total selections = 56
let's say that couple is always present in this committee of three.
This means that there are 4 ways to select 2 people of the committee. ( 4 couples and any one couple can be selected in 4 ways)
The third person can be selected out of remaining 6 people in 6 ways.
Therefore when couple exists there are: 4X6 = 24 ways
Thus no couple = (4X6) = 32