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Here is an interesting riddle on random matrices.  

(Rank of Random Binary Matrix). Let R(l,m,k) denote the number of binary matrices of dimension l \times m and rank k, so that by symmetry R(l,m,k)=R(m,l,k).  This is a repost of the solution that I have arrived at (certainly not the first!) and submitted as part of a homework (9) problem from the doctoral course  Modern coding theory (by Rudiger Urbanke) at  EPFL.  The sumbitted solution in PDF is available here.

Rank of a matrix G is essentially the number of nonzero rows when the matrix G is expressed in echelon form. So, we just need to compute the ways these matrices can be created with k non zero rows. Since the elements of the matrix are binary (from \mathbb{F}_{q=2}), we can simply do a counting.

It is trivial to compute R(l,m,k) for k=0 and k>l. For  k=0, only all zero matrix possible, and only one such matrix exist. Hence R(l,m,0)=1. For  l>k>0, since  k>\min(l,m), no matrix exist, which means R(l,m,k)=0 . 

Now we consider l=k>0.  How many ways? We have l=k  non zero rows of the l\times m  matrix, which means all rows must be nonzero. Without loss of generality, for counting, we could assume that, the rows are ordered. The last row (l^{th} row can be be done in 2^{m}-1,  since there anything other than all 0 vector (of size m) is allowed. On (l-1)-th row, anything other than that of row l is allowed. There are 2^{m}-2 ways here. l-2-th row can have anything except any linear combination of the rows l and l-1. This is nothing but 2^m-\left({\binom{2}{0}+\binom{2}{1}+\binom{2}{2}}\right)=2^m-2^2. Row l-3 then have 2^m-\left(\binom{3}{0}+\binom{3}{1}+\binom{3}{2}\right)=2^m-2^3 and so on. In all, Following the same procedure, we can have a total of  

= \left(2^m-1\right) \left(2^m-2^1\right)\left(2^m-2^2\right)\ldots \left(2^m-2^{l-1}\right)

=\left(2^m-1\right) 2^{1} \left(2^{m-1}-1\right) 2^{2} \left(2^{m-2}-1\right) \ldots 2^{l-1}\left(2^{m-l+1}-1\right)

=2^{0} 2^{1} 2^{2} \ldots 2^{l-1}\left(2^m-1\right)\left(2^{m-1}-1\right)\left(2^{m-2}-1\right)\ldots\left(2^{m-l+1}-1\right)

=\prod_{i=0}^{l-1}{{2^i}\left(2^{m-i}-1\right)}

=\prod_{i=0}^{l-1}{\left(2^{m}-2^{i}\right)}

=\prod_{i=0}^{l-1}{2^m \left(1-2^{i-m}\right)}

=2^{ml} \prod_{i=0}^{l-1}{ \left(1-2^{i-m}\right)}

 ways.  For l>k>0, we can construct a rank k matrix of size l \times m in any of the following ways:

  1.  Take a rank k-1 matrix of size (l-1) \times m and add an independent row.
  2.  Take a rank k matrix of size (l-1) \times m and add a dependent row.

For every (l-1) \times m matrix, 

2^{m}-1+\binom{k-1}{1}+\binom{k-1}{2}+\ldots +\binom{k-1}{k-1}=\left(2^m-2^{k-1}\right)

and hence,

R(l-1,m,k-1) \left(2^m-2^{k-1}\right)= R_{1}(l,m,k)

ways. (Essentially avoid all possible linear combinations of existing k-1 rows).  Using the second (item 2 above) method, we can have 1+\binom{k}{1}+\binom{k}{2}+\ldots +\binom{k}{k} = 2^k and 

R_{2}(l,m,k)= 2^k R(l-1,m,k) 

different ways a rank k matrix can be formed. Where,the first term (=1) is when the all zero row is picked as the new row. In\binom{k}{1} ways we can pick any one of the exisiting row as a dependent (new row). In general for 0\le j\le k we can have combination of j existing rows  out of k in \binom{k}{j} different ways to make a dependent (new) row.

So using (1) and (2) we get,

R(l,m,k)=2^k R(l-1,m,k)+\left(2^m-2^{k-1}\right)R(l-1,m,k-1)

Putting everything together,

R(l,m,k) = \begin{cases} 1, & k=0, \\2^{ml} \displaystyle \prod_{i=0}^{l-1}{ \left(1-2^{i-m}\right)} , & l=k>0 \\ 2^k R(l-1,m,k) + \left(2^m-2^{k-1}\right) R(l-1,m,k-1) &l>k>0 \\ 0 & k>l>0 \end{cases}

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