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−  <h1>Overview</h1>
 
   
−  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>A side note by aryabhata000:</h1>
 
−  This can also be done by stars and bars like so:
 
−  Let us call the <math>b</math> people <math>1, 2, ... b</math>
 
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−  Let the number of people before <math>1</math> in line be <math>y_1</math>, between <math>1, 2</math> be <math>y_2</math>, ... after <math>b</math> b3 <math>y_{b+1}</math>.
 
−  We have <cmath>y_1 + y_2 + y_3 + \dots y_{b+1} = ab</cmath>
 
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−  The number of ways to determine <math>y_1, y_2, \dots</math> is equivalent to the number of positive integer solutions to:
 
−  <cmath>x_1 + x_2 + .. + x_{b+1}</cmath> where <math>(x_2, ... x_b) = (y_2, ..., y_b) </math> and <math>(x_1, x_{b+1}) = (y_1 +1, y_{b+1})</math>.
 
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−  So, by stars and bars, the number of ways to determine <math>(y_2, ..., y_b) </math> is <cmath>F(a,b) = \dbinom{ab+1}{b} = \frac {(ab+1)!}{b!(a2b+1)!}</cmath>
 
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−  Furthermore, after picking positions for the people, we have <math>(ab)!</math> ways to order the <math>(ab)</math> people who can be together, and <math>b!</math> ways to order the <math>b</math> people who cannot be together. So for each <math>(y_1, y_2, ... y_{b+1}</math>, we have <math>b! (ab)!</math> orderings.
 
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−  Therefore, the final answer is <cmath>b! (ab)! F(a,b) = \frac{(ab)!(ab+1)!}{(a2b+1)!}</cmath>
 
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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>
 