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Clarissa will create her summer reading list by randomly choosing 4 bo

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Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 12 Jun 2017, 15:01
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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




The Official Guide for GMAT Review 2018

Practice Question
Problem Solving
Question No.: 140

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Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 12 Jun 2017, 16:11
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AbdurRakib 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

The Official Guide for GMAT Review 2018

Practice Question
Problem Solving
Question No.: 140

Using the fundamental counting principle (or permutation, because order matters -- she's listing them in their order chosen).

She chooses 4 books: __ * __ * __ * __

For her first choice she has 10 books to choose from; then she has nine books; then eight; and then seven.

10 * 9 * 8 * 7 = 5040

Answer D
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 27 Jun 2017, 15:26
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AbdurRakib 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


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:

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Re: Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 29 Jun 2017, 00:09
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4 books can be chosen as 10C4 ways. = 210 ways
Total list of 4 books = 4!
Therefore, total list of possible selection = 210 *4! = 210 *24 =5,040.
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 08 Aug 2017, 23:45
AbdurRakib 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




The Official Guide for GMAT Review 2018

Practice Question
Problem Solving
Question No.: 140


as the order is important here, so its a Permutation case. So arranging 4 books out of 10 books is
\(10_P_4\) = 5040

Option D
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 15 Nov 2017, 17:21
AbdurRakib 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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Re: Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 26 Nov 2017, 03:47
genxer123 wrote:
AbdurRakib 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

The Official Guide for GMAT Review 2018

Practice Question
Problem Solving
Question No.: 140

Using the fundamental counting principle (or permutation, because order matters -- she's listing them in their order chosen):

For her first book she has 10 to choose from, then nine, than eight, then seven.

10 * 9 * 8* 7 = 5040

Answer D


Hello. I dont understand one thing :? I did so 10! / 4! (10-6)! = 210 so I chose 210 why its D :?
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 26 Nov 2017, 03:54
ScottTargetTestPrep wrote:
AbdurRakib 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


and if the order didn't matter the answer would be 210 ? like this ? -- > 10! / 4!(10-4)! =210
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Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 26 Nov 2017, 07:03
dave13 wrote:
ScottTargetTestPrep wrote:
AbdurRakib 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


and if the order didn't matter the answer would be 210 ? like this ? -- > 10! / 4!(10-4)! =210


"She will list the books in the order in which they are chosen""--Order matters here ...So 10P4
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 30 Jul 2018, 22:27
generis wrote:
AbdurRakib 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

The Official Guide for GMAT Review 2018

Practice Question
Problem Solving
Question No.: 140

Using the fundamental counting principle (or permutation, because order matters -- she's listing them in their order chosen):

For her first book she has 10 to choose from, then nine, than eight, then seven.

10 * 9 * 8* 7 = 5040

Answer D


Hi,

so if order didn't matter the answer would be...

10!/4!(10-4)! = 210??
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 30 Jul 2018, 23:04

Solution



Given:
    • Clarissa will choose 4 books from the 10 books.
    • She will list the books in the order in which they are chosen.

To find:
    • The number of different possible lists.
Approach and Working:

    • She is first selecting 4 books and then listing them.
    • This is same as selectin and then arrangement.
    o Hence, this a permutation question.
    • Thus, total ways= \(^{10}P_4\)= 5040.

Hence, option D is the correct answer.

Answer: D
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 30 Jul 2018, 23:07
EgmatQuantExpert,

so if order didn't matter then the answer would be...10!/4!(10-4)! = 210??
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Clarissa will create her summer reading list by randomly choosing 4 bo  [#permalink]

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New post 30 Jul 2018, 23:12
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Hey ENEM,

Yes, if the order does not matter then the total ways will be 210.

You can actually read a lot on this topic in our Permutation and Combination article.

Here is the link of the article:Fool-proof method to Differentiate between Permutation & Combination Questions

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