I went about this one by first finding the number of different groups possible without any restrictions and then finding the number that involved a married couple and subtracting.

Number of possible groups

8 people, choose 3 = (8!)/(3!5!) = 56 (from formula for combinations)

Number of possible groups with a married couple

For a given couple, they can be involved together in 6 different groups. This is because they must be 2 of the members and there are then 6 possibilities for the other group member. Or, in terms of combinations

Groups with married couple = (2 C 2)(6 C 1) = 1*6 = 6

This is true of all 4 married couples, so in total there are 4*6 = 24 group combinations that involve a married couple. Subtracting 24 from 56 gives us 32 groups that will not involve a couple.

Lets say the couple are as follows :

Male 1 , Female 1
Male 2 , Female 2
Male 3 , Female 3
Male 4 , Female 4

For the 1st Member of the committee we have 8 choices (Assume we picked Male 1)

For the 2nd Member of the committee we have 6 choices (Already picked Male 1 and cant pick Female 1. Assume this time we picked Female 3 )

For the 3rd Member of the committee we have 4 choices (Already picked Male 1 and Female 3. Cant pick Female 1 and Male 3)

Now the committee can be arranged in 3! ways = 6

So the different committee are : (8*6*4)/6 = 32

HTH