The image displays four distinct graphs, each labeled accordingly:**Graph 1**: This is a bipartite graph with two sets of nodes. Each node in the first set is connected to every node in the second set. The left set has four nodes, and the right set has three nodes, resulting in twelve edges.**Graph 2**: This consists of two disconnected components. Each component is a triangle, indicating three nodes with each node connected to the other two.**Graph 3**: This features two disconnected components as well. The left component includes three nodes with two connected by an edge. The right component is similar to those in Graph 2—a triangle with each node connected to the other two.**Graph 4**: Shaped like an "H", this graph is composed of five nodes. The central node connects to both left and right nodes, which each also connect to an additional node, forming a symmetrical structure.These visual representations illustrate different configurations of nodes and edges, commonly used in studies of graph theory. **Transcription:**7) What is the number of Hamilton circuits in K₁₅?**Explanation:**This question asks about the number of Hamiltonian circuits in the complete graph K₁₅. A Hamiltonian circuit is a closed loop on a graph where every node (vertex) is visited exactly once before returning to the starting node.In a complete graph Kₙ, where there are n vertices, the number of Hamiltonian circuits can be calculated using the formula:\[\frac{(n-1)!}{2}\]For K₁₅:\[\frac{(15-1)!}{2} = \frac{14!}{2}\] This calculates the number of unique tours possible in the graph. Since each tour can be traversed in reverse, we divide by 2 to avoid counting each circuit twice.

We know that Hamiltonian circuit is a circuits that visits every vertex exactly once except the starting vertex which is the ending vertex also. Kn is the complete graph with n vertices in which each vertex is connected to every other vertex through an edge. for example, k2 is a line joining two points K3 is the triangle K3 has two Hamiltonian circuits ignoring the starting points . K4 has 4 vertices and 6 edges . So, number of Hamiltonian circuits =2×3=3!We know that number of Hamiltonian circuits in kn is n-1! In K15 , n=15 So, number of Hamiltonian circuits is 15-1!=14!

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Math

Advanced Math

What is the number of Hamilton circuits in K15?

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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