(related to Problem: The Nine Schoolboys)
The boys can walk out as follows:—
|1st Day||2nd Day||3rd Day||4th Day||5th Day||6th Day|
|A B C||B F H||F A G||A D H||G B I||D C A|
|D E F||E I A||I D B||B E G||C F D||E H B|
|G H I||C G D||H C E||F I C||H A E||I G F|
Every boy will then have walked by the side of every other boy once and once only.
Dealing with the problem generally, $12n + 9$ boys may walk out in triplets under the conditions on $9n + 6$ days, where n may be naught or any integer. Every possible pair will occur once. Call the number of boys $m.$ Then every boy will pair $m-1$ times, of which $(m- 1)/4$ times he will be in the middle of a triplet and $(m- 1)/2$ times on the outside. Thus, if we refer to the solution above, we find that every boy is in the middle twice (making $4$ pairs) and four times on the outside (making the remaining $4$ pairs of his $8$). The reader may now like to try his hand at solving the two next cases of $21$ boys on $15$ days, and $33$ boys on $24$ days. It is, perhaps, interesting to note that a school of $489$ boys could thus walk out daily in one leap year, but it would take $731$ girls (referred to in the solution to Three Men in a Boat) to perform their particular feat by a daily walk in a year of $365$ days.
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