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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Medium

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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 61-90 of 250 questions
61 Medium
61 of 250
Application +4 | -1
The number of natural numbers which are less than 2⋅1082 \cdot 10^8 and which can be written by means of the digit 1 and 2, is
View answer and explanation
Correct option: D
766
Option D (766) is correct because the sum of all possible numbers formed using digits 1 and 2, up to 8 digits, plus the 9-digit numbers starting with 1, equals 766.
62 Medium
62 of 250
Application +4 | -1
If 2n+1Pn−1:2n−1Pn=3:5{ }^{2 n+1} P_{n-1}:{ }^{2 n-1} P_n=3: 5, then nn is equal to
View answer and explanation
Correct option: A
4
Option A (4) is correct because solving the ratio of permutations leads to a quadratic equation with n=4 as the valid solution.
63 Medium
63 of 250
Analyzing +4 | -1
There are 20 questions in a question paper. If no two students solve the same combination of questions but solve equal number of questions, then the maximum number of students who appeared in the examination, is
View answer and explanation
Correct option: C
20C10{ }^{20} C_{10}
Option B (20C10{ }^{20} C_{10}) is correct because the maximum number of students is achieved when each student solves 10 questions, which corresponds to the largest binomial coefficient.
64 Medium
64 of 250
Analyzing +4 | -1
The total number of ways of arranging the letters AAAA BBB CC D E F in a row such that letters C are separated from one another is
View answer and explanation
Correct option: B
1386000
Option B (1386000) is correct because it accurately calculates the arrangements where the two C's are separated by first arranging the other letters and then placing the C's in the gaps.
65 Medium
65 of 250
Application +4 | -1
Out of 7 consonants and 4 vowels, the number of words (not necessarily meaningful) that can be made, each consisting of 3 consonants and 2 vowels, is
View answer and explanation
Correct option: C
25200
Option C (25200) is correct because we select 3 consonants out of 7 and 2 vowels out of 4, then arrange them in 5! ways.
66 Medium
66 of 250
Application +4 | -1
Sixteen men compete with one another in running swimming and riding. How many prize lists could be made if there were altogether 6 prizes of different values, one for running, 2 for swimming and 3 for riding?
View answer and explanation
Correct option: B
163×152×14{16^3} \times {15^2} \times 14
Option B (163×152×14{16^3} \times {15^2} \times 14) is correct because it represents the product of the number of ways to award each prize.
67 Medium
67 of 250
Application +4 | -1
A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions, so that exactly 10 predictions are correct, is equal to
View answer and explanation
Correct option: A
20C10⋅210{ }^{20} C_{10} \cdot 2^{10}
Option A (20C10⋅210{ }^{20} C_{10} \cdot 2^{10}) is correct because it represents choosing the 10 correct matches and then the possibilities for the incorrect ones.
68 Medium
68 of 250
Application +4 | -1
Number of ways is which 50 identical books can be distributed in 100 students if each student can get atmost one book is
View answer and explanation
Correct option: A
(51)(52)…….(100)50!\frac{(51)(52) \ldots \ldots .(100)}{50 !}
Option A ((51)(52)...(100)/50!) is correct because it represents the number of ways to choose 50 students out of 100 to receive the books.
69 Medium
69 of 250
Application +4 | -1
The number of words of four letters containing equal number of vowels and consonants, repetition allowed, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated number of words does not match any of the given options.
70 Medium
70 of 250
Application +4 | -1
The number of words which can be formed the letters of the word MAXIMUM, if two consonants cannot occur together, is
View answer and explanation
Correct option: A
4!
Option A (4!) is correct because the vowels must occupy alternate positions, and the consonants can then be arranged in the remaining spaces.
71 Medium
71 of 250
Application +4 | -1
A variable name in certain computer language must be either an alphabet or an alphabet followed by a decimal digit. The total number of different variable names that can exist in that language is equal to
View answer and explanation
Correct option: C
286
Option C (286) is correct because it correctly calculates the total number of variable names by considering alphabets and alphabets followed by digits.
72 Medium
72 of 250
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.
73 Medium
73 of 250
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.
74 Medium
74 of 250
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.
75 Medium
75 of 250
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.
76 Medium
76 of 250
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.
77 Medium
77 of 250
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.
78 Medium
78 of 250
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.
79 Medium
79 of 250
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'.
80 Medium
80 of 250
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.
81 Medium
81 of 250
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.
82 Medium
82 of 250
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.
83 Medium
83 of 250
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.
84 Medium
84 of 250
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.
85 Medium
85 of 250
Application +4 | -1
The value of ∑r=115r2(15Cr15Cr−1)\sum\limits_{{r=1}}^{15} r^2\left(\frac{{ }^{15} C_r}{{ }^{15} C_{r-1}}\right) is equal to:
View answer and explanation
Correct option: D
680
Option D (680) is correct because the summation simplifies to a manageable expression after simplifying the ratio of combinations, and the final calculation yields 680.
86 Medium
86 of 250
Analyzing +4 | -1
If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which C and S do not come together, is (6!)k(6 !) k, then kk is equal to
View answer and explanation
Correct option: B
5670
Option B (5670) is correct because it calculates the total number of arrangements of the letters in MATHEMATICS and subtracts the number of arrangements where C and S are together.
87 Medium
87 of 250
Application +4 | -1
The number of 5-digit numbers that can be made using the digits 1 and 2 and in which atleast one digit is different, is
View answer and explanation
Correct option: A
30
Option C (3×443 \times 4^4) is correct because it is the number of different matrices that can be formed with the given elements.
88 Medium
88 of 250
Analyzing +4 | -1
(2n)!22n(n!)2≤\frac{(2 n) !}{2^{2 n}(n !)^2} \leq
View answer and explanation
Correct option: A
13n+1\frac{1}{\sqrt{3 n+1}}
Option A (13n+1\frac{{1}}{{\sqrt{{3n+1}} }}) is correct because substituting n=1 into the expression results in equality, and for n>1 the inequality holds.
89 Medium
89 of 250
Application +4 | -1
  nPr=3024{\;^n}{P_r} = 3024 and   nCr=126{\;^n}{C_r} = 126, then rr is
View answer and explanation
Correct option: B
4
Option B (4) is correct because by dividing the permutation formula by the combination formula, we can solve for r.
90 Medium
90 of 250
Analyzing +4 | -1
Find the number of permutations x1,x2,x3,x4,x5x_{1}, x_{2}, x_{3}, x_{4}, x_{5} of numbers 1,2,3,4,51,2,3,4,5 such that the sum of five products x1x2x3+x2x3x4+x3x4x5+x4x5x1+x5x1x2x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+x_{4} x_{5} x_{1}+x_{5} x_{1} x_{2} is divisible by 3
View answer and explanation
Correct option: B
80
The correct answer is 80 because the sum of the products is divisible by 3 if x2+x4x_2 + x_4 is divisible by 3, leading to 8 combinations for x2x_2 and x4x_4, and then 10 permutations for the remaining numbers.
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