Amount of numbers with even zeros Given all the numbers between $10^{99}$ and $10^{100}$. How many of them has an even amount of zeros?
I tried by a sum of permutations, but I think I was totally mistaken.
 A: You can define a coupled pair of recurrences.  Let $E(n)$ be the number of $n$ digit numbers that have an even number of zeros and $O(n)$ the number of $n$ digit numbers that have an odd number of zeros.  You start with $O(1)=9,E(1)=0$ because you require that the first digit not be zero.  Then $O(n)=9E(n-1)+O(n-1)$ because you can add a non-zero to any number that has an odd number of zeros and still have an odd number or add a zero to any number that has an even number of zeros.  Similarly $E(n)=9O(n-1)+E(n-1)$.  Solve them and find $E(100)$.  Then worry about whether your between includes the endpoints or not.  This includes $10^{99}$ but not $10^{100}$  You will need an arbitrary precision package to compute the final answer.  It is about half of the numbers but there are more odds.
A: $10^{99}$ is a hundred digit number, all first digits are from [1-9], then have to have even number of zeroes in 99 spaces, non zeroes can be any [1-9].  10^100 has hundred zeroes, so it qualifies if the range is inclusive.  The 9 before the sum represents the 9 choices for the first digit
$$1 + 9 \sum_{i=0}^{49} {99 \choose 2i} 9^{99 - 2i}$$
I got 4.50000000114583E+99 even numbers - I lack immediate knowledge of languages such as Python which can handle large integers. 
