Question

有多少种方法可以把整数 $1$ 到 $14$ 分成 $7$ 对，使得每一对中较大的数至少是较小的数的 $2$ 倍？

How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?

$108$

$120$

$126$

$132$

$144$

答案：E

知识点：有限制的排列配对与分组乘法原理

难度评级：2390

解答：

数 $8$ 到 $14$ 之间不能互相配对，所以它们必须分别与 $1$ 到 $7$ 中的一个数配对。特别地， $7$ 必须与 $14$ 配对，因为没有其他可用的数至少是 $7$ 的两倍。

再看其余各数的搭档。 $8$ 和 $9$ 可与 $1-4$ 中任一数配对； $10$ 和 $11$ 可与 $1-5$ 中任一数配对； $12$ 和 $13$ 可与 $1-6$ 中任一数配对。

$8$ 有 $4$ 种选择，随后 $9$ 只剩 $3$ 种，因为 $8$ 已占用一个数。 $10$ 仍有 $3$ 种选择，因为虽已占用 $2$ 个选择，但它多出一个可选数 $(5)$ 。接着 $11$ 有 $2$ 种， $12$ 有 $2$ 种， $13$ 只有 $1$ 种。

相乘得到 $4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 144.$

所以正确答案是 E。

The numbers from $8$ through $14$ cannot be paired with one another, so they must be paired with the numbers from $1$ through $7.$ In particular, $7$ must be paired with $14,$ since no other available number is at least twice $7.$

Now let's look at what the other numbers can pair with. $8$ and $9$ can pair with any number $1-4.$ $10$ and $11$ can pair with any number $1-5,$ and $12$ and $13$ can pair with any number $1-6.$

$8$ can pair with $4$ numbers, but then $9$ only has $3$ options since $8$ took one. $10$ then has $3$ options, since $2$ choices are taken, but it has one more to choose from $(5).$ $11$ then has $2$ options, $12$ has $2$ options, and $13$ only has $1.$

Multiplying these together yields $4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 144.$

Thus, E is the correct answer.

2022 AMC 10A 第 14 题

答案：E

难度评级：2390

解答：

其他年份的第 14 题