# Difference between revisions of "2014 AIME II Problems/Problem 9"

Kevin38017 (talk | contribs) |
(→Solution 3 (Complementary Counting)) |
||

Line 16: | Line 16: | ||

− | Which method do you like best? | + | Which method do '''you''' like best? |

## Revision as of 22:45, 11 April 2014

## 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 3 chairs cannot contain 3 adjacent chairs. There are only 10 sets of 3 chairs so that they are all 3 adjacent. There are 10 subsets of 4 chairs where all 4 are adjacent, and 10 * 5 or 50 where there are only 3. If there are 5 chairs, 10 have all 5 adjacent, 10 * 4 or 40 have 4 adjacent, and 10 * 5c2 or 100 have 3 adjacent. With 6 chairs in the subset, 10 have all 6 adjacent, 10 * 3 or 30 have 5 adjacent, 10 * 4c2 or 60 have 4 adjacent, 10 * 3 / 2 or 15 have 2 groups of 3 adjacent chairs, and 10 * (5c2 - 3) or 70 have 1 group of 3 adjacent chairs. All possible subsets with more than 6 chairs have at least 1 group of 3 adjacent chairs, so we add 10c7 or 120, 10c8 or 45, 10c9 or 10, and 10c10 or 1. Adding, we get 10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = 581.

## 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, I claim that n chairs give , as confirmed above. This claim can be verified by PIE: there are ways to arrange 3 adjacent chairs, but then we subtract ways to arrange 4. Finally, we add 1 to account for the full subset of chairs. Thus, for n = 10 we get a first count of 641.

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 = .

## 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 .

Which method do **you** like best?