For the committee to include 2 females, we must have 3 males, so that there should be 5 members.
That is, (43)×(32)\binom{4}{3} \times \binom{3}{2}(34)×(23)
= 4!(4−3)! 3!×3!(3−2)! 2!\frac{4!}{(4-3)!\,3!} \times \frac{3!}{(3-2)!\,2!}(4−3)!3!4!×(3−2)!2!3!
= 4 × 3 = 12 ways
