Let $a,b \in \mathbb{N}$ arbitrary with $a > b$. Is that true that
$$\sum_{j=0}^b (-1)^j\binom{a}{j}\binom{b}{j} \neq 0$$
Till now I wasn't able to find a conterexample but also my attempts to prove it fail.
I get only $\sum_{j=0}^b (-1)^j\binom{a}{j}\binom{b}{j} = \sum_{j=0}^b (-1)^j \frac{a!b!}{(a-j)!(b-j)(j!)^2}$. Does anybody see the key how to cope with it?
This is less of an answer, but more of a guide. Because
$$\binom{b}{j}=\binom{b}{b-j}$$
you can rewrite your sum as
$$S_{a,b}=\sum_{j=0}^b(-1)^j\binom{a}{j}\binom{b}{b-j}$$
which is an alternating version of Chu-Vandermonde's identity
Also, this page might help you get further in your calculation. It looks like their might be similarities between your alternating sum...