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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Easy

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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 421-450 of 485 questions

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421 Easy
421 of 485
Application +4 | -1
Four boys picked 30 apples. The number of ways in which they can divide them if all the apples are identical, is
View answer and explanation
Correct option: C
5456
Option C (5456) is correct because the number of ways to divide identical items among people is calculated using combinations with repetition.
422 Easy
422 of 485
Application +4 | -1
The total number of ways can 15 identical Mathematics books be distributed among six students, is
View answer and explanation
Correct option: B
20!5!15!\frac{20 !}{5 ! 15 !}
Option B (20!5!15!\frac{{20 !}}{{{{5 ! 15 !}}}}) is correct because it represents the number of ways to distribute 15 identical books among 6 students, allowing blank boxes.
423 Easy
423 of 485
Application +4 | -1
Three dice are rolled. The number of possible outcomes in which at least one die shows 5 is
View answer and explanation
Correct option: D
91
Option D (91) is correct because the number of outcomes with at least one 5 is found by subtracting the number of outcomes with no 5s from the total number of outcomes.
424 Easy
424 of 485
Application +4 | -1
When a group photograph is taken, all the seven teachers should be in the first row and all the twenty students should be in the second row. If the two corners of the second row are reserved for the two tallest students, interchangeable only between them, and if the middle seat of the front row is reserved for the principal. The no. of such possible 'arrangements' is
View answer and explanation
Correct option: C
1440×181440 \times 18 !
Option C (1440×181440 \times 18 !) is correct because it correctly accounts for the arrangements of teachers, tallest students, and remaining students.
425 Easy
425 of 485
Application +4 | -1
In a plane, there are two sets of parallel lines, one of mm lines and the other of nn lines. If the lines of one set cut those of the other, then the number of different parallelograms that can be formed is
View answer and explanation
Correct option: C
nm(m−1)(n−1)4\frac{n m(m-1)(n-1)}{4}
Option C (nm(m−1)(n−1)4\frac{n m(m-1)(n-1)}{4}) is correct because it represents the number of ways to choose two lines from each set of parallel lines to form a parallelogram.
426 Easy
426 of 485
Application +4 | -1
mm men and ww women are to be seated in a row so that all women sit together. The number of ways in which they can be seated is
View answer and explanation
Correct option: A
(m+1)!w!(m+1) ! w !
Option A ((m+1)!w!(m+1) ! w !) is correct because the women are treated as a single block, and this block along with the men can be arranged in (m+1)! ways, with the women themselves arranged in w! ways.
427 Easy
427 of 485
Application +4 | -1
If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at 315th 315^{\text {th }} position in this arrangement is :
View answer and explanation
Correct option: C
NRAPGU
Option D (4) is correct because it is the value of the derivative of g(x) evaluated at x=0, calculated using the chain rule and product rule.
428 Easy
428 of 485
Application +4 | -1
The number of natural numbers less than one million that can be formed by using the digits, 0, 2, and 3 is
View answer and explanation
Correct option: A
728
Option A (728) is correct because it sums the number of possible natural numbers of lengths 1 to 6 using the digits 0, 2, and 3.
429 Easy
429 of 485
Application +4 | -1
There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. The number of ways in which the votes can be given is
View answer and explanation
Correct option: B
1024
Option B (1024) is correct because each of the 5 men has 4 choices, leading to 45=10244^5 = 1024 possible voting combinations.
430 Easy
430 of 485
Application +4 | -1
If there are 62 onto mapping from a set XX containing nn elements to the set Y={−1,1}Y=\{-1,1\}, then nn is equal to:
View answer and explanation
Correct option: C
6
Option C (6) is correct because the number of onto mappings from a set X with n elements to a set Y with 2 elements is 2n−22^n - 2, which equals 62 when n=6.
431 Easy
431 of 485
Application +4 | -1
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
View answer and explanation
Correct option: A
Atleast 1000
Option D (2) is correct because the sum of the coefficients of all odd degree terms simplifies to 2 after expansion and simplification.
432 Easy
432 of 485
Understanding +4 | -1
There are ' nn ' seats round a table numbered 1,2,3,…n1,2,3, \ldots n The numbers of in which m(≤n)m(\leq n) persons can take seats is
View answer and explanation
Correct option: A
nPm{ }^n P_m
Option A (nPm{ }^n P_m) is correct because the seats are numbered, indicating a linear arrangement, and thus the number of ways to seat m persons in n seats is given by the permutation nPm{ }^n P_m.
433 Easy
433 of 485
Application +4 | -1
The number of ways in which the letters of the word 'SUCCESSFUL' be arranged such that all the ' ss ' come together is
View answer and explanation
Correct option: C
8!2!2!\frac{8 !}{2 ! 2 !}
Option C (8!2!2!\frac{8 !}{2 ! 2 !}) is correct because when the 3 'S's are treated as one unit, there are 8 units to arrange, accounting for repetitions of 'U' and 'C'.
434 Easy
434 of 485
Application +4 | -1
The number of ways in which 7 people can be arranged at a round table, so that 2 particular persons may be together, is
View answer and explanation
Correct option: C
240
Option C (240) is correct because it treats the two people as one unit and accounts for their internal arrangement in a circular permutation.
435 Easy
435 of 485
Application +4 | -1
From 17 consonants and 5 vowels. The number of words containing 3 consonants and 2 vowels that can be made if all the letters are different, is
View answer and explanation
Correct option: A
816000
Option A (816000) is correct because it represents the number of words formed with 3 consonants and 2 vowels.
436 Easy
436 of 485
Application +4 | -1
Let x1,x2,x3,…,xkx_1, x_2, x_3, \ldots, x_k be the divisors of positive integer ' nn ' (including 1 and xx ). If x1+x2+…+xk=75x_1+x_2+\ldots+x_k=75, then ∑i=1k1xi\sum\limits_{i=1}^k \frac{1}{x_i} is equal to
View answer and explanation
Correct option: D
75n\frac{75}{n}
Option D (75/n) is correct because the sum of the reciprocals of the divisors is equal to the sum of the divisors divided by n.
437 Easy
437 of 485
Application +4 | -1
A man invites 6 non-vegeterian and 5 vegeterian friends for a dinner party. He arrange 6 non-vegeterian friends on one round table and 5 vegeterian friends along another round table. The number of ways this can be done is:
View answer and explanation
Correct option: C
2880
Option C (2880) is correct because arranging the non-vegetarians and vegetarians separately on round tables gives (6-1)! * (5-1)! = 2880 ways.
438 Easy
438 of 485
Application +4 | -1
The least positive integral value of nn which satisfies the inequality 10Cn−1>2⋅10Cn{ }^{10} {C}_{n-1}>2 \cdot{ }^{10} {C}_n is
View answer and explanation
Correct option: B
8
Option B (8) is correct because it is the smallest integer value of n that satisfies the given inequality.
439 Easy
439 of 485
Application +4 | -1
The number of arrangements of the letters of the word BANANA in which the two N's do not appear adjacently, is
View answer and explanation
Correct option: A
40
Option A (40) is correct because it subtracts the number of arrangements where the two N's are adjacent from the total number of arrangements of the word BANANA.
440 Easy
440 of 485
Application +4 | -1
The rank of the word MOTHER when the letters of the word are arranged alphabetically as in a dictionary, is
View answer and explanation
Correct option: C
309
Option C (309) is correct because it is the rank calculated by arranging the letters alphabetically and finding the number of words before MOTHER.
441 Easy
441 of 485
Understanding +4 | -1
A set contains (2n+1)(2 n+1) elements. The number of subsets of the set which contain at most nn elements is
View answer and explanation
Correct option: D
22n2^{2 n}
Option D (22n2^{2 n}) is correct because the sum of binomial coefficients from 0 to n in a set of (2n+1) elements equals half of the total subsets.
442 Easy
442 of 485
Application +4 | -1
Let AA be the set of 4-digit numbers a1a2a3a4a_1 a_2 a_3 a_4, where a1>a2>a3>a4a_1 > a_2 > a_3 > a_4, then n(A)n(A) is equal to
View answer and explanation
Correct option: C
210
Option C (210) is correct because it represents the number of ways to choose 4 distinct digits from 10.
443 Easy
443 of 485
Application +4 | -1
How many numbers lying between 999 and 10000 can be formed with the help of the digits 0, 2, 3, 6, 7, 8 when the digits are not be repeated?
View answer and explanation
Correct option: C
300
Option C (300) is correct because it calculates the number of 4-digit numbers formed with the given digits, excluding those starting with 0.
444 Easy
444 of 485
Application +4 | -1
If each of 10 points on a straight line be joined to each of 10 points on a parallel line then the total number of triangles that can be formed with the given points as vertices, is
View answer and explanation
Correct option: B
900
Option B (900) is correct because it represents the total number of triangles that can be formed with the given points as vertices.
445 Easy
445 of 485
Application +4 | -1
The value of 50C4+∑r=1656−rC3{ }^{50} C_4+\sum\limits_{{r=1}}^6{ }^{56-r} C_3 is
View answer and explanation
Correct option: B
56C4{ }^{56} C_4
Option B (56C4{ }^{56} C_4) is correct because the summation can be simplified using the identity nCr+nCr−1=n+1Cr{ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r repeatedly.
446 Easy
446 of 485
Application +4 | -1
Number of 4 digit numbers of the form N=abcdN=a b c d which satisfy following three conditions:

(i) 4000≤N<60004000 \leq N < 6000

(ii) NN is a multiple of 5

(iii) 3≤b<c≤63 \leq b < c \leq 6

is equal to:
View answer and explanation
Correct option: C
24
Option C (24) is correct because the number of possible 4-digit numbers is found by considering the constraints on each digit.
447 Easy
447 of 485
Application +4 | -1
There are nn different books and pp copies of each. The number of ways in which a selection can be made from them is
View answer and explanation
Correct option: C
(p+1)n−1{\left( {p + 1} \right)^n} - 1
Option C ((p+1)n−1{\left( {p + 1} \right)^n} - 1) is correct because each book can be selected in p+1 ways (0 to p copies), and we subtract 1 for the case where no book is selected.
448 Easy
448 of 485
Application +4 | -1
If x∈Nx \in {N} and (x+2)!(2x−1)!⋅(2x+1)!(x+3)!=727\frac{(x+2) !}{(2 x-1) !} \cdot \frac{(2 x+1) !}{(x+3) !}=\frac{72}{7}, then xx is equal to
View answer and explanation
Correct option: B
4
Option B (4) is correct because simplifying the given factorial equation leads to a quadratic equation, which has x=4 as a valid solution.
449 Easy
449 of 485
Application +4 | -1
The number of natural numbers smaller than 104,{10^4}, in the decimal notation of which all the digit are different is
View answer and explanation
Correct option: A
5274
The correct answer is A (5274) because it is the number of natural numbers smaller than 104{10^4} with all distinct digits.
450 Easy
450 of 485
Application +4 | -1
There are pp copies each of nn different books. The number of different ways in which a non-empty selection can be made from them, is
View answer and explanation
Correct option: A
(p+1)n−1(p+1)^n-1
Option A ((p+1)n−1(p+1)^n-1) is correct because it represents the number of non-empty selections from n different books with p copies each.
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