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There are $2n$ people. There are $2n!$ ways to arrange them.

The number of ways to arrange them such that couples are always together is $n! \cdot 2^n$

How do you calculate the number of ways to arrange them such that no couples are together?

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marked as duplicate by N. F. Taussig combinatorics Aug 18 '18 at 19:05

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    $\begingroup$ No, since it is possible that some couples are together while others are not. See Derangments $\endgroup$ – lulu Aug 18 '18 at 15:41

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