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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 601-630 of 742 questions

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601 Medium
601 of 742
Understanding +4 | -1
If n=2p−1(2p−1)n=2^{p-1}\left(2^p-1\right), where 2p−12^p-1 is a prime, then the sum of the divisors of nn is equal to
View answer and explanation
Correct option: B
2n2 \mathrm{n}
Option B is correct because the sum of divisors of n, where n is in the given form, is equal to 2n.
602 Easy
602 of 742
Application +4 | -1
∑r=0m  n+rCn\sum \limits_{r = 0}^m {\;^{n + r}}{C_n} is equal to
View answer and explanation
Correct option: A
  n+m+1Cn+1{\;^{n + m + 1}}{C_{n + 1}}
Option A (  n+m+1Cn+1{\;^{n + m + 1}}{C_{n + 1}}) is correct because it represents the sum of combinations using the Hockey Stick Identity.
603 Easy
603 of 742
Application +4 | -1
The number of permutations of 4 letters that can be made out of the letters of the word EXAMINATION is
View answer and explanation
Correct option: A
2454
The correct answer is A (2454) because it accounts for all possible permutations of 4 letters from EXAMINATION, considering repetitions of A, I, and N.
604 Easy
604 of 742
Application +4 | -1
The number of arrangements of the letters of the word "pencil" so that ee always comes earlier than ii, is
View answer and explanation
Correct option: C
360
Option C (360) is correct because it calculates the arrangements of the word PENCIL treating e and i as identical.
605 Easy
605 of 742
Application +4 | -1
Out of 5 apples, 10 mangoes and 15 oranges, the number of ways of distributing 15 fruits each to two persons, is n\mathrm{n} then n22\frac{\mathrm{n}}{22} is equal to
View answer and explanation
Correct option: A
3
Option A (3) is correct because the number of ways of distributing the fruits can be found using coefficient extraction, and then dividing by 22.
606 Medium
606 of 742
Understanding +4 | -1
Two players P1P_1 and P2P_2 play a series of '2n2 n' games. Each game can result in either a win or loss for P1P_1. Total number of ways in which P1P_1 can win the series of these games, is equal to
View answer and explanation
Correct option: A
12(22n−2nCn)\frac{1}{2}\left(2^{2 n}-{ }^{2 n} C_n\right)
Option A (12(22n−2nCn)\frac{1}{2}\left(2^{2 n}-{ }^{2 n} C_n\right)) is correct because it represents the sum of ways P1P_1 can win more than n games, which is half of the total ways minus the ways to win exactly n games.
607 Easy
607 of 742
Analyzing +4 | -1
Let f:{1,  2,  3,  4,  5}→{1,  2,  3,  4,  4,  5}f:\left\{ {1,\;2,\;3,\;4,\;5} \right\} \to \left\{ {1,\;2,\;3,\;4,\;4,\;5} \right\} that are onto and f(x)≠if\left( x \right) \ne i is equal to
View answer and explanation
Correct option: B
44
Option B (44) is correct because the number of onto functions with the given condition is equivalent to the number of derangements of 5 objects, which is 44.
608 Medium
608 of 742
Application +4 | -1
15 identical balls have to be put in 5 different boxes. Each box can contain any number of balls. Total number of ways of putting the balls into box, so that each box contains atleast 2 balls, is equal to
View answer and explanation
Correct option: A
9C4{ }^9 C_4
Option A (9C4{ }^9 C_4) is correct because it represents the number of ways to distribute the balls after giving each box the minimum required number of balls.
609 Easy
609 of 742
Application +4 | -1
The number of numbers lying between 40000 and 70000 that can be made from the digits 0,1,2,4,5,70,1,2,4,5,7, if digits can be repeated in the same number, is
View answer and explanation
Correct option: A
2591
Option A (2591) is correct because it calculates the number of 5-digit numbers between 40000 and 70000 formed using the given digits with repetition allowed, considering the restrictions on the ten-thousands place.
610 Medium
610 of 742
Application +4 | -1
number of ways in which 15 indistinguishable oranges can be distributed in 3 different boxes so that every has atmost 8 oranges is
View answer and explanation
Correct option: A
52
Option A (52) is correct because it represents the number of ways to distribute 15 indistinguishable oranges into 3 distinct boxes with a maximum of 8 oranges per box, calculated using stars and bars with inclusion-exclusion.
611 Medium
611 of 742
Application +4 | -1
Total number of 6 digit numbers in which all the odd digits and only odd digits appear is
View answer and explanation
Correct option: A
52∠6\frac{5}{2} \angle 6
Option A (52∠6\frac{{5}}{{2}} \angle 6) is correct because it represents the number of 6-digit numbers with only odd digits, where one odd digit is repeated.
612 Easy
612 of 742
Application +4 | -1
How many four-digit numbers can be formed using the digits 0,1,2,3,4,50,1,2,3,4,5, if repetition of digits is not allowed?
View answer and explanation
Correct option: C
300
Option C (300) is correct because it accounts for the restriction that the first digit cannot be 0 and then calculates the permutations for the remaining digits.
613 Medium
613 of 742
Application +4 | -1
The number of ordered pairs (m,n)(m, n), m,n∈{1,2,…,100}m, n \in\{1,2, \ldots, 100\} such that 7m+7n7^m+7^n is divisible by 5, is
View answer and explanation
Correct option: C
2500
Option C (2500) is correct because it represents the number of ordered pairs (m, n) such that 7^m + 7^n is divisible by 5, which is determined by analyzing the cyclical pattern of the last digit of 7^r.
614 Medium
614 of 742
Application +4 | -1
The number of word formed by using all letter of word 'RANDOM' in which vowels are in alphabetical order is
View answer and explanation
Correct option: B
360
Option B (360) is correct because it divides the total permutations by the number of ways to arrange the vowels to ensure alphabetical order.
615 Easy
615 of 742
Application +4 | -1
All letters of the word EAMCET are arranged in all possible ways. The number of such arrangement in which no two vowels are adjacent to each other, is
View answer and explanation
Correct option: C
72
Option C (72) is correct because the number of arrangements where no two vowels are adjacent is calculated by arranging consonants first and then placing vowels in the gaps.
616 Easy
616 of 742
Understanding +4 | -1
In a certain test, aia_i students gave wrong answers to atleast ii questions, where i=1,2,3,…,ki=1,2,3, \ldots, k. No student gave more than kk wrong answers. The total number of wrong answers given is
View answer and explanation
Correct option: A
a1+a2+…+aka_1+a_2+\ldots+a_k
Option A (a1+a2+…+aka_1+a_2+\ldots+a_k) is correct because it represents the sum of students who gave at least i wrong answers for each i from 1 to k, which equals the total number of wrong answers.
617 Easy
617 of 742
Application +4 | -1
Find the number of different necklaces that can be made from 17 identical pearls and 2 different diamonds.
View answer and explanation
Correct option: A
9
Option A (9) is correct because the 19 pearls (17 identical + 2 different) can be arranged in a circle in 18 ways, and since mutual arrangement of two different diamonds does not give a different necklace, the required number of different necklaces is 18/2 = 9.
618 Easy
618 of 742
Analyzing +4 | -1
Let S={1,2,3,…,n}S=\{1,2,3, \ldots, n\}. If XX denotes the set of all subsets of SS containing exactly two elements, then the value of ∑A∈X(min⁡A)\sum\limits_{{A \in X}}(\min A) is
View answer and explanation
Correct option: A
n+1C3{ }^{n+1} C_3
Option A (n+1C3{ }^{n+1} C_3) is correct because it accurately sums the minimum elements of all two-element subsets of S.
619 Easy
619 of 742
Application +4 | -1
Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is tt then t400\frac{t}{400} is.
View answer and explanation
Correct option: A
6
Option A (6) is correct because the number of ways to insert the blanks is calculated using combinations, and then multiplied by the number of permutations of the letters.
620 Easy
620 of 742
Analyzing +4 | -1
The sum of factors of 8!8 ! which are odd and are of the form 3m+23 m+2, where mm is a natural number, is
View answer and explanation
Correct option: A
40
Option A (40) is correct because it is the sum of the odd factors of 8! that are of the form 3m+2.
621 Easy
621 of 742
Application +4 | -1
The letters of the word MODESTY are written in all possible orders and these words are written out as in a dictionary, then the rank of the word MODESTY is
View answer and explanation
Correct option: C
1681
Option C (1681) is correct because the rank is calculated by counting the number of words that come before MODESTY in the dictionary order.
622 Medium
622 of 742
Application +4 | -1
In a single correct match the column question, column I contain 10 questions and Column II contain 10 answers written in some arbitrary order. If the number ways a student can answer this question so that exactly 6 of his matching are correct is k\mathrm{k}, then (sum of digits of k) /2/ 2 is equal to
View answer and explanation
Correct option: A
9
Option A (9) is correct because the sum of digits of k (1890) divided by 2 is equal to 9.
623 Medium
623 of 742
Application +4 | -1
Let a be a factor of 120 . Then the number of positive integral solutions of x1,x2,x3=a\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3=\mathrm{a} is k\mathrm{k} then k80\frac{\mathrm{k}}{80} is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the number of positive integral solutions of x1, x2, x3 = a is found to be 320, and then k/80 = 320/80 = 4.
624 Easy
624 of 742
Application +4 | -1
The number of ways in which two 10 paise, two 20 paise, three 25 paise and one 50 paise coins can be distributed among 8 children, so that each child gets only one coin, is
View answer and explanation
Correct option: B
1680
Option B (1680) is correct because it calculates the number of ways to distribute the coins considering repetitions.
625 Easy
625 of 742
Application +4 | -1
If n+5pn+1=12(11)(n−1)(n+3pn){ }^{n+5} p_{n+1}=\frac{1}{2}(11)(n-1)\left({ }^{n+3} p_n\right) then nn is equal to:
View answer and explanation
Correct option: D
6, 7
The correct answer is 6, 7 because substituting these values into the given equation satisfies the equality after simplifying the permutations.
626 Medium
626 of 742
Application +4 | -1
The number of ways in which the sum of upper faces of four distinct dices can be six, is
View answer and explanation
Correct option: A
10
Option A (10) is correct because it represents the number of ways to distribute the sum of 6 among four distinct dice, each having a minimum value of 1.
627 Easy
627 of 742
Application +4 | -1
The number of arrangements that can be formed out of GANESHPURI so that "The vowels occupy even places are
View answer and explanation
Correct option: B
5P4⋅6P6{ }^5 P_4 \cdot{ }^6 P_6
Option B (5P4⋅6P6{ }^5 P_4 \cdot{ }^6 P_6) is correct because it represents the number of ways to arrange the 4 vowels in the 5 even places, and then arrange the remaining 6 letters in the remaining 6 places.
628 Medium
628 of 742
Application +4 | -1
The number of ways in which 3 children can distribute 10 tickets out of 15 consecutively numbered tickets themselves such that they get consecutive blocks of 5, 3 and 2 tickets is
View answer and explanation
Correct option: C
8C5(3!)2{ }^8 {C}_5(3 !)^2
Option B (π+1\pi+1) is correct because it maximizes the sum of the inverse trigonometric functions by considering the range of each function and the given domain.
629 Easy
629 of 742
Analyzing +4 | -1
In the decimal system of numeration, the number of 6-digit numbers in which the sum of digits is divisible by 5, is
View answer and explanation
Correct option: A
180000
Option A (180000) is correct because it is the result of calculating the number of 6-digit numbers where the sum of the digits is divisible by 5.
630 Easy
630 of 742
Application +4 | -1
How many numbers between 5000 and 10,000 can be formed using the digits 1,2,3,4,5,6,7,8,9, each digit appearing not more than once in each number?
View answer and explanation
Correct option: A
5×  8P35 \times {\;^8}{P_3}
Option A (5×  8P35 \times {\;^8}{P_3}) is correct because there are 5 choices for the thousands digit (5, 6, 7, 8, 9) and then 8ˆP3{{\^8}P_3} ways to arrange the remaining 3 digits from the remaining 8 digits.
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