Question

If all permutations of the letters of the word AGAIN are arranged in the order as in a dictionary, what is the $49^{th}$ word?

• A. AAGIN
• B. NAAGI
• C. IAAGN
• D. GAAIN

Answer 1

Answer: (B)

Starting with letter $A$, and arranging the other four letters, there are $4!=24$ words. These are the first $24$ words. Then starting with $G$, and arranging $A,A,I$ and $N$ in different ways, there are $\frac{4!}{2!1!1!}=12$ words. Next the $37^{th}$ word starts with $I$. There are again $12$ words starting with $I$. This accounts up to the $48^{th}$ word. The $49^{th}$ word is $NAAGI$.