This position may be arrived at in as few as forty-six moves, as follows:  16—­21, 16—­22, 16—­23, 17—­16, 12—­17, 12—­22, 12—­21,7—­12, 7—­17, 7—­22, 11—­12, 11—­17, 2—­7, 2—­12, 6—­11, 8—­7, 8—­6, 13—­8, 18—­13, 11—­18, 2—­17, 18—­12, 18—­7, 18—­2, 13—­7, 3—­8, 3—­13, 4—­3, 4—­8, 9—­4, 9—­3, 14—­9, 14—­4, 19—­14, 19—­9, 3—­14, 3—­19, 6—­12, 6—­13, 6—­14, 17—­11, 12—­16, 2—­12, 7—­17, 11—­13, 16—­18 = 46 moves.  I am, of course, not able to say positively that a solution cannot be discovered in fewer moves, but I believe it will be found a very hard task to reduce the number.

345.—­THE TWO PAWNS.

Call one pawn A and the other B. Now, owing to that optional first move, either pawn may make either 5 or 6 moves in reaching the eighth square.  There are, therefore, four cases to be considered:  (1) A 6 moves and B 6 moves; (2) A 6 moves and B 5 moves; (3) A 5 moves and B 6 moves; (4) A 5 moves and B 5 moves.  In case (1) there are 12 moves, and we may select any 6 of these for A. Therefore 7x8x9x10x11x12 divided by 1x2x3x4x5x6 gives us the number of variations for this case—­that is, 924.  Similarly for case (2), 6 selections out of 11 will be 462; in case (3), 5 selections out of 11 will also be 462; and in case (4), 5 selections out of 10 will be 252.  Add these four numbers together and we get 2,100, which is the correct number of different ways in which the pawns may advance under the conditions. (See No. 270, on p. 204.)

346.—­SETTING THE BOARD.

The White pawns may be arranged in 40,320 ways, the White rooks in 2 ways, the bishops in 2 ways, and the knights in 2 ways.  Multiply these numbers together, and we find that the White pieces may be placed in 322,560 different 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 x 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 a board of n squared squares contains ((n squared + n) squared)/4 rectangles, of which (2n cubed + 3n squared + n)/6 are squares and (3n^4 + 2n cubed — 3n squared — 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 triangular 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.