Permutations & Combinations
Counting
Grade 11
Question:
<p>How many numbers can be made with the digits 3, 4, 5, 6, 7, 8 lying between 3000 and 4000, which are divisible by 5 while repetition of any digit is not allowed in any number?</p>
<p>(1) 60</p>
<p>(2) 12</p>
<p>(3) 120</p>
<p>(4) 24</p>
Step-by-Step Solution
Key Concept: A number between 3000 and 4000 must start with 3 (thousands digit fixed), and to be divisible by 5, it must end in 5 (since only 5 from the given digits is divisible by 5). This leaves 4 remaining digits for 2 middle positions.
<p><strong>Step 1:</strong> Identify the constraints. The number must:</p><ul><li>Lie between 3000 and 4000 → First digit (thousands) = 3 (FIXED)</li><li>Be divisible by 5 → Last digit (units) = 5 (FIXED, since 5 is the only digit divisible by 5)</li><li>No repetition allowed</li></ul><p><strong>Step 2:</strong> Structure of the 4-digit number: <strong>3 _ _ 5</strong></p><p><strong>Step 3:</strong> Fill the two middle positions (hundreds and tens places).</p><ul><li>After using digits 3 and 5, remaining digits available: {4, 6, 7, 8} = 4 digits</li><li>Hundreds place: Choose 1 from 4 remaining digits = 4 ways</li><li>Tens place: Choose 1 from 3 remaining digits = 3 ways</li></ul><p><strong>Step 4:</strong> Apply multiplication principle.</p><p>Total numbers = 4 × 3 = <strong>12</strong></p><p>∴ Answer: D</p>
Correct Answer: D