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− | =Definition= | + | Do not make false theorems. This may cause others to think of this as true and use it. ~[[User:cryptographer|<font color="#FF2998">crypto</font>]] ([[User talk:Cryptographer|<font color="#FF0000">talk</font>]]) |
− | The Georgeooga-Harryooga Theorem states that if you have <math>a</math> distinguishable objects and <math>b</math> of them cannot be together, then there are <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math> ways to arrange the objects.
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− | Created by George and Harry of The Ooga Booga Tribe of The Caveman Society
| + | This theorem is too op to delete - [[User:Onafets|<font color ="#0000FF">Onafets</font>]] |
− | =Proofs=
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− | ==Proof 1==
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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.
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− | Then we can organize our objects like so <math>\square1\square2\square3\square...\square a-b-1\square a-b\square</math>.
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− | We have <math>(a-b)!</math> ways to arrange the objects in that list.
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− | 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.
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− | By fundamental counting principal our answer is <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math>.
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− | Proof by RedFireTruck
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− | =Applications=
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− | ==Application 1==
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− | ===Problem===
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− | Zara has a collection of <math>4</math> 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?
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− | <math>\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24</math>
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− | (Source [[2020 AMC 8 Problems/Problem 10]])
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− | ===Solutions===
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− | ====Solution 1====
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− | By the Georgeooga-Harryooga Theorem there are <math>\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{\textbf{(C) }12}</math> way to arrange the <math>4</math> marbles.
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Revision as of 12:41, 2 March 2021
Do not make false theorems. This may cause others to think of this as true and use it. ~crypto (talk)
This theorem is too op to delete - Onafets
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