Question

The manager of an amateur baseball team has selected the nine players for an upcoming game. How many batting orders are possible if the pitcher must be last, the first baseman must bat sixth, and the catcher must bat second?

Answer 1

The number of ways to pick the pitcher, the first baseman and the catcher is the number of ordered selections of three elements from a set of 9 elements

$$\dfrac{9!}{(9-6)}=504$$

We can arrange (9-3)=6 players in $6!=720$ ways.

$$504(720)=362880$$

362880 batting orders are possible.