(related to Problem: Dominoes In Progression)

There are twenty-three different ways. You may start with any domino, except the $4—4$ and those that bear a $5$ or $6,$ though only certain initial dominoes may be played either way round. If you are given the common difference and the first domino is played, you have no option as to the other dominoes. Therefore all I need do is to give the initial domino for all the twenty-three ways, and state the common difference. This I will do as follows:—

With a common difference of $1,$ the first domino may be either of these: $0—0,$ $0—1,$ $1—0,$ $0—2,$ $1—1,$ $2—0,$ $0—3,$ $1—2,$ $2—1,$ $3—0,$ $0—4,$ $1—3,$ $2—2,$ $3—1,$ $1—4,$ $2—3,$ $3—2,$ $2—4,$ $3—3,$ $3—4.$ With a difference of $2,$ the first domino may be $0—0,$ $0—2,$ or $0—1.$ Take the last case of all as an example. Having played the $0—1,$ and the difference being $2,$ we are compelled to continue with $1—2,$ $2—3,$ $3—4,$ $4—5,$ $5—6.$ There are three dominoes that can never be used at all. These are $0—5,$ $0—6,$ and $1—6.$ If we used a box of dominoes extending to $9—9,$ there would be forty different ways.

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Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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