Difference between revisions of "2020 AIME II Problems/Problem 9"
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==Solution (Bash)== | ==Solution (Bash)== | ||
| − | There are <math>2^{5}-1</math> intersections that we must consider if we are to perform a PIE bash on this problem. Since we don't really want to think that hard, and bashing does not take that long for this problem, we can write down the all permutations that satisfy the conditions presented in the problem in "lexicographically next" order to keep track easily. We do this for all cases such that the first "person" is <math>1-3</math> | + | There are <math>2^{5}-1</math> intersections that we must consider if we are to perform a PIE bash on this problem. Since we don't really want to think that hard, and bashing does not take that long for this problem, we can write down the all permutations that satisfy the conditions presented in the problem in "lexicographically next" order to keep track easily. We do this for all cases such that the first "person" is <math>1-3</math>, and multiply by two, since the number of working permutations with <math>4-6</math> as the first number is the same as if it were <math>1-3</math>, hence, after doing such a bash, we get <math>45\times2=90</math> permutations that result in no consecutive letters being adjacent to each other. |
~afatperson | ~afatperson | ||
Revision as of 03:07, 8 June 2020
Problem
While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next to each other after the break. Find the number of possible seating orders they could have chosen after the break.
Solution (Bash)
There are 
intersections that we must consider if we are to perform a PIE bash on this problem. Since we don't really want to think that hard, and bashing does not take that long for this problem, we can write down the all permutations that satisfy the conditions presented in the problem in "lexicographically next" order to keep track easily. We do this for all cases such that the first "person" is 
, and multiply by two, since the number of working permutations with 
as the first number is the same as if it were , hence, after doing such a bash, we get 
permutations that result in no consecutive letters being adjacent to each other.
~afatperson
See Also
| 2020 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
