All the letters of the word 'EAMCOT' are arranged in different posible ways. The number of such arrangements in which no two vowels are adjacent to each other is :

Options

A. 360
B. 144
C. 72
D. 54

Answer

To solve the problem of arranging the letters in the word "EAMCOT" such that no two vowels are adjacent, we can follow these steps:

### Step 1: Identify the vowels and consonants
In the word "EAMCOT", the vowels are A, E, O and the consonants are M, C, T.

### Step 2: Arrange the consonants
First, we will arrange the consonants. The consonants M, C, and T can be arranged among themselves. The number of arrangements of 3 consonants is given by:
\[ 3! = 6 \]

### Step 3: Determine positions for the vowels
Once the consonants are arranged, we need to find the positions where we can place the vowels. When the consonants are arranged, they create gaps where the vowels can be placed. For example, if we arrange the consonants as MCT, the gaps for placing vowels are:
- _ M _ C _ T _

This gives us 4 gaps (before M, between M and C, between C and T, and after T) to place the vowels.

### Step 4: Choose positions for the vowels
We have 4 available gaps and we need to choose 3 of them to place the vowels. The number of ways to choose 3 gaps from 4 is given by:
\[ \binom{4}{3} = 4 \]

### Step 5: Arrange the vowels
The vowels A, E, O can be arranged among themselves in the chosen gaps. The number of arrangements of 3 vowels is given by:
\[ 3! = 6 \]

### Step 6: Calculate the total arrangements
Now, we can find the total number of arrangements where no two vowels are adjacent by multiplying the number of arrangements of consonants, the number of ways to choose gaps, and the arrangements of vowels:
\[ \text{Total arrangements} = (\text{Arrangements of consonants}) \times (\text{Ways to choose gaps}) \times (\text{Arrangements of vowels}) \]
\[ = 3! \times \binom{4}{3} \times 3! = 6 \times 4 \times 6 \]
\[ = 144 \]

Thus, the number of arrangements in which no two vowels are adjacent is **144**.