Answer
$4$
Work Step by Step
Since $\left(\begin{array}{c} n\\ r\end{array} \right)=\dfrac{n!}{r!(n-r)!}$, then, $\left(\begin{array}{c}
4 \\ 3
\end{array} \right)$ is equal to
\begin{array}{l}\require{cancel}
\dfrac{4!}{3!(4-3)!}
\\\\=
\dfrac{4!}{3!(1)!}
\\\\=
4
\end{array}