= 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 $6$, $\binom62=15$, and
$$
\frac{6!}{2^3\,3!}=15
$$
<synthemes>. 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 \b[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 $3\cdot3-3+1=7$ <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 gives
$$
\boxed{\#\text{points}=6,\quad\#\text{duads}=15,\quad\#\text{synthemes}=15,\quad\#\text{totals}=15\cdot2/5=6.}
$$
Two 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 \b[each pair of totals has exactly one common <syntheme>]. This incidence property drives the next construction.
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