Question

The manager of a baseball team has 15 players to choose from for his nine person batting order. How many different ways can he arrange the players in the lineup. A.5005. B.362880. C.3603600. D.1816214400

Answer 1

this is 15P9 Number of permutations of 9 in 15.

= 15! / (15-9)!

= 1816214400

Answer 2

Answer: D. 1816214400

Explanation:

When we select r things from n things in order we apply permutations and the number of ways to select r things =
`^nP_r=(n!)/((n-r)!)`

Given : Total player = 15

Required number of players for Batting order = 9

Then the number of different ways to select 9 person batting order so that he arrange the players in the lineup would be
`^(15)P_9=(15!)/((15-9)!)`

`=(15*14*13*12*11*10*9*8*7*6!)/(6!)`

`=1816214400`

∴ The number of different ways can he arrange the players in the lineup = 1816214400

Hence, the correct answer is D. 1816214400