Question

How many ways can a class of 25 students be assigned to three different lab sections if each lab section has at least 5 students?

Short answer

There are 66 ways to assign the 25 students to the lab sections.

Step-by-step solution

Understand the Constraints

We must distribute 25 students into three lab sections, each with at least 5 students. This means each section will have between 5 and 15 students, depending on how the remaining students are distributed.

Calculate Initial Distribution

The minimum requirement is 5 students per section. Assign the first 15 students (5 per section): \((5, 5, 5)\). This leaves us with 10 students left to distribute freely among the three sections, so no section has fewer than 5 students.

Distribute Remaining Students

We need to find the number of ways to distribute 10 remaining students into 3 sections. This is a classic "stars and bars" problem where we find the number of ways to partition 10 identical items (students) into 3 distinct groups (sections).

Apply Stars and Bars Formula

Using the stars and bars formula, the number of ways to distribute \(n\) identical items into \(k\) groups is \(\binom{n+k-1}{k-1}\). Here, \(n = 10\) and \(k = 3\). So, the calculation is \(\binom{10 + 3 - 1}{3 - 1} = \binom{12}{2}\).

Solve the Binomial Coefficient

Calculate \(\binom{12}{2}\), which gives us the number of ways to distribute the students as \(\frac{12 \times 11}{2 \times 1} = 66\). Therefore, there are 66 ways to assign the students.

Key concepts

Stars and Bars Method

The Stars and Bars method is a popular combinatorial technique used to solve problems related to distributing identical items into distinct groups. Imagine you have a total number of items (stars) and you need to divide them into several groups using dividers (bars). Each division suggests how many items go into each group.

In this exercise, we have 10 students (our stars) that need to be distributed across 3 lab sections (our groups). To determine the number of ways to distribute these stars, we use the formula involving bars, which are the dividers between the groups.

• We begin by lining up all the stars in a row.
• Then, we place the bars to create groups, while ensuring each group has at least its minimum requirement already fulfilled (in this case, 5 students per group before the distribution).
• The number of such setups can be calculated using the Stars and Bars formula, which is \(\binom{n+k-1}{k-1}\).

This method efficiently handles the task of partitioning identical items, leading us to the correct number of distributions.

Binomial Coefficient

The Binomial Coefficient is a fundamental component in combinatorics, representing the number of ways to choose a subset of items from a larger set. It is denoted as \(\binom{n}{k}\), where \(n\) is the total number of items, and \(k\) is the number of items to choose.

To compute the binomial coefficient, use the formula: \[\binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \(n!\) (n factorial) is the product of all positive integers up to \(n\).

In our exercise, the binomial coefficient appears when applying the Stars and Bars method. We are essentially calculating how many ways we can "insert" two bars in between a sequence of 12 items, representing the 10 students plus the additional two group dividers. This is calculated as \(\binom{12}{2}\), resulting in 66 possible distributions. This highlights how the Binomial Coefficient can help us deduce combinatorial distributions effectively using its simple yet powerful formula.

Partitioning Problems

Partitioning problems in combinatorics involve dividing a set or a number into parts that satisfy certain conditions. These problems often appear in different contexts, such as distributing resources, seating arrangements, or group assignments.

Our exercise is a typical partitioning problem. We aim to partition a class of 25 students into three lab sections, each requiring at least 5 students. An efficient technique to handle such partitioning is leveraging combinatorial methods like Stars and Bars combined with Binomial Coefficients.

• We first fulfill the constraint of having at least 5 students in each section, leaving us with 10 students to freely distribute.
• The task is then reduced to partitioning these remaining students among the sections, which is then resolved using the combinatorial tools mentioned before.
• This systematic approach ensures that all constraints are met, and all possible distributions are considered.

By understanding and applying these methods, partitioning problems that might initially seem challenging become manageable and straightforward to solve.