## BACHET'S SQUARE.

(

Chessboard Problems)

One of the oldest card puzzles is by Claude Caspar Bachet de Meziriac,

first published, I believe, in the 1624 edition of his work. Rearrange

the sixteen court cards (including the aces) in a square so that in no

row of four cards, horizontal, vertical, or diagonal, shall be found two

cards of the same suit or the same value. This in itself is easy enough,

but a point of the puzzle is to find in how many different ways this may

be done. The eminent French mathematician A. Labosne, in his modern

edition of Bachet, gives the answer incorrectly. And yet the puzzle is

really quite easy. Any arrangement produces seven more by turning the

square round and reflecting it in a mirror. These are counted as

different by Bachet.

Note "row of four cards," so that the only diagonals we have here to

consider are the two long ones.

## Answer:

[Illustration: 1]

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]

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.