15 players for a softball team show up for a game: (a) How many ways are there to choose 10 players to take the field? 10!/(5!) (b) How many ways are there to assign the 10 positions by selecting players from the 15 people who show up? (c) Of the 15 people who show up, 5 are women. How many ways are there to choose 10 players to take the field if at least one of these players must be women?

a) It is combination (order does not matter):

$C(15,10)=\frac{15!}{10!5!}=3003$

b) It is permutation (order is important):

$P(15,10)=\frac{15!}{15-10!}=\frac{15!}{5!}=10897286400$

c) Lets determine number of ways to choose 10 players to take the field such that none are a woman. Since 5 of 15 players are women, there are 15-5=10 men. Hence number of ways to select 10 men of ten players is 1. There are 3003 ways to choose the 10 players overall (part a)). Therefore

Number of ways with at least one woman = Number of ways – Number of ways without women = 3003 – 1 = 3002

Answer:

a) 3003 ways to choose 10 players to take the field.

b) 10897286400 ways to assign the 10 positions.

c) 3002 ways to choose 10 players such that at least one is a woman.