AMC10 2020 B

AMC10 2020 B · Q5

AMC10 2020 B · Q5. It mainly tests Combinations.

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.)

有 1 块棕色瓷砖、1 块紫色瓷砖、2 块绿色瓷砖和 3 块黄色瓷砖，从左到右排成一行，有多少种不同的排列方式？（相同颜色的瓷砖不可区分。）

(A) 210 210

(B) 420 420

(D) 840 840

(E) 1050 1050

Answer

Correct choice: (B)

正确答案：（B）

Solution

If all the tiles were distinguishable from each other, there would be $7! = 5040$ possible arrangements. However, this total must be divided by $2! = 2$ to account for the 2 different equivalent orderings of the 2 green tiles and also divided by $3! = 6$ to account for the 6 different equivalent orderings of the 3 yellow tiles. Therefore the answer is $$\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{2 \cdot 6} = 420.$$\n\nOR\n\nThere are $\binom{7}{3}$ ways to choose the positions for the yellow tiles, $\binom{4}{2}$ ways to choose the positions for the green tiles among the remaining positions, and then 2 ways to choose the positions for the brown and purple tiles. Therefore the answer is $$\binom{7}{3} \cdot \binom{4}{2} \cdot 2 = 420.$$

如果所有瓷砖都可区分，则有 $7! = 5040$ 种可能的排列。但是，需要除以 $2! = 2$ 来考虑 2 块绿色瓷砖的 2 种等价排序，并且除以 $3! = 6$ 来考虑 3 块黄色瓷砖的 6 种等价排序。因此答案是 $$\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{2 \cdot 6} = 420。$$\n\n或者\n\n有 $\binom{7}{3}$ 种方法选择黄色瓷砖的位置，在剩余位置中有 $\binom{4}{2}$ 种方法选择绿色瓷砖的位置，然后有 2 种方法选择棕色和紫色瓷砖的位置。因此答案是 $$\binom{7}{3} \cdot \binom{4}{2} \cdot 2 = 420。$$

Topics

CP.003 Combinations