An exam consisted of $28$ problems. Each student solved $7$ problems correctly. For every pair of problems solved, there are exactly $2$ students who solved them correctly. How many students took the exam?

I believe I have the correct answer which is $36$ students, but I don't know how to formally set it up to turn in.

  • $\begingroup$ This question is ambiguous. When you say "For every pair of problems solved", do you mean $14$ pair or $28 \choose 2$ = $378$ pair? $\endgroup$
    – David
    Oct 12, 2014 at 15:45

2 Answers 2


Note that each student solves $21$ pairs of problems.

There are $\cfrac {28\times 27}2=14\times 27=21\times 18$ pairs of problems in the set.

If every pair of problems is solved, then every pair is solved twice, giving $21\times 36$ pairs of solutions, which would take $36$ students. Any more than $36$ students and at least one of the problems would have been solved more than twice.

However the problem as stated does not tell us that every pair of problems has been solved at least once. Two students both solving the same seven problems would also satisfy the conditions as written.


Here is how i solved it then maybe it can give more understanding.

Let x=number of students lets count the number of pairs of problems that are solved correctly in 2 differenty ways since each students solved 7 correctly that gives us x(7c2)pairs pairs of correctly solved problems.

since there were 28 problems and for every pair there are exactly two students who solved this gives us 2(28c2) paris of correcte solved problems.

so x(7c2)=2(28c2) 21x=2*378 x=36


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