"Mary, the wife of John Goffe of Rochester, being afflicted with a long illness, removed to her father's house at West Mulling, about nine miles from her own. There she died on 4th June, this present year, 1691. "The day before her departur... Read more of The Dying Mother {101} at Scary Stories.caInformational Site Network Informational
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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.










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