Page:Amusements in mathematics.djvu/226

ent, each containing five squares, with one square piece of four squares.

294.—THE CHESSBOARD SENTENCE.

The pieces may be fitted together, as shown in the illustration, to form a perfect chessboard.

295.—THE EIGHT ROOKS.

Obviously there must be a rook in every row and every column. Starting with the top row, it is clear that we may put our first rook on any one of eight different squares. Wherever it is placed, we have the option of seven squares for the second rook in the second row. Then we have six squares from which to select the third row, five in the fourth, and so on. Therefore the number of our difierent ways must be 8×7×6×5×4×3×2×1=40,320 (that is |8), which is the correct answer.

How many ways there are if mere reversals and reflections are not counted as different has not yet been determined; it is a difficult problem. But this point, on a smaller square, is considered in the next puzzle.

296.—THE FOUR LIONS.

There are only seven different ways under the conditions. They are as follows: 1 2 3 4, 1 2 4 3, 1 3 2 4, 1 3 4 2, 1 4 3 2, 2 1 4 3, 2 4 1 3. Taking the last example, this notation means that we place a lion in the second square of first row, fourth square of second row, first square of third row, and third square of fourth row. The first example is, of course, the one we gave when setting the puzzle.

297.—BISHOPS—UNGUARDED.

This cannot be done with fewer bishops than eight, and the simplest solution is to place the bishops in line along the fourth or fifth row of

the board (see diagram). But it will be noticed that no bishop is here guarded by another, so we consider that point in the next puzzle.

298.—BISHOPS—GUARDED.

This puzzle is quite easy if you first of all give it a httle thought. You need only consider squares of one colour, for whatever can be done in the case of the white squares can always be repeated on the black, and they are here quite independent of one another. This equality, of course, is in consequence of the fact that the number of squares on an ordinary chessboard, sixty-four, is an even number. If a square chequered board has an odd number of squares, then there will always be one more square of one colour than of the other.

Ten bishops are necessary in order that every square shall be attacked and every bishop guarded by another bishop. I give one way of arranging them in the diagram. It will be noticed that the two central bishops in the group