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I am trying to do a combinatorial proof of ${n\choose p}{n\choose q}=\sum_{k=0}^{n} {n \choose k}{n-k \choose p-k}{n-k \choose q-k}$

For the left side. I thought of two urns with n red and n blue balls and choosing p-red balls and q-blue balls.

For the right side, i am not very sure, but I thought of make k the number of couples of red and blue balls. Making this is ${n \choose k}$ ways. Since it's the same counting ${n \choose k}$ or ${n \choose n-k}$. I choose ${n-k \choose p-k}$ red balls and the same way with blue.

But I do not think this is right, any help will be appreciated.

Emma Wool
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    First step: $\dbinom{n}{k} \dbinom{n-k}{p-k} = \dbinom{n}{p} \dbinom{p}{k}$. You can prove this easily by bijection. – darij grinberg Sep 24 '17 at 23:56
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    Actually, the identity is false. My bet is that some of the minuses should be pluses or some of the subscripts have been switched, as I recall seeing identities like that (e.g., parts (f) and (g) of Proposition 2.31 in detnotes). – darij grinberg Sep 24 '17 at 23:58
  • i was reading this one https://math.stackexchange.com/questions/1879541/prove-n-choose-k2-sum-i-0kn-choose-in-i-choose-k-in-k-choos?rq=1 which is pretty close about this one, but like you said i don't know if this problem (from a professor) is actually true, @darijgrinberg – Emma Wool Sep 25 '17 at 00:00
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    Actually, if you replace $\dbinom{n-k}{q-k}$ by $\dbinom{n-p}{q-k}$, you get a (correct) corollary of Vandermonde convolution. – darij grinberg Sep 25 '17 at 00:02
  • even your ball thought needs work because if the balls are indistinguishable in each set of n you can say that it only has one choice of p or q balls that work. –  Sep 25 '17 at 00:13

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The question should read \begin{eqnarray*} \binom{n}{\color{red}{p}} \binom{n}{\color{blue}{q}} = \sum_{k=0}^{n} \binom{n}{\color{purple}k} \binom{n-k}{\color{red}{p-k}} \binom{n-p}{\color{blue}{q-k}} \end{eqnarray*} as is pointed out in the comments by Darij Grinberg.

The combinatorial proof is then to place $\color{red}{p}$ red balls & $\color{blue}{q}$ blue balls into $n$ boxes, with the restriction that each box can only contain at most one ball of each colour. This clearly gives the LHS.

Now there are $\color{purple}k$ boxes that recieve balls of both colours ($ \binom{n}{k} $ ways). There are $\color{red}{p-k}$ red balls left to place in $n-k$ boxes ($ \binom{n-k}{p-k}$ ways). Finally there are $\color{blue}{q-k}$ blue balls left to place in the last $n-p$ boxes ($\binom{n-p}{q-k}$ ways.) $k$ can range from zero up to $n$.

Donald Splutterwit
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