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The identity

$$\sum_{k=a}^{n-b} \binom{k}{a} \binom {n-k}{b} = \binom{n+1}{a+b+1} \tag{1}$$

was given in this answer, and should be a restatement of the same identity

$$ \sum_{m=0}^{M} \binom{m+k}{k}\binom{M-m}{n} \tag{2} $$

proven in this past answer. I can follow each step of the proof to $(2)$. I am trying to convince myself that $(1)$ is a restatement of $(2)$, maybe just with a change of variables. It may not even be necessary to use any other combinatorial identities. Still, I still keep getting confused over the meanings of variables when I attempt a substitution. I was hoping someone could please spell out the steps, because I have been stuck here for a while.

Community
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NaN
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1 Answers1

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The original identity $(1)$ was misquoted and that may be the source of the confusion. To get from $(2)$ to $(1)$, substitute as follows: $$ k\mapsto a\\ n\mapsto b\\ M\mapsto n-a\\ m\mapsto k-a $$ This substitution yields $$ \sum_{k=a}^n \binom{k}{a} \binom {n-k}{b} = \binom{n+1}{a+b+1} $$ but note that for $n-b\lt k\le n$, $\binom{n-k}{b}=0$, therefore, we get $$ \sum_{k=a}^{n-b} \binom{k}{a} \binom {n-k}{b} = \binom{n+1}{a+b+1} $$

robjohn
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  • Ah I did misquote the identity $(1)$ and that was a typo in the question. I did mean to write $\binom{k=a}{n-b}$. – NaN Dec 23 '13 at 23:29
  • thanks for returning to the question, since your answer to my question and your proof $(2)$ already seemed thorough. I didn't want to bother you with this stuff again. – NaN Dec 23 '13 at 23:29