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Permutations and Combinations

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data

    7 distinct flags are hoisted in a post. Find the number of ways of arranging them if
    2 of the flags must be separated. (The answer is 3600)

    2. Relevant equations

    Permutation: n!
    Circular Permutation: (n-1)!

    3. The attempt at a solution

    1: Flag 1, 2: Flag 2:

    The arrangements should be:

    1 X 2 X X X X
    X 1 X 2 X X X ...

    2 X 1 X X X X
    X 2 X 1 X X X ...

    1 X X 2 X X X
    X 1 X X 2 X X ...

    2 X X 1 X X X
    X 2 X X 1 X X ...

    ......

    I don't know how to calculate the solution provided.

    Can anyone tell me how to solve the above problem?

    Thank you very much!
     
  2. jcsd
  3. Feb 28, 2009 #2

    CompuChip

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    Science Advisor
    Homework Helper

    I don't know if this is the way you are supposed to find, but for me a simple counting argument always works.

    Suppose that the first flag is not placed in the first or last position. How many possibilities are there? For each choice, how many possibilities are there to place the second flag? For each fixed placement of flag 1 and 2, how many possibilities are there to arrange the remaining 5?
     
  4. Feb 28, 2009 #3
    1) (6 x 1 x 5 x 4 x 3 x 2 x 1) x 5 ?

    2) Don't know

    3) 5! ?
     
  5. Feb 28, 2009 #4

    CompuChip

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    3) is right. For any placement of the two marked flags, you have 5! possibilities to arrange the rest. So this will give you 5! x (number of placements for the first two flags).

    For 1) you are thinking much too complicated. I have 7 empty places, of which there are five which have two neighbours (I am excluding the case where flag 1 is in the first or last position for a moment). I want to put the first flag in one of them. How many ways are there in which I can do that.
    Then how many spots out of the 7 are still allowed for the second flag?
     
  6. Feb 28, 2009 #5
    How many arrangements have flag 1 adjacent to flag 2?
     
  7. Feb 28, 2009 #6
    1) 5 ?
     
  8. Feb 28, 2009 #7
    Treating Flag 1 and Flag 2 as one unit, there are (5+1)! arrangements available.

    However, inside that unit, there are 2! arrangements available.

    Totally, we have (5+1)! x 2! arrangements?
     
  9. Feb 28, 2009 #8
    davieddy, is that the solution should be 7! - (5+1)! x 2! = 3600?

    Thank you very much!
     
  10. Mar 1, 2009 #9
    Well done.
    Compuchip was trying to guide you through the direct
    approach whiich you were trying originally. It gives:

    (5*4 +2*5) * 5! = 3600

    Not really too complicated, but messier.

    David
     
  11. Mar 1, 2009 #10
    Number of ways of locating flags 1 and 2 non adjacent:

    5*4 + 2*5 = 7*6 - 6*2 = 30

    Nothing in it really!
    Perhaps there are further different ways of counting it.

    David

    BTW * is used in computing/math instead of x for a reason
    too obvious to mention.
     
    Last edited: Mar 1, 2009
  12. Mar 1, 2009 #11
    Actually, I don't understand why you can get 5*4 + 2*5 OR 7*6 - 6*2.

    And why does the first flag not place in the first or last position?
     
  13. Mar 1, 2009 #12
    (7*6 - 6*2)*5! is what you worked out correctly earlier.

    If flag one is not at one end, then there are 4 places to put flag 2
    which are not adjacent.
    He (Compuchip) went on to ask how many ways can you place flag 2
    with flag 1 at an end. Answer 2"5.
     
  14. Mar 1, 2009 #13
    I got it!

    Thank CompuChip & davieddy very much!
     
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