1
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ 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? $\endgroup$ – whuber Oct 19 '12 at 14:09
1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.