🗺️ Grid Paths
Problems involving counting paths on a grid with various constraints and restrictions.
🎯 Recognition Cues
- Keywords: “path”, “walk”, “grid”, “lattice”, “right and up”, “diagonal”
- Setup: Moving on a grid from one point to another
- Constraints: Blocked squares, diagonal moves, specific directions
📋 Solution Template
Identify the path type:
- Basic lattice path: Only right and up moves
- Diagonal path: Includes diagonal moves
- Blocked path: Some squares are off-limits
Apply the appropriate method:
- Basic lattice: Use $\binom{m+n}{n}$ formula
- Diagonal: Use modified formula or recursion
- Blocked: Use inclusion-exclusion or dynamic programming
Check for symmetry:
- Reflection symmetry
- Rotation symmetry
- Diagonal symmetry
💡 Micro-Examples
Basic Lattice Path
- Problem: How many ways can you walk from (0,0) to (3,2) moving only right and up?
- Solution: $\binom{3+2}{2} = \binom{5}{2} = 10$ ways
Diagonal Path
- Problem: How many ways can you walk from (0,0) to (3,2) moving right, up, or diagonally up-right?
- Solution: Use recursion: $a_{i,j} = a_{i-1,j} + a_{i,j-1} + a_{i-1,j-1}$
Blocked Path
- Problem: How many ways can you walk from (0,0) to (3,2) avoiding the square (1,1)?
- Solution: Total paths - Paths through (1,1) = $\binom{5}{2} - \binom{2}{1} \cdot \binom{2}{1} = 10 - 4 = 6$
⚠️ Common Pitfalls & Variants
Pitfall: Forgetting to account for blocked squares
- Wrong: “Path from (0,0) to (3,2) avoiding (1,1)” = $\binom{5}{2}$
- Right: Total - Paths through (1,1) = $\binom{5}{2} - \binom{2}{1} \cdot \binom{2}{1}$
Pitfall: Confusing lattice paths with general paths
- Wrong: “Path from (0,0) to (3,2)” = $3! \cdot 2! = 12$ ways
- Right: $\binom{5}{2} = 10$ ways (lattice path formula)
Pitfall: Misapplying diagonal path formulas
- Wrong: “Diagonal path from (0,0) to (3,2)” = $\binom{5}{2}$
- Right: Use recursion or modified formula for diagonal moves
🏆 AMC-Style Worked Example
Problem: How many ways can you walk from (0,0) to (4,3) moving only right and up, avoiding the squares (1,1) and (2,2)?
Solution:
- Step 1: Total paths from (0,0) to (4,3): $\binom{7}{3} = 35$
- Step 2: Paths through (1,1): $\binom{2}{1} \cdot \binom{4}{2} = 2 \cdot 6 = 12$
- Step 3: Paths through (2,2): $\binom{4}{2} \cdot \binom{2}{1} = 6 \cdot 2 = 12$
- Step 4: Paths through both (1,1) and (2,2): $\binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} = 8$
- Step 5: By inclusion-exclusion: $35 - 12 - 12 + 8 = 19$
Key insight: Use inclusion-exclusion for multiple blocked squares!
🔗 Related Topics
- Permutations & Combinations - Foundation for path counting
- Inclusion-Exclusion - Essential for blocked paths
- Symmetry & Invariance - Symmetry in path problems