Combination and Permutation: Calculating Different Committee Compositions

In this article, we will explore how to determine the number of different committees that can be formed from a given group, using the principles of combination and permutation. Specifically, we will focus on a scenario where a committee of five members is to be formed from six females and four males, with the requirement that the committee consists of three females and two males. We will cover both the brute force solution and the application of permutation and combination theory.

Brute Force Solution Using J Programming Language

Consider the problem where a committee of five members is to be formed from 6 females and 4 males, with the requirement that the committee consists of 3 females and 2 males. We can solve this using the J programming language, a high-level, general-purpose programming language. Here is how we can approach it:

The code snippet provided uses the assignment operator to define a simple boolean array. The result, 120, represents the number of different committees that can be formed.

Code:

/m.2/

Output:

120

The answer is 120 different committees.

Listing All Committee Possibilities

If the committee manager wants to see all the committee possibilities, we can list them. Assuming the females are numbered 1 to 6 and the males are numbered 11 to 14, we can systematically list all the combinations. However, for brevity, we will not list all combinations here but explain the process.

Let's break this down into the steps of forming two independent committees: one for the females and one for the males.

Applying Combination and Permutation Theory

1. **Females Selection:** We need to choose 3 females out of 6. This can be calculated using the combination formula:

NCr frac{n!}{r!(n - r)!}

Here, n 6 (total number of females) and r 3 (number of females to be chosen).

Thus, the number of ways to choose 3 females out of 6 is:

C(6, 3) frac{6!}{3!(6-3)!} frac{6!}{3!3!} 20

2. **Males Selection: ** We need to choose 2 males out of 4. Using the combination formula again:

C(4, 2) frac{4!}{2!(4-2)!} frac{4!}{2!2!} 6

Since these are independent formations, the total number of ways to form the required committee is the product of these two numbers:

20 times 6 120

Therefore, the answer is 120 different committees.

Conclusion

By applying the principles of combination and permutation, we can determine the number of different committees that can be formed. Whether using a brute force approach with a programming language or applying mathematical formulas, the result remains the same. This method provides a systematic and efficient way to solve such problems, ensuring accuracy and reliability in committee formation.