Permutations & Combinations
Combinations and Permutations
Grade None
Question:
<p>If \(\dfrac{^{n+2}C_6}{^{n-2}P_2} = 11\), then the value of \(n\) is:</p>
<p>(1) 6</p>
<p>(2) 7</p>
<p>(3) 9</p>
<p>(4) 10</p>
Step-by-Step Solution
Key Concept: Expand the combination and permutation formulas separately, then simplify the ratio by canceling factorials. The permutation in the denominator will significantly reduce the expression, making algebraic solution feasible.
<p><strong>Step 1:</strong> Write out the formulas.</p><p>$^{n+2}C_6 = \frac{(n+2)!}{6!(n-4)!}$ and $^{n-2}P_2 = (n-2)(n-3)$</p><p><strong>Step 2:</strong> Simplify $^{n+2}C_6$.</p><p>$^{n+2}C_6 = \frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{6!} = \frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{720}$</p><p><strong>Step 3:</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 4:</strong> Cancel $(n-2)(n-3)$ from numerator and denominator.</p><p>$\frac{(n+2)(n+1)n(n-1)}{720} = 11$</p><p><strong>Step 5:</strong> Solve for $n$.</p><p>$(n+2)(n+1)n(n-1) = 7920$</p><p>Testing $n = 8$: $(10)(9)(8)(7) = 5040$ ✗</p><p>Testing $n = 9$: $(11)(10)(9)(8) = 7920$ ✓</p><p>∴ Answer: C (n = 9)</p>
Correct Answer: C