Is there a formula to state the number of 'sets' of 'ordered sets within ordered groups'?
Here the explanation for the numbers we found in the comments, which unfortunately do not match yours. If you have additional explanations I can delete this answer, or perhaps improve.

Goal: given $a$ students, make $b$ groups, each consisting of $c$ persons, $bc \le a$. What is the number of possibilities? 

My assumptions, the groups are unordered, so $(1,2,3)$ is the same group as $(3,1,2)$. Also the solution $(1,2)(3,4)$ is equal to $(3,4)(1,2)$

Put your students in a row: $a!$ possibilities. The first $b$ students form the first group, the next $b$ students the second group, etc. The order in the line-up between each group of students does not matter, so we have to divide by th1s number of possibilities which is $(c!)^b$. Again we divide by $(a-bc)!$ the number of permutations of the not chosen ones. Also, the order of the groups themselves does not matter, we additionally divide by $b!$.

Answer: $a!\;/\;{(c!)^b (a-bc)! \;b!}$

Another way of obtaining the same number. Choose a group of $c$ students from total $a$, then a group of $c$ students from the remaining $c-b$, etc. This can be done in ${a \choose c}{a-c\choose c}\dots{a-c(b-1)\choose c}$. Again I divide by the possible orderings of the groups which is $b!$.
Thus $\frac{a!}{c!(a-c)!}\frac{(a-c)!}{c!(a-2c)!}\dots \frac{(a-(b-1)c)!}{c!(a-bc)!} / b!$

Your first example $a=12$, $b=4$, $c=3$. You have 880, this number is 15400.

Second example  $a=12$, $b=3$, $c=2$. 13680, we agree.