## Thinking Mathematically (6th Edition)

${}_{n}P_{r}=\displaystyle \frac{n!}{(n-r)!},\quad {}_{n}C_{r}=\frac{n!}{(n-r)!r!}$ ------------- We see that ${}_{n}P_{r}=r!\cdot {}_{n}C_{r}$ The text gives: ${}_{n}P_{r}=6\cdot {}_{n}C_{r}$, so $r!=6=3!$ $r=3$ ${}_{n}P_{3}=3!\cdot {}_{n}C_{3}$ is true for ANY $n \geq 3$, So, we need more information to determine n.