Question

一副牌中的每张牌都画有一种图形：圆形、正方形或三角形；并且涂成三种颜色之一：红色、蓝色或绿色。此外，每种颜色还使用三种深浅之一：浅色、中等或深色。整副牌有 $27$ 张，每一种图形、颜色、深浅的组合各有一张。若三张牌组成的集合满足以下所有条件，则称为互补：

• 三张牌的图形要么各不相同，要么全都相同。

• 三张牌的颜色要么各不相同，要么全都相同。

• 三张牌的深浅要么各不相同，要么全都相同。

有多少个不同的互补三张牌集合？

Every card in a deck has a picture of one shape — circle, square, or triangle, which is painted in one of the three colors — red, blue, or green. Furthermore, each color is applied in one of three shades — light, medium, or dark. The deck has $27$ cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:

• Either each of the three cards has a different shape or all three of the cards have the same shape.

• Either each of the three cards has a different color or all three of the cards have the same color.

• Either each of the three cards has a different shade or all three of the cards have the same shade.

How many different complementary three-card sets are there?

答案

答案：117

知识点：数对计数双重计数

难度评级：2450

解答：

给定任意两张不同的牌，恰好有一张牌能把它们补成互补集合：在每个属性上，如果两张牌相同，第三张牌必须也取相同值；如果两张牌不同，第三张牌必须取剩下的那个值。补出的牌不同于原来的两张牌，因为原来的两张牌至少在某个属性上不同，而在这个属性上第三张牌与二者都不同。

因此 $\binom{27}{2} = 351$ 对牌各自延伸成一个互补集合，而每个互补集合由其中的 $\binom{3}{2} = 3$ 对牌产生。集合数为 $\frac{351}{3} = 117$ 。

Given any two distinct cards, there is exactly one card completing them to a complementary set: in each attribute, if the two cards agree, the third card must share that value, and if they differ, the third must take the one remaining value. The completing card is distinct from both (the two given cards differ somewhere, and in that attribute the third card differs from each).

So the $\binom{27}{2} = 351$ pairs of cards each extend to one complementary set, and each complementary set is produced by $\binom{3}{2} = 3$ of these pairs. The number of sets is $\frac{351}{3} = 117.$

1997 AIME 第 10 题

答案：117

难度评级：2450

解答：

其他年份的第 10 题