In how many ways `3` books of Mathematics, `4` books of Physics and `2` books of Chemistry can be arranged on a table if all the books of one subject placed together?

Answer

To solve the problem of arranging 3 Mathematics books, 4 Physics books, and 2 Chemistry books on a table with the condition that all books of one subject are placed together, we can follow these steps:

### Step-by-Step Solution:

1. **Group the Books by Subject**:
Since all books of one subject must be placed together, we can treat each subject as a single group. Therefore, we have three groups:
- Group 1: Mathematics (M)
- Group 2: Physics (P)
- Group 3: Chemistry (C)

2. **Arrange the Groups**:
The three groups (M, P, C) can be arranged among themselves. The number of ways to arrange these 3 groups is given by the factorial of the number of groups:
\[ \text{Ways to arrange groups} = 3! = 6 \]

3. **Arrange the Books Within Each Group**:
- For the Mathematics books, there are 3 books, which can be arranged in:
\[ 3! = 6 \text{ ways} \]
- For the Physics books, there are 4 books, which can be arranged in:
\[ 4! = 24 \text{ ways} \]
- For the Chemistry books, there are 2 books, which can be arranged in:
\[ 2! = 2 \text{ ways} \]

4. **Calculate the Total Arrangements**:
To find the total number of arrangements, we multiply the number of ways to arrange the groups by the number of arrangements within each group:
\[ \text{Total arrangements} = (\text{Ways to arrange groups}) \times (\text{Ways to arrange Mathematics}) \times (\text{Ways to arrange Physics}) \times (\text{Ways to arrange Chemistry}) \]
Substituting the values we calculated:
\[ \text{Total arrangements} = 3! \times 3! \times 4! \times 2! = 6 \times 6 \times 24 \times 2 \]

5. **Perform the Multiplication**:
\[ 6 \times 6 = 36 \]
\[ 36 \times 24 = 864 \]
\[ 864 \times 2 = 1728 \]

Thus, the total number of ways to arrange the books is:
\[ \boxed{1728} \]