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!
 A: 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..
A: 
$10$ 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 $i=2,i=3$.

$i = 2$
As you observed, the sample space has size $|\Omega| = 4^{10}$.
There are indeed $\binom{4}{2}$ ways to select the two versions which exactly three students will receive.  We can assign three of the ten students the lower numbered of the two selected versions in $\binom{10}{3}$ ways and three of the remaining seven students the higher numbered of the two selected version in $\binom{7}{3}$ ways.  You are also correct that there are eight ways to assign versions to the remaining students so that neither of the remaining versions is taken by exactly three students since there are $\binom{4}{2}$ ways to assign exactly two students to take the lower-numbered of the remaining two versions and one way to assign the remaining two students the remaining version and $2$ ways to assign all four students to take one of the remaining versions.
Therefore, the probability that there are exactly two versions which exactly three students receive is
$$\frac{\dbinom{4}{2}\dbinom{10}{3}\dbinom{7}{3}\left[\dbinom{4}{2} + \dbinom{2}{1}\right]}{4^{10}} \approx 0.1923$$
Where did you make your mistake?
You made two mistakes.
You meant to write that the number of ways of assigning exactly three of the remaining four students one of the two remaining versions is
$$\binom{2}{1}\binom{4}{3} = 8$$
since there are two ways to select the version three of the remaining four students will receive, $\binom{4}{3}$ to select the three students who will receive that version, and one way to assign the remaining student the remaining version.
Note that
$$\frac{4!}{(4 - 3)!} = \frac{4!}{1!} = \frac{4!}{1} = 24$$
Evidently, what you wrote is not what you meant.
Also, you made a computational error.
$$\Pr(A) = \frac{|A|}{|\Omega|} = \frac{\dbinom{4}{2}\dbinom{10}{6}\dbinom{6}{3}\cdot 8}{4^{10}} \approx 0.1923$$

$i = 3$
There are $\binom{4}{3}$ ways to select the three versions which exactly three students each will receive, $\binom{10}{3}$ ways to assign three of the ten students the lowest numbered of those versions, $\binom{7}{3}$ to assign three of the remaining seven students the next lowest numbered of the selected versions, and $\binom{4}{3}$ ways to assign three of the remaining four students the highest numbered of the selected versions.  The other student must receive the fourth version.  Hence, the probability that there are three versions which exactly three students each will receive is
$$\frac{\dbinom{4}{3}\dbinom{10}{3}\dbinom{7}{3}\dbinom{4}{3}}{4^{10}} \approx 0.0641$$
Where did you make your mistake?
After you selected which nine students would receive the three versions which exactly three students each would receive, the number of ways of distributing versions to students so that exactly three students each would receive them is
$$\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
Therefore, you should have obtained
$$\Pr(A) = \frac{|A|}{|\Omega|} = \frac{\dbinom{4}{3}\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{4^{10}} = 0.0641$$
Your term $3^9$ is the number of ways of distributing nine students to take those versions without restriction, which is why you obtained a much larger number.
