Difference between revisions of "2020 AIME I Problems/Problem 5"
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<math>12(345)6 \rightarrow 125346, 124536</math> | <math>12(345)6 \rightarrow 125346, 124536</math> | ||
| − | It's not hard to see that no overcount is possible, and that the cycle is either <math>1</math> "right" or <math>1</math> "left." Therefore, we consider how many elements we flip by. If we flip <math>1</math> or <math>2</math> such elements, then there is one way to cycle them. Otherwise, we have <math>2</math> ways. Therefore, the total number of ascending is <math>1 + | + | It's not hard to see that no overcount is possible, and that the cycle is either <math>1</math> "right" or <math>1</math> "left." Therefore, we consider how many elements we flip by. If we flip <math>1</math> or <math>2</math> such elements, then there is one way to cycle them. Otherwise, we have <math>2</math> ways. Therefore, the total number of ascending is <math>1 + 5 + 2(4 + 3 + 2 + 1) = 26</math>, and multiplying by two gives <math>\boxed{052}.</math> ~awang11 |
==See Also== | ==See Also== | ||
Revision as of 17:34, 12 March 2020
Note: Please do not post problems here until after the AIME.
Problem
Six cards numbered 
through 
are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.
Solution 1
Realize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two.
If we choose any of the numbers through , there are five other spots to put them, so we get 
. However, we overcount some cases. Take the example of 
. We overcount this case because we can remove the 
or the 
. Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract 
cases (namely, 
,) to get 
, but we have to add back one more for the original case, 
. Therefore, there are 
cases. Multiplying by gives the desired answer, 
.
-molocyxu
Solution 2 (Inspired by 2018 CMIMC combo round)
Similar to above, a 
correspondence between ascending and descending is established by subtracting each number from 
.
We note that the given condition is equivalent to "cycling" for a contiguous subset of it. For example,
It's not hard to see that no overcount is possible, and that the cycle is either "right" or "left." Therefore, we consider how many elements we flip by. If we flip or such elements, then there is one way to cycle them. Otherwise, we have ways. Therefore, the total number of ascending is 
, and multiplying by two gives 
~awang11
See Also
| 2020 AIME I (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. 
