<p>If all permutations of the letters of the word 'AGAIN' are arranged as in a dictionary, then 50th word is</p>
Step-by-Step Solution
Key Concept: Count permutations systematically by fixing prefixes; use factorial division for repeated letters.
<p>The word AGAIN has 5 letters with A appearing twice: A, G, A, I, N.</p><p>Total distinct permutations = $\frac{5!}{2!} = 60$.</p><p>Words starting with A: $\frac{4!}{1!} = 24$ (remaining letters G, A, I, N with one A)</p><p>Words starting with G: $\frac{4!}{2!} = 12$ (remaining letters A, A, I, N)</p><p>Words starting with I: $\frac{4!}{2!} = 12$ (remaining letters A, A, G, N)</p><p>Words starting with N: $\frac{4!}{2!} = 12$ (remaining letters A, A, G, I)</p><p>After 24 words (starting with A), we have words 25-36 (starting with G), 37-48 (starting with I), 49-60 (starting with N).</p><p>The 50th word is the 2nd word starting with N.</p><p>Words starting with NA: $\frac{3!}{1!} = 6$ (remaining A, G, I)</p><p>The 49th word is the 1st starting with N; the 50th is the 2nd starting with NA.</p><p>Words starting with NAA: Remaining G, I → NAAGI, NAAGI (1st is 49th overall)</p><p>Words starting with NAG: Remaining A, I → NAGAI, NAIGA (2nd of NA...)</p><p>Actually, NAA gives 2! = 2 words: NAAGI (49th), NAAIG (not 50th). NAG gives: NAGAI (50th), NAGIA.</p><p>The 50th word is NAGAI.</p>
Correct Answer: d