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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 391-420 of 485 questions

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391 Easy
391 of 485
Application +4 | -1
The number of even proper divisors of 1008 is
View answer and explanation
Correct option: A
23
Option A (23) is correct because it is the number of even proper divisors of 1008, calculated by considering the prime factorization and excluding the number itself.
392 Easy
392 of 485
Application +4 | -1
From 3 mangoes, 4 apples and 2 oranges. The number of selections of fruits that can be made, taking at least one of each kind is
View answer and explanation
Correct option: A
24
Option A (24) is correct because it represents the number of selections with at least one of each kind of fruit.
393 Easy
393 of 485
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: B
41
Option B (41) is correct because substituting r = 41 into the given permutation equation satisfies the condition.
394 Easy
394 of 485
Application +4 | -1
The number of 3 letters words, with or without meaning which can be formed out of the letters of the word 'NUMBER'.

Statement I: When repetition of letters is not allowed is 120.

Statement II: When repetition of letters is allowed is 216.

Choose the correct option.
View answer and explanation
Correct option: C
Both I and II are correct
Option C (Both I and II are correct) is correct because the number of 3-letter words without repetition is 120 and with repetition is 216.
395 Easy
395 of 485
Analyzing +4 | -1
We are required to form different words with the help of the letters of the word INTEGER. Let m1m_1 be the number of words in which I and N are never together and m2m_2 be the number of words which begin with I and end with R, then m1m2\frac{m_1}{m_2} is given by
View answer and explanation
Correct option: B
30
Option B (30) is correct because the ratio of the number of words with I and N not together to the number of words starting with I and ending with R is 1800/60 = 30.
396 Easy
396 of 485
Application +4 | -1
The number of five digit numbers that can be formed by using digits 1, 2, 3 only, such that three digits of the formed number are identical, is
View answer and explanation
Correct option: C
120
The correct answer is C (120) because it accounts for the cases where exactly three digits are identical and the remaining two are different, and where three digits are identical and the remaining two are also identical.
397 Easy
397 of 485
Application +4 | -1
The number of words that can be formed from the letters a,b,c,d,e,fa, b, c, d, e, f, taken 3 at a time, each word containing at least one vowel is
View answer and explanation
Correct option: A
96
Option A (96) is correct because it represents the number of 3-letter words formed with at least one vowel from the given letters.
398 Easy
398 of 485
Application +4 | -1
The range of the function f(x)=7−xPx−3f(x) = {}^{7-x} P_{x-3} is
View answer and explanation
Correct option: A
{1,2,3}\left\{ {1,2,3} \right\}
Option A ({1,2,3}\{ {1,2,3} \}) is correct because the range is found by determining the valid domain of x and then evaluating the function at those points.
399 Easy
399 of 485
Analyzing +4 | -1
The number of divisors of n=27⋅35⋅53n=2^7 \cdot 3^5 \cdot 5^3, which are of the form 4t+1(t∈N)4 t+1(t \in N), is
View answer and explanation
Correct option: C
11
Option C (11) is correct because it accurately counts the number of divisors of the given form 4t+1.
400 Easy
400 of 485
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.
401 Easy
401 of 485
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.
402 Easy
402 of 485
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.
403 Easy
403 of 485
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.
404 Easy
404 of 485
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.
405 Easy
405 of 485
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.
406 Easy
406 of 485
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.
407 Easy
407 of 485
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.
408 Easy
408 of 485
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.
409 Easy
409 of 485
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.
410 Easy
410 of 485
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.
411 Easy
411 of 485
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.
412 Easy
412 of 485
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.
413 Easy
413 of 485
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.
414 Easy
414 of 485
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.
415 Easy
415 of 485
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.
416 Easy
416 of 485
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.
417 Easy
417 of 485
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.
418 Easy
418 of 485
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.
419 Easy
419 of 485
Application +4 | -1
The number of all five digit numbers which are divisible by 4 that can be formed from the digits 0,1,2,3,4 (without repetition) is
View answer and explanation
Correct option: B
30
Option B (30) is correct because it accurately counts the number of five-digit numbers divisible by 4 that can be formed using the digits 0, 1, 2, 3, 4 without repetition.
420 Easy
420 of 485
Analyzing +4 | -1
The numbers of times the digits 3 will be written when listing the integers from 1 to 1000 is
View answer and explanation
Correct option: B
300
Option B (300) is correct because the digit 3 appears 300 times when listing integers from 1 to 1000.
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