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Hi I'd like to know if the following proof of Vandermonde's Identity is correct (is really easy):

Let $m,n,r$ be natural numbers such that $r\le \min \{m,n\}$. The Vandermonde's Identity gives us that $$ \binom{m+n}r = \sum_{k=0}^r \binom mk \binom n{r-k} $$

$\bf{Solution}$ Let $\gamma$ be the unit circle $\lvert z\lvert =1$. Now since $${m \choose r}= \frac{1}{2\pi i}\int _{\gamma}\frac{(1+z)^m}{z^{r+1}} dz$$

Then

\begin{align} 2\pi i \binom{m+n}r= \int _{\gamma}\frac{(1+z)^{m+n}}{z^{r+1}} dz &= \int _{\gamma}\frac{(1+z)^{n}}{z^{r+1}}\sum_{k=0}^m {m \choose k}z^k \,dz\\ &= \sum_{k=0}^m { m\choose k}\int_\gamma \frac{(1+z)^n}{z^{r-k+1}} dz \\ &= \sum_{k=0}^m 2\pi i { m\choose k} {n\choose r-k} \end{align}

For all $k>r$, we set ${n\choose r-k}=0$, i.e., the left-hand side is zero and imply the desired identity

$$ \binom{m+n}r = \sum_{k=0}^r \binom mk \binom n{r-k} $$

Thanks :)

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  • $\begingroup$ @gt6989b Do you think is correct? $\endgroup$ – Jose Antonio Sep 2 '15 at 3:18
  • $\begingroup$ yes, i think this is right $\endgroup$ – gt6989b Sep 2 '15 at 3:46
  • $\begingroup$ @gt6989b I like it because is incredible easy. Even more than the usual approach using generating functions. Thanks :) $\endgroup$ – Jose Antonio Sep 2 '15 at 3:59
  • $\begingroup$ @JoseAntonio Instead of using generating functions, I think an easy combinatoric proof is also possible. $\endgroup$ – PSPACEhard Sep 2 '15 at 5:29
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    $\begingroup$ Nice use of complex calculus. A combinatorial proof : Let A have m members and let B have n members with $ A\cap B = \phi$. For k from 0 to r , compute the number of ways to choose a set with r members with k of them in A and r-k of them in B. Now sum over k. $\endgroup$ – DanielWainfleet Sep 2 '15 at 8:16
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To keep the question off the unanswered list. Yes, your approach is correct.

See Complex Analysis proof of multinomial expression for something in the same spirit.

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  • $\begingroup$ Thanks. Is completely correct to set ${n \choose r-k}=0$ for $k>r$? $\endgroup$ – Jose Antonio Sep 2 '15 at 7:19
  • $\begingroup$ Wow I've never seen this approach. I think is amazing, thanks for the link :) $\endgroup$ – Jose Antonio Sep 2 '15 at 7:20

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