Forming a Committee with a Minimum of 4 Students
This article discusses the process of forming a committee of 10 members from 9 teachers and 6 students, ensuring the committee has at least 4 students. We will explore the problem using both direct calculation and the principle of complementary counting.
Case Breakdown for Direct Calculation
To solve the problem directly, we can break it down into cases based on the number of students in the committee. The possible cases are:
4 students and 6 teachers 5 students and 5 teachers 6 students and 4 teachersCase 1: 4 Students and 6 Teachers
The number of ways to choose 4 students out of 6 is given by the combination formula binom{6}{4}, which equals 15. The number of ways to choose 6 teachers out of 9 is binom{9}{6}, which equals 84. Therefore, the total number of ways for this case is:
15 times 84 1260
Case 2: 5 Students and 5 Teachers
The number of ways to choose 5 students out of 6 is binom{6}{5}, which equals 6. The number of ways to choose 5 teachers out of 9 is binom{9}{5}, which equals 126. Therefore, the total number of ways for this case is:
6 times 126 756
Case 3: 6 Students and 4 Teachers
The number of ways to choose 6 students out of 6 is binom{6}{6}, which equals 1. The number of ways to choose 4 teachers out of 9 is binom{9}{4}, which equals 126. Therefore, the total number of ways for this case is:
1 times 126 126
Total Ways to Form the Committee with at Least 4 Students
Adding the totals of all three cases, we get:
1260 756 126 2142
Thus, the total number of ways to form the committee with at least 4 students is 2142.
Using the Principle of Complementary Counting
Alternatively, we can solve the problem using the principle of complementary counting. First, we calculate the total number of ways to select the committee without any restrictions. Then, we find the number of ways to select the committee with less than 4 students and subtract this from the total number of combinations.
The total number of ways to select a committee of 10 members from 9 teachers and 6 students is given by binom{15}{10}, which equals 3003. The number of ways to select a committee with less than 4 students is the sum of the ways to select committees with 0, 1, 2, or 3 students.Case 1: Committee with 0 Students
The number of ways to select a committee with 0 students and 10 teachers is 0, since there are only 9 teachers.
Case 2: Committee with 1 Student
The number of ways to select a committee with 1 student and 9 teachers is given by binom{9}{9} times binom{6}{1} 6.
Case 3: Committee with 2 Students
The number of ways to select a committee with 2 students and 8 teachers is given by binom{9}{8} times binom{6}{2} 90.
Case 4: Committee with 3 Students
The number of ways to select a committee with 3 students and 7 teachers is given by binom{9}{7} times binom{6}{3} 630.
The total number of ways to select a committee with less than 4 students is:
0 6 90 630 726
Therefore, the number of ways to select a committee with at least 4 students is:
3003 - 726 2277
Therefore, using the principle of complementary counting, the number of ways to form the committee with at least 4 students is 2277. This method provides an alternative approach to solve the given problem.
Conclusion
In conclusion, we have discussed two methods for solving the problem of forming a committee with at least 4 students. The first method involves breaking down the problem into cases based on the number of students and calculating the combinations for each case. The second method uses complementary counting to subtract the number of invalid combinations from the total number of combinations. Both methods yield the same result, which is 2142 or 2277 ways, depending on the interpretation.