AMC10 2019 A

AMC10 2019 A · Q17

AMC10 2019 A · Q17. It mainly tests Combinations, Casework.

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

一个孩子使用相同形状但不同颜色的立方体搭建塔形。有2个红色、3个蓝色和4个绿色立方体，能搭建多少种高度为8个立方体的不同塔？（会剩下一个立方体。）

(A) 24 24

(B) 288 288

(D) 1260 1260

(E) 40320 40320

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Of the 9 cubes available, 1 cube will not be used. Because there are three different kinds of cubes and limited numbers of each kind, there are three different possibilities for the set of cubes that are used. One possibility is 1 red cube, 3 blue cubes, and 4 green cubes; the second possibility is 2 red cubes, 2 blue cubes, and 4 green cubes; and the third possibility is 2 red cubes, 3 blue cubes, and 3 green cubes. Cubes of the same color are indistinguishable. Hence the number of different towers is $$ \frac{8!}{1!\cdot 3!\cdot 4!}+\frac{8!}{2!\cdot 2!\cdot 4!}+\frac{8!}{2!\cdot 3!\cdot 3!}=1,260. $$

答案（D）：在现有的 9 个立方体中，有 1 个不会被使用。由于有三种不同颜色的立方体且每种数量有限，因此被使用的立方体集合有三种可能。第一种可能是 1 个红色、3 个蓝色和 4 个绿色；第二种可能是 2 个红色、2 个蓝色和 4 个绿色；第三种可能是 2 个红色、3 个蓝色和 3 个绿色。同色立方体不可区分。因此不同塔的数量为 $$ \frac{8!}{1!\cdot 3!\cdot 4!}+\frac{8!}{2!\cdot 2!\cdot 4!}+\frac{8!}{2!\cdot 3!\cdot 3!}=1,260. $$

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