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.

Please help. Greatly appreciated.

Thanks

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.

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