If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group? Choose the correct choice from the following options:
A) 32  B) 24  C) 56  D) 44  E) 40
The first member can be anybody, so 8 possibilities.   The second member cannot be the sibling of the previous member, therefore 6 possibilities (anybody from the remaining 3 pairs).   Finally, the third member, can be chosen from the remaining 2 pairs, so 4 possibilities.   This would give us 8*6*4 possibilities, but in this case we don't care about the order in which we choose them, we have to divide the product we obtained by 6(=3!), which is the number of possibilities we can chose the same three distinct members, say ABC (it can be BAC, CAB,...). You can count those possibilities for such a small number as 3, but why avoid permutations? Is it better to start listing the possibilities instead of just accepting the already proven result?   So, the correct answer is   8*4= 
<gadget id="calculator">  8*4</gadget>
<output>32</output>
32.   32nswer 32.
<result>A</result>
If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?