A committee of three people is to be chosen from four married couples. What is the number of different commities that can be chosen if two people who are married to each other cannot both serve on this committee?
16
24
26
30
32
Number of options for the 1st person = 8.
Number of options for the 2nd person = 6. (Of the 7 remaining people, anyone but the spouse of the first person selected.)
Number of options for the 3rd person = 4. (Of the 6 remaining people, anyone but the spouses of the first 2 people selected.)
To combine these options, we multiply:
8*6*4.
Since the ORDER of the selections doesn't matter -- ABC is the same committee as BCA -- the product above must be divided by the number of ways to ARRANGE the 3 people chosen (3!):
(8*6*4)/(3*2*1) = 32.
The correct answer is
E.
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