If two of the $13$ members of a computer programming team refuse to work together, how many groups of seven can be chosen to work on a project?

Question

How is this calculated? I know a possible way to calculate it would be to find the total number of combinations and subtract the number of combinations with only the two people in the group. None of the solutions I found are convincing. Please help!

Answer 1

The number of groups of $7$ where the two members in question work together is $\binom{11}{5}$. (This is the number of ways to choose 5 of the remaining 11 team members to work together with the two problematic team members, forming a group of 7.) Thus the number of acceptable groups of $7$ is $\binom{13}{7} - \binom{11}{5}$.

Answer 2

The total number of ways to choose the seven member is $\binom{13}{7}$. Let's count the combinations that we DON'T want, namely, that the two people are working together. If the two people are on the project, there are another 5 people needed to be chosen from the remaining 11 people, or $\binom{11}{5}$. So your answer is these two combinations subtracted from the other.

Answer 3

But, how can we take out a group of five people from a group of seven people? That's the thing driving me crazy.

I read your words as: You interpreted the symbol $\Large{11\choose 5}$ as: the result is a group of five people. This is wrong, as @N. F. Taussig pointed out in the comment that is a shorthand of

$${2\choose2}\cdot{11\choose5}=1\cdot{11\choose5}={11\choose5},$$

you should find that the first in the above has two numbers added up to $7$. When solving a combinatorics problem I would recommend you first write down each step you're doing clearly, e.g. the steps for the above is:

$$\textrm{(choose the two who fight into a group)}\cdot\textrm{(choose the other five people to join them)}.$$

Answer 4

Method 1: A direct count.

A group formed without the two team members who refuse to work together must contain neither of them or exactly one of them.

Neither of the two team members who refuse to work together is selected: Then seven of the remaining $13 - 2 = 11$ team members must be selected, which can be done in $$\binom{2}{0}\binom{11}{7}$$ ways.

Exactly one of the two team members who refuse to work together is selected: We must select one of the two team members who refuse to work together and six of the remaining eleven team members, which can be done in $$\binom{2}{1}\binom{11}{6}$$ ways.

Total: Since the two cases are mutually exclusive and exhaustive, the number of ways a group of seven team members can be selected if two of the team members refuse to work together is $$\binom{2}{0}\binom{11}{7} + \binom{2}{1}\binom{11}{6}$$

Method 2: An indirect count.

There are $$\binom{13}{7}$$ ways to select a group of seven of the thirteen team members. From these, we must subtract those selections that include both of the team members who refuse to work together.

If both team members who refuse to work together were selected, we would also have to select five of the other team members to complete the group, which can be done in $$\binom{2}{2}\binom{11}{5}$$ ways.

Hence, the number of ways seven team members can be selected for the project if two of the team members refuse to work together is $$\binom{13}{7} - \binom{2}{2}\binom{11}{5}$$