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swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

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

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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Source: Beat The GMAT — Problem Solving | Tue Nov 28, 2017 5:32 pm

EconomistGMATTutor: GMAT Instructor; Posts: 555; Joined: Wed Oct 04, 2017 4:18 pm; Thanked: 180 times; Followed by:12 members

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
We offer 70+ point score improvement money back guarantee.
Our average student improves 98 points.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

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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