If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)}^n$ is $\sqrt{6}:1$then n=

$\frac{{\mathrm{T}}_5\text{ from begining }}{{\mathrm{T}}_5\text{ from the end }}=\frac{\sqrt{6}}{1}\frac{{\mathrm{T}}_5}{{\mathrm{T}}_{\mathrm{n}-3}}=\frac{\sqrt{6}}{1}{\mathrm{T}}_5=\sqrt{6}\cdot {\mathrm{T}}_{\mathrm{n}-3}{\mathrm{T}}_{4+1}=\sqrt{6}\cdot {\mathrm{T}}_{(\mathrm{n}-4)+1}$
$\begin{array}{l}{ }^{\mathrm{n}}{\mathrm{C}}_4{\left(2^{\frac{1}{4}}\right)}^{\mathrm{n}-4}\cdot {\left(\frac{1}{3^{\frac{1}{4}}}\right)}^4=\sqrt{6}{\cdot }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-4}\cdot {\left(2^{\frac{1}{4}}\right)}^4\cdot {\left(\frac{1}{3^{\frac{1}{4}}}\right)}^{\mathrm{n}-4}\text{ But }\\ { }^{\mathrm{n}}{\mathrm{C}}_4{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-4}\left(\text{ so cancelling them }\right)\frac{2^{\frac{\mathrm{n}-4}{4}}}{3}=3^{\frac{1}{2}}\cdot 2^{\frac{1}{2}}\cdot \frac{2}{3^{\frac{\mathrm{n}-4}{4}}}\\ \Rightarrow (6{)}^{\frac{\mathrm{n}-4}{4}}=(6{)}^{\frac{1}{2}+1}\\ \frac{\mathrm{n}-4}{4}=\frac{3}{2}\therefore \mathrm{n}-4=\frac{12}{2}\Rightarrow \mathrm{n}=6+4\\ \therefore \mathrm{n}=10\end{array}$