Page:Amusements in mathematics.djvu/219

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solutions with a 9 in the corner. If, however, we substitute 8, the two corners in the same row and column may contain 0, 0, or 1, 1, or 0, 1, or 1, 0. In the case of B, ten different selections may be made for the fourth corner; but in each of the cases C, D, and E, only nine selections are possible, because we cannot use the 9. Therefore with 8 in the top left-hand corner there are 10+(3×9) = 37 different solutions. If we then try 7 in the corner, the result will be 10+27+40, or 77 solutions. With 6 we get 10+27+40+49=126; with 5, 10+27+40+49+54=180; with 4, the same as with 5, 55=235; with 3, the same as with 4, +52=287; with 2, the same as with 3, +45 = 332; with 1, the same as with 2, +34=366, and with nought in the top left-hand corner the number

of solutions will be found to be 10+27+40+49+54+55+52+45+34+19=385. As there is no other number to be placed in the top left-hand corner, we have now only to add these totals together thus, 10+37+77+126+180+235+287+332 + 366+385=2,035. We therefore find that the total number of ways in which tenants may occupy some or all of the eight villas so that there shall be always nine persons living along each side of the square is 2,035. Of course, this method must obviously cover all the reversals and reflections, since each comer in turn is occupied by every number in all possible combinations with the other two comers that are in line with it.

Here is a general formula for solving the puzzle: $$\frac{(n^2+3n+2)(n^2+3n+3).}{6}$$ Whatever may be the stipulated number of residents along each of the sides (which number is represented by $$n$$), the total number of different arrangements may be thus ascertained. In our particular case the number of residents was nine. Therefore (81+27+2)×(81+27+3) and the product, divided by 6, gives 2,035. If the number of residents had been 0, 1, 2, 3, 4, 5, 6, 7, or 8, the total arrangements would be 1, 7, 26, 70, 155, 301, 532, 876, or 1,365 respectively.

277.—COUNTER CROSSES. us first deal with the Greek Cross, There are just eighteen forms in which the numbers may be paired for the two arms. Here they are:—