For pennies you have the sequence $\langle 1,1,1,\dots\rangle$, which as you say correponds to the formal power series
$$1+x+x^2+\ldots=\sum_{n\ge 0}x^n\;.$$
That’s just a geometric series, so you know what the generating function is (even if you didn’t realize it right away:
$$\sum_{n\ge 0}x^n=\frac1{1-x}\;.$$
For nickels you can do almost the same thing. Your sequence is $\langle 1,0,0,0,0,1,0,\dots\rangle$, corresponding to another geometric series:
$$1+x^5+x^{10}+\ldots=\sum_{n\ge 0}x^{5n}=\sum_{n\ge 0}\left(x^5\right)^n=\frac1{1-x^5}\;.$$
For pennies and nickels you want
$$\left(1+x+x^2+\ldots\right)\left(1+x^5+x^{10}+\ldots\right)\;:\tag{1}$$
$(1)$ has one $x^n$ term for every way to write $n$ in the form $5m+k$ with $m,k\in\Bbb N$, which is precisely the number of ways to make the amount $n$ with nickels and pennies. Clearly, then, the generating function is just
$$\frac1{(1-x)(1-x^5)}\;.$$
I’ll leave the last part to you.
First observe that
$$\sum_{a+b+2c=k}(-1)^{b+c}\binom{65+a}a\binom{15+b}b\binom{15+c}c$$
is the coefficient of $x^k$ in the product
$$\left(\sum_{a\ge 0}\binom{65+a}ax^a\right)\left(\sum_{b\ge 0}(-1)^b\binom{15+b}bx^b\right)\left(\sum_{c\ge 0}(-1)^c\binom{15+c}cx^{2c}\right)\,.$$
In closed form this is the product
$$\begin{align*}
\frac1{(1-x)^{66}}\cdot\frac1{(1+x)^{16}}\cdot\frac1{(1+x^2)^{16}}&=\frac1{(1-x)^{66}(1+x)^{16}(1+x^2)^{16}}\\
&=\frac1{(1-x)^{50}(1-x^2)^{16}(1+x^2)^{16}}\\
&=\frac1{(1-x)^{50}(1-x^4)^{16}}\\
&=\frac1{(1-x)^{50}}\cdot\frac1{(1-x^4)^{16}}\,,
\end{align*}$$
which in turn is the closed form of
$$\left(\sum_a\binom{49+a}ax^a\right)\left(\sum_b\binom{15+b}bx^{4b}\right)\,.$$
The coefficient of $x^k$ in this product is
$$\sum_{a+4b=k}\binom{49+a}a\binom{15+b}b\,;$$
can you see why this is the number of ways to make change for $25k$ cents using these $66$ types of coins?
Best Answer
You can just read it off via: $x^{10}x^{90}+x^{20}x^{80}+\ldots$. So it amounts to figuring out how many terms you have.