TITLE: How many $N$ digits binary numbers can be formed where $0$ is not repeated
QUESTION [5 upvotes]: How many $N$ digits binary numbers can be formed where $0$ is not repeated.
Note - first digit can be $0$.
I am more interested on the thought process to solve such problems, and not just the answer.
If anyone can cite some resources for learning how to solve such problems would be great.

REPLY [6 votes]: First consider $B_{a}(n)$ as the number of "binary number" without 0 repeated of length $n$, which starts with $a$, with $a \in \left\{0, 1\right\}$.
Then:
$$B(n) = B_{0}(n) + B_{1}(n)$$
is the number of "binary number" without 0 repeated of length $n$.
We can know work on the $B_{a}(n+1)$. These number can build by adding $0$s or $1$s in front of a number of length $n$. In particular we have that:
$$B_{0}(n+1) = B_{1}(n) \\
B_{1}(n+1) = B_{0}(n) + B_{1}(n) = B(n)$$
Summing up, we have that:
$$B(n+1) = B(n) + B_{1}(n)$$
But it also clear from the previous relation that $B_1(n) = B(n-1)$ (since $B_{1}(n+1) = B(n)$ is true for $n$, then it is true even for $n-1$), so we finally have the recurrence relation:
$$B(n+1) = B(n) + B(n-1)$$
At this point you have to determinate the number $B(1)$ and $B(2)$ and then apply the recurrence relation we have derived before. 
We have $B(1) = 2$, since we have the sequences $[0]$ and $[1]$.
Also, $B(2) = 3$, since we have the sequences $[0,1]$, $[1,0]$ and $[1,1]$.
So we can evaluate $B(3) = B(2) + B(1) = 5$ and so on.
Note: the recurrence relation is the same of the Fibonacci sequence.