How many ways can you arrange the letters in the word "letters"
You have the letters l,e,t,r and s in 7 positions. The number of permutations as if the "t" and "e" are unique is $7!$. However, that you can swap the two "t"s in each of those permutations and it will not change the word. Similarly for "e".
$$ \frac{7!}{2!2!} = 1,260 $$Every morning 7 people enter an office. How many different orders can they enter the office?
You are scheduling exams in a single classroom. There are 8 exams that need to be scheduled and each lasts 2 hours. The classroom can be scheduled for up to 16 hours. How many different permutations of the exam schedules can you make?
You are scheduling exams in two classrooms. There are 8 exams that need to be scheduled and each lasts 2 hours. Each classroom can be scheduled for up to 8 hours. How many different permutations of the exam schedules can you make?
Imagine that you lie the exams in a line and assign the first four to classroom 1 and the second four to classroom 2. Take letters of the alphabet to be the exams:
$$ \underbrace{ABCD}_{\textrm{Classroom 1}}\underbrace{DEFG}_{\textrm{Classroom 2}} $$You can see that as you write out the permutations, there is no way to rearrange which go to classroom 1 or 2 without copying another permutation. Thus the number is the same.
$$ 8! = 40,320 $$You are scheduling exams in three classrooms. There are 8 exams that need to be scheduled and each lasts 2 hours. Each classroom can be scheduled for up to 8 hours. How many different permutations of the exam schedules can you make? Assume all gaps without exams are exactly two hours
Now we have a new exam option: "no exam". How many of these do we have? There are total 16 hours of exams and we have 24 hours of classroom time. That means we have 4 "no exams". However, these no exam gaps are interchangeable just like the problem above with two "t"s in the word letters.
$$ \frac{12!}{4!} = 19,958,400 $$