The number of arrangements of the word " IDIOTS " such that vowels are at the places which form three consecutive terms of an A.P. is :

Options

A. 36
B. 72
C. 24
D. 108

Answer

To solve the problem of arranging the word "IDIOTS" such that the vowels occupy positions that form three consecutive terms of an arithmetic progression (A.P.), we can follow these steps:

### Step 1: Identify the vowels and consonants
The word "IDIOTS" consists of the following letters:
- Vowels: I, I, O (3 vowels)
- Consonants: D, T, S (3 consonants)

### Step 2: Determine possible positions for vowels
We need to place the vowels in positions that form three consecutive terms of an A.P. in a 6-letter arrangement. The possible sets of positions for the vowels are:
1. (1, 2, 3)
2. (2, 3, 4)
3. (3, 4, 5)
4. (4, 5, 6)
5. (1, 3, 5) - common difference of 2
6. (2, 4, 6) - common difference of 2

Thus, we have a total of 6 valid arrangements for the vowels.

### Step 3: Arrange the vowels
The vowels we have are I, I, O. Since the letter I is repeated, the number of distinct arrangements of these vowels is given by the formula for permutations of multiset:
\[ \text{Arrangements of vowels} = \frac{3!}{2!} = 3 \]
where \(3!\) is the factorial of the total number of vowels and \(2!\) accounts for the repetition of the letter I.

### Step 4: Arrange the consonants
The consonants D, T, and S can be arranged among themselves in:
\[ 3! = 6 \text{ ways} \]

### Step 5: Calculate total arrangements
Now, we can calculate the total number of arrangements by multiplying the number of ways to arrange the vowels by the number of ways to arrange the consonants, and then by the number of valid vowel positions:
\[ \text{Total arrangements} = (\text{Number of ways to choose positions for vowels}) \times (\text{Arrangements of vowels}) \times (\text{Arrangements of consonants}) \]
\[ = 6 \times 3 \times 6 = 108 \]

### Final Answer
Thus, the total number of arrangements of the word "IDIOTS" such that the vowels are at positions that form three consecutive terms of an A.P. is **108**.

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