Arun sharma

Question

A mixed doubles tennis game is to be played between two teams (each team consists of
one male and one female). There are four married couples. No team is to consist of a husband and his
wife. What is the maximum number of games that can be played?
(a) 12 (b) 21
(c) 36 (d) 42

Comments

No .. it's 42

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is it 21?

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No 42 is given as answer . Option d

Answer 1

Let's  four couples are (A,a), (B,b), (C,c), and (D,d) where A,B,C,D are the husbands and a,b,c,d are the wife.
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.
so 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.