What is the number of ways to put n distinct balls into n+1 boxes, n of which are identical and the other of which is different (say, it is larger than the rest)?
I understand that this is a question regarding Bell numbers, but I am unsure how to include the box that is different in this. I have that there are n+1 ways to pick this box, and n+1 ways to pick the empty box, and the number of ways to partition [n] is B(n). So the total would be $B(n)(n+1)^2$
There are ${n\choose k}$ ways to choose the balls that go into the distinguished box for $k=0,1\dots,n.$ Then we put $n-k$ distinct balls into $n$ identical boxes, so we have $$\sum_{k=0}^n{n\choose k}P(n-k, n)$$ ways where $P(n,n-k)$ is the number of ways to put $n-k$ distinct objects into $n$ identical boxes. Since we don't care which boxes we use, this is simply the number of ways to partition a set of $n-k$ objects, or to put it another way, the number of equivalence classes on a set of $n-k$ objects. That is, $P(n-k,n)$ is just $B(n-k)$, the Bell number. The final answer is $$\sum_{k=0}^n{n\choose k}B(n-k)$$