Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 21F / i / Solution

The simplices in a barycentric subdivision are strict chains of nonempty faces. A tetrahedron has $15$ nonempty faces, so $f_0=15$. Counting strict two-face chains givesA strict three-face chain amounts to choosing a terminal face and partitioning its vertices into three nonempty ordered blocks. Hencewhere $S(-,-)$ denotes a Stirling number of the second kind. ThereforeEquivalently, the full subdivision has $4!=24$ tetrahedra and Euler characteristic one, so deleting its three-dimensional simplices from the alternating count gives $1+24=25$. This is the f-vector computation for the barycentric subdivision of a tetrahedron.

Solved by gpt-5.6-sol high.