<p>In how many different ways can three persons <em>A</em>, <em>B</em>, <em>C</em> having respectively 6, 7 and 8 one-rupee coins donate 10 rupees collectively?</p>
Step-by-Step Solution
Key Concept: Model the problem as finding non-negative integer solutions to a+b+c=10 where a≤6, b≤7, c≤8. Use generating functions where each person's constraint becomes a factor: (1+p+p²+...+p⁶)(1+p+p²+...+p⁷)(1+p+p²+...+p⁸), then extract the coefficient of p¹⁰.
<p><strong>Step 1:</strong> Set up the constraint equation. Let a, b, c be rupees donated by A, B, C respectively. We need: a+b+c=10 where 0≤a≤6, 0≤b≤7, 0≤c≤8.</p><p><strong>Step 2:</strong> Express as generating functions. Each person's contribution is represented by: A contributes (1+p+p²+...+p⁶), B contributes (1+p+p²+...+p⁷), C contributes (1+p+p²+...+p⁸).</p><p><strong>Step 3:</strong> Convert to closed form using geometric series: (1+p+...+p⁶)=(1-p⁷)/(1-p), (1+p+...+p⁷)=(1-p⁸)/(1-p), (1+p+...+p⁸)=(1-p⁹)/(1-p).</p><p><strong>Step 4:</strong> The generating function becomes: [(1-p⁷)(1-p⁸)(1-p⁹)]/[(1-p)³].</p><p><strong>Step 5:</strong> Rewrite as: (1-p⁷)(1-p⁸)(1-p⁹)(1-p)⁻³. The answer is the coefficient of p¹⁰ in this expansion.</p><p><strong>Step 6:</strong> Expand (1-p)⁻³ = Σ C(n+2,2)pⁿ. Then subtract contributions from terms p⁷, p⁸, p⁹ that violate constraints.</p><p>∴ Answer: coefficient of p¹⁰ in (1-p⁷)(1-p⁸)(1-p⁹)(1-p)⁻³</p>
Correct Answer: coefficient of \(p^{10}\) in \((1-p^7)(1-p^8)(1-p^9)(1-p)^{-3}\)