Chapter 11 - Section 11.5 - The Binomial Theorem - Exercise Set - Page 664: 51

$28$

Since $\left(\begin{array}{c} n\\ r\end{array} \right)=\dfrac{n!}{r!(n-r)!}$, then, $\left(\begin{array}{c} 8 \\ 2 \end{array} \right)$ is equal to \begin{array}{l}\require{cancel} \dfrac{8!}{2!(8-2)!} \\\\= \dfrac{8!}{2!(6)!} \\\\= 28 \end{array}