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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?

user267839
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    Do not know if this helps, but the sum you write will be the $x^a$-coefficient of $(1-x)^a(1+x)^b = (1-x^2)^b(1-x)^{a-b}$. – Hw Chu Oct 08 '18 at 20:18

1 Answers1

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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...

Eleven-Eleven
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