Well, let's see what do Mathematica say about this :
In:= Sum[(-1)^p*Binomial[a, p]*Binomial[n - p, r], {p, 0, a}] Out= Binomial[-a + n, -a + r]
It looks like your statement is not true... (EDIT but of course it is true !)
Let's go for a proof. Let $f_{n,a,p,r}$ denote the expression $(-1)^p \binom{a}{p} \binom{n-p}{r}$, and $S_u$, the shift w.r.t. the variable $u$. For example, $(S_a f)_{n, a, p, r} = f_{n, a+1, p, r}$.
Let $g$ denote the sum $\sum_{p=0}^a f_{n,a,p,r}$, and $g'$ the expression $\binom{n-a}{r-a}$. We want to proof $g=g'$.
You can check that $$ \left((1+r)S_r+p+r-n\right)f + (S_p - 1)\left(\frac{-p - n p + p^2}{1 + r}f\right) = 0$$ When you sum, you get (notice the telescoping sum) $$ \left((1+r)S_r+p+r-n\right)g + \underbrace{\left[\frac{-p - n p + p^2}{1 + r}f\right]_{p=0}^{a+1}}_{=0} = 0$$
This gives you a recurrence for $g$, and it is easy to check that $g'$ satisfies the same.
You have similar formulas for the variables $n$ ans $r$. And you check that $g$ satisfies the defining recurrence equations of $g'$. And so $g=g'$.
These formulas can be found with the method of creative telescoping, which is implemented in Mathematica by the package HolonomicFunctions of Christoph Koutschan : you can prove this equality in a algorithmic way ! strong text