| 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 $A L E,$ 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 $J E K,$ 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.
Solutions: 1
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