If all the words with or without meaning made using all the letters of the word "KANPUR" are arrange

Q: Wait, why do we use factorials like $5!$ and $4!$?

Great question! When we fix a letter at the beginning of a word, we are essentially locking one position. The remaining letters can be arranged in any order in the remaining slots. For $5$ remaining slots, the number of ways to arrange $5$ distinct letters is $5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$. It is the fundamental principle of counting permutations!

Q: How do we know when to stop and start listing words manually?

You stop when the next 'block' of words (calculated by the factorial of the remaining letters) would push your cumulative total beyond the target number. For example, if your current total is $432$ and the next block is $24$, the new total would be $456$. Since $456 > 440$, you know the target word lies within that block, so you 'zoom in' by fixing the next letter and repeating the process.

Q: What if the word had repeated letters, like 'BANANA'?

That is a fantastic observation! If there were repeated letters, we would have to divide the total permutations by the factorial of the frequency of each repeated letter. For example, if 'A' appeared $3$ times, we would divide by $3!$. Since 'KANPUR' has all distinct letters, we don't need to worry about that here, but always keep it in mind for other problems!

Question

If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at $440^{th}$ position in this arrangement, is :

Options:

(A) PRNAUK

(B) PRKANU

(C) PRKAUN

(D) PRNAKU

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the $440^{th}$ word in the dictionary of permutations of the word **KANPUR**, we first identify the unique letters involved: **K, A, N, P, U, R**.

Sorting these letters alphabetically provides our roadmap: **A, K, N, P, R, U**. There are $6$ distinct letters, meaning there are $6! = 720$ total permutations.

### The Macro Search

We calculate the number of words starting with each letter to narrow down the position of the $440^{th}$ word.

Words starting with **A**: $5! = 120$ words. (Cumulative: $120$)

Words starting with **K**: $5! = 120$ words. (Cumulative: $240$)

Words starting with **N**: $5! = 120$ words. (Cumulative: $360$)

Since $360 < 440$, we proceed to the next letter. Words starting with **P** will contain our target word, as the cumulative count would reach $480$.

### Narrowing the Focus

We now fix the first letter as **P** and determine the second letter using the remaining set: **{A, K, N, R, U}**.

Words starting with **PA**: $4! = 24$ words. (Cumulative: $360 + 24 = 384$)

Words starting with **PK**: $4! = 24$ words. (Cumulative: $384 + 24 = 408$)

Words starting with **PN**: $4! = 24$ words. (Cumulative: $408 + 24 = 432$)

If we included words starting with **PR**, the total would reach $456$. Because $432 < 440 < 456$, our target word must start with **PR**.

### The Final Stretch

With the prefix **PR** fixed, we determine the third letter using the remaining set: **{A, K, N, U}**.

Words starting with **PRA**: $3! = 6$ words. (Cumulative: $432 + 6 = 438$)

Words starting with **PRK**: $3! = 6$ words. (Cumulative: $438 + 6 = 444$)

Since $444 > 440$, our target word must start with **PRK**.

The $439^{th}$ word is the first permutation starting with **PRK**, which is **PRKANU** (arranging the remaining letters **A, N, U** in alphabetical order).

The $440^{th}$ word is the next permutation after **PRKANU**. By swapping the last two letters, we obtain **PRKAUN**.

The $440^{th}$ word is **PRKAUN**.

Animated Solution for Mathematics - Permutations and Combinations: If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at $44 0^{t h}$ position in this arrangement, is :

Select Answer:

Visualized Solution

Alphabetical Sorting

Words Starting with $A$

Words Starting with $K$

Words Starting with $N$

Narrowing Down to $P$

Words Starting with $P A$

Words Starting with $P K$

Words Starting with $P N$

Narrowing Down to $P R$

Words Starting with $P R A$

Narrowing Down to $P R K$

The $43 9^{th}$ Word

The $44 0^{th}$ Word

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Macro Search

Narrowing the Focus

The Final Stretch

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Select Answer:

Visualized Solution

Alphabetical Sorting

Words Starting with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Words Starting with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

Words Starting with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Narrowing Down to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Words Starting with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PA

Words Starting with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PK

Words Starting with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PN

Narrowing Down to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PR

Words Starting with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PRA

Narrowing Down to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PRK

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 439th Word

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 440th Word