Difference between revisions of "Georgeooga-Harryooga Theorem"
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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. | 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. | ||
Revision as of 15:32, 19 November 2020
Contents
ICE MARTIX 2 PLES STOP
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.
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 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
Applications
Application 1
Problem
Zara has a collection of 
marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
Source 2020 AMC 8 Problem 10
Solutions
Solution 1
By the Georgeooga-Harryooga Theorem there are 
way to arrange the marbles.
Solution 2
We can arrange our marbles like so 
.
To arrange the 
and 
we have 
ways.
To place the 
and 
in the blanks we have 
ways.
By fundamental counting principle our final answer is 
Solution 3
Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by 
and , respectively. If we ignore the constraint that and cannot be next to each other, we get a total of 
ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that and can be next to each other. If we place and next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. 
). However, we could also have placed and in the opposite order (i.e. 
). Thus there are 6 ways of placing and directly next to each other. Next, notice that for each of these placements, we have two open slots for placing and . Specifically, we can place in the first open slot and in the second open slot or switch their order and place in the first open slot and in the second open slot. This gives us a total of 
ways to place and next to each other. Subtracting this from the total number of arrangements gives us 
total arrangements 
.
We can also solve this problem directly by looking at the number of ways that we can place and such that they are not directly next to each other. Observe that there are three ways to place and (in that order) into the four slots so they are not next to each other (i.e. 
). However, we could also have placed and in the opposite order (i.e. 
). Thus there are 6 ways of placing and so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing and . Specifically, we can place in the first open slot and in the second open slot or switch their order and place in the first open slot and in the second open slot. This gives us a total of ways to place and such that they are not next to each other .
~ junaidmansuri
Testimonials
"Thanks for rediscovering our theorem [User:Redfiretruck RedFireTruck]" - George and Harry of The Ooga Booga Tribe of The Caveman Society
