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Difference between revisions of "Cooga Georgeooga-Harryooga Theorem"
Line 9: Line 9:
 
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.
 
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.
  
−
Then we can organize our objects like so <math>\square1\square2\square3\square...\square a-b-1\square a-b\square</math>.
+
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 <math>(a-b)!</math> ways to arrange the objects in that list.
 
We have <math>(a-b)!</math> ways to arrange the objects in that list.

Revision as of 19:33, 31 January 2021

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 $\frac{(a-b)!(a-b)!}{(a-2b)!}$ 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 fundamental counting principal our answer is $\frac{(a-b)!(a-b)!}{(a-2b)!}$.


Proof by RedFireTruck

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