There are 12 students in a class. Find the number of n ways so that the student can take tests if 4 students are to take each test. There are 12 students in a class.  Find the number of n ways so that the student can take tests if 4 students are to take each test.
I tried to answer this but I think that my answer is very wrong. My solution is
$$\frac{12!}{(12-3)!3!}\frac{12!}{(12-4)!4!}.$$
I did my best but I think this is not the correct answer, please help me with this

I think that your thought is good. First you choose $4$ students out of $12$, then for the second group you choose $4$ students out of $8$ and the at last $4$ students out of $4$. But as you've mentioned it can lead to double counting.
So we calculate:
$$\binom{12}{4} \times \binom{8}{4} \times \binom{4}{4}$$
But instead of dividing with $3$, you should divide by $6$, because the number of permutations of $3$ is $6$. Here's example:
Let studnets from $1-4$ be in the first group, $5-8$ in the second and $9-12$ in the third.
We have $6$ permutations.
$$A = {1,2,3,4} \quad B = {5,6,7,8} \quad C = {9,10,11,12}$$
$$A = {1,2,3,4} \quad B = {9,10,11,12} \quad C = {5,6,7,8}$$
$$A = {5,6,7,8} \quad B = {1,2,3,4} \quad C = {9,10,11,12}$$
$$A = {5,6,7,8} \quad B = {9,10,11,12} \quad C = {5,6,7,8}$$
$$A = {9,10,11,12} \quad B = {1,2,3,4} \quad C = {5,6,7,8}$$
$$A = {9,10,11,12} \quad B = {5,6,7,8} \quad C = {1,2,3,4}$$
This 6 example are completely the same, but using our formula for generating number of ways we obtain all of them, so we have to divide by $6$.
So the number of ways is:
$$n = \frac{\binom{12}{4} \times \binom{8}{4} \times \binom{4}{4}}{3!}$$
$$n = \frac{495 \times 70 \times 1}{6}$$
$$n = 5775\text{ ways}$$
