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

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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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

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

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

EgmatQuantExpert,

so if order didn't matter then the answer would be...10!/4!(10-4)! = 210??

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

Regards,
Ashutosh
e-GMAT

Here AB is not equal to BA because order matters.
Also, Repetition is not allowed so
Permutation formula will be applied. 10!/6! = 5040

Fast math!

___ ___ ___ ___
10 x 9 x 8 x 7

8x7=56
56x10=560
560x9=?

Now, look at the options, they are not close to one another!
560x9~560x10=5600

Which one is closest? D!

10C4 = 210

We're told that the books will be in the order they're chosen.

210 * 4! = 5,040.

Answer is D.

There are 10 books available for the first pick this 10
There are now 9 books available for the second pick
For the third pick only 8 books are available
For the fourth pick only 7

Thus 10*9*8*7 =5040 (You just applied the Fundamental counting principle)

Just a couple of reflections:
You realize though that when you pick a book you include ALL THE ARRANGEMENTS of that book. Books can be arranged in 4! ways.
In this context order matters because you pick the possible lists depending on ORDER. You don't just pick books (combination), you have to count also the arrangements.

Since when you apply combinations you don't care about order, you shoud realize that you can de-arrange things by 5040/4! and you obtain all the pickings WITHOUT the possible arrangements.

Answer: Option D

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Point to note in this question is "She will list the books in the order in which they are chosen."
so here order matters AB is not equal to BA.
We need to go for permutation 10P4 = 5040
or
10C4*4! = 5040

Ans D

1. She will first choose 4 books out of 10.
2. She will list those books in order.

Therefore, this is a Permutation question. 10!/4! = 5040.

PMO: this is picking 4 books from 10 books with the arrangement --> 10P4 = 5040

Correct option : D

Information required to attend the question for solution:
Combination : only arrangement
Permutation : Arrangement + selection

Question stem:
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 (arrangement) in the order in which they are chosen (Selection).

To Find : How many different lists are possible?
Permutation : 10P4 = 10 x 9 x 8 x 7 = 5,040 = Option B

here order matters so
10*9*8*7 = 5040

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