#### Answer

$40320$

#### Work Step by Step

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:
$8!=8\cdot7\cdot6\cdot...\cdot3\cdot2\cdot1=40320$