How many ways there are to arrange a boolean $2\times5$ matrix such that there won't be two zeros one above the other 
How many ways there are to arrange a boolean $2\times5$ matrix such that there won't be two zeros one above the other.

For example, this is not allowed
01110
01111

I can brute force it but it doesn't seem right, there are too many calculations to do, so I thought that this is the same as asking that there won't be any 1s one above the other, which is symmetry, so if I'll add that condition, then there are only 2 ways to arrange the matrix:
10101  01010
01010  10101

But dividing 2 by 2 (since it's the same condition twice) can't be the answer so could it be $2^{10}-1$?

The answer is $3^5$ since for each column there are three options: 
1 0 1
0 1 1

Thanks JMoravitz.
