# Combinatorics proof for Negative Binomial.

I'm trying to demostrate this for a proof about two random variables(Binomial Negative): $$\sum_{j=0}^k {j+r-1 \choose j}{k-j+s-1 \choose k-j}= {k+r+s-1 \choose k}$$

I know the basics about combinatory so: $$\sum_{j=0}^k {j+r-1 \choose j}\cdot{k-j+s-1 \choose k-j}= \frac{(j+r-1)!}{j!(r-1)!}\cdot\frac{(k-j+s-1)!}{(k-j)!(s-1)!}$$

But i don't know how to arrange them for the proof.

$\dbinom {j+r-1}{j}$ counts the ways to place $j$ indistinct balls into $r$ boxes.

$\dbinom {(k-j)+s-1}{k-j}$ counts the ways to place $k{-}j$ indistinct balls into $s$ boxes.

If we count the ways to take $0$ to $k$ such balls from $k$, place them in $r$ boxes and the rest into $s$ boxes, then we have counted the ways to place $k$ indistinct balls into $r+s$ boxes.

This is counted by $\dbinom{k+(r+s)-1}{k}$.

$\blacksquare$

$$\sum_{j=0}^k\dbinom{j+r-1}{j}\dbinom{(k-j)+s-1}{k-j} ~=~ \dbinom{k+(r+s)-1}{k}$$

• As I noted in my answer, Vandermonde's identity claims the $-1$ in the answer should be $-2$, however you also get $-1$ like the assignment. What causes this? Am I reading the identity incorrectly? – orlp May 3 '17 at 1:12
• Yes. @orip Note the $j$ and $k$ in the numerator of the binomials. Vandermonde's identity is $$\sum_{j=0}^k \binom{r-1}{j}\binom{s-1}{k-j}=\binom{r+s-2}{k}$$ – Graham Kemp May 3 '17 at 1:20
• I went off Vandermonde's identity off of Wikipedia, but for a second I forgot I am cancelling a bounded variable ($j$) outside of the scope of that bounded variable. In other words I was almost assuredly doing something wrong. My bad. – orlp May 3 '17 at 1:23
• when the convolution involves both the upper and the lower term, at the bottom you get the sum of the lower terms, at top the sum +1 of the upper terms – G Cab Aug 3 '17 at 15:13

From a negative binomial standpoint, note that

$$\binom{j+r-1}{j}=\left[z^j\right](1-z)^{-r}$$

and

$$\binom{k-j+s-1}{k-j}=\left[z^{k-j}\right](1-z)^{-s}$$

therefore

\begin{align}\binom{k+r+s-1}{k}&=\left[z^k\right](1-z)^{-(r+s)}\\&=\left[z^{k}\right](1-z)^{-r}(1-z)^{-s}\\&=\sum_{j=0}^{k}\left[z^j\right](1-z)^{-r}\left[z^{k-j}\right](1-z)^{-s}\\&=\sum_{j=0}^{k}\binom{j+r-1}{j}\binom{k-j+s-1}{k-j}\end{align}