How many vertices do the following graphs have if they contain 21 edges 3 vertices of degree 4 and others each of degree 3?
Table of Contents
- 1 How many vertices do the following graphs have if they contain 21 edges 3 vertices of degree 4 and others each of degree 3?
- 2 How many vertices will be there in a graph which contains 16 edges and all vertices of degree 4?
- 3 How many vertices graph have?
- 4 What is sum of the degree in all vertices of K 3?
- 5 Is it possible to construct a graph with 12 vertices such that two of the vertices have degree 3 and the remaining vertices have degree 4?
How many vertices do the following graphs have if they contain 21 edges 3 vertices of degree 4 and others each of degree 3?
∴ Number of vertices in the graph is 18.
How many vertices the graph have if it contains 21 edges 3 vertices?
A simple non directed graph contains 21 edges, 3 vertices of degree 4 and the other vertices are of degree 2. then the number of vertices in the graph is?…Subscribe to GO Classes for GATE CSE 2022.
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How many vertices will be there in a graph which contains 16 edges and all vertices of degree 4?
We see a graph that satisfies the requirements of 8 vertices, 16 edges and all vertices have degree 4 that certainly not planar.
How many vertices does a complete graph have with 21 edges?
A graph with 21 edges has seven vertices of degree 1, three of degree 2, seven of degree 3 and the rest of degree 4.
How many vertices graph have?
A graph is a diagram of points and lines connected to the points. It has at least one line joining a set of two vertices with no vertex connecting itself. The concept of graphs in graph theory stands up on some basic terms such as point, line, vertex, edge, degree of vertices, properties of graphs, etc.
How many vertices does the graph have with 21 edges?
What is sum of the degree in all vertices of K 3?
Handshaking Theorem states in any given graph, Sum of degree of all the vertices is twice the number of edges contained in it.
Does a 3 regular graph on 14 vertices exist?
If k 1 = 4 and k 2 = 4 , then is isomorphic to and hence, by Theorem 1.1, there is a 3-regular, 3-connected subgraph of on 14 vertices.
Is it possible to construct a graph with 12 vertices such that two of the vertices have degree 3 and the remaining vertices have degree 4?
So, in your case, you cannot have any vertex of degree ≥ 4, but there must be at least one vertex of degree ≥ 3, hence there must be a vertex of degree 3.