The number of ways in which a mixed double game can be arranged from amongst 9
married couples if no husband and wife play in the same game is?

Dear  s adhithya

In mixed doubles, there are 2 men and 2 women..

2 men out of 9 can be selected in ⁹c₂ ways.

now we can only set 2 women out of 7.bec two the selected two men's wife can't play in a match

so in ⁷c₂ we can select women

now there are 2 men and 2 women M1, M2, W1,W2

possible combinatoin:

M1W1  and M2W2

M1W2 and M2W1..

so 2 possible combination..

hence answer= ⁹c₂ *  ⁷c_2 *2

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Sahil Arora

Dear Sir,

I still have a doubt after seeing your reply. The answer you have given is what i arrived at myself on encountering the question in my bitsat sample paper, but on 2nd thoughts, i felt that besides the case of 2 men being chosen out of 9 and then 2 women being selected out of 7, there is also the possible case of 2 women being selected out of 9 and then  2 men being selected out of 7 and also the third case of 2 men being selected out of eight and then 2 women being selected out of 8 , i.e, a husband from 1 couple and a wife from another couple sit out. why aren't these 2 cases being considered?

Dear Student,
Please find below the solution to your problem.

In mixed doubles, there are 2 men and 2 women..
2 men out of 9 can be selected in9c2ways.
now we can only set 2 women out of 7.bec two the selected two men's wife can't play in a match
so in7c2we can select women
now there are 2 men and 2 women M1, M2, W1,W2
possible combinatoin:
M1W1 and M2W2
M1W2 and M2W1..
so 2 possible combination..
hence answer=9c2*7c2*2

Thanks and Regards