Difference between revisions of "Georgeooga-Harryooga Theorem"
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+ | =BEFORE YOU CONTINUE READING, THIS IS NOT A REAL THEOREM–IT ONLY EXISTS IN REDFIRETRUCK'S MIND= | ||
=Definition= | =Definition= | ||
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. | ||
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We can also solve this problem directly by looking at the number of ways that we can place <math>S</math> and <math>T</math> such that they are not directly next to each other. Observe that there are three ways to place <math>S</math> and <math>T</math> (in that order) into the four slots so they are not next to each other (i.e. <math>S\square T\square, \square S\square T, S\square\square T</math>). However, we could also have placed <math>S</math> and <math>T</math> in the opposite order (i.e. <math>T\square S\square, \square T\square S, T\square\square S</math>). Thus there are 6 ways of placing <math>S</math> and <math>T</math> so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing <math>A</math> and <math>B</math>. Specifically, we can place <math>A</math> in the first open slot and <math>B</math> in the second open slot or switch their order and place <math>B</math> in the first open slot and <math>A</math> in the second open slot. This gives us a total of <math>6\times 2=12</math> ways to place <math>S</math> and <math>T</math> such that they are not next to each other <math>\implies\boxed{\textbf{(C) }12}</math>.<br> | We can also solve this problem directly by looking at the number of ways that we can place <math>S</math> and <math>T</math> such that they are not directly next to each other. Observe that there are three ways to place <math>S</math> and <math>T</math> (in that order) into the four slots so they are not next to each other (i.e. <math>S\square T\square, \square S\square T, S\square\square T</math>). However, we could also have placed <math>S</math> and <math>T</math> in the opposite order (i.e. <math>T\square S\square, \square T\square S, T\square\square S</math>). Thus there are 6 ways of placing <math>S</math> and <math>T</math> so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing <math>A</math> and <math>B</math>. Specifically, we can place <math>A</math> in the first open slot and <math>B</math> in the second open slot or switch their order and place <math>B</math> in the first open slot and <math>A</math> in the second open slot. This gives us a total of <math>6\times 2=12</math> ways to place <math>S</math> and <math>T</math> such that they are not next to each other <math>\implies\boxed{\textbf{(C) }12}</math>.<br> | ||
~ junaidmansuri | ~ junaidmansuri | ||
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Revision as of 16:06, 19 November 2020
Contents
BEFORE YOU CONTINUE READING, THIS IS NOT A REAL THEOREM–IT ONLY EXISTS IN REDFIRETRUCK'S MIND
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.
An imaginary theorem created by the made up people George and Harry of The Ooga Booga Tribe of The Caveman Society. These people only exist in redfiretruck’s mind. ~icematrix
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 by RedFireTruck
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 by RedFireTruck
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