Question

Let $\frac{3n!}{k(n!{)}^3}$ be the number of different ways a set $A$ of $3n$ elements be partitioned into $3$ subsets of equal number of elements? (the subsets $P,Q,R$ form a partition if $P\cup Q\cup R=A,P\cap R=\varphi ,Q\cap R=\varphi ,R\cap P=\varphi ).$Find $k$ ?

Answer 1

The required number of ways $=$ the number of ways in which $3n$ different things can be divided in $3$ equal groups
$=$ The number of ways to distribute $3n$ different things equally among three persons
$=\frac{3n!}{3!(n!{)}^3}$
$=\frac{3n!}{6(n!{)}^3}$