TITLE: Arrangements of Chairs in a Circle
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: Another approach is to count the good configurations directly.  There is one subset of size $10$, ${10 \choose 9}=10$ of size $9$, ${10 \choose 8}=45$ of size $8$ and ${10 \choose 7}=120$ of size $7$.   All $176$ of these have three chairs in a row.   
There are $10$ subsets of size $3$ that have three in a row.  For subsets of size $4$ there are $10$ ways to choose the first of the three and six places for the other one where we exclude the one just before the three to avoid double counting four in a row.  That gives $60$.  For subsets of size $5$ there are $10$ ways to choose the three in a row and ${6 \choose 2}=15$ ways to choose the two others, again disallowing the one just before the set of three.  That gives $150$.  This paragraph totals $220$.  
Finally we have subsets of size $6$.  We follow the same approach as with $5$, choosing a block of three in $10$ ways, then choosing the three others in ${6 \choose 3}=20$ ways.  There are fifteen configurations with two isolated sets of three, which we have counted twice.  That gives $185$ subsets.  
The grand total is $176+220+185=581$