By way of enrichment I would like to point out that using the Polya
Enumeration Theorem the closed form is also given by
$$n! [z^k] Z(P_n)\left(\frac{1}{1-z}\right)$$
where $Z(P_n) = Z(A_n)-Z(S_n)$ is the difference between the cycle
index of the alternating group and the cycle index of the symmetric
group. This cycle index is known in species theory as the set operator
$\mathfrak{P}_{=n}$ and the species equation here is
$$\mathfrak{P}_{=n}\left(\mathcal{E} + \mathcal{Z} + \mathcal{Z}^2 +
\mathcal{Z^3} + \cdots\right).$$
Recall the recurrence by Lovasz for the cycle index $Z(P_n)$ of
the set operator $\mathfrak{P}_{=n}$ on $n$ slots, which is
$$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l})
\quad\text{where}\quad
Z(P_0) = 1.$$
This recurrence lets us calculate the cycle index $Z(P_n)$ very easily.
For example when $n=3$ as in the introduction to the problem the cycle
index is
$$Z(P_3) =
1/6\,{a_{{1}}}^{3}-1/2\,a_{{2}}a_{{1}}+1/3\,a_{{3}} $$
and the generating function becomes
$$1/6\, \left( 1-z \right) ^{-3}-1/2\,{\frac {1}{ \left( -{z}^{2}+1
\right) \left( 1-z \right) }}+1/3\, \left( -{z}^{3}+1 \right) ^
{-1} $$
which gives the sequence
$$0, 0, 6, 6, 12, 18, 24, 30, 42, 48, 60, 72, 84, 96,\ldots$$
which is six times OEIS A069905.
Similarly when $n=5$ we get the cycle index
$$Z(P_5) =
{\frac {{a_{{1}}}^{5}}{120}}-1/12\,a_{{2}}{a_{{1}}}^{3}+1/6\,a_{{
3}}{a_{{1}}}^{2}+1/8\,a_{{1}}{a_{{2}}}^{2}\\-1/4\,a_{{4}}a_{{1}}-1/
6\,a_{{2}}a_{{3}}+1/5\,a_{{5}}$$
and the generating function becomes
$${\frac {1}{120\, \left( 1-z \right) ^{5}}}-1/12\,{\frac {1}{
\left( -{z}^{2}+1 \right) \left( 1-z \right) ^{3}}}\\+1/6\,{
\frac {1}{ \left( -{z}^{3}+1 \right) \left( 1-z \right) ^{2}}}+1
/8\,{\frac {1}{ \left( -{z}^{2}+1 \right) ^{2} \left( 1-z
\right) }}-1/4\,{\frac {1}{ \left( -{z}^{4}+1 \right) \left( 1-
z \right) }}\\-1/6\,{\frac {1}{ \left( -{z}^{2}+1 \right) \left( -
{z}^{3}+1 \right) }}+1/5\, \left( -{z}^{5}+1 \right) ^{-1}$$
which gives the sequence
$$0, 0, 0, 0, 0, 0, 0, 0, 0, 120, 120, 240, 360, 600, \ldots$$
which is $120$ times OEIS A001401.
The prefix of zeroes (these two examples start at one)
compared to the two OEIS entries represents the fact that
the minimum value attainable with $n$ distinct summands in
$$[z^k] Z(P_n)\left(\frac{1}{1-z}\right)$$
is $0+1+2+\cdots+n-1 = 1/2\times n\times (n-1).$
These sequences match the formula by @bof, which is
$$n! p_n\left(k - \frac{1}{2} n (n - 3)\right).$$
There are many more related links at
MSE Meta on Burnside/Polya.
The Maple code for these was as follows.
p :=
proc(n, k)
option remember;
if k=1 then return 1 fi;
if n<k then return 0 fi;
p(n-1, k-1)+p(n-k,k)
end;
pet_cycleind_symm :=
proc(n)
local p, s;
option remember;
if n=0 then return 1; fi;
expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;
pet_cycleind_set :=
proc(n)
local p, s;
option remember;
if n=0 then return 1; fi;
expand(1/n*add((-1)^(l-1)*
a[l]*pet_cycleind_set(n-l), l=1..n));
end;
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
q1 :=
proc(n, k)
option remember;
local gf;
gf := pet_varinto_cind(1/(1-z), pet_cycleind_set(n));
n!*coeftayl(gf, z=0, k);
end;
q2 :=
proc(n, k)
option remember;
n!*p(k-n*(n-3)/2, n);
end;
Addendum. As per request we now give a mixed (combinatorial,
algebraic) proof of the identity
$$[z^n] Z(P_k)\left(\frac{1}{1-z}\right)
= p_k\left(n - \frac{1}{2} k(k-3)\right).$$
Observe that we have reverted to the standard convention of using $n$
for the sum of the partition and $k$ for the number of parts.
By the same construction as before (PET) we have
$$p_k\left(n - \frac{1}{2} k(k-3)\right) =
[z^{n- \frac{1}{2} k(k-3)}] Z(S_k)\left(\frac{z}{1-z}\right)$$
with $Z(S_k)$ being the cycle index of the symmetric group (unlabelled
multisets with operator $\mathfrak{M}_{=k}$.)
Using basic algebra this becomes
$$[z^{n- \frac{1}{2} k(k-3)}] z^k Z(S_k)\left(\frac{1}{1-z}\right)
= [z^{n- \frac{1}{2} k(k-3) -\frac{1}{2} 2k}]
Z(S_k)\left(\frac{1}{1-z}\right)
\\ = [z^{n- \frac{1}{2} k(k-1)}]
Z(S_k)\left(\frac{1}{1-z}\right).$$
But this is the species
$$\mathfrak{M}_{=k}\left(\mathcal{E} + \mathcal{Z} + \mathcal{Z}^2 +
\mathcal{Z^3} + \cdots\right),$$
i.e. partitions with empty constitutents being permitted and
constituents not necessarily distinct.
There is however a straighforward bijection between these and
partitions with potentially empty but distinct constituents. To go
from the former to the latter add $q$ circles on the left of every row
in the Ferrers diagram with row indices $q$ starting at zero. Now if
we had two adjacent rows with the first above the second with length
$b_1$ and $a_1$ where $b_1\ge a_1$ then the resulting pair is $b_1+q$
and $a_1+q-1.$ The difference between these is $b_1-a_1+1\ge 1$ so the
new pair is distinct and in non-decreasing order seen from below. To
go from the latter to the former remove $q$ circles from every row
(index is $q$), turning $b_2$ and $a_2$ where $b_2>a_2$ into $b_2-q$
and $a_2 -(q-1).$ The difference is $b_2-a_2-1\ge 0$ and the pair is
non-decreasing order but not necessarily distinct.
The number of circles being added/removed is
$$\sum_{q=0}^{k-1} q = \frac{1}{2} k(k-1).$$
We have shown that
$$ [z^{n- \frac{1}{2} k(k-1)}] Z(S_k)\left(\frac{1}{1-z}\right)
= [z^n] Z(P_k)\left(\frac{1}{1-z}\right).$$
Here we have $n\ge \frac{1}{2} k (k-1),$ both sides are zero otherwise.
This concludes the argument.
Best Answer
$$(b-a)+(d-c)=9$$
The number of red balls from the start till the first green ball gives you the value for $a$.
Imagine a line with 9 red balls and 3 green balls. Consider the following example,
The number of red balls from the start till the second green ball gives you the value for $b-a$.
The number of red balls from the second green ball till the third green ball gives you the value for $c$. The number of red balls from the second green ball till the end gives you the value for $d-c$.
This has $\binom{12}3=220$ solutions