To find the number of three-digit positive integers that have their hundreds, tens, and units digits in ascending order, we need to consider the possible combinations of digits.

1. The hundreds digit: Since we're looking for three-digit numbers, the hundreds digit can be any number from 1 to 9. So, there are 9 possibilities for the hundreds digit.

2. The tens digit: This digit must be greater than the hundreds digit. So, if the hundreds digit is 1, the tens digit can be any number from 2 to 9, giving us 8 possibilities. If the hundreds digit is 2, the tens digit can be any number from 3 to 9, giving us 7 possibilities. This pattern continues until the hundreds digit is 9, in which case there are no possibilities for the tens digit because there's no number greater than 9.

3. The units digit: This digit must be greater than the tens digit. So, the number of possibilities for the units digit depends on what the tens digit is. If the tens digit is 2, the units digit can be any number from 3 to 9, giving us 7 possibilities. If the tens digit is 3, the units digit can be any number from 4 to 9, giving us 6 possibilities. This pattern continues until the tens digit is 9, in which case there are no possibilities for the units digit because there's no number greater than 9.

So, to find the total number of three-digit positive integers that have their hundreds, tens, and units digits in ascending order, we need to add up all the possible combinations of digits. This can be calculated as follows:

For hundreds digit 1, there are 8 possibilities for the tens digit and for each of these, there are 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 possibilities for the units digit. So, there are 8 * 28 = 224 numbers with hundreds digit 1.

We repeat this calculation for hundreds digits 2 through 8 (we don't need to consider 9 because there are no possibilities for the tens digit in this case). This gives us:

For hundreds digit 2, there are 7 * 21 = 147 numbers.
For hundreds digit 3, there are 6 * 15 = 90 numbers.
For hundreds digit 4, there are 5 * 10 = 50 numbers.
For hundreds digit 5, there are 4 * 6 = 24 numbers.
For hundreds digit 6, there are 3 * 3 = 9 numbers.
For hundreds digit 7, there are 2 * 1 = 2 numbers.
For hundreds digit 8, there are 1 * 0 = 0 numbers.

Adding all these up, we find that there are 224 + 147 + 90 + 50 + 24 + 9 + 2 + 0 = 546 three-digit positive integers that have their hundreds, tens, and units digits in ascending order.

Question

To find the number of three-digit positive integers that have their hundreds, tens, and units digits in ascending order, we need to consider the possible combinations of digits.

1. The hundreds digit: Since we're looking for three-digit numbers, the hundreds digit can be any number from 1 to 9. So, there are 9 possibilities for the hundreds digit.

2. The tens digit: This digit must be greater than the hundreds digit. So, if the hundreds digit is 1, the tens digit can be any number from 2 to 9, giving us 8 possibilities. If the hundreds digit is 2, the tens digit can be any number from 3 to 9, giving us 7 possibilities. This pattern continues until the hundreds digit is 9, in which case there are no possibilities for the tens digit because there's no number greater than 9.

3. The units digit: This digit must be greater than the tens digit. So, the number of possibilities for the units digit depends on what the tens digit is. If the tens digit is 2, the units digit can be any number from 3 to 9, giving us 7 possibilities. If the tens digit is 3, the units digit can be any number from 4 to 9, giving us 6 possibilities. This pattern continues until the tens digit is 9, in which case there are no possibilities for the units digit because there's no number greater than 9.

So, to find the total number of three-digit positive integers that have their hundreds, tens, and units digits in ascending order, we need to add up all the possible combinations of digits. This can be calculated as follows:

For hundreds digit 1, there are 8 possibilities for the tens digit and for each of these, there are 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 possibilities for the units digit. So, there are 8 * 28 = 224 numbers with hundreds digit 1.

We repeat this calculation for hundreds digits 2 through 8 (we don't need to consider 9 because there are no possibilities for the tens digit in this case). This gives us:

For hundreds digit 2, there are 7 * 21 = 147 numbers.
For hundreds digit 3, there are 6 * 15 = 90 numbers.
For hundreds digit 4, there are 5 * 10 = 50 numbers.
For hundreds digit 5, there are 4 * 6 = 24 numbers.
For hundreds digit 6, there are 3 * 3 = 9 numbers.
For hundreds digit 7, there are 2 * 1 = 2 numbers.
For hundreds digit 8, there are 1 * 0 = 0 numbers.

Adding all these up, we find that there are 224 + 147 + 90 + 50 + 24 + 9 + 2 + 0 = 546 three-digit positive integers that have their hundreds, tens, and units digits in ascending order.

Knowee AI · Accepted Answer