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Consider the set of all the vectors in $C^8$ each of whose coordinates is either 0 or 1.

How many different bases does this set contain?

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    By C^8, do you mean $\Bbb C^8$, the $8$-dimensional vector space over the complex numbers? – 2017-01-23
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    no closed form is known for this, see here: http://mathoverflow.net/questions/18636/number-of-invertible-0-1-real-matrices – 2017-01-23
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    @JorgeFernándezHidalgo What does your comment mean? The answer will be a number. – 2017-01-23
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    @IgorRivin I mean that no formula has been found for a general $n$, of course it is easy to solve with brute force, the link I gave contains the link to the OEIS sequence, the answer for $n=8$ is $8286284310367538176$ – 2017-01-23
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    @JorgeFernándezHidalgo: No, that is the number of matrices that are _not_ invertible. You need to subtract that from $2^{64}$. – 2017-01-23
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    in hindsight brute force may not be the way to go for this one, but surely some combination of computers and intelligent casework. – 2017-01-23
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    @HenningMakholm oh yeah, of course. Sorry about that. – 2017-01-23
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    @JorgeFernándezHidalgo I am sure the numbers have not been obtained by brute force, since iterating through $2^{64}$ possibilities might take a while. The obvious hack of looking at the vectors in sorted order saves a factor of $8!$ gets you down to $2^{49},$ which is still largish. – 2017-01-23
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    @IgorRivin maybe you can get it down by stopping if the first $k$ rows are allready LD. Although there's probably more stuff involved, I don't know really xD. – 2017-01-23
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    I saw that you already had a comment in the link I posted. – 2017-01-23

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