Town Plan Shortest Paths

To take the shortest path from $A$ to $B$, Neelam must move through the diagonal road inside the park $P$. Let the entry point to the park diagonal be $E$ and the exit point be $F$.

• Number of ways from $A$ to $E$ (2 steps right, 2 steps down) $= \frac{(2+2)!}{2!2!} = 6$.
• Number of ways along the park diagonal $E \to F = 1$.
• Number of ways from $F$ to $B$ (4 steps right, 2 steps down) $= \frac{(4+2)!}{4!2!} = 15$.

Total shortest paths $= 6 \times 1 \times 15 = 90$.

From $A$ to $B$, there are 90 paths as calculated in Q9.

From $B$ to $C$ (moving left/up):

• Case 1: Via point $X$ (below region $D$): $\frac{(5+1)!}{5!1!} = 6$ ways to $X$, and $2$ ways from $X$ to $C$, giving $6 \times 2 = 12$ ways.
• Case 2: Via point $Y$ (around region $D$): 1 way. Total ways from $B$ to $C = 12 + 1 = 13$.

Total ways from $A$ to $C$ via $B = 90 \times 13 = 1170$.