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Probably this is a easy question, but I was unable to solve it. Let $n$ be a even natural number. Is true that the following number is natural for all $n$? $$\frac{n!}{\left(n/2\right)!\left(n/2\right)!}$$

I can see, for example, that $(n/2)!$ divides $n!$, but then I can't conclude by using this arguing.

Thank you

Tomás
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  • Yes, because the symmetric group on $n$ objects contains a direct product of two copies of the symmetric group on $n/2$ objects (those fixing the first $n/2$ objects and those fixing the last $n/2$ objects). – Tobias Kildetoft Jul 22 '13 at 11:24
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    It's a binomial coefficient, hence an integer. – Daniel Fischer Jul 22 '13 at 11:24
  • Also, $k!$ divides any product of $k$ consecutive integers. (which is due $\begin{pmatrix}n\k\end{pmatrix}$ being an integer, of course.) – zuggg Jul 22 '13 at 11:35

2 Answers2

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Yes, it is. Suppose that $n=2m$; then

$$\frac{n!}{(n/2)!(n/2)!}=\frac{(2m)!}{m!m!}=\binom{2m}m\;,$$

which is the number of $m$-element subsets of a set of $2m$ things. This clearly must be an integer. For more information, see the Wikipedia article on binomial coefficients.

Brian M. Scott
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Yes. This is just $\displaystyle {n \choose n/2}$.

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