Question:

Step 1: Understand the problem setup and constraints. We need to count paths from $A_0(0,0)$ to $(6,6)$ where: - Each step moves either horizontally by $a \in \{1,2\}$ or vertically by $b \in \{1,2\}$ - The path consists of exactly two perpendicular segments (meaning exactly one direction change) - This gives us two cases: horizontal-then-vertical (H-V) or vertical-then-horizontal (V-H) Step 2: Define the composition function. Let $f(n)$ denote the number of ways to write $n$ as an ordered sum of 1s and 2s (compositions of $n$ using parts from $\{1,2\}$). The recurrence relation is: $$f(n) = f(n-1) + f(n-2)$$ with base cases $f(1) = 1$ and $f(2) = 2$. This is because to form $n$, we can either add 1 to a composition of $(n-1)$, or add 2 to a composition of $(n-2)$. Step 3: Calculate $f(6)$. Computing the sequence: $$f(1) = 1$$ $$f(2) = 2$$ $$f(3) = f(2) + f(1) = 2 + 1 = 3$$ $$f(4) = f(3) + f(2) = 3 + 2 = 5$$ $$f(5) = f(4) + f(3) = 5 + 3 = 8$$ $$f(6) = f(5) + f(4) = 8 + 5 = 13$$ Step 4: Count paths for Case 1 (Horizontal then Vertical). For paths of the form H-V: - We need to partition the horizontal distance 6 into steps of size 1 or 2: $f(6) = 13$ ways - We need to partition the vertical distance 6 into steps of size 1 or 2: $f(6) = 13$ ways - Total for Case 1: $13 \times 13 = 169$ Step 5: Count paths for Case 2 (Vertical then Horizontal). For paths of the form V-H: - We need to partition the vertical distance 6 into steps of size 1 or 2: $f(6) = 13$ ways - We need to partition the horizontal distance 6 into steps of size 1 or 2: $f(6) = 13$ ways - Total for Case 2: $13 \times 13 = 169$ Step 6: Calculate the total number of paths. The total number of paths with exactly two perpendicular segments (one direction change) is: $$\text{Total} = \text{Case 1} + \text{Case 2} = 169 + 169 = 338$$ **Final Answer:** The number of paths from $(0,0)$ to $(6,6)$ consisting of exactly two perpendicular segments with $a, b \in \{1,2\}$ is $\boxed{338}$. This corresponds to **Option 3: 338**.