2014 AIME II Problems/Problem 9
Problem
Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.
Solution 1 (Casework)
We know that a subset with less than 
chairs cannot contain adjacent chairs. There are only 
sets of chairs so that they are all adjacent. There are subsets of 
chairs where all are adjacent, and 
or 
where there are only 
If there are 
chairs, have all adjacent, 
or 
have adjacent, and 
or 
have adjacent. With 
chairs in the subset, have all adjacent, 
or 
have adjacent, 
or 
have adjacent, 
or 
have 
groups of adjacent chairs, and 
or 
have 
group of adjacent chairs. All possible subsets with more than chairs have at least group of adjacent chairs, so we add 
or 
, 
or 
, 
or , and 
or 
Adding, we get 
Solution 2 (PIE)
Starting with small cases, we see that four chairs give 
, five chairs give 
, and six chairs give 
Thus, n chairs should give 
, as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are 
ways to arrange adjacent chairs, but then we subtract 
ways to arrange 
Finally, we add to account for the full subset of chairs. Thus, for 
we get a first count of 
However, we overcount cases in which there are two distinct groups of three or more chairs. Time to casework: we have cases for two groups of directly opposite each other, for two groups of four, 
for two groups of not symmetrically opposite, for a group of and a group of , and for a group of and a group of 
Thus, we have 
.
See also
| 2014 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
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