I'm struggling with proving the identity $$\sum_{k=p}^{q}\binom{l}{m+k}\binom{s}{n+k}=\binom{l+s}{l-m+n}$$ where $$p=-\min(m,n)~ \text{and}~q=\min(l-m,s-q).$$ It reminds me of Vandermonde's identity but still I can't get it right. I would appreciate an algebraic or combinatorial proof.
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What are your attempts? – user690234 Mar 13 '20 at 09:53
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What is the variation of $k$ the lowest and the smallest of it. Are there any restrictions on m,n, l and s! – Z Ahmed Mar 13 '20 at 09:58
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Let $l=s=m=n=1$ and sum over $k=0,1$ sum is 1, but the claimed result goves 2? – Z Ahmed Mar 13 '20 at 10:11
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$\sum_{k=p}^{q}{\left(\binom{l}{m+k}\binom{s}{n+k}\right)}$ is the coefficient of $x^{m+k}\cdot \left(\frac{1}{x}\right)^{n+k}=x^{m-n}$ from $\left(1+x\right)^{l}\left(1+\frac{1}{x}\right)^{s}=\frac{\left(1+x\right)^{l+s}}{x^{s}}$ which is $\binom{l+s}{m-n+s}=\binom{l+s}{l-m+n}.$ Here $p=-min(m,n)$ and $q=min(l-m,s-n).$
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Thank you! I tried to do something like this but I could't imagine the proper way to do this :) – Nerwena Mar 13 '20 at 10:01
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Let $l=s=m=n=1$ and sum over $k=0,1$ sum is 1, but the claimed result goves 2? – Z Ahmed Mar 13 '20 at 10:09
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@DrZafarAhmedDSc he forgot to include the limit of $k$ which should be from $-min(m,n)$ to $min(l-m,s-n)$ – acat3 Mar 13 '20 at 10:59
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I copied it from my lecturer site it looks excatly like I wrote it here. He's a lazy person he doesb't write limits at all so I don't know how it is supposed to look like. I've searched the Internet but I didn't find this equation. – Nerwena Mar 13 '20 at 11:55
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@DrZafarAhmedDSc in Your example it should be $\sum_{k=-2}^{5}$ not $\sum_{k=2}^{5}$ – acat3 Mar 13 '20 at 12:01