Maximum Paths of Length 6 in an Undirected Graph

Let $G$ be an arbitrary undirected graph with $n$ vertices. We want to find the maximum number of paths of length 6 between any pairs of vertices. /n/nConsider a pair of vertices $u$ and $v$ in $G$. Let $N(u)$ and $N(v)$ be the sets of neighbors of $u$ and $v$, respectively. Since the maximum length of a path is 6, any vertex on a path between $u$ and $v$ must be at most 3 edges away from $u$ or $v$. Therefore, any path of length 6 between $u$ and $v$ must consist of the form $u-x-y-z-w-v$, where $x,y,z,w$ are distinct vertices in $N(u)/cup N(v)$. /n/nLet $S=N(u)/cup N(v)$. Then, the number of paths of length 6 between $u$ and $v$ is equal to the number of ways to choose 4 distinct vertices from $S$, which is $/binom{|S|}{4}$. Therefore, the maximum number of paths of length 6 between any pairs of vertices in $G$ is $/binom{|S|}{4}$, where $S$ is the union of the neighborhoods of any pair of vertices in $G$. /n/nTo maximize $/binom{|S|}{4}$, we want to maximize $|S|$. By the pigeonhole principle, the union of the neighborhoods of any pair of vertices in $G$ has at most $2(n-1)$ vertices. Therefore, the maximum number of paths of length 6 between any pairs of vertices in $G$ is $/binom{2(n-1)}{4}$. /n/nIn summary, the maximum number of paths of length 6 between any pairs of vertices in an arbitrary undirected graph with $n$ vertices is $/binom{2(n-1)}{4}$.