Answer
The statement is false. The correct statement is: "if r is greater than 1, $_{r}{{P}_{n}}$ is less than $_{r}{{C}_{n}}$."
Work Step by Step
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.
