We have 6 people which we have to seat at the 2 tables . Each table has 3 chairs to sit on.
How many ways can we seat the people when we do not distinquish the order of tables? If each person has the same person by right and left hand , its the same way.
How could we calculate this using Stirling number of the first kind.
s(n, 1) = (n − 1)!, ∀n ≥ 1,
s(n, k) = 0, ∀k > n,
s(n + 1, k) = s(n, k − 1) + n · s(n, k), ∀k, n ∈ N, 1 ≤ k ≤ n.
Should be forulas , but how to modify it to reach the right answer?
Take an individual, who has $5$ candidates for the left-hand position and then $4$ candidates for the right-hand position. That then leaves three individuals for the other table: take one of those, who has $2$ candidates for the left-hand position and then $1$ candidate for the right-hand position
So there are $5\times 4 \times 2 \times 1=40$ possibilities
This is easily extended to $n$ indistinguishable round tables with $k$ people on each table giving a total of $\dfrac{(nk)!}{n!\,k^n}$ possibilities. So in this example you have $\dfrac{6!}{2! \times 3^2}=\dfrac{720}{2 \times 9}=40$
I do not see a connection with Stirling numbers of the first kind