Number of mixed doubles pairs such that no one plays with his/her spouse?  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.
 A: 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.
A: 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 $840$.  
A: 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.
A: My answer is 1854.
The total number of derangements of n letters to n envelopes is 
$$n!\left(\frac{(-1)^0}{0!}+\frac{(-1)^1}{1!}+\dots+\frac{(-1)^n}{n!}\right)$$
A: 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
