There may be a way to work out a final exact formula for this, but note that first of all you can assume that each min is 0 by subtracting the sum of mins from $N$, and subtracting each min from the corresponding max, so e.g. you can assume the count of $a$ is between $0$ and $max(a) - min(a)$, and so forth, and that the sum is $N - min(a) - min(b) - min(c)$.
Let $N_a$ be the count of $a$ and so forth for $b$ and $c$. If you choose $N_a$, then each valid choice of $N_b$ leads to one solution for $N_c$, which may either be in the valid range for count of $c$ or not. The choices of $N_b$ are between $0$ and $N - N_a$ but you must have the max possible value of $N_c$ is greater than or equal to $N - N_a - N_b$. So this gives you the minimum possible value for $N_b$, given $N_a$, and the maximum possible value of $N_b$ given $N_a$ is equal to the minimum of $N - N_a$ and the original maximum value for $N_b$. So this tells you how many choices you have for $N_b$ given $N_a$, such that you can find a valid assignment for $N_c$ to complete the total assignment. And the choice of $N_c$ is unique given the choices for $N_b$ and $N_c$. Thus, in $O(N)$ time you can just do a for loop over the values of $N_a$ and count the number of valid solutions in $O(1)$ time for each value of $N_a$, without need for further nested for loops. Of course, if it's possible to eliminate the need for looping over $N_b$ then it should be possible to similarly eliminate the need for originally looping over $N_a$ too, but I haven't worked through the messy details yet to come up with a final closed form formula.
