Filling k different sized boxes with n objects Is there a known mechanism to count the number of ways in which I can fill $k$ boxes of various sizes, $r_1,...,r_k$, with $n$ objects without counting permutations of objects within a box as a separate allocation.
e.g. Suppose $n=3$, $k=2$, $r_1=1$, and $r_2=2$. Then, there are 3 objects and 2 boxes. Box 1 can fit 1 object and box 2 can fit 2 objects. Also, if object 1 in box 1 and objects 2 and 3 in box 2 is an allocation that is counted, I would not like to additionally count object 1 in box 1 and objects 3 and 2 in box 2.
Any guidance would be greatly appreciated!
 A: I am assuming that $\sum_{i=1}^{k}r_i = n$, then the answer is
$$\prod_{i=1}^{k} C(n - \sum_{j=1}^{i-1}r_j,r_i)$$
where $C$ is the binomial coefficient. This can be further reduced to
$$\frac{n!}{\prod_{i=1}^{k}r_i!}$$
Explanation - Choose $r_1$ objects and out them into box 1, then choose $r_2$ from $N - r_1$ and put them into box 2 and so on.
A: From your description in the question, I take it that objects are distinct.
The answer is directly given by the multinomial coefficient
$\dbinom{n}{r_1, r_2, ... r_k}$, where $r_1+r_2+...+r_k=n$
Added
You might like to note a number of useful interpretations of the multinomial coefficient


*

*Ways to put $n$ labelled balls into $k$ labelled boxes with $r_1, r_2, .. r_k$ in boxes $1,2,...k$

*Selecting people for labelled groups (choosing $n_1$ to be labelled $1$, $n_2$ to be labelled $2$, etc

*Permutations with repetitions of indistinguishable objects, e.g. of MISSISSIPPI
For the last one, you would be familiar with $\dfrac{11!}{1!4!4!2!}$
Note that $\dfrac{11!}{1!4!4!2!} = \binom{11}1\binom{10}4\binom64\binom22= \dbinom{11}{1,4,4,2}$  
