Groups of $3$ out of $23$ s.t. all students cooperate exactly once

Question

A class consists of $23$ students. We want to create groups of $3,$
in a way  that all students cooperate (two students cooperate iff they coexist in some group) exactly once. How many groups of $3$ as above are there in total?

Of course $binom{23}{3}$ is the total number of ways to create groups of $3.$ But how do we exclude all the repetitions?
Thanks in advance.

Josh B. · Answer

Overall there are $frac{23*22}{2}=253$ pairs of students in the class. And a group of $3$ students will contain $3$ pairs of students. But $253$ is not divisible by $3$, so the desired assignments are impossible without some repetition.
See also: Steiner Triple Systems.

paw88789 · Answer

I may be missing something (or not understanding the problem), but it doesn't seem possible. Each of the $23$ individual students must participate in $11$ different groups of $3$ (so that they can work with each of the other $22$ students exactly once). This suggests $23cdot 11$ total groups, BUT that counted each group three times (once for each member). Thus the actual number of groups would be $frac{23cdot 11}{3}$. That's not a whole number!

Answered by paw88789 on February 11, 2021

Groups of $3$ out of $23$ s.t. all students cooperate exactly once

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