To solve the problem, we need to find the total number of distinct arrangements of the letters in the word "AGAIN" and then determine the 50th word when these arrangements are listed in dictionary order.
### Step 1: Count the total arrangements of the letters in "AGAIN"
The word "AGAIN" consists of 5 letters where:
- A appears 2 times
- G appears 1 time
- I appears 1 time
- N appears 1 time
The formula for the number of distinct arrangements of letters when there are repetitions is given by:
\[
\text{Total arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots \times p_k!}
\]
where:
- \( n \) is the total number of letters,
- \( p_1, p_2, \ldots, p_k \) are the frequencies of the repeated letters.
For "AGAIN":
\[
\text{Total arrangements} = \frac{5!}{2!} = \frac{120}{2} = 60
\]
### Step 2: List the arrangements in dictionary order
To find the 50th word, we need to list the words in alphabetical order. The letters in "AGAIN" in alphabetical order are: A, A, G, I, N.
1. **Fixing 'A' as the first letter:**
- Remaining letters: A, G, I, N
- Arrangements:
\[
\frac{4!}{1!} = 24 \text{ (words starting with A)}
\]
2. **Fixing 'G' as the first letter:**
- Remaining letters: A, A, I, N
- Arrangements:
\[
\frac{4!}{2!} = \frac{24}{2} = 12 \text{ (words starting with G)}
\]
3. **Fixing 'I' as the first letter:**
- Remaining letters: A, A, G, N
- Arrangements:
\[
\frac{4!}{2!} = \frac{24}{2} = 12 \text{ (words starting with I)}
\]
4. **Fixing 'N' as the first letter:**
- Remaining letters: A, A, G, I
- Arrangements:
\[
\frac{4!}{2!} = \frac{24}{2} = 12 \text{ (words starting with N)}
\]
### Step 3: Determine the 50th word
Now we can summarize the counts:
- Words starting with A: 24 (1 to 24)
- Words starting with G: 12 (25 to 36)
- Words starting with I: 12 (37 to 48)
- Words starting with N: 12 (49 to 60)
Since the 50th word falls in the range of words starting with N, we need to find the 2nd word starting with N.
#### Arrangements starting with N:
1. **Fixing 'N' as the first letter:**
- Remaining letters: A, A, G, I
- Arrangements in alphabetical order:
- A, A, G, I
- A, A, I, G
- A, G, A, I
- A, I, A, G
- G, A, A, I
- G, A, I, A
- I, A, A, G
- I, A, G, A
The first arrangement starting with N is "NAAIG".
The second arrangement starting with N is "NAAGI".
Thus, the 50th word in the dictionary order is:
\[
\text{50th word} = NAAIG
\]
### Final Answer
The number of distinct arrangements of the letters in "AGAIN" is 60, and the 50th word in dictionary order is **NAAIG**.
