From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed.
How many different committees are possible if
(c) 1 man and 1 woman refuse to serve together?

committee without the "problematic" woman ${6\choose 3}\cdot{7\choose 3}$
committee without the "problematic" man ${5\choose 3}\cdot{8\choose 3}$
Now it seems that I counted twice committes without the "problematic" people so overall it is ${6\choose 3}\cdot{7\choose 3} + {5\choose 3}\cdot{8\choose 3}-{5\choose 3}\cdot{7\choose 3}=910$.

Is there a way to calculate the committees without "over-counting"?

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    $\begingroup$ Note that neither of the "problematic" people refuse to serve on their own. Instead, count the total number of committees and subtract the ones that include both of the people who don't wish to work with each other. Correcting for this overcounting is called "inclusion-exclusion" and is completely standard. $\endgroup$ May 10, 2015 at 18:21

1 Answer 1


There are ${8 \choose 3} \cdot {6 \choose 3}$ committees in total, of which ${7 \choose 2} \cdot {5 \choose 2}$ contain the problematic pair.

This gives ${8 \choose 3} \cdot {6 \choose 3}-{7 \choose 2} \cdot {5 \choose 2}$ committees without the problematic pair.

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    $\begingroup$ why ${7\choose 2}\cdot {5\choose 2}$ count the problematic pair? $\endgroup$
    – gbox
    May 10, 2015 at 18:25
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    $\begingroup$ While this corrects the mistake in the question, it doesn't actually answer the question. $\endgroup$ May 10, 2015 at 18:25
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    $\begingroup$ @gbox Because if both of the problematic people are included, then you only get to choose 2 more men and 2 more women (of which there are 7 and 5 remaining, respectively). $\endgroup$ May 10, 2015 at 18:26
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    $\begingroup$ @gbox Yes, it is mistaken; this answer is correct. Somehow you have the right numerical answer, but your equation is not true (by a large factor). The lefthand side of your equation is 391650, not 910. $\endgroup$ May 10, 2015 at 18:30
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    $\begingroup$ I made the edit, it should show up soon, clearly you meant to add instead of multiply $\endgroup$
    – WW1
    May 10, 2015 at 20:19

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