How many ways can 5 people, A, B, C, D, and E, sit in a row at a movie theater D and E won't si

What I'm actually looking for is an explanation about how to resolve this statistics problem. How do I do the permutation?

How many ways can 5 people, A, B, C, D, and E, sit in a row at a movie theater D and E won't sit together?getting late

Start by ignoring the problems between D and E. If anyone sits anywhere, there are 5 choices for seat 1, 4 for seat 2, etc. So there are 5*4*3*2*1 = 5! = 120 total possible ways.

It's hard to find cases where D and E are next to each other out of the 120. Instead, group D and E as a single person, leaving 4 total seats and see how many there are now. That's 4! = 24. D and E are treated as a single location but there are 2 ways D and E could be seated in the groupings. Either D on the left or D on the right. This gives 48 possible ways for D and E to be together.

Your answer then is the total number of arrangements (120) minus the number where D and E are together (48) leaving 72 for your answer.