Stats seating arrangement problem

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kuahji
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How many ways can five people, A, B, C, D, E, sit in a row at a movie theater if D & E will not sit next to each other?

If everyone "would" sit next to each other, then it'd just be 5! or 120. However, without actually drawing out a picture, I'm not sure exactly how to work the problem with the idea that D & E won't sit next to each other.
 
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It's easier to count the number of ways they CAN sit next to each other, then subtract.
You must have DE or ED in positions 12 or 23 or ...; then, how many ways can the remaining people be arranged?
 
It would be 5!-4!2!. Total number of ways minus the ways they do sit together.
 
More specifically, to count the ways they CAN sit together, consider D and E as a single person. You now have 4 "people" and so 4! ways to seat them. Of course, D and E, while sitting together, can "swap" positions. There are 2!= 2 ways to do that so we get chaoseverlasting's "2!4!" ways they we can set them so that D and E are sitting together. Since you already know there are 5!= 120 ways to seat 5 people and now we know that there are 2!4!= 2(24)= 48 ways to seat them so that D and E are sitting together, there are 5!- 2!4!= 120-48= 72 ways to seat them so that D and E are not sitting together.
 
HallsofIvy said:
More specifically, to count the ways they CAN sit together, consider D and E as a single person.

Thank you for your response, this was the best for helping me to understand the problem.
 
Another approach is, first let ABC sit. They can be ordered in 3! ways. Among A,B,C, there are 4 places from among which two can be occupied. Arrange D and E in these four places, which can be done in 4P2 ways. You can choose any method, but the second problem let's you visualize what's going on, i.e. visualize the condition, not its negation. Once you understand that, you can follow any method that suits you.

Heres the diagrammatic representation,

* A * B * C *. "*" Represents places D and E can occupy, such that they're never together.

Regards,
Sleek.