<p>The total number of positive integral solutions of \(15 < x_1 + x_2 + x_3 \leq 20\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize that the inequality 15 < a₁ + a₂ + a₃ < 25 with positive integers requires converting to a standard stars-and-bars problem by substituting variables to shift the constraint from strict inequalities to a range count.
<p><strong>Step 1:</strong> Convert the constraint. We need 15 < a₁ + a₂ + a₃ < 25 where a₁, a₂, a₃ are positive integers.</p><p>This is equivalent to: 16 ≤ a₁ + a₂ + a₃ ≤ 24 (since we have integers)</p><p><strong>Step 2:</strong> Use substitution. Let bᵢ = aᵢ - 1, so bᵢ ≥ 0. Then:</p><p>a₁ + a₂ + a₃ = (b₁ + 1) + (b₂ + 1) + (b₃ + 1) = b₁ + b₂ + b₃ + 3</p><p><strong>Step 3:</strong> The inequality becomes: 16 ≤ b₁ + b₂ + b₃ + 3 ≤ 24</p><p>Which simplifies to: 13 ≤ b₁ + b₂ + b₃ ≤ 21</p><p><strong>Step 4:</strong> Count solutions by summing each case:</p><p>∑(k=13 to 21) C(k+2, 2) = C(15,2) + C(16,2) + C(17,2) + ... + C(23,2)</p><p>Using the hockey-stick identity: ∑(k=15 to 23) C(k, 2) = C(24, 3) = 2024</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A