Q11Discrete Mathematics

A complete graph of order $n$, mathematically denoted as $K_n$, is a highly interconnected simple undirected graph wherein every single distinct pair of vertices is physically connected by exactly one unique edge. We are explicitly analyzing $K_4$, which possesses exactly 4 vertices (let us label them $A,B,C,D$) and 6 edges. The problem demands calculating the exact number of paths of length 3 strictly between any two given vertices. In graph theory, the "length" of a path is formally defined as the total number of distinct edges traversed. A valid "path" strictly prohibits the revisiting of any vertex.

Combinatorial Path Construction

Let us isolate two specific, distinct vertices: the starting vertex $A$ and the destination vertex $B$. We require a path $A\to v_1\to v_2\to B$. Because the path length must be exactly 3, we must traverse exactly three edges, which fundamentally requires passing through exactly two intermediate vertices ($v_1$ and $v_2$).

Because the graph is $K_4$, it only contains 4 vertices in total. If $A$ is the start and $B$ is the end, the only available candidates for the intermediate vertices $v_1$ and $v_2$ are the remaining two vertices in the graph: $C$ and $D$.

Step 1: Selecting the First Intermediate Vertex ($v_1$) From the starting vertex $A$, we must travel to the first intermediate node. We cannot travel directly to $B$ (that would be a path of length 1). We have exactly two valid choices: we can travel to $C$ OR we can travel to $D$.

Step 2: Selecting the Second Intermediate Vertex ($v_2$) Once we have arrived at the first intermediate vertex, we must travel to the second. - Scenario 1: If we chose $C$ as $v_1$, we cannot go back to $A$ (no revisiting), and we cannot go to $B$ yet (that would create a path length of 2). Therefore, our only mathematically possible move is to travel to $D$. Thus, $v_2$ must be $D$. The path segment is $A\to C\to D$. - Scenario 2: If we chose $D$ as $v_1$, by the exact same logical constraint, our only possible move is to travel to $C$. Thus, $v_2$ must be $C$. The path segment is $A\to D\to C$.

Step 3: Reaching the Destination From the second intermediate vertex, we must take the final edge directly to the destination $B$. - Following Scenario 1: $A\to C\to D\to B$. (Length = 3 edges) - Following Scenario 2: $A\to D\to C\to B$. (Length = 3 edges)

Mathematical Conclusion

Through exhaustive combinatorial tracing, we have discovered exactly 2 unique paths. This logic holds universally for any two chosen vertices in a $K_4$ graph due to its perfect symmetry. Therefore, the total number of paths of length 3 between any two vertices in $K_4$ is exactly 2.