AMC10 2008 A

AMC10 2008 A · Q23

AMC10 2008 A · Q23. It mainly tests Basic counting (rules of product/sum), Combinations.

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?

要从集合$S = \{a, b, c, d, e\}$中选择两个子集，使得它们的并集是$S$，且交集恰好包含两个元素。有多少种方法可以做到这一点，假设选择子集的顺序无关紧要？

(A) 20 20

(B) 40 40

(D) 160 160

(E) 320 320

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let the two subsets be $A$ and $B$. There are $\binom{5}{2}=10$ ways to choose the two elements common to $A$ and $B$. There are then $2^3=8$ ways to assign the remaining three elements to $A$ or $B$, so there are 80 ordered pairs $(A,B)$ that meet the required conditions. However, the ordered pairs $(A,B)$ and $(B,A)$ represent the same pair $\{A,B\}$ of subsets, so the conditions can be met in $\frac{80}{2}=40$ ways.

答案（B）：设这两个子集为 $A$ 和 $B$。选择同时属于 $A$ 和 $B$ 的两个元素有 $\binom{5}{2}=10$ 种方法。接着将剩下的三个元素分配到 $A$ 或 $B$ 中有 $2^3=8$ 种方法，因此满足条件的有序对 $(A,B)$ 共有 80 个。但是，有序对 $(A,B)$ 和 $(B,A)$ 表示同一个子集对 $\{A,B\}$，所以满足条件的方法数为 $\frac{80}{2}=40$ 种。

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