Place $4$ identical $a's$ and $4$ identical $b's$ in $3$ circular tables such that two of these circular tables have $3$ seats and these two tables are identical, the rest one has $4$ seats. How many possible arrangements are there ?
My attempt: I made use of Stirling number of first kind in my calculation such that we have two empty seats, but lets think all of these $a's$,$b's$ and the empty seats as distinct objects. Then ,using the formula of stirling number of first kind : $$\frac{10!}{3^2 \times 2! \times 4}$$
However, the $a's$,$b's$ and the empty seats were identical, so $$\frac{10!}{3^2 \times 2! \times 4 \times (4!)^2 \times 2!}$$
I hesitated about my answer, because of there may be overcounting and Burnsides' lemma need to be implemented.
What is your thoughts about my answer ?