Permutations & Combinations
Grade 11
Question:
<p><span class="math-tex">\({ }^{n-1} {C}_{r}=\left(k^{2}-8\right)^{n} {C}_{r+1}\)</span> if and only if:</p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{2} \lt k \leq 3\)</span></p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{3} \lt k \leq 3 \sqrt{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{2} \lt k \lt 2 \sqrt{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{3} \lt k \lt 3 \sqrt{3}\)</span></p>
Step-by-Step Solution
Key Concept: Simplify the binomial coefficient equation ^(n-1)C_r = (k²-8)·^nC_(r+1) using factorial expansion to get (r+1)/n = k²-8, then apply the constraint that 0 ≤ (r+1)/n ≤ 1 for valid combinations to determine the range of k²-8, which is [0,1), yielding 8 ≤ k² < 9.
<p>Given, <span class="math-tex">${ }^{n-1} {C}_{r}=\left(k^{2}-8\right)^{n} {C}_{r+1}$</span><br />
<span class="math-tex">$\Rightarrow \frac{(n-1)!}{r!(n-r-1)!}=\left(k^{2}-8\right) \frac{n(n-1)!}{(n-r-1)!(r+1) r!}$</span><br />
<span class="math-tex">$\Rightarrow \frac{r+1}{n}=k^{2}-8$</span><br />
<span class="math-tex">$\Rightarrow 0 \lt k^{2}-8 \leq 1 \Rightarrow 8 \leq k^{2} \leq 9$</span><br />
<span class="math-tex">$\Rightarrow k \in(2 \sqrt{2}, 3]$</span></p>
Correct Answer: A