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I want to find out if the following sum has a closed form: $$\sum_{j=0}^N{j+k-1\choose k-1}{N-j+k-1\choose k-1}$$

I tried to rewrite the sum such that I could use the following form of the Vandermonde identity: $$\sum_{m=0}^n{m\choose j}{n-m\choose k-j}={n+1\choose k+1}$$ but I couldn't arrive at any result.

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Hint: You are doing good! Using what you put there you will need $n-(j+k-1)=N-j+k-1$ implies $n=N+2k-2$ and you also have that $2k-2-(k-1)=k-1$ So take the $k$ in the second equation to be $2k-2.$

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  • $\begingroup$ yea, but if I substitute $m=j+k-1$, the sum does not start from 0 afterwards. I can write it in terms of the sum that starts from 0, but a term appears that has the same form as the starting sum. $\endgroup$
    – Dinu Danut-Valentin
    Commented Nov 1, 2020 at 14:44
  • $\begingroup$ Notice that this is not very important because $\binom{n}{k}=0$ for $n<k$ so is like your LHS starts at $m=j$ $\endgroup$
    – Phicar
    Commented Nov 1, 2020 at 14:52

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