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This might be basic, but I'm still having a hard time understanding it.
Say 100 pieces of pie has to be distibuted to 100 different plates and the way they are distributed are random. (n=100, r=100)
Then what is the chance of all the pie pieces to be distributed evenly? I know that there are 100^100 (n^r) different ways of them to be distributed total, but don't really know how to find a formula for them not to "collide"
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[eqn] \prod_{k=1}^{100} \frac{101-k}{100} [/eqn]
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Try your problems with 3 pieces of cake.
It's a piece of cake.
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>>8571278
Alright. Let's say 3 pieces of cake has to be distributed on 10 plates then. There are 10^3 (1000) different ways that the pieces can be distributed.
I still don't know how to calculate the probability of then getting their own plate. I figured I have to divide with 1000, the total amount of combinations, but how?
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>>8571319
Just think I figured it out.
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[math]\frac{1}{\binom{n+r-1}{n}}[/math]
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>>8571336
p = number of plates / pies
There are p^p ways the pies can be distributed. The number of possible ways the pies can be distributed evenly is p!. The probability would be (p!)/p^p
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[math]\_frac{1}{\binom{n+r-1}{n}}[/math]
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[math]\frac_{1}{\binom{n+r-1}{n}}[/math]
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>>8571237
here's my attempt. the number of plates has to divide the number of pies otherwise they can't be evenly distributed
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Sorry everyone, forgot to post what my orignal solution was. I came to this after some thinking and calculation:
n = number of plates
r = number of pie pieces
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