Question

In how many different ways can a mixed doubles tennis game be organised between four married couples if no husband and wife play in the same game ?

Answer 1

Let's denote the four couples by (A,a), (B,b), (C,c), and (D,d) where A,B,C,D are the husbands and a,b,c,d are the wives.
For a games, let's choose first the husbands. We have 4C2=4*3/2=6 possibilities. Now we know that the final number of games will be a multiple of 6, so we are down to two choices, C and D.

Once husbands were chosen, say A and B, let's count how many possibilities we have to choose their partners under the given restrictions.
We can choose a and b and we have to pair them as (A,b) and (B,a) - 1 possibility.
We can choose one of the wives, a or b, but we have to pair her with the other husband and in addition, we have to choose another partner for the second husband.
This we can do in 2*2 = 4 ways, as there are two possibilities to choose from a and b, then we have 2 possibilities to choose the other wife, c or d - 4 possibilities
Finally, we can choose the other two wives, c and d, and we have two possibilities to team them up with the men, (A,c), (B,d) or (A,d), (B,c) - 2 possibilities.
In conclusion, for every pair of husbands, we have 1 + 4 + 2 = 7 possibilities to choose their partners for the game.
Total number of possibilities 6 * 7 = 42.
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