Cooga Georgeooga-Harryooga Theorem

Revision as of 11:09, 7 December 2021 by Jwelsh (talk | contribs) (Reverted edits by Cozzmo (talk) to last revision by Sosiaops)

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)