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1st combinatorial problem

  1. Jul 18, 2007 #1
    1. The problem statement, all variables and given/known data
    1. A carpenter has 3 identical hammers, 5 different screwdrivers, 2 identical mallets, 2 different saws and a tape-measure. She wishes to hang the tools in a row on a tool rack on the wall. In how many ways can this be done if the first and last positions on the rack are to be mallets and the hammers are not to be next to each other?


    2. Relevant equations
    The number of ways of arranging n objects which include 'a' identical objects of one type, 'b' identical objects of another type,.... is
    n!/(a!b!....)


    n objects divided into m groups with each group having G1, G2, ..., Gm objects respectively has m! * G1! * G2! * ... *Gm!



    3. The attempt at a solution
    Since the mallets are identical and there are only 2, we don't have to worry about them. We can reduce the problem to 11 objects to be arranged. Since out of the 11 objects, 3 are identical which are the hammers we have a total of 11!/3! ways of permuting the 11 objects. However we don't want the
    hammers to be next to each other.

    So calculate the ways they are next to each other. We have 9 groups of objects. As the hammers are identical and must all be next to each other in a threesome, we have 9! ways of permuting the 11 objects. So we subtract the cases when the hammers are next to each other.
    11!/3!-9!=6289920

    However the answers suggested 13063680 ways.

    I can't see what is wrong with my reasoning.
     
  2. jcsd
  3. Jul 18, 2007 #2

    VietDao29

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    Homework Helper

    Yeah, looks nearly correct to me. You forget to drop out the ways, in which only 2 hammers go together, and the 3rd one is not next to any hammer.

    Btw, what book are you using? I don't think you should trust the answer provided by that book. Since, in here, I see at least 2 of 4 answers the book proposed are wrong, i.e 50% wrong. :bugeye: So, don't trust it. o:)
     
  4. Jul 18, 2007 #3
    Very good. I didn't take that into accout. I'll do the calculation as total permutation minus all cases when two hammers are together since that will include cases when 3 hammers are together. When two hammers are next to each other , there are 10! different permutations so 11!/3!-10!=3024000
     
    Last edited: Jul 18, 2007
  5. Jul 18, 2007 #4

    VietDao29

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    Homework Helper

    Well, that's what I get, too. :)
     
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