**Previous exercise – Permutation and Combination – Exercise 4.2 **

#### Permutation and Combination – Exercise 4.3

- Convert the following products into factorials.

(i)1 x 2 x 3 x 4 x 5 x 6 x 7

(ii) 18 x 17 … x 3 x 2 x 1

(iii) 6 x 7 x 8 x 9

(iv) 2 x 4 x 6 x 8

- Evaluate

(i) 6!

(ii) 9!

(iii) 8! – 5!

(iv) ^{7!}/_{5!}

(v) ^{12!}/_{(9!)(3!)}

(vi) ^{30!}/_{80!}

- Evaluate

(i) ^{n!}/_{(n-r)!}

(ii) ^{n!}/_{(n – r)!r! } when n = 15 and r = 2

- Find the LCM of 4! , 5! , 6!
- If (n+1)! = 12(n – 1)!, find the value of n

#### Permutation and Combination – Exercise 4.3 – Solutions:

**Convert the following products into factorials.**

**(i)1 x 2 x 3 x 4 x 5 x 6 x 7**

**(ii) 18 x 17 … x 3 x 2 x 1**

**(iii) 6 x 7 x 8 x 9**

**(iv) 2 x 4 x 6 x 8**

Solution:

(i)1 x 2 x 3 x 4 x 5 x 6 x 7 = 7!

(ii) 18 x 17 … x 3 x 2 x 1 = 18!

(iii) 6 x 7 x 8 x 9 = ^{9!}/_{5!}

(iv) 2 x 4 x 6 x 8 = 384 = 16*24 = 16*4!

**Evaluate**

**(i) 6!**

**(ii) 9!**

**(iii) 8! – 5!**

**(iv) ^{7!}/_{5!}**

**(v) ^{12!}/_{(9!)(3!)}**

**(vi) ^{30!}/_{80!}**

Solution:

(i) 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720

(ii) 9! = 362880

(iii) 8! – 5! = 40200

(iv) ^{7!}/_{5!} = ^{1x2x3x4x5x6x7}/_{1x2x3x4x5} = 6 x 7 = 42

(v) ^{12!}/_{(9!)(3!)} = ^{1x2x3x4x5x6x7x8x9x10x11x12}/_{(1x2x3x4x5x6x7x8x9)(1x2x3)} = ^{10x11x12}/_{1x2x3} = 5x11x4 = 220

(vi) ^{30!}/_{28!} = ^{28! 29*30}/_{28!} = 29 * 30 = 870

**Evaluate**

**(i) ^{n!}/_{(n-r)!} when n = 15 and r = 2**

**(ii) ^{n!}/_{(n – r)!r! } when n = 15 and r = 2**

Solution:

(i) ^{n!}/_{(n-r)!} when n = 15 and r = 2

^{n!}/_{(n-r)!} = ^{15!}/_{(15 – 2)!} = ^{15!}/_{13!} = 210

(ii) ^{n!}/_{(n – r)!r! } when n = 15 and r = 2

^{n!}/_{(n – r)!r! } = ^{15!}/_{(15-2)!2!} = ^{15!}/_{(13)!2!} = 105

**Find the LCM of 4! , 5! , 6!**

Solution:

4! = 1 x 2 x 3 x 4

5! = 1 x 2 x 3 x 4 x 5

6! = 1 x 2 x 3 x 4 x 5 x 6

Therefore, LCM of 4! , 5! and 6! is 720

**If (n+1)! = 12(n – 1)!, find the value of n**

Solution:

(n+1)! = 12(n – 1)!

(n – 1)! x n x (n + 1) = 12 x (n – 1)!

n(n+1) = 12

n^{2} + n = 12

n^{2} + n – 12 = 0

n^{2} + 4n – 3n – 12 = 0

n(n + 4) – 3(n + 4) = 0

(n – 3)(n + 4) = 0

n – 3 = 0 or n + 4 = 0

n = +3 or n = – 4

**Next Exercise Permutation and Combination – Exercise 4.4 **