Forming Committees with a Minimum Number of Students
When faced with the task of forming a committee of 5 people from a group of 6 teachers and 4 students, the condition often specified is that the committee must include at least one student. How do you calculate the number of such possible committees? This question can be effectively answered through the use of combinatorial methods and the principle of complementary counting. Let's delve into the process step-by-step.
The Principle of Complementary Counting
To find the number of committees of 5 that can be formed from 6 teachers and 4 students, including at least one student, we can employ the principle of complementary counting. This involves calculating the number of committees that do not meet the criteria (in this case, committees with no students) and then subtracting this from the total number of possible committees.
Total Number of Committees Without Restrictions
The total number of ways to choose 5 members from a group of 10, where the group consists of 6 teachers and 4 students, is given by the combination formula:
[ C(n, k) frac{n!}{k!(n-k)!} ]Here, ( n 10 ) and ( k 5 ).
Calculating this, we get:
[ C(10, 5) frac{10!}{5!10-5!} frac{10!}{5!5!} 252 ]Committees with No Students Only Teachers
The number of ways to choose 5 teachers from 6 is:
[ C(6, 5) frac{6!}{5!6-5!} frac{6!}{5!1!} 6 ]Committees with At Least One Student
To find the number of committees that include at least one student, we subtract the number of all-teacher committees from the total number of committees:
[ text{Committees with at least one student} C(10, 5) - C(6, 5) 252 - 6 246 ]Thus, the total number of committees that can be formed including at least one student is 246.
Calculating Specific Committee Combinations
Alternatively, we can break down the problem into specific cases and calculate the number of committees for each:
1 teacher and 4 students:The number of ways to choose 1 teacher from 6 and 4 students from 4 is:
[ C(6, 1) times C(4, 4) 6 times 1 6 ] 2 teachers and 3 students:The number of ways to choose 2 teachers from 6 and 3 students from 4 is:
[ C(6, 2) times C(4, 3) 15 times 4 60 ] 3 teachers and 2 students:The number of ways to choose 3 teachers from 6 and 2 students from 4 is:
[ C(6, 3) times C(4, 2) 20 times 6 120 ] 4 teachers and 1 student:The number of ways to choose 4 teachers from 6 and 1 student from 4 is:
[ C(6, 4) times C(4, 1) 15 times 4 60 ] 5 teachers and 0 students:The number of ways to choose 5 teachers from 6 is:
[ C(6, 5) times C(4, 0) 6 times 1 6 ]The total number of different committees that can be formed is:
[ 6 60 120 60 6 252 ]Additional Scenarios and Probabilities
Now, let's explore a related scenario where we calculate the number of committees with at least one teacher, and the probability of having more teachers than students.
1 teacher and 4 students:The number of ways to choose 1 teacher from 6 and 4 students from 4 is:
[ C(6, 1) times C(4, 4) 6 times 1 6 ] 2 teachers and 3 students:The number of ways to choose 2 teachers from 6 and 3 students from 4 is:
[ C(6, 2) times C(4, 3) 15 times 4 60 ] 3 teachers and 2 students:The number of ways to choose 3 teachers from 6 and 2 students from 4 is:
[ C(6, 3) times C(4, 2) 20 times 6 120 ] 4 teachers and 1 student:The number of ways to choose 4 teachers from 6 and 1 student from 4 is:
[ C(6, 4) times C(4, 1) 15 times 4 60 ] 5 teachers and 0 students:The number of ways to choose 5 teachers from 6 is:
[ C(6, 5) times C(4, 0) 6 times 1 6 ]The total number of committees with at least one teacher is 246 as calculated previously. The number of committees where more teachers than students are present is:
[ 6 60 120 186 ]The probability of having more teachers than students is:
[ frac{186}{246} frac{11}{41} ]Conclusion
By using combinatorial methods and the principle of complementary counting, we can effectively calculate the number of committees that can be formed from a group of 6 teachers and 4 students, with at least one student included. The total number of such committees is 246, and the probability of having more teachers than students is 11/41.