I am attempting to solve the following probability problem but am having trouble conceptualizing the solution. I found one similar inquiry here (Complicated probability question) but was unable to successfully apply it to my situation.
I have 9 university courses, each of which can be registered for by 0 to 20 students. Each of 25 university students must select 3 of the 9 courses to register in.
Notes - Once a course fills (reaches registration of 20 students) it is effectively closed and can no longer be registered for. From that point, students have one less course from which to select. (e.g. If one course fills, students subsequently have only 8 courses (9-1) to choose from.)
I'm trying to figure out the probability that each of the 9 courses will be registered for by 5 or more students.
I understand that the number of possible registration choices for each student is 84 (9!/3!(9-3))...but I'm not even sure if that is relevant given the problem I am attempting to solve. I also understand that the sum total of registrants needs to be 75 (25x3) if my registration rules are strict (each student must register for 3 courses, no more/no less) but cannot figure out where this should be used either.
I would be most appreciative of any advice that pushes me in the correct direction.
A partial answer (or hint):
Consider a specific class. Each student has a 3/9 = 1/3 chance of registering for it. Hence (naively) the number of students in the class will follow a binomial distribution of $25$ students, each having chance $1/3$ of registering:
One can then easily calculate the probability the number of the students in the class is $5$ or more ($p = 0.953799$). And then you can easily calculate the probability that all classes have at least $5$ students ($p^{25}=0.6533$).
BUT this is not the correct answer, for two reasons.
Nevertheless... it is an approximation... a start!