Problem with 3 urns and balls of 3 different colors

Question

Suppose we have 4 red balls, 5 green balls and 6 blue balls. In how many ways can we distribute these balls into 3 numbered urns, $U_1, U_2$, and $ U_3$? Note that the balls can be distributed in any way, even if all balls end up in a single urn.

My attempt: Let $R_1, R_2, R_3$ denote the number of red balls in urns $U_1, U_2,$ and $U_3$, respectively. Similarly, define $G_1, G_2, G_3$ and $B_1, B_2, B_3$ for green and blue balls. Then, I need to find all the combinations $(R_1, R_2, R_3), (G_1, G_2, G_3)$, and $(B_1, B_2, B_3)$ such that $R_1+R_2+R_3=4$, $G_1+G_2+G_3=5$, and $B_1+B_2+B_3=6$. It is well-known that the number of solutions to the equation $x_1+x_2+...+x_k=n$ with $x_i \in \{0,1,2,...,n\}$ is given by $\binom{n+k-1}{k-1}$. Therefore, the total number of possible combinations is given by:

$$ \binom{6}{2}\cdot\binom{7}{2}\cdot\binom{8}{2}=8820. $$

Question: Is my solution correct? If so, are there simpler ways to solve this problem?

Answer 1

This depends upon some assumptions made:

• If the model is that a child is given these balls, and asked to make all possible numbered heaps of the colored balls, your answer is correct

• If, on the other hand, the model is that each ball is put uniformly at random in the urns, we have to use the distinct balls into distinct urns model, and then the # of ways will be
  $3^4\cdot3^5\cdot3^6 = 3^{15}$ ways

Answer 2

The analysis in the posted question is valid, precisely because the urns, being numbered, are to be considered distinguishable, rather than indistinguishable.

That is, with the urns distinguishable, the only way that two distributions given by [red-1,green-1,blue-1] and [red-2,green-2,blue-2] are distinct is if red-1:red-2 is distinct, or green-1:green-2 is distinct, or blue-1:blue-2 is distinct.

In the alternative problem where the urns are un-numbered (and therefore to be considered indistinguishable), the corresponding analysis would be invalid.

As a counter example, assume that exactly 2 blue balls are assigned to each of the 3 indistinguishable urns.

Further assume that in both distributions, all the red balls go into the same urn, and in both distributions, all the green balls go into the same urn.

Then, the two distributions pictured below are different:

Urn-1    Urn-2    Urn-3
red=4    red=0    red=0
grn=5    grn=0    grn=0
blu=2    blu=2    blu=2

Urn-1    Urn-2    Urn-3
red=4    red=0    red=0
grn=0    grn=5    grn=0
blu=2    blu=2    blu=2

From an intuitive point of view, just prior to the green balls being distributed, Urn-1 and Urn-2, which originally are to be considered indistinguishable, become distinguishable, by virtue of their each receiving a different number of red balls.