Question

A baseball team’s roster consists of 25 players with 13 position players and the rest being pitchers. The team’s batting order for a particular game consists of 9 players. The first 8 places in the batting order are taken by position players. The last place is taken by a pitcher. How many batting orders are possible? Use the formula for permutations to find your answer.

Answer 1

D. 622,702,080

Explanation:

Answer 2

The total number of permutations is the product of the number of options for each choice:

`$\sum_(i=0)^(n) O_{C_(i)}`

For the first position there are 13 choices, for the second 12 and so on. This pattern holds until the last spot which we are told must be one of the 12 pitchers. Thus:

`N_(BO)=13 * 12*11*10*9*8*7*6*12`

`N_(BO)=622,702,080`