At a party, n people are seated in a room with n+1 chairs. They go to another room for supper; when they return and sit down again, it is found that no one occupies the same chair as before. Show that this can occur in precisely Dn + Dn+1 ways.
When I tried to solve this, I got (n+1)*Dn + Dn+1 instead, and I'd like to know what I got wrong.
Here's my strategy: add a new "invisible person" to the n people, and consider the seating for the n+1 people & n+1 chairs.
Case 1) the invisible person sits in the same chair
Then the remaining n seats should make a derangement, so there are Dn possibilities.
Case 2) the invisible person also gets a new chair the second time
All n+1 chairs should make up a derangement, so there are Dn+1 possibilities.
But for Case 1, there are n+1 seats where the invisible person can occupy, so in total there are (n+1)*Dn + Dn+1 ways.
Why is the (n+1) factor unnecessary?