# 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?

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Of the possible committees how many contain at least 2 women?

Of the possible committees how many contain no married couple?

Of the possible committees how many contain at least 2 women?

Of the possible committees how many contain no married couple?

##### 1 Answer

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:

**Total number of ways**

**Exactly 2 women**

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

**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

4 Women

Summing up:

**No married couple**

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