Words From KANPUR in Dictionary Order: 440th Word

JEE Main 2025 — 29 January, Evening Shift. Previous Year Question.

Problem

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 the 440th position in this arrangement is:

Key insight. With $6! = 720$ possible arrangements, listing them out to find the 440th is impossible by hand — but since every arrangement starting with a fixed letter uses up a predictable block of positions ($5!=120$ of them), the target word can be built up one letter at a time by tracking which block of 120, then 24, then 6, the 440th position falls into.

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Approach

In dictionary order, the letters of KANPUR sort as A, K, N, P, R, U. Every word beginning with a particular letter accounts for a fixed block of $5! = 120$ positions (since the remaining 5 letters can be arranged in any order after it). So the strategy is to repeatedly ask “which starting letter’s block of 120 (or, once that letter is fixed, blocks of 24, then 6, then smaller) does position 440 fall inside?” — fixing one letter of the answer at each stage.

Solution

Step 1 — Fix the first letter

The letters in dictionary order are A, K, N, P, R, U.

• Words starting with A: positions 1–120
• Words starting with K: positions 121–240
• Words starting with N: positions 241–360

That accounts for 360 positions. The 440th word must start with the next letter, P, and falls at position $440-360=80$ within the P-block.

Step 2 — Fix the second letter

Within words starting with P, the remaining letters A, K, N, R, U sort in that order, and each choice of second letter fixes a block of $4!=24$ positions.

• P, then A: positions 1–24 (within the P-block)
• P, then K: positions 25–48
• P, then N: positions 49–72

That’s 72 positions used, leaving $80-72=8$ into the next block. So the second letter is R (the next letter after N in the remaining set A, K, N, R, U).

Step 3 — Fix the third letter

Within PR-words, the remaining letters A, K, N, U sort in that order, each fixing a block of $3!=6$ positions.

• PR, then A: positions 1–6

That uses 6 of the 8 remaining positions, leaving $8-6=2$ into the next block. So the third letter is K (the next letter after A in A, K, N, U).

Step 4 — Fix the remaining letters

Within PRK-words, the remaining letters are A, N, U, sorting in that order, each fixing a block of $2!=2$ positions.

• PRK, then A: positions 1–2

The target position, 2, falls exactly within this block. With A fixed as the fourth letter, the remaining letters N and U sort as N first, then U, so the two words in this block are, in order, PRKANU and PRKAUN — and position 2 of this block is the second one.

Step 5 — Read off the word

$$\text{PRKAUN}$$

Answer

Common mistakes

• Forgetting to update the pool of remaining letters at each step. Once a letter is fixed, it’s removed from consideration, and the next letter’s blocks are built from a shorter alphabetically-sorted list — mixing this up gives the wrong block sizes.
• Off-by-one slips when converting “position 440 overall” into “position within the current block.” Subtracting the letters already accounted for at each stage (360, then 72, then 6) needs care to land on the right remaining offset.

Practise next

• Find the 100th word in the dictionary arrangement of all the letters of CHAIR, by the same block-by-block method.