Animated Solution for Mathematics - Permutations and Combinations: 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 :

Unlocking the $315^{\text{th}}$ Word

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are embarking on a journey through the structured, elegant world of combinatorics.

Imagine you are standing before a massive library of words, all formed from the letters of the word $NAGPUR$. Our mission is to find the $315^{\text{th}}$ word in this dictionary. It sounds daunting, but with the right perspective, it becomes a beautiful dance of logic.

The Alphabetical Foundation

Before we can count, we must organize. The dictionary is a creature of order. The first rule of any dictionary problem is to sort the letters alphabetically.

Our letters are $N,A,G,P,U,R$. When we line them up in their proper, disciplined order, we get: $A,G,N,P,R,U$. This is our alphabet for the day. Every word we build will follow this sequence.

The Power of Blocks

Now, let us start counting. We want to reach the $315^{\text{th}}$ word. We will group words into 'blocks' based on their starting letter.

First, consider all words starting with $A$. If we fix $A$ at the first position, we have $5$ remaining slots to fill with the remaining $5$ letters. The number of ways to arrange these is:

So, the first $120$ words in our dictionary all start with $A$.

Next, we move to words starting with $G$. Fixing $G$ at the front, we again have $5!=120$ words. Our cumulative total is now:

We are making progress!

Then, we look at words starting with $N$. If we were to count all of them, we would add another $120$, bringing our total to $360$.

But wait! Our target is $315$. Since $240<315<360$, we know with absolute certainty that our target word starts with $N$. We have entered the $N$-block.

The Final Countdown

Inside the $N$-block, we continue our systematic search. We look at the second letter. The first available letter after $N$ is $A$.

If we fix $NA$, we have $4$ slots left, giving us $4!=24$ words. Our total is now:

Next is $NG$. Again, $4!=24$ words. Our total is:

Next is $NP$. Another $4!=24$ words. Our total is:

Stop! Take a breath. We are at $312$, and our target is $315$. We are only $3$ words away.

This is where the block method ends and the manual search begins. We are now in the $NR$ block. The remaining letters are $A,G,P,U$. We arrange them in alphabetical order to find the next words.

The $313^{\text{th}}$ word is $NRAGPU$. The $314^{\text{th}}$ word is $NRAGUP$.

Finally, for the $315^{\text{th}}$ word, we move to the next prefix after $NRAG$, which is $NRAP$. The remaining letters are $G$ and $U$. Arranging them alphabetically gives us $GU$.

Thus, our $315^{\text{th}}$ word is $NRAPGU$.

There it is. The $315^{\text{th}}$ word is $NRAPGU$.

It wasn't just about the calculation; it was about the patience to build the structure, the wisdom to know when to switch from formulas to manual listing, and the clarity to see the pattern. Keep this methodical approach in your toolkit, and no combinatorics problem will ever stand in your way again.