A class of 96 students must divide itself into 24 sets of 4 for an assignment. However, during the year there are 4 such assignments, and no student can ever work with any of the same partners. Prove that this is possible or impossible to do.
Note that, if for project 1, Anne, Bob, Chris, and David are working together, then for project 2 Anne cannot work with Bob, Chris, or David for any other project.
I can't come up with a combinatorial argument for the general case, but we can construct an example (and hence proving it is possible) as follows:
We label each student as $s_i$ for $i \in\{1,2,\ldots,96 \}.$ The first time we assign the groups as follows:
$$\begin{bmatrix}s_1& s_2& s_3 & s_4 & s_5 & s_6 &\cdots& s_{22} & s_{23} & s_{24}\\ s_{25}& s_{26}& s_{27} & s_{28} & s_{29} & s_{30} &\cdots& s_{46} & s_{47} & s_{48} \\ s_{49}& s_{50}& s_{51} & s_{52} & s_{53} & s_{54} &\cdots& s_{70} & s_{71} & s_{72}\\ s_{73}& s_{74}& s_{75} & s_{76} & s_{77} & s_{78} &\cdots& s_{94} & s_{95} & s_{96} \end{bmatrix} \tag{1}$$ where each column represents a group. Notice that there are $24$ groups, with $4$ students in each one. Now, for the second time, we proceed as follows: we leave the first row as it is, we shift the second row one place to the right, the third row two places to the right and the fourth row three places to the right, so we end up with $$\begin{bmatrix} s_1& s_2& s_3 & s_4 & s_5 & s_6 &\cdots& s_{22} & s_{23} & s_{24} \\ s_{48}& s_{25}& s_{26} & s_{27} & s_{28} & s_{29} &\cdots& s_{45} & s_{46} & s_{47} \\ s_{71}& s_{72}& s_{49} & s_{50} & s_{51} & s_{52} &\cdots& s_{68} & s_{69} & s_{70} \\ s_{94}& s_{95}& s_{96} & s_{73} & s_{74} & s_{75} &\cdots& s_{91} & s_{92} & s_{93} \end{bmatrix} \tag{2}$$
Starting from $(2)$, we do the same for the third assignment:
$$\begin{bmatrix} s_1& s_2& s_3 & s_4 & s_5 & s_6 &\cdots& s_{22} & s_{23} & s_{24} \\ s_{47}& s_{48}& s_{25} & s_{26} & s_{27} & s_{28} &\cdots& s_{44} & s_{45} & s_{46} \\ s_{69}& s_{70}& s_{71} & s_{72} & s_{49} & s_{50} &\cdots& s_{66} & s_{67} & s_{68} \\ s_{91}& s_{92}& s_{93} & s_{94} & s_{95} & s_{96} &\cdots& s_{88} & s_{89} & s_{90} \end{bmatrix} \tag{3}$$
and the fourth: $$\begin{bmatrix} s_1& s_2& s_3 & s_4 & s_5 & s_6 &\cdots& s_{22} & s_{23} & s_{24} \\ s_{46}& s_{47}& s_{48} & s_{25} & s_{26} & s_{27} &\cdots& s_{43} & s_{44} & s_{45} \\ s_{67}& s_{68}& s_{69} & s_{70} & s_{71} & s_{72} &\cdots& s_{64} & s_{65} & s_{66} \\ s_{88}& s_{89}& s_{90} & s_{91} & s_{92} & s_{93} &\cdots& s_{85} & s_{86} & s_{87} \end{bmatrix} \tag{4}$$
This ensures that each of the $4$ times no student meets with another previously met student. We can also see that there is more than one solution to this problem.