The Pigeonhole Principle

Example 1 In a room of 13 people, 2 or more people have their birthday in the same month. Proof: (by contradiction) 1. Assume the room has 13 people and no 2 people have their birthday in the same month. 2. There must be at least 13 different months. 3. Statement 2. is false, so the assumption is false.

Example 2 If 41 balls are chosen from a set of red, white, blue, garnet, and gold colored balls, then at least 12 red balls, 15 white balls, 4 blue, 10 garnet, or 4 gold balls chosen. Proof by contradiction: (use DeMorgan’s law) 1. Assume that 41 balls are chosen from this set and that at most 11 red, 14 white, 3 blue, 9 garnet, and 3 gold balls are chosen. 2. At most 40 balls were chosen, a contradiction.

The pigeonhole principle Let m1, m2, … , mn be positive integers. If m1 + m2 + . . . + mn - n + 1 objects are put into n boxes, Then either the 1st box has at least m1, or the 2nd box has at least m2, or …, or the nth box has at least mn objects.

Proof by Contradiction Assume m1 + m2 + . . . + mn - n + 1 objects are put into n boxes, and the 1st box has at most m1 - 1, and the 2nd box has at most m2 - 1, and …, and the nth box has at most mn - 1 objects. Then, at most m1 + m2 + . . . + mn - n objects are in the boxes, a contradiction.

Another Form of Pigeonhole Principle If A is the average number of pigeons/hole, then some hole contains at least A pigeons and some hole contains at most A pigeons.

Intuition A A A Cannot have all holes contain less than the average. Cannot have all holes contain more than the average.

Proof of Alternate Principle By contradiction: Assume A is the average number of pigeons/hole. Assume every hole contains at most A - 1 pigeons or every hole contains at least A + 1 pigeons. 3. Let n denote the number of holes. 4. Assume every hole contains at most A - 1 pigeons. 5. All holes contain at most n(A - 1 ) < nA pigeons, a contradiction.

5. Assume every hole contains at least A + 1 pigeons. 6. All holes contain at least n(A + 1) > nA pigeons, a contradiction. 7. Therefore, some hole contains at least A pigeons and some hole contains at most A pigeons.

Applications of pigeonhole principle If n + 1 pigeons are distributed among n holes, then some hole contains at least 2 pigeons. If 2n + 1 pigeons are distributed among n holes, then some hole contains at least 3 pigeons. If kn + 1 pigeons are distributed among n holes, then some hole contains at least k + 1 pigeons. The average number of pigeons/hole = k + 1/n and  k + 1/n  = k + 1.

Applications ... In any group of 367 people, there must be at least 1 pair with the same birthday. If 4 different pairs of socks are scrambled in the drawer, only 5 socks need to be selected to guarantee finding a matching pair. In a group of 61 people, at least 6 were born in the same month.

Applications ... If 401 letters were delivered to 50 houses, then some house received at most 8 letters. If x1, x2, …, x8 are distinct integers, then some pair of these have the same remainder when divided by 7.

Applications ... Given a set of 7 distinct integers, there are 2 whose sum or difference is divisible by 10. Set this up so that there are 6 pigeon holes. Partitioning the integers into equivalence classes according to their remainder when divided by 10 yields too many classes. Consider: {[0]}, {[1],[9]}, {[2],[8]}, {[3],[7]}, {[4],[6]}, {[5]}. If 2 integers are in the same set either their difference is divisible by 10 or their sum is divisible by 10.

Applications ... Suppose Then, 50 chairs are arranged in a rectangular array of 5 rows and 10 columns. 41 students are seated randomly in the chairs (1 student/chair). Then, some row contains at least 9 students some row contains at most 8 students some column contains at least 5 students some column contains at most 4 students.

Applications ... A patient has 45 pills, with instructions to take at least 1 pill/day for 30 days. Prove: there is a period of consecutive days in which the patient takes a total of 14 pills. 1. Let ai be the number of pills taken through the end of the ith day. 2. 1  a1 < a2 < . . . < a30  45. 3. a1 + 14 < a2 + 14 < . . . < a30 + 14  45 + 14 = 59

5. 2 of these integers must be the same. 4. We have: 60 integers: a1, a2 , . . . , a30 , a1 +14, a2 +14 , . . . , a30 +14 59 holes. 5. 2 of these integers must be the same. 6. They cannot both be in a1, a2 , . . . , a30 . 7. They cannot both be in a1 +14, a2 +14 , . . . , a30 +14. 8. One is in each: ai = aj + 14, for some i and j. 9. For that i and j, ai - aj = 14. 10. That is, aj+1+aj+2 + . . . + ai = 14.