Grouping students into groups of 3 I have 6 students.
I wanted to divide the students into discussion groups of size 3.
I wished to do this such that all possible groups of 3 students get to discuss together for exactly 1 round.
I could do this trivially in 10 rounds so that the assignment of students into groups (A and B) at different rounds were:
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Round& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\ \hline
Student & & &\\ \hline
1 & A & A & A & A & A & A & A & A & A & A\\ \hline
2 & A & A & A & A & B & B & B & B & B & B\\ \hline
3 & A & B & B & B & A & A & A & B & B & B\\ \hline
4 & B & A & B & B & A & B & B & A & A & B\\ \hline
5 & B & B & A & B & B & A & B & A & B & A\\ \hline
6 & B & B & B & A & B & B & A & B & A & A\\ \hline
\end{array}
What if I would have had 9 students?
How to then find the optimal partition of students over the time, so that all possible groups of size 3 would be covered in the minimum number of rounds? 

$$\frac1{3!}\binom9{3,3,3}=\frac1{3!}\frac{9!}{3!3!3!}=280$$
The second factor of LHS equals the number of arrangments when $9$ students are divided in $3$ distinghuishable groups of $3$. Since the groups are not distinghuishable we must repair the multiple counting. For that the first factor of LHS.
Applying it on $6$ (mentioned in your question) gives $\frac1{2!}\binom6{3}=10$ as it should.
