Permutations & Combinations
Arrangements with Restrictions
Grade 11
Question:
<p>There are \(n\) distinct white and \(n\) distinct black balls. If the number of ways of arranging them in a row so that neighboring balls are of different colors is 1152, then the value of \(n\) is ___.</p>
Step-by-Step Solution
Key Concept: For alternating color arrangement, fix one color's positions first (either all whites or all blacks in alternate slots), then permute whites and blacks separately. The total arrangements = 2 × n! × n! (factor of 2 accounts for starting with either color).
<p><strong>Step 1:</strong> For neighboring balls to have different colors, we must arrange them in alternating pattern: W-B-W-B-... or B-W-B-W-...</p><p><strong>Step 2:</strong> There are 2 ways to choose which color starts (either white or black).</p><p><strong>Step 3:</strong> Once the color pattern is fixed, we can arrange n distinct whites in n! ways and n distinct blacks in n! ways.</p><p><strong>Step 4:</strong> Total arrangements = 2 × n! × n!</p><p><strong>Step 5:</strong> Given: 2 × n! × n! = 1152</p><p>Therefore: n! × n! = 576</p><p><strong>Step 6:</strong> We need (n!)² = 576</p><p>Taking square root: n! = 24 = 4!</p><p><strong>Step 7:</strong> Therefore n = 4</p><p><strong>Verification:</strong> 2 × 4! × 4! = 2 × 24 × 24 = 1152 ✓</p><p>∴ Answer: 4</p>
Correct Answer: 4