Question

The numbers \(1, 2, 3, \ldots, m\) are arranged in random order. The number of ways this can be done, so that the numbers \(1, 2, \ldots, r \, (r < m)\) appear as neighbours is:

• A. \((m - r)!\)
• B. \((m - r + 1)!\)
• C. \((m - r)!r!\)
• D. \((m - r + 1)!r!\)

Hint:

When numbers are restricted to be together, treat them as a single block and arrange accordingly. Multiply by the number of ways to arrange the numbers within the block.

Answer 1

The Correct Option is B

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:

\[ (m - r + 1)! r! \]