Did you know? Put six people in one room. Assume every pair either knows each other or does not, and that knowing someone is mutual. No matter how those relationships are arranged, you will always find either three people who all know one another or three people who are all strangers to one another. Five people do not guarantee the same result.
For families researching a JC2 H3 Mathematics tutor in Bukit Timah, this is a memorable introduction to Ramsey theory and two-colour graph proofs. It uses the pigeonhole principle, considering cases, symmetry, proof by contradiction and combinatorial reasoning—skills explicitly included in the 2027 Singapore–Cambridge H3 Mathematics (9820) syllabus. The six-person theorem itself is useful enrichment, not a separately named H3 topic.
Quick answer: Colour each connection between six vertices either red or blue. Choose any vertex; its five connections include at least three of the same colour. If those three neighbours have even one connecting edge in that colour, you obtain a same-colour triangle with the chosen vertex. Otherwise their three connecting edges all have the other colour, giving a same-colour triangle among the neighbours. This proves R(3,3)≤6, and a five-cycle colouring proves that six is the smallest guaranteed number.
Quick navigation
- Six people explained in plain English
- Turn the puzzle into a coloured graph
- Complete proof for six vertices
- Why five vertices are not enough
- Meaning of the Ramsey number
- A general Ramsey recurrence
- The role of symmetry and the pigeonhole principle
- What changes with incomplete or directed relationships
- Practice and problem-solving clinics
- Common reasoning errors
- A seven-session learning plan
- Frequently asked questions and sources
The six-person party problem in plain English
Imagine six people at a table. Between every two people, either there is a mutual acquaintance or there is not. The problem says that some three form a mutual-acquaintance group, or some three form a mutual-stranger group. You do not know which three in advance.
The condition is not about everybody being friendly or about a majority agreeing on one opinion. It is about a local pattern among three people. A mutual-acquaintance triple means every pair inside the triple knows one another. A mutual-stranger triple means no pair inside it knows one another.
It is easy to find an example where the claim holds. If all six know each other, take any three. But a mathematical theorem is stronger: the claimed configuration must exist for every possible assignment of relationships to the fifteen distinct pairs.
Trying to check every possible relationship pattern would mean considering 2¹⁵=32768 colourings, each with its own possible triples. A proof by enumeration might be programmable, but it is not the clearest explanation.
The strategic observation is that one person has five relationships. Among five pairs, at least three must be of the same kind. That tiny local fact drives the entire proof, no matter how the rest of the graph is coloured.
This is the mathematical beauty of the problem: a large collection of choices still forces a small regular structure. The proof gives a reason beyond a sample calculation, and the reason transfers to many other combinatorial settings.
Translate the story into graph colouring
Represent each person by a vertex, or dot. Draw an edge between every pair of distinct vertices. Because we record exactly one relationship for every pair, the result is a complete graph on six vertices, written K₆.
Colour each edge red when the people know one another and blue when they do not. There are C(6,2)=15 edges because each unordered pair of vertices corresponds to exactly one edge.
A triangle in the graph consists of three vertices and all three edges among them. A monochromatic triangle is a triangle whose three edges share one colour. Red corresponds to three mutual acquaintances; blue corresponds to three mutual strangers.
The graph problem is therefore: prove that every red-blue colouring of the fifteen edges of K₆ contains a monochromatic triangle. This statement includes all possible arrangements and makes the roles of the two colours mathematically symmetric.
Be careful about the word ‘triangle’. The people do not need to sit in a geometric triangle and their dots do not need equal physical spacing. A triangle here is a three-vertex complete subgraph, determined only by pairwise connections.
Similarly, the picture’s crossing lines do not create additional vertices unless the graph definition explicitly says so. Graph-theoretic incidence concerns endpoints, not accidental crossings in a drawing.
This translation is an example of a general H3 problem-solving heuristic: restate the problem in a representation where its conditions become easier to exploit.
A complete proof of the six-person theorem
Step 1 — choose one vertex. Select any person A. There are five other people, so A has five edges connecting to them. Each edge is either red or blue.
Step 2 — apply pigeonhole. With five edges placed into two colour categories, at least one colour appears three times. Otherwise each colour would appear at most twice and the total would be at most four. By symmetry of the colours, suppose A has red edges to three vertices B,C,D.
Step 3 — examine the three connecting edges. The pairs B–C, B–D and C–D each have an edge. There are exactly two possibilities: at least one of these three edges is red, or all three are blue.
Case 1 — some connecting edge is red. Suppose B–C is red. Since A–B and A–C are also red, A,B,C form a red triangle. The desired conclusion is established.
Case 2 — no connecting edge is red. Since every edge is assigned either red or blue, the three edges B–C, B–D and C–D must all be blue. Hence B,C,D form a blue triangle.
Conclusion: Every red-blue colouring of K₆ contains a monochromatic triangle. Because the argument did not depend on a particular arrangement, it holds universally.
If the three edges from A had initially been blue rather than red, the same reasoning would work after swapping the colour names. That is a justified use of symmetry, not an unexplained ‘without loss of generality’.
Notice what the theorem does not assert. It does not promise that every group of three is monochromatic, that a specific trio works, or that both a red and blue triangle must occur. It guarantees at least one such triple of either colour.
This proof is a model of H3 exposition: it states a representation, uses a precise counting fact, divides all remaining possibilities into cases and reaches exactly the required conclusion.
Why five people do not give the same guarantee
To prove six is the smallest number with this property, we must also show that the guarantee fails for five. A single five-vertex colouring with no monochromatic triangle is enough; we do not need to classify every colouring of five vertices.
Label five vertices 1,2,3,4,5 around a cycle. Colour the cycle edges 1–2, 2–3, 3–4, 4–5 and 5–1 red. Colour the other five edges blue.
The red subgraph is a cycle of length five, so no three red edges form a triangle: no red chord closes a three-vertex loop. The blue edges form another five-cycle, following the order 1–3–5–2–4–1.
The blue five-cycle also has no triangle. Therefore this colouring of the complete graph K₅ contains neither a red nor a blue triangle. The guarantee cannot hold for every five-person arrangement.
Together with the six-vertex proof, this establishes R(3,3)=6, the exact two-colour Ramsey number for triangles.
The idea of a counterexample is important. A graph with five vertices in which a monochromatic triangle does occur does not show the threshold is five. To lower the guaranteed threshold, every possible colouring would need the required triangle; the five-cycle colouring refutes that universal statement.
A useful classroom exercise is to draw the red pentagon and its blue complement and ask the learner to identify why the two subgraphs are isomorphic. That visual symmetry makes the lower-bound example memorable without replacing its proof.
What the notation R(3,3) means
The Ramsey number R(s,t), in the two-colour graph setting, is the smallest integer N such that every red-blue colouring of the edges of K_N contains either a red complete subgraph with s vertices or a blue complete subgraph with t vertices.
For R(3,3), we require a same-colour triangle of either colour. The numbers in parentheses describe the desired clique sizes, not the numbers of red or blue edges.
The six-person theorem proves R(3,3)≤6. The pentagon construction proves R(3,3)>5. Together they give the exact equality six. A proof of only the first inequality would provide an upper bound but not minimality.
Another easy boundary is R(2,t)=t for t≥2. Among t vertices, if there is any red edge we already have a red K₂. If there is no red edge, every edge is blue and the t vertices form a blue K_t.
The symmetry R(s,t)=R(t,s) follows by exchanging red and blue in any colouring. It does not require that the red and blue subgraphs have equal numbers of edges.
These definitions broaden the lesson beyond one party. However, large Ramsey numbers are often difficult to determine exactly, so students should distinguish a constructive lower bound, a universal upper bound and a fully proved exact value.
The 2027 H3 Mathematics syllabus prescribes combinatorial arguments and the pigeonhole principle; the Ramsey notation is included here as enrichment so learners can see how those principles interact in a non-routine setting.
A general recurrence for Ramsey upper bounds
The six-person proof suggests a more general approach. To find a red K_s or a blue K_t, choose a vertex of a larger two-coloured complete graph and classify its adjacent edges by colour.
Let A=R(s−1,t) and B=R(s,t−1), with s,t≥3. Take a complete graph on N=A+B vertices and colour every edge red or blue.
Choose one vertex v. It has N−1=A+B−1 neighbours. Either at least A of its incident edges are red or at least B are blue. If both counts were below those thresholds, their sum would be at most (A−1)+(B−1)=N−2, a contradiction.
Suppose v has at least A red neighbours. By the definition of A=R(s−1,t), those neighbours contain a red K_{s−1} or a blue K_t. A red K_{s−1} joins with v to form a red K_s because every edge from v to those vertices is red. A blue K_t already satisfies the other target.
If instead v has at least B blue neighbours, the definition of B=R(s,t−1) yields either a red K_s or a blue K_{t−1}, which joins with v to make a blue K_t.
Hence R(s,t)≤R(s−1,t)+R(s,t−1). It is an upper-bound recurrence and need not be an equality in every case.
Using R(2,3)=R(3,2)=3 gives R(3,3)≤3+3=6, recovering the upper bound. The five-cycle construction is still needed to show the exact value is not smaller.
The recurrence is a powerful H3-style example because the proof structure contains a pigeonhole step, symmetry between colours and conditional use of smaller subproblems. It also shows why an upper bound and an exact value are distinct claims.
Why the pigeonhole and symmetry steps are valid
At a chosen vertex in K₆ there are five coloured edges. If red occurs fewer than three times and blue occurs fewer than three, each colour occurs at most twice, giving at most four edges. That contradicts the actual degree five.
This arithmetic is the pigeonhole principle in its smallest useful form. The ‘pigeons’ are five edges, and the ‘holes’ are two colours. The guaranteed number in one colour is ⌈5/2⌉=3.
After choosing a colour with at least three edges, the proof treats it as red. This is valid because the original conclusion is symmetric under exchanging the two colours. If the majority is blue, exchange labels and use the identical argument.
Inside the three selected neighbours, the statement ‘either some edge is red or all edges are blue’ covers every possible colouring of their mutual edges. No third category remains because the problem specified exactly two colours.
A student may accidentally claim that all three mutual edges must be the other colour. That is too strong. Only when none is red do all three become blue; otherwise just one red edge is enough to create the red triangle with the chosen vertex.
These distinctions matter in mathematical prose. The reader must see why there are three same-colour edges, why all relevant cases have been considered, and why each case creates the promised pattern.
The proof is short only because the representation and the cases were chosen carefully. Searching for a regular structure is an H3 problem-solving skill, not a memory trick for one graph.
What changes if the story’s assumptions change?
The graph model assumes every pair has exactly one of two outcomes, and that the outcomes are symmetric. If some relationships are unknown, some pairs have no colour; the K₆ proof no longer applies to that incomplete graph.
If ‘knows’ is directed—person A knows B, but B may not know A—then a single undirected red or blue edge may not describe the data. A different directed-graph question would need its own theorem.
If there are three possible edge colours, the five incident edges no longer guarantee three in one colour. The original pigeonhole count changes to a maximum of two in each of three groups, which could accommodate all five edges.
If the question asks for four mutual acquaintances or four mutual strangers, three red incident edges are no longer sufficient. A triangle argument cannot simply be relabelled as a four-vertex proof.
Even the seating chart is irrelevant unless distances or visibility are part of the assumptions. The graph theorem uses only pairwise relationships; geometric placement is a convenient picture, not a hidden condition.
These are productive variations for a JC2 tutorial. Change one assumption, ask the student which line of the original proof fails, and decide whether a new argument or counterexample is required.
A well-written mathematical answer states the version proved. It avoids overgeneralising from a striking special case and thereby demonstrates the precision expected in the 2027 H3 syllabus.
Try extending the argument to seven people
Suppose seven people meet under the same mutual acquaintance-or-stranger rules. Do we need an entirely new theorem? No. Choose any six of the seven people. The six-person theorem guarantees the desired triple inside that subset, so the triple also exists in the original group of seven.
The same logic holds for every number N≥6: any group with at least six members contains a six-member subset, and that subset contains a monochromatic triangle. This is a simple but important monotonicity property of Ramsey guarantees.
A student may claim that seven people guarantee an additional kind of structure, such as four mutual acquaintances, merely because there are more choices. That does not follow from the six-person theorem. A stronger conclusion needs its own proof or a different threshold.
This extension teaches an H3 habit: after proving a threshold works, ask whether the property persists for larger inputs and why. The subset argument supplies a rigorous reason without a new round of casework.
A three-colour counterexample that exposes the assumptions
Can the exact six-vertex argument be used if every pair is assigned one of three colours? Not automatically. At a chosen vertex, five incident edges can split into counts 2,2,1, so no colour must occur three times.
In fact, six vertices can be three-coloured with no monochromatic triangle. Begin with the five-vertex red-blue colouring based on the pentagon and its complement, which has no monochromatic triangle. Add a sixth vertex and colour all five edges connecting it to the existing vertices green.
Any triangle involving the new vertex has two green edges and one old red or blue edge, so it is not monochromatic. Any triangle entirely within the original five vertices was already nonmonochromatic. Thus the resulting three-colour K₆ has no monochromatic triangle.
This does not contradict Ramsey theory. It only shows that six is not a guaranteed threshold for three colours. The conditions of a theorem determine how and where it applies, and a correct counterexample reveals the boundary of the statement.
The exercise links the named H3 methods of counterexample, construction and considering cases. It makes the often-ignored number of colour categories a visible mathematical assumption.
A graph-colouring proof audit
Before using the party theorem, check that the graph is complete: every pair of distinct people has a known relationship. Check that each relationship has exactly one of two colours. Check that the relationship can be modelled by an undirected edge rather than two potentially different directed responses.
Next define precisely what configuration is sought. A red K₃ has three red edges between the chosen three vertices, not merely two red edges meeting at one vertex. The same holds for a blue K₃.
During the proof, name the chosen vertex and the three neighbours sharing a colour. Explain why at least three exist using the degree five and two colour categories. Then write the complete case split: some edge among the neighbours has the first colour, or none does.
Finally, distinguish an upper-bound proof from an exact threshold. The argument on six vertices guarantees existence, while the five-cycle construction proves that five cannot guarantee it. Only together do they establish R(3,3)=6.
A graph picture is helpful, but every sentence should remain true regardless of how the vertices are physically positioned. The argument depends on adjacency and colour, not artistic layout.
Original graph-colouring and proof clinics
1. How many edges does K₆ have?
Answer: It has C(6,2)=15 edges, one for each unordered pair of vertices.
Why it matters: Count each pair once. Treating A–B and B–A as different would double the edge total.
2. How many candidate triangles are in K₆?
Answer: There are C(6,3)=20 sets of three vertices.
Why it matters: A candidate triangle is determined by choosing three vertices, but the three edge colours may be mixed.
3. How many edges are in K₅?
Answer: There are C(5,2)=10 edges.
Why it matters: The red cycle in the extremal counterexample uses five edges and its blue complement uses the remaining five.
4. How many candidate triangles are in K₅?
Answer: There are C(5,3)=10 triples of vertices.
Why it matters: The five-cycle construction makes all ten candidate triangles nonmonochromatic.
5. What does monochromatic mean?
Answer: Every edge inside the chosen configuration has the same colour.
Why it matters: It is not enough for two edges of a triangle to match; all three must match.
6. What is the first pigeonhole step?
Answer: Five edges from one vertex are coloured red or blue; at least three share a colour.
Why it matters: If both colours appeared at most twice, there would be at most four edges.
7. Can four red edges also occur?
Answer: Yes. The pigeonhole statement guarantees at least three, not exactly three.
Why it matters: A proof may select any three same-colour neighbours if there are four or five.
8. What if A has three red neighbours?
Answer: If any edge among those neighbours is red, that edge and the two red edges from A form a red triangle.
Why it matters: The selected vertex supplies the other two needed edges.
9. What if none of those connecting edges is red?
Answer: All three edges within the selected neighbour triple must be blue, so those neighbours form a blue triangle.
Why it matters: This is the exhaustive second case when exactly two colours exist.
10. What if the majority colour was blue?
Answer: Exchange the colour names and apply the same argument.
Why it matters: The desired conclusion treats red and blue symmetrically, so the relabelling is justified.
11. Does the proof specify the exact trio beforehand?
Answer: No. It proves some trio exists in every colouring.
Why it matters: An existence theorem need not supply a predetermined set of vertices.
12. Does it guarantee both coloured triangles?
Answer: No. A completely red K₆ contains red triangles but no blue edges or blue triangles.
Why it matters: The conclusion is an OR, not an AND.
13. Does it guarantee every triple works?
Answer: No. A mixed-colour triple is possible even though some other triple must be monochromatic.
Why it matters: The theorem is existential, not universal over all triples.
14. Does a five-person group always contain such a triple?
Answer: No. The red pentagon and its blue complementary pentagon avoid all monochromatic triangles.
Why it matters: One valid colouring is a counterexample to the five-person universal claim.
15. What are the red edges in the five-cycle?
Answer: They are 1–2,2–3,3–4,4–5,5–1.
Why it matters: A cycle of length five has no triangle because it contains no chords.
16. What are the blue edges in that example?
Answer: They are 1–3,1–4,2–4,2–5,3–5.
Why it matters: They form another five-cycle in the order 1–3–5–2–4–1.
17. Why is six the exact Ramsey number?
Answer: Every K₆ colouring has a monochromatic triangle, while one K₅ colouring does not.
Why it matters: An exact minimum requires both an upper-bound proof and a lower-bound example.
18. What is R(2,3)?
Answer: It is three. If any edge in K₃ is red, there is a red K₂; if none is red, all of K₃ is blue.
Why it matters: A boundary condition supplies a simple Ramsey recurrence base.
19. What is R(3,2)?
Answer: It is three by exchanging the roles of the colours in R(2,3).
Why it matters: Ramsey numbers satisfy R(s,t)=R(t,s) under colour interchange.
20. What is R(2,4)?
Answer: It is four: any red edge works, otherwise all four vertices form a blue K₄.
Why it matters: A complete graph with fewer than four vertices cannot contain the required blue K₄ in the all-blue case.
21. What does R(3,3)≤6 say?
Answer: Six vertices guarantee a red or blue triangle.
Why it matters: It gives an upper bound; it does not alone prove six is the smallest possible.
22. What does R(3,3)>5 say?
Answer: Some two-colouring of K₅ avoids both kinds of monochromatic triangle.
Why it matters: The five-cycle example supplies that necessary lower bound.
23. What is the general recurrence?
Answer: R(s,t)≤R(s−1,t)+R(s,t−1) for integers s,t≥3.
Why it matters: Choose a vertex and apply pigeonhole to its red and blue neighbours.
24. Use the recurrence for R(3,3)
Answer: Since R(2,3)=R(3,2)=3, it gives R(3,3)≤6.
Why it matters: A recurrence upper bound still needs a separate lower-bound construction for equality.
25. What is an upper bound for R(3,4)?
Answer: The recurrence gives R(3,4)≤R(2,4)+R(3,3)=4+6=10.
Why it matters: This is a valid bound, not necessarily the exact Ramsey number.
26. What if there are seven people?
Answer: Select any six and apply the six-person theorem.
Why it matters: A configuration guaranteed in a subset is also present in its containing group.
27. Can a colouring of K₄ avoid monochromatic triangles?
Answer: Yes. Colour a four-cycle red and its two diagonals blue.
Why it matters: The red subgraph has no triangles and the blue subgraph has only two disjoint edges.
28. Can a K₃ colouring avoid a monochromatic triangle?
Answer: Yes. Colour two edges red and the remaining edge blue.
Why it matters: A triangle is monochromatic only when all three edges match.
29. Can an all-red K₅ avoid a monochromatic triangle?
Answer: No. Any three vertices form a red triangle.
Why it matters: A good colouring with the property does not prove every five-vertex colouring has it.
30. Does the three-colour version still work for six?
Answer: No. Extend the two-colour pentagon counterexample with a sixth vertex joined by green edges.
Why it matters: The resulting three-colour K₆ has no monochromatic triangle.
31. Why does three colours change pigeonhole?
Answer: Five edges distributed among three colours guarantee only two in one colour, not three.
Why it matters: The numerical threshold depends on the number of categories.
32. Does the graph need to be complete?
Answer: Yes, for the theorem as stated: every pair has exactly one coloured edge.
Why it matters: Unknown or missing relationships change the mathematical model.
33. Are directed relationships covered?
Answer: Not by this undirected two-colour proof without extra assumptions.
Why it matters: Mutual friendship can be represented by one undirected edge; one-way following cannot.
34. Can geometric edge crossings be ignored?
Answer: Yes. Crossing lines in a drawing are not graph vertices unless defined as such.
Why it matters: Graph topology depends on declared vertices and endpoints, not the drawing’s geometry.
35. Does acquaintance need to be transitive?
Answer: No. A knowing B and B knowing C does not have to imply A knows C.
Why it matters: The theorem works under arbitrary symmetric pair classifications.
36. Can the proof be checked by brute force?
Answer: Yes, K₆ has 15 edges and 2¹⁵=32768 red-blue assignments.
Why it matters: Enumeration can verify a finite claim, but the pigeonhole proof explains its structure far more clearly.
37. What is the relevant H3 syllabus principle?
Answer: The pigeonhole principle, proof by cases, symmetry, and combinatorial arguments are all explicitly named.
Why it matters: Ramsey theory here is an enrichment example applying those listed capabilities.
38. What is the key logical wording?
Answer: The theorem says ‘there exists a monochromatic triple’ in every two-colouring of K₆.
Why it matters: The nested universal-and-existential structure matters to the proof.
39. What would refute the six-person theorem?
Answer: A single red-blue edge colouring of K₆ with no monochromatic triangle.
Why it matters: The proof shows such a counterexample cannot exist.
40. How should the final argument end?
Answer: State that either case produces a same-colour triangle, so every colouring meets the requirement.
Why it matters: Conclude the claimed universal existence, not a stronger unsupported pattern.
A seven-session JC2 learning programme
Session one turns the party language into an undirected complete graph with two edge colours. Ask the student to identify which assumptions permit that representation and to count the fifteen edges of K₆.
Session two proves the six-person claim by pigeonhole. Require the student to name the chosen vertex, count five incident edges, identify three same-colour neighbours and write the two possible cases without skipping one.
Session three constructs the five-vertex counterexample. Draw the red pentagon, label every remaining edge blue and explain why each colour class is a triangle-free five-cycle.
Session four compares the upper-bound argument with the lower-bound construction. Ask why neither one by itself establishes R(3,3)=6 and what the word ‘smallest’ adds to the mathematical task.
Session five introduces Ramsey notation and easy boundaries R(2,t)=t. Derive the general recurrence by choosing a vertex and considering red or blue neighbourhoods.
Session six varies the assumptions: allow a third colour, remove an edge or direct the relationships. Have students locate precisely which line of the original proof no longer applies, then construct a counterexample where possible.
Session seven gives an unfamiliar two-colour proof. Ask the learner to choose a vertex or subset strategically, justify the case split and write a readable conclusion. The aim is a transferable proof method, not the memorisation of the number six.
Parents can ask whether every pair has one known relationship, why three same-colour edges occur and what happens if a connecting edge has that colour. A learner who can explain each step understands the reasoning.
The immutable eduKateSG three-student Mathematics reference near Sixth Avenue MRT is a Secondary 1 teaching example rather than evidence of current JC2 H3 classes. Contact eduKateSG for availability and advanced mathematics support.
Frequently asked questions
Is Ramsey theory a separately prescribed H3 9820 topic?
The official syllabus names the pigeonhole principle, symmetry and combinatorial proof, not Ramsey theory as a separate additional topic. The party theorem is a connected enrichment application.
Why are six people enough?
Choose one person and three same-colour connections among that person’s five relationships. Either a matching edge exists among the three neighbours or all their connections have the opposite colour.
Why are five people insufficient?
Colour a five-cycle red and its complement blue. Both colour classes are five-cycles without triangles, so no monochromatic triple is forced.
What does R(3,3)=6 mean?
It is the smallest number of vertices such that every red-blue edge colouring of a complete graph contains a red triangle or a blue triangle.
Must both a red and a blue triangle appear?
No. The theorem requires at least one triangle of either colour. An all-red complete graph gives a red triangle but no blue triangle.
Does the theorem apply to one-way social media following?
Not automatically. The proof models each pair by one undirected two-colour edge, while one-way following may be directed and require a different mathematical problem.
How does the pigeonhole principle enter?
Five incident edges divided into two colour classes guarantee at least three in one class, because two plus two could account for only four edges.
What is the significance of the Ramsey recurrence?
It gives upper bounds R(s,t)≤R(s−1,t)+R(s,t−1), derived by choosing a vertex and examining its coloured neighbours.
Can computer enumeration replace the proof?
It can check a finite collection of colourings, but the short pigeonhole argument gives the structural reason and supports generalisations.
How should a tutor check a student’s solution?
Ask for the assumptions, graph representation, exact pigeonhole count, exhaustive case split and distinction between proving an upper bound and proving an exact threshold.
Official syllabus and connected proof guides
The official 2027 H3 Mathematics (9820) syllabus explicitly names combinatorial arguments, the pigeonhole principle, symmetry, proof by cases and problem-solving heuristics. This article uses Ramsey’s theorem as original enrichment built on those capabilities.
Continue with induction and the pigeonhole principle, proof by cases and symmetry, and double counting. The lesson is to make a complicated set of pairwise possibilities yield to one carefully chosen local argument.