2014 AIME II Problems/Problem 9
Contents
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 
.
Solution 3 (Complementary Counting)
It is possible to use recursion to count the complement. Number the chairs 
If chair is not occupied, then we have a line of 
chairs such that there is no consecutive group of three. If chair is occupied, then we split into more cases. If chairs and are empty, then we have a line of 
If chair is empty but chair is occupied, then we have a line of chairs (because chair cannot be occupied); this is similar to when chair is occupied and chair is empty. Finally, chairs and cannot be simultaneously occupied. Thus, we have reduced the problem down to computing 
, where 
counts the ways to select a subset of chairs from a group of n chairs such that there is no group of chairs in a row.
Now, we notice that 
(representing the cases when the first, second, and/or third chair is occupied). Also, 
, and hence 
. Now we know the complement is 
, and subtracting from 
gives 
.
See also
| 2014 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
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