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For natural number $m, n$, i found out that $\{\{\{a_1, a_2, \cdots a_n\},\{a_{n+1}, a_{n+2}, \cdots a_{2n}\}, \cdots \{a_{mn-n+1}, \cdots a_{mn}\}\}| \{a_1,a_2, \cdots a_{mn}\}=\{1,2, \cdots mn\}\}$ has $\frac{(mn)!}{(n!)^mm!}$ elements. This would be combinatorial proof that $(n!)^mm!$ is divisor of $(mn)!$. However, I want to know the 'straight proof' using number theory. Any help?

tneserp
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jan 22 '23 at 07:52
  • I think this does help. Thank you – tneserp Jan 22 '23 at 08:28

1 Answers1

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There are $m$ runs of $n$ consecutive numbers, so $n!^m|(mn)!$.

Furthermore, the $k^{th}$ run (i.e. $kn+1,\dotsm (k+1)n$) leaves at least $k$ when divided by $n!$, which together multiply to $m!$.

JMP
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