Number of ways to pick out pairs of balls from an urn We have an urn with $2N$ balls, of four colors $(a,b,c,d)$. I want to know how many ways there are to pick pairs of balls until the urn is empty. Note that $(x,y)$ pair is equivalent to $(y,x)$. Also see Mark Fischler's comment below for additional equivalences.
For example, we have an urn with $(0, 2, 0, 8)$ balls, $2N=10$. 
First, we can pick a pair $(b,b)$, leaving us with $(0,0,0,8)$. Now we can only take $(d,d)$ pairs four times until the urn is empty. This sequence of moves leads to the outcome $1\times(b,b), 4\times(d,d)$.
Alternatively, we can pick $(b,d)$, leaving us with $(0,1,0,7)$. Now we have to pick another $(b,d)$ pair (since the leftover $b$ has to be picked eventually), leading to $(0,0,0,6)$. We empty the urn by taking $(d,d)$ three more times. This sequence leads to $2\times(b,d), 3\times(d,d)$.
Thus, starting from $(0,2,0,8)$, we have the following possible pair selections:
\begin{aligned}
1\times&(b,b)\\
2\times&(b,d)\\
(4+3)\times&(d,d)
\end{aligned}
Is there a name for this selection scheme? Is there a closed for solution for this? The best I was able to come up with so far has been an iterative algorithm that generates all the possible pair selections, and then counts them up.
 A: I have the answer in what could be considered "closed form"!
First some notation, following that in Concrete Mathematics: Let $P(a, b, \ldots , z)$ be some polynomial in $\{a, b, \ldots , z)\}$.  Then
$$
[a^\alpha b^\beta \cdots z^\omega] P
$$
means the coefficient in $P$ of the term $[a^\alpha b^\beta \cdots z^\omega]$.  For example, 
$$
[x^2y^4](x^2+x y + y^2)^3  = 6
$$ 
because $(x^2+x y + y^2)^3 = \mathbf{6}x^2 y^4 + 6 y^2 x^4 + x^6 + y^6 + 3 x^5 y + 3 y^5 x + 7x^3 y^3$.
Consider an urn with $2N$ balls of $C$ colors, and say that for each value of color $c$ the urn containing $n_c$ balls. 
(In your case, there are $2N$ balls, choosing among $4$ colors, so $n_1 + n_2 + n_3 + n_4 = 2N$.) 
Let $Q(x_1, x_2, \ldots ,x_C)$ be the most general homogeneous quadratic polynomial with unit coefficients:
$$
Q(x_1,  x_2, \ldots ,x_C) = \sum_i x_i^2 + \sum_{i\neq j} x_i x_j
$$
where the indices $i$ and $j$ range from $1$ to $C$.  In your 4 color example,
$$
Q(x_1,  x_2, x_3 ,x_4) = (x_1^2 + x_2^2 + x_3^2 + x_4^2 +x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)
$$
Then the number of distinct ways to choose a sequence of pairs from those balls is given by

$$
[x_1^{n_1} x_2^{n_2} \cdots x_C^{n_C}]\,\,Q^N
$$

As an example, say there are $5$ red balls, $3$ blue balls, $6$ green balls and $8$ yellow balls.  $N = 11$ in that case. Then there will be 
$$
[x^5y^3z^6w^8]\,\,  (x^2 + y^2+ z^2 + w^2 + xy+xz+xw+yz+yw+zw) ^{11}= 281953980
$$
ways to draw the 11 pairs.
Doing this with explicit recursion would take forever; using this polynomial expansion it is instantaneous in Mathematica and even plausible by hand.
You can see why this works, since each contribution to $Q^N$ selects $N$ pairs in some order and needs to consume each of the colors in exactly the right numbers.
A: If you just want to know how many, you can make a recurrence.  Let $A,B,C,D$ be the number of balls of each color.  Then
$$f(A,B,C,D)=f(A-2,B,C,D)+f(A-1,B-1,C,D)+\ldots$$
all the other similar terms.  The advantage of this approach is that you don't have to generate all the different orders.  This suggests a recursive algorithm in your favorite language.  I doubt it will be easy to get a closed form.
