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Question: There are $4$ couples . They decide to form a committee of $4$ people where no couple finds a place, then total number of ways are ??

My Attempt: What I did its simply $2^4=16$ as total ways of selecting men are ${4\choose 0}+..+{4\choose4}$ as ways of selecting women will be fixed on selecting the men as a woman whose husband is selected can't take place in it. Am I right??

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    $\begingroup$ Looks good. Another way to see it is to note that you have to choose exactly one individual from each married pair. $\endgroup$
    – lulu
    Jan 24 '16 at 11:22
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Your thought procedure is correct, no doubt. But it is a bit confusing and hazy, just like it has confused you yourself that you had to ask this question.

Better, I would suggest that you take the $4$ couples one by one and choose any one partner from each of the couples. In that way, the given condition is satisfied and the answer is also the same. Only the calculation varies.

Total no. of ways = $\binom{2}{1}\cdot \binom{2}{1}\cdot \binom{2}{1}\cdot \binom{2}{1} = 16$

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If I understand your question, it means that there can be only one person from the couple in the committee.

Since you choose $4$ people among $4$ couples, and no couple can be in the committee, you have to choose one person per couple. So yes, you choose $1$ per per couple four times, therefore the number of committees possible will be $2^4 = 16$.

Now imagine we had to choose $m$ people for the committee among $n \geq m$ couples, and the same criterion for choosing a committee as above. Then you have to simply choose $m$ couples among $n$, and for each couple you have two possibilities, either choose the man or the woman. Therefore the number of possibilities will be $2^m$$ {n}\choose{m}$.

In your example: $n = m = 4 \Rightarrow 2^4$${4}\choose{4}$$=16$.

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  • $\begingroup$ The exact wording is "no couple finds a place" which isn't the same as your interpretation of "only one person from each couple" $\endgroup$ Jan 24 '16 at 12:52
  • $\begingroup$ What does it mean then? $\endgroup$
    – K.A.
    Jan 24 '16 at 12:53
  • $\begingroup$ Isn't that the same? Saying that the two people of a couple can't be both in the committe? $\endgroup$
    – K.A.
    Jan 24 '16 at 13:02
  • $\begingroup$ Yes, it is the same. +1 for generalization of the problem. $\endgroup$ Jan 24 '16 at 13:29

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