So, there is actually an approach by hand (without calculating via a computer program or creating a $10 \times 10$ table of a recurrence relation).
Observe that to get a distance of $10$ by using planks of length $2$ and $3$:
We are using exactly $4$ planks, except a case of using $5$ $2$'s. If we are using $4$ planks, we will use exactly $2$ $2$'s and $2$ $3$'s, making $\binom{2+2}{2}=6$ combination cases.
Thus, from $(0,0)$ to $(10,10)$, the number of ways — including going into the whirlpool — is:
- Using $5$ planks on horizontal and $5$ planks on vertical: $\binom{5+5}{5} = 252$.
- Using $5$ planks on horizontal and $4$ planks on vertical: $\binom{5+4}{5} \times 6 = 756$.
- Using $4$ planks on horizontal and $5$ planks on vertical: $\binom{5+4}{4} \times 6 = 756$.
- Using $4$ planks on horizontal and $4$ planks on vertical: $\binom{4+4}{4} \times 6 \times 6 = 2520$.
Overall, there will be $252+756+756+2520=4284$ ways.
Now, we may then count how many ways of going from $(0,0)$ to $(10,10)$ and landing on $(5,5)$. It's simply:
Count the number of ways going from $(0,0)$ to $(5,5)$, multiplying the number of ways going from $(5,5)$ to $(10,10)$.
Again, using a similar observation to get a distance of $5$ and going from $(0,0)$ to $(5,5)$ (and also $(5,5)$ to $(10,10)$):
We are using exactly $2$ planks: $1$ $2$'s and $1$ $3$'s, making $2$ possible cases.
Using $2$ planks on horizontal and $2$ planks on vertical: $\binom{2+2}{2} \times 2 \times 2 = 24$ ways.
So, the final answer will be:
$4284 - (24 \times 24) = 3708$ ways.