0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Right you are. Must have punched the wrong number into the calculator. $\endgroup$ – Dunka Dec 6 '14 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.