Page:Amusements in mathematics.djvu/244

ways. The Black pieces may, of course, be placed in the same number of ways. Therefore the men may be set up in 322,560×322,560 = 104,044,953,600 ways. But the point that nearly everybody overlooks is that the board may be placed in two different ways for every arrangement. Therefore the answer is doubled, and is 208,089,907,200 different ways.

347.—COUNTING THE RECTANGLES.

There are 1,296 different rectangles in all, 204 of which are squares, counting the square board itself as one, and 1,092 rectangles that are not squares. The general formula is that aboard of $n^{2}$ contains ${\frac {n^{2}+n)^{2}}{4}}$ rectangles, of which ${\frac {2n^{3}+3n^{2}+n}{6}}$ are squares and ${\frac {3n^{4}+2n^{3}-3n^{2}+2n}{12}}$ are rectangles that are not squares. It is curious and interesting that the total number of rectangles is always the square of the tri- angular number whose side is $n$.

348.—THE ROOKERY.

The answer involves the little point that in the final position the numbered rooks must be in numerical order in the direction contrary to that in which they appear in the original diagram, otherwise it cannot be solved. Play the rooks in the following order of their numbers. As there is never more than one square to which a rook can move (except on the final move), the notation is obvious— 5, 6, 7, 5, 6, 4, 3, 6, 4, 7, 5, 4, 7, 3, 6, 7, 3, 5, 4, 3, 1, 8, 3, 4, 5, 6, 7, 1, 8, 2, 1, and rook takes bishop, checkmate. These are the fewest possible moves — thirty-two. The Black king's moves are all forced, and need not be given.

349.—STALEMATE.

Working independently, the same position was arrived at by Messrs. S. Loyd, E. N. Frankenstein, W. H. Thompson, and myself. So the following may be accepted as the best solution possible to this curious problem:—

White.		Black.
1.	P-Q 4	1.	P—K 4
2.	Q—Q 3	2.	Q—R 5
3.	Q—KKt 3	3.	B—Kt 5 ch
4.	Kt—Q 2	4.	P—QR 4
5.	P—R 4	5.	P—Q 3
6.	P—R 3	6.	B—K 3
7.	R—R 3	7.	P—KB 4
8.	Q—R 2	8.	P—B 4
9.	R—KKt 3	9.	B—Kt 6
10.	P—QB 4	10.	P—B 5
11.	P—B 3	11.	P—K 5
12.	P—Q 5	12.	P—K6

And White is stalemated.

We give a diagram of the curious position arrived at. It will be seen that not one of White's pieces may be moved.

350.—THE FORSAKEN KING.

Play as follows : —

White.		Black.
1.	P to K 4th	1.	Any move
2.	Q to Kt 4th	2.	Any move except on KB file (a)
3.	Q to Kt 7th	3.	K moves to royal row
4.	B to Kt 5th	4.	Any move
5.	Mate in two moves
		If 3, K other than to royal row
4.	P to Q 4th	4.	Any move
5.	Mate in two moves
		(a) If 2, Any move on KB file
3.	Q to Q 7th	3.	K moves to royal row
4.	P to Q Kt 3rd	4.	Any move
5.	Mate in two moves
	If 3, K other than to royal row
4.	P to Q 4th	4.	Any move
5.	Mate in two moves

Of course, by "royal row" is meant the row on which the king originally stands at the beginning of a game. Though, if Black plays badly, he may, in certain positions, be mated in fewer moves, the above provides for every variation he can possibly bring about.

351.—THE CRUSADER.

White.		Black.
1.	Kt to QB 3rd	1.	P to Q 4th
2.	Kt takes QP	2.	Kt to QB 3rd
3.	Kt takes KP	3.	P to KKt 4th
4.	Kt takes B	4.	Kt to KB 3rd
5.	Kt takes P	5.	Kt to K 5th
6.	Kt takes Kt	6.	Kt to B 6th
7.	Kt takes Q	7.	R to KKt sq
8.	Kt takes BP	8.	R to KKt 3rd
9.	Kt takes P	9.	R to K 3rd
10.	Kt takes P	10.	Kt to Kt 8th
11.	Kt takes B	11.	R to R 6th
12.	Kt takes R	12.	P to Kt 4th
13.	Kt takes P (ch)	13.	K to B 2nd