How many positive integers up to and including 1000 have no repeated digits?
Answer and solution
Correct answer: 738
Solution
Answer: 738
1. Natural numbers are positive integers for this question; start at 1, not 0.
2. One-digit choices: 1 through 9, giving 9.
3. Two-digit choices: first digit has 9 nonzero choices; second has 9 choices including zero but excluding the first digit. Count 81.
4. Three digits: 9 choices, then 9, then 8, giving 648.
5. The only four-digit candidate is 1000, which repeats zero. Total=9+81+648=738.
Why this method works:
The leading digit cannot be 0, but later digits may be 0 unless already used. Therefore the second digit of a three-digit number has 9 choices, not 8: only the first digit is excluded from all 10 digits.
Independent check and common trap:
The disjoint counts 9,81,648 cover all positive numbers below 1000. The sole remaining candidate 1000 repeats 0, so it contributes 0. Their sum 738 can also be checked by enumerating the integers 1 through 1000 and comparing each digit count with its number of distinct digits.
Apply: Count 1-,2-,3-digit numbers: 9+9×9+9×9×8=738. The endpoint 1000 repeats zero, so contributes nothing.
Check: The disjoint counts 9,81,648 cover all positive numbers below 1000. The sole remaining candidate 1000 repeats 0, so it contributes 0. Their sum 738 can also be checked by enumerating the integers 1 through 1000 and comparing each digit count with its number of distinct digits.
1. Natural numbers are positive integers for this question; start at 1, not 0.
2. One-digit choices: 1 through 9, giving 9.
3. Two-digit choices: first digit has 9 nonzero choices; second has 9 choices including zero but excluding the first digit. Count 81.
4. Three digits: 9 choices, then 9, then 8, giving 648.
5. The only four-digit candidate is 1000, which repeats zero. Total=9+81+648=738.
Why this method works:
The leading digit cannot be 0, but later digits may be 0 unless already used. Therefore the second digit of a three-digit number has 9 choices, not 8: only the first digit is excluded from all 10 digits.
Independent check and common trap:
The disjoint counts 9,81,648 cover all positive numbers below 1000. The sole remaining candidate 1000 repeats 0, so it contributes 0. Their sum 738 can also be checked by enumerating the integers 1 through 1000 and comparing each digit count with its number of distinct digits.
G Strategy shortcut
Spot: Permutations and Combinations.Apply: Count 1-,2-,3-digit numbers: 9+9×9+9×9×8=738. The endpoint 1000 repeats zero, so contributes nothing.
Check: The disjoint counts 9,81,648 cover all positive numbers below 1000. The sole remaining candidate 1000 repeats 0, so it contributes 0. Their sum 738 can also be checked by enumerating the integers 1 through 1000 and comparing each digit count with its number of distinct digits.
G Strategy: Spot: Permutations and Combinations. Apply: Count 1-,2-,3-digit numbers: 9+9×9+9×9×8=738. The endpoint 1000 repeats zero, so contributes nothing. Check: The disjoint counts 9,81,648 cover all positive numbers below 1000. The sole remaining candidate 1000 repeats 0, so it contributes 0. Their sum 738 can also be checked by enumerating the integers 1 through 1000 and comparing each digit count with its number of distinct digits.