<p>Let <i>n</i> denote the number of all <i>n</i>-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let \(b_n\) = the number of such <i>n</i>-digit integers ending with digit 1 and \(c_n\) = the number of such <i>n</i>-digit integers ending with digit 0.<br><br><b>Problem 1:</b> The value of \(b_6\) is</p>
Step-by-Step Solution
Key Concept: Recognize that b_n (numbers ending in 1) can be formed by appending 1 to any valid (n-1)-digit number, while c_n (numbers ending in 0) can only be formed by appending 0 to numbers ending in 1. This gives recurrence relations: b_n = b_{n-1} + c_{n-1} and c_n = b_{n-1}.
<p><strong>Step 1:</strong> Establish base cases. For 1-digit numbers: b₁ = 1 (only digit '1'), c₁ = 0 (digit '0' cannot start an n-digit number).</p><p><strong>Step 2:</strong> Set up recurrences. A valid n-digit number ending in 1 comes from any valid (n-1)-digit number + append 1. A valid n-digit number ending in 0 comes only from (n-1)-digit numbers ending in 1 + append 0 (no consecutive 0s). Thus: b_n = b_{n-1} + c_{n-1} and c_n = b_{n-1}.</p><p><strong>Step 3:</strong> Calculate successively:</p><p>n=1: b₁ = 1, c₁ = 0</p><p>n=2: b₂ = 1 + 0 = 1, c₂ = 1</p><p>n=3: b₃ = 1 + 1 = 2, c₃ = 1</p><p>n=4: b₄ = 2 + 1 = 3, c₄ = 2</p><p>n=5: b₅ = 3 + 2 = 5, c₅ = 3</p><p>n=6: b₆ = 5 + 3 = 8, c₆ = 5</p><p><strong>∴ Answer: b₆ = 8 (Option B)</strong></p>
Correct Answer: B