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A confusing combinatorics problem

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The 10 students in my physics class sit at a round table in the following manner: The students line up in alphabetical order. The student whose name is first sits anywhere she wants. Each subsequent student chooses a chair that is next to a student who is already seated. How many different seatings are possible, assuming that two seatings are the same if each student in both seatings has the same student to the left and the same student to the right?

Jul 13, 2022

#1
+118579
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Lets see,

The first stuent has only one place to chose from as they denot the 'begining of the circe'

The second student has 2 places to chose from

The third has 2 to shooe from

The ninth has 2 to chose from and the last has only one.

1*2*2*2*2*2*2*2*2*1

2^8 = 256

I am not certain that it is right though.

Jul 13, 2022
#2
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I believe your answer is correct. But your solution is "wrong". First off, the first student gets 10 choices and not one. And the answer is not 10 * 2^8 because we'll have to divide 10. Remember two seatings are the same if each studnet in both seats hsa the same sttudent to the left and to the right. Since there are 10 seats, we divide by 10 to get 2^8 as our answer.

Jul 13, 2022
#3
+118579
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If my answer is right then my logic, and therefore my solution,  is also right.

Yes there are ten chairs to choose from but they are all the same. It makes no differnece which is chosen. Take a chair, any chair, and it defines the beginning of the sequence.

I said 1*

you said 10/10 *

They are the same.

Melody  Jul 14, 2022
edited by Melody  Jul 14, 2022