2008 AMC 12A Problems/Problem 21

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

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

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2008 AMC 12A Problems/Problem 21

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