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 440th position in this arrangement is:

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For dictionary order problems, calculate each letter's positional contribution and count words accordingly.
Updated On: Nov 2, 2025
  • PRNAKU
  • PRKANU
  • PRKAUN
  • PRNAUK
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The Correct Option is C

Approach Solution - 1

Arranging the letters alphabetically: {A, K, N, P, R, U} 

Step 1: Words starting with A = 5! = 120

Step 2: Words starting with K = 5! = 120

Step 3: Words starting with N = 5! = 120

Step 4: Words starting with PA = 4! = 24

Step 5: Words starting with PK = 4! = 24

Step 6: Words starting with PN = 4! = 24

Step 7: Words starting with PRKA = 3! = 6

Step 8: PRKAN is the 439th word

Step 9: PRKAUN is the 440th word

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Approach Solution -2

Step 1: Write the letters of the word in alphabetical order.
The word is KANPUR.
Arrange the letters alphabetically:
A, K, N, P, R, U.
Total letters = 6.

Step 2: Total number of words that can be formed.
Since all letters are distinct, total possible arrangements = \( 6! = 720 \).

Step 3: Find how many words come before the words starting with each letter.
Each letter can be the first letter in \( 5! = 120 \) words.

- Words starting with A: 120 words (positions 1–120)
- Words starting with K: next 120 words (positions 121–240)
- Words starting with N: next 120 words (positions 241–360)
- Words starting with P: next 120 words (positions 361–480)

Since 440 lies between 361 and 480, the required word starts with P.

Step 4: Fix P as the first letter.
Remaining letters: A, K, N, R, U (arranged alphabetically).
Each of these 5 letters can be second letters with \( 4! = 24 \) words each.

- Second letter A: 24 words (361–384)
- Second letter K: next 24 words (385–408)
- Second letter N: next 24 words (409–432)
- Second letter R: next 24 words (433–456)

The 440th word lies between 433 and 456, so the second letter is R.

Step 5: Fix P and R as the first two letters.
Remaining letters: A, K, N, U.
Each can be the third letter with \( 3! = 6 \) words each.

- Third letter A: positions 433–438
- Third letter K: positions 439–444

The 440th position lies in this range, so the third letter is K.

Step 6: Fix P, R, K and list remaining letters.
Remaining: A, N, U.
Words in dictionary order:
1. PRKAUN
2. PRKANU
3. PRKNAU
4. PRKNUA
5. PRKUAN
6. PRKUNA

The first word in this set (position 439 + 1 = 440) is PRKAUN.

Final Answer:
\[ \boxed{\text{PRKAUN}} \]
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