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Fedor Petrov
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Combinatorial proof of a certain binomial identity

Let $n,p,q$ be non-negative integers. Then $$ \sum_{k=0}^n{2k+2p\choose k+p,k,p}{2(n-k)+2q\choose n-k+q,n-k,q}=4^n{2p\choose p}{2q\choose q}{n+p+q\choose n}.\quad\quad(\heartsuit) $$ In particular, for $p=q=0$ $(\heartsuit)$ reads as $$ \sum_{k=0}^n{2k\choose k}{2(n-k)\choose n-k}=4^n.\quad\quad\quad(\clubsuit) $$ There are known nice combinatorial (essentially bijective) proofs of $(\clubsuit)$, I wonder whether there is anything equally nice working for $(\heartsuit)$.

For what it worth, $(\heartsuit)$ is just Chu--Vandermonde identity $\sum {x\choose k}{y\choose n-k}={x+y\choose n}$ for $x=-p-1/2$, $y=-q-1/2$.

Fedor Petrov
  • 108.9k
  • 9
  • 264
  • 459