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 Post subject: Combinatorics & Probability XII - Page # 27 - A committee ofPosted: Sun Jan 23, 2011 10:56 pm

Joined: Sun Jan 23, 2011 10:38 pm
Posts: 2
A committee of 4 is to be chosen from 7 employees for a special project at ACME Corporation. 2 of the 7 employees are unwilling to work with each other. How many committees are possible if the 2 employees do not work together?

Solution: Total arrangements without restriction is 7!/4!(7-4)! = 35. Now the question I have is how to find the arrangements that violate the restrictions. According to the explanation in the book, the # of arrangements that violate the restrictions are 5!/2!(5-2)! = 10. I am unable to understand this explanation of 5 remaining employees fill 2 remaining spots. Below is how I interpreted the problem. Please help and tell me what I am doing wrong.

1) We have 4 spots to fill.
2) If we combine 2 employees as one unit, there would be total of 6 - 1 unit (2 employees) + 5 remaining employees - employees.
3) So the 6 employees fill 4 spots in 6!/4!(6-2)! = 15.

So the total arrangements would be 35-15 = 20.

Thanks

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 Post subject: Re: Combinatorics & Probability XII - Page # 27 - A committee ofPosted: Mon Jan 24, 2011 6:05 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
We have to find how many ways that the two employees could be on the committee of 4.

If we fill spots with the two "enemy" employees, there are 2 more spots left to fill.

There are 7 - 2 = 5 employees left from which to choose to fill these spot.

So, 5!/(2! *(5-2)! ) = number of ways to have the two "enemies" on the 4 person committee.

-- Veritas Help

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