<s>[INST] Probability of randomly assigning elements to groups 
10 students are tested in an exam with 4 different versions. Each
student is randomly assigned to one of the versions. What is the
probability that there are exactly $i$  versions in which exactly  > $3$ students were assigned. Answer separately for $i=2, i = 3$.

I've just started my probability course and it's been a while since I've touched combinatorics (and admittedly I was never good at it), and I'm having trouble getting it right..
My attempt:
Note that as we choose the cards randomly we have a symmetric probability space.
We have $10$ students, each student "chooses" one of $4$ versions. Order matters and repetition allowed, therefore
$|\Omega| = 4^{10}$
For $i = 2$ we first choose two versions which will have exactly $3$ students
${4 \choose 2}$
Then we choose $6$ students to be assigned to the versions we've chosen
${10 \choose 6}$
Then we count the number of different combinations for $6$ students in $2$ versions, where there are exactly $3$ students in each version. We choose $3$ students out of $6$ to place in one version and the remaining students will be in the second version.
${6 \choose 3}$
Now we are left with $4$ students to assign between the remaining two versions. We either have two versions with $2$ students each, or one version with $4$ students. We calculate total options to allocate students and subtract the options in which there are $3$ students in the same version.
$3$ Students in $1$ version - We choose the version ${2 \choose 1}$, allocate $3$ out of $4$ students to it with the remaining student in the last version $\frac{4!}{\left(4-3\right)!}$, so in total
${2 \choose 1}\frac{4!}{\left(4-3\right)!} = 8$
Total options to allocate $4$ students into $2$ versions without restrictions $2^{4} = 16$
Therefore in total $|A| = {4 \choose 2}{10 \choose 6}{6 \choose 3}(16-8)={4 \choose 2}{10 \choose 6}{6 \choose 3}8$ and we have
$P(A) = \frac{\left|A\right|}{\left|Ω\right|} = \frac{{4 \choose 2}{10 \choose 6}{6 \choose 3}8}{4^{10}} = 0.384$
For $i = 3$ we choose $3$ versions to have $3$ students.
${4 \choose 3}$
We choose $9$ students to allocate to these versions, with the remaining one on the other version.
${10 \choose 9}$
Total options to organize $9$ students in $3$ versions.
$3^{9}$
Therefore $|A| = {4 \choose 3}{10 \choose 9}3^{9}$ and we have
$P(A) = \frac{\left|A\right|}{\left|Ω\right|} = \frac{{4 \choose 3}{10 \choose 9}3^{9}}{4^{10}} = 0.750$
I can't seem to find any mistakes, but I would also be really grateful if you can comment whether my way of thinking is correct and maybe give a few tips on how to approach these types of questions!
 [/INST] For another approach and showing the mistake :
$1-)$ For $i=2$ , By your reasoning select firstly $2$ case which take exactly $3$ sutdents by $C(4,2)$ ,after that ,  choose $3$ students for the first and the second by $C(10,3) \times C(7,3)$ ,respectively .Now , we have $4$ students to disperse to $2$ versions , but they cannot have $3$ students. Lets find that case by exponential generating functions such that the exponential generating function for each version is $$\bigg(1 + x+ \frac{x^2}{2!}+\frac{x^4}{4!}\bigg)$$
Then , find $[x^4]$ in the expasion of $$\bigg(1 + x+ \frac{x^2}{2!}+\frac{x^4}{4!}\bigg)^2$$
So , $[x^4]=8$
Calculation of E.G.F :
Then , $$\frac{C(4,2) \times C(10,3) \times C(7,3) \times 8 }{4^{10}} = \frac{6 \times 120 \times 35 \times 8}{4^{10}}= 0,1922...$$
I guess you made mistake in $C(4,2)$ , you might have counted  it $12$ instead of $6$
$2-)$For , $i=3$ , Select $3$ versions taking exactly $3$ students by $C(4,3)$ ,  select $3$ students for the selectd versions by $C(10,3) \times C(7,3) \times C(4,3)$. The remainig will go to the remaining version automatically. Then , $$\frac{C(4,3) \times C(10,3) \times C(7,3) \times C(4,3) }{4^{10}}= \frac{4 \times 120 \times 35 \times 4}{4^{10}}=0,06408..$$
When you write $3^9$ in your answer  , you do not ensure that the selected version will have exactly $3$ students , it is the mistake..
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