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Using a pool of problems, 20 tests will be formed. -Every test should have the same number of problems. -Any problem should be included in at most 10 tests. -For every 5 tests, there should be at least 2 problems common to all of them. What can be the minimum number of problems in this pool?

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Not an answer, but an approach. Make groups of 10 tests and put two unique problems on each test of the group. There are ${20 \choose 5}=15504$ groups of tests that must share two problems. Each group of $10$ has ${10 \choose 5}=252$ five element subsets, so it will take at least $\frac {15504}{252}\approx 61.5$ groups of $10$ to cover all the five element subsets of $20$ in this way. This way we should be able to get by with not many more than $124$ questions. The fact that it doesn't divide evenly shows we can't have a perfect set of groups, so that each $5$ element subset of the $20$ is in only one group, but we should be able to get close. Each test will be in about half the groups, so the equal number of problems per test will be satisfied. Otherwise make a few more problems to equalize.

This doesn't prove that you can form the groups to minimize the overlap, but generally it doesn't take much "wiggle room" to make these possible.