How many ways to order vectors in $\{0,1\}^n$?

Question

How many different rankings can be produced for the vectors in $\{0,1\}^n$ that also respect the usual $\geqq$ ordering of vectors (defined below)?

I want to produce a complete ordering where, for $x\neq y$, if $x_i \geq y_i$ for all $i=1,\dots,n$ then $x$ is greater than $y$ according to that new vector ordering.

This leaves some freedom in the ordering where I could have $(1,0,0) > (0,1,1)$ as a possibility. $(1,0,0) < (0,1,1)$ is also allowed.

Example: For $n=2$, there are two rankings.

$(1,1) \geq (0,1) \geq (1,0) \geq (0,0)$ and $(1,1) \geq' (1,0) \geq' (0,1) \geq' (0,0)$.

Answer 1

You have to first rank the entries of the vectors. As in the first ranking of your example the second entry has 'higher importance' for the order (-> higher priority for the actual ranking of the whole set) than the first entry.

In your second ranking it is the first entry which dominates for the purpose of ranking.

How many possibilities are there to rank the $n$ entries there are?

The answer to this (and your) question is $n!=n(n-1)\cdots$

Answer 2

As observed by Brian M. Scott in the comments, these are linear extensions of the subset lattice (also known as the Boolean lattice).

So let us look them up, following Samuel's advice: compute some small instances in SageMath,

sage: for n in range(2,5): (n, posets.BooleanLattice(n).linear_extensions().cardinality())
(2, 2)
(3, 48)
(4, 1680384)

followed by OEIS lookup, which takes us to A046873. From there (and following links) we learn the next numbers:

5 14807804035657359360
6 141377911697227887117195970316200795630205476957716480
7 630470261306055898099742878692134361829979979674711225065761605059425237453564989302659882866111738567871048772795838071474370002961694720

and apparently that is how far they are currently known (the $n=7$ value computed in 2017 by Brouwer and Christensen). OEIS also gives pointers to asymptotic results.