Revision as of 10:05, 18 November 2020

Definition

The Georgeooga-Harryooga Theorem states that if you have $a$ distinguishable objects and $b$ of them cannot be together, then there are $\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$ 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 $a$ 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 $1\square2\square3\square...\square a-b-1\square a-b$

We have $(a-b)!$ ways to arrange the objects in that list.

Now we have $a-b+1$ blanks 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 final answer is $\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$ .

@@ Line 4: / Line 4: @@
 Created by George and Harry of The Ooga Booga Tribe of The Caveman Society
+=Proofs=
+=Proof 1=
+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>1\square2\square3\square...\square a-b-1\square a-b</math>
+We have <math>(a-b)!</math> ways to arrange the objects in that list.
+Now we have <math>a-b+1</math> blanks 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 final answer is <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math>.

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Difference between revisions of "Georgeooga-Harryooga Theorem"

Revision as of 10:05, 18 November 2020

Definition

Proofs

Proof 1