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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-wid... | 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-wid... | 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 me... | 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<... | 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 Explanat... | 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;... | 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>... | 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 Explanati... | 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 Comb... | 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 di... | 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-seri... | 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</... | [] | 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 $b_{2} b_{3} b_{4}$... | 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/><b... | 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... | 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/>
$$
\begin{aligned}
&\therefore(\mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g})-(\mathrm{b}+\mathrm{d}+\mathrm{f})=11 \mathrm{x} \\\\
&\mat... | integer | jee-main-2022-online-24th-june-evening-shift |
1l5w0pg9y | maths | permutations-and-combinations | conditional-permutations | <p>The number of 6-digit numbers made by using the digits 1, 2, 3, 4, 5, 6, 7, without repetition and which are multiple of 15 is ____________.</p> | [] | null | 360 | <p>A number is multiple of 15 when the number is divisible by 5 and sum of digits of the number is divisible by 3.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l65uqr2z/3aa8732e-2bc7-4c8a-a45b-f53f925f4291/0f4fb6b0-0ee7-11ed-a7de-eff776fdb55c/file-1l65uqr30.png?format=png" data-orsrc="https:/... | integer | jee-main-2022-online-30th-june-morning-shift |
1l6gjaknm | maths | permutations-and-combinations | conditional-permutations | <p>The number of 5-digit natural numbers, such that the product of their digits is 36 , is __________.</p> | [] | null | 180 | <p>Factors of 36 = 2<sup>2</sup> . 3<sup>2</sup> . 1</p>
<p>Five-digit combinations can be</p>
<p>(1, 2, 2, 3, 3) (1, 4, 3, 3, 1), (1, 9, 2, 2, 1)</p>
<p>(1, 4, 9, 11) (1, 2, 3, 6, 1) (1, 6, 6, 1, 1)</p>
<p>i.e., total numbers</p>
<p>$${{5!} \over {2!2!}} + {{5!} \over {2!2!}} + {{5!} \over {2!2!}} + {{5!} \over {3!}} ... | 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-1l97twac... | 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=... | 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 t... | 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... | 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/... | 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 separatel... | 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 $$0, 1, 2, 3, 4... | integer | jee-main-2023-online-15th-april-morning-shift |
1lgoy7m2y | maths | permutations-and-combinations | conditional-permutations | <p>Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits $$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) \... | 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 cann... | 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... | 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/>$$
\begin{array}{ll}
& \therefore n(p)=5! \times 1 \\\\
& \Rightarrow n(q)=4! \times 1 \\\\
& \Rightarrow n(p \cap q)=2!
\end{array}... | integer | jee-main-2023-online-10th-april-morning-shift |
1lgylnv0d | maths | permutations-and-combinations | conditional-permutations | <p>If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which $$\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 !}... | 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 $r$ t... | 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/>$$
\begin{aligned}
& ={ }^3 C_2 \times{ }^4 C_2 \times 4 ! \\\\
& ={ }^3 C_1 \t... | integer | jee-main-2023-online-6th-april-evening-shift |
1lguwuo8m | maths | permutations-and-combinations | dearrangement | <p>In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is _________.</p> | [] | null | 44 | This problem can be solved using the concept of derangements, which is a permutation of objects where no object appears in its original position. In this case, we have 5 students who should not sit in their allotted seats.
<br/><br/>The formula for calculating the number of derangements (also known as subfactorials) ... | integer | jee-main-2023-online-11th-april-morning-shift |
38m7NMVKw7MZYgve | maths | permutations-and-combinations | divisibility-of-numbers | The sum of integers from 1 to 100 that are divisible by 2 or 5 is : | [{"identifier": "A", "content": "3000"}, {"identifier": "B", "content": "3050"}, {"identifier": "C", "content": "3600"}, {"identifier": "D", "content": "3250"}] | ["B"] | null | According to this question, any number between 1 to 100 should be divisible by 2 or 5 but not by 2$$ \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 $$ Sum of the nu... | mcq | aieee-2002 |
vKUBJgogUkTqMHWSIq45x | maths | permutations-and-combinations | divisibility-of-numbers | The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is : | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "30"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "48"}] | ["B"] | null | Here number should be divisible by 3, that means sum of numbers should be divisible by 3.
<br><br>Possible 4 digits among 0, 1, 2, 3, 4 which are divisible by 3 are
<br><br>(1)$$\,\,\,\,$$ (0, 2, 3, 4) Sum of digits = 0 + 2 + 3 +4 = 9 (divisible by 3)
<br><br>(2) $$\,\,\,\,$$ (0, 1, 2, 3) Sum of digits = 0 + 1 + 2 + 3... | mcq | jee-main-2018-online-16th-april-morning-slot |
OjrcXgele0CMwAVPCy1kluvyg3b | maths | permutations-and-combinations | divisibility-of-numbers | A natural number has prime factorization given by n = 2<sup>x</sup>3<sup>y</sup>5<sup>z</sup>, where y and z are such <br/>that y + z = 5 and y<sup>$$-$$1</sup> + z<sup>$$-$$1</sup> = $${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 $$ (y $$-$$ ... | mcq | jee-main-2021-online-26th-february-evening-slot |
1krygm1ew | maths | permutations-and-combinations | divisibility-of-numbers | Let n be a non-negative integer. Then the number of divisors of the form "4n + 1" of the number (10)<sup>10</sup> . (11)<sup>11</sup> . (13)<sup>13</sup> is equal to __________. | [] | null | 924 | N = 2<sup>10</sup> $$\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... | 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... | 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>$$ = \sum\limits_{r = 1}^{... | mcq | jee-main-2016-online-10th-april-morning-slot |
jaoe38c1lscnqca5 | maths | permutations-and-combinations | factorial | <p>Let $$\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}$$\n"}, {"identifier": "D",... | ["A"] | null | <p>$$\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 !}{(... | 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>$$\begin{aligned}
& { }^6 \mathrm{C}_{\mathrm{m}}+2\left({ }^6 \mathrm{C}_{\mathrm{m}+1}\right)+{ }^6 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& { }^7 \mathrm{C}_{\mathrm{m}+1}+{ }^7 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& { }^8 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& \therefore \mathrm{m}... | mcq | jee-main-2024-online-31st-january-evening-shift |
txW7MxHrrC595ct0 | maths | permutations-and-combinations | number-of-combinations | If $${}^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... | 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 n... | 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 ... | [{"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 i... | ["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 ... | 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>$$ = {{10\,.\,9\,.\,8} \ov... | mcq | aieee-2011 |
lWJFreINJxGWYGy2 | maths | permutations-and-combinations | number-of-combinations | <br/> <b> Statement - 1: </b> The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is emply is $${}^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}$$. | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1."}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false, Statement - 2 is true."}, {"identi... | ["A"] | null | <p>Let XA<sub></sub>, X<sub>B</sub>, X<sub>C</sub> and X<sub>D</sub> represent number of balls present in box A, B, C and D respectively.</p>
<p>As no box can be empty so,</p>
<p>X<sub>A</sub> $$\ge$$ 1, X<sub>B</sub> $$\ge$$ 1, X<sub>C</sub> $$\ge$$ 1 and X<sub>D</sub> $$\ge$$ 1</p>
<p>$$\Rightarrow$$ X<sub>A</sub> $$... | 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 c... | 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> – (<s... | 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<s... | 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>$$ \therefore $$ $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{... | mcq | jee-main-2016-online-9th-april-morning-slot |
IYJ08gkoyHdlZ5bWWXi8X | maths | permutations-and-combinations | number-of-combinations | If $${{{}^{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 $$ (n + 2)! = 11.6! (n $$... | 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 par... | [{"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 La... | 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>$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \o... | mcq | jee-main-2019-online-10th-january-evening-slot |
d9rYjwxlxiAXsKNTjN8AV | maths | permutations-and-combinations | number-of-combinations | Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by n<sub>i</sub>, the label of the ball drawn from the i<sup>th</sup> box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n<sub>1</sub> < n<sub>... | [{"identifier": "A", "content": "164"}, {"identifier": "B", "content": "240"}, {"identifier": "C", "content": "82"}, {"identifier": "D", "content": "120"}] | ["D"] | null | Number of ways = <sup>10</sup>C<sub>3</sub> = 120 | mcq | jee-main-2019-online-12th-january-morning-slot |
wyrLSbxVQxkAIJHo7FrTT | maths | permutations-and-combinations | number-of-combinations | There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "11"}] | ["A"] | null | Let m-men, 2-women
<br><br><sup>m</sup>C<sub>2</sub> $$ \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}$$
<br><br>Possible values of r for integr... | mcq | jee-main-2020-online-7th-january-evening-slot |
Df2NYHYq0uAHcIQV3H7k9k2k5gjegt5 | maths | permutations-and-combinations | number-of-combinations | If a, b and c are the greatest value of <sup>19</sup>C<sub>p</sub>, <sup>20</sup>C<sub>q</sub>
and <sup>21</sup>C<sub>r</sub> respectively, then : | [{"identifier": "A", "content": "$${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 {1... | ["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... | 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 pres... | 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 o... | 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>$$ \Rightarrow {1 \over {r(n - r)!}} = {1 \... | mcq | jee-main-2021-online-25th-july-evening-shift |
1l6nox8pu | maths | permutations-and-combinations | number-of-combinations | <p>A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168 , then $$\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$$ and ... | integer | jee-main-2022-online-28th-july-evening-shift |
1l6p3l4h6 | maths | permutations-and-combinations | number-of-combinations | <p>The number of matrices of order $$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!}}$$</p>
<... | integer | jee-main-2022-online-29th-july-morning-shift |
ldoaldz7 | maths | permutations-and-combinations | number-of-combinations | Let $\mathrm{A}=\left[\mathrm{a}_{i j}\right], \mathrm{a}_{i j} \in \mathbb{Z} \cap[0,4], 1 \leq i, j \leq 2$.
<br/><br/>The number of matrices A such that the sum of all entries is a prime number $\mathrm{p} \in(2,13)$ is __________. | [] | null | 204 | $A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{13} & a_{14}\end{array}\right]$
<br/><br/>Such that $\Sigma a_{i i}=3,5,7$ or 11
<br/><br/>Then for sum 3, the possible entries are $(0,0,0,3)$, $(0,0,1,2),(0,1,1,1)$.
<br/><br/>Then total number of possible matrices
$$
=4+12+4
$$
$=20$
<br/><br/>For sum 5 the poss... | integer | jee-main-2023-online-31st-january-evening-shift |
1ldu5w6ww | maths | permutations-and-combinations | number-of-combinations | <p>$$\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+{ ... | 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,... | 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=... | 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 $$... | mcq | jee-main-2023-online-8th-april-morning-shift |
lsaoelst | maths | permutations-and-combinations | number-of-combinations | If $\mathrm{n}$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $\mathrm{n}$ is equal to : | [{"identifier": "A", "content": "47"}, {"identifier": "B", "content": "53"}, {"identifier": "C", "content": "51"}, {"identifier": "D", "content": "43"}] | ["C"] | null | <p>To determine the number of ways in which five different employees can be seated in four indistinguishable offices, we will use the concept of partitioning of an integer.</p>
<p>The problem is equivalent to partitioning the number 5 (representing the 5 different employees) into at most 4 parts (representing the 4 in... | mcq | jee-main-2024-online-1st-february-morning-shift |
lsapvsz6 | maths | permutations-and-combinations | number-of-combinations | The number of elements in the set $\mathrm{S}=\{(x, y, z): x, y, z \in \mathbf{Z}, x+2 y+3 z=42, x, y, z \geqslant 0\}$ equals __________. | [] | null | 169 | $$
x+2 y+3 z=42
$$
<br/><br/>$$
x, y, z \geq 0
$$
<br/><br/>as
<br/><br/>$\begin{array}{ll}z=0 & x+2 y=42 \Rightarrow 22 \text { cases } \\\\ z=1 & x+2 y=39 \Rightarrow 20 \text { cases } \\\\ z=2 & x+2 y=36 \Rightarrow 19 \text { cases } \\\\ z=3 & x+2 y=33 \Rightarrow 17 \text { cases } \\\\ z=4 & x+2 y=30 \Rightarro... | integer | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lsfk9w8g | maths | permutations-and-combinations | number-of-combinations | <p>Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "15"}] | ["D"] | null | <p>$$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) & ... | 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 ... | [] | 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 q... | 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 si... | [{"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="... | 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,... | 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>$$={ }^{15} C_3-4 \cdot{ }^9 C_3+6=1... | integer | jee-main-2024-online-5th-april-morning-shift |
lv9s208h | maths | permutations-and-combinations | number-of-combinations | <p>Let the set $$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$$. The maximum number of such possible parti... | [{"identifier": "A", "content": "1640"}, {"identifier": "B", "content": "1520"}, {"identifier": "C", "content": "1710"}, {"identifier": "D", "content": "1680"}] | ["D"] | null | <p>Given set $$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$$ is equal to :</p> | [{"identifier": "A", "content": "62"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "55"}, {"identifier": "D", "content": "50"}] | ["D"] | null | <p>Given $0 \leq r \leq n$. If $\binom{n+1}{r+1} : \binom{n}{r} : \binom{n-1}{r-1} = 55 : 35 : 21$, then we are to determine the value of $2n + 5r$.</p>
<h3>Step-by-Step Solution:</h3>
<ol>
<li>Write the given proportions involving binomial coefficients:</li>
</ol>
<p>$ \frac{n+1}{r+1} \times \binom{n}{r} : \binom... | mcq | jee-main-2024-online-6th-april-evening-shift |
MWbLRehjPiKaEgcg | maths | permutations-and-combinations | number-of-permutations | If the letter of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number | [{"identifier": "A", "content": "601"}, {"identifier": "B", "content": "600"}, {"identifier": "C", "content": "603"}, {"identifier": "D", "content": "602"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263736/exam_images/ru0tccwaposwg6xx5sll.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264497/exam_images/bawohr8zdetwd8ozm1dc.webp"><source media="(max-wid... | mcq | aieee-2005 |
KscQ5CpVa6h3aZAR | maths | permutations-and-combinations | number-of-permutations | From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is : | [{"identifier": "A", "content": "at least 500 but less than 750"}, {"identifier": "B", "content": "at least 750 but less than 1000"}, {"identifier": "C", "content": "at least 1000"}, {"identifier": "D", "content": "less than 500"}] | ["C"] | null | From 6 different novels 4 novels can be chosen = $${}^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}$$ ways
<br><br>Let 4 nove... | mcq | aieee-2009 |
cpWdidtToPeRE0b1 | maths | permutations-and-combinations | number-of-permutations | The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is: | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "72"}, {"identifier": "C", "content": "216"}, {"identifier": "D", "content": "192"}] | ["D"] | null | <p>For a four digit number the first place can be filled in 3 ways with 6 or 7 or 8 and the remaining four places in 4! ways i.e., 3 $$\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$$</p>
<p>Note that, ne... | mcq | jee-main-2016-offline |
4sIda7kghDbn5SBTSeZMx | maths | permutations-and-combinations | number-of-permutations | If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is : | [{"identifier": "A", "content": "44<sup>th</sup> "}, {"identifier": "B", "content": "45<sup>th</sup> "}, {"identifier": "C", "content": "46<sup>th</sup> "}, {"identifier": "D", "content": "47<sup>th</sup> "}] | ["C"] | null | <p>To find the position of the word QUEEN:</p>
<p>$$\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... | 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 nove... | 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 $$ \times $... | mcq | jee-main-2018-online-15th-april-morning-slot |
xe5ZfgVxYCdbpMKEGIuOF | maths | permutations-and-combinations | number-of-permutations | The number of four letter words that can be formed using the letters of the word <b>BARRACK</b> is : | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "144"}, {"identifier": "C", "content": "264"}, {"identifier": "D", "content": "270"}] | ["D"] | null | <b>Case 1 :</b>
<br><br>When all the four letters different then no of words
<br>= <sup>5</sup>C<sub>4</sub> $$ \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... | 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>$$... | mcq | jee-main-2020-online-7th-january-morning-slot |
qbRvgJn6hv5comatCj7k9k2k5hketjb | maths | permutations-and-combinations | number-of-permutations | The number of 4 letter words (with or without
meaning) that can be formed from the eleven
letters of the word 'EXAMINATION' is
_______. | [] | null | 2454 | 2A, 2I, 2N, E, X, M, T, O
<br><br>To form four letter words
<br><br><b>Case 1 :</b> All same ( not possible)
<br><br><b>Case 2 :</b> 1 different, 3 same (not possible)
<br><br><b>Case 3 :</b> 2 different, 2 same
<br><br>= <sup>3</sup>C<sub>1</sub> $$ \times $$ <sup>7</sup>C<sub>2</sub> $$ \times $$ $${{4!} \over {2!}}$... | integer | jee-main-2020-online-8th-january-evening-slot |
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