Why is $ \binom{-r}{z} = (-1)^z\binom{z+r-1}{r-1} $

Question

How do we prove this identity? I Can't find a simple solution.

Have you tried using the definition of the left hand side? – Mariano Suárez-Álvarez Apr 16 '17 at 06:19 — Mariano Suárez-Álvarez, Apr 16 '17 at 06:19

score 1 · Answer 1 · answered Apr 16 '17 at 06:23

1

$$\binom{-r}{z}=\frac{(-r)(-r-1)(-r-2)\cdots(-r-z+1)}{z!}\\=\frac{(-1)^z r(r+1)\cdots(z+r-1)}{z!}=(-1)^z\binom{z+r-1}{z}=(-1)^z\binom{z+r-1}{r-1}$$

answered Apr 16 '17 at 06:23

Shaswata

Thanks for the answer. Why is $ \frac{r(r+1)...(z+r-1)}{z!}=\binom{z+r-1}{z} $ – matt Apr 16 '17 at 06:57
Rewrite it as $\frac{(z+r-1)(z+r-2)\cdots r }{z!}=\frac{(z+r-1)(z+r-2)\cdots r , (r-1)!}{z! (r-1)!}=\frac{(z+r-1)!}{z!(r-1)!}$ – Shaswata Apr 16 '17 at 09:51

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