Q:

Can this be solved using a formula?

I've attempted this problem by drawing several possible paths from S to E given the above conditions - a process that is very tedious. I suspect that there is an easier way. Any advice/suggestions will be greatly appreciated.

A:

As other solutions show, it is easier to do this with a bit of forethought (and some combinatorics) rather than rely on a formula, but just in case you were truly looking for a direct formula, it can be done with generating functions.  The answer will be the coefficient of $x^3 y^3$ in the Maclaurin series expansion of
$$f(x,y) = \frac{1}{1 - y^2 - x - xy}.$$
The exponents on $x$ and $y$ represent the relative displacements in the horizontal and vertical directions: $x^3 y^3$ represents the steps needed to get from S to E, $y^2$ represents move A, etc.
We can extract this coefficient mechanically by taking the mixed partial derivative $\partial^6 f \over \partial x^3 \partial y^3$,  evaluating at $(0,0)$, and dividing out by $36 = 3! \cdot 3!$ (which is the same mixed partial applied to $x^3 y^3$).  This single-line formula
$$\frac{1}{3!3!}\frac{\partial^6}{ \partial x^3 \partial y^3}\left(\frac{1}{1 - y^2 - x - xy} \right)\bigg|_{x=0,y=0}$$
ultimately yields 13, after a great amount of tedium to compute it :).

