Question

一个孩子用形状相同但颜色不同的立方体搭塔。用 $2$ 个红色立方体、 $3$ 个蓝色立方体和 $4$ 个绿色立方体，可以搭出多少种高度为 $8$ 个立方体的不同塔？（会剩下一个立方体。）

A child builds towers using identically shaped cubes of different colors. How many different towers with a height $8$ cubes can the child build with $2$ red cubes, $3$ blue cubes, and $4$ green cubes? (One cube will be left out.)

$24$

$288$

$312$

$1,260$

$40,320$

答案：D

知识点：多重集排列双射

难度评级：1480

解答：

给定一个合法的高度为 $8$ 的塔，各颜色数量唯一确定了未使用的立方体；把该立方体放在顶端。反过来，从全部 $9$ 个立方体的任意排列中去掉顶端立方体，都会得到一个合法的高度为 $8$ 的塔。这两个操作互为逆操作，所以所求塔与全部 $9$ 个立方体的排列一一对应。

高度为 $9,$ 的塔共有 $9!$ 种排列，但同色立方体之间的交换会造成重复计算。

因此要分别除以红色立方体的 $2!$ 种排列、蓝色立方体的 $3!$ 种排列和绿色立方体的 $4!$ 种排列。

所以合法排列数为 $\dfrac{9!}{2! \cdot 3! \cdot 4!} = 1,260.$

所以正确答案是 D。

Given a valid height- $8$ tower, its color counts determine the one unused cube; place that cube on top. Conversely, removing the top cube from any arrangement of all $9$ cubes gives a valid height- $8$ tower. These operations are inverses, so the desired towers are in one-to-one correspondence with arrangements of all $9$ cubes.

There are $9!$ ways to make a tower of height $9,$ but we are overcounting since there are multiple cubes of the same color.

We have to divide through by $2!$ ways to arrange the red cubes, $3!$ for the blue cubes, and $4!$ for the green cubes.

Therefore, the number of valid arrangements is $\dfrac{9!}{2! \cdot 3! \cdot 4!} = 1,260.$

Thus, D is the correct answer.

2019 AMC 10A 第 17 题

答案：D

难度评级：1480

解答：

其他年份的第 17 题