Exponential Generating function Use an exponential generating function to determine how many ways there are to make an r-arrangement of pennies, nickels, dimes and quarters with at least one penny and an odd number of quarters.  coins of the same denomination are identical.
You need exponential generating functions if you want to consider the order of the coins. You have the following generating functions:


Pennies: At least one, $\frac{z}{1!} + \frac{z^2}{2!} + \ldots = e^z - 1$
Dimes: Any number, $1 + \frac{z}{1!} + \frac{z^2}{2!} + \ldots = e^z$
Quarters: An odd number, $\frac{z}{1!} + \frac{z^3}{3!} + \ldots 
                              = \frac{e^z - e^{-z}}{2}$


Thus the exponential generating function for $n$ coins is:
$$
(e^z - 1) \cdot e^z \cdot \frac{e^z - e^{-z}}{2}
  = \frac{1}{2} \left( 
                  e^{3 z} - e^{2 z} - e^z + 1
                \right)
$$
We want:
\begin{align}
n! [z^n] \frac{1}{2} \left(
                       e^{3 z} - e^{2 z} - e^z + 1
                     \right)
  &amp;= \frac{n!}{2} \left(
                    \frac{3^n}{n!} - \frac{2^n}{n!} - \frac{1}{n!} + [n = 0]
                  \right) \\
  &amp;= \frac{3^n - 2^n - [n \ne 0]}{2}
\end{align}
(Here $[n = 0]$ is Iverson's bracket, i.e., it is 1 if $n = 0$ and 0 otherwise).
I am going to use regular generating functions instead of exponential ones because I think that is simpler - I'm not sure why the question asks for exponential specifically. I will make series where the exponent $n$ in $x^n$ is the number of cents produced and the coefficient of $x^n$ is the number of possible ways to make those $n$ cents out of the coin(s) in question.

With at least one penny there is one way to make 1 cent, 1 way to make 2 cents etc... so the GF is $$(1)x+(1)x^2+\cdots=\frac{x}{1-x}$$ With an odd number of quarters there is one way to make 25 cents, one way to make 75 cents, etc... so the GF is $$(1)x^{25}+(1)x^{75}+(1)x^{125}+\cdots=\frac{x^{25}}{1-x^{50}}$$ With nickels and dimes the GF's are respectively $$1+x^5+x^{10}+\cdots=\frac{1}{1-x^5}$$ and $$1+x^{10}+x^{20}+\cdots=\frac{1}{1-x^{10}}$$ We multiply (multiplication gives the total number of possible ways to make some number of cents with all the possible combinations) these functions to get $$\frac{x^{26}}{(1-x)(1-x^5)(1-x^{10})(1-x^{50})}$$ So to find the number of ways to make $n$ cents you would take the coefficient of the $x^n$ term in that series: $$\text{#ways to make $n$ pennies}=[x^n]\left(\frac{x^{26}}{(1-x)(1-x^5)(1-x^{10})(1-x^{50})}\right)$$

You should probably use a CAS to write that as a power series, or else use partial fractions to decompose it into pieces.