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.
 A: For part (a), since we require exactly $5$ red balls, we know that the $7$ other balls must be composed of only white and green balls.  How many ways can we choose $5$ distinct red balls from $10$?  This is exactly what the symbol $\binom{10}{5}$ measures.  It's formula is given by:
$$\binom{10}{5}=\frac{10!}{5!(10-5)!}=\frac{10\cdot9\cdot8\cdot7\cdot6}{5\cdot4\cdot3\cdot2\cdot1}=630$$
Now, we have to ask ourselves how many ways can we make a collection of $7$ balls from $5$ identical white balls and $7$ identical green balls.  Since there are only $5$ white balls, we can have anywhere from $0$ to $5$ white balls, making $6$ possibilities.  This means, altogether, there are $6\times 630=3780$ possible different collections of $12$ balls with exactly $5$ red balls.
Notice that since the red balls are distinguishable, there are many more ways to choose them than there are to choose the identical white and green balls.
As is mentioned in the comments, part (b) is relatively straightforward.  See if you can do part (c) applying some of the concepts we used for part (a) (just do part (a) for exactly $6,7,8,9,$ and $10$ red balls, and add the results).
A: 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.
A: 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 $k=6,7,8,9,10$.
