Now, the present puzzle is a variation of the above. It is simply this: Instead of using the fifteen letters given, the reader is allowed to select any fifteen different letters of the alphabet that he may prefer. Then construct thirty-five groups in accordance with the conditions, and show as many good English words as possible. 272.--THE NINE SCHOOLBOYS. This is a new and interesting companion puzzle to the "Fifteen Schoolgirls" (see solution of No. 269), 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 week 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 ]