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Mathematics · Permutations and Combinations

JEE Main 2025 — 29 January, Evening Shift — Question 51

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 440th 440^{\text {th }} position in this arrangement, is

  1. Option A:

    PRNAKU

  2. Option B:

    PRKANU

  3. Option C:

    PRKAUN

    Correct
  4. Option D:

    PRNAUK

Answer: C

Step-by-step solution

A, K, N, P, R, U

A ……………=5!=120K ……………=5!=120N ……………=5!=120PA …………=4!=24PK …………=4!=24PN …………=4!=24PRA ………=3!=6PRKANU=1PRKAUN=1\begin{aligned} A &\ \ldots\ldots\ldots\ldots\ldots = 5! = 120 \\ K &\ \ldots\ldots\ldots\ldots\ldots = 5! = 120 \\ N &\ \ldots\ldots\ldots\ldots\ldots = 5! = 120 \\[4pt] PA &\ \ldots\ldots\ldots\ldots = 4! = 24 \\ PK &\ \ldots\ldots\ldots\ldots = 4! = 24 \\ PN &\ \ldots\ldots\ldots\ldots = 4! = 24 \\[4pt] PRA &\ \ldots\ldots\ldots = 3! = 6 \\[4pt] PRKANU &= 1 \\ PRKAUN &= 1 \end{aligned} Total=440\text{Total} = 440 ⇒440th word\Rightarrow \text{440}^{\text{th}} \text{ word}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
29 January, Evening Shift
Subject
Mathematics
Chapter
Permutations and Combinations
Topic
Permutations