Help With 1 Simple Discrete Math Problem? Please

here is the question:http://i.imgur.com/bpRCfjq.jpg

I think i got part A, but does someone have a good firm answer for part B?

For 1A:There are as many edges in K_n as there are subsets of both endpoints (2) that can be chosen from a set of n vertices. The number of ways to select 2 vertices from n vertices is n(n-1) / 2 aka n choose 2.

Thus the # of edges in K_n is n(n-1)/2

Is that how to "show" this?

Then for 1b.

the one part of graph has m vertices each with degree n an the other part has n vertices each with degree m..

i see that m*n = e, if make my own little graph but i don't quite see how that comes from what i said above.

Re: Help With 1 Simple Discrete Math Problem? Please

Quote:

Originally Posted by

**outkast32** here is the question:

http://i.imgur.com/bpRCfjq.jpgThen for 1b.

the one part of graph has m vertices each with degree n an the other part has n vertices each with degree m..

i see that m*n = e, if make my own little graph but i don't quite see how that comes from what i said above.

In $\mathcal{K}_{m\times n}$ you have two disjoint sets of vertices: one has $m$ elements and the other has $n$ elements,

The is an edge between any two points one from each set. There is no edge between two points in the same set.