Permutation/Combination Problem

I was thinking that this question sounds familiar... yes, it's pretty much the exact same set up as the question aatech linked to. My explanation for that one still stands here: just think about it logically. In my opinion, equations tend to get students in trouble.

One approach to solve this question is -

total no of ways of selecting 3 people from 8 = 8C3 = 56

Now if the siblings are present in the combination then 2 places are booked and only one place is left to be filled....

1*1*6 for each sibling pair..We have 4 pairs ...so total number of combinations in which sibling pair will be present is -

4*1*1*6 = 24

Hence the number of combinations in which the sibling pair will not be presnet = 56 - 24 = 32

2* 4C3 ( 3 boys or gals committe) + 2 * 4C2 * 2C1 ( 2 boys and one gal or 2 gals and one boy) = 32


Neo