Let $\varepsilon$ and $\eta$ be multilinear, skewsymmetric forms taking their values on $\mathbb R$. How can I write the sum:
$$(\varepsilon\wedge \eta)(a_1, \ldots, a_{k}, a_{k+1}, \ldots, a_{k+l})=\sum_{\sigma\in Sh(k, l)} \textrm{sign}(\sigma) \varepsilon(a_{\sigma(1)}, \ldots, a_{\sigma(k)})\eta(a_{\sigma(k+1)}, \ldots, a_{\sigma(k+l)})$$ without using shuffles?
That is, I'm look to write something like $$\sum_{1\leq j_1<\ldots< j_k \leq k+l\atop 1\leq i_1<\ldots<i_l\leq k+2} (-1)^{j_1+\ldots+j_k}(-1)^{i_1+\ldots+i_l} \varepsilon(a_{j_1}, \ldots, a_{j_k})\eta(a_{i_1}, \ldots, a_{i_l}).$$ The above sum is not correct though, because there are overlapping indices.
The point is that once we know $1\leq j_1<\ldots<j_k\leq k+l$ the remaining indices $i_1<\ldots<i_l$ are uniquely determined, but I don't know to use this information properly.
Thanks.
This is probably not a satisfying answer, but I give it a try.
Assume $\varepsilon$ and $\eta$ to be a $k$- and an $h$-form respectively. As you say, when you fix $a_{j_1},\dots,a_{j_k}$, with $j_1 < j_2 < \dots < j_k$, the remaining indices are uniquely determined and appear in the sequence of vectors in $\eta$ in increasing order due to the process of contraction. So there should be no sum over the $i$s, rather a remark that the $i$s are the indices which complete the sequence $j_1,\dots,j_k$ to $1,\dots,k+h$, and that they appear in increasing order.
Also, $$ \begin{align} (-1)^{j_1+\dots+j_k}(-1)^{i_1+\dots+i_h} & = (-1)^{1+\dots+(k+h)}\\ & = (-1)^{(k+h)(k+h+1)/2}. \end{align} $$ I did not check whether the signs are ok in general, but they look right in special cases where $h$ and $k$ are small.
Using contractions I end up with the expression you want to avoid, which is the only one I have always seen around in the general case.