Chapter 13.2, Problem 58AYU

To determine

To find: How many batting orders is it possible for a manager to use?

Answer to Problem 58AYU

There are 40320 possible batting orders.

Explanation of Solution

Given:

In the National Baseball League, the pitcher usually bats ninth.

Formula used:

$P\left(n,r\right)=n!/\left(n-r\right)!$

Calculation:

Number of players available is 9 and the pitcher usually bats ninth.

Hence,

Number of choices for the first player is $P\left(8,1\right)=8$.

Number of choices for the second player is $P\left(7,1\right)=7$.

Number of choices for the third player is $P\left(6,1\right)=6$.

Number of choices for the fourth player is $P\left(5,1\right)=5$.

Number of choices for the fifth player is $P\left(4,1\right)=4$.

Number of choices for the sixth player is $P\left(3,1\right)=3$.

Number of choices for the seventh player is $P\left(2,1\right)=2$.

Number of choices for the eighth player is $P\left(1,1\right)=1$.

Number of choices for the ninth player is $P\left(1,1\right)=1$.

Total number of choices is $8*7*6*5*4*3*2*1*1=40320$.

There are 40320 possible batting orders.