There are 10 students. Find the number of ways in which they can be divided into 3 groups such that each group has at least 1 student and third group has at most 3
We can make 3 separate cases where $G_3$ has $1$, $2$, $3$ students respectively.
As for the remaining two groups, there are no restrictions other than that no group can have $0$ students.
But I am not sure about how to divide the remaining students into $2$ groups. Clearly the stars and bars method won’t work because students are not identical objects.
Best Answer
Let's take the case where $G_3$ has one student. Pick the student there and you have nine left. Now, $G_1$ (say) can have $1$ to $8$ students. Pick the students for $G_1$ and then all of the others are on $G_2$.
So for this case you'll get
$$10\cdot \sum_{j=1}^8 {9 \choose j}$$
groups. And similarly for the other cases for $G_3$.