Let G be a complete graph with n vertices. Consider the case where n=2. With only 2 vertices it is clear that there will only be one edge. Now add one more vertex to get n = 3. We must now add edges between the two old vertices and the new one for a total of 3 vertices. We see that adding a vertex to a graph with n vertices gives us n more edges. We get the following sequence
Edges on a graph with n vertices: 0+1+2+3+4+5+...+n-1. Adding this to itself and dividing by two yields the following formula for the number of edges on a complete graph with n vertices: n(n-1)/2.
(n-1)*(n-2)/2
That number would be achieved if all but one vertex were connected to every other vertex but that one remained isolated.
N*(N - 1)/2
Given an undirected graph G=(V,E) and an integer k, find induced subgraph H=(U,F) of G of maximum size (maximum in terms of the number of vertices) such that all vertices of H have degree at least k
In a connected component of a graph with Mi vertices, the maximum number of edges is MiC2 or Mi(Mi-1)/2. So if we have k components and each component has Mi vertices then the maximum number of edges for the graph is M1C2+M2C2+...+MKC2. Of course the sum of Mi as i goes from 1 to k must be n since the sum of the vertices in each component is the sum of all the vertices in the graph which you gave as n. Where MC2 means choose 2 from M and there are M(M-1)/2 ways to do that.
It's a perfect square.
An odd number.
When the number is a square of a prime number. Of course, it will have a repeated factor, but it will still only have 3 distinct factors. 4, 9, 25, 49, 121 are some examples.
V*(V-1)/2
Given an undirected graph G=(V,E) and an integer k, find induced subgraph H=(U,F) of G of maximum size (maximum in terms of the number of vertices) such that all vertices of H have degree at least k
n * (n - 1) / 2 That would ignore the "acyclic" part of the question. An acyclic graph with the maximum number of edges is a tree. The correct answer is n-1 edges.
It is a true statement.
n - 1
n-1
Every prism has vertices. They have an even number of vertices, with a minimum of 6 and no maximum.
There is no such polyhedron. The numbers given in the question do not satisfy the Euler characteristic for simply connected polyhedra.Alternatively, the fact that it has eight faces means that is should be an octahedron. However, among the 257 topologically distinct convex octahedra, the maximum number of vertices is 12 and the maximum number of edges is 18. Both these numbers are well below what you require!
discuss the possible number of points of interscetion of two distinct circle
Six.
Six (6)
I believe that such an object cannot exist in normal 3-d space. If there are 6 vertices, the maximum number of edges is 12.