How many ways are there to pick five people for a committee if there are six
(different) men and eight (different) women and the selection must include at
least one man and one woman?
I know to solve this with very basic combinatorial analysis (though I know the principle is also basic). We find the total people we can choose for the committee and proceed with cases: 1 where we have all men on the committee, 1 where we have all women on the committee and then we subtract the two cases from our total. This gives us
$${14 \choose 5} – {8 \choose 5}-{6 \choose 5}.$$
How would I compute for Inclusion-Exclusion Principle, though? I've no idea where to begin with it other drawing a venn diagram and labeling set M and W…
Best Answer
You have more-or-less done it already.
For $i \in \{0,1,\ldots\}$, let $a_i$ be the number of committees with $\geq i$ absent genders.
We include $a_0=\binom{14}{5}$.
We exclude $a_1=\binom{8}{5}+\binom{6}{5}$.
We include $a_2=0$.
We exclude $a_3=0$.
And so on.