Consider the following problem:

How many different mixed gender committees of 3 people can be chosen from a group of 5 men and 5 women?

Wrong approach: These are committees of the form $M_1M_2W_1$ or $W_1W_2M_1$ (the ordering does not matter). There are $5\choose 2$$5\choose 1$ of the former and $5\choose 2$$5\choose 1$ of the latter for a total of $2{5\choose 2}{5\choose 1}=100$ possible committee.

Correct approach: Take the total number of committees of 3 possible and subtract from the number of non-mixed ones. There are ${10\choose 3}$ possible committees and ${3\choose 5}$ all men (and as many all women) ones. Therefore, one gets ${10\choose 3}-2{5\choose 3}=80$ possible committees.

Here is my question: The first approach over-counts 20 committees. I wonder of what form are these over counted committees. More generally, I fail to understand why the first approach gives an incorrect answer.


1 Answer 1


Both approaches are correct, but you have evaluated the result of the second approach incorrectly. It should be \begin{equation} \binom{10}{3} - 2\,\binom{5}{3} = 120 - 2\times 10 = 100, \end{equation} which agrees with the result of the first approach. So, this was just a calculation error.

  • $\begingroup$ Argh: sorry, the correct approach was from a book and of course, I didn't check their calculations (they put the expression in terms of combinatorial and the final answer only)! $\endgroup$
    – user603
    Commented Feb 28, 2016 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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