In the expansion,

\( (a+b)^n=\;^nC_0a^n+^nC_1a^{n-1}+^nC_2a^{n-2}b^2\)

\( +......+^nC_{n-1}+^nC_nb^n\)

Fifth term from the beginning \( ^nC_4a^{n-4}b^4\)

Fifth term from the end = \( ^nC_4a^4b^{4-4}\)

Therefore, it is evident that in the expansion of

\( (\sqrt[4]{2} +\dfrac{1}{\sqrt[4]{3} })^n\) are fifth term from the beginning is

\( ^nC_4(\sqrt[4]{2})^{n-4}(\dfrac{1}{\sqrt[4]{2}})^4\) and the fifth term from the end is

\( ^nC_{n-4}(\sqrt[4]{2})^4(\dfrac{1}{\sqrt[4]{3}})^{n-4}\)

\( ^nC_4(\sqrt[4]{2})^{n-4}(\dfrac{1}{\sqrt[4]{3}})^4\)

= \( ^nC_4\dfrac{(\sqrt[4]{2})^n}{(\sqrt[4]{2})^n}\times \dfrac{1}{3}\)

\( =\dfrac{n!}{6\times 4!(n-4)!}(\sqrt[4]{2})^n\)..........(1)

\( ^nC_{n-4}(\sqrt[4]{2})^n(\dfrac{1}{\sqrt[4]{2}})^{n-4}\)

= \( ^nC_{n-4}\dfrac{(\sqrt[4]{3})^4}{(\sqrt[4]{3})^n}\)

\( =^nC_{n-4}\times 2\times \dfrac{3}{(\sqrt[4]{3})^n}\)

= \( \dfrac{6!}{(n-4)!4!}\times \dfrac{1}{(\sqrt[4]{3})^n}\) ...........(2)

It is given that the ratio of the fifth term from the beginning to the fifth term from the end is \(\sqrt{ 6} :1.\) Therefore, from (1) and (2), we obtain

\( =\dfrac{n!}{6\times 4!(n-4)!}(\sqrt[4]{2})^n:\)

= \( \dfrac{6!}{(n-4)!4!}\times \dfrac{1}{(\sqrt[4]{3})^n}=\sqrt{ 6} :1\)

\(=\dfrac{(\sqrt[4]{2})^n}{6}:\dfrac{6}{(\sqrt[4]{3})^n}=\sqrt{ 6} :1\)

= \((\sqrt[4]{2})^n=36 \sqrt{ 6} \)

= \( 6^\dfrac{n}{4}=6^\dfrac{5}{2}\)

= \( \dfrac{n}{4}=\dfrac{5}{2}\)

\( n=4\times \dfrac{5}{2}=10\)

Thus, the value of n is 10.

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