<s>[INST] How many base $10$ numbers are there with $n$ digits and an even number of zeros? 
How many base $10$ numbers are there with 
  $n$ digits and an even number of zeros?

Solution:
Lets call this number $a_n$.
This is the number of $n-1$ digits that have an even number of zeros
times $9$ possibilities for the $n$th digit 
+ number of $n-1$ digits
that have an odd number of zeros and a zero for the $n$th digit.
$a_n = 9a_{n-1} + (10^{n-1} - a_{n-1})$
$a_n = 8a_{n-1} + 10^{n-1}$
We define
$a_0 = 1$
$a_1 = 9$
The generating function is
$G(x) = 1 + 9x + 82x^2 + 756x^3 + \cdots $
$G(x) = \sum_{n=0}^{\infty}a_n x^n$
$$\begin{align}
G(x) - 1 & = \sum_{n=1}^{\infty} ([8a_{n-1} + 10^{n-1}] x^n)\\
 & = \sum_{n=1}^{\infty} 8a_{n-1}x^n +
 \sum_{n=1}^{\infty}10^{n-1}x^n\\ 
 & =  8x\sum_{n=1}^{\infty} a_{n-1}x^{n-1} +
 x\sum_{n=1}^{\infty}10^{n-1}x^{n-1}\\
 & =  8x\sum_{n=0}^{\infty} a_{n}x^{n} +
 x\sum_{n=0}^{\infty}10^{n}x^{n}\\ 
 & = 8x G(x) +
 x\left(\frac{1}{1-10x}\right)
 &
\end{align}
$$
$(1-8x)G(x) = x\left(\frac{1}{1-10x}\right) + 1$
$G(x) = \frac{1-9x}{(1-8x)(1-10x)}$
$G(x) = \frac{1/2}{1-8x} + \frac{1/2}{1-10x}$
$\therefore$ $a_n=\frac{1}{2}(8^n+10^n)$
Is this solution/method valid?
Note that the way i have set up the solution, and defined $a_0$ and $a_1$, there are supposed to be $82$ numbers in $a_2$. I am including the $0$ numbers of zeros,  i.e. there are $9 \times 9 = 81$ numbers with $0$ zeros and $1$ number $00$. 
 [/INST] Here's another solution: clearly the number of $n$-digit numbers with $k$ zeroes is $f(n,k) = {n \choose k} 9^k$.  Then we have
$$f(n,0) + f(n,1) + \cdots + f(n,n) = \sum_{k=0}^n {n \choose k} 9^k$$
and by the binomial theorem this is $(9+1)^n = 10^n$.  On the other hand, 
$$f(n,0) - f(n,1) + \cdots + (-1)^n f(n,n) = \sum_{k=0}^n {n \choose k} 9^k (-1)^k$$
and this is, again by the binomial theorem, $(9-1)^n = 8^n$.  Adding the two equations together, we get
$$2 f(n,0) + 2 f(n,2) + \cdots + 2 f(n, n) = 8^n + 10^n$$
if $n$ is even, and 
$$2 f(n,0) + 2 f(n,2) + \cdots + 2 f(n, n-1) = 8^n + 10^n$$
if $n$ is odd.  Dividing through by 2 gives the result.
To be fair, this solution is not the first one that springs to mind unless you know the answer in advance.
</s>