# Number of ways to form a committee with men and women

1. Mar 26, 2013

### semidevil

1. The problem statement, all variables and given/known data
From a group of 5 women and 7 men, how many different committees con-
sisting of 2 women and 3 men can be formed? What if 2 of the men are
feuding and refuse to serve on the committee together?

Part 1:
my attempt:
for men, we have 7 choose 3
for woman, we have 5 choose 2.
multiply together, and we get 350 possible committees

Part 2 - what if 2 of the men are feuding.
I'm trying to grasp the concept of that and what it means and how to build an equation for it. So we have 7 men and any 2 can be feuding, so if A and B are feuding, A can still serve with C, D, E, F, or G. So that means 5 ways for A. B can do the same, so 5 ways for B, 5 ways for C, 5 ways for D, 5 ways for E which is 30 ways. The men can be swapped around in 30 ways.
Women can still be grouped in to 5 choose 2, so 30 x 10 = 300.

I was able to work out the solution manually thinking it out, but I want to know how the author did it for part 2.

author solution:

There are C(5; 2)C(7; 3) = 350 possible committees consisting of 2 women
and 3 men. Now, if we suppose that 2 men are feuding and refuse to serve
together then the number of committees that do not include the two men
is C(7; 3) - C(2; 2)C(5; 1) = 30 possible groups. Because there are still
C(5; 2) = 10 possible ways to choose the 2 women, it follows that there are
30 10 = 300 possible committees

what is C(2; 2) and C(5; 1)?? how did he come up with that?

2. Mar 26, 2013

### Ibix

He's using C(n;r) to mean n-choose-r, sometimes written nCr, nCr, or Cnr.

He has divided the men into two groups - the two feuders and the other five. If C(7;3) is all groups of three men, and C(7;3)-C(2;2)C(5;1) is all groups not including both feuders, what must C(2;2)C(5;1) be?