One of my teachers gives his students essay finals. First, he tells us three numbers, A, B, and C. He gives us A essay questions to study before the test. He picks B essay questions to put on the test, and we must pick out C of them to answer. He tells us to study A-B+C of the given questions if we want to pass the final.

As a procrastinator, I only studied the night before. Luckily, some other students had taken the test a day early, and could tell me which of the questions the teacher had picked. Of course, the teacher would pick a different combination of questions to give to the rest of the students. After hearing which questions were given, I realized I needed to study N less questions than was necessary before!

Find N, generalizing to all possible A, B, and C.

(In reply to Solution by Eric)

Consider the case A=4, B=3, C=2. A-B+C=3. The 3-subsets are 123, 124, 134, and 234. Supposing 234 given already and therefore not to be given again, we still have left 123, 124, and 134, but you must still study 3 of the 4. For if you consider just studying 12, 13, 14, 23, 24, or 34, there is always at least one of 123, 124, and 134 that does not contain both of the studied questions. Hence N=0 for this case.

Posted by Richard on 2005-04-09 00:55:11