1. **Stating the problem:** We need to count the number of ways to travel from point $X$ (bottom-left corner) to point $Y$ (top-right corner) on a $6 \times 6$ grid by moving only East or North along the gridlines. Additionally, the paths must not pass through points $A$, $B$, $C$, and $D$, which form a $2 \times 2$ block in the middle-left part of the grid. 2. **Total number of ways from $X$ to $Y$ without restrictions:** The grid has $6$ steps East and $6$ steps North. The total steps to reach $Y$ are $12$ with $6$ East and $6$ North moves. The number of distinct paths is given by combinations: $$\binom{12}{6} = 924.$$ 3. **Positions of points:** We clarify coordinates with $X$ at $(0,0)$, increasing East and North. - $X=(0,0)$ - $Y=(6,6)$ - Block points: $A=(2,3), B=(3,3), D=(2,2), C=(3,2)$. 4. **Number of ways passing through each forbidden point:** Use the multiplication rule: $$ \text{ways}(X \to P) \times \text{ways}(P \to Y) $$ with $$ \text{ways}(P \to Q) = \binom{steps}{East} $$ where $steps$ is total steps between $P$ and $Q$. Calculate ways through each point: - Through $A=(2,3)$: - $X \to A$: $2$ East, $3$ North: $\binom{5}{2} = 10$ - $A \to Y$: from $(2,3)$ to $(6,6)$ is $4$ East and $3$ North: $\binom{7}{4} = 35$ - Total: $10 \times 35 = 350$ - Through $B=(3,3)$: - $X \to B$: $3$ East, $3$ North: $\binom{6}{3} = 20$ - $B \to Y$: $3$ East, $3$ North: $\binom{6}{3} = 20$ - Total: $20 \times 20 = 400$ - Through $C=(3,2)$: - $X \to C$: $3$ East, $2$ North: $\binom{5}{3} = 10$ - $C \to Y$: $3$ East, $4$ North: $\binom{7}{3} = 35$ - Total: $10 \times 35 = 350$ - Through $D=(2,2)$: - $X \to D$: $2$ East, $2$ North: $\binom{4}{2} = 6$ - $D \to Y$: $4$ East, $4$ North: $\binom{8}{4} = 70$ - Total: $6 \times 70 = 420$ 5. **Number of ways passing through intersections of forbidden points (Inclusion-Exclusion Principle):** Calculate ways passing through pairs: - $A$ and $B$ (path must go $X \to A \to B \to Y$): - $A \to B$: $(3,3)-(2,3)$ = 1 East, 0 North $\Rightarrow \binom{1}{1} = 1$ - $X \to A$: 10 (from above) - $B \to Y$: 20 (from above) - Total: $10 \times 1 \times 20 = 200$ - $A$ and $C$ (no direct path from $A$ to $C$ moving only East/North since $C$ is south of $A$), so zero. - $A$ and $D$ (no path going through $A$ then $D$ because $D$ is south of $A$), zero. - $B$ and $C$ (no path $B\to C$ since $C$ is south of $B$), zero. - $B$ and $D$ (no path $B\to D$), zero. - $C$ and $D$ (no path $C\to D$), zero. Calculate ways passing through triples: No triples are possible in order East/North path as points would have to go backward. Calculate ways passing through all four: No path goes through all four for similar reasons. 6. **Summarize using Inclusion-Exclusion:** Number of forbidden paths = $$( ext{sum of single forbidden}) - ( ext{sum of pairs}) + ( ext{sum of triples}) - \dots$$ Sum of singles = $350 + 400 + 350 + 420 = 1520$ Sum of pairs = $200$ Sum of triples = 0 Forbidden paths = $1520 - 200 = 1320$ 7. **Wait, this is more than total paths (924)!** This suggests overcounting because the points $A,B,C,D$ create overlaps that cannot be ignored with naive inclusion-exclusion directly. We must use a careful Inclusion-Exclusion with ordered intersections. Because only pair $A$ and $B$ overlaps are possible in order. 8. **Use an alternative method - Count total paths passing through any forbidden point by inclusion-exclusion carefully with valid path order:** Forbidden set $S = \{A, B, C, D\}$. Let's consider sets: - $S_A = $ paths through $A$ - $S_B = $ paths through $B$ - $S_C = $ paths through $C$ - $S_D = $ paths through $D$ Pairs with possible ordered intersections (paths with points in increasing order): Only $A$ then $B$ possible. No other pairs have valid directed paths. Therefore, forbidden paths = $|S_A| + |S_B| + |S_C| + |S_D| - |S_A \cap S_B|$. From above: - $|S_A| = 350$ - $|S_B| = 400$ - $|S_C| = 350$ - $|S_D| = 420$ - $|S_A \cap S_B| = 200$ Forbidden = $350 + 400 + 350 + 420 - 200 = 1320 - 200 = 1120$. 9. **Check again against total (924):** This can't be correct as forbidden paths exceed total paths. There is an error because calculating $|S_A|$, $|S_B|$, $|S_C|$, $|S_D|$ counts paths passing through multiple forbidden points multiple times. Correct approach: We must use Inclusion-Exclusion only with mutually exclusive sets. Because forbidden points are in a block, paths passing through multiple forbidden points are counted multiple times. 10. **Alternative method: Count all paths from $X$ to $Y$ avoiding the block points.** We can treat $A,B,C,D$ as forbidden vertices and use the principle of restriction: Allowed paths = total paths - paths through these points. One approach is dynamic programming on grid with forbidden points. Calculate ways to each point: - At $X=(0,0)$ ways = 1 - For each point, ways = ways from left + ways from below, unless forbidden (0 ways). Calculate ways step-by-step: Row 0: all points 0 to 6 East (0,0 to 6,0): ways=1 each along bottom. Row 1: (0,1)=1 (from below) (1,1)=ways(0,1)+ways(1,0)=1+1=2 (2,1)=ways(1,1)+ways(2,0)=2+1=3 (3,1)=3+1=4 (4,1)=4+1=5 (5,1)=5+1=6 (6,1)=6+1=7 Row 2: (0,2)=1 (1,2)=1+2=3 (2,2)=$D$ forbidden=0 (3,2)=$C$ forbidden=0 (4,2)=0+5=5 (5,2)=5+6=11 (6,2)=11+7=18 Row 3: (0,3)=1 (1,3)=1+3=4 (2,3)=$A$ forbidden=0 (3,3)=$B$ forbidden=0 (4,3)=0+5=5 (5,3)=5+11=16 (6,3)=16+18=34 Row 4: (0,4)=1 (1,4)=1+4=5 (2,4)=5+0=5 (3,4)=5+0=5 (4,4)=5+5=10 (5,4)=10+16=26 (6,4)=26+34=60 Row 5: (0,5)=1 (1,5)=1+5=6 (2,5)=6+5=11 (3,5)=11+5=16 (4,5)=16+10=26 (5,5)=26+26=52 (6,5)=52+60=112 Row 6: (0,6)=1 (1,6)=1+6=7 (2,6)=7+11=18 (3,6)=18+16=34 (4,6)=34+26=60 (5,6)=60+52=112 (6,6)=112+112=224 Final answer: ways from $X$ to $Y$ avoiding $A,B,C,D$ = $224$. **Answer:** $$\boxed{224}.$$