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We have a set $S$ of $2N$ distinct elements. I want to partition it into $N$ parts each containing 2 elements. My motivation is partitioning a group of people into pairs.

What is the formula that gives the number of different partitions of $S$?

I prefer closed-form formula. If no such formula exists then asymptotic growth rate is fine.

EDIT: I received several answers. The formula seems hard to calculate. Hence, I am also interested in an accurate Stirling approximation of the formula.

I got this Stirling approximation: $\sqrt 2 (2N/e)^N$. Is this the most accurate approximation?

Peter Taylor
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Mohammad Al-Turkistany
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2 Answers2

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Why not repeatedly choose the 2 components for each partition in order, to get $$ \binom{2N}{2} \times \binom{2N-2}{2} \times \ldots \times \binom{4}{2} \binom{2}{2} = \prod_{k=1}^N \binom{2k}{2}, $$ which you can simplify since $\binom{n}{2} = n(n-1)/2$ to get $$ \prod_{k=1}^N \binom{2k}{2} = \prod_{k=1}^N \frac{2k(2k-1)}{2} = \frac{(2N)!}{2^N} $$

UPDATE To account for the fact that this is an ordered partition, divide by the overcount factor of $N!$ to get $$ \frac{(2N)!}{N! \cdot 2^N} = (2N-1)!!, $$ as was pointed out by Thomas Andrews in a comment.

gt6989b
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The desired number is

$$\frac{(2N)!}{N!\cdot2^N}$$

To get this, consider that there are $N!$ arrangements for every unordered partition.

Peter
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