Given $26$ balls - $8$ yellow, $7$ red and $11$ white - how many ways are there to select $12$ of them? I'm interested in knowing and  understanding the solution to the following problem: 
given $26$ balls - $8$ yellow, $7$ red and $11$ white - how many ways are there to select $12$ of them (all balls of the same colour are indistinguishable).
I've read something about generating functions but I hoped there was a more straightforward way to "see" the solution...I really can't figure it out.
Thanks in advance for your help!

Forget for a while about the fact that the numbers of each colour are limited. If there were at least $12$ of each colour, the problem would be straightforward Stars and Bars. The answer would be $\binom{12+3-1}{3-1}$. 
From this we must subtract the "bad" choices that involve using more balls of a given colour than are available. The bad choices are (i) too many yellow, (ii) too many red, and (iii) too many white.  Note that we can count these separately and add, since we will never need simultaneously too many balls of two or more colours. 
We count the number of choices with too many yellow, that is, $9$ or more. The number of ways to have $9$ yellow or more is the number of ways 
to choose $3$ balls from yellow, red, and/or white  to accompany $9$ yellow. This is easy Stars and Bars, but also can be done by explicit listing.
The number of choices with too many red is done the same way. The number of choices with too many white needs no machinery.
