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Difference between revisions of "2021 AMC 10A Problems/Problem 20"

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==See Also==
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Revision as of 02:59, 12 February 2021

Problem

In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing? $\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$

Solution (bashing)

We write out the $120$ cases. These cases are the ones that work: $13254,14253,14352,15243,15342,21435,21534,23154,24153,24351,25143,25341,\linebreak 31425,31524,32415,32451,34152,34251,35142,35241,41325,41523,42315,42513,\linebreak 43512,45132,45231,51324,51423,52314,52413,53412. \linebreak$ We count these out and get $\boxed{\text{D: }32}$ 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 $3!=6$ possibilities. Similarly, the fourth scenario also has 6 possibilities.

Together, Case (1) has $2+2+6+6=16$ possibilities. By symmetry, Case (2) also has $16$ possibilities. Together, the answer is $16+16=\boxed{\textbf{(D)} ~32}.$

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)

https://youtu.be/Fqak5BArpdc

~ 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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