<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 4-digit number into independent choices: fix the range constraint (a = 4-9), apply divisibility by 5 (d ∈ {0,5}), and apply the middle digit constraint (b ∈ {3,4,5,6,7,8,9}), then multiply using the multiplication principle.
<p><strong>Step 1:</strong> Analyze the constraint N ≥ 4000. Since N = abcd is a 4-digit number, we need a ≥ 4. Combined with a ≤ 9, we have a ∈ {4,5,6,7,8,9}, giving <strong>6 choices for a</strong>.</p><p><strong>Step 2:</strong> Apply divisibility by 5. A number is divisible by 5 if and only if its last digit is 0 or 5. Therefore d ∈ {0,5}, giving <strong>2 choices for d</strong>.</p><p><strong>Step 3:</strong> Apply the constraint 3 ≤ b. Since b is a digit, b ∈ {0,1,2,...,9}. Combined with 3 ≤ b, we have b ∈ {3,4,5,6,7,8,9}, giving <strong>7 choices for b</strong>.</p><p><strong>Step 4:</strong> Identify c's constraint. The problem statement has no explicit constraint on c, so c ∈ {0,1,2,...,9}, giving <strong>10 choices for c</strong>.</p><p><strong>Step 5:</strong> Apply the multiplication principle. Total count = (choices for a) × (choices for b) × (choices for c) × (choices for d) = 6 × 7 × 10 × 2 = <strong>840</strong>.</p><p>∴ Answer: C</p>
Correct Answer: C