AMC 8 1994 Problem 24
AMC8 1994 年第 24 题
基本计数原理★★★★☆
A 2 by 2 square is divided into four 1 by 1 squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares.
一个 2 × 2 的正方形被分成四个 1 × 1 的小正方形。每个小正方形要被涂成绿色或红色。共有多少种不同的涂色方式，使得没有绿色正方形与任何红色正方形共享其顶边或右边？小绿色正方形的数量可以从零到四个不等。
- A.4
- B.6
- C.7
- D.8
- E.16
正确答案 · Correct Answer
B
解析 · Solution
解法1
常规解法 · 按绿格数分类
绿格的上方、右侧都不能是红。0 绿：1 种；1 绿（须在右上角）：1 种；2 绿（上面一行或右边一列）：2 种；3 绿（红在左下角）：1 种；4 绿：1 种。合计 1+1+2+1+1=6。
巧解 · 阶梯形
规则等价于「绿格的上方与右方也必须是绿」，即绿色区域是向上、向右封闭的阶梯。2×2 共有 6 种这样的阶梯（从全红到全绿），故 6。
Solution 1
Elementary · split by number of green cells
A green cell may not have a red above or to its right. By the number of greens: 0 greens (1 way); 1 green, which must sit in the top-right (1 way); 2 greens, the top row or the right column (2 ways); 3 greens, with the red in the bottom-left (1 way); 4 greens (1 way). Total 1+1+2+1+1=6.
Slick · staircase shape
The rule means a green cell forces the cells above it and to its right to be green, so the green region is an 'up-and-right closed' staircase. The 2×2 grid has exactly 6 such staircases (from all-red to all-green), so 6.
继续练习