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The number of ways in which a mixed double game can be arranged from amongst 9married 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 * 7c2 *2
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best s adhithya !!! Regards, Askiitians Experts 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?
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