<p>How many 4-digit numbers \(N = \overline{abcd}\) are there such that \(4000 \leq N < 6000\), \(3 \leq b \leq 6\), and \(N\) is a multiple of 5?</p>
Step-by-Step Solution
Key Concept: The number N = abcd must satisfy 4000 ≤ N < 5000, which means a = 4 is fixed, and we need to count valid choices for b, c, d with the constraint that all digits are distinct.
<p><strong>Step 1:</strong> Identify the range constraint. Since 4000 ≤ N < 5000, we must have <strong>a = 4</strong> (first digit fixed).</p><p><strong>Step 2:</strong> For digits b, c, d to be distinct from each other and from a = 4:</p><ul><li>For <strong>b</strong>: Choose from {0, 1, 2, 3, 5, 6, 7, 8, 9} → <strong>9 choices</strong></li><li>For <strong>c</strong>: Choose from remaining digits (excluding 4 and b) → <strong>8 choices</strong></li><li>For <strong>d</strong>: Choose from remaining digits (excluding 4, b, and c) → <strong>7 choices</strong></li></ul><p><strong>Step 3:</strong> Apply multiplication principle: Total = 9 × 8 × 7 = <strong>504</strong></p><p><strong>Note:</strong> We use permutation logic P(9,3) = 9!/(9-3)! = 9 × 8 × 7 = 504, selecting 3 distinct digits from 9 available digits (all except 4) where order matters.</p><p>∴ Answer: <strong>504</strong> (Option C)</p>
Correct Answer: C