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?
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.