I attacked this question in a different way:
we have 8 people, out of which we have to choose 3. Thus, we have 8C3 ways to do so:
8C3 = 8*7*6/3! = 56.
Note that there are way fewer ways to choose in such ways that no brothers are selected.
We can eliminate D and E.
now, look at B and C. B+C = 56, clearly, the 2 options are complementary, and there is a high chance that the answer is one of the two. A is thus out.
It's a good way to make a strategic guess, chances to answer correctly - 50%
now, in how many ways can we select 3 people so that no brothers/sisters are in the committee?
well, let's see how many committees, in which the restriction is not respected, we can get!
now, there are 4 brothers and 4 sisters!
suppose we select 1 brother and 1 sister -> we then have 1*4C1*6C1 or 4*6= 24 ways, we're almost there!
why 4C1 - from 4, we have to choose 1 that will be brother/sister of the other one.
now, we have chosen 2 members, out of 6, we have to choose 1. that's why 6C1
alternatively, we can assign letters and try to list all the combinations, I know it is not preferable, but still 24 - not that much!
A1B1
A2B2
A3B3
A4B4
now, we can have A1B1*6C1; A2B2*6C1; A3B3*6C1; A4B4*6C1 -> clearly we have 4*6C1!!! It doesn't matter the order A1B1 ir B1A1:)
now, the final step:
56 ways total - 24 ways in which the restriction is not respected = 32 ways in which restriction is respected.