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

(Solution 2 (Casework on Two Consecutive Digits))
m (Reformatted: Make the boxes consistent and Solution 1 look nicer.)
Line 7: Line 7:
 
We write out the <math>120</math> cases.  
 
We write out the <math>120</math> cases.  
 
These cases are the ones that work:
 
These cases are the ones that work:
 +
 
<math>13254,14253,14352,15243,15342,21435,21534,23154,24153,24351,25143,25341,\linebreak  
 
<math>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
 
31425,31524,32415,32451,34152,34251,35142,35241,41325,41523,42315,42513,\linebreak
43512,45132,45231,51324,51423,52314,52413,53412. \linebreak</math>
+
43512,45132,45231,51324,51423,52314,52413,53412.</math>
We count these out and get <math>\boxed{\text{D: }32}</math> permutations that work. ~contactbibliophile
+
 
 +
We count these out and get <math>\boxed{\textbf{(D)} ~32}</math> permutations that work.  
 +
 
 +
~contactbibliophile
  
 
==Solution 2 (Casework on the Consecutive Digits)==
 
==Solution 2 (Casework on the Consecutive Digits)==
Line 60: Line 64:
 
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.
 
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 <math>2*2*(3!+2!)=4*(6+2)=32</math>.
+
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 <math>2*2*(3!+2!)=4*(6+2)=\boxed{\textbf{(D)} ~32}</math>.
  
 
~~Xhte
 
~~Xhte
Line 67: Line 71:
  
 
We only need to find the # of rearrangements when 5 is the 4th digit and 5th digit. Find the total, and multiply by 2. Then we can get the answer by adding the case when 5 is the third digit.
 
We only need to find the # of rearrangements when 5 is the 4th digit and 5th digit. Find the total, and multiply by 2. Then we can get the answer by adding the case when 5 is the third digit.
 
  
 
'''Case <math>1</math>:''' 5 is the 5th digit. __ __ __ __ 5
 
'''Case <math>1</math>:''' 5 is the 5th digit. __ __ __ __ 5
Line 76: Line 79:
  
 
__ __ 4 __ 5, then the 1st digit must be <math>2</math> or <math>3</math>, <math>2</math> gives us <math>1</math> way, and <math>3</math> gives us <math>2</math> ways. (Can't be <math>1</math> because the first digit would increasing). Therefore, <math>4</math> in the middle and <math>5</math> in the last would result in <math>3</math> ways.  
 
__ __ 4 __ 5, then the 1st digit must be <math>2</math> or <math>3</math>, <math>2</math> gives us <math>1</math> way, and <math>3</math> gives us <math>2</math> ways. (Can't be <math>1</math> because the first digit would increasing). Therefore, <math>4</math> in the middle and <math>5</math> in the last would result in <math>3</math> ways.  
 
  
 
'''Case <math>2</math>:''' <math>5</math> is the fourth digit. __ __ __ 5 __
 
'''Case <math>2</math>:''' <math>5</math> is the fourth digit. __ __ __ 5 __
  
 
Then the last digit can be all of the 4 numbers <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math>. Let's say if the last digit is <math>4</math>, then the 2nd digit would be the largest for the remaining digits to prevent increasing order or decreasing order. Then the remaining two are interchangeable, give us <math>2!</math> ways. All of the <math>4</math> can work, so case <math>2</math> would result in <math>2!+2!+2!+2!=8</math> ways.  
 
Then the last digit can be all of the 4 numbers <math>1</math>, <math>2</math>, <math>3</math>, and <math>4</math>. Let's say if the last digit is <math>4</math>, then the 2nd digit would be the largest for the remaining digits to prevent increasing order or decreasing order. Then the remaining two are interchangeable, give us <math>2!</math> ways. All of the <math>4</math> can work, so case <math>2</math> would result in <math>2!+2!+2!+2!=8</math> ways.  
 
  
 
'''Case <math>3</math>:''' <math>5</math> is in the middle. __ __ 5 __ __
 
'''Case <math>3</math>:''' <math>5</math> is in the middle. __ __ 5 __ __
Line 92: Line 93:
 
<math>8+3+2=13</math>, so the total ways for case 1 and case 2 with both increasing and decreasing would be <math>13*2=26.</math>
 
<math>8+3+2=13</math>, so the total ways for case 1 and case 2 with both increasing and decreasing would be <math>13*2=26.</math>
  
<math>26+6=\boxed{\textbf{(D)} ~32}.</math>
+
Finally, we have <math>26+6=\boxed{\textbf{(D)} ~32}.</math>
  
 
~Michael595
 
~Michael595

Revision as of 10:57, 18 June 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 1 (Bashing)

We write out the $120$ cases. These cases are the ones that work:

$13254,14253,14352,15243,15342,21435,21534,23154,24153,24351,25143,25341,\linebreak  31425,31524,32415,32451,34152,34251,35142,35241,41325,41523,42315,42513,\linebreak 43512,45132,45231,51324,51423,52314,52413,53412.$

We count these out and get $\boxed{\textbf{(D)} ~32}$ permutations that work.

~contactbibliophile

Solution 2 (Casework on the Consecutive Digits)

Reading the terms from left to right, we have two cases for the consecutive digits, where $+$ means increase and $-$ means decrease:

$\textbf{Case \#1: }\boldsymbol{+,-,+,-}$

$\textbf{Case \#2: }\boldsymbol{-,+,-,+}$

For $\text{Case \#1},$ note that for the second and fourth terms, one term must be $5,$ and the other term must be $3$ 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 $4,$ 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.

Finally, 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 $2*2*(3!+2!)=4*(6+2)=\boxed{\textbf{(D)} ~32}$.

~~Xhte

Solution 4: Symmetry

We only need to find the # of rearrangements when 5 is the 4th digit and 5th digit. Find the total, and multiply by 2. Then we can get the answer by adding the case when 5 is the third digit.

Case $1$: 5 is the 5th digit. __ __ __ __ 5

Then $4$ can only be either 1st digit or the 3rd digit.

4 __ __ __ 5, then the only way is that $3$ is the 3rd digit, so it can be either $231$ or $132$, give us $2$ results.

__ __ 4 __ 5, then the 1st digit must be $2$ or $3$, $2$ gives us $1$ way, and $3$ gives us $2$ ways. (Can't be $1$ because the first digit would increasing). Therefore, $4$ in the middle and $5$ in the last would result in $3$ ways.

Case $2$: $5$ is the fourth digit. __ __ __ 5 __

Then the last digit can be all of the 4 numbers $1$, $2$, $3$, and $4$. Let's say if the last digit is $4$, then the 2nd digit would be the largest for the remaining digits to prevent increasing order or decreasing order. Then the remaining two are interchangeable, give us $2!$ ways. All of the $4$ can work, so case $2$ would result in $2!+2!+2!+2!=8$ ways.

Case $3$: $5$ is in the middle. __ __ 5 __ __

Then there are only two cases: 1. $42513$, then 4 and 3 are interchangeable, which results in $2!*2!$. Or it can be $43512$, then 4 and 2 are interchangeable, but it can not be $23514$, so there can only be 2 possible ways: $43512$, $21534$.

Therefore, case 3 would result in $4+2=6$ ways.

$8+3+2=13$, so the total ways for case 1 and case 2 with both increasing and decreasing would be $13*2=26.$

Finally, we have $26+6=\boxed{\textbf{(D)} ~32}.$

~Michael595

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

https://youtu.be/Fqak5BArpdc

~ pi_is_3.14

Video Solution by Power of Logic (Using Idea of Symmetrically Counting)

https://youtu.be/ZLQ8KYtai_M

Video Solution by TheBeautyofMath

https://youtu.be/UZZoSYHBJlI

~IceMatrix

See Also

2021 AMC 10A (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS