Permutations & Combinations
Permutations and Combinations
Grade None
Question:
<p>If \(\dfrac{{}^{n+2}C_6}{{}^{n-2}P_2} = 11\), then \(n\) satisfies the equation:</p>
<p>\(n^2 + n - 110 = 0\)</p>
<p>\(n^2 + 2n - 80 = 0\)</p>
<p>\(n^2 + 3n - 108 = 0\)</p>
<p>\(n^2 + 5n - 84 = 0\)</p>
Step-by-Step Solution
Key Concept: Expand the combination and permutation formulas separately, then create a rational equation by recognizing that C(n+2,6) involves factorials up to (n+2)! and P(n-2,2) = (n-2)(n-3), allowing algebraic simplification.
<p><strong>Step 1:</strong> Write out the formulas explicitly.</p><p>$${}^{n+2}C_6 = \frac{(n+2)!}{6!(n-4)!} = \frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{720}$$</p><p>$${}^{n-2}P_2 = (n-2)(n-3)$$</p><p><strong>Step 2:</strong> Form the equation.</p><p>$$\frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{720(n-2)(n-3)} = 11$$</p><p><strong>Step 3:</strong> Cancel common factors (valid since n ≥ 4).</p><p>$$\frac{(n+2)(n+1)n(n-1)}{720} = 11$$</p><p><strong>Step 4:</strong> Simplify.</p><p>$$(n+2)(n+1)n(n-1) = 7920$$</p><p><strong>Step 5:</strong> Test values. For n = 8:</p><p>$$10 \times 9 \times 8 \times 7 = 5040 \text{ (too small)}$$</p><p>For n = 9:</p><p>$$11 \times 10 \times 9 \times 8 = 7920 ✓$$</p><p>∴ Answer: <strong>n = 9</strong></p>
Correct Answer: C