I'm working on this problem and am having some difficulties.
I know something important to keep in mind is that (as an example)
$1110001111$ is just one bit string despite it having two individual substrings with at least 3 consecutive ones.
I know there are $2^{10} = 1024$ total possible bit strings of length $n = 10$. We want to find the number of bit strings that contain at least $3$ consecutive ones.
I start by first counting the number of bit strings that contain $3$ consecutive ones.
There are ${10}\choose{3}$ possible ways to insert three ones in this bit string of length $10$, but that isn't what we want, we want to count the number of ways to have three consecutive ones in a bit string of length $10$. And this is where I'm getting stuck.
To solve this problem, I started with trying by exhaustively listing out the total possible strings, but quickly started to run into some problems and thought there was a more efficient way to do it. So then I was thinking about using the inclusion-exclusion principle, where my sets would be something like "$A_{1}$ is the set containing three consecutive ones; $A_{2}$ is the set containing four consecutive ones,..., $A_{10}$ is the set containing $10$ consecutive ones" but then I realized this would be a truly ginormous computation and that there must still be a better way to go about solving it.
So my question, is how to I count the number of bit strings of length $n = 10$ where I have at least $3 \leq k \leq 10$ consecutive ones?
