Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 109 / 1 / iii

The assertion is false. Take $n=6$, $r=3$, andThere are $1+3\cdot 3=10$ members. Its lower shadow consists of the three pairs inside $\{1,2,3\}$ and the nine pairs having one point there and one in $\{4,5,6\}$, so $|\partial \mathcal{A}|=12$. Its upper shadow consists of the twelve four-sets containing at least two points of $\{1,2,3\}$, so $|\nabla \mathcal{A}|=12$. The sum is $24$.

The initial lex segment of length ten is the star of all triples containing $1$; its lower and upper shadows have sizes $15$ and $10$. The initial colex segment consists of all triples in $[5]$; its lower and upper shadows have sizes $10$ and $15$. Both sums are $25$, so neither canonical segment minimizes the sum.

Solved by gpt-5.6-sol high.