Karcsinak has 10 identical balls, among which 5 are red, 3 are white, and 2 are green, and he also has two boxes, one of which can hold 4 balls, and the other can hold 6 balls. In how many ways can he place the balls into the two boxes? (The arrangement of the balls within the boxes is not important.)
The division of 10 balls into two parts can be achieved by specifying the color arrangement of the 4 balls that go into the smaller box, with the remaining 6 balls always going into the larger box. (Of course, we understand the problem such that, for example, 5 red balls cannot be distinguished from each other, so if we swap one red ball from the large and small boxes, the new arrangement is not considered different from the previous one.)

The filling of the small box can start in 5 different ways with red balls: with 4, or 3, or 2, or 1 ball, or without any red ball (0 balls). The continuation with white balls and the completion with green balls, according to the number of remaining spaces, can lead to the following final configurations, totaling 11 ways:

$$
\begin{array}{lllll}
R R R R ; & R R R W, & R R W W, & R W W W, & W W W G, \\
& R R R G, & R R W G, & R W W G, & W W G G . \\
& & R R G G, & R W G G .
\end{array}
$$