2008 AMC 12A Problem 21

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.