Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 168 / 3 / iv / Solution

Assume first that no two members of the uniform set family $\mathcal{A}\subseteq [n{]}^{(k)}$ meet in exactly one point. Fix $A_0\in \mathcal{A}$. Since $\mathcal{A}$ is an intersecting family, every $B\in \mathcal{A}$ then contains at least two elements of $A_0$. ConsequentlyThis proves the stated dichotomy.

If the small alternative holds, then for sufficiently large $n$ any fixed $i$ has all but at most $\epsilon \left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ members of $\mathcal{A}$ containing it. Otherwise choose $A,B\in \mathcal{A}$ with $A\cap B=\{i\}$. Every $C\in \mathcal{A}$ not containing $i$ must meet both $A\setminus \{i\}$ and $B\setminus \{i\}$, soFor sufficiently large $n$ this is at most $\epsilon \left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$, as required.