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

Options

A. a. `126xx 10!`
B. b. 125000
C. c. 357500
D. d. 123040

Answer

To find the number of ways in which 12 different books can be arranged on a shelf such that three particular books do not come together, we can follow these steps:

### Step 1: Calculate the total arrangements of 12 books
The total number of ways to arrange 12 different books is given by the factorial of the number of books. Therefore, the total arrangements can be calculated as:

\[ 12! = 479001600 \]

### Step 2: Calculate the arrangements where the three particular books are together
To find the arrangements where the three particular books (let's call them B1, B2, and B3) are together, we can treat these three books as a single unit or block. This means we now have 10 units to arrange (the block of 3 books + the other 9 books).

The number of ways to arrange these 10 units is:

\[ 10! = 3628800 \]

Additionally, within the block of three books, the three books can be arranged among themselves in:

\[ 3! = 6 \]

Thus, the total arrangements where the three particular books are together is:

\[ 10! \times 3! = 3628800 \times 6 = 21772800 \]

### Step 3: Subtract the arrangements where the three books are together from the total arrangements
Now, we subtract the number of arrangements where the three particular books are together from the total arrangements:

\[ \text{Arrangements where the three books are not together} = 12! - (10! \times 3!) \]

Substituting the values we calculated:

\[ = 479001600 - 21772800 = 457228800 \]

### Final Answer
Thus, the number of ways in which 12 different books can be arranged on a shelf so that three particular books shall not be together is:

\[ \boxed{457228800} \]

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