The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is
Question:
The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is
The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is
Options
Answer: (1) 1440
Explanation:
Step 1: Understand the problem We want integers greater than 2000 formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once. The integers can be 4, 5, or 6 digits (because digits 0–5, so max 6 digits). But let's check carefully: To be greater than 2000, a 4-digit number must have the thousands digit ≥ 2. The digits are 0, 1, 2, 3, 4, 5. Step 2: Count 4-digit numbers Let’s form numbers abcd, where a is the thousands digit. Case 1: a = 2, 3, 4, or 5 (to be >2000) a = 2: Remaining digits for b, c, d: 0,1,3,4,5 → choose 3 digits in 3! = 6 ways? Wait, careful: There are 5 remaining digits, and we need to arrange 3 of them: Number of arrangements = P(5,3) = 5 × 4 × 3 = 60 Similarly, a = 3, 4, 5 → each gives 60 numbers Total 4-digit numbers = 4 × 60 = 240 ✅ Step 2 done Step 3: Count 5-digit numbers Any 5-digit number with digits 0–5 and no repetition is automatically >2000, because first digit ≥1, but since we only exclude numbers ≤2000, all 5-digit numbers formed from these digits are >2000. Number of 5-digit numbers: First digit ≠ 0 → 5 options (1,2,3,4,5) Remaining 4 digits: choose from remaining 5 digits → arrange 4 digits = 5 × 4 × 3 × 2 = 120 Total 5-digit numbers = 5 × 120 = 600 Step 4: Count 6-digit numbers 6-digit numbers use all digits 0–5 → automatically >2000 Number of arrangements: first digit ≠ 0 → 5 options Remaining 5 digits can be arranged in 5! = 120 ways Total 6-digit numbers = 5 × 120 = 600 Step 5: Add them up 4-digit numbers: 240 5-digit numbers: 600 6-digit numbers: 600 Total = 240 + 600 + 600 = 1440
Explanation:
Step 1: Understand the problem We want integers greater than 2000 formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once. The integers can be 4, 5, or 6 digits (because digits 0–5, so max 6 digits). But let's check carefully: To be greater than 2000, a 4-digit number must have the thousands digit ≥ 2. The digits are 0, 1, 2, 3, 4, 5. Step 2: Count 4-digit numbers Let’s form numbers abcd, where a is the thousands digit. Case 1: a = 2, 3, 4, or 5 (to be >2000) a = 2: Remaining digits for b, c, d: 0,1,3,4,5 → choose 3 digits in 3! = 6 ways? Wait, careful: There are 5 remaining digits, and we need to arrange 3 of them: Number of arrangements = P(5,3) = 5 × 4 × 3 = 60 Similarly, a = 3, 4, 5 → each gives 60 numbers Total 4-digit numbers = 4 × 60 = 240 ✅ Step 2 done Step 3: Count 5-digit numbers Any 5-digit number with digits 0–5 and no repetition is automatically >2000, because first digit ≥1, but since we only exclude numbers ≤2000, all 5-digit numbers formed from these digits are >2000. Number of 5-digit numbers: First digit ≠ 0 → 5 options (1,2,3,4,5) Remaining 4 digits: choose from remaining 5 digits → arrange 4 digits = 5 × 4 × 3 × 2 = 120 Total 5-digit numbers = 5 × 120 = 600 Step 4: Count 6-digit numbers 6-digit numbers use all digits 0–5 → automatically >2000 Number of arrangements: first digit ≠ 0 → 5 options Remaining 5 digits can be arranged in 5! = 120 ways Total 6-digit numbers = 5 × 120 = 600 Step 5: Add them up 4-digit numbers: 240 5-digit numbers: 600 6-digit numbers: 600 Total = 240 + 600 + 600 = 1440
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