📚 Three-Set Problems | 三集合问题
Three-set problems in IGCSE Mathematics involve using Venn diagrams with three overlapping circles to solve counting problems. These problems test your ability to organise information, interpret set notation, and apply the principle of inclusion–exclusion.
IGCSE 数学中的三集合问题涉及使用包含三个相交圆形的维恩图来解决计数问题。这类题目考察你组织信息、解读集合符号以及应用容斥原理的能力。
1. What Are Three-Set Problems? | 什么是三集合问题
A three-set problem involves three sets, usually labelled A, B and C, that may overlap with one another. The universal set ξ contains all the elements being considered. A Venn diagram with three circles divides the universal set into 8 distinct regions.
三集合问题涉及三个通常标记为 A、B 和 C 的集合，这些集合可能相互重叠。全集 ξ 包含所有被考虑的元素。一个包含三个圆形的维恩图将全集划分为 8 个不同的区域。
These problems require you to work with counts (denoted by n(A), n(B), etc.) and to determine how many elements belong to various combinations of sets.
这类问题要求你处理集合的元素个数（记为 n(A)、n(B) 等），并确定有多少元素属于集合的各种不同组合。
2. The Three-Circle Venn Diagram | 三圆维恩图
The three-circle Venn diagram is drawn as three overlapping circles inside a rectangle that represents the universal set ξ. Each circle represents one set, and the overlaps represent intersections between sets.
三圆维恩图画在一个代表全集 ξ 的矩形内部，包含三个相互重叠的圆形。每个圆形代表一个集合，重叠部分代表集合之间的交集。
When drawing diagrams in exams, always make sure the three circles overlap symmetrically so that all 8 regions are clearly visible. Label each circle with the set name, and label the rectangle with ξ.
在考试中画图时，务必确保三个圆对称重叠，使全部 8 个区域清晰可见。在每个圆上标注集合名称，并在矩形上标注 ξ。
3. Notation and Symbols | 符号与记号
You must be familiar with the following notation for three-set problems:
你必须熟悉三集合问题中的以下符号：
- A ∪ B : the union of A and B — elements in A or B or both | A 与 B 的并集——属于 A 或 B 或两者的元素
- A ∩ B : the intersection of A and B — elements in both A and B | A 与 B 的交集——同时属于 A 和 B 的元素
- A′ : the complement of A — elements not in A | A 的补集——不属于 A 的元素
- n(A) : the number of elements in set A | 集合 A 中元素的个数
- ξ : the universal set | 全集
- ∈ : is an element of | 属于
- ∉ : is not an element of | 不属于
- A ∩ B ∩ C : elements in all three sets | 同时属于三个集合的元素
In three-set problems, A ∩ B means the region where A and B overlap only, not including the part that also belongs to C, unless A ∩ B ∩ C is specified.
在三集合问题中，A ∩ B 指的是 A 和 B 重叠的区域，但不包括同时也属于 C 的部分，除非特别指明 A ∩ B ∩ C。
4. The 8 Regions of a Three-Set Venn Diagram | 三圆维恩图的八个区域
It is essential to identify all 8 regions. Let a be the number of elements in A only, b in B only, c in C only, d in A ∩ B only, e in A ∩ C only, f in B ∩ C only, and g in A ∩ B ∩ C. The 8th region is the area outside all three circles, which belongs to ξ but to none of A, B or C.
识别全部 8 个区域至关重要。设 a 为仅属于 A 的元素个数，b 为仅属于 B 的个数，c 为仅属于 C 的个数，d 为仅属于 A ∩ B 的个数，e 为仅属于 A ∩ C 的个数，f 为仅属于 B ∩ C 的个数，g 为属于 A ∩ B ∩ C 的个数。第 8 个区域是三个圆之外的部分，属于 ξ 但不属于 A、B、C 中的任何一个。
n(ξ) = a + b + c + d + e + f + g + h
where h is the number of elements outside all three sets.
其中 h 是三个集合之外的元素个数。
This breakdown is the key to solving three-set problems: write down what each region represents, then fill in the numbers.
这种分解是解决三集合问题的关键：写出每个区域代表什么，然后填入数字。
5. The Inclusion–Exclusion Principle | 容斥原理
The most important formula for three-set problems is the inclusion–exclusion principle:
三集合问题最重要的公式是容斥原理：
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C)
This formula works by adding the three individual sets, then subtracting the overlaps that were counted twice, and finally adding back the triple overlap that was subtracted too many times.
这个公式的运作方式是：先将三个集合相加，然后减去被重复计算的重叠部分，最后加回被多减的三重交集部分。
If you are given n(ξ) and the number outside all three sets, you can also use:
如果你已知 n(ξ) 以及三个集合之外的元素个数，你还可以使用：
n(A ∪ B ∪ C) = n(ξ) − n(A ∪ B ∪ C)′
6. Applying the Formula | 应用公式
In exam questions, you are often given n(A), n(B), n(C), the various pairwise intersections, the triple intersection, and the number outside all sets. Your task is to find missing values.
在考试题目中，你通常会得到 n(A)、n(B)、n(C)、各成对交集、三重交集以及所有集合之外的个数。你的任务是求出缺失的值。
Work systematically: first place the number in the triple intersection g at the centre. Then work outwards: fill d = n(A ∩ B) − g, e = n(A ∩ C) − g, f = n(B ∩ C) − g. Then find a = n(A) − d − e − g, and similarly for b and c. Finally, verify that all 8 regions sum to n(ξ).
系统地进行计算：首先将三重交集 g 的数字放在中心。然后向外推算：d = n(A ∩ B) − g，e = n(A ∩ C) − g，f = n(B ∩ C) − g。接着求出 a = n(A) − d − e − g，b 和 c 同理。最后验证全部 8 个区域之和等于 n(ξ)。
7. Interpreting Word Problems | 理解文字题
Three-set problems in exams are usually presented as word problems. For example, a survey may ask students which of three sports they play: football, tennis and swimming. Some students play only one sport, some play two, some play all three, and some play none.
考试中的三集合问题通常以文字题形式呈现。例如，一项调查可能询问学生参加三种运动中的哪些：足球、网球和游泳。有些学生只参加一项运动，有些参加两项，有些三项都参加，还有些一项都不参加。
Key phrases to watch for: “all three” means A ∩ B ∩ C; “at least one” means A ∪ B ∪ C; “exactly two” means the sum of the three pairwise-only regions; “none” means the region outside all circles.
需要注意的关键短语：”三项都” 表示 A ∩ B ∩ C；”至少一项” 表示 A ∪ B ∪ C；”恰好两项” 表示三个仅属于成对交集区域之和；”一项都没有” 表示三个圆之外的部分。
Always read the question carefully to determine whether a given number refers to a single set, a pairwise intersection (including the triple) or exactly a pairwise intersection (excluding the triple).
务必仔细阅读题目，判断给定数字是指单个集合、成对交集（包含三重交集）还是仅仅指成对交集（不包含三重交集）。
8. Worked Example 1 | 例题一
In a class of 30 students, 15 play football, 12 play tennis, and 10 play swimming. 6 play both football and tennis, 5 play both tennis and swimming, 4 play both football and swimming, and 2 play all three. How many students play none of the three sports?
在一个 30 人的班级中，15 人踢足球，12 人打网球，10 人游泳。6 人既踢足球又打网球，5 人既打网球又游泳，4 人既踢足球又游泳，2 人三项都参与。有多少学生三项运动都没有参加？
Solution: Using the inclusion–exclusion principle:
解答：使用容斥原理：
n(F ∪ T ∪ S) = 15 + 12 + 10 − 6 − 5 − 4 + 2 = 24
Since n(ξ) = 30, the number who play none is 30 − 24 = 6.
由于 n(ξ) = 30，一项都不参加的人数为 30 − 24 = 6。
We can also verify this using the region method. Let g = 2. Then d (F ∩ T only) = 6 − 2 = 4, e (F ∩ S only) = 4 − 2 = 2, f (T ∩ S only) = 5 − 2 = 3. Then a (F only) = 15 − 4 − 2 − 2 = 7, b (T only) = 12 − 4 − 3 − 2 = 3, c (S only) = 10 − 2 − 3 − 2 = 3. Adding: 7 + 3 + 3 + 4 + 2 + 3 + 2 = 24, so 6 students play none.
我们也可以用区域法进行验证。设 g = 2。则 d（仅 F ∩ T）= 6 − 2 = 4，e（仅 F ∩ S）= 4 − 2 = 2，f（仅 T ∩ S）= 5 − 2 = 3。则 a（仅 F）= 15 − 4 − 2 − 2 = 7，b（仅 T）= 12 − 4 − 3 − 2 = 3，c（仅 S）= 10 − 2 − 3 − 2 = 3。相加：7 + 3 + 3 + 4 + 2 + 3 + 2 = 24，因此有 6 名学生一项运动都没参加。
9. Worked Example
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