# Proof of combination

• Feb 19th 2008, 06:19 AM
hockey777
Proof of combination
Prove the following equality

$\displaystyle \binom{-r}{k}$=$\displaystyle (-1)^{k}$$\displaystyle \binom{r+k-1}{k} Any help would greatly appreciated. • Feb 19th 2008, 07:46 AM topsquark Quote: Originally Posted by hockey777 Prove the following equality \displaystyle \binom{-r}{k}=\displaystyle (-1)^{k}$$\displaystyle \binom{r+k-1}{k}$

Any help would greatly appreciated.

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

-Dan
• Feb 19th 2008, 08:18 AM
Plato
In this case the meaning is $\displaystyle { - r \choose k} = \frac{{\left( { - r} \right)\left( { - r - 1} \right)\left( { - r - 2} \right) \cdots \left( { - r - k + 1} \right)}}{{k!}}$.
The proof is tedious. Expand both sides and compare.
• Feb 19th 2008, 09:15 AM
hockey777
Quote:

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

=$\displaystyle -1^k$ $\displaystyle \frac{n(n+1)...(n+k-1)}{k!}$

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

Would that work?