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

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

解答：

其他年份的第 23 题