The number of different permutations of all the letters of the word PERMUTATION such that any two consecutive letters in the arrangement are neither both vowels nor both identical is :

Options

A. ` 63 xx lfloor 6 xx lfloor5 `
B. ` 8 xx lfloor 6 xx lfloor5 `
C. ` 57 xx lfloor 5 xx lfloor 5 `
D. ` 7 xx lfloor 7 xx lfloor 5 `

Answer

To solve the problem of finding the number of different permutations of the letters in the word "PERMUTATION" such that no two consecutive letters are both vowels or both identical, we can follow these steps:

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

### Step 2: Count the Letters
The total number of letters in "PERMUTATION" is 12. Among these:
- Total consonants = 6 (P, R, M, T, T, N)
- Total vowels = 5 (E, U, A, I, O)

### Step 3: Arrange the Consonants
First, we arrange the consonants. The consonants include one repeated letter (T). The number of ways to arrange the consonants is given by the formula for permutations of multiset:

\[ \text{Arrangements of consonants} = \frac{6!}{2!} = \frac{720}{2} = 360 \]

### Step 4: Create Spaces for Vowels
When we arrange the consonants, we create spaces for the vowels. For 6 consonants, there are 7 possible spaces (before the first consonant, between consonants, and after the last consonant):

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

### Step 5: Place the Vowels
We need to select 5 out of these 7 spaces to place the vowels. The number of ways to choose 5 spaces from 7 is given by:

\[ \binom{7}{5} = \binom{7}{2} = 21 \]

### Step 6: Arrange the Vowels
Next, we arrange the vowels. Since all vowels are distinct, the number of arrangements of the vowels is:

\[ 5! = 120 \]

### Step 7: Calculate Total Arrangements
Now, we can calculate the total arrangements by multiplying the arrangements of consonants, the ways to choose spaces, and the arrangements of vowels:

\[ \text{Total arrangements} = \left(\frac{6!}{2!}\right) \times \binom{7}{5} \times 5! = 360 \times 21 \times 120 \]

Calculating this gives:

\[ 360 \times 21 = 7560 \]
\[ 7560 \times 120 = 907200 \]

### Step 8: Exclude Invalid Arrangements
Now, we need to exclude the arrangements where the two T's are together. If we treat the two T's as a single entity (TT), we have the following letters to arrange: P, R, M, TT, N (5 letters total). The arrangements of these letters are:

\[ 5! = 120 \]

Again, we create spaces for the vowels. With 5 consonants (including TT), we have 6 spaces:

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

The number of ways to choose 5 spaces from 6 is:

\[ \binom{6}{5} = 6 \]

The arrangements of the vowels remain:

\[ 5! = 120 \]

Thus, the total arrangements where the T's are together is:

\[ 120 \times 6 \times 120 = 86400 \]

### Step 9: Final Calculation
Finally, we subtract the invalid arrangements from the total arrangements:

\[ \text{Valid arrangements} = 907200 - 86400 = 820800 \]

### Conclusion
The number of different permutations of the letters in the word "PERMUTATION" such that no two consecutive letters are both vowels or both identical is **820800**.

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