# Suppose that a committee of 4 is to be chosen from 6 married couples. In how many ways can this be done? Of the possible committees how many contain exactly 2 women?

## Of the possible committees how many contain at least 2 women? Of the possible committees how many contain no married couple?

See below:

#### Explanation:

We have a committee of 4 people being chosen from a pool of 6 married couples. I'm going to assume that "married couple" means 1 man and 1 woman, and so we have 6 men and 6 women, or 12 people total.

We'll be using combinations because a committee with Andy, Barb, Clark, and Dave is the same as Dave, Clark, Barb, and Andy. The general formula is:

C_(n,k)=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

Total number of ways

C_(12,4)=(12!)/((4!)(12-4)!)=(12!)/(8!4!)=(12xx11xx10xx9xx8!)/(8!xx4xx3xx2)=495

Exactly 2 women

We'll pick 2 women from the 6 available and the same with the men, and so:

(C_(6,2))(C_(6,2))=(C_(6,2))^2=((6!)/(2!4!))^2=((6xx5xx4!)/(4!xx2))^2=15^2=225

At least 2 women

We can add in the numbers of ways to have 3 and 4 women on the committee to the calculation above:

3 Women

(C_(6,3))(C_(6,1))=((6!)/(3!3!))((6!)/(5!1!))=>

=>((6xx5xx4xx3!)/(3!xx3xx2))((6xx5!)/(5!))=20xx6=120

4 Women

C_(6,4)=(6!)/((4!)(6-4)!)=(6xx5xx4!)/(4!xx2)=15

Summing up:

$225 + 120 + 15 = 360$

No married couple

For this one, we can pick one person from 4 of the married couples and no one from the other 2:

(C_(2,1))^4(C_(2,0))^2=((2!)/((1!)(2-1)!))^4((2!)/((0!)(2-0)!))^2=>

$\implies {\left(2\right)}^{4} {\left(1\right)}^{2} = 16$