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# Departments A, B, and C have 10 employees each, and department D has

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Re: Departments A, B, and C have 10 employees each, and department D has [#permalink]
10c1 * 10c1* 10c1*20c2 = 10*10*10*190=190,000. Hence D is the correct answer.
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Re: Departments A, B, and C have 10 employees each, and department D has [#permalink]
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Number of ways to choose from Department A,B and C : 10c1 = 10 each

Number of ways to choose from Department D(2 of 20) : 20c2 = 20*19/2 = 190

Total number of task forces possible are : 190*10*10*10 = 190000(Option D)
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Re: Departments A, B, and C have 10 employees each, and department D has [#permalink]
Doesn't this question require long calculation.
Can GMAT test on this type of question?

Thought the Concept is easy.
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Departments A, B, and C have 10 employees each, and department D has [#permalink]
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Ratnaa19 the calculation is not too long. As long as you realize that you need to multiply 10*10*10*19*10 = 19*10^4 = 190,000 Moreover, this question is from one of the exam packs.

quick question: I agree with the possibilities for A, B, C (10*10*10). However, I would like to understand why we need to do D with Combinations and we cannot simply say: XY are two employees from department D (20 employees) so we have 20*19= 380 possibilities

update: I can answer myself: actually if you do 20*19, you would count employee XY and YX twice, but it's the same "pair" so that's why we use combinations
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Re: Departments A, B, and C have 10 employees each, and department D has [#permalink]
Bunuel 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 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

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Re: Departments A, B, and C have 10 employees each, and department D has [#permalink]
Hey guys! and Bunuel
I'm having trouble here with the interpretation of this question.
So, they're forming a task force that has 1 member of A, 1 member of B, 1 member of C, 2 members of D.
So total possible combinations = 10*10*10*20*19, and we have to consider if the order matters.
I saw multiple resolutions here considering that the order among A, B, and C does matter, while the order between D1 (1st D member) and D2 (2nd D member) doesn't; however since we're talking about a task force of People, the order, in my opinion, doesn't matter for all of the taskforce (A,B,C,D1 and D2).
Let's think of a case in which, for group A, Ana was selected.
For group B, Barbara was selected.
For group C, Carlos was selected
And for group D, Daniel and Donald were selected.
Alright, so we have the following task force: Ana, Barbara, Carlos, Daniel, and Donald.
And in my conception, it would be the same task force if we write: Barbara, Carlos, Ana, Donald and Daniel (just changed the order, but still the exact same people)
Thus, the order doesn't matter at all for the whole task force.
Thus, the solution of the problem must be (A*B*C*D*D-1)/6! = (10*10*10*20*19)/ 6! = 19000/36

What am I missing here? Why does the order among A, B, and C groups matter?

Thanks!
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Re: Departments A, B, and C have 10 employees each, and department D has [#permalink]
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Re: Departments A, B, and C have 10 employees each, and department D has [#permalink]
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