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?

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="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=>#

#=>(2)^4(1)^2=16#