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Manager
Joined: 08 Jan 2018
Posts: 129

Re: A committee of three people is to be chosen from four married couples.
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15 Sep 2019, 07:13
We have 4 couples. Let the couples be
AB CD EF GH
Number of ways to select ANY 3 out of 8 is 8C3 = 56
Number of ways we select the 3 such that there is at least one couple = Number of ways to select a couple = 4 ways Number of ways to select the third member = 6 ways (as after picking one couple, 6 will remain) Thus Number of ways to select the group such that there is at least one couple = 4*6 = 24
Hence, the number of ways to select the group such that there are no couples = 56  24 = 32
Answer E



Intern
Joined: 14 Sep 2016
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Re: A committee of three people is to be chosen from four married couples.
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16 Oct 2019, 06:38
For the reverse approach to this question, is it logically correct to proceed as follows:
1. 8C3 = 56 Total # ways 2. 8C1 = 8 ways to choose person not included in the couple selection 3. 3C1 = 3 ways to choose 1 couple from the 3 couples, omitting the 1 person chosen and his spouse?
56  (8*3) = 32
Wondering if this approach would work as well as i believe i just did the selection in a reverse order. Or, is this just a coincidence?
Appreciate the help!




Re: A committee of three people is to be chosen from four married couples.
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16 Oct 2019, 06:38



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