A committee to contain $5$ members and have at least one woman. There are $7$ women and $9$ men. I think that we can fix one woman as one of the $5$ members, and randomly select the rest from a combined pool of men and women. I think the answer should be $C(15, 4)$. I thought we might have to multiply this by $7$ because any one of the women can be fixed, but this is not necessary right?

  • $\begingroup$ Not sure but looks incorrect. $\endgroup$ – A---B Jun 16 '16 at 18:45
  • 1
    $\begingroup$ Your method would double count a lot of cases (if you multiply by $7$) or undercount otherwise. The undercount because there would be correct options that exclude your initial fixed choice, and the overcount because of the cases with more than one woman on committee. $\endgroup$ – Joffan Jun 16 '16 at 18:47
  • $\begingroup$ @Joffan Ah ok thanks! $\endgroup$ – Ovi Jun 16 '16 at 18:49

The only problem with multiplying by seven is that you could have one case where you fix one woman and choose other members in such a way so that the people on the committee are the same as in another case where you fix another woman. To make this clearer, let W1, W2, ..., W7 be the women who can be chosen to be on the committee, and let M1, M2, ..., M9 be the possible men. For example, you would be counting the case where you fix W1 and choose W2, W3, M1, and M2 separately from the case where you fix W2, and choose W1, W3, M1, and M2, even though they form the same committees.

I also don't think the correct answer is C(15,4), because the choice of which woman to fix is important as well. For example, counting the number of possibilities as C(15,4) would count all the cases where you choose one woman and the rest men as one case, even though these are in fact seven separate cases. I would rather simply count the total number of committees and then subtract the number that don't have any women (i.e. with all men) to get the number with at least one woman on the committee. This should come out with C(16,5)-C(9,5) as your answer.


Total number of combinations:


Number of combinations with no women:


Number of combinations with at least one woman:


  • $\begingroup$ Ah thanks makes sense! I'm not really sure what's wrong with $C(15, 4)$ though? $\endgroup$ – Ovi Jun 16 '16 at 18:46
  • $\begingroup$ @Ovi: You're welcome. I can't answer this question because I'm not really sure what's right with $\binom{15}{4}$ in this context. $\endgroup$ – barak manos Jun 16 '16 at 18:48

With problems like this one, (or problems in general) it is often useful to compute an answer by two different methods as a check. Problems of this nature are particularly amenable to doing that.

Method A: (total of all committees) - (all male committees) = (committees with at least one woman) (This is the quickest solution.)

Method B: For (w in [1,2,3,4,5]) count committees with (w women) and ( (5-w) men )

This would give C(7,1)*C(9,4) + C(7,2)*C(9,3) + C(7,3)*C(9,2) + C(7,4)*C(9,1) + C(7,5)*C(9,0) (Which is obviously much more laborious, and not very practical when the numbers are larger).

Determining that those two methods give the same answer boosts confidence in that answer, and often gives additional understanding into the problem as well.


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.