Let us use the letters A, K, Q, J, to denote ace, king, queen, jack; and D, S, H, C, to denote diamonds, spades, hearts, clubs.  In Diagrams 1 and 2 we have the two available ways of arranging either group of letters so that no two similar letters shall be in line—­though a quarter-turn of 1 will give us the arrangement in 2.  If we superimpose or combine these two squares, we get the arrangement of Diagram 3, which is one solution.  But in each square we may put the letters in the top line in twenty-four different ways without altering the scheme of arrangement.  Thus, in Diagram 4 the S’s are similarly placed to the D’s in 2, the H’s to the S’s, the C’s to the H’s, and the D’s to the C’s.  It clearly follows that there must be 24x24 = 576 ways of combining the two primitive arrangements.  But the error that Labosne fell into was that of assuming that the A, K, Q, J must be arranged in the form 1, and the D, S, H, C in the form 2.  He thus included reflections and half-turns, but not quarter-turns.  They may obviously be interchanged.  So that the correct answer is 2 x 576 = 1,152, counting reflections and reversals as different.  Put in another manner, the pairs in the top row may be written in 16 x 9 x 4 x 1 = 576 different ways, and the square then completed in 2 ways, making 1,152 ways in all.

305.—­THE THIRTY-SIX LETTER BLOCKS.

I pointed out that it was impossible to get all the letters into the box under the conditions, but the puzzle was to place as many as possible.

This requires a little judgment and careful investigation, or we are liable to jump to the hasty conclusion that the proper way to solve the puzzle must be first to place all six of one letter, then all six of another letter, and so on.  As there is only one scheme (with its reversals) for placing six similar letters so that no two shall be in a line in any direction, the reader will find that after he has placed four different kinds of letters, six times each, every place is occupied except those twelve that form the two long diagonals.  He is, therefore, unable to place more than two each of his last two letters, and there are eight blanks left.  I give such an arrangement in Diagram 1.

[Illustration:  1]

[Illustration:  2]

The secret, however, consists in not trying thus to place all six of each letter.  It will be found that if we content ourselves with placing only five of each letter, this number (thirty in all) may be got into the box, and there will be only six blanks.  But the correct solution is to place six of each of two letters and five of each of the remaining four.  An examination of Diagram 2 will show that there are six each of C and D, and five each of A, B, E, and F. There are, therefore, only four blanks left, and no letter is in line with a similar letter in any direction.

306.—­THE CROWDED CHESSBOARD.

[Illustration]