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  4. If all permutations of the letters of the word AGAIN are arranged in the order as in a dictionary, what is the 49th word?

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Q. 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?

Permutations and Combinations

Solution:

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 $NAAGI$.