最多报两门人数

Criado em: 21 de janeiro de 2025

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Pergunta

一个班级总共有30名学生。心理学、社会学和市场营销专业的学生人数分别是总人数的2/5、4/15和1/3。问：这三门课程中，最多有多少名学生同时报名了其中的两门课程？

Pensamento

Calculating student overlaps

I’m figuring out the overlap of students in psychology, sociology, and marketing courses using fractions to avoid guessing. This makes me think that analyzing the data precisely is essential.

Clarifying enrollment limits: I’m making sure to check exact student counts across psychology, sociology, and marketing. I’m also examining the ambiguity between simultaneous enrollment in two courses and their combined total, aiming to manage class sizes effectively.

Breaking down intersections

I'm trying to figure out the maximum number of students taking exactly two out of three courses. This leads me to ponder set notation and intersections, ensuring clarity in the interpretation.

Maximizing student overlap

I'm gathering info on the highest number of students enrolled in exactly two out of three courses, considering intersections and the overall total of 30 students.

Mapping out enrollment

I’m noting that the sets total 30 students. To maximize overlap, students should ideally enroll in exactly one course each, with no additional students in overlapping courses.

Applying set theory

Using the principle of inclusion-exclusion, I’m noting that if there’s overlap among psychology, sociology, and marketing groups, the total number of distinct students must be adjusted to maintain the total count.

Pinpointing key intersections

I'm realizing that unclear details might lead to students being counted multiple times. This highlights the need for careful consideration of pairwise and triple intersections.

Clarifying ambiguity

I'm working through whether the question is about students in exactly two or at least two courses, given the total student count of 30 and possible overlap.

Maximizing triple intersection

I'm maximizing the number of students in all three sets to manage the total count and ensure other sets have the needed students by reducing the triple intersection.

Assessing course inclusion

I'm thinking through whether every student is enrolled in at least one course and why the sums of the course-specific fractions align with the total student count.

Mapping course splits

OK, let me see. Ensuring the total count matches the course student fractions is key. This might mean some students skipped courses entirely, clarifying the ambiguity about dual course registrations.

Setting intersections

Piecing together, I'm thinking about 8 students in all three courses, 4 in psychology alone, and 2 in marketing alone. This setup explains 16 students not in any course.

Rethinking pairwise intersections

I'm noting that by potentially eliminating the triple intersection, we can explore the principle of inclusion-exclusion to find the maximum number of students in at least two sets.

Simplifying intersections

I’m focusing on the principle of inclusion-exclusion to determine the maximum number of distinct students in at least two sets, considering their intersections without repeatedly counting triple intersections.

Mapping intersections

I’m thinking through the number of distinct students in at least two sets, factoring in pairwise intersections and triple intersections, to ensure they are counted accurately.

Adjusting my approach

Mapping out the number of distinct students in at least two sets, I’m focusing on refining pairwise intersections and fine-tuning the calculations to ensure accuracy.

Adjusting my approach

I’m verifying student counts and their distribution across different categories to ensure accuracy and maximize the number of students in at least two sets by tweaking the constraints.

Maximizing student overlap

OK, I'm focusing on minimizing students in only one course by setting p0, s0, and m0 to zero. This approach increases overlap among students in multiple courses.

Pinpointing constraints

I'm working through the inequalities to ensure x', y', z', and w stay nonnegative integers, which helps in determining the maximum number of students in at least two sets.

Mapping intersections

I’m piecing together how x', y', and z' adjust with different w values. This involves adding and subtracting constraints to figure out the maximum students in at least two sets.

Translating constraints

I’m figuring out the relationship between x' and w to ensure we accurately understand how student counts link to the maximum possible students in at least two sets.

Ensuring nonnegative integers

I’m working through the constraints for x', y', and z' to ensure they stay nonnegative integers. This involves evaluating constraints and verifying the maximum number of students in at least two sets.

Ensuring even parity

I’m gathering info on w needing to match the parity of 10, 14, and 6 for x', y', and z' to stay nonnegative integers and meet constraints.

Testing scenarios

I’m examining possible values for w to ensure x', y', and z' meet the constraints. The goal is to maximize students in at least two sets while keeping all values nonnegative.

Assessing outcomes

I’m looking at different w values to find the maximum possible students in at least two sets, ensuring constraints are met and all student counts stay nonnegative.

Thought about课程重叠人数，持续 1m 6s

Resposta

先给出题目中的信息：

全班总人数 $N=30$
报心理学（P）的人数： $\frac{2}{5} \times 30 = 12$
报社会学（S）的人数： $\frac{4}{15} \times 30 = 8$
报市场营销（M）的人数： $\frac{1}{3} \times 30 = 10$

记：

$p_0$ ：只报心理学的人数
$s_0$ ：只报社会学的人数
$m_0$ ：只报市场营销的人数
$x' = |P\cap S| - |P\cap S\cap M|$ ：只同时报心理学和社会学的人数（正好两门）
$y' = |P\cap M| - |P\cap S\cap M|$ ：只同时报心理学和市场营销的人数（正好两门）
$z' = |S\cap M| - |P\cap S\cap M|$ ：只同时报社会学和市场营销的人数（正好两门）
$w = |P\cap S\cap M|$ ：同时报三门的人数
$n_0$ ：三门都不报的人数

这样，每门课的人数可以写成：

\begin{cases} |P| = p_0 + x' + y' + w = 12, \\ |S| = s_0 + x' + z' + w = 8, \\ |M| = m_0 + y' + z' + w = 10. \end{cases}

全班总人数则是

p_0 + s_0 + m_0 + x' + y' + z' + w + n_0 = 30.

我们要“最大化同时报名了两门课程的人数”。这里的“同时报名了两门”通常指的是“至少两门”还是“正好两门”，要根据题意及常见做法来判断。

如果题目要求“至少两门”，那么计数就是 $(x'+y'+z') + w$ 。
如果题目要求“正好两门”，那么计数就是 $x'+y'+z'$ 。

从题干“最多有多少名学生同时报名了其中的两门课程”这一表述来看，更常见的理解是“正好报了两门”的人数; 但很多竞赛/练习题也会把它理解成“至少两门”。在此类题目中，最常见的做法是：

若想要让“报至少两门”的人数尽可能多，往往会把三门交集（ $w$ ）尽量做大；
若想要让“正好两门”的人数尽可能多，往往会让三门交集尽量为 0，以免“同报三门”的人也被分摊进各对交集，从而“抢占”了两门交集的人数。

下面演示两种思路，但会发现想要“报两门”尽量多(不管“至少两门”还是“正好两门”)，结果都会倾向于令三门交集 $w=0$ 。原因在于本题给定的三个课程人数之和刚好等于全班人数 $12+8+10=30$ ，这使得增加三门交集会“挤占”其他两门交集的总和。

1. 若题目要的是“至少两门”的最大人数

令

x' = |P\cap S| - w, \quad y' = |P\cap M| - w, \quad z' = |S\cap M| - w.

则“至少两门”的总人数是

x' + y' + z' + w.

为简化和最大化重报，先尝试让只报一门的人数都为 0，即 $p_0 = s_0 = m_0 = 0$ 。这样各科人数公式变为

\begin{cases} x' + y' + w = 12,\\ x' + z' + w = 8,\\ y' + z' + w = 10. \end{cases}

把三式相加得

2(x' + y' + z') + 3w = 12 + 8 + 10 = 30.

记

X = x'+y'+z'.

则有

2X + 3w = 30 \quad \Longrightarrow \quad X = 15 - \frac{3}{2}w.

又因为每个 $x',y',z'$ 都必须是非负整数，进而要求 $(10 - w), (14 - w), (6 - w)$ 都为非负的偶数（这是由分别求出 $x',y',z'$ 的表达式可知）。经过枚举可得使 $x',y',z'\ge 0$ 的最大可行 $w$ 值不会让总和 $x'+y'+z'+w$ 超过某一极值。最终发现：

当 $w=0$ 时，可令 $\begin{cases} x' = 5, \\ y' = 7, \\ z' = 3, \end{cases} \quad \text{则 } x'+y'+z' = 15, \quad \text{“至少两门”} = 15.$ 并且此时只要再放 $n_0 = 30 - 15 = 15$ 个学生到“三门都不报”即可满足总人数为 30。
如果让 $w>0$ ，则会牺牲 $x'+y'+z'$ 的和，导致最终 $(x'+y'+z') + w$ 也不会超过 15。

因此，当题意是“至少两门”时，可使最多 $15$ 人同时报了两门或以上。

2. 若题目要的是“正好两门”的最大人数

如果题目更严格地要求“正好两门”，那么人数就是

x' + y' + z'.

从上面的分析可见，为了让 $x'+y'+z'$ 最大，应取三门交集 $w=0$ 并令只报一门的人数都为 0，这样就得到

x' = 5,\quad y' = 7,\quad z' = 3,\quad \text{总和 } x'+y'+z' = 15.

这意味着有 15 人各自正好报了两门课，另外有 15 人没有报任何课（或者报一门都可以视需要再做调整，不过为了不违背三个学科各自总数的限制，最简单就是都不报）。

3. 小结

无论是“至少两门”还是“正好两门”，在本题给定数据（ $12+8+10=30$ ）的情形下，能同时报两门的那部分人数上限都是 15。区别在于：

“至少两门”的情形：那 15 人中不一定全部是“正好两门”，也可以包括“三门都报”。但经过计算，若想总数达 15，必须让三门交集为 0，并在两门交集上达最大。
“正好两门”的情形：同理可得最大值也是 15，且必须让三门交集为 0。

因此本题答案为：最多可以有 15 人同时报了其中的两门课程。