This essentially boils down to the partitions of 2017 with just 3 terms.
So we are going from x2017y0z0 to x0y0z2017.
So, just looking at the powers:
2017-0-0
2016-1-0
2016-0-1
2015-2-0
2015-1-1
2015-0-2 etc etc....
so we have the sum of integers n(n+1)/2 = 2018x2019/2 = 2037171 (as we start with zero, not1)
That's a great way of doing it. My way of remembering it is, "How many ways can you divide 2017 pieces of candy among 3 children?" Where this is a combination problem. The little tricks are in making sure you choose 1 less arrangement, for example, you only need to choose 2, and you need 2019 items (candy + children - 1). So 2019 choose 2 is also the answer.
Yes, that's the same problem. Just as with the kids and candy, order is important here; the number partition problem can be done as permutations or combinations, depending on the problem.