1. **State the problem:** We want to find the number of ways to travel from point $X$ at the bottom-left corner to point $Y$ at the top-right corner of a $6 \times 6$ grid, moving only East or North along gridlines, without passing through points $A, B, C,$ or $D$ located in a square block near the center-left. 2. **Identify coordinates:** Let’s label the coordinates of points with $X$ at $(0,0)$ and $Y$ at $(6,6)$. Assuming the block $ABCD$ occupies points: $A=(2,2)$, $B=(3,2)$, $C=(3,3)$, and $D=(2,3)$. 3. **Calculate total paths without restrictions:** The number of ways to move from $(0,0)$ to $(6,6)$ moving only East and North is the combination: $$ \binom{12}{6} = 924 $$ 4. **Calculate paths passing through each forbidden point:** For each forbidden point $P$, calculate number of paths from $X$ to $P$ and from $P$ to $Y$, then multiply. - For $A=(2,2)$: $$ \text{paths}_{X \to A} = \binom{4}{2} = 6, \quad \text{paths}_{A \to Y} = \binom{8}{4} = 70 $$ $$ \text{total}_A = 6 \times 70 = 420 $$ - For $B=(3,2)$: $$ \text{paths}_{X \to B} = \binom{5}{3} = 10, \quad \text{paths}_{B \to Y} = \binom{7}{3} = 35 $$ $$ \text{total}_B = 10 \times 35 = 350 $$ - For $C=(3,3)$: $$ \text{paths}_{X \to C} = \binom{6}{3} = 20, \quad \text{paths}_{C \to Y} = \binom{6}{3} = 20 $$ $$ \text{total}_C = 20 \times 20 = 400 $$ - For $D=(2,3)$: $$ \text{paths}_{X \to D} = \binom{5}{2} = 10, \quad \text{paths}_{D \to Y} = \binom{7}{4} = 35 $$ $$ \text{total}_D = 10 \times 35 = 350 $$ 5. **Count paths passing through intersections of forbidden points:** Use Inclusion-Exclusion Principle. - Paths through both $A$ and $B$: Not possible because must go East or North; $B$ is to the East of $A$ but must pass through $A$ first: $$ \text{paths}_{X \to A \to B \to Y} = 0 $$ - Paths through both $A$ and $D$: $$ \text{paths}_{X \to A} = 6, \quad \text{paths}_{A \to D} = \binom{1+1}{1} = 2, \quad \text{paths}_{D \to Y} = 35 $$ $$ \text{total}_{A,D} = 6 \times 2 \times 35 = 420 $$ - Paths through both $A$ and $C$: Must go $A \to B \to C$ (since $C$ is diagonally above $B$), but $A$ and $C$ are not on a single path without passing $B$, so zero. - Paths through both $B$ and $C$: $$ \text{paths}_{X \to B} = 10, \quad \text{paths}_{B \to C} = \binom{1+1}{1} = 2, \quad \text{paths}_{C \to Y} = 20 $$ $$ \text{total}_{B,C} = 10 \times 2 \times 20 = 400 $$ - Paths through both $B$ and $D$: impossible, $D$ to the west. - Paths through both $C$ and $D$: Must go $D \to C$, which is east, then north, so $$ \text{paths}_{X \to D} = 10, \quad \text{paths}_{D \to C} = 1, \quad \text{paths}_{C \to Y} = 20 $$ $$ \text{total}_{C,D} = 10 \times 1 \times 20 = 200 $$ 6. **Paths through three or four forbidden points:** These will be zero or negligible as they require backtracking or impossible paths. 7. **Apply Inclusion-Exclusion:** $$ \text{Valid paths} = Total - (A+B+C+D) + (A\cap D + B\cap C + C\cap D) $$ $$ = 924 - (420 + 350 + 400 + 350) + (420 + 400 + 200) $$ $$ = 924 - 1520 + 1020 = 424 $$ **Final answer:** There are $\boxed{424}$ valid paths from $X$ to $Y$ avoiding points $A, B, C,$ and $D$.