I wish to find the number of ways in which at least two adjacent points can be chosen from a line of five points with the inclusion-exclusion principle.
We start with two adjacent points, of which there are 4 choices. Then, there are $2^3$ ways the other points can be chosen, so, our first count is $4\cdot2^3$.
Then, we count three adjacent points, of which there are 3 choices. There are $2^2$ ways the other points can be chosen. These three adjacent points were each counted 2 times by the previous count (of two adjacent points), so now we must subtract our count of $3\cdot2^2$ once. So, our running total is now $4\cdot2^3-3\cdot2^2$.
Then, we count four adjacent points, of which there are 2 choices. There are $2$ ways the other point can be chosen. These four adjacent points were each counted three times by the first count (of two adjacent points) and negative two times by the second count (of three adjacent points), so we do not count this, because we have already counted it $3-2=1$ net times. So, our answer is $4\cdot2^3-3\cdot2^2=32-12=20$.
By intuition, it seems as though I have counted something incorrectly or my inclusion-exclusion argument is incorrect. Is my solution vaild?
First, you made a small arithmetic mistake: $3\cdot2^2=12$, not $24$, so your argument actually yields $32-12=20$.
That’s still off by one, though. You’ve correctly counted the configurations with one run of at least two adjacent points and made sure that longer runs of adjacent points are counted exactly once. But you still counted the configuration where all points except the middle point are chosen twice, since it has two runs of adjacent points and you didn’t account for that. So the correct count is $19$.
You might be interested in Möbius inversion on partially ordered sets, which is a generalization of the standard inclusion–exclusion method that would also yield the coefficients you used for the runs of adjacent points.