(related to Problem: Bachet's Square)
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 — although 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 $24 \times 24 = 576$ 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 $2 \times 576 = 1,152,$ counting reflections and reversals as different. Put in another manner, the pairs in the top row may be written in $16 \times 9 \times 4 \times 1 = 576$ different ways, and the square then completed in $2$ ways, making $1,152$ ways in all.
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