Difference between revisions of "2019 AIME II Problems/Problem 5"
Mathkiddie (talk | contribs) m (→Solution 2) |
Mathkiddie (talk | contribs) (→Solution 2) |
||
| Line 20: | Line 20: | ||
#Place the other 3 teams in the chosen gaps (<math>6</math> permutations) | #Place the other 3 teams in the chosen gaps (<math>6</math> permutations) | ||
So, the total is <math>12\cdot35\cdot6=2520</math>. | So, the total is <math>12\cdot35\cdot6=2520</math>. | ||
| − | Therefore, the remainder is <math>\boxed{520}</math> | + | |
| + | Therefore, the remainder is <math>\boxed{520}</math>. | ||
~Mathkiddie | ~Mathkiddie | ||
Revision as of 19:49, 18 December 2023
Problem
Four ambassadors and one advisor for each of them are to be seated at a round table with 
chairs numbered in order 
to . Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are 
ways for the 
people to be seated at the table under these conditions. Find the remainder when is divided by 
.
Solution 1
Let us first consider the 
ambassadors and the 
even-numbered chairs. If we consider only their relative positions, they can sit in one of 
distinct ways: Such that the 
empty even-numbered chairs are adjacent, such that they are separated by one occupied even-numbered chair, and such that they are opposite each other. For each way, there are 
ways to assign the ambassadors to the selected seats.
In the first way, there are distinct orientations. The advisors can be placed in any of the 
odd-numbered chairs adjacent to the ambassadors, and for each placement, there is exactly one way to assign them to the ambassadors. This means that there are 
total seating arrangements for this way.
In the second way, there are distinct orientations. advisors can be placed in any of the chairs adjacent to the "chunk" of ambassadors, and advisor can be placed in either of the chairs adjacent to the "lonely" ambassador. Once again, for each placement, there is exactly one way to assign the advisors to the ambassadors. This means that there are 
total seating arrangements for this way.
In the third way, there are distinct orientations. For both "chunks" of ambassadors, advisors can be placed in any of the chairs adjacent to them, and once again, there is exactly one way to assign them for each placement. This means that there are 
total seating arrangements for this way.
Totalling up the arrangements, there are 
total ways to seat the people, and the remainder when 
is divided by is 
. ~emerald_block
Solution 2
In the diagram, the seats are numbered 1...12. Rather than picking seats for each person, however, each ambassador/assistant team picks a gap between the seats (A...L) and the ambassador sits in the even seat while the assistant sits in the odd seat. For example, if team 1 picks gap C then Ambassador 1 will sit in seat 2 while assistant 1 will sit in seat 3. No two teams can pick adjacent gaps. For example, if team 1 chooses gap C then team 2 cannot pick gaps B or D. In the diagram, the teams have picked gaps C, F, H and J. Note that the gap-gaps - distances between the chosen gaps - (in the diagram, 2, 1, 1, 4) must sum to 8. So, to get the number of seatings, we:
- Choose a gap for team 1 (
options) - Choose 3 other gaps around the table with positive gap-gaps. The number of ways to do this is the number of ways to partition 8 with 4 positive integers. This is the same as partitioning 4 with 4 non-negative integers, and using stars-and-bars, this is
- Place the other 3 teams in the chosen gaps (
permutations)
So, the total is 
.
Therefore, the remainder is .
~Mathkiddie
Solution 3
There are 
total ways to let everyone sit. However, this may lead to advisors sitting in the same chair, leading to awkward situations. So we find how many ways this happens. There are 
ways to choose which advisors end up sitting together, times 
ways to find neighboring even seats (order matters), and for the other two ambassadors, there are 
ways to sit and 4 ways for their advisors to sit. This results in 
ways for this to happen. However, we overcounted the case when all four of the advisors run out of room to sit, where there are 
to choose the places the ambassadors sit and 4! ways to arrange the ambassadors. 
ways to happen. So the answer is 
, which has a remainder of when divided by 1000.
~minor 
fixes by virjoy2001
Solution 4
We see that for every 2 adjacent spots on the table, there is exactly one way for an ambassador and his or her partner to sit. There are 2 cases:
- There is an ambassador at the 12th chair and his or her partner at the 1st chair
- There is no pair that has their chairs numbered as 12 and 1
For the first case, there are 3 pairs and 4 empty spaces, which results in 
ways to arrange the pairs.
For the second case, there are 4 pairs and 4 empty spaces, which results in 
ways to arrange the pairs.
Finally, we have to assign ambassadors and their advisors to the pairs, which has ways, so 
, or
~PenguinMoosey
See Also
| 2019 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
