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
 A: 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:
$$n = \frac{\binom{12}{4} \times \binom{8}{4} \times \binom{4}{4}}{3!}$$
$$n = \frac{495 \times 70 \times 1}{6}$$
$$n = 5775\text{ ways}$$
A: This looks to me like a multinomial problem: we can work with $\binom{12}{4,  4, 4}$ since three groups of four students (total 12) are testing.
The multinomial coefficient $$\binom{12}{4, 4, 4} = \dfrac{12!}{4!4!4!} = 34650$$ ways that 12 students can be grouped into groups of $4$ to take one of 3 specific tests: $4$ taking test 1, $4$ taking test 2, and $4$ taking test 3.
There is no double counting because each group is taking a different test: there are three distinct tests, after all.
A: ** It's an attempt only! **
12 students, 4 students each test, which means there are three tests. First you take 4 students for the first one (12 choose 4), and then (8 choose 4), and (4 choose 4).
Now that the groups are determined, we need to assign each group to a test, for that we have 3! = 6 options for each group selection.
Though in order to avoid double counting in the group selection, we divide by 3 because the groups can be chosen in a different order.
Finally, it's  $12\choose 4$ * $8\choose4 $* $4 \choose 4$ $ * 2 $
