What did you do wrong?
Note that designating particular balls to be the representative of that color in a particular box leads to overcounting whenever you have more than one ball of that color drawn from that box since such selections are counted once for each way you could have designated one of the balls of that color as the ball of that color.
To illustrate, suppose the red balls are numbered $r_{ij}$ and the blue balls are numbered $b_{ij}$, where $i$ denotes the box number and $j$ denotes the number in that box. Then you count the selection $b_{11}, b_{12}, b_{21}, b_{31}, b_{41}, r_{11}, r_{12}, r_{21}, r_{31}, r_{41}$ four times.
\begin{array}{c c c c c}
\text{box 1} & \text{box 2} & \text{box 3} & \text{box 4} & \text{additional balls}\\ \hline
b_{11}, r_{11} & b_{21}, r_{21} & b_{31}, r_{31} & b_{41}, r_{41} & b_{12}, r_{12}\\
b_{11}, r_{12} & b_{21}, r_{21} & b_{31}, r_{31} & b_{41}, r_{41} & b_{12}, r_{11}\\
b_{12}, r_{11} & b_{21}, r_{21} & b_{31}, r_{31} & b_{41}, r_{41} & b_{11}, r_{12}\\
b_{12}, r_{12} & b_{21}, r_{21} & b_{31}, r_{31} & b_{41}, r_{41} & b_{11}, r_{11}
\end{array}
Solution
Let's consider cases, depending on how many balls of each color are chosen and how many balls are selected from each box. Since at least one ball of each color must be selected from each box, there can be at most six balls of the same color.
Six red and four blue balls are selected
We must select a blue ball from each box, which can be done in $\binom{2}{1}^4$ ways. For the red balls, there are two possibilities. Either three red balls are selected from one box and one each is selected from the other three boxes, which can occur in $\binom{4}{1}\binom{3}{3}\binom{3}{1}^3$ ways, or two red balls each are selected from two of the four boxes and one red ball each is selected from the other two boxes, which can occur in $\binom{4}{2}\binom{3}{2}^2\binom{3}{1}^2$ ways. Hence, there are
$$\binom{2}{1}^4\left[\binom{4}{1}\binom{3}{3}\binom{3}{1}^3 + \binom{4}{2}\binom{3}{2}^2\binom{3}{1}^2\right]$$
such selections.
Five red and five blue balls are selected
There are two possibilities. We select two red balls and two blue balls from the same box and one ball of each color from each of the other boxes, which can occur in $\binom{4}{1}\binom{3}{2}\binom{2}{2}\binom{3}{1}^3\binom{2}{1}^3$ ways, or we can select two red balls from one box, one red ball each from the other three boxes, two blue balls from a box different than the box from which two red balls were selected, and one blue ball from each of the other three boxes, which can occur in $\binom{4}{1}\binom{3}{2}\binom{3}{1}^3\binom{3}{1}\binom{2}{2}\binom{2}{1}^3$ ways. Hence, there are
$$\binom{4}{1}\binom{3}{2}\binom{2}{2}\binom{3}{1}^3\binom{2}{1}^3 + \binom{4}{1}\binom{3}{2}\binom{3}{1}^3\binom{3}{1}\binom{2}{2}\binom{2}{1}^3$$
such selections.
Four red and six blue balls are selected
We must select one red ball from each box, which can occur in $\binom{3}{1}^4$ ways. We must select two blue balls each from two of the four boxes and one blue ball from each of the other two boxes, which can occur in $\binom{4}{2}\binom{2}{2}^2\binom{2}{1}^2$ ways. Hence, there are
$$\binom{3}{1}^4\binom{4}{2}\binom{2}{2}^2\binom{2}{1}^2$$
such selections.
Total: Since these cases are mutually exclusive and exhaustive, the number of possible selections is
$$\binom{2}{1}^4\left[\binom{4}{1}\binom{3}{3}\binom{3}{1}^3 + \binom{4}{2}\binom{3}{2}^2\binom{3}{1}^2\right] + \binom{4}{1}\binom{3}{2}\binom{2}{2}\binom{3}{1}^3\binom{2}{1}^3 + \binom{4}{1}\binom{3}{2}\binom{3}{1}^3\binom{3}{1}\binom{2}{2}\binom{2}{1}^3 + \binom{3}{1}^4\binom{4}{2}\binom{2}{2}^2\binom{2}{1}^2$$
Best Answer
Are the balls distinct? That is, do they have numbers on them such that it matters which balls are chosen, which balls are red, and which are blue? In that case, you need to consider "choice" to account for the different arrangements. There are $\binom{n}{k}$ ways to choose balls to color, then each ball can be colored one of two ways, for $2^k$ combinations. That is, you’ve already committed to painting the ball in hand, so is it going to be red, or is it going to be blue? With one ball, two possible outcomes. With two balls, 4, three gives 8 combinations and so on. The answer is the product of the choice and the how-to-paint-each-ball.
If not, the problem becomes much simpler. I envision a hopper of colorless, indistinguishable balls. The only restriction I see is that k of them must be colored. Therefore, select k balls out of the hopper--you only do this once--then consider the combinations of painting none of the balls blue (all red), one of the balls blue (k-1 red), two blue (k-2 red) and so on, up to k balls blue. Since there is no distinguishability among balls, each combination happens once.
Therefore, there are k+1 ways to do this.