There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?  Choose the correct choice from the following options:
A) 4
B) 8
C) 12
D) 16
E) 20
4 pairs = 4*2= 
<gadget id="calculator">4*2</gadget>
<output>8</output>
8 people total   8C4= 8!/4!4! = 70 total outcomes   Total – unfavorable = favorable   Unfavorable outcomes   Assuming one pair of twins in the committee, we have two spaces left. Since we plugged a pair of twins in the committee, we have 8-2= 
<gadget id="calculator">8-2</gadget>
<output>6</output>
6 people to fill 2 spaces.   6C2= 6!/2!4! = 15 ways to fill the two remaining slots   We only filled the slots with one pair, and we have to account for arrangements of the pairs. Now, we have 4 pairs. 4*15= 60 total arrangements   When we place members into the remaining slots, there may be an additional set of twins. There are 2 remaining slots to which we can fit a pair of twins. If it were one remaining slot, we cannot fit a pair of twins, so we wouldn’t have to account for duplicates.   Now we account for the number of duplicates.   # of duplicates = Total arrangements - # of unique combinations   To find the # of duplicates of twins, we need treat a pair of twins as one unit.   This means the 4 slots are really 2 slots.   Total ways of arranging four pairs of twins in two slots   4P2 = 4*3= 
<gadget id="calculator">4*3</gadget>
<output>12</output>
12 total ways   Total # of unique combinations   Choosing two pairs out of 4 pairs= 4C2   = 6   Therefore, # of duplicates = 12 - 6= 
<gadget id="calculator">12 - 6</gadget>
<output>6</output>
6 duplicates   70 – 60 + 6= 16   ANS:16
<result>D</result>
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?