Olympiad Combinatorics Problems Solutions Apr 2026

At a party, some people shake hands. Prove that the number of people who shake an odd number of hands is even.

Show that in any group of 6 people, there are either 3 mutual friends or 3 mutual strangers.

Take a classic problem like “Prove that in any set of 10 integers, there exist two whose difference is divisible by 9.” Apply the pigeonhole principle. You’ve just taken the first step into a larger world.

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

Whenever you see sums of numbers counting relationships, try counting the total number of pairs or triples in two ways. 4. Extremal Principle: Look at the Extreme Pick an object that maximizes or minimizes some quantity. Then show that if the desired condition isn’t met, you can find a contradiction by modifying that extreme object.

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.

In a tournament (every pair of players plays one game, no ties), prove there is a ranking such that each player beats the next player in the ranking. At a party, some people shake hands

But here’s the secret:

Color the board black and white in the usual pattern. A knight always moves from a black square to a white square and vice versa. For a closed tour, the knight must make an equal number of black and white moves, but there are 64 squares. Since 64 is even, a closed knight’s tour is possible in theory—but parity alone doesn’t guarantee it; it’s a starting point for deeper invariants.

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. Take a classic problem like “Prove that in

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.

If you’ve ever looked at an International Mathematical Olympiad (IMO) problem and felt your brain do a double backflip, chances are it was a combinatorics question. Unlike algebra or geometry, where formulas and theorems provide a clear roadmap, combinatorics problems often feel like puzzles wrapped in riddles.

This is equivalent to showing every tournament has a Hamiltonian path. Use induction: Remove a vertex, find a path in the remaining tournament, then insert the vertex somewhere.