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 "2008 AMC 12A Problems/Problem 21"

(→‎Solution)
m
 
(One intermediate revision by one other user not shown)
Line 4: Line 4:
 
<math>\mathrm{(A)}\ 36\qquad\mathrm{(B)}\ 40\qquad\textbf{(C)}\ 44\qquad\mathrm{(D)}\ 48\qquad\mathrm{(E)}\ 52</math>
 
<math>\mathrm{(A)}\ 36\qquad\mathrm{(B)}\ 40\qquad\textbf{(C)}\ 44\qquad\mathrm{(D)}\ 48\qquad\mathrm{(E)}\ 52</math>
  
−
==Solution 1==
+
==Solution 1 (Symmetry) ==
 
There are <math>5!=120</math> total permutations.  
 
There are <math>5!=120</math> total permutations.  
  
Line 29: Line 29:
  
 
Therefore, our answer is <math>8 + 12 + 8 + 12 + 8 = 48 \Rightarrow \boxed{D}</math>.
 
Therefore, our answer is <math>8 + 12 + 8 + 12 + 8 = 48 \Rightarrow \boxed{D}</math>.
 +
 +
-MP8148
  
 
== See also ==
 
== See also ==

Latest revision as of 00:21, 31 July 2023

Problem

A permutation $(a_1,a_2,a_3,a_4,a_5)$ of $(1,2,3,4,5)$ is heavy-tailed if $a_1 + a_2 < a_4 + a_5$. What is the number of heavy-tailed permutations?

$\mathrm{(A)}\ 36\qquad\mathrm{(B)}\ 40\qquad\textbf{(C)}\ 44\qquad\mathrm{(D)}\ 48\qquad\mathrm{(E)}\ 52$

Solution 1 (Symmetry)

There are $5!=120$ total permutations.

For every permutation $(a_1,a_2,a_3,a_4,a_5)$ such that $a_1 + a_2 < a_4 + a_5$, there is exactly one permutation such that $a_1 + a_2 > a_4 + a_5$. Thus it suffices to count the permutations such that $a_1 + a_2 = a_4 + a_5$.

$1+4=2+3$, $1+5=2+4$, and $2+5=3+4$ are the only combinations of numbers that can satisfy $a_1 + a_2 = a_4 + a_5$.

There are $3$ combinations of numbers, $2$ possibilities of which side of the equation is $a_1+a_2$ and which side is $a_4+a_5$, and $2^2=4$ possibilities for rearranging $a_1,a_2$ and $a_4,a_5$. Thus, there are $3\cdot2\cdot4=24$ permutations such that $a_1 + a_2 = a_4 + a_5$.

Thus, the number of heavy-tailed permutations is $\frac{120-24}{2}=48 \Rightarrow D$.

Solution 2 (Casework)

We use case work on the value of $a_3$.

Case 1: $a_3 = 1$. Since $a_1 + a_2 < a_4 + a_5$, $(a_1, a_2)$ can only be a permutation of $(2, 3)$ or $(2, 4)$. The values of $a_1$ and $a_2$, as well as the values of $a_4$ and $a_5$, are interchangeable, so this case produces a total of $2(2 \cdot 2) = 8$ solutions.

Case 2: $a_3 = 2$. Similarly, we have $(a_1, a_2)$ is a permutation of $(1, 3)$, $(1, 4)$, or $(1, 5)$, which gives a total of $3(2 \cdot 2) = 12$ solutions.

Case 3: $a_3 = 3$. $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 4)$, which gives a total of $2(2 \cdot 2) = 8$ solutions.

Case 4: $a_3 = 4$. $(a_1, a_2)$ is a permutation of $(1, 2)$, $(1, 3)$, or $(2, 3)$, which gives a total of $3(2 \cdot 2) = 12$ solutions.

Case 5: $a_3 = 5$. $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 3)$, which gives a total of $2(2 \cdot 2) = 8$ solutions.

Therefore, our answer is $8 + 12 + 8 + 12 + 8 = 48 \Rightarrow \boxed{D}$.

-MP8148

See also

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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=2008_AMC_12A_Problems/Problem_21&oldid=196274"