ALE FOE HOD BGN
    CAB HEN JOG KFM
    HAG GEM MOB BFH
    FAN KIN JEK DFL
    JAM HIM GCL LJH
    AID JIB FCJ NJD
    OAK FIG HCK MLN
    BED OIL MCD BLK
    ICE CON DGK

The above is the solution of a puzzle I gave in Tit-bits in the summer of 1896.  It was required to take the letters, A, B, C, D, E, F, G, H, I, J, K, L, M, N, and O, and with them form thirty-five groups of three letters so that the combinations should include the greatest number possible of common English words.  No two letters may appear together in a group more than once.  Thus, A and L having been together in ALE, must never be found together again; nor may A appear again in a group with E, nor L with E. These conditions will be found complied with in the above solution, and the number of words formed is twenty-one.  Many persons have since tried hard to beat this number, but so far have not succeeded.

More than thirty-five combinations of the fifteen letters cannot be formed within the conditions.  Theoretically, there cannot possibly be more than twenty-three words formed, because only this number of combinations is possible with a vowel or vowels in each.  And as no English word can be formed from three of the given vowels (A, E, I, and O), we must reduce the number of possible words to twenty-two.  This is correct theoretically, but practically that twenty-second word cannot be got in.  If JEK, shown above, were a word it would be all right; but it is not, and no amount of juggling with the other letters has resulted in a better answer than the one shown.  I should, say that proper nouns and abbreviations, such as Joe, Jim, Alf, Hal, Flo, Ike, etc., are disallowed.

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.