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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 151-180 of 485 questions

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151 Easy
151 of 485
Application +4 | -1
The exponent of 3 in 100 !, is
View answer and explanation
Correct option: C
48
Option A (72) is correct because it calculates the number of words formed with at least one vowel.
152 Easy
152 of 485
Understanding +4 | -1
There are 10 true-false questions. The number of ways, in which they can be answered is
View answer and explanation
Correct option: B
2102^{10}
Option B (2102^{10}) is correct because each of the 10 questions can be answered in 2 ways (true or false), leading to 2102^{10} total possibilities.
153 Easy
153 of 485
Understanding +4 | -1
The value of 10C5/11C6{ }^{10} C_5 /{ }^{11} C_6, when numerator and denominator takes their greatest value, is
View answer and explanation
Correct option: A
611\frac{6}{11}
Option A (611\frac{6}{11}) is correct because it is the simplified ratio of the two binomial coefficients.
154 Easy
154 of 485
Application +4 | -1
If all permutations of the letters of the word PENCIL are arranged as in a dictionary, then 413th word is
View answer and explanation
Correct option: D
LICPEN
Option D (LICPEN) is correct because after arranging the permutations of the letters in the word PENCIL in dictionary order, the 413th word is LICPEN.
155 Easy
155 of 485
Application +4 | -1
A seven-digit numbers divisible by 9 is to be formed by using 7 out of numbers {1,2,3,4,5,6,7,8,9}\{1,2,3,4,5, 6,7,8,9\}. The number of ways in which this can be done, is
View answer and explanation
Correct option: D
4⋅7!4 \cdot 7!
Option D (4⋅7!4 \cdot 7!) is correct because there are 4 pairs of digits that can be excluded to make the sum of the remaining 7 digits divisible by 9, and each set of 7 digits can be arranged in 7! ways.
156 Easy
156 of 485
Application +4 | -1
How many four digit numbers can be formed using the digits 1, 2, 3, 4, 5 such that at least one of the digit is repeated?
View answer and explanation
Correct option: D
54−5!{5^4} - 5!
Option D (54−5!{5^4} - 5!) is correct because it subtracts the number of four-digit numbers with no repetition from the total number of four-digit numbers that can be formed.
157 Easy
157 of 485
Application +4 | -1
A dictionary is printed consisting of 7 letters words only that can be made with a letters of the word CRICKET. If the words are printed at the alphabetical order as in an ordinary dictionary, then the number of words before the word CRICKET is
View answer and explanation
Correct option: A
530
Option A (530) is correct because it calculates the number of words before \"CRICKET\" by considering alphabetical order and accounting for repeated letters.
158 Easy
158 of 485
Application +4 | -1
The number of six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7, so that digits do not repeat and the terminal digits are even, is
View answer and explanation
Correct option: D
720
The correct answer is D (720) because it correctly calculates the number of six-digit numbers with even terminal digits from the given digits without repetition.
159 Easy
159 of 485
Application +4 | -1
The number of ways in which pp positive and nn negative signs may be placed in a row so that no two negative signs shall be together is
View answer and explanation
Correct option: A
p+1Cn{ }^{p+1} {C}_n
Option A (p+1Cn{}^{p+1} {C}_n) is correct because it represents the number of ways to place negative signs such that no two are together.
160 Easy
160 of 485
Analyzing +4 | -1
The number of 6 digits numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digit is repeated, is
View answer and explanation
Correct option: A
60
Option A (60) is correct because it is the total number of 6-digit numbers that can be formed using the given digits, divisible by 11, with no repetition.
161 Easy
161 of 485
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 A can be mapped to any of the k elements in A.
162 Easy
162 of 485
Application +4 | -1
If 10Cr−1>2(10Cr){ }^{10} C_{{r}-1} > 2\left({ }^{10} C_{\mathrm{r}}\right) then rr is equal to
View answer and explanation
Correct option: C
8, 9, 10
Option C (8, 9, 10) is correct because these values of r satisfy the inequality 10Cr−1>2(10Cr){}^{10} C_{{r}-1} > 2\left({}^{10} C_{\mathrm{r}}\right) after simplification.
163 Easy
163 of 485
Application +4 | -1
If x∈Nx \in {N} and (2x)!3!(2x−3)!:x!2!(x−2)=44:3\frac{(2 x) !}{3 !(2 x-3) !}: \frac{x !}{2 !(x-2)}=44: 3, then xx is equal to
View answer and explanation
Correct option: A
6
Option A (6) is correct because solving the given ratio of permutations leads to the equation 2x-1=11, which gives x=6.
164 Easy
164 of 485
Application +4 | -1
A basket ball team consists of 12 pairs of twin brothers. On the first day of training, all 24 players stand in a circle in such a way that all pairs of twin brothers are neighbours. Number of ways this can be done is :
View answer and explanation
Correct option: B
212(11!)2^{12}(11 !)
Option B (212(11!)2^{12}(11 !)) is correct because there are 11! ways to arrange the 12 pairs in a circle, and each pair can be arranged in 2 ways.
165 Easy
165 of 485
Analyzing +4 | -1
Consider the following statement:

1.The number of ways of arranging m different things taken all at a time in which p⩽mp \leqslant m perticular things are never together is m!−(m−p+1)!p!m! - \left( {m - p + 1} \right)!p!

2. A pack of 52 cards can be divided equally among four players in order in 52!(13!)4\frac{{52!}}{{{{\left( {13!} \right)}^4}}} ways

Which of these is/are correct?
View answer and explanation
Correct option: C
Both of these
Both statements are correct. Statement 1 correctly describes the number of arrangements where p particular things are never together, and statement 2 accurately represents the number of ways to divide a deck of cards equally among four players.
166 Easy
166 of 485
Application +4 | -1
The number of ways distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is
View answer and explanation
Correct option: B
21
Option B (21) is correct because it represents the number of ways to distribute 8 identical balls into 3 distinct boxes with no box empty, which is calculated using the stars and bars method.
167 Easy
167 of 485
Analyzing +4 | -1
A man has three friends. The number of ways he can invite one friend everyday for dinner on six successive nights, so that no friend is invited more than three times, is
View answer and explanation
Correct option: D
510
The correct answer is D (510) because it accounts for all possible ways the man can invite his friends, considering the constraint that no friend is invited more than three times.
168 Easy
168 of 485
Application +4 | -1
The number of arrangements which can be made out of the letters of the word ALGEBRA, without changing the relative order (positions) of vowels and consonants, is
View answer and explanation
Correct option: A
72
Option A (72) is correct because it accounts for arranging vowels and consonants separately while considering repeated vowels.
169 Easy
169 of 485
Application +4 | -1
Eleven books consisting of 5 Mathematics, 4 physics and 2 Chemistry are places on a shelf. The number of possible ways of arranging them on the assumption that the books of the same subject are all together, is
View answer and explanation
Correct option: C
5!4!3!2!5!4!3!2!
Option C (5!4!3!2!5!4!3!2!) is correct because it accounts for arranging the books within each subject and then arranging the subjects themselves.
170 Easy
170 of 485
Application +4 | -1
If the letters of the word ‘SACHIN’ are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number
View answer and explanation
Correct option: D
601
Option D (601) is correct because it represents the position of the word 'SACHIN' when all permutations are arranged alphabetically.
171 Easy
171 of 485
Application +4 | -1
The number of ways in which nn things of which rr are alike, can be arranged in a circular order, is
View answer and explanation
Correct option: B
(n−1)!r!\frac{(n-1) !}{r !}
Option B (48) is correct because we calculate the total permutations of 5-digit numbers and subtract those that start with 3 to ensure the number is greater than 50,000.
172 Easy
172 of 485
Application +4 | -1
Four married couples are to be seated in a row having 8 chairs. The number of ways so that spouses are seated next to each other, is
View answer and explanation
Correct option: C
384
Option C (384) is correct because it accounts for arranging the couples and then arranging the spouses within each couple.
173 Easy
173 of 485
Understanding +4 | -1
The value of ∑r=1n   nPrr!\sum \limits_{r = 1}^n \,\frac{{{\;^n}{P_r}}}{{r!}} is
View answer and explanation
Correct option: B
2n−1{2^n} - 1
Option B (2n−1{2^n} - 1) is correct because the given summation is equivalent to the binomial expansion of (1+1)n(1+1)^n excluding the first term.
174 Easy
174 of 485
Application +4 | -1
A man has 6 friends. In how many ways he can invite one or more of them at dinner?
View answer and explanation
Correct option: C
63
Option C (63) is correct because each friend can either be invited or not, leading to 262^6 possibilities, but we subtract 1 for the case where no friends are invited.
175 Easy
175 of 485
Application +4 | -1
Find the number of ways of arranging the letters of the word ENGINEERING
View answer and explanation
Correct option: D
11!3!3!2!2!\frac{11 !}{3 ! 3 ! 2 ! 2 !}
Option D (11!3!3!2!2!\frac{11 !}{3 ! 3 ! 2 ! 2 !}) is correct because it accounts for the repetitions of the letters E, N, G, and I in the word ENGINEERING.
176 Easy
176 of 485
Application +4 | -1
From 7 gentlemen and 4 ladies, a committee of 5 is to be formed. The number of ways in which this can be done so as to include atleast one lady is
View answer and explanation
Correct option: A
805
Option A (805) is correct because it sums the number of ways to form the committee with at least one lady.
177 Easy
177 of 485
Application +4 | -1
If one test (on screening paper basis) was conducted on Batch A, maximum number of marks is (90×3)=270.4(90 \times 3)=270.4 students get the marks lower than 80 . Coaching institute decide to inform their guardians, that is why their result card were sent to their home. The number of ways, in which all the letters were put in wrong envelopes, is
View answer and explanation
Correct option: A
9
Option A (9) is correct because the number of ways to put all 4 letters in the wrong envelopes is calculated using the derangement formula, which results in 9.
178 Easy
178 of 485
Application +4 | -1
How many 10 digit numbers can be written by using the digits 1 and 2?
View answer and explanation
Correct option: B
210{2^{10}}
Option B (210{2^{10}}) is correct because each of the 10 digits can be either 1 or 2, resulting in 2 options for each digit.
179 Easy
179 of 485
Application +4 | -1
If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent. The number of points in which they intersect each other is
View answer and explanation
Correct option: B
190
Option B (190) is correct because it calculates the number of intersection points formed by 20 lines, where no two are parallel and no three are concurrent.
180 Easy
180 of 485
Application +4 | -1
The number of ways 12 different books can be distributed equally among 4 persons, is
View answer and explanation
Correct option: C
12!(3!)4\frac{12 !}{(3 !)^4}
Option C (12!(3!)4\frac{{12 !}}{{{{\left( {3!} \right)}^4}}}) is correct because it represents the number of ways to distribute 12 distinct books equally among 4 people.
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