<s>[INST] No of matrices of rank 3 I was tackling this problem 
How many 4 x 3 matrices can be formed of rank 3 where entries are coming from a field  of 3 elements.
When I solved it as finding total no of linearly independent sets with cardinality 4 in F^4(F). Assuming that each column of the matrix is coming from F^4(F).
I am getting correct answer as follows 
(3^4-1)(3^4-3)(3^4-3^2)
 But if I am solving this problem as taking each row from F^3 (F). 
Then our task is to find matrices with rank 3. So any ordered basis can serve to form any three rows of A and remaining row can be anything 
So I am getting (3^3-1)(3^3-3)(3^3-3^2)3^3.
But it doesn't matches with that answer which I got using columns. Where I am doing mistake I am unable to identify. 
 [/INST] You're only counting matrices where the upper $3\times3$ part has full rank. You need to add three more cases: If dependence first occurs in the third row, we have $(3^3-1)(3^3-3)3^2(3^3-3^2)$ options; if it first occurs in the second row, we have $(3^3-1)3(3^3-3)(3^3-3^2)$ options, and if it first occurs in the first row, we have $1(3^3-1)(3^3-3)(3^3-3^2)$ options, for a total of
\begin{eqnarray*}
&&(3^3-1)(3^3-3)(3^3-3^2)3^3+(3^3-1)(3^3-3)3^2(3^3-3^2)+(3^3-1)3(3^3-3)(3^3-3^2)+1(3^3-1)(3^3-3)(3^3-3^2)
\\
&=&
(3^3-1)(3^3-3)(3^3-3^2)(3^3+3^2+3+1)
\\
&=&
(3^3-1)(3^3-3)(3^3-3^2)\frac{3^4-1}{3-1}
\\
&=&
(3^3-1)(3^3-3)\frac{3^3-3^2}{3-1}(3^4-1)
\\
&=&
(3^3-1)(3^3-3)3^2(3^4-1)
\\
&=&
(3^4-3)(3^4-3^2)(3^4-1)
\\
&=&
(3^4-1)(3^4-3)(3^4-3^2)\;.
\end{eqnarray*}
One more piece of advice: You would probably have found the error yourself if you'd tried the simpler case of a $2\times1$ matrix.
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