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 $A_0$. Consequently
$$
|\mathcal A|
\leq\binom{k}{2}\binom{n-2}{k-2}
=O(n^{k-2})
=o\left(\binom{n-1}{k-1}\right).
$$
This proves the stated dichotomy.

If the small alternative holds, then for sufficiently large $n$ any fixed $i$ has all but at most $\varepsilon\binom{n-1}{k-1}$ 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 $B\setminus\{i\}$, so
$$
|\{C\in\mathcal A:i\notin C\}|
\leq(k-1)^2\binom{n-2}{k-2}
=o\left(\binom{n-1}{k-1}\right).
$$
For sufficiently large $n$ this is at most $\varepsilon\binom{n-1}{k-1}$, as required.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-168/3/iv/solution