Basic inclusion exclusion problem - picking 24 balls out of four sets of 10 balls This is probably a basic problem, but I'm having a really difficult time solving it.
There are ten red balls, ten green balls, ten blue balls, and ten white balls. Balls of the same color are considered equal. How many ways are there to select 24 balls, regardless of order? (Sorry if the question was phrased awkwardly, I translated it)
The question is supposed to be solved using the inclusion-exclusion principle. 
I'd appreciate advice not only on how to solve this specific problem, but how to solve general inclusion-exclusion problems.

I usually start with a smaller problem and look for a pattern to scale up to the goal.
Let's start with 2 balls of each color (2,2,2,2) and we want to draw 5 (roughly over half, just like 24/40).  
On the first draw, all colors are equally likely, thus there are 4 possibilities (because all balls of the same color count the same).
On the second draw, I'm going to draw from a pool of 1,2,2,2 balls (one of the colors is lacking).  There are still 4 possibilities here (although one is less likely than the other 3).  However, since you said order doesn't matter, we've got to subtract off 4 chose 2 or 6 of those
So far, there are 4 * 4  - 16 = 10 possibilities, but you probably could have worked that out yourself.  The next draw, things get tricky.
I could be at two different states (1,1,2,2) or (0,2,2,2).  So, at this level are there 3 or 4 possibilities?  Both.  Of the 10 possibilities we have thus far, 4 are are the case in which we drew two of the same color and depleted that color leaving only 3 additional possiblities.  For the other 6, they have 4 colors to pick from.  So, at this point there are 4 * 3 + 6 * 4 = 36 combinations, but we have to account for duplicates again.  The only ways we can have duplicates is either 2 of one color and 1 of another or 3 distinct colors.  The latter is easy, 4 choose 3 or 4.  The former is a bit more tricky but there are 4 choose 2 pairs of colors and in each pair the first color could be the one with 2 or the second color could be, thus 2 * (4 choose 2) or 12 duplications.  
So, after three pulls we have 4 * 3 + 6 * 4 - 4 - 12 = 20 combinations.  You may be thinking "ugh!  This is all drudgework and we still have 2 more pulls to do!"  Fear not.  These sort of combinatorial problems usually have symmetry to them because you can just reverse the bag of balls and the table that you are setting the balls on, i.e. The combinations of 0 iterations are the same as n, 1 iterations is the same as n-1.  Since we have calculated 3 steps, that's the same as 5, so we are done.  
As far as scaling up your problem, you can easily parse through the first 10 steps, until you deplete a color.  Then, go 6 more steps and stop because the work done at 16 will be the same as 24, which you are looking for.  
Each iteration, find out how many total combinations there are for that step and then subtract off the duplicates.  You may wish to try another small scale of 3 balls of each color, just to get the hang of all the tricks you may encounter at the scale of 10 balls of each color.
