Permutations & Combinations
Arrangements with restrictions
Grade 11
Question:
<p>There are \((n+1)\) white and \((n+1)\) black balls, each set numbered 1 to \(n+1\). The number of ways in which the balls can be arranged in a row so that the adjacent balls are of different colors is</p>
<p>\((2n+2)!\)</p>
<p>\((2n+2)! \times 2\)</p>
<p>\((n+1)! \times 2\)</p>
<p>\(2[(n+1)!]^2\)</p>
Step-by-Step Solution
Key Concept: For alternating color arrangements, fix one color's pattern first, then the other color must follow the complementary pattern. With equal counts of both colors, we have exactly 2 valid alternating sequences (starting with white or black), and within each color, numbered balls can be permuted independently.
<p><strong>Step 1:</strong> For adjacent balls to be different colors, the arrangement must alternate: either W-B-W-B-... or B-W-B-W-...</p><p><strong>Step 2:</strong> There are exactly 2 valid color patterns.</p><p><strong>Step 3:</strong> Once the color pattern is fixed, the (n+1) numbered white balls can be arranged in their positions in (n+1)! ways.</p><p><strong>Step 4:</strong> Similarly, the (n+1) numbered black balls can be arranged in their positions in (n+1)! ways.</p><p><strong>Step 5:</strong> By multiplication principle: Total arrangements = 2 × (n+1)! × (n+1)! = 2[(n+1)!]²</p><p>∴ Answer: D</p>
Correct Answer: D