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Step-by-Step Solution
Step 1: Identify the letters and their frequency
The given word is "SMALL", which contains the letters S, M, A, L, L. Notice that the letter L is repeated twice, while S, M, and A appear once each.
Step 2: Arrange the letters in alphabetical order
Put the distinct letters in alphabetical order: A, L, L, M, S. We will count how many valid 5-letter permutations come before the word "SMALL" in dictionary (lexicographic) order.
Step 3: Count permutations starting with each letter that comes before 'S' in the first position
Words starting with A
The remaining letters to arrange are S, M, L, L. The total permutations of these 4 letters (with L repeated twice) is
$ \frac{4!}{2!} = 12 $.
Words starting with L
Since we used one L, the remaining letters to arrange are S, M, A, L (only one L left). The total permutations of these 4 letters is
$ 4! = 24 $.
Words starting with M
The remaining letters to arrange are S, A, L, L. The total permutations is
$ \frac{4!}{2!} = 12 $.
So far, the count of words up to (but not including) those starting with S is:
$ 12 + 24 + 12 = 48 $.
Step 4: Count permutations starting with 'S' up to "SMALL"
Now consider words that begin with S. The remaining letters are M, A, L, L. Order these letters alphabetically: A, L, L, M.
Second letter A (SA...)
Then the remaining letters are M, L, L. The number of permutations is
$ \frac{3!}{2!} = 3 $
(because there are two L's). These cover words #49, #50, #51 if we continue the counting.
Second letter L (SL...)
Use one L, leaving M, A, L. The number of permutations is
$ 3! = 6 $.
These cover words #52 through #57.
NEXT: Second letter M (SM...)
The next word in order, after counting the above, is "SMALL", which is the 58th word.
Step 5: Conclusion
Hence, the word "SMALL" is in the 58th position when all 5-letter arrangements of the letters of "SMALL" are listed in dictionary order.
