The Johnson graph $J(n,k)$ has as vertices the $k$-element subsets of an $n$-element set, where two vertices are adjacent whenever the corresponding subsets intersect in exactly $k - 1$ elements. I am wondering if the number of triangles on any vertex $v$ of $J(n,k)$ is a straight forward formula? This number is definitely independent of a vertex in $J(n,k)$.
The formula I come up with are all wrong, as I test them against the correct values using Sagemath codes. I seem to be missing something though I suspect its something easy. If $J(n,k)$ is difficult, will the case $J(2k,k)$ be easier?
We may assume that $v = \{1, 2, \dots, k\}$. Let's first count triangles using both $v$ and the vertex $w = \{2, 3, \dots, k+1\}$, which is one of the $k(n-k)$ neighbors of $v$.
A neighbor of both $v$ and $w$ could contain all of the elements of $v\cap w = \{2,3,\dots,k\}$, in which case there are $n-k-1$ ways to pick a $k^{\text{th}}$ element (not equal to $1$ or $k+1$) and get a vertex. Alternatively, a neighbor of $v$ and $w$ could be missing one of the elements of $v\cap w$; in this case, it must include both $1$ and $k-1$, and there are $k-1$ ways to choose the element of $v\cap w$ that it's missing. Altogether, we find $n-2$ common neighbors.
This would give us $k(n-k)(n-2)$ triangles containing $v$, except that we are double-counting: either one of the other vertices of the triangle could be the vertex we chose as $w$. Therefore $v$ is contained in $\frac12 k(n-k)(n-2)$ triangles.