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Clarissa will create her summer reading list by randomly

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Clarissa will create her summer reading list by randomly

by BTGmoderatorDC » Sun Oct 21, 2018 6:55 pm

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Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?

A. 6
B. 40
C. 210
D. 5,040
E. 151,200

OA D

Source: Official Guide
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by Jay@ManhattanReview » Sun Oct 21, 2018 10:17 pm
BTGmoderatorDC wrote:Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?

A. 6
B. 40
C. 210
D. 5,040
E. 151,200

OA D

Source: Official Guide
Since the reading list lays importance to order of the books, we will apply Permutation and not Combination. In other words, if Clarissa chooses A, B, C, and D as chosen 4 books, for example, ABCD and ADCB are two different lists.

So, the number of ways to choose 4 books (Order in important) = 10P4 = 10.9.8.7 = 5040.

The correct answer: D

Hope this helps!

-Jay
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by swerve » Mon Oct 22, 2018 9:18 am
We can try as follows,

4 books can be chosen as 10C4 ways = 210 ways.

The total list of 4 books = 4!

Therefore, the total list of possible selection = 210*4! = 210*24 = 5,040.

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by Brent@GMATPrepNow » Mon Oct 22, 2018 9:20 am
BTGmoderatorDC wrote:Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?

A. 6
B. 40
C. 210
D. 5,040
E. 151,200

OA D

Source: Official Guide
Take the task of creating the reading list and break it into stages.


Stage 1: Select a book to read 1st
There are 10 books to choose from. So, we can complete stage 1 in 10 ways

Stage 2: Select a book to read 2nd
There are 9 books remaining to choose from (since we already chose a book in stage 1).
So, we can complete stage 2 in 9 ways

Stage 3: Select a book to read 3rd
There are 8 books remaining to choose from. So, we can complete stage 3 in 8 ways

Stage 4: Select a book to read 4th
There are 7 books remaining to choose from. So, we can complete stage 4 in 7 ways

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create a reading list) in (10)(9)(8)(7) ways (= 5040 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should ... 67256.html
- https://www.beatthegmat.com/counting-pr ... 44302.html
- https://www.beatthegmat.com/picking-a-5 ... 73110.html
- https://www.beatthegmat.com/permutation ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatori ... 73194.html
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- https://www.beatthegmat.com/doubt-on-se ... 71047.html
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DIFFICULT
- https://www.beatthegmat.com/wonderful-p ... 71001.html
- https://www.beatthegmat.com/permutation ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladi ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Tue Oct 23, 2018 6:24 pm
BTGmoderatorDC wrote:Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?

A. 6
B. 40
C. 210
D. 5,040
E. 151,200
Since the order of the books in her list matters, we use permutation. Thus, the number of ways 4 books can be chosen and ordered from 10 books is 10P4 = 10!/(10-4)! = 10 x 9 x 8 x 7 = 5,040.

Answer: D

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