[Math] In how many ways can 20 persons be seated round a table if there are 9 chairs

Question

The problem given is in the title:

In how many ways can $20$ persons be seated round a table if there are $9$ chairs?

I tried solving it as follows:

I can fix one person to one of $9$ chairs. I can select any one of $20$ for this. Now I need to permute $8$ out of remaining $19$. So total count $=20\times {}^{19}P_8=609493224800$

But given solution was:

We can select $9$ out of $20$ person in ${}^{20}C_9$ ways. Those $9$ persons can seat around circular table in $8!$ ways. So total count ${}^{20}C_9\times 8!=6772147200$

So where did I made mistake in my approach?

Answer 1

For each way:

One time you fix me (for example) and then sit other,

another time fix yourself and then sit other,

and so on for every of 9 person,

so you count every way 9 times!

Also after fixing first person,

sitting others from left of first person to his right isn't different from inverse direction,

so doesn't make new way,

but you count it!