If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is:

We have to find the position (rank) of the word $$\text{QUEEN}$$ when every distinct arrangement of its letters is listed in strict English (lexicographic) order.

The letters present are $$Q,\;U,\;E,\;E,\;N$$ - five letters in total, with the letter $$E$$ repeated twice and the other three letters occurring once each.

Whenever we speak of the “rank”, we always start counting from $$1$$. Hence the rank of the very first word in the dictionary list will be $$1$$, the next will be $$2$$, and so on. Our final answer will therefore be

Rank = 1 + (number of words that appear before QUEEN).

To count the words that come before $$\text{QUEEN}$$, we proceed letter by letter, always using the remaining letters in alphabetical order. The alphabetical order of the distinct letters available to us is

$$E \lt N \lt Q \lt U.$$

Step 1: Fixing the first letter. The first letter of $$\text{QUEEN}$$ is $$Q$$. Before we reach $$Q$$, any word whose first letter is alphabetically smaller will obviously precede it. Those smaller first letters are $$E$$ and $$N.$$ We count the arrangements for each choice.

• First letter = E. After using one $$E$$ we still have the letters $$\{E,N,Q,U\}$$ left, all of them distinct. The number of ways to arrange these $$4$$ letters is

$$\frac{4!}{1!}=24.$$

• First letter = N. After fixing $$N$$ we are left with $$\{E,E,Q,U\}$$ - that is, $$4$$ letters with the letter $$E$$ repeated twice. The number of distinct arrangements is

$$\frac{4!}{2!}=12.$$

Adding these, the total number of words that begin with a letter smaller than $$Q$$ is

$$24 + 12 = 36.$$

Step 2: Fixing the second letter. Having now fixed the first letter as $$Q$$, we move to the second letter. The remaining multiset is $$\{U,E,E,N\}$$ and their alphabetical order is $$E \lt N \lt U.$$ The second letter of $$\text{QUEEN}$$ is $$U.$$ So we must count words whose second letter is alphabetically smaller than $$U$$, namely words with second letter $$E$$ or $$N.$$

• Second letter = E. We use one $$E$$, leaving $$\{E,N,U\}$$ which are all distinct. The number of their arrangements is

$$3! = 6.$$

• Second letter = N. We use $$N$$, leaving $$\{E,E,U\}$$ - three letters with two $$E$$’s. The number of distinct arrangements is

$$\frac{3!}{2!}=3.$$

Thus, with the first letter fixed as $$Q$$, the number of words that still come before $$\text{QUEEN}$$ thanks to a smaller second letter is

$$6 + 3 = 9.$$

Step 3: Fixing further letters. Up to this point we have accumulated

$$36 + 9 = 45$$

words that precede $$\text{QUEEN}$$. Now the first two letters are exactly $$Q$$ and $$U$$, matching our target word. The remaining letters are $$\{E,E,N\}$$ in that order. The third letter of $$\text{QUEEN}$$ is $$E,$$ which is the smallest available letter, so no earlier word can be produced by choosing a smaller third letter (because none exists). The same reasoning applies to the fourth letter, which is again $$E.$$ Finally, the fifth letter is forced to be $$N.$$

Therefore no additional words are counted, and the total number of words that appear before $$\text{QUEEN}$$ remains $$45$$.

Calculating the rank. Recalling that we start counting ranks from $$1$$, we add $$1$$ to the number of preceding words:

$$\text{Rank of QUEEN} \;=\; 45 + 1 = 46.$$

Hence, the correct answer is Option C.