Number of Complete Bipartite Graphs with n Vertices

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Homework Statement


How many complete bipartite graphs have n vertices?


Homework Equations





The Attempt at a Solution


I said, let the first bipartition set have x vertices and the second bipartition set have y vertices. So x+y=n.

I think that you would do C(n, x)*C(n-x, y)=C(n,x)*C(y,y)=C(n,x) but I feel like this is either wrong or not good since I made x and y up arbitrarily.

Help?
 
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So that's the number of complete bipartite graphs with the "first" partition having size x. What are the possible values for x?
 
There are n possible values for x... so is the answer n*C(n,x)?

Is there any problem with me having made x up without it being in the problem?

Also, would I instead want to do something like C(x,1)*C(y,1) [to get the number of ways to pick the vertices] *C(n,x) [to get the number of ways of making the 'first' bipartition set] *n?
 
Could it matter if there is an even/odd number of vertices in anyone bipartition set?
 
Any new ideas?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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