Probability question: From a group of 8 women, 6 men,

[Jamin2112]

Probability question: "From a group of 8 women, 6 men, ..."

Homework Statement

From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. How many different committees are possible if [...] 2 of the men refuse to serve together?

Homework Equations

n choose k = n!/((n-k)! * k!)

multiplication rule

The Attempt at a Solution

Clearly the number of total committees that can be formed from 3 of 8 women and 3 of 6 men is (8 choose 3) * (6 choose 3). I need to subtract the number of those committees in which some arbitrary pair of men would've served together. Could someone please explain (or hint at) the logic involved in finding this?

OT link removed

 

[HallsofIvy]

I would do this: call the 6 men a, b, c, d, e, and f and assume, without loss of generality, that the two men who refuse to serve on the same committee are a and b. Then there are three ways to choose the three men to serve on the commitee:
1) Choose a but not b. Then we must choose 2 more men from the four, c, d, e, and f. There are 4(3)= 12 ways to do that.
2) Choose b but not a. As before, there are 12 ways to do that.
3) Choose neither a nor b to be on the committee. That is, we choose 3 out of c, d, e, and f. There are 4(3)(2)= 24 ways to do that.

There are, therefore, 12+ 12+ 48= 96 ways to choose the three men.

There are, as you say. "8 choose 3"= 8(7)(6)= 336 ways to choose the three women.

 

[Strants]

1) Choose a but not b. Then we must choose 2 more men from the four, c, d, e, and f. There are 4(3)= 12 ways to do that.
2) Choose b but not a. As before, there are 12 ways to do that.
3) Choose neither a nor b to be on the committee. That is, we choose 3 out of c, d, e, and f. There are 4(3)(2)= 24 ways to do that.

There are, therefore, 12+ 12+ 48= 96 ways to choose the three men.

I disagree here; order doesn't matter for any group of men, and as such the choose function should show up here.

As far as the OP's elimination method goes, again, order does not matter. How many 'illegal' committees are there? (Hint: think about how many more men there are to choose from, and how big a committee is.)

Alternatively, using the letter system, it's actually not unrealistic to enumerate all possible 'legal' combinations of three men (though I'll grant it's rather tedious).