Answer
$12,870$
Work Step by Step
We notice that we have:
$$\binom{8}{8-k}=\dfrac{8!}{k!(8-k)!}=\binom{8}{k}.$$
Evaluate the sum:
$$\begin{align*}
\sum_{k=0}^8\binom{8}{k}\binom{8}{8-k}&=\sum_{k=0}^8\binom{8}{k}^2\\
&=\binom{8}{0}^2+\binom{8}{1}^2+\binom{8}{2}^2\\
&\phantom{=}+\binom{8}{3}^2+\binom{8}{4}^2+\binom{8}{5}^2\\
&\phantom{=}+\binom{8}{6}^2+\binom{8}{7}^2+\binom{8}{8}^2\\
&=1^2+8^2+28^2+56^2+70^2+56^2+28^2+8^2+1^2\\
&=12,870.
\end{align*}$$
