Number of routes not passing in 3 points 
Given a coordinates system in which you can move in unit steps right or up, how many possible routes are there from $(0, 0)$ to $(k, n)$ while also not passing in any of the points $(z,c), (y,b), (x,a)$?
$a<b<c<n,x<y<z<k$
all parameters are in $\mathbb{N}$.

I managed to solve it with inclusion-exclusion but the solution is extremely ugly. Looking for a hint for a more elegant solution.
 A: Overview
The number of paths of east and north steps (${\bf E}$ and ${\bf N}$) on the grid without constraints is $T = {k+n \choose k}$.  You can envision this as $k+n$ slots (steps) into which you place $k$ ${\bf E}$ steps, where the rest must be ${\bf N}$.
Consider the first forbidden point, at $(x,a)$, which we call $A$.  The number of the total paths that pass through $A$ is the product of the number of legal paths from $(0,0)$ to $(x,a)$ times the number of legal paths from $(x,a)$ to the endpoint $(k,n)$.  Those numbers, multiplied, are $N(A) = {a+x \choose a}{(k-x)+(n-a) \choose k-x}$.
So you subtract these from the total number to find the total number of paths that do not pass through the first forbidden point.
A similar calculation holds for the second and the third forbidden points, $B$ and $C$.
However, for the full problem you must consider paths that don't go through multiple such points.
This is a straightforward matter of counting all the path segments that do or do not pass through the points, using the general mathematical formula above.  This is the technique of inclusion/exclusion.  Call the total number of paths $T$ (as above) and the number passing through $A$ as $N(A)$, and likewise for point $B$ and point $C$.  Then the total number of ways that avoid $A$ and $B$ and $C$ is:
$$T - N(A) - N(B) -N(C) + N(A \cap B) + N(A \cap C) + N(B \cap C) - N(A \cap B \cap C)$$
