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 151-180 of 742 questions

Preparing practice and saved progress…

151 Easy
151 of 742
Application +4 | -1
The number of ways in which 6 different beads can be string into a necklace is
View answer and explanation
Correct option: A
60
Option A ((n−2)(n−1)!(n-2)(n-1)!) is correct because we calculate the total arrangements and subtract the arrangements where the two books are together.
152 Easy
152 of 742
Application +4 | -1
The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is
View answer and explanation
Correct option: C
77
Option C (77) is correct because the total number of integers is the sum of the possibilities from two cases: five 1's, one 2, one 3 and four 1's, three 2's.
153 Medium
153 of 742
Application +4 | -1
In a polygon, no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon is 70, then the number of diagonals of the polygon is
View answer and explanation
Correct option: A
20
Option D (None of these) is incorrect. Option A (20) is correct because the number of diagonals is calculated by nC2−n=20{}^nC_2 - n = 20 where n = 8.
154 Easy
154 of 742
Application +4 | -1
The number of ways in which you can put five beads of five different colours to form a necklace is:
View answer and explanation
Correct option: A
12
Option A (12) is correct because the number of ways to arrange the beads in a necklace is (n-1)!/2 due to rotational and reflectional symmetry.
155 Easy
155 of 742
Application +4 | -1
Five persons are to address a meeting. If a specified speaker is to speak before another specified speaker, find the number of ways in which this could be arranged. In how many of these arrangements will the first speaker come immediately before the second?
View answer and explanation
Correct option: A
60 , 24
Option A (60 , 24) is correct because it accurately calculates the number of arrangements with the specified speaker order and the number of arrangements where they speak immediately before each other.
156 Medium
156 of 742
Application +4 | -1
Four speakers will address a meeting, where speaker QQ will always speak after speaker PP. Then, the number of ways in which the order of speakers can be prepared is
View answer and explanation
Correct option: D
12
Option D (12) is correct because in half of the total arrangements, speaker Q will speak after speaker P, so the number of ways is 4!/2=124!/2 = 12.
157 Easy
157 of 742
Application +4 | -1
In how many ways a garland can be made from exactly 10 flowers?
View answer and explanation
Correct option: D
9!2\frac{{9!}}{2}
Option D (9!2\frac{{9!}}{2}) is correct because arranging flowers in a garland is a circular permutation, and since a garland can be flipped, we divide by 2.
158 Easy
158 of 742
Application +4 | -1
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then the number of such arrangement is
View answer and explanation
Correct option: C
At least 1000
Option C (At least 1000) is correct because the total number of arrangements is calculated by selecting the novels and dictionary, then arranging them with the dictionary in the middle, resulting in 1080.
159 Medium
159 of 742
Application +4 | -1
The number of ways of arranging letters of the word HAVANA so that VV and NN do not appear together is
View answer and explanation
Correct option: B
80
Option B (80) is correct because we calculate the total arrangements of HAVANA and subtract the arrangements where V and N are together.
160 Easy
160 of 742
Application +4 | -1
The number of ways in which 5 pictures can be hung from 7 picture nails on the wall is
View answer and explanation
Correct option: C
2520
Option C (2520) is correct because the number of ways to hang 5 pictures from 7 nails is a permutation problem, calculated as 7P5=7!2!=2520{7}{P_5} = \frac{{7!}}{{2!}} = 2520.
161 Medium
161 of 742
Application +4 | -1
From the vertices of a regular polygon of 10 sides, the number of ways of selecting three vertices such that no two vertices are consecutive is
View answer and explanation
Correct option: C
50
Option C (50) is correct because the number of ways to select three non-consecutive vertices is calculated by subtracting the number of ways to select consecutive vertices from the total number of ways to select three vertices.
162 Medium
162 of 742
Application +4 | -1
Total number of 3 letter words that can be formed from the letter of the word 'AAAHNPRRS' is equal to
View answer and explanation
Correct option: C
247
Option D (227) is the closest to the correct answer, but the provided solution steps are incorrect. The correct approach involves considering different cases of letter combinations.
163 Medium
163 of 742
Application +4 | -1
Find the number of 6 digits numbers whose sum is 10 using the digits 1,2,31,2,3 only
View answer and explanation
Correct option: C
90
Option C (90) is correct because it sums the arrangements for all possible cases where the sum of 6 digits using 1, 2, and 3 equals 10.
164 Easy
164 of 742
Application +4 | -1
How many nine digit numbers can be formed by using the digits 2,2,3,3,5,5,8,8,82,2,3,3,5,5,8,8,8 so that the odd digits occupy even positions?
View answer and explanation
Correct option: D
60
The correct answer is 60 because the odd digits must occupy the even positions, and the even digits must occupy the odd positions. The number of arrangements is calculated by multiplying the arrangements of odd digits in even places and even digits in odd places.
165 Easy
165 of 742
Analyzing +4 | -1
Total number of divisors of n=25⋅34⋅510⋅76n=2^5 \cdot 3^4 \cdot 5^{10} \cdot 7^6 that are of the form 4λ+2,λ≥14 \lambda+2, \lambda \geq 1, is equal to
View answer and explanation
Correct option: C
384
Option C (384) is correct because it accurately counts the number of divisors of the specified form 4λ+2.
166 Medium
166 of 742
Application +4 | -1
Seven different lecturers are to deliver lectures in seven periods of a class on a particular day. A,BA, B and CC are three of the lecturers. The number of ways in which a routine for the day can be made such that AA delivers his lecture before BB and BB before CC, is
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because the number of ways to arrange the lectures with A before B and B before C is 840.
167 Medium
167 of 742
Application +4 | -1
The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy B1B_{1} and a particular girl G1\mathrm{G}_{1} never sit adjacent to each other, is :
View answer and explanation
Correct option: A
5×65 \times 6 !
Option A (5 x 6!) is correct because it accounts for arranging the remaining boys and girls around the table after fixing B1 and G1's positions, ensuring they are not adjacent.
168 Easy
168 of 742
Application +4 | -1
The number of 4-digit even numbers that can be formed using 0,1,2,3,4,5,60,1,2,3,4,5,6 without repetition is
View answer and explanation
Correct option: C
420
Option C (420) is correct because it accounts for the restrictions on the first digit (cannot be 0) and the last digit (must be even) when forming the 4-digit even numbers.
169 Medium
169 of 742
Application +4 | -1
The number of ways in which 6 men can be arranged in a row so that three particular men are consecutive is
View answer and explanation
Correct option: B
4P4×3P3{ }^{4} P_{4} \times{ }^{3} P_{3}
Option B (4P4×3P3{ }^{4} P_{4} \times{ }^{3} P_{3}) is correct because it treats the three consecutive men as a single unit and then arranges them internally.
170 Medium
170 of 742
Application +4 | -1
The letters of the word GHOTU is arranged alphabetically as in a dictionary. The rank of the word TOUGH is
View answer and explanation
Correct option: D
89
Option A (π4ln⁡2\frac{\pi}{4}\ln 2) is incorrect. Option B (π2ln⁡2\frac{\pi}{2}\ln 2) is the correct answer because by using the substitution x = 1/t, the integral can be simplified and solved.
171 Easy
171 of 742
Application +4 | -1
Seven women and seven men are to sit round a circular table such that there is a man on either side of every women; the number of seating arrangements is
View answer and explanation
Correct option: C
6  !×7  !6\;! \times 7\;!
Option C (6  !×7  !6\;! \times 7\;!) is correct because it accounts for arranging the women around the table and then arranging the men in the spaces between them.
172 Medium
172 of 742
Application +4 | -1
A round table conference is to be held between 20 delegates of 2 countries. In how many ways can they be seated if two particular delegates are always together
View answer and explanation
Correct option: D
18!×2!18 ! \times 2 !
Option D (18!×2!18 ! \times 2 !) is correct because treating the two delegates as one unit leaves 19 units to arrange in a circle (18!), and the two delegates can switch places (2!).
173 Easy
173 of 742
Application +4 | -1
In a class of 10 students there are 3 girls students. The number of ways in which they can be arranged in a row, so that no two girls are consecutive is k(8!),k(8!), then kk is equal to
View answer and explanation
Correct option: D
42
Option D (42) is correct because arranging the boys and then placing the girls in the gaps results in k=42 when expressed as k(8!).
174 Easy
174 of 742
Application +4 | -1
Let AA be a set containing nn elements. The number of ways of choosing two subsets PP and QQ of AA such that P∩Q=ϕP \cap Q=\phi is
View answer and explanation
Correct option: B
3n3^n
Option B (3n3^n) is correct because each element has three possibilities: belonging to P, belonging to Q, or belonging to neither.
175 Easy
175 of 742
Understanding +4 | -1
The number of five digits numbers that can be formed without any restriction, is
View answer and explanation
Correct option: C
90000
Option C (90000) is correct because the first digit can be any number from 1 to 9 (9 options), and the remaining four digits can be any number from 0 to 9 (10 options each).
176 Medium
176 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 letters with atleast three between every two. The number of ways in which this can be done, is
View answer and explanation
Correct option: C
2400
Option C (2400) is correct because it represents the number of ways to insert 15 blanks between 5 letters with at least 3 blanks between each pair.
177 Easy
177 of 742
Application +4 | -1
Seven atheletes are participating in a race. In how many ways can the first three prizes be won?
View answer and explanation
Correct option: B
210
Option B (210) is correct because it represents the number of ways to arrange 7 athletes taken 3 at a time, calculated using permutations.
178 Easy
178 of 742
Analyzing +4 | -1
The number of ways in which the letters of the word ARRANGE be arranged so that "neither two A's nor two R's are together
View answer and explanation
Correct option: B
660
The correct answer is B (660) because it represents the number of arrangements where neither two A's nor two R's are together, calculated by subtracting arrangements with two R's together from those with two A's together.
179 Easy
179 of 742
Application +4 | -1
How many numbers of 6 digits can be formed from the digits of the number 112233?
View answer and explanation
Correct option: C
90
Option C (90) is correct because the number of 6-digit numbers formed from 112233 is calculated by dividing the total permutations by the repetitions of each digit.
180 Easy
180 of 742
Application +4 | -1
The number of ways in which 6 men and 5 women can dine at a round table, if no two women are to sit together, is
View answer and explanation
Correct option: A
6!×5!6! \times 5!
Option A, 6!×5!6! \times 5!, is correct because we first arrange the men around the table and then arrange the women in the spaces between them.
Revision notes and key takeaways

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

Showing 151-180 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.