<p>How many numbers of <em>n</em> digits can be made with the nonzero digits in which no two consecutive digits are the same?</p>
Step-by-Step Solution
Key Concept: Use the multiplication principle with a recurrence structure: the first digit has 9 choices (1-9), and each subsequent digit has 8 choices (any nonzero digit except the previous one). This gives 9 × 8^(n-1) total n-digit numbers.
<p><strong>Step 1:</strong> Identify constraints. We need n-digit numbers using only nonzero digits {1,2,3,4,5,6,7,8,9}, where no two consecutive digits are equal.</p><p><strong>Step 2:</strong> Count choices for the first digit. The first digit can be any of the 9 nonzero digits. <strong>Choices = 9</strong></p><p><strong>Step 3:</strong> Count choices for the second digit. The second digit must be a nonzero digit different from the first digit. Since there are 9 nonzero digits total and we must exclude 1 (the first digit), we have 8 choices.</p><p><strong>Step 4:</strong> Count choices for remaining digits. For each position after the first, the digit must be nonzero and different from the previous digit. This gives us 8 choices at each of the remaining (n-1) positions.</p><p><strong>Step 5:</strong> Apply multiplication principle. Total numbers = (choices for 1st digit) × (choices for 2nd) × (choices for 3rd) × ... × (choices for nth)</p><p style='font-size: 14px;'>= 9 × 8 × 8 × ... × 8 (with 8 appearing (n-1) times)</p><p style='font-size: 14px;'>= 9 × 8^(n-1)</p><p><strong>∴ Answer: </strong>\(9 \times 8^{n-1}\)</p>
Correct Answer: \(9 \times 8^{n-1}\)