The number of ways in which the letters of the word FRACTION be arranged so that no vowels are together is

Options

A. 14400
B. 16440
C. 17330
D. none of these

Answer

To solve the problem of arranging the letters of the word "FRACTION" such that no vowels are together, we can follow these steps:

### Step 1: Identify the Letters
The word "FRACTION" consists of 8 letters: F, R, A, C, T, I, O, N. Among these, the vowels are A, I, and O. Therefore, we have:
- Vowels: A, I, O (3 vowels)
- Consonants: F, R, C, T, N (5 consonants)

### Step 2: Arrange the Consonants
First, we will arrange the consonants. The consonants are F, R, C, T, N. The number of ways to arrange these 5 consonants is given by:
\[ 5! = 120 \]

### Step 3: Identify the Gaps for Vowels
Once the consonants are arranged, we can visualize the arrangement as follows:
- _ F _ R _ C _ T _ N _

This arrangement creates 6 gaps (represented by underscores) where the vowels can be placed.

### Step 4: Place the Vowels in the Gaps
To ensure that no vowels are together, we need to choose 3 out of these 6 gaps to place the vowels. The number of ways to choose 3 gaps from 6 is given by the combination formula:
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

### Step 5: Arrange the Vowels
The vowels A, I, and O can be arranged among themselves in:
\[ 3! = 6 \]

### Step 6: Calculate the Total Arrangements
Now, we can calculate the total number of arrangements where no vowels are together by multiplying the number of arrangements of consonants, the number of ways to choose gaps, and the arrangements of vowels:
\[ \text{Total arrangements} = (5!) \times \left(\binom{6}{3}\right) \times (3!) \]
\[ = 120 \times 20 \times 6 \]
\[ = 120 \times 120 \]
\[ = 14400 \]

### Final Answer
Thus, the total number of ways to arrange the letters of the word "FRACTION" such that no vowels are together is **14400**.

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