Arrangements of Chairs in a Circle 
Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.

Hints only please!
This is a confusing worded-problem. 
We could break it into, $7$ cases but that would take very long? 
Case 1: 3 chairs adjacent. 
Ways to do this: $$\binom{10}{3} \cdot \binom{7}{7} = 120$$
But I see that, for the next, $\binom{10}{4}$. 
So it will be:
$$1 + \binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} + \cdots + \binom{10}{9}$$
$$= 1 + 120 + 210 + 252+\cdots$$
But this isnt a legit method it looks like.
I am not sure how to use PIE/anything else here?

The number of sets of exactly three adjacent chairs is $10$.  Either they're chairs $1$, $2$, and $3$, or $2,3,4$, or $3,4,5$, etc.  After $8,9,10$ comes $9,10,1$ and then $10,1,2$.
The number of subsets of the remaining set of seven chairs is $2^7=128$.
However, if one just picks $10\times128=1280$, then one is counting some combinations more than once.  For example, if one has $1,2,3,4,6,7,8$, then one counts each of these:
\begin{align}
& \underbrace{1,2,3,}\ 4, 6, 7, 8 \\[10pt]
& 1,\ \underbrace{2,3,4,}\ 6,7,8 \\[10pt]
& 1,2,3,4,\ \underbrace{6,7,8}
\end{align}
But they're really all three the same combination and should be counted only once.  One should therefore expect a smaller number than $1280$.
So I would search for combinations in which no chairs are adjacent or in which exactly two chairs are adjacent, and subtract that from the total number of subsets.
