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Math Help - What to do?

  1. #1
    Junior Member Dragon's Avatar
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    What to do?

    After rolling the first ball of a frame in a game 10-pin bollowing, how many pin configurations are physically possible?
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  2. #2
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    Quote Originally Posted by Dragon View Post
    After rolling the first ball of a frame in a game 10-pin bollowing, how many pin configurations are physically possible?
    Consider the following diagram,
    Code:
       *
      **
     ***
    ****
    In each star there are 10, but after you use 1 pin there are 9 left. For the next star there are 9 choices and so on....
    Thus,
     10\cdot 9\cdot 8\cdot ... \cdot 2\cdot 1=10!
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  3. #3
    Member Rimas's Avatar
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    what?
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  4. #4
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    Me and dragon are friends, and he didnt post the correct problem.

    It should read:
    After rolling the first ball of a frame in a game of 10 pin bowling, how many different pin configurations can remain (assuming all configurations are physically possible)?

    Assuming ThePerfectHacker's solution is still correct, could somebody explain it a little more clearly?

    (I'd make a new thread but I'm not sure if it violates the identical thread rule.)
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  5. #5
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by ceasar_19134 View Post
    Me and dragon are friends, and he didnt post the correct problem.

    It should read:
    After rolling the first ball of a frame in a game of 10 pin bowling, how many different pin configurations can remain (assuming all configurations are physically possible)?

    Assuming ThePerfectHacker's solution is still correct, could somebody explain it a little more clearly?

    (I'd make a new thread but I'm not sure if it violates the identical thread rule.)
    Try this:

    Draw the diagram of the pins:
    Code:
    o o o o 
     o o o
      o o
       o
    Now knock over a pin:
    Code:
    o o o o 
     o o o
      x o
       o
    That's 1 combination, now put that pin back up and knock over another pin:
    Code:
    o o x o 
     o o o
      o o
       o
    That's another combination, since you can knock over ten individual pins there are 10 physical combinations of pins when 1 is knocked over...

    Now what if 2 were knocked over?

    Well first thing to do is to draw the the pins with 1 knocked over:
    Code:
    o o o o 
     o o o
      o o
       x
    Now, knock another one over:
    Code:
    o o o o 
     o o o
      x o
       x
    Put it back up and knock another one over:
    Code:
    o o o o 
     o o x
      o o
       x
    If you keep doing this you'll realize that there are nine combinations..

    Thus you have 9 combinations per pin, since you have 10 pins you have 9(10) combinations for the whole thing, which is 90

    However every one of those combinations is counted twice, notice that if I knock down a pin:
    Code:
    o o o o 
     o x o
      o o
       o
    and knocked down another pin:
    Code:
    o o o o 
     o x o
      x o
       o
    it is the same combination as if I knocked down this pin:
    Code:
    o o o o 
     o o o
      x o
       o
    and then knocked down the second pin:
    Code:
    o o o o 
     o x o
      x o
       o
    So the number of combinations for 2 pins is 90/2, which equals 45

    So far you have 10+45=55 combinations

    more to come (going to do it in another post)...
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  6. #6
    MHF Contributor Quick's Avatar
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    New England
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    Quote Originally Posted by Quick View Post
    So the number of combinations for 2 pins is 90/2, which equals 45
    Now onto 3 pins, knock over 2 pins:
    Code:
    o o o o
     o o o
      x o
       x
    Now knock over another one:
    Code:
    o o o x
     o o o
      x o
       x
    Reset and do it again:
    Code:
    o x o o
     o o o
      x o
       x
    You might realize that there are 8 possible combinations.

    Thus the number of combinations for three pins is the number of combinations for 2 pins times 8, which is 45*8=360

    But wait! each combination is counted three times! so you divide the number by 3, thus you have 360/3=120 combinations

    Add that to the amount of combinations you have so far to get: 10+45+120=175

    Now you try doing the amount of combinations for 4, 5, 6, 7, 8, 9, and 10 pins, and solve from there!
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  7. #7
    MHF Contributor Quick's Avatar
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    New England
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    Quote Originally Posted by ceasar_19134 View Post
    Assuming ThePerfectHacker's solution is still correct, could somebody explain it a little more clearly?

    (I'd make a new thread but I'm not sure if it violates the identical thread rule.)
    1) It would have violated the identical thread rule

    2) I don't think PH's solution is correct
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