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 .

  • 1
    $\begingroup$ 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

For any prime $p$, by Legendre's Theorem $$ \nu_p\left(\frac{(2n)!(2m)!}{n!m!(n+m)!}\right)=\sum_{h\geq 0}\left(\underbrace{\left\lfloor\frac{2n}{p^h}\right\rfloor+\left\lfloor\frac{2m}{p^h}\right\rfloor-\left\lfloor\frac{n}{p^h}\right\rfloor-\left\lfloor\frac{m}{p^h}\right\rfloor-\left\lfloor\frac{n+m}{p^h}\right\rfloor}_{\color{red}{\geq 0}.}\right)$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.