Suppose there are 15 people consisting of 8 women and 7 men.

Question

How many ways are there to form a committee with at least 3 women and at least 2 men?

My answer is $2^{15}- 2^7\binom{8}{0}-2^7\binom{8}{1}-2^7\binom{8}{2}-2^8\binom{7}{0}-2^8\binom{7}{1}$ = 25984 but the answer given is 26280. Can someone tell me where I got wrong? Thank you!

Answer 1

Consider a committee consisting of, for example, $2$ women and $1$ man. This is a committee you do not want, so you have to subtract it from the total; but if you look carefully you will see that you have actually subtracted it twice, once in the term $2^7\binom82$ and once in the term $2^8\binom71$.

Answer 2

Hint: You count some cases twice. When computing all cases of women with less than the number of required men, and then computing all cases of men with less than the number of required women, you will count intersection of these two computations twice.

Hence, you should add intersection of these two to the sum.

Answer 3

We want to choose $3$ women from $8$ and $2$ men from $7$ so we have \begin{eqnarray*} \binom{8}{3} \times \binom{7}{2} =? \times ? = \color{red}{?}. \end{eqnarray*}