In how many ways the letters of the word COMBINATORICS can be arranged if all vowel and all consonants are alphabetically ordered.

Explanation

To solve the problem of arranging the letters of the word "COMBINATORICS" with the condition that all vowels and all consonants must be in alphabetical order, we can follow these steps:

### Step 1: Identify the letters
The word "COMBINATORICS" consists of 13 letters. Let's separate them into vowels and consonants.

- **Vowels**: O, I, A, O, I (which includes A, I, I, O, O)
- **Consonants**: C, M, B, N, T, R, C, S (which includes B, C, C, M, N, R, S, T)

### Step 2: Count the vowels and consonants
- **Total vowels**: 5 (A, I, I, O, O)
- **Total consonants**: 8 (B, C, C, M, N, R, S, T)

### Step 3: Arrange vowels and consonants alphabetically
- **Vowels in alphabetical order**: A, I, I, O, O
- **Consonants in alphabetical order**: B, C, C, M, N, R, S, T

### Step 4: Determine the total arrangements
Since the vowels and consonants must be in alphabetical order, we can treat them as fixed groups. The problem now reduces to choosing positions for the vowels among the total letters.

### Step 5: Choose positions for vowels
We need to choose 5 positions out of the total 13 for the vowels. The remaining positions will automatically be for the consonants. The number of ways to choose 5 positions from 13 is given by the combination formula:

\[ \text{Number of ways} = \binom{13}{5} \]

### Step 6: Calculate the combinations
Now, we calculate \(\binom{13}{5}\):

\[ \binom{13}{5} = \frac{13!}{5!(13-5)!} = \frac{13!}{5! \cdot 8!} \]

Calculating this gives:

\[ \binom{13}{5} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287 \]

### Step 7: Final answer
Thus, the total number of ways to arrange the letters of the word "COMBINATORICS" with all vowels and consonants in alphabetical order is:

\[ \text{Total arrangements} = 1287 \]