If all the words (with or without meaning) having five letters, formed using the letters of the word $$SMALL$$ and arranged as in a dictionary; then the position of the word $$SMALL$$ is

We have the five-letter multiset $$\{S,\,M,\,A,\,L,\,L\}$$ in which the letter $$L$$ is repeated twice. Our task is to arrange all distinct five-letter words obtainable from these letters in ordinary dictionary (lexicographic) order and then locate the position of the specific word $$SMALL$$ among them.

Whenever some letters repeat, the number of different permutations of all the letters is given by the formula $$\text{Number of distinct words}= \dfrac{n!}{p_1!\,p_2!\,\dots}$$ where $$n$$ is the total number of letters and each $$p_i$$ is the frequency of a particular repeated letter. With $$n=5$$ and only the letter $$L$$ repeating twice, we have $$\dfrac{5!}{2!}= \dfrac{120}{2}=60$$ so exactly $$60$$ different five-letter words will appear in the dictionary list.

Dictionary ordering depends on the first letter, then the second, and so on. For clarity let us first write the five letters in alphabetical order: $$A \;<\; L \;<\; M \;<\; S$$ (there are two $$L$$’s, but alphabetically both are just $$L$$).

Now we count, step by step, how many words come before $$SMALL$$.

Step 1 - First letter strictly smaller than $$S$$

The letters smaller than $$S$$ are $$A,\,L,\,M$$. For each such first letter we arrange the remaining four letters (remembering the duplicate $$L$$). If the first letter is $$A$$, the leftover multiset is $$\{L,\,L,\,M,\,S\}$$. Using the formula, the number of words beginning with $$A$$ is $$\dfrac{4!}{2!}= \dfrac{24}{2}=12.$$

If the first letter is $$L$$, the remaining set is $$\{A,\,L,\,M,\,S\}$$ (only one $$L$$ left), so $$4! = 24$$ such words.

If the first letter is $$M$$, the remaining set is $$\{A,\,L,\,L,\,S\}$$, giving $$\dfrac{4!}{2!}=12$$ words.

Adding these three blocks, the number of words that definitely precede all words beginning with $$S$$ is $$12 + 24 + 12 = 48.$$

Step 2 - First letter $$S$$, second letter smaller than $$M$$

The word we want is $$S\!\,M\!\,A\!\,L\!\,L$$, whose second letter is $$M$$. Among the remaining letters $$\{A,\,L,\,L,\,M\}$$, those alphabetically smaller than $$M$$ are $$A$$ and $$L$$.

• Second letter $$A$$: leftover letters $$\{L,\,L,\,M\}$$ give $$\dfrac{3!}{2!}=3$$ words.

• Second letter $$L$$: leftover letters $$\{A,\,L,\,M\}$$ (each distinct) give $$3! = 6$$ words.

Total words beginning with $$S$$ whose second letter is less than $$M$$: $$3 + 6 = 9.$$

So far, the cumulative count of words before $$SMALL$$ is $$48 + 9 = 57.$$

Step 3 - First two letters $$S\,M$$, third letter smaller than $$A$$

After fixing $$S$$ and $$M$$, the remaining letters are $$\{A,\,L,\,L\}$$. There is no letter alphabetically smaller than $$A$$, so no words are added at this stage.

Step 4 - First three letters $$S\,M\,A$$, fourth letter smaller than $$L$$

The leftover letters $$\{L,\,L\}$$ are both $$L$$, so again there is no smaller choice. Likewise, the last letter is forced. Thus, the very next word after the 57 counted so far is precisely $$SMALL$$.

Therefore, the position of $$SMALL$$ in the complete dictionary list is $$57 + 1 = 58.$$

Hence, the correct answer is Option B.