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

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== 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}}
 
[[Category:Intermediate Geometry Problems]]
 

Revision as of 22:38, 19 February 2008

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?

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

Solution

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 $\frac{120-24}{2}=48 \Rightarrow D$.

See also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 12 Problems and Solutions