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Instead of using the Klein group, I used the symbols 1 and 0, and added mod 2. Assuming this is legitimate, I could derive the same result about possible positions when one marble is left. For example, if the center hole has a 1 then there will be an odd number of 1's in the remaining holes, meaning that the sum is odd. All holes labeled 0 are eliminated as possibilities. Using the same type of symmetry arguments, there are only 2 types of position that do not have 0.