Graph theory framing. Represent each team as a vertex and each game as an edge connecting the two teams that play. The condition “every team plays exactly 3 games” means every vertex has degree 3. We must count all such graphs on 6 labeled vertices.

A quick check: since each edge contributes 2 to the total degree sum, the number of games per schedule is (6 × 3) / 2 = 9. Figure 1 shows one example.

Solution 1 — Direct Count

We classify all possible 3-regular graphs on 6 vertices. There are exactly two structurally distinct types.

Step 1: The two schedule types

• Type 1 — The Prism. The 6 teams split into two groups of 3. Within each group the teams form a triangle (every pair in the group plays each other), and each team plays one cross-group game.
• Type 2 — K_3,3 (complete bipartite). The teams split into two groups of 3. Every team in Group A plays every team in Group B; no games occur within a group.

Why are these the only two types? Suppose some vertex (say team 1) belongs to a triangle with teams 2 and 3. Since each of teams 1, 2, 3 has degree 3 and already has 2 edges within the triangle, each must have exactly one edge leaving the triangle to some vertex in {4, 5, 6}. These three edges are distinct (they go to different vertices), so they form a perfect matching between {1, 2, 3} and {4, 5, 6}. Teams 4, 5, 6 each receive one cross-group edge, leaving each with 2 edges still needed — which must lie within {4, 5, 6}, forming a second triangle. This gives the Prism. Conversely, if no triangle exists, one can verify the only possibility is K_3,3.

Step 2: Counting Type 2 schedules (K_3,3)

We only need to choose which 3 teams form Group A (the remaining 3 form Group B automatically, and the schedule is then fully determined). Since the partition {A,B,C} vs {D,E,F} is the same schedule as {D,E,F} vs {A,B,C}, divide by 2.

Step 3: Counting Type 1 schedules (Prism)

A Prism schedule requires two independent choices:

• Partition: Choose which 3 teams form each triangle. As above, C(6,3)/2 = 10 ways.
• Matching: How does each team in one group get paired with an opponent in the other group? Team 1 (say) can be assigned to any of the 3 opponents (3 ways); team 2 gets either of the 2 remaining opponents (2 ways); team 3 is forced (1 way). Total: 3! = 6 matchings.

Different matchings produce genuinely different sets of games, so we multiply:

Step 4: Total

Every valid schedule is exactly one of these two types, so:

Solution 2 — Complement Approach

Instead of asking which games are played, we ask which games are omitted.

In a full round-robin among 6 teams, every pair would play — that is K₆, with C(6,2) = 15 games. Our schedule uses only 9 of these, so exactly 6 games are omitted per schedule. Since each team has 5 opponents in K₆ but plays only 3, each team skips exactly 2 games. The 6 omitted games form their own graph — the complement of the schedule — in which every vertex has degree exactly 2.

Step 1: Structure of 2-regular graphs

A 2-regular graph is always a disjoint union of cycles covering all vertices. (Following any path, you can never get stuck — degree 2 guarantees a next edge — and you can never branch, so every path must eventually loop back to its start.)

We must partition all 6 vertices into cycles, each of length at least 3 (no loops or double edges). The partitions of 6 into parts ≥ 3 are:

• Case 1: a single cycle of length 6 — one hexagon spanning all teams.
• Case 2: two cycles of length 3 — two disjoint triangles.

Other partitions are impossible: 4+2 and 5+1 fail because 2 and 1 are less than 3.

Step 2: Counting Case 1 — 6-cycle complement

How many distinct labeled 6-cycles exist on 6 vertices? There are 6! ways to arrange 6 teams in a sequence, but each cycle is counted 6 times by rotation and 2 times by the direction of traversal:

Each 6-cycle complement gives a unique schedule, so there are 60 schedules of this type.

Step 3: Counting Case 2 — two-triangle complement

Choose which 3 vertices form the first triangle: C(6,3) = 20 ways. The remaining 3 form the second triangle automatically. Since the two triangles play symmetric roles (neither is “first”), we divide by 2.

For each partition, there is exactly one triangle on any given 3 labeled vertices — by the same cycle-counting formula: (3−1)!/2 = 1. So each partition determines a unique complement, giving 10 schedules of this type.

Step 4: Total

Connection Between the Two Approaches

Both solutions partition all valid schedules into the same two cases, arrived at from opposite directions. The complement of a Prism schedule is a 6-cycle; the complement of a K_3,3 schedule is two triangles. The table below shows the exact correspondence.

Notice also the interplay of the two counting techniques:

• Solution 1 counts K_3,3 by choosing a partition (10 = C(6,3)/2), and counts the Prism by choosing a partition and a matching (60 = 10 × 6).
• Solution 2 counts two-triangle complements by choosing a partition (10 = C(6,3)/2), and counts 6-cycle complements by the cycle-counting formula (60 = (6−1)!/2 = 5!/2).

Both approaches ultimately rest on the same algebraic facts; the complement lens simply offers a cleaner path to the harder case. As a general problem-solving principle: when the structure of what is present is difficult to classify directly, it can help to classify what is absent instead.