Question:

<p><strong>Step 1:</strong> Write the combination formulas.</p><p>$${}^{2n}C_3 = \frac{(2n)!}{3!(2n-3)!} = \frac{(2n)(2n-1)(2n-2)}{6}$$</p><p>$${}^nC_3 = \frac{n!}{3!(n-3)!} = \frac{n(n-1)(n-2)}{6}$$</p><p><strong>Step 2:</strong> Form the ratio from the given condition.</p><p>$$\frac{{}^{2n}C_3}{{}^nC_3} = \frac{11}{1}$$</p><p>$$\frac{\frac{(2n)(2n-1)(2n-2)}{6}}{\frac{n(n-1)(n-2)}{6}} = 11$$</p><p>$$\frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} = 11$$</p><p><strong>Step 3:</strong> Simplify the left side.</p><p>$$\frac{2n \cdot (2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 11$$</p><p>$$\frac{4n(2n-1)(n-1)}{n(n-1)(n-2)} = 11$$</p><p>$$\frac{4(2n-1)}{n-2} = 11$$</p><p><strong>Step 4:</strong> Solve for n.</p><p>$$4(2n-1) = 11(n-2)$$</p><p>$$8n - 4 = 11n - 22$$</p><p>$$-4 + 22 = 11n - 8n$$</p><p>$$18 = 3n$$</p><p>$$n = 6$$</p><p><strong>Verification:</strong> For n = 6: ${}^{12}C_3 = 220$ and ${}^6C_3 = 20$, so 220:20 = 11:1 ✓</p><p>∴ <strong>Answer: n = 6</strong></p>