Find the number of distributions of 12 blue, 13 white, and 14 red balls into 4 different boxes, where each box is supposed to get at least two balls of each color. (Note: Balls of the same color are indistinguishable.)
So far I did $x_1 + x_2 + x_3 + x_4 = 39$ balls.
$y = x + 2$, $y_1 + y_2 + y_3 + y_4 = 31$ balls.
$$\binom{31 + 4 - 1}{4 -1}$$
Is this correct?
Give each box its required two balls of each color.
Thus there are $4$ remaining blue, $5$ remaining white, and $6$ remaining red balls that can be placed arbitrarily.
Can you take it from there?
You'll need a hypergeometric distribution.
Put two balls of each color in each of the four boxes. That leaves us with four blue balls, five white balls, and six red balls to distribute to the four boxes.
Distribute the remaining balls of each color separately.
The number of ways we can distribute the remaining blue balls is the number of solutions of the equation $$b_1 + b_2 + b_3 + b_4 = 4$$ in the nonnegative integers, where $b_i$ represents the number of blue balls placed in box $i$, $1 \leq i \leq 4$.
The number of ways we can distribute the remaining white balls is the number of solutions of the equation $$w_1 + w_2 + w_3 + w_4 = 5$$ in the nonnegative integers, where $w_i$ represents the number of white balls placed in box $i$, $1 \leq i \leq 4$.
The number of ways we can distribute the remaining red balls is the number of solutions of the equation $$r_1 + r_2 + r_3 + r_4 = 6$$ in the nonnegative integers, where $r_i$ represents the number of white balls placed in box $i$, $1 \leq i \leq 4$.
A particular solution of the equation $$x_1 + x_2 + x_3 + \cdots + x_n = k$$ in the nonnegative integers corresponds to the placement of $n - 1$ addition signs in a row of $k$ ones.
To use the example of the remaining red balls, $$1 1 + + 1 + 1 1 1$$ corresponds to the solution $r_1 = 2$, $r_2 = 0$, $r_3 = 1$, $r_4 = 3$.
The number of solutions of the equation $$x_1 + x_2 + x_3 + \cdots + x_n = k$$ in the nonnegative integers is $$\binom{k + n - 1}{n - 1}$$ since we must choose which $n - 1$ of the $k + n - 1$ positions required for $k$ ones and $n - 1$ addition signs will be filled with addition signs.
Since the three colors can be distributed independently, we multiply the number of ways of distributing the remaining balls of each color.
There are $$\binom{4 + 4 - 1}{4 - 1}$$ ways to distribute the remaining blue balls, $$\binom{5 + 4 - 1}{4 - 1}$$ ways to distribute the remaining white balls, and $$\binom{6 + 4 - 1}{4 - 1}$$ ways to distribute the remaining red balls. Hence, the number of ways of distributing the balls to four distinct boxes so that there are at least two balls of each color in each box is $$\binom{4 + 4 - 1}{4 - 1}\binom{5 + 4 - 1}{4 - 1}\binom{6 + 4 - 1}{4 - 1}$$