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The number of ways in which $4n$ students can be distributed equally among $4$ sections = ?

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

1 Answer

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$
$=\large\frac{(4n)!}{n!^{4}}$
answered Aug 14, 2013 by rvidyagovindarajan_1
 

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