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There are $12$ people: $8$ men and $4$ women. I want to divide them into $4$ groups of $2$ men and $1$ woman each. How many possible configurations do I have?

My solution to this question was:

"First of all, I suppose that each man and woman is distinguishable and I suppose the same for the four groups.

For the men:

I have ${8}\choose{2}$ possible combinations for the first group. After choosing a combination for the first group, I have ${6}\choose{2}$ possible combinations for the second group. After I choose the combinations for the first and second groups, I have ${4}\choose{2}$ possible combinations for the third group and after I choose the combination for the first, second and third group, I have ${2}\choose{2}$ possible combinations for the last group.Thus, if I choose a combination for the first $3$ groups, I automatically choose a combination for the last group.

I have a total of ${8}\choose{2}$$ {6}\choose{2}$ ${4}\choose{2}$${2}\choose{2}$ possible ways to separate the $8$ men into $4$ groups of $2$ men each.

But, I dont need to start choosing a combination of $2$ men for the first group and after that for the second, etc. I can start with the third group, for example. Then, I have a total of $4!$ permutations of the order of the groups, giving a total number of ways:

$4!$${8}\choose{2}$$ {6}\choose{2}$ ${4}\choose{2}$${2}\choose{2}$

Following the same argument, we have, for the women: $4!$${4}\choose{1}$$ {3}\choose{1}$ ${2}\choose{1}$${1}\choose{1}$ $=4!4!$ number of ways.

The total number of ways is, then: $4!4!4!$${8}\choose{2}$$ {6}\choose{2}$ ${4}\choose{2}$${2}\choose{2}$."

I'm not confident about this solution, specially about the $4!$ in front of the binomial coefficients.

Some of my friends believe that the correct answer is simply ${8}\choose{2}$ $\cdot$ ${4}\choose{1}$

I'm afraid that both solutions might be wrong. So, how would you solve this?

Thanks.

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    $\begingroup$ Since each group has precisely one woman in it, you can index each group by the woman it contains. Then you have to choose two me for each woman. $\endgroup$ Nov 4 '14 at 17:08
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You had the right solution, almost. Call the women A, B, C, D. Woman A can choose her partners in $\binom{8}{2}$ ways. For each such way, woman B can choose her partners in $\binom{6}{2}$ ways. And then woman C has $\binom{4}{2}$ choices. Multiply, and it's over.

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I think your reasons for multiplying by $4!$ are a little confused. An easier way to think of it is this:

As there are four women, and four groups each of which contains one woman, we can "label" each group by the woman in it.

Woman one has ${8}\choose{2}$ ways to choose her two partners.

Woman two has ${6}\choose{2}$ ways to choose hers.

Woman three has ${4}\choose{2}$ ways to choose hers.

Woman four is left with the remaining two men (${2}\choose{2}$ = $1$).

So there are ${8}\choose{2}$× ${6}\choose{2}$×${4}\choose{2}$× ${2}\choose{2}$ combinations.

If the order of the groups is important, then we can multiply by the $4!$ ways of ordering them.

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