Evaluate $\displaystyle\sum_{r=0}^8 (-1)^r \binom{20}{r} \binom{20}{8-r}$

Question

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$$\sum_{r=0}^8 (-1)^r \binom{20}{r} \binom{20}{8-r}$$

Answer 1

Evaluate $$\sum_{r=0}^8 (-1)^r \binom{20}{r} \binom{20}{8-r}$$

I would evaluate $$\sum_{r=0}^8 (-1)^r \binom{20}{r} \binom{20}{12+r}={20\choose 16}$$

that is the coefficient of $x^8$ in the binomial series of $(x^2-1)^{20}$ got by multiplying the binomial series below. \begin{align*} (1+x)^{20}&=\sum_{i=0}^{20}{20\choose k}x^i\\ (x-1)^{20}&=\sum_{j=0}^{20}(-1)^j{20\choose j}x^{20-j}\\ \end{align*}