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Difference between revisions of "Cooga Georgeooga-Harryooga Theorem"

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Now we have <math>a-b</math> blanks and <math>b</math> other objects so we have <math>_{a-b}P_{b}=\frac{(a-b)!}{(a-2b)!}</math> ways to arrange the objects we can't put together.
 
Now we have <math>a-b</math> blanks and <math>b</math> other objects so we have <math>_{a-b}P_{b}=\frac{(a-b)!}{(a-2b)!}</math> ways to arrange the objects we can't put together.
  
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By The Fundamental Counting Principal our answer is <math>\frac{(a-b)!(a-b)!}{(a-2b)!}</math>.
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By The Fundamental Counting Principal our answer is <math>\frac{(a-b)!^2}{(a-2b)!}</math>.
  
  
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Proof by [[User:RedFireTruck|RedFireTruck]]
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Proof by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 12:12, 1 February 2021 (EST)
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=Applications=
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==Application 1==
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===Problem===
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===Solutions===
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====Solution 1====
  
 
=Testimonials=
 
=Testimonials=
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"This is GREAT!!!" ~ hi..
 
"This is GREAT!!!" ~ hi..
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"This is a very nice theorem!" - [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FF0000">talk</font>]]) 10:53, 1 February 2021 (EST)
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Latest revision as of 17:41, 7 September 2024

Definition

The Cooga Georgeooga-Harryooga Theorem (Circular Georgeooga-Harryooga Theorem) states that if you have $a$ distinguishable objects and $b$ objects are kept away from each other, then there are $\frac{(a-b)!^2}{(a-2b)!}$ ways to arrange the objects in a circle.


Created by George and Harry of The Ooga Booga Tribe of The Caveman Society

Proofs

Proof 1

Let our group of $a$ objects be represented like so $1$, $2$, $3$, ..., $a-1$, $a$. Let the last $b$ objects be the ones we can't have together.

Then we can organize our objects like so [asy] label("$1$", dir(90)); label("BLANK", dir(60)); label("$2$", dir(30)); label("BLANK", dir(0)); label("$3$", dir(-30)); label("BLANK", dir(-60)); label("$\dots$", dir(-90)); label("BLANK", dir(-120)); label("$a-b-1$", dir(-150)); label("BLANK", dir(-180)); label("$a-b$", dir(-210)); label("BLANK", dir(-240)); [/asy]

We have $(a-b)!$ ways to arrange the objects in that list.

Now we have $a-b$ blanks and $b$ other objects so we have $_{a-b}P_{b}=\frac{(a-b)!}{(a-2b)!}$ ways to arrange the objects we can't put together.

By The Fundamental Counting Principal our answer is $\frac{(a-b)!^2}{(a-2b)!}$.


Proof by RedFireTruck (talk) 12:12, 1 February 2021 (EST)

Applications

Application 1

Problem

Solutions

Solution 1

Testimonials

"Thanks for rediscovering our theorem RedFireTruck" - George and Harry of The Ooga Booga Tribe of The Caveman Society

"This is GREAT!!!" ~ hi..

"This is a very nice theorem!" - RedFireTruck (talk) 10:53, 1 February 2021 (EST)

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