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.