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by BigFisch » Mon Mar 17, 2008 6:42 am
Hello everyone, i really appreciate that if anyone could tell me how to solve this question. Is there any formula for this type of question? Is this a permutation question? Thank you so much guys.

(Question 2) A committee of three people is to be chosen from four 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?

(a) 16
(b) 24
(c) 26
(d) 30
(e) 32
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by jonnyais » Mon Mar 17, 2008 7:33 am
I thought about it as the difference between the total number of combinations and the number of combinations that contain couples.

Total combination of couples: this is simply 8 choose 3. If you haven't memorized the formula for the binomial coefficient yet, visit this site:
https://mathworld.wolfram.com/BinomialCoefficient.html

The number of combinations with couples: I think this is the product of the 4 couples multiplied by the remaining members that could be the third person on the committee, with after the couple is selected means there are 6 people left = 4*6.

I think the answer is 8-choose-3 minus 4*6 = 56-24 = 32
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by BigFisch » Mon Mar 17, 2008 9:38 am
jonnyais, thank you so much for the explaination, i really appreciate for your help. :D
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