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

Revision as of 04:58, 14 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:

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

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

( $+$ 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

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

https://youtu.be/Fqak5BArpdc

~ pi_is_3.14

@@ Line 14: / Line 14: @@
 Reading the terms from left to right, we have two cases:
-(1) <math>+,-,+,-</math>
+<math>\text{Case \#1: }+,-,+,-</math>
-(2) <math>-,+,-,+</math>
+<math>\text{Case \#2: }-,+,-,+</math>
 (<math>+</math> stands for increase and <math>-</math> 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:
+For <math>\text{Case \#1},</math> note that for the second and fourth terms, one of which must be a <math>5,</math> and the other one must be a <math>3</math> or <math>4.</math> We have four sub-cases:
-_3_5_
+<math>(1) \ \underline{\hspace{3mm}}3\underline{\hspace{3mm}}5\underline{\hspace{3mm}}</math>
-_5_3_
+<math>(2) \ \underline{\hspace{3mm}}5\underline{\hspace{3mm}}3\underline{\hspace{3mm}}</math>
-_4_5_
+<math>(3) \ \underline{\hspace{3mm}}4\underline{\hspace{3mm}}5\underline{\hspace{3mm}}</math>
-_5_4_
+<math>(4) \ \underline{\hspace{3mm}}5\underline{\hspace{3mm}}4\underline{\hspace{3mm}}</math>
-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 <math>(1),</math> the first two blanks must be <math>1</math> and <math>2</math> in some order, and the last blank must be a <math>4,</math> for a total of <math>2</math> possibilities. Similarly, <math>(2)</math> also has <math>2</math> possibilities.
-For the third scenario, there are no restrictions for the numbers 1, 2, and 3. So, we have <math>3!=6</math> possibilities. Similarly, the fourth scenario also has 6 possibilities.
+For <math>(3),</math> there are no restrictions for the numbers <math>1, 2,</math> and <math>3.</math> So, we have <math>3!=6</math> possibilities. Similarly, <math>(4)</math> also has <math>6</math> possibilities.
-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, <math>\text{Case \#1}</math> has <math>2+2+6+6=16</math> possibilities. By symmetry, <math>\text{Case \#2}</math> also has <math>16</math> possibilities. Together, the answer is <math>16+16=\boxed{\textbf{(D)} ~32}.</math>
 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

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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