I got this question which i couldn't figure out:
Given $m,n,k \geq 1$ such that $k \ge n-m$. in how many ways can we put $k$ diffrent balls in $n$ boxes, that excatly $m$ boxes will be empty?
i thought about $n \choose m$ for the choosing the empty boxes, and ${n-m-k-1} \choose {n-m}$ for putting the balls in the remainig boxes. The result will be the multiplication of these two.
Am i right?