For instance, if $k=m\ge n,$ you are asking for the number of surjections from $[m]$ to $[n],$ a classical counting problem. Like the problem of counting surjections, your more general question has an answer in the form of an inclusion-exclusion formula:
$$\sum_{u=0}^m\sum_{v=0}^n(-1)^{u+v}\binom mu\binom nv\binom{(m-u)(n-v)}k\tag1$$
or equivalently (setting $r=m-u,s=n-v$)
$$\sum_{r=0}^m\sum_{s=0}^n(-1)^{m+n+r+s}\binom mr\binom ns\binom{rs}k.\tag2$$
Example. Consider the case $m=2,n=3,k=3.$ The answer is clearly $2^3-2=6.$ With $m=2$ and $n=3$ in formula (2), the only nonzero terms occur when $rs\ge3,$ that is, when $(r,s)$ is $(1,3)$ or $(2,2)$ or $(2,3),$ and we get
$$-\binom21\binom33\binom33-\binom22\binom32\binom43+\binom22\binom33\binom63=-2-12+20=6.$$
Explanation. Let $E$ be the set of all $m\times n$ matrices of zeros and ones with exactly $k$ ones. For $i\in[m]$ let $A_i$ be the set of all matrices in $E$ with no ones in the $i^{\text{th}}$ row. For $j\in[n]$ let $B_j$ be the set of all matrices in $E$ with no ones in the $j^{\text{th}}$ column. The expression (1) comes from the inclusion-exclusion formula for
$$|E\setminus(\bigcup_{i\in[m]}A_i\cup\bigcup_{j\in[n]}B_j)|\tag3$$
by simplifying it using the fact that, for $I\in\binom{[m]}u$ and $J\in\binom{[n]}v,$
$$|\bigcap_{i\in I}A_i\cap\bigcap_{j\in J}B_j|=\binom{(m-u)(n-v)}k.\tag4$$