How many students can be given the set of unique exam problems?

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    2015
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SUMMARY

The discussion centers on determining the maximum number of students that can be assigned unique sets of exam problems from a pool of 8 problems, where each student receives 3 problems and no two students share more than one problem. The solution involves combinatorial mathematics, specifically the use of combinations and the principle of inclusion-exclusion. The conclusion reached is that a maximum of 8 students can be accommodated under these constraints.

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A set of 8 problems was prepared for an examination. Each student was given 3 of them. No two students received more than one common problem.

What is the largest possible number of students?
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No one answered last week's problem. :( You can find the proposed solution below:

Label the problems by a, b, c, d, e, f, g and h, then 8 possible problems set are abc, ade, afg, bdg, bfh, cdh, cef, egh.

Hence, there could be 8 students.

Suppose that some problem (e.g. b) was given to 4 students. Then each of these 4 students should receive 2 different supplementary problems, and there should be at least 9 problems, which leads to a contradiction. Therefore, each problem was given to at most 3 students, and there were at most $8(3)=24$ awarding of problems.

As each students was awarded 3 problems, there were at most 8 students.
 

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