eek days
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:--

A B C
D E F
G H I

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.

273.--THE ROUND TABLE.

Seat the same n persons at a round table on

(n - 1)(n - 2)
--------------
2

occasions so that no person shall ever have the same two neighbours
twice. This is, of course, equivalent to saying that every person must
sit once, and once only, between every possible pair.

274.--THE MOUSE-TRAP PUZZLE.

[Illustration

6 20 2 19
13 21
7 5
3 18
17 8
15 11
14 16
1 9
10 4 12

]

This is a modern version, with a difference, of an old puzzle of the
same name. Number twenty-one cards, 1, 2, 3, etc., up to 21, and place
them in a circle in the particular order shown in the illustration.
These cards represent mice. You start from any card, calling that card
"one," and count, "one, two, three," etc., in a clockwise direction, and
when your count agrees with the number on the card, you have made a
"catch," and you remove the card. Then start at the next card, calling
that "one," and try again to make another "catch." And so on. Supposing
you start at 18, calling that card "one," your first "catch" will be 19.
Remove 19 and your next "catch" is 10. Remove 10 and your next "catch"
is 1. Remove the 1, and if you count up to 21 (you must never go
beyond), you cannot make another "catch." Now, the ideal is to "catch"
all the twenty-one mice, but this is not here possible, and if it were
it would merely require twenty-one different trials, at the most, to
succeed. But the reader may make any two cards change places before he
begins. Thus, you can change the 6 with the 2, or the 7 with the 11, or
any other pair. This can be done in several ways so as to enable you to
"catch" all the twenty-one mice,