How many ways can $28$ black balls and $10$ white ones be put in $3$ urns? I am trying to figure out in how many ways, one can put $28$ black balls and $10$ white ones in $3$ urns. Can you assist please? Thank you.
 A: Here I am assuming that the urns in your question are distinguishable.
Now, since the urns are identical all we need to do is to treat the case of black and white balls independently. 
Putting 28 black balls in three urns is equivalent to finding three non-negative integers whose sum is 28 i.e.
\begin{align}
x_1+x_2+x_3=28
\end{align}
Here $x_1,x_2,x_3$ are non-negative integers.
This is a problem of "combinations with repetitions", also known as the "stars and bars" problem. The number of ways of making s selections from among r distinguishable possibilities, where the order does not matter and repetitions are allowed is
\begin{align}
{r+s-1 \choose s}
\end{align}
Hence the number of possibilities in our case is ${30 \choose 28}$.
Similarly in the case of white balls it is ${12 \choose 10}$.
Hence, the total number of ways is  ${12 \choose 10}{30 \choose 28}$.
P.S.: You can find a question similar to yours here 
A: If each urn is distinct then the white and black balls can be considered separately.
There are 66 $(\space =\space^{12}C_10)$ ways to distribute 10 white balls between the 3 urns.
There are 435 $(\space =\space^{30}C_28)$ ways to distribute 28 black balls between the 3 urns.
So there is 28710 total ways to distribute 38 balls assuming that each ball is distinct.
