Revision as of 19:34, 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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Difference between revisions of "Cooga Georgeooga-Harryooga Theorem"

Revision as of 19:34, 31 January 2021

Contents

Definition

Proofs

Proof 1

Testimonials

@@ Line 11: / Line 11: @@
 Then we can organize our objects like so <asy>
 label("$1$", dir(90));
-label("_", dir(60));
+label("BLANK", dir(60));
 label("$2$", dir(30));
-label("_", dir(0));
+label("BLANK", dir(0));
 label("$3$", dir(-30));
-label("_", dir(-60));
+label("BLANK", dir(-60));
 label("$\dots$", dir(-90));
-label("_", dir(-120));
+label("BLANK", dir(-120));
 label("$a-b-1$", dir(-150));
-label("_", dir(-180));
+label("BLANK", dir(-180));
 label("$a-b$", dir(-210));
-label("_", dir(-240));
+label("BLANK", dir(-240));
 </asy>