Count restricted committees using cases and complements

There are 18 people altogether, so the number is $\binom{18}{4}=3060$.

With no boys, choose 4 girls: $\binom84=70$. With one boy, choose 1 of 10 boys and 3 of 8 girls: $\binom{10}{1}\binom83=10\times56=560$. Add the disjoint cases:70+560=630. Multiplication counts the choices within the one-boy case; addition joins the alternatives. Reversing the selection stages gives $\binom83\binom{10}{1}=56\times10=560$, the same count. Changing the order of these counting stages does not assign roles to committee members.

The four disjoint cases are:

Add:960+1260+560+70=2850.

Choose any 4 people, then subtract all-boy committees: $\binom{18}{4}-\binom{10}{4}=3060-210=2850$. The source recommends no restriction minus complement cases when the direct approach creates too many cases.

Here or is inclusive: both A and B may serve. Committees containing A number $\binom{17}{3}=680$; those containing B also number 680. Those containing both number $\binom{16}{2}=120$. Therefore the required number is $\binom{17}{3}+\binom{17}{3}-\binom{16}{2}=680+680-120=1240$. The Venn diagram follows the Teacher note: subtract the intersection once because the first two counts include it twice.

A is included and B excluded: $\binom{16}{3}=560$. B is included and A excluded: $\binom{16}{3}=560$. Both are included: $\binom{16}{2}=120$. Thus 560+560+120=1240. In the second case, both source copies print Boy B; this is a name slip for Boy A, since the distinguished people are Boy A and Girl B.