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Feb 25, 2012 at 18:01 comment added Alexander Chervov To comment just above --- if "K" is very big e.g. extreme case K=N - so there is only one "subspace" which is space itself, then of course number of weight "k" vectors is C_n^k, so if "K" is something near it we may expect similar result...
Feb 25, 2012 at 17:54 vote accept Alexander Chervov
Feb 24, 2012 at 20:21 comment added Seva One more comment to interpret the result. We have shown that the proportion of vectors of weight $k$ among all vectors of a random $K$-dimensional subset is about $2^{-N}\binom Nk$, which is the proportion of vectors of weight $k$ among all vectors of $F_2^K$. In other words, lying in a $K$-dimensional subspace is independent from having weight $k$. Pretty expectable!
Feb 24, 2012 at 19:57 comment added Seva What we computed is the average (=expected) number of vectors of weight $k$ in a randomly chosen $K$-dimensional subspace. Summing over $k$, we get the total number of vectors in a random $K$-dimensional subspace, which is $2^K$. Makes sense?
Feb 24, 2012 at 19:21 comment added Alexander Chervov if sum k=1...N I expect to get 1+O(1), but we get 2^K... is it correct ?
Feb 24, 2012 at 19:13 comment added Seva I appended two lines to my answer to address this.
Feb 24, 2012 at 19:11 history edited Seva CC BY-SA 3.0
added 130 characters in body; added 3 characters in body
Feb 24, 2012 at 18:09 comment added Alexander Chervov @Seva WOW !!! Can we understand what distribution we get when K,N are "large" ? It might be just something simple related to binomial distribution tending to normal distribution ...
Feb 24, 2012 at 15:39 history edited Seva CC BY-SA 3.0
more typos and corrections
Feb 24, 2012 at 15:16 history answered Seva CC BY-SA 3.0