Difference between revisions of "2020 AIME I Problems/Problem 5"
(→Solution) |
|||
Line 4: | Line 4: | ||
== Solution == | == 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 <math>1</math> through <math>6</math>, there are five other spots to put them, so we get <math>6 \cdot 5 = 30</math>. However, we overcount some cases. Take the example of <math>132456</math>. We overcount this case because we can remove the <math>3</math> or the <math>2</math>. Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract <math>5</math> cases (namely, <math>213456, 132456, 124356, 123546, 123465</math>,) to get <math>30-5=25</math>, but we have to add back one more for the original case, <math>123456</math>. Therefore, there are <math>26</math> cases. Multiplying by <math>2</math> gives the desired answer, <math>\boxed{052}</math>. | ||
+ | |||
+ | -molocyxu | ||
==See Also== | ==See Also== |
Revision as of 17:12, 12 March 2020
Note: Please do not post problems here until after the AIME.
Problem
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.