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 3 students on the committee
To form a committee of 5 people with 3 students, we need to choose 3 students from the 12 available and 2 teachers from the 6 available.
The number of ways to choose 3 students from 12 is calculated as 12 choose 3, which can be written as 12C3 or denoted mathematically as ${12 \choose 3}$.
Using the combination formula,
${12 \choose 3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!} = \frac{12\times11\times10}{3\times2\times1} = 220$.
To choose 2 teachers from 6, we can use the same formula as above:
${6 \choose 2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6\times5}{2\times1} = 15$.
To calculate the total number of possible committees of 5 people with 3 students, we multiply the number of ways to choose the students by the number of ways to choose the teachers:
$220 \times 15 = 3300$.
Therefore, there can be 3300 committees of 5 people that include 3 students.
To form a committee of 5 people with 3 students, you need to choose 3 students from the 12 available and 2 people from the remaining 6 teachers.
The number of ways to choose 3 students from 12 is given by the combination formula:
C(12, 3) = 12! / (3! * (12-3)!) = 12! / (3! * 9!) = (12 * 11 * 10) / (3 * 2 * 1) = 220
Similarly, the number of ways to choose 2 teachers from 6 is given by:
C(6, 2) = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
Therefore, the total number of committees that can be formed is:
220 * 15 = 3300 committees.