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Say I have a group of 20 people, and I want to split them to pairs, I know that the number of different ways to do it is $\frac{(2n)!}{2^n \cdot n!}$ But let's say that I have to pair a boy with a girl? I got confused because unlike the first option the number of total elements (pairs) isn't $(2n)!$, and I failed to count it. I think that the size of each equivalence class is the same - $2^n \cdot n!$, I am just missing the number of total elements. Any ideas? Thanks!

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If you have $n$ boys and $n$ girls, give them each a number and sort the pairs by wlog the boy's number. Then there are $n!$ possible orderings for the girls, so $n!$ ways of forming the pairs.

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So simple! thanks – yotamoo Mar 11 '12 at 12:51

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