$\begin{array}{1 1} \large\frac{(4n)!}{n!} \\ \large\frac{(4n)!}{4!n!} \\ \large\frac{(4n)!}{n!^{4}} \\ \large\frac{(4n)!}{4!n!^{4}} \end{array}$

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Each section has $n$ students.

$1^{st} $ section can be formed in $^{4n}C_n$ ways.

$2^{nd}$ section can be formed in $^{3n}C_n$ ways.

$3^{rd} $ section can be formed in $^{2n}C_n$ ways and

$4^{th}$ section is formed in $^nC_n$ ways.

The required no. of ways = $^{4n}C_n.^{3n}C_n.^{2n}C_n.^nC_n$ ways.

$=\large\frac{(4n)!}{(3n)!n!}.\frac{(3n)!}{(2n)!.n!},\frac{(2n)!}{n!.n!}$$.1$

$=\large\frac{(4n)!}{n!^{4}}$

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