How many $4\times 3$ matrices of rank three are there over a finite field of three elements? 
Let $M$ be the space of all $4\times 3$ matrices with entries in the finite field of three elements. Then the number of matrices of rank three in $M$ is
A. $(3^4 - 3)(3^4 - 3^2)(3^4-3^3)$
B. $(3^4 - 1)(3^4 - 2)(3^4 - 3)$
C. $(3^4-1)(3^4-3)(3^4-3^2)$
D. $3^4(3^4 - 1)(3^4 - 2)$

Is there any specific formula to solve this type of problem? Is there any specific formula to calculate the number of matrices? I have no idea how to start this problem. Any guidance please.

Sorry for the previous post.the answer I am getting is c
$(3^4-1)(3^4-3)(3^4-3^2)$
For the $1$st column we have $4$ places and $3$ elements to fill it.So $3^4$ choices but the elements can't be all zero.S0 we have $3^4-1$ choices .For the second column  we have $3^4$ choices but the second column cant be linearly dependent with the $1$st .So $(3^4-3)$ choices .Similarly the $3$rd column is not linearly dependent to both $1$st and $2$ nd cloumn
