Question

Thirteen people on a softball team show up for a game. a)How many ways are there to choose 10 players to take the field? b)How many ways are there to assign the 10 positions by selecting players from the 13 people who show up? c)Of the 13 people who show up, three are women. How many ways are there to choose 10 players to take the field if at least one of these players must be a woman?

Answer 1

Answer:

a) 286

b) 1,037,836,800

c) 285

Step-by-step explanation:

a) Since order is not important, the total possible number of ways to choose 10 players out of 13 is the following combination:

$C(13,10)=\frac{13!}{(13-10)!10! } \\C(13,10)=\frac{13*12*11*10!}{(3*2*1)10! }\\C(13,10) = 286$

b) The total number of possibilities to assign positions by the selecting 10 players is the permutation of 13 players for 10 positions:

$P(13,10) = \frac{13!}{(13-10)!}\\P(13,10) = 13*12*11*10*9*8*7*6*5*4\\P(13,10) =1,037,836,800$

c) The number of ways to pick 10 players including at least one woman is equal to the total number of ways to pick 10 players (found in item a) minus the the number of ways to pick 10 players without picking a single woman.

Since there 10 male players for 10 positions, there is only one possible way to pick a team without women, therefore:

$P=286-1 =285$