9OiX | maths | permutations-and-combinations | number-of-combinations | There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "66"}, {"identifier": "C", "content": "108"}, {"identifier": "D", "content": "3"}] | ["C"] | null | <p>Thus number of ways $$ = ({}^3{C_2}) \times ({}^9{C_2}) = 3 \times {{9 \times 8} \over 2} = 108$$</p> | mcq | aieee-2010 |
hZ0WqGlvxb9cOBdE | maths | permutations-and-combinations | number-of-combinations | These are 10 points in a plane, out of these 6 are collinear, if N is the number of triangles formed by joining these points. then: | [{"identifier": "A", "content": "$$N \\le 100$$ "}, {"identifier": "B", "content": "$$100 < N \\le 140$$ "}, {"identifier": "C", "content": "$$140 < N \\le 190\\,$$ "}, {"identifier": "D", "content": "$$N > 190$$ "}] | ["A"] | null | <p>We need 3 points to create a triangle. With 10 points number of triangle possible $${}^{10}{C_3}$$</p>
<p>Here 6 points are on the same line so we can't make any triangle with those 6 points.</p>
<p> So subtract $${}^{6}{C_3}$$.</p>
<p>$$\therefore$$ $$N = {}^{10}{C_3} - {}^6{C_3}$$</p>
<p>$$ = {{10\,.\,9\,.\,8} \over {1\,.\,2\,.\,3}} - {{6\,.\,5\,.\,4} \over {1\,.\,2\,.\,3}}$$</p>
<p>$$ = 120 - 20$$</p>
<p>$$ = 100$$</p> | mcq | aieee-2011 |
lWJFreINJxGWYGy2 | maths | permutations-and-combinations | number-of-combinations | <br/> <b> Statement - 1: </b> The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is emply is $${}^9{C_3}$$.
<br/> <b> Statement - 2: </b> The number of ways of choosing any 3 places from 9 different places is $${}^9{C_3}$$. | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1."}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false, Statement - 2 is true."}, {"identifier": "D", "content": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1."}] | ["A"] | null | <p>Let XA<sub></sub>, X<sub>B</sub>, X<sub>C</sub> and X<sub>D</sub> represent number of balls present in box A, B, C and D respectively.</p>
<p>As no box can be empty so,</p>
<p>X<sub>A</sub> $$\ge$$ 1, X<sub>B</sub> $$\ge$$ 1, X<sub>C</sub> $$\ge$$ 1 and X<sub>D</sub> $$\ge$$ 1</p>
<p>$$\Rightarrow$$ X<sub>A</sub> $$-$$ 1 $$\ge$$ 0, $$\Rightarrow$$ X<sub>B</sub> $$-$$ 1 $$\ge$$ 0, $$\Rightarrow$$ X<sub>C</sub> $$-$$ 1 $$\ge$$ 0 and $$\Rightarrow$$ X<sub>D</sub> $$-$$ 1 $$\ge$$ 0</p>
<p>t<sub>A</sub> $$\ge$$ 0, t<sub>B</sub> $$\ge$$ 0, t<sub>C</sub> $$\ge$$ 0 and t<sub>D</sub> $$\ge$$ 0</p>
<p>According to the question,</p>
<p>X<sub>A</sub> + X<sub>B</sub> + X<sub>C</sub> + X<sub>D</sub> = 10</p>
<p>$$\Rightarrow$$ (X<sub>A</sub> $$-$$ 1) + (X<sub>B</sub> $$-$$ 1) + (X<sub>C</sub> $$-$$ 1) + (X<sub>D</sub> $$-$$ 1) = 6</p>
<p>$$\Rightarrow$$ t<sub>A</sub> + t<sub>B</sub> + t<sub>C</sub> + t<sub>D</sub> = 6</p>
<p>Now question becomes, box A, B, C, and D can have none or one or more balls and total balls are 6</p>
<p>From formula we know, n things can be distributed among r people in $${}^{n + r - 1}{C_{r - 1}}$$ ways where each people can have either 0 or more things.</p>
<p>$$\therefore$$ 6 balls can be distributed among 4 boxes in $${}^{6 + 4 - 1}{C_{4 - 1}} = {}^9{C_3}$$ ways where each box can have either 0 or more balls.</p>
<p>Therefore, Statement 1 is correct. The number of ways of choosing any 3 places from 9 different places is $${}^9{C_3}$$ ways. But Statement - 2 is not the correct explanation of Statement - 1.</p> | mcq | aieee-2011 |
FrmiixXJ0JcQkToz | maths | permutations-and-combinations | number-of-combinations | Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is: | [{"identifier": "A", "content": "880"}, {"identifier": "B", "content": "629"}, {"identifier": "C", "content": "630"}, {"identifier": "D", "content": "879"}] | ["D"] | null | <p>For alike n objects, number of ways we can select zero or more objects = n + 1 and number of ways we can select at least one object = n</p>
<p>Given 10 identical white balls, 9 identical green balls and 7 identical black balls.</p>
<p>To find number of ways for selecting atleast one ball.</p>
<p>Number of ways to choose zero or more white balls = (10 + 1) [since, all white balls are mutually identical]</p>
<p>Number of ways to choose zero or more green balls = (9 + 1) [since, all green balls are mutually identical]</p>
<p>Number of ways to choose zero or more black balls = (7 + 1) [since, all black balls are mutually identical]</p>
<p>Hence, number of ways to choose zero or more balls of any colour = (10 + 1) (9 + 1) (7 + 1)</p>
<p>Also, number of ways to choose a total of zero balls = 1</p>
<p>Hence, the number, if ways to choose at least one ball (irrespective of any colour) = (10 + 1) (9 + 1) (7 + 1) $$-$$ 1 = 879</p> | mcq | aieee-2012 |
CMaU4ZiR4nN6vcs3UI1qpqahkk8g5fe1k | maths | permutations-and-combinations | number-of-combinations | Let A and B be two sets containing 2 elements and
4 elements respectively. The number of subsets of
A $$ \times $$ B having 3 or more elements is : | [{"identifier": "A", "content": "219"}, {"identifier": "B", "content": "211"}, {"identifier": "C", "content": "256"}, {"identifier": "D", "content": "220"}] | ["A"] | null | A $$ \times $$ B will have 2 $$ \times $$ 4 = 8 elements.
<br><br>The number of subsets having atleast 3 elements
<br><br>= <sup>8</sup>C<sub>3</sub> + <sup>8</sup>C<sub>4</sub> + <sup>8</sup>C<sub>5</sub> + <sup>8</sup>C<sub>6</sub> + <sup>8</sup>C<sub>7</sub> + <sup>8</sup>C<sub>8</sub>
<br><br>= 2<sup>8</sup> – (<sup>8</sup>C<sub>0</sub> + <sup>8</sup>C<sub>1</sub> + <sup>8</sup>C<sub>2</sub>) = 256 – 1 – 8 – 28 = 219 | mcq | jee-main-2013-offline |
foj68NgT56Ij10bv | maths | permutations-and-combinations | number-of-combinations | Let $${T_n}$$ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If $${T_{n + 1}} - {T_n}$$ = 10, then the value of n is : | [{"identifier": "A", "content": "7 "}, {"identifier": "B", "content": "5 "}, {"identifier": "C", "content": "10 "}, {"identifier": "D", "content": "8"}] | ["B"] | null | Number of possible triangle using n vertices = <sup>n</sup>C<sub>3</sub>
<br><br>$$ \therefore $$ T<sub>n</sub> = <sup>n</sup>C<sub>3</sub>
<br><br>then T<sub>n + 1</sub> = <sup>n + 1</sup>C<sub>3</sub>
<br><br>Given, $${T_{n + 1}} - {T_n}$$ = 10
<br><br>$$ \Rightarrow $$ <sup>n + 1</sup>C<sub>3</sub> - <sup>n</sup>C<sub>3</sub> = 10
<br><br>$$ \Rightarrow $$ $${{\left( {n + 1} \right)n\left( {n - 1} \right)} \over 6} - {{n\left( {n - 1} \right)\left( {n - 2} \right)} \over 6}$$ = 10
<br><br>$$ \Rightarrow $$ 3n(n - 1) = 60
<br><br>$$ \Rightarrow $$ n(n - 1) = 20
<br><br>$$ \Rightarrow $$ n<sup>2</sup> - n - 20 = 0
<br><br>$$ \Rightarrow $$ (n - 5)(n + 4) = 0
<br><br>$$ \therefore $$ n = 5 | mcq | jee-main-2013-offline |
JQOZHrRPZakaNrrEKuSe1 | maths | permutations-and-combinations | number-of-combinations | The value of $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}}} \right)$$ is equal to : | [{"identifier": "A", "content": "560 "}, {"identifier": "B", "content": "680"}, {"identifier": "C", "content": "1240"}, {"identifier": "D", "content": "1085"}] | ["B"] | null | We know,
<br><br>$${{{}^n{C_r}} \over {{}^n{C_{r - 1}}}} = {{n - r + 1} \over r}$$
<br><br>$$ \therefore $$ $${{^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}} = {{15 - r + 1} \over r} = {{16 - r} \over r}$$
<br><br>$$ \therefore $$ $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {^{15}{C_{r - 1}}}}} \right)$$
<br><br>$$=$$ $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{16 - r} \over r}} \right)$$
<br><br>$$ = \sum\limits_{r = 1}^{15} {\left( {16r - {r^2}} \right)} $$
<br><br>$$ = 16\sum\limits_{r = 1}^{15} {r - \sum\limits_{r = 1}^{15} {{r^2}} } $$
<br><br>$$ = 16 \times {{15 \times 16} \over 2} - {{15 \times 16 \times 31} \over 6}$$
<br><br>$$=$$ 120 $$ \times $$ 16 $$-$$ 40 $$ \times $$ 31
<br><br>$$=$$ 680 | mcq | jee-main-2016-online-9th-april-morning-slot |
IYJ08gkoyHdlZ5bWWXi8X | maths | permutations-and-combinations | number-of-combinations | If $${{{}^{n + 2}C{}_6} \over {{}^{n - 2}{P_2}}}$$ = 11, then n satisfies the
equation : | [{"identifier": "A", "content": "n<sup>2</sup> + 3n \u2212 108 = 0"}, {"identifier": "B", "content": "n<sup>2</sup> + 5n \u2212 84 = 0\n"}, {"identifier": "C", "content": "n<sup>2</sup> + 2n \u2212 80 = 0"}, {"identifier": "D", "content": "n<sup>2</sup> + n \u2212 110 = 0"}] | ["A"] | null | $${{{}^{n + 2}{C_6}} \over {{}^{n - 2}{P_2}}} = 11$$
<br><br>$$ \Rightarrow $$ $${{\left( {n + 2} \right)!} \over {6!\,\left( {n - 4} \right)!}} = 11\,.\,{{\left( {n - 2} \right)!} \over {\left( {n - 4} \right)!}}$$
<br><br>$$ \Rightarrow $$ (n + 2)! = 11.6! (n $$-$$ 2)!
<br><br>$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11.6!
<br><br>$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11 . 6 . 5 . 4 . 3 . 2 . 1
<br><br>$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11 . 10 . 9 . 8
<br><br>$$ \therefore $$ n = 9
<br><br>This value of n satisfy the equation,
<br><br>n<sup>2</sup> + 3n $$-$$ 108 = 0 | mcq | jee-main-2016-online-10th-april-morning-slot |
hPoq3OSkej5kavXb | maths | permutations-and-combinations | number-of-combinations | A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are
ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X
and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in
this party, is: | [{"identifier": "A", "content": "468"}, {"identifier": "B", "content": "469"}, {"identifier": "C", "content": "484"}, {"identifier": "D", "content": "485"}] | ["D"] | null | <table class="tg">
<tbody><tr>
<th class="tg-0lax"></th>
<th class="tg-s6z2" colspan="2"><span style="font-weight:bold">X(7 Friends)</span></th>
<th class="tg-s6z2" colspan="2"><span style="font-weight:bold">Y(7 Friends)</span></th>
</tr>
<tr>
<td class="tg-0lax"></td>
<td class="tg-s6z2">4 Ladies</td>
<td class="tg-s6z2">3 Men</td>
<td class="tg-s6z2">3 Ladies</td>
<td class="tg-baqh">4 Men</td>
</tr>
<tr>
<td class="tg-0lax"><span style="font-weight:bold">Case 1</span></td>
<td class="tg-s6z2">3</td>
<td class="tg-s6z2">0</td>
<td class="tg-s6z2">0</td>
<td class="tg-baqh">3</td>
</tr>
<tr>
<td class="tg-0lax"><span style="font-weight:bold">Case 2</span></td>
<td class="tg-s6z2">0</td>
<td class="tg-s6z2">3</td>
<td class="tg-s6z2">3</td>
<td class="tg-baqh">0</td>
</tr>
<tr>
<td class="tg-0lax"><span style="font-weight:bold">Case 3</span></td>
<td class="tg-s6z2">2</td>
<td class="tg-s6z2">1</td>
<td class="tg-s6z2">1</td>
<td class="tg-baqh">2</td>
</tr>
<tr>
<td class="tg-0lax"><span style="font-weight:bold">Case 4</span></td>
<td class="tg-s6z2">1</td>
<td class="tg-s6z2">2</td>
<td class="tg-s6z2">2</td>
<td class="tg-baqh">1</td>
</tr>
</tbody></table>
<br><br>In Case 1, Case 2, Case 3 and Case 4, total 6 friends are present and 3 from X and 3 from Y and among those 6 friend 3 are ladies and 3 are men in every case.
<br><br>$$\therefore$$ No of ways 6 friends can be invited =
<br><br>$$({}^4{C_3} \times {}^3{C_0} \times {}^3{C_0} \times {}^4{C_3})$$ + $$({}^4{C_0} \times {}^3{C_3} \times {}^3{C_3} \times {}^4{C_0})$$ + $$\left( {{}^4{C_2} \times {}^3{C_1} \times {}^3{C_1} \times {}^4{C_2}} \right)$$ + $$\left( {{}^4{C_1} \times {}^3{C_2} \times {}^3{C_2} \times {}^4{C_1}} \right)$$
<br><br>= 16 + 1 + 324 + 144 = 485 | mcq | jee-main-2017-offline |
nopfCkMOXIUlIZhNggMvD | maths | permutations-and-combinations | number-of-combinations | If $$\sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r}.{}^{50 - r}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} ,\,\,$$ then K is equal to : | [{"identifier": "A", "content": "2<sup>24</sup> "}, {"identifier": "B", "content": "2<sup>25</sup>$$-$$ 1"}, {"identifier": "C", "content": "2<sup>25</sup>"}, {"identifier": "D", "content": "(25)<sup>2</sup>"}] | ["C"] | null | $$\sum\limits_{r = 0}^{25} {^{50}} {C_r}.{}^{50 - r}{C_{25 - r}}$$
<br><br>$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {r!\left( {50 - r} \right)!}}} \times {{\left( {50 - r} \right)!} \over {\left( {25} \right)!\left( {25 - r} \right)!}}$$
<br><br>$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \over {\left( {25 - r} \right)!\left( {r!} \right)}}} $$
<br><br>$$ = {}^{50}{C_{25}}\sum\limits_{r = 0}^{25} {^{25}} {C_r} = \left( {{2^{25}}} \right){}^{50}{C_{25}}$$
<br><br>$$ \therefore $$ $$K = {2^{25}}$$ | mcq | jee-main-2019-online-10th-january-evening-slot |
d9rYjwxlxiAXsKNTjN8AV | maths | permutations-and-combinations | number-of-combinations | Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by n<sub>i</sub>, the label of the ball drawn from the i<sup>th</sup> box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n<sub>1</sub> < n<sub>2</sub> < n<sub>3</sub> is : | [{"identifier": "A", "content": "164"}, {"identifier": "B", "content": "240"}, {"identifier": "C", "content": "82"}, {"identifier": "D", "content": "120"}] | ["D"] | null | Number of ways = <sup>10</sup>C<sub>3</sub> = 120 | mcq | jee-main-2019-online-12th-january-morning-slot |
wyrLSbxVQxkAIJHo7FrTT | maths | permutations-and-combinations | number-of-combinations | There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "11"}] | ["A"] | null | Let m-men, 2-women
<br><br><sup>m</sup>C<sub>2</sub> $$ \times $$ 2 = <sup>m</sup><sup></sup>C<sub>1</sub> <sup>2</sup>C<sub>1</sub> . 2 + 84
<br><br>m<sup>2</sup> $$-$$ 5m $$-$$ 84 = 0 $$ \Rightarrow $$ (m $$-$$ 12) (m + 7) = 0
<br><br>m = 12 | mcq | jee-main-2019-online-12th-january-evening-slot |
IiAhsoFZAqNRGsldWG3rsa0w2w9jx5e78h8 | maths | permutations-and-combinations | number-of-combinations | The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21
are distinct, is : | [{"identifier": "A", "content": "2<sup>20</sup> - 1"}, {"identifier": "B", "content": "2<sup>20</sup>"}, {"identifier": "C", "content": "2<sup>20</sup> + 1"}, {"identifier": "D", "content": "2<sup>21</sup> "}] | ["B"] | null | <sup>21</sup>C<sub>0</sub> + <sup>21</sup>C<sub>1</sub> + <sup>21</sup>C<sub>2</sub> + ....... + <sup>21</sup>C<sub>10</sub> = $${{{2^{21}}} \over 2} = {2^{20}}$$ | mcq | jee-main-2019-online-12th-april-morning-slot |
mfec6lSyyVk6WkwqUq7k9k2k5fifl9j | maths | permutations-and-combinations | number-of-combinations | The number of ordered pairs (r, k) for which <br/>6.<sup>35</sup>C<sub>r</sub>
= (k<sup>2</sup> - 3). <sup>36</sup>C<sub>r + 1</sub>, where k is an integer, is :
| [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}] | ["D"] | null | 6.<sup>35</sup>C<sub>r</sub>
= (k<sup>2</sup> - 3). <sup>36</sup>C<sub>r + 1</sub>
<br><br>$$ \Rightarrow $$ 6.<sup>35</sup>C<sub>r</sub>
= (k<sup>2</sup> - 3).$${{36} \over {r + 1}}$$<sup>35</sup>C<sub>r</sub>
<br><br>$$ \Rightarrow $$ k<sup>2</sup> - 3 = $${{r + 1} \over 6}$$
<br><br>Possible values of r for integral values of k, are
<br><br>r = 5, 35
<br><br>number of ordered pairs are 4
<br><br>(5, 2), (5, –2), (35, 3), (35, 3) | mcq | jee-main-2020-online-7th-january-evening-slot |
Df2NYHYq0uAHcIQV3H7k9k2k5gjegt5 | maths | permutations-and-combinations | number-of-combinations | If a, b and c are the greatest value of <sup>19</sup>C<sub>p</sub>, <sup>20</sup>C<sub>q</sub>
and <sup>21</sup>C<sub>r</sub> respectively, then : | [{"identifier": "A", "content": "$${a \\over {11}} = {b \\over {22}} = {c \\over {21}}$$"}, {"identifier": "B", "content": "$${a \\over {10}} = {b \\over {22}} = {c \\over {21}}$$"}, {"identifier": "C", "content": "$${a \\over {10}} = {b \\over {11}} = {c \\over {42}}$$"}, {"identifier": "D", "content": "$${a \\over {11}} = {b \\over {22}} = {c \\over {42}}$$"}] | ["D"] | null | (
<sup>19</sup>C<sub>p</sub>)<sub>max</sub> =
<sup>19</sup>C<sub>9</sub> or
<sup>19</sup>C<sub>10</sub> = a
<br><br>(
<sup>20</sup>C<sub>p</sub>)<sub>max</sub> =
<sup>20</sup>C<sub>10</sub> = b
<br><br>(
<sup>21</sup>C<sub>r</sub>)<sub>max</sub> =
<sup>21</sup>C<sub>10</sub> or
<sup>21</sup>C<sub>11</sub> = c
<br><br>1 = $${a \over {^{19}{C_{10}}}} = {b \over {^{20}{C_{10}}}} = {c \over {^{21}{C_{10}}}}$$
<br><br>$$ \Rightarrow $$ $${a \over {11}} = {b \over {22}} = {c \over {42}}$$
| mcq | jee-main-2020-online-8th-january-morning-slot |
TgQzv9xzFjU3klQ0D27k9k2k5h0eczd | maths | permutations-and-combinations | number-of-combinations | An urn contains 5 red marbles, 4 black marbles
and 3 white marbles. Then the number of ways
in which 4 marbles can be drawn so that at the
most three of them are red is ___________. | [] | null | 490 | Here 5 red marbels and 7 non red marbels presents.
<br><br>No of ways 4 marbels can be chosen where atmost 3 red marbels can be present.
<br><br><b>Case 1:</b> When 3 red marbels present
<br><br>No of ways = <sup>5</sup>C<sub>3</sub> $$ \times $$ <sup>7</sup>C<sub>1</sub>
<br><br><b>Case 2:</b> When 2 red marbels present
<br><br>No of ways = <sup>5</sup>C<sub>2</sub> $$ \times $$ <sup>7</sup>C<sub>2</sub>
<br><br><b>Case 3:</b> When 1 red marbels present
<br><br>No of ways = <sup>5</sup>C<sub>1</sub> $$ \times $$ <sup>7</sup>C<sub>3</sub>
<br><br><b>Case 4:</b> When 0 red marbels present
<br><br>No of ways = <sup>5</sup>C<sub>0</sub> $$ \times $$ <sup>7</sup>C<sub>4</sub>
<br><br>$$ \therefore $$ Total number of ways
<br><br> = <sup>5</sup>C<sub>3</sub> $$ \times $$ <sup>7</sup>C<sub>1</sub> + <sup>5</sup>C<sub>2</sub> $$ \times $$ <sup>7</sup>C<sub>2</sub> + <sup>5</sup>C<sub>1</sub> $$ \times $$ <sup>7</sup>C<sub>3</sub> + <sup>5</sup>C<sub>0</sub> $$ \times $$ <sup>7</sup>C<sub>4</sub>
<br><br>= 70 + 210 + 175 + 35
<br><br>= 490 | integer | jee-main-2020-online-8th-january-morning-slot |
SLjj7ptO9uQcIkLlhTjgy2xukfakunkv | maths | permutations-and-combinations | number-of-combinations | A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is __________. | [] | null | 135 | Select any 4 questions in <sup>6</sup>C<sub>4</sub>
ways which are
correct.
<br><br>Answering right option for each question is possible in 1 way.
<br><br>So ways of choosing right option for 4 questions = 1.1.1.1 = (1)<sup>4</sup>
<br><br>Number of ways of choosing wrong option for each question = 3
<br><br>So ways of choosing wrong option for 2 questions = (3)<sup>2</sup>
<br><br>$$ \therefore $$ Required number of ways = <sup>6</sup>C<sub>4</sub>.(1)<sup>4</sup>.(3)<sup>2</sup> = 135 | integer | jee-main-2020-online-4th-september-evening-slot |
FlAz2HmdDhqlvaBRdA1kmjbf0bo | maths | permutations-and-combinations | number-of-combinations | Team 'A' consists of 7 boys and n girls and Team 'B' has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}] | ["C"] | null | Total matches between boys of both team = $${}^7{C_1} \times {}^4{C_1} = 28$$<br><br>Total matches between girls of both team = $${}^n{C_1}\,{}^6{C_1} = 6n$$<br><br>Now, 28 + 6n = 52<br><br>$$ \Rightarrow $$ n = 4 | mcq | jee-main-2021-online-17th-march-morning-shift |
1krzq59sa | maths | permutations-and-combinations | number-of-combinations | If $${}^n{P_r} = {}^n{P_{r + 1}}$$ and $${}^n{C_r} = {}^n{C_{r - 1}}$$, then the value of r is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["C"] | null | $${}^n{P_r} = {}^n{P_{r + 1}} \Rightarrow {{n!} \over {(n - r)!}} = {{n!} \over {(n - r - 1)!}}$$<br><br>$$ \Rightarrow (n - r) = 1$$ .....(1)<br><br>$${}^n{C_r} = {}^n{C_{r - 1}}$$<br><br>$$ \Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}$$<br><br>$$ \Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}$$<br><br>$$ \Rightarrow n - r + 1 = r$$<br><br>$$ \Rightarrow n + 1 = 2r$$ ..... (2)<br><br>From (1) and (2), $$ 2r - 1 - r = 1 \Rightarrow r = 2$$ | mcq | jee-main-2021-online-25th-july-evening-shift |
1l6nox8pu | maths | permutations-and-combinations | number-of-combinations | <p>A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168 , then $$\mathrm{b}+3 \mathrm{~g}$$ is equal to ____________.</p> | [] | null | 17 | <p>$${}^b{C_3}\,.\,{}^g{C_2} = 168$$</p>
<p>$$ \Rightarrow {{b(b - 1)(b - 2)} \over 6}\,.\,{{g(g - 1)} \over 2} = 168$$</p>
<p>$$ \Rightarrow b(b - 1)(b - 2)\,\,\,\,\,\,g(g - 1) = {2^5}{.3^2}.7$$</p>
<p>$$ \Rightarrow b(b - 1)(b - 2)\,\,\,\,\,\,g(g - 1) = 6\,.\,7\,.\,8\,.\,3\,.\,2$$</p>
<p>$$\therefore$$ $$b = 8$$ and $$g = 3$$</p>
<p>$$\therefore$$ $$b + 3g = 17$$</p> | integer | jee-main-2022-online-28th-july-evening-shift |
1l6p3l4h6 | maths | permutations-and-combinations | number-of-combinations | <p>The number of matrices of order $$3 \times 3$$, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is __________.</p> | [] | null | 282 | <p>In a $$3\times3$$ order matrix there are $$9$$ entries.</p>
<p>These nine entries are zero or one.</p>
<p>The sum of positive prime entries are $$2, 3, 5$$ or $$7$$.</p>
<p>Total possible matrices $$ = {{9!} \over {2!\,.\,7!}} + {{9!} \over {3!\,.\,6!}} + {{9!} \over {5!\,.\,4!}} + {{9!} \over {7!\,.\,2!}}$$</p>
<p>$$ = 34 + 84 + 126 + 36$$</p>
<p>$$ = 282$$</p> | integer | jee-main-2022-online-29th-july-morning-shift |
ldoaldz7 | maths | permutations-and-combinations | number-of-combinations | Let $\mathrm{A}=\left[\mathrm{a}_{i j}\right], \mathrm{a}_{i j} \in \mathbb{Z} \cap[0,4], 1 \leq i, j \leq 2$.
<br/><br/>The number of matrices A such that the sum of all entries is a prime number $\mathrm{p} \in(2,13)$ is __________. | [] | null | 204 | $A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{13} & a_{14}\end{array}\right]$
<br/><br/>Such that $\Sigma a_{i i}=3,5,7$ or 11
<br/><br/>Then for sum 3, the possible entries are $(0,0,0,3)$, $(0,0,1,2),(0,1,1,1)$.
<br/><br/>Then total number of possible matrices
$$
=4+12+4
$$
$=20$
<br/><br/>For sum 5 the possible entries are $(0,0,1,4)$, $(0,0,2,3),(0,1,2,2),(0,1,1,3)$ and $(1,1,1,2)$.
<br/><br/>$\therefore $ Total possible matrices $=12+12+12+12+4=52$
<br/><br/>For sum 7 the possible entries are $(0,0,3,4)$, $(0,2,2,3),(0,1,2,4),(0,1,3,3),(1,2,2,2)$, $(1,1,2,3)$ and $(1,1,1,4)$.
<br/><br/>$\therefore $ Total possible matrices $=80$
<br/><br/>For sum 11 the possible entries are $(0,3,4,4)$, $(1,2,4,4),(2,3,3,3),(2,2,3,4)$.
<br/><br/>$\therefore $ Total number of matrices $=52$
<br/><br/>$\therefore $ Total required matrices $=20+52+80+52$
$$
=204
$$ | integer | jee-main-2023-online-31st-january-evening-shift |
1ldu5w6ww | maths | permutations-and-combinations | number-of-combinations | <p>$$\sum\limits_{k = 0}^6 {{}^{51 - k}{C_3}} $$ is equal to :</p> | [{"identifier": "A", "content": "$$\\mathrm{{}^{51}{C_4} - {}^{45}{C_4}}$$"}, {"identifier": "B", "content": "$$\\mathrm{{}^{51}{C_3} - {}^{45}{C_3}}$$"}, {"identifier": "C", "content": "$$\\mathrm{{}^{52}{C_3} - {}^{45}{C_3}}$$"}, {"identifier": "D", "content": "$$\\mathrm{{}^{52}{C_4} - {}^{45}{C_4}}$$"}] | ["D"] | null | $$
\begin{aligned}
& \sum_{\mathrm{k}=0}^6{ }^{51-\mathrm{k}} \mathrm{C}_3 \\\\
& ={ }^{51} \mathrm{C}_3+{ }^{50} \mathrm{C}_3+{ }^{49} \mathrm{C}_3+\ldots+{ }^{45} \mathrm{C}_3 \\\\
& ={ }^{45} \mathrm{C}_3+{ }^{46} \mathrm{C}_3+\ldots \ldots+{ }^{51} \mathrm{C}_3 \\\\
& ={ }^{45} \mathrm{C}_4+{ }^{45} \mathrm{C}_3+{ }^{46} \mathrm{C}_3+\ldots . .+{ }^{51} \mathrm{C}_3-{ }^{45} \mathrm{C}_4 \\\\
& \left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}\right) \\\\
& ={ }^{52} \mathrm{C}_4-{ }^{45} \mathrm{C}_4
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldv2uyem | maths | permutations-and-combinations | number-of-combinations | <p>Let $$x$$ and $$y$$ be distinct integers where $$1 \le x \le 25$$ and $$1 \le y \le 25$$. Then, the number of ways of choosing $$x$$ and $$y$$, such that $$x+y$$ is divisible by 5, is ____________.</p> | [] | null | 120 | Let $x+y=5 \lambda$
<br/><br/>Possible cases are
<br/><br/>$\begin{array}{llc}x & y & \text { Number of ways } \\ 5 \lambda(5,10,15,20,25) & 5 \lambda(5,10,15,20,25) & 20 \\ 5 \lambda+1(1,6,11,16,21) & 5 \lambda+4(4,9,14,19,24) & 25 \\ 5 \lambda+2(2,7,12,17,22) & 5 \lambda+3(3,8,13,18,23) & 25 \\ 5 \lambda+3(3,8,13,18,23) & 5 \lambda+2(2,7,12,17,22) & 25 \\ 5 \lambda+4(4,9,14,19,24) & 5 \lambda+1(1,6,11,16,21) & 25\end{array}$
<br/><br/>Total number of ways $=20+25+25+25+25=120$
<br/><br/><b>Note :</b> In first case total number of ways = 20 as in the question given that the chosen $$x$$ and $$y$$ are distinct integers each time. So when you choose x as 5 then you can't choose y as 5. Possible values of y are (10, 15, 20, 25). So, here four possible pairs of (x, y) possible { (5, 10), (5, 15), (5, 20), (5, 25)}.
<br/><br/>Similarly, four possible pairs of (x, y) possible each time when x = 10, 15, 20 and 25.
<br/><br/>$$ \therefore $$ Total number of ways in the first case = 5 $$ \times $$ 4 = 20. | integer | jee-main-2023-online-25th-january-morning-shift |
1lgvpequ0 | maths | permutations-and-combinations | number-of-combinations | <p>Eight persons are to be transported from city A to city B in three cars of different makes. If each car can accommodate at most three persons, then the number of ways, in which they can be transported, is :</p> | [{"identifier": "A", "content": "560"}, {"identifier": "B", "content": "1680"}, {"identifier": "C", "content": "3360"}, {"identifier": "D", "content": "1120"}] | ["B"] | null | Let $C_1, C_2$ and $C_3$ be the three cars in which 8 person are to be transported from city $A$ to city $B$.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnj292pt/52c17b6b-83cf-405a-ba54-57beea816528/4aa6a710-66ba-11ee-bc75-8500166e483f/file-6y3zli1lnj292pu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnj292pt/52c17b6b-83cf-405a-ba54-57beea816528/4aa6a710-66ba-11ee-bc75-8500166e483f/file-6y3zli1lnj292pu.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Evening Shift Mathematics - Permutations and Combinations Question 32 English Explanation">
<br>Let the number of persons transported in cars $C_1, C_2$ and $C_3$ are 3,3 and 2 respectively.
<br><br>There are total $\frac{8 !}{3 ! 3 ! 2 ! 2 !}$ group
<br><br>$\therefore$ They can travel $\frac{8 !}{3 ! 3 ! 2 ! 2 !} \times 3$ ! ways
or
1680 ways | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgzzp7q6 | maths | permutations-and-combinations | number-of-combinations | <p>Let the number of elements in sets $$A$$ and $$B$$ be five and two respectively. Then the number of subsets of $$A \times B$$ each having at least 3 and at most 6 elements is :</p> | [{"identifier": "A", "content": "782"}, {"identifier": "B", "content": "772"}, {"identifier": "C", "content": "752"}, {"identifier": "D", "content": "792"}] | ["D"] | null | <p>First, let's determine the number of elements in the Cartesian product $$A \times B$$. If set $$A$$ has 5 elements and set $$B$$ has 2 elements, then the number of elements in $$A \times B$$ is:</p>
<p>
<p>$$|A \times B| = |A| \times |B| = 5 \times 2 = 10$$</p>
</p>
<p>We need to find the number of subsets of $$A \times B$$ with at least 3 elements and at most 6 elements. The total number of elements in $$A \times B$$ is 10, so we must consider the subsets containing 3, 4, 5, and 6 elements.</p>
<p>The number of ways to choose $$r$$ elements from a set of 10 elements is given by the binomial coefficient:</p>
<p>
<p>$$\binom{10}{r} = \frac{10!}{r!(10-r)!}$$</p>
</p>
<p>We will find the sum of the binomial coefficients for $$r = 3, 4, 5, 6$$:</p>
<p>First, $$\binom{10}{3}$$:</p>
<p>
<p>$$\binom{10}{3} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$</p>
</p>
<p>Next, $$\binom{10}{4}$$:</p>
<p>
<p>$$\binom{10}{4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$</p>
</p>
<p>Next, $$\binom{10}{5}$$:</p>
<p>
<p>$$\binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$$</p>
</p>
<p>Next, $$\binom{10}{6}$$:</p>
<p>
<p>$$\binom{10}{6} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$</p>
</p>
<p>Finally, we sum these values:</p>
<p>
<p>$$\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} = 120 + 210 + 252 + 210 = 792$$</p>
</p>
<p>Therefore, the number of subsets of $$A \times B$$ each having at least 3 and at most 6 elements is:</p>
<p>Option D - 792</p> | mcq | jee-main-2023-online-8th-april-morning-shift |
lsaoelst | maths | permutations-and-combinations | number-of-combinations | If $\mathrm{n}$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $\mathrm{n}$ is equal to : | [{"identifier": "A", "content": "47"}, {"identifier": "B", "content": "53"}, {"identifier": "C", "content": "51"}, {"identifier": "D", "content": "43"}] | ["C"] | null | <p>To determine the number of ways in which five different employees can be seated in four indistinguishable offices, we will use the concept of partitioning of an integer.</p>
<p>The problem is equivalent to partitioning the number 5 (representing the 5 different employees) into at most 4 parts (representing the 4 indistinguishable offices) where each part represents the number of employees in each office.</p>
<p>The partitions of 5 into at most 4 parts are as follows :</p>
<ol>
<li>$(5, 0, 0, 0)$ – One office has 5 employees, and the other three have none.</li><br>
<li>$(4, 1, 0, 0)$ – One office has 4 employees, another one has 1, and the other two have none.</li><br>
<li>$(3, 2, 0, 0)$ – One office has 3 employees, another one has 2, and the other two have none.</li><br>
<li>$(3, 1, 1, 0)$ – One office has 3 employees, two have 1 each, and one has none.</li><br>
<li>$(2, 2, 1, 0)$ – Two offices have 2 employees each, one has 1, and one has none.</li><br>
<li>$(2, 1, 1, 1)$ – One office has 2 employees, and the other three have 1 each.</li>
</ol>
<p>However, since the offices are indistinguishable, we must not count partitions that differ only by the order of the parts. This means that partitions like $(5, 0, 0, 0)$, $(0, 5, 0, 0)$, $(0, 0, 5, 0)$, and $(0, 0, 0, 5)$ all represent the same scenario and thus are counted as one.</p>
<p>So we end up with 6 distinct partition scenarios.</p>
<p>Now, for each partition, we need to find the number of ways to assign the employees according to each partition :</p>
<ol>
<li>$(5, 0, 0, 0)$ – All employees are in one office. There is 1 way to do this because the offices are indistinguishable.</li><br>
<li>$(4, 1, 0, 0)$ - For the group with 4 employees, there are $${}^5{C_4}$$ ways to choose which 4 out of the 5 will be together. Then the remaining employee is automatically assigned to the second office. So, there are $${}^5{C_4}$$ ways for this partition.</li><br>
<li>$(3, 2, 0, 0)$ - For the group with 3 employees, there are ${}^5{C_3}$ ways to choose them, and for the group with 2 employees, there are $${}^2{C_2}$$ ways to choose them (which is essentially 1 way since the last 2 employees are automatically grouped). But the 2 offices these groups can occupy are indistinguishable, so we do not multiply by 2. So, there are $${}^5{C_3}$$ ways for this partition.</li><br>
<li>$(3, 1, 1, 0)$ - For the group with 3 employees, there are $${}^5{C_3}$$ ways. Then we have two indistinguishable offices, each with 1 employee. The order does not matter as offices are indistinguishable. So, there are simply $${}^5{C_3}$$ ways.</li><br>
<li>$(2, 2, 1, 0)$ – For the first group of 2 employees, there are $${}^5{C_2}$$ ways to choose them. For the next group of 2 employees, there are $${}^3{C_2}$$ ways from the remaining 3. Then, the last employee is alone, and since the offices are indistinguishable, there are no further combinations to consider. However, we have double-counted since the two groups of two are indistinguishable. To adjust for this, we divide by 2. This gives us $\frac{{}^5{C_2} \cdot {}^3{C_2}}{2}$ ways.</li><br>
<li>$(2, 1, 1, 1)$ - For the group with 2 employees, there are ${}^5{C_2}$ ways to choose them. The other three employees are each in their own office, with no further combinations since the offices are indistinguishable, so there are ${}^5{C_2}$ ways for this partition.</li>
</ol>
<p>Let's compute these cases :</p>
<ul>
<li>$(5, 0, 0, 0)$ corresponds to $1$ way.</li><br>
<li>$(4, 1, 0, 0)$ corresponds to ${}^5{C_4} = 5$ ways.</li><br>
<li>$(3, 2, 0, 0)$ corresponds to ${}^5{C_3} = 10$ ways.</li><br>
<li>$(3, 1, 1, 0)$ corresponds to ${}^5{C_3} = 10$ ways.</li><br>
<li>$(2, 2, 1, 0)$ corresponds to $\frac{{}^5{C_2} \cdot {}^3{C_2}}{2} = \frac{10 \cdot 3}{2} = 15$ ways.</li><br>
<li>$(2, 1, 1, 1)$ corresponds to ${}^5{C_2} = 10$ ways.</li>
</ul>
<p>Adding up all the ways we get :</p>
<p>$$ 1 + 5 + 10 + 10 + 15 + 10 = 51 $$</p>
<p>Therefore, the number of ways five different employees can sit into four indistinguishable offices ($\mathrm{n}$) is 51, which corresponds to Option C.</p> | mcq | jee-main-2024-online-1st-february-morning-shift |
lsapvsz6 | maths | permutations-and-combinations | number-of-combinations | The number of elements in the set $\mathrm{S}=\{(x, y, z): x, y, z \in \mathbf{Z}, x+2 y+3 z=42, x, y, z \geqslant 0\}$ equals __________. | [] | null | 169 | $$
x+2 y+3 z=42
$$
<br/><br/>$$
x, y, z \geq 0
$$
<br/><br/>as
<br/><br/>$\begin{array}{ll}z=0 & x+2 y=42 \Rightarrow 22 \text { cases } \\\\ z=1 & x+2 y=39 \Rightarrow 20 \text { cases } \\\\ z=2 & x+2 y=36 \Rightarrow 19 \text { cases } \\\\ z=3 & x+2 y=33 \Rightarrow 17 \text { cases } \\\\ z=4 & x+2 y=30 \Rightarrow 16 \text { cases } \\\\ z=5 & x+2 y=27 \Rightarrow 14 \text { cases }\end{array}$
<br/><br/>$z=6 \quad x+2 y=24 \Rightarrow 13$ cases
<br/><br/>$z=7 \quad x+2 y=21 \Rightarrow 11$ cases
<br/><br/>$z=8 \quad x+2 y=18 \Rightarrow 10$ cases
<br/><br/>$z=9 \quad x+2 y=15 \Rightarrow 8$ cases
<br/><br/>$z=10 x+2 y=12 \Rightarrow 7$ cases
<br/><br/>$z=11 x+2 y=9 \Rightarrow 5$ cases
<br/><br/>$z=12 x+2 y=6 \Rightarrow 4$ cases
<br/><br/>$z=13 x+2 y=3 \Rightarrow 2$ cases
<br/><br/>$z=14 x+2 y=0 \Rightarrow 1$ case
<br/><br/>Total = 169 | integer | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lsfk9w8g | maths | permutations-and-combinations | number-of-combinations | <p>Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "15"}] | ["D"] | null | <p>$$3 \text { Shelf empty: }(8,0,0,0) \rightarrow 1 \text { way }$$</p>
<p>$$\left.2 \text { shelf empty: } \begin{array}{c}
(7,1,0,0) \\
(6,2,0,0) \\
(5,3,0,0) \\
(4,4,0,0)
\end{array}\right] \rightarrow 4 \text { ways }$$</p>
<p>$$\left.1 \text { shelf empty: } \begin{array}{cc}
(6,1,1,0) & (3,3,2,0) \\
(4,2,1,0) & (4,2,2,0) \\
(4,3,0) &
\end{array}\right] \rightarrow 5 \text { ways }$$</p>
<p>$$\left.0 \text { Shelf empty : } \begin{array}{ll}
(1,2,3,2) & (5,1,1,1) \\
(2,2,2,2) \\
(3,3,2,1,1) \\
(4,2,1,1) &
\end{array}\right] \rightarrow 5 \text { ways }$$</p>
<p>Total $$=15$$ ways</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsg4z01p | maths | permutations-and-combinations | number-of-combinations | <p>In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections : $$A, B$$ and $$C$$. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section $$A$$ has 8 questions, section $$B$$ has 6 questions and section $$C$$ has 6 questions, then the total number of ways a student can select 15 questions is __________.</p> | [] | null | 11376 | <p>The problem involves choosing 15 questions out of a total of 20 available questions, with the constraint that at least 4 questions must be chosen from each of the three sections A, B, and C. To evaluate the total number of ways a student can select these questions, we need to consider every possible combination of questions from sections A, B, and C that sum up to 15 questions while respecting the constraints.</p><p>Section A has 8 questions, Section B and C each have 6 questions. The student must choose at least 4 questions from each section, which satisfies the minimum requirement. However, since the student is to attempt a total of 15 questions, there are several combinations to consider, as outlined below:</p><ul><li>Choosing 4 questions from A, 5 from B, and 6 from C</li><li>Choosing 4 questions from A, 6 from B, and 5 from C</li><li>Choosing 7 questions from A, 4 from B, and 4 from C</li><li>Choosing 6 questions from A, 5 from B, and 4 from C</li><li>Choosing 6 questions from A, 4 from B, and 5 from C</li><li>Choosing 5 questions from A, 5 from B, and 5 from C</li><li>Choosing 5 questions from A, 6 from B, and 4 from C</li><li>Choosing 5 questions from A, 4 from B, and 6 from C</li></ul>
<br/>$$
\begin{array}{|c|c|c|l|c|}
\hline \mathrm{A} & \mathrm{B} & \mathrm{C} & \Rightarrow & \begin{array}{c}
\text { No. of } \\
\text { question }
\end{array} \\
\hline 4 & 5 & 6 & \rightarrow & { }^8 C_4{ }^6 C_5{ }^6 C_6 \\
\hline 4 & 6 & 5 & \rightarrow & { }^8 C_4{ }^6 C_6{ }^6 C_5 \\
\hline 7 & 4 & 4 & \rightarrow & { }^8 C_7{ }^6 C_4{ }^6 C_4 \\
\hline 6 & 5 & 4 & \rightarrow & { }^8 C_6{ }^6 C_5{ }^6 C_4 \\
\hline 6 & 4 & 5 & \rightarrow & { }^8 C_6{ }^6 C_4{ }^6 C_5 \\
\hline 5 & 5 & 5 & \rightarrow & { }^8 C_5{ }^6 C_5{ }^6 C_5 \\
\hline 5 & 6 & 4 & \rightarrow & { }^8 C_5{ }^6 C_6{ }^6 C_4 \\
\hline 5 & 4 & 6 & \rightarrow & { }^8 C_5{ }^6 C_4{ }^6 C_6 \\
\hline
\end{array}
$$
<br/><br/>Total ways of select $=11376$ | integer | jee-main-2024-online-30th-january-evening-shift |
lv0vxc34 | maths | permutations-and-combinations | number-of-combinations | <p>There are 5 points $$P_1, P_2, P_3, P_4, P_5$$ on the side $$A B$$, excluding $$A$$ and $$B$$, of a triangle $$A B C$$. Similarly there are 6 points $$\mathrm{P}_6, \mathrm{P}_7, \ldots, \mathrm{P}_{11}$$ on the side $$\mathrm{BC}$$ and 7 points $$\mathrm{P}_{12}, \mathrm{P}_{13}, \ldots, \mathrm{P}_{18}$$ on the side $$\mathrm{CA}$$ of the triangle. The number of triangles, that can be formed using the points $$\mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{18}$$ as vertices, is:</p> | [{"identifier": "A", "content": "751"}, {"identifier": "B", "content": "776"}, {"identifier": "C", "content": "796"}, {"identifier": "D", "content": "771"}] | ["A"] | null | <p>Number of points on side $$A B=5$$</p>
<p>Number of points on side $$B C=6$$</p>
<p>Number of points on side $$A C=7$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwk6ko9r/1d69134a-fee4-473a-84cb-efe47fa9ec3c/ecc716f0-1985-11ef-a7bd-376696e028ce/file-1lwk6ko9s.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwk6ko9r/1d69134a-fee4-473a-84cb-efe47fa9ec3c/ecc716f0-1985-11ef-a7bd-376696e028ce/file-1lwk6ko9s.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Morning Shift Mathematics - Permutations and Combinations Question 11 English Explanation"></p>
<p>Number of ways selecting three points from side</p>
<p>$$A B={ }^5 C_3$$</p>
<p>Number of ways selecting three points from side</p>
<p>$$B C={ }^6 C_3$$</p>
<p>Number of ways selecting three points from side</p>
<p>$$A C={ }^7 C_3$$</p>
<p>Total number of triangle possible formed using the points $$P_1 P_2 \ldots P_{18}$$</p>
<p>$$\begin{aligned}
& ={ }^{18} C_3-{ }^5 C_3-{ }^6 C_3-{ }^7 C_3 \\\\
& =816-10-20-35 \\\\
& =751
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv2er3yn | maths | permutations-and-combinations | number-of-combinations | <p>There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is ________.</p> | [] | null | 5626 | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-0lax{text-align:left;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 596px">
<colgroup>
<col style="width: 150px">
<col style="width: 139px">
<col style="width: 307px">
</colgroup>
<thead>
<tr>
<th class="tg-0lax">Group A</th>
<th class="tg-0lax">Group B</th>
<th class="tg-0lax">Ways</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0lax">$$4m$$<br><br>$$3m+1w$$<br><br>$$2m+2w$$<br><br>$$1m+3w$$<br><br>$$4w$$<br></td>
<td class="tg-0lax">$$4w$$<br><br>$$1m+3w$$<br><br>$$2m+2w$$<br><br>$$3m+w$$<br><br>$$4m$$<br></td>
<td class="tg-0lax">$${ }^4 C_4 \cdot{ }^4 C_4 \quad=1$$<br><br>$${ }^4 C_1 \cdot{ }^5 C_1 \cdot{ }^5 C_1^4 C_3 \quad=400$$<br><br>$${ }^4 C_2 \cdot{ }^5 C_2{ }^5 C_2{ }^4 C_2 \quad=3600$$<br><br>$${ }^4 C_1{ }^5 C_3{ }^5 C_3{ }^4 C_1 \quad=1600$$<br><br>$${ }^5 C_4{ }^5 C_4 \quad=25$$<br></td>
</tr>
</tbody>
</table></p>
<p>$$\begin{aligned}
\text { Total ways } & =1+400+3600+1600+25 \\
& =5626
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-evening-shift |
lv7v47sx | maths | permutations-and-combinations | number-of-combinations | <p>The number of ways of getting a sum 16 on throwing a dice four times is ________.</p> | [] | null | 125 | <p>Number of ways $$=$$ coefficient of $$x^{16}$$ in $$\left(x+x^2+\ldots+\right.$$ $$\left.x^6\right)^4$$</p>
<p>$$=$$ coefficient of $$x^{16}$$ in $$\left(1-x^6\right)^4(1-x)^{-4}$$</p>
<p>$$=$$ coefficient of $$x^{16}$$ in $$\left(1-4 x^6+6 x^{12} \ldots\right)(1-x)^{-4}$$</p>
<p>$$={ }^{15} C_3-4 \cdot{ }^9 C_3+6=125$$</p> | integer | jee-main-2024-online-5th-april-morning-shift |
lv9s208h | maths | permutations-and-combinations | number-of-combinations | <p>Let the set $$S=\{2,4,8,16, \ldots, 512\}$$ be partitioned into 3 sets $$A, B, C$$ with equal number of elements such that $$\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=\mathrm{S}$$ and $$\mathrm{A} \cap \mathrm{B}=\mathrm{B} \cap \mathrm{C}=\mathrm{A} \cap \mathrm{C}=\phi$$. The maximum number of such possible partitions of $$S$$ is equal to:</p> | [{"identifier": "A", "content": "1640"}, {"identifier": "B", "content": "1520"}, {"identifier": "C", "content": "1710"}, {"identifier": "D", "content": "1680"}] | ["D"] | null | <p>Given set $$S=\left\{2^1, 2^2, \ldots 2^9\right\}$$ which consist of 9 elements.</p>
<p>Maximum number of possible partitions (in set $$A, B$$ and $$C$$)</p>
<p>$$={ }^9 C_3 \cdot{ }^6 C_3 \cdot{ }^3 C_3=1680$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
lvb294ut | maths | permutations-and-combinations | number-of-combinations | <p>Let $$0 \leq r \leq n$$. If $${ }^{n+1} C_{r+1}:{ }^n C_r:{ }^{n-1} C_{r-1}=55: 35: 21$$, then $$2 n+5 r$$ is equal to :</p> | [{"identifier": "A", "content": "62"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "55"}, {"identifier": "D", "content": "50"}] | ["D"] | null | <p>Given $0 \leq r \leq n$. If $\binom{n+1}{r+1} : \binom{n}{r} : \binom{n-1}{r-1} = 55 : 35 : 21$, then we are to determine the value of $2n + 5r$.</p>
<h3>Step-by-Step Solution:</h3>
<ol>
<li>Write the given proportions involving binomial coefficients:</li>
</ol>
<p>$ \frac{n+1}{r+1} \times \binom{n}{r} : \binom{n}{r} : \frac{r}{n} \times \binom{n}{r} = 55 : 35 : 21 $</p>
<ol>
<li>Simplify the proportions:</li>
</ol>
<p>$ \frac{n+1}{r+1} = \frac{55}{35} \quad \text{and} \quad \frac{n}{r} = \frac{35}{21} $</p>
<ol>
<li>From the simplified ratios, we establish the following two equations:</li>
</ol>
<p>$ \frac{n+1}{r+1} = \frac{11}{7} \quad \Rightarrow \quad 7(n+1) = 11(r+1) \quad \Rightarrow \quad 7n - 11r = 4 \quad \text{.... (1)} $</p>
<p>$ \frac{n}{r} = \frac{5}{3} \quad \Rightarrow \quad 3n = 5r \quad \Rightarrow \quad 3n - 5r = 0 \quad \text{.... (2)} $</p>
<ol>
<li>Solve equations (1) and (2) simultaneously:</li>
</ol>
<ul>
<li>From equation (2), solve for $n$:</li>
</ul>
<p>$ 3n = 5r \quad \Rightarrow \quad n = \frac{5r}{3} $</p>
<ul>
<li>Substitute $n$ into equation (1):</li>
</ul>
<p>$ 7 \left(\frac{5r}{3}\right) - 11r = 4 \quad \Rightarrow \quad \frac{35r}{3} - 11r = 4 $</p>
<p>$ \frac{35r - 33r}{3} = 4 \quad \Rightarrow \quad \frac{2r}{3} = 4 \quad \Rightarrow \quad 2r = 12 \quad \Rightarrow \quad r = 6 $</p>
<ul>
<li>Substituting $r$ back to find $n$:</li>
</ul>
<p>$ n = \frac{5 \times 6}{3} = 10 $</p>
<ol>
<li>Compute $2n + 5r$:</li>
</ol>
<p>$ 2n + 5r = 2 \times 10 + 5 \times 6 = 20 + 30 = 50 $</p>
<p>Thus, the value of $2n + 5r$ is 50.</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
MWbLRehjPiKaEgcg | maths | permutations-and-combinations | number-of-permutations | If the letter of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number | [{"identifier": "A", "content": "601"}, {"identifier": "B", "content": "600"}, {"identifier": "C", "content": "603"}, {"identifier": "D", "content": "602"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263736/exam_images/ru0tccwaposwg6xx5sll.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264497/exam_images/bawohr8zdetwd8ozm1dc.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264658/exam_images/ku4u7xqus8aqacpzqeks.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266868/exam_images/udrmfpmvl6yghk6upkr8.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265634/exam_images/trl438xafzvaoxkw1zrx.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="AIEEE 2005 Mathematics - Permutations and Combinations Question 170 English Explanation 1"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267385/exam_images/ibozet9suohl1dwpkuih.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264951/exam_images/haqurugpgxhpkdjhjioo.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265218/exam_images/lmrb2thuizvxcjaivnbp.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265445/exam_images/mxzfpaabesix90enqhto.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266793/exam_images/m8iisj7kowslkqzbmaox.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="AIEEE 2005 Mathematics - Permutations and Combinations Question 170 English Explanation 2"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265618/exam_images/ivzvumhobexplzmok9t6.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265587/exam_images/a25zofjc4urwvsibdct8.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266872/exam_images/ozgit8mdsewsutv0ht28.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265586/exam_images/yd4iraelcfqxnujronks.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265597/exam_images/xu3tqfawcekq0xckw7ej.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264419/exam_images/tgwhspuv73d1hl6mxgsx.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="AIEEE 2005 Mathematics - Permutations and Combinations Question 170 English Explanation 3"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266063/exam_images/ydx4sjeubbgl11t7fahx.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267532/exam_images/rs8ed4birhdkcyra9bno.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267308/exam_images/yv6tg2jkqw5e5zkfhpjn.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263758/exam_images/eas1iwl5pvqf8clrko2u.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267734/exam_images/vnt57cckvux7ausi0knw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="AIEEE 2005 Mathematics - Permutations and Combinations Question 170 English Explanation 4"></picture> | mcq | aieee-2005 |
KscQ5CpVa6h3aZAR | maths | permutations-and-combinations | number-of-permutations | From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is : | [{"identifier": "A", "content": "at least 500 but less than 750"}, {"identifier": "B", "content": "at least 750 but less than 1000"}, {"identifier": "C", "content": "at least 1000"}, {"identifier": "D", "content": "less than 500"}] | ["C"] | null | From 6 different novels 4 novels can be chosen = $${}^6{C_4}$$ ways
<br><br>And from 4 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways
<br><br>$$\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \times {}^3{C_1}$$ ways
<br><br>Let 4 novels are N<sub>1</sub>, N<sub>2</sub>, N<sub>3</sub>, N<sub>4</sub> and 1 dictionary is D<sub>1</sub>.
<br><br> Dictionary should be in the middle. So the arrangement will be like this
<br><br>_ _ D<sub>1</sub> _ _
<br><br>On those 4 blank places 4 novels N<sub>1</sub>, N<sub>2</sub>, N<sub>3</sub>, N<sub>4</sub> can be placed. And 4 novels can be arrange $$4!$$ ways.
<br><br>$$\therefore$$ Total no of ways = $${}^6{C_4} \times {}^3{C_1}$$$$ \times 4!$$ = 1080 | mcq | aieee-2009 |
cpWdidtToPeRE0b1 | maths | permutations-and-combinations | number-of-permutations | The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is: | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "72"}, {"identifier": "C", "content": "216"}, {"identifier": "D", "content": "192"}] | ["D"] | null | <p>For a four digit number the first place can be filled in 3 ways with 6 or 7 or 8 and the remaining four places in 4! ways i.e., 3 $$\times$$ 4! = 72.</p>
<p>For a five digit number it can be arranged in 5! ways,</p>
<p>$$\therefore$$ total number of integers = (72 + 120) = 192.</p> | mcq | jee-main-2015-offline |
IplDhMsmdCoRdapu | maths | permutations-and-combinations | number-of-permutations | If all the words (with or without meaning) having five letters,formed using the letters of the word SMALL and arranged as in a dictionary, then the position of the word SMALL is : | [{"identifier": "A", "content": "$${46^{th}}$$ "}, {"identifier": "B", "content": "$${59^{th}}$$"}, {"identifier": "C", "content": "$${52^{nd}}$$"}, {"identifier": "D", "content": "$${58^{th}}$$"}] | ["D"] | null | <p>Clearly, number of words start with $$A = {{4!} \over {2!}} = 12$$</p>
<p>Number of words start with $$L = 4! = 24$$</p>
<p>Number of words start with $$M = {{4!} \over {2!}} = 12$$</p>
<p>Number of words start with $$SA = {{3!} \over {2!}} = 3$$</p>
<p>Number of words start with $$SL = 3! = 6$$</p>
<p>Note that, next word will be "SMALL"</p>
<p>Hence, the position of word "SMALL" is 58th.</p> | mcq | jee-main-2016-offline |
4sIda7kghDbn5SBTSeZMx | maths | permutations-and-combinations | number-of-permutations | If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is : | [{"identifier": "A", "content": "44<sup>th</sup> "}, {"identifier": "B", "content": "45<sup>th</sup> "}, {"identifier": "C", "content": "46<sup>th</sup> "}, {"identifier": "D", "content": "47<sup>th</sup> "}] | ["C"] | null | <p>To find the position of the word QUEEN:</p>
<p>$$\bullet$$ The number of words starting with E is 4! = 24.</p>
<p>$$\bullet$$ The number of words starting with N is $${{4!} \over 2} = 12$$.</p>
<p>$$\bullet$$ The number of words starting with QE is 3! = 6.</p>
<p>$$\bullet$$ Number of words starting with QN is $${{3!} \over 2} = 3$$.</p>
<p>Therefore, the position of the word QUEEN is next to the sum, 24 + 12 + 6 + 3 = 45.</p>
<p>That is, the word 'QUEEN' will be on 46th position.</p> | mcq | jee-main-2017-online-8th-april-morning-slot |
dE3mDjQalcUd2nH6 | maths | permutations-and-combinations | number-of-permutations | From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and
arranged in a row on a shelf so that the dictionary is always in the middle. The number of such
arrangements is : | [{"identifier": "A", "content": "at least 750 but less than 1000"}, {"identifier": "B", "content": "at least 1000"}, {"identifier": "C", "content": "less than 500"}, {"identifier": "D", "content": "at least 500 but less than 750"}] | ["B"] | null | From 6 different novels 4 novels can be chosen = $${}^6{C_4}$$ ways
<br><br>And from 3 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways
<br><br>$$\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \times {}^3{C_1}$$ ways
<br><br>Let 4 novels are N<sub>1</sub>, N<sub>2</sub>, N<sub>3</sub>, N<sub>4</sub> and 1 dictionary is D<sub>1</sub>.
<br><br> Dictionary should be in the middle. So the arrangement will be like this
<br><br>_ _ D<sub>1</sub> _ _
<br><br>On those 4 blank places 4 novels N<sub>1</sub>, N<sub>2</sub>, N<sub>3</sub>, N<sub>4</sub> can be placed. And 4 novels can be arrange $$4!$$ ways.
<br><br>$$\therefore$$ Total no of ways = $${}^6{C_4} \times {}^3{C_1}$$$$ \times 4!$$ = 1080 | mcq | jee-main-2018-offline |
XLd4i0VsYtBkRaHaQmAcd | maths | permutations-and-combinations | number-of-permutations | n$$-$$digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "9"}] | ["B"] | null | In n digit number first place can be filled with any one of 2, 5, 7. So no of ways first digit can be filled = 3<br><br>
Similarly,<br>
no of ways 2nd digit can be filled = 3 ways<br>
.<br>
.<br>
.<br>
.<br>
- - - - - - nth - - - - - - - = 3 ways<br>
<br>
$$ \therefore $$ Total numbers = 3 $$ \times $$ 3 $$ \times $$ 3 .... n times<br>
= 3<sup>n</sup><br>
$$ \therefore $$ According to question, for smallest value of n, <br><br>
3<sup>n</sup> > 900<br><br>
3<sup>6</sup> = 729 < 900<br><br>
3<sup>7</sup> = 2187 > 900<br><br>
$$ \therefore $$ n = 7 | mcq | jee-main-2018-online-15th-april-morning-slot |
xe5ZfgVxYCdbpMKEGIuOF | maths | permutations-and-combinations | number-of-permutations | The number of four letter words that can be formed using the letters of the word <b>BARRACK</b> is : | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "144"}, {"identifier": "C", "content": "264"}, {"identifier": "D", "content": "270"}] | ["D"] | null | <b>Case 1 :</b>
<br><br>When all the four letters different then no of words
<br>= <sup>5</sup>C<sub>4</sub> $$ \times $$4!
<br><br><b>Case 2 :</b>
<br><br>When out of four letters two letters are R and other two different letters are chosen from B, A, C, K then the no of words <br>= <sup>4</sup>C<sub>2</sub> $$ \times $$ $${{4!} \over {2!}}$$ = 72
<br><br><b>Case 3 :</b>
<br><br>When out of four letters two letters are A and other two different letters are chosen from B, R, C, K then the no of words <br>= <sup>4</sup>C<sub>2</sub> $$ \times $$ $${{4!} \over {2!}}$$ = 72
<br><br><b>Case 4 :</b>
<br><br>When word is formed using two R and two A then number of words<br> = $${{4!} \over {2!2!}}$$ = 6
<br><br>So, total number of 4 letters words possible = 120 + 72 + 72 + 6 = 270 | mcq | jee-main-2018-online-15th-april-evening-slot |
lrAD8xOCEl2vEKUBAe7k9k2k5e2s565 | maths | permutations-and-combinations | number-of-permutations | Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is : | [{"identifier": "A", "content": "$${5 \\over 2}\\left( {6!} \\right)$$"}, {"identifier": "B", "content": "$${6!}$$"}, {"identifier": "C", "content": "5<sup>6</sup>"}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {6!} \\right)$$"}] | ["A"] | null | Here none number repeats more than once.
<br><br>We can choose the number which repeats more than once among 1, 3, 5, 7, 9 in <sup>5</sup>C<sub>1</sub> ways.
<br><br>Let number 3 repeats more than once. So six digits are 1, 3, 3, 5, 7, 9.
<br><br>We can arrange those six digits in $${{6!} \over {2!}}$$ ways.
<br><br>$$ \therefore $$ Total six digit numbers = <sup>5</sup>C<sub>1</sub> $$ \times $$ $${{6!} \over {2!}}$$ = $${5 \over 2}\left( {6!} \right)$$ | mcq | jee-main-2020-online-7th-january-morning-slot |
qbRvgJn6hv5comatCj7k9k2k5hketjb | maths | permutations-and-combinations | number-of-permutations | The number of 4 letter words (with or without
meaning) that can be formed from the eleven
letters of the word 'EXAMINATION' is
_______. | [] | null | 2454 | 2A, 2I, 2N, E, X, M, T, O
<br><br>To form four letter words
<br><br><b>Case 1 :</b> All same ( not possible)
<br><br><b>Case 2 :</b> 1 different, 3 same (not possible)
<br><br><b>Case 3 :</b> 2 different, 2 same
<br><br>= <sup>3</sup>C<sub>1</sub> $$ \times $$ <sup>7</sup>C<sub>2</sub> $$ \times $$ $${{4!} \over {2!}}$$ = 756
<br><br><b>Case 4 :</b> 2 same of one kind, 2 same same of other kind
<br><br>= <sup>3</sup>C<sub>2</sub> $$ \times $$ $${{4!} \over {2!2!}}$$ = 18
<br><br><b>Case 5 :</b> All letters are different
<br><br>= <sup>8</sup>C<sub>4</sub> $$ \times $$ 4! = 1680
<br><br>$$ \therefore $$ Total ways = 1680 + 756 + 18 = 2454
| integer | jee-main-2020-online-8th-january-evening-slot |
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