math
posted by Friends on .
My study group is lost on this one also..
Suppose you are managing 25 employees, and you need to form three teams to work on three different projects.
Assume that all employees will work on exactly one team. Also, each employee has the same qualifications/skills, so any employee can serve on any team.
The number of employees on each project are as follows: 8 on Team A, 3 on Team B, and 14 on Team C
(a) How many ways can Team A be selected from the available employees?
(b) Then, how many ways can Team B be selected from the remaining available employees?
(c) Then, how many ways can Team C be selected from the remaining available employees?
(d) Then, how many ways can all teams be selected?
(e) What is the probability that three workers randomly selected from all employees will all be from team A?
We know it's something like this but we don't know how to calculate it.
1. number of team A = C(25,8)
2. number of team B = C(17,3)
3. number of team C = C(14,14)
4. multiply them
5. C(22,5) x C(17,3) x C(14,14) / (answer to 4)
C(25,8)meaning choosing 8 people from a group of 25 but that's where we are lost because we don't know how to do that part.
any help..?

I had answered this question here
http://www.jiskha.com/display.cgi?id=1292393115
The notation C(n,r) is essential to the type of question you asked
by definition it is equal to n!/((nr)!r!)
so for something like C(5,2)
it would be 5!/(3!2!) = 5x4x3x2x1/(2x1x3x2x1)
notice we can cancel
= 5x4/(2x1) = 20/2 = 10
you should have a key on your calculator that does this,
often it is called _{n} C _{r}
to do C(5,2)
enter 5
press 2nd _{n} C _{r}
press 2
press =
you should get 10
for your C(25,8) it means choose 8 people from 25.
Your are dealing with combinations, where the order in which the people are chosen does not matter.
See if you can make out the different answers I gave you and how they match up with the question.
e.g. for e), since we know 3 definite people are in the group we only need 5 more for team A, but the choice for that is from the other 22 available.
that leaves 17, and we need 3 of those for team B
finally leaving the remaining 14 to form team C
good luck