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How does one prove $$\sum_{v} (-1)^v \binom {a} {v} \binom {n-v} {r}=\binom {n-a} {n-r}$$ where $n,r$ are positive integers, $a$ is arbitary real ,

using the given two identites. $$ \binom {-a} {v}=(-1)^v\binom {a+v-1} {v} $$where $a>0$ is real and $$\binom {m} {0}\binom {l} {r}+\binom {m} {1}\binom {l} {r-1}+....\binom {m} {r}\binom {l} {0}=\binom {m+l} {r}$$

where $m,l$ are arbitary numbers and $r$ is a positive integer. $$ $$I have been trying for a long time now. I can prove it using comparing degrees of certain terms in polynomials. But I don't see a way using the given identities.

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  • $\begingroup$ Are you sure that $\binom {-a} {v}=(-1)^v\binom {a+v-1} {v}$ for $a>0$? $\endgroup$ – Monadologie Jul 23 at 15:28
  • $\begingroup$ @Monadologie Yes, isn't it? Am I overlooking something? $\endgroup$ – saulspatz Jul 23 at 15:40
  • $\begingroup$ @ Monadologie yes. Its from Feller vol 1. $\endgroup$ – jnyan Jul 23 at 16:25
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We obtain \begin{align*} \color{blue}{\sum_{\nu}}&\color{blue}{ (-1)^{\nu}\binom{a}{\nu}\binom{n-\nu}{r}}\\ &=\sum_{\nu} (-1)^{\nu} \binom{a}{\nu}\binom{n-\nu}{n-\nu-r}\tag{1}\\ &=(-1)^{n-r}\sum_{\nu} \binom{a}{\nu}\binom{-r-1}{n-\nu-r}\tag{2}\\ &=(-1)^{n-r}\binom{a-r-1}{n-r}\tag{3}\\ &\,\,\color{blue}{=\binom{n-a}{n-r}}\tag{4} \end{align*} and the claim follows.

Comment:

  • In (1) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

  • In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

  • In (3) we use the Chu-Vandermonde Identity $\sum_{\nu}\binom{p}{\nu}\binom{q}{n-\nu}=\binom{p+q}{n}$.

  • In (4) we use again the binomial identity as we did in (2).

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  • $\begingroup$ thank you very much $\endgroup$ – jnyan Jul 25 at 5:56
  • $\begingroup$ @jnyan: You're welcome. $\endgroup$ – Markus Scheuer Jul 25 at 6:22

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