Page:Amusements in mathematics.djvu/222

210 dominoes. Take your pencil and follow out the following route, starting at the 4: 41 304210234. 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 264×32 = 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, 8,448×15 gives 126,720 different ways as the correct answer to the puzzle.

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.

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.

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 i, 5, 6, 10, or 2, 5, 6, 9, in fourteen ways. Thus there are sixty-five ways in all.

286.—PAINTING THE DIE.

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.

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.

are 255 different ways of cutting the board into two pieces of exactly the same size