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Suppose there is a group of 24 people, 12 men and 12 women. I’d like to arrange them into groups of three, so that each group contains at least one man and at least one woman. How many ways are there to do this?

Here is how I thought one might approach the problem: Since there will be 8 groups of three, first let’s choose 8 women. Then we’ll choose 8 men to ‘pair up’ with the 8 women. This will leave 4 men and 4 women, which we’ll distribute randomly onto the pairs. So, formally, this amounts to: $${12 \choose 8}\cdot{12 \choose 8}\cdot8!\cdot8!$$ (The first 8! is for pairing the 8 men with the 8 women. The second 8! is for distributing the 8 remaining people onto the pairs.)

The problem is that we are counting some triples more than once. E.g., the method yields the triple ‘Jill-Jack-Mary’, and also ‘Mary-Jack-Jill’. But they are the same triple, of course. Note that the process does NOT yield any of the following: ‘Jack-Jill-Mary’, ‘Jack-Mary-Jill’, ‘Jill-Mary-Jack’, ‘Mary-Jill-Jack’. That’s because the process picks a woman first and then a man (and then either a man or a woman). So, although we are arranging into triples, I think we need to divide by 2, and not by 3!, to compensate for the over-counting. That gives us:

$$\frac{{12 \choose 8}\cdot{12 \choose 8}\cdot8!\cdot8!}{2}$$

Am I at all close to the right answer? Thanks!

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    $\begingroup$ Helps, I think, to note that you must have $4$ each of $WWM,WMM$ triples. $\endgroup$
    – lulu
    Jun 17 '17 at 12:39
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I'd start by choosing the sets of $4$ women and $4$ men who will be the unique members of their gender in the groups. Having chosen these, each of these unique individuals can be assigned a pair of opposite gender members to complete the groups.

The order of those two "opposite gender members" does not matter, so we have:

$$ \binom{12}{4}^2 \cdot \left( \frac{8!}{2!2!2!2!} \right)^2 $$

ways to assemble the eight groups.

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  • $\begingroup$ Thanks – that makes sense. I don’t see the mistake(s) in my reasoning yet, though. $\endgroup$
    – MarkOxford
    Jun 17 '17 at 13:02
  • $\begingroup$ At the end of your calculation you divide by two, but you only do it once. There is actually a factor two that needs to be divided out for each of the eight groups. $\endgroup$
    – hardmath
    Jun 17 '17 at 13:09

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