In how many different ways can a set A of 3n elements be partitioned into 3 subsets of equal number of elements?
My approach :
The number of permutations of n objects with n1 identical objects of type 1, n2 identical objects of type 2, , and n3 identical objects of type is n!/n1!n2!n3!
Here ans could be (3n)!/(n!)^3
but given is 3n)!/6* (n!)^3
How did again 6 come in denominator , why is he arranging again in 6 i..e., 3! ways ??