Notes on Norman Wildberger's Six: A mathematical exploration — lecture 2 of an elementary course in pure mathematics built around the combinatorics of six objects. This page works through one specific claim from the lecture: that there are exactly 45 meets, and why the 15 × 3 construction generates all of them without overlap.

A meet is a 2-set of non-incident edges — two edges that share no node — on six nodes. The claim is that there are exactly 45 meets. The argument has two parts:

  1. Sorting. Sort every meet into one of 15 buckets, one per edge.
  2. Counting. Show each bucket has exactly 3 meets.

If the sorting puts every meet into exactly one bucket — no meet skipped, no meet in two buckets — then the total is just 15 × 3 = 45.

The sorting rule

Given any meet e₁ · e₂, the two edges use 4 different nodes. The remaining 2 nodes form a third edge, which I'll call the complement. That complement is the bucket label.

Example: the meet 1-2 · 3-4 uses nodes {1,2,3,4}. The leftover nodes are {5,6}, so the complement edge is 5-6. This meet goes in bucket 5-6.

meet 1-2 · 3-4
complement: 5-6

Why each meet lands in exactly one bucket

The complement is unique. Given a meet, the 4 nodes it uses are fixed, so the 2 leftover nodes are fixed, so the complement edge is fixed. There is no choice involved. One meet → one complement → one bucket.

This means:

  • No meet is in two buckets — its complement is uniquely determined, so it goes to only that bucket.

Why the construction misses no meets

The argument above rules out one failure mode (a meet appearing in two buckets). The other failure mode is a meet appearing in zero buckets — being missed entirely. To rule that out, take an arbitrary meet and trace it through the construction.

Pick any meet — say 2-5 · 3-6. Where does it show up?

meet 2-5 · 3-6
complement: 1-4
  1. This meet uses nodes {2, 3, 5, 6}.
  2. The leftover nodes are {1, 4}, so its complement edge is 1-4.
  3. The construction visits all 15 edges, so it visits 1-4. When it does, it generates the 3 pair-ups of the 4 leftover nodes {2, 3, 5, 6}:
2-3 · 5-6
2-5 · 3-6 ←
2-6 · 3-5

Our meet 2-5 · 3-6 is the middle one. The construction produces it — we didn't have to know about it in advance, and we didn't have to search for it. It fell out automatically when we processed bucket 1-4.

The general argument. Take any meet M:

  1. M consists of two disjoint edges, so M uses exactly 4 of the 6 nodes.
  2. The 2 leftover nodes form an edge e — the complement.
  3. The construction visits e (it visits all 15 edges, no exceptions).
  4. When it visits e, it generates every way to pair up the 4 leftover nodes into 2 pairs.
  5. M is one of those pair-ups — because that's literally what M is: two disjoint edges on those 4 leftover nodes.

So M must be produced. Since M was an arbitrary meet, every meet is produced.

Concern Why it can't happen
A meet ends up in two buckets (counted twice) A meet has only one complement, so only one bucket matches.
A meet ends up in zero buckets (missed) A meet always has a complement, and the bucket for that complement always generates it.

Together: every meet appears in exactly one bucket. The buckets partition the meets.

Why each bucket has exactly 3 meets

Fix a bucket — say bucket 5-6. The meets in it use only the 4 leftover nodes {1,2,3,4}. So the question becomes: in how many ways can we split 4 nodes into 2 pairs?

Pick node 1. It must pair with one of {2, 3, 4} — that's 3 choices. The remaining 2 nodes are then forced into the second pair. So 3 splits:

1-2 · 3-4
1-3 · 2-4
1-4 · 2-3

This is the same calculation no matter which edge is the bucket label — it always reduces to "split 4 leftover nodes into 2 pairs," which always has 3 answers.

All 15 buckets — the receipts

Each card is a bucket. The bucket label (the dashed gray edge) is the complement. The two solid colored edges are the actual meet.

What this argument is not doing

This is not a "list everything and count" proof. It's a structural proof: it pairs up two ways of counting the same set.

  • One way: walk through all meets, find the count directly. (Hard — that's what we're trying to avoid.)
  • The other way: group meets by their complement edge, count each group, then multiply by the number of groups. 15 × 3.

Both ways count the same thing — the set of meets — so they must give the same answer. The structural way is easier because the local question ("how many meets share this complement?") has a clean answer (3) that we can compute without enumerating everything.

The key step is convincing yourself the sorting rule is well-defined: every meet has a complement, and only one. Once you believe that, the partition is automatic, and 15 × 3 = 45 follows.