Question

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? 3003 (b) How many ways are there to assign the 10 positions by selecting players from the 15 people who show up? 3003 (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 a woman?

Answer 1

Answer:

(a) 3003 ways

(b) 10897286400 ways

(c) 3002 ways

Step-by-step explanation:

Given

$n = 15$ --- 15 players

Solving (a) Ways of selecting 10 players.

This implies combination.

So, we have:

$r=10$

Using:

$^nC_r = \frac{n!}{(n-r)!r!}$

We have:

$^{15}C_{10} = \frac{15!}{(15-10)!10!}$

$^{15}C_{10} = \frac{15!}{5!10!}$

Simplify

$^{15}C_{10} = \frac{15*14*13*12*11*10!}{5!10!}$

$^{15}C_{10} = \frac{15*14*13*12*11}{5!}$

$^{15}C_{10} = \frac{15*14*13*12*11}{5*4*3*2*1}$

$^{15}C_{10} = \frac{360360}{120}$

$^{15}C_{10} = 3003$

Solving (b) Ways of assigning positions to 10 players.

This implies permutation.

So, we have:

$r=10$

Using:

$^nP_r = \frac{n!}{(n-r)!}$

We have:

$^{15}P_{10} = \frac{15!}{(15-10)!}$

$^{15}P_{10} = \frac{15!}{5!}$

Solve each factorial

$^{15}P_{10} = \frac{1307674368000}{120}$

$^{15}P_{10} = 10897286400$

Solving (c) Ways of choosing at least 1 woman

We have:

$Men = 10$

$Women = 5$

Ways of selecting 10 players is: (a) 3003 ways

Since the number of men are 10, there is 1 way of selecting 10 men (i.e. selection without women)

Using complement rule:

At least 1 woman = Total - No woman

$At\ least\ 1\ woman = 3003 - 1$

$At\ least\ 1\ woman = 3002$