Intuitive understanding of the formula $\frac{(m+n+p)!}{m!n!p!}$ for dividing $m+n+p$ things into three groups of sizes $m,n$ and $p$ The number of ways in which $m+n+p$ things can be divided into three unequal groups containing $m,n$ and $p$ things is $\dfrac{(m+n+p)!}{m!n!p!}$
I need help understanding this formula intuitively and its proof. Moreover, I don't get why this formula has no multiplication by $3!$ since there are $3!$ permutations of those 3 groups possible. 

Since all group sizes occur only once, they can be thought of as indices of the groups, i.e. instead of ordering groups by $1,2,3$, we can order them by their sizes $m,n,p$, say, in nondecreasing order, and the identification is unique. If some group sizes occurred more than once, those groups could be permuted, and we would need to divide by more factors than those in your question.
In general, suppose we divide the set of $N$ elements into groups, where the multiset of sizes is $S=\{s_1^{r_1},\dots,s_k^{r_k}\}$ (i.e. each distinct size $s_i$ occurs $r_i>0$ times), so $N=\sum_{i=1}^{k}{s_ir_i}$.
Given a list of those $N$ elements, they can be permuted in $N!$ ways. For each way of partitioning this set of $N$ elements into subsets with the multiset of sizes given by $S$, some permutations preserve the size of the part to which each element belongs. Such permutations are those that can do only two things:


*

*permute elements within each part,

*permute parts of equal sizes.


Elements in a part of size $s_i$ can be permuted in $s_i!$ ways. The $r_i$ parts of the same size $s_i$ can be permuted in $r_i!$ ways. Thus, the number of distinct partitions of $N$ elements into parts with the multiset of sizes $S=\{s_1^{r_1},\dots,s_k^{r_k}\}$ is
$$
\frac{N!}{\prod_{i=1}^{k}{\Bigl(\left(s_i!\right)^{r_i}r_i!\Bigr)}}.
$$
In our case, $k=3$, $s_1=m$, $s_2=n$, $s_3=p$, $r_1=r_2=r_3=1$, since all $m,n,p$ are distinct.
