Learn more about the Pigeonhole Principle and other powerful techniques for combinatorics problems
in our Intermediate Counting
& Probability textbook by USA Math Olympiad winner (and MIT PhD) David Patrick.
Difference between revisions of "Pigeonhole Principle"
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Also known as the [[Dirichlet]] box principle, | Also known as the [[Dirichlet]] box principle, | ||
the basic '''pigeonhole principle''' states that if there are <math>n</math> boxes, and more than <math>{n}</math> objects, then one box must contain two or more objects. | the basic '''pigeonhole principle''' states that if there are <math>n</math> boxes, and more than <math>{n}</math> objects, then one box must contain two or more objects. |
Revision as of 12:11, 5 July 2008
This is an AoPSWiki Word of the Week for July 4-July 10 |
Also known as the Dirichlet box principle, the basic pigeonhole principle states that if there are boxes, and more than objects, then one box must contain two or more objects.
The extended version of the pigeonhole principle states that for boxes, and more than objects, some box must contain at least objects. Although this theorem seems obvious, many challenging olympiad problems can be solved by applying the pigeonhole principle.
Contents
Examples
Introductory Problems
Intermediate Problems
- Show that in any group of five people, there are two who have an identical number of friends within the group. (Solution) (Mathematical Circles)
- Six distinct positive integers are randomly chosen between 1 and 2006, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of 5? (Solution)
Olympiad Problems
- Seven line segments, with lengths no greater than 10 inches, and no shorter than 1 inch, are given. Show that one can choose three of them to represent the sides of a triangle. (Solution) (Manhattan Mathematical Olympiad 2004)
- Prove that having 100 whole numbers, one can choose 15 of them so that the difference of any two is divisible by 7. (Solution) (Manhattan Mathematical Olympiad 2005)
- Prove that from any set of one hundred whole numbers, one can choose either one number which is divisible by 100, or several numbers whose sum is divisible by 100. (Solution) (Manhattan Mathematical Olympiad 2003)
- Prove that among any ten points located on a circle with diameter 5, there exist at least two at a distance less than 2 from each other. (Solution) (Japan 1997)
- Every point in a place is either red, green, or blue. Prove that there exists a rectangle in the plane such that all of its vertices are the same color. (Solution) (USAMTS Year 18 - Round 1 - Problem 4)
- There are 51 senators in a senate. The senate needs to be divided into committees such that each senator is on exactly one committee. Each senator hates exactly three other senators. (If senator A hates senator B, then senator B does 'not' necessarily hate senator A.) Find the smallest such that it is always possible to arrange the committees so that no senator hates another senator on his or her committee. (Solution) (Red MOP lecture 2006)
- Show that for any and positive integer , there exists a rational number with such that (Solution) (the classical Rational Approximation Theorem)