Find sum of given binomial series for $n>3$

Question

We are given that $n>3$ and we have to find the sum of the series given by: $$S=xyz\binom{n}{0}-(x-1)(y-1)(z-1)\binom{n}{1}+...+(-1)^n(x-n)(y-n)(z-n)\binom{n}{n}$$

I figured out that the general term is $$t(r)=(-1)^r(x-r)(y-r)(z-r)\binom{n}{r}$$ but I see no obvious manipulations between the terms nor does any particular series strike my mind.

Can someone provide an approach? Any help would be appreciated.

Answer 1

The given sum is zero for $n>3$. Note that $$(-1)^r(x-r)(y-r)(z-r)\binom{n}{r}$$ is a linear combination of $\binom{n}{r}r^k(-1)^r$ with $k=0,1,2,3$. Moreover, for $0\leq k<n$ we have that $$\sum_{r=0}^n\binom{n}{r}r^k(-1)^r=0$$ (for a detailed proof see Simplifying $\sum_{r = 0}^{n} {{n}\choose{r}}r^k(-1)^r$ ).