# Math Help - Clarification on enumeration

1. ## Clarification on enumeration

In a group of 15 people, 7 are women. They must make a committee of 5 with at least 1 woman on it, how many possible commitees are there?

I get stuck on doing 15C5 to get the total possible number of comitees to be 3003, but I'm not sure exactly how to go from there. Any help is appreciated, thanks!

In a group of 15 people, 7 are women. They must make a committee of 5 with at least 1 woman on it, how many possible commitees are there?

I get stuck on doing 15C5 to get the total possible number of comitees to be 3003, but I'm not sure exactly how to go from there. Any help is appreciated, thanks!
at least one woman means one or more. thus, you want the sum of the number of committees of each type below.

1 woman, 4 men
2 women, 3 men
3 women, 2 men
4 women, 1 man
5 women

note, this is not a probability problem. it is a combinatorics problem. in general, there is no need to find the total possible outcomes. unless you want to do the compliment perhaps. we can get the same answer by finding the numer of committees that have no women, and subtract them from the total number of possible committees. can you finish?

In a group of 15 people, 7 are women.
They must make a committee of 5 with at least 1 woman on it.
How many possible committees are there?

I get 15C5 to get the total possible number of committees to be 3003.

This is correct, but do you know why we might need that number?
You don't have any game plan whatsoever?

One way is to COUNT the committees! . . .
(What a concept!)

"At least one woman" means that we have committees with:
. . one woman, two women, three women, four women, or five women.

How many have one woman (and four men)?
There are: . $_7C_1 = 7$ ways to pick one woman.
There are: . $_8C_4 = 70$ ways to pick 4 men.
. . Hence, there are: . $7\cdot70 \:=\:{\color{blue}490}$ committees with one woman.

How many have two women (and three men)?
There are: . $_7C_2 = 21$ ways to pick 2 women.
There are: . $_8C_3 = 56$ ways to pick 3 men.
, , Hence, there are: . $21\cdot56\:=\:{\color{blue}1176}$ committees with two women.

How many have three women (and two men)?
There are: . $_7C_3 = 36$ ways to pick 3 women.
There are: . $_8C_2 = 28$ ways to pick 2 men.
. . Hence, there are: . $35\cdot28 \:=\:{\color{blue}980}$ committees with three women.

How many have four women (and one man)?
There are: . $_7C_4 = 35$ ways to pick 3 women.
There are: . $_8C_1 = 8$ ways to pick 1 man.
. . Hence, there are: . $35\cdot8 \:=\:{\color{blue}280}$ commiittees with four women.

How many have five women (and no men)?
There are: . $_7C_5 = 21$ ways to pick 5 women.
. . Hence, there are ${\color{blue}21}$ committees with five women.

Therefore: . $490 + 1176 + 980 + 280 + 21 \;=\;\boxed{2947}$ committees with at least one woman.

Now I'll someone else show you the fast way to solve this problem.