Abhijit K wrote:If a committee of 3 people is to be selected from among 5 married couples so that the
committee does not include two people who are married to each other, how many such
committees are possible?
A. 20
B. 40
C. 50
D. 80
E. 120
Number of options for the first person = 10. (Any of the 10 people.)
Number of options for the second person = 8. (Of the 9 remaining people, anyone but the spouse of the first person selected.)
Number of options for the third person = 6. (Of the 8 remaining people, anyone but the spouses of the first two people selected.)
To combine these options, we multiply:
10*8*6.
Since the ORDER of the three people doesn't matter -- ABC is the same committee as CAB -- we divide by the number of ways the three people can be arranged (3!):
(10*8*6)/(3*2*1) = 80.
The correct answer is
D.
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