Number of ways to choose 3 couples from 5 couples = 5C3 = 10
Now each of these 3 couples can send two persons (husband or wife), number of ways of doing this = 2*2*2 = 2^3 = 8
Therefore, total number of ways: 10*8 = 80
The correct answer is D.
Refer to the post here for a similar problem >> https://www.beatthegmat.com/committe-of- ... tml#601345
Probability 2
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Total selections possible = 10C3 = 120
Ways to select with 1 couple, 1 non-couple = 5C1 * 8 = 40. Since there are 5 couples we can select any one couple in 5 ways. that fills 2 seats out of 3 and the third person can be any one out of 8 remaining.
Therefore no of selections such that no couple is selected = 120 - 40 = 80
hence D
Ways to select with 1 couple, 1 non-couple = 5C1 * 8 = 40. Since there are 5 couples we can select any one couple in 5 ways. that fills 2 seats out of 3 and the third person can be any one out of 8 remaining.
Therefore no of selections such that no couple is selected = 120 - 40 = 80
hence D
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Number of options for the first person = 10. (Any of the 10 people.)dvalenz 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?
20
50
50
80
120
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 selections doesn't matter -- ABC is the same committee as BCA -- we divide by the number of ways to ARRANGE the 3 people selected (3!):
(10*8*6)/(3*2*1) = 80.
The correct answer is D.
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