One of the questions in my textbook is asking the following: in how many ways can we split an unspecified number of non empty groups, have each group sit into a circular table and then serve them either soda, water, neither or both.
So I'm pretty sure that we must use composition of generating functions to solve this problem. What I have done so far: Lets call the composite function that solves this problem $H(x) = B(A(x))$ where $A(x)$ is the number of ways that we can arrange the groups in a circular table. So $A(x) = log(\frac{1}{1-x})$ now my trouble is finding the generating function for $B(x)$ which is the number of ways to serve each table soda, water, neither or both. If there are k tables, then would $B(x) = \sum_{k\geq0}4^k \frac{x^k}{k!} = e^{4x}$? where at each table we choose one of the four options or am I missing something here?