THE FIFTEEN DOMINOES.
(
Combination and Group Problems)
In this case we do not use the complete set of twenty-eight dominoes to
be found in the ordinary box. We dispense with all those dominoes that
have a five or a six on them and limit ourselves to the fifteen that
remain, where the double-four is the highest.
In how many different ways may the fifteen dominoes be arranged in a
straight line in accordance with the simple rule of the game that a
number must always be placed against a similar number--that is, a four
against a four, a blank against a blank, and so on? Left to right and
right to left of the same arrangement are to be counted as two different
ways.
Answer:
The reader may have noticed that at each end of the line I give is a
four, so that, if we like, we can form a ring instead of a line. It can
easily be proved that this must always be so. Every line arrangement
will make a circular arrangement if we like to join the ends. Now,
curious as it may at first appear, the following diagram exactly
represents the conditions when we leave the doubles out of the question
and devote our attention to forming circular arrangements. Each number,
or half domino, is in line with every other number, so that if we start
at any one of the five numbers and go over all the lines of the pentagon
once and once only we shall come back to the starting place, and the
order of our route will give us one of the circular arrangements for the
ten dominoes. Take your pencil and follow out the following route,
starting at the 4: 41304210234. You have been over all the lines once
only, and by repeating all these figures in this way,
41--13--30--04--42--21--10--02--23--34, you get an arrangement of the
dominoes (without the doubles) which will be perfectly clear. Take other
routes and you will get other arrangements. If, therefore, we can
ascertain just how many of these circular routes are obtainable from
the pentagon, then the rest is very easy.
Well, the number of different circular routes over the pentagon is 264.
How I arrive at these figures I will not at present explain, because it
would take a lot of space. The dominoes may, therefore, be arranged in a
circle in just 264 different ways, leaving out the doubles. Now, in any
one of these circles the five doubles may be inserted in 2^5 = 32
different ways. Therefore when we include the doubles there are 264 x 32
= 8,448 different circular arrangements. But each of those circles may
be broken (so as to form our straight line) in any one of 15 different
places. Consequently, 8,448 x 15 gives 126,720 different ways as the
correct answer to the puzzle.
[Illustration:
-----
| |
/ | |
/ -----
/ . .
----- . . -----
| | . . | o o |
| o | -.--------.--- | |
| | . . . | o o |
----- . . .. -----
. . . . /
----- .. -----
| o | . . |o |
| | --------- | o |
| o |. .| o|
----- -----
]
I purposely refrained from asking the reader to discover in just how
many different ways the full set of twenty-eight dominoes may be
arranged in a straight line in accordance with the ordinary rules of the
game, left to right and right to left of any arrangement counting as
different ways. It is an exceedingly difficult problem, but the correct
answer is 7,959,229,931,520 ways. The method of solving is very complex.
Informational
