In how many ways can 20 books be arranged on a shelf so that a particular pair of books shall not come together?

Answer

To solve the problem of arranging 20 books on a shelf such that a particular pair of books (let's call them B1 and B2) do not come together, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate Total Arrangements of 20 Books:**
The total number of ways to arrange 20 books without any restrictions is given by the factorial of the number of books.
\[ \text{Total arrangements} = 20! \]

2. **Consider B1 and B2 as a Single Entity:**
To find the arrangements where B1 and B2 are together, we can treat them as a single entity or "block." This means we now have 19 entities to arrange (the B1-B2 block and the other 18 books).
\[ \text{Arrangements with B1 and B2 together} = 19! \]

3. **Account for the Internal Arrangement of B1 and B2:**
Within the block, B1 and B2 can be arranged in 2 ways (B1 first or B2 first).
\[ \text{Internal arrangements of B1 and B2} = 2! \]

4. **Calculate Total Arrangements with B1 and B2 Together:**
Therefore, the total arrangements where B1 and B2 are together is:
\[ \text{Total arrangements with B1 and B2 together} = 19! \times 2! \]

5. **Calculate Arrangements Where B1 and B2 Do Not Come Together:**
To find the number of arrangements where B1 and B2 do not come together, we subtract the arrangements where they are together from the total arrangements.
\[ \text{Arrangements where B1 and B2 do not come together} = 20! - (19! \times 2!) \]

6. **Simplify the Expression:**
Since \(2! = 2\), we can simplify:
\[ \text{Arrangements where B1 and B2 do not come together} = 20! - 2 \times 19! \]

7. **Factor Out 19!:**
We can factor out \(19!\) from the expression:
\[ = 19! \times (20 - 2) = 19! \times 18 \]

### Final Answer:
Thus, the number of ways to arrange the 20 books such that the particular pair of books does not come together is:
\[ \text{Final Answer} = 18 \times 19! \]