Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Georgeooga-Harryooga Theorem"
Line 36: Line 36:
 
Solution by [[User:Redfiretruck|RedFireTruck]]
 
Solution by [[User:Redfiretruck|RedFireTruck]]
 
<hr>
 
<hr>
−
<strong>ALL THINGS ABOVE EXCEPT FOR THE OVERVIEW TAB IS MADE BY RedFireTruck.</strong>
+
<strong>ALL THINGS ABOVE EXCEPT FOR THE OVERVIEW TAB AND THE PROBLEM IS MADE BY RedFireTruck </strong>

Revision as of 20:45, 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 AND THE PROBLEM IS MADE BY RedFireTruck

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Georgeooga-Harryooga_Theorem&oldid=138426"