JAMB - Mathematics (2019 - No. 62)

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?

144 ways

15 ways

185 ways

12 ways

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