Hopefully all of the evil things Mathematica's LaTeX conversion tool did to my expressions. I should warn the reader that Mozilla is not rendering this correctly for me.
In an effort to understand the component definition given by Ciufolini and Wheeler in Gravitation and Inertia of the exterior product
\begin{align*} \alpha_{a_{1}\ldots a_{p+q}} & =\left(\tilde{\theta}\wedge\tilde{\omega}\right)_{a_{1}\ldots a_{p+q}}=\frac{\left(p+q\right)!}{p!q!}\theta_{\left[a_{1}\ldots a_{p}\right.}\omega_{\left.a_{p+1}\ldots a_{p+q}\right]} \end{align*}
I created the following example:
\begin{align*} \theta_{ij} & =\theta_{[ij]},\\ \omega_{ijk} & =\omega_{[ijk],}\\ \tilde{\theta} & =\frac{1}{2}\theta_{[ij]}d\mathit{x}^{i}\wedge d\mathit{x}^{j},\\ \tilde{\omega} & =\frac{1}{6}\omega_{[ijk]}d\mathit{x}^{i}\wedge d\mathit{x}^{j}\wedge d\mathit{x}^{k},\\ \tilde{\alpha} & =\tilde{\theta}\wedge\tilde{\omega},\\ & =\frac{1}{2}\frac{1}{6}\theta_{ab}\omega_{cde}d\mathit{x}^{a}\wedge d\mathit{x}^{b}\wedge d\mathit{x}^{c}\wedge d\mathit{x}^{d}\wedge d\mathit{x}^{e}\\ & =\frac{1}{\left(2+3\right)!}\alpha_{abcde}d\mathit{x}^{a}\wedge d\mathit{x}^{b}\wedge d\mathit{x}^{c}\wedge d\mathit{x}^{d}\wedge d\mathit{x}^{e},\\ \alpha_{abcde} & =\frac{6!}{12}\theta_{[ab}\omega_{cde]}. \end{align*}
This is my question: would it be incorrect to write
\begin{align*} \alpha_{abcde} & =\frac{6!}{12}\theta_{ab}\omega_{cde}, \end{align*}
because there's nothing in the simple product of scalar components to ensure we have
\begin{align*} \theta_{ab}\omega_{cde} & =-\theta_{ac}\omega_{bde}? \end{align*}
I'm confident that a proper p-form will have the same meaning when expressed on a standard tensor basis. That is, assuming $\tilde{\alpha}$ to be a 5-form, we have
\begin{align*} \tilde{\alpha} & =\alpha_{abcde}d\mathit{x}^{a}\otimes d\mathit{x}^{b}\otimes d\mathit{x}^{c}\otimes d\mathit{x}^{d}\otimes d\mathit{x}^{e}\\ & =\frac{1}{5!}\alpha_{abcde}d\mathit{x}^{a}\wedge d\mathit{x}^{b}\wedge d\mathit{x}^{c}\wedge d\mathit{x}^{d}\wedge d\mathit{x}^{e} \end{align*}
That's because the components already have the property $\alpha_{abcde}=\alpha_{\left[abcde\right]}.$ When either expression of the tensor acts on an argument list of five vectors, it will produce the alternating sum of the tensor product components, which is the definition of the determinant. That is built into the wedge product basis, but not into the tensor product basis. So, for the two expressions to behave identically, the components have to be antisymmetric.