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If I understand it correctly, a binary linear $(n,k)$-code C is a code where the Alphabet is $\{0,1\}$ and $C=f(\mathbb F^k_2 )$ for some injective vectorspacehomomorphism and $f : \mathbb F^n_2 \rightarrow \mathbb F^k_2 $ such that $C$ is a linear subspace of $ \mathbb F^n_2$ with $k$ dimensions.

I want to know how many binary linear $(10,8)$-codes there are.

Is this just the number of linear subspaces of $F^{10}_2$ with $8$ dimensions - which would be $ 10 \choose 8$ if I am not mistaken, or do I need to consider something else?

willwissen
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    After you figured out you'd just be counting subspaces, you should have typed "counting subspaces of dimension" into the search box, and you would have probably seen the post I linked or a duplicate linked to it. – rschwieb Jun 21 '21 at 15:17

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Well, yes and no.

It looks like you mean the usual binomial coefficient $n\choose k$, but this obviously doesn't work. To choose a smaller example, suppose we wanted to count how many $1$-dimensional subspaces of $F_2\times F_2$ there are. Without breaking a sweat you say "three." But that is not $2\choose 1$.

However, it turns out there is something called the Gaussian binomial coefficient as mentioned at this solution.

Using that formula, it comes out to ${10\choose 8}_2=174251$, but $10\choose 8$ is just 45.

If you think about it, $10 \choose 8$ just counts the number of $8$ dimensional subspaces you can extract from a single basis of $10$ elements by picking some of the 10 elements. It does not take into account which of the (numerous) bases you start with.

rschwieb
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