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.

This involves counting the same set or property in two different ways to establish an identity or inequality.

In a competition, there are ( m ) contestants and ( n ) judges, where ( n \ge 3 ) is odd. Each judge rates each contestant as either “pass” or “fail”. Suppose for any two judges, their ratings coincide for at most ( k ) contestants. Prove that [ \frackm \ge \fracn-12n. ]

Start with “overcount” then divide by the number of equivalent representations.

Many combinatorial problems—about friendships, tournaments, networks, or matchings—are secretly graph problems.

This principle states that if ( n ) items are placed into ( m ) boxes and ( n > m ), at least one box contains two items. While trivial at first glance, its applications are profound.