Step 1: We are given the numbers \( 1, 2, 3, \ldots, m \), and we need to arrange them such that the numbers \( 1, 2, \ldots, r \) appear as neighbours.
Step 2: Treat the numbers \( 1, 2, \ldots, r \) as a single block. By doing this, we reduce the problem to arranging \( m - r + 1 \) objects: the block and the remaining \( m - r \) numbers.
Step 3: The number of ways to arrange these \( m - r + 1 \) objects is:
\[ (m - r + 1)! \]
Step 4: Within the block, the \( r \) numbers can be arranged in \( r! \) different ways.
Step 5: Therefore, the total number of arrangements is the product of the two:
\[ (m - r + 1)! \times r! \]
Conclusion:
The total number of ways the numbers can be arranged with the numbers \( 1, 2, \ldots, r \) as neighbours is:
\[ (m - r + 1)! r! \]
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: