Question

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

Answer 1

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