1. Start with the given relation

   $\binom{n-1}{r}=(k^2-8)\binom{n}{r+1}$

2. Use factorial form / standard identity

   $\binom{n-1}{r}=\frac{(n-1)!}{r!(n-r-1)!}$ $\binom{n}{r+1}=\frac{n!}{(r+1)!(n-r-1)!}$

   Hence,

   \frac{\binom{n-1}{r}}{\binom{n}{r+1}}=rac{(n-1)!}{r!(n-r-1)!}\cdot \frac{(r+1)!(n-r-1)!}{n!}

   $=\frac{r+1}{n}$

   Therefore,

   $k^2-8=\frac{r+1}{n}$

3. Find the possible range of $\dfrac{r+1}{n}$

   Since $\binom{n-1}{r}$ is defined, we must have

   $0\le r\le n-1$

   So,

   $1\le r+1\le n$

   Dividing by $n>0$,

   $\frac{1}{n}\le \frac{r+1}{n}\le 1$

   Thus,

   $0<k^2-8\le 1$

   (strictly positive because $r+1\ge 1$)

   So,

   $8<k^2\le 9$

4. Convert to range of $k$

   Assuming $k$ is positive (as in the options),

   $2\sqrt{2}<k\le 3$

5. Match with the options

   This corresponds to:

   Option B: $2\sqrt{2}<k\le 3$

6. Check other options briefly

   • A excludes $k=3$, but $k=3$ is possible when $k^2-8=1$, i.e. $r=n-1$.
   • C and D give values larger than $2\sqrt{3}$, which do not satisfy $k^2\le 9$.

Therefore, the correct answer is B.