Question

A committee must be formed with 3 teachers and 6 students. If there are 8 teachers to choose from, and 15 students, how many different ways could the committee be made?

Answer 1

In order to find the number of different ways the committee can be made, first let's calculate the possibilities for the teachers.

Since we have 8 teachers and need to choose 3, we have a combination of 8 choose 3.

A combination of n choose p is calculated with the formula below:

`C(n,p)=(n!)/(p!(n-p)!)`

So, for n = 8 and p = 3, we have:

`C(8,3)=(8!)/(3!(8-3)!)=(8!)/(3!\cdot5!)=(8\cdot7\cdot6\cdot5!)/(3\cdot2\cdot5!)=56`

Now, for the students, we have 15 students and need to choose 6, so we have a combination of 15 choose 6:

`C(15,6)=(15!)/(6!9!)=(15\cdot14\cdot13\cdot12\cdot11\cdot10\cdot9!)/(6\cdot5\cdot4\cdot3\cdot2\cdot9!)=5005`

Multiplying both numbers of possibilities, we have the final result:

`56\cdot5005=280280`

Therefore there are 280,280 different ways of making the committee.