Question

If all permutations of the letters of the word $AGAIN$ are arranged as in dictionary, the forty-ninth word is

• A. $NAAGI$
• B. $NAGAI$
• C. $NAAIG$
• D. $NAIAG$

Answer 1

The correct option is A $NAAGI$

Stating with the 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, the $37th$ word starts with $I$ and there are $12$ Words Starting with $I$.

Total goes up to $48$ words .

Now, $49th$ word is$NAAGI.$