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−  <h1>Overview</h1>
 +  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>]]) 
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−  This is not a legit theorem  
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−  <i>@Sugar rush</i> Even though this is not a real theorem, it could be useful to use this, so I will bring parts of it back:
 
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−  <h1>Definition</h1>
 
−  The GeorgeoogaHarryooga Theorem states that if you have <math>a</math> distinguishable objects and <math>b</math> are kept away from each other, then there are <math>\frac{(ab)!(ab+1)!}{(a2b+1)!}</math> ways to arrange the objects.
 
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−  <h1>Proof</h1>
 
−  Let our group of <math>a</math> objects be represented like so <math>1</math>, <math>2</math>, <math>3</math>, ..., <math>a1</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>\square1\square2\square3\square...\square ab1\square ab\square</math>.
 
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−  We have <math>(ab)!</math> ways to arrange the objects in that list.
 
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−  Now we have <math>ab+1</math> blanks and <math>b</math> other objects so we have <math>_{ab+1}P_{b}=\frac{(ab+1)!}{(a2b+1)!}</math> ways to arrange the objects we can't put together.
 
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−  By fundamental counting principal our answer is <math>\frac{(ab)!(ab+1)!}{(a2b+1)!}</math>.
 
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−  Proof by [[User:RedfiretruckRedFireTruck]]
 
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−  <h1>Application</h1>  
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−  Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream. Fred and George are identical twins, so they are indistinguishable. Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line.
 
−  With these conditions, how many different ways can you arrange these kids in a line?
 
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−  Problem by Math4Life2020
 
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−  <h2>Solution</h2>
 
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−  If Eric and Fred were distinguishable we would have <math>\frac{(83)!(83+1)!}{(82\cdot3+1)!}=14400</math> ways to arrange them by the GeorgeoogaHarryooga Theorem. However, Eric and Fred are indistinguishable so we have to divide by <math>2!=2</math>. Therefore, our answer is <math>\frac{14400}2=\boxed{7200}</math>.
 
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−  Solution by [[User:RedfiretruckRedFireTruck]]
 
−  <hr>  
−  <strong>ALL THINGS ABOVE EXCEPT FOR THE OVERVIEW TAB AND THE PROBLEM IS MADE BY RedFireTruck </strong>
 
Latest revision as of 15:54, 20 February 2021
Do not make false theorems. This may cause others to think of this as true and use it. ~crypto (talk)