# How to prove this combinatorial identity

I am wondering how to prove the following identity:
$$\sum_{k=0}^r {r-k \choose m} {s \choose k-t} (-1)^{k-t} = {r-t-s \choose r-t-m}$$
It seems that I can negating the upper index of ${s \choose k-t} (-1)^{k-t}$,but I can not find a way to continue.

t, r, m are nonnegative integers, this question comes from TAOCP vol 1, page 59, (24)

I have solved it myself.But I am not sure whether it's true
$$\sum_{k=0}^r {r-k \choose m} {s \choose k-t} (-1)^{k-t}$$ $$= \sum_{k=0}^r {r-k \choose r-k-m} {s \choose k-t} (-1)^{k-t}$$ $$= \sum_{k=0}^r {-m-1 \choose r-k-m} {s \choose k-t} (-1)^{r-t-m}$$ $$= {s-m-1 \choose r-t-m} (-1)^{r-t-m}$$ $$= {r-t-s \choose r-t-m}$$

and is there other solutions?

-
What have you tried? Where does this problem come from? The values $r$,$s$,$t$ and $m$ belongs to what? Are they related? – user37238 Feb 21 '14 at 9:15
How did you get the step where the summation disappears? – ShreevatsaR Feb 21 '14 at 14:46
@ShreevatsaR vandermonde convolution – Matsune Feb 21 '14 at 14:49
Ah I see. Your solution is right, then. – ShreevatsaR Feb 21 '14 at 16:48