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Can you help me with this problem?

There are $7$ married couples. What will be the number of mixed double pairs in tennis such that no one plays with his/her spouse?

Can some one help me with this? The answer is $840$.

Thanks!

EDIT : The question asks for the number of teams possible in which no wife is teamed up with her husband.

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1  
I don't think this problem is clear enough. I thought a "mixed doubles pair" was just a male and female pairing, but there are only $42$ such pairings. Do you mean that no wife plays with her husband as a partner, or that no wife plays on the opposing side against her husband, or that no wife plays on either side, either with or against her husband? Given the answer is 840, I think the question means the latter. –  Thomas Andrews Oct 25 '11 at 19:03
    
(Whoops, editing time ran out.) Insert before my questions the sentence: "Presumably you want the number of pairs of pairs that can participate in match. Do you want ..." –  Thomas Andrews Oct 25 '11 at 19:10
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I don't think not playing with your spouse is a very reasonable constraint for a tennis tournament. But I can conceive other social events where it makes sense... –  PseudoNeo Mar 22 '12 at 16:22

5 Answers 5

If no married couple can play together, but are allowed to oppose one another, you can pick the first pair in $7\cdot 6=42$ ways. Then you can pick the second woman in one way as the wife of the first man, in which case she can pair with six other men, or five ways as not the wife of the first man, in which case she can pair with five other men. In total we have $7\cdot 6 (1\cdot 6+5\cdot 5)=1302$ possibilities if the order of pairs matters. If the order of pairs does not matter, it is half this, $651$ possibilities.

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The problem can be interpreted in more than one way. For the sake of simplicity, we think of one of each couple as female, and of the other as male.

We interpret the problem as follows: How many ways are there to pick $4$ people, $2$ of each sex, so that no two people picked are a "couple," and then divide these $4$ people into two gender-mixed teams of $2$?

The females can be picked in $\binom{7}{2}$ ways. For each of these ways, we can pick the males in $\binom{5}{2}$ ways, for a total of $\binom{7}{2}\binom{5}{2}$. Once we have done the picking of the $4$ people, they can be divided into two mixed teams of two in $2$ ways, giving a total of $$2\binom{7}{2}\binom{5}{2}.$$ This turns out to be $420$. To make the answer of $840$ correct, we would need to find another interpretation of the problem. The interpretation that having a couple in the foursome is OK, as long as they are not on the same side, gives a number well below $840$. We do get $840$ if the only thing we forbid is two couples, but that would be a very strange interpretation of the wording of the question.

Edit: In response to the edited version, line up the "females" in order of height, or social Insurance number. Then line up the males opposite them. Producing the $7$ teams is equivalent to producing the derangements of $7$ objects. This number can be computed in various ways. It is substantially larger than $840$.

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And the answer is $1854$, not $840$. –  Brian M. Scott Oct 25 '11 at 19:00
    
The question is unclear, but I cannot think of an interpretation where the answer is the number of derangements of $7$ items. Any hints? :) –  Srivatsan Oct 25 '11 at 19:06
    
@Srivatsan Narayanan: I do not play tennis, so saw $7$ couples dancing. –  André Nicolas Oct 25 '11 at 19:40
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I can get another factor $2$ by saying you pick one team in 7*6=42 ways, then pick another in 5(pick a woman not married to the man in the previous team)*4(pick a man not married to either woman selected)=20 ways. This corresponds to caring which order the pairs are picked compared to your solution. It gets the requested $840$, but I like yours better. –  Ross Millikan Oct 25 '11 at 19:40
    
@Ross Millikan: If one team is to be distinguished in some way (like getting the nice end of the court, or serving first) then doubling the $420$ would make sense, but that I think is a less natural interpretation. My favourite interpretation is that it is really $8$ couples. –  André Nicolas Oct 25 '11 at 19:47

My answer is 1854.

The total number of derangements of n letters to n envelopes is

$$n!\left(\frac{(-1)^0}{0!}+\frac{(-1)^1}{1!}+\dots+\frac{(-1)^n}{n!}\right)$$

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Have a look here –  drhab Sep 23 at 8:30

I got the answer 840.here is how First,out of 7 women,the no of permutations in arranging two women out of seven is $ ^7P_2 $ and for arrangement of two men out of the remaining five(who are not the husbands of either eomen) is $ ^5P_2 $ .hence total no of ways is $ 7.6.5.4 = 840 $

Now,if any of people got other answers even with correct methods,then it means you are also the considering arrangements of the men and women in the court,like arranging them in the right or the left,or playing one side or the other side of the court.

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882-210-21 = 651

ttl mixed double pairs possible - no. of one married couple on either side - no. of exactly one married couple & one mixed couple

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ttl? Isn't that the counter of how many routers a ping packet goes through, or something like that? –  Asaf Karagila Mar 22 '12 at 17:15
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@Asaf: I'd guess "total", not "time to live". –  Willie Wong Mar 23 '12 at 8:32
    
@Willie: I'd guess that sarcastic remarks on writing these kind of shortcuts on this site don't go well through text. –  Asaf Karagila Mar 23 '12 at 8:47
    
@Pawan Please go through the edit available by clicking on the timestamp above my name. I am not sure about your answer however... –  user21436 Mar 23 '12 at 11:39
    
@Kannappan: I reverted the edit because you changed his meaning to something I am all but certain was not the OP had in mind. –  Willie Wong Mar 23 '12 at 15:27

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