Number of Paths from Point A to Point B Is there a closed formula for determining how many paths there are
from A to B, if we are only allowed to move to the right,
down, or down and to the right diagonally? I've looked at a graph theory approach  to this problem but it leads me no where to finding a closed form expression.

 A: There are $m+n-l$ steps where $m$ is the number down from A to B, $n$ is number to the right from A to B, and $l$ is the number of diagonal steps you take. From here you know the number of ways should be:
\begin{equation}
\sum_{l=0} ^{\min\{m,n\}} \binom{m+n-l}{m-l,n-l,l}
\end{equation}
In your case this would be:
\begin{equation}
\sum_{l=0} ^3 \binom{6-l}{3-l,3-l,l}
\end{equation}
For an $n$ steps right by $n$ steps down this simplifies to:
\begin{equation}
\binom{2 n}{n} {}_2F_1(-n, -n, -2 n, -1)
\end{equation}
If you want to check my work I get 63 for your case.
A: Name the nodes (intersections) of the grid as follows:


*

*$a_0,a_1,a_2,a_3$ for the first line

*$b_0,b_1,b_2,b_3$ for the second line

*$c_0,c_1,c_2,c_3$ for the third line

*$d_0,d_1,d_2,d_3$ for the fourth line


And let $A(n),B(n),C(n),D(n)$ denote the number of paths from $A$ to $a_n$, $b_n$, $c_n$, $d_n$, respectively. Thus, we want to compute $D(3)$.
$A(n),B(n),C(n),D(n)$ verify the following recursive relations:
\begin{cases}
A(n)=A(n-1)\\
B(n)=B(n-1)+A(n-1)+A(n)\\
C(n)=C(n-1)+B(n-1)+B(n)\\
D(n)=D(n-1)+C(n-1)+C(n)
\end{cases}
with $A(0)=B(0)=C(0)=D(0)=1$.
A little bit of algebra yields
$$
\boxed{
D(n)=\frac{4}{3}n^3+2n^2+\frac{8}{3}n+1
}
$$
So the answer is $D(3)=63$, which matches the answer proposed by @Kitter Catter.
