(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.
This eBook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this edition or online at http://www.gutenberg.org. If you are not located in the United States, you'll have to check the laws of the country where you are located before using this ebook.