The number of different words that can be formed from the letters of the word ‘INTERMEDIATE’ such that two vowels never come together is ________.

Answer

To find the number of different words that can be formed from the letters of the word "INTERMEDIATE" such that no two vowels come together, we can follow these steps:

### Step 1: Identify the letters and their frequencies
The word "INTERMEDIATE" consists of the following letters:
- Vowels: I, E, E, I, A, E (Total: 6 vowels)
- Consonants: N, T, R, M, D, T (Total: 6 consonants)

The frequencies of the letters are:
- Vowels: I (2), E (3), A (1)
- Consonants: N (1), T (2), R (1), M (1), D (1)

### Step 2: Arrange the consonants
First, we will arrange the consonants. The consonants are N, T, R, M, D, T. The arrangement of these consonants can be calculated as follows:

\[ \text{Number of arrangements of consonants} = \frac{6!}{2!} \]

Here, \(6!\) accounts for the total arrangements of the consonants, and \(2!\) accounts for the repetition of the letter T.

Calculating this gives:
\[ 6! = 720 \quad \text{and} \quad 2! = 2 \]
\[ \text{Number of arrangements of consonants} = \frac{720}{2} = 360 \]

### Step 3: Create spaces for vowels
Once the consonants are arranged, we can create spaces for the vowels. Arranging 6 consonants creates 7 possible spaces (before the first consonant, between consonants, and after the last consonant):

\[ \_ C \_ C \_ C \_ C \_ C \_ C \_ \]

### Step 4: Place the vowels in the spaces
We need to select 6 spaces out of the 7 available spaces to place the vowels. The number of ways to choose 6 spaces from 7 is given by:

\[ \binom{7}{6} = 7 \]

### Step 5: Arrange the vowels
Next, we need to arrange the vowels I, E, E, I, A, E. The arrangement of these vowels can be calculated as follows:

\[ \text{Number of arrangements of vowels} = \frac{6!}{3! \cdot 2!} \]

Here, \(6!\) accounts for the total arrangements of the vowels, \(3!\) accounts for the repetition of the letter E, and \(2!\) accounts for the repetition of the letter I.

Calculating this gives:
\[ 6! = 720, \quad 3! = 6, \quad 2! = 2 \]
\[ \text{Number of arrangements of vowels} = \frac{720}{6 \cdot 2} = \frac{720}{12} = 60 \]

### Step 6: Calculate the total arrangements
Finally, we multiply the number of arrangements of consonants, the number of ways to choose spaces, and the number of arrangements of vowels:

\[ \text{Total arrangements} = (\text{Arrangements of consonants}) \times (\text{Ways to choose spaces}) \times (\text{Arrangements of vowels}) \]
\[ \text{Total arrangements} = 360 \times 7 \times 60 \]

Calculating this gives:
\[ 360 \times 7 = 2520 \]
\[ 2520 \times 60 = 151200 \]

Thus, the total number of different words that can be formed from the letters of the word "INTERMEDIATE" such that no two vowels come together is **151200**.