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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Mixed difficulty

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Practice format and chapter quiz

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 1-30 of 742 questions
1 Medium
1 of 742
Application +4 | -1
How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?
View answer and explanation
Correct option: C
60
Option C (60) is correct because it represents the number of ways to arrange the digits such that the odd digits occupy the even positions, calculated using permutations.
2 Easy
2 of 742
Application +4 | -1
The number of ways in which rr letters can be posted in nn letterboxes in a town is
View answer and explanation
Correct option: A
nrn^{r}
Option A (nrn^{r}) is correct because each of the rr letters can be posted in any of the nn letterboxes, giving nn choices for each letter.
3 Medium
3 of 742
Application +4 | -1
The number of four-digit numbers that can be made with the digits 1,2,3,41,2,3,4 and 5 in which atleast two digits are identical, is
View answer and explanation
Correct option: B
505
Option B (505) is correct because it calculates the number of four-digit numbers with at least two identical digits by subtracting the number of numbers with distinct digits from the total number of possible numbers.
4 Easy
4 of 742
Application +4 | -1
The number of ways in which a pack of 52 cards can be divided equally into four groups, is
View answer and explanation
Correct option: B
52!(13!)4×4!\frac{{52!}}{{{{\left( {13!} \right)}^4} \times 4!}}
Option B (52!(13!)4×4!\frac{{52!}}{{{{\left( {13!} \right)}^4} \times 4!}}) is correct because it represents the number of ways to divide 52 cards into four equal groups without considering the order of the groups.
5 Medium
5 of 742
Application +4 | -1
Five digited numbers with distinct digits are formed by using the digits, 5, 4, 3, 2, 1, 0. The number of those numbers which are multiples of 3, is
View answer and explanation
Correct option: C
216
Option C (216) is correct because the numbers formed must have a digit sum divisible by 3, considering both cases: including and excluding 0.
6 Easy
6 of 742
Application +4 | -1
A man invites a party to (m+n)\left( {m + n} \right) friends to dinner and places m at one round table and nn at another. The number of ways of arranging the guests is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the given options do not match the correct formula for arranging guests at two round tables.
7 Easy
7 of 742
Application +4 | -1
There are 5 gentlemen and 4 ladies to dine at a round table. The number of ways in which they can seat themselves so that no two ladies are together is
View answer and explanation
Correct option: B
2880
Option B (2880) is correct because it calculates the number of arrangements where men are seated first, then ladies are arranged in the gaps.
8 Medium
8 of 742
Application +4 | -1
The number of four letter words that can be formed using the letters of the word BARRACK is
View answer and explanation
Correct option: D
270
Option D (270) is correct because it accounts for all possible four-letter words formed using the letters of BARRACK, considering cases with different letters, two R's, and two A's.
9 Easy
9 of 742
Analyzing +4 | -1
If n>1n>1, then
View answer and explanation
Correct option: C
(2n)!(n!)2>4n2n+1\frac{(2 n) !}{(n !)^2}>\frac{4 n}{2 n+1}
Option C ((2n)!(n!)2>4n2n+1\frac{{(2n)!}}{{(n!)^2}} > \frac{{4n}}{{2n+1}}) is correct because substituting n=2 into the expression confirms the inequality.
10 Easy
10 of 742
Application +4 | -1
Let AA and BB be finite sets containing respectively 3 and 2 elements. Find the number of functions that can be defined from A to B
View answer and explanation
Correct option: A
8
Option A (8) is correct because the number of functions from A to B is 23 = 8.
11 Easy
11 of 742
Application +4 | -1
If nPr=2520{ }^{{n}} P_{{r}}=2520 and nCr=21{ }^{{n}} C_{{r}}=21 then nn is equal to:
View answer and explanation
Correct option: B
7
Option B (7) is correct because by using the given values of nPr{}^{{n}} P_{{r}} and nCr{}^{{n}} C_{{r}}, we can find r=5 and then solve for n, which results in n=7.
12 Easy
12 of 742
Application +4 | -1
The number of ways in which ten candidates A1,A2,A3,A4,⋯ ,A10A_1, A_2, A_3, A_4, \cdots, A_{10} can be arranged in a row "If A1A_1 is always above A2A_2 and A2A_2 is above A3A_3 is
View answer and explanation
Correct option: A
10!3!\frac{10 !}{3 !}
Option A (10!3!\frac{10 !}{3 !}) is correct because it represents the total arrangements divided by the arrangements of the three specific candidates.
13 Medium
13 of 742
Application +4 | -1
Number of ways in which four prize can be given away to 8 students when a students can get any number of prize is
View answer and explanation
Correct option: B
848^4
Option D (None of these) is correct. There appears to be an error in the provided solution. The correct approach involves derangements, and the number of functions f: A -> B such that f(i) != a_i is given by the number of derangements of 5 elements, which is 44. Therefore, none of the given options are correct.
14 Easy
14 of 742
Application +4 | -1
A boy has 3 library tickets and 9 books of his interest in the library. Out of these 8, he does not want to borrow Chemistry part II, unless Chemistry part I is also borrowed. The number of ways in which he can choose the three books to be borrowed, is
View answer and explanation
Correct option: A
41
Option A (41) is correct because it considers both cases: including and excluding Chemistry part II.
15 Easy
15 of 742
Application +4 | -1
The number of wards can be formed with the letters of the word PATALIPUTRA without changing the relative positions of vowels and consonants is 400n400 \mathrm{n} then n=\mathrm{n}=
View answer and explanation
Correct option: A
9
Option A (9) is correct because the number of arrangements of the letters in PATALIPUTRA, keeping the relative positions of vowels and consonants unchanged, is 3600, and 3600/400 = 9.

Explore Other Chapters in MATHEMATICS

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
16 Easy
16 of 742
Application +4 | -1
The number of ways in which a committee of 3 women and 4 men be chosen from 8 women and 7 men if Ms Bryn refuses to serve on the committee if Mr. Arnav is a member of the committee (Ms. Bryn and Mr. Arnav are among 8 women & 7 men respectively) is :
View answer and explanation
Correct option: C
1540
Option C (5050π5050 \pi) is correct because it is the sum of the total area of the green regions, calculated by summing the areas of alternating annuli.
17 Medium
17 of 742
Application +4 | -1
The total number of numbers that are less than 3⋅1083 \cdot 10^8 and can be formed using the digits 1,2,31,2,3, is equal to
View answer and explanation
Correct option: C
12(7⋅38−3)\frac{1}{2}\left(7 \cdot 3^8-3\right)
Option C (12(7⋅38−3)\frac{1}{2}\left(7 \cdot 3^8-3\right)) is correct because it represents the total numbers that can be formed using the digits 1, 2, and 3, and are less than 3⋅1083 \cdot 10^8.
18 Easy
18 of 742
Understanding +4 | -1
Let SS be the set of all functions from the set AA to the set AA. If n(A)=kn(A)=k, then n(S)n(S) is
View answer and explanation
Correct option: B
kkk^k
Option B (kkk^k) is correct because each of the k elements in set A can be mapped to any of the k elements in set A, resulting in kkk^k possible functions.
19 Easy
19 of 742
Application +4 | -1
Let x1,x2,x3x_1, x_2, x_3 ______ xnx_n be nn observation such that Σ(xi)2=400\Sigma\left(x_i\right)^2=400 and Σxi=40\Sigma x_i=40 then a possible value of nn among the following is
View answer and explanation
Correct option: B
4
Option B (4) is correct because the condition derived from AM-GM inequality requires n > 4, making 4 the smallest possible integer value for n.
20 Easy
20 of 742
Application +4 | -1
Anil have tiled his square bathroom wall with congruent square tiles. All the tiles are red, except those along the two diagonals, which are all blue. If he used 121 blue tiles, then the number of red tiles used are
View answer and explanation
Correct option: C
3600
Option C (3600) is correct because with 121 blue tiles, the square is 61x61, and subtracting the blue tiles leaves 3600 red tiles.
21 Easy
21 of 742
Application +4 | -1
The total number of arrangements of the letters in the expressiona3b2c4{a^3}{b^2}{c^4} when written at full length, is
View answer and explanation
Correct option: A
1260
Option A (1260) is correct because the total number of arrangements is calculated by dividing the total factorial by the factorials of the repeated letters.
22 Medium
22 of 742
Application +4 | -1
No. of ways in which nn different prizes can be distributed among mm - persons (m<n)(m<n) if each is entitled to receive at most (n−1)(n-1) prizes is
View answer and explanation
Correct option: D
mn−mm^n-m
Option D (mn−mm^n-m) is correct because it subtracts the cases where one person receives all prizes from the total possible distributions.
23 Easy
23 of 742
Application +4 | -1
Let f(n)f(n) denotes the number of different ways the positive integer nn can be expressed as the sum of 1's and 2's. For example f(4)=5f(4)=5, since 4=1+1+1+1,1+1+2,1+2+1,2+1+1,2+24=1+1+1+1,1+1+2,1+2+1,2+1+1,2+2 (Note Order of 1's and 2's is important). Then, f{f(6)}f\{f(6)\} is
View answer and explanation
Correct option: C
f(13)f(13)
Option C (f(13)f(13)) is correct because f(6)f(6) is calculated to be 13, and therefore f(f(6))=f(13)f(f(6)) = f(13).
24 Easy
24 of 742
Application +4 | -1
A candidate is required to answer 7 questions out of 12 questions which are divided into two groups each containing 6 questions. He is not permitted to attempt more than 5 questions from each group. The number of ways in which he can choose the 7 questions is
View answer and explanation
Correct option: A
780
Option A (780) is correct because it sums the number of ways to choose questions from each group under the given constraints.
25 Easy
25 of 742
Application +4 | -1
The number of proper divisors of 2p⋅6q⋅15r2^p \cdot 6^q \cdot 15^r is
View answer and explanation
Correct option: B
(p+q+1)(q+r+1)(r+1)−2(p+q+1)(q+r+1)(r+1)-2
Option B ((p+q+1)(q+r+1)(r+1)−2(p+q+1)(q+r+1)(r+1)-2) is correct because it calculates the total number of divisors and then subtracts 2 to exclude 1 and the number itself, giving the number of proper divisors.
26 Medium
26 of 742
Application +4 | -1
A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. It internal arrangement inside the car does not matter then the number of ways in which they can travel, is
View answer and explanation
Correct option: C
126
Option C (126) is correct because it calculates the number of ways to divide 8 students into two cars, one seating 5 and the other seating 4 (but only needing 3), or one seating 4 and the other seating 5 (but only needing 4).
27 Easy
27 of 742
Application +4 | -1
The number of ways in which 10 identical balls can be distributed in 4 distinct boxes, so that none of the boxes remain empty, is
View answer and explanation
Correct option: D
None of these
None of these is correct because the correct answer is 84, calculated using stars and bars with the condition that no box is empty.
28 Easy
28 of 742
Understanding +4 | -1
Which of the following statement is false?
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because all the statements A, B, and C are true; it is possible to find consecutive natural numbers that are not prime.
29 Medium
29 of 742
Application +4 | -1
Total number of positive integral solutions of 15<x1+x2+x3≤2015 < x_1+x_2+x_3 \leq 20 is equal to
View answer and explanation
Correct option: D
685
Option D (685) is correct because it calculates the total number of positive integral solutions for the given inequality by summing the solutions for each possible sum value.
30 Easy
30 of 742
Application +4 | -1
The maximum possible number of points of intersection of 7 straight lines and 5 circles is:
View answer and explanation
Correct option: A
111
Option A (111) is correct because it represents the sum of the maximum number of intersection points between lines, circles, and lines and circles.
Revision notes and key takeaways

Review PERMUTATION AND COMBINATION in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 742 single-correct-answer MCQs on PERMUTATION AND COMBINATION.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
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