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

Given:

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

To find:

In how many ways can this be done when the committee consists of at least 6 women?

Solution:

We are given that a committee of 10 members has to be formed consisting of at least 6 women (i.e., women should be 6 or more)

So, the selection can be done as:

• 6 women and 4 men
• 7 women and 3 men

To solve the given problem we will use the following formula of combinations:

$\boxed{\bold{^nC_r = \frac{n!}{r! (n-r)!} }}$

Therefore,

The total no. of ways of selecting the members in the committee are,

= $[^7C_6\times ^1^5C_4] + [^7C_7\times ^1^5C_3]$

= $[\frac{7!}{6!}\times\frac{15!}{4!\times 11!} ] + [\frac{7!}{7!} \times \frac{15!}{3!\times 12!} ]$

= $[\frac{7\times 6!}{6!}\times\frac{15\times 14\times 13\times 12\times 11!}{4\times 3\times 2\times 11!} ] + [1 \times \frac{15\times 14\times 13\times 12!}{3\times 2\times 12!} ]$

= $[7\times\frac{32760}{24} ] + [1 \times \frac{2730}{6} ]$

= $9555 + 455$

= $\bold{10010}$

Thus, this can be done in 10010 ways when the committee consists of at least 6 women.

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