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 :

The word "KANPUR" has 6 distinct letters: A, K, N, P, R, U.

Arranging in alphabetical order: A, K, N, P, R, U.

Total permutations = 6! = 720.

Words starting with each letter = 5! = 120.

Words starting with A: positions 1-120

Words starting with K: positions 121-240

Words starting with N: positions 241-360

Words starting with P: positions 361-480

440th word starts with P. Position within P-words: 440 - 360 = 80.

After P, remaining letters: A, K, N, R, U (alphabetical order).

Words starting with PA: 4! = 24 (positions 1-24)

Words starting with PK: 4! = 24 (positions 25-48)

Words starting with PN: 4! = 24 (positions 49-72)

Words starting with PR: 4! = 24 (positions 73-96)

80th word starts with PR. Position within PR-words: 80 - 72 = 8.

After PR, remaining letters: A, K, N, U.

Words starting with PRA: 3! = 6 (positions 1-6)

Words starting with PRK: 3! = 6 (positions 7-12)

8th word starts with PRK. Position within PRK-words: 8 - 6 = 2.

After PRK, remaining letters: A, N, U.

Words starting with PRKA: 2! = 2 (positions 1-2)

2nd word starting with PRKA: remaining letters N, U arranged as UN.

PRKAUN (position 2 within PRKA-words)

Wait: 1st position is PRKANU, 2nd position is PRKAUN.

So the 440th word is PRKAUN.

The correct answer is Option 3: PRKAUN.