For the numbers from 100 through 199, how many have different digits in increasing order?

123, 124, 125, 126, 127, 128, 129, 134, 135....

Uh what

To find the numbers from 100 through 199 that have different digits in increasing order, we can follow these steps:

1. Start with the first number, 100.
2. Check if each digit in the number is greater than the previous digit.
a. In the number 100, 1 is not greater than 0, so it does not meet the criteria.
3. Move on to the next number, 101.
a. In this number, 1 is greater than 0, but 0 is not greater than 1, so it also does not meet the criteria.
4. Continue this process for the remaining numbers from 100 to 199.
a. Keep track of the numbers that meet the criteria.

Following these steps, we find that the numbers from 100 to 199 that have different digits in increasing order are:

123, 124, 125, 126, 127, 128, 129, 134, 135, 136, 137, 138, 139, 145, 146, 147, 148, 149, 156, 157, 158, 159, 167, 168, 169, 178, 179, 189.

So, there are 28 numbers from 100 through 199 with different digits in increasing order.

To find the answer, we need to count the numbers between 100 and 199 (inclusive) that have different digits in increasing order.

We can break down the problem into three cases:

1. Numbers with three distinct digits in increasing order.
2. Numbers with two distinct digits, where the second digit is greater than the first.
3. Numbers with two identical digits, followed by a third digit greater than the first.

Case 1: Numbers with three distinct digits in increasing order:
For this case, we can choose any three distinct digits from 1 to 9, and put them in ascending order. Since there are 9 choices for the first digit (1-9), 8 choices for the second digit (excluding the first digit), and 7 choices for the third digit, there are a total of 9 * 8 * 7 = 504 numbers in this case.

Case 2: Numbers with two distinct digits, where the second digit is greater than the first:
In this case, we need to choose one digit from 1 to 9 as the first digit, and another digit greater than the first digit as the second digit. There are 9 choices for the first digit (1-9), and 8 choices for the second digit (2-9). So there are a total of 9 * 8 = 72 numbers in this case.

Case 3: Numbers with two identical digits, followed by a third digit greater than the first:
For this case, we need to choose one digit from 1 to 9 as the first digit, and another digit greater than the first digit as the third digit. There are 9 choices for the first digit (1-9), and 8 choices for the third digit (2-9). So there are a total of 9 * 8 = 72 numbers in this case.

Adding up the numbers from each case, we have 504 + 72 + 72 = 648 numbers between 100 and 199 that have different digits in increasing order.