2008 AMC 10A 第 23 题

Question

要从集合

S = \{a, b, c, d, e\}

中选择两个子集，使它们的并集为

S

，且交集恰好包含两个元素。如果选择两个子集的顺序不重要，共有多少种方法？

Two subsets of the set

S = \{a, b, c, d, e\}

are to be chosen so that their union is

S

and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?

20

40

60

160

320

答案：B

知识点：子集组合乘法原理

难度评级：1770

小提示：

先选择属于两个子集共有的两个元素。

First choose the two elements that belong to both subsets

大提示：

其余每个元素恰好进入一个子集；最后对无序子集对进行修正。

Each of the remaining elements goes to exactly one subset; then correct for the unordered pair

解答：

共有元素可用

\binom{5}{2} = 10

种方式选择。

剩余

3

个元素必须恰好属于一个子集，有

2^3 = 8

种分配方式，所以有

80

个有序子集对。

因为两个子集的顺序不重要，除以

2

，得到

\dfrac{80}{2} = 40

。

所以正确答案是 B。

Choose the two common elements in

\binom{5}{2} = 10

ways.

Each of the remaining

3

elements must lie in exactly one subset, giving

2^3 = 8

assignments, for

80

ordered pairs.

Since the order of the two subsets does not matter, divide by

2

to get

\dfrac{80}{2} = 40.

Thus, the correct answer is B.

2008 AMC 10A 第 23 题

答案：B

难度评级：1770

小提示：

大提示：

解答：

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