Find of the total number of words which are formed by using the all the letters of the word "SUCCESS" such the no two alike letters are together.

Answer

To solve the problem of finding the total number of words formed by using all the letters of the word "SUCCESS" such that no two alike letters are together, we can follow these steps:

### Step 1: Identify the letters and their frequencies
The word "SUCCESS" consists of:
- S: 3
- U: 1
- C: 2
- E: 1

### Step 2: Calculate the total arrangements without restrictions
First, we calculate the total arrangements of the letters in "SUCCESS" without any restrictions. The formula for arrangements of letters where some letters are repeated is given by:

\[ \text{Total arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdots} \]

Where \( n \) is the total number of letters, and \( n_1, n_2, n_3, \ldots \) are the frequencies of each distinct letter.

In our case:
- Total letters, \( n = 7 \) (3 S's, 2 C's, 1 U, 1 E)
- Frequencies: \( n_S = 3, n_C = 2, n_U = 1, n_E = 1 \)

Thus, the total arrangements are:

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

### Step 3: Calculate arrangements with restricted letters (C's together)
Next, we consider the scenario where the two C's are together. We treat the two C's as a single entity (CC). The letters we now have are:
- S: 3
- U: 1
- CC: 1 (as a single entity)
- E: 1

This gives us a total of 6 entities: S, S, S, U, CC, E.

The arrangements of these 6 entities are:

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

### Step 4: Calculate arrangements with restricted letters (S's together)
Now, we consider the scenario where the three S's are together. We treat the three S's as a single entity (SSS). The letters we have now are:
- SSS: 1 (as a single entity)
- U: 1
- C: 2
- E: 1

This gives us a total of 5 entities: SSS, U, C, C, E.

The arrangements of these 5 entities are:

\[ \text{Arrangements with SSS together} = \frac{5!}{2! \cdot 1! \cdot 1! \cdot 1!} = \frac{120}{2} = 60 \]

### Step 5: Calculate arrangements with both C's and S's together
Now we consider the scenario where both the S's and C's are together. We treat SSS as one entity and CC as another entity. The letters we have now are:
- SSS: 1
- U: 1
- CC: 1
- E: 1

This gives us a total of 4 entities: SSS, U, CC, E.

The arrangements of these 4 entities are:

\[ \text{Arrangements with SSS and CC together} = \frac{4!}{1! \cdot 1! \cdot 1! \cdot 1!} = 24 \]

### Step 6: Apply the principle of inclusion-exclusion
Now, we can use the principle of inclusion-exclusion to find the total arrangements where no two alike letters are together:

\[ \text{Total arrangements with restrictions} = \text{Total arrangements} - (\text{Arrangements with CC together} + \text{Arrangements with SSS together} - \text{Arrangements with both together}) \]

Substituting the values we calculated:

\[ \text{Total arrangements with restrictions} = 420 - (120 + 60 - 24) = 420 - 156 = 264 \]

### Final Answer
The total number of words formed by using all the letters of the word "SUCCESS" such that no two alike letters are together is **264**.