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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 241-270 of 742 questions

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241 Easy
241 of 742
Analyzing +4 | -1
Consider the following statement:

1.The number of ways of arranging m different things taken all at a time in which p⩽mp \leqslant m perticular things are never together is m!−(m−p+1)!p!m! - \left( {m - p + 1} \right)!p!

2. A pack of 52 cards can be divided equally among four players in order in 52!(13!)4\frac{{52!}}{{{{\left( {13!} \right)}^4}}} ways

Which of these is/are correct?
View answer and explanation
Correct option: C
Both of these
Both statements are correct. Statement 1 correctly describes the number of arrangements where p particular things are never together, and statement 2 accurately represents the number of ways to divide a deck of cards equally among four players.
242 Easy
242 of 742
Application +4 | -1
The number of ways distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is
View answer and explanation
Correct option: B
21
Option B (21) is correct because it represents the number of ways to distribute 8 identical balls into 3 distinct boxes with no box empty, which is calculated using the stars and bars method.
243 Medium
243 of 742
Analyzing +4 | -1
We are to form different words with the letters of the word INTEGER. Let m1{m_1} be the number of words in which I and N are never together and m2{m_2} be the number of words which begin with I and end with R, then m1/m2{m_1}/{m_2} is equal to
View answer and explanation
Correct option: A
30
Option A (30) is correct because it is the ratio of the number of words where I and N are not together to the number of words starting with I and ending with R.
244 Medium
244 of 742
Application +4 | -1
If 56Pr+6:54Pr+3=30800:1{ }^{56} P_{r+6}:{ }^{54} P_{r+3}=30800: 1, then the value of rr is
View answer and explanation
Correct option: C
41
Option C (41) is correct because solving the equation 56Pr+654Pr+3=30800\frac{{ }^{56} P_{r+6}}{{ }^{54} P_{r+3}}=30800 for rr gives r=41r=41.
245 Hard
245 of 742
Analyzing +4 | -1
Ten persons numbered 1,2,3,…,101,2,3, \ldots, 10 play a chess tournament, each player playing against every player exactly one game. It is known that no game ends in a draw. Let w1,w2,w3,…,w10w_1, w_2, w_3, \ldots, w_{10} be the number of games won by player 1,2,3,…,101,2,3, \ldots, 10 respectively and l1,l2,l3,…l_1, l_2, l_3, \ldots, l10l_{10} be the number of games lost by the players 1,2,3,…1,2,3, \ldots, 10 respectively, then
View answer and explanation
Correct option: C
Σwi2=Σli2\Sigma w_i^2=\Sigma l_i^2
Option C (Σwi2=Σli2\Sigma w_i^2=\Sigma l_i^2) is correct because in a tournament where each game has a winner and a loser, the sum of the squares of the wins equals the sum of the squares of the losses.
246 Medium
246 of 742
Application +4 | -1
The sum of the digits in the unit place of all the numbers formed with the help of 3,4,5,63,4,5,6 taken all at a time is
View answer and explanation
Correct option: B
108
Option B (108) is correct because it calculates the sum of the digits in the unit place of all numbers formed using the given digits.
247 Medium
247 of 742
Application +4 | -1
Number of numbers greater than 24000 that can be formed by using 1,2,3,4,51,2,3,4,5 when no digit being repeated is
View answer and explanation
Correct option: C
84
Option C (84) is correct because it accounts for numbers starting with 3, 4, 5, and those starting with 24 or 25, without repetition.
248 Medium
248 of 742
Application +4 | -1
Let A\mathrm{A} be the set of all positive prime integers less than 30 . The number of different rational numbers whose numerator and denominator belongs to A\mathrm{A} is
View answer and explanation
Correct option: A
90
Option A (90) is correct because the number of different rational numbers is the number of permutations of choosing two distinct primes from the set A.
249 Medium
249 of 742
Application +4 | -1
The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48 , isThe number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48 , is
View answer and explanation
Correct option: A
432
Option A (432) is correct because it is calculated by finding the number of 3-digit numbers divisible by 3 or 4 and subtracting those divisible by 48 to avoid overcounting.
250 Medium
250 of 742
Application +4 | -1
nn different toys have to be distributed among nn children. Total number of ways in which these toys can be distributed, so that exactly one child gets no toy, is equal to
View answer and explanation
Correct option: B
n!⋅nC2n ! \cdot{ }^{n} C_2
Option B (n!⋅nC2n! \cdot{ }^{n} C_2) is correct because it represents the ways to choose two children, give one an extra toy, and then distribute the toys.
251 Medium
251 of 742
Application +4 | -1
If letters of the word 'KUBER' are written in all possible orders and arranged as in a dictionary, then rank of the word 'KUBER' will be
View answer and explanation
Correct option: A
67
Option A (67) is correct because it is the correct lexicographical rank of the word 'KUBER'.
252 Medium
252 of 742
Application +4 | -1
The one's and ten's digits in 1!+2!+3!+…+491 !+2 !+3 !+\ldots+49! are respectively
View answer and explanation
Correct option: B
3 and 1
Option B (3 and 1) is correct because the sum of the factorials from 1! to 49! ends in 13, so the one's digit is 3 and the ten's digit is 1.
253 Easy
253 of 742
Analyzing +4 | -1
A man has three friends. The number of ways he can invite one friend everyday for dinner on six successive nights, so that no friend is invited more than three times, is
View answer and explanation
Correct option: D
510
The correct answer is D (510) because it accounts for all possible ways the man can invite his friends, considering the constraint that no friend is invited more than three times.
254 Medium
254 of 742
Application +4 | -1
The no. of ways in which 5 boys and 4 girls sit around a circular table so that no two girls sit together is
View answer and explanation
Correct option: A
5!4!5 ! 4 !
Option A (5!4!5 ! 4 !) is correct because it arranges the boys in a circle and then arranges the girls in the spaces between them.
255 Easy
255 of 742
Application +4 | -1
The number of arrangements which can be made out of the letters of the word ALGEBRA, without changing the relative order (positions) of vowels and consonants, is
View answer and explanation
Correct option: A
72
Option A (72) is correct because it accounts for arranging vowels and consonants separately while considering repeated vowels.
256 Medium
256 of 742
Application +4 | -1
8 players S1, S2,…...S8\mathrm{S}_1, \mathrm{~S}_2, \ldots . . . \mathrm{S}_8, play in tournament. Number of ways in which they can be grouped into 4 pairs so that S1S_1 and S2S_2 are in different group is
View answer and explanation
Correct option: B
90
Option B (90) is correct because it represents the number of ways to group the players into pairs such that S1 and S2 are in different groups.
257 Easy
257 of 742
Application +4 | -1
Eleven books consisting of 5 Mathematics, 4 physics and 2 Chemistry are places on a shelf. The number of possible ways of arranging them on the assumption that the books of the same subject are all together, is
View answer and explanation
Correct option: C
5!4!3!2!5!4!3!2!
Option C (5!4!3!2!5!4!3!2!) is correct because it accounts for arranging the books within each subject and then arranging the subjects themselves.
258 Easy
258 of 742
Application +4 | -1
If the letters 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
View answer and explanation
Correct option: D
601
Option D (601) is correct because it represents the position of the word 'SACHIN' when all permutations are arranged alphabetically.
259 Easy
259 of 742
Application +4 | -1
The number of ways in which nn things of which rr are alike, can be arranged in a circular order, is
View answer and explanation
Correct option: B
(n−1)!r!\frac{(n-1) !}{r !}
Option B (48) is correct because we calculate the total permutations of 5-digit numbers and subtract those that start with 3 to ensure the number is greater than 50,000.
260 Easy
260 of 742
Application +4 | -1
Four married couples are to be seated in a row having 8 chairs. The number of ways so that spouses are seated next to each other, is
View answer and explanation
Correct option: C
384
Option C (384) is correct because it accounts for arranging the couples and then arranging the spouses within each couple.
261 Easy
261 of 742
Understanding +4 | -1
The value of ∑r=1n   nPrr!\sum \limits_{r = 1}^n \,\frac{{{\;^n}{P_r}}}{{r!}} is
View answer and explanation
Correct option: B
2n−1{2^n} - 1
Option B (2n−1{2^n} - 1) is correct because the given summation is equivalent to the binomial expansion of (1+1)n(1+1)^n excluding the first term.
262 Easy
262 of 742
Application +4 | -1
A man has 6 friends. In how many ways he can invite one or more of them at dinner?
View answer and explanation
Correct option: C
63
Option C (63) is correct because each friend can either be invited or not, leading to 262^6 possibilities, but we subtract 1 for the case where no friends are invited.
263 Medium
263 of 742
Application +4 | -1
Total no. of ways in which six ' + ' and four '-' signs can be arranged in a row so that no two '-' signs to be together, is
View answer and explanation
Correct option: A
35
Option A (35) is correct because arranging the '+' signs creates gaps where the '-' signs can be placed such that no two '-' signs are together, and the number of ways to do this is 35.
264 Easy
264 of 742
Application +4 | -1
Find the number of ways of arranging the letters of the word ENGINEERING
View answer and explanation
Correct option: D
11!3!3!2!2!\frac{11 !}{3 ! 3 ! 2 ! 2 !}
Option D (11!3!3!2!2!\frac{11 !}{3 ! 3 ! 2 ! 2 !}) is correct because it accounts for the repetitions of the letters E, N, G, and I in the word ENGINEERING.
265 Easy
265 of 742
Application +4 | -1
From 7 gentlemen and 4 ladies, a committee of 5 is to be formed. The number of ways in which this can be done so as to include atleast one lady is
View answer and explanation
Correct option: A
805
Option A (805) is correct because it sums the number of ways to form the committee with at least one lady.
266 Easy
266 of 742
Application +4 | -1
If one test (on screening paper basis) was conducted on Batch A, maximum number of marks is (90×3)=270.4(90 \times 3)=270.4 students get the marks lower than 80 . Coaching institute decide to inform their guardians, that is why their result card were sent to their home. The number of ways, in which all the letters were put in wrong envelopes, is
View answer and explanation
Correct option: A
9
Option A (9) is correct because the number of ways to put all 4 letters in the wrong envelopes is calculated using the derangement formula, which results in 9.
267 Easy
267 of 742
Application +4 | -1
How many 10 digit numbers can be written by using the digits 1 and 2?
View answer and explanation
Correct option: B
210{2^{10}}
Option B (210{2^{10}}) is correct because each of the 10 digits can be either 1 or 2, resulting in 2 options for each digit.
268 Medium
268 of 742
Application +4 | -1
The number of ways 6 different flowers can be given to 10 girls, if each can receive any number of flowers is
View answer and explanation
Correct option: B
10610^{6}
Option B (10610^{6}) is correct because each flower can be given to any of the 10 girls, resulting in 10 choices for each flower.
269 Easy
269 of 742
Application +4 | -1
If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent. The number of points in which they intersect each other is
View answer and explanation
Correct option: B
190
Option B (190) is correct because it calculates the number of intersection points formed by 20 lines, where no two are parallel and no three are concurrent.
270 Easy
270 of 742
Application +4 | -1
The number of ways 12 different books can be distributed equally among 4 persons, is
View answer and explanation
Correct option: C
12!(3!)4\frac{12 !}{(3 !)^4}
Option C (12!(3!)4\frac{{12 !}}{{{{\left( {3!} \right)}^4}}}) is correct because it represents the number of ways to distribute 12 distinct books equally among 4 people.
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