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Amusements in Mathematics
Henry Ernest Dudeney
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next above it and the cell in the bottom right-hand corner. The answer for six couples will
be the same as the number of ways in which you can place six rooks (not using the
cancelled cells) so that no rook shall ever attack another rook. It will be found that the six
rooks may be placed in eighty different ways, which agrees with the above table.
262.—THOSE FIFTEEN SHEEP.
A certain cyclopædia has the following curious problem, I am told: "Place fifteen sheep
in four pens so that there shall be the same number of sheep in each pen." No answer
whatever is vouchsafed, so I thought I would investigate the matter. I saw that in dealing
with apples or bricks the thing would appear to be quite impossible, since four times any
number must be an even number, while fifteen is an odd number. I thought, therefore,
that there must be some quality peculiar to the sheep that was not generally known. So I
decided to interview some farmers on the subject. The first one pointed out that if we put
one pen inside another, like the rings of a target, and placed all sheep in the smallest pen,
it would be all right. But I objected to this, because you admittedly place all the sheep in
one pen, not in four pens. The second man said that if I placed four sheep in each of three
pens and three sheep in the last pen (that is fifteen sheep in all), and one of the ewes in
the last pen had a lamb during the night, there would be the same number in each pen in
the morning. This also failed to satisfy me.
The third farmer said, "I've got four hurdle pens down in one of my fields, and a small
flock of wethers, so if you will just step down with me I will show you how it is done."
The illustration depicts my friend as he is about to demonstrate the matter to me. His
lucid explanation was evidently that which was in the mind of the writer of the article in
the cyclopædia. What was it? Can you place those fifteen sheep?
263.—KING ARTHUR'S KNIGHTS.
King Arthur sat at the Round Table on three successive evenings with his knights—
Beleobus, Caradoc, Driam, Eric, Floll, and Galahad—but on no occasion did any person
have as his neighbour one who had before sat next to him. On the first evening they sat in
alphabetical order round the table. But afterwards King Arthur arranged the two next
sittings so that he might have Beleobus as near to him as possible and Galahad as far
away from him as could be managed. How did he seat the knights to the best advantage,
remembering that rule that no knight may have the same neighbour twice?
264.—THE CITY LUNCHEONS.
Twelve men connected with a large firm in the City of London sit down to luncheon
together every day in the same room. The tables are small ones that only accommodate
two persons at the same time. Can you show how these twelve men may lunch together
on eleven days in pairs, so that no two of them shall ever sit twice together? We will
represent the men by the first twelve letters of the alphabet, and suppose the first day's
pairing to be as follows—
(A B) (C D) (E F) (G H) (I J) (K L).