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The head of a large company has 9 close friends.

a) In how many ways can he invite six of them to dinner?

b) Repeat a) if two of his friends are divorced, hate each other, and cannot both be invited.

c) Repeat a) if the friends consist of three single people and three married couples and if a husband or wife is invited, the spouse must be invited, too.

Part a is relatively straight forward. It would simply be $C(9,6)$ since the order in which he invites them does not matter.

Part b and c are a little more complicated and I'm not sure how to do them. Any help would be appreciated.

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For part b), how many ways can you choose the divorced couple AND four of the other seven individuals. So if the divorced couple is invited, then the remaining four choices are restricted to the remaining seven potential guests. Since we are excluding this case, the number of ways to invite six people is: $C(9,6) - C(7,4)$.

HINT: For part c), you either choose two singletons and 2 couples OR no singletons and 3 couples.

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  • $\begingroup$ @michaelwal: Please see the hint above. If you're still stuck, I'm willing to help as required. $\endgroup$ – Chris K Dec 4 '13 at 22:54
  • $\begingroup$ I'm not sure... $\endgroup$ – micheal wal Dec 4 '13 at 23:14
  • $\begingroup$ @michaelwal: I'll add some details to show you how it's done. $\endgroup$ – Chris K Dec 4 '13 at 23:15

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