## Precalculus (6th Edition) Blitzer

Let us consider the left hand side expression of the provided equation, $_{7}{{P}_{3}}$ Recall, $_{n}{{\operatorname{P}}_{r}}=\frac{n!}{\left( n-r \right)!}$ Therefore, \begin{align} & _{7}{{P}_{3}}=\frac{7!}{\left( 7-3 \right)!} \\ & =\frac{7!}{4!} \\ & =\frac{7\times 6\times 5\times 4!}{4!} \\ & =210 \end{align} Also, consider the left hand side expression of the provided equation, $3!\cdot \left( _{7}{{C}_{3}} \right)$ Recall, $_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ Then, \begin{align} & 3!\cdot \left( _{7}{{C}_{3}} \right)=3!\times \frac{7!}{3!\left( 7-3 \right)!} \\ & =3!\times \frac{7!}{3!\left( 4 \right)!} \\ & =\frac{7!}{\left( 4 \right)!} \\ & =\frac{7\times 6\times 5\times 4!}{4!} \end{align} So, $3!\cdot \left( _{7}{{C}_{3}} \right)=210$ Thus, $_{7}{{P}_{3}}=3!\cdot \left( _{7}{{C}_{3}} \right)$