Skip to main content
Back to PERMUTATION AND COMBINATION

Your Progress

0 / 742 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 181-210 of 742 questions

Preparing practice and saved progress…

181 Easy
181 of 742
Application +4 | -1
The sum of all 4 digited numbers that can be formed using the digits 1,2,4,5,61,2,4,5,6 without repetition is
View answer and explanation
Correct option: A
479952
Option A (479952) is correct because it calculates the sum of all 4-digit numbers formed using the given digits without repetition.
182 Medium
182 of 742
Application +4 | -1
If NN is the number of positive integral solution of x1x2x3x4=770{x_1}{x_2}{x_3}{x_4} = 770, then the value of NN is
View answer and explanation
Correct option: D
256
Option D (256) is correct because each prime factor of 770 can be assigned to any of the four variables, resulting in 44 possible solutions.
183 Easy
183 of 742
Application +4 | -1
If   12Pr=  11P6+6.  11P5{\;^{12}}{P_r} = {\;^{11}}{P_6} + 6.{\;^{11}}{P_5} then rr is equal to
View answer and explanation
Correct option: A
6
Option A (6) is correct because by using the given relation and simplifying, we find that r = 6.
184 Medium
184 of 742
Application +4 | -1
If seven women and seven men are to be seated around a circular table such that there is a man on either side of every woman, then the number of seating arrangements is
View answer and explanation
Correct option: A
6!76 ! 7 !
Option A (6!7!) is correct because 7 women can be arranged in 6! ways around a circular table, and then the 7 men can be arranged in 7! ways in the spaces between the women.
185 Medium
185 of 742
Application +4 | -1
There are 3 set of parallel lines containing respectively pp lines, qq lines and rr lines respectively. The greatest number of parallelograms that can be formed by the system
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the formula provided does not correctly calculate the number of parallelograms.
186 Easy
186 of 742
Application +4 | -1
The number of lines drawn through 6 points lying on a circle is
View answer and explanation
Correct option: B
15
Option B (15) is correct because it calculates the number of ways to choose 2 points out of 6 to form a line.
187 Easy
187 of 742
Understanding +4 | -1
How many 10 digits numbers can be written by using digits 9 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 a 9 or a 2, giving 2 options for each digit.
188 Medium
188 of 742
Application +4 | -1
The exponent of 12 in 50! is
View answer and explanation
Correct option: A
22
Option A (22) is correct because the exponent of 12 in 50! is determined by the minimum of the exponent of 3 and half the exponent of 2 in the prime factorization of 50!.
189 Easy
189 of 742
Application +4 | -1
Total number of three letter words that can be formed from the letters of the word 'SAHARANPUR' is equal to
View answer and explanation
Correct option: C
247
Option C (247) is correct because it accurately counts the number of three-letter words that can be formed from the letters of 'SAHARANPUR'.
190 Medium
190 of 742
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.
191 Easy
191 of 742
Application +4 | -1
In a circus there are ten cages for accommodating ten animals. Out of these four cages are so small that five out of 10 animals cannot enter into them. In how many ways will it be possible to accommodate ten animals in these ten cages?
View answer and explanation
Correct option: B
86400
Option B (86400) is correct because it first arranges the animals that can't go into the small cages, then arranges the remaining animals in the remaining cages.
192 Easy
192 of 742
Application +4 | -1
If the permutations of a,b,c,d,ea,b,c,d,e taken all together be written down in alphabetical order as in dictionary and numbered, then the rank of the permutation debacdebac is
View answer and explanation
Correct option: D
93
Option D (93) is correct because the rank of the permutation 'debac' is determined by counting the permutations that come before it in lexicographical order.
193 Medium
193 of 742
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.
194 Easy
194 of 742
Analyzing +4 | -1
Let pp be a prime number such that p≥3p \geq 3. Let n=p!+1.n=p !+1. The number of primes in the list n+1,n+2,n+3,....,n+p−1n + 1, n + 2, n + 3, ...., n + p − 1 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because for each kk in the list, n+kn+k is divisible by k+1k+1, meaning none of the numbers in the list are prime.
195 Easy
195 of 742
Application +4 | -1
How many numbers divisible by 5 and lying between 3000 and 4000 that can be formed from the digits 1, 2, 3, 4, 5, 6 (repetition of digits is not allowed)?
View answer and explanation
Correct option: C
  4P2{\;^4}{P_2}
Option A (216) is incorrect. Option B (375) is correct because it correctly calculates the total number of four-digit odd numbers that can be formed using the given digits.
196 Hard
196 of 742
Analyzing +4 | -1
lim⁡n→∞∑1⩽i∑⩽j⩽n(i+1)(j+1)n4\lim\limits_{n \to \infty } \frac{{\sum\limits_{1 \leqslant i} {\sum\limits_{ \leqslant j \leqslant n} {\left( {i + 1} \right)\left( {j + 1} \right)} } }}{{{n^4}}} is equal to
View answer and explanation
Correct option: C
18\frac{1}{8}
Option C (18\frac{1}{8}) is the correct answer because the limit of the given expression as n approaches infinity is 18\frac{1}{8}.
197 Easy
197 of 742
Application +4 | -1
If [y][y] denote the greatest integer ≤y\leq y, and 2[x8]2+3[x8]=202\left[\frac{x}{8}\right]^2+3\left[\frac{x}{8}\right]=20, then xx lies in the smallest interval [a,b)[a, b) where b−ab-a is equal to
View answer and explanation
Correct option: D
8
Option D (LICPEN) is correct because by arranging the permutations in dictionary order, the 413th word is LICPEN.
198 Medium
198 of 742
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.
199 Easy
199 of 742
Application +4 | -1
How many different arrangements can be made out of the letters in the expansion A2B3C4A^2 B^3 C^4, when written in full?
View answer and explanation
Correct option: B
9!2!3!4!\frac{9 !}{2 ! 3 ! 4 !}
Option B (9!2!3!4!\frac{{9 !}}{{2 ! 3 ! 4 !}}) is correct because it correctly applies the formula for permutations with repetitions.
200 Easy
200 of 742
Application +4 | -1
If   nPr=30240{\;^n}{P_r} = 30240 and   nCr=252{\;^n}{C_r} = 252, then the ordered pair (n,  r)\left( {n,\;r} \right) is equal to
View answer and explanation
Correct option: B
(10,5)
Option B (10,5) is correct because it satisfies both the permutation and combination equations given in the problem.
201 Easy
201 of 742
Application +4 | -1
In an examination a minimum is to be secured in each of 5 subjects for a pass. The number of ways in which a student can fail is
View answer and explanation
Correct option: A
31
Option A (31) is correct because it represents the total number of ways a student can fail in at least one subject out of 5, calculated using combinations.
202 Easy
202 of 742
Application +4 | -1
A1,A2…AnA_1, A_2 \ldots A_n are points on the line y=xy=x lying in the positive quadrant such that OAn=nOAn−1,OO A_n=n O A_{n-1}, O being the origin. If OA1=1O A_1=1 and the coordinates of AnA_n are (25202,25202)(2520 \sqrt{2}, 2520 \sqrt{2}), then n=n=
View answer and explanation
Correct option: C
7
Option C (7) is correct because OAn=n!OA_n = n! and solving for nn with the given coordinates of AnA_n gives n=7n = 7.
203 Easy
203 of 742
Application +4 | -1
Let Pm{P_m} stand for   mPm.{\;^m}{P_m}. Then, 1+P1+2  P2+3  P3+…+n.Pn1 + {P_1} + 2\;{P_2} + 3\;{P_3} + \ldots + n.{P_n} is equal to
View answer and explanation
Correct option: C
(n+1)  !−1\left( {n + 1} \right)\;! - 1
Option C ((n+1) !−1\left( {n + 1} \right)\,! - 1) is correct because it represents the sum of the series 1+P1+2  P2+3  P3+…+n.Pn1 + {P_1} + 2\;{P_2} + 3\;{P_3} + \ldots + n.{P_n}.
204 Easy
204 of 742
Application +4 | -1
The number of ways in which the letters of the word "STRANGE" can be arranged so that the vowels may appear in the odd places, is
View answer and explanation
Correct option: A
1440
Option A (1440) is correct because the 2 vowels can be arranged in 4 odd places in P(4,2){P}(4,2) ways and the 5 consonants can be arranged in the remaining 5 places in P(5,5){P}(5,5) ways.
205 Hard
205 of 742
Application +4 | -1
The number of ways of selecting nn objects out of 3n3 n objects, of which nn are alike and rest are different, is
View answer and explanation
Correct option: A
22n−1+(2n−1)!n!(n−1)!2^{2 n-1}+\frac{(2 n-1) !}{n !(n-1) !}
Option A (22n−1+(2n−1)!n!(n−1)!2^{2 n-1}+\frac{{(2 n-1) !}}{{n !(n-1) !}}) is correct because it represents the number of ways to select n objects out of 3n objects, where n are alike and the rest are different.
206 Easy
206 of 742
Application +4 | -1
The number of all possible words that can be formed using the letters of the word “MATHEMATICS” is
View answer and explanation
Correct option: A
11  !2  !2  !2  !\frac{{11\;!}}{{2\;!2\;!2\;!}}
Option A (11  !2  !2  !2  !\frac{{11\;!}}{{2\;!2\;!2\;!}}) is correct because it accounts for the repetitions of the letters M, A, and T in the word MATHEMATICS.
207 Easy
207 of 742
Application +4 | -1
Find the number of ways of arranging the letters of the word 'BANANA'
View answer and explanation
Correct option: A
60
Option A (60) is correct because it accounts for the repetitions of the letters A and N in the word BANANA.
208 Easy
208 of 742
Application +4 | -1
The number of ways in which 20 different pearls of two colours can be set alternately on a necklace, there being 10 pearls of each colour, is
View answer and explanation
Correct option: B
5×(9!)25 \times {\left( {9!} \right)^2}
Option B (5×(9!)25 \times {\left( {9!} \right)^2}) is correct because it accounts for arranging the pearls of one color in a circle and then arranging the pearls of the other color in the spaces.
209 Medium
209 of 742
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.
210 Medium
210 of 742
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.
Revision notes and key takeaways

Review PERMUTATION AND COMBINATION in your textbook or class notes, then use this set to check your understanding.

Showing 181-210 of 742 questions

Your practice

0 / 742 answered · 742 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.