A fresh look at an old problem

A while ago I did the following problem:
Consider all 5 digit numbers (from 10,000 to 99,999). Call a 5 digit number weakly increasing if the first two digits (ten-thousands and thousands) are strictly increasing from left to right, and the last 3 digits (hundreds, tens. and ones) are strictly increasing from left to right. How many weakly increasing 5 digit numbers are there

If you read my solution, you'll notice that it was tedious and had a LOT of casework. However, I'm a bit more experienced now and have a much simpler solution.

Notice that for any arrangement of digits, there is exactly one way to arrange them in increasing order. So for the first two digits, (only 9 digits because we can't use 0, it will always be the first digit and that makes it a 4 digit number), and for the last three digits, . So our final answer is

The lesson to be learned from this is, even when you've solved a problem, try to solve it in a different way. If you have time, it's the best method to check your work, and it can give you new insights too.