There are n distinctive balls to be distributed to two distinctive urns. One urn can hold a maximum of three balls while the other can hold a maximum of six balls.
The task is to find an exponential generating function for this problem.
I don't quite understand how to find the exponential generating function $C(x) = \sum\limits_{n=0}^{\infty} \frac{c_n x^n}{n!} $ without a series $c_{n}$.
I started off by writing down a generating function for each of the urns and multiplying them:
\begin{align} &A(x) = (x^0+x^1+x^2+x^3) = \sum\limits_{i=0}^{3}x^i \\ &B(x) = (x^0+x^1+x^2+x^3+x^4+x^5+x^6)= \sum\limits_{i=0}^{6}x^i \\ &C(x) = A(x) B(x) = \left(\sum\limits_{i=0}^{3}x^i \right) \left(\sum\limits_{i=0}^{6}x^i \right)\end{align}
I think though that I might be totally off (as I know $A(x)$ would be used to express e.g. "there are 3 1$-notes"). Can someone provide a good example of how this is done and an explanation?
Let $c_n$ be the number of ways to distribute the $n$ balls to the two distinct urns. Then note that $$ \begin{align} \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\right)\left(1+x+\frac{x^2}{2!}+\dotsb+\frac{x^6}{6!}\right)&=\sum_{n\geq 0}\left(\sum_{k_1+k_2=n;\, 0\leq k_1\leq 2;\, 0\leq k_2\leq 6}\frac{n!}{k_1!k_2!}\right)\frac{x^n}{n!}\\ &=\sum_{n\geq 0}c_n\frac{x^n}{n!} \end{align} $$