If a committee of 3 is to be selected from a group of 15 students and 6 teachers, then the
number of different possible committees that have at most 1 teacher is...
I got 1260 by doing (6C1)(15C1)(14C1), but the correct answer is supposed to be 1085. How did they get this?
At most 1 teacher ----> 1 teacher or no teacher
so C(6,1)*C(15,2) + C(6,0)*C(15,3)
= 6(105) + 1(455)
= 1085
Well, it seems like your calculations are as accurate as a bullseye shot from a rubber band gun! However, the correct answer of 1085 might have been achieved like this:
To find the number of committees with at most 1 teacher, we can calculate the following:
Number of committees with 0 teachers = (6C0) * (15C3)
Number of committees with 1 teacher = (6C1) * (15C2)
So, the total number of committees with at most 1 teacher would be the sum of these two calculations:
Total = (6C0) * (15C3) + (6C1) * (15C2)
Now, let's do the math and cross our fingers that the answer matches your expected result of 1085!
To find the number of different possible committees with at most 1 teacher, we need to consider two cases:
Case 1: Committee with no teacher (all students)
In this case, we need to select 3 students from the 15 students available. The number of ways to do this is given by the combination formula C(15, 3) = (15!)/(3!(15-3)!) = 455.
Case 2: Committee with 1 teacher
In this case, we choose 1 teacher and 2 students.
The number of ways to choose 1 teacher from 6 teachers is C(6, 1) = 6.
The number of ways to choose 2 students from 15 students is C(15, 2) = (15!)/(2!(15-2)!) = 105.
Therefore, the total number of different possible committees with at most 1 teacher is 455 + 6 * 105 = 455 + 630 = 1085.
So, the correct answer is indeed 1085.
To find the number of different possible committees that have at most 1 teacher, we need to consider two cases separately: committees with 0 teachers and committees with 1 teacher.
Case 1: Committees with 0 teachers
To form a committee with 0 teachers, we need to select 3 students from the group of 15 students. The number of ways to do this can be calculated using the combination formula:
C(15,3) = 15! / (3! * (15 - 3)!) = 455
Case 2: Committees with 1 teacher
To form a committee with 1 teacher, we need to select 1 teacher from the group of 6 teachers and 2 students from the group of 15 students. The number of ways to do this can be calculated as:
C(6,1) * C(15,2) = (6! / (1! * (6 - 1)!)) * (15! / (2! * (15 - 2)!)) = 6 * 105 = 630
Now, we need to calculate the total number of committees by adding the number of committees from each case:
Total number of committees = Case 1 + Case 2 = 455 + 630 = 1085
Therefore, the correct answer is 1085, not 1260.