<p>In how many ways can we get a sum of at most 17 by throwing six distinct dice? In how many ways can we get a sum greater than 17?</p>
Step-by-Step Solution
Key Concept: Use stars-and-bars with inclusion-exclusion to count solutions to x₁+x₂+...+x₆ ≤ 17 where 1 ≤ xᵢ ≤ 6, then subtract from total 6⁶ for the complement.
<p><strong>Step 1:</strong> Transform variables. Let yᵢ = xᵢ - 1 where xᵢ is the value on die i. Then 0 ≤ yᵢ ≤ 5 and we need y₁ + y₂ + ... + y₆ ≤ 11 (since sum ≤ 17 means 6 + (y₁+...+y₆) ≤ 17).</p><p><strong>Step 2:</strong> Introduce slack variable z ≥ 0: y₁ + y₂ + ... + y₆ + z = 11 with 0 ≤ yᵢ ≤ 5.</p><p><strong>Step 3:</strong> Apply stars-and-bars without upper bound constraints: <sup>17</sup>C₁₁ solutions (distributing 11 among 7 variables).</p><p><strong>Step 4:</strong> Apply inclusion-exclusion for upper bound violations. If yᵢ ≥ 6 for some i, substitute yᵢ' = yᵢ - 6, giving y₁' + ... + y₆' + z = 5, which has <sup>11</sup>C₅ solutions. Exactly 6 dice can violate the bound (no two simultaneously), so subtract 6·<sup>11</sup>C₅.</p><p><strong>Step 5:</strong> Sum at most 17: <sup>17</sup>C₁₁ - 6·<sup>11</sup>C₅ = 12376 - 6(462) = 12376 - 2772 = <strong>9604</strong></p><p><strong>Step 6:</strong> Total ways to throw 6 dice = 6⁶ = 46656. Sum greater than 17: 46656 - 9604 = <strong>37052</strong></p><p>∴ Answer: Sum at most 17: <sup>17</sup>C₁₁ - 6·<sup>11</sup>C₅ (= 9604); Sum greater than 17: 6⁶ - (<sup>17</sup>C₁₁ - 6·<sup>11</sup>C₅) (= 37052)</p>
Correct Answer: Sum at most 17: \({}^{17}C_{11} - 6 \cdot {}^{11}C_5\); Sum greater than 17: \(6^6 - ({}^{17}C_{11} - 6\cdot{}^{11}C_5)\)