Revision as of 09:36, 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 $\square1\square2\square3\square...\square a-b-1\square a-b\square$

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

@@ Line 13: / Line 13: @@
 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
+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.
-By fundamental counting principal our final answer is <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math>.
+By fundamental counting principal our answer is <math>\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}</math>.
 Proof by RedFireTruck

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Difference between revisions of "Georgeooga-Harryooga Theorem"

Revision as of 09:36, 18 November 2020

Definition

Proofs

Proof 1