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?