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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 GeorgeoogaHarryooga Theorem states that if you have <math>a</math> distinguishable objects and <math>b</math> objects are kept away from each other, then there are <math>\frac{(ab)!(ab+1)!}{(a2b+1)!}</math> ways to arrange the objects in a line.
 
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−  Created by George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]
 
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−  =Proofs=
 
−  ==Proof 1==
 
−  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 The Fundamental Counting Principle our answer is <math>\frac{(ab)!(ab+1)!}{(a2b+1)!}</math>.
 
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−  Proof by [[User:RedFireTruck<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck<font color="#FF0000">talk</font>]]) 12:07, 1 February 2021 (EST)
 
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−  ==Proof 2==
 
−  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> be <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} = ab + 2</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} + 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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−  Proof by Aryabhata000
 
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−  =Applications=
 
−  ==Application 1==
 
−  ===Problem===
 
−  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.
 
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−  With these conditions, how many different ways can you arrange these kids in a line?
 
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−  Problem by [https://artofproblemsolving.com/community/c4h2342517p18900597 Math4Life2020]
 
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−  ===Solutions===
 
−  ====Solution 1====
 
−  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:RedFireTruck<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck<font color="#FF0000">talk</font>]]) 12:07, 1 February 2021 (EST)
 
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−  ==Application 2==
 
−  ===Problem===
 
−  Zara has a collection of <math>4</math> marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
 
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−  <math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math>
 
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−  Problem by [[2020 AMC 8 Problems/Problem 10The Mathematical Association of America's American Mathematics Competitions]]
 
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−  ===Solutions===
 
−  ====Solution 1====
 
−  By the [[GeorgeoogaHarryooga Theorem]] there are <math>\frac{(42)!(42+1)!}{(42\cdot2+1)!}=\boxed{\textbf{(C) }12}</math> way to arrange the marbles.
 
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−  Solution by [[User:RedFireTruck<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck<font color="#FF0000">talk</font>]]) 12:07, 1 February 2021 (EST)
 
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−  ====Solution 2====
 
−  We can arrange our marbles like so <math>\square A\square B\square</math>.
 
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−  To arrange the <math>A</math> and <math>B</math> we have <math>2!=2</math> ways.
 
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−  To place the <math>S</math> and <math>T</math> in the blanks we have <math>_3P_2=6</math> ways.
 
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−  By fundamental counting principle our final answer is <math>2\cdot6=\boxed{\textbf{(C) }12}</math>
 
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−  Solution by [[User:RedFireTruck<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck<font color="#FF0000">talk</font>]]) 12:07, 1 February 2021 (EST)
 
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−  ====Solution 3====
 
−  Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by <math>A,B,S,</math> and <math>T</math>, respectively. If we ignore the constraint that <math>S</math> and <math>T</math> cannot be next to each other, we get a total of <math>4!=24</math> ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that <math>S</math> and <math>T</math> can be next to each other. If we place <math>S</math> and <math>T</math> next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. <math>ST\square\square, \square ST\square, \square\square ST</math>). However, we could also have placed <math>S</math> and <math>T</math> in the opposite order (i.e. <math>TS\square\square, \square TS\square, \square\square TS</math>). Thus there are 6 ways of placing <math>S</math> and <math>T</math> directly next to each other. Next, notice that for each of these placements, we have two open slots for placing <math>A</math> and <math>B</math>. Specifically, we can place <math>A</math> in the first open slot and <math>B</math> in the second open slot or switch their order and place <math>B</math> in the first open slot and <math>A</math> in the second open slot. This gives us a total of <math>6\times 2=12</math> ways to place <math>S</math> and <math>T</math> next to each other. Subtracting this from the total number of arrangements gives us <math>2412=12</math> total arrangements <math>\implies\boxed{\textbf{(C) }12}</math>.<br>
 
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−  We can also solve this problem directly by looking at the number of ways that we can place <math>S</math> and <math>T</math> such that they are not directly next to each other. Observe that there are three ways to place <math>S</math> and <math>T</math> (in that order) into the four slots so they are not next to each other (i.e. <math>S\square T\square, \square S\square T, S\square\square T</math>). However, we could also have placed <math>S</math> and <math>T</math> in the opposite order (i.e. <math>T\square S\square, \square T\square S, T\square\square S</math>). Thus there are 6 ways of placing <math>S</math> and <math>T</math> so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing <math>A</math> and <math>B</math>. Specifically, we can place <math>A</math> in the first open slot and <math>B</math> in the second open slot or switch their order and place <math>B</math> in the first open slot and <math>A</math> in the second open slot. This gives us a total of <math>6\times 2=12</math> ways to place <math>S</math> and <math>T</math> such that they are not next to each other <math>\implies\boxed{\textbf{(C) }12}</math>.<br>
 
−  ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
 
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−  ====Solution 4====
 
−  Let's try complementary counting. There <math>4!</math> ways to arrange the 4 marbles. However, there are <math>2\cdot3!</math> arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, <cmath>4!2\cdot3!=\boxed{12 \textbf{(C)}}</cmath>
 
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−  ====Solution 5====
 
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−  We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a "super marble." There are <math>2!</math> ways to arrange Steelie and Tiger within this "super marble." Then there are <math>3!</math> ways to arrange the "super marble" and Zara's two other marbles in a row. Since there are <math>4!</math> ways to arrange the marbles without any restrictions, the answer is given by <math>4!2!\cdot 3!=\textbf{(C) }12</math>
 
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−  franzliszt
 
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−  ====Solution 6====
 
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−  We will use the following
 
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−  <math>\textbf{GeorgeoogaHarryooga Theorem:}</math> The [[GeorgeoogaHarryooga 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{(ab)!(ab+1)!}{(a2b+1)!}</math> ways to arrange the objects.
 
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−  <math>\textit{Proof. (Created by AoPS user RedFireTruck)}</math>
 
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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>a1</math>, <math>a</math>. Let the last <math>b</math> objects be the ones we can't have together.
 
−   
−  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.
 
−   
−  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:RedFireTruck<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck<font color="#FF0000">talk</font>]]) 12:07, 1 February 2021 (EST)
 
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−  Back to the problem. By the [[GeorgeoogaHarryooga Theorem]], our answer is <math>\frac{(42)!(42+1)!}{(42\cdot2+1)!}=\textbf{(C) }12</math>.
 
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−  franzliszt
 
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−  ====Solution 7====
 
−  https://youtu.be/pB46JzBNM6g
 
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−  ~savannahsolver  
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−  ==Application 3==
 
−  ===Problem===
 
−  The <math>7</math> members of The Ooga Booga Tribe, Lord Ooga Booga, Ooga, Booga, Foogle, Hoogle, George, and Harry, are about to perform a ritual. They have invited <math>2</math> priests, Agoob and Agoo, from a neighboring tribe. In this ritual they will line up in a row and sit down. The <math>2</math> priests must sit next to each other. Lord Ooga Booga, Ooga, and Booga just had a family argument so they must stay away from each other. In how many ways can The Ooga Booga Tribe perform their ritual.
 
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−  Problem by [https://artofproblemsolving.com/community/c3h2319596p19094529 RedFireTruck]
 
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−  ===Solutions===
 
−  ====Solution 1====
 
−  Let Agoob and Agoo be one object called Agooboo. Then, by the [[GeorgeoogaHarryooga Theorem]] there are <math>\frac{(83)!(83+1)!}{(82\cdot3+1)!}=14400</math> ways to arrange the <math>7</math> members and Agooboo. However, there are <math>2</math> ways to "split" Agooboo. So, by the Fundamental Counting Principle, our answer is <math>14400\cdot2=\boxed{28800}</math>.
 
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−  Solution by [[User:RedFireTruck<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck<font color="#FF0000">talk</font>]]) 12:07, 1 February 2021 (EST)
 
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−  =Testimonials=
 
−  "Thanks for rediscovering our theorem [[User:RedfiretruckRedFireTruck]]"  George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]
 
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−  "Wow! George and Harry are alive???" ~ samrocksnature
 
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−  "Hi" ~ jasperE3
 
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−  "I used this theorem on the AMC 8 and got a 25. Very useful!"  [[User:RedFireTruck<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck<font color="#FF0000">talk</font>]]) 11:01, 1 February 2021 (EST)
 
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−  "This is very complicated, but great."  Jiseop55406
 
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−  I used this theorem on the AMC 8 too! ~ ilp
 
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−  "Very nice theorem" ~ IdkHowToAddNumbers
 
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−  "This theorem is really worth of being used in the AMC 8! Very useful!" ~ [[User:Aopsg5gethsemanea2Aopsg5gethsemanea2]] ([[User talk:Aopsg5gethsemanea2talk]]) 20:48, 21 December 2020 (EST)
 
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−  "I have realized the way" ~ hi..
 
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−  "This theorem even works on AMC10 and 12"  TryhardMathlete
 
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−  "This is very cool"~Physicsisfun123
 
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−  "Interesting!" ~justin6688
 
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−  I used this on AMC 8 problem 10 and I got it right! THANK YOU! Onafets {Satire}
 