In how many ways we could put $n+5$ indistinguishable balls into $n$ distinguishable boxes and at least $2$ boxes have to be empty?
This is my answer: ${n}\choose{2} $$\cdot$$ {n+5+n-2-1}\choose{n+5}$ because first of all we have to choose $2$ empty boxes and after that we put $n+5$ balls into $n-2$ boxes. Is it correct?