(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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