Different words are being formed by arranging the letters of the word "SUCCESS".
The probability of choosing a word in which the two C are together but no two S are together is

Options

A. `2/35`
B. `2/7`
C. `24/105`
D. 1

Answer

To solve the problem of finding the probability of forming words from the letters of "SUCCESS" where the two C's are together and no two S's are together, we will follow these steps:

### Step 1: Calculate the total number of arrangements of the letters in "SUCCESS"

The word "SUCCESS" consists of the letters S, U, C, E, S, S, C. The counts of each letter are:
- S: 3
- C: 2
- U: 1
- E: 1

The total number of letters is 7. The formula for the number of arrangements of letters where some letters are repeated is given by:

\[ \text{Total arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!} \]

where \( n \) is the total number of letters and \( n_1, n_2, \ldots, n_k \) are the frequencies of the repeated letters.

Using this formula, we have:

\[ \text{Total arrangements} = \frac{7!}{3! \cdot 2! \cdot 1! \cdot 1!} = \frac{5040}{6 \cdot 2 \cdot 1 \cdot 1} = \frac{5040}{12} = 420 \]

### Step 2: Calculate the number of arrangements where the two C's are together

To treat the two C's as a single unit, we can denote this unit as "CC". The letters we now have are S, S, S, U, E, and CC. This gives us the new arrangement:

- S: 3
- U: 1
- E: 1
- CC: 1

Now we have a total of 6 units (3 S's, 1 U, 1 E, and 1 CC). The number of arrangements of these 6 units is:

\[ \text{Arrangements with CC together} = \frac{6!}{3! \cdot 1! \cdot 1!} = \frac{720}{6} = 120 \]

### Step 3: Ensure that no two S's are together

To ensure that no two S's are together, we will first place U, E, and CC. These three letters can be arranged in:

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

Now, we have arranged U, E, and CC, which creates gaps where we can place the S's. The arrangement looks something like this:

- _ U _ E _ CC _

This creates 4 gaps (before U, between U and E, between E and CC, and after CC). We need to choose 3 out of these 4 gaps to place the S's, ensuring that no two S's are together.

The number of ways to choose 3 gaps from 4 is given by:

\[ \binom{4}{3} = 4 \]

Now, we can multiply the arrangements of U, E, and CC by the ways to choose gaps for S's:

\[ \text{Total arrangements with CC together and no two S's together} = 6 \cdot 4 = 24 \]

### Step 4: Calculate the probability

The probability of choosing a word in which the two C's are together and no two S's are together is given by the ratio of the favorable outcomes to the total outcomes:

\[ \text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{24}{420} = \frac{2}{35} \]

### Final Answer

Thus, the probability of choosing a word in which the two C's are together but no two S's are together is:

\[ \frac{2}{35} \]