How many ways are there to pick 12 balls from large piles of (identical) red, white and blue balls? Would the below be correct? Since there are $3$ types of balls from $12$, you would do $12$ choose $9$ and then multiply by $3$?
$$\binom{12}{9} \cdot 3$$ 
 A: No.
You (effectively) wish to count the ways to fill three boxes with 12 identical balls, and then the balls get coloured according to which box they are put in.
Are you familiar with the "Stars and Bars" theorem?

 $$\binom{12+3-1}{12} = 91$$

A: Let $x_r$, $x_w$, and $x_b$ denote, respectively, the number of red, white, and blue balls selected.  Since a total of twelve balls are selected, the number of ways to select $12$ balls from large piles of identical red, white, and blue balls is the number of solutions of the equation
$$x_r + x_w + x_b = 12$$
in the non-negative integers.  A particular solution corresponds to the placement of two addition signs in a row of twelve ones.  For instance,
$$1 1 1 1 1 + 1 1 1 1 + 1 1 1$$
corresponds to the solution $x_r = 5$, $x_w = 4$, and $x_b = 3$, while 
$$1 1 1 1 + + 1 1 1 1 1 1 1 1$$
corresponds to the solution $x_4 = 4$, $x_2 = 0$, and $x_8 = 8$.  Thus, the number of solutions of the equation in the non-negative integers is 
$$\binom{12 + 2}{2} = \binom{14}{2}$$
since we must choose which two of the fourteen symbols (twelve ones and two addition signs) will be addition signs.
