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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.

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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.

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  • $\begingroup$ 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 Feb 28 '16 at 11:27

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