Question

If all the words, with or without meaning, made using all the letters of the word ``RANCHI'' are arranged as in a dictionary, then the word at \(560^{\text{th}}\) position is:

• A. \text{NICHRA}
• B. \text{NICAHR}
• C. \text{NICARH}
• D. \text{NICHAR}

Hint:

For dictionary order problems, always subtract 1 from the given position and use factorial grouping to identify each letter step-by-step.

Answer 1

The Correct Option is C

Concept:
The number of arrangements of \(n\) distinct letters is \(n!\).
Dictionary (lexicographical) order depends on alphabetical arrangement.
To find the word at a given position, we use the factorial number system.
Step 1: Arrange letters alphabetically The letters of the word RANCHI arranged alphabetically are: \[ A,\ C,\ H,\ I,\ N,\ R \] Total number of words: \[ 6! = 720 \]
Step 2: Convert position to zero-based index \[ 560 - 1 = 559 \]
Step 3: Determine letters position-wise
Each block of words starting with a fixed first letter has \(5! = 120\) words. \[ 559 \div 120 = 4 \quad (\text{remainder } 79) \] The \(5^{\text{th}}\) letter is N. Remaining letters: \(A, C, H, I, R\)
Each block now has \(4! = 24\) words. \[ 79 \div 24 = 3 \quad (\text{remainder } 7) \] The \(4^{\text{th}}\) letter is I. Remaining letters: \(A, C, H, R\)
Each block now has \(3! = 6\) words. \[ 7 \div 6 = 1 \quad (\text{remainder } 1) \] The \(2^{\text{nd}}\) letter is C. Remaining letters: \(A, H, R\)
Each block now has \(2! = 2\) words. \[ 1 \div 2 = 0 \quad (\text{remainder } 1) \] The \(1^{\text{st}}\) letter is A. Remaining letters: \(H, R\)
Each block now has \(1! = 1\) word. \[ 1 \div 1 = 1 \] The \(2^{\text{nd}}\) letter is R. The last remaining letter is H.
Step 4: Required word \[ \boxed{\text{NICARH}} \]