Reduce the following decision problem to CNF-SAT Input:


$X$ = {$x_1$,$x_2$,$x_3$,...,$x_n$}
$Y$ = {$y_1$,$y_2$,$y_3$,...,$y_m$}
$k$, where, $k$ $\leq$ $m$


Output (Yes/No):
Satisfying the following condition, can all the elements in set $X$ be assigned to elements in set $Y$, such that, no elements in $X$ is left unassigned?

Condition: At least $k$ elements in $Y$ is needed to access all the elements in $X$. All the elements assigned to $y_i \in Y$ are said to be accessible using $y_i$.

Reduce this decision problem to CNF-SAT.

Thoughts:

I have considered $n \times m$ no. of variables. Each variables for a pair of element of set $X$ and set $Y$. i.e., $a_{ij}$ : for all $x_i$ and $y_j$. 

For ensuring that no elements in $X$ is left unassigned the CNF expression will be. 
$(a_{11} \vee a_{12} \vee ... \vee a_{1m}) \wedge (a_{21} \vee a_{22} \vee ... \vee a_{2m})... (a_{n1} \vee a_{n2} \vee ... \vee a_{nm})$

I am not able to form the CNF expression for ensuring that at least $k$ elements from $Y$ are required to access all the elements in $X$. It seems that there will be an exponential number of clauses in the expression.

Thanks for helping.
Yes, you certainly can reduce it to SAT: the problem you state is obviously in NP, and thanks to Cook's theorem, everything in NP can be reduced to SAT.

I suspect that wasn't what you actually meant to ask.  Actually, your problem is in P: Hall's theorem says that such an assignment exists if and only if the corresponding bipartite graph has a perfect matching.  You can test whether a bipartite graph has a perfect matching in polynomial time using standard algorithms.