This is a new and interesting companion puzzle to the "Fifteen Schoolgirls" (see the solution of Three Men in a Boat), and even in the simplest possible form in which I present it, there are unquestionable difficulties. Nine schoolboys walk out in triplets on the six weekdays so that no boy ever walks side by side with any other boy more than once. How would you arrange them?
If we represent them by the first nine letters of the alphabet, they might be grouped on the first day as follows:— \[\begin{array}{ccc} A&B&C\\ D&E&F\\ G&H&I \end{array}\]
Then A can never walk again side by side with $B,$ or $B$ with $C,$ or $D$ with $E,$ and so on. But $A$ can, of course, walk side by side with $C.$ It is here not a question of being together in the same triplet, but of walking side by side in a triplet. Under these conditions they can walk out on six days; under the "Schoolgirls" conditions they can only walk on four days.
Solutions: 1
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