Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2012 AMC 12B Problems/Problem 18"
(→‎Solution)
Line 5: Line 5:
 
<math>\textbf{(A)}\ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ 362,880</math>
 
<math>\textbf{(A)}\ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ 362,880</math>
  
−
== Solution ==
+
== Solution 1==
 
Let <math>1\leq k\leq 10</math>. Assume that <math>a_1=k</math>. If <math>k<10</math>, the first number appear after <math>k</math> that is greater than <math>k</math> must be <math>k+1</math>, otherwise if it is any number <math>x</math> larger than <math>k+1</math>, there will be neither <math>x-1</math> nor <math>x+1</math> appearing before <math>x</math>. Similarly, one can conclude that if <math>k+1<10</math>, the first number appear after <math>k+1</math> that is larger than <math>k+1</math> must be <math>k+2</math>, and so forth.
 
Let <math>1\leq k\leq 10</math>. Assume that <math>a_1=k</math>. If <math>k<10</math>, the first number appear after <math>k</math> that is greater than <math>k</math> must be <math>k+1</math>, otherwise if it is any number <math>x</math> larger than <math>k+1</math>, there will be neither <math>x-1</math> nor <math>x+1</math> appearing before <math>x</math>. Similarly, one can conclude that if <math>k+1<10</math>, the first number appear after <math>k+1</math> that is larger than <math>k+1</math> must be <math>k+2</math>, and so forth.
  
Line 16: Line 16:
 
<cmath>\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} = 2^9=512 ...... \framebox{B}</cmath>
 
<cmath>\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} = 2^9=512 ...... \framebox{B}</cmath>
  
 +
== Solution 2 (Noticing Stuff)==
 +
If there is only 1 number, the number of ways to order would be 1.
 +
If there are 2 numbers, the number of ways to order would be 2.
 +
If there are 3 numbers, by listing out, the number of ways is 4.
 +
We can then make a conjecture that the problem is simply powers of 2.
 +
<math>2^(10-1)=512</math>.
 +
:)
  
 
== See Also ==
 
== See Also ==

Revision as of 00:31, 1 September 2015

Problem 18

Let $(a_1,a_2, \dots ,a_{10})$ be a list of the first 10 positive integers such that for each $2 \le i \le 10$ 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

Let $1\leq k\leq 10$. Assume that $a_1=k$. If $k<10$, the first number appear after $k$ that is greater than $k$ must be $k+1$, otherwise if it is any number $x$ larger than $k+1$, there will be neither $x-1$ nor $x+1$ appearing before $x$. Similarly, one can conclude that if $k+1<10$, the first number appear after $k+1$ that is larger than $k+1$ must be $k+2$, and so forth.

On the other hand, if $k>1$, the first number appear after $k$ that is less than $k$ must be $k-1$, and then $k-2$, and so forth.

To count the number of possibilities when $a_1=k$ is given, we set up $9$ spots after $k$, and assign $k-1$ of them to the numbers less than $k$ and the rest to the numbers greater than $k$. The number of ways in doing so is $9$ choose $k-1$.

Therefore, when summing up the cases from $k=1$ to $10$, we get

\[\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} = 2^9=512 ...... \framebox{B}\]

Solution 2 (Noticing Stuff)

If there is only 1 number, the number of ways to order would be 1. If there are 2 numbers, the number of ways to order would be 2. If there are 3 numbers, by listing out, the number of ways is 4. We can then make a conjecture that the problem is simply powers of 2. 

$2^(10-1)=512$.
)

See Also

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 12 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_18&oldid=71902"