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 muwt be at least one teacher and at least one student on the committee
To find the number of committees with 5 people, we need to determine how many ways we can choose 1 teacher and 4 students, or 2 teachers and 3 students, or 3 teachers and 2 students, etc.
First, let's consider choosing 1 teacher and 4 students:
- Choose 1 teacher from 6: 6C1 = 6 ways
- Choose 4 students from 12: 12C4 = 495 ways
The total number of committees with 1 teacher and 4 students = 6C1 * 12C4 = 6 * 495 = 2,970 ways.
Next, let's consider choosing 2 teachers and 3 students:
- Choose 2 teachers from the remaining 5 (after removing the 1 teacher chosen above): 5C2 = 10 ways
- Choose 3 students from the remaining 12: 12C3 = 220 ways
The total number of committees with 2 teachers and 3 students = 5C2 * 12C3 = 10 * 220 = 2,200 ways.
We need to consider all possible combinations with different numbers of teachers and students until there are not enough teachers or students left.
Adding up the total number of committees from each case:
2,970 (1 teacher and 4 students) + 2,200 (2 teachers and 3 students) = 5,170
Therefore, there can be 5,170 different committees of 5 people if there must be at least one teacher and at least one student on the committee.
To find the total number of committees, we need to consider the different cases.
Case 1: 1 teacher and 4 students
We can choose 1 teacher out of 6 teachers in $\binom{6}{1}$ ways and 4 students out of 12 students in $\binom{12}{4}$ ways.
So the total number of committees with 1 teacher and 4 students is $\binom{6}{1} \cdot \binom{12}{4}$.
Case 2: 2 teachers and 3 students
We can choose 2 teachers out of 6 teachers in $\binom{6}{2}$ ways and 3 students out of 12 students in $\binom{12}{3}$ ways.
So the total number of committees with 2 teachers and 3 students is $\binom{6}{2} \cdot \binom{12}{3}$.
Case 3: 3 teachers and 2 students
We can choose 3 teachers out of 6 teachers in $\binom{6}{3}$ ways and 2 students out of 12 students in $\binom{12}{2}$ ways.
So the total number of committees with 3 teachers and 2 students is $\binom{6}{3} \cdot \binom{12}{2}$.
Case 4: 4 teachers and 1 student
We can choose 4 teachers out of 6 teachers in $\binom{6}{4}$ ways and 1 student out of 12 students in $\binom{12}{1}$ ways.
So the total number of committees with 4 teachers and 1 student is $\binom{6}{4} \cdot \binom{12}{1}$.
Adding all the cases together, the total number of committees is
$\binom{6}{1} \cdot \binom{12}{4} + \binom{6}{2} \cdot \binom{12}{3} + \binom{6}{3} \cdot \binom{12}{2} + \binom{6}{4} \cdot \binom{12}{1}$
$= 6 \cdot 495 + 15 \cdot 220 + 20 \cdot 66 + 15 \cdot 12$
$= 2970 + 3300 + 1320 + 180$
$= \boxed{7830}$ committees.