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Eight dogs are in a pen when the owner comes to walk some

by M7MBA » Sun Feb 18, 2018 4:56 am
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 the option A.

Is there a fast and easy way to solve this PS question? Experts, may you show me how you'd solve it? Thanks in advanced.
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by [email protected] » Sun Feb 18, 2018 10:34 am
Hi M7MBA,

We're told that 8 dogs are in a pen and the owner lets five dogs out of the pen (one at a time). We're asked for the number of different variations in the line of dogs leaving the pen. This question is a fairly standard version of a Permutation question, so you would probably find it easiest to just write out the numbers and do whatever calculations are necessary.

Since we have 8 dogs and we're going to deal with 5 of them (one at a time - in a line), there would be....
8 options for the 1st dog
7 options for the 2nd dog
6 options for the 3rd dog
5 options for the 4th dog
4 options for the 5th dog

(8)(7)(6)(5)(4) total options

While you could calculate the exact result, the answer choices are 'spaced out' enough that you can estimate that calculation....
(8x7) = about 60
(6x5) = 30
(60)(30) = 1800 and THEN you still have to multiply by 4...
(1800)(4) = about 7000
There's only one answer that's close...

Final Answer: A

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by Jeff@TargetTestPrep » Thu Feb 22, 2018 8:42 am
M7MBA 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 "ordered 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!/3! = 8 x 7 x 6 x 5 x 4 = 6,720

Answer: A

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by Brent@GMATPrepNow » Thu Feb 22, 2018 9:33 am
M7MBA 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
Take the task of arranging the 5 dogs in a line and break it into stages.

Stage 1: Select a dog to be 1st in the line
There are 8 dogs to choose from, so, we can complete stage 1 in 8 ways

Stage 2: Select a dog to be 2nd in the line
There are 7 remaining dogs from which to choose, so we can complete this stage in 7 ways.

Stage 3: Select a dog to be 3rd in the line
There are 6 remaining dogs from which to choose, so we can complete this stage in 6 ways.

Stage 4: Select a dog to be 4th in the line
There are 5 remaining dogs from which to choose, so we can complete this stage in 5 ways.

Stage 5: Select a dog to be 5th in the line
There are 4 remaining dogs from which to choose, so we can complete this stage in 4 ways.

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus arrange 5 dogs in a line) in (8)(7)(6)(5)(4) ways (= 6720 ways)

Answer: A
--------------------------

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Brent Hanneson - Creator of GMATPrepNow.com
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