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I am trying to figure out if the following expression

$$\frac{(n^2 - n)! }{ n! ((n-1)!)^n }$$

is an integer for all positive integer $n.$

I tried the induction, but induction case is running into problem. So I was looking at it from permutation/combination problems. But so far, couldn't come up with a convincing argument.

Does anybody has any thoughts on this to share ?

Thank you very much.

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closed as off topic by Todd Trimble, Anthony Quas, Dan Petersen, Emil Jeřábek, quid Jan 17 '13 at 17:43

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This site is not the right place for your question; this site is for research interests of professional mathematicians. Try mathstackexchange.com instead. –  Todd Trimble Jan 17 '13 at 16:44

2 Answers 2

Todd was right, this is not appropriate for MO. Just an easy explanation: Let $\Sigma_n$ be the permutation group on $n$ letters. Then $\Sigma_m\wr\Sigma_n=(\Sigma_m)^n\rtimes \Sigma_n$ is a subgroup of $\Sigma_{mn}$. The above is a special case.

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More generally, $\frac{(nm)!}{n!(m!)^n}$ counts the partitions of a set of $nm$ elements into $n$ classes of $m$ elements.

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