AMC8 2022

AMC8 2022 · Q23

AMC8 2022 · Q23. It mainly tests Basic counting (rules of product/sum), Inclusion–exclusion (basic).

A $\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$-by-$3$ grid. Shown below is a sample configuration with three $\triangle$s in a line. How many configurations will have three $\triangle$s in a line and three $\bigcirc$s in a line?

在3×3网格的九个方格中每个放置一个$\triangle$或$\bigcirc$。下面是一个示例配置，有一行三个$\triangle$。有多少种配置既有行三个$\triangle$又有行三个$\bigcirc$？

(A) 39 39

(B) 42 42

(D) 84 84

(E) 96 96

Answer

Correct choice: (D)

正确答案：（D）

Solution

We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are $3$ ways to choose a column with all $\bigcirc$'s and $2$ ways to choose a column with all $\triangle$'s. The third column can be filled in $2^3=8$ ways. Therefore, we have a total of $3\cdot2\cdot8=48$ cases. However, we overcounted the cases with $2$ complete columns of with one symbol and $1$ complete column with another symbol. This happens in $2\cdot3=6$ cases. $48-6=42$. However, we have to remember to double our answer, giving us $\boxed{\textbf{(D) }84}$ ways to complete the grid.

我们只考虑列的情况（三个相同符号在同一列），最后再乘以2因为行也一样。有3种选择全$\bigcirc$的列，2种选择全$\triangle$的列。第三列有$2^3=8$种填充方式。因此，总数$3\cdot2\cdot8=48$。但多计了有2列一种符号1列另一种的情况，这种情况有$2\cdot3=6$。所以$48-6=42$。记得乘2，得$\boxed{\textbf{(D) }84}$。

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