Probability question

perusal

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Ten books are placed in random order on a bookshelf. Find the probability of three given books being side by side.

I can't seem to arrive at the right answer. Could someone please walk me through this question?

Thank you
 

Subhotosh Khan

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Ten books are placed in random order on a bookshelf. Find the probability of three given books being side by side.

I can't seem to arrive at the right answer. Could someone please walk me through this question?

Thank you
Take those three books and tie those together to pretend you have one book.

Now you have 8 books arrange.

How many ways can you arrange those?
 

Jomo

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Ten books are placed in random order on a bookshelf. Find the probability of three given books being side by side.

I can't seem to arrive at the right answer. Could someone please walk me through this question?

Thank you
Sure we could walk you through the problem. But 1st you need to tell us exactly where you are stuck so that we can give you leading hints to continue. It is you who will solve this problem, not us.
 

perusal

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Sure we could walk you through the problem. But 1st you need to tell us exactly where you are stuck so that we can give you leading hints to continue. It is you who will solve this problem, not us.

Thanks.

Well, my understanding is that there are 10!/3!*7! possible ways to arrange the books so that 3 given books are together. I make this 120 possible combinations.

I then figure that you can select the 3 given books in 3!/1!*1!*! ways or 6 ways.

So with N(A)/N I have 6/120 == 1/20

The textbook informs me that the answer is 1/15 so I know I know my reasoning is incorrect.
 

perusal

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Take those three books and tie those together to pretend you have one book.

Now you have 8 books arrange.

How many ways can you arrange those?
I would make that 8! However, the order in which the 3 books are arranged with respect to each other is inconsequential, or so I believe. Would that mean the I then have and an equation of 3!/1!*1!*1! to fit in somewhere? My understanding is that you can arrange 3 distinct objects in 6 different combinations and they will still be next to one another.
 

perusal

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I think i got it

Sure we could walk you through the problem. But 1st you need to tell us exactly where you are stuck so that we can give you leading hints to continue. It is you who will solve this problem, not us.
I think I may have worked it out.

You can shift a block of 3 books to the right 8 times to get them from one end of the shelf to the other. With each iteration, there are 7! possible locations for the other books. Therefore 8 * 7! possibilities or 8! I then multiply 8! by 3! to allow for the possible positions the 3 given books may be in and still be side by side. My N(A) = 8!*3!

My N is simply equal to 10! the number of ways 10 distinct objects can be arranged.

N(A)/N = 1/15

Does this look correct?
 
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