Question

A committee of 10 members has to be formed from 15 men and 7 women. In how
many ways can this be done when the committee consists of at least 6 women ?
od 16​

Answer 1

Answer:

Number of ways = 10010

Step-by-step explanation:

Given:

• A committee of 10 members is to be formed from 15 men and 7 women.

To Find:

• Number of ways the committee can be formed if it consists of atleast 6 women

Solution:

By given the committee must not exceed 10 members and should contain atleast 6 women.

Thus the committee can be selected in the following ways:

6 women and 4 men or 7 women and 3 men

This can be represented as,

$\tt Total\:number\:of\:ways=\:^7C_6\times \:^{15}C_4+\: ^7C_7+\: ^{15}C_3$

We know that,

$\boxed{\tt ^nC_r=\dfrac{n!}{r!(n-r)!} }$

Therefore we get,

$\tt Total\:number\:of\:ways=\dfrac{7!}{6!}\times \dfrac{15!}{4!\times 11!} +\dfrac{7!}{7!} \times \dfrac{15!}{3!\times 12!}$

$\tt \implies 7\times \dfrac{15\times 14\times 13\times 12}{4\times 3\times 2} +1\times \dfrac{15\times 14\times 13}{3\times 2}$

$\tt \implies 7\times1365+ 455$

$\tt \implies 9555+455$

$\tt \implies 10010$

Hence the committee can be formed in 10010 ways.

Answer 2

required Answer :-

☯ GIVEN ,

A committee of 10 members has to be formed from 15 men and 7 women.

✿ Solution ,

$= \frac{7!}{6!} \times \frac{15!}{4! \times 11!} + \frac{7!}{7!} \times \frac{15!}{3! \times 21!}$

$= 7 \times \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2} + 1 \times \frac{15 \times 14 \times 13}{3 \times 2}$

⟹$now \: 7 \times 1365 + 455$

⟹$9555 + 445$

⟹$10010$

Hence , 10010 is the Answer :)