# Difference between revisions of "Georgeooga-Harryooga Theorem"

Redfiretruck (talk | contribs) (Undo revision 140416 by Sugar rush (talk)) (Tag: Undo) |
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− | + | <h1>Overview</h1> | |

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+ | This is not a legit theorem | ||

+ | <i>@Sugar rush</i> 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> | |

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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. | ||

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

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+ | <h1>A side note by aryabhata000:</h1> | ||

+ | This can also be done by stars and bars like so: | ||

Let us call the <math>b</math> people <math>1, 2, ... b</math> | Let us call the <math>b</math> people <math>1, 2, ... b</math> | ||

− | Let the number of people before <math>1</math> in line be <math>y_1</math>, between <math>1, 2</math> be <math>y_2</math>, ... after <math>b</math> | + | Let the number of people before <math>1</math> in line be <math>y_1</math>, between <math>1, 2</math> be <math>y_2</math>, ... after <math>b</math> b3 <math>y_{b+1}</math>. |

We have <cmath>y_1 + y_2 + y_3 + \dots y_{b+1} = a-b</cmath> | We have <cmath>y_1 + y_2 + y_3 + \dots y_{b+1} = a-b</cmath> | ||

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Therefore, the final answer is <cmath>b! (a-b)! F(a,b) = \frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</cmath> | Therefore, the final answer is <cmath>b! (a-b)! F(a,b) = \frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</cmath> | ||

+ | <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. | |

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− | 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. | ||

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

+ | Problem by Math4Life2020 | ||

− | + | <h2>Solution</h2> | |

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− | Solution | ||

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− | <math>\ | + | 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]] | Solution by [[User:Redfiretruck|RedFireTruck]] | ||

− | + | <hr> | |

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

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## Revision as of 14:01, 30 December 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

# A side note by aryabhata000:

This can also be done by stars and bars like so: Let us call the people

Let the number of people before in line be , between be , ... after b3 . We have

The number of ways to determine is equivalent to the number of positive integer solutions to: where and .

So, by stars and bars, the number of ways to determine is

Furthermore, after picking positions for the people, we have ways to order the people who can be together, and ways to order the people who cannot be together. So for each , we have orderings.

Therefore, the final answer is

# 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 AND THE PROBLEM IS MADE BY RedFireTruck **