If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at 315^ th position in this arrangement is :

Permutations and Combinations PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q45 with Solution

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Correct answer: C

The word is NAGPUR with letters: $A, G, N, P, R, U$ All letters are distinct.

So all permutations are arranged lexicographically using: $A < G < N < P < R < U$

Each fixed first letter gives: $5! = 120$ words.

Group the words by first letter:

Since: $241 \le 315 \le 360$ The $315^{\text{th}}$ word starts with $N$ .

Now find its position within the $N$ -group: $315 - 240 = 75$ So we need the $75^{\text{th}}$ word among permutations of: $A, G, P, R, U$

With first letter fixed as $N$ , each choice of second letter gives: $4! = 24$ words.

Possible second letters in order: $A, G, P, R, U$

Blocks:

Since $75$ lies in $73$ to $96$ , the second letter is $R$ .

Now relative position within the $NR$ -block: $75 - 72 = 3$ So we need the $3^{\text{rd}}$ word among permutations of: $A, G, P, U$

For fixed prefix $NR$ , each third letter gives: $3! = 6$ words.

Possible third letters in order: $A, G, P, U$

Blocks:

Since $3$ lies in $1$ to $6$ , the third letter is $A$ .

Now position within the $NRA$ -block is still $3$ . We now arrange remaining letters: $G, P, U$

For fixed prefix $NRA$ , each fourth letter gives: $2! = 2$ words.

Possible fourth letters in order: $G, P, U$

Blocks:

Since $3$ lies in $3$ to $4$ , the fourth letter is $P$ .

Now relative position inside $NRAP$ block: $3 - 2 = 1$ So we need the $1^{\text{st}}$ arrangement of remaining letters: $G, U$

The permutations of $G, U$ in dictionary order are:

So the first one is $GU$ .

Thus the required word is: $\boxed{NRAPGU}$

Hence the correct option is: $\boxed{C}$