How many positive integer solutions does the equation $a+b+c=100$ have if we require $aHow many positive integer solutions does the equation $a+b+c=100$ have if we require $a<b<c$?
I know how to solve the problem if it was just $a+b+c=100$ but the fact it has the restriction $a<b<c$ is throwing me off.
How would I solve this?

$a<b<c$ can be translated by $b=a+x$ and $c=b+y=a+(x+y)$, where $a,x,y>0$.
$$a + b + c = 100\iff a+a+x+a+x+y=100 \iff3a+2x+y=100$$
Let $a_k$ the number of solutions of the diophantine equation $3a+2x+y=k$, where $a,x,y>0$. 
$$\sum_{d=0}^\infty a_k t^k=\sum_{k=0}^\infty(\sum_{3a+2x+y=k\\a,x,y>0}t^k)=(\sum_{k=1}^\infty t^{3k})(\sum_{k=1}^\infty t^{2k})(\sum_{k=1}^\infty t^{k})=\frac{t^6}{(1-t^3)(1-t^2)(1-t)}$$
We will calculate explecitely $a_k$ now :
$$\frac{t^6}{(1-t^3)(1-t^2)(1-t)}\\=\frac{t + 2}{9 (t^2 + t + 1))} - \frac{89}{72 (t - 1)} + \frac{1}{8 (t + 1)} - \frac{3}{4 (t - 1)^2} -\frac{1}{6 (t - 1)^3}-1\\=\sum_{n=1}^∞ t^n\frac{1}{72}  (47 + 9 (-1)^n + 8 e^{-(2 i n π)/3} + 8 e^{(2 i n π)/3} + 6 n^2-36n)\\=\sum_{n=1}^∞ t^n\frac{1}{72}  (47 + 9 (-1)^n + 16\cos(2\pi \frac{n}{3}) + 6 n^2-36n)$$
So : $$a_{100}=\frac{1}{72}  (47+9 -8 +60000-3600)=784$$
