# Permutation-Combination Exercise 2

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- Most Important Multiple Choice Questions
- Online Permutation and Combination Exercise with Correct Answer Key and Solutions
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- Question 1 of 20
##### 1. Question

In how many ways can six different rings be worn on four fingers of one hand?

##### Hint

Required number of ways = ways of selecting 4 objects out of 6 given objects

= ⁶C₄ = (6 × 5)/2 = 15

- Question 2 of 20
##### 2. Question

There are three prizes to be distributed among five students. If no student gets more than one prize, then this can be done in :

##### Hint

It is a question of arrangement without repetitions.

Required no. of ways = 5 × 4 × 3 = 60

- Question 3 of 20
##### 3. Question

Find the number of ways in which 8064 can be resolved as the product of two factors?

##### Hint

First of all we will prime factorize 8064.

8064 = 2 × 4032

= 2² × 2016

= 2³ × 1008 = 2⁴ × 504

= 2⁵ × 252 = 2⁶ × 126 = 2⁷ × 63

= 2⁷ × 3² × 7¹

Required no. of ways

= (7 + 1) (2 + 1) . 1

= 8 × 3 = 24

- Question 4 of 20
##### 4. Question

In how many ways can twelve girls be arranged in a row if two particular girls must occupy the end places?

##### Hint

Two particular girls can be arranged in 2! ways and remaining 10 girls can be arranged in 10! ways.

Required no. of ways = 2! × 10!

- Question 5 of 20
##### 5. Question

In how many ways can a selection of 5 letters be made out of 5 A’s, 4 B’s, 3 C’s, 2D’s and 1 E’s?

##### Hint

5 letters out of 15 letters can be selected in ¹⁵C₅ ways.

- Question 6 of 20
##### 6. Question

The number of four digit numbers with distinct digits is :

##### Hint

The thousandth place can be filled up in 9 ways with any one of the digits 1, 2, 3, …., 9.

After that the other three places can be filled up in ⁹P₃ ways, with any one of the remaining 9 digits including zero.

Hence, the number of four digit numbers with distinct digits = 9 × ⁹P₃

- Question 7 of 20
##### 7. Question

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

##### Hint

The first and the last (terminal) digits are even and there are three even digits.

This arrangement can be done in ³P₂ ways.

For any one of these arrangements, two even digits are used; and the remaining digits are 5 (4 odd and 1 even) and the four digits in the six digits (leaving out the terminal digits) may be arranged using these 5 digits in ⁵P₄ ways.

The required number of numbers is

³P₂ × ⁵P₄ = 6 × 120 = 720.

- Question 8 of 20
##### 8. Question

The number of numbers greater than a million that can be formed with the digits 2, 3, 0, 3, 4, 2 and 3 is

##### Hint

Any number greater than a million must be of 7 or more than 7 digits.

Here number of given digits is seven, therefore we have to form numbers of seven digits only.

Now there are seven digits of which 3 occurs thrice and two occurs twice.

∴ number of numbers formed

= = 420

But this also includes those numbers of seven digits whose first digit is zero and so in fact they are only six digit numbers.

Number of numbers of seven digits having zero in the first place = 60.

Hence required number

= 420 – 60 = 360

- Question 9 of 20
##### 9. Question

Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed) is

##### Hint

Required number of numbers

= 3 × 5 × 5 × 5 = 375

- Question 10 of 20
##### 10. Question

The sum of integers from 1 to 100 that are divisible by 2 or 5 is

##### Hint

Required sum

= (2 + 4 + 6 + … + 100) + (5 + 10 + 15 + …+ 100) – (10 + 20 + … + 100)

= 2550 + 1050 – 530 = 3070.

- Question 11 of 20
##### 11. Question

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

##### Hint

X – X – X – X – X.

The four digits 3, 3, 5,5 can be arranged at (-) places in

= 6 ways.

The five digits 2, 2, 8, 8, 8 can be arranged at (X) places in

ways = 10 ways

Total no. of arrangements

= 6 × 10 = 60 ways

- Question 12 of 20
##### 12. Question

If , then the value of r is

##### Hint

⇒

⇒

⇒

⇒

⇒

⇒ 10 – r = 5

⇒ r = 5

- Question 13 of 20
##### 13. Question

The number of integers satisfying the inequality is

##### Hint

The inequality is

We must have n + 1 ≥ 3 and n + 1 ≥ 2

⇒ n ≥ 2 and n ≥ 1

⇒ n ≥ 2 and also

⇒ (n + 1) n(n – 4) ≤ 600

By trial the values of n satisfying this are 2, 3, 4, 5, 6, 7, 8, 9 which are eight in number.

- Question 14 of 20
##### 14. Question

If , then least value of m is

##### Hint

⇒ m > 6

The least value of m is 7.

- Question 15 of 20
##### 15. Question

The number of values of r satisfying the equation is

##### Hint

⇒ r² = 3r

⇒ r² = 40 – 3r

⇒ r =0, 3 or -8, 5

3 and 5 are the values as the given equation is not defined by r = 0 and r = –8.

Hence, the number of values of r is 2.

- Question 16 of 20
##### 16. Question

If

^{n}P_{r}=^{n}P_{r + 1}and^{n}C_{r}=^{n}C_{r – 1}, then the values of n and r are##### Hint

We have,

^{n}P_{r}=^{n}P_{r+1}⇒ n – r = 1 …(1)

Also,

^{n}C_{r}=^{n}C_{r–1}⇒ r + r – 1 = n

⇒ 2r – n = 1 …(2)

Solving (1) and (2), we get r = 2 and n = 3

- Question 17 of 20
##### 17. Question

If

^{n}P_{r}= 720^{n}C_{r}, then r is equal to##### Hint

^{n}P_{r}= 720^{n}C_{r}⇒

⇒ r! = 720 = 1 × 2 × 3× 4 × 5 × 6!

⇒ r = 6

- Question 18 of 20
##### 18. Question

If and

^{k}P₃ = 9240, then##### Hint

We have

=

=

=

⇒ m = 10, n = 12

Given, kP₃ = 9240

⇒ k(k – 1) (k – 2) = 9240 = 22 × 21 × 20

⇒ k = 22

∴ m + n – k = 10 + 12 – 22 = 0

- Question 19 of 20
##### 19. Question

If = , then the value of r and the minimum value of n are

##### Hint

Given,

=

….. ….. ….. ….. ….. ….. ….. ….. ….. ….. ….. …..

⇒ r² – r – 132 = 0

⇒ (r – 12) (r + 11) = 0

⇒ r = 12

⇒ n ≥ 12

So, minimum value of n = 12.

- Question 20 of 20
##### 20. Question

The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon is

##### Hint

Number of all possible triangles

= Number of selections of 3 points from 8 vertices

= ⁸C₃ = 56

Number of triangle with one side common with octagon

= 8 × 4 = 32

(Consider side A₁A₂. Since two points A₃, A₈ are adjacent, 3rd point should be chosen from remaining 4 points.)

Number of triangles having two sides common with octagon : All such triangles have three consecutive vertices, viz., A₁A₂A₃, A₂A₃A₄, ….. A₈A₁A₂.

Number of such triangles = 8

∴ Number of triangles with no side common

= 56 – 32 – 8 = 16.