Answer
$35$
Work Step by Step
Using $\left( \array{n\\r} \right)=\dfrac{n!}{r!(n-r)!},$ the given expression, $\left( \array{
7\\3
} \right)$ evaluates to
\begin{array}{l}\require{cancel}
=\dfrac{7!}{3!(7-3)!}
\\\\=
\dfrac{7!}{3!4!}
\\\\=
\dfrac{7(6)(5)(4!)}{3(2)(1)(4!)}
\\\\=
\dfrac{7(\cancel{6})(5)(\cancel{4!})}{\cancel{3(2)}(1)(\cancel{4!})}
\\\\=
\dfrac{35}{1}
\\\\=
35
\end{array}
