Question

All the words (with or without meaning) formed using all the five letters of the word GOING are arranged as in a dictionary. Then the word OGGIN occurs at the place which is:

• A. \(48^{\text{th}}\)
• B. \(49^{\text{th}}\)
• C. \(50^{\text{th}}\)
• D. \(51^{\text{th}}\)

Hint:

For rank problems with repeated letters: always divide by factorial of repetitions and count only letters \textbf{strictly smaller} at each position.

Answer 1

The Correct Option is B

Step 1: Arrange letters in alphabetical order Letters of the word GOING are: \[ G,\ O,\ I,\ N,\ G \] Alphabetical order: \[ G,\ G,\ I,\ N,\ O \] Step 2: Find rank of the word OGGIN We calculate the number of words that come before OGGIN. First letter: O Letters smaller than \(O\) are \(G, G, I, N\). Number of permutations using remaining 4 letters (with two G’s): \[ \frac{4!}{2!}=12 \] Number of such letters before \(O\): \(4\) \[ \Rightarrow 4 \times 12 = 48 \] Step 3: Remaining letters After fixing \(O\), the remaining word is GGIN. This is already the smallest possible arrangement of these letters. So, no additional words are added. Step 4: Final rank \[ \text{Rank} = 48 + 1 = 49 \] \[ \boxed{49^{\text{th}}} \]