Here's the Question,
12 employees are eligible for working on a company's project. The team must consist of 4 people. However, two employees had a fight and refuse to work together. How many ways can the team be formed?
So I am having a hard time checking/confirming if my thinking behind this is correct, even when looking at exclusion-inclusions principles. So I will explain what I think is right and hopefully someone can confirm the logic.
1) $^{12}C_4$ is the total number of ways of selecting a team of 4.
$^{10}C_2$ is the total number of ways to exclude the 2 people being together. $^{12}C_4 – ^{10}C_2 =450$ possible ways.
2) Excluding 1 of the two leaves $^{11}C_4$ possible choices.
Excluding the other 1 instead is also $^{11}C_4$ possible choices. However, this over counts the 10 other people in both cases which is $^{10}C_4$ choices. Therefore $2 \times ^{11}C_4 – ^{10}C_4 = 450$ possible ways.
I know the answer of 450 is right, but I am unsure if my reasoning behind these calculations is correct. Maybe there is another way to look at this problem that is easier to grasp?
Many Thanks
Ken
Best Answer
Your first method is an elegant way to solve the problem.
Here is another approach. Observe that either exactly one of the two feuding employees is placed on the team or neither one is.
Exactly one of the two feuding employees is placed on the team: We choose which of the two feuding employees is placed on the team and which three of the other ten employees is placed on the team with that person, which can be done in $$\binom{2}{1}\binom{10}{3}$$ ways.
Neither of the two feuding employees is placed on the team: We must choose which four of the other ten employees is placed on the team, which can be done in $$\binom{10}{4}$$ ways.
Total: Since these two cases are mutually exclusive, the number of ways the team of four employees can be chosen without placing the two feuding employees together is $$\binom{2}{1}\binom{10}{3} + \binom{10}{4} = 450$$ as you found.