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Difference between revisions of "2012 AMC 10B Problems/Problem 22"

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==Problem 22=
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==Problem==
 
Let (<math>a_1</math>, <math>a_2</math>, ... <math>a_{10}</math>) be a list of the first 10 positive integers such that for each <math>2\le</math> <math>i</math> <math>\le10</math> either <math>a_i + 1</math> or <math>a_i-1</math> or both appear somewhere before <math>a_i</math> in the list. How many such lists are there?
 
Let (<math>a_1</math>, <math>a_2</math>, ... <math>a_{10}</math>) be a list of the first 10 positive integers such that for each <math>2\le</math> <math>i</math> <math>\le10</math> either <math>a_i + 1</math> or <math>a_i-1</math> or both appear somewhere before <math>a_i</math> in the list. How many such lists are there?
  
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<math>\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880</math>
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==Solution 1==
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If we have 1 as the first number, then the only possible list is <math>(1,2,3,4,5,6,7,8,9,10)</math>.
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If we have 2 as the first number, then we have 9 ways to choose where the <math>1</math> goes, and the numbers ascend from the first number, <math>2</math>, with the exception of the <math>1</math>.
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For example, <math>(2,3,1,4,5,6,7,8,9,10)</math>, or <math>(2,3,4,1,5,6,7,8,9,10)</math>. There are <math>\dbinom{9}{1}</math> ways to do so.
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If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are <math>\dbinom{9}{2}</math> ways to do this.
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In the same way, the total number of lists is:
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<math>\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}</math>
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By the [[binomial theorem]], this is <math>2^{9}</math> = <math>512</math>, or <math>\boxed{\textbf{(B)}}</math>
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==Solution 2==
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Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is <math>(1,2,3,4,5...10)</math>. There are 9 arrows, so the answer is <math>2^{9}</math> = <math>512</math> <math>\boxed{\textbf{(B)}}</math>
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NOTE:
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Solution cited from: http://www.artofproblemsolving.com/Videos/external.php?video_id=269.
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==Solution 3==
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Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, <math>(1, 2)</math> and <math>(2, 1)</math>. If we replace it with 3, there are four lists, <math>(1, 2, 3), (2, 1, 3), (2, 3, 1),</math> and <math>(3, 2, 1)</math>. Since 2 and 4 are both powers of 2, it is likely that the number of lists is <math>2^{n-1}</math>, where <math>n</math> is the length of the lists. <math>2^{10-1}=512=\boxed{\textbf{(B)}}</math>
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==Solution 4 (Recursion)==
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If <math>a_1=10</math>, the sequence must be <math>10, 9, 8,7,6,5,4,3,2,1</math>. If <math>a_2=10</math>, then <math>a_1=9</math>, and the sequence is <math>9, 10, 8, 7, 6, 5,4,3,2,1</math>. If <math>a_3=10</math>, then the possible sequences are <cmath>9,8,10,7,6,5,4,3,2,1 \text{ and}</cmath><cmath>8,9,10,7,6,5,4,3,2,1.</cmath> In general, for an <math>n</math>-length sequence, if <math>a_i=n</math>, then <math>a_1</math> through <math>a_{i-1}</math> can be filled in <math>f(i-1)</math> ways with <math>n-i+1</math> through <math>n-1</math>, and <math>a_{i+1}</math> through <math>a_{n}</math> must be sorted in decreasing order with the remaining numbers (<math>1</math> through <math>n-i</math>), in one way. Thus <math>f(n) = \sum_{i=0}^{n-1} f(i)</math>, where <math>f(0)=1</math>.
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We can see (or prove by induction) that <math>f(n)=2^{n-1} ~\forall~ n \ge 1</math>. Hence, <math>f(10)=2^9=\boxed{\textbf{(B) }512}</math>.
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- ColtsFan10
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==Solution 5==
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Solution 3 states that <math>f(n)=2^{n-1} ~\forall~ n \ge 1</math> without formal proof.  Solution 4 gives a formal proof.  Here is another formal proof:
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<math>f(1)=1</math>.  When the list goes from <math>n-1</math> numbers to <math>n</math> numbers, there are <math>2</math> ways to make the new lists:
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Case 1: append <math>n</math> to the end of lists with <math>n-1</math> numbers to make a new list, the number of the new lists is <math>f(n-1)</math>;
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Case 2: put number <math>1</math> at the end of the new lists, the way to arrange <math>(2,3,...,n-1,n)</math> as the first <math>n-1</math> items is the same as to arrange <math>(1,2,...,n-2,n-1)</math>, by subtracting 1 from each of the elements, so the number of the new lists is also <math>f(n-1)</math>.
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So <math>f(n)=f(n-1)+f(n-1)=2f(n-1)=2^{n-1} ~\forall~ n \ge 1</math>
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-[https://artofproblemsolving.com/wiki/index.php/User:Junche junche]
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==Video Solution by Richard Rusczyk==
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https://artofproblemsolving.com/videos/amc/2012amc10b/269
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~dolphin7
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==Video Solution by TheBeautyofMath==
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https://youtu.be/bXPSv93GVbg
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~IceMatrix
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== See Also ==
  
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<math>\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880</math>
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{{AMC10 box|year=2012|ab=B|num-b=21|num-a=23}}
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{{MAA Notice}}

Latest revision as of 10:19, 21 April 2024

Problem

Let ($a_1$, $a_2$, ... $a_{10}$) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?

$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880$

Solution 1

If we have 1 as the first number, then the only possible list is $(1,2,3,4,5,6,7,8,9,10)$.

If we have 2 as the first number, then we have 9 ways to choose where the $1$ goes, and the numbers ascend from the first number, $2$, with the exception of the $1$. For example, $(2,3,1,4,5,6,7,8,9,10)$, or $(2,3,4,1,5,6,7,8,9,10)$. There are $\dbinom{9}{1}$ ways to do so.

If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are $\dbinom{9}{2}$ ways to do this.

In the same way, the total number of lists is: $\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}$

By the binomial theorem, this is $2^{9}$ = $512$, or $\boxed{\textbf{(B)}}$

Solution 2

Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is $(1,2,3,4,5...10)$. There are 9 arrows, so the answer is $2^{9}$ = $512$ $\boxed{\textbf{(B)}}$

NOTE: Solution cited from: http://www.artofproblemsolving.com/Videos/external.php?video_id=269.

Solution 3

Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, $(1, 2)$ and $(2, 1)$. If we replace it with 3, there are four lists, $(1, 2, 3), (2, 1, 3), (2, 3, 1),$ and $(3, 2, 1)$. Since 2 and 4 are both powers of 2, it is likely that the number of lists is $2^{n-1}$, where $n$ is the length of the lists. $2^{10-1}=512=\boxed{\textbf{(B)}}$

Solution 4 (Recursion)

If $a_1=10$, the sequence must be $10, 9, 8,7,6,5,4,3,2,1$. If $a_2=10$, then $a_1=9$, and the sequence is $9, 10, 8, 7, 6, 5,4,3,2,1$. If $a_3=10$, then the possible sequences are \[9,8,10,7,6,5,4,3,2,1 \text{ and}\]\[8,9,10,7,6,5,4,3,2,1.\] In general, for an $n$-length sequence, if $a_i=n$, then $a_1$ through $a_{i-1}$ can be filled in $f(i-1)$ ways with $n-i+1$ through $n-1$, and $a_{i+1}$ through $a_{n}$ must be sorted in decreasing order with the remaining numbers ($1$ through $n-i$), in one way. Thus $f(n) = \sum_{i=0}^{n-1} f(i)$, where $f(0)=1$.

We can see (or prove by induction) that $f(n)=2^{n-1} ~\forall~ n \ge 1$. Hence, $f(10)=2^9=\boxed{\textbf{(B) }512}$.

- ColtsFan10

Solution 5

Solution 3 states that $f(n)=2^{n-1} ~\forall~ n \ge 1$ without formal proof. Solution 4 gives a formal proof. Here is another formal proof:

$f(1)=1$. When the list goes from $n-1$ numbers to $n$ numbers, there are $2$ ways to make the new lists:

Case 1: append $n$ to the end of lists with $n-1$ numbers to make a new list, the number of the new lists is $f(n-1)$;

Case 2: put number $1$ at the end of the new lists, the way to arrange $(2,3,...,n-1,n)$ as the first $n-1$ items is the same as to arrange $(1,2,...,n-2,n-1)$, by subtracting 1 from each of the elements, so the number of the new lists is also $f(n-1)$.

So $f(n)=f(n-1)+f(n-1)=2f(n-1)=2^{n-1} ~\forall~ n \ge 1$

-junche

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc10b/269

~dolphin7

Video Solution by TheBeautyofMath

https://youtu.be/bXPSv93GVbg

~IceMatrix

See Also

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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

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