Coupon collector without replacement Given a urn containing balls of $k$ distinct color such that $i$th color has $c_i$ number of balls. How many balls in expectation one needs to sample in order to to get at least one ball of each color. Problem looks very similar to coupon collector(each color being coupon) but without replacement.
I am able to solve 2 color case. For example if urn contain r red balls and b blue balls then: 
Let S be the event that first ball is red and we have 
$$E[X]=P(S)E[X|S]+P(\bar S)E[X|\bar S]$$
Now $E[X|S]=1+(\frac{r-1}{b+1}+1)$, where $\frac{r-1}{b+1}$ is the expected number of red ball one needs to sample to get blue ball from bag that contatin $r-1$ red balls and $b$ blue balls. Similarly $E[X|\bar S]=1+(\frac{b-1}{r+1}+1)$, using these we can get $E[X]$.
But I don't know how to generalize this to k color case. Any hints?  
 A: I will use $n_i$ in place of $c_i$ for the number of balls of the $i^{th}$ color, and let $n=n_1+\dots+n_k$ be the total number of balls.
For each $i\in \{1,\dots,k\}$, let $T_i$ be the number of draws it takes to see the $i^{th}$ color. The variable you are interested in is
$$
T=\max(T_1,\dots,T_k),
$$
which is the number of draws it takes to get all colors.
Note that $E[T_i]=(n+1)/(n_i+1)$. This is by the same logic for the two color example in your post. If you imagine lining up all the balls in the order they are drawn, then the $n_i$ balls of color $i$ will divide the other balls into $n_i+1$ sections, each with an expected length of $(n-n_i)/(n_i+1)$, to which you must add one.
By the same logic, for any subset of colors $S\subseteq \{1,\dots,k\}$,
$$
E\left[\min_{i\in S} T_i\right]=\frac{n+1}{\left(\sum_{i\in S}n_i\right)+1}
$$
The other piece we need is this equation which relates the max of $k$ numbers to the min's of their subsets:
$$
\max(x_1,\dots,x_k)=\sum_{S\subseteq \{1,\dots,k\}, S\neq \varnothing}(-1)^{|S|-1}\min_{i\in S} x_i
$$
valid for real numbers $x_1,\dots,x_k$. Applying this with $x_1=T_1,\dots,x_k=T_k$, we conclude
$$
\boxed{E[T]
=\sum_{S\subseteq \{1,\dots,k\}, S\neq \varnothing}(-1)^{|S|-1}\frac{n+1}{\left(\sum_{i\in S}n_i\right)+1}}
$$
In the special case where there are the same number $c$ of each color of ball, this simplifies to
$$
E[T]=\sum_{i=1}^k (-1)^{i-1} \binom{k}{i} \frac{ck+1}{ci+1}
$$
