Difference between revisions of "2014 AIME II Problems/Problem 9"
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However, we overcount cases in which there are two distinct groups of three or more chairs. Time to casework: we have <math>5</math> cases for two groups of <math>3</math> directly opposite each other, <math>5</math> for two groups of four, <math>20</math> for two groups of <math>3</math> not symmetrically opposite, <math>20</math> for a group of <math>3</math> and a group of <math>4</math>, and <math>10</math> for a group of <math>3</math> and a group of <math>5.</math> Thus, we have <math>641 - 60 = \boxed{581}</math>. | However, we overcount cases in which there are two distinct groups of three or more chairs. Time to casework: we have <math>5</math> cases for two groups of <math>3</math> directly opposite each other, <math>5</math> for two groups of four, <math>20</math> for two groups of <math>3</math> not symmetrically opposite, <math>20</math> for a group of <math>3</math> and a group of <math>4</math>, and <math>10</math> for a group of <math>3</math> and a group of <math>5.</math> Thus, we have <math>641 - 60 = \boxed{581}</math>. | ||
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== See also == | == See also == |
Revision as of 14:55, 7 March 2015
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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