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Eight dogs are in a pen when the owner comes...

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by swerve » Tue Nov 28, 2017 5:32 pm
Eight dogs are in a pen when the owner comes to walk some of them. The owner lets five dogs out of the pen one at a time. How many different variations in the line of dogs leaving the pen are possible?

A. 6,720
B. 3,360
C. 1,680
D. 560
E. 56

The OA is A.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.
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by EconomistGMATTutor » Thu Nov 30, 2017 1:48 pm
Hello swerve.

Let's take a look at your question.

We have to notice that the order is important. So, we are talking about a permutation.

The first dog can be chosen from 8 ---------- 8 options.
The second dog can be chosen from 7 -------7 options.
The third dog can be chosen from 6 ---------6 options.
The fourth dog can be chosen from 5 -------5 options.
The fifth dog can be chosen from 4 ----------4 options.

So, the total different variations are: 8*7*6*5*4=6,720.

So, the correct answer is A .

I hope this explanation may help you.

I'm available if you'd like a follow up.

Regards,
GMAT Prep From The Economist
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by Scott@TargetTestPrep » Wed Dec 13, 2017 12:17 pm
swerve wrote:Eight dogs are in a pen when the owner comes to walk some of them. The owner lets five dogs out of the pen one at a time. How many different variations in the line of dogs leaving the pen are possible?

A. 6,720
B. 3,360
C. 1,680
D. 560
E. 56
We are given that there are 8 dogs and the owner needs to arrange 5 of them. Since we have an "order arrangement," order matters. So we have a permutation problem. Thus, the number of ways to arrange 5 dogs from 8 is:

8P5 = 8!/(8-5)! = 8 x 7 x 6 x 5 x 4 = 6,720

Answer: A

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