# Difference between revisions of "Georgeooga-Harryooga Theorem"

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+ | <h1>Overview</h1> | ||

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This is not a legit theorem | This is not a legit theorem | ||

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+ | [@Sugar rush] Even though this is not a real theorem, it could be useful to use this, so I will bring parts of it back: | ||

+ | |||

+ | <h1>Definition</h1> | ||

+ | The Georgeooga-Harryooga Theorem states that if you have <math>a</math> distinguishable objects and <math>b</math> are kept away from each other, then there are <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math> ways to arrange the objects. | ||

+ | |||

+ | <h1>Proof</h1> | ||

+ | 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+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. | ||

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+ | By fundamental counting principal our answer is <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math>. | ||

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+ | Proof by [[User:Redfiretruck|RedFireTruck]] | ||

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+ | <h1>Application</h1> | ||

+ | |||

+ | Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream. Fred and George are identical twins, so they are indistinguishable. Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line. | ||

+ | With these conditions, how many different ways can you arrange these kids in a line? | ||

+ | |||

+ | Problem by Math4Life2020 | ||

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+ | <h2>Solution</h2> | ||

+ | |||

+ | If Eric and Fred were distinguishable we would have <math>\frac{(8-3)!(8-3+1)!}{(8-2\cdot3+1)!}=14400</math> ways to arrange them by the Georgeooga-Harryooga Theorem. However, Eric and Fred are indistinguishable so we have to divide by <math>2!=2</math>. Therefore, our answer is <math>\frac{14400}2=\boxed{7200}</math>. | ||

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+ | |||

+ | Solution by [[User:Redfiretruck|RedFireTruck]] | ||

+ | <hr> | ||

+ | <strong>ALL THINGS ABOVE EXCEPT FOR THE OVERVIEW TAB IS MADE BY RedFireTruck.</strong> |

## Revision as of 20:42, 24 November 2020

# Overview

This is not a legit theorem

[@Sugar rush] Even though this is not a real theorem, it could be useful to use this, so I will bring parts of it back:

# Definition

The Georgeooga-Harryooga Theorem states that if you have distinguishable objects and are kept away from each other, then there are ways to arrange the objects.

# Proof

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 and other objects so we have ways to arrange the objects we can't put together.

By fundamental counting principal our answer is .

Proof by RedFireTruck

# Application

Alice, Bob, Carl, David, Eric, Fred, George, and Harry want to stand in a line to buy ice cream. Fred and George are identical twins, so they are indistinguishable. Alice, Bob, and Carl had a serious disagreement in 6th grade, so none of them can be together in the line. With these conditions, how many different ways can you arrange these kids in a line?

Problem by Math4Life2020

## Solution

If Eric and Fred were distinguishable we would have ways to arrange them by the Georgeooga-Harryooga Theorem. However, Eric and Fred are indistinguishable so we have to divide by . Therefore, our answer is .

Solution by RedFireTruck

**ALL THINGS ABOVE EXCEPT FOR THE OVERVIEW TAB IS MADE BY RedFireTruck.**