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First, you have to select the pairs whose members will be elected to committee. That is 4C3 = 4 distinct ways.
Now for each pair there are 2 ways to select a person.

Total, there are 8C3 ways to choose a group of 3 = 56 ways

To calculate the # of ways to choose a group WITH siblings:

Assume that there is a sibling in the group - 4 pairs of siblings = 4 choices
For each pair of sibling that's chosen, there are 6 choices for the third member of the group.

Therefore, there are 4x6 = 24 groups with sibling pairs.

Number of groups without sibling pairs = 56-24 = 32.

I don't think there's a particular way to tackling comb/permutation questions. The key is to really understand the concepts behind these questions so you can manipulate them when you need to...this is esp important since these questions come in all shapes/forms.