Number of permutations in lawn tennis so no husband and wife play together. In how many ways can a lawn tennis mixed doubles be made up from seven married couples if no husband and wife play in the same set? Please explain the logic.
 A: I take the question as given in the detailed part (no husband and wife play in the same set)
We can choose two females in $\dbinom72$ ways, males who are not their spouses in $\dbinom52$ ways, and the $4$ can be paired in just $2$ different ways.
Putting it all together, we get $\dbinom72\cdot\dbinom52\cdot2 = 420$ 
A: In a mixed doubles match 4 players play out of which 2 are male and 2 are female.
CASE 1[when males are selected first]
we can select 2 males out of 7 which is 7C2
Now since the two men have already been selected their wifes cannot be a part of the match hence available options are 7-2 = 5 therefore we select 2 women out of 5 i.e 5C2
Total ways = 7C2 * 5C2 = 210
CASE 2[ when females are selected first]
now we do exactly the opposite
We select 2 women from 7 and 2 men from 5 
Total ways = 7C2 * 5C2 = 210
GRAND TOTAL = 210 +210 = 420 AS PER MY LOGIC 
HOWEVER ANSWER AS PER RD SHARMA IS 840 
A: There are 7 males and 7 females. Out of these, you have got to chose 2 males and 2 females, such that no husband and wife play in the same set (I believe the cause of doubt might be confusing set with team).
You can start by selecting a male (or alternatively a female). This can be done in 7 ways. Now chose another male. This can be done in 6 ways. Now to chose a female, you have 5 options because you cannot chose the wives of the two males you've already chosen. So chose this female from the 5 options you have and the next female from the remaining 4 options you have got left with yourself after chosing the first female. By principle of multiplication, total number of ways to do this task is 7*6*5*4=840.
Hope this helps. Please comment in case it is not clear.
