## Precalculus (6th Edition) Blitzer

The statement is false. The correct statement is: "if r is greater than 1, $_{r}{{P}_{n}}$ is less than $_{r}{{C}_{n}}$."
Let us consider, $r=2,n=4$ And recall, $_{n}{{\operatorname{P}}_{r}}=\frac{n!}{\left( n-r \right)!}$ Therefore, \begin{align} & _{4}{{P}_{2}}=\frac{4!}{\left( 4-2 \right)!} \\ & =\frac{4!}{2!} \\ & =\frac{4\times 3\times 2!}{2!} \\ & =12 \end{align} Recall, $_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ So, \begin{align} & _{4}{{C}_{2}}=\frac{4!}{2!\left( 4-2 \right)!} \\ & =\frac{4!}{2!\left( 2 \right)!} \\ & =\frac{4\times 3\times 2}{2\times 2} \\ & =6 \end{align} We know that 12 is greater than 6. Hence, the statement is false for all value of n when r is greater than 1.