2021 AMC 10A Problems/Problem 20

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 1 (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:

$\text{Case \#1: }+,-,+,-$

$\text{Case \#2: }-,+,-,+$

( $+$ stands for increase and $-$ stands for decrease.)

For $\text{Case \#1},$ note that for the second and fourth terms, one of which must be a $5,$ and the other one must be a $3$ or $4.$ We have four sub-cases:

$(1) \ \underline{\hspace{3mm}}3\underline{\hspace{3mm}}5\underline{\hspace{3mm}}$

$(2) \ \underline{\hspace{3mm}}5\underline{\hspace{3mm}}3\underline{\hspace{3mm}}$

$(3) \ \underline{\hspace{3mm}}4\underline{\hspace{3mm}}5\underline{\hspace{3mm}}$

$(4) \ \underline{\hspace{3mm}}5\underline{\hspace{3mm}}4\underline{\hspace{3mm}}$

For $(1),$ 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, $(2)$ also has $2$ possibilities.

For $(3),$ there are no restrictions for the numbers $1, 2,$ and $3.$ So, we have $3!=6$ possibilities. Similarly, $(4)$ also has $6$ possibilities.

Together, $\text{Case \#1}$ has $2+2+6+6=16$ possibilities. By symmetry, $\text{Case \#2}$ 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

Solution 3 (similar to solution 2)

Like Solution 2, we have two cases. Due to symmetry, we just need to count one of the cases. For the purpose of this solution, we will be doing $-,+,-,+$ . Instead of starting with 5, we start with 1.

There are two ways to place it:

_1_ _ _

_ _ _1_

Now we place 2, it can either be next to 1 and on the outside, or is place in where 1 would go in the other case. So now we have another two "sub case":

_1_2_(case 1)

21_ _ _(case 2)

There are 3! ways to arrange the rest for case 1, since there is no restriction.

For case 2, we need to consider how many ways to arrange 3,4,5 in a a>b<c fashion. It should seem pretty obvious that b has to be 3, so there will be 2! way to put 4 and 5.

Now we find our result, times 2 for symmetry, times 2 for placement of 1 and times (3!+2!) for the two different cases for placement of 2. This give us $2*2*(3!+2!)=4*(6+2)=32$ .

~~Xhte

== Solution 4: Symmetry

Due to symmetry, we only need to find the number of ways that no three adjacent digits are increasing.

Therefore, we only need to find # of rearrangements when 5 is the 3rd digit, 4th digit, or 5th digit. Find the total, and multiply by 2.

Case 1: 5 is in the middle __ __ 5 __ __

__ __ 5 __ __ 2 2

Then there are only three cases: 1. 42513, then 4 and 3 are interchangable, which results in 2!*2!. Or it can be 43512, then 4 and 2 are interchangable, but it can not be 23514, so there can only be 2 possible ways: 43512, 21534. The last case would be

Video Solution by OmegaLearn (Using PIE - Principle of Inclusion Exclusion)

https://youtu.be/Fqak5BArpdc

~ pi_is_3.14

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
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