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

Take the task of creating the task force and break it into stages.

Stage 1: Select one person from department A
There are 10 people to choose from, so we can complete stage 1 in 10 ways

Stage 2: Select one person from department B
There are 10 people to choose from, so we can complete stage 2 in 10 ways

Stage 3: Select one person from department C
There are 10 people to choose from, so we can complete stage 3 in 10 ways

Stage 4: Select 2 people from department D
Since the order in which we select the 2 people does not matter, we can use combinations.
We can select 2 people from 20 people in 20C2 ways (190 ways)
So, we can complete stage 4 in 190 ways

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create a task force) in (10)(10)(10)(190) ways (= 190,000 ways)

Answer:

If anyone is interested, here's a video on calculating combinations (like 20C2) in your head

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\(\begin{align}&\text{Department A, B, C}&&10\text{ Employees}&&\text{select }1\\\\
&\text{Department D}&&20\text{ Employees}&&\text{select }2\\
\end{align}\)

\(\text{Total }= {10 \choose 1}^3 \times {20 \choose 2} =\\\\
10^3 \times \frac{20!}{18!\times 2!} =\\\\
1,000 \times \frac{20 \times 19 \times 18!}{18! \times 2 \times 1} =\\ 1,000 \times \frac{20 \times 19}{2} =\\ 1,000 \times 190 = 190,000\\\\
\text{Answer: (D) 190,000}\)

10c1 * 10c1* 10c1*20c2 = 10*10*10*190=190,000. Hence D is the correct answer.
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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)

Doesn't this question require long calculation.
Can GMAT test on this type of question?

Thought the Concept is easy.

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

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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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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