Clarification of a concept in Permutation Statement 1
No. of ways in which $(m+n+p)!$ different things can be divivded into different groups containing m,n & p things respectively. is $(m+n+p)!/m!n!p!$ 
Statement 2
If $m=n=p$ and the groups have identical qualititive characteristic then the no. of groups=$(3n)!/(n!)^3*3!)$
How this '$3!$' arrives? Why doesnot substituting the values work in statement 2?
 A: It is because the groups are identical in statement $1$ and identical in statement $2$ that the difference arises.
The difference can be seen in the following example:
There is exactly $1$ way to split a group of three people into three groups of size $1$, $\frac{3!}{1!1!1!3!}$. However if each group is distinct there are $6$ ways to split them, $\frac{3!}{1!1!1!}$
A: We are distributing distinct objects to basically identical groups which may be distinguished/identified if group sizes differ.  
Statement 1: Suppose 6 objects ABCDEF are distributed in 3-2-1 pattern, e.g. ABC | DE | F
Obviously there are 6! permutations, but we need to remove the permutation within each group, thus $\dfrac{6!}{3!2!1!}$ 
The groups get distinguished/identified by the size of the group.
Statement 2: Now suppose we instead distribute in 2-2-2 pattern, e.g. AB | CD | EF 
Again there are 6! permutations, and we need to divide by $(2!)^3$ in the previous pattern.
But apart from that, there is no difference between AB | CD | EF and, say, CD| EF | AB , i.e. the groups canot be distinguished/identified.
So we also need to remove permutations between groups, hence division by 3!
NOTE
As a further aid to understanding, suppose the distribution was in the pattern 4-1-1, we would not be able to distinguish/identify only 2 of the groups, and the answer would be $\dfrac{6!}{4!1!1*2!}$
