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| − | =Definition=
| + | Do not make false theorems. This may cause others to think of this as true and use it. ~[[User:cryptographer|<font color="#FF2998">crypto</font>]] ([[User talk:Cryptographer|<font color="#FF0000">talk</font>]]) |
| − | The Georgeooga-Harryooga Theorem states that if you have <math>a</math> distinguishable objects and <math>b</math> of them cannot be together, then there are <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math> ways to arrange the objects.
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| − | Created by George and Harry of The Ooga Booga Tribe of The Caveman Society
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| − | =Proofs=
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| − | =Proof 1=
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| − | Let our group of <math>a</math> objects be represented like so <math>1</math>, <math>2</math>, <math>3</math>, ..., <math>a-1</math>, <math>a</math>. Let the last <math>b</math> objects be the ones we can't have together.
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| − | Then we can organize our objects like so <math>1\square2\square3\square...\square a-b-1\square a-b</math>
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| − | We have <math>(a-b)!</math> ways to arrange the objects in that list.
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| − | Now we have <math>a-b+1</math> blanks so we have <math>_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}</math> ways to arrange the objects we can't put together
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| − | By fundamental counting principal our final answer is <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math>.
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| − | Proof by RedFireTruck
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Revision as of 16:54, 20 February 2021
Do not make false theorems. This may cause others to think of this as true and use it. ~crypto (talk)
