Dividing 30 Balls

Distribute the white balls and the black balls in any fashion. Since they add up to $15$ balls, it's impossible for a box to be overfilled, and hence every arrangement of them is valid. The red balls fill in the rest, completing each box to $15$ balls each.

Now, there are $6$ ways to distribute the white balls (one way for each of $0,1,2,\ldots,5$ balls in box A), and similarly $11$ ways to distribute the black balls (one way for each of $0,1,2,\ldots,10$ balls in box A), so in total there are $6 \cdot 11 = \boxed{66}$ ways to distribute the balls.

The question can be solved using generating functions without painstaking calculations . We can associate the generating functions $A_x=1+x+...+x^5=\frac{x^6-1}{x-1}$ $B_x=1+x+...+x^{10}=\frac{x^{11}-1}{x-1}$ $C_x=1+x+...+x^{15}=\frac{x^{16}-1}{x-1}$ to the sets of balls. Solving the question comes down to finding the coefficient of $x^{15}$ since deciding the configuration of box A determines box B's one (there are only 30 balls).

We will write the product $A_xB_xC_x= \frac{(x^{16}-1)(x^{11}-1)(x^6-1)}{(x-1)^3}$ in a nicer way. Multiplying out the numerator gives $x^{11}+x^6-1$ , ignoring the terms of degree higher than $15$ . The denominator can be written as $(1-x)^{-3}= \sum_{k=0}^{\infty} {3+k-1 \choose k} \cdot (-1)^k \cdot (-x)^k = {2 \choose 2} + {3 \choose 2} x + { 4 \choose 2 } x^2 + \cdots + { 15+2 \choose 2 } x^ {15} + \cdots .$ Since this has to be multiplied by $x^{11}+x^6-1$ , we are only concerned with $15 x^4+55x^9+136x^{15}$ : taking the product of those two the term of degree $15$ coefficient's is $|15+55-136|=66.$

Note: expressing $(1-x)^{-3}$ as a power series is particularly nice since ${n \choose 2}$ is the $n-th$ triangular number, so you just have to keep summing $n+1$ to the coefficient of the previous term while writing it out: $(1-x)^{-3}=1+3x+6x^2+10x^3+15x^4+21x^5+ \dots .$

Here we can also use generating functions , we can write the equation as................................................................ (x^0+x^1+...+x^5) (x^0+x^1+...+x^10) (x^0+x^1+...+x^15) Solving this (1-x^6) (1-x^11) (1-x^16) (1-x)^-3 ... From this we can get the coefficient of x^15 to get the answer which is 66. This solution requires knowledge of Generating functions for Combinatorics which it is simple yet powerful.

Using stars and bars method: $w + b + r = 15$

In total $^{17}C_2 = 136$ ways to arrange them

We need to remove cases which have either $w >= 6$ or $b > = 11$ or $r >= 16$ .

Using PIE, for the above, you get total number of invalid combinations as $^{11}C_2 + ^{6}C_2 = 70$ .

Thus, final answer $=66$