Fifth term from the beginning is $T_{4+1}$ i.e., $T_5$ = $\binom{n}{4} (\sqrt[4]{2})^{n-4} \left(\frac{1}{\sqrt[4]{3}}\right)^4$ = $\binom{n}{4} (2)^{\frac{n-4}{4}} (3)^{-\frac{4}{4}}$ Fifth term from the end is $T_{n-5+2}$ i.e., $T_{n-3}$ ∴ $T_{n-3}$ = $\binom{n}{n-4} (\sqrt[4]{2})^4 \left(\frac{1}{\sqrt[4]{3}}\right)^{n-4}$ = $\binom{n}{4} (2)^{\frac{4}{4}} (3)^{\frac{4-n}{4}}$ (Since $\binom{n}{n-4}$ = $\binom{n}{4}$) According to given condition, we have $\frac{\binom{n}{4} (2)^{\frac{n-4}{4}} (3)^{-1}}{\binom{n}{4} (2) (3)^{\frac{4-n}{4}}}$ = $\frac{\sqrt{6}}{1}$ ⇒ $2^{\frac{n-4}{4} - 1} 3^{-1 - \frac{4-n}{4}}$ = $\sqrt{6}$ ⇒ $2^{\frac{n-8}{4}} 3^{\frac{n-8}{4}}$ = $\sqrt{2} \sqrt{3}$ = $(2)^{\frac{1}{2}} (3)^{\frac{1}{2}}$ On comparing, we get ⇒ $\frac{n-8}{4}$ = $\frac{1}{2}$ ⇒ 2n - 16 = 4 ⇒ n = 10