<p>If \({}^nC_3 + {}^nC_4 > {}^{n+1}C_3\), then</p>
Step-by-Step Solution
Key Concept: Use Pascal's identity (^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1}) to simplify ^nC_3 + ^nC_4, then solve the resulting inequality systematically.
<p><strong>Step 1:</strong> Apply Pascal's identity: ^nC_3 + ^nC_4 = ^{n+1}C_4</p><p><strong>Step 2:</strong> The inequality becomes ^{n+1}C_4 > ^{n+1}C_3</p><p><strong>Step 3:</strong> For ^{n+1}C_4 > ^{n+1}C_3, we need the ratio of consecutive binomial coefficients:</p><p>^{n+1}C_4 / ^{n+1}C_3 = (n+1-3)/(4) = (n-2)/4 > 1</p><p><strong>Step 4:</strong> This gives n - 2 > 4, so n > 6</p><p><strong>Step 5:</strong> Also, n ≥ 4 for ^nC_4 to be defined</p><p><strong>Step 6:</strong> Combining constraints: n > 6, which means n ∈ {7, 8, 9, ...}</p><p>∴ Answer: n > 6 (or n ≥ 7 for integer values)</p>
Correct Answer: A