Find the number of ways in which 12 different books can be arranged on a shelf so that two particular books shall not be together

Options

A. a. 12!
B. b. `1320 xx 9!`
C. c. 387892
D. d. 39916800

Answer

To solve the problem of arranging 12 different books on a shelf such that two particular books (let's call them B11 and B12) are not together, we can follow these steps:

### Step 1: Calculate the total arrangements of 12 books
The total number of arrangements of 12 different books is given by the factorial of the number of books.

\[ \text{Total arrangements} = 12! = 479001600 \]

### Step 2: Calculate the arrangements where the two particular books are together
To find the arrangements where B11 and B12 are together, we can treat them as a single unit or block. This means we now have 11 units to arrange (the block of B11 and B12 plus the other 10 books).

\[ \text{Arrangements with B11 and B12 together} = 11! \]

Since B11 and B12 can be arranged within their block in 2 ways (B11 first or B12 first), we multiply by 2:

\[ \text{Total arrangements with B11 and B12 together} = 11! \times 2 = 3991680 \times 2 = 7983360 \]

### Step 3: Calculate the arrangements where the two particular books are not together
To find the number of arrangements where B11 and B12 are not together, we subtract the arrangements where they are together from the total arrangements:

\[ \text{Arrangements where B11 and B12 are not together} = 12! - (11! \times 2) \]

Substituting the values we calculated:

\[ \text{Arrangements where B11 and B12 are not together} = 479001600 - 7983360 = 471018240 \]

### Final Answer
Thus, the number of ways to arrange the 12 different books such that the two particular books are not together is:

\[ \boxed{471018240} \]