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两门课程最大人数

बनाया गया: 24 जनवरी 2025

का उपयोग करके उत्तर दिया गया O1 द्वारा Chat01

प्रश्न

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25、415和13\frac{2}{5} 、 \frac{4}{15} 和 \frac{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？

विचार

Calculating distribution

I'm focusing on the number of students in each course: psychology has 12, sociology has 8, and marketing has 10. Next, I'll figure out the overlap to determine the maximum possible students enrolled in exactly two courses.

Interpreting ambiguity

OK, let me see. The question is curious about the maximum number of students enrolled in any two courses, considering the ambiguity and potential enrollment in all three.

Applying the approach

I'm using the principle of inclusion-exclusion to determine the maximum number of students in at least two courses, which includes calculating sums and overlaps for the given course enrollments.

Prioritizing unique overlaps

OK, let me see. I’m focusing on reducing triple overlaps to maximize the count of distinct students across multiple courses, aiming for the highest possible number in at least two sets.

Mapping student sets

I’m putting together the numbers of students in psychology, sociology, and marketing, considering overlaps in pairs and all three courses to maximize the count of students in two or more sets.

Understanding course overlaps

Considering the ambiguity, I’m leaning towards "exactly two courses" to pinpoint students in psychology, sociology, and marketing. This involves interpreting language precision and typical puzzle structures.

Enhancing course engagement

I’m working through maximizing students enrolled in exactly two courses, like psychology, sociology, or marketing, by balancing partial overlaps and ensuring all students are allocated uniquely.

Tweaking the distribution

I’m aiming to minimize overlap in all three classes while balancing students enrolled in just one or two classes, ensuring the total number of students aligns with the 30 seats available.

Crunching the numbers

OK, I’m working out the count of students in exactly two classes, factoring in the total of 30 seats, individual class enrollments, and overlaps.

Piecing together constraints

Reassessing the principles, I’m ensuring they align with given constraints and avoiding contradictions. This involves subtracting membership counts for each set and reassessing intersections.

Crunching intersection sums

Subtracting equation (2') from (1'), I figure out that x + 2y must be zero. This means both x and y are zero, confirming no double intersections.

Confirming feasibility

OK, so it’s clear that assigning each student to one set, with the total cardinality equaling the population, works. This approach avoids any forced overlaps and confirms the feasibility.

Evaluating options

I’m digging into ways to maximize students in exactly two sets, while balancing total seats and student numbers, to ensure optimal allocation. This makes me think about balance and efficiency.

Rearranging for better results

OK, let’s see. By shuffling student assignments, we can boost those in exactly two courses while keeping the original set sizes intact. This uses our available seats to full advantage.

Maximizing students

With 30 students and 30 seats, each student must be in one set. This ensures the desired allocation, maximizing efficiency.

Maximizing student impact

I'm working on efficiently aligning students with set sizes, aiming to maximize the number of students in exactly two courses without exceeding the total of 30 seats.

Balancing student assignments

I’m working through distributing 29 students among 28 seats, considering each can take at most 3 seats, ensuring every student is assigned at least one seat.

Identifying potential gaps

OK, let's think this through: it's possible that some students might not be enrolled in the mentioned courses, given there's no implication that all 30 students are pursuing them.

Determining course participation

OK, so this means not all 30 students are in psychology, sociology, or marketing, and the actual count is likely below 30.

Maximizing student allocation

OK, so I’m thinking about how many students can be in exactly two courses, considering overlaps and ensuring the total doesn’t exceed 30. This involves calculating and adjusting for overlaps.

Maximizing exactly two enrollments

I'm noting the set sizes and individual enrollments to ensure the sum of students in exactly two sets is maximized, without exceeding the total of 30 students.

Allocating precisely

OK, let me see. I’m piecing together the constraints to ensure the focus remains on maximizing students involved in exactly two courses, with careful attention to overlaps and specific exclusions.

Optimizing constraints

OK, let me see. I'm figuring out how to maximize students in exactly two courses by tweaking constraints and using variables like a, b, c, d, e, f, and h.

Pulling together expressions

I’m combining terms to simplify the constraints, aiming to maximize students in exactly two courses while adhering to the total of 30 students, considering overlaps and exclusions.

Balancing the equation

I’m thinking h should match the sum of d, e, and f. This ensures space for students not taking any courses, keeping the total count consistent.

Piecing together options

I'm thinking through how to maximize students in exactly two courses, ensuring each term's constraints are satisfied for d, e, f, and their sums.

Maximizing course engagement

OK, let me see. I’m piecing together variables to maximize students in exactly two courses, ensuring the total is 30 and all constraints are satisfied. This requires careful balancing.

Checking constraints

I’m verifying that d,e,f≥0d, e, f \geq 0 with p,q,rp, q, r set to their maximum values, ensuring all constraints are satisfied. The goal is to maximize XX.

Confirming feasibility

I’m verifying that the solution is feasible, with students in exactly two courses, no courses, or specific courses, and all constraints are satisfied. The solution aligns with the student count.

Thought about students in multiple courses，持续 3m 20s

उत्तर

分析思路如下：

  1. 已知信息：

    • 全班共有 3030 名学生；
    • 选修心理学（P）的有 25×30=12\frac{2}{5}\times 30 = 12 人；
    • 选修社会学（S）的有 415×30=8\frac{4}{15}\times 30 = 8 人；
    • 选修市场营销（M）的有 13×30=10\frac{1}{3}\times 30 = 10 人。

    我们记

    ∣P∣=12,∣S∣=8,∣M∣=10.|P| = 12,\quad |S| = 8,\quad |M| = 10.

  2. 问题理解：
    题目“最多有多少名学生同时报名了其中的两门课程”通常理解为：在满足三个科目人数各自为 12、8、10 的前提下，最多能有多少学生“恰好”在两个科目里（不在三个科目里），即我们要尽量让“只选两门”的人数最大。

  3. 集合与变量设定：
    令

    • a=a = 只选心理学（P）的人数；
    • b=b = 只选社会学（S）的人数；
    • c=c = 只选市场营销（M）的人数；
    • d=d = 只选了心理学和社会学（P∩S，但不选M）的人数；
    • e=e = 只选了心理学和市场营销（P∩M，但不选S）的人数；
    • f=f = 只选了社会学和市场营销（S∩M，但不选P）的人数；
    • g=g = 同时选了三门课（P∩S∩M）的人数；
    • h=h = 三门都没选的人数（可能存在，有些学生什么课都没选）。

    则有总人数约束：

    a+b+c+d+e+f+g+h=30.a + b + c + d + e + f + g + h = 30.

    三个学科人数分别给出以下约束：

    {∣P∣=a+d+e+g=12,∣S∣=b+d+f+g=8,∣M∣=c+e+f+g=10.\begin{cases} |P| = a + d + e + g = 12,\\ |S| = b + d + f + g = 8, \\ |M| = c + e + f + g = 10. \end{cases}

    我们想要最大化的是只选了两门课的人数之和

    d+e+f.d + e + f.

    为了尽量“腾出”人数让更多的人落在“两门”里，往往会让三门都选的 gg 变为 0（因为三门都选并不能增加“恰好两门”的人数），并尽可能减少只选一门的人数 a,b,ca,b,c。

  4. 在 g=0g=0 的情况下进行推导：
    如果令 g=0g=0，则

    ∣P∣=a+d+e=12,∣S∣=b+d+f=8,∣M∣=c+e+f=10.|P|=a+d+e=12,\quad |S|=b+d+f=8,\quad |M|=c+e+f=10.

    同时，有

    a+b+c+d+e+f+h=30.a + b + c + d + e + f + h = 30.

    我们把只选一门的人数 a,b,ca,b,c 用 d,e,fd,e,f 表达：

    {a=12−d−e,b=8−d−f,c=10−e−f.\begin{cases} a = 12 - d - e, \\ b = 8 - d - f, \\ c = 10 - e - f. \end{cases}

    把它们带回到总人数公式：

    (12−d−e)+(8−d−f)+(10−e−f)  +  d+e+f  +  h  =  30.(12 - d - e) + (8 - d - f) + (10 - e - f) \;+\; d + e + f \;+\; h \;=\; 30.

    化简可得：

    12+8+10=30,消去后得到 h=d+e+f.12+8+10 = 30,\quad \text{消去后得到 } h = d + e + f.

    也就是说，只要选了两门课的人数（d+e+fd+e+f）是多少，便有同样数量的人 hh 可以什么课都不选，从而不影响三门课的总人数要求。

  5. 进一步的约束：

    • a=12−d−e≥0  ⇒  d+e≤12.a=12-d-e \ge 0 \;\Rightarrow\; d+e \le 12.
    • b=8−d−f≥0  ⇒  d+f≤8.b=8 -d-f \ge 0 \;\Rightarrow\; d+f \le 8.
    • c=10−e−f≥0  ⇒  e+f≤10.c=10-e-f \ge 0 \;\Rightarrow\; e+f \le 10.
    • 以及 d,e,f≥0d,e,f \ge 0。

    我们要最大化

    X=d+e+fX = d+ e+ f

    在满足

    d+e≤12,d+f≤8,e+f≤10d+e \le 12,\quad d+f \le 8,\quad e+f \le 10

    的前提下，让 XX 尽量大。

  6. 寻找可行最大值：
    直观做法是尝试让

    d+e=12,d+f=8,e+f=10d+e = 12,\quad d+f = 8,\quad e+f = 10

    同时成立。如果它们都能同时满足非负性，就能把 d,e,fd,e,f “推”到可能的最大值。于是：

    d=(d+e)+(d+f)−(e+f)2=12+8−102=102=5,d = \frac{(d+e) + (d+f) - (e+f)}{2} = \frac{12 + 8 - 10}{2} = \frac{10}{2} = 5, e=(d+e)+(e+f)−(d+f)2=12+10−82=142=7,e = \frac{(d+e) + (e+f) - (d+f)}{2} = \frac{12 + 10 - 8}{2} = \frac{14}{2} = 7, f=(d+f)+(e+f)−(d+e)2=8+10−122=62=3.f = \frac{(d+f) + (e+f) - (d+e)}{2} = \frac{8 + 10 - 12}{2} = \frac{6}{2} = 3.

    这样得到

    d=5,e=7,f=3,d=5,\quad e=7,\quad f=3,\quad

    对应的

    d+e+f=5+7+3=15.d+e+f = 5 + 7 + 3 = 15.

    检查是否一切非负：

    a=12−d−e=12−5−7=0,b=8−d−f=8−5−3=0,c=10−e−f=10−7−3=0,a = 12 - d - e = 12 -5 -7=0, \quad b = 8 - d - f = 8 -5 -3=0, \quad c = 10 - e - f = 10 -7 -3=0,

    都没有出现负数，说明这是可行解。再看剩余学生人数

    h=d+e+f=15,h = d+e+f = 15,

    表示有 15 个人“两门都不选”（即选 0 门）。

    在该分配下：

    • PP 中共有 d+e=5+7=12d+e = 5+7=12 人；
    • SS 中共有 d+f=5+3=8d+f = 5+3=8 人；
    • MM 中共有 e+f=7+3=10e+f = 7+3=10 人；
    • 且没有人只选一门或三门，最大化了“恰好选两门”的人数。

  7. 结论：
    满足题目条件（各科人数分别为 12、8、10）时，“最多有多少名学生同时报名了其中的两门课程？”的答案是

    15.\boxed{15}.

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