Permutations and Combinations
220 Questions
Start JEE Mains Test
2021
Q151
JEE Mains
MCQ
14 Mar 2026
If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :
A.
240
B.
360
C.
333
D.
364
2021
Q152
JEE Mains
MCQ
14 Mar 2026
Team 'A' consists of 7 boys and n girls and Team 'B' has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n is equal to :
A.
5
B.
2
C.
4
D.
6
2021
Q153
JEE Mains
MCQ
14 Mar 2026
Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then ($\beta$ $-$ $\alpha$) is equal to :
A.
717
B.
795
C.
1890
D.
1173
2021
Q154
JEE Mains
MCQ
14 Mar 2026
A natural number has prime factorization given by n = 2x3y5z, where y and z are such
that y + z = 5 and y$-$1 + z$-$1 = ${5 \over 6}$, y > z. Then the number of odd divisions of n, including 1, is :
that y + z = 5 and y$-$1 + z$-$1 = ${5 \over 6}$, y > z. Then the number of odd divisions of n, including 1, is :
A.
11
B.
6
C.
12
D.
6x
2021
Q155
JEE Mains
MCQ
14 Mar 2026
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 :
A.
35
B.
42
C.
82
D.
77
2021
Q156
JEE Mains
MCQ
14 Mar 2026
The total number of positive integral solutions (x, y, z) such that xyz = 24 is :
A.
36
B.
24
C.
45
D.
30
2021
Q157
JEE Mains
MCQ
14 Mar 2026
A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at
least 2 Indians and double the number of foreigners as Indians. Then the number of ways,
the committee can be formed, is :
A.
1050
B.
575
C.
560
D.
1625
2020
Q158
JEE Mains
Numerical
14 Mar 2026
The number of words (with or without meaning)
that can be formed from all the letters of the
word “LETTER” in which vowels never come
together is ________ .
Correct Answer: 120
Explanation:
Consonants $ \to $ LTTR
Vowels $ \to $ EE
Total No of words = ${{6!} \over {2!2!}}$ = 180
Total no of words if vowels are together
= ${{5!} \over {2!}}$ = 60
$ \therefore $ Total no of words where
vowels never come together = 180 – 60 = 120.
Vowels $ \to $ EE
Total No of words = ${{6!} \over {2!2!}}$ = 180
Total no of words if vowels are together
= ${{5!} \over {2!}}$ = 60
$ \therefore $ Total no of words where
vowels never come together = 180 – 60 = 120.
2020
Q159
JEE Mains
Numerical
14 Mar 2026
Four fair dice are thrown independently 27 times. Then the expected number of times, at
least two dice show up a three or a five, is _________.
Correct Answer: 11
Explanation:
4 dice are independently thrown. Each die has probability to show 3 or 5 is
$P = {2 \over 6} = {1 \over 3}$
$ \therefore $ $q = 1 - {1 \over 3} = {2 \over 3}$ (not showing 3 or 5)
Experiment is performed with 4 dices independently
$ \therefore $ Their binomial distribution is
${(q + p)^4} = {(q)^4} + {}^4{C_1}{q^3}p + {}^4{C_2}{q^2}{p^2} + {}^4{C_3}q{p^3} + {}^4{C_4}{P^4}$
$ \therefore $ In one throw of each dice probability of showing 3 or 5 at least twice is
= ${p^4} + {}^4{C_3}q{p^3} + {}^4{C_2}{q^2}{p^2}$
$ = {{33} \over {81}}$
Given such experiment performed 27 times
$ \therefore $ So expected outcomes = np
= ${{33} \over {81}} \times 27$
= 11
$P = {2 \over 6} = {1 \over 3}$
$ \therefore $ $q = 1 - {1 \over 3} = {2 \over 3}$ (not showing 3 or 5)
Experiment is performed with 4 dices independently
$ \therefore $ Their binomial distribution is
${(q + p)^4} = {(q)^4} + {}^4{C_1}{q^3}p + {}^4{C_2}{q^2}{p^2} + {}^4{C_3}q{p^3} + {}^4{C_4}{P^4}$
$ \therefore $ In one throw of each dice probability of showing 3 or 5 at least twice is
= ${p^4} + {}^4{C_3}q{p^3} + {}^4{C_2}{q^2}{p^2}$
$ = {{33} \over {81}}$
Given such experiment performed 27 times
$ \therefore $ So expected outcomes = np
= ${{33} \over {81}} \times 27$
= 11
2020
Q160
JEE Mains
Numerical
14 Mar 2026
The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the letters of the word ’SYLLABUS’ such that two letters are distinct and two letters are alike, is :
Correct Answer: 240
Explanation:
In 'SYLLABUS' word
1. Two S letters
2. Two L letters
3. One Y letter
4. One A letter
5. One B letter
6. One U letter
Number of ways we can select two alike letters = 2C1
Then number of ways we can select two distinct letters = 5C2
Then total arrangement of selected letters = ${{4!} \over {2!}}$
So total number of words, with or without meaning, that can be formed
= 2C1 $ \times $ 5C2 $ \times $ ${{4!} \over {2!}}$ = 240
1. Two S letters
2. Two L letters
3. One Y letter
4. One A letter
5. One B letter
6. One U letter
Number of ways we can select two alike letters = 2C1
Then number of ways we can select two distinct letters = 5C2
Then total arrangement of selected letters = ${{4!} \over {2!}}$
So total number of words, with or without meaning, that can be formed
= 2C1 $ \times $ 5C2 $ \times $ ${{4!} \over {2!}}$ = 240
2020
Q161
JEE Mains
Numerical
14 Mar 2026
A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is __________.
Correct Answer: 135
Explanation:
Select any 4 questions in 6C4
ways which are
correct.
Answering right option for each question is possible in 1 way.
So ways of choosing right option for 4 questions = 1.1.1.1 = (1)4
Number of ways of choosing wrong option for each question = 3
So ways of choosing wrong option for 2 questions = (3)2
$ \therefore $ Required number of ways = 6C4.(1)4.(3)2 = 135
Answering right option for each question is possible in 1 way.
So ways of choosing right option for 4 questions = 1.1.1.1 = (1)4
Number of ways of choosing wrong option for each question = 3
So ways of choosing wrong option for 2 questions = (3)2
$ \therefore $ Required number of ways = 6C4.(1)4.(3)2 = 135
2020
Q162
JEE Mains
Numerical
14 Mar 2026
The total number of 3-digit numbers, whose
sum of digits is 10, is __________.
Correct Answer: 54
Explanation:
Let xyz is 3 digits number.
Given that sum of digits = 10
$ \therefore $ x + y + z = 10 ......(1)
Also x can't be 0 as if x = 0 then it will become 2 digits number.
So, x $ \ge $ 1, y $ \ge $ 0, z $ \ge $ 0
As x $ \ge $ 1
$ \Rightarrow $ x $-$ 1 $ \ge $ 0
Let x $-$ 1 = t
$ \therefore $ t $ \ge $ 0
From equation (1)
(x $-$ 1) + y + z = 9
$ \Rightarrow $ t + y + z = 9
Now this problem becomes, distributing 9 things among 3 people t, y, z.
Number of ways we can do that
= ${}^{9 + 3 - 1}{C_{3 - 1}} = {}^{11}{C_2} = 55$
Now when 3 digit number is 900 then t = 9, y = 0, z = 0.
And when t = 9, then
x $-$ 1 = 9
$ \Rightarrow $ x = 10
But we can't take x = 10 in a 3 digits number. So, we have to remove this case.
$ \therefore $ Total number of 3 digit numbers = 55 $-$ 1 = 54.
Given that sum of digits = 10
$ \therefore $ x + y + z = 10 ......(1)
Also x can't be 0 as if x = 0 then it will become 2 digits number.
So, x $ \ge $ 1, y $ \ge $ 0, z $ \ge $ 0
As x $ \ge $ 1
$ \Rightarrow $ x $-$ 1 $ \ge $ 0
Let x $-$ 1 = t
$ \therefore $ t $ \ge $ 0
From equation (1)
(x $-$ 1) + y + z = 9
$ \Rightarrow $ t + y + z = 9
Now this problem becomes, distributing 9 things among 3 people t, y, z.
Number of ways we can do that
= ${}^{9 + 3 - 1}{C_{3 - 1}} = {}^{11}{C_2} = 55$
Now when 3 digit number is 900 then t = 9, y = 0, z = 0.
And when t = 9, then
x $-$ 1 = 9
$ \Rightarrow $ x = 10
But we can't take x = 10 in a 3 digits number. So, we have to remove this case.
$ \therefore $ Total number of 3 digit numbers = 55 $-$ 1 = 54.
2020
Q163
JEE Mains
Numerical
14 Mar 2026
If the letters of the word 'MOTHER' be permuted
and all the words so formed (with or without
meaning) be listed as in a dictionary, then the
position of the word 'MOTHER' is ______.
Correct Answer: 309
Explanation:
2020
Q164
JEE Mains
Numerical
14 Mar 2026
The number of 4 letter words (with or without
meaning) that can be formed from the eleven
letters of the word 'EXAMINATION' is
_______.
Correct Answer: 2454
Explanation:
2A, 2I, 2N, E, X, M, T, O
To form four letter words
Case 1 : All same ( not possible)
Case 2 : 1 different, 3 same (not possible)
Case 3 : 2 different, 2 same
= 3C1 $ \times $ 7C2 $ \times $ ${{4!} \over {2!}}$ = 756
Case 4 : 2 same of one kind, 2 same same of other kind
= 3C2 $ \times $ ${{4!} \over {2!2!}}$ = 18
Case 5 : All letters are different
= 8C4 $ \times $ 4! = 1680
$ \therefore $ Total ways = 1680 + 756 + 18 = 2454
To form four letter words
Case 1 : All same ( not possible)
Case 2 : 1 different, 3 same (not possible)
Case 3 : 2 different, 2 same
= 3C1 $ \times $ 7C2 $ \times $ ${{4!} \over {2!}}$ = 756
Case 4 : 2 same of one kind, 2 same same of other kind
= 3C2 $ \times $ ${{4!} \over {2!2!}}$ = 18
Case 5 : All letters are different
= 8C4 $ \times $ 4! = 1680
$ \therefore $ Total ways = 1680 + 756 + 18 = 2454
2020
Q165
JEE Mains
Numerical
14 Mar 2026
An urn contains 5 red marbles, 4 black marbles
and 3 white marbles. Then the number of ways
in which 4 marbles can be drawn so that at the
most three of them are red is ___________.
Correct Answer: 490
Explanation:
Here 5 red marbels and 7 non red marbels presents.
No of ways 4 marbels can be chosen where atmost 3 red marbels can be present.
Case 1: When 3 red marbels present
No of ways = 5C3 $ \times $ 7C1
Case 2: When 2 red marbels present
No of ways = 5C2 $ \times $ 7C2
Case 3: When 1 red marbels present
No of ways = 5C1 $ \times $ 7C3
Case 4: When 0 red marbels present
No of ways = 5C0 $ \times $ 7C4
$ \therefore $ Total number of ways
= 5C3 $ \times $ 7C1 + 5C2 $ \times $ 7C2 + 5C1 $ \times $ 7C3 + 5C0 $ \times $ 7C4
= 70 + 210 + 175 + 35
= 490
No of ways 4 marbels can be chosen where atmost 3 red marbels can be present.
Case 1: When 3 red marbels present
No of ways = 5C3 $ \times $ 7C1
Case 2: When 2 red marbels present
No of ways = 5C2 $ \times $ 7C2
Case 3: When 1 red marbels present
No of ways = 5C1 $ \times $ 7C3
Case 4: When 0 red marbels present
No of ways = 5C0 $ \times $ 7C4
$ \therefore $ Total number of ways
= 5C3 $ \times $ 7C1 + 5C2 $ \times $ 7C2 + 5C1 $ \times $ 7C3 + 5C0 $ \times $ 7C4
= 70 + 210 + 175 + 35
= 490
2020
Q166
JEE Mains
MCQ
14 Mar 2026
Two families with three members each and one family with four members are to be seated in a row.
In how many ways can they be seated so that the same family members are not separated?
A.
2! 3! 4!
B.
(3!)3.(4!)
C.
3! (4!)3
D.
(3!)2.(4!)
2020
Q167
JEE Mains
MCQ
14 Mar 2026
There are 3 sections in a question paper and
each section contains 5 questions. A candidate
has to answer a total of 5 questions, choosing
at least one question from each section. Then
the number of ways, in which the candidate
can choose the questions, is :
A.
2250
B.
2255
C.
3000
D.
1500
2020
Q168
JEE Mains
MCQ
14 Mar 2026
The value of (2.1P0
– 3.2P1 + 4.3P2 .... up to
51th term)
+ (1! – 2! + 3! – ..... up to 51th term) is equal to :
+ (1! – 2! + 3! – ..... up to 51th term) is equal to :
A.
1
B.
1 + (51)!
C.
1 – 51(51)!
D.
1 + (52)!
2020
Q169
JEE Mains
MCQ
14 Mar 2026
Let n > 2 be an integer. Suppose that there are
n Metro stations in a city located along a
circular path. Each pair of stations is connected
by a straight track only. Further, each pair of
nearest stations is connected by blue line,
whereas all remaining pairs of stations are
connected by red line. If the number of red lines
is 99 times the number of blue lines, then the
value of n is :
A.
201
B.
199
C.
101
D.
200
2020
Q170
JEE Mains
MCQ
14 Mar 2026
If the number of five digit numbers with distinct
digits and 2 at the 10th place is 336 k, then k
is equal to :
A.
6
B.
8
C.
4
D.
7
2020
Q171
JEE Mains
MCQ
14 Mar 2026
If a, b and c are the greatest value of 19Cp, 20Cq
and 21Cr respectively, then :
A.
${a \over {11}} = {b \over {22}} = {c \over {21}}$
B.
${a \over {10}} = {b \over {22}} = {c \over {21}}$
C.
${a \over {10}} = {b \over {11}} = {c \over {42}}$
D.
${a \over {11}} = {b \over {22}} = {c \over {42}}$
2020
Q172
JEE Mains
MCQ
14 Mar 2026
The number of ordered pairs (r, k) for which
6.35Cr = (k2 - 3). 36Cr + 1, where k is an integer, is :
6.35Cr = (k2 - 3). 36Cr + 1, where k is an integer, is :
A.
6
B.
3
C.
2
D.
4
2020
Q173
JEE Mains
MCQ
14 Mar 2026
Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is :
A.
${5 \over 2}\left( {6!} \right)$
B.
${6!}$
C.
56
D.
${1 \over 2}\left( {6!} \right)$
2019
Q174
JEE Mains
MCQ
14 Mar 2026
A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can
randomly be selected from this group such that there is at least one boy and at least one girl in each team, is
1750, then n is equal to :
A.
24
B.
25
C.
27
D.
28
2019
Q175
JEE Mains
MCQ
14 Mar 2026
The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21
are distinct, is :
A.
220 - 1
B.
220
C.
220 + 1
D.
221
2019
Q176
JEE Mains
MCQ
14 Mar 2026
Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the
top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total
number of beams is :
A.
180
B.
210
C.
170
D.
190
Any two non-adjacent pillers are joined by beams
2019
Q177
JEE Mains
MCQ
14 Mar 2026
The number of 6 digit 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 :
A.
36
B.
60
C.
72
D.
48
2019
Q178
JEE Mains
MCQ
14 Mar 2026
A committee of 11 members is to be formed from
8 males and 5 females. If m is the number of ways
the committee is formed with at least 6 males and
n is the number of ways the committee is formed
with at least 3 females, then :
A.
n = m – 8
B.
m = n = 78
C.
m + n = 68
D.
m = n = 68
2019
Q179
JEE Mains
MCQ
14 Mar 2026
The number of four-digit numbers strictly greater
than 4321 that can be formed using the digits
0,1,2,3,4,5 (repetition of digits is allowed) is :
A.
306
B.
288
C.
310
D.
360
2019
Q180
JEE Mains
MCQ
14 Mar 2026
All possible numbers are formed using the digits
1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number
of such numbers in which the odd digits occupy
even places is :
A.
175
B.
162
C.
160
D.
180
2019
Q181
JEE Mains
MCQ
14 Mar 2026
There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is :
A.
12
B.
9
C.
7
D.
11
2019
Q182
JEE Mains
MCQ
14 Mar 2026
Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is :
A.
164
B.
240
C.
82
D.
120
2019
Q183
JEE Mains
MCQ
14 Mar 2026
If $\sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r}.{}^{50 - r}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} ,\,\,$ then K is equal to :
A.
224
B.
225$-$ 1
C.
225
D.
(25)2
2019
Q184
JEE Mains
MCQ
14 Mar 2026
Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is :
A.
9
B.
18
C.
36
D.
32
2019
Q185
JEE Mains
MCQ
14 Mar 2026
The number of natural numbers less than 7,000 which can be formed by using the digits 0, 1, 3, 7, 9 (repitition of digits allowed) is equal to :
A.
374
B.
372
C.
375
D.
250
2019
Q186
JEE Mains
MCQ
14 Mar 2026
Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can
be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same
team, is :
A.
500
B.
350
C.
200
D.
300
2018
Q187
JEE Mains
MCQ
14 Mar 2026
The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is :
A.
24
B.
30
C.
36
D.
48
2018
Q188
JEE Mains
MCQ
14 Mar 2026
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and
arranged in a row on a shelf so that the dictionary is always in the middle. The number of such
arrangements is :
A.
at least 750 but less than 1000
B.
at least 1000
C.
less than 500
D.
at least 500 but less than 750
2018
Q189
JEE Mains
MCQ
14 Mar 2026
The number of four letter words that can be formed using the letters of the word BARRACK is :
A.
120
B.
144
C.
264
D.
270
2018
Q190
JEE Mains
MCQ
14 Mar 2026
n$-$digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is :
A.
6
B.
7
C.
8
D.
9
2017
Q191
JEE Mains
MCQ
14 Mar 2026
The number of ways in which 5 boys and 3 girls can be seated on a round table if a
particular boy B1 and a particular girl G1 never sit adjacent to each other, is :
A.
5 $ \times $ 6!
B.
6 $ \times $ 6!
C.
7!
D.
5 $ \times $ 7!
2017
Q192
JEE Mains
MCQ
14 Mar 2026
If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is :
A.
44th
B.
45th
C.
46th
D.
47th
2017
Q193
JEE Mains
MCQ
14 Mar 2026
A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are
ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X
and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in
this party, is:
A.
468
B.
469
C.
484
D.
485
2016
Q194
JEE Mains
MCQ
14 Mar 2026
If ${{{}^{n + 2}C{}_6} \over {{}^{n - 2}{P_2}}}$ = 11, then n satisfies the
equation :
A.
n2 + 3n − 108 = 0
B.
n2 + 5n − 84 = 0
C.
n2 + 2n − 80 = 0
D.
n2 + n − 110 = 0
2016
Q195
JEE Mains
MCQ
14 Mar 2026
The sum $\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $ is equal to :
A.
(11)!
B.
10 $ \times $ (11!)
C.
101 $ \times $ (10!)
D.
11 $ \times $ (11!)
2016
Q196
JEE Mains
MCQ
14 Mar 2026
The value of $\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}}} \right)$ is equal to :
A.
560
B.
680
C.
1240
D.
1085
2016
Q197
JEE Mains
MCQ
14 Mar 2026
If the four letter words (need not be meaningful ) are to be formed using the
letters from the word “MEDITERRANEAN” such that the first letter is R and the fourth letter is E, then the total number of all such words is :
A.
${{11!} \over {{{\left( {2!} \right)}^3}}}$
B.
110
C.
56
D.
59
2016
Q198
JEE Mains
MCQ
14 Mar 2026
If all the words (with or without meaning) having five letters,formed using the letters of the word SMALL and arranged as in a dictionary, then the position of the word SMALL is :
A.
${46^{th}}$
B.
${59^{th}}$
C.
${52^{nd}}$
D.
${58^{th}}$
2015
Q199
JEE Mains
MCQ
14 Mar 2026
The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is:
A.
120
B.
72
C.
216
D.
192
2013
Q200
JEE Mains
MCQ
14 Mar 2026
Let ${T_n}$ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If ${T_{n + 1}} - {T_n}$ = 10, then the value of n is :
A.
7
B.
5
C.
10
D.
8