Page:Amusements in mathematics.djvu/234

222 entirely convinced myself of the truth of the statement. There are at least five different ways of arranging the queens so as to leave eleven squares unattacked.

317.—A PUZZLE WITH PAWNS.

pawns may be placed so that no three shall be in a straight line in any possible direction, as in the diagram. We regard, as the conditions required, the pawns as mere points on a plane.

318.—LION-HUNTING.

There are 6,480 ways of placing the man and the lion, if there are no restrictions whatever except that they must be on different spots. This is obvious, because the man may be placed on any one of the 81 spots, and in every case there are 80 spots remaining for the lion; therefore 81×80=6,480. Now, if we deduct the number of ways in which the lion and the man may be placed on the same path, the result must be the number of ways in which they will not be on the same path. The number of ways in which they may be in line is found without much difficulty to be 816. Consequently, 6,480-816=5,664, the required answer.

The general solution is this: $$\frac{1}{2}(n-1) (3n^2-n+2)$$. This is, of course, equivalent to saying that if we call the number of squares on the side of a "chessboard" $$n$$, then the formula shows the number of ways in which two bishops may be placed without attacking one another. Only in this case we must divide by two, because the two bishops have no distinct individuality, and cannot produce a different solution by mere exchange of places.

319.— THE KNIGHT-GUARDS.

The smallest possible number of knights with which this puzzle can be solved is fourteen.