2020 AMC 10B 第 5 题

Question

将

1

块棕色砖、

1

块紫色砖、

2

块绿色砖和

3

块黄色砖从左到右排成一行，有多少种可区分的排列？同色砖不可区分。

How many distinguishable arrangements are there of

1

brown tile,

1

purple tile,

2

green tiles, and

3

yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)

210

420

630

840

1050

先把七块砖都看作不同来排列。

First arrange the seven tiles as if they were all different

再除以相同绿色砖和相同黄色砖的排列数。

Divide by the orders of the identical green and yellow tiles

砖块总数为

1+1+2+3=7

。若七块都不同有

7!

种排列。两块绿色砖不可区分，三块黄色砖不可区分，所以要除以

2!

和

3!

。

因此可区分的排列数为

\frac{7!}{2!3!}=420\text{。}

所以正确答案是 B。

There are

1+1+2+3=7

total tiles. If all seven tiles were distinct, there would be

7!

arrangements. The two green tiles are indistinguishable, and the three yellow tiles are indistinguishable, so we divide by

2!

and

3!.

Thus the number of arrangements is

\frac{7!}{2!3!}=420.

Thus, B is the correct answer.

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