In this section we look at a method of counting that uses a correspondence between two sets. We can think of one set as a set of objects and the other as a set of boxes. The name of the principle comes from thinking of a set of letters and pigeonholes, which is what they used to call rows of letterboxes. Though it is also fun to think about correspondences between pigeons and pigeonholes.
A function from one finite set to a smaller finite set cannot be one-to-one. There must be at least two elements in the domain that have the same image in the codomain.
The Pigeonhole Principle really just says that if there are more objects (pigeons) than boxes (pigeonholes) then there must be at least one box with more than one object.
In a drawer with 5 pairs of red socks and 5 pairs of blue socks, how many socks do you need to randomly pull out of the drawer to be certain that you have a pair?
It is important to note that pigeonhole problems are not about probability. We want to know when we are guaranteed to have a pair of socks, not when it is likely that we have a pair.
Since we are looking for sums of 11, our boxes can be pairs of numbers that sum to 11: \(\{1, 10\}, \{2, 9\}, \{3, 8\}, \{4, 7\}, \{5, 6\}\text{.}\) Then our objects are the numbers themselves. We need the number of numbers to be greater than the number of boxes, which is 5. Thus, if we choose 6 numbers from \(A\text{,}\) we are guaranteed to have 2 from the same subset. This means with 6 numbers we must have two numbers that sum to 11.
Let \(A=\{1, 2, 3, 4, 5, 6, 7, 8\}\text{.}\) If 5 integers are selected from \(A\text{,}\) must at least one pair have a sum of 9? If 4 integers are selected from \(A\text{,}\) must at least one pair have a sum of 9?
How many integers from 0 to 60 must you choose in order to get at least one that is odd? How many integers from 0 to 60 must you choose in order to get at least one that is even?
If there are \(n\) objects mapped to \(m\) boxes, and for some \(k\text{,}\) we have \(n>km\text{,}\) then at least one box contains \(k+1\) or more objects.
Let \(X, Y\) be finite sets with the same number of elements. Suppose \(f\) is a function from \(X\) to \(Y\text{.}\) Then \(f\) is one-to-one if and only if \(f\) is onto.
The Pigeonhole Principle can be used to prove the theorem, but it is more technical than we need. If you try a few examples of functions from a set of \(n\) elements to another set of \(n\) elements, you can probably convince yourself that a one-to-one function must be onto and vice versa.
What is the fewest number of digits you need to randomly select from \(A=\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}\) to guarantee you have either two even or two odd numbers?
What is the fewest number of digits you need to randomly select from \(A=\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}\) to guarantee you have two with the same remainder when divided by 3?
If \(n+1\) integers are chosen from the set \(\{1, 2, 3,\ldots 2n\}\text{,}\) where \(n\) is a positive integer, must at least one of them be even? Why?
Suppose six pairs of similar-looking boots are thrown together in a pile. How many individual boots must you pick to be sure of getting a matched pair? Why?