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Difference between revisions of "2008 AMC 12A Problems/Problem 21"

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<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>
  
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==Solution==
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==Solution 1 (Symmetry) ==
 
There are <math>5!=120</math> total permutations.  
 
There are <math>5!=120</math> total permutations.  
  
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There are <math>3</math> combinations of numbers, <math>2</math> possibilities of which side of the equation is <math>a_1+a_2</math> and which side is <math>a_4+a_5</math>, and <math>2^2=4</math> possibilities for rearranging <math>a_1,a_2</math> and <math>a_4,a_5</math>. Thus, there are <math>3\cdot2\cdot4=24</math> permutations such that <math>a_1 + a_2 = a_4 + a_5</math>.  
 
There are <math>3</math> combinations of numbers, <math>2</math> possibilities of which side of the equation is <math>a_1+a_2</math> and which side is <math>a_4+a_5</math>, and <math>2^2=4</math> possibilities for rearranging <math>a_1,a_2</math> and <math>a_4,a_5</math>. Thus, there are <math>3\cdot2\cdot4=24</math> permutations such that <math>a_1 + a_2 = a_4 + a_5</math>.  
  
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Thus, the number of <u>heavy-tailed</u> permutations is <math>\frac{120-24}{2}=48 \Rightarrow D</math>.  
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Thus, the number of <u>heavy-tailed</u> permutations is <math>\frac{120-24}{2}=48 \Rightarrow D</math>.
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==Solution 2 (Casework)==
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We use case work on the value of <math>a_3</math>.
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Case 1: <math>a_3 = 1</math>. Since <math>a_1 + a_2 < a_4 + a_5</math>, <math>(a_1, a_2)</math> can only be a permutation of <math>(2, 3)</math> or <math>(2, 4)</math>. The values of <math>a_1</math> and <math>a_2</math>, as well as the values of <math>a_4</math> and <math>a_5</math>, are interchangeable, so this case produces a total of <math>2(2 \cdot 2) = 8</math> solutions.
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Case 2: <math>a_3 = 2</math>. Similarly, we have <math>(a_1, a_2)</math> is a permutation of <math>(1, 3)</math>, <math>(1, 4)</math>, or <math>(1, 5)</math>, which gives a total of <math>3(2 \cdot 2) = 12</math> solutions.
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Case 3: <math>a_3 = 3</math>. <math>(a_1, a_2)</math> is a permutation of <math>(1, 2)</math> or <math>(1, 4)</math>, which gives a total of <math>2(2 \cdot 2) = 8</math> solutions.
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Case 4: <math>a_3 = 4</math>. <math>(a_1, a_2)</math> is a permutation of <math>(1, 2)</math>, <math>(1, 3)</math>, or <math>(2, 3)</math>, which gives a total of <math>3(2 \cdot 2) = 12</math> solutions.
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Case 5: <math>a_3 = 5</math>. <math>(a_1, a_2)</math> is a permutation of <math>(1, 2)</math> or <math>(1, 3)</math>, which gives a total of <math>2(2 \cdot 2) = 8</math> solutions.
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Therefore, our answer is <math>8 + 12 + 8 + 12 + 8 = 48 \Rightarrow \boxed{D}</math>.
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-MP8148
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2008|num-b=20|num-a=22|ab=A}}
 
{{AMC12 box|year=2008|num-b=20|num-a=22|ab=A}}
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[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

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
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All AMC 12 Problems and Solutions

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