Probability problem from math competition I was training for the math competitions when I've found this problem. I'll try to translate it from italian:
A bad chemist arrives in a lab where 4 groups are working , each consisting of 3 people. Everyone of these people holds a flask with distilled water, each flask with the same amount of water. The bad chemist goes to the first group and puts $a_1$ drops  of sulfuric acid in the first flask, $a_2$ in the second and $a_3$ in the third, so that $a_1+a_2+a_3=4$. In the same way, he goes to the second group and puts $b_1$ drops  of sulfuric acid in the first flask, $b_2$ in the second and $b_3$ in the third, so that $b_1+b_2+b_3=3$. He does the same with the third group, putting $c_1, c_2, c_3$ drops respectively, so that $c_1+c_2+c_3=2$, and with the fourth group, putting $d_1, d_2, d_3$ drops, so that $d_1+d_2+d_3=1$. Assuming that the drops are all of the same size and that the chemist puts the sulfuric acid drops in the flasks randomly, what is the percentage probability that the flask with the most acid solution belongs to the first group (most acid of all the flasks, no ties).
The answer to this problem is $36\%$ (I don't know if it's rounded). I tried to solve the problem in the following way. The possible configurations are:
First group: $4$-$0$-$0$ ($3x$), $3$-$1$-$0$ ($6x$), 2-2-0 (3x), 2-1-1 (3x) -> 15 configurations
Second group: $3$-$0$-$0$ ($3x$), $2$-$1$-$0$ ($6x$), 1-1-1 (1x) -> 10 configurations
Third group: $2$-$0$-$0$ ($3x$), $1$-$1$-$0$ ($3x$) -> 6 configurations
Fourth group: $1$-$0$-$0$ ($3x$) -> 3 configurations
Thus, all the possible configurations are $15\cdot 10\cdot 6\cdot 3=2700$. The "good" ones are:
$4$-$0$-$0$ for the first group:  all of the second, third, fourth group configurations are "good"-> $3\cdot 10\cdot 6\cdot 3=540$
$3$-$1$-$0$ for the first group:  2-1-0 and 1-1-1 of the second group, all of the third and fourth are "good" -> $6\cdot 7\cdot 6\cdot 3=756$
$2$-$1$-$1$ for the first group:  1-1-1 of the second group, 1-1-0 of the third, all of the fourth are "good"-> $3\times 1\times 3 \times 3=27$
$2$-$2$-$1$ for the first group:  no configurations, because we have a tie
The probability become $$\dfrac{540+756+27}{2700}=0,49=49\%$$
I can't figure out what I am doing wrong. Could anyone help me?
 A: Following the suggestion of @JMoravitz, I solved the problem. The mistake in my first solution was that I considered all the "configurations" equally likely to occur. Like suggested, you have to put attention also to the order of placing the drops in the set of flasks. Doing that, all the possible configuarations are:
First group: 4-0-0 (3x), 3-1-0 (24x), 2-2-0 (18x), 2-1-1 (36x) -> $3^4=81$ configurations
Second group: 3-0-0 (3x), 2-1-0 (18x), 1-1-1 (6x) -> $3^3=27$ configurations
Third group: 2-0-0 (3x), 1-1-0 (6x) -> $3^2=9$ configurations
Fourth group: 1-0-0 (3x) -> $3$ configurations
Thus, all the possible configurations are $3^4\cdot 3^3\cdot 3^2\cdot 3= 3^{10}=59049$. The "good" ones are:
4-0-0 for the first group: all of the second, third, fourth group configurations are "good"-> $3\cdot 27\cdot 9\cdot 3=2187$
3-1-0 for the first group: 2-1-0 and 1-1-1 of the second group, all of the third and fourth are "good"-> $24\cdot 24\cdot 9\cdot 3=15552$
2-1-1 for the first group: 1-1-1 of the second group, 1-1-0 of the third, all of the fourth are "good"-> $36\times 6\times 6 \times 3=3888$
2-2-1 for the first group: no configurations, because we have a tie
The probability become $$\dfrac{2187+15552+3888}{59049}=\dfrac{21627}{59049}=\dfrac{89}{243}\approx 36\%$$
Thank you all again.
