Problem 83 - Algebra for JEE ADVANCED

Question

The number of ways in which we can distribute \(m n\) students equally among \(m\) sections is given by a. \(\frac{(m n) !}{n !}\) b. \(\frac{(m n) !}{(n !)^{m}}\) c. \(\frac{(m n) !}{m ! n !}\) d. \((m n)^{m}\)

Step-by-Step Solution

Answer

The number of ways is \(\frac{(m n)!}{(n!)^m}\), which is option b.

Step 1: Understand the Problem

We have \(m n\) students and we need to distribute them equally among \(m\) sections. This implies each section will contain \(n\) students, since the total number of students is \(m \times n\). The task is to find the number of ways to make this distribution.

Step 2: Choose Students for the First Section

Out of \(m n\) total students, we need to select \(n\) students for the first section. The number of ways to choose \(n\) students from \(m n\) students is given by the binomial coefficient: \[ \binom{m n}{n} \] This is calculated as \( \frac{(m n)!}{n! (m n - n)!} \).

Step 3: Choose Students for the Remaining Sections

After choosing \(n\) students for the first section, we have \(m n - n\) students left to distribute. For the second section, choose another \(n\) students from the remaining \(m n - n\) students, which is \[ \binom{m n-n}{n} = \frac{(m n-n)!}{n! (m n - 2n)!} \].Continue this process for all \(m\) sections, until you're left with \(n\) students for the last section. At each step, the pattern repeats.

Step 4: Combine All Choices

To get the total number of ways to distribute the students, multiply the choices for each section. This results in: \[ \frac{(m n)!}{n! (m n - n)!} \cdot \frac{(m n-n)!}{n! (m n - 2n)!} \cdot \ldots \cdot \frac{n!}{n! 0!} \] By simplifying this, we see that most factors cancel out. The result is: \[ \frac{(m n)!}{(n!)^m} \] Therefore, the correct answer is option b.

Key Concepts

Binomial Coefficient

In combinatorial mathematics, the binomial coefficient is a fundamental tool used for counting combinations. It helps us determine how many ways we can choose a specific number of elements (k) from a larger set (n) without considering the order of selection.

The formula for the binomial coefficient is given by:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

Here, \(n!\) represents the factorial of \(n\), which is the product of all positive integers up to \(n\).

Factorials allow us to account for all potential arrangements of a set. This tool is crucial for various problems in combinatorial analysis, including those involving distributions, selections, and arrangements of objects. Understanding binomial coefficients can greatly simplify complex problems by breaking them down into manageable parts.

Permutation and Combination

Permutations and combinations are key concepts in combinatorics that allow us to explore different ways of arranging or choosing elements from a set. These two concepts apply based on whether the order of selection is important.

• Permutation: When order matters, a permutation is used. It refers to the arrangement of \(n\) elements taken \(r\) at a time. The formula is: \[ P(n, r) = \frac{n!}{(n-r)!} \]
• Combination: Unlike permutations, combinations disregard order. It's simply about selecting \(r\) elements from a total of \(n\). This is where the binomial coefficient comes into play: \[ C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

In our exercise, the appropriate concept is combinations, since the order in which students are placed into sections doesn't matter. We're interested in the number of ways to choose, not arrange, students for each section from the total pool.

Combinatorial Analysis

Combinatorial analysis is a branch of mathematics focused on counting, arranging, and grouping objects. It is essential in fields such as computer science, cryptography, and even biology for resource optimization and problem-solving.

It comprises techniques like permutations, combinations, and binomial coefficients to systematically count large sets, groups, or arrangements.

• It extends beyond simple problems and includes tasks like finding the number of ways to distribute objects into distinct groups, akin to our exercise of distributing students.
• In such distribution tasks, understanding the principle of multiplication is crucial, where the product of choices at each stage gives the total number of distributions.

The exercise at hand uses these principles to determine how to equally distribute students across sections, ensuring every section is full and the count is exhaustive. With these tools in your mathematical toolkit, solving complex organizational problems becomes less daunting.