In how many ways can 13 different alphabets (a, b, c, ... m) be arranged so that the alphabets f and g never come together?

Options

A. 13 ! – 12 !
B. 13 ! – 12! / 2!
C. 13 ! – 2 `xx` 12 !
D. None of these

Answer

To solve the problem of arranging 13 different alphabets (a, b, c, ..., m) such that the alphabets f and g never come together, we can follow these steps:

### Step 1: Calculate the total arrangements of 13 alphabets
First, we need to find the total number of arrangements of the 13 different alphabets. The number of ways to arrange n distinct objects is given by n factorial (n!).

For 13 alphabets, the total arrangements are:
\[ 13! \]

### Step 2: Consider f and g as a single unit
Next, we consider the case where f and g are together. We can treat f and g as a single unit or block. This means that instead of 13 individual alphabets, we now have 12 units to arrange (11 individual alphabets + 1 block of f and g).

### Step 3: Calculate arrangements with f and g together
The number of ways to arrange these 12 units (11 individual alphabets + 1 block) is:
\[ 12! \]

However, within the block of f and g, f and g can be arranged in 2 different ways (fg or gf). Therefore, we need to multiply the arrangements of the 12 units by the arrangements within the block:
\[ 12! \times 2 \]

### Step 4: Calculate the arrangements where f and g are not together
To find the arrangements where f and g are not together, we subtract the arrangements where they are together from the total arrangements:
\[ \text{Arrangements where f and g are not together} = 13! - (12! \times 2) \]

### Step 5: Simplify the expression
Now we can simplify the expression:
\[ = 13! - 2 \times 12! \]
Since \( 13! = 13 \times 12! \), we can rewrite it as:
\[ = 13 \times 12! - 2 \times 12! \]
\[ = (13 - 2) \times 12! \]
\[ = 11 \times 12! \]

### Final Answer
Thus, the number of ways to arrange the 13 different alphabets such that f and g never come together is:
\[ 11 \times 12! \]