All the letters of the word 'AGAIN' be arranged and the words thus formed are known as 'Simple Words'. If a vowel appears in between two similar letters, the number of simple words is

Options

A. 12
B. 6
C. 36
D. 14

Explanation

To solve the problem of arranging the letters of the word "AGAIN" such that a vowel appears between the two similar letters (the two A's), we can follow these steps:

### Step 1: Identify the letters in the word "AGAIN"
The letters in the word "AGAIN" are A, G, A, I, N. Here, we have:
- Vowels: A, A, I
- Consonants: G, N

### Step 2: Understand the condition
We need to arrange the letters such that a vowel appears between the two similar letters (the two A's). The only vowel that can be placed between the two A's is I, as it is the only other vowel present.

### Step 3: Treat the AAI as a single unit
If we consider the arrangement of AAI as a single unit, we can represent it as (AAI). Therefore, we can think of our arrangement as:
- (AAI), G, N

### Step 4: Count the total units
Now, we have three units to arrange:
1. (AAI)
2. G
3. N

### Step 5: Calculate the arrangements
The total number of ways to arrange these three units is given by the factorial of the number of units:
\[ \text{Total arrangements} = 3! = 6 \]

### Step 6: Conclusion
Thus, the total number of simple words formed under the given condition is 6.

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