In how many ways can 10 books on English and 8 books on physics be placed in a row on a shelf so that two books on physics may not be together?

Options

A. 165
B. 195
C. 155
D. 135

Answer

To solve the problem of arranging 10 English books and 8 Physics books on a shelf such that no two Physics books are adjacent, we can follow these steps:

### Step 1: Arrange the English Books
First, we arrange the 10 English books. The number of ways to arrange 10 distinct English books is given by the factorial of the number of books:

\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800 \]

### Step 2: Identify the Gaps for Physics Books
Once the English books are arranged, we can identify the gaps where the Physics books can be placed. When we arrange 10 English books, they create gaps where the Physics books can be placed.

For 10 English books, the gaps are as follows:
- Before the first English book
- Between the English books (9 gaps)
- After the last English book

This gives us a total of \(10 + 1 = 11\) gaps.

### Step 3: Choose Gaps for Physics Books
Now, we need to choose 8 gaps from the 11 available gaps to place the Physics books. The number of ways to choose 8 gaps from 11 is given by the combination formula:

\[ \binom{11}{8} = \frac{11!}{8!(11-8)!} = \frac{11!}{8! \cdot 3!} \]

Calculating this, we have:

\[ \binom{11}{8} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \]

### Step 4: Arrange the Physics Books
Finally, we can arrange the 8 Physics books in the chosen gaps. The number of ways to arrange 8 distinct Physics books is given by:

\[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320 \]

### Step 5: Calculate the Total Arrangements
Now, we can find the total number of arrangements by multiplying the number of ways to arrange the English books, the number of ways to choose the gaps, and the number of ways to arrange the Physics books:

\[ \text{Total arrangements} = 10! \times \binom{11}{8} \times 8! \]

Substituting the values we calculated:

\[ \text{Total arrangements} = 3,628,800 \times 165 \times 40,320 \]

Calculating this gives:

\[ \text{Total arrangements} = 3,628,800 \times 165 \times 40,320 = 24,576,000,000 \]

Thus, the total number of ways to arrange the books such that no two Physics books are together is **24,576,000,000**.