Ball-counting problem (Combinatorics) I would like some help on this problem, I just can't figure it out.
In a box there are 5 identical white balls, 7 identical green balls and 10 red balls (the red balls are numbered from 1 to 10). A man picks 12 balls from the box. How many are the possibilities, in which:
a) exactly 5 red balls are drawn --
b) a red ball isn't drawn --
c) there is a white ball, a green ball and at least 6 red balls
Thanks in advance.

Hints: (a) How many ways can we choose $5$ numbers from $1,2,...,9,10$? (This will tell you how many different collections of $5$ red balls he may draw.) How many distinguishable collections of $7$ balls can he draw so that each of the seven is either green or white? Note that the answers to those two questions do not depend on each other, so we'll multiply them together to get the solution to part (a).
(b) Don't overthink it. How many ways can this happen?
(c) You can split this into $5$ cases (depending on the number of red balls drawn) and proceed in a similar way to what we did in part (a) for each case (bearing in mind that we've already drawn one green ball and one white ball). Then, add up the numbers of ways each case can happen.
