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Difference between revisions of "Georgeooga-Harryooga Theorem"

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==Proof 1==
 
==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.
 
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 <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.
 
We have <math>(a-b)!</math> ways to arrange the objects in that list.
 +
 
Now we have <math>a-b+1</math> blanks and <math>b</math> other objects so we have <math>_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}</math> ways to arrange the objects we can't put together.
 
Now we have <math>a-b+1</math> blanks and <math>b</math> other objects so we have <math>_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}</math> ways to arrange the objects we can't put together.
 +
 
By fundamental counting principal our answer is <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math>.
 
By fundamental counting principal our answer is <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math>.
  
  
 
Proof by RedFireTruck
 
Proof by RedFireTruck

Revision as of 10:36, 18 November 2020

Definition

The Georgeooga-Harryooga Theorem states that if you have $a$ distinguishable objects and $b$ of them cannot be together, then there are $\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$ ways to arrange the objects.


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 $a-b+1$ blanks and $b$ other objects so we have $_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}$ ways to arrange the objects we can't put together.

By fundamental counting principal our answer is $\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$.


Proof by RedFireTruck

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