THE CITY LUNCHEONS.
(
Combination and Group Problems)
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).
Then give any pairing you like for the next day, say--
(A C) (B D) (E G) (F H) (I K) (J L),
and so on, until you have completed your eleven lines, with no pair ever
occurring twice. There are a good many different arrangements possible.
Try to find one of them.
Answer:
The men may be grouped as follows, where each line represents a day and
each column a table:--
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.
