, 2024

Solution:

On her first turn, Sally cannot continue moving down the middle row. She must turn either to the bottom row or the top row. WLOG, she turns to the top row, and enters the cell $(3,1)$ and we will multiply by 2 later. Then, we can see that the path must finish in $(1,1)$. So, we will follow these two branches of the path, one for the start and one for the end. These branches must both move one unit up, and then one of the paths must move into the center row. Both branches move up one unit, and then the path in the middle row must go back to fill the corner. After this, we have exactly the same scenario as before, albeit with two fewer rows. So, for each additional two rows, we have a factor of two and thus there are $2^{12}=4096$ paths.

Solution:

We solve this problem for a general 3 by $2 n$ grid.
On her first turn, Sally cannot continue moving down the middle row. She must turn either to the topmost row or the bottommost row. WLOG, she turns to the top row.
Suppose Sally returns to the middle row $k$ times. There are $k$ "blocks". However, $2 k$ of the squares are already occupied by Sally's row shifts. Thus, we are solving
$$2\left(x_{1}+x_{2}+\ldots x_{k}\right)=2(n-k) .$$
There are
$$\sum_{k=1}^{n}\binom{n-k+k-1}{k-1}=2^{n-1}$$
solutions. We multiply by 2 to get $2^{n}$.
For $n=12$, this evaluates to $2^{12}=4096$.