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.

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.