Pick one person, say Alex. Among the other 5, either at least 3 are friends with Alex or at least 3 are strangers to Alex. By focusing on that group of 3, you apply the pigeonhole principle again to force a monochromatic triangle in the friendship graph.
Happy counting! 🧩 Do you have a favorite Olympiad combinatorics problem or a clever solution that blew your mind? Share it in the comments below! Olympiad Combinatorics Problems Solutions
Count the total number of handshakes (sum of all handshake counts divided by 2). The sum of degrees is even. The sum of even degrees is even, so the sum of odd degrees must also be even. Hence, an even number of people have odd degree. Pick one person, say Alex
When a problem involves moves or transformations, look for what doesn’t change modulo 2, modulo 3, or some clever coloring. 3. Double Counting: Two Ways to Tell the Same Story One of the most elegant weapons in the Olympiad arsenal. Count the same set of objects in two different ways to derive an identity. Happy counting
Let’s break down the most common types of Olympiad combinatorics problems and the strategies to solve them. The principle is deceptively simple: If you put (n) items into (m) boxes and (n > m), at least one box contains two items.
When stuck, ask: “What’s the smallest/biggest/largest/minimal possible …?” 5. Graph Theory Modeling: Turn the Problem into Vertices & Edges Many combinatorial problems—about friendships, tournaments, networks, or matchings—are secretly graph problems.
At a party, some people shake hands. Prove that the number of people who shake an odd number of hands is even.