# Probability

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)

from Kaplan

I tried solving it with combination formula because the order doesnot matter (12!/3!*8! )*3! . The 3! is the number of team. What am I missing?

This requires the multinomial formula, not the combination formula. When you are trying to separate n object into n groups of n the correct formula is :

N!/ n1 ! * n2 * ! n3 * ! nn !

Where N is the total number of objects and n is the number of objects per groups. So in your case it would be

12! / 4! 4! 4! = 34,650

How many ways can you choose the first team?

12! / (4! × 8!) = 495

How many ways can you choose the second team?

8! / (4! × 4!) = 70

How many ways can you choose the third team?

1

How many ways can you choose the three teams if the order of choosing teams matters?

12! / (8! × 4!) × 8! / (4! × 4!) × 1 = 12! / (4! × 4! × 4!) = 34,650

And if the order of choosing the teams doesn’t matter?

12! / (4! × 4! × 4!) / 3! = 5,775

The question makes it sound as if the order of choosing the teams doesn’t matter. The answer choices offered make it sound as if the order of choosing the teams does matter.

It’s a poorly worded question.