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If $m,n \in \mathbb{W}=\{0,1,2,3,...\}$ prove that $$\frac{(2n)!(2m)!}{n!m!(m+n)!}\in \mathbb{W} $$
I tried to divide into cases $$(1) :m=n\\(2):n>m\\(3):m>n $$ (2),(3) are the same .
(1):$$m=n \to\frac{(2n)!(2n)!}{n!n!(n+n)!}=\frac{(2n)!}{n!n!}=\begin{pmatrix} 2n \\ n \end{pmatrix} \in \mathbb{W} $$ (2):$$n>m \to \frac{(2n)!(2m)!}{n!m!(m+n)!}=\\ \frac{(2n)!(2m)!}{n!m!(m+n)!} $$ at this step ,i get stuck to show $ \in \mathbb{W}$ how can I conclude ?
Is there another idea to prove (like combinational proof) ?
thanks for any help in advance .

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    $\begingroup$ I'm not sure if this is an exact duplicate of this or not since yours demands that it needs to be a non-negative integer. $\endgroup$ – John Doe Oct 12 '17 at 22:49
  • $\begingroup$ if $m$ and $n$ are non-negative, that expression will never be negative, so it is a duplicate $\endgroup$ – Nick Pavlov Oct 12 '17 at 22:52
  • $\begingroup$ Is there some particular reason for renaming $\mathbb{N}$ as $\mathbb{W}$? $\endgroup$ – Jack D'Aurizio Oct 13 '17 at 2:00
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For any prime $p$, by Legendre's Theorem $$ \nu_p\left(\frac{(2n)!(2m)!}{n!m!(n+m)!}\right)=\sum_{h\geq 0}\left(\underbrace{\left\lfloor\frac{2n}{p^h}\right\rfloor+\left\lfloor\frac{2m}{p^h}\right\rfloor-\left\lfloor\frac{n}{p^h}\right\rfloor-\left\lfloor\frac{m}{p^h}\right\rfloor-\left\lfloor\frac{n+m}{p^h}\right\rfloor}_{\color{red}{\geq 0}.}\right)$$

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  • $\begingroup$ It might be good to prove the inequality. $\endgroup$ – marty cohen Oct 14 '17 at 6:03

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