<p>A man parks his car among <em>n</em> cars standing in a row, his car not being parked at an end. On his return he finds that exactly <em>m</em> of the <em>n</em> cars are still there. What is the probability that both the cars parked on two sides of his car have left?</p>
<p>\(\dfrac{(m-1)(m-2)}{(n-1)(n-2)}\)</p>
<p>\(\dfrac{(n-m)(n-m-1)}{(n-1)(n-2)}\)</p>
<p>\(\dfrac{m(m-1)}{(n-1)(n-2)}\)</p>
<p>\(\dfrac{(n-m-1)(n-m-2)}{(n-1)(n-2)}\)</p>
Step-by-Step Solution
Key Concept: The man's car position is fixed (not at an end), so we need to find the probability that both neighboring cars have left given that exactly m cars remain from the original n cars. This requires counting favorable outcomes where both adjacent positions are empty among the m remaining cars.
<p><strong>Step 1:</strong> The man's car is parked at some position (not at an end), dividing the row into left and right sides. Since his car remains, we focus on the other (n-1) cars.</p><p><strong>Step 2:</strong> Exactly m cars remain total, so exactly m cars total are present (including his car). This means (n-m) cars have left from the original n cars.</p><p><strong>Step 3:</strong> Among the (n-1) other cars, exactly (m-1) remain and (n-m) have left. The total ways to choose which (m-1) cars remain from (n-1) cars is C(n-1, m-1).</p><p><strong>Step 4:</strong> For both adjacent cars to have left, we need: both the car on the left and car on the right of his car are among the (n-m) cars that left. This means these 2 specific positions must be empty.</p><p><strong>Step 5:</strong> If both neighbors left, then from the remaining (n-3) cars (excluding his car and the 2 neighbors), exactly (m-1) cars must remain. The number of favorable ways is C(n-3, m-1).</p><p><strong>Step 6:</strong> Probability = C(n-3, m-1) / C(n-1, m-1) = [C(n-3, m-1)] / [C(n-1, m-1)]</p><p><strong>Step 7:</strong> Using the formula: C(n-3, m-1) / C(n-1, m-1) = [(n-3)!(m-1)!(n-m)!] / [(m-1)!(n-m-2)!(n-1)!] = [(n-m)(n-m-1)] / [(n-1)(n-2)]</p><p>∴ Answer: <strong>B</strong> [The probability is (n-m)(n-m-1)/[(n-1)(n-2)]]</p>
Correct Answer: B