Let $X$ be an abstract simplicial complex. Recall that the usual simplicial chain complex for $X$ is defined as follows. Let $C_k(X)$ be the quotient of the free abelian group on formal symbols $[v_0,\ldots,v_k]$ where $\{v_0,\ldots,v_k\}$ is a $k$-simplex of $X$ which identifies $[v_0,\ldots,v_k]$ and $(-1)^{|\sigma|}[v_{\sigma(0)},\ldots,v_{\sigma(k)}]$ for all permutations $\sigma$ of $\{0,\ldots,k\}$. The differentials are defined in the usual way.

My question is what happens when you instead do the following. Let $D_k(X)$ be the free abelian group on formal symbols $[v_0,\ldots,v_k]_D$ where $\{v_0,\ldots,v_k\}$ is a simplex of $X$ of dimension at most $k$ (so possibly there are repetitions among the $v_i$). We can define a differential on $D_{\ast}(X)$ in the usual way.

To summarize : The difference between $D_k(X)$ and $C_k(X)$ are that

In $D_k(X)$, the ordering matters (not just the orientation); for instance, if $v_0$ and $v_1$ are joined by an edge in $X$, then $[v_0,v_1]_D$ is unrelated to $[v_1,v_0]_D$, while in $C_1(X)$ we would have $[v_0,v_1]=-[v_1,v_0]$.

In $D_k(X)$, we allow "degenerate" simplices that have repetitions in their simplices in our chain complex, while this is not allowed in $C_k(X)$.

Does $D_{\ast}(X)$ also compute the homology of $X$? I can verify that it works in degree $0$ (trivial) and $1$ (easy; but this would not work if we did not allow repetitions). It seems similar to stuff I've seen with simplicial sets, but I don't know the literature there enough to extract it from there.