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Combination of Bowling Pins

  1. Aug 14, 2008 #1
    Hi! Me and some guys at work has a little argue about how many combinations you can have after that you have throwed away the first globe in bowling. There are 10 pins and 10 postions for them to stand. Are there any quations that works out for this?

    Here's an example of all the 10 pins, rised at all 10 positons:
    x x x x
    .x x x
    ..x x

    (I hope this is the right thread caus its not an homework)
    Last edited: Aug 14, 2008
  2. jcsd
  3. Aug 14, 2008 #2


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    If each pin is allowed to be 'up' or 'down', and

    . . . x
    .. . x
    ... .

    is different from, say,

    . . x .
    .. x .
    ... .

    then all you need to know is that each position has two states, up or down, and there are 10. This gives 2^10 = 1024 positions.

    If you combine certain arrangements, like the two above, you'll have fewer. If you additionally disallow 'impossible' arrangements like

    x x x x
    .x . x
    ..x x

    then you'll have fewer still.
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