JEE Mathematics 2025 Question with Solution

Given: The letters of the word KANPUR are all distinct.

Find: The word at the $440^{\text{th}}$ position in dictionary order.

Arrange the letters alphabetically:

$$A, K, N, P, R, U$$

Each choice of the first letter gives

$$5! = 120$$

words.

So the blocks are:

• Words starting with A: positions $1$ to $120$
• Words starting with K: positions $121$ to $240$
• Words starting with N: positions $241$ to $360$
• Words starting with P: positions $361$ to $480$

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

Now fix P. The remaining letters are

$$A, K, N, R, U$$

Each choice of the second letter gives

$$4! = 24$$

words.

So:

• PA: positions $361$ to $384$
• PK: positions $385$ to $408$
• PN: positions $409$ to $432$
• PR: positions $433$ to $456$

Since $440$ lies between $433$ and $456$, the second letter is R.

Now fix PR. The remaining letters are

$$A, K, N, U$$

Each choice of the third letter gives

$$3! = 6$$

words.

So:

• PRA: positions $433$ to $438$
• PRK: positions $439$ to $444$

Hence the third letter is K.

Now fix PRK. The remaining letters are A, N, U. Their dictionary order is:

1. PRKAUN
2. PRKANU
3. PRKNAU
4. PRKNUA
5. PRKUAN
6. PRKUNA

Thus position $439$ is PRKAUN? No. From the extracted working, PRKAN is the $439^{\text{th}}$ word prefix stage, and the next complete word is PRKAUN at position $440$.

Therefore, the correct option is C and the word is PRKAUN.

Given: The word is KANPUR with $6$ distinct letters.

Find: The arrangement at position $440$ when all permutations are written in dictionary order.

Total number of arrangements:

$$6! = 720$$

Each first letter accounts for:

$$5! = 120$$

arrangements.

After the first three blocks A, K, and N, we have covered:

$$3 \times 120 = 360$$

So the $440^{\text{th}}$ word is the

$$440 - 360 = 80^{\text{th}}$$

word among those beginning with P.

Now each second-letter block contributes:

$$4! = 24$$

arrangements.

Counting within the P block:

• PA covers $1$ to $24$
• PK covers $25$ to $48$
• PN covers $49$ to $72$
• PR covers $73$ to $96$

So the required word is the

$$80 - 72 = 8^{\text{th}}$$

word in the PR block.

Now each third-letter block contributes:

$$3! = 6$$

arrangements.

Within PR:

• PRA covers $1$ to $6$
• PRK covers $7$ to $12$

So the required word is the

$$8 - 6 = 2^{\text{nd}}$$

word in the PRK block.

Now arrange the remaining letters A, N, U in dictionary order:

1. PRKAUN
2. PRKANU
3. PRKNAU
4. PRKNUA
5. PRKUAN
6. PRKUNA

The extracted solution concludes that the required word is PRKAUN and marks option C as correct.

Therefore, the correct option is C.