The number of all natural numbers up to 1000 with non-repeating digits is Option 1: 738 Option 2: 648 Option 3: 504 Option 4: 585

Answer: 738. Solution: We need to count all natural numbers from 1 to 1000 where no digit appears more than once in the number. Valid examples: 123, 456, 789 (all digits are different) Invalid examples: 112, 233, 1001 (digits repeat) The number 1000 has three zeros, so digits repeat. Any number above 1000 would be 4+ digits, which we don't need to consider. We'll use the fundamental counting principle: if we have multiple choices to make, we multiply the number of options for each choice. Single-digit numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9 Count = 9 numbers We don't include 0 because natural numbers start from 1. For a two-digit number: _ _ First digit (tens place): - Can be: 1, 2, 3, 4, 5, 6, 7, 8, 9 - Cannot be 0 (otherwise it wouldn't be a two-digit number) - 9 choices Second digit (units place): - Can be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 - Cannot be the same as the first digit (no repetition allowed) - 9 choices (10 total digits - 1 already used) Total two-digit numbers = 9 x 9 = 81 For a three-digit number: _ _ _ First digit (hundreds place): - Can be: 1, 2, 3, 4, 5, 6, 7, 8, 9 - Cannot be 0 (otherwise it wouldn't be a three-digit number) - 9 choices Second digit (tens place): - Can be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 - Cannot be the same as the first digit - 9 choices (10 total digits - 1 already used) Third digit (units place): - Can be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 - Cannot be the same as the first OR second digit - 8 choices (10 total digits - 2 already used) Total three-digit numbers = 9 x 9 x 8 = 648 The only four-digit number ≤ 1000 is 1000 itself. But 1000 = 1, 0, 0, 0 has repeating digits (three zeros), so it doesn't qualify. Count = 0 Total numbers = Single-digit + Two-digit + Three-digit Total = 9 + 81 + 648 = 738 When counting arrangements with restrictions: 1. We identify the positions to fill 2. We count choices for each position (considering restrictions) 3. We apply the fundamental counting principle 4. We remember: Once a digit is used, it reduces options for subsequent positions This systematic approach works for any similar counting problem.

Modern Math

Permutation & Combination

Medium

The number of all natural numbers up to $1000$ with non-repeating digits is

$738$

$648$

$504$

$585$

✅ Correct Option: 1

We need to count all natural numbers from 1 to 1000 where no digit appears more than once in the number.

Valid examples: 123, 456, 789 (all digits are different)

Invalid examples: 112, 233, 1001 (digits repeat)

The number 1000 has three zeros, so digits repeat. Any number above 1000 would be 4+ digits, which we don't need to consider.

We'll use the fundamental counting principle: if we have multiple choices to make, we multiply the number of options for each choice.

Single-digit numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9

Count = 9 numbers

We don't include 0 because natural numbers start from 1.

For a two-digit number: _ _

First digit (tens place):

Can be: 1, 2, 3, 4, 5, 6, 7, 8, 9
Cannot be 0 (otherwise it wouldn't be a two-digit number)
9 choices

Second digit (units place):

Can be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Cannot be the same as the first digit (no repetition allowed)
9 choices (10 total digits - 1 already used)

Total two-digit numbers = $9 \times 9 = 81$

For a three-digit number: _ _ _

First digit (hundreds place):

Can be: 1, 2, 3, 4, 5, 6, 7, 8, 9
Cannot be 0 (otherwise it wouldn't be a three-digit number)
9 choices

Second digit (tens place):

Can be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Cannot be the same as the first digit
9 choices (10 total digits - 1 already used)

Third digit (units place):

Can be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Cannot be the same as the first OR second digit
8 choices (10 total digits - 2 already used)

Total three-digit numbers = $9 \times 9 \times 8 = 648$

The only four-digit number ≤ 1000 is 1000 itself.

But 1000 = 1, 0, 0, 0 has repeating digits (three zeros), so it doesn't qualify.

Count = 0

Total numbers = Single-digit + Two-digit + Three-digit

Total = $9 + 81 + 648 = 738$

When counting arrangements with restrictions:

We identify the positions to fill
We count choices for each position (considering restrictions)
We apply the fundamental counting principle
We remember: Once a digit is used, it reduces options for subsequent positions

This systematic approach works for any similar counting problem.

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