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?

Since $a$ is the smallest, the largest number it can be is $32$, so $a$ ranges from $1$ to $32$. The remaining sum $b+c$ must be equal to $100 - a$ and since $b$ is smaller than $c$ then $b$ ranges from $a+1$ to $\lfloor \frac{100-a}{2}\rfloor $. So for a given $a$ there are $\lfloor \frac{100-a}{2}\rfloor - a -1$ different $b$s one can choose. Therefore the possible permutations are 
$$\sum_{a=1}^{32} \lfloor \frac{100-a}{2}\rfloor - a -1$$
$$=\sum_{a=1}^{32} \lfloor \frac{100-a}{2}\rfloor - \sum_{a=1}^{32}a -32 $$
$$ = 2\sum_{a=34}^{49}a - \sum_{a=1}^{32}a -32$$
$$=49(50) - 34(33) - \frac{32(33)}{2} -32= 768$$
EDIT: 
Following the comments of Vik, it is quite clear that since we are including the case $b = a +1$ the expression should change to 
$$\sum_{a=1}^{32} \lfloor \frac{100-a}{2}\rfloor - a $$
However as Vik also mentioned this is an over estimate as it would also give us the combination where we have $b = c= (100-a)/2$ the case where $a$ is even. Since $b$ is strictly less than $c$, we need to remove $16$ off of our expression as there are 16 cases of when $a$ is even. Therefore, we get $$\sum_{a=1}^{32}( \lfloor \frac{100-a}{2}\rfloor - a) - 16= 784$$
