Key Concepts and Formulas

• Lexicographical Order: Ordering words as they appear in a dictionary.
• Permutation: The number of ways to arrange $n$ distinct objects is $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.
• If we have $n$ distinct objects and want to arrange $r$ of them, the number of permutations is $P(n, r) = \frac{n!}{(n-r)!}$. If we want to arrange all $n$ distinct objects, then $r=n$ and $P(n,n) = n!$.

Step-by-Step Solution

Step 1: Arrange the Letters Alphabetically

The letters in the word "KANPUR" are A, K, N, P, R, U. These are already distinct. Arranging them alphabetically gives us: A, K, N, P, R, U. This order will be the basis for constructing words in lexicographical order.

Step 2: Count Words Starting with 'A'

If the first letter is 'A', the remaining 5 letters (K, N, P, R, U) can be arranged in $5!$ ways. $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. So, there are 120 words that start with 'A'.

Step 3: Count Words Starting with 'K'

If the first letter is 'K', the remaining 5 letters (A, N, P, R, U) can be arranged in $5!$ ways. $5! = 120$. So, there are 120 words that start with 'K'.

Step 4: Count Words Starting with 'N'

If the first letter is 'N', the remaining 5 letters (A, K, P, R, U) can be arranged in $5!$ ways. $5! = 120$. So, there are 120 words that start with 'N'.

Step 5: Count Words Starting with 'P'

If the first letter is 'P', the remaining 5 letters (A, K, N, R, U) can be arranged in $5!$ ways. $5! = 120$. So, there are 120 words that start with 'P'.

Step 6: Calculate the Cumulative Count

So far, the number of words starting with A, K, N, and P is $120 + 120 + 120 + 120 = 480$. Since we are looking for the 440th word, it must start with A, K, N, or P. Since 440 < 480, the 440th word must start with one of these letters. Also, since $3 \times 120 = 360 < 440$ and $4 \times 120 = 480 > 440$, the 440th word must start with 'P'. We need to find the (440 - 360) = 80th word starting with 'P'.

Step 7: Count Words Starting with 'PA'

If the first two letters are 'PA', the remaining 4 letters (K, N, R, U) can be arranged in $4!$ ways. $4! = 4 \times 3 \times 2 \times 1 = 24$. So, there are 24 words that start with 'PA'.

Step 8: Count Words Starting with 'PK'

If the first two letters are 'PK', the remaining 4 letters (A, N, R, U) can be arranged in $4!$ ways. $4! = 24$. So, there are 24 words that start with 'PK'.

Step 9: Count Words Starting with 'PN'

If the first two letters are 'PN', the remaining 4 letters (A, K, R, U) can be arranged in $4!$ ways. $4! = 24$. So, there are 24 words that start with 'PN'.

Step 10: Count Words Starting with 'PR'

If the first two letters are 'PR', the remaining 4 letters (A, K, N, U) can be arranged in $4!$ ways. $4! = 24$. So, there are 24 words that start with 'PR'.

Step 11: Calculate the Cumulative Count (Starting with 'P')

The number of words starting with PA, PK, PN, and PR is $24 + 24 + 24 + 24 = 96$. Since we are looking for the 80th word starting with 'P', and $3 \times 24 = 72 < 80$ and $4 \times 24 = 96 > 80$, the 80th word starting with 'P' must start with 'PR'. We need to find the (80 - 72) = 8th word starting with 'PR'.

Step 12: Count Words Starting with 'PRA'

If the first three letters are 'PRA', the remaining 3 letters (K, N, U) can be arranged in $3!$ ways. $3! = 3 \times 2 \times 1 = 6$. So, there are 6 words that start with 'PRA'.

Step 13: Count Words Starting with 'PRK'

If the first three letters are 'PRK', the remaining 3 letters (A, N, U) can be arranged in $3!$ ways. $3! = 6$. So, there are 6 words that start with 'PRK'.

Step 14: Calculate the Cumulative Count (Starting with 'PR')

The number of words starting with PRA and PRK is $6 + 6 = 12$. Since we are looking for the 8th word starting with 'PR', and $6 < 8$ and $12 > 8$, the 8th word starting with 'PR' must start with 'PRK'. We need to find the (8 - 6) = 2nd word starting with 'PRK'.

Step 15: Count Words Starting with 'PRKA'

If the first four letters are 'PRKA', the remaining 2 letters (N, U) can be arranged in $2!$ ways. $2! = 2 \times 1 = 2$. The words are PRKANU and PRKAU.

Since we are looking for the 2nd word starting with 'PRK', it is PRKAU. We are looking for the 2nd word starting with PRK, which is the 2nd lexicographical arrangement of the remaining letters A, N, and U, after PRK. Sorting A, N, U alphabetically gives A, N, U. Then the arrangements of A,N,U are: ANU, AUN, NAU, NUA, UAN, UNA. We seek the second arrangement of these letters after PRK, so we seek the second word starting with PRK. After fixing PRKA, the remaining letters are N and U, so the arrangements are NU and UN, so the 1st word is PRKANU and the 2nd is PRKAUN.

Therefore, the 440th word is PRKAUN.

Common Mistakes & Tips

• Alphabetical Order: Always arrange the letters alphabetically at the beginning. This is crucial for lexicographical ordering.
• Tracking the Remainder: Remember to subtract the number of words already counted from the target position to find the position within the next prefix.
• Factorial Calculation: Be careful while calculating the factorial values. Double-check your calculations.

Summary

To find the 440th word, we systematically counted the number of words starting with each letter. By repeatedly fixing prefixes and calculating the number of permutations of the remaining letters, we narrowed down the possibilities until we arrived at the 440th word, which is PRKAUN.

Final Answer

The final answer is \boxed{PRKAUN}, which corresponds to option (B).