# How to prove that $\frac{(mn)!}{m!(n!)^m}$ is an integer?

$\forall m,n\in\mathbb Z$ , $m\ge1$ and $n\ge1$ how to prove that $$\frac{(mn)!}{m!(n!)^m}$$ is an integer?

Thanks in advance.

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There is a nice proof of this using group theory. Which course are you taking? –  Geoff Robinson Feb 23 '13 at 18:44
@Geoff Robinson:i know concept of group theory.passed one years ago. but i didn't this question in group theory –  Maisam Hedyelloo Feb 23 '13 at 18:52
There is a nice subgroup of the Symmetric group $S_{mn}$ whose order is $m!(n!)^{m}$. If you can see this, you can apply Lagrange's theorem. –  Geoff Robinson Feb 23 '13 at 18:54

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