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Here is the wording of my question:

In how many ways can a class with 20 students (12 boys and 8 girls elect a class president, vice president, and secretary if each student is willing to serve in any of the offices and no student can be elected to more than one office, and with at least one boy and one girl each among the students selected?

There are two ways that I am thinking to do this. The first way would be to take the total number of cases and subtract off the ones that only have boys and only have girls elected. So that would be:

$$20\cdot19\cdot18-\binom{8}{3}-\binom{12}{3}$$

Which equals $6564$.

The other way I thought to do it was to select first a boy, then a girl, then one final student from the pool left and multiply it by the 6 ways you can arrange the 3 positions, so:

$$\binom{12}{1}\cdot\binom{8}{1}\cdot\binom{18}{1}\cdot3!$$

Which equals $10368$.

In my mind, both ways of doing this should be equivalent. I have checked my computation several times over but it is possible I am continuing to make a mistake. Can anyone offer an explanation as to which way is the correct way and explain why the other way is wrong? Maybe both of them are wrong...

Thank you very much!

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Neither attempt is correct.

The first attempt undersubtracts: it should subtract ordered ways of arranging the students, so it should be

$$20 \times 19 \times 18 - \left(\binom{8}{3}+\binom{12}{3}\right) \times 3!=5184.$$

The second attempt overcounts, since we can e.g. pick Alice using the first "choose" then Eve using the second "choose", or pick Eve using the first "choose" then Alice using the second "choose". Thus, we've counted this situation twice.

It could be repaired by choosing two boys one girl, and two girls one boy, then summing the result and multiplying by $3!$:

$$\left(\binom{12}{2}\binom{8}{1}+\binom{12}{1}\binom{8}{2}\right) \times 3!=5184.$$

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  • $\begingroup$ I see, I was forgetting to order the combinations of only boys and girls. Thanks! $\endgroup$ – Rolando Martino Feb 3 '14 at 3:51

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