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Combinatorics problem

  1. Sep 14, 2014 #1
    Hi, just a simple combinatorics problem I cant figure out how to do!

    We want to place n books, of which m are broken, on a bookshelf so that there are at least 2 consecutive broken books. The broken books are indissociable from one another and so are the good book, how many ways can we do this

    Thank you in advance!
     
  2. jcsd
  3. Sep 14, 2014 #2

    mathman

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    What does this word mean? I suggest you rewrite the last sentence.
     
  4. Sep 14, 2014 #3
    I am sorry, I am French-Canadian and didnt know how to write this word in English (thought it was the same thing :P), I meant that the broken books are all exactly the same and cant be told apart from each other and so are the books in good condition!
     
  5. Sep 15, 2014 #4

    mathman

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    Please clarify the original question. Do you mean any arrangement as long as at least two broken books are together? If m > (n+1)/2, all arrangements will work.
     
  6. Sep 16, 2014 #5
    Ok I will rewrite it as an exact translation of the original question: We want to place n books, of which m are broken, on a shelf so that there are at least 2 consecutive broken books. The borken books cannot be told apart one from another, and that is also the case for the books in good condition. In how many ways can we do this?

    Now this is a problem given to me by my teacher's assistant who doesn't explain as well as the teacher. She gave us the answer, but I can't figure out how she got to it, this is it

    n!/(m!(n-m))! - ((n-m+1) choose m) or n!/(m!(n-m))! - (n-m+1)!/(m!(n-2m+1)!)

    The only thing I know is that the +1 at the end of the second term is for if we placed a broken book at the edge of the shelf because she said this in class. Otherwise, isn't n choose m (the first term) the different ways to select m books from n, but what does that signify in this case as the m books will all be on the shelf anyways? So that's that thank you in advance!
     
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