It depends on the range of values for $i$ and $j$, and some initial values for $c_{ij}$.
The common way to solve a linear recurrence of this sort is to use generating functions.
Assume your initial conditions are on $c_{i0}$ and $c_{0j}$, and $i,j$ are supposed to be non-negative integers. Define:
$f(x,y) = \sum_{i,j} c_{ij} x^i y^j$
Then:
$(x+y) f(x,y) = f(x,y) - c_{00} - \sum_{i>0}(c_{i0}-c_{i-1,0})x^i - \sum_{j>0} (c_{0j}-c_{0,j-1}) y^j$
For the particular $c_{0j}=c_{i0}=1$, this resolves to the nice formula:
$f(x,y) = \frac{1}{1-(x+y)}$
But $\frac{1}{1-(x+y)} = \sum_m (x+y)^m = \sum_{i,j} {{i+j}\choose i} x^iy^j$
So $c_{ij} = {{i+j}\choose i}$.