The number of all positive integers up to $500$ with non-repeating digits is

The number of all positive integers up to $500$ with non-repeating digits is

A number has non-repeating digits if no digit appears more than once. For example, 123 has non-repeating digits while 122 has repeating digits (the digit 2 appears twice).

Since we want numbers up to 500, we need to consider 1-digit, 2-digit, and 3-digit numbers separately. Each category has different constraints for the digits we can choose.

---

1-digit numbers

The 1-digit positive integers are: 1, 2, 3, 4, 5, 6, 7, 8, 9

Since each number has only one digit, there's no way for digits to repeat. Every single 1-digit number automatically satisfies our condition.

Count for 1-digit numbers: 9

---

2-digit numbers

A 2-digit number has the form $AB$ where $A$ is the tens digit and $B$ is the units digit.

For non-repeating digits: $A \neq B$ (the two digits must be different)

Choose the tens digit $(A)$:

$A$ can be any digit from 1 to 9. It has 9 choices

$A$ cannot be 0, otherwise it wouldn't be a 2-digit number (it would be a 1-digit number)

Choose the units digit $(B)$:

$B$ can be any digit from 0 to 9, except it cannot equal $A$

We have 10 total digits (0,1,2,3,4,5,6,7,8,9), but since $A$ already used one digit, $B$ has 9 remaining choices

Example: If $A = 3$, then $B$ can be any of {0,1,2,4,5,6,7,8,9} - that's 9 choices.

Total 2-digit numbers: $9 \times 9 = 81$

---

3-digit numbers

A 3-digit number has the form $ABC$. Since we want numbers $\leq 500$, the hundreds digit $A$ can only be 1, 2, 3, or 4.

Choose the hundreds digit $(A)$:

$A$ can be 1, 2, 3, or 4

So $A$ has 4 choices

Choose the tens digit $(B)$:

$B$ can be any digit from 0 to 9, except it cannot equal $A$

Since $A$ used one digit, $B$ has 9 remaining choices

Choose the units digit $(C)$:

$C$ can be any digit from 0 to 9, except it cannot equal $A$ or $B$

Since $A$ and $B$ already used two different digits, $C$ has 8 remaining choices

Total 3-digit numbers: $4 \times 9 \times 8 = 288$

---

Adding all cases together:

1-digit numbers: 9

2-digit numbers: 81

3-digit numbers: 288

Total: $9 + 81 + 288 = 378$