It will be seen that I have played six dominoes, in the illustration, in accordance with the ordinary rules of the game, $4$ against $4,$ $1$ against $1,$ and so on, and yet the sum of the spots on the successive dominoes, $4, 5, 6, 7, 8, 9,$ are in arithmetical progression; that is, the numbers taken in order have a common difference of $1.$ In how many different ways may we play six dominoes, from an ordinary box of twenty-eight, so that the numbers on them may lie in arithmetical progression? We must always play from left to right, and numbers in decreasing arithmetical progression (such as $9, 8, 7, 6, 5, 4)$ are not admissible.
Solutions: 1
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