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.

It seems that I misunderstood the problem. I assumed that the balls are distinguishable objects, as one would do when counting probabilities of the corresponding events. If green and white balls are not supposed to be distinguishable, then the below is wrong.
Because there seems to be nothing to gain from deleting a solution, I'm leaving it with the above disclaimer.

For a) you essentially want to choose $5$ out of $10$ red balls, and then $7$ of the remaining $12$ balls. Altogether, ${10 \choose 5} \cdot {12 \choose 7}$
For b) you just want to draw $12$ non-red balls. It so happens that there are $12$ non-red balls in the box, hence the answer is just $1$.
For c), fix the number of red balls you draw, call it $k$. Then, there are ${10 \choose k}$ ways to draw $k$ red balls, and $12 \choose 12 - k$ ways to draw the rest. We want to exclude the situations in which only white or only green balls are drawn apart from the red ones. This happens in, respectively, $5 \choose k$ and $7 \choose k$ ways. Hence, for given $k$, there are:
$${10 \choose k} \cdot \left( {12 \choose 12 - k} - {5 \choose k} - {7 \choose k} \right)$$
I am afraid there is no other way than laboriously sum this for $k=6,7,8,9,10$.
