6 white and 6 black balls in 10 urns, without empty urns 6 white and 6 black balls of the same size are distributed among 10 urns so that there is at least one ball in each urn. What is the number of distributions of balls?
 A: Hint/outline:
Case 1: One urn has $3$ balls and each of nine urns has one ball.  In this case, there are four possibilities, all based on the balls in the urn with $3$ balls: WWW, WWB, WBB, BBB. (Since the urns are indistinguishable, we don't differentiate between which urns get which of the remaining colors.)
Case 2: Two urns each have $2$ balls and each of eight urns has one ball.  As in the previous case, you just need to determine the number of possible distributions for the urns with more than $1$ ball.

Update for distinguishable urns: 
Let's look at a couple of cases:
Case: Suppose we have an urn with WWB and $9$ urns with one ball:
Choose an urn for three balls: $10$ ways.
Choose $4$ of remaining $9$ urns each for $1$ white ball: ${9\choose 4}$ ways.
Total for this case: $10\cdot{9\choose 4}$ ways.
Case: Suppose we have an urn with WW and an urn with WB and eight urns with one ball each:
Choose an urn for WW ($10$ ways)
Choose a remaining urn for WB ($9$ ways)
Choose $3$ of remaining $8$ urns for a white ball (${8\choose 3}$ ways).
Total for this case: $10\cdot 9\cdot {8\choose 3}$ ways.
Note: Try to work out the other cases similarly.
