Stats seating arrangement problem

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Homework Help Overview

The problem involves determining the number of seating arrangements for five individuals, A, B, C, D, and E, in a row at a movie theater with the condition that D and E cannot sit next to each other.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss counting the arrangements where D and E sit together and then subtracting that from the total arrangements. Some suggest treating D and E as a single entity to simplify the counting process.

Discussion Status

Multiple approaches are being explored, including direct counting and visualization techniques. Some participants express concern about providing complete solutions, while others offer guidance that aids understanding of the problem without resolving it fully.

Contextual Notes

There is a mention of homework rules regarding the level of assistance that can be provided, indicating a need to balance guidance with the original poster's learning process.

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.
 
I thought we weren't supposed to solve the problems completely for the OP ...
 
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.
 

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