## Precalculus (6th Edition) Blitzer

The required solution is $\frac{{}_{7}{{P}_{3}}}{3!}-{}_{7}{{C}_{3}}=0$ .
The permutation and combination formulas respectively are: ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ And ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ Applying these formulas into the provided expression, we get: \begin{align} & \frac{{}_{7}{{P}_{3}}}{3!}-{}_{7}{{C}_{3}}=\frac{7!}{3!\left( 7-3 \right)!}-\frac{7!}{3!\left( 7-3 \right)!} \\ & =0 \end{align} $\frac{{}_{7}{{P}_{3}}}{3!}-{}_{7}{{C}_{3}}=0$