Departments A, B, and C have 10 employees each . . .

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Departments A, B, and C have 10 employees each, and department D has 20 employees. Departments A, B, C, and D have no employees in common. A task force is to be formed by selecting 1 employee from each of departments A, B, and C and 2 employees from department D. How many different task forces are possible?

A. 19,000
B. 40,000
C. 100,000
D. 190,000
E. 400,000

The OA is D.

What formulas should I set here in order to solve this PS question? Experts, may you help me? Thanks.

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by [email protected] » Tue Dec 05, 2017 10:51 am
Hi Vincen,

We're told that Departments A, B, and C have 10 employees each, Department D has 20 employees and the Departments have no employees in common. A task force is to be formed by selecting 1 employee from each of departments A, B, and C and 2 employees from department D. We're asked for the number of different task forces that are possible.

With the first 3 Departments, we have (10)(10)(10) = 1,000 possible groups. With Department D though, we have to choose 2 people from 20 total people, so we have to use the Combination Formula:

20!/(2!)(18!) = (20)(19)/(2)(1) = 190

Thus, we have (1,000)(190) = 190,000 possible task forces

Final Answer: D

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by Scott@TargetTestPrep » Sun Sep 29, 2019 6:08 pm
Vincen wrote:Departments A, B, and C have 10 employees each, and department D has 20 employees. Departments A, B, C, and D have no employees in common. A task force is to be formed by selecting 1 employee from each of departments A, B, and C and 2 employees from department D. How many different task forces are possible?

A. 19,000
B. 40,000
C. 100,000
D. 190,000
E. 400,000

The OA is D.

What formulas should I set here in order to solve this PS question? Experts, may you help me? Thanks.
The number of ways to select 1 employee from each of departments A, B, and C and 2 employees from department D is

10C1 x 10C1 x 10C1 x 20C2 = 10 x 10 x 10 x (20 x 19)/2 = 10 x 10 x 10 x 10 x 19 = 190,000

Answer: D

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