# Why are there 5 groups that contain the 2 enemies?

Suppose I must form a committee of 3 people chosen out of 7. 2 of the 7 are enemies of each other. These 2 people would wreak havoc if they are both chosen to be on the committee. How may ways are there to form committees containing both of them?

Why is the answer $\binom{2}{2}\binom{5}{1} = 5$? Why am I picking 2 out of 2 people and then 1 out of 5 people?

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Think of the pool as consisting of two groups, the awkward two and the other five. To form one of these useless committees you must pick both of the members of the awkward group, which you can do in $\binom22$ ways, and then you can combine them with any one member of the other group, whom you can choose in $\binom51$ ways. You’re really just observing that in order to get a useless committee, you have to include both of the awkward folks and one of the others, so the only choice involved is in which one of the $5$ others you pick.

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You wish to know how you can choose $3$ people out of $7$, when $2$ of them have to be chosen. This means that you have to choose $2$ out of the $2$ enemies, and another from the remaining five, that would be $\binom{5}{1}$.

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You're picking three members.

First you take the two enemies (since that's one requirement), for the first two members (there is one combination of two possible from two enemies, a.k.a $\binom{2}{2}$)

Then there are 5 neutral people to choose from for the third member (there are five combinations of one possible from five neutral people $\binom{5}{1}$).

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