# How to calculate the number of m-partitions of an N-partite quantum state?

I am reading the structure of multipartite entanglement from section 3.3. It is stated there that the number of possible partitions of $$N$$ parties into $$m$$ parts is given by $$\frac{m^N}{m!}$$.

I could not understand how they are doing the calculation. Can anyone please help as to how should I proceed?

This is a purely combinatorics problem. Let us index our registers by $$1,\cdots,N$$. We have $$m$$ "bags" $$P_1,\cdots,P_m$$ into which which we want to put our indexes.
• For the first index ($$1$$), we have $$m$$ choices: we can put it in any of the $$P_i$$.
• For the second index ($$2$$), well, it's still the case, so we also have $$m$$ choices.
• For the last index ($$N$$), we still have $$m$$ choices.
All in all, this results in $$m^N$$ possibilities. However, we mustn't count the order of the partitions. For instance, if $$N=3$$ and $$m=2$$, $$P_1=\{1\}$$ and $$P_2=\{2, 3\}$$ corresponds to the same partition as $$P_1=\{2, 3\}$$ and $$P_2=\{1\}$$.
That is, the way we counted our sets, we have counted each configuration along with all its permutations. There are $$m!$$ of them, so we must divide by this number in order to get the final result.
Note that the way they counted, they allow for $$P_i$$ to be nil sets.