A total of $2018$ tickets, numbered $1$, $2$, $3$, $\cdots$, $2014$, $2015$ are placed in an empty bag. Alfrid removes ticket $a$ from the bag. Bernice then removes ticket $b$ from the bag. Finally, Charlie removes ticket $c$ from the bag. They notice that $a < b < c$ and $a + b + c = 2018$. In how many ways could this happen?
This is equivalent to counting the number of ordered and distinct positive integer solutions to the equation $$a+b+c=2018$$
The total count, without any restriction, is $$\binom{2018-1}{3-1}=\binom{2017}{2}$$
Because $2018$ is not divisible by $3$, therefore $a$, $b$, and $c$ cannot be all equal. Hence, there are two possibilities: (1) all the numbers are distinct, and (2) two of them are the same.
The $2^{nd}$ case is easy to count because the same number can only be $1$, $2$, $\cdots$, $1008$. Consequently, the total number of possibilities where two numbers are equal is $$1008\cdot\binom{3}{1}$$
In the $1^{st}$ case, when $a$, $b$ and $c$ are distinct, each triplet $(a, b, c)$ can form $3!=6$ different solutions of which only $1$ is ordered. This means only $\frac{1}{6}$ of the cases in this category satisfies the condition $a < b < c$.
Hence, the final answer to the original question is $$\frac{1}{6}\cdot\left(\binom{2017}{2} - 3\times 1008\right)=\boxed{338352}$$