2008 AMC 12A Problem 21

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

By symmetry,

a_1 + a_2 \lt a_4 + a_5

and

a_1 + a_2 \gt a_4 + a_5

occur equally often

Count the balanced permutations

a_1 + a_2 = a_4 + a_5;

here

a_3 \in \{1, 3, 5\}

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.

Problem 21 in other years