Difference between revisions of "2020 AIME I Problems/Problem 5"
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== Problem == | == Problem == | ||
| + | Six cards numbered <math>1</math> through <math>6</math> 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 == | == Solution == | ||
Revision as of 16:18, 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
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
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. 
