Allocating objects to pigeon holes

Can someone please check my attempted solution to a very old past exam question that I came across:

12 members of a tennis club, 6 men and 6 women, are to be divided into 3 groups, each group to play mixed doubles. In how many ways can this be done?

The 6 men can be allocated to the 3 groups of 2 in $6\mathrm C2 \times 4\mathrm C2 \times 2\mathrm C2 = 90 \text{ ways}$. Similarly, the 6 women can be allocated in 90 ways. Hence there are $90\times 90=8\mathord,100$ ways of allocating mixed doubles teams into 3 separate groups.

Is this correct?

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I do not think it is quite right. You are counting the number of ways of assigning two men to the first court, two women to the first court, then two men and two women to the second court and so on.

I would interpret the question though so that two solutions are the same if everyone is playing with the same people. In that case your answer will be $\frac{8100}{3!}= 1350.$

Note that we have still not assigned people into pairs. This would double the number of possibilities.

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thanks, I agree with you assuming there is no order to the 3 required groups to be formed –  TryinHard Nov 20 '12 at 7:34