So here's the context. I'm reading a paper on...real moduli spaces (as opposed to fake ones, ya know?), and the author says that $B_n\langle S_3\rangle$ is combinatorially equivalent to the product of three simplices $\Delta_i\times \Delta_j\times\Delta_k$ where $i + j + k = n$. $S_3$ here denotes a circle with three distinct fixed points, and $B_n\langle M\rangle$ is defined for a (real) manifold $M$ as the closure in $M^n$ (or possibly in $M^n/\mathbb{S}_n$) of $(M^n\setminus\Delta)/\mathbb{S}_n$ where $\Delta$ is the so-called "large diagonal," i.e. the set of $n$-tuples which have any two coordinates coinciding, and $\mathbb{S}_n$ is the symmetric group on $n$ points, i.e. the group of permutations of the $n$ coordinates so that ordering of the coordinates is irrelevant here.
I'm trying to understand what "combinatorially equivalent" means in this context. My guess is that it means you can find a triangulation of each manifold such that the triangulations are like...graph isomorphic or something?
Aha! I seem to have found the answer in Milnor's...article (I suppose) "On the Relationship between Differentiable Manifolds and Combinatorial Manifolds" (published 1956).
"Two triangulated spaces $(K_1,f_x,X_1)$ and $(K_2,f_2,X_2)$ are isomorphic $K_1$ is isomorphic to $K_2.$ They are combinatorially equivalent if they have isomorphic subdivisions."
"A triangulation $(K,f)$ of a space $X$ consists of a simplicial complex $K$, together with a homeomorphism $f$ of $|K|$ onto $X$."
So there it is! :D