Can someone indicate me how to count the numbers of $m$-simplices in the barycentric subdivision of an $n$-simplex (m $\leq n$) in an efficient way?
For $m = n$, I have come up with the following reasoning:
The number of $n$ simplices in the barycentric subdivision of an $n$ simplex is (calculated recursively as) $(n + 1)!$, since
1) the number of $(n - 1)$ simplices in an $n$-simplex is equal to $n + 1$,
2) the barycentric subdivision of an $n$-simplex comprises the cones formed by the barycentre and the barycentric subdivision of each $(n - 1)$ simplex.
I tried to go on, but I got stuck. I think there must be recursive formulae to calculate those numbers, but I have no idea. The "count them all and then remove duplications" method does not seem to work well.