There are $N$ different balls and $K$ numbered cells, $K<N$. I need to calculate the number of options to distribute the balls into exactly $L$ cells ($L<K$) so each cell will have at least 1 ball. I tried doing this: I assume all the balls are identical and arrange all the balls in a line, there are $\binom{N-1}{L-1}$ ways to distribute them to L cells so each cell has at least 1 ball, now because the balls are different I have $N!$ ways to arrange them in the line so in total I have $\binom{N-1}{L-1}\cdot N!$ ways to distribute the balls into cells. I have $\binom{K}{L}$ ways to choose the cells that have balls, so the total number of options is $$\binom{K}{L}\cdot \binom{N-1}{L-1}\cdot N!$$ but it is apparently wrong but I don't understand what did I do wrong here
EDIT: after the comment I tried doing this instead: I choose $\binom{K}{L}$ cells and put 1 ball in each cell, I choose the $L$ balls in $\binom{N}{L}$ ways and because the balls are different I have $L!$ ways to distribute them between the $L$ chosen cells, the remaining $N-L$ balls can be distributed in $L^{N-L}$ ways so in total $$\binom{K}{L}\cdot \binom{N}{L} \cdot L! \cdot L^{N-L}$$ but again I got the wrong answer, I assume because there are intersections I missed, at this point I'm not sure how I'm supposed to approach this