Committee Formation: Combinatorial Mathematics in Group Selection

Introduction

Combining different subsets to form a desired group is a common problem in combinatorial mathematics and has numerous applications in various fields. This article explores the process of forming a committee of 4 boys and 4 girls from 4 classes, where each class contains 2 boys and 2 girls. We will delve into the mathematical principles and techniques to solve such problems, providing a detailed step-by-step analysis for clarity and comprehensiveness.

Selection of Boys and Girls

The problem at hand involves selecting a committee of 4 boys and 4 girls, with the requirement that 2 members must be chosen from each of the 4 available classes.

Method 1: Sequential Selection

To choose the boys, we first select 2 boys from 4, which can be done in 4C2 6 ways. Similarly, to choose 2 girls from 4, we can also do it in 4C2 6 ways. Since the choice of boys is independent of the choice of girls, we multiply the two results to get the total number of ways to form the committee:

6 x 6 36 ways.

Method 2: Direct Combinatorial Calculation

We can also approach the problem by directly calculating the combinations. For selecting 2 boys out of 4, we use the combination formula:

(4C2) 4! / (2! 2!) 6

Similarly, for selecting 2 girls out of 4:

(4C2) 4! / (2! 2!) 6

Combining the two, we get:

6 x 6 36 ways.

Method 3: Combination of Independent Events

We can also consider the selection process as a combination of independent events. First, select 2 boys out of 4, which can be done in 4C2 6 ways. Then, select 2 girls out of 4, also in 4C2 6 ways. Again, the total number of ways to form the committee is:

6 x 6 36 ways.

Alternative Problem Scenarios

Let's now consider a slightly more complex scenario involving a combination of selecting boys and girls from different classes. We'll analyze the following cases:

1. One Boy and One Girl from Each Class

For each class, there are 2C1 2 ways to choose one boy and one girl. Since we have 4 classes, the total number of ways to choose one boy and one girl from each class is:

2^4 16 ways.

Note: The original 256 ways mentioned in the example might have considered multiple permutations of choosing multiple boys or girls, which we do not consider here.

2. Two Boys or Two Girls from One Class and One Boy and One Girl from the Other Classes

For each class, choosing 2 boys or 2 girls can be done in 2C2 1 way. Choosing one boy and one girl from the other 3 classes can be done in:

(2C1)^3 2^3 8 ways.

Since there are 4 classes, we have:

4 x 1 x 8 32 ways.

Note: The original 1024 ways mentioned in the example might have overcounted the permutations.

3. Two Boys or Two Girls from Two Classes and One Boy and One Girl from the Other Classes

Choosing 2 boys or 2 girls from 2 classes can be done in:

(2C2)^2 1^2 1 way.

Choosing one boy and one girl from the remaining 2 classes can be done in:

(2C1)^2 2^2 4 ways.

Since we can choose 2 classes out of 4 in:

(4C2) 4! / (2! 2!) 6 ways,

the total number of ways is:

6 x 1 x 4 24 ways.

Note: The original 1586 ways mentioned in the example might have overcounted the permutations.

Summary and Conclusion

In conclusion, the problem of forming a committee of 4 boys and 4 girls from 4 classes, where each class has 2 boys and 2 girls, can be solved using various combinatorial methods. The total number of ways to form such a committee is 36, considering the independent selection of boys and girls from each class. Additionally, for more complex scenarios, we can extend our analysis to consider other combinations of selecting boys and girls from different classes, as demonstrated in the alternative problem scenarios.

Through this exploration, we have seen how combinatorial mathematics can be applied to solve real-world problems such as committee formation, highlighting the importance of systematic and logical approaches in problem-solving.