**Previous exercise – Permutation and Combination – Exercise 4.5 **

**Permutation and Combination – Exercise 4.6**

- Evaluate (i)
^{10}C_{3}(ii)^{60}C_{60 }(iii)^{100}C_{97}

(i) if ^{n}C_{4} = ^{n}C_{7} find n

(ii) IF ^{n}P_{r} = 840, ^{n}C_{r} = 35 find n

- If
^{2n}C_{3}:^{n}C_{3}= 11 : 1 , find n - Verify that
^{8}C_{4}+^{8}C_{5}=^{9}C_{4} - Prove that
^{nCr}/_{n-1Cr-1}=^{n}/_{r}when 1 ≤ r ≤ n

#### Permutation and Combination – Exercise 4.6 – Solutions:

**Evaluate**

**(i) ^{10}C_{3}**

**(ii) ^{60}C_{60 }**

**(iii) ^{100}C_{97}**

Solution:

(i) ^{10}C_{3}

We know, ^{n}C_{r} = ^{n!}/_{(n – r)!r!}

^{10}C_{3} = ^{10!}/_{(10 – 3)!3!}

= ^{10!}/_{7! x 3!}

= ^{7! x 8 x 9 x 10}/_{7! x 2 x 3}

= ^{8 x 9 x 10}/_{2 x 3}

= 4 x 3 x 10

= 120

(ii)^{60}C_{60 }

We know, ^{n}C_{r} = ^{n!}/_{(n – r)!r!}

^{60}C_{60} = ^{60!}/_{(60 – 60)!60!}

= ^{60!}/_{0! x 60!}

= ^{60!}/_{1 x 60!}

= 1

(iii)^{100}C_{97}

We know, ^{n}C_{r} = ^{n!}/_{(n – r)!r!}

^{100}C_{97} = ^{100!}/_{(100 – 97)!97!}

= ^{10!}/_{3! x 97!}

= ^{97! x 98 x 99 x 100}/_{97! x 2 x 3}

= ^{98 x 99 x 100}/_{2 x 3}

= 98 x 33 x 50

= 161700

**2. (i) if ^{n}C_{4} = ^{n}C_{7} find n**

**(ii) IF ^{n}P_{r} = 840, ^{n}C_{r} = 35 find n**

Solution:

(i) if ^{n}C_{4} = ^{n}C_{7} , we have to find n

^{n}C_{4} = ^{n}C_{7}

^{n}C_{4} = ^{n}C_{n – 7}

4 = n – 7

n = 4 + 7 = 11

(ii) If ^{n}P_{r} = 840, ^{n}C_{r} = 35, we have to find n

We know, ^{n}P_{r} =r! ^{n}C_{r}

840 = r! x 35

r! = ^{840}/_{35} = 24

r! = 1 x 2 x 3 x 4 = 4!

Therefore, r = 4

**If**^{2n}C_{3}:^{n}C_{3}= 11 : 1 , find n

Solution:

^{2n}C_{3} : ^{n}C_{3} = 11 : 1

^{2nC3}/_{nC3} = ^{11}/_{1}

^{2n}C_{3} = 11 x ^{n}C_{3}

^{(2n)!}/_{(2n-3)! 3!} = 11 x ^{n!}/_{(n – 3)!3!}

^{(2n-3)! (2n – 2)(2n – 1) 2n}/_{(2n – 3)! 3!} = 11 x ^{(n-3)!(n-2)(n-1)n}/_{(n-3)! 3!}

^{(2n – 2)(2n – 1) 2n}/_{ 3!} = 11 x ^{(n-2)(n-1)n}/_{3!}

(2n – 2)(2n – 1) 2n = 11 x (n-2)(n-1)n

2(n-1) x 2 x n x(2n-1) = 11n(n-2)(n-1)

4(2n-1) = 11(n – 2)

8n – 4 = 11n – 22

8n – 11n = 4 – 22

-3n = – 18

n = ^{18}/_{3} = 6

**Verify that**^{8}C_{4}+^{8}C_{5}=^{9}C_{4}

Solution:

LHS = ^{8}C_{4} + ^{8}C_{5}

= ^{8!}/_{(8-4)!4!} + ^{8!}/_{(8-5)!5!}

= ^{8!}/_{4!4!} +^{8!}/_{3!5!}

= ^{4! x 5 x 6 x 7 x 8}/_{4! x 1 x 2 x 3 x 4} + ^{5! x 6 x 7 x 8}/_{5! x 1 x 2 x 3 }

= 5 x 7 x 2 + 7 x 8

= 70 + 56

RHS = ^{9}C_{4}

= ^{9!}/_{(9-4)!4!}

= ^{9!}/_{5!4!}

= ^{5! x 6 x 7 x 8 x 9}/_{5! x 1 x 2 x 3 x 4}

= 7 x 2 x 9

= 126

LHS = RHS

Hence proved.

**Next exercise Permutation and Combination – Exercise 4.7 **