<p>If \({}^n P_r = {}^n P_{r+1}\) and \({}^n C_r = {}^n C_{r-1}\), then the value of \(n + r\) is ___.</p>
Step-by-Step Solution
Key Concept: From ⁿPᵣ = ⁿPᵣ₊₁, deduce that r = n-r (using the permutation formula), and from ⁿCᵣ = ⁿCᵣ₋₁, use the binomial coefficient property that equal combinations occur at symmetric positions or when r = (n-1)/2.
<p><strong>Step 1:</strong> Solve ⁿPᵣ = ⁿPᵣ₊₁</p><p>Using ⁿPᵣ = n!/(n-r)!, we have:</p><p>n!/(n-r)! = n!/(n-r-1)!</p><p>This gives (n-r-1)! = (n-r)!, which means n-r-1 = 0</p><p>Therefore: <strong>n = r + 1</strong></p><p><strong>Step 2:</strong> Solve ⁿCᵣ = ⁿCᵣ₋₁</p><p>For binomial coefficients, ⁿCₚ = ⁿCᵩ when p = q or p + q = n</p><p>So either r = r-1 (impossible) or r + (r-1) = n</p><p>Therefore: <strong>n = 2r - 1</strong></p><p><strong>Step 3:</strong> Solve the system simultaneously</p><p>From Step 1: n = r + 1</p><p>From Step 2: n = 2r - 1</p><p>Setting equal: r + 1 = 2r - 1</p><p>Therefore: r = 2 and n = 3</p><p><strong>Verification:</strong> ³P₂ = 6 = ³P₃ ✓ and ³C₂ = 3 = ³C₁ ✓</p><p>∴ Answer: n + r = 3 + 2 = <strong>5</strong></p>
Correct Answer: 3