Question

All letters of the word `AGAIN' are permuted in all possible ways, and the words so formed (with or without meaning) are written as in a dictionary. Then the $50^{th}$ word is

• A. IAANG
• B. INAGA
• C. NAAIG
• D. NAAGI

Hint:

For dictionary order permutations with repeated letters, calculate total permutations using $\frac{n!}{k!}$, group by first letter, and list systematically to find the $k^{th}$ term.

Answer 1

The Correct Option is C

The word `AGAIN' has letters A, A, G, I, N (5 letters, two A’s identical). Total permutations: \[ \frac{5!}{2!} = \frac{120}{2} = 60 \] Arrange in dictionary (alphabetical) order: A, G, I, N. Count permutations by first letter:
- A: Remaining letters A, G, I, N (4 letters, one A repeated): $\frac{4!}{2!} = 12$
- G: Remaining letters A, A, I, N: $\frac{4!}{2!} = 12$
- I: Remaining letters A, A, G, N: $\frac{4!}{2!} = 12$
- N: Remaining letters A, A, G, I: $\frac{4!}{2!} = 12$
Cumulative count: A (1–12), G (13–24), I (25–36), N (37–48). The $50^{th}$ word starts with N (since $49, 50>48$). List words starting with N:
- NA: Remaining A, G, I: $\frac{3!}{1!} = 6$ (NAAGI, NAAIG, NAGAI, NAGIA, NAIGA, NAIAG)
- NAA: Words are NAAIG, NAAGI (2 permutations)
- NAAG: Words are NAAGI, NAAGG (but NAAGG is invalid)
- NAAI: Words are NAAIG, NAAIA (but NAAIA is invalid)
Order: NAAIG, NAAGI, NAGAI, NAGIA, NAIGA, NAIAG. The $49^{th}$ is NAAGI, and the $50^{th}$ is NAAIG. Option (3) is correct. Options (1), (2), and (4) are not the $50^{th}$ word.