👤

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 :

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:

[tex]C(n,p)=\frac{n!}{p!(n-p)!}[/tex]

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

[tex]C(8,3)=\frac{8!}{3!(8-3)!}=\frac{8!}{3!\cdot5!}=\frac{8\cdot7\cdot6\cdot5!}{3\cdot2\cdot5!}=56[/tex]

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

[tex]C(15,6)=\frac{15!}{6!9!}=\frac{15\cdot14\cdot13\cdot12\cdot11\cdot10\cdot9!}{6\cdot5\cdot4\cdot3\cdot2\cdot9!}=5005[/tex]

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

[tex]56\cdot5005=280280[/tex]

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