How many combinations can be made with 10 bowling pins in 10 positions?

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SUMMARY

The discussion centers on calculating the number of combinations for 10 bowling pins in 10 positions, where each pin can either be 'up' or 'down'. The formula used is 2^10, resulting in 1024 possible combinations. However, when considering specific arrangements and disallowing 'impossible' configurations, the total number of valid combinations decreases. This highlights the importance of defining constraints in combinatorial problems.

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  • Understanding of binary states (up/down) in combinatorial contexts
  • Familiarity with basic combinatorial mathematics
  • Knowledge of constraints in arrangement problems
  • Ability to visualize and represent arrangements
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  • Research combinatorial mathematics and its applications in real-world scenarios
  • Explore the concept of permutations and combinations in greater detail
  • Learn about constraints in combinatorial problems and how they affect outcomes
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This discussion is beneficial for mathematicians, game designers, and anyone interested in combinatorial analysis or optimization problems involving arrangements and constraints.

oNesterud
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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
...x

(I hope this is the right thread caus its not an homework)
 
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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
...x

then you'll have fewer still.
 

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