Difference between revisions of "Georgeooga-Harryooga Theorem"
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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> | + | 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. | ||
Revision as of 09:15, 18 November 2020
Definition
The Georgeooga-Harryooga Theorem states that if you have 
distinguishable objects and 
of them cannot be together, then there are 
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 objects be represented like so 
, 
, 
, ..., 
, . Let the last objects be the ones we can't have together.
Then we can organize our objects like so 
We have 
ways to arrange the objects in that list.
Now we have 
blanks so we have 
ways to arrange the objects we can't put together
By fundamental counting principal our final answer is .
Proof by RedFireTruck
