同时选课人数最大值

Creato il: 11 gennaio 2025

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Domanda

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？

Pensiero

Breaking down the numbers

I found that 12 students are in psychology, 8 in sociology, and 10 in marketing. The aim is to determine the maximum number of students enrolled in exactly two courses simultaneously.

Clarifying the question

I’m interpreting the question to mean determining the maximum number of students taking exactly two out of the three courses, likely through set theory concepts.

Working through set theory

I'm piecing together the number of students concurrently enrolled in exactly two of the three courses by exploring pairwise intersections and solving for students taking all three courses.

Mapping out student numbers

I'm focusing on the maximum number of students enrolled in exactly two courses by subtracting triple intersections from the sum of pairwise intersections.

Maximizing exact enrollments

I'm examining the constraints to maximize students enrolled in exactly two courses by minimizing the triple intersection and carefully distributing students among the pairwise intersections.

Evaluating feasibility

Check if having 8 students in psychology, 10 in sociology, and 8 in marketing, with no overlap, covers all 30 students. This approach seems a bit constraining.

Crunching numbers

OK, let me see. I’m working out the maximum number of students in exactly two courses. Taking a closer look at pairwise intersections helps avoid double counting.

Balancing intersections

I'm thinking through the formula to ensure a + b + c - 3x equals 30, keeping exact enrollments maximized while avoiding double counting of triple intersections.

Tallying constraints

I’m setting the lowest value for x to maximize students in exactly two courses, ensuring total distinct students remain 30 without going below zero for any single course.

Mapping feasible values

I’m figuring out the minimum x to boost the sum of students in exactly two courses while ensuring all constraints and total distinct students stay intact.

Piecing together constraints

I'm working through maximizing a + b + c by focusing on saturation of constraints like b + c ≤ 10 and a + c ≤ 8, while adhering to all other rules.

Balancing constraints

I’m ensuring the sum of a + b + c stays within limits by balancing constraints like a + b ≤ 12, b + c ≤ 10, and a + c ≤ 8.

Ensuring count integrity

I'm figuring out how to keep the total student count at 30 while maximizing students in exactly two courses, checking that all constraints align without overcounting.

Laying out the constraints

OK, I’m checking that each course enrollment is distinct. The goal is to hit the 30-student mark by tweaking the pairwise intersections to steer clear of double counting.

Ensuring student diversity

I’m piecing together how to maintain exactly 30 distinct students by carefully managing pairwise intersections. This keeps the number of students in exactly two courses maximized.

Clarifying overlap

I’m digging into the overlaps between course enrollments to ensure all 30 students are represented accurately, avoiding any double or triple counting of students.

Assessing intersections

I’m carefully analyzing the intersections of psychology, sociology, and marketing courses to ensure the total unique student count is 30, considering overlaps and avoiding any discrepancies.

Considering course alignment

I’m thinking through the course distribution to ensure it matches the 30 students criterion, factoring in overlaps and the potential for students being enrolled in more than one course.

Ensuring alignment

I’m working through how 30 students are spread over three courses, noting overlaps and possible multiple enrollments. This leads me to understand the total effectively.

Balancing intersections

I'm mapping out the total course enrollments, ensuring each student is enrolled in at least one course and balancing the intersections to prevent any discrepancies or leftover students.

Assessing feasibility

Mapping out if x can be 0 without conflicts, leading to no overlap and 30 distinct students, while ensuring the maximum number covers students in exactly two courses.

Happy to clarify

I’m piecing together that the total course enrollments add up to 30, reflecting individual course participation rates. This ensures progress in understanding unique student distribution.

Clarifying course overlap

I’m considering the maximum number of students enrolled in exactly two courses by minimizing those in single courses and any triple intersections. This seems the most straightforward approach.

Maximizing pairwise intersections

I’m thinking through how to boost the count of students in exactly two courses by simplifying and focusing on pairwise intersections, minimizing any overlap or triple intersections.

Evaluating feasibility

Setting psm=0, I'm working through equations to see if maximizing students in exactly two courses is feasible by evaluating constraints and simplifying intersections.

Evaluating constraints

Piecing together the variables, I’m pulling together constraints to see if maximizing students in exactly two courses within a 30-student class is doable.

Piecing together constraints

I’m figuring out how to keep students in exactly two courses maximized by carefully evaluating and balancing constraints, ensuring fairness and avoiding double counting.

Evaluating feasibility

I’m focusing on how feasible it is to allocate all 30 students to courses, considering overlap and their accurate representation in the class without any being outside the courses given.

Piecing together course enrollments

I'm thinking about the possibility of students being counted in multiple courses, conflicting with the total class size of 30 and course enrollment sums. This creates an interesting paradox.

Balancing student enrollments

I'm exploring the balance between distinct student counts and total enrollments, figuring out how to best distribute students across courses to capture the desired overlaps.

Maximizing course overlap

I'm pulling together the overlap of course enrollments to ensure the total remains 30, which reveals the maximum number of students in exactly two courses.

Maximizing student overlap

I'm mapping out how to maximize students in exactly two courses by analyzing total enrollments and individual course participation, ensuring the total hits 30 without exceeding enrollments.

Maximizing course enrollments

I’m optimizing student enrollments, ensuring the total is 30. By defining T1_P, T1_S, T1_M, T2_{PS}, T2_{PM}, T2_{SM}, and T3, I aim to maximize students in exactly two courses.

Maximizing students

I'm sorting out how to maximize T2, the count of students in exactly two courses, by tweaking the variables and ensuring all constraints are satisfied. Progress is steady and thoughtful.

Maximizing student distribution

I’m figuring out the best way to allocate students to maximize T2 (students in exactly two courses) and ensure the total counts and constraints are balanced. Progress is steady.

Maximizing constraints

OK, let me see. By setting T3 to 0, I’m piecing together constraints to ensure the total number of students in exactly two courses is maximized.

Coinciding variables

Interestingly enough, if T3 equals zero, T0 and T2 align. This setup maximizes students in exactly two courses while ensuring the total and constraints are satisfied.

Maximizing student participation

I’m digging into boosting students in exactly two courses by mapping out constraints and piecing together x, y, and z to hit the sweet spot. This ensures alignment with all rules.

Balancing constraints

I’m working through balancing constraints like x, y, and z to ensure all rules are aligned while maximizing students in exactly two courses. Progress is steady and thoughtful.

Balancing constraints

I’m thinking about balancing constraints to maximize students in exactly two courses. Progressing through various x, y, and z combinations, ensuring consistency with all rules.

Optimizing student enrollment

I'm mapping out student enrollments, aiming to maximize those in exactly two courses while ensuring a consistent total of 30 students across all scenarios. Progress is steady and strategic.

Thought about 课程报名重叠，持续 1m 44s

Risposta

这道题如果只看各科人数之和 $12 + 8 + 10 = 30$ ，似乎“刚好等于全班 30 人”，容易让人以为“所有同学都各选了一门课、且三门课之间无重叠”。但题目问的是“最多有多少名学生同时报了其中的两门课”，显然允许存在重叠（选多门课）的情形。

要点是：“人数分别是总人数的几分之几”指的是某门课的“选课人数”之和，并不要求这 30 个“选课名额”必须正好覆盖 30 个不同的学生。因为一个学生如果同时选了两门课，就会在“总选课人次”里被算两次；如果有人没选任何课，则在这三门课的人次统计里根本没出现。

1. 设定记号

设全班共有 30 人，分别可能选修心理学（P）、社会学（S）、市场营销（M），也可能一门都不选。为了方便，记：

$T_1$ ：只选了 1 门课的学生数。
$T_2$ ：同时选了 2 门课（恰好两门）的学生数。
$T_3$ ：同时选了 3 门课的学生数。
$T_0$ ：一门都不选的学生数。

这样，全班总人数的分解就是

T_0 \;+\; T_1 \;+\; T_2 \;+\; T_3 \;=\; 30.

而“总的选课人次”则是

1 \times T_1 \;+\; 2 \times T_2 \;+\; 3 \times T_3 \;=\; \underbrace{12 + 8 + 10}_{=30}.

题目给出的每门课的人数加起来恰好是 30，所以

T_1 + 2T_2 + 3T_3 \;=\; 30.

我们要找的是“ $T_2$ 的最大可能值”。

2. 先直觉判断：尽量多的学生同时选两门

如果想让“同时选两门课的人”尽可能多，那么有两个直觉：

尽量减少同时选三门课的人数（因为选三门并不会增加“正好选两门”的人数）。在极端情况下，可以先考虑 $T_3 = 0$ 。
只要总选课人次固定（=30），当有人一下子占了 3 人次，等于“用一个人贡献了 3 个选课名额”；不如让更多人各自贡献 2 个选课名额，这样可以在不变的“30 人次”之内多凑出更多的“选两门”。

因此先假设 $T_3=0$ ，看能否做出可行分配，并尽量让 $T_2$ 变大。

3. 建立方程并求解

在 $T_3 = 0$ 时，

T_1 \;+\; 2 T_2 = 30,

且全班总人数

T_0 \;+\; T_1 \;+\; T_2 \;=\; 30.

从第一式可得

T_1 = 30 - 2T_2.

把它代入总人数那式：

T_0 + (30 - 2T_2) + T_2 \;=\; 30 \;\;\Longrightarrow\;\; T_0 = T_2.

也就是说，在不出现三门同修的情况下，如果想让 $T_2$ 再多一些，就会相应地多出一些“什么都不选”的同学 $T_0$ 。只要最后不出现负数，就说明可行。

3.1 还要满足各科具体人数的限制

心理学（P）共 12 人次
社会学（S）共 8 人次
市场营销（M）共 10 人次

若记

$x =$ 只选 $\{P,S\}$ 的人数，
$y =$ 只选 $\{P,M\}$ 的人数，
$z =$ 只选 $\{S,M\}$ 的人数，

那么

T_2 \;=\; x + y + z.

同时各门课的人数还要满足

选了 P 的总人数 $= 12$ 。它来自 “只选 P” + “PS” + “PM” + “三门”，但我们假设三门为 0，则 $(\text{只选P}) + x + y = 12.$
选了 S 的总人数 $= 8$ 。它来自 “只选 S” + “PS” + “SM”，无三门则 $(\text{只选S}) + x + z = 8.$
选了 M 的总人数 $= 10$ 。它来自 “只选 M” + “PM” + “SM”，无三门则 $(\text{只选M}) + y + z = 10.$

此外，总选课人次 $= 30$ 也等价于

(\text{只选P}) + (\text{只选S}) + (\text{只选M}) \;+\; 2(x+y+z) \;=\; 30.

但这与上面三条相加本质是同一个限制。

化简做“只选某一门”的人数

只选 P 的人数 $= 12 - x - y$ 。
只选 S 的人数 $= 8 - x - z$ 。
只选 M 的人数 $= 10 - y - z$ 。

只要它们都不为负，就说明分配可行。

要最大化 $x+y+z$

同时还要满足：

\begin{cases} x+y \;\le\; 12,\\ x+z \;\le\; 8,\\ y+z \;\le\; 10,\\ x, y, z \;\ge\; 0. \end{cases}

令 $S = x+y+z$ ，我们就想让 $S$ 尽量大。

一个常见做法是“尽量让不等式都顶住上限”来试探。试着令

\begin{cases} x + z = 8,\\ y + z = 10, \end{cases}

则可得

y - x = 2.

若再让 $x + y = 12$ 同时顶住，则

x + (x+2) = 12 \;\;\Longrightarrow\;\; 2x=10\;\;\Longrightarrow\;\; x=5,\quad y=7.

再由 $x+z=8$ 得 $z=3$ 。检查是否都满足：

$x+y = 5+7=12$ （不超 12）
$x+z = 5+3=8$ （不超 8）
$y+z = 7+3=10$ （不超 10）

都刚好“等号”，无冲突。于是

S = x+y+z = 5 + 7 + 3 = 15.

对应的“只选 P/S/M”人数：

只选 P = $12 - 5 - 7 = 0$ 。
只选 S = $8 - 5 - 3 = 0$ 。
只选 M = $10 - 7 - 3 = 0$ 。

都不为负，刚好全是零。换言之，所有选课的同学都恰好是“二门并修”；这样一来他们贡献的选课总人次是

T_2 \times 2 = 15 \times 2 = 30

正好覆盖了那 30 个“选课名额”。这时候 $T_1=0$ ，再加上班里还有 $T_0$ 人完全不选课。

因为全班共 30 人，而现在只有 15 人在选课（这 15 人各选了 2 门），因此还有

T_0 = 30 - 15 = 15

人什么都没选。这样总人数也对上了。

选课的 15 人：每人同时选了 2 门课

不选课的 15 人：一门都没报

一共 30 人

由此可见，在不违背各科人数之和为 12、8、10 的前提下，最多可以让 15 个同学“同时报了两门课”。

4. 答案

\boxed{15}

也就是说，这三门课里，“同时选了其中两门”的学生，最多可以达到 15 人。

补充：为什么会出现“一半人不选课”？

很多同学可能会疑惑：题目不是说“心理学有 12 人，社会学 8 人，市场营销 10 人”吗？怎么还会有大批同学没选任何课？

这是因为“选课总人次 = 30” 并不必然等于“有 30 个不同学生都选了至少一门”。一个人选了 2 门，就在总人次里被计算了 2 次；如果要“尽量多的人同时选两门”，就势必要有一些人“腾出来”不选任何课(这样才不至于把某一门课的人数冲得太高，无法满足另一门的限制)。

这类计数/集合问题的常见陷阱就在于：“某几组统计数之和 = 总人数”，并不自动代表“全班所有人正好一人一份地分摊到这些组里，没有重叠也没有空缺”。题目正是要我们去利用重叠(多门并修)的可能性，来最大化某个目标。

结论就是：最多可以有 15 个同学同时选了两门课。

Domanda

Pensiero

Risposta

1. 设定记号

2. 先直觉判断：尽量多的学生同时选两门

3. 建立方程并求解

3.1 还要满足各科具体人数的限制

化简做“只选某一门”的人数

要最大化 x+y+zx+y+zx+y+z

4. 答案

补充：为什么会出现“一半人不选课”？

Condividi questa Q&A

要最大化 $x+y+z$