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 $a$ distinguishable objects and $b$ are kept away from each other, then there are $\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$ ways to arrange the objects.

Proof

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

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

Solution by RedFireTruck

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

@@ Line 1: / Line 1: @@
+<h1>Overview</h1>
 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:
+<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.
+By fundamental counting principal our answer is <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math>.
+Proof by [[User:Redfiretruck|RedFireTruck]]
+<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
+<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>.
+Solution by [[User:Redfiretruck|RedFireTruck]]
+<hr>
+<strong>ALL THINGS ABOVE EXCEPT FOR THE OVERVIEW TAB IS MADE BY RedFireTruck.</strong>

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