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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.