(This is because permuting rows and permuting columns doesn t change anything.) Therefore, for this case we may assume that 6ħ every single column has exactly 2 blue squares and 2 white squares. First, note that the argument above would hold even if we assumed that another column had 3 blue squares. Case 2: 2 blue squares in the first column. This means that we ll have a subboard all of whose corners are white, as required. Indeed, there are only 4 ways to color in the first 3 rows of a column by using at most 1 blue square here they are: By the pigeonhole principle, since there are only 4 potential colorings, and 6 columns to color in, some two columns will agree on the first 3 rows. As noted above, if any of the remaining columns have blue squares in 2 of those rows, we d have a subboard with matching corners for example, if column 5 has 2 blue squares in the first 3 rows, here s what can happen: Therefore, we need to assume that each of the remaining 6 columns each have at least 2 white squares in the first 3 rows here s an example of such a coloring: (note that for now, we re ignoring the fourth row, hence the light blue squares) As should be clear, we re now having an issue with having a subboard all of whose corners are white. Since permuting the rows doesn t change anything, let s assume that the first column has blue squares in rows 1, 2 and 3. Case 1: 3 blue squares in the first column. Either the first column has 2 of each color, or we have at least 3 squares of 5Ħ one color (either blue or white.) Since there s no difference between blue and white in the context of the problem, let s start with the case where we have at least 3 blue squares in the first column. Let s consider the first column of the checkerboard. Let us proceed by contradiction: assume that there exists a coloring such that no subboard has all four corners of the same color. Note that we get a subboard whose corners are the same color if two columns agree on two squares: for example, if column 2 and column 5 both have blue squares in the 1nd and 4th rows, then we d get the following (the squares for which we don t know the colors are colored in light blue): As another example, in the coloring given in the question, columns 1 and 6 both have blue squares in the 3rd and 4th row, which similarly leads to a desired subboard. Here s an example of a coloring: see if you can find a subboard with four corners of the same color! Solution: Let us first discuss how to approach this. Show that there is a subboard all of whose corners are blue or all of whose corners are white. Here s an example of a subboard, with squares shaded in red: Now, suppose that each of the 28 squares is colored either blue or white. A subboard of a checkerboard is a board you can cut-out of the checkerboard by only taking the squares which are between a specified pair of rows and a specified pair of columns. Show that if we take n + 1 numbers from the set then a 1 + a 3 + a 4 + a 5 = a 3 + a 5 + a 7 + a 8 a 1 + a 4 = a 7 + a 8 Thus, this will allow us to find two disjoint subsets of the 10 numbers with the same sum. Together we will work through countless problems and see how the pigeonhole principle is such a simple but powerful tool in our study of combinatorics.1 Pigeonhole Principle Solutions 1. Consequently, using the extended pigeonhole principle, the minimum number of students in the class so that at least six students receive the same letter grade is 26.
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