AMC12 2008 A

AMC12 2008 A · Q21

AMC12 2008 A · Q21. It mainly tests Basic counting (rules of product/sum), Permutations.

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?

$(1,2,3,4,5)$ 的一个排列 $(a_1,a_2,a_3,a_4,a_5)$ 若满足 $a_1 + a_2 < a_4 + a_5$，则称为重尾排列。重尾排列的个数是多少？

(A) 36 36

(B) 40 40

(D) 48 48

(E) 52 52

Answer

Correct choice: (D)

正确答案：（D）

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

共有 $5!=120$ 个排列。对于每一个满足 $a_1 + a_2 < a_4 + a_5$ 的排列 $(a_1,a_2,a_3,a_4,a_5)$，恰好存在一个排列满足 $a_1 + a_2 > a_4 + a_5$。因此只需统计满足 $a_1 + a_2 = a_4 + a_5$ 的排列个数。只有 $1+4=2+3$、$1+5=2+4$、$2+5=3+4$ 这三种数组合能满足 $a_1 + a_2 = a_4 + a_5$。共有 $3$ 种数组合；等式两边哪一边是 $a_1+a_2$、哪一边是 $a_4+a_5$ 有 $2$ 种；并且 $a_1,a_2$ 与 $a_4,a_5$ 各自内部交换共有 $2^2=4$ 种。因此满足 $a_1 + a_2 = a_4 + a_5$ 的排列共有 $3\cdot2\cdot4=24$ 个。所以重尾排列的个数为 $\frac{120-24}{2}=48 \Rightarrow D$。

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