There are $8$ people lining up for a photo. Brianna refuses to stand next to Kendall, but she must stand next to Amanda. How many ways can they line up?
The answer says that it is $8640$, but I have no idea how I would get such an answer.
There are $8$ people lining up for a photo.
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$\begingroup$
combinatorics
discrete-mathematics
permutations
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1Try it with fewer people than 8 and try and find the pattern. Try 3 first, then four. – 2017-01-26
3 Answers
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Count the number of ways that Brianna and Amanda are next to others $(2 \times 7!)$
Subtract the ways that Both Kendall and Amanda are next to Brianna $(2 \times 6!)$
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Take Kendall and Amenda as 1. We have 6 persons + 1 group.
So they can be arrange in 7! ways. But Amenda and kendall can be further arranged in group in 2 ways. So we have 2 × 7!.
Now take amenda in between kendall and briana as 1 group. We have 5 persons + 1 group.
So they can be arrange in 6! ways. But amenda can be in between briana and kendall. We have 2 ways.
So we have 2 × 6! ways.
Total cases = 2 × 7! - 2 × 6! = 8640
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Count Amanda and Brianna as a single unit. This leaves us with 6 people that includes Kendall as well. Remove Kendall from this set of 6 people.That leaves us with 5 people.Now arrange these $5 $people+$1$unit. This can be done in $6!$×$2$ ways. These are all the arrangements where Brianna and Amanda are together.Now for each and every arrangement Kendall has only 6 choices to sit at.
So the answer is $6$$(6!)$$×$$2$