-----
| |
/ | | \
/ ----- \
/ . . \
----- . . -----
| | . . | 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.
284.--THE CROSS TARGET.
[Illustration:
-- --
(CD)( )
-- --
(AE)(A )
-- -- -- -- -- --
(CE)(E )(A )(AB)(C )(D )
-- -- -- -- -- --
(D )( )(B )(E )(EB)( )
-- -- -- -- -- --
(C )(B )
-- --
( )(ED)
-- --
]
Twenty-one different squares may be selected. Of these nine will be of
the size shown by the four A's in the diagram, four of the size shown by
the B's, four of the size shown by the C's, two of the size shown by the
D's, and two of the size indicated by the upper single A, the upper
single E, the lower single C, and the EB. It is an interesting fact that
you cannot form any one of these twenty-one squares without using at
least one of the six circles marked E.
285.--THE FOUR POSTAGE STAMPS.
Referring to the original diagram, the four stamps may be given in the
shape 1, 2, 3, 4, in three ways; in the shape 1, 2, 5, 6, in six ways;
in the shape 1, 2, 3, 5, or 1, 2, 3, 7, or 1, 5, 6, 7, or 3, 5, 6, 7, in
twenty-eight ways; in shape 1, 2, 3, 6, or 2, 5, 6, 7, in fourteen ways;
in shape 1, 2, 6, 7, or 2, 3, 5, 6, or 1, 5, 6, 10, or 2, 5, 6, 9, in
fourteen ways. Thus there are sixty-five ways in all.
286.--PAINTING THE DIE.
The 1 can be marked on any one of six different sides. For every side
occupied by 1 we have a selection of four sides for the 2. For every
situation of the 2 we have two places for the 3. (The 6, 5, and 4 need
not be considered, as their positions are determined by the 1, 2, and
3.) Therefore 6, 4, and 2 multiplied together make 48 different
ways--the correct answer.
287.--AN ACROSTIC PUZZLE.
There are twenty-six letters in the alphabet, giving 325 different
pairs. Every one of these pairs may be reversed, making 650 ways. But
every initial letter may be repeated as the final, producing 26 other
ways. The total is therefore 676 different pairs. In other words, the
answer is the square of the number of letters in the alphabet.
288.--CHEQUERED BOARD DIVISIONS.
Public-domain text, read in full here on John Shaqi.
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