Determine exponential generating function of the case given

Question

I have a problem to determine EGF that show how many ways we can distribute r different people into n different rooms such that every room has at least 2 people and no more than 5 people. Of course, we know that this is a permutation case (and that's why it leads to EGF).

In ordinary generating function, we can manipulate the expression $(x^2 + x^3 + x^4 + x^5)$ as $x^2(1 + x + x^2 + x^3) = x^2(1-x^4)(1 + x + x^2 + \cdots)$, but how about in exponential generating function which lets to $\left(\dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!}\right)$?

Could anyone help me? Your help will be appreciated. Thank you.

Answer 1

Using the coefficient of operator $[x^j]$ to denote the coefficient of $x^j$ in a series the wanted number is \begin{align*} r![x^r]&\left(\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}\right)^n\\ &=r![x^r]\frac{x^{2n}}{5!}\left(60+20x+5x^2+x^3\right)^n\\ &=\frac{r!}{5!}[x^{r-2n}]\left(60+20x+5x^2+x^3\right)^n\tag{1}\\ \end{align*} Here we have $r\geq 2n$ since we need at least $2n$ people to arrange them into $n$ rooms.

Comment: In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

Hint:

• The paper Bracket notation for the "coefficient of" operator by D.E. Knuth might be instructive.

• This MSE post provides an easy application of the coefficient of operator.

• ... and this one a somewhat more demanding.