## 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.