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There are 11 chairs around a circular table. In how many ways can we arrange 10 people in these seats?


If
(i) There are 11 identical chairs placed equally apart around the table

My solution is:

First fix the one person in any of the chair.
Then permute other 9 people in remaining 10 seats.
i.e $10P9$ which is same as $10!$


(ii) If there are 11 distinctly coloured chairs placed equally apart around the table.

My solution is:

All the chairs are distinct.
So arrangements like $A_1A_2A_3A_4A_5A_6A_7A_8A_9A_{10}$ and $A_{10}A_1A_2A_3A_4A_5A_6A_7A_8A_9$ would be different. So it is like linear arrangement.
Arrangement of 10 people in 11 chairs can be done in $11P10$, ways which is same as $11!$


(iii) If there are 10 identically coloured chair and 1 chair is distinctly coloured.

My solution is:

First fix one person in the coloured seat.
Then permute remaining 9 people in remaining 10 seats.
i.e $10P9$ which is same as $10!$


The answer key says
(i)$9!$
(ii)$11!$
(iii)$11!$

Consider that I am a semi-beginner and learning this subject. What am I doing wrong?

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In part I, Fix the empty chair and $10$ people can be seated in 10 identical chairs in $10P10= 10!$ ways.

In part II, by fixing one chair, you have made the chairs in a linear fashion. Then there are 11 ways a person can be seated, 10 ways the second person can be seated and so on until the last person can be seated in 2 ways to give $(11*10*9...2) = 11!$ ways.

In part III, similar reasoning, by fixing the coloured chair, you have made the chairs in a linear fashion. Then there are 10 ways a person can be seated in the colored chair and the rest of 9 could be seated in the 10 identical chairs in $10P9$ ways + You can keep the colored chair empty and seat the 10 in 10 identical chairs in $10P10$ ways to a total of $(10.10!+10!) = 11!$ ways.

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    $\begingroup$ But the OP says there are $11$ chairs around the table, not $10$. So it seems to me that there are $10!$ possible arrangements (and the answer key is wrong), thinking of the $11$ chairs as seating $10$ people plus one "dummy". $\endgroup$
    – awkward
    Nov 13, 2019 at 15:33
  • $\begingroup$ Simply one chair is empty, I thought it does not matter that there are 11 chairs for the first part. These chairs are around a circular table and there are 10 people to be arranged in 10 chairs around the table. $\endgroup$ Nov 13, 2019 at 15:35
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    $\begingroup$ I think I edited my comment while you were responding. It seems to me the way to treat the problem is that there are $10$ people plus one "dummy", for a total of $11$. Then $11$ objects can placed around a circular table in $10!$ ways. $\endgroup$
    – awkward
    Nov 13, 2019 at 15:38
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    $\begingroup$ In part I, fix the empty chair. Then the ten people can be arranged in the remaining ten chairs in $10!$ ways. $\endgroup$ Nov 13, 2019 at 15:46
  • $\begingroup$ For (iii) I can make someone sit in the distinct chair and permute other 9 people in 10 identical chairs in 10P9 ways OR make no one sit in distinct chair and permute 10 people in 10 identical chairs in 10P10 ways. The presence of an empty distinct chair make the entire arrangement linear, so no need to fix someone on any identical chair. Overall ways of arrangement: 10P9+10P10. Is this correct? $\endgroup$
    – rsonx
    Nov 13, 2019 at 16:28

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