If the letters of the word ASSASSIN are written down in a row, the probability that no two S's occur together, is

Options

A. `(1)/(35)`
B. `(1)/(21)`
C. `(1)/(14)`
D. `(1)/(28)`

Answer

To solve the problem of finding the probability that no two S's occur together in the arrangement of the letters of the word "ASSASSIN", we will follow these steps:

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

The word "ASSASSIN" consists of:
- 4 S's
- 2 A's
- 1 I
- 1 N

The total number of letters is 8.

The formula for the total arrangements of letters when there are repetitions is given by:

\[ \text{Total arrangements} = \frac{n!}{n_1! \times n_2! \times n_3! \times \ldots} \]

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

Thus, we have:

\[ \text{Total arrangements} = \frac{8!}{4! \times 2! \times 1! \times 1!} \]

### Step 2: Calculate the arrangements where no two S's are together.

To ensure that no two S's are together, we first arrange the letters that are not S (A, A, I, N).

The letters A, A, I, N can be arranged as follows:

\[ \text{Arrangements of A, A, I, N} = \frac{4!}{2! \times 1! \times 1!} = 12 \]

Now, when we arrange A, A, I, N, we create gaps where S's can be placed. The arrangement looks like this:

\[ \_ A \_ A \_ I \_ N \_ \]

This creates 5 gaps (before the first letter, between letters, and after the last letter) where we can place the S's.

### Step 3: Choose gaps to place S's.

We need to choose 4 gaps from these 5 to place the S's. The number of ways to choose 4 gaps from 5 is given by:

\[ \binom{5}{4} = 5 \]

### Step 4: Calculate the total arrangements with the S's placed.

Thus, the total arrangements where no two S's are together is:

\[ \text{Total arrangements with S's not together} = \text{Arrangements of A, A, I, N} \times \text{Ways to choose gaps} \]

\[ = 12 \times 5 = 60 \]

### Step 5: Calculate the probability.

The probability that no two S's occur together is given by the ratio of the favorable outcomes to the total outcomes:

\[ P(\text{no two S's together}) = \frac{\text{Total arrangements with S's not together}}{\text{Total arrangements}} \]

Substituting the values we calculated:

\[ P(\text{no two S's together}) = \frac{60}{\frac{8!}{4! \times 2! \times 1! \times 1!}} \]

Calculating \( 8! \):

\[ 8! = 40320 \]
Calculating \( 4! \times 2! \times 1! \times 1! \):

\[ 4! = 24, \quad 2! = 2 \quad \Rightarrow \quad 4! \times 2! = 24 \times 2 = 48 \]

Thus,

\[ \text{Total arrangements} = \frac{40320}{48} = 840 \]

Now substituting back into the probability formula:

\[ P(\text{no two S's together}) = \frac{60}{840} = \frac{1}{14} \]

### Final Answer:

The probability that no two S's occur together is:

\[ \boxed{\frac{1}{14}} \]