Not every collection of choices between roots $\alpha$ and $-\alpha$ is "allowed". Yes, there is a partition of the set of all roots into positive $\Phi$ and negative $-\Phi$, but, also, $\Phi$ must be closed under addition. In the case of $GL(n)$, the Weyl group (permutation matrices, if you like) acts simply-transitively on the set of such choices, and/so there are exactly $n!$ such choices, rather than the $2^{n(n-1)/2}$ choices of sign.

Not every collection of choices between roots $\alpha$ and $-\alpha$ is "allowed". Yes, there is a partition of the set of all roots into positive $\Phi$ and negative $-\Phi$, but, also, $\Phi$ must be closed under addition. In the case of $GL(n)$, the Weyl group (permutation matrices, if you like) acts simply-transitively on the set of such choices, and/so there are exactly $n!$ such choices, rather than the $2^{n(n-1)/2}$ choices of sign.

[Edit:] In response to @Brad H-D's comment/query: yes, I was a bit glib... Indeed, it is not the case that all possible (iterated) sums of positive roots are roots, since this would (erroneously) require that there be infinitely-many, etc. Rather, as Brad H-D leadingly-asked, it means that either the sum of two "positive" roots is a "positive" root, or is not a root at all. (Thx, Brad H-D.)