Problem #DES-3

Problem

Today we will be solving problems using the pigeonhole principle. What is it? Simply put, we are asked to place pigeons in pigeonholes, but the number of pigeons is larger than the number of pigeonholes. No matter how we try to do that, at least one pigeonhole will have to contain at least 2 pigeons. By ”pigeonholes” we can mean any containers and by ”pigeons” we mean any items, which are placed in these containers. This is a simple observation, but it is helpful in solving some very difficult problems. Some of these problems might seem obvious or intuitively true. Pigeonhole principle is a useful way of formalising things that seem intuitive but can be difficult to describe mathematically.

There is also a more general version of the pigeonhole principle, where the number of pigeons is more than \(k\) times larger than the number of pigeonholes. Then, by the same logic, there will be one pigeonhole containing \(k+1\) pigeons or more.

A formal way to prove the pigeonhole principle is by contradiction - imagine what would happen if each pigeonhole contained only one pigeon? Well, the total number of pigeons could not be larger than the number of pigeonholes! What if each pigeonhole had \(k\) pigeons or fewer? The total number of pigeons could be \(k\) times larger than the number of pigeonholes, but not greater than that.