AMC10 2007 B

AMC10 2007 B · Q20

AMC10 2007 B · Q20. It mainly tests Basic counting (rules of product/sum), Combinations.

A set of 25 square blocks is arranged into a 5 × 5 square. How many different combinations of 3 blocks can be selected from that set so that no two are in the same row or column?

25个正方形方块排成5×5正方形。从中选3个方块的不同组合有多少种，使得没有两个在同一行或同一列？

(A) 100 100

(B) 125 125

(D) 2300 2300

(E) 3600 3600

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): After one of the 25 blocks is chosen, 16 of the remaining blocks do not share its row or column. After the second block is chosen, 9 of the remaining blocks do not share a row or column with either of the first two. Because the three blocks can be chosen in any order, the number of different combinations is $$ \frac{25 \cdot 16 \cdot 9}{3!} = 25 \cdot 8 \cdot 3 = 600. $$

答案（C）：选定 25 个方块中的一个后，剩下的方块中有 16 个与它不在同一行或同一列。选定第二个方块后，剩下的方块中有 9 个与前两个方块都不在同一行或同一列。由于这三个方块可以按任意顺序选取，不同组合的数量为 $$ \frac{25 \cdot 16 \cdot 9}{3!} = 25 \cdot 8 \cdot 3 = 600. $$

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