If different words are formed with all the letters of the word 'AGAIN' and are arranged alphabetically amond themselvs as in a dictionary, the word at the 50th place will be

Options

A. NAAGI
B. NAAIG
C. IAAGN
D. IAANG

Answer

To find the 50th word formed by the letters of the word "AGAIN" when arranged in alphabetical order, we will follow a systematic approach using permutations and combinations.

### Step-by-Step Solution:

1. **Identify the Letters**: The letters in "AGAIN" are A, A, G, I, N.

2. **Arrange the Letters Alphabetically**: The alphabetical order of the letters is A, A, G, I, N.

3. **Calculate Total Permutations**:
The total number of distinct permutations of the letters can be calculated using the formula for permutations of multiset:
\[ \text{Total permutations} = \frac{n!}{p_1! \times p_2! \times \ldots \times p_k!} \]
where \( n \) is the total number of letters, and \( p_1, p_2, \ldots, p_k \) are the frequencies of each letter.
Here, we have:
- Total letters (n) = 5
- Frequency of A = 2, G = 1, I = 1, N = 1
\[ \text{Total permutations} = \frac{5!}{2! \times 1! \times 1! \times 1!} = \frac{120}{2} = 60 \]

4. **Count Words Starting with A**:
If we fix A as the first letter, we have the letters A, G, I, N left. The number of permutations of these letters is:
\[ \text{Permutations} = \frac{4!}{1! \times 1! \times 1!} = 24 \]
So, the first 24 words start with A.

5. **Count Words Starting with AG**:
If we fix AG as the first two letters, we have A, I, N left. The number of permutations of these letters is:
\[ \text{Permutations} = \frac{3!}{1! \times 1!} = 6 \]
So, the next 6 words (25th to 30th) start with AG.

6. **Count Words Starting with AI**:
If we fix AI as the first two letters, we have A, G, N left. The number of permutations of these letters is:
\[ \text{Permutations} = \frac{3!}{1! \times 1!} = 6 \]
So, the next 6 words (31st to 36th) start with AI.

7. **Count Words Starting with AN**:
If we fix AN as the first two letters, we have A, G, I left. The number of permutations of these letters is:
\[ \text{Permutations} = \frac{3!}{1! \times 1!} = 6 \]
So, the next 6 words (37th to 42nd) start with AN.

8. **Count Words Starting with AIG**:
If we fix AIG as the first three letters, we have A, N left. The number of permutations of these letters is:
\[ \text{Permutations} = \frac{2!}{1!} = 2 \]
So, the next 2 words (43rd to 44th) start with AIG.

9. **Count Words Starting with AIN**:
If we fix AIN as the first three letters, we have A, G left. The number of permutations of these letters is:
\[ \text{Permutations} = \frac{2!}{1!} = 2 \]
So, the next 2 words (45th to 46th) start with AIN.

10. **Count Words Starting with AGI**:
If we fix AGI as the first three letters, we have A, N left. The number of permutations of these letters is:
\[ \text{Permutations} = \frac{2!}{1!} = 2 \]
So, the next 2 words (47th to 48th) start with AGI.

11. **Count Words Starting with AN**:
If we fix AN as the first two letters, we have A, G, I left. The number of permutations of these letters is:
\[ \text{Permutations} = \frac{3!}{1! \times 1!} = 6 \]
So, the next 6 words (49th to 54th) start with AN, which includes the 50th word.

12. **Determine the 50th Word**:
The 49th word is "ANAIG", and the 50th word is "ANAGI".

### Final Answer:
The word at the 50th place is **"ANAGI"**.

---