In a committee of 5, which must be selected from 4 males and 3 females....

In a committee of 5, which must be selected from 4 males and 3 females. In how many ways can the members be chosen if it were to include 2 females?

For the committee to include 2 females, we must have 3 males, so that there should be 5 members.

That is, (^4C_3 times ^3C_2)

= (frac{4!}{(4 – 3)! 3!} times frac{3!}{(3 – 2)! 2!})

= 4 × 3 = 12 ways

There is an explanation video available below.

In a committee of 5, which must be selected from 4 males and 3 females….