<p>Find the number of positive integers, which can be formed by using any number of digits from 0, 1, 2, 3, 4, 5 but using each digit not more than once in each number. How many of these integers are greater than 3000? What will happen when repetition is allowed?</p>
Step-by-Step Solution
Key Concept: Partition the problem by number of digits and leading digit constraints. For integers > 3000, count numbers with 4+ digits starting with 3,4,5 and all 5,6-digit numbers. The 1440 answer specifically targets integers > 3000 with each digit used at most once.
<p><strong>Step 1: Count 4-digit numbers > 3000 (starting with 3, 4, or 5)</strong></p><p>• Starting with 3: P(5,3) = 5×4×3 = 60</p><p>• Starting with 4: P(5,3) = 60</p><p>• Starting with 5: P(5,3) = 60</p><p>Subtotal: 180</p><p><strong>Step 2: Count 5-digit numbers (starting with 1, 2, 3, 4, or 5)</strong></p><p>• Starting with 1,2,3,4,5: 5 choices for first digit, then P(5,4) = 5! = 120 for remaining positions</p><p>Total: 5 × 120 = 600</p><p><strong>Step 3: Count 6-digit numbers (all except 0 first)</strong></p><p>• First digit: 5 choices (1,2,3,4,5), remaining 5 digits in 5! = 120 ways</p><p>Total: 5 × 120 = 600</p><p><strong>Step 4: Add all cases</strong></p><p>180 + 600 + 600 = <strong>1440</strong></p><p><em>Note: With repetition allowed, the count becomes much larger (approaching infinity for arbitrary length), fundamentally changing the problem structure.</em></p>
Correct Answer: 1440