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: Given numbers \( 1, 2, 3, \ldots, m \), arrange them so \( 1, 2, \ldots, r \) are neighbors.

Step 2: Consider \( 1, 2, \ldots, r \) as one block. This simplifies to arranging \( m - r + 1 \) items: the block and the other \( m - r \) numbers.

Step 3: The arrangements for these \( m - r + 1 \) objects is:

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

Step 4: The \( r \) numbers within the block can be arranged in \( r! \) ways.

Step 5: Thus, the total arrangements equal the product:

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

Conclusion:

The total arrangements with \( 1, 2, \ldots, r \) as neighbors is:

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