The words that can be formed using alphabets \[A, H, L, U,\] and \[R\] are written as in a dictionary (no alphabet is repeated). Then find the rank of the word \[RAHUL\].
A. \[71\]
B. \[72\]
C. \[73\]
D. \[74\]

Hint In the given question, 5 alphabets are given. By using the factorial function, we will find the number of words formed using the given alphabets and the rank of the word \[RAHUL\].

Formula used
Factorial: The factorial of a non-negative integer is the product of all positive integers less than or equal to that number.
\[n! = 1 \cdot 2 \cdot 3 \cdot ....\left( {n - 1} \right) \cdot n\]

Complete step by step solution:
The given alphabets are \[A, H, L, U,\] and \[R\].
In the dictionary, the order of alphabets is \[A, H, L, R,\] and \[U\].
The words formed before \[RAHUL\] will starts with the letters \[A, H, L\] or \[R\].

Let’s calculate the number of words that comes before the word \[RAHUL\].
A number of words are formed when the first letter is \[A\]:
If \[A\] is the first letter and the other 4 letters are placed without repetition, then
A number of words start with \[A\]: \[4! = 24\]

A number of words are formed when the first letter is \[H\]:
If \[H\] is the first letter and the other 4 letters are placed without repetition, then
A number of words start with \[H\]: \[4! = 24\]

A number of words are formed when the first letter is \[L\]:
If \[L\] is the first letter and the other 4 letters are placed without repetition, then
A number of words start with \[L\]: \[4! = 24\]

In alphabetic manner, the only word starts with \[R\] and comes before \[RAHUL\] is \[RAHLU\].
So, the number of words comes before \[RAHUL\] in the dictionary is
Number of words \[ = 24 + 24 + 24 + 1\]
\[ \Rightarrow \] Number of words \[ = 73\]

Therefore, the rank of the word \[RAHUL\] is \[74\].
Hence the correct option is D.

Note: Students often get confused about the calculation of repetition and without repetition conditions. For without repetition, we will use factorial and for repetition, we will use permutation formula.