Given, the letters of the word AGAIN. Now, we arrange them in alphabetical order: A, A, G, I, N. Step 1: Counting arrangements starting with 'A' If the word starts with 'A', the remaining four letters (A, G, I, N) can be arranged in: \[ 4! = 24 \text{ ways} \] Step 2: Counting arrangements starting with 'G' If the word starts with 'G', the remaining four letters (A, A, I, N) include two A's, so the number of distinct arrangements is: \[ \frac{4!}{2!} = \frac{24}{2} = 12 \text{ ways} \] Step 3: Counting arrangements starting with 'I' If the word starts with 'I', the remaining four letters (A, A, G, N) also include two A's, so the number of distinct arrangements is: \[ \frac{4!}{2!} = 12 \text{ ways} \] Step 4: Finding the 49th word So far, we have counted: \[ 24 + 12 + 12 = 48 \text{ words} \] Thus, the 49th word will start with 'N' and will be NAAGI. \[ \boxed{\text{NAAGI}} \]