Partitioning $n$ objects into $k$ boxes I know that if $n$ is a positive integer and $n_1,n_2,...n_k$ are positive integers such that $n=n_1+n_2+...+n_k$. Then the number of ways to partition $n$ objects into k boxes is: 
If $n_1=n_2=....=n_k$ and the boxes are unlabeled, then it is equal to $\frac{n!}{n_1!\cdot n_2!\cdot...\cdot n_k!}$
But what if $n_j$'s are not necessarily the same? I could not find a formula for that. Can you give me a hand?
 A: I’m assuming that the $n$ objects are distinguishable and that $n_1,\dots,n_k$ are the specified sizes of the pieces of the partition. The number of such partitions is 
$$\frac{n!}{n_1!n_2!\dots n_k!}$$
regardless of whether the $n_i$’s are the same or different. This number is a multinomial coefficient, a generalization of the more familiar binomial coefficient; it can be written 
$$\binom{n}{n_1,n_2,\dots,n_k}=\frac{n!}{n_1!n_2!\dots n_k!}\;.$$
Added 13 April 2020:
As a couple of commenters have noted, the original question is a bit confused: alev speaks of the boxes being unlabelled but gives a formula that is correct for labelled boxes. My answer was based on the assumption that ‘unlabelled’ was an error and that the formula was consistent with alev’s actual intent; if the boxes really are supposed to be unlabelled, the problem is of course a bit harder in general.
Suppose that there are $n$ unlabelled boxes, $n_1,\ldots,n_m$ are the distinct box sizes, and that for $i=1,\ldots,m$ there are $r_i$ boxes of size $n_i$, so that $\sum_{i=1}^mr_i=k$ and $\sum_{i=1}^mr_in_i=n$. Then the number of partitions is
$$\frac{n!}{\prod_{i=1}^mr_i!(n_i!)^{r_i}}\;,$$
which reduces to the multinomial coefficient only when $m=k$.
