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.
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.
882-210-21 = 651

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