In how many ways, can the letters of the word 'DIRECTOR' be arranged, so that the three vowels are never together?

Answer

To solve the problem of arranging the letters of the word "DIRECTOR" such that the three vowels (I, E, O) are never together, we can follow these steps:

### Step 1: Calculate the total arrangements of the letters in "DIRECTOR"
The word "DIRECTOR" consists of 8 letters where 'R' appears twice. The formula for the total arrangements of letters where some letters are repeated is given by:

\[ \text{Total arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots} \]

Here, \( n \) is the total number of letters, and \( p_1, p_2, \ldots \) are the frequencies of the repeated letters.

For "DIRECTOR":
- Total letters (n) = 8
- Repeated letter (R) = 2

Thus, the total arrangements are:

\[ \text{Total arrangements} = \frac{8!}{2!} \]

Calculating this gives:

\[ 8! = 40320 \quad \text{and} \quad 2! = 2 \]

So,

\[ \text{Total arrangements} = \frac{40320}{2} = 20160 \]

### Step 2: Calculate arrangements where the three vowels are together
To find the arrangements where the vowels (I, E, O) are together, we can treat the three vowels as a single unit or block. Thus, we can consider the block of vowels as one letter.

Now, we have the following letters to arrange:
- Vowel block (IEO)
- D
- R
- C
- T
- R

This gives us a total of 6 units (the vowel block + D + R + C + T + R).

The arrangement of these 6 units, where R is repeated twice, is given by:

\[ \text{Arrangements with vowels together} = \frac{6!}{2!} \]

Calculating this gives:

\[ 6! = 720 \quad \text{and} \quad 2! = 2 \]

So,

\[ \text{Arrangements with vowels together} = \frac{720}{2} = 360 \]

### Step 3: Calculate arrangements of the vowels within the block
The vowels (I, E, O) can be arranged among themselves in:

\[ 3! = 6 \text{ ways} \]

### Step 4: Total arrangements where vowels are together
Now, we multiply the arrangements of the blocks by the arrangements of the vowels:

\[ \text{Total arrangements with vowels together} = 360 \times 6 = 2160 \]

### Step 5: Calculate arrangements where vowels are not together
To find the arrangements where the vowels are not together, we subtract the arrangements where they are together from the total arrangements:

\[ \text{Arrangements where vowels are not together} = \text{Total arrangements} - \text{Total arrangements with vowels together} \]

Substituting the values we calculated:

\[ \text{Arrangements where vowels are not together} = 20160 - 2160 = 17900 \]

### Final Answer
Thus, the number of ways to arrange the letters of the word "DIRECTOR" such that the three vowels are never together is:

\[ \boxed{17900} \]