Permutations & Combinations – Rank of Word – JEE Main 29 Jan 2025 Shift 2

Question ID: #561
JEE Main29 January Shift 2, 2025Algebra

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:

  • (1) PRNAKU
  • (2) PRKANU
  • (3) PRKAUN
  • (4) PRNAUK

Solution:


The letters of the word “KANPUR” in alphabetical order are: A, K, N, P, R, U.

We need to find the word at the $440^{th}$ position.

Words starting with A: $5! = 120$

Words starting with K: $5! = 120$

Words starting with N: $5! = 120$

Total words so far = $120 + 120 + 120 = 360$.

We need to reach 440, so the $440^{th}$ word must start with P.

Words starting with PA: $4! = 24$. (Cumulative Total = $360 + 24 = 384$)

Words starting with PK: $4! = 24$. (Cumulative Total = $384 + 24 = 408$)

Words starting with PN: $4! = 24$. (Cumulative Total = $408 + 24 = 432$)

The next letters in alphabetical order after N is R. So, we check words starting with PR.

Words starting with PRA: $3! = 6$. (Cumulative Total = $432 + 6 = 438$)

We are now very close to 440. The next word starts with PRK. The remaining letters are A, N, U.

$439^{th}$ word: P R K A N U (Alphabetical order of remaining letters)

$440^{th}$ word: P R K A U N

Ans. (3)

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