Difference between revisions of "2014 AIME II Problems/Problem 9"
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==Solution 2 (PIE)== | ==Solution 2 (PIE)== | ||
− | Starting with small cases, we see that four chairs give 4 + 1 = 5, five chairs give 5 + 5 + 1 = 11, and six chairs give 6 + 6 + 6 + 6 + 1 = 25. Thus, | + | Starting with small cases, we see that four chairs give <math>4 + 1 = 5</math>, five chairs give <math>5 + 5 + 1 = 11</math>, and six chairs give <math>6 + 6 + 6 + 6 + 1 = 25.</math> Thus, n chairs should give <math>n 2^{n-4} + 1</math>, as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are <math>n 2^{n-3}</math> ways to arrange <math>3</math> adjacent chairs, but then we subtract <math>n 2^{n-4}</math> ways to arrange <math>4.</math> Finally, we add <math>1</math> to account for the full subset of chairs. Thus, for <math>n = 10</math> we get a first count of <math>641.</math> |
− | However, we overcount cases in which there are two distinct groups of three or more chairs. Time to casework: we have 5 cases for two groups of 3 directly opposite each other, 5 for two groups of four, 20 for two groups of 3 not symmetrically opposite, 20 for a group of 3 and a group of 4, and 10 for a group of 3 and a group of 5. Thus, we have 641 - 60 = | + | 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>. |
==Solution 3 (Complementary Counting)== | ==Solution 3 (Complementary Counting)== |
Revision as of 21:19, 20 May 2014
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 1, 2, 3, ..., 10. If chair 1 is not occupied, then we have a line of 9 chairs such that there is no consecutive group of three. If chair 1 is occupied, then we split into more cases. If chairs 2 and 10 are empty, then we have a line of 7. If chair 2 is empty but chair 10 is occupied, then we have a line of 6 chairs (because chair 9 cannot be occupied); similarly for chair 2 occupied and chair 10 empty. Finally, chairs 2 and 10 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 3 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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