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.
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Difference between revisions of "Pigeonhole Principle"

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=== Pigeonhole Principle ===
 
=== Pigeonhole Principle ===
 
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Also known as [[Dirichlet principle]],
The basic pigeonhole principle says that if there are <math>n</math> holes, and <math>n+k</math> pigons (k>1), then one hole MUST contain two or more pigeons. The extended version of the pigeonhole principle states that for n holes, and <math>{nk+j}</math> pigeons, j>1, some hole must contain k+1 pigeons. If you see a problem with the numbers n, and nk+1, think about pigeonhole.
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the basic pigeonhole principle says 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 extended version of the pigeonhole principle states that for n boxes, and more than <math>{nk}</math> objects, some box must contain at least <math>k+1</math> objects. If you see a problem with the numbers <math>{n}</math>, and <math>nk+1</math>, think about pigeonhole.
  
 
=== Examples ===
 
=== Examples ===
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*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.(Japan 1997)
 
*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.(Japan 1997)
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*Show that for any <math>{x\in\mathbb R}</math> and positive integer <math>{n}</math>, there exists a rational number <math>{\frac pq}</math> with <math>1\le q\le n</math> such that <math>\left|x-\frac pq\right|<\frac 1{nq}</math> (the classical Rational Approximation Theorem)

Revision as of 18:41, 18 June 2006

Pigeonhole Principle

Also known as Dirichlet principle, the basic pigeonhole principle says that if there are $n$ boxes, and more than ${n}$ objects, then one box MUST contain two or more objects. The extended version of the pigeonhole principle states that for n boxes, and more than ${nk}$ objects, some box must contain at least $k+1$ objects. If you see a problem with the numbers ${n}$, and $nk+1$, think about pigeonhole.

Examples

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You could paste in these... (maybe, just a suggestion)

  • Show that in any group of five people, there are two who have an identical number of friends within the group. (Mathematical Circles)


  • 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. (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. (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. (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.(Japan 1997)
  • Show that for any ${x\in\mathbb R}$ and positive integer ${n}$, there exists a rational number ${\frac pq}$ with $1\le q\le n$ such that $\left|x-\frac pq\right|<\frac 1{nq}$ (the classical Rational Approximation Theorem)