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Problem Solving With Permutations

  1. Feb 20, 2012 #1
    Hi, I have the following questions

    1) A coin is tossed 9 times. In how many ways could the results be six heads and three tails?

    2) A man bought two vanilla ice cream cones, three chocolate ones, four strawberry, and one butterscotch. In how many ways could he distribute them among his 10 children?

    3) Juan's soccer team played a total of 14 games this season. Their record was eight wins, four losses, and two ties. In how many orders could this have happened?

    I think for the above problems you take the total factorial and divide by the duplicate factorials. For ex for 1) 9! / 6! x 3! = 84

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Feb 20, 2012 #2

    kai_sikorski

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    To try to explain the formula think about it like this. Lets say the coins were labeled 1 to 9. Lets say we were lining them up on a table and on the table we marked out three slots for tails and six slots for heads.

    t t t | h h h h h h

    How many ways are there to arrange coins into these slots? Well you have 9 choices for the first slot, 8 for the next, etc. So there are 9! ways.

    But now we notice that for example

    1 2 3 | 4 5 6 7 8 9

    and

    2 1 3 | 4 5 6 7 8 9

    Correspond to the same coins being heads and the same set of coins being tails.

    Okay so say that you have a fixed set of 3 tail-coins, and a fixed set of 6 tail coins. How many permutations does that correspond to? Well line up the tail coins first, there are 3! ways to do that and then there are 6! possibilities for the head coins.

    Okay so

    TOTAL_PERMUTATIONS = NUMBER_OF_POSSIBLE_SETS * PERMUTATIONS_PER_SET

    9! = NUMBER_OF_POSSIBLE_SETS * (6!3!)
    NUMBER_OF_POSSIBLE_SETS = (9!)/(6!3!)

    So yes your formula is correct.
     
  4. Feb 20, 2012 #3

    kai_sikorski

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    Err sorry this was wrong
     
    Last edited: Feb 20, 2012
  5. Feb 20, 2012 #4

    tiny-tim

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    welcome to pf!

    hi koolkidx45! welcome to pf! :wink:
    yes, that's correct :smile:

    (and for three types, you have three factorials on the bottom instead of two)
     
  6. Feb 20, 2012 #5
    Thanks for all the help guys! So for my two other problems i would do the same thing? But the denominators would be different factorials? For ex.

    3) 14!/8!*4!*2!
     
  7. Feb 20, 2012 #6

    tiny-tim

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