Let $$b_n=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\binom{m+k}{m-r}$$ By Vandermonde's identity, there has $$\binom{m+n}{m-r}=\binom{n+m}{n+r}=\sum_{k=0}^n\binom{n}{k}\binom{m}{k+r}$$ so $$b_n=\binom{m}{n+r}$$ I wonder if it is possible to directly calculate $$\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\binom{m+k}{m-r}$$ to get $$\binom{m}{n+r}$$
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1You may look at Binomial inversion formula .. https://math.stackexchange.com/questions/55659/combinatorial-interpretation-of-binomial-inversion – r9m Sep 25 '21 at 13:42
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@r9m Yeah,but I have calculated it using the binomial inversion formula, I wonder how to directly calculate it. – YulinYang Sep 26 '21 at 00:35