10. Place 11 identical balls into six distinct boxes so that at most three boxes are empty. The number of ways to do this is $\qquad$.
10. 4212 .

If there are no empty boxes, there are $\mathrm{C}_{10}^{5}$ ways to place them; if there is one empty box, there are $\mathrm{C}_{6}^{1} \mathrm{C}_{10}^{4}$ ways to place them; if there are two empty boxes, there are $\mathrm{C}_{6}^{2} \mathrm{C}_{10}^{3}$ ways to place them; if there are three empty boxes, there are $\mathrm{C}_{6}^{3} \mathrm{C}_{10}^{2}$ ways to place them. Therefore, the total number of ways to place them with at most three empty boxes is $\mathrm{C}_{10}^{5}+\mathrm{C}_{6}^{1} \mathrm{C}_{10}^{4}+\mathrm{C}_{6}^{2} \mathrm{C}_{10}^{3}+\mathrm{C}_{6}^{3} \mathrm{C}_{10}^{2}=4212$