A baseball team consists of nine players. How many different batting orders are possible? How many are possible if the pitcher bats last?

Since the pitcher bats last, there is only one choice for the 9th slot. For the 1st slot, there are 8 choices remaining. For the 2nd slot, there are 7 choices remaining, etc.

So the number of possible batting orders is:

8*7*6*5*4*3*2*1 = 8! = 40,320