Difference between revisions of "2021 AMC 10A Problems/Problem 20"
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Together, Case (1) has <math>2+2+6+6=16</math> possibilities. By symmetry, Case (2) also has <math>16</math> possibilities. Together, the answer is <math>16+16=\boxed{\textbf{(D)} ~32}.</math> | Together, Case (1) has <math>2+2+6+6=16</math> possibilities. By symmetry, Case (2) also has <math>16</math> possibilities. Together, the answer is <math>16+16=\boxed{\textbf{(D)} ~32}.</math> | ||
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+ | 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 | ~MRENTHUSIASM |
Revision as of 00:51, 12 February 2021
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