There is a circular table with $10$ seating , we want to place $6$ people to this circular table.How many ways are there ?
My try: I thought the empty seats as identical object that will be placed by $6$ distinct object (i.e $6$ people) , so my answer is $$\frac{(10-1)!}{4!}=15,120$$
IS MY SOLUTION CORRECT ?
Secondly , i want to learn how to solve this problem using $\color{red}{\text{polya counting method}}$ such that "How many possible way are there arrange $6$ distinct and $4$ identical object in
a-) circular table (i.e $2-D$)
b-) necklace (i.e, $3-D$ )
IMPORTANT NOTE : @user2661923 corrected my answer , but my main purpose is to see application of Polya theorem in this question . I mentioned my this aim in "Secondly" part. Hence , i am waiting for the solutions of part "a" and "b". Thanks in advance !
I am ignorant of Polya Theory and I came up with the same answer by a different approach.
With one of the people arbitrarily designated as the head of the table (AKA the twelve-oclock position), the remaining $(5)$ people can be permuted in $(5!)$ ways.
Then, the $(6)$ regions of (potentially) empty spaces (between the people) can be filled by the number of non-negative integer solutions to
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 4 ~: ~\binom{9}{5} ~\text{solutions.} \tag1 $$
$\displaystyle (5!) \times \binom{9}{5}~$ agrees with your computation.
See also this Stars and Bars article and this one.
Edit
It could be argued that Polya Theory would become necessary for the following related problem:
Suppose that you have $(6)$ groups of identical triplets ($18$ people). Further suppose that you are seating them at a circular table with $(30)$ chairs. Further suppose that each set of triplets is to be regarded as indistinguishable from each other, but distinguishable from the other $(15)$ people, with respect to enumerating distinct seating arrangements.
Then, how many rotationally distinct ways are there of seating the $(18)$ people around the table?