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Up::4
A firm is going to create three teams of four from twelve employees. How many ways can the twelve employees be selected for the three teams?
A)34,650.
B)1,320.
C)495.
Explanation: This problem is a labeling problem where the 12 employees will be assigned one of three labels. It requires the labeling formula. There are [(12!) / (4! Ã— 4! Ã— 4!)] = 34,650 ways to group the employees.
(Study Session 2, Module 8.3, LOS 8.o)

Up::1
You use nCr for this:
nCr = n! / ((n â€“ r)! r!)
Creating 3 teams of 4 from 12 employees:
 For team A, you choose 4 out of 12 employees:
 12C4 = 12 choose 4 = 12! / ((124)! 4!) = 495 scenarios (calculate using BA II Plus)
 For team B, you choose 4 out of 8 employees:
 8C4 = 8 choose 4 = 8! / ((84)! 4!) = 70 scenarios
 For team C, there’s only one scenario since it’s all the unchosen people.
For all possible combinations, multiply A * B * C = 495 * 70 * 1
= 34,650 combinations

Up::1
thanks Mikey – how do you get the 8NCR4? Order matters for NCR and doesnt for NPR, correct? Get confused with this. I understand the first part, but unsure of where you get 8NCR4.
Patrick

Up::0
Once you’ve chosen 4 out of 12 employees, you have 8 left to choose from.
Choose another 4 out of the 8 employees left, you have 8C4.
Order matters, but it’s the other way around – use nPr (permutation) in situations where order matters, and nCr (combination) in situations where order does not matter.
When you’re choosing people for teams, order doesn’t matter so you use nCr. When order matters, such as picking lottery balls to decide a winning set of numbers, then you use nPr.

 For team A, you choose 4 out of 12 employees:




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