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Bill has a set of encyclopedias with 26 volumes, one per letter of the

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Bill has a set of encyclopedias with 26 volumes, one per letter of the  [#permalink]

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New post 27 Nov 2019, 01:58
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Bill has a set of encyclopedias with 26 volumes, one per letter of the alphabet. He has a special shelf built for them with 26 slots in a row, each labeled alphabetically. After moving to a new house, Bill is faced with the task of putting the books back in their proper slots. He decides to do it in such a way that, during the process, there is never a gap between any of the books that are on the shelf. That is, each book he puts back after the first one is adjacent to a book that is already on the shelf. If he can start with any of the 26 books, how many ways can Bill accomplish his task?

A. \(26!\)

B. \(2^{25}\)

C. \(24!\)

D. \(26^2\)

E. \(14!12!\)


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Re: Bill has a set of encyclopedias with 26 volumes, one per letter of the  [#permalink]

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New post 15 Jan 2020, 06:06
1
ShankSouljaBoi wrote:
nick1816 wrote:
Total number of ways
= \(\frac{(n-1)!}{(n-1)!}+\frac{(n-1)!}{1!(n-2)!}+........+\frac{(n-1)!}{(n-2)!1!}+\frac{(n-1)!}{(n-1)!}\)
= \(2^{n-1}\)
=\(2^{25}\)



Bunuel wrote:
Bill has a set of encyclopedias with 26 volumes, one per letter of the alphabet. He has a special shelf built for them with 26 slots in a row, each labeled alphabetically. After moving to a new house, Bill is faced with the task of putting the books back in their proper slots. He decides to do it in such a way that, during the process, there is never a gap between any of the books that are on the shelf. That is, each book he puts back after the first one is adjacent to a book that is already on the shelf. If he can start with any of the 26 books, how many ways can Bill accomplish his task?

A. \(26!\)

B. \(2^{25}\)

C. \(24!\)

D. \(26^2\)

E. \(14!12!\)


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Hi chetan2u ,

Could you please help with this one.


Thanks :)


Hi..

Such questions have to have a pattern..
(I)
1) If there are 2 books -- 2 ways.\(=2^1=2^{2-1}\)
2) If 3 books, there are 4 ways --\(=2^2=2^{3-1}\) choose the corner ones A or C, then 1 way each AND when choose the middle one, two ways BCA or BAC.
3) Therefore, for 26 books \(2^{26-1}=2^{25}\)
(II)
There would be a length process, when you choose the first book initially or the second book or so on. calculate the number of ways
(III)
Logically if we look it other way, it may become a bit simpler
Way to place should be SAME as removing it. So if you place all 26, first book can be either of the corners, A or Z, so 2 ways. The moment we pick the first one, we again have two in the corners, so again TWO ways.
This will continue till last one is left, that can be picked in 1 way.
Total ways = \(2*2*2..(25 times)*1=2^{25} \)

B

(IV) I have not tried, but should work.
I can take the left ones as step upwards and right ones as step downwards, and now we have to move from one corner to opposite corner, where total steps are 25.
for example say we start from E, so 4 ( A,B,C,D) upwards and remaining sideways. E is in one corner and you have to take shortest route to the opposite corner
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Bill has a set of encyclopedias with 26 volumes, one per letter of the  [#permalink]

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New post 27 Nov 2019, 06:03
Total number of ways
= \(\frac{(n-1)!}{(n-1)!}+\frac{(n-1)!}{1!(n-2)!}+........+\frac{(n-1)!}{(n-2)!1!}+\frac{(n-1)!}{(n-1)!}\)
= \(2^{n-1}\)
=\(2^{25}\)



Bunuel wrote:
Bill has a set of encyclopedias with 26 volumes, one per letter of the alphabet. He has a special shelf built for them with 26 slots in a row, each labeled alphabetically. After moving to a new house, Bill is faced with the task of putting the books back in their proper slots. He decides to do it in such a way that, during the process, there is never a gap between any of the books that are on the shelf. That is, each book he puts back after the first one is adjacent to a book that is already on the shelf. If he can start with any of the 26 books, how many ways can Bill accomplish his task?

A. \(26!\)

B. \(2^{25}\)

C. \(24!\)

D. \(26^2\)

E. \(14!12!\)


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Re: Bill has a set of encyclopedias with 26 volumes, one per letter of the  [#permalink]

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New post 16 Dec 2019, 12:54
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1
nick1816 wrote:
Total number of ways
= \(\frac{(n-1)!}{(n-1)!}+\frac{(n-1)!}{1!(n-2)!}+........+\frac{(n-1)!}{(n-2)!1!}+\frac{(n-1)!}{(n-1)!}\)
= \(2^{n-1}\)
=\(2^{25}\)



Bunuel wrote:
Bill has a set of encyclopedias with 26 volumes, one per letter of the alphabet. He has a special shelf built for them with 26 slots in a row, each labeled alphabetically. After moving to a new house, Bill is faced with the task of putting the books back in their proper slots. He decides to do it in such a way that, during the process, there is never a gap between any of the books that are on the shelf. That is, each book he puts back after the first one is adjacent to a book that is already on the shelf. If he can start with any of the 26 books, how many ways can Bill accomplish his task?

A. \(26!\)

B. \(2^{25}\)

C. \(24!\)

D. \(26^2\)

E. \(14!12!\)


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Can you please explain your solution?I didn't understand it.

Thanks,
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Re: Bill has a set of encyclopedias with 26 volumes, one per letter of the  [#permalink]

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New post 14 Jan 2020, 04:57
nick1816 wrote:
Total number of ways
= \(\frac{(n-1)!}{(n-1)!}+\frac{(n-1)!}{1!(n-2)!}+........+\frac{(n-1)!}{(n-2)!1!}+\frac{(n-1)!}{(n-1)!}\)
= \(2^{n-1}\)
=\(2^{25}\)



Bunuel wrote:
Bill has a set of encyclopedias with 26 volumes, one per letter of the alphabet. He has a special shelf built for them with 26 slots in a row, each labeled alphabetically. After moving to a new house, Bill is faced with the task of putting the books back in their proper slots. He decides to do it in such a way that, during the process, there is never a gap between any of the books that are on the shelf. That is, each book he puts back after the first one is adjacent to a book that is already on the shelf. If he can start with any of the 26 books, how many ways can Bill accomplish his task?

A. \(26!\)

B. \(2^{25}\)

C. \(24!\)

D. \(26^2\)

E. \(14!12!\)


Are You Up For the Challenge: 700 Level Questions

Hi chetan2u ,

Could you please help with this one.


Thanks :)
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Bill has a set of encyclopedias with 26 volumes, one per letter of the  [#permalink]

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New post 14 Jan 2020, 10:49
Bunuel wrote:
Bill has a set of encyclopedias with 26 volumes, one per letter of the alphabet. He has a special shelf built for them with 26 slots in a row, each labeled alphabetically. After moving to a new house, Bill is faced with the task of putting the books back in their proper slots. He decides to do it in such a way that, during the process, there is never a gap between any of the books that are on the shelf. That is, each book he puts back after the first one is adjacent to a book that is already on the shelf. If he can start with any of the 26 books, how many ways can Bill accomplish his task?

A. \(26!\)

B. \(2^{25}\)

C. \(24!\)

D. \(26^2\)

E. \(14!12!\)


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here we have to place books such that no space is there between them

so first select a book and place it in the rack (since the position is determined with respect to the first book placed)
so now after placing the first book from second book we have to places for every book ie either extreme right or extreme left
hence 2^25(as there are 25 books left)
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Re: Bill has a set of encyclopedias with 26 volumes, one per letter of the  [#permalink]

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New post 14 Jan 2020, 14:34
I have a question.

It seems to me that the task will be completed in 26 steps.

Step 1: Select the first book. This can be done in 26 ways
Step 2: There are only 2 books you can select that leave no gaps, so the second book can be selected 2 ways
.
.
.
Step 26: You are at the last book, there is only 1 way this one can be selected

So why isn't the answer:

26*2^24*1

or written differently

13*2^25

?

Thank you in advance
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Re: Bill has a set of encyclopedias with 26 volumes, one per letter of the  [#permalink]

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New post 16 Jan 2020, 07:27
1
Bunuel wrote:
Bill has a set of encyclopedias with 26 volumes, one per letter of the alphabet. He has a special shelf built for them with 26 slots in a row, each labeled alphabetically. After moving to a new house, Bill is faced with the task of putting the books back in their proper slots. He decides to do it in such a way that, during the process, there is never a gap between any of the books that are on the shelf. That is, each book he puts back after the first one is adjacent to a book that is already on the shelf. If he can start with any of the 26 books, how many ways can Bill accomplish his task?

A. \(26!\)

B. \(2^{25}\)

C. \(24!\)

D. \(26^2\)

E. \(14!12!\)


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Solution:

Instead of starting with 26 volumes, let’s look at some easier examples that will allow us to develop a pattern which will be applicable for the entire set of 26 volumes. For example, let’s assume the set of encyclopedias has only 3 volumes: A, B, and C. If Bill starts with A, then he can only put them as A-B-C. If he starts with B, then he can put them as B-A-C or as B-C-A. If he starts with C, then he can only put them as C-B-A. We see that if there are 3 volumes, there are 4 ways to put the books onto the shelf. (Note: The three letter arrangements with hyphens, B-A-C for example, means the order the books are put onto the shelf. In this case, B is the first book to put onto the shelf, A is the second and C is the third. It doesn’t mean the arrangement of the books on the shelf is BAC since, after all, when all the books are put onto the shelf, the arrangement, from left to right, should be ABC.)

Now let’s say there are 4 volumes: A, B, C and D. If Bill starts with A, then he can only put them as A-B-C-D. If he starts with B, then he can put them as B-A-C-D, B-C-A-D or as B-C-D-A. If he starts with C, then he can put them as C-B-A-D, C-B-D-A, C-D-B-A. Finally, if he starts with D, then he can only put them as D-C-B-A.We see that if there are 4 volumes, there are 8 ways to put the books onto the shelf.

From this, we can see a pattern. Let n be the number of volumes. If n = 3, the total number of ways is 4 = 2^2 = 2^(3 - 1). If n = 4, the total number of ways is 8 = 2^3 = 2^(4 - 1). Therefore, if n = 26, the total number of ways is 2^(26 - 1) = 2^25.

Alternate Solution:

Notice that the number of ways Bill can shelvethe books is the same as the number of ways he can unshelve the books without leaving any gaps. To see that, suppose B1 - B2 - B3 - … - B26 is one of the allowable ways of putting the books on shelves. Then, we can simply reverse the order and remove books from the shelf in the order B26 - B25 - … - B1. Using this fact, we will count the number of ways he can unshelve the books.

Beginning with the 26 books on the shelf, he can remove either the leftmost book or the right most book on the shelf in order to avoid leaving any gaps. He has two choices for the first book.

Once the first book is removed, he can remove either the leftmost book or the right most book from the remaining 25 books. He has two choices for the second book.

Continuing the pattern, he will have 2 choices for all but the last book. When there is only one book left at the shelf, he will have only one choice. Thus, the number of ways Bill can accomplish the task is 2^25.

Answer: B
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Re: Bill has a set of encyclopedias with 26 volumes, one per letter of the   [#permalink] 16 Jan 2020, 07:27
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