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I recently noticed that for a triple of integers $k \geq 2$, $k \geq m \geq t \geq 1$, the following identity seems to hold

$\sum_{j=0}^{m-t} (-1)^{m-t-j}{k \choose j}{m-1-j \choose t-1}={k-t \choose m-t}.$

Is this a known binomial identity? If yes, is there any reference where I can find the proof?

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closed as off-topic by Gjergji Zaimi, Joonas Ilmavirta, Chris Godsil, Alexey Ustinov, Peter Humphries Nov 13 '15 at 5:31

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    $\begingroup$ It is. A version of Chu-Vandermonde. $\endgroup$ – Wadim Zudilin Nov 12 '15 at 9:58
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I don't see how this can be obtained directly from Vandermonde's identity, but it is an easy induction over $t$ using ${k \choose j}={k-1 \choose j-1}+{k-1 \choose j}$. Grouping neighboring terms together, you will see that the LHS is just the sum of iterated differences of neighboring binomial coefficients, and iterating them $t$ times, it all boils down to ${k-t \choose m-t}$. Rather than writing it formally (who will ever read it?), I suggest you try it out with values like $k=10,m=8,t=3$.

If we put $n:=m-t$, we get as a by-product for $r=1,...,n$ $$\sum_{j=0}^{n} (-1)^{j}{m-r \choose j}{m-j-1 \choose m-n-1}=0.$$

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  • $\begingroup$ Hint: $(-1)^{m-t-j}\binom{m-1-j}{t-1}=\binom{-t}{m-t-j}$. $\endgroup$ – Gjergji Zaimi Nov 12 '15 at 19:57
  • $\begingroup$ @GjergjiZaimi Oh yes, sure enough. I never think of those negative "numerators". $\endgroup$ – Wolfgang Nov 12 '15 at 20:23

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