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Prove that: $$\forall n\in \mathbb{N}, \ \binom{2n}{n-1}=\sum_{i=0}^{n-1}\binom{n}{i}\binom{n}{n-1-i}$$

I've no idea how to approach this.

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  • $\begingroup$ See en.wikipedia.org/wiki/Cauchy_product $\endgroup$
    – Robert Z
    Commented Sep 18, 2017 at 13:27
  • $\begingroup$ You could try generating series, or proving a recurrence, or induction, perhaps? $\endgroup$
    – Pedro
    Commented Sep 18, 2017 at 13:27
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    $\begingroup$ Hint: Vandermonde convolution. $\endgroup$ Commented Sep 18, 2017 at 13:33
  • $\begingroup$ @SeanRoberson best hint...it's direct with Vandermonde identity +1 $\endgroup$ Commented Sep 18, 2017 at 13:43
  • $\begingroup$ @SeanRoberson The combinatorial proof for Vandermonde's identity was most understandable for me. Thanks! $\endgroup$ Commented Sep 18, 2017 at 14:01

1 Answer 1

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$\forall n\in \mathbb{N}, \ \binom{2n}{n-1}=\sum_{i=0}^{n-1}\binom{n}{i}\binom{n}{n-1-i}$

posing $n=m$ and $r=n-1$

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

Vandermonde 's Identity

https://en.wikipedia.org/wiki/Vandermonde%27s_identity

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