# Thread: Proof of combination

1. ## Proof of combination

Prove the following equality

$\binom{-r}{k}$= $(-1)^{k}$ $\binom{r+k-1}{k}$

Any help would greatly appreciated.

2. Originally Posted by hockey777
Prove the following equality

$\binom{-r}{k}$= $(-1)^{k}$ $\binom{r+k-1}{k}$

Any help would greatly appreciated.
How are you defining ${-r \choose k}$? Normally the combinitorial function is not defined for negative values.

-Dan

3. In this case the meaning is ${ - r \choose k} = \frac{{\left( { - r} \right)\left( { - r - 1} \right)\left( { - r - 2} \right) \cdots \left( { - r - k + 1} \right)}}{{k!}}$.
The proof is tedious. Expand both sides and compare.

4. Originally Posted by Plato
In this case the meaning is ${ - r \choose k} = \frac{{\left( { - r} \right)\left( { - r - 1} \right)\left( { - r - 2} \right) \cdots \left( { - r - k + 1} \right)}}{{k!}}$.
The proof is tedious. Expand both sides and compare.
= $-1^k$ $\frac{n(n+1)...(n+k-1)}{k!}$

= $-1^k$ $\binom{n+k-1}{k}$

Would that work?