2021 AMC 10A Problems/Problem 20
Contents
Problem
In how many ways can the sequence 
be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
Solution (bashing)
We write out the 
cases.
These cases are the ones that work:
We count these out and get 
permutations that work. ~contactbibliophile
Solution 2 (Casework)
Reading the terms from left to right, we have two cases:
(1) 
(2) 
(
stands for increase and 
stands for decrease.)
For Case (1), note that for the 2nd and 4th terms, one of which must be a 5, and the other one must be a 3 or 4. We have four scenarios:
_3_5_
_5_3_
_4_5_
_5_4_
For the first scenario, the first two blanks must be 1 and 2 in some order, and the last blank must be a 4, for a total of 2 possibilities. Similarly, the second scenario also has 2 possibilities.
For the third scenario, there are no restrictions for the numbers 1, 2, and 3. So, we have 
possibilities. Similarly, the fourth scenario also has 6 possibilities.
Together, Case (1) has 
possibilities. By symmetry, Case (2) also has 
possibilities. Together, the answer is 
This problem is a little similar to the 2004 AIME I Problem 6: https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_6
~MRENTHUSIASM
Video Solution by OmegaLearn (Using PIE - Principal of Inclusion Exclusion)
~ pi_is_3.14
See Also
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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