<s>[INST] Generating Functions help Using generating functions, ﬁnd the number of ways to make change for a $\$100$
bill using only dollar coins and $\$1,\$2$, and $\$5$ bills.
Thanks for the help.
 [/INST] Your generating function will have the form $$f(x)=\sum_{n\ge 0}a_nx^n\;,$$ where $a_n$ is the number of ways to make a total of $n$ dollars using the prescribed coin and bills. Each $\$1$ coin must therefore add $1$ to the exponent, as must each $\$1$ bill; each $\$2$ bill must add $2$ to the exponent, and each $\$5$ bill must add $5$. Thus, 
$$f(x)=\underbrace{(1+x+x^2+\ldots)}_{\$1\text{ coins}}\underbrace{(1+x+x^2+\ldots)}_{\$1\text{ bills}}\underbrace{(1+x^2+x^4+\ldots)}_{\$2\text{ bills}}\underbrace{(1+x^5+x^{10}+\ldots)}_{\$5\text{ bills}}\;,$$
or more compactly, 
$$f(x)=\left(\sum_{n\ge 0}x^n\right)^2\left(\sum_{n\ge 0}x^{2n}\right)\left(\sum_{n\ge 0}x^{5n}\right)\;.$$
Now use the basic geometric generating function to rewrite this as
$$f(x)=\left(\frac1{1-x}\right)^2\left(\frac1{1-x^2}\right)\left(\frac1{1-x^5}\right)=\frac1{(1-x)^2(1-x^2)(1-x^5)}\;,$$
which can be further simplified to $$f(x)=\frac1{(1-x)^3(1+x)(1-x^5)}\;,$$ if you wish.
The answer to the question is now the coefficient of $x^{100}$ in $f(x)$.
Added: Suppose that you want the number of $\$5$ bills to be at least $2$ and at most $6$. Then the factor that accounts for the $\$5$ bills would be 
$$\begin{align*}
x^{2\cdot5}+x^{3\cdot5}+x^{4\cdot5}+x^{5\cdot5}+x^{6\cdot5}&=x^{10}+x^{15}+x^{20}+x^{25}+x^{30}\\
&=x^{10}\left(1+x^5+x^{10}+x^{15}+x^{20}\right)\\
&=\frac{x^{10}\left(1-x^{25}\right)}{1-x^5}\;.
\end{align*}$$
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