There are $5$ letters in the word $AGAIN$, in which $A$ appears $2$ times. Therefore, the required number of words $=\frac{5 !}{2 !}=60$

To get the number of words starting with $A$, we fix the letter $A$ at the extreme left position, we then rearrange the remaining $4$ letters taken all at a time. There will be as many arrangements of these $4$ letters taken $4$ at a time as there are permutations of $4$ different things taken $4$ at a time. Hence, the number of words starting with $A=4 !=24 .$ Then, starting with $G$, the number of words $=\frac{4 !}{2 !}=12$ as after placing $G$ at the extreme left position, we are left with the letters $A , A , I$ and $N$. Similarly, there are $12$ words starting with the next letter $I$. Total number of words so far obtained $=24+12+12=48$

The $49^{\text {th }}$ word is $NAAGI$. The $50^{\text {th }}$ word is $NAAIG$.