6 teachers and 12 students volunteer for a committee to discuss extra curricular activities. How many committees of 5 people can be made if there must be no more than 2 teachers on the committee .
We can solve this problem using combinations.
If there are no teachers on the committee, we need to choose 5 students from 12 students. The number of ways to do this is denoted by the combination "12 choose 5" and can be calculated as:
(12 choose 5) = 12! / (5!(12-5)!) = (12*11*10*9*8) / (5*4*3*2*1) = 792.
If there is 1 teacher on the committee, we need to choose 4 more people from 6 teachers and 12 students. The number of ways to do this is the product of the number of ways to choose 1 teacher from 6 teachers and the number of ways to choose 4 people from 12+6-1=17 total people. Thus, the number of committees with 1 teacher is:
(6 choose 1) * (17 choose 4) = 6 * (17! / (4!(17-4)!) = 6 * (17*16*15*14) / (4*3*2*1) = 2380.
If there are 2 teachers on the committee, we need to choose 3 more people from 6 teachers and 12 students. The number of ways to do this is the product of the number of ways to choose 2 teachers from 6 teachers and the number of ways to choose 3 people from 12+6-1=17 total people. Thus, the number of committees with 2 teachers is:
(6 choose 2) * (17 choose 3) = (6! / (2!(6-2)!)) * (17! / (3!(17-3)!)) = (6*5 / (2*1)) * (17*16*15 / (3*2*1)) = 1800.
Therefore, the total number of committees can be made is:
792 + 2380 + 1800 = 4972 committee.
To find the number of committees, we need to consider the different combinations of teachers and students.
1. Selecting 0 teachers:
- Selecting 5 students from 12: C(12, 5) = 792 combinations
2. Selecting 1 teacher:
- Selecting 1 teacher from 6: C(6, 1) = 6 combinations
- Selecting 4 students from 12: C(12, 4) = 495 combinations
- Total combinations: 6 * 495 = 2,970 combinations
3. Selecting 2 teachers:
- Selecting 2 teachers from 6: C(6, 2) = 15 combinations
- Selecting 3 students from 12: C(12, 3) = 220 combinations
- Total combinations: 15 * 220 = 3,300 combinations
To get the total number of committees that satisfy the given conditions, we add the results from each scenario:
792 + 2,970 + 3,300 = 7,062
Therefore, there are 7,062 different committees that can be formed.