The total number of words that can be made by writing the letters of the word PERMUTATION so that no vowel occupies any space between two consonants is:

Options

A. `(lfloor(7))/(lfloor(2))xxlfloor(5)`
B. `""^(7)C_(2)xx(lfloor(5))^(2)`
C. `(lfloor(6))/(lfloor(2))xxlfloor(6)`
D. `""^(6)C_(4)xxlfloor(4)`

Answer

To solve the problem of finding the total number of words that can be formed from the letters of the word "PERMUTATION" such that no vowel occupies any space between two consonants, we can follow these steps:

### Step 1: Identify the letters in "PERMUTATION"
The word "PERMUTATION" consists of the following letters:
- Consonants: P, R, M, T, T, N (6 consonants)
- Vowels: E, U, A, I, O (5 vowels)

### Step 2: Arrange the consonants
First, we will arrange the consonants. Since the consonant 'T' is repeated, we need to account for this in our arrangements.

The number of ways to arrange the consonants is given by the formula:
\[ \text{Number of arrangements of consonants} = \frac{n!}{p_1! \times p_2! \times \ldots} \]
where \( n \) is the total number of consonants and \( p_1, p_2, \ldots \) are the frequencies of the repeated consonants.

For our consonants:
- Total consonants = 6 (P, R, M, T, T, N)
- The letter T is repeated 2 times.

Thus, the number of arrangements of the consonants is:
\[ \text{Arrangements of consonants} = \frac{6!}{2!} \]

### Step 3: Determine the positions for the vowels
When we arrange the consonants, they create gaps where vowels can be placed. For 6 consonants, there are 7 possible gaps (before the first consonant, between consonants, and after the last consonant).

The arrangement will look like this:
\[ \_ C \_ C \_ C \_ C \_ C \_ C \_ \]
Where C represents a consonant.

### Step 4: Choose gaps for the vowels
We need to select 5 out of these 7 gaps to place the vowels. The number of ways to choose 5 gaps from 7 is given by:
\[ \binom{7}{5} = \binom{7}{2} = 21 \]

### Step 5: Arrange the vowels
The vowels (E, U, A, I, O) can be arranged in the selected gaps. The number of arrangements of the 5 vowels is:
\[ 5! = 120 \]

### Step 6: Calculate the total arrangements
Now, we can find the total number of arrangements by multiplying the number of arrangements of consonants, the number of ways to choose gaps, and the arrangements of vowels:
\[ \text{Total arrangements} = \left(\frac{6!}{2!}\right) \times \binom{7}{5} \times 5! \]

Calculating each part:
- \( 6! = 720 \)
- \( 2! = 2 \)
- \( \frac{6!}{2!} = \frac{720}{2} = 360 \)
- \( \binom{7}{5} = 21 \)
- \( 5! = 120 \)

Now, substituting these values:
\[ \text{Total arrangements} = 360 \times 21 \times 120 \]

Calculating this gives:
\[ 360 \times 21 = 7560 \]
\[ 7560 \times 120 = 907200 \]

### Final Answer
The total number of words that can be formed from the letters of the word "PERMUTATION" such that no vowel occupies any space between two consonants is **907200**.