# Proving that the number of ways $mn$ different objects can be divided into $m$ groups of $n$ objects is $\frac{1}{m!} \cdot \frac{(mn)!}{(n!)^m}$

Please can you explain how these two theorems are derived?

The number of ways in which $mn$ different objects can be divided equally into $m$ groups each group containing $n$ objects and the order of the groups is not important is

$$\left[\frac{(mn)!}{(n!)^m}\right] \cdot \frac{1}{m!}$$

Also, please explain how to derive a formula when the order IS important.

I tried to reason it the following way , is this correct ? When $mn$ different things are divided equally into $n$ groups , each containing $n$ objects then the number of objects in a particular group is identical and hence the number of permutations decrease by a factor of $n!n!n!.....(m times)$ since there are $m$ groups. The number of permutations further decreases by $m!$ because the groups are not ordered and distributing the first group of $n$ things to a person is the same as distributing another group of $n$ things . Hence we divide by the number of groups , that is $m$. Please correct my reasoning.

• When you pose a question here, it is expected that you share your own thoughts on the problem. You should edit your question to explain what you have attempted and where you are stuck so that you receive responses that address the specific difficulties you are encountering. – N. F. Taussig Oct 16 '17 at 12:32
• Hello , thanks for the suggestion . I have made the required edits – Aditi Oct 16 '17 at 12:33

When order is not important, number of permutations of $mn$ different objects is $(mn)!$, but we have to have them in $m$ groups with size $n$ each, so we have to divide by permutations of the $m$ groups and the $n$ balls in each group, i.e. $m!(n!)(n!)... (n!)$, terms $n!$ occur $m$ times, so we get the required formula
When order IS important, just multiply above formula by $(n!)^m$ as now order of balls in each group is important, order of groups is still not important.