Combinatorics of weighted vector compositions I am interested in a problem related to the one of enumerating all vector compositions (as described in chapter 4.3 from "The Theory of Partitions" from G.E.Andrews):
Vector (2,1) can be decomposed into following patterns: 
$$(2,1)$$
$$(2,0)+(0,1)$$
$$(0,1)+(2,0)$$
$$(1,1)+(1,0)$$
$$(1,0)+(1,1)$$
$$(1,0)+(1,0)+(0,1)$$
$$(1,0)+(0,1)+(1,0)$$
$$(0,1)+(1,0)+(1,0)$$
And following Andrew's notation:
$$c(2,1;1)=1$$
$$c(2,1;2)=4$$
$$c(2,1;3)=3$$
a total of $$c(2,1)=8$$
Its analytic calculation is done through theorem's 4.7 and 4.8 as a sum of product of binomial numbers.
I would like to introduce a weight in each of the described compositions interpreting them as multinomial numbers:
$$(2,1) \rightarrow 3 $$
$$(2,0)+(0,1) \rightarrow 1*1=1 $$
$$(0,1)+(2,0) \rightarrow 1*1=1 $$
$$(1,1)+(1,0) \rightarrow 2*1=2 $$
$$(1,0)+(1,1) \rightarrow 2*1=2 $$
$$(1,0)+(1,0)+(0,1) \rightarrow 1*1*1=1 $$
$$(1,0)+(0,1)+(1,0) \rightarrow 1*1*1=1 $$
$$(0,1)+(1,0)+(1,0) \rightarrow 1*1*1=1 $$
which lead to (using analogous "f" in the notation):
$$f(2,1;1)=3$$
$$f(2,1;2)=6$$
$$f(2,1;3)=3$$
a total of $$f(2,1)=12$$
I would like a similar expresion than mentioned theorems (sum of product of binomial numbers) but any help on that would be appreciated. 
I already searched for some examples with vectors that sum N=3 and N=4 into oeis but no success on that.
 A: I originally wrote this for vectors of length $2$, but it generalized to vectors of any length easily, and that generalization is at the end.

Let $f(a,b;k)$ denote your weighted sum into $k$ parts. I claim that $f(a,b;k)$ has the following combinatorial interpretation. There are $a$ red cards, $b$ blue cards, and $k-1$ white cards. Then $f(a,b;k)$ is the number of ways to stack the deck so that every while card is sandwiched between two non-white cards. The decomposition $(x_1,y_1)+(x_2,y_2)+\dots+(x_k,y_k)$ means the following; the white cards divide the deck into $k$ sections, and in the $i^{th}$ section, there are $x_i$ red cards and $y_i$ blue cards. There are then $\binom{x_i+y_i}{x_i}$ ways to arrange the colored cards within each section.
The number of ways to shuffle the deck is $\binom{a+b+k-1}{a,b,k-1}=\frac{(a+b+k-1)!}{a!b!(k-1)!}$. However, we must subtract out the bad shufflings where one of the gaps does not have any colored cards. Effectively, there is one fewer white cards to deal with (either it must appear on the top or bottom, or it must appear after a particular white card). So we subtract out $\binom{k}1\cdot \binom{a+b+k-2}{a,b,k-2}$. But we must add back in the doubly subtracted shufflings, then subtract the triply subtracted shufflings, etc. Fortunately, it turns out that when counting the shuffles where any $i$ gaps are empty, it is equivalent to count shuffles which have $k-1-i$ white cards. Take some time to think about this, for it is a nice coincidence. For example, if $i=2$ and the two gaps are adjacent, then there must be a block of three white cards $``WWW"$ which behaves like a single white cards, while if the two blocks are separate, then there are two pairs of adjacent white cards $``WW"$ and $``WW"$. Either way, there are effectively two fewer white cards. 
The result of this application of the principle of inclusion exclusion is
$$
f(a,b;k)=\sum_{i=0}^{k-1}(-1)^i\binom{k}i\binom{a+b+k-1-i}{a,b,k-1-i}
$$
For example,
$$
f(2,1;2)=\binom{2}0\frac{4!}{2!1!1!}-\binom{2}1\frac{3!}{2!1!0!}=12-2\cdot 3=6
$$
$$
f(2,1,3)=\binom{3}0\frac{5!}{2!1!2!}-\binom{3}1\frac{4!}{2!1!1!}+\binom{3}2\frac{3!}{2!1!0!}=30-3\cdot 12+3\cdot 3=3
$$
In general, if $L=(l_1,l_2,\dots,l_m)$, then
$$
f(L;k)=\sum_{i=0}^k (-1)^i\binom{k}i\binom{l_1+l_2+\dots+l_m+k-1-i}{l_1,l_2,\dots,l_m,k-1-i}
$$
