What are the number of ways of distributing n objects(mixture of distinct and identical objects) into r groups of different sizes? Example:
lets say each object can be identified by a unique numerical I.D.
What are the number of ways of distributing 1,1,1,1,1,0,0,2,2,2,3 into 4 groups of size 4,2,2 and 3.
 A: Your question is equivalent to the counting the matrices with prescribed sums of rows and columns, which can be formulated as:
Given two vectors $M=(M_1,M_2,\dots, M_m)$
and $N=(N_1,N_2,\dots, N_n)$ with positive integer elements $M_i,N_i>0$, such that $$\sum_{i=1}^mM_i=\sum_{j=1}^nN_j$$ how many $m\times n$ matrices with non-negative integer elements $A_{ij}\ge0$ do exist, such that 
$$\sum_{j=1}^n A_{ij}=M_i\text{ and } \sum_{i=1}^m A_{ij}=N_j.$$
The matrices are often referred to as 2-way contingency tables. In your problem $m$ and $n$ are the number of different object types and that of groups, respectively.
Unfortunately the above question has no simple answer. For example in a quite recent paper the authors state:

The enumeration of integer-matrices has been the subject of considerable study and it is unlikely that a simple formula exists.

On the other hand for small matrices the recursive calculation of the number of possible distributions does not represent a problem. The following Mathematica code:
count[a_,b_]:=Module[{c},
  If[Length[b]==1,1,c=Select[Combinatorica`Compositions[b[[1]],Length[a]],Min[a-#]>=0&];
  Sum[count[a-c[[i]],b[[2;;]]],{i,Length[c]}]]
]

gives for your example the answer:
count[{1,2,3,5},{2,2,3,4}]
431

A: By Polya.
Suppose we have four sorts of objects denoted by $$e, i, h, l$$ that come in the quantities of  1, 2, 3 respectively 5 items.
then a group may contain (without the size restriction)
$$ group := (1+we)(1+wi+w^2i^2)(1+wh+w^2h^2+w^3h^3)(1+wl+\cdots+w^5l^5) $$
where $w$ is a counter (weight) for the number of items, independently of their sort.
First group has 4 items so we are interested in
$$ group_1 := coeff (group, w^4) $$
and for the rest of the groups
$$ group_2 := coeff (group, w^2) $$
$$ group_3 := coeff (group, w^2) $$
$$ group_4 := coeff (group, w^3) $$
The whole partition is described by
$$partition := group_1\cdot group_2\cdot group_3\cdot group_4$$
We are now interested in
$$coeff (e^1 i^2h^3l^5, partition)$$
which indicates the number of configurations that contains 1,2,3 and 5 given items and it is $431$
See also Spelling Bee - real combinatorics example
