Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 3 / 4 / a / Solution

A point is an element of $H$. A duad is an unordered two-element subset. A syntheme is a partition of $H$ into three duads, and a total of synthemes is a collection of five synthemes whose duads partition all fifteen duads. In graph terms these are vertices, edges, perfect matchings and one-factorization of $K_6$.

The first counts are $6$, $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$, andsynthemes. To count totals, first observe that two edge-disjoint synthemes have union a six-cycle. Its complement in $K_6$ is a triangular prism: two triangles on alternate cycle vertices, joined by the three remaining cross edges. Its perfect matchings are the matching using all three cross edges and three matchings using one cross edge each. The all-cross matching cannot be used in a factorization, because the remaining two odd triangles cannot be matched. The other three matchings partition the prism edges. Therefore every pair of disjoint synthemes extends to a unique total.

Fix a syntheme $s$. Each of its three duads belongs to three synthemes. Inclusion-exclusion shows that $3\cdot 3-3+1=7$ synthemes share a duad with $s$, including $s$ itself. Thus eight are disjoint from $s$. A total containing $s$ uses four of these, and each disjoint syntheme determines exactly one such total. Hence $s$ belongs to $8/4=2$ totals. Counting incidences givesTwo different totals share at most one syntheme, by the unique-completion assertion. There are fifteen pairs of totals and fifteen synthemes each belonging to two totals; consequently each pair of totals has exactly one common syntheme. This incidence property drives the next construction.