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DOMINOES IN PROGRESSION.

(Problems Concerning Games.)
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.


Answer:

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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