TITLE: What is wrong with my combinatorics method?
QUESTION [5 upvotes]: Suppose I want to select a team of $7$ from a pool of $10$ from $A$, $8$ from $B$ and $5$ from $C$. However, I want to make sure that I have at least one from each group.
My idea was to do the following $$10 \times 8 \times 5 \times \binom{20}{4}$$
the $10$, $8$ and $5$ are to choose the one member from each team, who is 'guaranteed' a spot. The $\binom{20}{4}$ is to choose the remaining $4$ team members from the remaining pool of $20$ people.
The problem is that this value is larger than $\binom{23}{7}$, which should be the largest possible value.

REPLY [4 votes]: I think you may have to use inclusion / exclusion for this one. Then you get
$${23\choose 7}-\left({13\choose 7}+{15\choose 7}+{18\choose 7}\right)+\left({{10\choose 7}+{8\choose 7}+0}\right)=205\ 310.$$
So the middle terms are the cases where no one is from A, B, or C respectively, and the rightmost terms are the cases where no one is from B or C, A or C, or A or B respectively.