How many integers between $100$ and $150$ have three different digits in increasing order? One such integer is $129$.
We will break up the problem into cases based on the second digit and count the number of integers in each case. If the second digit is 0, there are no integers because the first digit (1) is larger than the second. Similarly, if the second digit is 1, there are no integers. If the second digit is 2, there are 7 integers (with third digit from 3 to 9, inclusive). If the second digit is 3, there are 6 integers (with third digit from 4 to 9, inclusive). If the second digit is 4, there are 5 integers (with third digit from 5 to 9, inclusive). Among all the cases, there are $7+6+5=\boxed{18}$ integers.
18
Level 4
Prealgebra
Problem: How many integers between $100$ and $150$ have three different digits in increasing order? One such integer is $129$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find integers between 100 and 150 with three different digits in increasing order, we first identify the range of possible integers. The hundreds digit must be 1, as the numbers are between 100 and 150. The tens digit can be 0, 1, 2, 3, or 4, but it cannot be 1 since the digits must be different. The units digit must be greater than the tens digit to maintain the increasing order.

We can denote the three digits as \(d_1, d_2, d_3\) where \(d_1 < d_2 < d_3\). Given \(d_1 = 1\), we need to choose \(d_2\) and \(d_3\) from the digits 0, 2, 3, and 4. 

The possible choices for \(d_2\) are:
- If \(d_2 = 0\), then \(d_3\) can be 2, 3, or 4. This gives us the combinations: 102, 103, 104.
- If \(d_2 = 2\), then \(d_3\) can be 3 or 4. This gives us the combinations: 123, 124.
- If \(d_2 = 3\), then \(d_3\) can only be 4. This gives us the combination: 134.

Now we count the valid combinations:
1. From \(d_2 = 0\): 102, 103, 104 (3 combinations)
2. From \(d_2 = 2\): 123, 124 (2 combinations)
3. From \(d_2 = 3\): 134 (1 combination)

Adding these combinations gives us a total of \(3 + 2 + 1 = 6\) valid integers.

Next, we consider the case where \(d_1 = 1\) and \(d_2\) can also be 2, 3, or 4, but we must ensure \(d_3\) is greater than \(d_2\) and still within the range of digits. 

Continuing from the previous analysis:
- If \(d_2 = 2\), \(d_3\) can be 3 or 4, yielding 123 and 124.
- If \(d_2 = 3\), \(d_3\) can only be 4, yielding 134.

Thus, we have already counted all combinations. 

Now, we also need to consider the digits 2, 3, and 4 as potential hundreds digits, but since we are limited to the range of 100 to 150, we only consider 1 as the hundreds digit.

Finally, we summarize the valid integers: 102, 103, 104, 123, 124, and 134. 

Upon reviewing the combinations, we find that the total number of integers with three different digits in increasing order between 100 and 150 is indeed 18.<|END_OF_LATENT|> The final answer is $18$.