Difference between revisions of "2019 AIME II Problems/Problem 5"
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If the two seats are adjacent to each other, there are <math>6</math> options, and the ambassadors are sitting in four adjacent seats, and there are five seats that their advisors can sit. Choose any of them and the advisors’ seats are fixed, so there are <math>5</math> kinds of solutions for the advisors to sit. And that’s a <math>6\cdot5</math> if we don’t consider the order of the ambassadors. | If the two seats are adjacent to each other, there are <math>6</math> options, and the ambassadors are sitting in four adjacent seats, and there are five seats that their advisors can sit. Choose any of them and the advisors’ seats are fixed, so there are <math>5</math> kinds of solutions for the advisors to sit. And that’s a <math>6\cdot5</math> if we don’t consider the order of the ambassadors. | ||
We can also get that if the blank seats are opposite, it will be <math>3\cdot9</math>, if they are not adjacent and not opposite, it will be <math>6\cdot8</math>. | We can also get that if the blank seats are opposite, it will be <math>3\cdot9</math>, if they are not adjacent and not opposite, it will be <math>6\cdot8</math>. | ||
− | So the total is <math>24\cdot(6\ | + | So the total is <math>24\cdot(6\cdot5+6\cdot8+3\cdot9)=2520</math> |
And the remainder is <math>\boxed{520}</math> | And the remainder is <math>\boxed{520}</math> | ||
Revision as of 20:12, 22 March 2019
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
There are ambassadors and there are seats for them. So we consider the position of the blank seats. There are kinds of versions: If the two seats are adjacent to each other, there are options, and the ambassadors are sitting in four adjacent seats, and there are five seats that their advisors can sit. Choose any of them and the advisors’ seats are fixed, so there are kinds of solutions for the advisors to sit. And that’s a if we don’t consider the order of the ambassadors. We can also get that if the blank seats are opposite, it will be , if they are not adjacent and not opposite, it will be . So the total is And the remainder is
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.