Answer
$56$
Work Step by Step
Recall that $\left(\begin{array}{c}n\\ r\end{array}\right)=C(n,r)$ and denotes the number of combinations of a set of $n$ objects taken $r$, which can be computed as:
$$\left(\begin{array}{c}n\\ r\end{array}\right)=C(n,r)=\dfrac{n!}{(n-r)! r !}$$
using the formula above with $n=8$ and $r=3$ yields:
$$\left(\begin{array}{c}8\\ 3\end{array}\right)=\dfrac{8!}{(8-3)! 3 !}=\dfrac{8 \times 7 \times 6 \times 5!}{5! \cdot 3 \times 2 \times 1 }=56$$
