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Find the number of ways of forming a group of $2k$ people from $n$ couples,where $n,k \in \mathbb{N}$ with $2k \le n$, in each of the following cases: (i) There are $k$ couples in such a group; (ii) No couples are included in such a group: (iii) At least one couple is in included in such a group; (iv) Exactly two couples are included in such a group.

I know how to answer (i), and don't know how to answer (ii) and (iv), (iii) can be deduced from (ii) easily. Please help to answer (ii) and (iv). With great thanks!

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  • $\begingroup$ Hint for (ii): first choose the $2k$ couples. Then choose one member of each selected couple. $\endgroup$ – lulu Apr 21 '17 at 13:26
  • $\begingroup$ For case (ii), you can first select $2k$ couples and then select one person out of each couple. As such, the answer is ${{n}\choose{2k}} 2^{2k}$. For case (iv), first select 2 couples of which you will select both partners, then select $2k-4$ couples of which you will select one person. The total number of combinations is then ${{n}\choose{2}} {{n-2}\choose{2k-4}} 2^{2k-4}$. $\endgroup$ – jvdhooft May 3 '17 at 11:13

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