**Chapter 3 Permutations and Combination Ex 3.2**

## Chapter 3 Permutations and Combination Ex 3.2

**Question 1.Evaluate:(i) 8!Solution:**
8!

= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 40320

**(ii) 10!Solution:10!**

= 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 3628800

**(iii) 10! – 6!Solution:**

10! – 6!

= 10 × 9 × 8 × 7 × 6! – 6!

= 6! (10 × 9 × 8 × 7 – 1)

= 6! (5040 – 1)

= 6 × 5 × 4 × 3 × 2 × 1 × 5039

= 3628080

**(iv) (10 – 6)!Solution:**

(10 – 6)!

= 4!

= 4 × 3 × 2 × 1

= 24

**Question 2.Compute:**

= 12 × 11 × 10 × 9 × 8 × 7

= 665280

= 2!

= 2 × 1

= 2

**(iii) (3 × 2)!Solution:**

(3 × 2)!

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

**(iv) 3! × 2!Solution:**

3! × 2!

= 3 × 2 × 1 × 2 × 1

= 12

**Question 3.Write in terms of factorials(i) 5 × 6 × 7 × 8 × 9 × 10Solution:**

5 × 6 × 7 × 8 × 9 × 10 = 10 × 9 × 8 × 7 × 6 × 5

Multiplying and dividing by 4!, we get

**(ii) 3 × 6 × 9 × 12 × 15Solution:**
3 × 6 × 9 × 12 × 15

= 3 × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5)

= () (5 × 4 × 3 × 2 × 1)

= (5!)

**(iii) 6 × 7 × 8 × 9Solution:**
6 × 7 × 8 × 9 = 9 × 8 × 7 × 6

Multiplying and dividing by 5!, we get

**(iv) 5 × 10 × 15 × 20Solution:**

5 × 10 × 15 × 20

= (5 × 1) × (5 × 2) × (5 × 3) × (5 × 4)

= () (4 × 3 × 2 × 1)

= () (4!)

**Question 4.(i) n = 8, r = 6(ii) n = 12, r = 12(iii) n = 15, r = 10(iv) n = 15, r = 8Solution:**

**Question 5.Find n, ifSolution:**

Solution:

Solution:

**(iv) (n + 1)! = 42 × (n -1)!Solution:**
(n + 1)! = 42(n – 1)!

∴ (n + 1) n (n – 1)! = 42(n – 1)!

∴ + n = 42

∴ + n – 42 = 0

∴ (n + 7)(n – 6) = 0

∴ n = -7 or n = 6

But n ≠ -7 as n ∈ N

∴ n = 6

**(v) (n + 3)! = 110 × (n + 1)!Solution:**

(n + 3)! = (110) (n + 1)!

∴ (n + 3)(n + 2)(n + 1)! = 110(n + 1)!

∴ (n + 3) (n + 2) = (11) (10)

Comparing on both sides, we get

n + 3 = 11

∴ n = 8

**Question 6.Find n, if:Solution:**

∴ (17 – n) (16 – n) (15 – n) = 6 × 5 × 4

Comparing on both sides, we get

17 – n = 6

∴ n = 11

Solution:

∴ 12 = (n – 3)(n – 4)

(n – 3)(n – 4) = 4 × 3

Comparing on both sides, we get

n – 3 = 4

∴ n = 7

Solution:

∴ 120 = (n – 3)(n – 4) (n – 5)(n – 6)

∴ (n – 3)(n – 4) (n – 5)(n – 6) = 5 × 4 × 3 × 2

Comparing on both sides, we get

n – 3 = 5

∴ n = 8

Solution:

∴ (2n – 1)(2n – 3)(2n – 5) = 9 × 7 × 5

Comparing on both sides. We get

∴ 2n – 1 = 9

∴ n = 5

**Question 7.Solution:**

**Question 8.Solution:**

**Question 9.Solution:**

**Question 10.SimplifySolution:**

**Solution:**

Solution:

**(iv) n[n! + (n – 1)!] + (n – 1)! + (n + 1)!Solution:**

Solution:

Solution:

**Solution:**

Solution: