## Precalculus (6th Edition) Blitzer

The required solution is $\frac{{}_{20}{{P}_{2}}}{2!}-{}_{20}{{C}_{2}}=0$.
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{{}_{20}{{P}_{2}}}{2!}-{}_{20}{{C}_{2}}=\frac{20!}{2!\left( 20-2 \right)!}-\frac{20!}{2!\left( 20-2 \right)!} \\ & =0 \end{align} $\frac{{}_{20}{{P}_{2}}}{2!}-{}_{20}{{C}_{2}}=0$