Forming a Committee with Specific Teacher-Student Ratios

In this article, we explore the detailed process of forming a committee of 5 members from a group of 11 individuals, consisting of 4 teachers and 7 students. The committee must include at least 3 teachers, and we will delve into the various methods to achieve this specific ratio. Additionally, we will compare different approaches for forming the committee to determine the number of possible combinations in each scenario.

Introduction to the Problem

The problem in question is: How many ways can a committee of 5 be formed from a group of 11 people, consisting of 4 teachers and 7 students, if the committee must include at least 3 teachers? To solve this, we will break down the scenarios into distinct parts and use combinations (ashorthand for the combination notation, ( binom{n}{r} ) which represents the number of ways to choose ( r ) items from ( n ) items without regard to the order) to find the number of possible committees.

Solution and Analysis

Total Ways to Form Committees

There are two primary approaches to solve this problem: splitting into cases and direct counting. We will elaborate on both methods.

Case 1: Exactly 2 Teachers

If the committee includes exactly 2 teachers, then it must include exactly 3 students. The number of ways to choose 2 teachers from 4 and 3 students from 7 is calculated as follows:

4C2 * 7C3 6 * 35 210 ways

Case 2: At Least 3 Teachers

To include at least 3 teachers, the committee can either have 3 or 4 teachers. We will consider each possibility: 3 Teachers: 4C3 ways to choose 3 teachers and 7C2 ways to choose 2 students. 4 Teachers: 4C4 ways to choose 4 teachers and 7C1 ways to choose 1 student. The total number of ways for the committee to include at least 3 teachers is therefore:

4C3 * 7C2 4C4 * 7C1 4 * 21 1 * 7 84 7 91 ways

Direct Counting Approach

Alternatively, we can directly count the number of possible committees for each case:

Exactly 2 Teachers: This scenario necessitates 3 students, resulting in ( binom{4}{2} times binom{7}{3} ) combinations. Exactly 3 Teachers: Here, we have ( binom{4}{3} times binom{7}{2} ) combinations. Exactly 4 Teachers: This requires ( binom{4}{4} times binom{7}{1} ) combinations. The total number of combinations for a committee with at least 3 teachers is:

4C3 * 7C2 4C4 * 7C1 91 ways

Conclusion

Through the detailed exploration of both-case analysis and direct counting, we determine that the number of possible ways to form a committee of 5 with at least 3 teachers from a group of 11 people, consisting of 4 teachers and 7 students, is 91. This approach helps in understanding the flexibility and multiplicity of practical problem-solving in combinatorics and committee formation.