A committee of three people is to be chosen from 3 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?
Options:
A) 16 B) 24 C) 26 D) 30 E) 8
Total no of ways of choosing a committee = 6C3 = 20   No of ways to choose first couple = 3C1 = 3   No of ways to choose 1 person from remaining 4 = 4C1 = 4   Total number of ways of choosing a married couple to be in the same committee = 3*4= 
<gadget id="calculator">3*4</gadget>
<output>12</output>
12   So, no of ways not choosing a married couple= 20-12= 
<gadget id="calculator">20-12</gadget>
<output>8</output>
8   ANS:E. 8
<result>E</result>
A committee of three people is to be chosen from 3 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?