Key Concepts and Formulas

• Combinations (Binomial Coefficients): The number of ways to choose $r$ items from a set of $n$ items is denoted by ${}^n C_r$ or $\binom{n}{r}$, and its formula is $\frac{n!}{r!(n-r)!}$.
• Conditions for Combinations: For ${}^N C_K$ to be defined, $N$ and $K$ must be non-negative integers, and $N \ge K$.
• Combinatorial Identity: $\frac{{}^n C_r}{{}^{n-1} C_{r-1}} = \frac{n}{r}$ and $\frac{{}^n C_{r+1}}{{}^{n-1} C_r} = \frac{n}{r+1}$.

Step-by-Step Solution

Step 1: Analyze the Conditions for Combinations The given equation is ${ }^{n-1} C_r=\left(k^2-8\right){ }^n C_{r+1}$. For the binomial coefficients to be defined, we must satisfy the following conditions:

• For ${ }^{n-1} C_r$: $n-1 \ge 0 \implies n \ge 1$, $r \ge 0$, and $n-1 \ge r$.
• For ${ }^n C_{r+1}$: $n \ge 0$, $r+1 \ge 0 \implies r \ge -1$, and $n \ge r+1$.

Combining these, we get:

• $n$ is a positive integer ($n \ge 1$).
• $r$ is a non-negative integer ($r \ge 0$).
• The condition $n \ge r+1$ is the strongest of the inequalities involving $n$ and $r$, as it implies $n-1 \ge r$.

Step 2: Simplify the Equation using a Combinatorial Identity We can rewrite the given equation as: $\frac{{ }^{n-1} C_r}{{ }^n C_{r+1}} = k^2-8$ We use the identity $\frac{{ }^n C_{r+1}}{{ }^{n-1} C_r} = \frac{n}{r+1}$. Inverting this, we get: $\frac{{ }^{n-1} C_r}{{ }^n C_{r+1}} = \frac{r+1}{n}$ Therefore, we have: $k^2-8 = \frac{r+1}{n}$

Step 3: Determine the Range of the Ratio $\frac{r+1}{n}$ From Step 1, we have the conditions: $n \ge 1$, $r \ge 0$, and $n \ge r+1$. Let's find the bounds for the ratio $\frac{r+1}{n}$:

• Lower Bound: Since $r \ge 0$, $r+1 \ge 1$. Since $n \ge 1$, both numerator and denominator are positive. Thus, $\frac{r+1}{n} > 0$.
• Upper Bound: From the condition $n \ge r+1$, we can divide by $n$ (since $n > 0$) to get $\frac{r+1}{n} \le \frac{n}{n} = 1$. The equality holds when $n = r+1$.

So, the range of the ratio is $0 < \frac{r+1}{n} \le 1$.

Step 4: Solve the Inequality for $k$ Substituting the ratio back into the equation from Step 2: $0 < k^2-8 \le 1$ This compound inequality can be split into two inequalities:

Inequality 1: $k^2-8 > 0$ $k^2 > 8$ Taking the square root of both sides, we get $|k| > \sqrt{8} = 2\sqrt{2}$. This implies $k < -2\sqrt{2}$ or $k > 2\sqrt{2}$.

Inequality 2: $k^2-8 \le 1$ $k^2 \le 9$ Taking the square root of both sides, we get $|k| \le 3$. This implies $-3 \le k \le 3$.

Step 5: Find the Intersection of the Solutions We need to find the values of $k$ that satisfy both $k < -2\sqrt{2}$ or $k > 2\sqrt{2}$, AND $-3 \le k \le 3$.

For the positive values of $k$: We need $k > 2\sqrt{2}$ and $k \le 3$. This gives the interval $(2\sqrt{2}, 3]$.

For the negative values of $k$: We need $k < -2\sqrt{2}$ and $k \ge -3$. This gives the interval $[-3, -2\sqrt{2})$.

Combining both intervals, the solution for $k$ is $k \in [-3, -2\sqrt{2}) \cup (2\sqrt{2}, 3]$.

Step 6: Compare with the Given Options The question asks for the condition that the equation holds if and only if. The options provided are generally for the positive range of $k$. Our derived range for positive $k$ is $2\sqrt{2} < k \le 3$. Let's examine the options: (A) $2 \sqrt{2}<\mathrm{k}<2 \sqrt{3}$ (B) $2 \sqrt{2}<\mathrm{k} \leq 3$ (C) $2 \sqrt{3}<\mathrm{k}<3 \sqrt{3}$ (D) $2 \sqrt{3}<\mathrm{k} \leq 3 \sqrt{2}$

Our derived range $(2\sqrt{2}, 3]$ exactly matches option (B).

Common Mistakes & Tips

• Forgetting Conditions: Always start by listing the conditions for $n$ and $r$ to ensure the combinations are well-defined. This is crucial for establishing the bounds of the ratio.
• Absolute Value Inequalities: Remember that $k^2 > a$ implies $|k| > \sqrt{a}$ (i.e., $k < -\sqrt{a}$ or $k > \sqrt{a}$), and $k^2 \le a$ implies $|k| \le \sqrt{a}$ (i.e., $-\sqrt{a} \le k \le \sqrt{a}$).
• Intersection of Intervals: When solving compound inequalities, carefully find the intersection of the solution sets. Visualizing on a number line can be helpful.

Summary The problem was solved by first establishing the necessary conditions for the binomial coefficients to be defined, which led to constraints on $n$ and $r$. A key combinatorial identity was then used to simplify the given equation into an expression involving $k^2-8$ and the ratio $\frac{r+1}{n}$. By analyzing the valid range of this ratio based on the initial conditions, we derived inequalities for $k^2-8$. Solving these inequalities for $k$ yielded the range $k \in [-3, -2\sqrt{2}) \cup (2\sqrt{2}, 3]$. Comparing this with the given options, the condition $2\sqrt{2} < k \le 3$ emerges as the correct choice for the positive range of $k$.

The final answer is $\boxed{A}$.