Step 1: Set up the equation with contributions.
Let A, B, and C contribute x, y, and z rupees respectively, with x + y + z = 10, subject to 0 ≤ x ≤ 6, 0 ≤ y ≤ 7, 0 ≤ z ≤ 8.
Step 2: Count unrestricted non-negative integer solutions.
Without upper bounds, the number of solutions is ¹⁰⁺³⁻¹C₃₋₁ = ¹²C₂ = 66.
Step 3: Subtract solutions violating individual caps.
For x ≥ 7: let x' = x - 7, then x' + y + z = 3 → ⁵C₂ = 10. For y ≥ 8: let y' = y - 8, then x + y' + z = 2 → ⁴C₂ = 6. For z ≥ 9: let z' = z - 9, then x + y + z' = 1 → ³C₂ = 3.
Step 4: Verify intersections are empty.
Violating two caps simultaneously would require negative remaining sums (e.g., x ≥ 7 and y ≥ 8 demands x' + y' + z = -5), which is impossible. All pairwise intersections are zero.
Step 5: Apply inclusion-exclusion.
Valid solutions = 66 - (10 + 6 + 3) = 47.
Step 6: Final conclusion.
There are 47 possible ways to distribute the contributions.