Permutations & Combinations
Distribution of identical objects
Grade 11
Question:
<p>Number of ways equals number of solutions of \(x_1 + x_2 + x_3 = 8\); \(x_i \geq 1\) or \((x_1 - 1) + (x_2 - 1) + (x_3 - 1) = 5\); \(x_i \geq 1\) or \(y_1 + y_2 + y_3 = 5\); \(y_i \geq 0\), which is \({}^{5+2}C_2 = \dfrac{7 \times 6}{2} = 21\). The number of ways is:</p>
<p>15</p>
<p>18</p>
<p>20</p>
<p>21</p>
Step-by-Step Solution
Key Concept: Transform a constrained distribution problem (each variable ≥ 1) into an unconstrained one (each variable ≥ 0) using substitution yi = xi - 1, then apply stars and bars formula C(n+k-1, k-1).
<p><strong>Step 1:</strong> Recognize the constraint x₁ + x₂ + x₃ = 8 with xᵢ ≥ 1 means each variable must be at least 1.</p><p><strong>Step 2:</strong> Apply substitution yᵢ = xᵢ - 1, so yᵢ ≥ 0. This transforms the equation to:<br/>(y₁ + 1) + (y₂ + 1) + (y₃ + 1) = 8<br/>⟹ y₁ + y₂ + y₃ = 5, where yᵢ ≥ 0</p><p><strong>Step 3:</strong> Apply stars and bars formula: The number of non-negative integer solutions is C(n + k - 1, k - 1) where n = 5 (sum) and k = 3 (variables).</p><p><strong>Step 4:</strong> Calculate C(5 + 3 - 1, 3 - 1) = C(7, 2) = (7 × 6)/2 = 21</p><p>∴ Answer: <strong>21</strong></p>
Correct Answer: D