Key Concepts and Formulas

• Combinations ($^n C_r$): The number of ways to choose $r$ distinct items from a set of $n$ distinct items, where order doesn't matter: $^n C_r = \frac{n!}{r!(n-r)!}$

• Permutations ($^n P_r$): The number of ways to arrange $r$ distinct items chosen from a set of $n$ distinct items, where order matters: $^n P_r = \frac{n!}{(n-r)!}$

• Factorial: For a non-negative integer $n$, $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$. Also, $0! = 1$.

Step-by-Step Solution

Step 1: State the given equation and identify domain restrictions

We are given the equation: $\frac{{^{n + 2}C{}_6}}{{{^{n - 2}{P_2}}}} = 11$ Before proceeding, we need to establish the domain restrictions on $n$ based on the definitions of combinations and permutations. For $^{n+2}C_6$ to be defined, we must have $n+2 \ge 6$, which implies $n \ge 4$. For $^{n-2}P_2$ to be defined, we must have $n-2 \ge 2$, which implies $n \ge 4$. Also, $n$ must be an integer. Therefore, $n$ must be an integer and $n \ge 4$.

Step 2: Apply the formulas for combinations and permutations

Substitute the formulas for $^{(n+2)}C_6$ and $^{(n-2)}P_2$ into the given equation: $^{n+2}C_6 = \frac{(n+2)!}{6!(n+2-6)!} = \frac{(n+2)!}{6!(n-4)!}$ $^{n-2}P_2 = \frac{(n-2)!}{(n-2-2)!} = \frac{(n-2)!}{(n-4)!}$ Now, substitute these into the original equation: $\frac{\frac{(n+2)!}{6!(n-4)!}}{\frac{(n-2)!}{(n-4)!}} = 11$

Step 3: Simplify the expression

To simplify the complex fraction, multiply the numerator by the reciprocal of the denominator: $\frac{(n+2)!}{6!(n-4)!} \times \frac{(n-4)!}{(n-2)!} = 11$ Since $n \ge 4$, $(n-4)!$ is well-defined and non-zero, so we can cancel it: $\frac{(n+2)!}{6!(n-2)!} = 11$

Step 4: Expand the factorials and isolate terms

Multiply both sides by $6!$: $\frac{(n+2)!}{(n-2)!} = 11 \cdot 6!$ Expand $(n+2)!$ until we reach $(n-2)!$: $(n+2)! = (n+2)(n+1)(n)(n-1)(n-2)!$ Substitute this back into the equation: $\frac{(n+2)(n+1)(n)(n-1)(n-2)!}{(n-2)!} = 11 \cdot 6!$ Cancel out $(n-2)!$ from both the numerator and the denominator: $(n+2)(n+1)(n)(n-1) = 11 \cdot 6!$

Step 5: Calculate the numerical value

Calculate $6!$: $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ Then, multiply by 11: $11 \cdot 6! = 11 \cdot 720 = 7920$ So the equation becomes: $(n+2)(n+1)(n)(n-1) = 7920$

Step 6: Solve for n

We are looking for four consecutive integers whose product is 7920. We can estimate $n$ by taking the fourth root of 7920, which is slightly less than 10. Let's try $n=9$: $(9+2)(9+1)(9)(9-1) = (11)(10)(9)(8) = 110 \times 72 = 7920$ Thus, $n=9$ is a solution.

Step 7: Verify n against domain restrictions Since $n=9$ is an integer and $9 \ge 4$, our solution is valid.

Step 8: Identify the correct quadratic equation

We need to check which of the given options is satisfied by $n=9$.

• (A) $n^2 + 3n - 108 = 0$ Substituting $n=9$: $9^2 + 3(9) - 108 = 81 + 27 - 108 = 108 - 108 = 0$ This equation is satisfied by $n=9$.

• (B) $n^2 + 5n - 84 = 0$ Substituting $n=9$: $9^2 + 5(9) - 84 = 81 + 45 - 84 = 126 - 84 = 42 \ne 0$

• (C) $n^2 + 2n - 80 = 0$ Substituting $n=9$: $9^2 + 2(9) - 80 = 81 + 18 - 80 = 99 - 80 = 19 \ne 0$

• (D) $n^2 + n - 110 = 0$ Substituting $n=9$: $9^2 + 9 - 110 = 81 + 9 - 110 = 90 - 110 = -20 \ne 0$

Only option (A) is satisfied by $n=9$.

Common Mistakes & Tips

• Forgetting Domain Restrictions: Always determine the valid range for $n$ based on the definitions of permutations and combinations to avoid extraneous solutions.
• Incorrect Factorial Expansion: Be careful when expanding factorials, especially when dealing with variables.
• Arithmetic Errors: Double-check all calculations, as arithmetic errors can easily lead to incorrect answers.

Summary

This problem requires applying the formulas for combinations and permutations, simplifying the resulting expression, and solving for $n$. The key is to expand the factorials strategically and to be mindful of the domain restrictions. The value of $n$ that satisfies the given equation is $n=9$, and this value satisfies the quadratic equation $n^2 + 3n - 108 = 0$.

Final Answer: The final answer is \boxed{n^2 + 3n - 108 = 0}, which corresponds to option (A).