# How many ways can you choose a committee if for a married couple, you must choose both or neither

The full question is:

How many ways can you choose a committee of $5$ people from a group of $3$ married couples and $6$ single people, if for any married couple you must pick either both spouses or neither?

I'm having a lot of trouble separating out the ways to choose both or neither--especially because that affects how many you choose from the group of singles.

My work so far has gotten me the answer $\frac12 \Big( \frac {C(12,5)}{ 3!}\Big)$--where the $\frac12$ takes care of the neither or both, the $3!$ takes care of choosing a group, and other than that, you are choosing $5$ from a group of $12$.

Thanks for any help in advance!

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