Stars & Bars Question: Identical Balls in Distinct Boxes I am terrible at combinatorics so any and all help would be appreciated. 
20 identical balls are put into 10 distinct boxes so that at most 3 boxes are empty. In how many ways can this be done?
Thanks
 A: We can split this into $4$ cases based on the number of empty boxes, and use stars and bars for each case.

Case 1: No empty boxes.
We place $1$ ball in each box, and we imagine $9$ dividers and the remaining $10$ balls, so the number of possible ways to arrange the $19$ objects is $$\binom{19}{9}$$
Case 2: One empty box.
There are $10$ ways to choose the empty box. We place $1$ ball in each of $9$ boxes, and imagine $8$ dividers with the remaining $11$ balls, so the number of ways to arrange the $19$ objects is $$10\binom{19}{8}$$
Case 3: Two empty boxes.
There are $\binom{10}{2}=45$ ways to choose which boxes are empty. We place $1$ ball in each of $8$ boxes, and imagine $7$ dividers with the remaining $12$ balls, which gives $$45\binom{19}{7}$$
Case 4: Three empty boxes.
There are $\binom{10}{3}=120$ ways to choose which boxes are empty. We place $1$ ball in each of $7$ boxes, and imagine $6$ dividers with the remaining $13$ balls, which gives $$120\binom{19}{6}$$

Our final answer is
$$120\binom{19}{6}+45\binom{19}{7}+10\binom{19}{8}+\binom{19}{9}=\boxed{6371498}$$
A: Hint:  Can you do the case where no boxes are empty?  That is a standard stars and bars.  Then pick $9,8,7$ of the boxes (how many ways each?) and do the case where no boxes are empty.  Add them up.
