Number of bases of $F_2^3$:
Any ideas in which to express this?.
Is the formula $V=F_P^3$ in any way applicable?
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1And the more general case How to count number of bases and subspaces of a given dimension in a vector space over a finite field? – Arnaud D. Mar 30 '17 at 13:18
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Let's first find the number of ordered bases and then divide by 6. For the first vector, one can choose all vectors except for $0$, so this gives $2^3-1=7$ choices for the first one. Given such a choice, we can choose as a second vector any other vector that is not a multiple of the first one. There are 2 possible multiples of the first one, so there are $2^3-2=6$ possible choices for the second vector, given some choice for the first one. Now given a choice of two such vectors, we may choose the third one freely among all vectors that are not linear combinations of the preceding two, that is, we may choose freely from $2^3-2^2=4$ vectors. This gives a total of $7*6*4$ ordered bases, so a total of $7*4=28$ bases.
M. Van
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