# 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 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 $\frac{8100}{3!}= 1350.$