<p><strong>265.</strong> Number of 4 digit numbers of the form \(N = abcd\) which satisfy following three conditions:<br>(i) \(4000 \leq N < 6000\)<br>(ii) \(N\) is a multiple of 5<br>(iii) \(3 \leq b < c \leq 6\)<br>is equal to:</p>
Step-by-Step Solution
Key Concept: Break the problem into independent constraints: first digit a ∈ {4,5,6,7,8,9}, last digit d ∈ {0,5} for divisibility by 5, and middle digit b ∈ {3,4,5,6,7,8,9}. The remaining digit c has no restrictions.
<p><strong>Step 1:</strong> Analyze constraint (i): 4000 ≤ N < 10000</p><p>This means the first digit a ∈ {4, 5, 6, 7, 8, 9}. So <strong>6 choices for a</strong>.</p><p><strong>Step 2:</strong> Analyze constraint (ii): N is a multiple of 5</p><p>The last digit d must be 0 or 5. So <strong>2 choices for d</strong>.</p><p><strong>Step 3:</strong> Analyze constraint (iii): 3 ≤ b < 10</p><p>The second digit b ∈ {3, 4, 5, 6, 7, 8, 9}. So <strong>7 choices for b</strong>.</p><p><strong>Step 4:</strong> Find choices for c (third digit)</p><p>No restriction on c, so c ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. So <strong>10 choices for c</strong>.</p><p><strong>Step 5:</strong> Apply multiplication principle</p><p>Since all constraints are independent, total count = 6 × 7 × 10 × 2 = <strong>840</strong></p><p>∴ Answer: C</p>
Correct Answer: C