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A librarian has a set of ten books, including four different

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A librarian has a set of ten books, including four different

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A librarian has a set of ten books, including four different books about Abraham Lincoln. The librarian wants to put the ten books on a shelf with the four Lincoln books next to each other, somewhere on the shelf among the other six books. How many different arrangements of the ten books are possible?
(A) (10!)/(4!)
(B) (4!)(6!)
(C) (4!)(7!)
(D) (4!)(10!)
(E) (4!)(6!)(10!)

OA C

Source: Magoosh

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Quote:
A librarian has a set of ten different books, including four books about Abraham Lincoln. The librarian wants to put the ten books on a shelf with the four Lincoln books next to each other, somewhere on the shelf among the other six books. How many different arrangements of the ten books are possible?
(A) (10!)/(4!)
(B) (4!)(6!)
(C) (4!)(7!)
(D) (4!)(10!)
(E) (4!)(6!)(10!)
Source: Magoosh
$$?\,\,\,:\,\,\,\# \,\,\,{\rm{possibilities}}\,{\rm{,}}\,\,{\rm{Abe}}\,\,{\rm{books}}\;\,{\rm{together}}$$
$$10\,\,{\rm{different}}\,\,{\rm{books,}}\,\,{\rm{4}}\,\,{\rm{about}}\,{\rm{Abe}}\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,1\,\,{\rm{multiple - block}}\,\,\,\left( {4\,\,{\rm{Abe}}\,\,{\rm{books}}} \right) \hfill \cr
\,6\,\,{\rm{single}}\,{\rm{blocks}} \hfill \cr} \right.$$
$$\left. \matrix{
{P_7} = 7!\,\,\,{\rm{permutation}}\,\,{\rm{of}}\,\,{\rm{all}}\,\,{\rm{blocks}} \hfill \cr
{{\rm{P}}_{\rm{4}}} = 4!\,\,\,{\rm{permutation}}\,\,{\rm{of}}\,\,{\rm{Abe}}\,\,{\rm{books}}\,\,\,\, \hfill \cr} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,?\,\,\, = \,\,{P_7} \cdot {P_4}\,\, = \,\,7!4!\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{C}} \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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BTGmoderatorDC wrote:
A librarian has a set of ten books, including four different books about Abraham Lincoln. The librarian wants to put the ten books on a shelf with the four Lincoln books next to each other, somewhere on the shelf among the other six books. How many different arrangements of the ten books are possible?
(A) (10!)/(4!)
(B) (4!)(6!)
(C) (4!)(7!)
(D) (4!)(10!)
(E) (4!)(6!)(10!)

OA C

Source: Magoosh
Consider the 4 Lincoln books as one book for the purposes of arranging on the shelf, since they have to be together.

Along with the 6 other books, you can see that there are then 7 positions occupied. With the idea that order matters, there are then 7! ways to arrange the books on the shelf.

Going back to the 4 Lincoln books, since the problem stated that they are "different", we are being told that order matters, so the number of ways to arrange the 4 LIncoln books is 4!.

Total ways to arrange the books is therefore C, 7!x4!

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BTGmoderatorDC wrote:
A librarian has a set of ten books, including four different books about Abraham Lincoln. The librarian wants to put the ten books on a shelf with the four Lincoln books next to each other, somewhere on the shelf among the other six books. How many different arrangements of the ten books are possible?
(A) (10!)/(4!)
(B) (4!)(6!)
(C) (4!)(7!)
(D) (4!)(10!)
(E) (4!)(6!)(10!)
Take the task of arranging the 10 books and break it into stages.

Stage 1: Arrange the 4 books about Abe Lincoln in a row
We can arrange n objects in n! ways.
So, we can arrange the 4 books in 4! ways

IMPORTANT: Now we'll "glue" the 4 Abe Lincoln books together to form 1 SUPER BOOK (this will ensure that the 4 Abe Lincoln books remain together)
So, we now have 1 Abe Lincoln SUPER BOOK, along with 6 non-Abe Lincoln books (for a total of 7 "books")

Stage 2: Arrange the 7 "books"
We can arrange n objects in n! ways.
So, we can arrange the 7 books in 7! ways

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all of the books) in [color=blue](4!)(7!) ways

Answer: C
--------------------------

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: http://www.gmatprepnow.com/module/gmat-counting/video/775

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

Then you can try solving the following questions:

EASY
- http://www.beatthegmat.com/what-should-be-the-answer-t267256.html
- http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
- http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html
- http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
- http://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
- http://www.beatthegmat.com/arabian-horses-good-one-t150703.html
- http://www.beatthegmat.com/sub-sets-probability-t273337.html
- http://www.beatthegmat.com/combinatorics-problem-t273180.html
- http://www.beatthegmat.com/digits-numbers-t270127.html
- http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
- http://www.beatthegmat.com/combinatorics-problem-t267079.html


DIFFICULT
- http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
- http://www.beatthegmat.com/permutation-and-combination-t273915.html
- http://www.beatthegmat.com/permutation-t122873.html
- http://www.beatthegmat.com/no-two-ladies-sit-together-t275661.html
- http://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

_________________
Brent Hanneson – Creator of GMATPrepNow.com
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Let's first "glue" the 4 Lincoln books together to create one SUPER BOOK (this will ensure that the 4 books remain together)
We now have 7 books: 6 regular books and 1 super book
We can arrange these 7 books in 7! ways.

KEY: For each of the 7! arrangements, we can take the 4 Lincoln books (that comprise the SUPER BOOK) and arrange them in 4! ways.
So, the TOTAL number of arrangements = (7!)(4!)

Hence, the correct answer is C

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