How to solve this combination problem quickly? In how many ways can 3 men and their wives be made stand in a line such that none of the 3 men stand in a position that is ahead of his wife?
What is the best way to tackle such problems?
 Here is how I am tackling this problem

    M1 M2 M3  -> Men
    W1 W2 W3  -> Wives

    a.
     1  2   3   4    5    6
     W1 M1  W2  M2   W3   M3     ->  


    With the above in mind, W1 M1 is a pair and W2 M2 is a pair, W3 M3 is a pair.
    so, there can be 3 interchanges - if we assume that W1 M1 stick to each other.

    b. 

     W1 W2  M1  M2   [W3   M3]

W1 W2 can interchange in 2 ways. 
M1 M2 can interchange in 2 ways 

4 ways..* 2 ( when W3 M3 will be at first) = 8 ways.  



    c.  W1 W2  W3  M1   M2   M3

    W1 W2 W3 can be arranged in 3 ways
    M1 M2 M3 can be arranged in 3 ways.
    total ways are 9 ways. 

    -----------------

    so adding all -  20 ways.


However, the answer is not 20 ways. What is wrong in my thought process here?

 A: Choose two positions for man 1 and woman 1 (husband-wife pair).  $6\choose 2$ ways.  Put woman 1 in the earlier of these two positions.
Choose two remaining positions for man 2 and woman 2.  $4\choose 2$ ways.  Put woman 2 in the earlier of these two positions.  
Put man 3 and woman 3 into the two remaining slots, with woman 3 in the earlier position.
${6\choose 2}{4\choose 2}$ ways to do the process.
A: A good, if not the best, way to tackle this problem is pick two positions for the first couple to occupy, then another two for the second couple to occupy, which will leave just two positions for the third couple.  This can be done in
$${6\choose2}{4\choose2}{2\choose2}=15\cdot6\cdot1=90$$
ways.  Note that if you remove the condition that each man must stand behind his wife, the number of ways increases by a factor of $2\cdot2\cdot2=8$, which would give $90\cdot8=720$ ways, which makes sense since $720=6!$, which counts the number of ways of arranging $6$ people in a line with no restrictions.
A: I can see two slips.
Firstly, you forgot WMWWMM. [edit, sorry, you didn't.  However, in b, there are three choices for which pair is together - you have W3M3 together, but it might be W1M1 or W2M2.]
Secondly, WMWMWM can be arranged in six ways, not three:
W1M1W2M2W3M3,
W1M1W3M3W2M2,
W2M2W1M1W3M3,
W2M2W3M3W1M1,
W3M3W1M1W2M2,
W3M3W2M2W1M1.
Similarly for WWWMMM.
