Scheduling gym class My cousin came to me with this problem yesterday: 
She has 8 students in her gym class. In tomorrows class she has planned 4 different activities to rotate them through, each of which requires exactly 2 students. She wants every student to do every activity only once, and to have a different opponent after each rotation. That is, after each rotation every student will do a new activity and play a new person. There will be 4 rotations.
I don't think she can do it, but my only argument is through exhausting the possibilities. Is it truly impossible and is there some nice pigeon-hole kind of argument?
 A: You can think like this: separate the class in two groups of 4 people each. Then, each participant of group 1 will have as opponent one person of group 2 in each activity! 
A: As I noted in my comment above, the question can be interpreted in (at least) two different ways: Either there are four rounds, and each round consists of four pairs of people all doing the same activity at the same time, or each round has four pairs of people each pair doing a different activity per round.
If what is meant is the first case, then Luísa's solution is in fact valid. In the other case, the situation is somewhat more challenging, since the activities need to "rotate" as well.
What you are looking for here is a set of two so called "Mutually Orthogonal Latin Squares" (MOLS) or "Graeco-Latin Squares" of order 4. Luckily for you, such beasts exist - see e.g. http://math.ucdenver.edu/~wcherowi/courses/m6406/csln3.html.
From the solution given there, you can easily construct a solution to your problem for rounds I, II, III and IV, activities A, B, C and D and participants $1,\ldots 8$:
$$
\begin{array}{c|cccc}
& A & B & C & D \\
\hline
I   & 1:5 & 4:7 & 2:8 & 3:6 \\
II  & 2:6 & 3:8 & 1:7 & 4:5 \\
III & 3:7 & 2:5 & 4:6 & 1:8 \\
IV  & 4:8 & 1:6 & 3:5 & 2:7
\end{array}
$$
