total elements in group = 3n , it has to be partitioned / divided into 3 equal subsets /group let eeach of three group contains n elements all of equal sizes so according to division formula it will be $\frac{(3n!)}{(n!)^3 (3!)}$ 6 = 3! came due to power of n! which is 3
If question would have been : how many different ways can a set A of 3n elements be partitioned into 3 subsets of unequal number of elements
since groups are not of same size
your doubt : n!/n1!n2!n3! why no 3! ?? n1n2n3 can be arrranged in 3! ways ryt ??
no need of doing it in question he is not asking for permutations , only this is we have 3 equal groups and we have to divide it