JEE Main

Explore popular questions from Permutations and Combinations for JEE Main. This collection covers Permutations and Combinations previous year JEE Main questions hand picked by experienced teachers.

Mathematics

Permutations and Combinations

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Q 1. In a shop there are five types of ice creams available. A child buys six ice creams.
Statement 1: The number of different ways the child can buy the six ice creams is {tex} ^ { 10 } \mathrm { C } _ { 5 } {/tex}.
Statement 2: The number of different ways the child can buy the six ice creams is equal to the number of different ways of arranging {tex}6 A ^ { \prime } s {/tex} and {tex}4 B ^ { \prime }s {/tex} in a row.

Statement 1 is false, Statement 2 is true

B

Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

C

Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1

D

Statement 1 is true, Statement 2 is false

Explanation

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Q 2. 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. Then the number of such arrangements is

A

less than 500 .

B

at least 500 but less than 750 .

C

at least 750 but less than 1000 .

at least 1000

Explanation

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Q 3. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

A

36

B

66

108

D

3

Explanation

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Q 4. Statement {tex} 1 : {/tex} The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is {tex} ^ { 9 } \mathrm { C } _ { 3 } {/tex}. Statement 2: The number of ways of choosing any 3 places from 9 different places is {tex} ^ { 9 } \mathrm { C } _ { 3 } {/tex} .

Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1

B

Statement 1 is true, Statement 2 is false

C

Statement 1 is false, Statement 2 is true

D

Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Explanation

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Q 5. Let {tex} T _ { n } {/tex} be the number of all possible triangles formed by joining vertices of an {tex} n {/tex} -sided regular polygon. If {tex} T _ { n + 1 } - T _ { n } = 10 {/tex} , then the value of {tex} n {/tex} is

5

B

10

C

8

D

7

Explanation

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Q 6. Let {tex} A {/tex} and {tex} B {/tex} be two sets containing 2 elements and 4 elements, respectively. The number of subsets of {tex} A \times B {/tex} having 3 or more elements is

A

220

219

C

211

D

256

Explanation

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Q 7. The sum of the digits in the unit's place of all the {tex}4{/tex} -digit numbers formed by using the numbers {tex} 3,4,5 {/tex} and {tex} 6 , {/tex} without repetition is

A

432

108

C

36

D

18

Explanation

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Q 8. The number of integers greater than {tex}6000{/tex} that can be formed using the digits {tex} 3,5,6,7 {/tex} and {tex}8{/tex} without repetition is

192

B

120

C

72

D

216

Explanation

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Q 9. The value of {tex} \underset {r=1}{ \overset {15} \sum} r ^ { 2 } \left( \frac { ^ { 15 } C _ { r } } { ^ { 15 } C _ { r - 1 } } \right) {/tex} is equal to

A

1240

B

560

C

1085

680

Explanation

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Q 10. 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 {tex} R {/tex} and the fourth letter is {tex} E {/tex} , then the total number of all such words is

A

{tex}\small 110{/tex}

{tex}\small 59{/tex}

C

{tex}\large \frac { 11 ! } { ( 2 ! ) ^ { 3 } } {/tex}

D

{tex}\small 56{/tex}

Explanation

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Q 11. If {tex} \frac { ^{n + 2}C_6 } { ^{n - 2}P_2 } = 11 , {/tex} then {tex} n {/tex} satisfies the equation:

A

{tex} n ^ { 2 } + n - 110 = 0 {/tex}

B

{tex} n ^ { 2 } + 2 n - 80 = 0 {/tex}

{tex} n ^ { 2 } + 3 n - 108 = 0 {/tex}

D

{tex} n ^ { 2 } + 5 n - 84 = 0 {/tex}

Explanation

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Q 12. Everybody in a room shakes hand with everybody else. The total number of handshakes is equal to {tex} 153 . {/tex} The total number of persons in the room is equal to

18

B

19

C

17

D

16

Explanation

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Q 13. Number of triangles that can be formed joining the angular points of decagon is

A

30

B

20

C

90

120

Explanation

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Q 14. A class contains three girls and four boys. Every Saturday, five students go on a picnic, a different group of students is being sent each week. During the picnic, each girl in the group is given a doll by the accompanying teacher. All possible groups of five have gone once. The total number of dolls the girls have got is

A

21

45

C

27

D

24

Explanation

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Q 15. The number of all the odd divisors of 3600 is

A

45

B

4

C

18

9

Explanation

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Q 16. There are 'n' numbered seats around a round table. Total number of ways in which {tex} n _ { 1 } \left( n _ { 1 } < n \right) {/tex} persons can sit around the round table is equal to

A

{tex} ^ { n } C _ { n _ { 1 } } {/tex}

{tex} ^ { n } P _ { n_1} {/tex}

C

{tex} ^ { n } \mathrm { C } _ { n _ { 1 } - 1 } {/tex}

D

{tex} ^ { n } P _ { n _ { 1 } - 1 } {/tex}

Explanation

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Q 17. The total number of three-digit numbers, the sum of whose digits is even, is equal to

450

B

350

C

250

D

325

Explanation

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Q 18. A teacher takes three children from her class to the zoo at a time as often as she can, but she doesn't take the same set of three children more than once. She finds out that she goes to the zoo 84 times more than a particular child goes to the zoo. Total number of students in her class is equal to

A

12

B

14

10

D

11

Explanation

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Q 19. Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. The number of ways in which we can place the balls in the boxes (order is not considered in the box) so that no box remains empty is

150

B

300

C

200

D

None of these

Explanation

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Q 20. Total number of ways of selecting two numbers from the set {tex} \{ 1,2,3,4 , \ldots , 3 n \} , {/tex} so that their sum is divisible by {tex}3{/tex} is equal to

A

{tex}\large \frac { 2 n ^ { 2 } - n } { 2 } {/tex}

{tex}\large \frac { 3 n ^ { 2 } - n } { 2 } {/tex}

C

{tex} 2 n ^ { 2 } - n {/tex}

D

{tex} 3 n ^ { 2 } - n {/tex}

Explanation

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Q 21. The total number of selections of fruit which can be made from 3 bananas, 4 apples and 2 oranges, is

A

39

B

315

C

512

None of these

Explanation

We have,

Required number of ways = (2+1)(3+1)(4+10−1) = 59

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Q 22. If m = nC2, then mC2 is equal to

A

3 nC4

B

n + 1C4

3.n + 1C4

D

3. n + 1C3

Explanation

Given, $m = {\ ^{n}C}_{2} = \frac{n!}{2!(n - 2)!} = \frac{n(n - 1)}{2}$

Now, ${\ ^{n}C}_{2} = \frac{m!}{2!(m - 2)!} = \frac{m(m - 1)}{2}$

$= \frac{\frac{n(n - 1)}{2}.\left( \frac{n^{2} - n - 2}{2} \right)}{2}$

$= \frac{\left( n + 1 \right)n\left( n - 1 \right)(n - 2)}{8}$

= 3. n + 1C4

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Q 23. Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many ways can we place the balls so that no box remains empty?

A

50

B

100

150

D

200

Explanation

Let the boxes be marked as A, B, C. We have to ensure that no box remains empty and all five balls have to put in. There will be two possibilities.

(i) Any two box containing one ball each and 3rd box containing 3 balls. Number of ways

= A(1)B(1)C(3)

= 5C1.4C1.3C3 = 5.4.1 = 20

Since, the box containing 3 balls could be any of the three aboxes A, B, C. Hence, the required number of ways 20 × 3 = 60

(ii) Any two box containing 2 balls each and 3rd containing 1 ball, the number of ways

= A(2)B(2)C(1) = 5C2. 3C2. 1C1

= 10 × 3 × 1 = 30

Since, the box containing 1 ball could be any of the three boxes A, B, C. Hence, The required number of ways

= 30 × 3 = 90

Hence, total number of ways = 60 + 90 = 150

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Q 24. The interior angles of a regular polygon measure 160 each. The number of diagonals of the polygon are

A

97

B

105

135

D

146

Explanation

Let n be the number of sides of the polygon

n.160 = (n−2).180

⇒ 20.n = 360

∴ n = 18

Then number of diagonals = 18C2 − 18 = 153 − 18 = 135

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Q 25. There are n number of sets and m number of people have to be seated, then how many ways are possible to do this (m<n)?

nPm

B

nCm

C

nCn × (m−1)!

D

n − 1Pm − 1

Explanation

Required number of ways = nCm × m! = nPm