Distributing $3$ blue and $4$ green and $9$ red balls into $3$ distinct urns 
You have $16$ balls, $3$ blue, $4$ green, and $9$ red. You also have $3$ distinct urns. For each of the balls, you select an urn at random and put the ball into it. What is the probability that each urn contains all three colors?

Put $1$ each $B, G, R$ in all urns. Put remaining $6 \, R$ in $3$ urns in $^{6+3-1} C_{3-1}$ ways. Similarly for $B, \, G$. The required probability is $$\frac{^{3} C_{2} \times {^{8}C_{2}}}       {^{5} C_{2}\times {^{6} C_{2}}\times {^{11} C_{2}}}$$. 
Please tell where is the mistake. Thanks.
 A: The question is equivalent to finding the probability that while distributing the balls of a color there is no empty urn. At the end the probabilities should be multiplied so that the probability in question is:
$$
\prod_{i=R,G,B}\frac {3!}{3^{n_i}}{n_i \brace 3}=\frac{48400}{531441}\approx0.091,
$$
where ${p\brace q}$ stays for the Stirling number of the second kind.
A: Here is an answer that considered different approaches, some of them wrong, and I am curious to have someone point out the flaws in my reasoning...
First of all, it is $3$ separate issues for the three colors, as each is assigned independently and so we should be able to simply multiply the probabilities for each.
Considering the blue balls as distinct, there are $3^3$ ways they can go, and only $3! = 6$ of those options lands one ball in each.  So that is $6/27 = 2/9$.  Alternatively, P(1st ball okay) = $1$, P(2nd blue ball is okay) = $2/3$, and $P(3$rd ball is okay$) = 1/3$, giving $2/9$ again.
Counting the green arrangements: there are ${6 \choose 2} = 15$ ways to distribute the identical balls, but only $3$ of those ways are successes. So $P($green in all three$) = 3/15 = 1/5$. This answer is incorrect!
Alternate method for green: a tree, branching on the number of urns containing green at each step.  Working that way I find $4/9$ instead of $1/5$.
A third method for green: consider all $81$ possible outcomes, labeled by urn number.  $(1,1,2,3)$ has $12$ permutations, and likewise $(1,2,2,3)$ and $(1,2,3,3)$, for $36/81 = 4/9$ again, so it appears the first approach is miscounting somehow.
For the red: placing $9$ balls in $3$ urns is too many to count every option. The method that failed for green proceeds as follows:
$11 \choose 2$ ways to arrange the balls and bin boundaries, and if we consider $3$ placed, that leaves $6$ to place, so ${8 \choose 2}$ ways to arrange successfully. That gives $56/110 = 28/55$.  Somehow that is apparently a miscount.
Another approach would be to use inclusion exclusion: the probability of avoiding urn $1$ with all $9$ balls would be $(\frac{2}{3})^9$, and likewise for the second and third urns.  If I total those, I have overcounted by the number of ways to have all $9$ end up in a single urn, which is $3^{-8}$, giving $\frac{512\cdot3-3}{3^9} \approx 0.077884$, and so the probability of reaching all $3$ urns should be $\approx 0.9221155$
Putting it all together, I find $$\frac{2}{9}\cdot\frac{4}{9}\cdot\big(1 - \frac{1533}{3^9}\big) \approx 0.09107313...$$ which according to @user is the correct answer.
So where did the stars and bars argument go wrong?  I'm guessing that some things I thought equally likely aren't.
