Solving circular arrangement problem with k identical and m distinct positions.

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 $$10!$$

(i)$$9!$$
(ii)$$11!$$
(iii)$$11!$$

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

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.
• 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". Nov 13, 2019 at 15:33
• 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. Nov 13, 2019 at 15:38
• In part I, fix the empty chair. Then the ten people can be arranged in the remaining ten chairs in $10!$ ways. Nov 13, 2019 at 15:46