A group of 10 students participates in a mathematics competition formed by teams of 4 students. We know that any two of the teams have exactly one student in common.

a) What is the maximum number of teams a student can participate in? Provide an example of a distribution of 10 students where this maximum number can be verified.

b) Can the competition have 8 teams?
Solution

a) Consider a student $A$ who is part of the most teams and say that he is in a team with three other students $B, C$ and $D$. Any other team that also includes $A$ must contain three other students who are not in the set $\{B, C, D\}$. Since there are only $10-1=9$ students different from $A$, the maximum number of teams that can include $A$ is $9 / 3=3$. An example of distributing 10 students, represented by the letters of the set $\{A, B, C, D, E, F, G, H, I, J\}$, is

| A | B | C | D |
| :--- | :--- | :--- | :--- |
| A | E | F | G |
| A | H | I | J. |

Each row indicates a team, and all of them have only student $A$ in common.

b) Suppose, for the sake of contradiction, that there could be 8 teams. Since each team has 4 students, we would have at least $8 \cdot 4=32$ student participations, counted with repetitions. Given that there are only 10 students and $32 / 10>3$, at least one student would have to participate in 4 teams. This contradicts the previous item, and this contradiction shows that we cannot have 8 teams.