Number of ways to sum up to 12, using only 6 summands (including 0), and order matters This relates to a spatial model, each summand represent a particular location ($IxJ$ grid) where $N$ agents can go. In this case, there are only 6 ($2x3$ grid) possible locations to go and 12 ($N$) agents.
For example, 
$$
1 + 1 + 2 + 2 + 2 + 2 \leq 12,
$$
is different than 
$$
2 + 2 + 1 + 1 + 1 + 1 \leq 12.
$$
In the first case, just one agent goes to each of the first two locations and two agents go to each of the other locations. Two agents decide not to go to any location. In the second case, two agents go to each of the two first locations, and just one agent go to each of the other locations, and again, two agents decide not go anywhere.
Agents can decide not go anywhere, so that 
$$
0 + 0 + 0 + 0 + 0 + 0
$$ 
or 
$$
3 + 2 + 1 + 1 + 1 + 0
$$
are both valid arrangements.
I think this translates well into number theory, in particular combinatorics, but I have not found any way to solve this problem. Composition seems to be restricted to equality and including zero adds more complexity to the problem. I already know the answer to the specific case by checking in matlab using hard-wired code, but I rather have a way to compute the number analytically to generalize the model.
Here is the code, the size of the third dimension of "options" is what I am looking for.
for a1 = 0:N
    for a2 = 0:(N-a1)
        for a3 = 0:(N-a1-a2)
            for a4 = 0:(N-a1-a2-a3)
                for a5 = 0:(N-a1-a2-a3-a4)
                    for a6 = 0:(N-a1-a2-a3-a4-a5)
                        options(:,:,end+1) = [a1, a2; a3 a4; a5, a6];
                    end
                end
            end
        end
    end
end

options = options(:,:,2:end);   % remove first row of zeros

 A: Firstly, instead of looking at the six-tuples $(x_1,x_2,x_3,\dots,x_6)$ where $x_1+x_2+\dots+x_6\leq 12$ and each $x_i\geq 0$ let us add an additional dummy variable $y$ where $y=12-x_1-x_2-\dots-x_6$.  Given the restriction on the $x_i$'s this implies the restriction that $y\geq 0$.
This allows us to instead talk about finding seven-tuples $(x_1,x_2,\dots,x_6,y)$ where $x_1+x_2+\dots+x_6+y=12$, thus taking care of the problem caused by the inequality instead of equality.
Now, it should be a well known problem format where we can apply stars-and-bars to solve.  There are $7$ dinstinguishable bins and $12$ indistinguishable balls to place.

 This then amounts to $\binom{12+7-1}{7-1}=\binom{18}{6}=18564$ arrangements.

Using a different method, we could look instead at the expansion of $(1+x+x^2+x^3+\dots)^6$ and sum the entries of the coefficients of $x^n$ with $n$ less than or equal to $12$, or using a similar argument as before, look at the coefficient of $x^{12}$ in the expansion of $(1+x+x^2+\dots)^7$.  To calculate this with a computer, we can make the simplification that we stop in each parentheses at $x^{12}$.
This gives us the following expansion $1+7x+28x^2+84x^3+210x^4+462x^5+\dots$

$\dots+18564x^{12}+\dots$

A: Consider doing the following.  Ask the first agent how many she moved.  Write on a piece of paper that many "x" marks.  Then mark an "|" to indicate moving on to the next agent.  Then do the same for the next agent and then the next.  If after the six agents you haven't gotten $12$, mark off the remaining "x" to get $12$ "x" marks.
You will have written on your paper $12$ "x" marks and $6$ "|" marks.  
How many ways are there to do that?
You have made $18$ marks and of them $6$ of them are "|"s and the remainder of them are "x"s.  How many ways are there to choose $6$ marks out of $18$ to be "|"s?
There are ${18 \choose 6} = \frac {18!}{6!(18-6)!} = \frac {18!}{6!12!} = \frac {13*14*15*16*17*18}{6!} =  18564$
