Let $m$ be an integer greater than 3. In a party with more than $m$ participants, every group of $m$ people has exactly one common friend. How many friends does the person with the most friends have?
The number sought is clearly greater than $m$: take a group of $m$ people, then the common friend to the people in this group has at least $m$ friends. We will show that this maximum is in fact equal to $m$. Let $P$ be the person with the most friends, and $S$ the set of their friends. Consider a subset $S^{\prime}$ of $S$ with $m-1$ elements. Call $Q_{S^{\prime}}$ the common friend to the elements of $S^{\prime}$ and to $P$, it is a friend of $P$ and is therefore an element of $S$.

Let $S_{1}$ and $S_{2}$ be two subsets of $S$ with $m-1$ elements, distinct. Then $S_{1} \cup S_{2}$ has at least $m$ elements, and thus $Q_{S_{1}}$ is different from $Q_{S_{2}}$ (otherwise, by taking a subset with $m$ elements from $S_{1} \cup S_{2}$, the elements of this subset would have at least two common friends, $Q_{S_{1}}$ and $P$, which contradicts the assumptions).

We have thus shown that the set of subsets with $m-1$ elements of $S$ injects into $S$. This imposes

$$
\binom{|S|}{m-1} \leq|S|
$$

and therefore $|S| \leq m$, which is the desired result.