Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Cooga Georgeooga-Harryooga Theorem"

(→‎Definition)
Line 4: Line 4:
  
 
Created by George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]
 
Created by George and Harry of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]
 +
 +
=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>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>.
 +
 +
We have <math>(a-b)!</math> ways to arrange the objects in that list.
 +
 +
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.
 +
 +
By fundamental counting principal our answer is <math>\frac{(a-b)!(a-b)!}{(a-2b)!}</math>.
 +
 +
 +
Proof by [[User:RedFireTruck|RedFireTruck]]

Revision as of 19:25, 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 $\square1\square2\square3\square...\square a-b-1\square a-b\square$.

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Cooga_Georgeooga-Harryooga_Theorem&oldid=144198"