Question

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

$36$

$40$

$44$

$48$

$52$

Solution:

Call a permutation balanced if $a_1 + a_2 = a_4 + a_5.$ Reversing the entries swaps the two strict cases, so heavy-tailed and heavy-headed permutations are equally numerous.

The total $1 + 2 + 3 + 4 + 5 = 15$ is odd, so in a balanced permutation $a_3$ must be odd, one of $1, 3, 5.$ For each choice, the remaining four numbers split uniquely into two equal-sum pairs.

Any of the four can be $a_1$ (fixing $a_2$ ), and either remaining number can be $a_4$ (fixing $a_5$ ), giving $3 \cdot 4 \cdot 2 = 24$ balanced permutations.

The other $120 - 24 = 96$ permutations split evenly, so there are $\tfrac{96}{2} = 48$ heavy-tailed permutations.

Thus, D is the correct answer.

2008 AMC 12A Problem 21

Solution:

Problem 21 in Other Years