To solve the problem of finding the number of natural numbers from 1 to 1000 that have none of the digits repeated, we can break it down into three cases: one-digit numbers, two-digit numbers, and three-digit numbers.
### Step 1: Count One-Digit Numbers
The one-digit natural numbers are from 1 to 9.
- **Count**: There are 9 one-digit numbers (1, 2, 3, 4, 5, 6, 7, 8, 9).
**Total for one-digit numbers**: 9
### Step 2: Count Two-Digit Numbers
For two-digit numbers, the first digit can be any digit from 1 to 9 (since it cannot be 0), and the second digit can be any digit from 0 to 9, excluding the first digit.
- **First digit choices**: 9 options (1-9)
- **Second digit choices**: 9 options (0-9, excluding the first digit)
**Total for two-digit numbers**:
\[
9 \times 9 = 81
\]
### Step 3: Count Three-Digit Numbers
For three-digit numbers, the first digit can again be any digit from 1 to 9, the second digit can be any digit from 0 to 9 excluding the first digit, and the third digit can be any digit from 0 to 9 excluding the first and second digits.
- **First digit choices**: 9 options (1-9)
- **Second digit choices**: 9 options (0-9, excluding the first digit)
- **Third digit choices**: 8 options (0-9, excluding the first and second digits)
**Total for three-digit numbers**:
\[
9 \times 9 \times 8 = 648
\]
### Step 4: Calculate the Total
Now, we add the totals from all three cases:
\[
\text{Total} = \text{One-digit numbers} + \text{Two-digit numbers} + \text{Three-digit numbers}
\]
\[
\text{Total} = 9 + 81 + 648 = 738
\]
### Conclusion
Thus, the total number of natural numbers from 1 to 1000 that have none of the digits repeated is 738, which does not match the given statement of 691.
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