How many different rearrangements of the letters in the word CONTEST start with the two vowels ?

Options

A. 120
B. 60
C. 10
D. 5

Explanation

To find the number of different rearrangements of the letters in the word "CONTEST" that start with the two vowels, we can follow these steps:

### Step 1: Identify the vowels and consonants
The word "CONTEST" consists of the letters:
- Vowels: O, E
- Consonants: C, N, T, S, T

### Step 2: Fix the first two letters
Since we want the arrangements to start with the two vowels, we can fix the first two letters as "OE" or "EO".

### Step 3: Count the remaining letters
After fixing the first two letters, we have the following letters left to arrange:
- Remaining letters after "OE": C, N, T, S, T
- Remaining letters after "EO": C, N, T, S, T

### Step 4: Calculate the arrangements of the remaining letters
The remaining letters consist of 5 letters: C, N, T, S, T. Here, T appears twice, so we need to account for this repetition when calculating the arrangements.

The formula for the arrangements of n items where there are repetitions is given by:
\[ \text{Number of arrangements} = \frac{n!}{p_1! \times p_2! \times \ldots \times p_k!} \]
where \( n \) is the total number of items, and \( p_1, p_2, \ldots, p_k \) are the frequencies of the repeated items.

In our case:
- Total letters = 5 (C, N, T, S, T)
- T is repeated 2 times.

Thus, the number of arrangements is:
\[ \text{Arrangements} = \frac{5!}{2!} \]

### Step 5: Calculate the factorial values
Calculating the factorials:
\[ 5! = 120 \quad \text{and} \quad 2! = 2 \]
So,
\[ \text{Arrangements} = \frac{120}{2} = 60 \]

### Step 6: Account for both vowel arrangements
Since we can start with either "OE" or "EO", we multiply the number of arrangements by 2:
\[ \text{Total arrangements} = 2 \times 60 = 120 \]

### Final Answer
Thus, the total number of different rearrangements of the letters in the word "CONTEST" that start with the two vowels is **120**.

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