Thread: Geometery Formula for number of edges

HELLO, I REALLY NEED HELP TO FIND THE GEOMETERY FORMULA FOR THIS EQUATION. I FOULD OUT IT HAS TO BE AN EXPONOTIONAL FORMULA. HERE ARE THE SPEFICS...


I have a figure, if it has one vertex, it has 0 edges,so...
Vertex-----------edge
1-----------------0
2-----------------1
3-----------------3
4-----------------6
5----------------10
6----------------15
7----------------28(i think, im not sure)
8----------------(do not know, could not get)


please help to find a formula where i could find the number of edges with the number of vertexes and vice versa. I am toatally lost. this is an un directed graph. If possible please include graphs so i may undersatnd. also is a formaul and a function the same cause it says to find the function, thanxs!!!

This is funny, my class just went over this.

Connect all the vertices. Tell me, if there are 5 vertices, how many line segments touch each vertex? the answer is 4.

If there are 6 vertices, then the answer is 5.

If there are 7, then the answer is 6.

Do you understand what I did (I have not answered the problem yet, I've merely gone through the first, most important part)? If not, then I will go into more detail...

i have understood somewat of wat u said, please explain the whole, the formula please, thanxs

Originally Posted by OnMyWayToBeAMathProffesor

i have understood somewat of wat u said, please explain the whole, the formula please, thanxs

Draw out three vertices and connect them.

Notice that when there are 3 vertices there are 2 segments for each vertex.

So why isn't the rule 2x3, or 6 segments?

Because you are counting each segment twice, so the answer is really (3x2)/(2), or 3 segments.

Now draw a diagram with 4 vertices, there are 3 segments from each vertex, but each segment got counted twice, so divide by 2 and you'll find there are in all (4x3)/(2) segments, or 6 segments.

Continuing on you can say that a diagram with n vertices would have n-1 segments from each vertex, but each segment got counted twice, so divide by 2 and you'll find there are n(n-1)/2=(n^2-n)/2 line segments.

test it:

(5^2-5)/2=(25-5)/2=20/2=10 that checks out.

(6^2-6)/2=(36-6)/2=30/2=15 that checks out.

So the expression works!

Q.How many ways are to draw edges?
A.Since an edge is determined by 2 vertices then,
nC2
Which is,
n!/2!(n-2)!=n(n-1)/2 which is what Quick has.

This is the number of edges in a complete graph.
It is not possible to determine the number edge in a general graph, since it can be anything.

i got the part before the equal sign but wat about after the equal sign, how do u explain that?thanxs (this is for Quick)

wat are the ! for? thanxs

Originally Posted by OnMyWayToBeAMathProffesor

wat are the ! for? thanxs

That is the factorial.

n!=n(n-1)(n-2)...(2)(1)
For example,
4!=4*3*2*1=24

The formula I mentioned.
Is the number of combinations.
Thus, I was looking for the number of combination involving 2 verticies.

If the question was a digraph (which you purposely omit) then it can go both ways. Thus twice as much.
n(n-1)

THANXS TO EVERY ONE THAT HELPED ME!!!!!!!!!!!!!!

Originally Posted by OnMyWayToBeAMathProffesor

THANXS TO EVERY ONE THAT HELPED ME!!!!!!!!!!!!!!

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