Combinatorics of link choices in a fully connected network Four routers are fully interconnected with point-to-point links. If each pair of routers can be connected by exactly one of three line types (high, medium, or low speed), how many distinct link-type assignment topologies are possible?

Introduction / Context:
Network design often involves selecting link capacities between every pair of nodes. When the physical connectivity is fixed (complete graph), the design problem reduces to counting the number of ways to assign link types. This is a straightforward application of combinatorics to a networking scenario.

Given Data / Assumptions:

• Number of routers (nodes) = 4.
• Topology is fully connected (complete graph).
• Each link must be exactly one of three types: high, medium, or low speed.

Concept / Approach:
A fully connected undirected network of N nodes has E = N * (N - 1) / 2 distinct links. For each independent link, there are 3 choices. The multiplication rule for independent choices gives total configurations = 3^E. Here, N = 4, so E = 4 * 3 / 2 = 6; thus total = 3^6.

Step-by-Step Solution:
Compute number of links: E = C(4, 2) = 6.Assign choices per link: 3 types available per link.Apply product rule: total configurations = 3^6.Evaluate 3^6: 3^2 = 9, 3^3 = 27, 3^6 = 27 * 27 = 729.

Verification / Alternative check:
List smaller cases to verify the pattern: For N = 3, E = 3, total = 3^3 = 27. Extending to N = 4 yields 3^6 = 729, confirming consistency.

Why Other Options Are Wrong:
12: equals the number of directed arcs for a 4-node complete digraph without self-loops, not the count of assignments.81: equals 3^4, which would correspond to only 4 links.48: not a power of 3; does not match combinatorial structure.

Common Pitfalls:

• Forgetting the network is undirected and using N * (N - 1) instead of N * (N - 1) / 2.
• Confusing “number of topologies” with “number of physical graphs”; connectivity is fixed here, only link types vary.
• Assuming order matters between line types on the same link; it does not.

Final Answer:
729