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!



Consider separately the cases when you have 0, 1 or 2 couples on the committee.

  • $\begingroup$ ok, but then how do you put it together? so, 0 couples->C(6,5), 1 couple->C(6,3), 2 -> C(6,1), but you can't just do C(6,5)+C(3,1)*C(6,3)+C(3,2)*C(6,1). So how do you make it so that there is one statement that deals with all of them? $\endgroup$ – BadAtGraphs Nov 16 '12 at 8:57
  • 1
    $\begingroup$ @BadAtGraphs: Why can't you "just do" that? $\endgroup$ – ShreevatsaR Nov 16 '12 at 8:59
  • 2
    $\begingroup$ @BadAtGraphs You can just add them because they are mutually exclusive. You can't have 0 couples and 1 couple and 2 couples at the same time. $\endgroup$ – Daryl Nov 16 '12 at 10:21

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