A School Needs To Send 5 Students As Representatives; It Was Decided That Two Students Should Be From 10th Grade And 3 From 11th Grade. In 10th Grade, There Are 5 Qualified Students, And In 11th Grade, There Are 4. How Many Different Groups Can Be Formed?

Jun 18, 2025 by ADMIN 256 views

Forming Student Representative Groups: A Combinatorial Challenge

Introduction

In many educational institutions, the selection of student representatives is a crucial process, often involving specific criteria to ensure a diverse and capable group. This article delves into a scenario where a school aims to form a student delegation comprising representatives from different grade levels. The problem presented involves a combinatorial challenge: determining the number of distinct groups that can be formed given a set of eligible students from each grade. Specifically, the school needs to send five students as representatives, with the stipulation that two students must be from the 10th grade and three from the 11th grade. There are five qualified students in the 10th grade and four in the 11th grade. Our task is to calculate the number of different groups that can be formed under these conditions.

This problem falls under the domain of combinatorics, a branch of mathematics that deals with counting, arrangement, and combination of objects. The key concept here is combinations, which involves selecting items from a set without regard to the order of selection. This is particularly relevant because the order in which the students are chosen does not affect their role as representatives. We will utilize the combination formula to determine the number of ways to choose students from each grade level and then combine these results to find the total number of possible groups. Understanding and solving such problems is essential in various real-world scenarios, from forming committees to selecting teams, highlighting the practical application of combinatorial principles.

The process of solving this problem involves several steps. First, we need to determine the number of ways to select two students from the 10th grade out of the five qualified students. This is a combination problem, and we will use the combination formula, denoted as C(n, k) or "n choose k", where n is the total number of items, and k is the number of items to choose. The formula is given by C(n, k) = n! / (k!(n-k)!), where "!" denotes the factorial function. Applying this formula will give us the number of ways to form the 10th-grade contingent. Next, we will repeat this process for the 11th grade, calculating the number of ways to select three students from the four qualified students. Again, we will use the combination formula, this time with n = 4 and k = 3. Finally, to find the total number of different groups that can be formed, we will multiply the number of ways to choose the 10th-grade students by the number of ways to choose the 11th-grade students. This is because each combination of 10th-grade students can be paired with each combination of 11th-grade students, giving us the total number of possible groups. By systematically breaking down the problem into smaller parts and applying the appropriate combinatorial principles, we can arrive at the solution and understand the underlying mathematical concepts involved.

Determining the Number of Ways to Select 10th Grade Representatives

The first step in solving this combinatorial problem is to determine the number of ways to select two students from the 10th grade. As mentioned earlier, there are five qualified students in the 10th grade, and we need to choose two of them to be part of the representative group. This is a classic combination problem, where the order of selection does not matter. The combination formula, C(n, k) = n! / (k!(n-k)!), is the appropriate tool to calculate this. In this case, n is the total number of students in the 10th grade (5), and k is the number of students we need to select (2).

Applying the formula, we have C(5, 2) = 5! / (2!(5-2)!). Let's break this down step by step. The factorial of a number is the product of all positive integers up to that number. So, 5! (5 factorial) is 5 × 4 × 3 × 2 × 1 = 120. Similarly, 2! (2 factorial) is 2 × 1 = 2, and 3! (3 factorial) is 3 × 2 × 1 = 6. Substituting these values into the combination formula, we get C(5, 2) = 120 / (2 × 6) = 120 / 12 = 10. This result indicates that there are 10 different ways to choose two students from the five qualified students in the 10th grade.

Each of these 10 ways represents a unique pair of students who can represent the 10th grade in the student delegation. For example, if we label the students as A, B, C, D, and E, the possible pairs are AB, AC, AD, AE, BC, BD, BE, CD, CE, and DE. Each of these pairs is distinct, and no other combination is possible given the constraint of choosing two students from five. This calculation is crucial because it forms one part of the overall solution. To find the total number of possible groups, we need to combine this result with the number of ways to select students from the 11th grade. Understanding the combination formula and its application in this context is fundamental to solving this type of problem. The careful application of mathematical principles allows us to systematically determine the number of possibilities, ensuring that we have a complete and accurate solution. In the next step, we will apply the same principles to calculate the number of ways to select students from the 11th grade.

Determining the Number of Ways to Select 11th Grade Representatives

Having calculated the number of ways to select the 10th-grade representatives, the next step is to determine the number of ways to select the 11th-grade representatives. In this case, there are four qualified students in the 11th grade, and we need to choose three of them to be part of the representative group. Similar to the 10th-grade selection, this is also a combination problem, as the order of selection does not matter. We will again use the combination formula, C(n, k) = n! / (k!(n-k)!), but this time with different values for n and k. Here, n is the total number of students in the 11th grade (4), and k is the number of students we need to select (3).

Applying the formula, we have C(4, 3) = 4! / (3!(4-3)!). Let's break this down as we did before. 4! (4 factorial) is 4 × 3 × 2 × 1 = 24. 3! (3 factorial) is 3 × 2 × 1 = 6, and 1! (1 factorial) is simply 1. Substituting these values into the combination formula, we get C(4, 3) = 24 / (6 × 1) = 24 / 6 = 4. This result indicates that there are 4 different ways to choose three students from the four qualified students in the 11th grade.

These 4 ways represent the unique groups of three students who can represent the 11th grade in the student delegation. For instance, if we label the students as P, Q, R, and S, the possible groups are PQR, PQS, PRS, and QRS. Each of these groups is distinct, and no other combination is possible given the constraint of choosing three students from four. This calculation is just as crucial as the previous one, as it provides the other piece of the puzzle needed to find the total number of possible groups. By understanding the combination formula and its application, we can systematically determine the number of possibilities for the 11th-grade selection. Now that we have calculated the number of ways to select students from both the 10th and 11th grades, the final step is to combine these results to find the total number of different groups that can be formed. This will give us the ultimate answer to the problem posed at the beginning of the article.

Calculating the Total Number of Different Groups

With the number of ways to select the 10th-grade representatives and the 11th-grade representatives calculated, we can now determine the total number of different groups that can be formed. To do this, we will use the fundamental principle of counting, which states that if there are m ways to do one thing and n ways to do another, then there are m × n ways to do both. In this context, choosing the 10th-grade representatives is one task, and choosing the 11th-grade representatives is another. We have already established that there are 10 ways to choose the 10th-grade representatives and 4 ways to choose the 11th-grade representatives.

Applying the fundamental principle of counting, we multiply the number of ways to choose the 10th-grade students by the number of ways to choose the 11th-grade students. This gives us a total of 10 × 4 = 40 different groups. This means that there are 40 unique combinations of students that can be formed, each consisting of two students from the 10th grade and three students from the 11th grade. This final calculation provides the answer to the original problem and demonstrates how combinatorial principles can be applied to solve practical scenarios.

This solution is comprehensive because it considers all possible combinations of students from both grade levels while adhering to the specified constraints. Each of the 40 groups represents a unique composition of student representatives, ensuring that the school has a variety of options to choose from when forming the delegation. The systematic approach of breaking down the problem into smaller parts, calculating the combinations for each grade level separately, and then combining the results using the fundamental principle of counting is a testament to the power and elegance of combinatorial mathematics. This method can be applied to a wide range of similar problems, making it a valuable tool for decision-making and problem-solving in various contexts. Understanding and mastering these principles not only enhances mathematical proficiency but also provides a framework for tackling real-world challenges that involve selection and arrangement.

Conclusion

In conclusion, the problem of forming a student representative group with specific requirements from different grade levels has been systematically solved using combinatorial principles. By breaking down the problem into smaller parts, calculating the combinations for each grade level separately, and then combining the results, we have determined that there are 40 different groups that can be formed. This solution is comprehensive and considers all possible combinations of students while adhering to the given constraints.

The application of the combination formula, C(n, k) = n! / (k!(n-k)!), allowed us to calculate the number of ways to select students from each grade level. For the 10th grade, there were 10 ways to choose two students from five, and for the 11th grade, there were 4 ways to choose three students from four. By multiplying these results, we arrived at the total number of possible groups, which is 40. This demonstrates the fundamental principle of counting and its utility in solving combinatorial problems.

This type of problem-solving approach is not only valuable in mathematical contexts but also has practical applications in various real-world scenarios. From forming committees and teams to making selections and arrangements, the principles of combinatorics provide a framework for decision-making. Understanding and mastering these principles enhances mathematical proficiency and equips individuals with the tools to tackle challenges involving selection and arrangement. The systematic approach used in this article can be applied to a wide range of similar problems, making it a valuable skill for problem-solving and decision-making in diverse contexts. The ability to break down complex problems into smaller, manageable parts and apply appropriate mathematical principles is a key attribute of effective problem solvers. The solution presented here serves as a clear example of how such skills can be developed and applied to achieve a comprehensive and accurate solution.