# Question about combinations involving a caveat/exclusion

Working through an example in the Ross Probability textbook. Can someone explain the reasoning behind the answer?

From a group of 7 men, how many committees consisting of 3 men can be formed? what if 2 men are feuding and refuse to serve on the committee together?

Answer, part 1: First part is clear for me. with n=7 and r=3. Result = $\binom{7}{3}=35$.

Answer, part 2: Textbook gives the answer as 35 - 5 = 30, with 5 a result of $\binom{2}{2}\binom{5}{1}$.

Can someone please explain the reasoning behind the Part 2 answer?

• The reasoning might become clearer if you made this problem larger. If there were $7$ men and $3$ were feuding, then how would you go about selecting a committee to exclude from the valid possibilities?
– whuber
Commented Oct 19, 2012 at 14:09

## 1 Answer

Sure. The logic of part 2 is "create all possible committees, then chuck those that contain the feuders". If John and Bill are feuding, then there are 5 "bad" committees: those with John, Bil and any other person from amongst the remaining five.

That's what $\binom{2}{2} \binom{5}{1}$ means. The first bit: $\binom{2}{2}$ is 1 - namely how do you pick 2 people from 2 people. Clearly, there is only one way of doing that. Then you have to pick 1 person from the remaining 5 - and there are 5 ways of doing that.