If there are four distinct pairs of brothers and sisters, then in how

We need to divide \(8*6*4=192\) by the factorial of the # of people - 3! to get rid of duplications \(8*6*4=192\) contains ---> \(\frac{192}{3!}=32\) - correct answer.

Consider example with smaller number: there are two couples and we want to choose 2 people not married to each other.
Couples: \(A_1\), \(A_2\) and \(B_1\), \(B_2\). Committees possible:

\(A_1,B_1\);
\(A_1,B_2\);
\(A_2,B_1\);
\(A_2,B_2\).

Only 4 such committees are possible.

If we do as proposed in the solution you posted:
The first person on the committee can be anyone of the 4.
The second person on the committee can be only one out of 2 (the first person with her or his sibling excluded).

So we'll get: 4*2=8, so more than 4, which means that 8 contains some duplications. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

It's because if you pick A1 for the first pick and than pick B1 for the second you'll get the committee {A1, B1} but if you'll pick B1 for the first pick and then A1 you'll get the exact same committee {A1, B1} (dividing by the factorial of the # of people in committee you'll exclude this double countings).

Hope it helps.