<p>Find the number of ways in which we can get a score of 11 by throwing three dice.</p>
Step-by-Step Solution
Key Concept: Convert the problem to finding non-negative integer solutions of x + y + z = 11 where 1 ≤ x, y, z ≤ 6, then use substitution and stars-and-bars with constraint checking.
<p><strong>Step 1:</strong> Let the three dice show x, y, z. We need x + y + z = 11 where 1 ≤ x, y, z ≤ 6.</p><p><strong>Step 2:</strong> Substitute x' = x - 1, y' = y - 1, z' = z - 1 where 0 ≤ x', y', z' ≤ 5. Then (x'+1) + (y'+1) + (z'+1) = 11, giving x' + y' + z' = 8.</p><p><strong>Step 3:</strong> Without upper bound, solutions = C(8+3-1, 3-1) = C(10,2) = 45.</p><p><strong>Step 4:</strong> Apply inclusion-exclusion for constraint x', y', z' ≤ 5. Cases where one variable exceeds 5:</p><p>• If x' ≥ 6: Let x' = 6 + a. Then a + y' + z' = 2, giving C(4,2) = 6 solutions.</p><p>• By symmetry: 3 × 6 = 18 solutions to subtract.</p><p><strong>Step 5:</strong> Cases where two variables exceed 5 would need x', y' ≥ 6, requiring sum ≥ 12 > 8, so 0 such cases.</p><p>∴ Answer: 45 - 18 = <strong>27</strong></p>
Correct Answer: 27