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.
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 .
1 answer