Q. There are 35000 students at the university. Each of them takes four(distinct) courses. The university offers 999 different courses. When a student who has taken a course in discrete mathematics learned that the largest classroom holds only 135 students, she realized that there is problem. What is the problem?

Sol. There are 999 different courses and 35000 students, then
ceiling(35000/999) = 36, must take same course.
Now, since students have to take 4 distinct courses, there are at-least 144 students in a set(of 4 different courses).
And so, the problem is, when Discrete Mathematics class scheduled, there will not be enough room in the class for student to sit.

But, I'm not convinced, here is another argument(which i am going to present by a smaller set of data), which makes me think my above argument is false,

Example: Let, there are 50 students and a student takes 4 distinct courses out of 6 courses.
Now, total number of different set of 4 distinct courses which can be formed are, C(6,4) = 15.
Now, by Pigeonhole Principle, there are at-least 3 students which will take a set of courses.
Now, total number of sets which contains Discrete mathematics are, C(5,3) = 10.
Which assert that, there are at-least 30 students which converges, when Discrete mathematics class scheduled.

Now, the things is, what if, there are only 3 or 4 or, in general less than 6 students?
That means, we cannot predict that what is the least value of student attending the class.
Which is actually the case in the original question. If i try to solve it by the approach mentioned in example.

  • $\begingroup$ There seems to be a slight misunderstanding here. The question is not asking about the number of students in the discrete mathematics class; that is both unknown and unknowable. The issue is with the largest classroom relative to the most popular course. $\endgroup$ – Joffan May 12 '18 at 19:25
  • $\begingroup$ @Joffan Understand it in this way, let there are 5 processes and 6 resources and processes needs 2 or more for completion, now here it is ensured that, there exists a sequence in which process will complete, which ensures no deadlock. $Part-I$ $\endgroup$ – commonSense May 13 '18 at 19:57
  • $\begingroup$ @Joffan And, I think the word largest classroom is somewhat kind of same thing here. $1.)$ It is saying at-least $141$ $seats$, which means there could be $thousands$ $of$ $seats$, but that is not the point, coz you only know the exact number of student in a course, after every student is enrolled, which by-the-way changes every year. The point is $specifically$ for this question, $more-than-average$ number of students in a course which then $trade-offs$ with infrastructure. Correct me if i'm wrong. :) $Part-II$ $\endgroup$ – commonSense May 13 '18 at 19:57

There is $35000\times 4=140000$ elections. Now, we have $140000/999=140,14014$ students in at least one course, in particular, the greater.

  • $\begingroup$ Is there something wrong in the argument which i posed in my example? $\endgroup$ – commonSense May 6 '18 at 19:18
  • $\begingroup$ The Pigeonhole Principle assures you that you have something at least once, but you do not know what it is. So thinking about intersections or "multiplications" does not make sense $\endgroup$ – Martín Vacas Vignolo May 6 '18 at 19:24
  • 1
    $\begingroup$ in your solution, "there are at-least 144 students in a set" is totally wrong, because you know that one course is taken by at least 36 students, but about the other 3 courses of this students you don't have information with your argument $\endgroup$ – Martín Vacas Vignolo May 6 '18 at 19:28
  • $\begingroup$ i was going to ask about my Solution. Thanks. $\endgroup$ – commonSense May 6 '18 at 19:34

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