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