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Say I have 5 computers. How many permutations are available?

  1. Apr 28, 2004 #1
    Say I have 5 computers. How many permutations are available?
     
  2. jcsd
  3. Apr 28, 2004 #2

    matt grime

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    Erm. that needs more information. permutations of what?
     
  4. Apr 28, 2004 #3
    Five computers are to be wired in a linear network.
     
  5. Apr 28, 2004 #4

    matt grime

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    And? That doesn't answer the question.
     
  6. Apr 28, 2004 #5
    permutations of the five computers
     
  7. Apr 28, 2004 #6

    matt grime

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    So you want to know how many ways there are to order five objects ? Well, that is very elementary, and it is 120, If you want to know more try reading about permutations and such and factorials.
     
  8. Apr 28, 2004 #7
    Thanks. I guess the "linear" thing threw me off.

    So would I be correct in assuming, that if I had twenty different programs, and could only access four of them at a time, that I would multiply 20 X 19 X 18 X 17; which equals 16,279,200 to find out many different ways I could arrange them?

    If so, I think I got the hang of it.
     
  9. Apr 28, 2004 #8

    matt grime

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    half if not all of the point of my reply was that in saying you have a linear network does not mean the slightest thing to me mathematically. My answer has no bearing on whether or not they are linear or arranged in the shape of a quincunx.
     
  10. May 6, 2004 #9

    Gokul43201

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    I believe the linear part is probably useful to distinguish the arrangement from say, a 'tree network', where you may have to designate masters and slaves; or a circular network, where you would have only 24 arrangements. I'm sure 'linear' was meant in a homotopic sense.
     
  11. May 6, 2004 #10

    matt grime

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    as a tree and a straight line are homotopic, to a point, I don't think so.
     
  12. May 6, 2004 #11

    Gokul43201

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    Yeah, that's right. Okay, so that's not what I meant, but you get the idea. 'Linear' is like n-pentane, and 2,2 -dimethylpropane is a 'tree', while cyclopentane is a 'circle'.
     
  13. Oct 9, 2007 #12
    About the programs question (it may be alittle late now sorry) but you have to use a equation for combination. the equation is C(n,r)=n!/(n-r)!r!. If you not familiar with this the ! mean factorials. N represents total number of thing and in this case it would be 20 and r represents the restriction which would be 4.

    You new equation will be C(20,4)=20!/(20-4)!4!

    Go from there and if you need any help email me
     
  14. Oct 9, 2007 #13

    HallsofIvy

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    No, the OP specifically asked about permutations. He was correct that he needed to use 20!/(20-4)!= 20!/16!= 20(19)(18)(17). Dividing by 4! would count number of different ways without including permutations of the same 4 programs.
     
  15. Oct 10, 2007 #14
    It asked for Per. on the first Question but on the program Q. it wouldnt matter which program started with which one. It would just want tto find out how many different 4 combination could be used in the the process.
     
  16. Oct 10, 2007 #15

    matt grime

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    Well, that makes no sense, but since this is a THREE year old post I don't suppose it matters.
     
  17. Oct 10, 2007 #16
    Ya it doesnt matter but i had the Question on my test the other day and got it right so i thought i would share it. Its kinda hard to explain on the thread.
     
  18. Oct 11, 2007 #17

    Shooting Star

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    You are quite correct. The two questions were different.

    And giving the correct answer to a problem, however old it may be, is always good.
     
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