If all permutations of the letters of the word AGAIN are arranged as in dictionary, then fiftieth word is

Options

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

Answer

To find the 50th word in the dictionary arrangement of the letters in the word "AGAIN", we will follow these steps:

### Step 1: Identify the letters and their frequencies
The word "AGAIN" consists of the letters A, G, I, A, N. The frequency of each letter is:
- A: 2
- G: 1
- I: 1
- N: 1

### Step 2: Calculate the total permutations
The total number of distinct permutations of the letters can be calculated using the formula for permutations of multiset:
\[ \text{Total permutations} = \frac{n!}{n_1! \times n_2! \times \ldots} \]
Where \( n \) is the total number of letters, and \( n_1, n_2, \ldots \) are the frequencies of each distinct letter.

For "AGAIN":
\[ \text{Total permutations} = \frac{5!}{2! \times 1! \times 1! \times 1!} = \frac{120}{2} = 60 \]

### Step 3: Arrange permutations in dictionary order
The letters in alphabetical order are: A, A, G, I, N.

#### Count permutations starting with 'A':
- If the first letter is A, the remaining letters are A, G, I, N.
- The number of permutations of A, G, I, N:
\[ \frac{4!}{1! \times 1! \times 1!} = 24 \]
So, there are 24 permutations starting with A.

#### Count permutations starting with 'G':
- If the first letter is G, the remaining letters are A, A, I, N.
- The number of permutations of A, A, I, N:
\[ \frac{4!}{2! \times 1! \times 1!} = \frac{24}{2} = 12 \]
So, there are 12 permutations starting with G.

#### Count permutations starting with 'I':
- If the first letter is I, the remaining letters are A, A, G, N.
- The number of permutations of A, A, G, N:
\[ \frac{4!}{2! \times 1! \times 1!} = \frac{24}{2} = 12 \]
So, there are 12 permutations starting with I.

### Step 4: Calculate the cumulative count
- Total permutations starting with A: 24
- Total permutations starting with G: 12 (Total so far: 36)
- Total permutations starting with I: 12 (Total so far: 48)

Thus, the first 48 permutations are accounted for.

### Step 5: Find the 49th and 50th permutations
The 49th permutation will start with N (since we have exhausted A, G, and I).

Now, consider the letters A, A, G, I:
- If the first letter is N, the remaining letters are A, A, G, I.
- The number of permutations of A, A, G, I:
\[ \frac{4!}{2! \times 1! \times 1!} = \frac{24}{2} = 12 \]
The permutations starting with N are:
1. NAAIG
2. NAAGI
3. NAGAI
4. NAGIA
5. NAIGA
6. NAIGG
7. NGAAI
8. NGAIA
9. NGAIA
10. NIAAG
11. NIAAG
12. NIAAG

Thus, the 49th word is NAAIG, and the 50th word is NAAGI.

### Final Answer
The 50th word is **NAAGI**.