$\suṃ̣_{k=m}^{n} (-1)^k {n \choose k} {k \choose m}$ for all positive integers such that n > m.
My attempt, since k=m. I get $\suṃ̣_{k=m}^{n} (-1)^k {n \choose k} {m \choose m} = \suṃ̣_{k=m}^{n} (-1)^k {n \choose k}$.
What do I do from this step?
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1No, $k=m$ is only true for the first term of the sum, then the second term you have $k=m+1$ followed by $k=m+2$, and so on. – abiessu Dec 19 '19 at 00:06
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4Does this answer your question? $\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m}=0, n>m\geq 0$ – Maximilian Janisch Dec 19 '19 at 01:06
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See also https://math.stackexchange.com/questions/1276332/evaluate-sum-limits-k-mn-1k-n-choose-k-k-choose-m – Maximilian Janisch Dec 19 '19 at 01:06
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Use the identity $$ \binom{n}{k}\binom{k}{m}=\binom{n}{m}\binom{n-m}{k-m} $$ to rewrite the sum as $$ \binom{n}{m}\sum_{k=m}^n (-1)^k\binom{n-m}{k-m}= \binom{n}{m}(-1)^m\sum_{u=0}^{n-m}(-1)^{u}\binom{n-m}{u} =\binom{n}{m}(-1)^m(1-1)^{n-m}=0 $$ where in the first equality we make the change of indices $u=k-m$ and in the penultimate equality we use the binomial theorem and the fact that $n>m$.
Sri-Amirthan Theivendran
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