A. How many POSITIVE INTEGER solutions does the following

I think what "they're saying" is that all variables must be positive integers.

If not, infinity would be the answer to each question.
So I say 19 for the first one,

And for the second:

Ways to make three positive integers sum up to-
0 0
1 0
2 0
3 1
4 3-----------1,1,2 three permutations
5 6-----------1,2,2 three permutations 3,1,1 three permutations
6 10-----------1,1,4 (3 perm) 1,2,3 (6 perm) 2,2,2 (1 perm)
7 15----------1,1,5 (3p) 1,2,4 (6p) 1,3,3 (3p) 2,2,3 (3p)
8 21----------1,1,6 (3) 1,2,5 (6) 1,3,4 (6) 2,2,4 (3) 2,3,3(3)

A pattern emerges!
But I can't think of the formula!!!


There must be a smarter way!

Each variable must be between 1 and 98, inclusive.
if X is 1, then Y+Z must equal 99, giving us 98 possible answers for Y, and one answer for Z under each possible Y.
if X is 2, then Y+Z must equal 98, giving us 97 possible answers for Y, and one answer for Z under each possible Y.
if X is 3, then Y+Z must equal 97, giving us 96 possible answers for Y, and one answer for Z under each possible Y.
...
if X is 97, then Y+Z must equal 3, giving us 2 possible answers for Y, and one answer for Z under each possible Y.
if X is 98, then Y+Z must equal 2, giving us 1 possible answer for Y, and one answer for Z .

So the answer is 98+97+96...+2+1, or 49.5*98, or 4851.