Pigeon-Hole Principle ( PHP ) A Simple Idea in Combinatorics But Very Powerfull and Usefull

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Theorem Let n and k be positif integer, and n > k. Suppose we have to place n identical balls into k identical boxes. Then there will be at least one box in which we place at least two balls.

For warmup we begin with simple problems without solution Among three persons, there are two of the same sex. Among 13 persons, there are two born in the same month. Among 400 persons, there are a least two have the same birthday.

Example 1 In tournament with n players , everybody plays with everybody else exactly once. Prove that during the game there are always two players who have played the same number of game.

Proof The difficulties in proving with pigeon-hole principle is the identification of the ball and the box . Then the question is . . . What is the box? What is the ball?

Proof The number of players is n. And the number of possibilities for the number of game finished by any one player are 0, 1, 2, . . . , n-1. But the fact , however 0 and n-1 cannot both occur among the numbers of games finished by the players at any one time.

Proof Then, take The number of players ( n ) as balls and The number of game finished by one player ( n-1 ) as boxes. By PHP we get, there are two players that have the same number of the game.

Example 2 (ONMIPA 2010) Let S is a set of n positive integer. Prove that there exist an element in S or sum of elements in S that divisible by n.

{ a1, a2, a3, . . ., an } not necessary distinct Proof Suppose the elements in S are { a1, a2, a3, . . ., an } not necessary distinct If there is an element in S divisible by n, we are done. Otherwise, none of them is divisible by n. Consider the Set T = { s1, s2, s3, . . ., sn } Where, sn = a1 + a2 + . . . + an

Proof If there is i, 1≤ i ≤ n such that si is divisible by n, we are finished. So, assum all of si is not divisible by n. Then we get, the remainder if si divided by n are 1, 2, 3, . . . , n-1 There are n-1 possibilities but as we have n numbers, by PHP there are two numbers that have the same remainder.

Proof Without loss of generality, assum these numbers are si and sj where i < j. Then sj – si = ai + ai+1 + . . . + aj is divisible by n as desired.

Example 3 (ONMIPA 2009) We select n+1 different integers from the set { 1, 2, 3, … 2n }. Prove that there always be two among the selected integers such that one is divisible by other.

Hint. Write the numbers 1, 2, 3, . . . , 2n in the form 2p.a where, a is odd number and p ≥ 0, p integer Then used PHP to prove the statement.

Problem Solving can be learn ONLY by Solving Problems REMEMBER Thanks

This presentation is wrote by Tutur Widodo, In April 2010 Hi…hi…hi…!