It can be seen as distributing n unique objects above m groups
The students $S_1, S_2,…S_{10}$ are divided into 3 groups A, B and C such that each group has at least one students and C has at most 3. Find possibilities of forming groups
$$a+b+c=10$$
The number of positive integral solutions $(\ge 1)$ is
$$\binom {10-1}{3-1}=36$$
I don’t know how to account for the cases where C has at most 3 students
Also I know this method of solving is probably wrong but I don’t know how else to do it.
Best Answer
Leaving aside group $C$, each student has $2$ choices for groups
Group $C$ can receive a student in $\binom{10}1$ ways.
The remaining students could be received by the other two groups in $2^9$ ways, but this could result in one group receiving no student.
Any of the two groups could be empty, so we subtract $2$ to get legal ways as
$\binom{10}1\cdot (2^9-2)$, and proceeding similarly with group $C$ having $2$ or $3$ students, and adding up, we get
$\binom{10}1\cdot (2^9-2) + \binom{10}2\cdot(2^8-2) +\binom{10}3\cdot(2^7-2) = 31650$