Question

The manager of a baseball team has 15 players to choose from for his 9 person batting order

Answer 1

Given the question:
The manager of a baseball team has 15 players to choose from for his 9-person batting order. How many different ways can he arrange the players in the lineup?
5,005    362,880    3,603,600    1,816,214,400

The number of was in which a smaller group is choosing from a bigger group is given by the bigger number combination the smaller number.
$^nC_r= \frac{n!}{(n-r)!r!}$
Here, we want to choose a 9-person batting order from a team of 15 players. This is given by:
$^{15}C_9= \frac{15!}{(15-9)!9!}=\frac{15\times14\times13\times12\times11\times10\times9!}{6!\times9!}= \frac{3,603,600}{720} =5,005$

Answer 2

Answer:

The answer is 1,816,214,400 not 5005. Just took the quiz

Step-by-step explanation: