There are n job candidates that are being considered for a job and must complete a certain task to move forward. They are divided randomly into 2 groups, A and B. The interviewees in group A must take a test in one of 4 languages while those in group B must arrange themselves around a circular table and discuss one of 3 given topics (two circular arrangements are considered to be the same if each student has the same person sitting on his/her left in the two arrangements).
I am wondering how to go about translating this problem into a product of generating functions?
Any help here? I know that the number of ways to arrange people around a circular table is $(n-1)!$ but how to combine all the pieces above to come up with a product is not so clear to me...
Edit: Since the problem is poorly worded, I have edited out the last component and left it as is so the answer below is satisfactory.
Since the candidates are divided randomly, you want exponential generating functions. For the group A task that is $\sum_{n\ge 0}4^n\frac{x^n}{n!}=e^{4x}$, if we're interpreting the problem correctly. For the group B task it seems to be
$$1+\sum_{n\ge 1}3(n-1)!\frac{x^n}{n!}=1+3\sum_{n\ge 1}\frac{x^n}n=1-3\ln(1-x)\;.$$
The product is $e^{4x}\big(1-\ln(1-x)\big)$, and the coefficient of $x^n$ is
$$4^n+3\sum_{k=0}^{n-1}\binom{n}k4^k(n-k-1)!=4^n+3\sum_{k=1}^n\binom{n}k4^{n-k}(k-1)!\;.$$
I don't at the moment see either a series expansion of the product or a nice recurrence, and it's 3:30 a.m. here, so I'll post this in case it's of some use and take another look tomorrow.