<p>Find number of non-negative integral solutions of the equation \(x + y + z + 2w = 20\).</p>
Step-by-Step Solution
Key Concept: Transform the constraint equation by substituting 2w = 2w' to separate even and odd contributions, then use stars and bars on the modified equation x + y + z + w' = 10 where w' ranges over non-negative integers, recognizing that each value of w' corresponds to selecting 2 items from decreasing sets.
<p><strong>Step 1:</strong> Recognize that w can range from 0 to 10 (since 2w ≤ 20, so w ≤ 10).</p><p><strong>Step 2:</strong> For each fixed value of w = k, we need x + y + z = 20 - 2k where x, y, z ≥ 0.</p><p><strong>Step 3:</strong> By stars and bars, the number of solutions for x + y + z = 20 - 2k is C(20 - 2k + 2, 2) = C(22 - 2k, 2) = <sup>(22-2k)(21-2k)</sup>⁄₂.</p><p><strong>Step 4:</strong> Sum over all valid values of w from 0 to 10:</p><p>Total = C(22,2) + C(20,2) + C(18,2) + ... + C(2,2)</p><p>= ²²C₂ + ²⁰C₂ + ¹⁸C₂ + ... + ²C₂</p><p><strong>Step 5:</strong> Using the hockey stick identity: Σ C(2n, 2) for n = 1 to 11 = C(23, 3) = 1771</p><p>∴ <strong>Answer: ²²C₂ + ²⁰C₂ + ¹⁸C₂ + ... + ²C₂ = 1771</strong></p>
Correct Answer: ²²C₂ + ²⁰C₂ + ¹⁸C₂ + ... + ²C₂