This is not a homework. I just ask this question myself and thought it would be easy to figure out. But I did not get the solution.
Let $\mathbb{F}$ be a finite field with $|\mathbb{F}|=q$. Consider the $\mathbb{F}$-vectorspaces $V_1=\mathbb{F}^n,V_2=\mathbb{F}^m$ with dimensions $n<m$.
How many injective, surjective and bijectiv linear functions $f\colon \mathbb{F}^n \to \mathbb{F}^m$ exists?
My approach is: We have any basis $b_1,\ldots,b_n$ of $V_1$ and $c_1,\ldots,c_m$ of $V_2$. It clear that we only have to treat the function on this basis and there cannot be any bijective linear functions since $n<m$. Counting the functions must be similar to count the possibilities to do a injective map from $b_1,\ldots,b_n$ of $V_1$ to $c_1,\ldots,c_m$ of $V_2$.
How do I get the number of injective functions?
If I understand correctly, you already know that there are no surjective or bijective functions since $|\mathbb F^n|\lt|\mathbb F^m|$.
To count the injective functions, choose a non-zero vector in $\mathbb F^m$ to map $b_1$ to – there are $q^m - 1$ choices. Now choose a linearly independent vector to map $b_2$ to – there are $q^m-q$ choices. Continuing like this, you have
$$ \prod_{k=1}^n\left(q^m-q^{k-1}\right) $$
possibilities in all.