Solution Verification for How Many Class Rooms Are Needed

Question

The Question

There are 38 different time periods during which classes at a university can be scheduled. If there are 677 different classes, how many different rooms will be needed?

My Work

There are $677$ pigeons and $38$ holes. By the generalized pigeon hole principle, there is at least $1$ time slot with $[677/38]$ classes scheduled. This means we need $19$ rooms to avoid having $2$ classes being held in the same room.

It seems right, but it's an important question so I need to be sure.

Answer 1

That looks almost right. You can check by computing $18 * 38 = 684$. So with 18 classrooms and $38$ slots, you get a total of $684$ room-slots, which is more than enough for your $677$ classes. When you compute $677/38$, you should have gotten $17.815...$, so that you know that there was a time-slot with 18 classes scheduled, not 19.

Answer 2

You can also use the below formula for these type of questions.

[(N-1)/M]+1

Where N = classes(pigeon) M = time period(pigeonhole)

Therefore answer will be [(677-1)/38]+1 Answer is 18.