## Precalculus (10th Edition)

$40320$
The number of ways arranging $n$ distinct objects is $n!=n\cdot(n-1)\cdot(n-2)\cdot...\cdot3\cdot2\cdot1.$ Since the pitcher will bat ninth, then there are only $8$ players to arrange so $n=8$. Hence, the number of possible batting orders is: $8!=8\cdot7\cdot6\cdot...\cdot3\cdot2\cdot1=40320$