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.
Solutions: 1
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