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.