Elementary combinatorics... non-intuitive nPr/nCr? problem. Of course, when I say non-intuituve, it's completely subjective.
Suppose there are 12 woman and 10 men on the faculty. How many ways are there to pick a committee of 6?
For this portion of the problem, I simply did $\binom{22}{6} = 74613$ different ways to arrange a committee of 6 from 22 people.
But suppose some woman Jane and some woman Janet will not serve together. How would I apply this constraint? Or the constraint that at least one woman must be chosen.

The easiest way would just be to do complimentary counting and do $\binom{22}{6} - \binom{20}{4}$. The other way would be to do cases.
Take three cases. One where Janet is on and Jane is off and vice versa and the case where none are on. 
Case $1$: Janet on and Jane off: There are $20$ people left on the committee and we already have one spot taken, so $\binom{20}{5}$.
Case $2$: Janet off and Jane on: Same as case $1$, $\binom{20}{5}$.
Case $3$: Both Janet and Jane off: We have $20$ people to choose from, so $\binom{20}{6}$.
Thus the answer is $2\binom{20}{5}+\binom{20}{6} = 69768$.
