Expected number of colors in a sampling of colored balls without replacement, Suppose there is a box containing differently colored balls. There are $G$ colors, each color having $n$ balls of this color, i.e. $G \times n$ balls.
What is the expected number of different colors in a sample of size $s$ without replacement?
The problem is similar to Expected number of different colors, but without replacement. 
In case of $G = 2$ it becomes pretty easy, so with $s$ values of 1 and 2. But in general case I can't seem to avoid double counting the combinations.
 A: We have from first principles that the PGF in $u$ with the coefficient
on  $[u^q]$ representing  the probability  of $q$  different colors  /
coupons not being seen in a sample of size $s$ is given by
$$\frac{1}{s!} {nG\choose s}^{-1} s! [z^s]
\left(u + \sum_{k=1}^n \frac{n!}{(n-k)!} \frac{z^k}{k!}\right)^G.$$
This simplifies to
$${nG\choose s}^{-1} [z^s]
\left(u+\sum_{k=1}^n {n\choose k} z^k\right)^G
= {nG\choose s}^{-1} [z^s] (u-1+(1+z)^n)^G.$$
As  a sanity check we indeed have on setting $u=1$
$$ {nG\choose s}^{-1} [z^s] (1+z)^{nG} = 1.$$
For example, with four colors and four instances each we get for
six draws the distribution
$${16\choose 6}^{-1} [z^6] (u-1+(1+z)^4)^4
= {\frac {3\,{u}^{2}}{143}}
+{\frac {60\,u}{143}}+{\frac {80}{143}}.$$
where e.g. the last term gives the probability that none of the colors
are missing. We  cannot have three colors missing  because that leaves
only one  color to cover all  six draws, we have  only four instances,
however.   With  this  PGF  we  can answer  the  question  about  the
probability that $q$ colors are missing in a draw of $s$ items, which
is
$${nG\choose s}^{-1} [z^s] [u^q] (u-1+(1+z)^n)^G
= {nG\choose s}^{-1} [z^s]
{G\choose q} (-1+(1+z)^n)^{G-q}
\\ = {nG\choose s}^{-1} [z^s]
{G\choose q}
\sum_{p=0}^{G-q} {G-q\choose p} (-1)^{G-q-p} (1+z)^{np}.$$
This yields for the probability
$$\bbox[5px,border:2px solid #00A000]{
{nG\choose s}^{-1}
{G\choose q} 
\sum_{p=0}^{G-q} {G-q\choose p} (-1)^{G-q-p} {np\choose s}}$$
which is inclusion-exclusion.
 Returning to the main question we thus have for the expectation of
coupons that did not occur
$${nG\choose s}^{-1} \left.\frac{\partial}{\partial u}
[z^s]  (u-1+(1+z)^n)^G
\right|_{u=1}
\\ = {nG\choose s}^{-1} [z^s]
\left. G (u-1+(1+z)^n)^{G-1}
\right|_{u=1}
= {nG\choose s}^{-1}
[z^s] G (1+z)^{n(G-1)}.$$
We get for the number of coupons that did occur
$$\bbox[5px,border:2px solid #00A000]{
G - G {nG\choose s}^{-1} {nG-n \choose s}.}$$
E.g. when we draw one coupon we obtain
$$G - G \frac{1}{nG} (nG-n)
= G - G + G\frac{n}{nG} = 1$$
as expected. Also  note that we obtain the value  $G$ when $s\gt nG-n$
(second binomial  coeffcient is  zero).  This  is because  the maximum
coverage with $G-1$ colors is $nG-n$  and with the next sample we must
use the last missing color.
