AB CD EF GH IJ KL
AE DL GK FI CB HJ
AG LJ FH KC DE IB
AF JB KI HD LG CE
AK BE HC IL JF DG
AH EG ID CJ BK LF
AI GF CL DB EH JK
AC FK DJ LE GI BH
AD KH LB JG FC EI
AL HI JE BF KD GC
AJ IC BG EK HL FD
Note that in every column (except in the case of the A's) all the
letters descend cyclically in the same order, B, E, G, F, up to J, which
is followed by B.
265.--A PUZZLE FOR CARD-PLAYERS.
In the following solution each of the eleven lines represents a sitting,
each column a table, and each pair of letters a pair of partners.
A B -- I L | E J -- G K | F H -- C D
A C -- J B | F K -- H L | G I -- D E
A D -- K C | G L -- I B | H J -- E F
A E -- L D | H B -- J C | I K -- F G
A F -- B E | I C -- K D | J L -- G H
A G -- C F | J D -- L E | K B -- H I
A H -- D G | K E -- B F | L C -- I J
A I -- E H | L F -- C G | B D -- J K
A J -- F I | B G -- D H | C E -- K L
A K -- G J | C H -- E I | D F -- L B
A L -- H K | D I -- F J | E G -- B C
It will be seen that the letters B, C, D ...L descend cyclically. The
solution given above is absolutely perfect in all respects. It will be
found that every player has every other player once as his partner and
twice as his opponent.
266.--A TENNIS TOURNAMENT.
Call the men A, B, D, E, and their wives a, b, d, e. Then they may play
as follows without any person ever playing twice with or against any
other person:--
First Court. Second Court.
1st Day | A d against B e | D a against E b
2nd Day | A e " D b | E a " B d
3rd Day | A b " E d | B a " D e
It will be seen that no man ever plays with or against his own wife--an
ideal arrangement. If the reader wants a hard puzzle, let him try to
arrange eight married couples (in four courts on seven days) under
exactly similar conditions. It can be done, but I leave the reader in
this case the pleasure of seeking the answer and the general solution.
267.--THE WRONG HATS.
The number of different ways in which eight persons, with eight hats,
can each take the wrong hat, is 14,833.
Here are the successive solutions for any number of persons from one to
eight:--
1 = 0
2 = 1
3 = 2
4 = 9
5 = 44
6 = 265
7 = 1,854
8 = 14,833
To get these numbers, multiply successively by 2, 3, 4, 5, etc. When the
multiplier is even, add 1; when odd, deduct 1. Thus, 3 x 1 - 1 = 2; 4 x
2 + 1 = 9; 5 x 9 - 1 = 44; and so on. Or you can multiply the sum of the
number of ways for n - 1 and n - 2 persons by n - 1, and so get the
solution for n persons. Thus, 4(2 + 9) = 44; 5(9 + 44) = 265; and so on.
268.--THE PEAL OF BELLS.
The bells should be rung as follows:--
Public-domain text, read in full here on John Shaqi.
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