Number of ways in which 15 different books can be arraged on a shelf so that two particular books shall not be together is

Options

A. `14xx 15!`
B. `13xx14!`
C. 14! `xx` 15!
D. `(15!)^(2)`

Answer

To find the number of ways in which 15 different books can be arranged on a shelf such that two particular books are not together, we can follow these steps:

### Step 1: Calculate the total arrangements of 15 books
The total number of ways to arrange 15 different books is given by the factorial of the number of books. Therefore, the total arrangements are:

\[ 15! \]

### Step 2: Calculate the arrangements where the two particular books are together
To find the arrangements where the two specific books (let's call them A and B) are together, we can treat them as a single unit or block. This means we now have 14 units to arrange (the block of A and B plus the other 13 books).

The number of arrangements of these 14 units is:

\[ 14! \]

Since A and B can be arranged within their block in 2 different ways (AB or BA), we multiply by 2:

\[ \text{Arrangements with A and B together} = 14! \times 2! \]

### Step 3: Calculate the arrangements where the two particular books are not together
To find the arrangements where A and B are not together, we subtract the number of arrangements where they are together from the total arrangements:

\[ \text{Arrangements where A and B are not together} = 15! - (14! \times 2!) \]

### Step 4: Simplify the expression
We can factor out \(14!\) from the expression:

\[ 15! = 15 \times 14! \]
\[ 2! = 2 \]

Thus, we can rewrite the equation:

\[ \text{Arrangements where A and B are not together} = 15 \times 14! - 2 \times 14! \]

Factoring out \(14!\):

\[ = 14! \times (15 - 2) = 14! \times 13 \]

### Final Answer
The number of ways in which 15 different books can be arranged on a shelf so that the two particular books are not together is:

\[ \text{Required number of ways} = 13 \times 14! \]

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