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

by BTGmoderatorDC » Thu Jan 10, 2019 11:39 pm

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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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by fskilnik@GMATH » Fri Jan 11, 2019 6:00 am
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.

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Fabio.
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by regor60 » Fri Jan 11, 2019 6:48 am
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 [spoiler]C, 7!x4![/spoiler]
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by Brent@GMATPrepNow » Fri Jan 11, 2019 7:12 am
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 (4!)(7!) ways

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

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Cheers,
Brent
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by swerve » Fri Jan 11, 2019 8:36 am
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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by Scott@TargetTestPrep » Mon Jan 21, 2019 5:34 pm
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!)
Since the 4 Lincoln books must be together, we can, for now, treat them as just 1 book. Since there are 6 other books, there are a total of 1 + 6 = 7 books, and hence there are 7! arrangements. However, within these 4 Lincoln books, there are 4! ways to arrange them. Therefore, the total number of arrangements of all 10 books, with the 4 Lincoln books together, is:

4! * 7!

Answer: C

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