KenKen Conundrum – Solution of Permutations on a 4×4 Grid

by Bill Lombard on 17 January, 2009

kenken-latin-squares-3-and-4

We have the solution to the KenKen Conundrum posed recently. I asked my students to solve the following problem, and several of them succeeded. The original question was, “How many different arrangements (permutations) of the integers 1, 2, 3, 4 are possible in a 4×4 grid such as KenKen?

Several people commented to the post, and I also received an email from Tetsuya Miyamoto, the creator of KenKen. The problem posed for my students is a classical permutation exercise, and to make it a little more accessible for solution, I restricted the 4×4 grid to only those arrangements showing 1, 2, 3, 4 as the top row. According to an article on Latin Squares in Wikipedia, there are 576 ways to arrange 1, 2, 3, 4 in a 4×4 grid. Since 1, 2, 3, 4 is one permutation of these integers, and 242 = 576, then there must be 24 ways to complete the grid. Here is the solution for the homework exercise; the digits are arranged in numerical order from top to bottom.

kenken-4x4-latin-squares

My students noticed a lot of symmetry among the grids; a neat extension of this exercise is to color all the “1″ squares one color, the “2″ squares a different color, and so on. It makes a great wall display.

A simpler homework or classwork exercise is arranging all the permutations of the digits 1, 2, 3 for a 3×3 grid. There are only 12 ways to do this. It seems remarkable that going from a 3×3 to a 4×4 grid should permit so many more arrangements.

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