Find the value of n if (i)2nC3 : nC3 = 12 : 1 (ii) 2nC3 : nC3 = 11:1

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Please find below the solution to the asked query:
(i)
$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \frac{{}^{2\mathrm{n}}{\mathrm{C}}_3}{{}^{\mathrm{n}}{\mathrm{C}}_3}=\frac{12}{1} \\ \Rightarrow \frac{{\displaystyle \frac{2\mathrm{n}\left(2\mathrm{n}-1\right)\left(2\mathrm{n}-2\right)}{3.2.1}}}{{\displaystyle \frac{\mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)}{3.2.1}}}=12 \\ \Rightarrow \frac{4\mathrm{n}\left(2\mathrm{n}-1\right)\left(\mathrm{n}-1\right)}{\mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)}=12 \\ \Rightarrow \frac{4\left(2\mathrm{n}-1\right)}{\left(\mathrm{n}-2\right)}=12 \\ \Rightarrow \frac{\left(2\mathrm{n}-1\right)}{\left(\mathrm{n}-2\right)}=3 \\ \Rightarrow 2\mathrm{n}-1=3\mathrm{n}-6 \\ \Rightarrow 3\mathrm{n}-2\mathrm{n}=6-1 \\ \Rightarrow \boxed{\mathbf{n}\mathbf{=}\mathbf{5}} \end{gathered}$$

(ii)

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \frac{{}^{2\mathrm{n}}{\mathrm{C}}_3}{{}^{\mathrm{n}}{\mathrm{C}}_3}=\frac{11}{1} \\ \Rightarrow \frac{{\displaystyle \frac{2\mathrm{n}\left(2\mathrm{n}-1\right)\left(2\mathrm{n}-2\right)}{3.2.1}}}{{\displaystyle \frac{\mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)}{3.2.1}}}=11 \\ \Rightarrow \frac{4\mathrm{n}\left(2\mathrm{n}-1\right)\left(\mathrm{n}-1\right)}{\mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)}=11 \\ \Rightarrow \frac{4\left(2\mathrm{n}-1\right)}{\left(\mathrm{n}-2\right)}=11 \\ \Rightarrow \frac{4\left(2\mathrm{n}-1\right)}{1}=11\left(\mathrm{n}-2\right) \\ \Rightarrow 8\mathrm{n}-4=11\mathrm{n}-22 \\ \Rightarrow \boxed{\mathbf{n}\mathbf{=}\mathbf{6}} \end{gathered}$$

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